Statistical inference based on divergence measures

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Statistical Inference Based on Divergence Measures

© 2006 by Taylor & Francis Group, LLC

STATISTICS: Textbooks and Monographs D. B. Owen Founding Editor, 1972–1991

Associate Editors Statistical Computing/ Nonparametric Statistics Professor William R. Schucany Southern Methodist University Probability Professor Marcel F. Neuts University of Arizona

Multivariate Analysis Professor Anant M. Kshirsagar University of Michigan Quality Control/Reliability Professor Edward G. Schilling Rochester Institute of Technology

Editorial Board Applied Probability Dr. Paul R. Garvey The MITRE Corporation

Statistical Process Improvement Professor G. Geoffrey Vining Virginia Polytechnic Institute

Economic Statistics Professor David E. A. Giles University of Victoria

Stochastic Processes Professor V. Lakshmikantham Florida Institute of Technology

Experimental Designs Mr. Thomas B. Barker Rochester Institute of Technology

Survey Sampling Professor Lynne Stokes Southern Methodist University

Multivariate Analysis Professor Subir Ghosh University of California–Riverside

Time Series Sastry G. Pantula North Carolina State University

Statistical Distributions Professor N. Balakrishnan McMaster University


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Statistical Inference Based on Divergence Measures

Leandro Pardo Complutense University of Madrid Spain

Boca Raton London New York


Published in 2006 by Chapman & Hall/CRC Taylor & Francis Group 6000 Broken Sound Parkway NW, Suite 300 Boca Raton, FL 33487-2742 © 2006 by Taylor & Francis Group, LLC Chapman & Hall/CRC is an imprint of Taylor & Francis Group No claim to original U.S. Government works Printed in the United States of America on acid-free paper 10 9 8 7 6 5 4 3 2 1 International Standard Book Number-10: 1-58488-600-5 (Hardcover) International Standard Book Number-13: 978-1-58488-600-6 (Hardcover) Library of Congress Card Number 2005049685 This book contains information obtained from authentic and highly regarded sources. Reprinted material is quoted with permission, and sources are indicated. A wide variety of references are listed. Reasonable efforts have been made to publish reliable data and information, but the author and the publisher cannot assume responsibility for the validity of all materials or for the consequences of their use. No part of this book may be reprinted, reproduced, transmitted, or utilized in any form by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information storage or retrieval system, without written permission from the publishers. For permission to photocopy or use material electronically from this work, please access www.copyright.com (http://www.copyright.com/) or contact the Copyright Clearance Center, Inc. (CCC) 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400. CCC is a not-for-profit organization that provides licenses and registration for a variety of users. For organizations that have been granted a photocopy license by the CCC, a separate system of payment has been arranged. Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe.

Library of Congress Cataloging-in-Publication Data Pardo, Leandro. Statistical inference based on divergence measures / Leandro Pardo p. cm. -- (Statistics, textbooks and monographs ; v. 185) Includes bibliographical references and index. ISBN 1-58488-600-5 1. Divergent series. 2. Entropy (Information theory) 3. Multivariate analysis. 4. Statistical hypothesis testing. 5. Asymptotic expansions. I. Title. II. Series. QA295.P28 2005 519.5'4--dc22

2005049685

Visit the Taylor & Francis Web site at http://www.taylorandfrancis.com Taylor & Francis Group is the Academic Division of Informa plc.


and the CRC Press Web site at http://www.crcpress.com

This book is dedicated to my wife, Marisa


Preface The main purpose of this book is to present in a systematic way the solution to some classical problems of statistical inference, basically problems of estimation and hypotheses testing, on the basis of measures of entropy and divergence, with applications to multinomial (statistical analysis of categorical data) and general populations. The idea of using functionals of Information Theory, such as entropies or divergences, in statistical inference is not new. In fact, the so-called Statistical Information Theory has been the subject of much statistical research over the last forty years. Minimum divergence estimators or minimum distance estimators (see Parr, 1981) have been used successfully in models for continuous and discrete data. Divergence statistics, i.e., those ones obtained by replacing either one or both arguments in the measures of divergence by suitable estimators, have become a very good alternative to the classical likelihood ratio test in both continuous and discrete models, as well as to the classical Pearson—type statistic in discrete models. It is written as a textbook, although many methods and results are quite recent. Information Theory was born in 1948, when Shannon published his famous paper “A mathematical theory of communication.” Motivated by the problem of efficiently transmitting information over a noisy communication channel, he introduced a revolutionary new probabilistic way of thinking about communication and simultaneously created the first truly mathematical theory of entropy. In the cited paper, two new concepts were proposed and studied: the entropy, a measure of uncertainty of a random variable, and the mutual information. Verd´ u (1998), in his review paper, describes Information Theory as follows: “A unifying theory with profound intersections with Probability, Statistics, Computer Science, and other fields. Information Theory continues to set the stage for the development of communications, data storage and processing, and other information technolo-


x

Statistical Inference based on Divergence Measures

gies.” Many books have been written in relation to the subjects mentioned by Verd´ u, but the usage of tools arising from Information Theory in problems of estimation and testing has only been described by the book of Read and Cressie (1988), when analyzing categorical data. However, the interesting possibility of introducing alternative test statistics to the classical ones (like Wald, Rao or Likelihood ratio) in general populations is not yet found in any book, as far as I am concerned. This is an important contribution of this book to the field of Information Theory. But the following interesting question arises: Where exactly can be situated the origin of the link between Information Theory and Statistics? Lindley (1956) tries to answer our question, with the following words with reference to the paper of Shannon (1948), “The first idea is that information is a statistical concept” and “The second idea springs from the first and implies that on the basis of the frequency distribution, there is an essentially unique function of the distribution which measures the amount of the information.” This fact provided Kullback and Leibler (1951) the opportunity of introducing a measure of divergence, as a generalization of Shannon’s entropy, called the Kullback-Leibler divergence. Kullback, later in 1959, wrote the essential book “Information Theory and Statistics.” This book can be considered the beginning of Statistical Information Theory, although it has been necessary to wait a more few years for the statisticians to return to the problem. The contents of the present book can be roughly separated in two parts. The first part is dedicated to make, from a statistical perspective, an overview of the most important measures of entropy and divergence introduced until now in the literature of Information Theory, as well as to study their properties, in order to justify their application in statistical inference. Special attention is paid to the families of φ-entropies as well as on the φ-divergence measures. This is the main target of Chapter 1. Chapter 2 is devoted to the study of the asymptotic behavior of measures of entropy, and the use of their asymptotic distributions to solve different statistical problems. An important fact studied in this chapter is the behavior of the entropy measures as diversity indexes. The second part of the book is dedicated to two important topics: statistical analysis of discrete multivariate data in Chapters 3, 4, 5, 6, 7 and 8, and testing in general populations in Chapter 9. The statistical analysis of discrete multivariate data, arising from experiments


Preface

xi

where the outcome variables are the number of individuals classified into unique nonoverlapping categories, has received a great deal of attention in the statistical literature in the last forty years. The development of appropriate models for those kind of data is the common subject of hundreds of references. In these references, papers and books, the model is tested with the traditional Pearson goodness-of-fit test statistic or with the traditional loglikelihood ratio test statistic, and the unknown parameters are estimated using the maximum likelihood method. However, it is well known that this can give a poor approximation in many circumstances, see Read and Cressie (1988), and it is possible to get better results by considering general families of test statistics, as well as general families of estimators. We use the word “general” in the sense that these families contain as particular cases the Pearson and loglikelihood ratio test statistics, for testing, as well as the maximum likelihood estimator, for estimating. In Chapter 3, the problem of testing goodness-of-fit with simple null hypothesis is studied on the basis of the φ-divergence test statistics under different situations: Fixed number of classes, number of classes increasing to infinity, quantile characterization, dependent observations and misclassified data. The results obtained in this chapter are asymptotic and consequently valid just for large sample sizes. In Chapter 4, some methods to improve the accuracy of test statistics, in those situations where the sample size can not be assumed large, are presented. Chapter 5 is addressed to the study of a wide class of estimators suitable for discrete data, either when the underlaying distribution is discrete, or when it is continuous, but the observations are classified into groups: Minimum φ-divergence estimators. Their asymptotic properties are studied as well as their behavior under the set up of a mixture of normal populations. A new problem of estimation appears if we have some functions that constrain the unknown parameters. To solve this problem, the restricted minimum φ-divergence estimator is also introduced and studied in Chapter 5. These results will be used in Chapter 8, where the behavior of φ-divergences test statistics in contingency tables is discussed. Chapter 6 deals with the problem of goodness-of-fit with composite null hypothesis. For this problem, we consider φ-divergence test statistics in which the unknown parameters are estimated by minimum φ-divergence estimators. In addition to the classical problem, with fixed number of classes, the following nonstandard cases are also treated: φ-divergence test statistics when the unknown parameters are estimated by maximum likelihood estimator, φ-divergence test statistics with quantile characterizations, φ-divergence test statistics when parameters are estimated from an independent sample, φ-divergence test statistics with dependent


xii


observations and φ-divergence test statistics when there are some constraints on the parameters. Chapter 7 covers the important problem of testing in loglinear models by using φ-divergence test statistics. In this chapter, some of the most important results appeared in Cressie and Pardo (2000, 2002b), and Cressie et al . (2003) are presented. The properties of the minimum φ-divergence estimators in loglinear models are studied and a new family of test statistics based on them is introduced for the problems of testing goodness-of-fit and for testing a nested sequence of loglinear models. Pearson’s and likelihood ratio test statistics are members of the new family of test statistics. This chapter finishes with a simulation study, in which a new test statistic, placed “between” Pearson’s chisquare and likelihood ratio test statistics, emerged as a good choice, considering its valuable properties. Chapter 8 presents a unified study of some classical problems in contingency tables using the φ-divergence test statistic as well as the minimum φ-divergence estimator. We consider the problems of independence, symmetry, marginal homogeneity and quasi-symmetry in a two-way contingency table and also the classical problem of homogeneity. The domain of application of φ-divergence test statistics goes far beyond that of multinomial hypothesis testing. The extension of φ-divergence statistics to testing hypotheses in problems where random samples (one or several) obey distributional laws from parametric families has also given nice and interesting results in relation to the classical test statistics: likelihood ratio test, Wald test statistic or Rao statistic. This topic is considered and studied in Chapter 9. The exercises and their solutions included in each chapter form a part of considerable importance of the book. They provide not only practice problems for students, but also some additional results as complementary materials to the main text. I would like to express my gratitude to all the professors who revised parts of the manuscript and made some contributions. In particular, I would like to thank Professors Arjun Gupta, Nirian Mart´ın, Isabel Molina, Domingo Morales, Truc Nguyen, Julio Angel Pardo, Maria del Carmen Pardo and Kostas Zografos. My gratitude, also, to Professor Juan Francisco Padial for his support in the technical development of the book. Special thanks to Professor Arjun Gupta for his invitation to visit the De-


Preface

xiii

partment of Mathematics and Statistics of Bowling Green State University as Distinguished Lukacs Professor. Part of the book was written during my stay there. My final acknowledgment is to my wife, Professor Maria Luisa Menéndez, who read many times the early drafts of the manuscript. She gave me valuable advice and suggested many improvements. Her enthusiasm sustained me during the period spent in writing the manuscript, and this book is dedicated to her. Leandro Pardo Madrid


Contents

Preface

ix

1 Divergence Measures: Definition and Properties

1

1.1. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1

1.2. Phi-divergence Measures between Two Probability Distributions: Definition and Properties . . . . . . . . . . . . . . . . . . . . . . .

3

1.2.1. Basic Properties of the Phi-divergence Measures . . . . . .

8

1.3. Other Divergence Measures between Two Probability Distributions 18 1.3.1. Entropy Measures . . . . . . . . . . . . . . . . . . . . . . . 19 1.3.2. Burbea and Rao Divergence Measures . . . . . . . . . . . . 25 1.3.3. Bregman’s Distances . . . . . . . . . . . . . . . . . . . . . . 27 1.4. Divergence among k Populations . . . . . . . . . . . . . . . . . . . 27 1.5. Phi-disparities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 1.6. Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 1.7. Answers to Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . 34 2 Entropy as a Measure of Diversity: Sampling Distributions

55

2.1. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55


xvi


2.2. Phi-entropies. Asymptotic Distribution . . . . . . . . . . . . . . . 58 2.3. Testing and Confidence Intervals for Phi-entropies . . . . . . . . . 64 2.3.1. Test for a Predicted Value of the Entropy of a Population (Diversity of a Population) . . . . . . . . . . . . . . . . . . 64 2.3.2. Test for the Equality of the Entropies of Two Independent Populations (Equality of Diversities of Two Populations) . . 65 2.3.3. Test for the Equality of the Entropies of r Independent Populations . . . . . . . . . . . . . . . . . . . . . . . . . . . 65 2.3.4. Tests for Parameters . . . . . . . . . . . . . . . . . . . . . . 70 2.3.5. Confidence Intervals . . . . . . . . . . . . . . . . . . . . . . 74 2.4. Multinomial Populations: Asymptotic Distributions . . . . . . . . 75 2.4.1. Test of Discrete Uniformity . . . . . . . . . . . . . . . . . . 84 2.5. Maximum Entropy Principle and Statistical Inference on Condensed Ordered Data . . . . . . . . . . . . . . . . . . . . . . . . . . 87 2.6. Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93 2.7. Answers to Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . 98 3 Goodness-of-fit: Simple Null Hypothesis

113

3.1. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113 3.2. Phi-divergences and Goodness-of-fit with Fixed Number of Classes 116 3.3. Phi-divergence Test Statistics under Sparseness Assumptions . . . 125 3.4. Nonstandard Problems: Tests Statistics based on Phi-divergences . 132 3.4.1. Goodness-of-fit with Quantile Characterization . . . . . . . 132 3.4.2. Goodness-of-fit with Dependent Observations . . . . . . . . 135 3.4.3. Misclassified Data . . . . . . . . . . . . . . . . . . . . . . . 140 3.4.4. Goodness-of-fit for and against Order Restrictions . . . . . 144


CONTENTS

xvii

3.5. Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146 3.6. Answers to Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . 150 4 Optimality of Phi-divergence Test Statistics in Goodness-of-fit 165 4.1. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165 4.2. Asymptotic Efficiency . . . . . . . . . . . . . . . . . . . . . . . . . 166 4.2.1. Pitman Asymptotic Relative Efficiency . . . . . . . . . . . . 166 4.2.2. Bahadur Efficiency . . . . . . . . . . . . . . . . . . . . . . . 168 4.2.3. Approximations to the Power Function: Comparisons . . . 174 4.3. Exact and Asymptotic Moments: Comparison . . . . . . . . . . . . 175 4.3.1. Under the Null Hypothesis . . . . . . . . . . . . . . . . . . 176 4.3.2. Under Contiguous Alternative Hypotheses . . . . . . . . . . 185 4.3.3. Corrected Phi-divergence Test Statistic . . . . . . . . . . . 189 4.4. A Second Order Approximation to the Exact Distribution . . . . . 190 4.5. Exact Powers Based on Exact Critical Regions . . . . . . . . . . . 194 4.6. Small Sample Comparisons for the Phi-divergence Test Statistics . 198 4.7. Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203 4.8. Answers to Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . 204 5 Minimum Phi-divergence Estimators

213

5.1. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213 5.2. Maximum Likelihood and Minimum Phi-divergence Estimators . . 215 5.2.1. Minimum Power-divergence Estimators in Normal and Weibull Populations . . . . . . . . . . . . . . . . . . . . . . 220 5.3. Properties of the Minimum Phi-divergence Estimator . . . . . . . . 224


xviii


5.3.1. Asymptotic Properties . . . . . . . . . . . . . . . . . . . . . 227 5.3.2. Minimum Phi-divergence Functional Robustness . . . . . . 233 5.4. Normal Mixtures: Minimum Phi-divergence Estimator . . . . . . . 235 5.5. Minimum Phi-divergence Estimator with Constraints: Properties . 244 5.6. Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 247 5.7. Answers to Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . 249 6 Goodness-of-fit: Composite Null Hypothesis

257

6.1. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 257 6.2. Asymptotic Distribution with Fixed Number of Classes . . . . . . 259 6.3. Nonstandard Problems: Test Statistics Based on Phi-divergences . 271 6.3.1. Maximum Likelihood Estimator Based on Original Data and Test Statistics Based on Phi-divergences . . . . . . . . 271 6.3.2. Goodness-of-fit with Quantile Characterization . . . . . . . 276 6.3.3. Estimation from an Independent Sample . . . . . . . . . . . 278 6.3.4. Goodness-of-fit with Dependent Observations . . . . . . . . 279 6.3.5. Goodness-of-fit with Constraints . . . . . . . . . . . . . . . 281 6.4. Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 282 6.5. Answers to Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . 284 7 Testing Loglinear Models Using Phi-divergence Test Statistics 297 7.1. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 297 7.2. Loglinear Models: Definition . . . . . . . . . . . . . . . . . . . . . 305 7.3. Asymptotic Results for Minimum Phi-divergence Estimators in Loglinear Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . 307 7.4. Testing in Loglinear Models . . . . . . . . . . . . . . . . . . . . . . 309


CONTENTS

xix

7.5. Simulation Study . . . . . . . . . . . . . . . . . . . . . . . . . . . . 325 7.6. Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 336 7.7. Answers to Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . 340 8 Phi-divergence Measures in Contingency Tables

351

8.1. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 351 8.2. Independence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 352 8.2.1. Restricted Minimum Phi-divergence Estimator . . . . . . . 354 8.2.2. Test of Independence . . . . . . . . . . . . . . . . . . . . . . 360 8.2.3. Multiway Contingency Tables . . . . . . . . . . . . . . . . . 368 8.3. Symmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 368 8.3.1. Test of Symmetry . . . . . . . . . . . . . . . . . . . . . . . 373 8.3.2. Symmetry in a Three-way Contingence Table . . . . . . . . 378 8.4. Marginal Homogeneity . . . . . . . . . . . . . . . . . . . . . . . . . 382 8.5. Quasi-symmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . 388 8.6. Homogeneity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 394 8.7. Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 399 8.8. Answers to Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . 400 9 Testing in General Populations

407

9.1. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 407 9.2. Simple Null Hypotheses: Wald, Rao, Wilks and Phi-divergence Test Statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 408 9.2.1. Confidence Regions . . . . . . . . . . . . . . . . . . . . . . . 420 9.3. Composite Null Hypothesis . . . . . . . . . . . . . . . . . . . . . . 420


xx


9.4. Multi-sample Problem . . . . . . . . . . . . . . . . . . . . . . . . . 430 9.5. Some Topics in Multivariate Analysis . . . . . . . . . . . . . . . . . 436 9.6. Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 437 9.7. Answers to Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . 440 References


459

1

Divergence Measures: Definition and Properties 1.1.

Introduction

Let X be a random variable taking values on a sample space X (usually X will be a subset of Rn , n-dimensional Euclidean space). Suppose that the distribution function F of X depends on a certain number of parameters, and suppose further that the functional form of F is known except perhaps for a finite number of these parameters; we denote by θ the vector of unknown parameters associated with F. Let (X , βX , Pθ )θ∈Θ be the statistical space associated with the random variable X, where βX is the σ-field of Borel subsets A ⊂ X and {Pθ }θ∈Θ a family of probability distributions defined on the measurable space (X , βX ) with Θ an open subset of RM0 , M0 ≥ 1. In the following the support of the probability distribution Pθ is denoted by SX . We assume that the probability distributions Pθ are absolutely continuous with respect to a σ-finite measure µ on (X , βX ) . For simplicity µ is either the Lebesgue measure (i.e., satisfying the condition Pθ (C) = 0, whenever C has zero Lebesgue measure), or a counting measure (i.e., there exists a finite or countable set SX with the property Pθ (X -SX ) = 0). In the following   if µ is the Lebesgue measure,  fθ (x) dPθ fθ (x) = (x) = Prθ (X = x) = pθ (x) if µ is a counting measure,  dµ  (x ∈SX ) © 2006 by Taylor & Francis Group, LLC

2


denotes the family of probability density functions if µ is the Lebesgue measure, or the family of probability mass functions if µ is a counting measure. In the first case X is a random variable with absolutely continuous distribution and in the second case it is a discrete random variable with support SX . Let h be a measurable function. Expectation of h (X) is denoted by

Eθ [h(X)] =

 Z   h(x)fθ (x)dx      

XX

h(x)pθ (x)

if µ is the Lebesgue measure, if µ is a counting measure.

x∈SX

Since Mahalanobis (1936) introduced the concept of distance between two probability distributions, several coefficients have been suggested in statistical literature to reflect the fact that some probability distributions are “closer together” than others and consequently that it may be “easier to distinguish” between a pair of distributions which are “far from each other” than between those which are closer. Such coefficients have been variously called measures of distance between two distributions (see Adhikari and Joshi, 1956), measures of separation (Rao, 1949, 1954), measures of discriminatory information (Chernoff, 1952, Kullback, 1959) and measures of variation-distance (Kolmogorov, 1963). Many of the currently used tests, such as the likelihood ratio, the chi-square, the score and Wald tests, can in fact be shown to be defined in terms of appropriate distance measures. While the cited coefficients have not all been introduced for exactly the same purpose, they have the common property of increasing as the two distributions involved are “further from each other”. In the following, a coefficient with this property will be called divergence measure between two probability distributions. Before introducing the families of divergence measures that will be used in later chapters for studying different statistical problems, we consider two classical and important distances: the Kolmogorov and Lévy distances. Our aim is to illustrate the important role that distance measures play in Probability and Statistics. Given two probability measures Pθ 1 and Pθ 2 with associated unidimensional distribution functions Fθ 1 and Fθ 2 , respectively, the Kolmogorov distance, introduced by Kolmogorov (1933), between Fθ 1 and Fθ 2 (or between Pθ 1 and Pθ 2 ) is


Divergence Measures: Definition and Properties

3

given by K1 (Fθ 1 , Fθ 2 ) = sup |Fθ 1 (x) − Fθ 2 (x)| .

(1.1)

x∈R

It is the well known Glivenko-Cantelli Theorem, based on the previous distance, which states that the empirical distribution function is a uniformly strongly consistent estimate of the true distribution function; i.e., given a random sample X1 , ..., Xn from a population with distribution function Fθ 0 , for any ε > 0 it holds lim Pr {K1 (Fn , Fθ 0 ) > ε} = 0, n→∞

where Fn is the empirical distribution function, i.e., n

Fn (x) =

1X I (xi ) , n i=1 (−∞,x]

and IA is the indicator function of the set A. On the other hand, the Lévy distance is K2 (Fθ 1 , Fθ 2 ) = inf {ε > 0 : Fθ 1 (x − ε) ≤ Fθ 2 (x) ≤ Fθ 1 (x + ε) , for all x} ; it assumes values on [0, 1] and it is not easy to compute. It is interesting to note that convergence in the Lévy-metric implies weak convergence for distribution function in R (Lukacs, 1975, p. 71). It is shift invariant, but not scale invariant. This metric was introduced by Lévy (1925, pp. 199-200). Some other results about probability metrics, relationships between K1 and K2 as well as other interesting results can be seen in Gibbs and Su (2002).

1.2.

Phi-divergence Measures between Two Probability Distributions: Definition and Properties

In this section we shall introduce different divergence measures; in all the cases it must be understood provided the integral exists. Kullback-Leibler divergence measure, between the probability distributions Pθ 1 and Pθ 2 , is ¶¸ · µ Z f (X) f (x) DKull (θ1 , θ2 ) = fθ 1 (x) log θ 1 dµ (x) = Eθ 1 log θ 1 , (1.2) fθ 2 (x) fθ 2 (X) X


4


which was introduced and studied by Kullback and Leibler (1951) and Kullback (1959). Jeffreys (1946) used a symmetric version of (1.2), J (θ1 , θ2 ) = DKull (θ1 , θ2 ) + DKull (θ2 , θ1 ) , as a measure of divergence between two probability distributions. This divergence measure is also called J-divergence. Rényi (1961) presented the first parametric generalization of (1.2), Z 1 Dr1 (θ1 , θ2 ) = r−1 log fθ 1 (x)r fθ 2 (x)1−r dµ(x) X

1 r−1

=

log Eθ 1

"µ

fθ 1 (X ) fθ 2 (X )

¶r−1 #

,

r > 0, r 6= 1.

Later, Liese and Vajda (1987) extended it for all r 6= 1, 0, by Z 1 Dr1 (θ1 , θ2 ) = r(r−1) log fθ 1 (x)r fθ 2 (x)1−r dµ(x) X

=

1 r(r−1)

log Eθ 1

"µ

fθ 1 (X ) fθ 2 (X )

¶r−1 #

,

(1.3)

r 6= 0, 1.

In the following, expression (1.3) will be referred as Rényi divergence. The cases r = 1 and r = 0 are defined by D11 (θ1 , θ2 ) = lim Dr1 (θ1 , θ2 ) = DKull (θ1 , θ2 ) r→1

and D01 (θ1 , θ2 ) = lim Dr1 (θ1 , θ2 ) = DKull (θ2 , θ1 ) , r→0

respectively. The divergence measure DKull (θ2 , θ1 ) is called the Minimum discrimination information between the probability distributions Pθ 1 and Pθ 2 . Other two well known parametric generalizations of (1.2) are the one called r-order and s-degree divergence measure, and the other called 1-order and s-degree divergence measure. They were given by Sharma and Mittal (1977), by    s−1 r−1 Z  1  Drs (θ1 , θ2 ) = (s−1) fθ 1 (x)r fθ 2 (x)1−r dµ(x) − 1  =


Ã X "  s−1 ¶r−1 #! r−1 µ fθ 1 (X ) 1  Eθ 1 − 1 , (s−1) fθ 2 (X )

(1.4)

Divergence Measures: Definition and Properties

for r, s 6= 1 and D1s (θ1 , θ2 ) = =

1 (s−1)





exp (s − 1)

Z





f (x) fθ 1 (x) log fθθ 1 (x) dµ(x) − 1 2

X · µ µ µ ¶¸¶ ¶ fθ 1 (X ) 1 −1 , (s−1) exp (s − 1) Eθ 1 log f (X )

5

(1.5)

θ2

for s 6= 1. It can be easily shown that i) lim Drs (θ1 , θ2 ) = rDr1 (θ1 , θ2 ) s→1

ii) lim Drs (θ1 , θ2 ) = D1s (θ1 , θ2 ) r→1

iii) lim D1s (θ1 , θ2 ) = DKull (θ1 , θ2 ), lim Dr1 (θ1 , θ2 ) = DKull (θ1 , θ2 ) . s→1

r→1

The Kullback-Leibler divergence measure is the most famous special case of the φ-divergence family of divergence measures defined simultaneously by Csiszár (1963) and Ali and Silvey (1966) . Definition 1.1 The φ-divergence measure between the probability distributions Pθ 1 and Pθ 2 is defined by Z

´ x) fθ 2 (x) dµ(x) X · µ ¶¸ fθ 1 (X ) = Eθ 2 φ f ( X ) , φ ∈ Φ∗

Dφ (Pθ 1 , Pθ 2 ) = Dφ (θ1 , θ2 ) =

fθ 2 (x)φ

³f

θ1(

(1.6)

θ2

where Φ∗ is the class of all convex functions φ (x) , x ≥ 0, such that at x = 1, φ (1) = 0, at x = 0, 0φ (0/0) = 0 and 0φ (p/0) = limu→∞ φ (u) /u. Remark 1.1 Let φ ∈ Φ∗ be differentiable at x = 1, then the function ψ (x) ≡ φ (x) − φ0 (1) (x − 1)

(1.7)

also belongs to Φ∗ and has the additional property that ψ0 (1) = 0. This property,


6


together with the convexity, implies that ψ (x) ≥ 0, for any x ≥ 0. Further, µ µ µ ¶ ¶¶ Z fθ 1 (x) fθ 1 (x) 0 Dψ (θ1 , θ2 ) = fθ 2 (x) φ − φ (1) −1 dµ(x) fθ 2 (x) fθ 2 (x) ¶ µ ZX fθ 1 (x) dµ(x) fθ 2 (x)φ = fθ 2 (x) X

= Dφ (θ1 , θ2 ) . Since the two divergence measures coincide, we can consider the set Φ∗ to be equivalent to the set © ª Φ ≡ Φ∗ ∩ φ : φ0 (1) = 0 . Kullback-Leibler divergence measure is obtained for ψ (x) = x log x − x + 1 or φ (x) = x log x. We can observe that ψ (x) = φ (x) − φ0 (1)(x − 1). We shall denote by φ any function belonging to Φ or Φ∗ . In the following table we present some important measures of divergence studied in the literature which are particular cases of the φ-divergence. More examples can be seen in Arndt (2001), Pardo, L. (1997a) and Vajda (1989, 1995). φ-function

Divergence

x log x − x + 1 − log x + x − 1 (x − 1) log x 2 1 2 (x − 1) (x−1)2 (x+1)2 −xs +s(x−1)+1 , 1−s

³

−r

Balakrishnan and Sanghvi (1968)

s 6= 1, ´−1/r ,r>0

1−x 1+x 2 − 2 (1−x)2 0≤a≤1 2(a+(1−a)x) , ax log x−(ax+1−a) log(ax+1−a) , a(1−a) xλ+1 −x−λ(x−1) , λ 6= 0, −1 λ(λ+1) a 1/a

|1 − x |

Kullback-Leibler (1959) Minimum Discrimination Information J-Divergence Pearson (1900), Kagan (1963)

,0 0, i = 1, ..., M − 1 . i=1

Let Pθ be a probability measure on (X , βX ) such that for every θ ∈ Θ, Pθ ({xi }) = pi if i = 1, ..., M − 1

Pθ ({xM }) = 1 −

M−1 X

pi = pM .

i=1

Let µ be the counting measure on (X , βX ) attributing mass one to every xi ∈ X . It is clear that Pθ is absolutely continuous with respect to the measure µ for all θ ∈ Θ, and dPθ fθ (xi ) = i = 1, ..., M. (xi ) = pi , dµ In this context the φ-entropy is given by φ

φ

H (p) = H (θ) =

M X

φ(pi )

i=1

being p = (p1 , ..., pM−1 , pM )T ∈ 4M .

Let Y1 = y1 , ..., Yn = yn be a random sample from the population Pθ . The likelihood associated with this random sample is given by fθ (y1 , ..., yn ) = pn1 1 ...pnMM being ni the number of times that Yj = xi , j = 1, ..., n. The maximum likelihood estimator of the probabilities pi is given by pbi = ni /n, i = 1, ..., M. Then the maximum likelihood estimators of θ and p are given respectively by: b = (b b = (b θ p1 , ..., pbM−1 )T and p p1 , ..., pbM )T .

In the next result we compute the Fisher information matrix for the considered multinomial model. Proposition 2.1 The Fisher information matrix in the multinomial model is given by IF (p) = IF (θ) = (i(r,s) )r,s=1,...,M−1 , where  1 1  + if r = s    pr pM r, s = 1, ..., M − 1. i(r,s) =   1   if r 6= s pM © 2006 by Taylor & Francis Group, LLC

Entropy as a Measure of Diversity: Sampling Distributions

77

Proof. The (r, s)th-element i(r,s) , of the Fisher information matrix, for r 6= s, is Z

∂ log fθ (x) ∂ log fθ (x) fθ (x) dµ (x) ∂θr ∂θs X µ ¶ ¶ µ M−1 M−1 P P ∂ log 1 − pk ∂ log 1 − pk M−1 P ∂ log pk ∂ log pk k=1 k=1 pk + pM = ∂p ∂p¶s ∂pr ∂ps r k=1 ¶µ µ 1 1 1 − pM = = − , r 6= s, r, s = 1, ..., M − 1, pM pM pM

i(r,s) =

and the element (r, r) has the expression i(r,r)

µ

¶ ∂ log fθ (x) 2 = fθ (x) dµ(x) ∂θr X µ ¶ ¶¶2 µ µ 2 M−1 M−1 P P ∂ ∂ log pk pk + log 1 − pk pM = ∂pr k=1 ¶ ∂pr µ k=1 µ ¶ 1 2 1 2 1 1 = pr + − pM = + . pr pM pr pM Z

In the following Proposition the inverse of the Fisher information matrix is obtained. Proposition 2.2 The inverse of the Fisher information matrix is given by IF (θ)−1 = diag (θ) − θθT being θ = (p1 , ..., pM−1 )T . Proof. By multiplying the matrix IF (p)−1 = IF (θ)−1 with elements i−1 (r,s)

=

(

pr (1 − pr ) if r = s r, s = 1, ..., M − 1, if r 6= s −pr ps

by the matrix IF (p) = IF (θ) , with elements


78


i(r,s) =

 1 1  +    pr pM     1 pM

if r = s r, s = 1, ..., M − 1 , if r 6= s

we get the identity matrix. To check this equality we note that the (r, s)thelement of the product is given by ¶ µ M−1 P 1 1 1 = 1, if r = s, (−pr pj ) + pr (1 − pr ) + pM ps pM j=1 j6=r

M−1 P

1 1 (−pr pi ) + pr (1 − pr ) − pr ps pM pM i=1

i6=r,s

µ

1 1 + ps pM

¶

= 0, if r 6= s.

Then IF (θ)−1 IF (θ) = I (M−1)×(M−1) and IF (θ)−1 is the inverse of IF (θ) . Theorem 2.3 The analogical estimator, H φ (b p), obtained by replacing the p0i s by its relative frequencies, pbi , in a random sample of size n, verifies ´ ¡ ¢ √ ³ φ L n H (b p) − H φ (p) −→ N 0, σφ2 (p) n→∞

whenever σφ2 (p) > 0, being

σφ2 (p) = S T Σp S =

M X i=1

s2i pi −

ÃM X i=1

si pi

!2

,

with S = (s1 , ..., sM )T = (φ0 (p1 ), ..., φ0 (pM ))T and Σp = diag(p) − ppT . Proof. By Theorem 2.1, we have σφ2 (p) = T T IF (θ)−1 T , where T = (t1 , ..., tM−1 )T = (φ0 (p1 ) − φ0 (pM ), ..., φ0 (pM−1 ) − φ0 (pM ))T . Now we must show that T T IF (θ)−1 T = S T Σp S. In fact,


(2.7)


T T IF (θ)−1 T

79

= (φ0 (p1 ), ..., φ0 (pM−1 ))IF (θ)−1 (φ0 (p1 ), ..., φ0 (pM−1 ))T − 2(φ0 (p1 ), ..., φ0 (pM−1 ))IF (θ)−1 (φ0 (pM ), ..., φ0 (pM ))T + (φ0 (pM ), ..., φ0 (pM ))IF (θ)−1 (φ0 (pM ), ..., φ0 (pM ))T M−1 M−1 P P M−1 P = pi φ0 (pi )2 − pi pj φ0 (pi )φ0 (pj ) i=1 i=1 j=1 µ ¶ M−1 M−1 P P 0 0 0 0 − 2 φ (pM ) pi φ (pi ) − φ (pM )(1 − pM ) pi φ (pi ) i=1

+ (1 − pM )φ0 (pM )2 − φ0 (pM )2 =

M−1 P i=1

pi φ0 (pi )2 − M−1 P

i=1

pi pj

i=1 j=1

pi pj φ0 (pi )φ0 (pj )

i=1 j=1

pi φ0 (pi ) + pM φ0 (pM )2 − p2M φ0 (pM )2 i=1 µM ¶2 M P P 0 2 0 = pi φ (pi ) − pi φ (pi ) = S T Σp S. −

2φ0 (pM )pM

M−1 P M−1 P

M−1 P M−1 P

i=1

i=1

Remark 2.4 For the (h, φ)-entropies we have √ ³ ´ ¡ ¢ n L φ p) − Hhφ (p) −→ N 0, σφ2 (p) , µM ¶ Hh (b n→∞ P h0 φ(pi ) i=1

where σφ2 (p) was given in (2.7).

A complete study of the statistics based on (h, φ)-entropies, in simple random sampling as well as in stratified random sampling, can be seen in Salicr´ u et al . (1993). Remark 2.5 Consider an experiment whose outcomes belong to one of M (M ≥ 2) mutually exclusive and exhaustive categories, E1 , ..., EM and let pi (0 < pi < 1) be the probability that the outcome belongs to the ith category (i = 1, ..., M). Here PM i=1 pi = 1. Suppose that the experiment is performed n times, Y1 , ..., Yn , and that the n outcomes are independent. Furthermore, let Nj denote the number of P these outcomes belonging to category Ej (j = 1, ..., M), with M j=1 Nj = n. The random variable (N1 , ..., NM ) has a multinomial distribution with parameters n and p =(p1 , ..., pM )T .


80


If we define (i Tj

(i

(

=

1 if Yi ∈ Ej , i = 1, ..., n, j = 1, ..., M, 0 otherwise

(i

the vectors (T1 , ..., TM ), i = 1, ..., n, are functions of Yi and take values on the set {(1, 0, ..., 0) , ..., (0, ..., 0, 1)} . We also have: (1

(1

Y1 → (T1 , ..., TM ) (2 (2 Y2 → (T1 , ..., TM ) ... ... ... (n (n Yn → (T1 , ..., TM ) | {z } Ã n X |

i=1

(i T1 , ...,

n X i=1

≡ ≡ ≡ (i TM

{z (N1 , ..., NM )

! }

M(1; p1 , ..., pM ) M(1; p1 , ..., pM ) ... M(1; p1 , ..., pM )

≡ M (n; p1 , ..., pM ),

where ≡ is used to denote “distributed as” and M(n; p1 , ..., pM ) denotes the multinomial distribution with parameters n and p1 , ..., pM . On the other hand, given the sequence of M-dimensional random variables {U n }n∈N with U n = (U1n , ..., UMn )T where E [U n ] = µ

and

Cov [U n ] = Σ

the Central Limit Theorem states that the random vector   n n X X 1 1 Un =  U1j , ..., UMj  n n j=1

verifies

j=1

¢ L √ ¡ n U n − µ −→ N (0, Σ) . n→∞


∀n,


If we denote A=

µ

NM − npM N1 − np1 √ √ , ..., n n

¶

81

,

we have A =

µ

1 √ n

µ

µn ¶ ¶¶ P (i 1 (i T1 − np1 , ..., √ TM − npM n i=1 i=1 n P

µ ¶ ¶¶ µ n µ n √ √ 1 P (i 1 P (i = n T − p1 , ..., n T − pM n i=1 1 n i=1 M

and applying the Central Limit Theorem we have µ ¶ ¢ ¡ N1 − np1 NM − npM L √ √ −→ N 0, Σp , , ..., n→∞ n n where

Σp = diag (p) − ppT . Therefore,

¢ ¡ √ L n (b p − p) −→ N 0, Σp . n→∞

Theorem 2.4 If we assume that σφ2 (p) = 0, the analogical estimator H φ (b p), obtained by 0 replacing the pi s by their relative frequencies, pbi , obtained from a random sample of size n, verifies r ³ ´ X L φ φ 2n H (b p) − H (p) −→ βi Zi2 n→∞

i=1

where Z1 , ..., Zr are i.i.d. normal random variables with mean zero and variance 1, r = rank(Σp BΣp ) and βi0 s are the eigenvalues of the matrix BΣp being B the M × M matrix B = diag(φ00 (pi )i=1,...,M ).

Proof. In Theorem 2.2 it was established that the random variables ³ ´ b − H φ (θ) 2n H φ (θ) © 2006 by Taylor & Francis Group, LLC

and

√ b √ b n(θ − θ)T A n(θ − θ)

82


have the same asymptotic distribution. The elements of the matrix A are given by aij = φ00 (pi )δij + φ00 (pM ) i, j = 1, ..., M − 1 because aij =

=

Z

X

M−1 P k=1

×

µ ¶ ∂fθ (x) ∂fθ (x) ∂ 2 fθ (x) 0 00 φ (fθ (x)) + φ (fθ (x)) dµ (x) ∂θi ∂θj ∂θi ∂θj µ ¶ M−1 P ∂pk ∂pk ∂ 2 pk 0 φ00 (pk ) + φ (pk ) + φ00 (1 − pk ) ∂pi ∂pj ∂pi ∂pj k=1

µ ¶ µ ¶ M−1 M−1 P P ∂ 1− pk ∂ 1 − pk k=1

∂pi

k=1

∂pj

∂2 +

µ ¶ M−1 P 1− pk k=1

∂pi ∂pj

φ0

µ ¶ M−1 P 1− pk . k=1

b − θ) and denoting 1 = (1, .., 1)T , we have b − θ)T A(θ Then, for L = (θ

b − θ)T diag(φ00 (p1 ), ..., φ00 (pM−1 ))(θ b − θ) L = (θ T 00 T b b + (θ − θ) φ (pM )11 (θ − θ) M−1 M−1 M−1 P 00 P P φ (pi )(b pi − pi )2 + φ00 (pM ) (b pi − pi ) (b pj − pj ) = i=1

=

M P

i=1

i=1

j=1

φ00 (pi )(b pi − pi )2 = (b p − p)T B(b p − p),

being B = diag(φ00 (pi )i=1,...,M ). By applying Remark 2.5 we obtain √ L n(b p − p) −→ N(0, Σp ). n→∞ Then using the same argument as in Theorem 2.2 we have the desired result.

Remark 2.6 In the rest of the book the following result is important (Lemma 3, p. 57, in Ferguson, 1996): Let X be a k-variate normal random variable with mean vector 0 and variance-covariance matrix Σ. Then X T X is distributed chi-squared with r degrees of freedom if and only if Σ is a projection of rank r. Since Σ is symmetric and squared it is a projection if Σ2 = Σ, and therefore rank(Σ) =trace(Σ).


83


Corollary 2.2 If we consider u = (1/M, ..., 1/M)T , we get ´ 2nM ³ φ L φ (b p ) − H (u) −→ χ2 . H n→∞ M−1 φ00 (1/M)

(2.8)

Proof. In this case we have Σu = diag(u) − uuT . The random variables ³ ´ 2nM H φ (b p) − H φ (u) and (b p − u)T Mn diag(φ00 (1/M))(b p − u)

have the same asymptotic distribution, then ´ √ √ 2nM ³ φ φ (b p ) − H (u) and n(b p − u)T diag(u−1 ) n(b p − u) H 00 φ (1/M) have the same asymptotic distribution. But √ √ n(b p − u)T diag(u−1 ) n(b p − u) = X T X, where

√ p − u) X = n diag(u−1/2 )(b

and the asymptotic distribution of X is normal with mean vector 0 and variancecovariance matrix Σ∗ = diag(u−1/2 )Σu diag(u−1/2 ). Now we are going to establish that the matrix Σ∗ is a projection of rank M − 1. It is clear that ¢ ¡ Σ∗ = diag(u−1/2 )Σu diag(u−1/2 ) = I − diag(u−1/2 )uuT diag u−1/2 ,

where I denotes the M × M identity matrix. Then we have

Σ∗ Σ∗ = I − diag(u−1/2 )uuT diag(u−1/2 ) − diag(u−1/2 )uuT diag(u−1/2 ) × diag(u−1/2 )uuT diag(u−1/2 )diag(u−1/2 )uuT diag(u−1/2 ). But uT diag(u−1/2 )diag(u−1/2 )u = 1, then Σ∗ Σ∗ = I − diag(u−1/2 )uuT diag(u−1/2 ) = Σ∗ .


84


Now rank (Σ∗ ) = trace(diag(u−1/2 )Σu diag(u−1/2 )) = trace(diag(u−1 )Σu ), and

−1

diag(u

Therefore,



   )Σu =   

1 1 ) −M (1 − M 1 1 −M (1 − M ) ... ... ... ... 1 1 −M −M

1 ... ... −M 1 ... ... −M ... ... ... 1 1 −M ... (1 − M ) 1 1 ... −M (1 − M )

trace(diag(u−1 )Σu ) = M(1 −

1 M)



   .  

= M − 1.

Applying Remark 2.6 we get that the asymptotic distribution of the random variable X T X is chi-square with M − 1 degrees of freedom.

Remark 2.7 In the case of Shannon’s entropy we have L

2n (log M − H(b p)) −→ χ2M−1 , n→∞

and for the (h, φ)-entropies, Salicr´ u et al. (1993), ³ ´ 2nM L φ φ H (b p ) − H (u) −→ χ2M−1 . h h 00 0 n→∞ φ (1/M)h (Mφ(1/M)) Corollary 2.2 as well as Remark 2.7 permit constructing tests of goodness of fit to one distribution.

2.4.1.

Test of Discrete Uniformity

If we want to test H0 : p1 = ... = pM =

1 M

versus

H1 : ∃ i, j ∈ {1, ..., M} such that pi 6= pj

we can use the test statistic Z6 ≡


´ 2nM ³ φ φ (b p ) − H (u) H φ00 (1/M)

(2.9)


85

whose asymptotic distribution, by Corollary 2.2, is chi-square with M − 1 degrees of freedom. We should reject the null hypothesis if Z6 > χ2M−1,α . It is also possible to get the asymptotic power of the test for p∗ =(p∗1 , ..., p∗M )T 6= (1/M, ..., 1/M)T . This power is given by ´ ³ ¡ φ ¢ φ (u) > χ2 ∗ H (b p ) − H /H : p = p βφ,n (p∗ ) = Pr φ002nM 1 M−1,α (1/M) ¶¶ µ √ µ 00 φ (1/M)χ2M−1,α n φ (u) − H φ (p∗ ) = 1 − Φn σφ (p∗ ) + H , 2nM for a sequence of distributions Φn (x) tending uniformly to the standard normal distribution Φ (x). The expression of σφ (p∗ ) is given in Theorem 2.3. It is clear that lim βφ,n (p∗ ) = 1 − Φ (−∞) = 1,

n→∞

i.e., the test is consistent. The power obtained above can be used to determine approximately the sample size n∗ required to achieve a desired power β ∗ against a given alternative p∗ =(p∗1 , ..., p∗M )T 6= (1/M, ..., 1/M)T . The power is approximately Ã √ !! Ã φ00 (1/M) χ2M−1,α n 1−Φ . + H φ (u) − H φ (p∗ ) σφ (p∗ ) 2nM If we wish the power to be equal to β ∗ we must solve the equation Ã Ã √ !! 00 (1/M ) χ2 φ n M−1,α β∗ = 1 − Φ + H φ (u) − H φ (p∗ ) . σφ (p∗ ) 2nM It is not difficult to check that the sample size, n∗ , is the solution of the following quadratic equation ³ ´2 n2 H φ (u) − H φ (p∗ ) −nσφ (p∗ )(Φ−1 (1−β ∗ ))2 +2Sφ (M) (H φ (u)−H φ (p∗ )) = 0, where

Sφ (M) =


φ00 (1/M)χ2M−1,α 2M

.

86


The solution is given by ¶ µ q 1 2 2 B ± B − 4A Sφ (M) n = 2A ∗

where B = σφ (p∗ )(Φ−1 (1 − β ∗ ))2 and A = H φ (u) − H φ (p∗ ). The test of discrete uniformity given in this section permits testing the goodness of fit of any distribution, i.e., H0 : F (x) = F0 (x)

∀x ∈ R,

through testing H0 : pi = 1/M, i = 1, ..., M

versus

H1 : ∃ i, j / pi 6= pj ,

by simply partitioning the range of the random variable into M intervals with equal probabilities under F0 , and then to test if the observations are from the given discrete uniform distribution. We use the test statistic Z6 given in (2.9) and we should reject the null hypothesis if Z6 > χ2M−1,α . The particularization of the results obtained in the previous Section for general populations to multinomial populations is vital in order to study the behavior of the diversity among one or various populations. Based on last results we will have the possibility to test: i) if the diversity of a population is some specified value, ii) if the diversities of two populations are equal, iii) if the diversities of several populations are equal, and finally, iv) if the population is homogeneous (discrete uniformity) or not. Remark 2.8 It is interesting to observe that different diversity measures (entropy measures) may give different diversity ordering. For instance, if we consider two populations characterized by the following probability vectors p = (0.3, 0.24, 0.22, 0.12, 0.09, 0.03)T and q = (0.36, 0.21, 0.16, 0.12, 0.08, 0.07)T , we have HGS (p) = 0.7806 > HGS (q) = 0.7750



87

while H(p) = 1.61315 < H(q) = 1.631338.

Hence it appears the necessity for the definition of ordering of discrete probability distributions. In this sense we have the following results: Definition 2.1 The probability distribution p is majorized by the probability distribution q, and we denote, p ≺m q, iff r X i=1

p(i) ≤

r X i=1

q(i) , r = 1, ..., M − 1

where p(1) ≥ p(2) ≥ ... ≥ p(M) and q(1) ≥ q(2) ≥ ... ≥ q(M) . Roughly speaking, the sentence “p is majorized by the probability distribution q”, means that q is less “spread out” than p and the population represented by p has more diversity than the population represented by q. Definition 2.2 A real function f defined on the simplex 4M is said to be schur-concave on 4M if p ≺m q on 4M =⇒ f(p) ≥ f(q),

and f is strictly schur-concave on 4M if strict inequality holds whenever p ≺m q and p is not a permutation of q. Some entropy measures are schur-concave functions on 4M . Some examples are: Shannon, Havrda and Charvat, Arimoto, Rényi, etc. These results could be seen in Nayak (1983) as well as in Marshall and Olkin (1979).

2.5.

Maximum Entropy Principle and Statistical Inference on Condensed Ordered Data

The Maximum Entropy Principle (MEP) stated for the first time by Jaynes (1957) has had very important applications in Statistical Mechanics, Statistics, Geography, Spatial and Urban Structures, Economics, Marketing, Systems


88


Analysis, Actuarial Science, Finance, Computer Science, Spectrum Analysis, Pattern Recognition, Search Theory, Operations Research, etc. (e.g., Kapur (1982, 1989)). In Statistics many problems have been studied on the basis of the MEP, but perhaps one of the most important results can be found in the nonparametric density function estimation. This problem was first introduced by Theil and Laitinen (1980) and has been developed by Theil and O’Brien (1980), Theil and Fiebig (1981), Theil and Kidwai (1981a, b), Theil and Lightburn (1981), Rodr´ıguez and Van Ryzin (1985), etc. In relation with the problem of parametric estimation two interesting papers are: Kapur and Kesaven (1992) and Jiménez and Palacios (1993). These authors present the problem in the following way. Let Y1 , . . . , Yn be independent random variables with a common distribution Fθ ∈ {Fθ }θ∈Θ , being Θ an open subset of RM0 . Let Y(1) ≤ Y(2) ≤ . . . ≤ Y(n) be the ordered sample, define Y(0) = −∞, Y(n+1) = ∞ and consider p1 (θ) = Fθ (Y(1) ), pi (θ) = Fθ (Y(i) ) − Fθ (Y(i−1) ),

i = 2, . . . , n,

pn+1 (θ) = 1 − Fθ (Y(n) ).

e ∈ Θ, verifying In this context they propose to estimate θ by means of the value θ ! Ã n+1 n+1 X X e log pi (θ) e = max − − pi (θ) pi (θ) log pi (θ) . (2.10) θ∈Θ

i=1

i=1

This method can be justified by the fact that the order statistics Fθ (Y(1) ), . . . , Fθ (Y(n) )

divide the interval (0, 1) into (n + 1) random subintervals with equal expected length. Therefore, maximizing (2.10) is equivalent to choosing θ so that the 1 random variables pi (θ) are close as possible to their expected value n+1 . Also, Ranneby (1984) introduced an interesting method to estimate θ by using a statistical information theory approach. He proposes to minimize a Kullback—Leibler b of θ which maximizes divergence, or, equivalently, to select the value θ Sn (θ) =

1 n+1

n+1 X

log(pi (θ)(n + 1)),

i=1

and it is called the maximum spacing estimate.



89

The transformed spacings pi (θ) may be used to get test statistics for testing H0 : F = F0 ,

(2.11)

see, e.g., Greenwood (1946), Pyke (1965), Cressie (1976) and Hall (1986). Gebert and Kale (1969) indicate that test statistics of this type are useful for detecting departures from hypothesized density functions. If we assume that the above hypothesis is simple, applying the probability integral transformation F0 to the sample values Y1 , . . . , Yn permits assuming, without loss of generality, that the null hypothesis (2.11) specifies the uniform distribution on the unit interval. Then we wish to test H0 : G(u) = u, 0 ≤ u ≤ 1,

(2.12)

where G(u) is the distribution of Ui = F (Yi ), i = 1, . . . , n. If U(1) < U(2) < . . . < U(n) are the order statistics from the transformed sample, the spacings are defined by Vi = U(i) − U(i−1) and it is possible to define statistics to test (2.12). See, for instance, Kimball, (1947), Darling (1953), Kale and Godambe (1967), Kirmani and Alam (1974), Pyke (1965), Kale (1969), Sethuraman and Rao (1970) and Rao and Sethuraman (1975). Tests based symmetrically on spacings, namely, P of the form Tn = n1 ni=1 h(nVi ) are more common among these. Kuo and Rao (1984) demonstrated that among a wide class of such tests, the Greenwood test statistic obtained with h(x) = x2 has maximum efficacy. Jammalamadaka et al . (1989) established the asymptotic distribution for h(x) = x log x under the null hypothesis given in (2.12) and the alternatives given by the densities fn (x) = 1 + n−1/4 l(x), 0 ≤ x ≤ 1 where l(·) is assumed to be square integrable and continuously differentiable on [0, 1]. The above tests are known in the literature as test statistics based on firstorder spacings. Several authors have proposed generalizations of first-order uniform spacings. Cressie (1976) considers tests statistics of the form Sn(m)

=

n+2−m X i=1

g

³

(m) (n + 1)Vi (m)

´

,

m ≤ n + 1,

where g is a “smooth” function, and Vi = U(i−1+m) −U(i−1) , i = 1, . . . , n+2−m, are the mth-order spacings or mth-order gaps. See, e.g., Hall (1986), Stephens (1986), Cressie (1979), Del Pino (1979) among others. In this sense, the results for testing normality and uniformity, obtained by Vasicek (1976) and Dudewicz


90


and Van der Meulen (1981) respectively, are important. These authors propose the following test statistics based on Shannon’s entropy Hn,m =

1 n

n X

log

i=1

³ n ´ (Y(i+m) − Y(i−m) ) . 2m

Two modified versions of the above test statistic are given by Ebraimi et al. (1994). It is important to remark that previous methods for estimation and testing use all the sample information; however it is very hard to find the asymptotic distribution of the estimators even in the case of specific parametric models. This is obviously not the case for the maximum likelihood principle or the method of moments. Now we propose a method to decrease the amount of sample information in order to be able to have estimators with known asymptotic distribution when the MEP is used. We consider a population with probability density function fθ (x) . Suppose that we have a random sample Y1 , . . . , Yn , and we only observe the order statistics Y([ Mn ]+1) , Y([ 2n ]+1) , . . . , Y([ (M−1)n ]+1) , M

(2.13)

M

where [x] is the integer part function. We form M cells having boundaries −∞ = c0,n < c1,n < . . . < cM−1,n < cM,n = ∞, where ci,n = Y([ in ]+1) . The observed frequency Nin in the ith-cell (ci−1,n , ci,n ] is M

in ]−[ (i−1)n nonrandom, Nin = [ M M ]. The vectors of sample and population quantiles 1 2 M−1 of orders M , M , . . . M are cn = (c1,n , . . . , cM−1,n )T and c = (c1 , . . . , cM−1 )T , respectively. The ith “estimated cell probability” is therefore

pi (cn ; θ) = Fθ (ci,n ) − Fθ (c(i−1),n )

(2.14)

where θ = (θ1 , . . . , θM0 )T and Fθ is the probability distribution function associated with fθ (x). These are random, unlike the cell frequencies. For θ = θ0 (true 1 value of the parameter), we write pi = pi (c; θ0 ) = Mi − i−1 M = M , i = 1, . . . , M. In this context in Menéndez et al . (1997a) the following definition was given. Definition 2.3 Let Y1 , . . . , Yn be a random sample from a population described by the random



91

variable X with probability density function fθ (x) . The MEP estimator based on the order statistics given in (2.13) is defined by bSh = arg max H(p(cn ; θ)), θ θ∈Θ

where

H(p(cn ; θ)) = −

M X

(2.15)

pi (cn ; θ) log pi (cn ; θ),

i=1

and pi (cn ; θ), i = 1, ..., M, are defined in (2.14). We remark that the approach that takes some but not all the sample quantiles has been proposed by Bofinger (1973) for problems of testing goodness of fit with Pearson test statistic. Menéndez et al. (1997a) adapted this idea to problems of point estimation. Note also that the approach taken here is applied also to “multiple type II censoring”, in which observations between several sets of sample percentiles are unavailable. It is necessary only to take each unobserved interpercentile group as a cell. This is conceptually quite similar to the generality of censoring allowed in the procedures of Turnbull and Weiss (1978). Another interesting approach in which only a relatively small number of order statistics are used in testing for goodness-of-fit is given in Weiss (1974). He establishes that U([nδ ]) , U(2[nδ ]) , . . . , U(k(n)[nδ ]) , with δ ∈ (3/4, 1) and k(n) = [n1−δ ], are asymptotically sufficient and can be assumed to have a joint normal distribution for all asymptotic purposes. Under some assumptions, that can be seen in Menéndez et al . (1997a), it can bSh , verifying the condition (2.15), verifies the be established that the estimator, θ following properties, bSh converges in probability to θ0 . a) θ

bSh is asymptotically efficient, i.e., b) θ µ ¶ ∂ log L (θ) −1 b θSh = θ0 + n B 0 + oP (n−1/2 ), ∂θ θ=θ 0

where B 0 is a matrix of constants which may depend on θ0 and log L (θ) is given by M X log L (θ) = Nin log pi (cn ; θ). i=1


92


L bSh − θ0 ) −→ c) n1/2 (θ N(0, IF (θ0 )−1 ). n→∞

Based on sample quantiles, the MEP can be used to test if data come from a given parametric model. In this sense in the cited paper of Menéndez et al. (1997a) it was established that ³ ´ L bSh )) −→ Tn = 2n log M − H(p(cn ; θ χ2M−M0 −1 . n→∞



2.6.

93

Exercises

1. Let X = {C1 , ..., CM } be a set of M categories and let p =(p1 , ..., pM )T be a probability measure defined on X . Find the expected distance between two categories drawn at random when the distance is defined as zero if they belong to the same category and unity otherwise. 2. Find the asymptotic variance of the entropies of Rényi, Havrda-Charvat, Arimoto and Sharma-Mittal when the probability vector p =(p1 , ..., pM )T is b=(b replaced by the estimator p p1 , ..., pbM )T based on a simple random sample of size n. 3. Consider the hypothesis H0 : p1 = ... = pM = 1/M and the test statistics based on the entropies appearing in Exercise 2. Find, in each case, the corresponding expression of the test statistic.

4. The following data correspond to the occupational status by race and year in Walton County, Florida. Occupational Status

(Observed proportions) White White Black Black (1880) (1885) (1880) (1885)

Professional Manager, clerical, proprietor Skilled Unskilled Laborer Farmer

0.029 0.019 0.086 0.053 0.455 0.358

0.093 0.099 0.040 0.073 0.517 0.178

0.00 0.007 0.007 0.046 0.776 0.164

0.00 0.00 0.00 0.049 0.896 0.055

Sample size

209

151

152

144

Source: Nayak, T. K. (1983), Ph. D. Dissertation, University of Pittsburgh.

a) Compute the diversity indexes based on Shannon’s entropy, as well as the asymptotic variances, for the four populations. b) Test uniformity in every population with significance level α = 0.05. c) From data in the previous table one can observe that diversity for the black population has decreased over time. Check this conjecture by using a test statistic with significance level α = 0.05 and a 95% confidence interval. d) What happens for the white population? Do a similar study to the one given in c).


94


5. We have obtained from welding dive, four random samples corresponding to bounded areas of 1 squared meter from 5 meters deep. The sampling stations are found in the Spanish coast of Catalonia, from south to the north, from the delta of Ebro river (1) through the area of Maresme (2 and 3) to the Rosas bay (4). The two stations in the area of Maresme have a granite sandy bottom (thick); while the stations situated in delta of Ebro river and in the Rosas bay have a chalky sandy bottom (thin). The descriptions of the communities are listed in Tables 2.4, 2.5, 2.6 and 2.7. Sampling station 1: Riomar (Delta of Ebro river) date: 30/06/91 Type of sand: Thin Table 2.4 Species

Frequency

Lentidium mediterraneum Tellina tennis Spisula subtruncata Donax trunculus Mactra corallina Dosina lupinus Carastoderma flancum Otherwise

216 189 69 51 45 40 37 76

Total

723

Source: Pardo, L., Calvet, C. and Salicr´ u, M. (1992).

Sampling station 2: Canet de Mar (El Maresme) Date: 25/03/91 Type of sand: Granite Table 2.5 Species

Frequency

Spisula sultrancata Glycymeris glycymeris Alcanthocardia tuberculada Donax variegatus Donacilla cornea Chamelea gallina Callista chione Otherwise

345 52 36 36 34 28 18 71

Total

620




95

Sampling station 3: Malgrat de Mar (El Maresme) Date: 17/04/91 Type of sand: Granite Table 2.6 Species

Frequency

Chamelea gallina Spisula subtruncata Callista chione Donax variegatus Glycymeris glycymeris Dosina exoleta Corbula gibba Otherwise

440 377 38 35 33 31 8 41

Total

1003


Sampling station 4: Rosas (Rosas bay) Date: 09/03/91 Type of sand: Thin Table 2.7 Species

Frequency

Ceratoderma glaucum Spisula subtruncata Tapes decussatus Venerupis aurea Loripes lacteus Chamelea gallina Acanthoc. tuberculate Otherwise

134 93 69 59 52 33 32 135

Total

607


a) Compute the indexes of diversity based on the Gini-Simpson index for the four stations. b) Test the uniformity in every station with significance level α = 0.05. c) Give a 95% confidence interval for the diversity in the four populations.


96


6. Let X1 , ..., Xn a random sample of a normal population, with mean µ and variance σ2 . By using asymptotic properties of Shannon’s entropy statis√ L tic, prove the following relation 2n log σσb → N (0, 1) , where σ b is the n→∞ maximum likelihood estimator of σ.

7. Two classes of drugs, which are supposed to have some effect in a disease, were tested on 288 individuals. The individuals were classified in four groups according to class of drug they had received. Groups 1 and 2: different drug, group 3: both drugs, group 4: no drug. The records are presented in the following table. Group 1 12 10 13 13 12 10

Si2 =

1 ni −1

Pni

j=1 (xji

− xi )2

1.867

Group 2 12 4 11 7 8 10 12 5 9.696

Group 3 13 14 14 17 11 11 14 14 3.714

Group 4 1 8 9 9 9 4 0 1 16.411

Use the test statistic SSha given in (2.4) to test the equality of variances in the four groups with significance level α = 0.05. 8. The following values correspond to a simple random sample from a population with distribution function F and support the real line: -7.238, -0.804, -0.44, 0.18, -0.02, -1.08, 1.327, 1.98, -0.73, -0.27, -0.32, 0.58, -2.308, -4.37, 0.307, 4.72, 0.32, 0.124, -4.41, 1.98, -0.73, -1.27, -0.32, 0.58, -2.34, -8.19, 12.99, 1.51, 1.09, -4.37. Suppose that F is the Cauchy distribution function, whose probability density function is f(x) =

1 1 π 1 + x2

x∈R

and consider the partition, (−∞, −1], (−1, 0], (0, 1], (1, ∞). a) Calculate Shannon’s entropy associated with the previous partition. b) Based on Shannon’s entropy test the hypothesis that the previous observations come from a Cauchy population with significance level α = 0.05.



97

4 4 3 5 T c) Calculate the asymptotic power at the point p∗ = ( 16 , 16 , 16 , 16 ) . 2

9. Let X be a random variable whose density function is f (x) = |x| e−x , x ∈ R. a) Find an equiprobable partition with four classes in R. Use this partition to give a test statistic, based on Shannon’s entropy, for testing if the observations -1.1, 0.42, -0.25, 4.73, -7.2, 12.3, -0.22, 1.8, -0.7, -1.9, 4.77, 2.75, 0.01, -4.2, -0.1, -0.01, 3.23, -0.97, -0.12, -3.2, -0.15, -12.3, 0.75, -0.4, -7.4, 0.27, 1.51, -2.4, -2.67, 0.32 are from the population described by X with significance level α = 0.05. b) Find the asymptotic power at the point p∗ = (1/2, 1/4, 1/8, 1/8)T . 10. We have the following observations from a population with distribution function F and support the real line: 1.491, 2.495, 3.445, 1.108, 3.916, 1.112, 3.422, 2.278, 1.745, 3.131, 0.889, 2.099, 2.693, 2.409, 1.030, 1.437, 0.434, 1.655, 2.130, 1.967, 1.126, 3.113, 2.711, 0.849, 1.904, 1.570, 3.313, 2.599, 2.263, 2.208, 1.6771, 3.173, 1.235, 2.034, 4.007, 2.653, 2.269, 1.774, 4.077, 0.733, 0.061, 1.961, 1.916, 2.607, 2.060, 1.444, -0.357, 0.211, 2.555, 1.157. a) Test the hypothesis, with a significance level α = 0.05, if the previous observations come from a normal distribution with mean µ = 2 and variance σ 2 = 1.1, using Shannon’s entropy and a partition with 6 equiprobable classes. 1 2 3 3 3 3 T b) Obtain the asymptotic power at the point p∗ = ( 15 , 15 , 15 , 15 , 15 , 15 ) .

11. Let X1 , ..., Xn be a random sample of size n from an exponential distribution of parameter θ. Using Shannon’s entropy, give an asymptotic test statistic to test H0 : θ = θ0 versus H1 : θ 6= θ0 . 12. Using the entropy of Havrda and Charvat, H s (θ) . ´ √ ³ b − H s (θ) , where θ b is a) Find the asymptotic distribution of n H s (θ) the maximum likelihood estimator of θ. √ b) Obtain the asymptotic distribution of n (H s (b p) − H s (p)) , as a particular case of the result obtained in a). 13. We consider Laplace’s distribution f (x, θ1 , θ2 ) =


1 exp(−θ2−1 |x − θ1 |), 2θ2

−∞ < x < ∞, θ1 ∈ R, θ2 > 0

98


whose variance is 2θ22 . a) Find the expression of the entropy of Havrda and Charvat for this distribution and derive from it the expression of Shannon’s entropy. b) Test, using the entropy of Havrda and Charvat, H0 : θ2 = θ∗ versus H1 : θ2 6= θ∗ .

c) Derive a procedure for testing that the variances of s Laplace independent populations are equal. Give a procedure based on the entropy of Havrda and Charvat. 14. We consider the population given by the following density function fθ (x) =

θ2θ xθ+1

x ≥ 2,

θ > 0.

a) Find the test based on Shannon’s entropy for testing H0 : θ = θ0

against H1 : θ 6= θ0 .

b) Using the test obtained in a) and the observations 2.2408, 5.8951, 6.0717, 3.6448, 2.8551, 4.4065, 14.4337, 3.0338, 2.0676, 2.6155, 2.7269, 5.1468, 2.2178, 2.0141, 2.3339, 2.6548, 5.0718, 2.8124, 2.0501, 13.6717, test H0 : θ = 2 versus H1 : θ 6= 2 with significance level α = 0.05. 15. Find the acceptance region of SSha given in (2.4) for equality of two variances and compare it with the acceptance region of the test statistics given by Bartlett and Lhemann.

2.7.

Answers to Exercises

1. If we consider the distance, ( 1, Ci 6= Cj , Pr(d(Ci , Cj ) = 1) = pi pj , d (Ci , Cj ) = 0 Ci = Cj Pr (d (Ci , Ci ) = 0) = p2i we have Ep [d] =

M M X X i=1 j=1 i6=j


pi pj = 1 −

M X i=1

p2i = HGS (p).


99

2. It is a simple exercise to get from Remark 2.4 the following table Measure Havrda-Charvat

Rényi

Sharma-Mittal (1, s)

Asymptotic variance µ M ¶2 ! M P P s 2s−1 pi − pi

³

s s−1

´2

³

1 r−1

´2 µ P M

i=1

i=1

1 (t−1)2

Arimoto

Ã

Ãµ M P

i=1

i=1

pri

pi

pi2r−1

i=1

2 t −1

¶µ M P

i=1

1 t

pi

−

µM P

³

r r−1

pri

i=1

¶2(t−1)

−

µ ¶ M P pi log pi exp 2 (s − 1) i=1 Ã µM ¶2 ! M P P 2 × pi (log pi ) − pi log pi i=1

Sharma-Mittal (r, s)

¶−2 Ã M P

¶2 !

µM P

i=1

1 t

pi

¶2t !

i=1

´2 µ P M

i=1

pri

Ã ¶ 2(s−r) r−1 M P

i=1

p2r−1 i

−

µM P

i=1

pri

¶2 !

3. From Corollary 2.2 the expression of the statistic given in (2.8) for the different entropy measures is Measure Havrda-Charvat

Rényi

Statistics 2n (H s (b p) − H s (u)) (−s) M 1−s ¢ 2n ¡ 1 Hr (b p) − Hr1 (u) (−r)

Arimoto

2n (t H (b p) −t H (u)) (−t−1 ) M t−1

Sharma-Mittal (1, s)

2n (H1s (b p) − H1s (u)) −M 1−s

Sharma-Mittal (r, s)

2n (Hrs (b p) − Hrs (u)) −rM 1−s


100


4. a) First, we obtain the sample entropies, H(b pi ), i = 1, ..., 4, as well as their 2 asymptotic variances, σ (b pi ), i = 1, ..., 4, for the four populations. Based on Theorem 2.3 and taking into account that in the case of Shannon’s entropy φ(x) = −x log x, we have 2 σH (p) =

M X i=1

pi (log pi )2 −

ÃM X

pi log pi

i=1

!2

.

In the case of the first population we get H(b p1 ) = −0.029 log 0.029 − ... − 0.358 log 0.358 = 1.2707 2 σH (b p1 ) = 0.029(log 0.029)2 + ... + 0.358(log 0.358)2 − H(b p1 )2 = 0.6822. In a similar way we obtain the results for the other populations:

White (1880) White (1885) Black (1880) Black (1885)

H(b pi ) 1.2707 1.4179 0.7053 0.4060

2 (b σH pi ) 0.6822 0.7133 0.8706 0.7546

b) Now we have to test H0 : p1 = p2 = p3 = p4 = (1/6, ..., 1/6)T . We know that χ25,0.05 = 11.07 and, on the other hand,

White (1880) White (1885) Black (1880) Black (1885)

b ) H (p i 1.2707 1.4179 0.7053 0.4060

χ2

log M − M−1,α 2ni 1.7653 1.7551 1.7553 1.7533

Therefore we should reject the null hypothesis. c) In this case we must test H0 : H(p3 ) = H(p4 )


versus H1 : H(p3 ) 6= H(p4 ),


101

and we use the test statistic ³ ´ √ b ) p3 ) − H (p n3 n4 H(b 4 Z2 = q 2 (b 2 (b n4 σH p3 ) + n3 σH p4 )

whose value is Z2 = 2.857. On the other hand the critical region is given by (−∞, −1.96) ∪ (1.96, ∞) and we should reject the null hypothesis. A 95% confidence interval for H(p3 ) − H(p4 ) is given by (H(b p3 ) − H(b p4 )) ± z0.025

µ

p3 ) σ2 (b p4 ) σ2 (b + n3 n4

¶1/2

.

After some calculations the following interval is obtained (0.0934, 0.5039). This interval does not contain the value zero and then we can conclude that the change is significant. d) In a similar way we get that the value of the test statistic is Z2 = −1.64 and the confidence interval is (−2.5072, 2.128). We can not conclude that the change is significant. 5. a) The expression of Gini-Simpson index is HGS (p) = 1 −

M X

p2i .

i=1

Then 189 2 69 2 2 HGS (b p1 ) = 1 − ( 216 723 ) − ( 723 ) − ( 723 ) − 40 2 37 2 76 2 − ( 723 ) − ( 723 ) − ( 723 ) = 0.8078.

¡

¢ 51 2 723

45 2 − ( 723 )

In a similar way we can get the expression of the Gini-Simpson index for the other populations HGS (b p2 ) = 0.6576


HGS (b p3 ) = 0.6598

HGS (b p4 ) = 0.8428.

102


Using Exercise 2 (Havrda-Charvat with s = 2) the expression of the asymptotic variance is given by  Ã M !2  M X X 2 σGS (p) = 4  p3i − p2i  ; i=1

i=1

then,

2 2 2 2 σGS (b p1 ) = 0.0419, σGS (b p2 ) = 0.2312, σGS (b p3 ) = 0.0882, σGS (b p4 ) = 0.016.

b) From Exercise 3 we have nM

ÃM X i=1

pb2i − M −1

The numerical value of the test statistic following table: Station Riomar Canet del Mar Malgrat de Mar Rosas

!

L

−→ χ2M−1 .

n→∞

for each population appears in the Statistic 389.15 1078.40 1726.40 155.97

But χ27,0.05 = 14.076 and we should reject the homogeneity in each one of the four populations. c) A 100 (1 − α) % confidence interval for the Gini-Simpson index in each one of the four populations is given by µ ¶ σ b(b p) σ(b p) HGS (b p) − √ zα/2 , HGS (b p) + √ zα/2 . n n

In the following table we have the confidence intervals for each one of the four populations Station Riomar Canet del Mar Malgrat de Mar Rosas


Confidence Interval (0.8047, 0.8108) (0.6394, 0.6758) (0.6543, 0.6652) (0.8415, 0.8440)


103

On the basis of the confidence intervals obtained we have the following relations for the diversities: Rosas diversity is greater than Riomar diversity and Riomar diversity is greater than Canet diversity and Canet diversity equal to Malgrat de Mar diversity. Then in our case the stations with a bottom of thin sand present greater diversity than the stations with a bottom of granitic sand. A study of this problem was made by Pardo, L. et al. (1992). 6. For Shannon’s entropy we have H(σ 2 ) ≡ H(N(µ, σ 2 )) = log(σ2 2πe)1/2 , then ³ ´ ¡ ¢¢ L √ ¡ n H(b σ2 ) − H σ2 −→ N 0, T T IF (θ)−1 T , n→∞

where T

T

= (t1 , t2 ) is given in the example of Section 2.3.4 and θ =(µ, σ2 ), t1 =

∂H(σ2 ) = 0, ∂µ

t2 =

and 2 −1

IF (µ, σ )

=

Ã

∂H(σ2 ) 1 = 2, 2 ∂σ 2σ

σ2 0 0 2σ4

!

.

Then T T IF (θ)−1 T = 1/2 and ¡ ¢¢ L √ ¡ n H(b σ2 ) − H σ2 −→ N (0, 1/2) . n→∞

Substituting H(σ2 ) and H(ˆ σ2 ) by their expressions we have the statement. 7. Based on (2.4), we reject the null hypothesis H0 : σ12 = σ22 = σ32 = σ42 if r

1X Z4 = nj 2 j=1

In our case we have

Ã

log σ bj2

r ¡ ¢ nj Q − log σ bj2 N j=1

!2

> χ2r−1,α .

b22 = 8.4840, σ b32 = 3.25, σ b42 = 14.359, σ b12 = 1.5556, σ

and Z4 = 10.32. On the other hand we have χ23,0.05 = 7.82 and then we should reject the null hypothesis.


104


8. a) It is immediate to show that p1 = Pr((−∞, −1]) = p2 = Pr((−1, 0]) = p3 = Pr((0, 1]) = p4 = Pr((1, ∞)) = 1/4 and H(u) = log 4 = 1.3863, where u =(1/4, 1/4, 1/4, 1/4)T b) For testing H0 : The observations come from a Cauchy population, we could test H0 : p1 = p2 = p3 = p4 = 1/4. The critical region is 2n (log 4 − H(b p)) > χ2M−1,0.05 . In our case, pb1 = 0.333, pb2 = 0.267, pb3 = 0.2, pb4 = 0.2

and then H(b p) = 1.3625. Therefore,

p)) = 1.4268. Z6 = 2n (log 4 − H(b On the other hand χ23,0.05 = 7.815. Then the null hypothesis should be not rejected. ¡ 4 4 3 5 ¢T c) If we denote p∗ = 16 , 16 , 16 , 16 , we have H(p∗ ) = 1.3705 and σ2 (p∗ ) = 0.0308. ¡ 4 4 3 5 ¢T , 16 , 16 , 16 is given by The power at p∗ = 16 ¢ ¡ β30 (p∗ ) = Pr 2n(log 4 − H(b p)) > χ23,0.05 ¢ ¡ = Pr Ã−2nH(b p) > χ23,0.05 − 2n log 4 ! √ −χ23,0.05 + 2n(log 4 − H(p∗ )) n ∗ √ p) − H(p )) < = Pr σ(p∗ ) (H(b 2 nσ(p∗ ) µ ¶ −7.815 + 83.1776 − 82.2301 ≈ Pr Z < = 0.0002, 2 × 301/2 × 0.1755

where Z is a standard normal variable. 9. a) First we obtain the equiprobable partition. The first value will be the solution of the equation Z a1 1 2 − xe−x dx = , 4 −∞ © 2006 by Taylor & Francis Group, LLC


105

and this is given by a1 = −0.8305. The second point is obviously a2 = 0 and the third one, by symmetry, is a3 = 0.8325. Then the partition is given by (−∞, −0.8305] , (−0.8305, 0] , (0, 0.8305] , (0.8305, ∞) . On the basis of this partition we have 1 8 5 7 pb1 = , pb2 = , pb3 = , pb4 = . 3 30 30 30

Then

H(b p) = − and

4 X i=1

pbi log pbi = 1.3569

2n (log 4 − H(b p)) = 2 × 30 (log 4 − 1.3569) = 1.7636. But χ23,0.05 = 7.8115 and there is not statistical evidence to reject the null hypothesis. b) The power at p∗ = (1/2, 1/4, 1/8, 1/8)T is given by ³ √ ¡ ¢´ ∗) − H(p = 0.6591 β30 (p∗ ) ≈ Φ σ(p30∗ ) log 4 − 7.8115 60

where

1 1 1 1 1 1 1 1 H(p∗ ) = − log − log − log − log = 1.2130 2 2 4 4 8 8 8 8 and 2

∗

σ (p ) =

4 X

p∗i (log p∗i )2

i=1

−

Ã 4 X

p∗i log p∗i

i=1

!2

= 0.3303.

10. a) First we obtain the equiprobable partition. The first value, a1 , must verify 1 Pr (−∞ < N (2, 1.1) < a1 ) = ; 6 this is equivalent to Pr(−∞ < Then a1 verifies Φ(


a1 − 2 1 X −2 < )= . 1.1 1.1 6 a1 − 2 ) = 1/6, 1.1

106


where Φ denotes the distribution function of the standard normal distribution. Then a1 = 2 − 0.97 × 1.1 = 0.933 and the first interval is I1 = (−∞, 0.933] ; the interval I2 is given by I2 = (0.933, a2 ] where a2 is obtained in such a way that the probability of I2 is 16 . Then a2 = 1.527. In the same way we obtain I3 = (1.527, 2] , I4 = (2, 2.473] , I5 = (2.473, 3.067] and I6 = (3.067, ∞) . It is easy to get pb1 = 0.14, pb2 = 0.18, pb3 = 0.18, pb4 = 0.18, pb5 = 0.14 and pb6 = 0.18.

Then

H(b p) = −

6 X i=1

and

pbi log pbi = 1.7852,

Z6 = 2n(log M − H(b p)) = 1.1259. On the other hand χ25,0.05 = 11.07 and we should reject the null hypothesis. 1 2 3 3 3 3 T b) It is clear that the power at p∗ = ( 15 , 15 , 15 , 15 , 15 , 15 ) is

Ã√ 50 ¡ log 6 − β50 (p∗ ) ≈ Φ σ(p∗ )

11.07 100

! ¢ − H(p∗ ) = 0.0901

where 1 1 log 15 − H(p∗ ) = − 15

2 15

2 3 3 log 15 − 4 × ( 15 log 15 ) = 1.7367, n = 50

and σ(p∗ )

=

Ã

6 P

pi (log pi

i=1

)2

−

µ

6 P

pi log pi

i=1

¶2 !1/2

= 0.2936.

11. Shannon’s entropy for the exponential distribution is given by H (θ) = 1 − log θ, whereas the maximum likelihood estimator and the Fisher information are n θb = Pn

i=1 Xi


=X

−1

and IF (θ) = θ−2 .


107

We know that ´ ¡ ¢ √ ³ b L n H(θ) − H (θ0 ) −→ N 0, σ2 (θ0 ) , n→∞

with 2

σ (θ0 ) = IF (θ0 ) Then

−1

µ

∂H (θ0 ) ∂θ

¶2

= θ02 θ0−2 = 1.

√ L n(1 − log θb − 1 + log θ0 ) −→ N (0, 1) , n→∞

and

¢ L ¡ √ n log Xθ0 −→ N (0, 1) , n→∞

i.e., the null hypothesis should be rejected if ¯√ ¢¯ ¡ ¯ n log Xθ0 ¯ > zα/2 .

12. a) A first order Taylor expansion gives b = H s (θ0 ) + H s (θ)

M0 X ∂H s (θ0 ) i=1

∂θi

°´ ³° ° °b (θbi − θi0 ) + o °θ − θ0 ° .

°´ ³° √ b √ L ° °b But n(θ − θ0 ) −→ N(0, IF (θ0 )−1 ), then n o °θ − θ0 ° = oP (1) . n→∞ Therefore, the random variables ´ √ b √ ³ s b n H (θ) − H s (θ0 ) and T T n(θ − θ0 )

have the same asymptotic distribution, where T T = (t1 , ..., tM0 ) with Z ∂fθ 0 (x) ∂H s (θ0 ) 1 = sfθ 0 (x)s−1 dµ(x) s 6= 1. ti = ∂θi 1−s X ∂θi Therefore, ´ √ ³ s b L n H (θ) − H s (θ0 ) −→ N(0, T T IF (θ0 )−1 T ). n→∞

b) In this case we have µM−1 ¶ P s 1 s pi − 1 + H (p) = 1−s i=1

TT

= (t1 , ..., tM−1 ) =


1 s 1−s (pM

s−1 s 1−s (p1

− 1),

s−1 s−1 − ps−1 M , ..., pM−1 − pM ).

108


Then T T IF (p)−1 T

³ −1 s−1 s−1 s−1 T (ps−1 1 , ..., pM−1 )IF (p) (p1 , ..., pM−1 )

=

s2 (1−s)2

−

−1 s−1 s−1 s−1 2(ps−1 1 , ..., pM−1 )IF (p) (pM , ..., pM )

+

´ −1 s−1 s−1 s−1 (ps−1 , ..., p )I (p) (p , ..., p ) . F M M M M

In a similar way to Theorem 2.3 we have  Ã M !2  M 2 X X s  T T IF (p)−1 T = p2s−1 − psi  . i 2 (1 − s) i=1 i=1 13. a) The entropy of Havrda and Charvat for Laplace distribution is given by µZ ¶ 1 1 s H s (θ2 ) = exp(− |x − θ |)dx − 1 1 s 1−s θ2 R (2θ2 ) 1 = s 6= 1, s > 0, (21−s θ21−s s−1 − 1), 1−s and Shannon’s entropy by 1 (21−s θ21−s s−1 − 1) = log 2θ2 + 1. s→1 1 − s

lim H s (θ2 ) = lim

s→1

b) We want to test H0 : θ2 = θ∗ versus H1 : θ2 6= θ∗ and this is equivalent to test H0 : i.e.,

´ ¢ 1 ¡ 1−s 1−s −1 1 ³ 1−s ∗ 1−s −1 2 θ2 s − 1 = (θ ) s −1 , 2 1−s 1−s H0 : H s (θ2 ) = H s (θ ∗ ) .

For this purpose we use the test statistic ´ √ ³ s b n H (θ2 ) − H s (θ ∗ ) L −→ N (0, 1) Z1 = ∗ n→∞ σ (θ ) where σ2 (θ ∗ ) = T T IF (θ∗ )−1 T



and T

T

=

µ

∂H s (θ2 ) ∂H s (θ2 ) , ∂θ1 ∂θ2

¶

θ2 =θ∗

¶ µ 21−s ∗ −s (θ ) . = 0, s

Therefore, σ2 (θ∗ ) = T T IF (θ ∗ )−1 T Ã !µ µ ¶ ¶T 21−s ∗ −s 21−s ∗ −s (θ∗ )2 0 = 0, (θ ) (θ ) 0, s s 0 (θ∗ )2 22(1−s) ∗ −2s+2 = (θ ) . s2 Let us observe that −1

IF (θ2 )

=

Ã

θ22 0 0 θ22

!

because log fθ1 ,θ2 (x) ∂ log fθ1 ,θ2 (x) ∂θ2 ∂ 2 log fθ1 ,θ2 (x) ∂θ22 Then,

= − log 2θ2 −

,

1 |x − θ1 | θ2

1 1 + 2 |x − θ1 | θ2 θ2 ´ 1 ³ = 2 1 − θ22 |x − θ1 | . θ2 = −

¸ µ ¶ ∂ 2 log fθ1 ,θ2 (X) 1 2 1 IF (θ2 ) = −E = − 2 1 − E [|X − θ1 |] = 2 2 θ2 ∂θ2 θ2 θ2 ·

because E [|X − θ1 |] = θ2 . On the other hand we have ∂ log fθ1 ,θ2 (x) 1 x − θ1 =− ; ∂θ1 θ2 |x − θ1 | therefore,

It is also clear that


"µ ¶ # 1 θ1 − X 2 1 E − = 2. θ2 |X − θ1 | θ2 ¸ ∂ 2 log fθ1 ,θ2 (X) = 0, E ∂θ1 ∂θ2 ·

109

110


hence IF (θ2 ) =

Ã

θ2−2 0 0 θ2−2

!

.

It is not difficult to get that  Ã ! 1−s b √ 1  θ2 Z1 = n − 1 . 1−s θ∗

The null hypothesis should be rejected if ¯ ¯ Ã ! 1−s ¯ ¯ b ¯√ ¯ ¯ n 1  θ2  − 1 ¯¯ > zα/2 ¯ ∗ 1−s θ ¯ ¯ where

with x(1/2)

¯ ¯ ¯¢ 1 ¡¯¯ x1 − x(1/2) ¯ + ... + ¯xn − x(1/2) ¯ , θb2 = n the sample median.

c) In this case we want to test H0 : σ12 = ... = σs2 , i.e., (1)

(s)

H0 : 2(θ2 )2 = ... = 2(θ2 )2 ,

³ ´ (i) (i) where θ1 , θ2 represent the parameters in the ith-population.

For testing (2.16) we are going to test (1)

(s)

H0 : H s (θ2 ) = ... = H s (θ2 ). In this case the test statistic is Z4 =

s X i=1

being D=

´2 ni ³ s b(i) L ( θ ) − D −→ χ2s−1 , H 2 (i) n→∞ σ2 (θb ) 2

s X ni 1 (i) H s (θb2 ). s (i) P ni 2 b i=1 σi (θ2 ) b(i) ) i=1 σ 2 (θ 2


(2.16)


111

14. a) First, we obtain Shannon’s entropy associated with the considered population. We have Z ∞ θ2θ 1 θ2θ θ H (θ) = − log dx = − log + 1 + . θ+1 θ+1 x x 2 θ 2 Then we can test H0 : θ = θ0 by testing H0 : H (θ) = H (θ0 ) and in this case we know that √ ³ ´ n L b − H (θ0 ) −→ H(θ) N (0, 1) Z1 = n→∞ σ (θ0 )

with σ2 (θ0 ) = t2 IF (θ0 )−1 , being µ ¶ ∂H (θ) (1 + θ0 ) t= =− , ∂θ θ02 θ=θ0 We have

IF (θ0 ) = θ0−2 .

n . θb = P n log x2i i=1

Then we should reject the null hypothesis if ¯ √ ³ ´¯¯ ¯ n b ¯ ¯ ¯ σ (θ0 ) H(θ) − H (θ0 ) ¯ > zα/2 .

b) In this case we have θb = 1.7 and √ ³ ´ n b − H (θ0 ) = 2.9904. H(θ) σ (θ0 )

As z0.025 = 1.96 we should reject the null hypothesis.

15. We are going to obtain the acceptance region SSha given in (2.4) for r = 2. In this case, we have n ¡ ¢2 SSha = 12 n1 log σ b12 − nN2 log σ b22 b12 − nN1 log σ ¡ ¢2 o b22 − nN1 log σ + n2 log σ b12 − nN2 log σ b22 ³ ´ ³ ´ σ b2 2 σ b2 2 = 12 (n1 nN2 log σb12 + n2 nN1 log σb12 ) 2 2 µ 2 ¶2 σ b 1 n2 = n2N log 12 . σ b2 © 2006 by Taylor & Francis Group, LLC

112


Therefore the acceptance region is 0
χ2M−1,α / H1 : p = p∗ βn,φ (p∗1 , ..., p∗M ) = Pr Tnφ (b ³ ³ 00 ´´ √ 2 ∗ , p0 ) = 1 − Φn σ1 (1p? ) φ2√(1) χ − nD (p , φ n M−1,α

where Φn (x) tends uniformly to the standard normal distribution function Φ (x) and σ12 (p∗ ) is given in (3.10). This completes the proof. Based on this result an approximation of the power function of the test, with decision rule given in (3.9), at p∗ = (p∗1 , ..., p∗M )T , is βn,φ (p∗1 , ..., p∗M ) ' 1 − Φ

³

1 σ1 (p∗ )

³

φ00 (1) 2 √ χ 2 n M−1,α

−

where Φ is the standard normal distribution function.

´´ √ nDφ (p∗ , p0 ) ,

It is clear that limn→∞ βn,φ (p∗1 , ..., p∗M ) = 1, i.e., the test is consistent. Corollary 3.2 In a similar way to the previous theorem it is possible to establish that ¢ L √ ¡ b) − Dφ (p0 , p∗ ) −→ N (0, σ2 (p∗ )) n Dφ (p0 , p n→∞

where σ2 (p∗ ) is given by

σ22 (p∗ )

=

M X i=1


p∗i s2i

−

ÃM X i=1

p∗i si

!2

,

122


with

µ

p0 si = φ i∗ pi

¶

p0 − i∗ φ0 pi

µ

p0i p∗i

¶

, i = 1, ..., M.

Corollary 3.3 a) In the case of the Kullback-Leibler divergence measure, we have µ ¶ ¶2 µM P ∗ p∗i 2 p∗i log 0 = − pi log 0 , pi pi i=1 ¡ ¢ i=1 M p0 2 P i σ22 (p∗ ) = ∗ − 1. p i=1 i M P

σ12 (p∗ )

p∗i

b) In the case of (h, φ)-divergences we have

³ ¡ ¢ ³ p∗ ´´2 p∗i h0 Dφ (p∗ , p0 ) φ0 pi0 i i=1 ¶ µ ³ M ¢ 0 p∗i ´ 2 P ∗ 0¡ ∗ 0 − pi h Dφ (p , p ) φ p0

σ12 (p∗ ) =

M P

i

i=1

and σ22 (p∗ ) = −

M P

p∗i

i=1 µM P

i=1

³¡ ¡ ¢¢ ³ ³ p0i ´ h0 Dφ (p0 , p∗ ) φ p∗ − i

p0i 0 p∗i φ

³ 0 ´´´2 pi p∗i

³ 0 ´´ ¢¢ ³ ³ p0i ´ ¡ ¡ p pi h0 Dφ (p0 , p∗ ) φ p∗ − φ0 p∗i i

i

¶2

.

Proof. Part a) is a simple exercise. We prove part b). If we consider the function ÃM µ ¶! X yi g (y1 , ..., yM ) = h p0i φ 0 pi i=1 b, we get and its first Taylor’s expansion around p∗ = (p∗1 , ..., p∗M )T at p g (b p1 , ..., pbM ) = g (p∗1 , ..., p∗M ) +

¶ M µ X ∂g i=1

∂yi y=p∗

(b pi − p∗i ) + o (kb p − p∗ k) .

Therefore the random variables

M √ X √ n (g (b p1 , ..., pbM ) − g (p∗1 , ..., p∗M )) and n i=1


Ãµ

! ¶ ∂g ∗ (b pi − pi ) ∂yi y=p∗

Goodness-of-fit: Simple Null Hypothesis

123

have the same asymptotic distribution, i.e., M ´ √ ³ h √ X 0 h ∗ 0 n Dφ (b p, p ) − Dφ (p , p ) and n t∗i (b pi − p∗i ) i=1

have the same asymptotic distribution, where µ ¶ ¡ ¢ 0 p∗i ∗ 0 ∗ 0 , i = 1, ..., M. ti = h Dφ (p , p ) φ p0i In a similar way to Theorem 3.1 one can get σ22 (p∗ ).

In order to produce a nontrivial asymptotic power, Cochran (1952) suggested using a set of local alternatives contiguous to the null hypothesis as n increases. Consider the multinomial probability vector √ pn ≡ p0 + d/ n,

P where d = (d1 , ..., dM )T is a fixed M × 1 vector such that M j=1 dj = 0, and recall that n is the total-count parameter of the multinomial distribution. As n → ∞ the sequence of probability vectors {pn }n∈N converge to the probability vector ¢ ¡ p0 in the null hypothesis at the rate O n−1/2 . We say that √ H1,n : p = pn ≡ p0 + d/ n

(3.11)

is a sequence of contiguous alternative hypotheses, here contiguous to the null hypothesis p0 . Our interest is to study the asymptotic behavior of the test power under contiguous alternative hypotheses, i.e., ´ ³ φ 0 2 (3.12) p, p ) > χM−1,α /H1,n : p = pn . βn,φ (pn ) = Pr Tn (b

In what follows we show that under the alternative hypotheses H1,n , as n → ∞, Tnφ (b p, p0 ) converges in distribution to a noncentral chi-square random variable with noncentrality parameter δ, where δ is given in Theorem 3.3, and M − 1 de¡ ¢ grees of freedom χ2M−1 (δ) . Lehmann (1959) argues that contiguous alternative hypotheses are the only alternative hypotheses of interest.

It is interesting to observe that if we consider a point p∗ 6= p0 , we can write ¡√ ¡ ¢¢ √ = p0 + n−1/2 n p∗ − p0 , and if we define pn ≡ p0 + d/ n with d = ¢ √ ¡ ∗ n p − p0 we can use the expression given in (3.12) to get an approximation of the power function at p∗ . p∗


124


Theorem 3.3 The asymptotic distribution of the φ-divergence test statistic Tnφ (b p, p0 ), under the contiguous alternative hypotheses (3.11), is noncentral chi-square with M − 1 degrees of freedom and noncentrality parameter δ given by ³¡ ¢ ´ −1 d. δ = dT diag p0 Proof. We can write ¢ ¢ √ √ ¡ √ ¡ √ b − p0 = n p n (b p − pn ) + n pn − p0 = n (b p − pn ) + d, and under the hypothesis H1,n : p = pn ≡ p0 + √dn , we have ´ ³ √ L n (b p − pn ) −→ N 0, Σp0 n→∞

and

´ ³ ¢ L √ ¡ 0 b n p − p −→ N d, Σp0 . n→∞

By (3.8) in Theorem 3.1 we have Tnφ (b p, p0 ) = Then

³° ° ´ √ 2n 0 0 T √ 0 0 °2 ° b . (b p , p ) = n(b p − p ) C n(b p − p ) + o p − p D φ φ00 (1)

´ ´ ³¡ ¢ ³¡ ¢ ¢´T √ ³ ¢´ √ ³ −1/2 ¡ −1/2 ¡ b − p0 b − p0 n diag p0 n diag p0 p p °¢2 °¢2 ¡° ¡° b − p0 ° = X T X+o °p b − p0 ° + o °p

Tnφ (b p, p0 ) =

being³ the asymptotic distribution of X multivariate normal with mean vector ¡ 0 ¢−1/2 ´ d and variance-covariance matrix diag p diag

³¡ ¢ ´ ´ ³¡ ¢ −1/2 −1/2 p0 Σp0 diag p0 .

Applying Lemma (Ferguson 1996 p. 63) “Suppose that X is N (µ, Σ) . If Σ is idempotent of rank M and Σµ = µ, the distribution of X T X is noncentral chi-square with M −1 degrees of freedom and noncentrality parameter δ = µT µ”, the result follows if we establish that ´ ´ ³¡ ¢ ´ ³¡ ¢ ´ ³¡ ¢ ³¡ ¢ −1/2 −1/2 −1/2 −1/2 Σp0 diag p0 diag p0 d = diag p0 d, diag p0 © 2006 by Taylor & Francis Group, LLC


125

because in Theorem 3.1 it was proved that ³¡ ¢ ´ ´ ³¡ ¢ −1/2 −1/2 L = diag p0 Σp0 diag p0 ³¡ ¢ ´ ¡ ¢ ´ ³¡ ¢ −1/2 T −1/2 = I−diag p0 p0 p0 diag p0 , is a projection of rank M − 1.

We denote by U ³¡ ¢ ´ ´ ³¡ ¢ ´ ³¡ ¢ ´ ³¡ ¢ −1/2 −1/2 −1/2 −1/2 diag p0 Σp0 diag p0 diag p0 d = Ldiag p0

and we have ´ ¡ ¢ ´´ ³¡ ¢ ´ ³¡ ¢ ³ ³¡ ¢ −1/2 T −1/2 −1/2 p0 p0 diag p0 diag p0 d U = I−diag p0 ³¡ ¢ ´ ³¡ ¢ ´ ¡ ¢ ³¡ ¢ ´ −1/2 −1/2 T −1 = diag p0 d−diag p0 p0 p0 diag p0 d ³¡ ¢ ´ −1/2 = diag p0 d,

³¡ ¢ ´ ¡ ¢T −1 d =0. since p0 diag p0

The asymptotic distribution of the φ-divergence test statistic in a stratified random sampling was obtained in Morales et al . (1994) .

3.3.

Phi-divergence Test Statistics under Sparseness Assumptions

In the previous Section we have established the asymptotic distribution of the φ-divergence test statistic assuming M fixed and letting the sample size n tend to infinity. A different approach lets both n and M tend to infinity, but at the same rate so that n/M remains constant. In Bishop et al . (1975, p. 410) the following can be seen “...One reason for looking at this special type of asymptotic comes from practical considerations. Typically, multinomial data arrive in the form of a cross-classification of discrete variables. In many situations there are a large number of variables which can be used to cross-classify each observation, and if all of these variables are used the data would be spread too thinly over the cells in the resulting multidimensional contingency table. Thus if the investigator uses a subset of the variables to keep the average number of observations from becoming too small, he is in effect choosing M so that n/M


126


is moderate.” Holst (1972) pointed out the following “it is rather unnatural to keep M fixed when n → ∞; instead we should have that limn→∞ M = ∞.” For equiprobable cells Hoeffding (1965, p. 372) showed that Pearson test statistic is much more powerful than likelihood ratio test statistic against near alternative ¯ ¯ ¢ ¡ p satisfying maxi ¯pi − M −1 ¯ = O M −1 , for moderate significance levels, with n/M moderate and M large. In this Section we assume that the partition size M depends on the sample size n, i.e., M = Mn with limn→∞ Mn = ∞. The intervals or classes depend, in general, on n. For this reason we denote the partition by Pn and its elements by Enj , j = 1, ..., Mn . Then we have Pn = {En1 , ..., EnMn } , with 1 < Mn ≤ ∞. We denote γn = n/Mn and we assume that lim γn = γ ∈ (0, ∞) .

n→∞

(3.13)

Let pn = (pn1 , ..., pnMn )T be the vector of probabilities verifying pni = Pr (Eni ) , bn = (b i = 1, ..., Mn , and let p pn1 , ..., pbnMn )T be the relative frequency vector based on a random sample of size n, Y1 , ..., Yn , i.e., pbni = Nni /n being Nni the number of elements in the class Eni , i = 1, ..., Mn . We write N n = (Nn1 , ..., NnMn ) to denote the vector of absolute frequencies. Assumption (3.13) is realistic in goodness-of-fit testing, where partitions Pn are usually specified so that all observed frequencies Nni = nb pni were approximately the same and relatively large. If we denote the desired level of cell frequencies by γ then we obtain from the condition Nni = γ + OP (n) , and from the law of large numbers condition pbni = pni + oP (1) , that the ratio n/Mn , the expected number of observations in each cell, must be close to γ. On the other hand, (3.13) with small γ means that many cells are sparsely frequented. Thus (3.13) is also known as the sparseness assumption; see Section 4.3 in Read and Cressie (1988) devoted to testing under this assumption. The first results of this kind were published by Holst (1972). He developed a Poissonization technique leading under (3.13) to asymptotic distribution of Pearson test statistic and likelihood ratio test statistic. The so called Poissonization technique is originating from the fact that the vector Z = (Z1 , ..., ZM ) of independent Poisson random variables with E [Z] = (λ1 , ..., λM )T is under the condition Z1 + ... + ZM = n multinomially distributed with parameters n and p = (λ1 /n, ..., λM /n)T . Thus if M = Mn and the expectation of Z = Z n is npn


127


then the conditional distribution of Z n given Zn1 + ... + ZnMn = n coincides with the unconditional distribution of above defined N n . Morris (1975) derived a central limit law for a sequence of X 2 and G2 test statistics, each measuring the fit of a known multinomial to a count data set with M cells. He gave conditions under which X 2 and G2 , suitably standardized, approach normality as M tends to infinity. The closeness of these distributions to the normal, for selected sparse multinomial, was examined in a simulation study by Koehler and Larntz (1980). Dale (1986) obtained the asymptotic distribution of X 2 and G2 test statistics for product-multinomial model and later Morales et al . (2003) obtained, for this model, the asymptotic distribution of the φ-divergence test statistic. Holst (1972) considered the test statistic Sn =

Mn X

Φn (Nni , i/Mn )

i=1

where Φn : [0, ∞) × [0, 1] →R is a measurable function satisfying the condition |Φn (u, v)| ≤ c1 ec2 u for some c1 , c2 ∈ R not depending on n. He proved that the conditions: i) lim sup max n pin < ∞ n

1≤i≤Mn

ii) 0 < lim inf n

σn2 σ2 ≤ lim sup n < ∞ n n n

imply Sn − µn L −→ N(0, 1), n→∞ σn being µn =

Mn X

E [Φn (Zni , i/Mn )]

i=1

and σn2

=

Mn X i=1

−1

V ar [Φn (Zni , i/Mn )] − n


ÃM n X i=1

!2

Cov [Zni , Φn (Zni , i/Mn )]

,

128


where Zni , i = 1, ..., Mn are Poisson random variables with parameters npni (E [Zni ] = npni ) . If we consider q n = (qn1 , ..., qnMn )T with qni > 0, 1 ≤ i ≤ Mn and the continuous piecewise linear function gn : [0, 1] →R defined by gn (0) = 1 and gn (i/Mn ) = n qni

if 1 ≤ i ≤ Mn

and we define for u ∈ [0, ∞) and v ∈ [0, 1]

µ

u Φn (u, v) = gn (v) φ gn (v)

¶

,

then µ

Mn X

Nni nDφ (b pn , q n ) = n qni φ n qni i=1

¶

µ

Mn X

Nni = gn (i/Mn ) φ g (i/M n n) i=1

¶

= Sn .

To use the limit theorem of Holst it is necessary to establish that there exist c1 , c2 ∈ R not depending on n and verifying ¯ ¶¯ µ ¯ ¯ u ¯ ≤ c1 ec2 u . ¯ |Φn (u, v)| = ¯gn (v) φ gn (v) ¯

We know that φ (t) ≥ 0 and let us suppose φ(0) = lim φ(t) < ∞ and t↓0

log φ(t) < ∞. t→∞ t lim

(3.14)

Since φ(t) is convex in the domain 0 ≤ t < ∞, the function f (t) = (φ(t) − φ(0))/t is nondecreasing. Hence if t0 > 0 then 0 ≤ φ(t) ≤ φ(0) + t

φ(t0 ) − φ(0) t0

0 ≤ t ≤ t0 .

Further, by (3.14) there exist a positive value t0 and c ∈ R such that φ(t) ≤ ec t

t > t0 . ∗

Then there exist c∗1 , c∗2 ∈ R verifying 0 ≤ φ(t) ≤ c∗1 ec2 t Therefore for each 0< τ < ∞ we have the relation ³u´ ∗ ≤ τ c∗1 ec2 u/τ . τφ τ © 2006 by Taylor & Francis Group, LLC

∀ t ≥ 0.

129


Since φu (τ ) = τ φ(u/τ ), 0 ≤ u < ∞, are convex functions of τ , we have for each 0 < τ1 < τ2 < ∞ and 0 ≤ u < ∞ µ ¶ µ ¶¾ ½ ¯ ³ u ´¯ u u ¯ ¯ sup ¯τ φ , τ . φ φ ≤ max τ ¯ 1 2 τ τ τ 1 2 τ1 ≤τ ≤τ2 The assumption

0 < lim inf min n qni ≤ lim inf max n qni < ∞ n

n

1≤i≤Mn

1≤i≤Mn

(3.15)

implies the existence of 0 < τ1 < τ2 < ∞ and n0 with the property τ1 ≤ n qni ≤ τ2 for all 1 ≤ i ≤ Mn and n > n0 . It follows from here that under (3.15) the function Φn (u, v) satisfies for all n > n0 and c1 = max{τ1 c∗1 , τ2 c∗1 }, c2 = max{c∗2 /τ1 , c∗2 /τ2 } that |Φn (u, v)| ≤ c1 ec2 u . 2 for It is easy to check that in this case µn = n µφ,n and σn2 = n σφ,n

µφ,n =

Mn X i=1

¶¸ · µ Zni E qni φ n qni

and 2 σφ,n

=n

Mn X i=1

¶¸ ÃX ¶¸!2 · µ · µ Mn Zni Zni − V ar qni φ Cov Zni , qni φ , n qni n qni i=1

where Zni are Poisson random variables with E [Zni ] = n pni . Now the condition ii) given previously is equivalent to 2 2 ≤ lim sup σφ,n < ∞. 0 < lim inf σφ,n n

n

pn , q n ). Then we have established the following result for Dφ (b Theorem 3.4 If the null hypothesis H0 : p = pn satisfies lim sup max n pin < ∞, n


1≤i≤Mn

130


the conditions: i) φ(0) = lim φ(t) < ∞

and

t↓0

lim

t→∞

log φ(t) < ∞, t

ii) Given qn = (qn1 , ..., qnMn )T with qni > 0, ∀ 1 ≤ i ≤ Mn 0 < lim inf min n qni ≤ lim sup max n qni < ∞ n

1≤i≤Mn

1≤i≤Mn

and iii) 2 2 0 < lim inf σφ,n ≤ lim sup σφ,n −1. For nonuniform pn they established a similar result, but only for λ > −1 and integer-valued. For the proof they referred to Corollary 4.1 and Theorem 5.2 of Morris (1975), who employed a Poissonization idea alternative to Holst’s, leading to a similar result under slightly weaker assumptions about pn , but only for integer orders λ > −1. They formulated a conjecture that their asymptotic result can be extended to every λ > −1. The theorem presented in this Section, established in Menéndez et al . (1998b), provides asymptotic distributions also for test statistics not admitted in the theory of Morris, i.e., for the power-divergence of all orders λ > −1, since the corresponding functions φ(λ) (x) satisfy the condition φ(λ) (0) < ∞ if and only if λ > −1. This in particular means that the previous theorem confirms the conjecture of Read and Cressie.


132


3.4.

Nonstandard Problems: Tests Statistics based on Phi-divergences

3.4.1.

Goodness-of-fit with Quantile Characterization

The quantile characterization is an alternative method for testing goodnessp, p0 ) as discussed of-fit and, perhaps, has some advantage over that using Tnφ (b previously in Bofinger (1973), Durbin (1978), Menéndez et al . (2001b, 2001c) and others. The hypothetical and empirical quantile functions are defined as F0−1 (π) = inf {x : F0 (x) > π} and Fn−1 (π) = inf {x : Fn (x) > π} , respectively, for every π ∈ (0, 1) . Fn (x) is the empirical distribution function defined in Section 1 of Chapter 1. π 6

1 F (·)

F (c1 ) = π1

-

c1

c

Figure 3.1. Distribution function. Data are reduced by considering a partition π = (π1 , ..., πM−1 ) ∈ (0, 1)M−1 with π0 = 0 < π1 < ... < πM−1 < 1 = πM ,

(3.17)

and by applying the functions F0−1 and Fn−1 to π. (See Figures 3.1 and 3.2.) Let Y1 , ..., Yn be a random sample of size n. Hypothetical and empirical quantile vectors are calculated, respectively, as follows ¡ ¢ c = (c1 , ..., cM−1 ) = F0−1 (π1 ) , ..., F0−1 (πM−1 ) , ¢ ¡ −1 ¡ ¢ Y n = Yn1 , ..., YnM−1 = Fn (π1 ) , ..., Fn−1 (πM−1 ) , © 2006 by Taylor & Francis Group, LLC

133


where Yni = Y(ni ) (ni = [nπi ] + 1, i = 1, ..., M − 1) is the ni th-order statistic. c 6

6

F −1 (·)

F −1 (π1 ) = c1

-

0

π1

1

π

??

Figure 3.2. Quantile function. Hypothetical and empirical probability vectors, q 0 and p(Y n ), are calculated by 0 T q 0 =(q10 , ..., qM ) = (F0 (cj )−F0 (cj−1 ) : 1 ≤ j ≤ M)T = (πj −πj−1 : 1 ≤ j ≤ M)T ,

and ¡ ¢T p(Y n ) = (p1 (Y n ), ..., pM (Y n ))T = F0 (Ynj ) − F0 (Ynj−1 ) : 1 ≤ j ≤ M ,

where n0 = 0, nM = +∞, Yn0 = −∞ and YnM = +∞.

Once we have calculated the probability vectors q 0 and p(Y n ), different test statistics can be used to test H0 : F = F0 . (3.18)


134


The φ-divergence test statistic is given by Tnφ (p(Y n ), q 0 ) =

2n φ00 (1)

Dφ (p(Y n ), q0 ), φ ∈ Φ∗ .

In relation with this family of φ-divergence test statistics in Menéndez et al. (1998a, 2001a) the following result was obtained: Theorem 3.5 Let F0 (x) be continuous and increasing in the neighborhood of each cj = −1 F0 (πj ), j = 1, ..., M − 1. i) The decision rule “ Reject H0 if Tnφ (p(Y n ), q0 ) > χ2M−1,α ” defines a test for testing (3.18) with significance level α. ii) Let G be a distribution function with G 6= F0 and consider ∗ T ) = (G(cj ) − G(cj−1 ) : 1 ≤ j ≤ M)T . q ∗ = (q1∗ , ..., qM

The power βn,φ of the test with decision rule given in part i) satisfies, at the alternative q∗ , µ ´¶ √ 1 ³ φ00 (1) 2 ∗ ∗ ∗ 0 √ χ βn,φ (q1 , ..., qM ) = 1 − Φn − nDφ (q , q ) , σφ (q ∗ ) 2 n M−1,α where σφ2 (q ∗ ) is given by σφ2 (q ∗ )

=

M P

i=1

qi∗

µ µ ∗ ¶¶2 µ M µ ¶¶2 P ∗ 0 qi∗ qi 0 φ − qi φ , qi0 qi0 i=1

and Φn (x) is a sequence of distribution functions tending uniformly to the standard normal distribution function Φ (x) . iii) The test given in part i) is consistent in the sense of Fraser (1957), that is, for every alternative q ∗ 6= q0 , ∗ lim βn,φ (q1∗ , ..., qM ) = 1 for all α ∈ (0, 1) .

n→∞

iv) If we consider contiguous alternative hypotheses ¢ 1 ¡ H1,n : q n = q 0 + √ q ∗ − q 0 , n © 2006 by Taylor & Francis Group, LLC


135

the asymptotic distribution of the φ-divergence test statistic, Tnφ (p(Y n ), q0 ), given in part i), is noncentral chi-square with M − 1 degrees of freedom and noncentrality parameter δ given by

δ=

³ ´2 M qj∗ − qj0 X j=1

qj0

.

The proof of this Theorem is given in Menéndez et al . (2001a). An example for comparing the procedure given in this Section to the procedure given in Section 3.2 can be seen in this paper.

3.4.2.

Goodness-of-fit with Dependent Observations

In this Section we study the φ-divergence test statistic for testing the stationary distribution as well as the matrix of transition probabilities in a Markov chain.

Stationary Distribution Methods of statistical inference established for stationary independent data are often applied to dependent data. Investigations into the effects of Markov dependence seem to have been initiated by Bartlett (1951), who showed that such tests need no longer have the “usual” asymptotic distribution. Later this problem has been considered in many papers, see for instance Moore (1982), Glesser and Moore (1983a, 1983b), Molina et al . (2002) and references therein. Tavaré and Altham (1983) presented for irreducible aperiodic Markov chains a goodness-of-fit test for the stationary distribution, under simple null hypothesis, based on Pearson test statistic. In this Section we present a methodology, studied in Menéndez et al . (1997b, 1999a), for specification of critical values and powers of the φ-divergence test statistic in the framework of general statistical models with stationary data. The general methodology is illustrated in the model considered by Tavaré and Altham (1983). Let Y = {Yk , k ≥ 0} be a stationary sequence of random variables taking on values in the sample space X ⊂ R, and F the distribution function of components Yk on X . We consider the problem of testing the hypothesis H0 : F = F0


136


based on a realization of length n from Y . We consider the fixed partition P = Pn {Ei }i=1,...,M of X . Let Ni = j=1 IEi (Yj ) be the number of observations in Ei , i = 1, ..., M. In other words, we consider the classical goodness-of-fit tests b = (b for vectors p p1 , . . . , pbM )T , pbi = Ni /n, of the observed cell frequencies and vectors of the theoretical cell frequencies p = (p1 , . . . , pM )T , where pi = PrF (Ei ), i = 1, ..., M. The hypothesis F0 is indicated by writing p0 = (p01 , . . . , p0M )T , with p0i = PrF0 (Ei ), i = 1, ..., M, and it is assumed that all the components of p0 are nonzero. In this context Menéndez et al . (1997b) established the following result. Theorem 3.6 If the model satisfies the regularity assumptions: i) Under the null hypothesis H0 : p = p0 , for n → ∞, pbi = p0i + oP (1) , for all 1 ≤ i ≤ M. ii) The autocorrelation structure of the model verifies ¢ ¡ ¢´ L √ ³ 0 −1/2 ¡ n (p1 ) −→ N(0, V ), pb1 − p01 , ..., (p0M )−1/2 pbM − p0M n→∞

where V is a given matrix.

Then, the φ-divergence test statistic p, p0 ) = Tnφ (b

2n φ00 (1)

p, p0 ) Dφ (b

converges in law to a random variable X, defined by X≡

M X

ρi Zi2 ,

(3.19)

i=1

where ρi are the eigenvalues of the matrix V and Zi , i = 1, ..., M, are independent standard normal variables. Remark 3.2 Based on the previous theorem we should reject the null hypothesis F = F0 if Tnφ (b p, p0 ) > Qα , where Qα is the 100 (1 − α)-percentile of the random variable X given in (3.19). The matrix V and, consequently, the eigenvalues ρi may not be specified uniquely



137

by the null hypothesis F0 (uniquely the marginal distribution of components Yi is specified ). If V depends continuously on the model parameters which remain free under F0 , and there exist consistent estimates of these parameters leading to the estimate V n of the matrix V , then we can use the tests p, p0 ) > Qnα , Tnφ (b where Qnα is the 100 (1 − α)-percentile of X n≡

M X

ρni Zi2 ,

i=1

and ρni are the eigenvalues of the matrix V n . The continuity argument leads to the conclusion that ρni estimates consistently ρi , i.e., Qnα estimates consistently Qα . Theorem 3.6 and Remark 3.2 assert that all members of the family of φp, p0 ) are asymptotically equivalent from the point divergence test statistics Tnφ (b of view of the test size ´ ³ p, p0 ) > Qnα /p = p0 α = Pr Tnφ (b and preferences between them are based on the test powers. The power for p∗ 6= p0 is ³ ´ βn,φ (p∗ ) = Pr Tnφ (b p, p0 ) > Qnα /p = p∗ .

p, p0 ), can be applied to stationThe previous φ-divergence test statistic, Tnφ (b ary irreducible aperiodic Markov chains. We consider a random sample of size n from a stationary irreducible aperiodic Markov chain Y = {Yk , k ≥ 0} with state space {1, ..., M} . By P = (p(i, j))i,j=1,...,M we denote the matrix of transition probabilities of this chain and by p0 the stationary distribution, i.e., solution of the equation p0 = P (p0 )T . We assume that P is from the class of irreducible aperiodic stochastic matrices with one ergodic class so that the solution is unique. The irreducibility means that there are no transient states, i.e., pi 6= 0 for all 1 ≤ i ≤ M . The decomposition, considered in the previous theorem, may be defined by Ei = {i} , i = 1, ..., M and the assumptions i) and ii) hold (cf., e.g., Billingsley (1961a)). Since the probability distributions associated with the distribution functions F and F0 , in this case, coincide with the vectors p and p0 , we consider the hypothesis H0 : p = p0 about the stationary distribution


138


of the chain matrix P . Since no states are transient, the hypothesis satisfies the condition that all the components of p0 are nonzero. In the model under consideration the goodness-of-fit test of the null hypothesis H0 : p = p0 based on the Pearson’s test statistic has been considered by Tavaré and Altham (1983). This result was extended by Menéndez et al. (1997b, 1999a) to the φ-divergence test statistic. These authors established that in this case the φ-divergence test statistic Tnφ (b p, p0 ) converges in law to a random variable X, defined by M−1 X 1 + λi Z 2, (3.20) X≡ 1 − λi i i=1

where λ1 , ..., λM−1 are the nonunit eigenvalues of the chain matrix P .

If P = (p(i, j))i,j=1,...,M has identical rows then it is reversible and all its nonunit eigenvalues are zero. Thus Theorem 3.6 implies that if data Y1 , ..., Yn are independent then all φ-divergence test statistics Tnφ (b p, p0 ) are asymptotically distributed chi-squared with M − 1 degrees of freedom. More generally, if P = (1 − π) I M×M + π1(p0 )T , where 0 < π ≤ 1 and 1 = (1, ..., 1)T , then the nonunit eigenvalues of P are all equal 1 − π. Therefore all φ-divergence test staπ tistics 2−π Tnφ (b p, p0 ) tend in law to a chi-square distribution with M − 1 degrees of freedom. If the matrix P is not known we can use the relative frequencies pb(i, j) = υij /υi∗ , where υij =

n X k=2

I{(i,j)} (Yk−1 , Yk ) and υi∗ =

M X

υij ,

j=1

to estimate the transition probabilities p(i, j) of the matrix P consistently (cf., Billingsley (1961a)). By I{(i,j)} (Yk−1 , Yk ) we are denoting the function defined as ( 1 if Yk−1 = i and Yk = j . I{(i,j)} (Yk−1 , Yk ) = 0 otherwise Since the eigenvalues λi considered in (3.20) are continuous functions of the elements p(i, j) of P , the substitution p(i, j) = pb(i, j) in these functions leads to bi of λi . consistent estimates λ

The binary Markov model, considered in Exercise 13, was studied in Menéndez et al . (1997b) and they found that for the power-divergence test statistics the



139

value λ = −2 (Modified chi-square test statistic) is optimal in the sense of the power. It is interesting to observe that in the case of independent data φ Tn (−2) (b p, p0 ) is rarely optimal in this sense.

Chain Markov and Order We consider a random sample of size n from an irreduccible homogeneous Markov chain, Y = {Yk , k ≥ 0} , with state space {1, ..., M} and matrix of transition probabilities given by P = (p(i, j))i,j=1,...,M . Billingsley (1961a, 1961b) considered the problem of testing the hypothesis ¡ ¢ H0 : P =P 0 = p0 (i, j) i,j=1,...,M

on the basis of the likelihood ratio test statistic and chi-square test statistic. These results were extended in Menéndez et al . (1999b) by considering the φdivergence test statistic. They considered the family of φ-divergence test statistics given by ¶ µ M M X υi∗ X 0 pb(i, j) b ,P 0 ) = 2n Tnφ (P , (3.21) p (i, j)φ φ00 (1) i=1 n j=1 p0 (i, j)

and established that its asymptotic distribution is chi-square degrees n with c − M o 0 of freedom, where c is the number of elements in C0 = (i, j) : pij > 0 . The test statistics given by Billingsley appear as a particular case of the φ-divergence test statistic given in (3.21). Interesting papers applicable to stationary finitestate irreducible Markov chains have been written by Azalarov and Narkhuzhaev (1987, 1992), Mirvalev and Narkhuzhaev (1990), Ivchenko and Medvedev (1990), Basawa and Prakasa Rao (1980) and Rousas (1979). In Markov chains, the future evolution of the chain is conditionally independent of the past given the present state. It is, however, possible that the dependence relation is more complicated. In a second order Markov chain, the future evolution of the chain over times n + 1, n + 2, ... is independent of the past given the states at times n and n − 1. In a rth order Markov chain, r consecutive states must be conditioned upon for the future to be independent of the past. In a rth order Markov chain, the transition probabilities are r + 1 dimensional, r for the present and one for the future. The case r = 0 corresponds to a sequence of independent trials, while r = 1 corresponds to the usual Markov chain. It is important to be able to determine the order of the chain and for simplicity


140


to find the lowest acceptable value of r. This problem has been studied using the likelihood ratio test statistic and the chi-square test statistic by Billingsley (1961a, 1961b) and using φ-divergence test statistics by Menéndez et al. (2001d).

3.4.3.

Misclassified Data

The theory of goodness-of-fit tests, in the analysis of categorical data, has been developed extensively. One of the difficulties, often encountered in practice, is the possibility of a false classification of one or more individuals into the respective categories or classes. This problem was first discussed by Bross (1954) for the case of two categories. Bross established that the sample proportion is a biased estimate of the proportion and the bias is a function of the amount of misclassification of the data. Mote and Anderson (1965) studied the effect of misclassification on Pearson’s test statistic. If errors of misclassification are ignored, the size of the test will increase and the asymptotic power will be reduced. If we consider the family of φ-divergence test statistics Tnφ (b p, p0 ), a similar study to the one realized by Mote and Anderson (1965), gives analogous results. Then for the goodness-of-fit tests, the usual test requires modification when there are misclassification errors. In order to solve the difficulties involved in inference from a sample of categorical data, obtained by using a fallible classifying mechanism, Tenenbein (1970, 1971, 1972) presented double sampling methods for estimating the multinomial proportions in the presence of misclassification errors. Hochberg (1977) extended the use of Tenenbein’s double sample schemes for modeling and testing hypotheses on the true parameters from general multidimensional contingency tables with misclassification errors. Cheng et al. (1998) used also the Tenenbein’s double sampling scheme for introducing an adjusted chi-square test and the likelihood ratio test. In Pardo, L. and Zografos (2000), the family of φ-divergence test statistics for testing goodness-of-fit when the categorical data are subject to misclassification was considered. The double sample scheme is used in the context of the following experimental situation. Suppose that we have two methods of collecting the data: one which is error-free but expensive and the second which is fallible but inexpensive. An obvious dilemma results for the researcher, specially when funds are limited. Should



141

they sacrifice accuracy for quantity? Diamond and Lilienfield (1962) discuss an experimental situation in public health research where the true classification device is the physician’s examination whereas the fallible classifier is a questionnaire completed by the patient. The Tenenbein’s double sample scheme gives to the researcher another alternative which incorporates a balance between both measurement methods and their respective cost. The scheme suggests that, at the first stage, a sample of n units is drawn and the true and fallible classifications are obtained for each unit, and at the second stage a sample of N − n units is drawn and the fallible classification is obtained for each unit. Then, there are a total of n units in the sample which have been classified by both the true and fallible devices. The multinomial proportions can be estimated from the available data without going to the extreme of obtaining the true classification for all N units in the sample. We denote by Y the random variable associated with the true measurement, taking on the value “i” if the sampling unit belongs in fact to category Ei , i = 1, ..., M, and by Y0 the random variable associated with the fallible measurement, taking on the value “j” if the sampling unit is classified by the fallible device as being in category Ej , j = 1, ..., M. Let us denote the marginal probabilities of Y and Y 0 , by pi = Pr(Y = i), πj = Pr(Y 0 = j), i, j = 1, ..., M, P PM respectively, with M i=1 pi = j=1 πj = 1. To describe misclassification we define θij to be the probability that a unit, which, in fact, belongs to the category Ei , is classified in the category Ej . Thus θij = Pr(Y 0 = j|Y = i), i, j = 1, ..., M, and it is clear that M X j=1

M X θij = 1 and πj = pi θij , i, j = 1, ..., M. i=1

In this situation a double sampling scheme can be described as follows: i) A sample of n units is drawn and the true and fallible classifications, denoted by Y1 , ..., Yn and Y10 , ..., Yn0 , respectively, are obtained for each unit. We denote by nij the number of units in the sample whose true category is Ei , i = 1, ..., M, and whose fallible category is Ej , j = 1, ..., M, and by ni∗ = PM PM j=1 nij , n∗j = i=1 nij . © 2006 by Taylor & Francis Group, LLC

142


ii) A further sample of N − n units is drawn and the fallible classifications 0 , ..., Y 0 , are obtained for each unit. We denote by Yn+1 N mj =

N X

k=n+1

I{j} (Yk0 ), j = 1, ..., M,

the number of units whose fallible category is Ej , j = 1, ..., M, and by (m1 , ..., mM )T the vector of frequencies associated with the random sample 0 , ..., Y 0 . By I 0 Yn+1 {j} (Yk ) we are denoting N ( 1 if Yk0 = j . I{j} (Yk0 ) = 0 otherwise The joint likelihood function associated with the observed data 0 (Y1 , Y10 ), ..., (Yn , Yn0 ), Yn+1 , ..., YN0 ,

is given by M M Y M Y Y nij n∗j −nij (pi θij ) (πj − pi θij ) πkmk , L(p, Θ) = i=1j=1

k=1

with p = (p1 , ..., pM )T , Θ = (θij )i,j=1,..,M (c.f., Cheng et al. (1998)). Then the maximum likelihood estimators (cf., Tenenbein (1972)) are M X (mj + n∗j )nij (mj + n∗j )nij pbi = and θbij = , i, j = 1, ..., M. Nn∗j Nn∗j pbi j=1

Using the above expression for pbi , and assuming that n/N → f > 0, as N → ∞, the asymptotic distribution of (b p1 , ..., pbM−1 ) is √ L N(b p1 − p1 , ..., pbM−1 − pM−1 ) → N(0, Σ), N→∞

where the asymptotic variance-covariance matrix is defined (cf., Tenenbein (1972), Cheng et al . (1998)) by  pi qi  i=j  f [1 − (1 − f)Ki ], ³ ´ P Σ = (σij )i,j=1,..,M , σij = , M   1 − f1 λik λjk πk − pi qj , i = 6 j k=1


(3.22)


¯ ¡ £ ¤¢2 pi with qi = 1 − pi , Ki = Corr I{i} (Y ) , E[I{i} (Y ) ¯Y 0 ] = qi θ p i ij , for i, j = 1, ..., M − 1. and λij = E[I{i} (Y ) |Y 0 = j] = πj

143

µM 2 ¶ P θik −1 k=1 πk

Under the simple null hypothesis H0 : p=p0 = (p01 , ..., p0M )T , Cheng et al. (1998) established that N

M−1 X M−1 X i=1 j=1

L

(b pi − p0i )b τij (b pj − p0j ) −→ χ2M−1 , N→∞

with Σ−1 = (τij )i,j=1,..,M , and τbij the maximum likelihood estimator of τij , i, j = 1, ..., M − 1. Next result regarding the φ-divergence test statistic for misclassified data was established in the cited paper of Pardo, L. and Zografos (2000). Theorem 3.7 b=(b Based on Tenenbein’s (1972) double sampling scheme, let p p1 , ..., pbM )T be PM (mj + n∗j )nij with pbi = j=1 , and assume that n/N → f > 0, as N → ∞. Nn∗j Then we have i) Under the hypothesis H0 : p=p0 = (p01 , ..., p0M )T TNφ (b p,p0 )

r X 2N L 0 = 00 Dφ (b p,p ) −→ λi Zi2 , N→∞ φ (1) i=1

where Z1 , ..., Zr are independent and identically distributed normal random variables with mean zero and variance 1 and λi , i = 1, ..., r are the eigenval¡ ¢−1 ues of the matrix AΣ, being A the diagonal matrix with elements p0i , i = 1, ..., M −1, Σ the asymptotic variance-covariance matrix of the random √ vector N(b p−p0 ) given in (3.22) and r = rank(AΣA). ii) Let p∗ = (p∗1 , . . . , p∗M )T be a probability distribution with p∗ 6= p0 , then Ã √ µ ¶! N kα 00 ∗ ∗ 0 βN,φ (p ) = 1 − ΦN φ (1) − Dφ (p , p ) , σφ (p∗ ) 2N


144


where σφ2 (p∗ ) = T T ΣT , T = (t1 , ..., tM )T and tj =

µ

¶ ∂Dφ (p, p0 ) . ∂pj p=p∗

The value kα verifies ´ ³ Pr TNφ (b p,p0 ) > kα / H0 : p = p0 = α, and ΦN (x) is a sequence of distribution functions that tends uniformly to the standard normal distribution Φ (x).

3.4.4.

Goodness-of-fit for and against Order Restrictions

In some situations the probability vector p =(p1 , ..., pM )T exhibits a trend. If, for example, the classes {Ei }i=1,...,M have the same length in R and the original probability density function is unimodal, there is a positive integer k such that p1 ≤ ... ≤ pk−1 ≤ pk ≥ ... ≥ pM−1 . Some interesting examples in which the probability vector p exhibits a trend can be seen in Robertson (1978), Lee (1987), Robertson et al . (1988) and many others cited there. Statistical inference concerning a set of multinomial parameters under order restrictions has been studied since Chacko (1966) considered the maximum likelihood estimation of multinomial parameters subject to a simple order restriction. He also obtained the asymptotic null distribution of a chi-square type test statistic for testing homogeneity of a set of multinomial parameters against the simple order. The asymptotic null distribution of this test statistic is a mixture of chisquare distributions, which is called a chi-bar-square distribution. The mixing coefficients, which are called level probabilities, depend upon the multinomial parameter set as well as the order restriction. Robertson (1966) found maximum likelihood estimates of multinomial parameters subject to a partial order restriction and also Robertson (1978) generalized Chacko’s result to the one-sample likelihood ratio test of the equality of two multinomial parameters (one is known) against a partial order restriction. He showed that the asymptotic null distribution of this test statistic is chi-bar-square



145

and the level probabilities depend upon the known multinomial parameter only through the sets on which the known parameter is constant. He also considered the likelihood ratio test of an order restriction as a null hypothesis. He showed that the asymptotic null distribution of this test statistic is also chi-bar-square and that the level probabilities depend upon the true parameter only through the sets on which the true parameter is constant. Chi-square type tests have been studied by several researchers. Lee (1987) considered chi-square type tests for and against an order restriction on a set of multinomial parameters. He compared three test procedures, namely: i) the likelihood ratio test statistic, ii) the Pearson chi-square test statistic and iii) the modified chi-square test statistic. He showed that all three test statistics have the same asymptotic null distribution which is of chi-bar-square type. Menéndez et al. (2002) considered the φ-divergence test statistic for testing the equality of two multinomial parameters, one is known, against a partial order restriction and also for testing an order restriction as a null hypothesis. They established that the φ-divergence test statistics, under the null hypothesis, are asymptotically chibar-squared distributed. Other interesting results in this area, using φ-divergence test statistics, can be seen in Menéndez et al . (2003a, 2003c).


146

3.5.


Exercises

1. The following data represent the number of injured players in a random sample of 200 soccer matches: Injured players 0 1 2 3 ≥ 4 Soccer matches 82 90 20 7 1 a) Test the hypothesis that the distribution is Poisson with parameter λ = 0.775. Take significance level α = 0.05 and Freeman-Tukey test statistic. b) Find the power of the test at p∗ = (0.45, 0.35, 0.1, 0.05, 0.05)T . 2. Find the expression of the test statistic for goodness-of-fit based on divergence measures of Pearson, Matusita (a = 1/2), Balakrishnan-Sanghvi, Rathie-Kanappan, Power-divergence, Rukhin and Rényi. 3. Consider the divergence measure Dsr (p, q). Find the asymptotic variance for the statistic Dsr (b p, p0 ) and as a special case the asymptotic variance of the statistic associated with Rényi divergence. 4. Find the asymptotic variance of the estimated entropy of order r and degree s as a special case of the estimated divergence of order r and degree s. 5. Let u = (1/M, ..., 1/M)T . Find the asymptotic distribution of the estimated entropy of order r and degree s as a special case of the result obtained in Remark 3.1. 6. Let φ be a concave function with continuous second derivative. a) Find the asymptotic distribution of the test statistic 8nRφ (b p, p0 ) being Rφ the divergence measure introduced in Chapter 1 and p0 = (p01 , ..., p0M )T . b) Let p0 = (1/M, ..., 1/M)T . Show that p, p0 ) = − Sφ (b

8nM L p, p0 ) −→ χ2M−1 , Rφ (b n→∞ φ00 (1/M)

provided φ00 (1/M) < 0. c) Let φ (x) = −x log x. Find the asymptotic distribution for 8nRφ (b p, p0 ).



147

7. For the study of an urban system in a given region we suppose that a certain function of service f is more implanted relatively in the cities of major dimensions than in the smaller ones. For this purpose a spatial index is used. This index is given by If =

Af i Af r , Ati Atr

where: Af i represents the number of addresses of the function f in the city i. Af r represents the number of addresses of the function f in the region r. Ati represents the total number of addresses of the function f in the city i. Atr represents the total number of addresses of the function f in the region r. The cities are ordered in six levels taking into account If the largest (6) to the smallest (1). From previous studies the expected frequencies of each type of city according to its dimension are known. The theoretical and observed frequencies into the six categories are as follow: Levels 1 2 3 4 5 6 T otal T heoretical 1 2 5 7 15 20 50 Observed 1 3 6 10 17 13 50 Using the test statistic given in c) of the previous exercise, analyze if the presence of the function f is associated with the city dimension, taking as significance level α = 0.01. 8. We want to find a model to predict the probability of winning at a greyhound race track in Australia. Data collected on 595 races give the starting numbers of the eight dogs included in each race ordered according to the race finishing positions (the starting numbers are always the digits 1,...,8; 1 denotes the dog started on the fence, 2 denotes the second from the fence, etc.). We assume throughout that the initial positions are assigned at random to each of the eight dogs. We group the results into eight cells according to which starting number comes in first. Now we want to test


148


the hypothesis that all starting numbers have an equal chance of coming in first regardless of the positions of the other seven dogs, that is, H0 : pi = 1/8; i = 1, ..., 8. For this purpose use the test statistic given in Exercise 6 part b) with the function ( 1 a 1−a (x − x) si a 6= 1 φa (x) = −x log x si a = 1 for a = 1, 13/7 and 2 and significance level α = 0.05. The theoretical and observed frequencies into the eight categories are as follows: Dog i Observed Expected 1 0.175 0.125 2 0.16 0.125 3 0.111 0.125 4 0.106 0.125 5 0.104 0.125 6 0.097 0.125 7 0.101 0.125 8 0.146 0.125 Source: Haberman (1978, p. 2).

9. Let pn = (1/Mn , ..., 1/Mn )T and we consider the notation established in Section 3.3. Find the asymptotic distribution of the test statistic µ ¶ Mn X Nni 2 2 χ (b pn , pn ) = pni 1 − . npni i=1 10. Let pn = qn . Suppose that the following inequality holds 0 < lim inf min n qni ≤ lim sup max n qni < ∞. n

n

1≤i≤Mn

1≤i≤Mn

Show that for φ (x) = (1 − x)2 , µφ,n = and 2 σφ,n


Mn = n

Ã

Mn n

! Mn 1 X 1 −1 − γn + 2 . Mn n pni i=1


149

11. Consider the power divergence family. Find the expression of the mean and asymptotic variance corresponding to the family of test statistics based on it for qn = pn = (1/Mn , ..., 1/Mn )T and λ > −1. 12. We consider the Markov chain with states 1 and 2 and stochastic matrix P of transition probabilities given by Ã ! 1−β β P= ; 0 < β, γ ≤ 1, β + γ < 2. γ 1−γ Find the expression of the φ-divergence test statistic for testing H0 : p = p0 = (1/2, 1/2)T . 13. Find the expression of the φ-divergence test statistic given in Theorem 3.7 if p0 = (p0 , q0 )T , with q0 = 1 − p0 , 0 < p0 < 1. 14. Obtain the expression of the φ-divergence test statistic given in Theorem 3.7 for 1 φ (x) = (x − 1)2 . 2 15. A sample of n = 100 units is doubly classified by true and fallible methods and a second random sample of 400 measurements is taken and classified by the fallible method. The sample sizes n and N are respectively 100 and 500. The following table shows the obtained data.

True Device

0 1

Fallible 0 61 1 62

Device 1 7 31 38

68 32 100

First Sample

218

182

400

Second Sample

Test H0 : p = p0 = (1/2, 1/2)T by using the test statistic based on the power-divergence family with λ = 1, −1 and 0 and significance level α = 0.05. 16. With the notation of Section 3.3 establish that if Nni = γ + OP (n) and pbni = pni + oP (1) , then γ = npni + OP (1) .


(3.23)

150


17. We consider the population given by the distribution function ( 0 if x ≤ 2 . F0 (x) = 1 − x42 if x > 2 Use the procedure given in Section 3.4.1 to study if the observations 2.2408 5.8951 6.0717 3.6448 2.8551 4.4065 14.4337 3.0338 2.0676 2.6155 2.7269 5.1468 2.2178 2.0141 2.3339 2.6548 5.0718 2.8124 2.0501 13.6717 are from F0 (x) on the basis of the power-divergence test statistic for λ = −2, −1, −1/2, 0, 2/3 and 1 and significance level α = 0.05. 18. Find the expression of the asymptotic variance of ¢ √ ¡ n Dφ (b p, p0 ) − Dφ (p∗ , p0 ) ,

with p0 6= p∗ , using the divergence measures of Pearson, Matusita (a = 1/2), Balakrishnan-Sanghvi, Rathie-Kanappan, Power-divergence and Rukhin.

3.6.


1. If we consider the partition of the sample space given by E1 = {0} , E2 = {1} , E3 = {2} , E4 = {3} and E5 = {x ∈ N : x ≥ 4} , we have p0 = (0.4607, 0.3570, 0.1383, 0.0357, 0.0083)T . Now our problem consists of testing H0 : p = p0 , using the test statistic obtained from the power-divergence test statistics with λ = −1/2. The test statistic obtained in this case is the Freeman-Tukey’s test statistic and we should reject the null hypothesis if Ã ! 5 q X b p0 ) = 8n 1 − Tn−1/2 (p, p0i pbi > χ2M−1,α . i=1

b = (82/200, 90/200, 20/200, 7/200, 1/200)T , and From the data we get p then b p0 ) = 8.2971. Tn−1/2 (p,



151

On the other hand χ24, 0.05 = 9.4877. Therefore we should not reject the null hypothesis. The asymptotic power at the point p∗ = (0.45, 0.35, 0.1, 0.05, 0.05)T is given by Ã√ !! Ã 00 φ(−1/2) (1) χ2M−1,α 200 . β200,φ(−1/2) (p∗ ) ' 1−Φ − Dφ(−1/2) (p∗ , p0 ) σ1 (p∗ ) 400 It is not difficult to establish that  Ã 5 q !2  X p p0 p∗  , σ1 (p∗ ) = 4 1 − i

i

i=1

then

σ1 (p∗ ) = 0.2955, Dφ(−1/2) (p∗ , p0 ) = 0.0 4391 and φ00(−1/2) (1) = 1. With these values we get β200,φ(−1/2) (p∗ ) ' 0.8330. 2. In Theorem 3.1 it was established that b p0 ) Tnφ (p,

µ ¶ M 2n X 0 pbi L = 00 pi φ 0 −→ χ2M−1 . n→∞ φ (1) pi i=1

(x − 1)2 , then ¢2 M ¡ X Ni − p0i 2 φ 0 b p )= X ≡ Tn (p, . 0 np i i=1

For Pearson divergence, φ (x) =

1 2

This is the classical chi-square test statistic. √ 2 For Matusita (a = 1/2), φ (x) = (1 − x) , then b p0 ) Tnφ (p,

M µq X p ¶2 0 = 4n pi − pbi . i=1

It is not difficult to establish that this test statistic coincides with the Freeman-Tukey test statistic. For Balakrishnan-Sanghvi, φ (x) = b p0 ) Tnφ (p, © 2006 by Taylor & Francis Group, LLC

(x−1)2 , (x+1)2

= 4n

then

¶2 M µ X pbi − p0 i

i=1

pbi + p0i

.

152


For Rathie-Kanappan, φ (x) = b p0 ) Tns (p,

xs −x s−1 ,

then

2n = s (s − 1)

ÃM ! X (b pi )s ¡ 0 ¢s−1 − 1 . i=1 pi

For φ (x) = φ(λ) (x) , where φ(λ) (x) was given in the first chapter, then b p0 ) Tnλ (p,

2n = λ (λ + 1)

For Rukhin,

ÃM ! X (b pi )λ+1 ¡ 0 ¢λ − 1 , λ 6= 0, −1. pi i=1

(1 − x)2 , 2 (a + (1 − a) x)

φa (x) = and

b p0 ) = n Tnφa (p,

M X j=1

(p0j − pbj )2 ap0j + (1 − a)b pj

a ∈ [0, 1] , 0 ≤ a ≤ 1.

(3.24)

Finally, for Rényi divergence we have h (x) =

log (r (r − 1) x + 1) (xr − r (x − 1) − 1) and φ (x) = r (r − 1) r (r − 1)

and the family of Rényi test statistics is given by   M X 2n b p0 ) = log  (p0j )1−r (b pj )r − 1 . Tnr (p, r (r − 1) j=1

3. The divergence measure Drs (p, q) is a (h, φ)-divergence, with ´ s−1 1 ³ h (x) = (1 + r (r − 1) x) r−1 − 1 ; s, r 6= 1, s−1 and

φ (x) =

xr − r (x − 1) − 1 ; r (r − 1)

r 6= 0, 1.

b p0 ) is given by By Corollary 3.3, the asymptotic variance of Drs (p, r2

Ã

M P

i=1

r p∗i

1−r p0i

( )( ) (r−1)2

! 2 s−r

r−1

 ÃM !2  M X X ¡ ¢ ¡ ¢ 2(1−r) 1−r  , (p∗i )2r−1 p0i − (p∗i )r p0i i=1

i=1

(3.25)



because

153

s−r

h0 (x) = r (1 + r (r − 1) x) r−1 , ÃM ! X ¡ 0 ¢1−r 1 ∗ 0 ∗ r (pi ) pi −1 Dφ (p , p ) = r (r − 1) i=1

and

s−r ÃM ! r−1 X ¡ ¡ ¢ ¢ 1−r h0 Dφ (p∗ , p0 ) = r (p∗i )r p0i .

i=1

We know that

p, p0 ) = rDr1 (b p, p0 ), lim Drs (b

s→1

therefore for Rényi test statistic the asymptotic variance is µM ¶−2 P ∗ r ¡ 0 ¢1−r  ÃM !2  (pi ) pi M X X ¡ ¢ ¡ ¢ 2(1−r) 1−r i=1  . (p∗i )2r−1 p0i − (p∗i )r p0i (r − 1)2 i=1 i=1 4. Denoting the uniform distribution by u = (1/M, ..., 1/M)T , it is immediate to show that Drs (p, u) = M 1−s (Hrs (u) − Hrs (p)) , where Hrs (p) =

1 1−s

 Ã s−1 ! r−1 M X  pri − 1 i=1

is the entropy of r-order and s-degree (Entropy of Sharma and Mittal). From Exercise 3, we have ¡ ¢ √ L n (Drs (b p, u) − Drs (p∗ , u)) −→ N 0, σ12 (p∗ ) n→∞

where σ12 (p∗ ) is obtained from (3.25) replacing p0 by u. Therefore, ¡ ¢ √ L 2 n (Hrs (p∗ ) − Hrs (b p)) −→ N 0, σr,s (p∗ ) , n→∞

2 (p∗ ) = σ 2 (p∗ )/M 2(1−s) and where σr,s 1

σ12 (p∗ ) =

r2 (r − 1)2


ÃM X i=1

s−r  ÃM !2 r−1 !2  M X X  (p∗i )r (p∗i )2r−1 − (p∗i )r  .

i=1

i=1

154


5. By Remark 3.1(b), we have 2n h0 (0) φ00 (1)

L

p, u) −→ χ2M−1 . Dφh (b n→∞

But in our case we have h0 (0) = r and φ00 (1) = 1. Then, 2n s L D (b p, u) −→ χ2M−1 . n→∞ r r On the other hand, Drs (p, u) = M 1−s (Hrs (u) − Hrs (p)) , therefore, 2nM s−1 L p)) −→ χ2M−1 . (Hrs (u) − Hrs (b n→∞ r 6. a) A second order Taylor expansion gives that the random variables p, p0 ) ≡ 8nRφ (b p, p0 ) Snφ (b and

¢T √ ¡ ¢ √ ¡ b − p0 A n p b − p0 , n p

with A = diag (−φ00 (p1 ) , ..., −φ00 (pM )) , have the same asymptotic distribution. Then r X L 8nRφ (b p, p0 ) −→ λi Zi2 n→∞

i=1

where Z1 , ..., Zr are independent, identically distributed normal ´ random ³ variables with mean zero and variance 1, r = rank Σp0 AΣp0 and βi0 s are the eigenvalues of the matrix AΣp0 . b) In this case we have ¡ ¢ A = diag −φ00 (1/M) , ..., −φ00 (1/M) .

p, p0 ) and Then Snφ (b

¡ ¢T ¢√ ¡ ¢ √ ¡ b − p0 diag u−1 n p b − p0 , n p

with u = (1/M, ..., 1/M)T , have the same asymptotic distribution. This asymptotic distribution, see Theorem 3.1, is chi-square with M − 1 degrees of freedom.



c) In this case we have A = diag fore

155

³¡ ¢ ´ −1 , because φ00 (p0i ) = 1/p0i . Therep0 L

8nRφ (b p, p0 ) −→ χ2M−1 . n→∞

7. We have 400Rφ

(b p, p0 )

= 400 = 400

µ

6 P

³ i=1

pbi log pbi +p0i log p0i 2

−

0.02 log 0.02+0.02 log 0.02 2

+

.........

+

0.26 log 0.26+0.4 log 0.4 2

= 4.89182

−

6 P

i=1

−

pbi +p0i 2

¡ 0.02+0.02 2

¡ 0.26+0.4 2

log

pbi +p0i 2

¶

log 0.02+0.02 2

log 0.26+0.4 2

¢´

¢

and χ25, 0.01 = 15.086. Then the null hypothesis H0 should not be rejected. 8. In the following table we have computed the test statistics for the different values of a, a Sφa (b p, p0 )

1 29.1768

13/7 30.5175

2 . 30.788

But χ27, 0.05 = 14.07, and the null hypothesis should be rejected. pn , pn ) is obtained from Dφ (b pn , pn ) with φ (x) = 9. The expression of χ2 (b 2 (x − 1) . Applying Proposition 3.2 we get h ³ í´ √ ³ 2 n χ (b pn , pn ) − E φ Zγnn L r −→ N (0, 1), h h ³ í ³ í n→∞ γn V ar φ Zγnn − Cov2 Zn , φ Zγnn where Zn is a Poisson random variable with parameter γn . Then,

µφ,n


· µ ¶¸ ¸2 · Zn 1 Mn Zn =E φ −1 = = =E γn γn γn n

156


and 2 σφ,n

= =

= =

· · µ ¶¸ µ ¶¸ Zn Zn 2 − Cov Zn , φ γn V ar φ γn γn   · ¸4 Ã · ¸2 !2 Zn Zn  −1 − E −1 γn E γn γn Ã " µ ¶2 #!2 Zn − E (Zn − γn ) −1 γn Ã µ ¶2 ! · ¸2 γn + 3γn2 1 1 γn − − γn4 γn γn 1 3 1 1 2 2Mn + − − 2 = = , 2 γn γn γn γn γn n

because if Z is a Poisson variable with E [Z] = λ then for every a > 0, E E

£Z a

¤ −1 =

h¡ Z

−1

h¡ Z

−1

a

¢2 i

=

¢4 i

=

h ¡ ¢2 i = E (Z − λ) Za − 1 E

a

λ−a a λ+(λ−a)2 a2 λ+2λ(λ−a) a2 λ+3λ2 +4λ(λ−a)+6λ(λ−a)2 +(λ−a)4 . a4

Therefore, n √ 2Mn

¶ µ Mn L 2 χ ( bpn , pn ) − −→ N(0, 1). n→∞ n

10. By Theorem 3.4 and using the previous formulae, we have

µφ,n =

Mn X i=1


¶¸ X · µ Mn 1 Zni Mn = E pni φ pni = npni n pni n i=1


157

and 2 σφ,n =

= = =

¶2 ! µM ¶2 µ 2 Pn n p + 3(n p ) 1 1 ni ni 2 n pni − pni − (n pni )4 n pni n pni i=1 i=1 ¶ µ ¶ µ 2 2 M M Pn Pn 1 1 3n 1 n + 4 − 2 − 3 n n i=1 n pni i=1 n µ ¶2 M n 1 P 2Mn Mn 1 + − n2 p n n i=1 ¶ µ ni M n P 1 Mn 1 − γn−1 + 2 . n Mn n i=1 pni M Pn

Ã

11. In this case we have for λ > −1 and λ 6= 0 " Ãµ ¶ !# µ ¶¸ · Zn Zn λ+1 Zn 1 = E λ(λ+1) − µφ(λ) ,n = E φ(λ) γn γn γn =

1 λ(λ+1)

Ã "µ ¶ # ! Zn λ+1 E −1 γn

and σφ2 (λ) ,n

"

!# ¶ Zn λ+1 Zn = γn V ar − γn γn " Ãµ ¶ !# λ+1 Zn Zn 1 2 − − Cov Zn , λ(λ+1) γn γn Ã "µ ¶ # " µ ¶λ+1 #! λ+1 Z Z Zn n n 1 , = γn λ2 (λ+1) V ar − γn Cov2 . 2 γn γn γn 1 λ(λ+1)

Ãµ

For λ = 0, we get µφ(0) ,n

·

Zn Zn =E log γn γn

¸

and σφ2 (0) ,n

µ · ¸ · ¸¶ Zn Zn Zn 2 Zn Zn − γn Cov , = γn V ar log , log γn γn γn γn γn

where Zn is a Poisson random variable with mean γn . In the case of a zero observed cell frequency, Dφ(λ) ( bpn , pn ) is undefined for λ ≤ −1, since it requires taking positive powers of pni /b pni where Nni = 0. © 2006 by Taylor & Francis Group, LLC

158


12. The class of possible Markov matrices P is given by Ã ! 1−β β , 0 < β, γ < 1, α + γ < 2 γ 1−γ and the subclass of possible Markov matrices satisfying the condition pT = pT P is ! Ã ¶ µ p2 1−β β , β < 1, , 0 ≤ β ≤ min 1, p1 p1 β/p2 1 − p1 β/p2 where p =(p1 , p2 )T and the subclass satisfying (p0 )T = (p0 )T P is given by Ã ! 1−β β , 0 < β < 1. β 1−β The nonunit eigenvalue of this matrix is λ = 1 − 2β. Then we have L

p, p0 ) −→ Tnφ (b

n→∞

where Z is a standard normal variable.

1+λ 2 Z , 1−λ

Therefore, L

p, p0 ) −→ Tnφ (b

n→∞

1−β 2 Z . β

But if we consider the matrix Ã ! pb (1, 1) 1 − pb (1, 1) , 1 − pb (2, 2) pb (2, 2)

b = −1 + pb (2, 2) + pb (1, 1) . Then we have the nonunit eigenvalue is λ

and

b pb (2, 2) + pb (1, 1) 1+λ = b 2 − pb (2, 2) + pb (1, 1) 1−λ

2 − pb (2, 2) + pb (1, 1) L −→ χ2 . pb (2, 2) + pb (1, 1) n→∞ 1 Therefore we should reject the null hypothesis if p, p0 ) Tnφ (b

Tnφ (b p, p0 )

2 − pb (2, 2) + pb (1, 1) > χ21,α . pb (2, 2) + pb (1, 1)

A simulation study to choose the best value of the parameter λ in the power-divergence test statistics can be seen in Menéndez et al. (1997b).


159


13. In this case the null hypothesis is H0 : p = p0 , where p is the probability of having outcome 1 for the binary observation and q = 1 − p is the probability of having outcome 0. The misclassification probabilities are θ = Pr(Y 0 = 0|Y = 1) and ψ = Pr(Y 0 = 1|Y = 0). It is immediate that π = Pr(Y 0 = 1) = p(1 − θ) + qψ. Denote by nij the number of units in the validation subsample whose true category is i and fallible category is j; i, j = 0, 1, and mk =

N X

j=n+1

¡ ¢ I{k} Yj0 , k = 0, 1.

In this context, the maximum likelihood estimators of the probabilities p, n∗0 +m0 11 n∗1 +m1 10 n∗0 +m0 θ and ψ are respectively pb = nn∗1 + nn∗0 , θb = nn10 and N N N pb ∗0 n01 n∗1 +m1 b . ψ= n∗1 N(1−b p)

The matrix AΣ, appearing in Theorem 3.7 Ã ! −1 V ar(b p ) p Cov(b p , 1 − p b ) p−1 0 0 AΣ = q0−1 Cov(1 − pb, pb) q0−1 V ar(1 − pb) ! Ã −1 V ar(b p ) −p V ar(b p ) p−1 0 0 , = −q0−1 V ar(b p) q0−1 V ar(b p) with p0 q0 V ar(b p) = f

µ 1 − (1 − f)

¶ p0 q0 2 (1 − θ − ψ) , q0 = 1 − p0 , π(1 − π)

and f is given by limit as N → ∞ of n/N with n = n∗0 + n∗1 . It can be easily seen that the unique nonzero eigenvalue of AΣ is ¶ µ 1 p0 q0 2 µ= 1 − (1 − f) (1 − θ − ψ) . f π(1 − π) Then we have

2N L p, p0 ) −→ χ21 . Dφ (b 00 N→∞ µφ (1) ³ ´ p q 0 0 b 2 , then If we denote µ b = f1 1 − (1 − f ) π(1−π) (1 − θb − ψ) µ ¶ µ ¶¸ · pb qb 2N L −→ χ21 , p0 φ + q0 φ N→∞ µ b φ00 (1) p0 q0 with q0 = 1 − p0 and qb = 1 − pb.


(3.26)

160


14. For φ(x) = 12 (x − 1)2 , we get 1 p, p0 ) = (b p − p0 )2 /p0 q0 , Dφ (b 2 and 2N Dφ (b p, p0 ) = µ b φ00 (1)

N(b p − p0 )2 µ ¶. p0 q0 1 2 b b p0 q0 f 1 − (1 − f ) (1 − θ − ψ) π(1 − π)

This test statistic has been studied in Cheng et al . (1998).

15. For testing the null hypothesis H0 : p = (1/2, 1/2)T on the basis of a random sample of size N −n, and ignoring the possibility of misclassification, it holds 2 (N − n) L p, p0 ) −→ χ2M−1 . Dφ(λ) (b N→∞ φ00(λ) (1) The expression of the family of test statistics given in Theorem 3.7, for the power-divergence test statistics, can be written as  ³ λ+1 ´ pb qbλ+1 2N  + − 1 , λ 6= −1, 0  λ λ  µbλ(λ+1)µ p0 q0 ¶   ´ ³ ´ ³ p b q b 2N pb qb 2N Dφ(λ) (b , λ=0 p, p0 ) = , (3.27) µ b log p0 q0  µ b  ³³ ´ ³ ´ ´  p0 q0  pb qb  − 2N log , λ = −1 µ b p0 q0 p0 q0 b 2 ). b = f1 (1 − (1 − f) π(1−π) (1 − θb − ψ) with q0 = 1 − p0 , qb = 1 − pb and µ

From the data we have

pb = 0.368, θb = 0.0245, ψb = 0.129 and π b = 0.44.

If we do not consider the misclassification and we use only the second sample (218, 182), we have that the sampling proportions are 0.545 and 0.455 b1 = (b respectively, denoted by p p1 , qb1 = 1 − pb1 )T = (0.545, 0.455)T .

For testing the hypothesis H0 : p = (1/2, 1/2)T , we have  ´ 800 ³ pb12 qb12  + − 1 = 3.240, λ=1    2 1/2 µ³ 1/2  ´pb1 ³ ´qb1 ¶  pb1 qb1 = 3.244, λ = 0 , 800 log p, p0 ) = 2(N − n)Dφ(λ) (b 1/2 1/2  ¶ µ  1 1 ´ ³ ´ ³   pb1 2 qb1 2   −800 log = 3.253, λ = −1 1/2 1/2 © 2006 by Taylor & Francis Group, LLC


161

and χ21,0.05 = 3.841. Then for λ = 1, λ = 0 and λ = −1, the null hypothesis H0 should not be rejected if the size of the test is α = 0.05. Now, we are going to get the expression of the family of test statistics, given in Section 3.4.3, when we have a double sample data. First we obtain the expression for µ b, µ ¶ 1 p0 q0 2 b b µ b= 1 − (1 − f) (1 − θ − ψ) = 2.0919. f π b(1 − π b)

Then, based on (3.27), we have  ³ 2 ´ pb qb 2 2×500  + − 1 = 16.659, λ=1  2×2.0919 µ 1/2 1/2   ¶  ´ ³ ´ ³ pb qb  2×500 pb qb 2N = 16.858, λ=0 , p, p0 ) = Dφ(λ) (b 2.0919 log 1/2 1/2  µ b ¶ µ  ³ ´1 ³ ´1   2 2 pb qb  = 17.267, λ = −1 log  − 2×500 2.0919 1/2 1/2

because in this case pb = 0.368 and qb = 0.632. Thus if we consider double sample data, H0 should be rejected. Therefore we can see that if one uses only the fallible data then one may be led to the wrong conclusion to accept H0 .

16. To establish (3.23) we have to prove if for η > 0 there exists a constant k (η) such as if n ≥ n (η) , then Pr (|γ − npni | > k (η)) < η. Denoting q = Pr (|γ − npni | > k (η)) , we have q = Pr (|γ − Nni + Nni − npni | > k (η)) ´ ³ ´ ³ k(η) + Pr |N ≤ Pr |γ − Nni | > k(η) − np | > ni ni 2 2 ¯ ¯ ³¯ ´ ³¯ ´ ¯ γ Nni ¯ k(η) ¯ Nni npni ¯ k(η) = Pr ¯ n − n ¯ > 2n + Pr ¯ n − n ¯ > 2n ¯ ¯ ³¯ ´ ³¯ ´ ¯ ¯ Nni ¯ ¯ k(η) ≤ Pr ¯ nγ − Nnni ¯ > k(η) + Pr . > − p ¯ ¯ ni 2 n 2

Using the assumption Nni = γ + OP (n) , we have that given η/2 there exist ¡ ¢ a constant k1 η2 and n∗ = n (η) such that for all n ≥ n∗ , ¯ µ¯ ³ η ´¶ η ¯ γ Nni ¯ ¯ > k1 Pr ¯¯ − ≤ . n n ¯ 2 2 © 2006 by Taylor & Francis Group, LLC

162


¡ ¢ Using the assumption pbni = pni + oP (1) we have that given k1 η2 and ¡ ¡η ¢ ¢ η ∗∗ ∗∗ 2 > 0 there exists a number n = n k1 2 , η such that for n ≥ n , ¯ µ¯ ³ η ´¶ η ¯ Nni ¯ ¯ ¯ Pr ¯ − pni ¯ > k1 ≤ . n 2 2 ¡ ¢ If we consider k (η) = k1 η2 and n = max (n∗ , n∗∗ ) , we have that for all η > 0, η η Pr (|γ − npni | > k (η)) ≤ + . 2 2 17. We wish to test H0 : F = F0 using the family of power-divergence test statistics Ãµ ! ¶ M X 2n pi (Y n ) λ λ 0 Tn (p(Y n ), q ) = pi (Y n ) −1 , λ(λ + 1) qi0 i=1 where pi (Y n ) and qi0 , i = 1, ..., M must be calculated. We assume M = 4 and consider the partition of the unit interval defined by π0 = 0, π1 = 1/4, π2 = 2/4, π3 = 3/4, π4 = 1. We have ¢T ¡ q = q10 , ..., q40 = (F0 (cj ) − F0 (cj−1 ) : 1 ≤ j ≤ 4)T = 0

µ

1 1 1 1 , , , 4 4 4 4

¶T

,

and

p(Y n ) = (F0 (Y(6) ), F0 (Y(11) )−F0 (Y(6) ), F0 (Y(16) )−F0 (Y(11) ), 1−F0 (Y(16) ))T , because Yni = Y(ni ) and ni = [20 × πi ] + 1, i = 1, 2, 3. We have p(Y n ) = (0.26566, 0.24364, 0.3397, 0.151)T , because Y(6) = 2.3339, Y(11) = 2.8551 and Y(16) = 5.1468. In the following table we report the values of the power-divergence test statistics Tnλ (p(Y n ), q0 ) for different values of λ. Tnλ

λ = −2 1.7936

λ = −1 1.6259

λ = −1/2 1.5643

λ=0 1.5153

λ = 2/3 1.4676

λ=1 . 1.4506

On the other hand χ23, 0.05 = 7.815. Therefore we should not reject the null hypothesis.



163

18. The expression of the asymptotic variance is given by σ12 (p∗ ) Then we have Divergence Pearson Matusita (a=1/2)

Balakrishnan

Rathie-Kanappan Power-divergence Rukhin


=

M P

i=1

p∗i

µ µ ∗ ¶¶2 µ M µ ¶¶2 P ∗ 0 p∗i pi 0 − pi φ . φ p0i p0i i=1

Asymptotic variance µ M ∗ 2 ¶2 M (p∗ )3 P P (pi ) i − 0 0 2 i=1 (pi ) i=1 pi ¶2 µM q Pp ∗ 1− pi p0i i=1 µ ∗ ¶2 M P 4(pi − p0i )(p0i )2 ∗ pi (p∗i + p0i )3 i=1 ¶ µM P ∗ 4(p∗i − p0i )(p0i )2 2 − pi (p∗i + p0i )3 i=1Ã ¶2 ! µM M P P 2 s (p∗i )2s−1 (p0i )2(1−s) − (p∗i )s (p0i )1−s (1−s)2 i=1 i=1 Ã ¶2 ! µM M P P 1 (p∗i )2λ+1 (p0i )−2λ − (p∗i )λ+1 (p0i )−λ λ2 i=1 i=1 ¡ ¢2 M P (p0i − p∗i )2 p0i (a + 1) + (a − 1) p∗i ∗ pi (ap0i + (1 − a) p∗i )4 i=1Ã ¢ !2 ¡ 0 − p∗ ) p0 (a + 1) + (a − 1) p∗ M P (p i i i − p∗i i 0 + (1 − a) p∗ )2 (ap i=1 i i

4

Optimality of Phi-divergence Test Statistics in Goodness-of-fit 4.1.

Introduction

In the previous chapter we have studied the family of φ-divergence test stab p0 ), for the problem of goodness-of-fit. If we denote by FT φ (pb,p0 ) (t) tistics, Tnφ (p, n

b p0 ), for fixed φ, we established that the exact distribution of Tnφ (p, FTnφ (pb,p0 ) (t) = Fχ2

M−1

(t) + o (1)

as n → ∞,

(4.1)

under the null hypothesis

H0 : p = p0 .

(4.2)

Based on (4.1) we considered for the problem of goodness-of-fit given in (4.2) the decision rule b p0 ) > χ2M−1,α ”. “Reject, with significance level α, H0 if Tnφ (p,

(4.3)

Now in this chapter we shall present some criteria to choose the best function φ in some sense. In Section 4.2, Pitman asymptotic efficiency (contiguous alternative hypotheses), Bahadur efficiency and some asymptotic approximations of the power function for the φ-divergence test statistic are studied.


166


Result (4.1) is asymptotic and it is only valid for large sample sizes but in finite samples is frequently assumed to hold approximately in order to calculate critical regions for the φ-divergence test statistic in goodness-of-fit tests. Then it will be important to give some methods to improve the accuracy of the φ-divergence test statistic in those situations where the sample size cannot be assumed large. We shall study with the φ-divergence test statistic the same procedures studied previously by Read and Cressie (1988) with the power-divergence family of test statistics. In Section 4.3 we investigate the criterion based on the speed of conb p0 ), to its vergence of the exact moments of the φ-divergence test statistic, Tnφ (p, asymptotic moments. The exact distribution of every member of the family of b p0 ) (see (4.1)) differs from chi-square by o (1) . φ-divergence test statistics, Tnφ (p, In Section 4.4 a closer approximation to the exact distribution is obtained by extracting the φ-dependent second order component from the o (1) term. In Section 4.5 comparisons of exact power based on exact critical regions are presented and finally in Section 4.6 comparisons between the exact distribution of the φb p0 ), and the different asymptotic approximations divergence test statistics, Tnφ (p, are studied.

Throughout the chapter we shall assume that φ ∈ Φ∗ is 4 times continuously differentiable in the neighborhood of 1 and φ00 (1) 6= 0.

4.2.

Asymptotic Efficiency

Power functions are usually difficult to evaluate and we mostly have to be content with approximations based on limit results. In this Section we consider three approaches in order to choose an optimal φ-divergence test statistic: Pitman efficiency, Bahadur efficiency and comparisons based on some approximations to the power function.

4.2.1.

Pitman Asymptotic Relative Efficiency

For a probability vector p∗ 6= p0 , we established that ´ ³ b p0 ) > χ2M−1,α / H1 : p = p∗ = 1. lim βn,φ (p∗ ) = Pr Tnφ (p, n→∞

Hence to get a limit value less than 1, we must consider a sequence of contiguous alternative hypotheses


Optimality of Phi-divergence Test Statistics in Goodness-of-fit

167

√ H1,n : pn = p0 + d/ n, P where d =(d1 , ..., dM )T is a fixed M × 1 vector such that M j=1 dj = 0, because in this case ¡ ¢ lim βn,φ (pn ) = 1 − Gχ2 (δ) χ2M−1,α , (4.4) n→∞

M −1

where Gχ2 (δ) is the distribution function of a noncentral chi-square random M−1 variable with M − 1 degrees of freedom and noncentrality parameter ³¡ ¢ ´ −1 δ = dT diag p0 d. (4.5)

In this context we can consider the Pitman asymptotic relative efficiency to compare the behavior of two test statistics. Let us consider two φ-divergence test b p0 ) and Tnφ2 (p, b p0 ) and let us suppose that for a given n and a statistics Tnφ1 (p, significance level α there exists a number Nn such that ¡ ¢ βn,φ1 (pn ) = βNn ,φ2 pNn → β (any asigned value) < 1, n→∞

that is, the powers are equal and that Nn → ∞ as n → ∞. By pn and pNn we are denoting contiguous alternative hypotheses, p √ pn = p0 + d/ n and pNn = p0 + d/ Nn .

Definition 4.1 b p0 ) with respect Tnφ1 (p, b p0 ) The Pitman asymptotic relative efficiency of Tnφ2 (p, is given by n lim . n→∞ Nn For more details about the Pitman asymptotic efficiency see Rao (1973, pp. 467-470), Read and Cressie (1988, pp. 54-55) or Pitman (1979, pp. 56-62). Based on (4.4) the asymptotic power functions βn,φ1 (pn³) and βNń ,φ2 (pN ) will be equal because both of them are based in Gχ2 (δ) χ2M−1,α , with δ M −1 given on (4.5). For this reason the Pitman asymptotic efficiency is 1 for any φb p0 ) and Tnφ2 (p, b p0 ). We can conclude that using divergence test statistics Tnφ1 (p, Pitman asymptotic relative efficiency is not possible to discriminate between the φ-divergence test statistics.


168

4.2.2.


Bahadur Efficiency

Let {Yk }k∈N be a sequence of independent and identically distributed random variables from F and we denote by s = (y1 , y2 , .....) a possible outcome of the previous sequence. We consider the problem of goodness-of-fit H0 : F = F0 by partitioning the range of data in disjoint intervals, {Ei }i=1,..,M , by testing the hypothesis H0 : p = p0 , i.e., we consider the notation introduced in Section P 3.1. We denote Ni = nj=1 IEi (Yj ). We can see that Ni is based on the first P n components of the sequence {Yk }k∈N . We denote by ni (s) = nj=1 IEi (yj ) b the associated probability vector with the first n components of the and by p(s) b p0 ) for testing the null hypothesis outcome “s”. We consider a test statistic Hn (p, H0 : p = p0 , and its outcome based on (y1 , y2 , ....., yn ) given by Hn (pb (s) , p0 ). b p0 ) denotes a general family of test statistics, not necessarily the Here Hn (p, b p0 ) considered in same as the family of the φ-divergence test statistics Tnφ (p, (4.1). b p0 ) under the We denote by FHn (pb,p0 ) (t) the distribution function of Hn (p, b p0 ) is defined by null hypothesis H0 : p = p0 . The level attained by Hn (p, ¢ ¡ Ln (s) = 1 − FHn (pb,p0 ) Hn (pb (s) , p0 ) .

Bahadur (1971) pointed out that in typical cases Ln is asymptotically uniform distributed over (0, 1), under the null hypothesis H0 : p = p0 and Ln → 0 exponentially fast (with probability one) under p 6= p0 . Now we define the exact Bahadur slope. Definition 4.2 We consider p 6= p0 and p ∈ ∆+ M , where 4+ M

½ ¾ M P T = p = (p1 , ..., pM ) : pi > 0, i = 1, ..., M, pi = 1 . i=1

ª © b p0 ) n∈N has exact Bahadur slope c (p) , We shall say that the sequence Hn (p, 0 < c (p) < ∞, if 1 lim log Ln (s) = − c (p) , (4.6) n→∞ 2 with probability one when n → ∞.



169

This definition is motivated by Bahadur (1971) in the following terms “Consider the Fisherian transformation Vn (s) = −2 log Ln (s). Then, in typical cases, Vn → χ22 in distribution in the null case. Suppose now that a non-null p is obtained and that (4.6) holds, with 0 < c (p) < ∞. Suppose we plot, for a given s, the sequence of points {(n, Vn (s)) : n = 1, 2, ...} in the uv-plane. It follows from (4.6) that, for almost all s, this sequence of points moves out to infinity in the direction of a ray from the origin, the angle between the ray and the u-axis, on which axis the sample size n is being plotted, being tan−1 c (p) ”. Given ε > 0, 0 < ε < 1, and s, let N = N (ε, s) be the smallest integer m such that Ln (s) < ε for all n ≥ m and let N = ∞ if no such m exists. Then N © ª b p0 ) n∈N in order to become is the sample size required for the sequence Hn (p, significant (and remains significant) at the level ε. But the most important fact pointed out by Bahadur (1971) is that, for small ε, N is approximately inversely proportional to the exact Bahadur slope, i.e., if (4.6) holds and 0 < c (p) < ∞, then lim N (ε, s) /2 log (1/ε) = c (p)−1 ,

ε→0

(4.7)

with probability one when ε → 0 (see Theorem 7.1 in Bahadur (1971)). © ª © ª b p0 ) n∈N and Hn2 (p, b p0 ) n∈N are two sequences of test Suppose that Hn1 (p, b p0 ) has exact Bahadur slope ci (p) , 0 < ci (p) < ∞, statistics such that Hni (p, b p0 ) i = 1, 2. From (4.7) if N (i) (ε, s) is the sample size required to make Hni (p, significant at level ε, we have N (2) (ε, s) c1 (p) → , c2 (p) N (1) (ε, s)

with probability one when ε → 0. Consequently c1 (p)/c2 (p) is a measure of the b p0 ) relative to Hn2 (p, b p0 ). asymptotic efficiency of Hn1 (p, Based on this relation we can give the following definition:

Definition 4.3 1 b p0 ) relative to Let p ∈4+ M . A measure of the asymptotic efficiency of Hn (p, b p0 ) is given by c1 (p)/c2 (p). Hn2 (p, The following result given in Bahadur (1971) (Theorem 7.2) in general populations and adapted to multinomial populations in Lemma 3.1 in Cressie and Read (1984) will be important to establish the main theorem in this Section.


170


Lemma 4.1 b p0 ) is a test statistic for the simple null hypothesis H0 : p = p0 , based If Hn (p, on the first n observations of the sequence {Yk }k∈N , verifying: i) There exists a function b : ∆+ M → R with −∞ < b (p) < ∞, such that b p0 ) → b (p) n−1/2 Hn (p,

0 with probability one as n → ∞ for each p ∈4+ M and p 6= p .

ii) For each t in an open interval I, there exists a continuous function f , verifying ª © √ b p0 ) ≥ nt / p = p0 → −f (t) as n → ∞ n−1 log Pr Hn (p, © ª and b (p) /p 6= p0 ⊂ I.

b p0 ) is given by c(p) = 2f (b(p)) , for Then, the exact Bahadur slope for Hn (p, + 0 each p ∈4M and p 6= p .

Theorem 4.1 describes a useful method of finding the exact Bahadur slope for b p0 ). It is necessary to give some concepts the φ-divergence test statistic, Tnφ (p, before establishing its proof. Definition 4.4 Let A ⊂ 4+ M and An = 4M,n ∩ A, where ( 4M,n =

T

v = (v1 , ..., vM ) : vj = ij /n, ij ∈ N− {0} , j = 1, ..., M,

Let us say that A is p0 -regular, p0 ∈ 4+ M , if

M P

j=1

)

ij = n .

lim inf DKull (v, p0 ) = inf DKull (v, p0 ). v ∈A

n→∞ v ∈An

Some properties of this definition are given in the following proposition; its proof is in Bahadur (1971). Lemma 4.2 0 0 Let A ⊂ 4+ M and A the closure of A (interior of A). We have: i) If A ⊂ A0 (e.g., if A is open) A is p0 -regular



171

b ∈ 4+ ii) If A is p0 -regular and p M , there exists a positive constant, δ (M), depending only on M, such that µ ¶ − M−1 0 δ (M) n 2 exp −n inf DKull (v, p ) ≤ Prp0 {b p ∈ A} v ∈An µ ¶ ≤ (n + 1)M exp −n inf DKull (v, p0 ) . v ∈An Theorem 4.1 describes a practical way to get the exact Bahadur slope for the φ-divergence test statistic and it is based on Lemmas 4.1 and 4.2. Theorem 4.1 Let φ continuously differentiable and verifying φ (r) < ∞. r→∞ r

φ (0) + lim Then, the sequence of test statistics

verifies

ª © b p0 ) n∈N = Hn (p,

½q ¾ φ 0 bp ) Tn (p,

n∈N

0 i) Under p ∈4+ M and p 6= p ,

s q b p0 ) → n−1/2 Tnφ (p,

2 φ00 (1)

Dφ (p, p0 )

with probability 1 as n → ∞.

ii) Under the null hypothesis H0 : p = p0 , ¾ ½q φ −1 1/2 0 0 b p ) ≥ n t / p = p → − inf DKull (v, p0 ) n log Pr Tn (p, v ∈Aφ,t as n → ∞ for each t in an open interval with s ( ) 2 Aφ,t = v : v ∈ ∆+ DKull (v, p0 ) ≥ t . M and φ00 (1)

b p0 ) is given iii) The exact Bahadur slope of the φ-divergence test statistic Tnφ (p, by cφ (p) = inf 2DKull (v, p0 ), v ∈Bφ


172


where

© ª 0 0 Bφ = v : v ∈4+ M and Dφ (v, p ) ≥ Dφ (p, p ) ,

0 p ∈4+ M and p 6= p .

Proof. We shall establish the proof step by step. c.s.

b −→ p. Now the result follows by the continuity i) We know that under p 6= p0 , p n→∞

of Dφ (v, p0 ) with respect to the first argument. Condition i) of Lemma 4.1 is verified with s 2 b(p) = Dφ ( p, p0 ). 00 φ (1)

ii) It is well known that 0 < Dφ (v, p0 ) ≤ φ (0) + lim

r→∞

therefore if we denote by γ1 =

r

2

φ (r) ; r

³ φ (0) + lim

φ(r) r→∞ r

range of b, as p varies over 4+ M q φ 1/2 0 b p ) ≥ n t is equivalent to the event Tn (p, s ( pb ∈ Aφ,t =

v:v

´

we have that the © 0ª − p , is I = (0, γ1 ). For t ∈ I, the event φ00 (1)

∈4+ M

and

) 2 Dφ (v, p0 ) ≥ t . φ00 (1)

The probability vector pb takes its values on the lattice 4M,n . We consider the set An = Aφ,t ∩ 4M,n .

The continuity of Dφ (v, p0 ) in v implies that Aφ,t is the closure of its interior and by Lemma 4.2, p0 -regular for any v ∈4M,n . By ii) in Lemma 4.2, we have under p ∈4+ M ª © b ∈ Aφ,t /p = p0 = − inf DKull (v, p0 ), n−1 log Pr p v ∈Aφ,t i.e.,

−1

n

¶ µq √ φ 0 log Pr Tn (v, p0 ) ≥ nt/p = p = − inf DKull (v, p0 ). v ∈Aφ,t

Therefore we have the condition ii) of Lemma 4.1 with f(t) = inf DKull (v, p0 ), t ∈ I = (0, γ1 ) v∈Aφ,t



173

¡ ¢ and f is a continuous function, by the continuity of DKull v, p0 , in I and of ª © 0 . course I ⊃ b (p) /p ∈4+ M and p 6= p q iii) Applying Lemma 4.1 we have that the exact Bahadur slope for Tnφ (v, p0 ), with φ fixed, is 2f (p); i.e., cφ (p) ≡ 2f(p) = 2 inf v ∈Bφ DKull (v, p0 ), where Bφ = Aφ,f (p) . Since the function g(x) = x2 is a strictly monotonic increasing function of b p0 ) will x > 0, then the level obtained by the φ-divergence test statistic Tnφ (p, b p0 ) for every nnand s. Hence be the same as the one attained by Hn (p, o the exact ª © φ 0 0 b p ) n∈N and Tn (p, bp ) will be Bahadur slopes of both sequences Hn (p, n∈N

the same as Bahadur (1971, see Remark 2, p. 27) pointed out.

Remark 4.1 It is necessary to consider Hn instead of Tnφ due to condition i) in Lemma 4.1. Remark 4.2 If we consider φ (x) = x log x − x + 1 we have

and then

© ª 0 0 BKull = v : v ∈4+ M and DKull (v, p ) ≥ DKull (p, p ) 0 cKull (p) = 2DKull (p, p0 ), p ∈4+ M and p 6= p .

Therefore cφ (p) = inf 2DKull (v, p0 ) ≤ 2DKull (p, p0 ) = cKull (p) v ∈Bφ because p ∈Bφ . Then we have 0 cφ (p)/cKull (p) ≤ 1, for all v ∈4+ M p 6= p .

The likelihood ratio test obtained for φ (x) = x log x − x + 1 has maximal Bahadur efficiency among all the φ-divergence test statistics. However, in the same way as Cressie and Read (1984) pointed out with the power-divergence test statistic, other family members can be equally efficient if there does not exist a probability vector v satisfying both DKull (v, p0 ) < DKull (p, p0 ) and Dφ (v, p0 ) ≥ Dφ (p, p0 ).


174


Computing measures of large sample performance often leads to different conclusions. We have seen that, when M is fixed, the likelihood ratio test statistic is superior to any φ-divergence test statistic in terms of exact Bahadur efficiency, but all φ-divergence test statistics are equivalent in terms of Pitman efficiency. There are interesting results in relation to the chi-square test statistic and likelihood ratio test statistic when the number of cells is not fixed. Quine and Robinson (1985) studied the problem when Mn → ∞ and they found that chi-square test statistic has Pitman efficiency 1 and Bahadur efficiency 0 relative to likelihood ratio test statistic. But if n/Mn → γ ∈ (0, ∞) , i.e., sparseness assumption, chisquare test statistic is strictly superior to likelihood ratio test statistic in the Pitman sense but still has Bahadur efficiency 0. Similar results were found by Kallenberg et al . (1985). An interesting overview of this problem can be seen in Drost et al . (1989). The results presented by Quine and Robinson (1985) can be extended to the family of φ-divergence test statistics. This is an open problem.

4.2.3.

Approximations to the Power Function: Comparisons

In Chapter 3 we obtained two approximations to the power function, associated with the decision rule given in (4.3), via approximations to the limiting alternative distribution of the φ-divergence test statistic, one for contiguous alternative hypotheses and another for fixed alternatives. Given the contiguous alternative hypothesis pn we established that ¡ ¢ lim βn,φ (pn ) = 1 − Gχ2 (δ) χ2M−1,α , M−1

n→∞

where δ was given in (4.5).

Then every φ-divergence test statistic has the same asymptotic distribution. The distribution Gχ2 (δ) gives a fair approximation to the power function of M−1 the chi-square test statistic when n ≥ 100. The approximation is not as good for the likelihood ratio test statistic; see Kallenberg et al . (1985) and Broffitt and Randles (1977). Drost et al . (1989) pointed out that in the case of the power-divergence family the approximation for values of λ 6= 1 is very poor. The approach based on Gχ2 (δ) was considered for the first time by Patnaik (1949). M−1 Slakter (1968) simulated the power of the chi-square test statistic in many cases and compared it to the approach given by Gχ2 (δ) . He was very pessimistic. An M−1 interesting study is also given in Haber (1980).



175

The other asymptotic approximation has the expression βn,φ (p∗ ) ≈ 1 − Φ

³

1 σ1 (p∗ )

³

φ00√(1) 2 χ 2 n M−1,α

−

´´ √ nDφ (p∗ , p0 ) ,

where σ1 (p∗ ) is given in (3.10). Broffit and Randles (1977) compared the two previous approximations, in the case of the chi-square test statistic, and they conclude that for large values of the true power the normal approximation is better, but for moderate values of the power, the approximation Gχ2 (δ) is better. Menéndez et al . (2001a) considered M−1 a model in which the normal approximation, βn,φ (p∗ ), of the power function is very poor for the family of the power-divergence test statistics. Drost et al . (1989) proposed two new approximations to the power function for the power-divergence test statistic. Both the computation and results on asymptotic error rates suggest that the new approximations are greatly superior to the traditional power approximations. Sekiya et al. (1999) proposed a normal approximation based on the normalizing transformation of the power-divergence test statistic. Sekiya and Taneichi (2004) using multivariate Edgeworth expansion for a continuous distribution showed how the normal approximation can be improved. Obtaining these approximations for the φ-divergence test statistic is an open problem.

4.3.

Exact and Asymptotic Moments: Comparison

The speed of convergence of the exact moments to the asymptotic moments, in the family of φ-divergence test statistics, gives us information about the speed of convergence of the exact distribution to its asymptotic distribution. We shall consider a second order Taylor expansion of the first three exact b p0 ), and we compare them to moments of the φ-divergence test statistic, Tnφ (p, the corresponding moments of a chi-square distribution with M − 1 degrees of freedom. The sizes of the correction terms will give information about the errors that we are doing when we use the asymptotic distribution instead of the exact distribution. The method was used by Cressie and Read (1984). We therefore omit its justification or motivation. We consider the problem under the null hypothesis as well as under contiguous alternative hypotheses.


176


4.3.1.

Under the Null Hypothesis

We denote b p0 )) µβ (Tnφ (p,

=E

·³

´β ¸

b p0 ) Tnφ (p,

, β = 1, 2, 3,

and we shall establish, in Propositions 4.1, 4.2 and 4.3 that ³ ´ ¡ ¢ 1 b p0 ) = µβ χ2M−1 + fφβ + O(n−3/2 ), β = 1, 2, 3, µβ Tnφ (p, n

where µβ (χ2M−1 ) = E

h¡ h¡ ¢β i ¢β i . We point out that E χ2M−1 = M − 1, χ2M−1

M 2 − 1 or M 3 + 3M 2 − M − 3, if β = 1, 2 or 3, respectively.

Then fφβ controls the speed at which the first exact three moments, about b p0 ), converge to the first three the origin, of the φ-divergence test statistic Tnφ (p, moments of a chi-square distribution with M −1 degrees of freedom. The function φ for which fφβ = 0, β = 1, 2, 3, will be the best. In the next propositions we shall obtain the second order Taylor expansion of the first three moments, with respect to the origin, of the φ-divergence test statistic. Proposition 4.1 It holds

where

h i 1 b p0 ) = M − 1 + fφ1 + O(n−3/2 ), E Tnφ (p, n

fφ1 = and S =

φIV (1) φ000 (1) (2 − 3M + S) + (1 − 2M + S) 3φ00 (1) 4φ00 (1)

PM

0 −1 j=1 (pj ) .

Proof. Let us denote ¢ 1 ¡ Wj = √ Nj − np0j , j = 1, ..., M. n © 2006 by Taylor & Francis Group, LLC

(4.8)

177


b p0 ) around p0 gives A fourth order Taylor expansion of Dφ (p, µ ¶ ´ M M ³ P ∂Dφ (p,p0 ) ∂ 2 Dφ (p,p0 ) Wj 1 P 0 √ bp ) = + Dφ (p, ∂pj ∂p2j p=p0 n 2! j=1 j=1 p=p0 µ ¶ M 3 3 0 P W ∂ Dφ (p,p ) √j + 3!1 n n ∂p3j 0 j=1 µ ¶p=p M Wj4 ∂ 4 Dφ (p,p0 ) 1 P + 4! + OP (n−5/2 ). 4 n2 ∂pj j=1 p=p0 But ³

Wj2 n

µ µ ¶¶ p = φ0 p0j = φ0 (1), j p=p0 0 p=p ¶ µ ¶¶ µ µ ∂ 2 Dφ (p,p0 ) 1 00 pj = = p10 φ00 (1), 0φ 0 ∂p2j p p j j j p=p0 p=p0 ¶ µ ¶¶ µ µ ∂ 3 Dφ (p,p0 ) pj 1 000 = = 10 2 φ000 (1), 2φ ∂p3j p0j p0j ) ( (pj ) 0 0 p=p p=p µ ¶ µ ¶¶ µ ∂ 4 Dφ (p,p0 ) pj 1 IV = φ = 10 3 φIV (1). 0 0 3 ∂p4j p p j ( ) (pj ) j p=p0 p=p0 ∂Dφ (p,p0 ) ∂pj

´

Then b p0 ) = Tnφ (p,

+

2n

φ00 (1)

b p0 ) = Dφ (p,

φIV (1) 12nφ00 (1)

M P

Wj4

3

0 j=1 (pj )

M P

j=1

Wj2 p0j

+ OP

+

φ000 (1) √ 3 nφ00 (1)

(n−3/2 ).

M P

Wj3 2

0 j=1 (pj )

(4.9)

By (4.9) we can write h i b p0 ) = E Tnφ (p,

+

since

M P

j=1

E [Wj2 ] p0j

φIV (1) 12nφ00 (1)

+

φ000 (1) √ 3 nφ00 (1)

M P

j=1

E [Wj3 ]

(p0j )

2

M E [W 4 ] P j + O(n−3/2 ), 0 3 j=1 (pj )

i h E OP (n−3/2 ) = O(n−3/2 ).

The moment-generating function of a multinomial random variable, N = (N 1 , ..., NM )T ,


(4.10)

178


with parameters n and p0 is ¡ ¢n MN (t) = E[exp(tT N )] = p01 exp(t1 ) + ... + p0M exp(tM )

with t = (t1 , ..., tM )T . The moment generating function of the M-dimensional ¡ ¢ random variable W = √1n N − np0 is thus given by ¡ ¢ √ √ MW (t) = E[exp(tT W )] = E[exp(tT N / n − np0 )] √ √ = exp(− ntT p0 )E[exp(tT N / n)] √ √ = exp(− ntT p0 )M(t/ n),

and the ath-moment of Wj about the origin by ! Ã a M (t) ∂ W E[Wja ] = ∂taj t=0 for j = 1, ..., M,

(4.11)

(4.12)

a = 1, 2, ....

From (4.11) and (4.12) we have E[Wj2 ] = −(p0j )2 + p0j ³ ´ E[Wj3 ] = n−1/2 2(p0j )3 − 3(p0j )2 + p0j

)4 − 6(p0j )3 + 3(p0j )2 E[Wj4 ] = 3(p0j ³ ´ + n−1 −6(p0j )4 + 12(p0j )3 − 7(p0j )2 + p0j ,

and substituting these expressions in (4.10) the proof is complete. Proposition 4.2 It holds where

i h 1 b p0 )2 = M 2 − 1 + fφ2 + O(n−3/2 ), E Tnφ (p, n

¡ ¢ ¢ 000 (1) ¡ 2 2 − 2M − M 2 + S + 2φ 3φ00 (1) 10 − 13M − 6M (M + 8)S ³ 000 ´2 ¡ ¢ (1) 4 − 6M − 3M 2 + 5S + 13 φφ00 (1) ¢ IV (1) ¡ 3 − 5M − 2M 2 + (M + 3) S + φ2φ00 (1)

fφ2 =

and S = Proof.

PM

0 −1 j=1 (pj ) .


(4.13)

179


Squaring and taking expectations in (4.9) we get E

·³

´2 ¸

b p0 ) Tnφ (p,

M E [W 2 W 2 ] P j i + p0j p0i ( ) j=1 Ã j6=i ! M E [W 5 ] M E [W 2 W 3 ] 000 P P 2φ (1) j j i + 3√nφ00 (1) + 0 3 0 0 2 j=1 (pj ) j6=i pj (pi ) Ã Ã ! M E [W 6 ] M E [W 2 W 4 ] IV (1) P P φ j j i + + n1 6φ00 (1) (p0j )4 p0j (p0i )3 j=1 j6=i Ã !! ³ 000 ´2 P M E [W 6 ] M E [W 3 W 3 ] P φ (1) j j i + + O(n−3/2 ). + 3φ00 (1) (p0 )4 (p0 )2 (p0 )2

=

M P

E [Wj4 ] 2 p0j

j=1

j

j

j6=i

i

(4.14)

By (4.11) we have

E[Wja Wib ]

=

Ã

∂ a+b MW ( t) ∂taj ∂tbi

!

(4.15)

t=0

for j, i = 1, ..., M and a, b = 1, 2, ... . Then ³ ´ E[Wj5 ] = n−1/2 −20(p0j )5 + 50(p0j )4 − 40(p0j )3 + 10(p0j )2 ³ ´ + n−3/2 24(p0j )5 − 60(p0j )4 + 50(p0j )3 − 15(p0j )2 + p0j , and

0 6 0 5 0 4 0 3 E[Wj6 ] = −15(p ³ j ) + 45(pj ) − 45(pj ) + 15(pj ) ´ + n−1 130(p0j )6 − 390(p0j )5 + 415(p0j )4 − 180(p0j )3 + 25(p0j )2

+ O(n−2 ).

For j 6= i both fixed, we define pab = (p0j )a (p0i )b ; then E[Wj2 Wi2 ] = 3p22 − p21 − p12 + p11 + n−1 (−6p22 + 2p21 + 2p12 − p11 ) , E[Wj2 Wi3 ] = n−1/2 (−20p23 + 5p13 + 15p22 − 6p12 − p21 + p11 ) , E[Wj2 Wi4 ] = −15p24 + 18p23 + 3p14 − 3p22 − 6p13 + 3p12 + n1 (130p24 − 156p23 − 26p14 + 41p22 + 42p13 − p21 − 17p12 + p11 ) + O(n−2 ), E[Wj3 Wi3 ] = −15p33 + 9p32 + 9p23 − 9p22 + n−1 (130p33 − 78p32 − 78p23 + 63p22 + 5p31 + 5p13 − 6p21 − 6p12 + p11 ) + O(n−2 ),


180


and substituting these expressions in (4.14) and simplifying we have E where

·³

´2 ¸

b p0 ) Tnφ (p,

= M2 − 1 +

1 2 f + O(n−3/2 ), n φ

¡ ¢ 2 − 2M − M 2 + S ¢ 2φ000 (1) ¡ 2 3φ00 (1) 10 − 13M − 6M + (M + 8)S ³ 000 ´2 ¡ ¢ (1) 4 − 6M − 3M 2 + 5S + 13 φφ00 (1) ¢ IV (1) ¡ 3 − 5M − 2M 2 + (M + 3) S . + φ2φ00 (1)

fφ2 = +

To obtain the previous expression we have used the following relations, M M M X X ¡ 0 ¢2 X pj + p0j p0i = 1; p0j = M − 1 j=1

j6=i

j6=i

and M X p0j j6=i

p0i

= S − M.

Proposition 4.3 It holds ·³ ´3 ¸ b p0 ) E Tnφ (p, = M 3 + 3M 2 − M − 3 + n1 fφ3 + O(n−3/2 ) where

¡ ¢ 26 − 24M − 21M 2 − 3M 3 + (19 + 3M ) S ¢ ¢ ¡ φ000 (1) ¡ 2 3 2 S φ00 (1) 70 − 81M − 64M − 9M + 65 + 18M + M ³ 000 ´2 ¡ ¢ φ (1) 20 − 26M − 21M 2 − 3M 3 + (25 + 5M) S + φ00 (1) ¢ ¢ ¡ IV (1) ¡ 2 3 2 S + 3φ 4φ00 (1) 15 − 22M − 15M − 2M + 15 + 8M + M

fφ3 = +

and S = Proof.

PM

0 −1 j=1 (pj ) .


(4.16)


181

Cubing and taking expectations in (4.9) we have h i b p0 )3 = E Tnφ (p,

M E [W 6 ] P j

M E [W 4 W 2 ] M E [W 2 W 2 W 2 ] P P j i j i k + 2 0 p0 p0j p0i p0k p ( ) j i j=1 j6 = i j6 = i6 = k Ã M E [W 7 ] M E [W 4 W 3 ] M E [W 2 W 5 ] P P P φ000 (1) j j i j i + √nφ00 (1) + + 2 4 2 2 0 0 0 0 p0 3 p p p p ( ) ( ) ( ) ( ) j i j i j=1 j6=i Ã j6=i j Ã ¶ ³ ´ M 2 2 3 8 2 000 (1) P E [Wj ] E [Wj Wi Wk ] + n1 13 φφ00 (1) + 2 0 5 p0j p0i (p0k ) j=1 (pj ) ! M E [W 6 W 2 ] M E [W 3 W 5 ] M E [W 3 W 3 W 2 ] P P P j i j i j i k + +2 + 0 4 p0 0 2 0 3 0 2 0 2 0 j) i j6=i (pÃ j6=i (pj ) (pi ) j6=i6=k (pj ) (pi ) pk M E [W 8 ] M E [W 4 W 4 ] M E [W 2 W 6 ] IV (1) P P P j j i j i + φ4φ00 (1) 5 + 2 0 3 +2 0 0 0 p0 4 p p p p ( ) ( ) ( ) ( ) j j i j i j=1 j6=i !!j6=i M E [W 2 W 2 W 4 ] P j i k + + O(n−3/2 ). 0 p0 p0 3 p ( ) j i k j6=i6=k (4.17) Using again (4.11) we get ³ ´ E[Wj7 ] = n−1/2 210(p0j )7 − 735(p0j )6 + 945(p0j )5 − 525(p0j )4 + 105(p0j )3 p30j

+3

+ O(n−3/2 ),

E[Wj8 ] = 105(p0j )8 − 420(p0j )7 + 630(p0j )6 − 420(p0j )5 + 105(p0j )4 + O(n−1 ). For j 6= i 6= k all fixed, we define pab = (p0j )a (p0i )b and pabc = (p0j )a (p0i )b (p0k )c . Then E[Wj4 Wi3 ] = n−1/2 (210p43 − 105p42 − 210p33 + 3p41 + 144p32 + 36p23 − 6p31 − 39p22 + 3p21 ) + O(n−3/2 ), E[Wj4 Wi4 ] = 105p44 − 90 (p43 + p34 ) + 9 (p24 + 3p42 ) + 108p33 − 18(p32 + p23 ) + 9p22 + O(n−1 ), E[Wj5 Wi2 ] = n−1/2 (210p52 − 35p51 − 350p42 + 80p41 + 150p32 − 55p31 − 10p22 + 10p21 ) + O(n−3/2 ), E[Wj5 Wi3 ] = 105p53 − 45p52 − 150p43 + 90p42 + 45p33 − 45p32 + O(n−1 ), E[Wj6 Wi2 ] = 105p62 − 15p61 − 225p52 + 45p51 + 135p42 − 45p41 − 15p32 + 15p31 + O(n−1 ),


182


E[Wj2 Wi2 Wk2 ] = −15p222 + 3 (p122 + p212 + p221 ) − (p112 + p121 + p211 ) + p111 + n−1 (130p222 − 26 (p122 + p212 + p221 ) +7 (p112 + p121 + p211 ) − 3p111 ) + O(n−2 ), E[Wj2 Wi2 Wk3 ] = n−1/2 (210p223 − 105p222 − 35(p123 + p213 ) + 8p113 + 24(p112 + p212 ) + 3p221 − 9p112 − (p211 + p121 ) + p111 ) + O(n−3/2 ), E[Wj2 Wi3 Wk3 ] = 105p233 − 15p133 − 45(p223 + p232 ) + 9 (p132 + p123 ) + 27p222 − 9p122 + O(n−1 ), E[Wj2 Wi2 Wk4 ] = 105p224 − 15 (p124 + p214 ) − 90p223 + 3p114 + 18 (p123 + p213 ) + 9p222 − 6p113 − 3 (p122 + p212 ) + 3p112 + O(n−1 ), and by (4.17) we have ·³ ´3 ¸ 1 φ 0 bp ) = M 3 + 3M 2 − M − 3 + fφ3 + O(n−3/2 ), E Tn (p, n

where

¡ ¢ 26 − 24M − 21M 2 − 3M 3 + (19 + 3M ) S ¡ ¢ ¢ φ000 (1) ¡ 2 3 2 S φ00 (1) 70 − 81M − 64M − 9M + 65 + 18M + M ³ 000 ´2 ¡ ¢ φ (1) 20 − 26M − 21M 2 − 3M 3 + (25 + 5M ) S + φ00 (1) ¢ ¢ ¡ IV (1) ¡ 2 − 2M 3 + 15 + 8M + M 2 S . 15 − 22M − 15M + 3φ 00 4φ (1)

fφ3 = +

To obtain the previous expression we have used the following relations: M ¡ 0 ¢2 M X X p0j p0i pi + = S − 2M + 1 p0i p0k j6=i

j6=i6=k

and M M M X X X ¡ 0 ¢3 ¡ 0 ¢2 0 pj + 3 pj pi + p0j p0i p0k = 1. j=1

j6=i

j6=i6=k

Remark 4.3 It is clear that fφi , i = 1, 2, 3, control the speed at which the first three exact b p0 ), converge moments, about the origin, of the φ-divergence test statistic, Tnφ (p, to the three first moments, about the origin, of a chi-square random variable with M − 1 degrees of freedom.



183

Let us consider a function φ ∈ Φ∗ depending on a parameter “a”. In the following we shall denote it by φ ≡ φa . In this context as optimal in the sense of the moments of order β we consider the values of “a” from the set Rβ of roots of the equations fφi a = 0, i = 1, 2, 3. Strictly speaking these expansions are valid only as n → ∞ for M < ∞ fixed. If now we consider the null hypothesis H0 : p = p0 = (1/M, ..., 1/M)T , we PM 0 −1 = M 2 . However for M increasing the roots of the equation have j=1 (pj ) fφ1a = 0 converge to the roots of the equation IV 4φ000 a (1) + 3φa (1) = 0,

for φ00a (1) 6= 0,

(4.18)

since the equation (4.8) can be written as µ ¶ 2 − 3M + M 2 4φ000 (1) + 3φIV a a (1) = 0. 1 − 2M + M 2 Then the roots of this last equation converge to the roots of the equation (4.18) as M → ∞. In relation with the equations fφi a = 0, i = 2, 3, it is possible to apply similar arguments. If we consider the family of the power-divergence test statistics, φ ≡ φ(λ) , we have that the roots of the equation (4.18) are λ = 1 and λ = 2/3. These values were found directly by Read and Cressie (1988). Example 4.1 We consider the family of φ-divergence measures given by φa (x) =

(1 − x)2 , 2 (a + (1 − a) x)

a ∈ [0, 1] ,

(4.19)

i.e., the family of Rukhin’s divergence measures. This family was introduced by Rukhin (1994). The associated family of test statistics has the expression b p0 ) = n Tnφa (p,

M X j=1

(p0j − pbj )2

ap0j + (1 − a)b pj

0 ≤ a ≤ 1.

(4.20)

Some properties of this test statistic can be seen in Pardo, M. C. and J. A. Pardo b p0 ) is the modified chi-square test (1999). It is observed immediately that Tnφ0 (p, φ1 0 b p ) is the chi-square test statistic. statistic and Tn (p, For a ∈ [0, 1] we have

3

3(1−a) φ000 a (x) = − (a+(1−a)x)4


and

φIV a (x) =

12(1−a)4 (a+(1−a)x)5

184


and the equation (4.18) becomes 36a2 − 60a + 24 = 0. Then the roots of the equation (4.18) are a = 1 and a = 2/3. It seems interesting to know how large M has to be for using the roots a = 1 and a = 2/3. If M is not large we must use the roots of the equations fφi a = 0, i = 1, 2, 3 given in (4.8), (4.13) and (4.16). These solutions are given in Tables 4.1, P 0 −1 changes. 4.2, 4.3 and 4.4 as the number of classes M increases and M j=1 (pj ) PM In particular we have considered, for j=1 (p0j )−1 , the values M 2 , M 3 , M 4 and M 5.

fφ1a fφ2a fφ3a

M a1 a2 a1 a2 a1 a2

2 3 1.00 1.00 1.00 0.83 1.45 1.20 0.55 0.59 1.31 1.09 0.69 0.67

4 1.00 0.78 1.13 0.60 1.05 0.66

5 1.00 0.75 1.10 0.61 1.03 0.65

10 1.00 0.70 1.05 0.63 1.01 0.65

20 1.00 0.68 1.02 0.64 1.00 0.66

40 1.00 0.67 1.01 0.65 1.00 0.66

50 1.00 0.67 1.01 0.66 1.00 0.66

Table 4.1. Roots (a1 > a2 ) for fφi a = 0, i = 1, 2, 3, for

fφ1a fφ2a fφ3a

M a1 a2 a1 a2 a1 a2

2 3 1.00 1.00 0.73 0.70 1.06 1.03 0.55 0.57 0.93 0.93 0.62 0.62

4 1.00 0.68 1.02 0.58 0.93 0.62

5 1.00 0.68 1.01 0.59 0.93 0.62

10 1.00 0.67 1.00 0.61 0.95 0.63

20 1.00 0.67 1.00 0.63 0.97 0.64

40 1.00 0.67 1.00 0.65 0.98 0.65

fφ1a fφ2a fφ3a

2 3 1.00 1.00 0.69 0.67 1.02 1.01 0.55 0.57 0.91 0.91 0.61 0.61

4 1.00 0.67 1.00 0.58 0.92 0.61

5 1.00 0.67 1.00 0.58 0.93 0.61

10 1.00 0.67 1.00 0.61 0.95 0.63

20 1.00 0.67 1.00 0.63 0.96 0.64

40 1.00 0.67 1.00 0.65 0.98 0.65

0 −1 j=1 (pj )

50 1.00 0.67 1.00 0.65 0.98 0.65

100 1.00 0.67 1.00 0.65 0.99 0.66

= M 2.

200...∞ 1.00...1.00 0.67...2/3 1.00...1.00 0.66...2/3 0.99...1.00 0.66...2/3

PM

50 1.00 0.67 1.00 0.65 0.98 0.65



200...∞ 1.00...1.00 0.67... 2/3 1.00...1.00 0.66... 2/3 1.00...1.00 0.66... 2/3

PM


M a1 a2 a1 a2 a1 a2

100 1.00 0.67 1.00 0.66 1.00 0.66

0 −1 j=1 (pj )

100 1.00 0.67 1.00 0.66 0.99 0.66 PM

= M 3.

200...∞ 1.00...1.00 0.67...2/3 1.00...1.00 0.66...2/3 0.99...1.00 0.66...2/3

0 −1 j=1 (pj )

= M 4.


fφ1a fφ2a fφ3a

M a1 a2 a1 a2 a1 a2

2 3 1.00 1.00 0.68 0.67 1.01 1.00 0.55 0.57 0.91 0.91 0.60 0.60

4 1.00 0.67 1.00 0.57 0.92 0.61

5 1.00 0.67 1.00 0.59 0.92 0.61

10 1.00 0.67 1.00 0.61 0.95 0.63

20 1.00 0.67 1.00 0.63 0.96 0.64

40 1.00 0.67 1.00 0.65 0.98 0.65

50 1.00 0.67 1.00 0.65 0.98 0.65


100 1.00 0.67 1.00 0.66 0.99 0.66

185

200...∞ 1.00...1.00 0.67...2/3 1.00...1.00 0.66...2/3 0.99...1.00 0.66...2/3

PM

0 −1 j=1 (pj )

= M 5.

For M ≥ 20 all roots are within ±0.05 of the limiting roots a = 1 and a = 2/3. Therefore for M > 20, choosing a = 1 or a = 2/3 the convergence of the first three moments to those of a random variable chi-square with M − 1 degrees of freedom is faster. For M ≥ 4, most of the roots are within ±0.1 of the limiting roots a = 1 and a = 2/3. This suggests that the “a”-range [0.6, 1] is optimal for all M not too small. For M < 4 we should use the previous table.

4.3.2.

Under Contiguous Alternative Hypotheses

Under the contiguous alternative hypotheses, H1,n given in Section 3.2, it was established that the asymptotic distribution of the φ-divergence test stab p0 ), is noncentral chi-square with M − 1 degrees of freedom and tistic, Tnφ (p, P 2 0 noncentrality parameter δ = M j=1 dj /pj . In the same way as under the null hypothesis we shall establish in this case h i ¢ ¡ 1 b p0 ) = M − 1 + δ + √ gφ1 + O n−1 E Tnφ (p, n

i h ¢ ¡ 1 b p0 )2 = M 2 − 1 + 2(M − 1)δ + δ2 + √ gφ2 + O n−1 . E Tnφ (p, n

(4.21)

(4.22)

Based on (4.21) and (4.22) it is possible to give conditions to the function φ for improving the approximations of the exact moments to the asymptotic moments for the different values of M, p0 and d. The convergence speed of the exact moments to the asymptotic moments gives information about the convergence speed of the exact distribution to the asymptotic distribution. We are going to get the expressions of gφi , i = 1, 2. We shall omit the third ¢ ¡ moment and the expansion is only considered until the order O n−1/2 . Note © 2006 by Taylor & Francis Group, LLC

186


that b p0 ) = Tnφ (p,

M W2 M W3 P 2n φ000 (1) P j j φ 0) = −1 √ b D ( p, p + n 0 0 )2 + OP (n ), 00 (1) φ00 (1) 3 nφ p (p j=1 j j=1 j

where Wj = n−1/2 (Nj − np0j ), j = 1, ..., M. Consider the random variable

Vj = n1/2 (b pj − pj ), j = 1, ..., M, where pbj = Nj /n and N = (N1 , . . . , NM ) is a multinomial random variable with parameters n and p = (p1 , . . . , pM )T , with pj = p0j + n−1/2 dj , j = 1, ..., M. Then we have that Wj = Vj + dj , j = 1, ..., M, and the moments of Wj can be obtained from the moments of Vj , j = 1, ..., M . We know that

and

h i h i 2 i h E W 000 E Wj3 M M j P φ (1) P φ 0 bp ) = + √ 00 + O(n−1 ) E Tn (p, 3 nφ (1) j=1 (p0j )2 p0j j=1

(4.23)

E[Wj2 ] = E[(Vj + dj )2 ] = −(p0j )2 + p0j + d2j + n−1/2 (dj − 2p0j dj ) + O(n−1 ), E[Wj3 ] = E[(Vj + dj )3 ] = d3j − 3dj (p0j )2 + 3dj p0j + O(n−1/2 ). Substituting the expressions of E[Wj2 ] and E[Wj3 ] in (4.23) we have h i M d2 P j b p0 ) = M − 1 + E Tnφ (p, p0j Ã j=1 Ã !! M M M 3 P 000 P P 1 d dj dj φ (1) j + √ +3 + O(n−1 ) 0 + 3φ00 (1) 0 2 p0 n j=1 pj j=1 (pj ) j=1 j = M −1+δ+

√1 g 1 n φ

where gφ1 =


  M M X X d3j dj   . + 00 +3 3φ (1) j=1 (p0j )2 p0j p0 j=1 j

M X dj j=1

¢ ¡ + O n−1 ,

φ000 (1)

(4.24)


187

On the other hand, ·³ ´2 ¸ φ 0 bp ) = E Tn (p,

M P

M E [W 2 W 2 ] P j i + p0j p0i ( ) j=1 Ã j6=i ! M E [W 5 ] M E [W 2 W 3 ] 000 P P 2φ (1) j j i + 3√ + + O(n−1 ); 0 3 0 p0 3 nφ00 (1) p p ( ) ( ) j j i j=1 j6=i (4.25) and by using the previous procedure we have E [Wj4 ] 2 p0j

E[Wj4 ] = 3(p0j )4³− 6(p0j )3 + 3(p0j )2 + 6p0j d2j − 6(p0j )2 d2j + d4j

´ + n−1/2 20(p0j )3 dj − 30(p0j )2 dj + 10p0j dj + 6d3j − 12p0j d3j ,

E[Wj2 Wi2 ] = + − + +

3p22 − p21 − p12 + p11 + (p01 − p02 ) d2j − 4p11 dj di (p10 − p20 ) d2j + d2j d2i + n−1/2 (6p21 di + 6p12 dj − p20 di 2p11 dj − p02 dj − 2p11 di + p01 dj + p10 di + 2(2p12 − p11 )dj (di − 2p01 di )d2j + 2(2p21 − p11 )di − 4(p10 dj + p01 dj )dj di ¢ (dj − 2p10 dj )d2i ,

E[Wj5 ] = 5(3p40j − 6p30j + 3p20j )dj + 10(p0j − p20j )d3j + d5j + O(n−1/2 ),

E[Wj2 Wi3 ] = 6(p13 − p12 )dj + 3(3p22 − p21 − p12 + p11 )dj + 3(p01 − p02 )d2j di − 6p11 dj d2i + (p10 − p20 )d3i + d2j d3i + O(n−1/2 ), where pab = (p0j )a (p0i )b . Substituting the previous expressions in (4.25) we get Ã !2 ·³ ´2 ¸ M d2 M d2 P P φ j j 0 2 bp ) + E Tn (p, = M − 1 + 2(M − 1) p0 p0 j=1 j j=1 j ÃÃ M M M 2 P P dj dj P dj + √1n + 2 2(M + 3) 0 0 p p p0 j=1 j j=1 j j=1 j Ã ! M M P P φ000 (1) d3j dj + 4 2(M + 3) + 2 0 0 00 p p φ (1) j=1 ( j ) j=1 j M d2 P M M 2 M P P d3j dj P dj j + 2 + 23 2 p0j p0j p0j p0 j=1 j=1 j=1 !! j=1 ( j ) M P d3j + 23 (M + 5) + O(n−1 ) 0 2 p ( ) j=1 j ¢ ¡ = M 2 − 1 + 2(M − 1)δ + δ 2 + √1n gφ2 + O n−1 , © 2006 by Taylor & Francis Group, LLC

(4.26)

188


where gφ2 is given by ÃÃ

M P

gφ2 =

d

Ã

M P

d

!Ã

M P

d2j p0j

!

M P

d3j

j j 2(M + 3) +2 +4 2 p0 p0 (p0 ) j=1 j j=1 j j=1 Ã Ã ! Ã ! j=1 j M M M d2 000 (1) P P P dj dj j + φφ00 (1) +2 2(M + 3) p0j p0j p0j j=1 Ã j=1 Ã ! Ã j=1 ! !!! M d2 M M 3 P P P dj d3j j 2 2 + 3 (M + 5) . + 3 0 2 0 2 p0 j=1 j j=1 (pj ) j=1 (pj )

!

From (4.24) and (4.26) we observe that the first two moments under the contiguous alternative hypotheses coincide with the moments of a noncentral chisquare distribution with M − 1 degrees of freedom and noncentrality parameter P d2j 1 2 −1/2 ) that depend on δ= M j=1 p0j plus terms gφ and gφ of order O(n φ, M, plus a term of order O(n−1 ).

M M M X X d3j dj X d2j , and , p p p2 j=1 0j j=1 0j j=1 0j

The term of order O(n−1/2 ) is cancelled, in the first moment, if we choose φ verifying  −1  M M M 3 000 X X X d φ (1)  dj   dj  j = −3 0 0 )2 + 3 0 φ00 (1) p (p p j j=1 j j=1 j=1 j and for the second moment if Ã Ã !Ã ! ! M M M d2 M 3 P P P P φ000 (1) d dj dj j j = − 2(M + 3) +4 p0j + 2 0 2 p0j p0j p φ00 (1) ( ) j j=1 j=1 Ã j=1! Ã j=1! Ã M M M d2 P P P dj dj j +2 × 2(M + 3) p0j p0j p0j j=1 j=1 j=1 !!−1 Ã !Ã M M 3 2 P dj P dj + (M + 5) . + 23 2 0 p0 j=1 (pj ) j=1 j

Example 4.2 If we consider the family of divergence measures given by (4.19) and p0j = ¢ ¡ 1/M , j = 1, ..., M, the O n−1/2 correction factors in the first and second moment are a = 1 and 2 a=1− , M P 2 M dj + M + 5 j=1



189

P 2 respectively. Since M ≥ 1 and M j=1 dj ≥ 0 this value belongs to the interval (0, 1) and, when M increases, tends to 1. This result hints at the Pearson’s test statistic (a = 1) having closest distribution to the approximate noncentral chi-square under contiguous alternative hypotheses. The same happens with the power-divergence family. See Cressie and Read (1984, p. 454).

4.3.3.

Corrected Phi-divergence Test Statistic

We know that £ £ ¤ ¤ µ = E χ2M−1 = M − 1 and σ 2 = V ar χ2M−1 = 2(M − 1).

b p0 ) (we shall denote the corrected φ-divergence test statistic We can modify Tnφ (p, b p0 )), in such a way that by c Tnφ (p, b p0 )] = µ + o(n−1 ) E[c Tnφ (p,

and

b p0 ] = σ2 + o(n−1 ). V ar[c Tnφ (p,

We know that

b p0 )] = µ + fφ1 /n + o(n−1 ) E[Tnφ (p,

and

b p0 )] = σ2 + bφ /n + o(n−1 ) V ar[Tnφ (p,

where fφ1 was defined in (4.8) and bφ is given by

¡ ¢ ¢ 000 (1) ¡ 2 2 − 2M − M 2 + S + 2φ φ00 (1) 4 − 6M − M + 3S ³ 000 ´2 ¡ ¢ (1) + 13 φφ00 (1) 4 − 6M − 3M 2 + 5S

bφ =

+

where S =

2φIV (1) φ00 (1)

(1 − 2M + S) ,

PM

0 −1 j=1 (pj ) .

We define

c

in such a way that

b p0 )= Tnφ (p,

b p0 ) − γφ Tnφ (p, p δφ

b p0 )] = µ + o(n−1 ) E[c Tnφ (p, © 2006 by Taylor & Francis Group, LLC

190


and b p0 )] = σ2 + o(n−1 ). V ar[c Tnφ (p,

To do this it is necessary to consider ³ p ´ γφ = µ 1 − δφ + fφ1 /n i.e.,

+ + and

δφ = 1 + bφ /nσ2 , ¢

1 2 − 2M − M 2 + S 2(M−1)n ¡ 000 2φ (1) 2 φ00 (1) 4 − 6M − M + 3S) ³ 000 ´2 ¡ ¢ 1 φ (1) 4 − 6M − 3M 2 + 5S 3 φ00 (1) ´ 2φIV (1) (1 − 2M + S) , φ00 (1)

δφ = 1 + +

¡¡

and

³ 000 ¡ p ¢ φ (1) γφ = (M − 1) 1 − δφ + n1 3φ 00 (1) (2 − 3M + S) ´ IV (1) + φ4φ00 (1) (1 − 2M + S) .

Example 4.3 We consider the family of divergence measures given in (4.19) and p0j = 1/M, j = 1, ..., M. It is immediate to get ´ 1 ´ ³ ³ ¡ ¢ γa = (M − 1) 1 − δa1/2 + (1 − a) − 2 − 3M + M 2 + 3(1 − a) (M − 1)2 n

and

δa = 1 +

4.4.

¢ 1¡ −1 − 6(1 − a) (M − 2) + 3(1 − a)2 (5M − 6) . n

A Second Order Approximation to the Exact Distribution

In this Section we shall present an approximation of the exact distribution b p0 ) extracting the φ-dependent second order component from the o (1) of Tnφ (p, term in (4.1). This second order component was obtained by Yarnold (1972) for the chi- square test statistic under the null hypothesis (the approximation consists of a term of multivariate Edgeworth expansion for a continuous distribution and a discontinuous term), by Siatoni and Fujikoshi (1984) for the likelihood ratio



191

test statistic and Freeman-Tukey test statistic, by Read (1984a) for the powerdivergence test statistic and by Menéndez et al . (1997c) for the φ-divergence test b p0 ). statistic Tnφ (p, Let (N1 , ..., NM ) be a M-dimensional random vector multinomially distributed with parameters n and p0 . We consider the (M − 1)-dimensional random vector f w = (W1 , ..., WM−1 ) , defined by ¢ √ ¡ Wj = n Nj /n − p0j , j = 1, ..., M − 1.

The random vector f w takes values on the lattice n ´ o √ ³ e0 , m ∈ K , L= f w = n m/n − p w = (w1 , ..., wM−1 )T : f

e0 = (p01 , ..., p0M−1 )T and K is given by where p n f K= m = (n1 , ..., nM−1 )T : nj is a nonnegative integer, j = 1, ..., M − 1 o PM−1 with n ≤ n . j j=1

Siatoni and Fujikoshi (1984) established, under the null hypothesis, the following asymptotic expansion for the probability mass function of the random vector f w, ¶ µ ³ ´ ´ ³ 1 1 Pr g W =f w = n−(M−1)/2 g (f , w) 1 + √ h1 (f w) + h2 (f w) + O n−3/2 n n

where

µ ¶ ¯ ¯−1/2 ¯ T −1 1 exp − 2 f w Σe0 f w ¯Σp e0¯ p

−(M−1)/2 ¯

· g (f w) = (2π) · h1 (f w) = − 12 · h2 (f w) =

1 2

M P

j=1

wj p0j

+

1 6

(h1 (f w))2 +

being

M P

j=1 1 12

wj3 , (p0j )2

(1 − S) +

1 4

M P

j=1

wj2 (p0j )2

−

1 12

M P

j=1

wj4 (p0j )3

M−1 ³ ´T ³ ´ X 0 0 0 e e e − p and w = − wj . Σp p 0 = diag p M e j=1

This expansion provides a local Edgeworth approximation for the probability of f w at each point f w in L. If f w had a continuous distribution function it would © 2006 by Taylor & Francis Group, LLC

192


be possible to use the standard Edgeworth expansion to calculate the probability of any set B, by µ ¶ Z Z ´ ³ 1 1 Pr (f w∈ B) = · · · g (f w) 1 + √ h1 (f w) + h2 (f w) df w + O n−3/2 . n n B

(4.27) But f w has a lattice distribution and Yarnold (1972) indicated that in this case that expression is not valid and established for “extended convex sets” (convex set whose sections parallel to each coordinate axis are all intervals) B the following result ³ ´ ³ ´ Pr g W ∈ B = J1 + J2 + J3 + O n−3/2 ,

where J1 is the Edgeworth’s expansion for a continuous distribution given in (4.27), while J2 is a term to account for the discontinuity of f w. This term is ¡ −1/2 ¢ ¡ −1 ¢ and J3 term is O n and it has a very complicated expression (see O n Siatoni and Fujikoshi (1984)). ´T ³ P g e0 , w We denote p∗ = √wn + p eM = w√Mn + p0M , wM = − M−1 i=1 wi , and we consider the extended convex set n o f B φ (b) = (4.28) w: f w ∈ L and Tnφ (p∗ , p0 ) < b . We have (see Menéndez et al . (1997c)), under the null hypothesis, that the disb p0 ) can be expressed by tribution function of Tnφ (p,

being

³ ´ ³ ´ b p0 ) < b) = Pr g W ∈B φ (b) = J1φ + J2φ + J3φ + O n−3/2 Pr(Tnφ (p,

J1φ = Pr(χ2M−1 < b) +

3 ´ ³ ¢ 1 X φ ¡ 2 rj Pr χM−1+2j < b + O n−3/2 , 24n j=0

where · r0φ = 2(1 − S)

³ IV ´ ¢ 000 (1) ¡ (1) S − M 2 − 3 φφ00 (1) (S − 2M + 1) · r1φ = 6 φφ00 (1) ³ 000 ´2 ¡ ¢ φ (1) + 5S − 3M 2 − 6M + 4 + 3(3S − M 2 − 2M) φ00 (1) © 2006 by Taylor & Francis Group, LLC


193

³ 000 ´ ³ IV ´ (1) (1) (8S − 6M 2 − 6M + 4) + 3 φφ00 (1) (S − 2M + 1) · r2φ = −2 φφ00 (1) ³ 000 ´2 ¡ ¢ ¡ ¢ (1) − 2 φφ00 (1) 5S − 3M 2 − 6M + 4 − 6 −M 2 + 2S − 2M + 1

and

· r3φ =

³

φ000 (1) φ00 (1)

´2 + 1 (5S − 3M 2 − 6M + 4).

¢ ¡ W and J2φ is a discontinuous O n−1 term to account for the discontinuity in g can be approximated to first order by  1/2 , M ³ ´ Y Jb2φ = N φ (b) − n(M−1)/2 V φ (b) e−b/2 (2πn)(M−1) p0j  , j=1

where N φ (b) is the number of lattice points in B φ (b) , and (πb)(M−1)/2 V (b) = Γ((M + 1)/2) φ

ÃM Y i=1

p0j

!1/2 µ

¶ b (l1 − l2 ) , 1+ 24(M + 1)n

where ¶ ¶ µ IV ¢ φ000 (1) 2 ¡ φ (1) 2 (S − 2M + 1) , − 6M + 4 and l = 3 5S − 3M 2 φ00 (1) φ00 (1) P 1 bφ being S = M j=1 p0 . This term J2 does not look so complicated but it is very hard l1 =

µ

j

to obtain it when n and M are not small, because of the difficulty for getting V φ (b). ¢ ¡ Finally, J3φ = O n−1 and its expression is too complicated. Neglecting the J3φ term Menéndez et al. (1997c) proposed, in the same way as Read (1984a), the approximation b p0 ) < b) ' J1φ + Jb2φ . Pr(Tnφ (p, The proof of this result can be seen in Menéndez et al. (1997c). A similar result has been established in Taneichi et al . (2001a, b) under contiguous alternative hypotheses. Read (1984a) studied the usefulness of this approximation to the exact distribution for the power-divergence test statistics and established that it is externally close to the exact distribution for n so small as 10, and furthermore it provides a substantial improvement over the chi-square first order approximation.


194


Example 4.4 For the family of divergence measures given in (4.19), we have for p0j = 1/M, j = 1, ..., M φa b p0 ) < b) ' J1a + Jb2a FEd (b) ≡ Pr(Tnφ (p, being

J1a = Pr(χ2M−1 < b) +

3 ³ ´ ¢ 1 X a ¡ 2 rj Pr χM−1+2j < b + O n−3/2 , 24n j=0

where · r0a = 2(1 − M 2 ) · r1a = 6M(M − 1)(1 − 3(1 − a)2 · r2a = 12(1 − a)(M 2 − 3M + 2) + 36(1 − a)2 (M − 1) − 6(M − 1)2 and · r3a = 2(3a − 2)2 (M 2 − 3M + 2). The term Jb2a has the expression

.³ ´ 1/2 ³ ´ Jb2a = N a (b) − n(M−1)/2 V a (b) e−b/2 (2πn)(M−1) (1/M)M

where N a (b) is the number of lattice points in B φa (b), B φa (b) is defined in the same way as B φ (b) in (4.28) by replacing φ by φa , and ³ ´ (πb)(M−1)/2 ¡ 1 ¢M/2 3b 2 (M − M 2 ) 1 + V a (b) = Γ((M+1)/2) (1 − a) M 4(M+1)n + O(n−3/2 ).

4.5.

Exact Powers Based on Exact Critical Regions

We consider a function φ, some specified alternative hypotheses and a significance level α and we are going to get the exact power function, using the exact critical value (without any reference to asymptotic) for the φ-divergence b p0 ). We shall use this exact power function to compare the test statistic Tnφ (p, b p0 ). This approach members of the family of φ-divergence test statistics Tnφ (p, © 2006 by Taylor & Francis Group, LLC


195

is similar to one used by West and Kempthorne (1972) and Haber (1984) to compare the power of the chi-square test statistic and likelihood ratio test statistic, by Read (1984b) to compare the power-divergence test statistic and by Pardo, M. C. and Pardo, J. A. (1999) to compare the family of Rukhin test statistics given in (4.20). We restrict our attention to the equiprobable null model H0 : p0 = (1/M, ..., 1/M)T and we consider alternative models where one of the M probabilities is perturbed, and the rest are adjusted so that they still sum to 1,  M −1−δ   M(M − 1) H1 : pi = 1 + δ   M

if i = 1, ..., M − 1

(4.29)

if i = M

where −1 ≤ δ ≤ M − 1. We are going to justify a little bit the equiprobable null model. Sturges (1926) initiated the study of the choice of cell and recommended that the cell would be chosen to have equal probabilities with M = 1 + 2.303 log10 n. Mann and Wald ¢1/5 ¡ where zα denotes the (1942) for a sample size n recommended M = 4 2n2 zα−1 100(1−α) percentile of the standard normal distribution. Schorr (1974) confirmed that the “optimum” M is smaller than the value given by Mann and Wald and he suggested using M = 2n2/5 . Greenwood and Nikulin (1996) suggested using M ≤ min (1/α, log n) . Cohen and Sackrowitz (1975) proved that the tests which P lead to reject the above hypothesis if M i=1 hi (xi ) > c, where c is positive, hi , i = 1, ..., M, are convex functions and xi ≥ 0, i = 1, ..., M, are unbiased for equal cell probabilities. Bednarski and Ledwina (1978) stated that if tests of fit are based on continuous functions, then in general they are biased for testing an arbitrary simple hypothesis. b p0 ), In order to get the exact power for each φ-divergence test statistic Tnφ (p, it is necessary to choose a significance level α and calculate the associated critical region. To do this, we will see first the way to get exact 100 (1 − α) percentiles, b p0 ). tφn,M,α , corresponding to the exact distribution of Tnφ (p, is

b p0 ) under the null hypothesis H0 : p = p0 The distribution function of Tnφ (p, ³ ³ ´ ´ b p0 ) ≤ t = 1 − Pr p0 Tnφ (p, b p0 ) > t , FTnφ (pb,p0 ) (t) = Pr p0 Tnφ (p,


196


where ´ ³ b p0 ) > t = Pr p0 Tnφ (p,

and

X

Pr p0 (N1 = n1 , . . . , NM = nM ) ,

(n1 ,...,nM )∈An M,t

n o b p0 ) > t AnM,t = (n1 , . . . , nM ) ∈ (N ∪ {0})M /n1 + . . . + nM = n, Tnφ (p, Pr p0 (N1 = n1 , . . . , NM = nM ) =

n! (p0 )n1 . . . (p0M )nM . n1 ! . . . nM ! 1

b p0 ) is The set of upper tail probabilities of Tnφ (p, ³ n ´ o φ b p0 ) > t = α . Un,M = α ∈ (0, 1) : ∃t > 0 with Pr p0 Tnφ (p,

φ b p0 ) are obtained for any α ∈ Un,m through 100(1 − α) percentiles tφn,M,α of Tnφ (p, ³ ´ b p0 ) > tφn,M,α . In general it will not be possible to the equation α = Pr p0 Tnφ (p, get an exact percentile and therefore we shall consider an approximate 100(1 − α) φ percentile that we are going to define. If α ∈ (0, 1) − Un,M , we consider

³ o n ´ φ 0 b α1 = max α0 ∈ (0, α] : ∃t > 0 with Pr p0 Tn (p, p ) > t = α0 ,

so that tφn,M,α1 is defined as the approximate 100(1 − α) percentile. We calculate the approximate percentiles for α, M, n and φ all fixed. This process can be divided into four steps: 1. Generate all the elements xM = (n1 , . . . , nM ) of n o AnM = (n1 , . . . , nM ) ∈ (N ∪ {0})M /n1 + . . . + nM = n

and calculate the corresponding probabilities Prp0 (N1 = n1 , . . . , NM = nM ).

b p0 ). 2. For each xM ∈ AnM , calculate the test statistics Tnφ (p,

b p0 ) and Pr p0 (N1 = n1 , . . . , NM = nM ) in increasing order with 3. Put Tnφ (p, b p0 ). respect to the values of Tnφ (p, 4. Calculate the approximate 100(1 − α) percentile tφn,M,α1 .



197

φ We use randomized tests in order to decide with probability γn,M,α the rejecT 0 tion of the hypothesis H³0 : p = (1/M, ´ ..., 1/M) when the test statistic takes on φ φ b p0 ) be a function giving the probability of rejectthe value tn,M,α1 . Let ϕ Tn (p,

ing H0 when the φ-divergence test statistic is defined by the formula  if  1 ´  ³ φ φ 0 bp ) = ϕ Tn (p, γn,M,α if   0 if We have

b p0 ) is observed. This function Tnφ (p, b p0 ) > tφn,M,α1 Tnφ (p, b p0 ) = tφn,M,α1 . Tnφ (p, b p0 ) < tφn,M,α1 Tnφ (p,

(4.30)

h ³ í b p0 ) α = Ep0 ϕ Tnφ (p, ³ ´ ³ ´ φ b p0 ) > tφn,M,α1 + γn,M,α b p0 ) = tφn,M,α1 , Prp0 Tnφ (p, = 1 × Prp0 Tnφ (p,

and

φ γn,M,α

´ ³ b p0 ) > tφn,M,α1 α − Pr Tnφ (p, ´ . ³ = b p0 ) = tφn,M,α1 Pr Tnφ (p,

T Let us consider p ³= (p1 , ..., pM ´ ) with pi given in (4.29). The exact power b p0 ) , defined in (4.30), at p is function of the test ϕ Tnφ (p,

h ³ í b p0 ) βφ,n (p) = Ep ϕ Tnφ (p, ´ ´ ³ ³ φ b p0 ) > tφn,M,α1 + γn,M,α b p0 ) = tφn,M,α1 . = Prp Tnφ (p, Prp Tnφ (p,

A nice and extensive study about this problem can be seen in Marhuenda (2003). Now we present a practical example using the family of divergence measures given in (4.19). The study corresponding to the power-divergence test statistic can be seen in Cressie and Read (1988). Example 4.5 b p0 ), given in (4.20). We consider the family of Rukhin test statistics, Tnφa (p, Table 4.5 presents exact powers for the randomized test (4.30) based on α = 0.05 for different values of δ and “a”, n = 10, 20 and M = 5. For δ < 0 the power decreases as “a” increases. For δ > 0 the reverse occurs. For alternatives with δ > 0 we should choose “a” as large as other restrictions will allow in order to obtain the best power. For alternatives with δ < 0 we should choose “a” as


198


small as possible. If we wish to choose a test with reasonable power against these alternatives, for every δ value, we should choose a ∈ [0.6, 0.7]. The reason is that there is a marked reduction in power as “a” moves from 0.7 to 1 and a marked increase in power as “a” moves from 0 to 0.6.

a 0 .1 .2 .3 .4 .5 .6 2/3 .7 .8 .9 1

-0.9 .1670 .1637 .1620 .1670 .1517 .1517 .1516 .1429 .1367 .1313 .1372 .1247

(n = 10, M = 5) δ -0.5 0.5 1 .0819 .0707 .1392 .0805 .0719 .1412 .0797 .0728 .1426 .0818 .0711 .1400 .0766 .0755 .1570 .0766 .0755 .1570 .0774 .0797 .1749 .0772 .0753 .1677 .0758 .0833 .1977 .0746 .0844 .2028 .0767 .0833 .2009 .0737 .0858 .2099

1.5 .2600 .2617 .2630 .2607 .3053 .3053 .3417 .3353 .4056 .4056 .4032 .4212

a 0 .1 .2 .3 .4 .5 .6 2/3 .7 .8 .9 1

-.9 .5903 .5853 .5658 .5688 .5264 .4485 .3842 .3785 .3738 .3218 .2853 .2731

(n = 20, M = 5) δ -.5 .5 1 .1281 .0784 .1470 .1272 .0787 .1473 .1239 .0794 .1521 .1269 .0840 .1864 .1265 .0942 .2453 .1224 .1023 .2770 .1182 .1105 .3165 .1177 .1121 .3241 .1170 .1121 .3242 .1121 .1186 .3551 .1087 .1205 .3627 .1068 .1229 .3722

1.5 .2573 .2574 .2836 .3853 .5159 .5627 .6245 .6366 .6367 .6805 .6891 .7004

Table 4.5. Exact power functions for the randomized size α = 0.05 test of the symmetric hypothesis.

4.6.

Small Sample Comparisons for the Phi-divergence Test Statistics

The literature contains many simulation studies concerning the accuracy of using the chi-square distribution tail function Fχ2M−1 as an approximation to FT φ (pb,p0 ) for the chi-square test statistic and the likelihood ratio test statistic G2 n (e.g., Good et al. 1970; Roscoe and Byars, 1971; Tate and Hyer, 1973; Margolin and Light, 1974; Radlow and Alf, 1975; Chapman, 1976; Larntz, 1978; Kotze and Gokhale, 1980; Lawal, 1984; Kallenberg et al . 1985; Hosmane, 1986; Koehler, 1986; Rudas, 1986). Much of what follows in this section generalizes these studies. Two criteria used by Read (1984b) for comparing the family of power-divergence test statistics are proposed here for small n. A study in relation to Lin test statistic can be seen in Menéndez et al . (1997d) and in relation with Rukhin test statistic in Pardo, M. C. and Pardo, J. A. (1999).



199

We have obtained four different asymptotic approximations for the exact dis1 1 T b p0 ). We shall assume that p0 = ( M tribution of the test statistic Tnφ (p, , ..., M ) . The first one and the second one were obtained in Chapter 3. The first one is derived under the assumption that the number of classes is finite, Fχ2 (b) = M−1 ¢ ¡ Pr χ2M−1 ≤ b and the second one, FN (b) = Pr (N(0, 1) ≤ (b − µφ,n )/σφ,n ) , when the number of classes increases to infinity. We must remark that in this case √ ³ ´ n L b p0 ) − µφ,n −→ N(0, 1) Dnφ (p, n→∞ σφ,n where µφ,n and σφ,n are defined in Proposition 3.2. In this Chapter we have presented two other approaches. In Section 2 we have considered ´ ³ 1/2 FC (b) = Pr χ2M−1 ≤ (b − γφ )/δφ and finally in Section 3 we have considered FEd (b) ' J1φ + Jb2φ .

Two criteria, considered by Cressie and Read (1988), are presented to compare these four asymptotic approximations for small n. Criterion 1 consists of recording the maximum approximation error incurred by each of the four approxb p0 ). imations to the exact distribution, FT φ (pb,p0 ) (b) , of the test statistic Tnφ (p, n We calculate ¯ ³ ³ ´ ´¯ ¯ ¯ b p0 ) − Fi Tnφ (p, b p0 ) ¯ , max ¯FTnφ (pb,p0 ) Tnφ (p, pb for φ fixed and i = χ2M−1 , N, C, Ed. The sign associated with the maximum difference is also recorded. So, we know if the maximum error is an overestimate or an underestimate.

Criterion 2 consists of assessing the accuracy for the approximation in calculating the size of a test. We use the standard approximation Fχ2 to give a test M−1 with approximate significance level α. We choose cα such that 1−Fχ2 (cα ) = α, M−1

b p0 ), N, C, Ed, and i.e., cα = χ2M−1,α . We calculate 1 − Fi (χ2M−1,α ), i = Tnφ (p, assess how they vary for different functions φ. There are two reasons to take the critical cα to be the (1 − α) percentile of a chi-square distribution with M − 1 degrees of freedom. On the one hand, this is the most commonly used for the tests based on Pearson and likelihood ratio test. On the other hand, the critical region obtained from this approximation is independent of φ. In the next example we present a practical study based on the divergence measures given in (4.19).


200


Example 4.6 b p0 ), with p0 = (1/M, We consider the family of Rukhin test statistics, Tnφa (p, ..., 1/M )T , given in (4.20). Figures 4.1 and 4.2 il lustrate the maximum approximation error resulting from using the approximations Fχ2M−1 , FC , FEd and FN b p0 ) which are labelled Apr1, Apr2, Apr3 and for the exact distribution of Tnφ (p, Apr4 respectively on the graphs. The results are illustrated for specific values of “a” in the range [0,1], number of classes M = 5 and sample sizes n = 10 and n = 20. The approach FN is obtained in Exercise 1. At first glance, it is clear that the optimal parameter values are between 0.6 and 1 for the cases considered here. As n increases from 10 to 20, the error curves flatten over this range; furthermore, the size of the maximum errors decreases overall except for Apr4. But this behavior of the approximation FN is not surprising when one recalls that it relies on M increasing with n. In general the maximum error associated with the approximations Fχ2 , FC and FEd can be seen to be negative and of M−1 a similar order. However the normal approximation FN has larger maximum error than the others and of the opposite sign. Note that FN is not defined for a = 0 for the reason given in Exercise 1. Secondly, we assess the accuracy of the approximation in calculating the size of a test. We use the standard approximation Fχ2 to give a test with approximate significance level α, i.e., choose cα M−1

such that 1 − Fχ2M−1 (cα ) = α, i.e., cα = χ2M−1,α . We calculate 1 − Fi (χ2M−1,α ),

b p0 ), N, C, Ed, and assess how they vary for different values of “a”. i = Tnφa (p, The results are il lustrated in Figures 4.3 and 4.4 again for the specific values of “a”, sample sizes and number of classes used for criterion 1, and α = 0.1. There is a close agreement between the exact and nominal levels obtained for FC and FEd than for Fχ2M −1 . In general, the normal approximation is clearly poor in comparison to the other two approximations and tends to overestimate the true level for all “a” values considered. But this result must be expected because this approximation is an asymptotic result in n and M. Due to the value of reference the chi-square approximation, in the pictures, is represented by a line in 0.1. The exact and nominal levels for the approximation Fχ2 are quite similar for all M−1 a ∈ [0.6, 1] and it is not too similar outside this interval. If we want to use the approximation given by Fχ2 we should use a ∈ [0.6, 1] for small M. From the M −1 result obtained in relation with the criterion based on the speed of the convergence of the exact moment, when M increases, the optimal values were a = 2/3 and a = 1. This suggests that the range a ∈ [0.6, 1] is also optimum for big values of M. If we wish to consider values outside the interval [0.6, 1] the approximation



201

based on FC is a good alternative to the approximation FEd since the approximation FC is easier to get than the approximation FEd .

Figure 4.1

Figure 4.2


202


Figure 4.3

Figure 4.4



4.7.

203

Exercises

b p0 ), where φa is given 1. Consider the family of divergence measures, Dnφa (p, T in (4.19) and p0 = (1/M, ..., 1/M) . Find the asymptotic distribution obtained in Proposition 3.2 when the number of classes Mn verifies lim

n→∞

n = γ ∈ (0, ∞) . Mn

2. Consider the family of test statistics associated with Lin divergence measure (Lin, 1991) defined by φa (x) =

1 (ax log x − (ax + (1 − a)) log (ax + 1 − a)) , a ∈ (0, 1) , a (1 − a)

and φ0 (x) = lim φa (x) = x log x − x + 1, φ1 (x) = lim φa (x) = − log x − 1 + x. a→0

a→1

Is there some optimum value of “a” in goodness-of-fit according to the moment criterion given in (4.18)? 3. Consider the family of test statistics associated with the “harmonic mean divergence” given by ¡ ¢−1/r φr (x) = (1 − x) /2 − 21/r 1 + x−r , r > 0.

Find the optimum values of “r” in goodness-of-fit according to the moment criterion given in (4.18).

4. Consider the family of divergence measures given in Exercise 2 and the following alternatives  M −1−δ   if i = 1, ..., M − 1 M(M − 1) . H1 : πi = 1+δ   if i = M M Find the exact power for α = 0.05, n = 20 and M = 5.

5. Let φ : (0, ∞) → R be a concave function with continuous fourth derivative verifying φ00 (1/M) < 0. Find what conditions the function φ must verify b p0 ), given in Exercise 6 of Chapter 3, to be optimum the statistic Snφ (p, according to the moment criterion with p0 = (1/M, ..., 1/M)T . © 2006 by Taylor & Francis Group, LLC

204


6. Consider the family of Rφ -divergences given by ( 1 a 1−a (x − x) if a 6= 1 . φa (x) = −x log x if a = 1 b p0 ), p0 = Find the values of “a” for which the test statistic Snφa (p, (1/M, ..., 1/M)T is optimum according to the moment criterion as M → ∞.

7. Consider the power-divergence family of test statistics. Find the expression of the corrected test statistic given in Subsection 4.3.3 of this chapter.

8. Consider the family of test statistics given in Exercise 2. Find the expression of the corrected test statistic given in Subsection 4.3.3 of this chapter, with p0 = (1/M, ..., 1/M)T . b p0 ), with p0 = (1/M, ..., 1/M)T . Using the 9. Consider the test statistic Snφ (p, results given in Exercise 5, find the expression of the corrected test statistic given in Subsection 4.3.3 of this chapter. 10. Find the expression of the approximation based on Edgeworth’s expansion associated with Lin test statistic in the case of equiprobable hypothesis. 11. Find the expression of the approximation based on Edgeworth’s expansion associated with Pearson-test statistic. 12. Find the expression of the approximation based on Edgeworth’s expansion associated with the power-divergence test statistics for λ 6= −1 and λ 6= 0.

4.8.


b p0 ) the family of divergence measures obtained with 1. We denote by Dnφa (p, the functions defined in (4.19). By Proposition 3.2 we have √ ³ ´ n L b p0 ) − µn,a −→ N (0, 1) Dnφa (p, n→∞ σn,a where

µn,a

# " · µ ¶¸ (γn − Zn )2 Zn , a ∈ [0, 1] =E = E φa γn 2γn (γn a + (1 − a)Zn )



205

and 2 σn,a

# # " (γn − Zn )2 (γn − Zn )2 2 −Cov Zn , , = γn V ar 2γn (γn a + (1 − a)Zn ) 2γn (γn a + (1 − a)Zn ) "

a ∈ [0, 1] . Therefore, ´ ¢ ¡ √ ³ φa L b p0 ) − µa −→ N 0, σa2 n Dn (p, n→∞

where

"

(γ − Z)2 µa = E 2γ (γa + (1 − a)Z)

#

and σa2

"

# # " (γ − Z)2 (γ − Z)2 2 = γV ar −Cov Z, , a ∈ [0, 1] 2γ (γa + (1 − a)Z) 2γ (γa + (1 − a)Z)

where Z is a Poisson random variable with parameter γ. Finally, under the null hypothesis of equiprobability we have FN (b) = Pr(N(0, 1) ≤ (b − µa )/σa ).

¤ £ We can observe that µ0 and σ02 do not exist because E Z −1 does not exist. For this reason this approximation is valid only for a ∈ (0, 1]. 2. We can observe that for a → 0 we have the likelihood ratio test statistic and for a → 1 the modified likelihood ratio test statistic. It is clear that for a = 0 and a = 1, the equation (4.18) does not hold. Now we consider IV 2 a ∈ (0, 1) . We have φ000 a (1) = −(1 + a) and φa (1) = 2(a + a + 1) and the equation (4.18) becomes ¡ ¢ −4 (1 + a) + 6 a2 + a + 1 = 0. This equation does not have any real-valued solution for a ∈ (0, 1) , so there is not an optimum value according to this criterion.

IV 3. It is immediate to check that φ000 r (1) = −3/8 − 3/8r and φr (1) = 15/16 + 7/8r − 3/16r2 − 1/8r3 , then the equation (4.18) becomes µ ¶ ¶ µ 3 3 15 7 3 2 1 3 4 − − r +3 + r − r − r = 0. 8 8 16 8 16 8 √ The positive root of the equation (4.18) is r = − 14 + 14 57.


206


4. Using the procedure given in Section 4.4, we have computed the exact powers for the randomized test (4.30) with α = 0.05 and for different values of δ and a. Taking M = 5 and n = 10, we have obtained the powers

a 0 .1 .2 .3 .4 .5 .6 .7 .8 .9 1

(n = 20, M = 5) δ −.5 .5 1 .1213 .1076 .3056 .1244 .1012 .2765 .1252 .1001 .2718 .1267 .0953 .2521 .1268 .0933 .2424 .1281 .0886 .2205 .1272 .0849 .1971 .1251 .0821 .1759 .1282 .0800 .1640 .1270 .0790 .1503 .1262 .0789 .1474

−.9 .4463 .4841 .4948 .5260 .5336 .5627 .5689 .5670 .5844 .5849 .5803

1.5 .6103 .5678 .5588 .5279 .5128 .4705 .4222 .3646 .3361 .2792 .2574

For alternatives with δ > 0, we must choose “a” as larger as other restrictions permit getting the best power. On the contrary, for alternatives δ < 0 we must choose “a” as smaller as possible. Given a fixed “a” the power rises as |δ| increases. If we want to choose a test with a reasonable power against the given alternatives for any value of δ, we should choose a ∈ [0.3, 0.6] . 5. Using the procedure given for the family of φ-divergences we have ´ h i ³ 1 b p0 ) = M − 1 + fφ1 + O n−3/2 E Snφ (p, n ·³ ´2 ¸ ³ ´ 1 b p0 ) E Snφ (p, = M 2 − 1 + fφ2 + O n−3/2 n ·³ ´ ´3 ¸ ³ 1 3 φ 0 3 2 −3/2 b = M + 3M − M − 3 + fφ + O n E Sn (p, p ) n where

fφ1 =

φ000 (1/M) 2φ00 (1/M)

¡

2 M

¢ −3+M +

7φIV (1/M) 16φ00 (1/M)

¡

1 M2

−

¡ 10 ¢ 000 (1/M) fφ2 = −2M + 2 + M − 13 + 2M + M 2 φφ00 (1/M) ³ ´2 ¡ ¢ 1 φ000 (1/M) 12 18 + 4 φ00 (1/M) M2 − M + 6 ¢ IV (1/M) ¡ 3 5 + 7φ 8φ00 (1/M) M 2 − M + 1 + M © 2006 by Taylor & Francis Group, LLC

2 M

¢ +1


207

and ¢ φ000 (1/M) ¡ 210 2 3 fφ3 = 26 − 24M − 2M 2 + 2φ 00 (1/M) M − 243 + 3M + 27M + 3M ¢ IV (1/M) ¡ 45 66 2 + 7φ 16φ00 (1/M) M 2 − M + 18M + 3M ³ 000 ´ ¢ (1/M) 2 ¡ 180 − 234 + 14 φφ00 (1/M) M + 36 + 18M . M2

Therefore, the functions φ under which the asymptotic moments are closer to the exact moments, for fixed M, are those for which it holds fφi = 0, i = 1, 2, 3. This happens because the second order expansions of the first three moments of Snφ are the same as the first three moments of a chi-square dis¡ ¢ tribution χ2M−1 plus the correction factor of order O n−1 , fφi , i = 1, 2, 3, respectively. 6. We have to solve the equations fφi = 0, i = 1, 2, 3 given in the previous exercise for the given function φa as M → ∞. Solving the first equation, ¡ ¡ ¢ 7 2 fφ1a = 48 3 − 6M + 3M 2 a2 + − 81 48 M + ¢ ¡ 13 2 18 5 + 8 M − 8 M + 8 =0

and making M → ∞, we have a=

81 ±

138 48 M

−

57 48

¢

a

√ 812 − 6552 ; 42

therefore, the solutions of the equation are a = 2 and a = 13/7. On the other hand, solving the second equation fφ2a = 0, we have ¡ ¡ ¢ ¢ 45 − 71M + 19M 2 + 7M 3 a2 + −27M 3 − 67M 2 + 215M − 121 a ¡ ¢ + 26M 3 + 58M 2 − 162M + 78 = 0

and making M → ∞, we get

a= i.e., a = 2 and a = 13/7.


27 ±

√ 272 − 728 , 14

208


Finally to approximate the third asymptotic moment to the exact one, we have ¢ ¡ fφ3a = 1035 − 1398M + 144M 2 + 198M 3 + 21M 4 a2 ¡ ¢ + −81M 4 − 702M 3 − 552M 2 + 4110M − 2775 a ¡ ¢ + 78M 4 + 612M 3 + 496M 2 − 3012M + 1826 = 0 and as M → ∞, we obtain

√ 812 − 6552 ; a= 42 therefore the solutions of the equation are a = 2 and a = 13/7. 81 ±

This result is valid for large M. If M is small we should use the next table, which has the roots “a” of the equations fφi a = 0, i = 1, 2, 3, for fixed values of M that increase to ∞. fφ1a fφ2a fφ3a

M a1 a2 a1 a2 a1 a2

2 3.0 2.0 3.34 1.65 3.69 1.3

3 2.42 2.0 2.52 1.68 2.62 1.41

4 2.23 2.0 2.31 1.7 2.37 1.47

5 2.14 2.0 2.21 1.71 2.27 1.51

10 2.0 2.98 2.07 1.76 2.10 1.62

20 2.0 1.91 2.02 1.8 2.04 1.72

40 2.0 1.88 2.0 1.83 2.01 1.78

50 2.0 1.88 2.0 1.83 2.01 1.79

100 2.0 1.86 2.0 1.84 2.0 1.82

200 2.0 1.86 2.0 1.85 2.0 1.84

500 2.0 1.85 2.0 1.85 2.0 1.85

Values of the roots (a1 >a2 ) of fφi a = 0, i = 1, 2, 3.

In this table we observe that for M > 20 we can use the previous result since the first order factors of the three first moments are closer to 0 for a = 2 and a = 13/7. For M ≤ 20, it would be reasonable to choose one test statistic Snφa with a ∈ [1.5, 2] . For more details see Pardo, M. C. (1999). 7. In this case, the corrected test statistic is given by φ

c

φ b p0 ) Tn (λ) (p,

b p0 ) − γφ(λ) Tn (λ) (p, q ≡ δφ(λ)

where γφ(λ) and δφ(λ) are obtained from the theoretical results given in Subsection 4.3.3. IV Since φ00(λ) (1) = 1, φ000 (λ) (1) = λ − 1, φ(λ) (1) = (λ − 1) (λ − 2) we have ¢ ¡¡ 1 2 − 2M − M 2 + S δφ(λ) = 1 + 2(M−1)n + (λ − 1) (8 − 12M − 2M 2 +6S) ¡ ¢ + 13 (λ − 1)2 4 − 6M − 3M 2 + 5S + (λ − 1) (λ − 2) (2 − 4M + 2S))



209

and ´ ³ q ¡ ¢ γφ(λ) = (M − 1) 1 − δφ(λ) + n1 13 (λ − 1) (2 − 3M + S) (λ − 1) (λ − 2) + (1 − 2M + S) . 4 8. Similarly to the previous exercise and taking into account that

and

PM

IV 2 φ00a (1) = 1, φ000 a (1) = −(1 + a), φa (1) = 2(a + a + 1)

0 j=1 pj

γa

and

= M 2 , for a ∈ (0, 1) , we have ³ ´ ¡ ¢ 1/2 1 = (M − 1) 1 − δa (a + 1) 2 − 3M + M 2 − 3n ¡ ¢ 2 1 − 2M + M 2 + a +a+1 2n

¢ 1 2(a+1) ¡ 2 − 3M + M 2 + − (M−1)n n ¡ ¢ a2 +a+1 + 2 n(M−1) 1 − 2M + M 2 .

δa = 1 −

(a+1)2 3n(M−1)

¡ ¢ 2 − 3M + M 2

For a = 0, we have the likelihood ratio test statistic and for a = 1 the modified likelihood ratio test statistic. 9. We know that i h ¢ ¡ b p0 ) = M − 1 + aφ /n + o n−1 E Snφ (p,

and

h i ¡ ¢ φ 0 b V ar Sn (p, p ) = 2 (M − 1) + bφ /n + o n−1

where

φ000 (1/M) aφ = 00 2φ (1/M )

µ

2 −3+M M

¶

7φIV (1/M ) + 16φ00 (1/M)

µ

¶ 2 1 − +1 M2 M

and ¢ 1 ³ φ000 (1/M) ´2 000 (1/M) ¡ 12 bφ = −2M + 2 + φφ00 (1/M) − 18 + 6M + 2 φ00 (1/M) M ¡ ¡ 6 ¢ ¢ IV (1/M) 9 1 2 × M2 − M + 3 + 7φ 2φ00 (1/M) M 2 − M + 1 .

We define

c φ b p0 ) Sn (p, © 2006 by Taylor & Francis Group, LLC

=

b p0 ) − γφ Snφ (p, p δφ

210


in such a way that E and V ar

h

h

i

c φ b p0 ) Sn (p,

i

c φ b p0 ) Sn (p,

¡ ¢ = M − 1 + o n−1

¢ ¡ = 2 (M − 1) + o n−1 .

For this purpose it is necessary to consider ³ p ´ γφ = (M − 1) 1 − δφ + aφ /n and δφ = bφ /n2 (M − 1) , i.e.,

and

³ 000 ¢ φ (1/M) ¡ 6 1 δφ = 1 − n1 + n(M−1) φ00 (1/M) M − 9 + 3M ³ 000 ´ ¢ IV (1/M) ¡ (1/M) 2 ¡ 6 9 1 −M + 3 + 7φ + 14 φφ00 (1/M) 4φ00 (1/M) M 2 − M2 ¡ p ¢ γφ = (M³− 1) 1 − δφ ¢ φ000 (1/M) ¡ 2 1 + 2n φ00 (1/M) M − 3 + M +

7φIV (1/M) 8φ00 (1/M)

¡

1 M2

−

2 M

2 M

¢´ +1

¢´ +1 .

10. It is immediate to show that Lin test statistic is µ M φa pbi 1 P 0 b p ) = 2n 1−a Tn (p, pbi log abp +(1−a)p 0 i i i=1 ¶ M 0 P p i , a ∈ (0, 1) . + a1 p0i log abp +(1−a)p 0 i=1

i

i

The expression of the approximation based on Edgeworth’s expansion associated with this test statistic for p0 = (1/M, ..., 1/M)T is

being

b p0 ) < b) ' J1a + Jb2a Pr(Tnφa (p,

J1a = Pr(χ2M−1 < b) +

3 ´ ³ ¢ 1 X a ¡ 2 rj Pr χM−1+2j < b + O n−3/2 , 24n j=0

where · r0a = 2(1 − M 2 )

· r1a = -6(a2 + a + 1)(M − 1)2 +2(1 + a)2 (M 2 − 3M + 2)+6(M 2 − M)


211


¡ ¢ ¡ ¢ · r2a = -4a(a + 1) M 2 − 3M + 2 +6 a2 + a + 1 (M − 1)2 -6(M 2 − M)

and

· r3a = 2 (a + 2)2 (M 2 − 3M + 2).

The term Jb2a has the expression

.³ ´ 1/2 ³ ´ (2πn)(M−1) (1/M)M , Jb2a = N a (b) − n(M−1)/2 V a (b) e−c/2

where N a (b) is the number of points (w1 , . . . , wM−1 ) satisfying ¶ µ √ 1 Ni − , Ni = 0, 1, 2, . . . wi = n n M P φa b p0 ) < b, and such that M i=1 Ni = n and Tn (p, ³ ¡ (πc)(M −1)/2 c V a (b) = Γ((M+1)/2) (1/M)M/2 1 + 12(M+1)n (1 + a)2 (M 2 − 3M + 2) ¢¢ − 3(a2 + a + 1)(M − 1)2 + O(n−3/2 ).

11. In this case φ (x) = (x − 1)2 /2, then for the test statistic 2

X =n

M X (b pi − p0 )2 i

i=1

p0i

we have Pr(X 2 < b) ' J1φ + Jb2φ , being

¡ 1 Pr(χ2M−1 < b)2(1 − S) J1φ = Pr(χ2M−1 < b) + 24n ¢ + Pr(χ2M+5 < b)(5S − 3M 2 − 6M + 4)

and Jb2φ

where

³ ´ = N φ (b) − n(M−1)/2 V φ (b) e−b/2

N φ (b)

such that

(M−1)

(2πn)

M Y

p0j

i=1

is the number of points (w1 , . . . , wM−1 ) satisfying µ ¶ √ Ni 1 − , Ni = 0, 1, 2, . . . wi = n n M

PM

being S =

,Ã

i=1 Ni

PM

= n and X 2 < b and V φ (b) =

0 −1 j=1 (pj ) .


(πb)(M−1)/2 Γ((M + 1)/2)

µM Q

i=1

p0j

¶1/2

! 1/2

,

212


12. Using similar arguments to that used in the previous exercises and taking into account that φ00(λ) (1) = 1, φ000 (1) = λ − 1, φIV (λ) (1) = (λ − 1) (λ − 2) ³ φ ´ (λ) b p0 ) < b ' J1λ + Jb2λ being we have Pr Tn (λ) (p, J1λ

=

Pr(χ2M−1

3 ´ ³ ¢ 1 X λ ¡ 2 < b) + rj Pr χM−1+2j < b + O n−3/2 , 24n j=0

where · r0λ = 2(1 − S)

¢ ¡ · r1λ = 6 (λ − 1) S − M 2 − 3 (λ − 1) (λ − 2) (S − 2M + 1) ¢ ¡ + (λ − 1)2 5S − 3M 2 − 6M + 4 + 3(3S − M 2 − 2M) ¡ ¢ · r2λ = (λ − 1) −2(8S − 6M 2 − 6M + 4) + 3 (λ − 2) (S − 2M + 1) ¢ ¡ ¢ ¡ − 2 (λ − 1)2 5S − 3M 2 − 6M + 4 − 6 −M 2 + 2S − 2M + 1

and

· r3λ = λ2 (5S − 3M 2 − 6M + 4).

The term Jb2λ has the expression

Ã !1/2 M ³ ´ Y p0j Jb2λ = N λ (b) − n(M−1)/2 V λ (b) e−b/2 (2πn)(M−1) i=1

φ

b p0 ) < b and where N λ (b) is the number of X-values such that Tn (λ) (p, (πb)(M−1)/2 V φ (b) = Γ((M + 1)/2)

where

i=1

p0j

!1/2 µ

¶ b (l1 − l2 ) , 1+ 24(M + 1)n

¢ ¡ l1 = (λ − 1)2 5S − 3M 2 − 6M + 4

and being S =

ÃM Y

PM

l2 = 3 (λ − 1) (λ − 2) (S − 2M + 1) ,

0 −1 j=1 (pj ) .


For more details see Cressie and Read (1988).

5

Minimum Phi-divergence Estimators 5.1.

Introduction

In this chapter we consider a wide class of estimators which can be used when the data are discrete, either the underlying distribution is discrete or is continuous but the observations are classified into groups. The latter situation can occur either by experimental reasons or because the estimation problem without grouped data is not easy to solve; see Fryer and Robertson (1972). Some examples in which it is not possible to find the maximum likelihood estimator based on the original data can be seen in Le Cam (1990). For example, when we consider distributions resulting from the mixture of two normal populations, whose probability density function is given by Ã Ã µ µ ¶ ! ¶ ! 1 1 x − µ1 2 1 x − µ2 2 1 fθ (x) = w √ exp − exp − +(1−w) √ , 2 σ1 2 σ2 2πσ1 2πσ2 the likelihood function is not a bounded function. In this situation θ = (µ1 , µ2 , σ1 , σ2 , w), µ1 , µ2 ∈ R, σ1 , σ2 > 0 and w ∈ (0, 1), and the likelihood function for a random sample of size n, x1 , ..., xn is given by L(θ; x1 , . . . , xn ) =

n Y

j=1


fθ (xj ).

214


√ If we consider µ2 = µ1 = xi for some i (i = 1, . . . , n), then fθ (xi ) > w( 2πσ1 )−1 and Ã ¶ ! µ √ 1 xj − xi 2 −1 fθ (xj ) > (1 − w)( 2πσ2 ) exp − for j 6= i. 2 σ2 Therefore

 µ ¶2 n X xj − xi  1 −(n−1) L(θ; x1 , . . . , xn ) > (2π)−n/2 w(1−w)n−1 σ1−1 σ2 exp − 2 σ2 

j=1,j6=i

and choosing σ1 sufficiently small it is possible to do L as big as it is desired. Then there are not values w, σ1 , σ2 , µ1 and µ2 that maximize L.

Several authors have paid attention to estimation of the unknown parameters of a mixture of two unspecified normal densities. There are mainly three different approaches: moments, maximum likelihood and minimum distance. The moment solution to the problem of estimating the five parameters of an arbitrary mixture of two unspecified normal densities was studied by Karl Pearson (1894). Although many random phenomena have subsequently been shown to follow this distribution until 1966 this estimation problem was not considered. Important applications of mixture modeling occur in satellite remote-sensing of agricultural characteristics: specifically, the use of spectral measurements of light intensity to determine crop types, distributions of wind velocities and distributions of physical dimensions of various mass produced items. Hassenblad (1966) seems to have been the first to reopen the question. Since then the problem has also attracted the attention of Cohen (1967) who showed how the computation of Pearson’s moment method can be lightened to some extent. Since the likelihood function is not a bounded function, the objective in the maximum likelihood approach is to find an appropriate local maximum. Since closed-form solutions of the likelihood equations do not exist, they must be solved by using iterative techniques. Day (1969) and Behboodian (1970) find an appropriate local maximum of the likelihood function by using iterative techniques. Minimum distance estimation (MDE), in general, was presented for the first time by Wolfowitz (1957) and it provides a convenient method of consistently estimating unknown parameters. An extensive bibliography for minimum distance estimators is in Parr (1981). In the mixture model setting, Choi and Bulgren


Minimum Phi-divergence Estimators

215

(1968) and MacDonald (1971) estimated the mixture proportions (assuming the component distributions were known) by minimizing the sum-of-squares distance between the empirical and theoretical distribution functions. Quandt and Ramsey (1978) estimated the parameters in the mixture model by minimizing the sum-of-squares distance between the empirical and theoretical moment generating functions. Kumar et al. (1979), however, showed that this technique is highly sensitive to starting values. Bryant and Paulson (1983) examined the empirical characteristic function in this setting. Fryer and Robertson (1972) considered the MDE for grouped data and finally by using the families of Rφ -divergences and φ-divergences in Pardo, M. C. (1997b, 1999b) the problem of estimating the parameters of a mixture of normal distributions was considered. In this chapter we study the minimum φ-divergence estimator, considered in Morales et al . (1995), for grouped data.

5.2.

Maximum Likelihood and Minimum Phi-divergence Estimators

Let (X , βX , Pθ )θ∈Θ be the statistical space associated with the random variable X, where βX is the σ-field of Borel subsets A ⊂ X and {Pθ }θ∈Θ is a family of probability distributions defined on the measurable space (X , βX ) with Θ an open subset of RM0 , M0 ≥ 1. Let P = {Ei }i=1,...,M be a partition of X . The formula Prθ (Ei ) = pi (θ), i = 1, . . . , M, defines a discrete statistical model. Let Y1 , . . . , Yn be a random sample from the population described by the random P variable X, let Ni = nj=1 IEi (Yj ) and pbi = Ni /n, i = 1, . . . , M. Estimating θ by maximum likelihood method, under the discrete statistical model, consists of maximizing for fixed n1 , . . . , nM , Pr θ (N1 = n1 , . . . , NM = nM ) =

n! p1 (θ)n1 × . . . × pM (θ)nM n1 ! . . . nM !

(5.1)

or, equivalently, p, p(θ)) + k log Pr θ (N1 = n1 , . . . , NM = nM ) = −nDKull (b

(5.2)

b = (ˆ where p p1 , . . . , pˆM )T , p(θ) = (p1 (θ), . . . , pM (θ))T and k is independent of

θ. The equality (5.2) is easy to check because, if we denote l (θ) = log Pr θ (N1 =


216


n1 , . . . , NM = nM ), we have M P n! pbi log pi (θ) +n n1 ! . . . nM ! i=1 M M M P P P 1 n! −n +n pbi log pbi log pbi − n pbi log pbi = log n1 ! . . . nM ! pi (θ) i=1 i=1 i=1 M M P P n! pbi = log pbi log pbi log pbi −n +n n1 ! . . . nM ! pi (θ) i=1 i=1 M P pbi = −n pbi log p, p(θ)) + k. + k = −nDKull (b pi (θ) i=1

l (θ) = log

Then, estimating θ with the maximum likelihood estimator of the discrete model is equivalent to minimizing the Kullback-Leibler divergence on θ ∈ Θ ⊆ RM0 . Since Kullback-Leibler divergence is not the unique divergence measure, we can choose as estimator of θ the value ˜ θ verifying ˜ = D(b p, p(θ))

inf

θ∈Θ⊆RM0

D(b p, p(θ)),

(5.3)

where D is a divergence measure. In the following we assume that there exists a function p(θ) = (p1 (θ), . . . , pM (θ))T that maps each θ = (θ1 , . . . , θM0 )T into a point in ∆M , where ∆M was defined in Chapter 2, Section 2.1. As θ ranges over the values of Θ, p(θ) ranges over a subset T of ∆M . When we assume that a given model is “correct”, we are just assuming that there exists a value θ0 ∈ Θ such that p(θ0 ) = π, where π is the true value of the multinomial probability, i.e., π ∈ T . Definition 5.1 Let Y1 , . . . , Yn be a random sample from a population described by the random variable X with associated statistical space (X , βX , Pθ )θ∈Θ . The minimum φbφ ∈ Θ verifying divergence estimator of θ0 is any θ bφ )) = Dφ (b p, p(θ

inf

θ∈Θ⊆RM0

Dφ (b p, p(θ)).

In other words, the minimum φ-divergence estimator satisfies the condition bφ = arg θ © 2006 by Taylor & Francis Group, LLC

inf

θ∈Θ⊆RM0

Dφ (b p, p(θ)).

(5.4)


217

b in the sense of the φ-divergence This method chooses the point of T closest to p chosen.

Remark 5.1 If we consider the family of the power-divergence measures we obtain the minimum power-divergence estimator studied by Cressie and Read (1984). This is given by the condition b(λ) = arg θ

where

inf

θ∈Θ⊆RM0

Dφ(λ) (b p, p(θ)),

M

X 1 Dφ(λ) (b p, p(θ)) = pbi λ(λ + 1) i=1

Ãµ

pbi pi (θ)

¶λ

(5.5)

!

−1 .

For λ → 0 we obtain the maximum likelihood estimator, for λ = 1 the minimum chi-square estimator, for λ = −2 the minimum modified chi-square estimator (or minimum Neyman modified estimator), for λ → −1 the minimum modified likelihood estimator (or minimum discrimination information estimator), for λ = −0.5 Freeman-Tukey estimator and for λ = 2/3 Cressie-Read estimator. We know that Kullback-Leibler divergence measure can be obtained from the power-divergence measure with λ = 0. For this reason in the rest of the chapter in order to distinguish between the MLE based on original data and the MLE b(0) the maximum likelihood associated with the discrete model we shall denote by θ b the maximum likelihood estimator based estimator in the discrete model and by θ on the original data. We present an example to clarify all the notation and concepts introduced until now. Example 5.1 Suppose that n independent and identical distributed Poisson variables with mean θ (θ > 0) are observed, and let the observations be truncated at x = 2. Let N1 , N2 and N3 be the number of observations taking on the values 0, 1 and 2 or more, respectively. Then N =(N1 , N2 , N3 ) has the trinomial distribution (n; p1 (θ), p2 (θ), p3 (θ)), where p1 (θ) = Prθ (X = 0) = e−θ , p2 (θ) = Prθ (X = 1) = θe−θ and p3 (θ) = Prθ (X ≥ 2) = 1 − (1 + θ)e−θ . If we consider Cressie-Read estimator, we have to get the minimum in θ of


218


the function Dφ(2/3) (b p, p(θ)) =

9 10

+ pb3

¶ ¶ · µ³ µ³ ´2/3 ´2/3 pb1 pb2 − 1 + pb2 −1 pb1 exp(−θ) θ exp(−θ) µ³

pb3 1−(1+θ) exp(−θ)

´2/3

¶¸ −1 .

b = (0.2, 0.3, 0.5)T we obtain θb(2/3) = Now if we assume, for instance, that p 1.6595. Geometrically, ∆3 is the triangle side ABC depicted in Figure 5.1, that we represent in the plane through the triangle of Figure 5.2. p3 C = (0, 0, 1)

·@ @ · @ · @ · @ · B = (0, 1, 0) @ · ³ ³ ´ · ³³ ´ p2 · ´ ³³³ ³ ³ ·´´ ³ ³ ´ ·³ ´ ´ A = (1, 0, 0) ´ ´

p1

Figure 5.1. Set of probability distributions for Example 5.1. As θ varies over R+ = [0, ∞), p(θ) = (e−θ , θe−θ , 1 − (1 + θ)e−θ )T traces out a curve in ∆3 . This curve is the subset T . When θ → 0, p(θ) → (1, 0, 0)T , and when θ → ∞, p(θ) → (0, 0, 1)T . Thus, the boundary points of θ in this example correspond to the boundary points of ∆3 . Figure 5.2 shows the b in this example. If the Poisson model is relationships between ∆3 , T, π and p incorrect, then the true value of π does not generally lie on the curve, although © 2006 by Taylor & Francis Group, LLC


219

in principle it can. Because of the discreteness of the multinomial distribution, b does not lie on T (as is the case in the figure). The it often happens that p estimation method based on the minimum distance leads to a point in T closest b in the sense of the distance chosen. to p C = (0, 0, 1) ·TT · T · T · T · πX T z X · T · pˆ T · T · T · TH T j H · T · T · T ·· T

A = (1, 0, 0)

B = (0, 1, 0)

b for Example 5.1. Figure 5.2. Relation between ∆3 , T, π and p From a historical point of view the maximum likelihood estimator in grouped data or the minimum Kullback-Leibler divergence estimator was considered for the first time by Fisher (1925), the minimum chi-square estimator and the minimum modified chi-square estimator by Neyman (1949). Matusita (1955), using the measure of divergence that has his name, studied the corresponding estimator for a = 1/2 (Freeman-Tukey estimator). Read and Cressie (1988) considered the minimum power-divergence estimators and finally Morales et al. (1995) studied the minimum φ-divergence estimators. The results obtained by Read and Cressie (1988) as well as the results obtained in Morales et al . (1995) are generalizations of the results given by Birch (1964) for the maximum likelihood estimator. The results we present in this chapter follow the line of that given by Pardo, M. C. (1997a) for the Rφ -divergences, which are a generalization of the procedure given by Cox (1984) in relation to the maximum likelihood estimator.


220


5.2.1.

Minimum Power-divergence Estimators in Normal and Weibull Populations

In this Section we consider the minimum power-divergence estimator, given in (5.5), of the Normal and Weibull parameters for different values of λ. Moreover, we compare the results obtained with the maximum likelihood estimator based on the original data and the estimator based on minimizing the Kolmogorov distance (Kolmogorov estimator). A similar study was carried out for Weibull distribution by Pardo, M.C. (1997a) using the family of Rφ -divergences to define a minimum distance estimator. Definition 5.2 Kolmogorov estimator for the parameter θ, of a distribution family {Fθ }θ∈Θ , bKo ∈ Θ verifying is defined as the value θ bKo ) = min Dn (θ), Dn (θ θ∈Θ

for

Dn (θ) = sup{| Fn (x) − Fθ (x) |} = max{Dn+ (θ), Dn− (θ)}, x∈R

where Dn+ (θ) = sup{Fn (x) − Fθ (x)} = max{0, maxi=1,...,n { ni − Fθ (x(i) )}}, x∈R

Dn− (θ) = sup{Fθ (x) − Fn (x)} = max{0, maxi=1,...,n {Fθ (x(i) ) − x∈R

i−1 n }},

Fn is the empirical distribution function of the sample x1 , . . . , xn and x(1) ≤ x(2) ≤ . . . ≤ x(n) are the order statistics. It is well known that a random variable X has a Weibull distribution, W e (b, c) , with parameters θ = (b, c) , b, c, b > 0, c > 0, if the distribution function is given by Fθ (x) = 1 − exp (− (x/b)c ) , x ≥ 0. The general scheme for calculating the minimum power-divergence estimator is as follows: Step 1: We fix



221

a) sample size (n), b) number of classes in the partition (M), c) number of simulated samples (N): We consider the partition of the sample space, {Ei }i=1,...,M , Ei = (ai−1 , ai ], i = 1, . . . , M, where the values a0 , ..., aM are obtained from Z

ai

fθ (x)dx = 1/M,

i = 1, ..., M.

ai−1

Step 2: Given λ fixed, do for i = 1 to N a) Generate a random sample of size n, b) Calculate the relative frequencies, pbl , of El = (al−1 , al ], l = 1, ..., M,

b(λ),i =(θb1 , ..., θbM0 ) p, p(θ)). Estimator θ c) Minimize on θ the function Dφ(λ) (b (λ),i (λ),i is obtained. The minimum of the function Dφ(λ) (b p, p(θ)) has been obtained using the subroutine ZXMIN of the package IMSL. In other parts of the book we use the Newton-Raphson method to get minimum φ-divergence estimators. ˆ(λ) = (θˆ1 , ..., θˆM0 ) be the mean of the values obtained by Step 3: Let θ (λ) (λ) minimizing the function Dφ(λ) in step 2(c) for all the samples and mse the mean squared error of the estimated parameters, i.e.,

j θˆ(λ) =

1 N

N X

θˆφj (λ) ,i , mse(θj ) =

i=1


1 N

N ³ X i=1

θˆφj (λ) ,i − θj

´2

and mse =

1 M0

M0 X j=1

mse (θj ) .

222


W e(1, 1) ˆ θ bb b c mse b b θKo b b c mse b ˆ θ(−2) b b c mse b ˆ θ(−1) b b c mse ˆ(−0.5) bb θ b c mse ˆ(0) bb θ b c mse ˆ(1) bb θ b c mse

n = 20 .998783 1.063596 .055893 .984651 1.565195 1.023289 1.008785 1.423182 .776172 1.013693 1.406272 .756604 .992916 1.380089 .739701 .976527 1.369282 .739195 .974568 1.405006 .765451

n = 40 .994317 1.029655 .025969 1.006958 1.137134 .121828 1.010294 1.084708 .106446 1.006479 1.067542 .097466 1.000374 1.069138 .103279 .995429 1.052883 .093768 .982324 1.06289 .098793

n = 60 .994150 1.019258 .014047 .978376 1.185521 .108134 .985119 1.101808 .080867 .980422 1.087102 .078229 .977708 1.081392 .07735 .9735 1.085483 .077515 .967026 1.083276 .080175

Table 5.1. Estimators for parameters of a Weibull (1,1).

Tables 5.1 and 5.2, show the maximum likelihood estimator with original data, bKo , and the minimum power-divergence estimator θ ˆ ˆ(λ) θ, Kolmogorov estimator, θ for Normal and Weibull populations with parameters b = 1, c = 1, for Weibull populations and µ = 0, σ = 1, for Normal populations. These values have been calculated by computer simulation for 1000 samples, number of classes M = 6 and sample sizes n = 20, 40 and 60. The sums of the mean squared errors of the two parameters also appear in both tables. bKo and the θ ˆ(λ) need initial estimates. For The programs which calculate the θ Weibull populations these estimates have been calculated by the Dannenbring (1997) method, i.e., b b = x([.6321n]+1) and b c=

where x(1/2) is the sample median.


log(log 2) log(x(1/2) /b b)


223

In the case of normal populations we have taken as initial values the maximum likelihood estimators based on original data. As we expected, the mean squared error (mse) associated with the maximum likelihood estimator based on the original Weibull and Normal values, ˆ θ, is smaller ˆ(λ) . On than that associated with the minimum-power divergence estimators, θ bKo is greater the other hand the mse associated with the Kolmogorov estimator θ ˆ(λ) , for than the mse associated with all minimum power-divergence estimators, θ bKo , is based on the Weibull populations, although the Kolmogorov estimator, θ ˆ(λ) , classify the original data and the minimum power-divergence estimators, θ original data into classes.

ˆ θ

bKo θ ˆ(−2) θ

ˆ(−1) θ

ˆ θ(−0.5)

ˆ(0) θ

ˆ(1) θ

N(0, 1) µ b σ b mse µ b σ b mse µ b σ b mse µ b σ b mse µ b σ b mse µ b σ b mse µ b σ b mse

n = 20 -.011351 .960474 .038515 -.014687 .972221 .043739 -.012443 1.020231 .059777 -.010154 1.002308 .048961 -.010375 .991389 .044468 -.009577 .984915 .044277 -.009477 .970271 .039804

n = 40 -.004091 .978162 .018526 -.004947 .985606 .021340 -.005091 1.007078 .02851 -.005765 .995969 .024424 -.005316 .989804 .023099 -.006597 .986656 .022511 -.00711 .979003 .021464

n = 60 -.002958 .984269 .012468 -.003634 .987717 .014099 -.005469 1.001058 .017253 -.003588 .993877 .015981 -.002832 .992492 .015205 -.003736 .989282 .01498 -.002969 .981616 .014725

Table 5.2. Estimators for parameters of a Normal (0,1). bKo coincides, more or less, In Normal populations the mse associated with θ © 2006 by Taylor & Francis Group, LLC

224


ˆ(1) . For each sample size, θ ˆ(1) is the best of the with the mse associated with θ ˆ minimum power-divergence estimators θ(λ) for all sample sizes. It does not happen the same with Weibull populations because it depends on the sample size. ˆ(0) is the best estimator and for n = 60, the For n = 20 and 40 it seems that θ ˆ(−0.5) . best is θ From the comparison study we carried out, the new family of estimators we have introduced is a good alternative when it is necessary to classify the data. In Weibull populations, every member of the family of the minimum powerdivergence estimator is even better than Kolmogorov estimator which is based on original data. In relation to Normal populations the results are still better, ˆ(1) is as good as both estimators based on original data θ ˆ and θ ˆK0 . since θ

5.3.

Properties of the Minimum Phi-divergence Estimator

Throughout the Section, we assume that the model is correct, so that π = p(θ0 ), and M0 < M − 1. Furthermore, we restrict ourselves to unknown parameters θ0 satisfying the regularity conditions 1-6 introduced by Birch (1964): 1. θ0 is an interior point of Θ. 2. πi = pi (θ0 ) > 0 for i = 1, . . . , M. Thus π = (π1 , . . . , πM )T is an interior point of the set ∆M . 3. The mapping p : Θ → ∆M is totally differentiable at θ0 so that the partial derivatives of pi (θ0 ) with respect to each θj exist at θ0 and pi (θ) has a linear approximation at θ0 given by: M X ∂pi (θ0 ) pi (θ) = pi (θ0 ) + (θj − θ0j ) + o(kθ − θ0 k) ∂θj j=1

where o(kθ − θ0 k) denotes a function verifying lim

θ→θ 0

4. The Jacobian matrix J(θ0 ) =


µ

o(kθ − θ0 k) = 0. kθ − θ0 k

∂p (θ) ∂θ

¶

θ=θ 0

=

µ

∂pi (θ0 ) ∂θj

¶

i=1,...,M j=1,..,M0

(5.6)


225

is of full rank (i.e., of rank M0 ). 5. The inverse mapping p−1 : T → Θ is continuous at p(θ0 ) = π. 6. The mapping p : Θ → ∆M is continuous at every point θ ∈ Θ. We first derive the Fisher information matrix for the multinomial model. In some parts of the book it will be necessary to distinguish between the Fisher information matrix corresponding to the original model and the Fisher information matrix corresponding to an associated discretized model, i.e., the Fisher information matrix corresponding to a multinomial model. For this reason we denote IF (θ) the Fisher information matrix associated with the original model and I F (θ) the Fisher information matrix associated with the multinomial model. Proposition 5.1 We consider the multinomial model defined in (5.1) based on a random samP ple of size n, with n = M i=1 ni . Then the M0 × M0 Fisher information matrix associated with the random sample of size n is given by ¡ ¢ (n) I F (θ) = i(r,s) (θ) r,s=1,...,M 0 Ã ! M P 1 ∂pj (θ) ∂pj (θ) = n ∂θs j=1 pj (θ) ∂θr r,s=1,...,M0

= nA (θ)T A (θ)

where A (θ) is a M × M0 matrix given by ´ ³ A (θ) = diag p (θ)−1/2 J(θ).

(5.7)

(1)

For n = 1, I F (θ) ≡ I F (θ) = A (θ)T A (θ) .

Proof. We have log Pr θ (N1 = n1 , . . . , NM = nM ) = k +

M X

nj log pj (θ),

j=1

where k is independent of θ. Therefore, M

∂ log Pr θ (N1 = n1 , . . . , NM = nM ) X nj ∂pj (θ) = , ∂θr pj (θ) ∂θr j=1


226


and ·

∂ log Pr θ (.) ∂ log Pr θ (.) i(r,s) (θ) = E ∂θr ∂θs "

M P

¸

Ni ∂pi (θ) Nj ∂pj (θ) = E i,j=1 pi (θ) ∂θr pj (θ) ∂θs

#

  ·M ¸ M P Ni2 ∂pi (θ) ∂pi (θ) Ni ∂pi (θ) Nj ∂pj (θ)  P = E + E .  2 ∂θ ∂θ p r s i=1 pi (θ) i,j=1 i (θ) ∂θr pj (θ) ∂θs i6=j

But

E [Ni Nj ] = then we have

 2 2   npi (θ) (1 − pi (θ)) + n pi (θ)

if i = j

  −np (θ) p (θ) + n2 p (θ) p (θ) if i 6= j, i j i j

M np (θ) (1 − p (θ)) + n2 p (θ)2 ∂p (θ) ∂p (θ) P i i i i i i(r,s) (θ) = 2 (θ) ∂θ ∂θ p r s i=1 i

+

M −np (θ) p (θ) + n2 p (θ) p (θ) ∂p (θ) ∂p (θ) P i j i j i j p (θ)p (θ) ∂θ ∂θ i j r s i,j=1 i6=j

= n

− n

M P

M ∂p (θ) ∂p (θ) P 1 ∂pi (θ) ∂pi (θ) i i − n (1 − n) p (θ) ∂θ ∂θ ∂θ ∂θ r s r s i=1 i i=1 M ∂p (θ) ∂p (θ) M ∂p (θ) ∂p (θ) P P i j i j + n2 ∂θs ∂θs i,j=1 ∂θr i,j=1 ∂θr i6=j

= n

= n

i6=j

M ∂p (θ) ∂p (θ) ¢ P 1 ∂pi (θ) ∂pi (θ) ¡ 2 j i + n −n ∂θs ∂θs i=1 pi (θ) ∂θr i,j=1 ∂θr M P M P

1 ∂pi (θ) ∂pi (θ) . ∂θs i=1 pi (θ) ∂θr



By (5.7)

´ ³ A (θ) = diag p (θ)−1/2 J(θ),

therefore

A (θ)T A (θ) =

5.3.1.

227

Ã

M P

1 ∂pj (θ) ∂pj (θ) ∂θs j=1 pj (θ) ∂θr

!

r=1,...,M0 s=1,...,M0

.

Asymptotic Properties

In order to prove the following theorem we use the Implicit Function Theorem: Let F = (F1 , ..., FM0 ) : RM+M0 → RM0 be continuously differentiable in an open set D ⊂ RM+M0 , containing the point ¡ ¡ ¡ ¢ ¢¢ 0 x0 = x01 , ..., x0M , y0 = y10 , ..., yM 0

for which F (x0 , y0 ) = 0. Further, suppose that the matrix ¶ µ ∂Fi JF = i=1,...,M0 ∂yj j=1,...,M 0

is nonsingular at (x0 , y0 ) . Then there exists a neighborhood U of (x0 , y0 ) such that the matrix J F is nonsingular for all (x, y) ∈ U, an open set A ⊂ RM containing x0 and a continuously differentiable function g = (g1 , ..., gM0 ) : A → RM0 such that {(x, y) ∈ U : F (x, y) = 0} = {(x, g (x)) : x ∈ A} (see, e.g., p. 148 in Fleming (1977)). Theorem 5.1 Let φ ∈ Φ∗ be a twice continuously differentiable function in x > 0 with φ00 (1) > 0 and π =p (θ0 ). Under the Birch regularity conditions and assuming that the function p: Θ−→4M has continuous second partial derivatives in a neighborhood of θ0 , it holds ´ ³ bφ = θ0 + I F (θ0 )−1 A(θ0 )T diag π−1/2 (b b − π k), p − π) + o(k p θ © 2006 by Taylor & Francis Group, LLC

228


bφ is unique in a neighborhood of θ0 . where θ

Proof.

Let lM be the interior of the unit M-dimensional cube with ∆M ⊂ lM . Let V be a neighborhood of θ0 on which p : Θ → ∆M has continuous second partial derivatives. Let F = (F1 , ..., FM0 ) : lM × V → RM0 be defined by p1 , ..., peM ; θ1 , ..., θM0 ) = Fj (e

It holds

∂Dφ (e p, p (θ)) , ∀j = 1, ..., M0 . ∂θj

Fj (π1 , ..., πM ; θ01 , ..., θ0M0 ) = 0, ∀j = 1, ..., M0 due to µ ¶ ¶ ¶ M µ µ ∂Dφ (e p, p (θ)) X pel pel pel ∂pl (θ) 0 φ = −φ ∂θj pl (θ) pl (θ) pl (θ) ∂θj l=1

∀j = 1, ..., M0 .

Since ∂ ∂θr

µ

∂Dφ (e p, p (θ)) ∂θj

¶

= − +

M P

l=1

M P M P

l=1

µ

φ00

µ

φ0

µ

l=1

+

φ0

pel pl (θ)

pel pl (θ)

pel pl (θ)

¶

¶

¶

pel ∂pl (θ) ∂pl (θ) pl (θ)2 ∂θr ∂θj

pel ∂pl (θ) ∂pl (θ) pel pl (θ)2 ∂θr ∂θj pl (θ) pel ∂pl (θ) ∂pl (θ) pl (θ)2 ∂θr ∂θj

¶ ¶ ¶ µ µ µ M ∂ 2 p (θ) P pel pel pel l 0 + −φ , φ pl (θ) pl (θ) pl (θ) l=1 ∂θj ∂θr the M0 × M0 matrix J F associated with the function F at the point (p (θ0 ) , θ0 )



229

is given by ∂F ∂θ0

=

µ

¶ ∂F ∂θ (p e ,θ )=(π1 ,...,πM ;θ01 ,...,θ0M ) 0

Ãµ

! µ ¶¶ ∂Dφ (e p, p (θ)) ∂ = j=1,...,M0 ∂θr ∂θj r=1,...,M0 (p e ,θ )=(π1 ,...,πM ;θ01 ,...,θ0M ) 0 ¶ µM P 1 (θ ) ∂p (θ ) ∂p 0 0 l l 00 = φ (1) r=1,...,M0 ∂θr ∂θj l=1 pl (θ 0 ) j=1,...,M0 = φ00 (1) A (θ0 )T A (θ0 ) . We recall that if B is a p × q matrix with rank p and C is a q × s matrix with rank (C) = q then rank (BC) = p. By taking ´ ³ B = J(θ0 )TM0 ×M and C = diag p (θ0 )−1/2 M×M

it follows that A (θ0 )T = BC has rank M0 by condition 4 of Birch. Also,

´ ´ ³ ³ rank A (θ0 )T A (θ0 ) = rank A (θ0 ) A (θ0 )T = rank (A (θ0 )) = M0 . Therefore, the M0 × M0 matrix

∂F is nonsingular at (π1 , ..., πM ; θ01 , ..., θ0M0 ) . ∂θ0

Applying the Implicit Function Theorem there exists a neighborhood U of (p (θ0 ) , θ0 ) such that the matrix J F is nonsingular (in our case J F in (p (θ0 ) , θ0 ) is positive definite and then J F is positive definite for all (e p,θ) ∈ U, because F is continuously differentiable). Also, there exists a continuously differentiable ˜ :A ⊂ lM → RM0 such that p (θ0 ) ∈ A and function θ n o e p)) : p e∈ A . {(e p,θ) ∈ U : F (e p,θ) = 0} = (e p,θ(e (5.8) e (p (θ0 )) is an arg inf of We can observe that θ

Ψ (θ) = Dφ (p (θ0 ) , p (θ))

because p (θ0 ) ∈ A and then

³ ³ ´´ e (p (θ0 )) ³ ´ ∂Dφ p (θ0 ) , p θ e (p (θ0 )) = F p (θ0 ) , θ =0 ∂θ


230


e and ³ also by (5.8),´(p (θ0 ) , θ (p (θ0 ))) ∈ U and then J F is positive definite at e (p (θ0 )) . Therefore, p (θ0 ) , θ ³ ³ ´´ e (p (θ0 )) = inf Dφ (p (θ0 ) , p (θ)) Dφ p (θ0 ) , p θ θ∈Θ

e (p (θ0 )) = θ0 . and by the φ-divergence properties θ Applying the chain rule, we get

e=π, and, for p

e p) e p)) ∂F (e e p)) ∂ θ(e ∂F (e p, θ(e p, θ(e + =0 e p) ∂e p ∂e p ∂ θ(e ∂F ∂F ∂θ0 + = 0. ∂π ∂θ0 ∂π

Further we know that

∂F = φ00 (1)A(θ0 )T A(θ0 ) ∂θ0

∂F is ∂π ´ ³ ∂F = −φ00 (1)A(θ0 )T diag p (θ0 )−1/2 . ∂π

and we shall establish later that the M0 × M matrix

Therefore the M0 × M matrix

(5.9)

∂θ0 is ∂π

´ ³ ∂θ0 = (A(θ0 )T A(θ0 ))−1 A(θ0 )T diag p (θ0 )−1/2 . ∂π e around π yields The Taylor expansion of the function θ e p) = θ(π) e θ(e +

Ã

e (e ∂θ p) ∂e p

e As θ(π) = θ0 we obtain from here

!

e =π p

e − π k). (e p − π) + o(k p

´ ³ ¡ ¢ e p) = θ0 + A(θ0 )T A(θ0 ) −1 A(θ0 )T diag p (θ0 )−1/2 (e e − π k). p − π) + o(k p θ(e

c.s. e p) is the b −→ π =p(θ0 ), so that p b ∈ A and, consequently, θ(b We know that p n→∞ unique solution of the system of equations



³ ´ e p)) b, p(θ(b ∂Dφ p ∂θj

= 0,

231

j = 1, ..., M0

³ ´ e p) ∈ U. Therefore, θ(b e p) is the minimum φ-divergence estimator, b, θ(b and also p bφ , satisfying the relation θ ´ ³ ¡ ¢ bφ = θ0 + A(θ0 )T A(θ0 ) −1 A(θ0 )T diag p (θ0 )−1/2 (b p − p(θ0 )) θ +o(kb p − p(θ0 )k).

Finally, we establish the formula (5.9). We calculate the (i, j)th-element of the ∂F M0 × M matrix , ∂π ÃM µ µ ! ¶ ¶ ¶ µ ¶ µ X ∂Dφ (p, p (θ)) ∂ pl ∂pi (θ) ∂ pl pl 0 −φ = φ ∂pi ∂θj ∂pi pl (θ) pl (θ) pl (θ) ∂θj l=1 µ µ ¶¶ pi pi 00 ∂pi (θ) 1 − φ = pi (θ) pi (θ) pi (θ) ∂θj and for (π1 , ..., πM ; θ01 , ..., θ0M0 ) we have µ ¶¶ µ ∂Dφ (p, p (θ0 )) 1 ∂pi (θ0 ) ∂ =− . φ00 (1) ∂pi ∂θj pi (θ0 ) ∂θj ´ ³ Since A(θ0 ) = diag p (θ0 )−1/2 J (θ0 ) then (5.9) holds. Theorem 5.2 Under the assumptions of Theorem 5.1, it holds √ b L n(θφ − θ0 ) −→ N(0, I F (θ0 )−1 ). n→∞

Proof. Applying the Central Limit Theorem, we get √ L n(b p − p(θ0 )) −→ N(0, Σp(θ 0 ) ) n→∞

being Σp(θ 0 ) the M × M matrix Σp(θ 0 ) = diag (p (θ0 )) − p (θ0 ) p (θ0 )T .


232


Consequently √ b L n(θφ − θ0 ) −→ N (0, Σ∗ ) , n→∞

where Σ∗ =

´ ³ ¡ ¢−1 A(θ0 )T A(θ0 ) A(θ0 )T diag p (θ0 )−1/2 Σp(θ 0 )

´ ³ ¡ ¢−1 × diag p (θ0 )−1/2 A(θ0 ) A(θ0 )T A(θ0 ) ,

bφ has the expression because θ

´ ³ ¡ ¢−1 b − p(θ0 ) k). θ0 + A(θ0 )T A(θ0 ) p − p(θ0 )) + o(k p A(θ0 )T diag p (θ0 )−1/2 (b For the M × M matrix

³ ´ ´ ³ B = diag p (θ0 )−1/2 Σp(θ 0 ) diag p (θ0 )−1/2 ,

we have ´³ ´ ³ ´ ³ B = diag p (θ0 )−1/2 diag (p (θ0 )) − p (θ0 ) p (θ0 )T diag p (θ0 )−1/2 ´ ´ ³ ³ = diag p (θ0 )−1/2 diag (p (θ0 )) diag p (θ0 )−1/2

´ ´ ³ ³ − diag p (θ0 )−1/2 p (θ0 ) p (θ0 )T diag p (θ0 )−1/2 = I M×M − Therefore, Σ∗ = −

q p p (θ0 ) p (θ0 )T .

¡ ¢−1 A(θ0 )T A(θ0 )

q p ¡ ¢−1 ¡ ¢−1 A(θ0 )T A(θ0 ) A(θ0 )T p (θ0 ) p (θ0 )T A(θ0 ) A(θ0 )T A(θ0 ) .

Finally we establish that


q p (θ0 )T A(θ0 ) = 01×M0 .


233

We have 

then

    A (θ0 ) =     

p1 (θ0 )−1/2

∂p1 (θ0 ) ∂θ1

. . . pM (θ0 )−1/2

... ... ... ...

∂pM (θ0 ) ∂θ1

∂p1 (θ0 )   ∂θM0   .   .   .  ∂p (θ ) 

p1 (θ0 )−1/2

... pM (θ0 )−1/2

M

0

∂θM0

q ³ ´ p (θ0 )T A(θ0 ) = p1 (θ0 )1/2 , ..., pM (θ0 )1/2 A(θ0 )

µM ¶ M ∂p (θ ) P ∂pi (θ0 ) P i 0 = , ..., = 01×M0 . ∂θ1 i=1 i=1 ∂θM0

Finally

¡ ¢−1 Σ∗ = A(θ0 )T A(θ0 ) = I F (θ0 )−1 .

Based on the last two theorems we have that the minimum φ-divergence estimator is a BAN (Best Asymptotically Normal) estimator.

5.3.2.

Minimum Phi-divergence Functional Robustness

In this section we consider deviations from the discrete model, p(θ) = (p1 (θ), . . . , pM (θ))T , given by pε (θ) = (1 − ε)p(θ) + εp for ε > 0, θ ∈ Θ and p ∈ ∆M . Let θεφ (p) be the vector that minimizes the function ¶ µ M X pi gε (p, θ) = , pi (θ, ε)φ pi (θ, ε) i=1 where pε (θ) = (p1 (θ, ε), . . . , pM (θ, ε))T . In order to guarantee the robustness of θφ (p), we have to verify that slight deviations of p(θ) correspond to slight deviations of θεφ (p) or, analytically, that limε→0 θεφ (p) = θφ (p). The following theorem gives conditions that guarantee the functional robustness.


234


Theorem 5.3 Let the assumptions of Theorem 5.1 be fulfilled and in addition Θ is a compact set. Then lim θεφ (p) = θφ (p). ε→0

Proof. Let {εn } be an arbitrary sequence of positive numbers verifying lim εn = n→∞

0. Since φ is continuous and pi (θ, ²n ) −→ pi (θ), i = 1, . . . , M, we get that εn →0

gεn (p, θ) −→ g0 (p, θ), ∀θ ∈ Θ. Since Θ is compact the pointwise convergence εn →0

implies the uniform convergence. Consequently limεn →0 supθ∈Θ | gεn (p, θ) − g0 (p, θ) |= 0 which implies that lim | inf gεn (p, θ) − inf g0 (p, θ) |= 0,

εn →0

θ∈Θ

θ∈Θ

or equivalently lim | gεn (p, θεφn (p)) − g0 (p, θφ (p)) |= 0.

εn →0

So, we have proved that lim gεn (p, θεφn (p)) = g0 (p, θφ (p)).

εn →0

(5.10)

If limεn →0 θεφn (p) 6= θφ (p), the compactness of Θ guarantees the existence of a subsequence {θδφn (p)} ⊂ {θεφn (p)} such that limδn →0 θδφn (p) = θ∗ 6= θφ (p). By (5.10), g0 (p, θ∗ ) = g0 (p, θφ (p)) for θ∗ 6= θφ (p) which contradicts the assumed uniqueness of θφ (p). The statement of theorem follows from here since the sequence {εn } can be chosen arbitrarily. A more general way of studying the robustness is to assume that the true distribution π ∈ ∆M verifies kπ − p(θ)k < ε for some θ ∈ Θ and to prove that if ε is small, the θφ (π) value is near to θφ (p(θ)) = θ. Theorem 5.4 Let the assumptions of Theorem 5.1 hold and let π ∈ ∆M . Then lim

kπ−p(θ)k→0

θφ (π) = θφ (p(θ)) = θ.

Proof. It is immediate because θφ is a continuous function.



5.4.

235

Normal Mixtures: Minimum Phi-divergence Estimator

In this section we report the results of a simulation experiment designed to b(λ) that can compare empirically some minimum power-divergence estimators θ be seen in Pardo, M.C. (1999b). For the parameters of a mixture of two normal populations we analyze the efficiency as well as the robustness of the chosen estimators. We consider estimators corresponding to λ = −2, −1, −0.5, 0, 2/3 and 1. Simulations reported in this section are based on mixing proportions 0.25, 0.5 and 0.75. For each of these mixing proportions, firstly, we consider mixtures of the densities f1 (x) and f2 (x) where f1 (x) is the density for the random variable X = aY and f2 (x) is the density associated with X = Y + b where a > 0, b > 0 and Y is standard normal. Secondly, we assume that Y is a Student’s t distribution with two or four degrees of freedom, or double exponential, to study the robustness under symmetric departures from normality. Thus, “a” is the √ ratio of scale parameters which we take to be 1 and 2 while “b” is selected to provide the overlap desired between the two distributions. We consider “overlap” as the probability of misclassification, using the rule: Classify an observation x as being from population 1 if x < xc and from population 2 if x ≥ xc , where xc is theZunique point between µ1 and µ2 such that wf1 (xc ) = (1 − w)f2 (xc ) with µi =

xi fi (x) dx, i = 1, 2. The overlaps examined in the current study are 0.03

R

and 0.1. For each set of considered configurations, 5000 samples of size n = 100 were generated from the corresponding mixture distribution, and for each conb(−2) , θ b(−1) , θ b(−1/2) , sidered sample the minimum power-divergence estimators θ b(2/3) and θ b(1) were obtained. Our implementation of the minimum powerb(0) , θ θ b(λ) , employed the IMSL subroutine ZXMIN which minidivergence estimator, θ mizes a function of various variables. Although all these estimation procedures provide estimators of all five parameters, Tables 5.3, 5.5, 5.7 and 5.9 present only estimations for w; however, in Table 5.11 we present estimations for all the parameters. For either of the estimators proposed in the previous section to be used in practice, one must have starting values for the iterative procedures. We choose an ad hoc quasi-clustering technique used by Woodward et al . (1984) that is easy


236


to implement. They allow as possible values for the initial estimate of w only the values 0.1,...,0.9. For each of these values of w, the sample is divided into two subsamples, Y1 , Y2 , ..., Yn1 and Yn1 +1 , Yn1 +2 , ..., Yn , where Yi is the ith-order b is that value at statistic and n1 is “nw” rounded to the nearest integer. So, w which w(1 − w)(x1(1/2) − x2(1/2) )2 is maximized, and

b2 = x2(1/2) , σ b12 = ((x1(1/2) − x1(0.25) )/.6745)2 µ b1 = x1(1/2) , µ σ b22 = ((x2(0.75) − x2(1/2) )/.6745)2 ,

where xj(q) is the q th-percentile from the j th-subsample. In Table 5.3 we present summary results of the simulation carried out to compare the performance of the estimators for mixtures of normal components. Estimators of the bias and mean squared error (mse) based on the simulations are given by ns ns X 1 X [= 1 Bias (w bi − w) and mse d = (w bi − w)2 , ns ns i=1

i=1

bi denotes an estimate of w for the ithwhere ns is the number of samples and w sample. It should be noted that nmse d is the quantity actually given in the tables. b(λ) , We also table empirical measures of the relative efficiencies of each estimator, θ b(0) . The relative efficiency is given with respect to the MLE for grouped data θ by b(0) ) mse(6 d =θ b= E . b(0) ) mse( d θ We use a minimax criterion to find the best estimator. Let S = {−2, − 1, − .5, 0, 2/3, 1} b(λ) studied. Let βMIN (Pj ) = inf λ∈S nmse d λ (Pj ) , be the values of λ considered for θ j = 1, ..., 10 where Pj , j = 1, ..., 10 is each one of the mixture of normal under consideration. If we define the inefficiency function as iλ (Pj ) = nmse d λ (Pj ) − βMIN (Pj ) , j = 1, ..., 10, then, η(λ) = maxj (iλ (Pj )) can be considered as a inb(λ) , λ ∈ S. We efficiency measure of the minimum power-divergence estimator, θ consider the two best estimators under this criterion. Analyzing the results of b(1) , Table 5.3, we can see in Table 5.4 that the minimum chi-square estimator, θ b(−2) , are the best. and minimum modified chi-square estimator, θ



w .25

a 1

.5

1

.25

√ 2

.5

√ 2

.75

√ 2

Table

237

.1 Overlap .03 Overlap b b [ [ Estimator Bias n mse d E Bias n mse d E b θ(−2) .062 1.997 1.02 .037 .767 1.02 b θ(−1) .059 1.934 .99 .038 .797 1.06 b(−.5) θ .063 1.937 .99 .037 .767 1.02 b(0) θ .062 1.960 .036 .752 b θ(2/3) .062 2.053 1.05 .035 .771 1.03 b(1) θ .061 1.918 .98 .035 .760 1.01 b θ(−2) -.015 1.579 1.05 -.012 .707 .86 b(−1) θ -.015 1.565 1.04 -.014 .757 .92 b θ(−.5) -.015 1.455 .97 -.012 .719 .87 b(0) θ -.014 1.504 .011 .826 b(2/3) θ -.018 1.594 1.06 -.010 .800 .97 b(1) θ -.016 1.529 1.02 -.012 .781 .95 b θ(−2) -.023 .947 1.01 -.001 .588 1.05 b(−1) θ -.025 1.021 1.09 -.002 .573 1.03 b(−.5) θ -.025 .949 1.01 -.002 .567 1.02 b θ(0) -.026 .939 -.001 .557 b(2/3) θ -.026 .920 .98 -.002 .567 1.02 b(1) θ -.024 .951 1.01 -.001 .564 1.01 b(−2) θ -.102 2.135 1.01 -.057 1.003 .95 b θ(−1) -.100 2.176 1.02 -.062 1.057 1.00 b(−.5) θ -.101 2.152 1.01 -.061 1.079 1.02 b θ(0) -.101 2.123 -.060 1.054 b(2/3) θ -.101 2.120 1.00 -.063 1.124 1.07 b θ(1) -.098 2.043 .96 -.061 1.079 1.02 b(−2) θ -.168 4.563 .98 -.080 1.475 .95 b θ(−1) -.167 4.700 1.01 -.078 1.658 1.07 b(−.5) θ -.167 4.628 .99 -.080 1.620 1.04 b(0) θ -.166 4.667 -.076 1.553 b θ(2/3) -.167 4.704 1.01 -.073 1.379 .89 b(1) θ -.165 4.653 1.00 -.075 1.478 .95 5.3. Simulation results for mixtures of normal components.

η(.)

b θ (−2) 0.124

b(−1) θ 0.279

b(−.5) θ 0.241

b(0) θ 0.174

Table 5.4. Inefficiencies.

b(2/3) θ 0.141

b(1) θ 0.099

In Table 5.5 we display the results for the nonnormal components (double


238


exponential components). In this case (see Table 5.6), in accordance with our b(0) and θ b(2/3) . criterion, the best are θ

w .25

.5

.25

.5

.75

Table 5.5.

a

Estimator 1

b(−2) θ b θ(−1) b(−.5) θ b(0) θ b(2/3) θ b(1) θ b(−2) θ b(−1) θ b(−.5) θ b(0) θ b(2/3) θ b(1) θ b θ(−2) b(−1) θ b(−0.5) θ b(0) θ b(2/3) θ b(1) θ b(−2) θ b(−1) θ b(−.5) θ b(0) θ b(2/3) θ b(1) θ b(−2) θ b(−1) θ b(−.5) θ b(0) θ b(2/3) θ b(1) θ

.1 [ Bias

.056 .055 .057 .054 .054 .053 1 -.007 -.007 -.002 -.004 -.003 -.002 √ 2 .013 .012 .013 .014 .012 .015 √ 2 .052 -.049 -.051 -.048 -.050 -.050 √ 2 -.106 -.104 -.096 -.098 -.095 -.094 Simulation results for


Overlap n mse d

b E

.03 [ Bias

Overlap n mse d

b E

1.000 1.00 .057 .780 1.05 .975 .98 .056 .780 1.05 1.052 1.06 .055 .771 1.04 .995 .055 .740 .961 .97 .055 .776 1.05 .996 1.00 .053 .754 1.02 .682 .96 -.002 .425 1.07 .716 1.00 -.002 .381 .96 .703 .99 -.002 .391 .99 .713 .000 .397 .738 1.04 -.001 .390 .98 .712 1.00 .001 .405 1.02 .669 1.00 .036 .535 .97 .638 .96 .034 .517 .93 .659 .99 .036 .552 1.00 .666 .036 .554 .656 .99 .036 .562 1.02 .662 .99 .036 .555 1.00 1.157 1.07 -.029 .547 1.01 1.070 .99 -.027 .526 .97 1.111 1.03 -.029 .541 1.00 1.079 -.027 .541 1.039 .96 -.026 .558 1.03 1.124 1.04 -.028 .591 1.09 2.191 1.14 -.078 1.167 1.03 2.135 1.11 -.078 1.185 1.04 1.841 .96 -.078 1.181 1.04 1.921 -.075 1.136 1.908 .99 -.075 1.163 1.02 1.952 1.02 -.072 1.071 .94 mixtures of double exponential components.


b θ (−2) 0.35

η(.)

b(−1) θ 0.294

b(−.5) θ 0.091

b(0) θ 0.08


b(2/3) θ 0.067

239

b(1) θ 0.111

In Table 5.7 we present the results for the nonnormal components (t-Student with 4 degrees of freedom, t(4)). w .25

.5

.25

.5

.75

Table

a

Estimator 1

b(−2) θ b θ(−1) b(−.5) θ b(0) θ b(2/3) θ b(1) θ b(−2) θ b(−1) θ b(−0.5) θ b(0) θ b(2/3) θ b(1) θ b θ(−2) b(−1) θ b(−.5) θ b(0) θ b(2/3) θ b(1) θ b θ(−2) b(−1) θ b(−.5) θ b(0) θ b(2/3) θ b(1) θ b(−2) θ b(−1) θ b(−0.5) θ b(0) θ b(2/3) θ b(1) θ

.1 [ Bias

.084 .083 .081 .080 .077 .081 1 -.008 -.008 -.008 -.010 -.011 -.005 √ 2 -.008 -.007 -.007 -.007 -.010 -.007 √ 2 -.093 -.098 -.099 -.099 -.099 -.097 √ 2 -.189 -.186 -.187 -.181 -.183 -.182 5.7. Simulation results


Overlap n mse d

b E

.03 [ Bias

Overlap n mse d

b E

2.699 1.00 .049 .947 1.04 2.663 .99 .047 .901 .99 2.642 .98 .047 .913 1.00 2.695 .049 .911 2.576 .96 .046 .890 .98 2.604 .97 .048 .931 1.02 1.697 1.02 -.008 .827 1.04 1.657 .99 -.010 .793 1.00 1.598 .96 -.012 .838 1.06 1.668 .006 .794 1.745 1.05 -.008 .837 1.06 1.735 1.04 -.006 .809 1.02 1.007 1.01 .008 .582 1.03 1.055 1.06 .008 .554 .98 1.033 1.03 .008 .564 1.00 .999 .006 .565 .979 .98 .007 .553 .98 .967 .97 .007 .574 1.01 2.296 .92 -.060 1.191 1.01 2.539 1.01 -.061 1.18 .999 2.594 1.04 -.060 1.151 .98 2.504 -.058 1.177 2.48 .99 -.061 1.232 1.05 2.531 1.01 -.060 1.188 1.01 5.818 1.06 -.100 1.989 1.06 5.748 1.05 -.102 2.105 1.13 5.778 1.053 -.098 1.876 1.00 5.487 -.096 1.870 5.699 1.04 -.098 1.86 .99 5.541 1.01 -.099 1.893 1.01 for mixtures of t-Student (4) components.

240


η(.)

b(−2) θ 0.331

b θ (−1) 0.261

b(−0.5) θ 0.298

b(0) θ 0.208


b(2/3) θ 0.212

b(1) θ 0.235

b(0) and θ b(2/3) are again the best. Now we For t-Student(4) we have that θ present the results for a t-Student with 2 degrees of freedom, t(2). w .25

a

Estimator 1

b(−2) θ b θ(−1) b(−.5) θ b(0) θ b(2/3) θ b(1) θ b(−2) θ b(−1) θ b(−.5) θ b(0) θ b(2/3) θ b(1) θ b(−2) θ b(−1) θ b(−.5) θ b(0) θ b(2/3) θ b(1) θ b(−2) θ b θ(−1) b(−.5) θ b(0) θ b(2/3) θ b(1) θ b(−2) θ b θ(−1) b(−.5) θ b(0) θ b(2/3) θ b(1) θ

.1 [ Bias

.111 .113 .110 .112 .107 .109 .5 1 -.012 -.017 -.016 -.017 -.016 -.017 √ .25 2 .038 .037 .037 .037 .032 .028 √ .5 2 -.119 -.126 -.124 -.127 -.126 -.126 √ .75 2 -.229 -.234 -.234 -.231 -.236 -.230 Table.5.9. Simulation results


Overlap n mse d

b E

.03 [ Bias

6.742 1.00 .059 6.963 1.03 .060 6.755 1.00 .057 6.732 .057 6.746 1.00 .058 6.839 1.02 .058 3.900 .98 -.011 3.876 .98 -.009 3.903 .99 -.011 3.960 -.006 3.874 .98 -.007 3.995 1.01 -.005 4.736 .99 .011 4.588 .96 .011 4.715 .98 .009 4.797 .013 4.740 .99 .016 4.520 .94 .011 4.497 .93 -.072 4.829 1.00 -.070 4.551 .94 -.071 4.839 -.071 4.827 1.00 -.069 4.795 .99 -.069 11.064 .99 -.121 11.112 1.00 -.117 11.374 1.02 -.117 11.145 -.113 11.463 1.03 -.111 11.102 1.00 -.114 for mixtures of t-Student

Overlap n mse d

b E

1.468 1.00 1.509 1.02 1.493 1.01 1.473 1.491 1.01 1.495 1.01 1.193 1.07 1.136 1.02 1.136 1.02 1.112 1.006 .995 1.138 1.02 .832 1.03 .869 1.07 .836 1.03 .809 .815 1.01 .802 .99 1.874 1.01 1.794 .96 1.796 .96 1.862 1.799 .97 1.702 .91 3.353 1.06 3.177 1.00 3.096 .98 3.171 3.075 .97 3.184 1.00 (2) components.


b(−2) and θ b(1) , although θ b(1) has less bias. For t-Student(2), the best are θ b(−2) θ 0.278

η(.)

b(−1) θ 0.332

b(−.5) θ 0.31

b(0) θ 0.342

b(2/3) θ 0.399


b(1) θ 0.298

.1 Overlap µ1 σ1 µ2 σ2 w Normal b(−2) .25 1 θ 1.02 1.04 1.03 .99 1.02 b θ(−1) 1.31 1.05 1.02 1.00 .99 b(−.5) θ .95 1.01 1.02 .99 .99 b(2/3) θ 1.08 1.05 1.02 .98 1.05 b θ(1) 1.00 1.07 .99 .99 .98 b(−2) .5 1 θ 1.01 1.09 1.03 .97 1.05 b(−1) θ 1.83 1.04 1.04 .99 1.04 b(−.5) θ 1.00 1.05 .99 .96 .97 b θ(2/3) 1.04 1.03 1.04 1.01 1.06 b(1) θ 1.00 1.01 .98 .97 1.02 √ b .25 2 θ(−2) 1.00 .95 1.03 1.06 1.01 b(−1) θ .98 1.01 1.10 1.03 1.09 b(−.5) θ .97 1.02 1.07 1.06 1.01 b(2/3) θ .98 1.11 1.00 .99 .98 b θ(1) .97 1.00 1.01 1.04 1.01 √ b(−2) .5 2 θ 1.01 1.08 1.02 1.02 1.01 b(−1) θ .99 1.06 1.03 1.05 1.02 b θ(−.5) .96 1.01 1.02 .97 1.01 b(2/3) θ .99 .99 .97 1.02 1.00 b(1) θ 1.00 1.02 .96 .91 .96 √ b(−2) .75 2 θ 1.00 1.00 .96 .97 .98 b θ(−1) 1.11 1.10 .96 1.00 1.00 b(−.5) θ 1.06 1.09 .93 .98 .99 b θ(2/3) 1.00 1.02 .93 1.05 1.01 b(1) θ 1.02 1.03 .95 .98 1.00 Table 5.11. Estimated relative efficiencies of the estimators relative to the MLE for mixture. w

a


Estimator

241

242


w

a

.25

1

.5

1

.25

√ 2

.5

√ 2

.75

√ 2

.03 Overlap µ1 σ1 µ2 σ2 Normal b(−2) θ .93 1.03 .98 1.00 b θ (−1) .92 .89 1.04 1.01 b(−0.5) θ .94 .94 .98 .98 b θ (2/3) 1.03 1.02 .99 .97 b(1) θ .95 1.11 1.01 .96 b θ (−2) .98 1.04 .99 1.04 b(−1) θ .97 1.01 .93 1.16 b(−.5) θ .96 .99 .96 .98 b θ (2/3) .99 .99 .99 .99 b(1) θ 1.01 1.01 .99 .94 b(−2) θ 1.01 1.04 1.04 1.02 b(−1) θ 1.15 1.21 1.05 .98 b θ (−5) .98 .93 1.00 .98 b(2/3) θ 1.06 1.04 1.01 1.03 b(1) θ 1.10 1.03 1.00 1.00 b(−2) θ .96 1.07 1.03 1.12 b θ (−1) .99 1.04 1.20 1.07 b(−.5) θ 1.04 1.00 .97 .95 b θ (2/3) 1.04 .99 1.03 1.00 b(1) θ 1.05 1.02 1.09 1.01 b(−2) θ .95 .93 .90 .92 b θ (−1) .97 .93 1.03 1.05 b(−.5) θ 1.04 .97 .99 .99 b θ (2/3) .99 .96 .93 .98 b(1) θ .95 .96 .99 1.01 Table 5.11 (Continuation). Estimator

w 1.02 1.06 1.02 1.03 1.01 .86 .92 1.02 .97 .95 1.05 1.03 1.02 1.02 1.01 .95 1.00 1.02 1.07 1.02 .95 1.07 1.04 .89 .95

Although our emphasis has been put on the estimation of the mixing proportion, the estimation routines used here obtain estimations for all five parameters. So, it seems obvious to question about whether the results shown for w are similar for the rest of the parameters µ1 , σ1 , µ2 and σ2 . In Table 5.11 we display empirical relative efficiencies for all the parameters for normal and t(4) mixtures. From the table we see that the results for the other parameters with t(4) do not exhibit in general patterns similar to those shown for w. The Freeman-Tukey b(−.5) , seems to be the most efficient for µ1 , σ1 , µ2 and σ2 in Normal estimator, θ and t(4) mixtures.



w

a

.25

1

.5

1

.25

√ 2

.5

√ 2

.75

√ 2

.1 Overlap µ1 σ1 µ2 σ2 t-Student(4) b(−2) θ 1.01 1.01 1.03 1.00 b θ(−1) 1.03 1.07 1.03 1.00 b(−.5) θ 1.07 1.07 1.03 1.01 b θ(2/3) 1.02 1.33 1.00 1.03 b(1) θ 1.05 1.05 1.01 1.01 b θ(−2) .96 1.03 .99 1.07 b(−1) θ .96 .97 .99 1.01 b(0) θ .93 1.00 .96 1.03 b(2/3) θ .95 1.00 1.05 1.05 b(1) θ 1.05 1.14 1.01 .99 b θ(−2) .99 1.18 1.03 1.02 b(−1) θ 1.00 .91 1.02 1.05 b θ(−.5) .97 1.02 1.03 1.02 b(2/3) θ 1.02 .91 .98 .99 b θ(1) 1.02 .96 1.00 .95 b(−2) θ .97 .92 .90 .89 b θ(−1) .99 1.27 .97 1.03 b(−.5) θ 1.03 .84 .90 .94 b θ(2/3) 1.07 .88 .93 .93 b(1) θ 1.03 .85 .93 .90 b θ(−2) 1.05 .88 1.09 1.07 b(−1) θ 1.05 .84 1.03 1.00 b(−.5) θ 1.05 .85 1.05 1.03 b θ(2/3) 1.07 1.04 1.04 1.03 b(1) θ 1.04 .98 1.01 .97 Table 5.11. (Continuation). Estimator

243

w 1.00 .99 .98 .96 .97 1.02 .99 .96 1.05 1.04 1.01 1.06 1.03 .98 .97 .92 1.01 1.04 .99 1.01 1.06 1.05 1.05 1.04 1.01

To sum up, the Freeman-Tukey estimator seems to be, in general, the most efficient for all the parameters except the proportion parameter. For the proporb(2/3) , Cressie-Read estimator. It is the best tion parameter we recommend to use θ alternative for the component distributions double exponential and t(4), almost as good as the best at the true model. It performs only worse for t(2) when the departure from normality is more extreme. Finally, the minimum chi-square estimator is preferable for extreme departures from normality.


244

5.5.


w

a

.25

1

.5

1

.25

√ 2

.5

√ 2

.75

√ 2

.03 Overlap µ1 σ1 µ2 σ2 t-Student(4) b(−2) θ 1.01 .96 .99 1.02 b θ (−1) .98 .97 .99 1.01 b(−.5) θ 1.02 .97 1.02 1.08 b(2/3) θ 1.03 1.19 .96 .96 b(1) θ 1.07 1.16 .98 .99 b θ (−2) 1.03 1.04 1.14 1.11 b(−1) θ 1.04 1.36 1.01 1.08 b θ (−.5) 1.02 1.01 .96 1.04 b(2/3) θ 1.03 1.06 .94 .99 b(1) θ 1.06 1.13 .91 .94 b θ (−2) .86 .92 1.02 1.06 b(−1) θ .99 .97 .98 1.04 b θ (−.5) .93 .96 1.03 1.00 b(2/3) θ .92 1.38 1.01 1.01 b(1) θ .91 .91 1.01 .98 b(−2) θ .95 .86 1.15 1.16 b(−1) θ .99 .79 1.10 1.15 b(−.5) θ 1.04 .80 .99 1.01 b θ (2/3) 1.08 .77 1.05 1.01 b(1) θ 1.05 .77 1.13 1.06 b θ (−2) 1.03 1.16 1.09 .99 b(−1) θ 2.95 1.13 1.22 1.18 b(−.5) θ .98 1.12 .99 1.19 b θ (2/3) 1.00 1.03 1.01 .98 b(1) θ 1.02 1.02 1.26 .95 Table 5.11. (Continuation). Estimator

w 1.04 .99 1.00 .98 1.02 1.04 1.00 1.06 1.06 1.02 1.03 .98 1.00 .98 1.01 1.01 1.00 .98 1.05 1.01 1.06 1.13 1.00 .99 1.01

Minimum Phi-divergence Estimator with Constraints: Properties

A new problem appears if we suppose that we have ν real-valued functions f1 (θ) , ..., fν (θ) which constrain the parameter θ : fm (θ) = 0, m = 1, 2, ..., ν.



245

The problem to obtain the maximum likelihood estimator subject to constraints was considered, for the first time, by Aitchison and Silvey (1958) in the context of a random variable whose distribution function F depends on M0 parameters θ1 , ..., θM0 which are not mathematically independent but satisfy ν functional relationships. The Lagrangian multiplier method is used to find the maximum likelihood estimator of the parameters, which are found numerically by a process of iteration. The estimator is shown to have an asymptotic normal distribution. Silvey (1959) further discussed the Lagrangian method and the mathematical conditions for the existence of the maximum likelihood estimators. Diamond et al . (1960) considered the restricted maximum likelihood estimator, but in multinomial populations. Haber and Brown (1986) proposed a two-step algorithm for obtaining maximum likelihood estimators of the expected frequencies in loglinear models with expected frequencies subject to linear constraints. Problems of this type in multinomial models with the log-linear parameterization have been described by Bhapkar (1979), Bonett (1989), Gokhale (1973), Haber (1985), Haberman (1974) and Wedderburn (1974). In this Section we consider the minimum φ-divergence estimator subject to (r) some constraints: The restricted minimum φ-divergence estimator, θbφ , which is seen to be a generalization of the maximum likelihood estimator, subject to constraints, was studied for the first time in Pardo, J. A. et al . (2002). We suppose that we have ν (ν < M0 ) real-valued functions f1 (θ) , ..., fν (θ) that constrain the parameter θ ∈ Θ ⊂ RM0 , fm (θ) = 0, m = 1, ..., ν, and such that i) Every fm (θ) has continuous second partial derivatives, ii) The ν × M0 matrix B (θ) =

µ

∂fm (θ) ∂θk

¶

m=1,...,ν k=1,...,M0

is of rank ν. Under these assumptions the restricted minimum φ-divergence estimator of b(r) ∈ Θ satisfying the condition θ is θ φ (r)

b )) = p, p(θ Dφ (b φ


Dφ (b inf p, p (θ)). {θ∈Θ⊆RM0 :fm (θ)=0,m=1,...,ν }

246


In the cited paper of Pardo, J. A. et al. (2002) it was established, under the conditions of Theorem 5.1 and assuming the previous conditions i) and ii) about the functions fm (θ) = 0, m = 1, ..., ν, b(r) has the following expansion that θ φ

´ ³ θ0 + H (θ0 ) I F (θ0 )−1 A (θ0 )T diag p (θ0 )−1/2 (b p − p (θ0 )) + o (kb p − p (θ0 )k) (r)

b is unique in a neighborhood of θ0 and the M0 × M0 matrix H (θ0 ) is where θ φ defined by ³ ´−1 H (θ0 ) = I M0 ×M0 − I F (θ0 )−1 B (θ0 )T B (θ0 ) I F (θ0 )−1 B (θ0 )T B (θ0 ) .

In the cited paper of Pardo, J. A. et al . (2002) it is also established that √ b(r) L n(θφ − θ0 ) → N (0, Σ∗ ) , n→∞

(5.11)

where the M0 × M0 matrix Σ∗ is given by ¶ µ ³ ´−1 −1 T −1 T I F (θ0 ) B (θ0 ) I F (θ0 )−1 . I M0 ×M0 − B (θ0 ) B (θ0 ) I F (θ0 ) B (θ0 ) In Chapters 6 and 8 we describe some important applications of this result.



5.6.

247

Exercises

1. Let X, Y and Z be Lognormal, Gamma and Weibull random variables. Let gθ (x) be the p.d.f. obtained from the mixture of the three random variables, gθ (x) = π1 fµ,σ (x) + π2 fa,p (x) + π3 fc,d (x) , P with 0 < πi < 1, i = 1, 2, 3, 3i=1 πi = 1, and θ =(π1 , π2 , µ, σ, a, p, c, d). We consider a random sample of size n from the mixture gθ (x) . Prove that the corresponding likelihood function is not a bounded function. 2. Let X be a random variable with probability mass function Pr θ (X = x1 ) =

1 2 − θ, Pr θ (X = x2 ) = − θ and Pr θ (X = x3 ) = 2θ, 3 3

with 0 < θ < 13 . Find the minimum φ-divergence estimator of θ, with µ ¶ 1 1 φ (x) = +x−2 . 2 x 3. Let Y1 , ..., Yn be a sample from the population X with probability mass function p1 (θ) = Pr (X = 1) = 14 (2 + θ) p2 (θ) = Pr (X = 2) = 14 (1 − θ) p3 (θ) = Pr (X = 3) = 14 (1 − θ) p4 (θ) = Pr (X = 4) = 14 θ, θ ∈ (0, 1) . a) Find the minimum φ-divergence estimator with φ (x) = x log x − x + 1. b) Find its asymptotic distribution.

4. Let Y1 , ..., Yn be a sample from a Bernoulli random variable of parameter θ. Find the minimum φ-divergence estimator of θ with φ (x) = 12 (x − 1)2 as well as its asymptotic distribution. 5. Let X be a random variable with associated statistical space given by (X ,βX , Pθ )θ∈Θ⊂RM0 with X = Rd . We consider the discretized model given by the partition P = {Ei }i=1,...,M and pi (θ) = Prθ (Ei ) , i = 1, .., M. Find the asymptotic


248


³ ´T bφ ) = p1 (θ bφ ), ..., pM (θ bφ ) , being distribution of the probability vector p(θ bφ the minimum φ-divergence estimator. θ

bφ )). 6. Find the asymptotic distribution of Shannon’s entropy, H(p(θ

7. Let Y1 , ..., Yn be a sample from a Bernoulli random variable of parameter θ. Using the result given in Exercise 6 find the asymptotic distribution of Shannon’s entropy associated with the Bernoulli model when the parameter θ is estimated by the minimum φ-divergence estimator. 8. Solve Exercise 7 without using Exercise 6. 9. Let Y1 , ..., Yn be a sample of a random variable, X, with probability mass function, p1 (θ) = Pr (X = −1) = θ2 p2 (θ) = Pr (X = 0) = (1 − θ)2 p3 (θ) = Pr (X = 1) = 2θ (1 − θ) where θ ∈ (0, 1) . Find the minimum φ-divergence estimator of θ with φ (x) = x log x − x + 1 as well as its asymptotic distribution. b = (b 10. We consider a 2 × 3 × 2 contingency table and let p p111 , ..., pb232 )T be the nonparametric estimator of the unknown probability vector p = (p111 , ..., p232 )T where pijk = Pr (X = i, Y = j, K = k) . We consider the parameter space © ª Θ = θ : θ =(pijk ; i = 1, 2, j = 1, 2, 3, k = 1, 2; (i, j, k) 6= (2, 3, 2)) T and we denote by p (θ) = (p111 , ..., p232 )T = p the probability vector characterizing the model, with X p232 = 1 − pijk . i=1,2; j=1,2,3; k=1,2 (i,j,k)6=(2,3,2)

We assume the following constraints about the parameter θ f1 (p111 , ..., p231 ) = p111 p212 − p112 p211 = 0 f2 (p111 , ..., p231 ) = p121 p222 − p122 p221 = 0 . f3 (p111 , ..., p231 ) = p131 p232 − p132 p231 = 0 Find the asymptotic distribution of the minimum φ-divergence estimator with the previous constraints.



5.7.

249


1. We have L(θ; x1 , . . . , xn ) =

n Y

gθ (xj ) =

j=1

n Y

(π1 fµ,σ (xj ) + π2 fa,p (xj ) + π3 fc,d (xj )) ,

j=1

with θ = (π1 , π2 , µ, σ, a, p, c, d) . If we consider the particular point of θ given by (1/3, 1/3, log x1 , σ, a, p, c, d) , where x1 is the first sample value, we have ³ ´ ³ ´ 2 1 1 −log x1 ) + fa,p (x1 ) + fc,d (x1 ) exp −(log x2σ L(θ; x1 , . . . , xn ) = 13 √2πσx 2 1 n Q (π1 fµ,σ (xj ) + π2 fa,p (xj ) + π3 fc,d (xj )) × j=2

and choosing σ sufficiently small it is possible to do L as large as it is desired.

2. We consider a random sample of size n, y1 , ..., yn , and we denote by ni , i = 1, 2, 3, the number of values in the sample that coincide with xi . We have to obtain the minimum, at θ, of the function ¶ µ 3 3 n P pj (θ) pbj 1 P − 2 = 12 pj (θ) + (pj (θ) − pbj )2 , g (θ) = 2 p b p (θ) n j j j=1 j=1 j where pj (θ) = Prθ (X = xj ), j = 1, 2, 3. Differentiating with respect to θ and setting the result equal to zero, we have µ µ ¶ ¶ n3 ´ n 1 n1 n 2 n2 2n ³ 0 g (θ) = 2θ − −θ− (−1) + −θ− (−1) + , n1 3 n n2 3 n n3 n and from g 0 (θ) = 0 we obtain

1 2 + θbφ (y1 , ..., yn ) = 13n1 1 3n24 . + + n1 n2 n3 It is easy to prove that this point corresponds to a minimum. 3. a) It will be necessary to obtain the minimum, at θ, of the function g (θ) = D(b p, p (θ)) pb1 pb2 pb3 pb4 = pb1 log + pb2 log + pb3 log + pb4 log p1 (θ) p2 (θ) p3 (θ) p4 (θ) = −b p1 log (2 + θ) − pb2 log (1 − θ) − pb3 log (1 − θ) − pb4 log θ + c. © 2006 by Taylor & Francis Group, LLC

250


Differentiating with respect to θ and setting the result equal to zero, we have pb1 pb2 pb3 pb4 g0 (θ) = − + + − = 0. 2+θ 1−θ 1−θ θ Then θ2 (b p1 + pb2 + pb3 + pb4 ) + θ (2b p2 + 2b p3 − pb1 + pb4 ) − 2b p4 = 0,

i.e.,

θ2 + θ (1 − 2b p1 + pb2 + pb3 ) − 2b p4 = 0.

Therefore, θbφ (y1 , ..., yn ) =

´1/2 ³ p1 − pb2 − pb3 )2 + 8b p4 (−1 + 2b p1 − pb2 − pb3 ) + (−1 + 2b 2

´ ³ b) In this case the matrix diag p (θ)−1/2 is given by     

and

2 (2 + θ)−1/2 0 0 0 −1/2 0 0 0 2 (1 − θ) −1/2 0 0 0 2 (1 − θ) −1/2 0 0 0 2θ



  , 

∂p1 (θ) 1 ∂p2 (θ) 1 ∂p3 (θ) 1 ∂p4 (θ) 1 = , =− , = − and = . ∂θ 4 ∂θ 4 ∂θ 4 ∂θ 4 Therefore A (θ) is given by ³

2−1 (2 + θ)−1/2 , −2−1 (1 − θ)−1/2 , −2−1 (1 − θ)−1/2 , 2−1 θ−1/2

Then

A (θ)T A (θ) = =

´T

−1 −1 −1 1 1 1 −1 (2 µ + θ) + 4 (1 − θ)¶ + 4 (1 − θ) + 4 θ . 2θ + 1 1 2 1 1 1 + + = 4 2+θ 1−θ θ 2 (2 + θ) (1 − θ) θ 1 4

Finally, µ ¶ ´ √ ³ ) (1 − θ ) θ 2 (2 + θ L 0 0 0 n θbφ (Y1 , ..., Yn ) − θ0 −→ N 0, . n→∞ 2θ0 + 1 © 2006 by Taylor & Francis Group, LLC

.

.


251

4. We have to minimize, at θ, the function ¶2 ni npi (θ) i=1 2 (np (θ) − n )2 P i i = 2 p (θ) n i i=1 (np1 (θ) − n1 )2 (np2 (θ) − n2 )2 + = n2 p1 (θ) n2 p2 (θ) 2 (np1 (θ) − n1 )2 (np1 (θ) − n1 ) + = n2 p1 (θ) n2 p2 (θ) 2 (np1 (θ) − n1 ) = 2 n p1 (θ) p2 (θ) 1 . = (nθ − n1 )2 2 n θ (1 − θ)

g (θ) =

2 P

µ pi (θ) 1 −

It is clear that θbφ (y1 , ..., yn ) = n1 /n.

Now the asymptotic distribution is given by µ ³ ´ ´−1 ¶ √ ³ L T b . n θφ (Y1 , ..., Yn ) − θ0 −→ N 0, A (θ0 ) A (θ0 ) n→∞

But

´ ³ diag p (θ)−1/2 =

and

Ã

∂p1 (θ) = 1, ∂θ

θ−1/2 0 0 (1 − θ)−1/2

!

∂p2 (θ) = −1, ∂θ

then A (θ) =

Ã

θ−1/2 0 0 (1 − θ)−1/2

!Ã

1 −1

!

=

Ã

θ −1/2 − (1 − θ)−1/2

!

and A (θ)T A (θ) =

³

θ−1/2 , − (1 − θ)−1/2

´

Ã

θ−1/2 − (1 − θ)−1/2

!

Therefore √ b L n(θφ (Y1 , ..., Yn ) − θ0 ) −→ N (0, θ0 (1 − θ0 )) . n→∞


=

1 . θ (1 − θ)

252


This result could be obtained using directly the Central Limit Theorem ¢ √ ¡ N1 n n − θ0 L −→ N (0, 1) . θ0 (1 − θ0 ) n→∞ 5. We have bφ ) = p (θ0 ) + p(θ

µ

∂pi (θ0 ) ∂θj

¶

i=1,...,M j=1,...,M0

°´ ³° √ ° °b n o °θ − θ 0 ° = oP (1), therefore φ

But

°´ ³° bφ − θ 0 ) + o ° bφ − θ 0 ° (θ ° . °θ

´ √ ³ b L n p(θφ ) − p (θ0 ) −→ N (0, Σ∗ ) , n→∞

¡ ¢−1 with Σ∗ = J(θ0 ) A(θ0 )T A(θ0 ) J(θ0 )T , but

´ ³ A (θ0 ) = diag p (θ0 )−1/2 J(θ0 );

´ √ ³ b therefore we have that n p(θφ ) − p (θ0 ) converges to a normal distribution with mean vector zero and variance-covariance matrix given by ³ ´ ´ ³ ¡ ¢−1 diag p (θ0 )1/2 A (θ0 ) A(θ0 )T A(θ0 ) A(θ0 )T diag p (θ0 )1/2 .

6. We have

¶ µ ∂H (p (θ0 )) b bφ ) − p (θ0 )) H(p(θφ )) = H (p (θ0 )) + (p(θ ∂p (θ) i i=1,...,M °´ ³° ³ ´ ° ° b − p (θ ) . + o °p θ ° 0 φ

On the other hand

with

°´ ³° ° °b b b p(θφ ) = p (θ0 ) + J (θ0 ) (θφ − θ0 ) + o °θφ − θ0 ° ´ √ ³ b L n p(θφ ) − p (θ0 ) −→ N (0, Σ∗ ) , n→∞

being


¡ ¢−1 Σ∗ = J(θ0 ) A(θ0 )T A(θ0 ) J(θ0 )T .

253


Therefore

´ √ ³ bφ )) − H(p (θ0 )) n H(p(θ

converges in law to a normal distribution with mean vector zero and variancecovariance matrix Ãµ !T ¶ ¶ µ ∂H (p (θ0 )) ∂H (p (θ0 )) Σ∗ . ∂pi (θ) ∂pi (θ) i=1,...,M i=1,...,M 7. We know H (p (θ)) = −p1 (θ) log p1 (θ) − p2 (θ) log p2 (θ) . On the other hand ∂H (p (θ)) = −1 − log p2 (θ) = −1 − log (1 − θ) ∂p2 (θ) ∂H (p (θ)) = −1 − log p1 (θ) = −1 − log θ ∂p1 (θ) and

∂p1 (θ) ∂p2 (θ) = 1, = −1, ∂θ ∂θ

then we have Σ∗ = J(θ)(A(θ)T A(θ))−1 J(θ)T = Ã ! 1 −1 = θ (1 − θ) −1 1 because

Ã

1 −1

!

θ (1 − θ)

³

1 −1

´

¡ ¢−1 A(θ)T A(θ) = θ (1 − θ) .

Therefore, denoting Ãµ !T ¶ ¶ µ ∂H (p (θ)) ∂H (p (θ)) Σ∗ , L= ∂pi (θ) ∂pi (θ) i=1,2 i=1,2 we have L =

³

−1 − log θ, −1 − log (1 − θ) Ã ! −1 − log θ × −1 − log (1 − θ) µ ¶2 θ = θ (1 − θ) log , 1−θ


´

θ (1 − θ)

Ã

1 −1 −1 1

!

254


i.e., ´ √ ³ L bφ )) − H (p (θ0 )) −→ n H(p(θ N n→∞

Ã

µ 0, θ0 (1 − θ0 ) log

θ0 1 − θ0

¶2 !

.

8. We have H (p (θ)) = − (1 − θ) log (1 − θ) − θ log θ. Then

but

µ b H(p(θφ )) − H (p (θ)) = − log

θ (1 − θ)

¶

¯´ ³¯ ¯b ¯ b (θφ − θ) + o ¯θφ − θ ¯

√ b L n(θφ (Y1 , ..., Yn ) − θ0 ) −→ N (0, θ0 (1 − θ0 )) n→∞

and now the result is immediate.

9. We have to minimize, at θ, the function pb1 pb2 pb3 + pb2 log + pb3 log p1 (θ) p2 (θ) p3 (θ) pb1 log pb1 − pb1 log p1 (θ) + pb2 log pb2 − pb2 log p2 (θ) + pb3 log pb3 pb3 log p3 (θ) c − pb1 log p1 (θ) − pb2 log p2 (θ) − pb3 log p3 (θ) c − 2b p1 log θ − pb2 2 log (1 − θ) − pb3 log (2θ (1 − θ)) .

g (θ) = pb1 log = − = =

But

g0 (θ) = then we have

We know that

p2 θ − pb3 (1 − 2θ) − (1 − θ) 2b p1 + 2b , θ (1 − θ) 2b p1 + pb3 . θbφ (y1 , ..., yn ) = 2

´ √ ³ L n θbφ (Y1 , ..., Yn ) − θ0 −→ N n→∞


µ ³ ´−1 ¶ T 0, A (θ0 ) A (θ0 ) ,


255

but ³ ´ A (θ) = diag p (θ)−1/2 J (θ)  θ−1 0  0 (1 − θ)−1 =  

 = 

0

2 −2

√2(1−2θ) 2θ(1−θ)

0 

  0 2θ   0   −2 (1 − θ)  1 2 (1 − 2θ) (2θ (1 − θ))− 2

 ,

then we have A (θ)T A (θ) = and therefore

2 θ (1 − θ)

´ ¡ ¢ √ ³b L n θφ (Y1 , ..., Yn ) − θ0 −→ N 0, 2−1 θ0 (1 − θ0 ) . n→∞

10. We can observe that the constraints given can be written in the way p111 p212 =1 p112 p211

p121 p222 =1 p122 p221

and

p131 p232 = 1, p132 p231

i.e., using the odds ratios. These odds ratios characterize the model of conditional independence of the first and third variables given the second variable. In this case we know that the maximum likelihood estimator without constraints is given by pbijk =

nijk n

where nijk is the number of elements in the random sample of size n in the cell (i, j, k) . We denote by

and

b = (b θ p111 , pb112 , pb121 , pb122 , pb131 , pb132 , pb211 , pb212 , pb221 , pb222 , pb231 )T , Σθ = diag (θ) − θθT .

It is well known that √ b L n(θ − θ0 ) → N (0, Σθ 0 ) , n→∞


256


and the restricted maximum likelihood estimator is given by pb∗ijk =

where pbi∗k =

3 X j=1

pb∗ijk , pb∗jk =

2 X

b(r) with components The vector θ φ (r)

(r)

(r)

(r)

(r)

i=1

pbi∗k pb∗jk , pb∗∗k pb∗ijk and pb∗∗k =

(r)

(r)

(r)

3 2 X X i=1 j=1

(r)

pb∗ijk . (r)

(r)

(θbφ,111 , θbφ,112 , θbφ,121 , θbφ,122 , θbφ,131 , θbφ,132 , θbφ,211 , θbφ,212 , θbφ,221 , θbφ,222 , θbφ,231 )T

denotes the restricted minimum φ-divergence estimator. Using (5.11), we have √ L b(r) − θ0 ) → n(θ N (0, Σ∗ ) , φ n→∞

where Σ∗ is given by ³ ´ I F (θ0 )−1 I − B (θ0 )T (B (θ0 ) I F (θ0 )−1 B (θ0 )T )−1 B (θ0 ) I F (θ0 )−1 ,

because Σθ 0 = I F (θ0 )−1 . The matrix B (θ) is given by B (θ) = (A1 , A2 ) , where   p212 −p211 0 0 0 0   A1 =  0 0 p222 −p221 0 0  −p131 −p131 −p131 −p131 1 − 2p131 −p131 − p231

and



−p112  A2 =  0 −p131


p111 0 −p131

0 −p122 −p131

0 p121 −p131

 0  0 . −p131 − p132

6

Goodness-of-fit: Composite Null Hypothesis 6.1.

Introduction

In Chapter 3 we considered a random sample Y1 , . . . , Yn from F and we studied the problem of goodness-of-fit when F is completely known, i.e., H0 : F = F0 . It is common to wish to test the composite hypothesis that the distribution function F is a member of a parametric family {Fθ }θ∈Θ , where Θ is an open subset in RM0 , i.e., we are interested in testing H0 : F = Fθ .

(6.1)

An approach to this problem is to consider a discrete statistical model associated with the original model. In order to do this we consider a partition P = {Ei }i=1,...,M of the original sample space. Now the probabilities of the elements of the partition, Ei , i = 1, ..., M, depend on the unknown parameter θ, i.e., Z pi (θ) = Prθ (Ei ) =

dFθ , i = 1, ..., M.

Ei

The hypothesis given in (6.1) can be tested by the hypotheses

H0 : p = p(θ0 ) ∈ T for some unknown θ0 ∈ Θ versus H1 : p ∈∆M − T,


(6.2)

258

with


© ª T = p (θ) = (p1 (θ) , ..., pM (θ))T ∈ ∆M : θ ∈Θ ,

Θ ⊂ RM0 open and M0 < M − 1.

Pearson recommended (see D’Agostino and Stephens (1986, p. 65)) estimate based on Y1 , . . . , Yn and testing (6.2), by using the ing θ by an estimator θ chi-square test statistic, M ³ ´ X e 2 (Ni − npi (θ)) e ≡ . X2 θ e np ( θ) i i=1

(6.3)

e was a consistent estimator of θ, the asymptotic disPearson thought that if θ tribution of the test statistic (6.3) would coincide with the distribution of the test statistic, X 2 , given in Chapter 3 for testing the null hypothesis (3.1). He was wrong. Fisher (1924) established that the asymptotic distribution of the test statistic (6.3) was not chi-square with M − 1 degrees of freedom and also that depends on the method of estimation. Fisher pointed out that the right method of estimation was the method of the maximum likelihood for the discretized model or equivalently the maximum likelihood estimator based on grouped data. In b(0) , coincides with the previous chapter, it was established that this estimator, θ the minimum Kullback-Leibler divergence estimator. It is interesting to note that Fisher also established that the likelihood ratio test statistic, M ³ ´ X b(0) ≡ 2 G2 θ Ni log i=1

³

Ni , b(0) ) npi (θ

(6.4)

´ b(0) and that the minimum chi-square estiis asymptotically equivalent to X 2 θ b(1) , is asymptotically equivalent to θ b(0) . Finally, he established that the mator, θ ³ ´ ³ ´ ³ ´ b(0) , X 2 θ b(1) , G2 θ b(0) , asymptotic distribution of the test statistics X 2 θ ³ ´ b(1) is the same and is chi-square with M − M0 − 1 degrees of freedom. It is G2 θ interesting to note that Neyman (1949) established that the minimum modified b(−2) , is also asymptotically equivalent to θ b(0) . chi-square estimator, θ In this chapter we consider, for testing (6.2), the family of φ-divergence test statistics ³ ´ ³ ´ bφ ) , bφ ≡ 2n Dφ p b Tnφ1 θ , p( θ 2 2 φ001 (1) 1 © 2006 by Taylor & Francis Group, LLC

Goodness-of-fit: Composite Null Hypothesis

259

and we study its asymptotic distribution under different situations. We observe 2 1 that³ for φ ´1 (x) = 2 (x − 1) and φ2 (x) = x log x − x + 1 we get the test statistic b(0) ; for φ1 (x) = φ2 (x) = x log x − x + 1 we have the likelihood ratio X2 θ ³ ´ ³ ´ b(0) . The test statistic X 2 θ b(1) is obtained for φ1 (x) = test statistic, G2 θ ³ ´ b(1) for φ1 (x) = x log x − x + 1 φ2 (x) = 12 (x − 1)2 and the test statistic G2 θ and φ2 (x) =

1 2

(x − 1)2 .

We shall assume that φ1 , φ2 ∈ Φ∗ are twice continuously differentiable in x > 0 with the second derivatives φ001 (1) 6= 0 and φ002 (1) 6= 0.

6.2.

Asymptotic Distribution with Fixed Number of Classes

In the same way as in the case of the simple null hypothesis, we state three theorems. The first one gives the asymptotic distribution under the null hypothesis given in (6.2), the second one under an alternative hypothesis different from the hypothesis given in (6.2) and finally the third one under contiguous alternative hypotheses. Theorem 6.1 Under the null hypothesis given in (6.2), and assuming the regularity conditions given in Theorem 5.1, we have ³ ´ bφ = Tnφ1 θ 2

where φ1 , φ2 ∈ Φ∗ .

³ ´ 2n L b b ) −→ χ2 , D p, p(θφ2 φ n→∞ M−M0 −1 φ001 (1) 1

Proof. First, we obtain the asymptotic distribution of the random vector ´ √ ³ bφ ) . b − p(θ n p 2

We know that

bφ − θ 0 = θ 2

´ ³ ¡ ¢−1 p − p(θ0 )) A(θ0 )T A(θ0 ) A(θ0 )T diag p (θ0 )−1/2 (b

+ o (kb p − p(θ0 ))k)


260


and

°´ ³° bφ ) − p (θ0 ) = J (θ0 ) (θ bφ − θ 0 ) + o ° bφ − θ 0 ° p(θ θ ° , ° 2 2 2

where J (θ0 ) and A(θ0 ) were defined in (5.6) and (5.7), respectively. Therefore ³ ´ ¡ ¢ bφ ) − p (θ0 ) = J (θ0 ) A(θ0 )T A(θ0 ) −1 A(θ0 )T diag p (θ0 )−1/2 p(θ 2 °´ ³° ° °b p − p(θ0 ))k) . × (b p − p(θ0 )) + o °θφ2 − θ0 ° + o (kb

In the following we denote by L the M × M matrix

´ ³ ¡ ¢−1 A(θ0 )T diag p (θ0 )−1/2 L = J (θ0 ) A(θ0 )T A(θ0 )

and by I the M × M identity matrix. We have Ã

b − p(θ0 ) p bφ ) − p (θ0 ) p(θ 2

!

=

2M×1

+

Ã Ã

because

Ã

!

b − p(θ0 ) p bφ ) − p (θ0 ) p(θ 2

2M×M

(b p − p(θ0 ))M×1

0M×1 ´°´ ³°³ ° ° b p − p(θ0 ))k) o ° θ φ2 − θ 0 ° + o (kb

Then √ n

!

I L

L

−→ N

n→∞

2M×1

Ã Ã 0,

I L

!

!

.

! ¡ ¢ T Σp(θ 0 ) I, L ,

¢ ¡ √ L n (b p − p(θ0 )) −→ N 0, Σp(θ 0 ) , n→∞

i.e., √ n

Ã

b − p(θ0 ) p b p(θφ2 ) − p (θ0 )

!

2M×1

L

−→ N

n→∞

Ã Ã 0,

Σp(θ 0 ) Σp(θ 0 ) LT LΣp(θ 0 ) LΣp(θ 0 ) LT

!!

.

Therefore ´ ¡ ¢ √ ³ L bφ ) −→ b − p(θ n p N 0, Σp(θ 0 ) − Σp(θ 0 ) LT − LΣp(θ 0 ) + LΣp(θ 0 ) LT . 2 n→∞



A second order Taylor expansion of D³φ1 (p, q) around (p (θ0 ) , p (θ0 )) at ´ bφ ) , by bφ )) is given, denoting l = Dφ p b, p(θ (b p, p(θ 2

l= +

1

1 2

M ³ ∂ 2 D (p,q ) ´ P φ1

i,j=1 M P

1 22

2

∂pi ∂pj

³

(p(θ 0 ),p(θ 0 ))

´

(b pi − pi (θ0 ))(b pj − pj (θ0 ))

∂ 2 Dφ1 (p,q ) (b pi ∂pi ∂qj (p(θ 0 ),p(θ 0 ))

i,j=1 M ³ ∂2D P

p q) ´

φ1 ( , ∂qi ∂qj

bφ ) − pj (θ0 )) − pi (θ0 ))(pj (θ 2

bφ ) − pi (θ0 ))(pj (θ bφ ) − pj (θ0 )) (pi (θ 2 2 µ° °2 ¶ ´ ³ ° b ° + o kb p − p(θ0 )k2 + o °p(θ . φ2 ) − p(θ 0 )°

+

1 2

i,j=1

(p(θ 0 ),p(θ 0 ))

But

∂Dφ1 (p, q) = φ01 ∂pi

µ

pi qi

¶

,

∂Dφ1 (p, q) = φ1 ∂qi

and µ

¶ ∂ 2 Dφ1 (p, q) ∂pi ∂pj

(p(θ 0 ),p(θ 0 ))

In the same way µ

∂ 2 Dφ1 (p, q) ∂pi ∂qj

¶

(p(θ 0 ),p(θ 0 ))

Finally,

∂ 2 Dφ1 (p, q) ∂qi ∂qj and µ

¶ ∂ 2 Dφ1 (p, q) ∂qi ∂qj


µ

pi qi

¶

pi − φ01 qi

  φ00 (1) 1 1 pi (θ0 ) =  0

  −φ00 (1) 1 1 pi (θ0 ) =  0

µ ¶   φ00 pi pi 1 qi qi2 =  0

(p(θ 0 ),p(θ 0 ))

µ

i=j

pi qi

.

i 6= j

i=j

.

i 6= j

i=j i 6= j

  φ00 (1) 1 1 pi (θ0 ) =  0

i=j i 6= j

.

¶

261

262


Therefore M P

1 (b pi − pi (θ0 ))2 i=1 pi (θ 0 ) M P 1 bφ ) − pi (θ0 )) (b pi − pi (θ0 ))(pi (θ − 12 φ001 (1) 2 2 p (θ 0) i=1 i ´ ³ M P 1 bφ ) − pi (θ0 ))2 + o kb + 12 φ001 (1) (pi (θ p − p (θ0 )k2 2 µ° i=1 pi (θ0 ) ° ¶ ° b °2 + o °p(θ φ2 ) − p(θ 0 )°

l=

1 © (b pi − pi (θ0 ))2 p (θ ) i 0 i=1 o ¡ ¢ bφ ) − pi (θ0 )) + (pi (θ bφ ) − pi (θ0 ))2 + oP n−1 − 2(b pi − pi (θ0 ))(pi (θ 2 2 M ¡ ¢ P 1 bφ ))2 + oP n−1 (b pi − pi (θ = 12 φ001 (1) 2 i=1 pi (θ 0 )

=

and

1 00 2 φ1 (1)

1 00 2 φ1 (1)

M P

³ ´ bφ ) b, p(θ 2Dφ1 p 2 φ001 (1)

¡ ¢ 1 bφ ))2 + oP n−1 (b pi − pi (θ 2 i=1 pi (θ 0 ) ¡ ¢ bφ )) + oP n−1 bφ ))T C(b p − p(θ = (b p − p(θ 2 2

=

M P

´ ³ being C = diag p (θ0 )−1 . On the one hand

¡ ¢ √ L bφ )) −→ n(b p − p(θ N 0, Σp(θ 0 ) − Σp(θ 0 ) LT − LΣp(θ 0 ) + LΣp(θ 0 ) LT 2 n→∞

and on the other hand ³ ´ 2n bφ ))T C(b bφ )) + oP (1) b b ) = n(b p − p(θ p − p(θ D p , p( θ φ φ2 2 2 φ001 (1) 1 T = X X + oP (1) , where X=

´ ³ √ bφ )) p − p(θ n diag p (θ0 )−1/2 (b 2

is a M-variate normal random variable with mean vector 0 and variance-covariance matrix T ∗ given by ´ ´ ³ ³ (6.5) diag p (θ0 )−1/2 W (θ0 ) diag p (θ0 )−1/2 , © 2006 by Taylor & Francis Group, LLC


263

where W (θ0 ) = Σp(θ 0 ) − Σp(θ 0 ) LT − LΣp(θ 0 ) + LΣp(θ 0 ) LT . Therefore X T X (see Remark 2.6) is chi-squared distributed with r degrees of freedom if T ∗ is a projection of rank r. The matrix T ∗ can be written as ´¡ ³ ¢ T ∗ = diag p (θ0 )−1/2 Σp(θ 0 ) − Σp(θ 0 ) LT − LΣp(θ 0 ) + LΣp(θ 0 ) LT ´ ³ × diag p (θ0 )−1/2 ´ ´ ³ ³ = diag p (θ0 )−1/2 Σp(θ 0 ) diag p (θ0 )−1/2 ³ ´ ´ ³ − diag p (θ0 )−1/2 Σp(θ 0 ) LT diag p (θ0 )−1/2 ³ ´ ´ ³ − diag p (θ0 )−1/2 LΣp(θ 0 ) diag p (θ0 )−1/2 ´ ´ ³ ³ + diag p (θ0 )−1/2 LΣp(θ 0 ) LT diag p (θ0 )−1/2 . Let S denote the M × M matrix ³ ´ ´ ³ S = diag p (θ0 )−1/2 Σp(θ 0 ) diag p (θ0 )−1/2 .

We have

³ ´ ´ ³ S = diag p (θ0 )−1/2 (diag (p (θ0 )) − p (θ0 ) p (θ0 )T )diag p (θ0 )−1/2 ´ ´ ³ ³ = I − diag p (θ0 )−1/2 p (θ0 ) p (θ0 )T diag p (θ0 )−1/2 ³ ´T = I − p (θ0 )1/2 p (θ0 )1/2 . We define the matrix B ´ ´ ³ ³ B = diag p (θ0 )−1/2 LΣp(θ 0 ) diag p (θ0 )−1/2

and we are going to express it in a simpler way, ´ ³ ´ ³ ´ ³ B = diag p (θ0 )−1/2 L diag (p (θ0 )) − p (θ0 ) p (θ0 )T diag p (θ0 )−1/2 ³ ´ ³ ´ = diag p (θ0 )−1/2 L diag p (θ0 )1/2 ´ ´ ³ ³ − diag p (θ0 )−1/2 Lp (θ0 ) p (θ0 )T diag p (θ0 )−1/2 and

q ´ ³ p (θ0 ) = p (θ0 )T diag p (θ0 )1/2 . T


264


Then we get ´ ³ ´ ³ B = diag p (θ0 )−1/2 L diag p (θ0 )1/2 q ´ ³ ´p ³ − diag p (θ0 )−1/2 L diag p (θ0 )1/2 p (θ0 ) p (θ0 )T q ´ ³ ´ ³ p −1/2 1/2 L diag p (θ0 ) (I − p (θ0 ) p (θ0 )T ) = diag p (θ0 ) = K S,

being K the M × M matrix

Therefore

´ ³ ´ ³ K = diag p (θ0 )−1/2 L diag p (θ0 )1/2 . T ∗ = S − KS − SK T + KSK T

where

´ ³ ´ ³ K = diag p (θ0 )−1/2 L diag p (θ0 )1/2 ´ ´ ³ ³ = diag p (θ0 )−1/2 J (θ0 ) (A(θ0 )T A(θ0 ))−1 A(θ0 )T diag p (θ0 )−1/2 ³ ´ × diag p (θ0 )1/2 . But

then and

´ ³ A (θ0 ) = diag p (θ0 )−1/2 J (θ0 ) ,

¡ ¢−1 K = A (θ0 ) A(θ0 )T A(θ0 ) A(θ0 )T

¡ ¢−1 T ∗ = S − A (θ0 ) A(θ0 )T A(θ0 ) A(θ0 )T S − SK ¡ ¢ ¡ ¢−1 −1 )T A(θ0 ) A(θ0 )T S A (θ0 ) A(θ0 )T A(θ0 ) A(θ0 )T + A (θ0 ) A(θ0q p = I − p (θ0 ) p (θ0 )T µ ¶ q p ¡ ¢−1 T T T − A (θ0 ) A(θ0 ) A(θ0 ) A(θ0 ) I − p (θ0 ) p (θ0 ) ¶ µ q p ¡ ¢−1 A(θ0 )T − I − p (θ0 ) p (θ0 )T A (θ0 ) A(θ0 )T A(θ0 ) µ ¶ q p ¡ ¢−1 T T T + A (θ0 ) A(θ0 ) A(θ0 ) A(θ0 ) I − p (θ0 ) p (θ0 ) ¡ ¢ −1 A(θ0 )T . × A (θ0 ) A(θ0 )T A(θ0 ) © 2006 by Taylor & Francis Group, LLC


We know that

q p (θ0 )T A (θ0 ) = 01×M0

and hence T∗ = I −

265

q p ¡ ¢−1 p (θ0 ) p (θ0 )T − A (θ0 ) A(θ0 )T A(θ0 ) A(θ0 )T .

Now we are going to establish that the matrix T ∗ is idempotent, q p ¡ ¢−1 (T ∗ )2 = I − p (θ0 ) p (θ0 )T − A (θ0 ) A(θ0 )T A(θ0 ) A(θ0 )T q q q p p p − p (θ0 ) p (θ0 )T + p (θ0 ) p (θ0 )T p (θ0 ) p (θ0 )T q p + p (θ0 ) p (θ0 )T A (θ0 ) (A(θ0 )T A(θ0 ))−1 A(θ0 )T ¡ ¢−1 − A (θ0 ) A(θ0 )T A(θ0 ) A(θ0 )T q p ¡ ¢−1 + A (θ0 ) A(θ0 )T A(θ0 ) A(θ0 )T p (θ0 ) p (θ0 )T ¡ ¢−1 ¡ ¢−1 + A (θ0 ) A(θ0 )T A(θ0 ) A(θ0 )T A (θ0 ) A(θ0 )T A(θ0 ) A(θ0 )T , but

q p ¡ ¢−1 A (θ0 ) A(θ0 )T A(θ0 ) A(θ0 )T p (θ0 ) p (θ0 )T = 0M×1 q p p (θ0 )T p (θ0 ) = 1,

then

q p ¡ ¢−1 p (θ0 ) p (θ0 )T − A (θ0 ) A(θ0 )T A(θ0 ) A(θ0 )T q p ¡ ¢−1 = I − p (θ0 ) p (θ0 )T − A (θ0 ) A(θ0 )T A(θ0 ) A(θ0 )T = T ∗.

(T ∗ )2 = I −

In relation to rank of T ∗ we have µ ¶ q p T p (θ0 ) p (θ0 ) rank(T ) = trace(T ) = trace(I) − trace ´ ³ ¡ ¢−1 A(θ0 )T − trace A (θ0 ) A(θ0 )T A(θ0 ) ³¡ ¢−1 ¡ ¢´ A(θ0 )T A(θ0 ) = M − 1 − trace A(θ0 )T A(θ0 ) ∗

∗

= M − 1 − M0 ,

because

¡ ¢−1 ¡ ¢ A(θ0 )T A(θ0 ) = I M0 ×M0 . A(θ0 )T A(θ0 ) ³ ´ bφ is asymptotically distributed chiThen the φ-divergence test statistic Tnφ1 θ 2 squared with M − 1 − M0 degrees of freedom. © 2006 by Taylor & Francis Group, LLC

266


Based on the previous result we should reject the null hypothesis given in (6.1), with a significance level α, if ³ ´ bφ > χ2M−M −1,α . Tnφ1 θ (6.6) 2 0

We present, in the sequel, an approximation for the power function of the previous test statistic. Let q =(q1 , ..., qM )T be a point at the alternative hypothesis. If b tends, in probability, to q. the alternative hypothesis q is true, we have that p We denote by θa the point on Θ verifying θa = arg min Dφ2 (q, p (θ)), θ∈Θ

bφ ) tends, in probability, to p (θa ). In the next theorem, we and we have that p(θ 2 use the following assumption, ´ ´ √ ³³ L bφ ) − (q, p (θa )) −→ b, p(θ n p N (0,Σ) , (6.7) 2 n→∞

under the alternative hypothesis q, where Ã ! Σ11 Σ12 Σ= , Σ11 = diag (q) − qq T and Σ12 = Σ21 . Σ21 Σ22

Theorem 6.2 The asymptotic power for the test given by (6.6) at the alternative hypothesis q, assuming condition (6.7), is given by ¶¶ µ µ 00 √ 1 φ1 (1) 2 √ χ − nDφ1 (q, p (θa )) , βn,φ1 (q) = 1 − Φn σφ1 (q) 2 n M−M0 −1,α where σφ2 1 (q) = ZT Σ11 Z + 2ZT Σ12 S + ST Σ22 S, being

(6.8)

µ

¶ ∂Dφ1 (v, p (θa )) Z = , ∂v v =q µ ¶ ∂Dφ1 (q, w) T , S = ∂w w=p(θ a ) T

and Φn (x) is a sequence of distribution functions tending uniformly to the standard normal distribution function Φ (x) .



267

Proof. A first-order Taylor expansion of the divergence measure gives ³ ´ bφ ) − p (θa )) + oP (n−1/2 ). bφ ) = Dφ (q, p (θa )) + ZT (b b, p(θ p − q) + ST (p(θ Dφ1 p 2 1 2

Then

³ ´ ´ ¡ ¢ √ ³ L bφ ) − Dφ (q, p (θa )) −→ b, p(θ n Dφ1 p N 0,σφ2 1 (q) , 2 1 n→∞

for σφ21 (q) given in (6.8).

The previous result is not easy to apply. For this reason we are going to consider a sequence of contiguous alternatives that approach the null hypothesis ¢ ¡ at rate of O n−1/2 . Consider the contiguous alternative hypotheses,

1 H1,n : pn = p (θ0 ) + √ d, (6.9) n P where d ≡ (d1 , ..., dM )T is a fixed vector such that M j=1 dj = 0, with pn 6= p (θ 0 ) , θ0 unknown and θ0 ∈Θ. Then we have the following result:

Theorem 6.3 ³ ´ bφ , unThe asymptotic distribution of the φ1 -divergence test statistic Tnφ1 θ 2 der the contiguous alternative hypotheses given in (6.9), is noncentral chi-square with M − M0 − 1 degrees of freedom and noncentrality parameter δ given by ¶ ³ ³ ´−1 ´µ −1/2 T T T A (θ0 ) δ = d diag p (θ0 ) I − A (θ0 ) A (θ0 ) A (θ0 ) ´ ³ × diag p (θ0 )−1/2 d. Proof. We have √ √ n (b p − p (θ0 )) = n (b p − pn + pn − p (θ0 )) √ √ = n (b p − pn ) + n (pn − p (θ0 )) √ = n (b p − pn ) + d, then under the alternatives (6.9), ¡ ¢ √ L n(b p − p (θ0 )) −→ N d, Σp(θ 0 ) n→∞

because under the alternative hypotheses given in (6.9), ¢ √ L ¡ n(b p − pn ) −→ 0, Σp(θ 0 ) . n→∞


268


But ´−1 ³ ´ √ b √ ³ n(θφ2 − θ0 ) = n A (θ0 )T A (θ0 ) A (θ0 )T diag p (θ0 )−1/2 × (b p − p (θ0 )) + oP (1)

and

√ √ n(b p − p (θ0 )) = n(b p − pn ) + d,

then ´ ´−1 ³ √ ³ √ b n(θφ2 − θ0 ) = n A (θ0 )T A (θ0 ) A (θ0 )T diag p (θ0 )−1/2 ¢ ¡ × (b p − pn ) + n−1/2 d + oP (1) ´ ´−1 ³ √ ³ p − pn ) n A (θ0 )T A (θ0 ) A (θ0 )T diag p (θ0 )−1/2 (b = ´ ³ ´−1 ³ + A (θ)T A (θ0 ) A (θ)T diag p (θ0 )−1/2 d + oP (1) . bφ ) ≡ Therefore the random vector l(θ 2

´ √ ³ b n p(θφ2 ) − p (θ0 ) can be written as

bφ − θ0 ) + oP (1) bφ ) = J (θ0 ) √n(θ l(θ 2 2 ´ ³ ´−1 ³ √ p − pn ) A (θ0 )T diag p (θ0 )−1/2 (b = nJ (θ0 ) A (θ0 )T A (θ0 ) ´ ³ ´−1 ³ +J (θ0 ) A (θ0 )T A (θ0 ) A (θ0 )T diag p (θ0 )−1/2 d + oP (1) .

If we denote

´ ³ ´−1 ³ A (θ0 )T diag p (θ0 )−1/2 , L = J (θ0 ) A (θ0 )T A (θ0 )

and by I the M × M identity matrix, we have ´ √ √ ³ b n p(θφ2 ) − p (θ0 ) = nL(b p − pn ) + Ld+oP (1)

and

√ n

Ã

Hence √ n

Ã

b − p (θ0 ) p bφ ) − p (θ0 ) p(θ 2 b − p (θ0 ) p bφ ) − p (θ0 ) p(θ 2


!

√ = n

!

L

Ã

−→ N

n→∞

I L

ÃÃ

!

(b p − pn ) +

d Ld

! Ã ,

Ã

d Ld

!

+

Ã

0 oP (1)

Σp(θ 0 ) LT Σp(θ 0 ) LΣp(θ 0 ) LΣp(θ 0 ) LT

!

.

!!


269

and ¢ ¡ √ L bφ )) −→ n(b p−p(θ N (I − L)d, Σp(θ 0 ) − Σp(θ 0 ) LT − LΣp(θ 0 ) + LΣp(θ 0 ) LT . 2 n→∞

On the other hand ³ ´ ´ ³ 2n T diag p (θ )−1 (b bφ ))+oP (1) b b b p − p(θ ) = n(b p − p( θ )) D p , p( θ 0 φ φ2 φ2 00 2 φ1 (1) 1 = X T X + oP (1)

with

´ ³ √ −1/2 bφ )) (b p − p(θ X = ndiag p (θ0 ) 2

and

´ ´ ³ ³ L −1/2 ∗ (I − L) d, T −→ N diag p (θ ) X 0 n→∞

where T ∗ is defined in (6.5).

By Theorem 6.1 the rank of the matrix T ∗ is M − M0 − 1. If we prove that T µ = µ, where ´ ³ µ = diag p (θ0 )−1/2 (I M×M − L) d, ∗

³ ´ bφ is asymptotically distribwe will have established that the test statistic θ 2 uted as a noncentral chi-square distribution with M − M0 − 1 degrees of freedom and noncentrality parameter δ = µT µ. ´ ³ We know diag p (θ0 )1/2 A (θ0 ) = J (θ0 ) , then Tnφ1

´ ³ ´ ³ ´ ³ ³ diag p (θ0 )−1/2 d − diag p (θ0 )−1/2 diag p (θ0 )1/2 A (θ0 ) ´¶ ´−1 ³ ³ T T −1/2 d A (θ0 ) diag p (θ0 ) × A (θ0 ) A (θ0 ) µ ´ ³ ´−1 ³ = diag p (θ0 )−1/2 − A (θ0 ) A (θ0 )T A (θ0 ) ´´ ³ − A (θ0 )T diag p (θ0 )−1/2 d

µ =

By Theorem 6.1 the matrix T ∗ has the expression q ³ ´−1 p T = I − p (θ0 ) p (θ0 )T − A (θ0 ) A (θ0 )T A (θ0 ) A (θ0 )T . ∗


270


Then we have ¶ µ q ³ ´−1 p T T T ∗ I − p (θ0 ) p (θ0 ) − A (θ0 ) A (θ0 ) A (θ) A (θ0 ) µ T µ= and ∗

T µ= − − + ×

q ³ ³ ´ p µ − p (θ0 ) p (θ0 )T diag p (θ0 )−1/2 ´¶ ³ ´−1 ³ d A (θ0 ) A (θ0 )T A (θ0 ) A (θ0 )T diag p (θ0 )−1/2 ´ ³ ´−1 ³ A (θ0 ) A (θ0 )T A (θ0 ) A (θ0 )T diag p (θ0 )−1/2 d ³ ´−1 ³ ´−1 A (θ0 ) A (θ0 )T A (θ0 ) A (θ0 )T A (θ0 ) A (θ0 )T A (θ0 ) ´ ³ A (θ0 )T diag p (θ0 )−1/2 d.

Therefore ∗

T µ= + × − + =

¶ µ ´ ³ ³ ´−1 T T I − A (θ0 ) A (θ0 ) A (θ0 ) A (θ0 ) diag p (θ0 )−1/2 d q ³ ´−1 p p (θ0 ) p (θ0 )T A (θ0 ) A (θ0 )T A (θ0 ) ´ ³ A (θ0 )T diag p (θ0 )−1/2 d ´ ³ ´−1 ³ A (θ0 ) A (θ0 )T A (θ0 ) A (θ0 )T diag p (θ0 )−1/2 d ³ ´−1 ³ ´ A (θ0 ) A (θ0 )T A (θ0 ) A (θ0 )T diag p (θ0 )−1/2 d ¶ µ ´ ³ ´−1 ³ T T A (θ0 ) diag p (θ0 )−1/2 d = µ, I − A (θ0 ) A (θ0 ) A (θ0 )

and finally

³ ´ 2n L bφ ) −→ b, p(θ χ2M−M0 −1 (δ) Dφ1 p 2 n→∞ φ1 (1) 00

being δ = µT µ and

¶ µ ´ ³ ³ ´−1 T T A (θ) diag p (θ0 )−1/2 d. µ = I − A (θ0 ) A (θ0 ) A (θ0 ) ³ ´−1 If we denote U = I − A (θ0 ) A (θ0 )T A (θ0 ) A (θ0 )T , we have © 2006 by Taylor & Francis Group, LLC


271

³ ´−1 A (θ0 )T U T U = I − A (θ0 ) A (θ0 )T A (θ0 ) ³ ´−1 − A (θ0 ) A (θ0 )T A (θ0 ) A (θ0 )T ³ ´−1 + A (θ0 ) A (θ0 )T A (θ0 ) A (θ0 )T A (θ0 ) ³ ´−1 × A (θ0 )T A (θ0 ) A (θ0 )T ³ ´−1 = I − A (θ0 ) A (θ0 )T A (θ0 ) A (θ0 )T . Therefore ¶ ³ ³ ´−1 ´µ −1/2 T T I − A (θ0 ) A (θ0 ) A (θ0 ) A (θ0 ) δ = µ µ = d diag p (θ0 ) ´ ³ ×diag p (θ0 )−1/2 d. T

T

The result presented in this theorem was obtained for φ (x) = 12 (x − 1)2 , chisquare test statistic, by Diamond et al . (1960). For the φ-divergence test statistic the result was obtained by Menéndez et al . (2001a).

6.3.

Nonstandard Problems: Test Statistics Based on Phi-divergences

6.3.1.

Maximum Likelihood Estimator Based on Original Data and Test Statistics Based on Phi-divergences

If we only have the number Ni , i = 1, ..., M, of observations falling into the ith of the M cells, it is clear that we have to estimate the unknown parameters using a minimum φ-divergence estimator. If the original observations Y1 , ..., Yn are available, one is tempted to use more efficient estimators, such as the maxb from the likelihood function Qn fθ (yi ), i.e., the imum likelihood estimator, θ, i=1 maximum likelihood estimator based on the original data. One may reasonably expect this procedure to provide more powerful tests than those based on the grouped data; at the same time the estimates usually are simpler and easier to obtain.


272


Before proceeding to the main result we shall consider an example. Example 6.1 Let Y1 , ..., Yn be a random sample from a normal population with mean θ and variance 1. If we consider the partition of R given by E1 = (−∞, 0] and E2 = (0, ∞) , we have

! Ã 1 (x − θ)2 √ exp − dx Prθ (E1 ) = p1 (θ) = 2 2π −∞ Ã ! Z ∞ 1 (x − θ)2 √ exp − dx. Prθ (E2 ) = p2 (θ) = 1 − p1 (θ) = 2 2π 0 Z

0

It is well known that the maximum likelihood estimator of θ, based on the original observations, is given by θb = Y , (6.10)

and the maximum likelihood estimator based on grouped data is obtained maximizing n1 !n2 ! Prθ (N1 = n1 , N2 = n2 ) = p1 (θ)n1 p2 (θ)n2 . n! We consider 1 φ (x) = (1 − x)2 , 2 and we are ³going to study the asymptotic distribution of the φ-divergence test ´ φ b b statistic Tn θ , where θ is the maximum likelihood estimator given in (6.10), i.e., the maximum likelihood estimator based on the original data. We have Ã¡ ¡ ¡ ¢¢2 ¡ ¢¢2 ¡ ¡ ¢¢2 ! ³ ´ 2 P pbi − pi Y pb1 − p1 Y pb2 − p2 Y φ b ¡ ¢ ¡ ¢ ¡ ¢ =n + =n Tn θ pi Y p1 Y p2 Y i=1 Ã¡ ¡ ¡ ¢¢2 ¡ ¡ ¢¢2 ! ¡ ¢¢2 pb1 − p1 Y pb1 − p1 Y pb1 − p1 Y ¡ ¢ ¡ ¢ + =n ¡ ¢ ¡ ¢ . =n p1 Y p2 Y p1 Y p2 Y But

¡ ¢ ¡ ¢ P p1 Y p2 Y −→ p1 (θ) p2 (θ) , n→∞

³ ´ and therefore (see Ferguson 1996, p. 39) the asymptotic distribution of Tnφ θb coincides with the asymptotic distribution of ¡ ¢¢2 ¡ ³ ´ pb1 − p1 Y φ b . Rn θ = n p1 (θ) p2 (θ) © 2006 by Taylor & Francis Group, LLC


We have

273

³ ´ ¡ ¢ ¢ ∂p1 (θ) ¡ −1/2 , p1 Y = p1 (θ) + Y − θ + oP n ∂θ

but ∂p1 (θ) ∂θ

¶ µ 1 1 2 √ (x − θ) exp − (x − θ) dx = 2 2π −∞ µ ¶¶0 µ µ 2¶ 1 1 1 θ 2 = − √ exp − (x − θ) = − √ exp − , 2 2 2π 2π −∞ Z

0

then we get µ 2¶ √ ¢ ¢ √ ¡ ¡ ¢ n θ ¡ n p1 Y − p1 (θ) = − √ exp − Y − θ + oP (1) . 2 2π

Now we are going to establish the asymptotic distribution of the random variable ¢T √ ¡ n pb1 − p1 (θ) , Y − θ

that is a bivariate normal random variable with mean vector (0, 0)T and variancecovariance matrix with elements √ 1 V ar [ n (b p1 − p1 (θ))] = nV ar [b p1 ] = n 2 V ar [N1 ] n 1 = n 2 np1 (θ) p2 (θ) = p1 (θ) p2 (θ) n ¢¤ £ ¤ £ ¤ £√ ¡ V ar n Y − θ = nV ar Y − θ = nV ar Y 1 = n V ar [Yi ] = 1. n n ¡ ¢ 1 P I(−∞,0] (Yi ) − p1 (θ) , we have n i=1 · ¸ n £√ ¡ ¢¤ 1 P Cov n pb1 − p1 (θ) , Y − θ = nCov T, n (Yi − θ) i=1 ¤ £ = Cov I(−∞,0] (Yi ) − p1 (θ) , Yi − θ ¤ £ = E (I(−∞,0] (Yi ) − p1 (θ)) (Yi − θ) ¤ £ − E I(−∞,0] (Yi ) − p1 (θ) E [Yi − θ] £¡ ¢ ¤ = E I(−∞,0] (Yi ) (Yi − θ) Z 0 ´ ³ √1 (x − θ) exp − 1 (x − θ)2 dx = 2 2π −∞ ¡ 1 2¢ 1 = − √2π exp − 2 θ .

Denoting by T =


274


Therefore  Ã !  exp(−θ2 /2) √ p1 (θ) p2 (θ) − ¢T L √ ¡ 0 2π  . ,  exp(−θ2 /2) n pb1 − p1 (θ) , Y − θ −→ N  n→∞ 0 √ − 1 2π On the other hand √ n

¢T ¡ ¢ √ ¡ n pb1 − p1 (θ) , p1 Y − p1 (θ) can be written as

Ã

1

0 −

0 exp(−θ2 /2) √ 2π

!Ã

pb1 − p1 (θ) Y −θ

!

+ oP (1) ,

and denoting by   Ã ! exp(−θ2 /2) 1 0 p1 (θ) p2 (θ) − √2π  and A = Σ =  exp −θ2 /2 , exp(−θ2 /2) (√ ) √ 0 − − 1 2π 2π we have

Then



p1 (θ) p2 (θ) Σ∗ = AΣAT =  exp(−θ2 ) 2π

exp(−θ2 ) 2π exp(−θ2 ) 2π

¡ ¢ ¢T L √ ¡ n pb1 − p1 (θ) , p1 Y − p1 (θ) −→ N n→∞

Ã

Finally

¡ ¢¢ √ ¡ n pb1 − p1 Y

q p1 (θ) p2 (θ) − ¡ ¡ ¢¢2 L n pb1 − p1 Y −→

n→∞

and


exp(−θ2 )

0 0

!

∗

,Σ

!

L

−→ N (0, 1) .

n→∞

2π

¢! ¡ exp −θ2 χ21 p1 (θ) p2 (θ) − 2π

Ã

¡ ¢¢2 ¡ n pb1 − p1 Y L −→ λχ2 p1 (θ) p2 (θ) n→∞ 1

¢ ¡ exp −θ2 with λ = 1 − . 2πp1 (θ) p2 (θ)

.

¢! ¡ exp −θ2 0, p1 (θ) p2 (θ) − 2π

¡ ¢¢ L √ ¡ n pb1 − p1 Y −→ N n→∞

and

ÃÃ



,


275

Let us prove that λ ≥ 0. As the matrix Σ∗ is nonnegative definite, then ¢Ã ¢! ¡ ¡ exp −θ2 exp −θ2 ∗ 0 ≤ det (Σ ) = p1 (θ) p2 (θ) − 2π 2π Ã ¡ ¢ ! ¡ 2¢ exp −θ 2 p1 (θ) p2 (θ) = exp −θ . 1− 2π 2πp1 (θ) p2 (θ)

Therefore 0 ≤ λ ≤ 1.

Now we present the theorem ³ ´ that states the asymptotic distribution of the b . The proof can be seen in Morales et al . (1995) φ-divergence test statistic Tnφ θ for φ-divergence measures in general and in Chernoff and Lehman (1954) for the particular case φ1 (x) = 12 (x − 1)2 . Theorem 6.4 Under the conditions given in Morales et al . (1995) the asymptotic distribu³ ´ φ b b is the maximum likelihood tion of the φ-divergence test statistic Tn θ , where θ estimator based on the original data, verifies M

0 X 2n L 2 b −→ (b p , p( θ)) χ + (1 − λj ) Zj2 , D φ n→∞ M−M0 −1 φ00 (1)

j=1

where Zj are independent and normally distributed random variables with mean zero and unit variance, and the λj , 0 ≤ λj ≤ 1, are the roots of the equation det (I F (θ0 ) − λIF (θ0 )) = 0, where IF (θ0 ) and I F (θ0 ) are the Fisher information matrices from the original and discretized models, respectively. Remark 6.1 ¡ ¢ Let F1 F1 ≡ χ2M−M0 −1 be the asymptotic distribution function of the φdivergence test statistic

Ã

³ ´ bφ = Tnφ1 θ 2

and F2 F2 ≡ χ2M−M0 −1 +

M P0

j=1

2n

Dφ φ001 (1) 1

(1 − λj ) χ21j

!

³

bφ ) b, p(θ p 2

be the asymptotic distribution func-

tion of the φ-divergence test statistic ³ ´ b b = 2n Dφ (b p, p(θ)). Tnφ θ φ00 (1) © 2006 by Taylor & Francis Group, LLC

´

276


We have F2 (x) ≤ F1 (x)

∀x ≥ 0.

This result ³ ´ indicates that the decision rule “reject H0 , with a significance level α, φ b if Tn θ > χ2M−M0 −1,α ”, will lead to a probability of rejection greater than the desired level of significance when the hypothesis is true. In symbols ³ ´ b > χ2 α∗ = PrH0 φ002n(1) Dφ1 (b p, p(θ)) M−M0 −1,α ³1 ´ ³ ´ 2 = 1 − F2 χM−M0 −1,α ≥ 1 − F1 χ2M−M0 −1,α ³ ³ ´ ´ bφ ) > χ2 b, p(θ = PrH0 φ002n(1) Dφ1 p 2 M−M0 −1,α 1

= α.

Then when we consider the approximation F1 ' F2 , the probability of rejecting the null hypothesis increases and then we are raising the probability of type I error. However, a numerical investigation of a few special cases indicates that, at least in the Poisson models (see Chernoff and Lehman (1954)) this excess of probability of type I error will be so small as not to be serious. The situation appears to be not quite so favorable in the normal case.

6.3.2.

Goodness-of-fit with Quantile Characterization

In this section we consider the problem studied in Section 3.4.1, but here we assume that F ∈ {Fθ }θ∈Θ and Θ is an open set in RM0 . We consider the values πi defined in (3.17) and the vector q 0 = (πj − πj−1 : 1 ≤ j ≤ M)T . We are going to test H0 : F = Fθ by testing H0 : q = p(Y n , θ),

(6.11)

where ¡ ¢T p(Y n , θ) = (p1 (Y n , θ), ..., pM (Y n , θ))T = Fθ (Ynj ) − Fθ (Ynj−1 ) : 1 ≤ j ≤ M ,

n0 = 0, nM = +∞, Yni = Y(ni ) (ni = [nπi ] + 1) is the ni th-order statistic, Yn0 = −∞ and YnM = +∞. If we consider G ∈ / {Fθ }θ∈Θ an alternative hypothesis to the null hypothesis given in (6.11) is ¡ ¢T q(Y n ) = (q1 (Y n ), ..., qM (Y n ))T = G(Ynj ) − G(Ynj−1 ) : 1 ≤ j ≤ M . (6.12) © 2006 by Taylor & Francis Group, LLC

277


For testing (6.11) we can consider the φ-divergence test statistic defined by

where

³ ´ bφ = Tnφ1 θ 2 bφ = arg θ 2

³ ´ 2n bφ ), q 0 , p(Y , θ D n φ 2 φ001 (1) 1 inf

θ∈Θ⊂RM0

(6.13)

¡ ¢ Dφ2 p(Y n , θ), q0 .

Under some regularity assumptions Menéndez et al . (1998a) established that ³ ´ √ P L L bφ → bφ − θ 0 ) → bφ θ θ0 , n(θ N(0, I F (θ0 )−1 ) and Tnφ1 θ → χ2M−M0 −1 . 2 2 2 n→∞

n→∞

n→∞

We can consider the decision rule: “Reject the null hypothesis given in (6.11), with significance level α, if bφ ) > χ2 Tnφ1 (θ M−M0 −1,α ”. 2

(6.14)

Let θa a point in the parameter space verifying, under the alternative hybφ = θa + oP (1) . We denote pothesis given in (6.12), θ 2 c∗j = G−1 (πj ), j = 1, ..., M,

and

¡ ¡ ¡ ¢ ¢ ¢ p(c∗ , θa ) = Fθ a G−1 (πj ) − Fθ a G−1 (πj−1 ) : j = 1, ..., M .

Under the assumption

´ √ ³ L bφ ) − p(c∗ , θa ) −→ n p(Y n , θ N (0,Σ) , 2 n→∞

∗

q0 ,

the power of the φ-divergence test for some matrix Σ, and if p(c , θa ) 6= statistic (6.13) for testing (6.11), under the alternative hypothesis given in (6.12), satisfies µ ¶ ³ 00 √ ¡ ∗ ¢´ φ1 (1) 2 n 0 βn,φ1 (q(Y n )) = 1 − Φn σ (q(Y )) 2n χM−M0 −1,α − Dφ1 p(c , θa ), q , n φ1

where Φn (x) is a sequence of distribution functions tending uniformly to the standard normal distribution Φ(x) and Ã ! µ ¶ M X pj (c∗ , θa ) pk (c∗ , θa ) 1 1 2 0 0 σφ1 (q(Y n )) = σjk φ1 , φ1 qj0 qk0 qj0 qk0 j,k=1

for Σ = (σij )i,j=1,...,M . Other interesting results in this direction can be seen in Menéndez et al . (2001a, b, c).


278


6.3.3.

Estimation from an Independent Sample

Consider the problem in which we wish to obtain the limiting distribution of when θ is estimated using information from a second multinomial sample that is independent of Yn = (Y1 , . . . , Yn ). Then in addition to Yn = (Y1 , . . . , Yn ) we consider the samples Yn∗ ∗ = (Y1∗ , . . . , Yn∗∗ ) = (Yn+1 , ..., Yn+n∗ ) and Yn∗∗∗∗ = ∗ (Y1∗∗ , . . . , Yn∗∗ ∗∗ ) = (Y1 , . . . , Yn+n∗ ). Obviously, the samples Yn and Yn∗ are independent and Yn , Yn∗ ∗ are subsamples of Yn∗∗∗∗ . We assume that n∗ depends b∗ , θ b∗∗ be the minimum φ2 on n in such a way that τ = lim nn∗ exists. Let θ φ φ2 2 n→∞ divergence estimators based on the random samples Yn∗ ∗ and Yn∗∗∗∗ , respectively b∗ ) and p(θ b∗∗ ) be the parametric distribution estimators. Then we and let p(θ φ2 φ2 have for every τ > 0 that the asymptotic distributions of the φ-divergence test statistics ´ ´ ³ ³ 2n 2n b∗ ) b∗∗ ) b b D D and p , p( θ p , p( θ φ φ φ2 φ2 φ001 (1) 1 φ001 (1) 1 are 1+τ 2 τ χ2M−M0 −1 + χM0 and χ2M−M0 −1 + χ2 τ 1 + τ M0 respectively, where χ2M−M0 −1 and χ2M0 are independent chi-squared distributed Tnφ1

random variables with M − M0 − 1 and M0 degrees of freedom, respectively. This result has been established in Morales et al. (1995). This is an important generalization of the classical result with φ2 (x) = x log x− x + 1 and φ1 (x) = 12 (1 − x)2 , that is to say the classical chi-square test statistic. In that case the first result was given by Murthy and Gafarian (1970) and the second one Chase (1972). ˆ∗ and θ ˆ∗∗ the MLE based on the original data Yn∗ ∗ = Now we denote by θ b the relative frequency vector (Y1∗ , ..., Yn∗∗ ) and Yn∗∗∗∗ = (Y1∗∗ , ..., Yn∗∗ ∗∗ ) and by p based on the sample Yn = (Y1 , . . . , Yn ). Under some regularity conditions given in Morales et al. (1995) the φ-divergence test statistics ³ ´ 2n b∗ ) b D p , p( θ φ 1 φ001 (1)

are asymptotically distributed as χ2M−M0 −1 +

and

³ ´ 2n ˆ∗∗ ) b D p , p( θ φ φ001 (1) 1

M0 M0 X X 2 2 (1 + λj /τ )Zj and χM−M0 −1 + (1 − λj /(1 + τ ))Zj2 , j=1


j=1


279

respectively, where Zj , j = 1, ..., M0 , are independent standard normal variables and the λj , 0 ≤ λj ≤ 1, are the roots of the equation det (I F (θ0 ) − λIF (θ0 )) = 0, where IF (θ0 ) and I F (θ0 ) are the Fisher information matrices of the original and discretized models, respectively.

6.3.4.

Goodness-of-fit with Dependent Observations

In this Section we extend some results considered in Chapter 3, related to stationary irreducible aperiodic Markov chains Y = {Yk , k ≥ 0} with state space {1, ..., M} , to the situation in which the stationary distribution depend on an unknown parameter θ ∈ Θ and Θ is an open subset of RM0 . We denote by p (θ) the stationary distribution and by P θ the set of all matrices P such that their stationary distribution p coincides with p (θ) and each element of P θ by P (θ) . A basic statistical problem is how to estimate in a consistent and asymptotically normal way the unknown true parameter θ0 ∈ Θ by using a random sample of size n from Y = {Yk , k ≥ 0} about the states of the chain, i.e., how to find a e taking on values in Θ such that measurable mapping θ P e −→ i) θ θ0 n→∞

ii)

√ e L n(θ − θ0 ) −→ N (0, V 0 ) n→∞

and how to evaluate the M0 × M0 matrix V 0 . In this context the minimum φ-divergence estimator was obtained in Menéndez et al . (1999a) and is given by bφ = arg θ 2

inf

θ∈Θ⊂RM0

Dφ2 (b p, p (θ)) ,

b is the nonparametric estimator based on the sample of size n. where p

Under some regularity assumptions in the cited paper of Menéndez et al. the following statements were established: a.s. bφ −→ θ0 a) θ 2 n→∞

´ ³ ´−1 ³ bφ = θ0 + A (θ0 )T A (θ0 ) p − p (θ0 )) b) θ A(θ0 )T diag p (θ0 )−1/2 (b 2 b − p (θ0 ) k) +o(k p


280

c)


¢ ¡ √ b L n(θφ2 − θ0 ) −→ N 0,∆T0 B 0 ∆0 , n→∞

where

³ ´−1 · ∆0 = A(θ0 ) A (θ0 )T A (θ0 ) ´ ´ ³ ³ · B 0 = diag p (θ0 )−1/2 S 0 diag p (θ0 )−1/2

and

· S 0 = diag (p (θ0 )) C 0 +C T0 diag (p (θ0 ))−diag (p (θ0 ))−p (θ0 ) p (θ0 )T , ´−1 ³ and 1 = (1, ..., 1)T is the where C 0 = I M×M − P (θ0 ) + 1p (θ0 )T column vector of M units. After estimating the unknown parameter θ0 we are interested in testing H0 : p=p (θ0 ) . Regarding this problem of testing, in the cited paper of Menéndez et al . (1999a), it was established that M ´ ³ X 2n L b b D ) −→ ρi Zi2 , p , p( θ φ1 φ2 n→∞ φ001 (1) j=1

where Zi are independently normally distributed random variables with mean zero and unit variance and ρi are the eigenvalues of the matrix ´ ´ ³ ³ L0 = diag p (θ0 )−1/2 (I − W + ) S 0 (I − W − ) diag p (θ0 )−1/2 , where

and

´ ´ ³ ³ W + = diag p (θ0 )1/2 Σ0 diag p (θ0 )−1/2 , ³ ³ ´ ´ W − = diag p (θ0 )−1/2 Σ0 diag p (θ0 )1/2 , ´−1 ³ A(θ0 ). Σ0 = A(θ0 )T A (θ0 )T A (θ0 )

Another interesting problem is to test the transition matrix of the chain, i.e., to test H0 : P = P (θ0 ) = (pij (θ0 ))i,j=1,...,M , for some unknown θ0 ∈ Θ. Menéndez et al . (1999b) considered the φ-divergence test statistic Ã ! M M ³ ´ X X p b (i, j) υ 2n i∗ bφ )φ1 bφ = Tnφ1 θ pij (θ , 2 2 bφ ) φ00 (1) n pij (θ 2 i=1 j=1 © 2006 by Taylor & Francis Group, LLC


where pb(i, j) was defined in Section 3.4.2, and bφ = arg θ 2

inf

θ∈Θ⊂RM0

b , P (θ)) = arg Dφ2 (P

inf

θ∈Θ⊂RM0

M M X υi∗ X i=1

n

pij (θ) φ2

j=1

µ

281

pb(i, j) pij (θ)

¶

.

Its asymptotic distribution is chi-square with c − M − M0 degrees of freedom, where c is the number of elements of the set, C = {(i, j) : pij (θ0 ) > 0} . The results presented in this section extend the ideas of Glesser and Moore (1983a, 1983b).

6.3.5.

Goodness-of-fit with Constraints

With the notation introduced in Section 5.5, Pardo, J. A. et al . (2002), established that given ν (ν < M0 ) real valued functions f1 (θ) , ..., fν (θ) that constrain the parameter θ ∈ Θ ⊂ RM0 , fm (θ) = 0, m = 1, ..., ν and under the null hypothesis given in (6.2) it holds 2n φ001 (1)

´ ³ L b(r) ) −→ b, p(θ Dφ1 p χ2M−M0 +ν−1 . φ2 n→∞

The importance of this result will be seen in Chapter 8.


282

6.4.


Exercises

1. Consider the following random sample of size n = 20; 1, 1, 1, 2, 2, 2, 2, 1, 1, 1, 1, 3, 3, 3, 4, 4, 4, 1, 1, 1. We wish to test if these observations fit the distribution given in Exercise 3 of Chapter 5, using φ1 (x) = φ2 (x) = x log x − x + 1 and significance level α = 0.05. 2. Consider the following random sample of size n = 20; -1, 1, 1,-1, 1, 1, 1, 1, 1, 0, 0, 1, 0, -1, -1, 0, 0, 1, 0, 1. We wish to test if these observations fit the distribution given in Exercise 9 of Chapter 5, using φ1 (x) = 12 (1 − x)2 , φ2 (x) = x log x − x + 1 and significance level α = 0.05. 3. A genetic model indicates that the distributions of the population between men or women and colour-blind or normal have the probabilities Men Normal Colour-blind

θ 2 (1−θ) 2

Women ¡ ¢ θ 1 − θ2 (1−θ)2 2

a) Find the minimum φ2 -divergence estimator with φ2 (x) = x log x − x + 1. b) Obtain its asymptotic distribution. c) We consider a random sample of size n = 2000 and we obtain the following results

Normal Colour-blind

Men 894 81

Women . 1015 10

We want to know if the previous random sample is from the genetic model considered using φ2 (x) = 12 (x − 1)2 , φ1 (x) = x log x − x + 1 and significance level α = 0.05. 4. Solve Example 6.1 using Theorem 6.4. 5. Let Y1 , ..., Yn be a random sample from a exponential population of parameter θ. We consider the partition E1 = (0, 1) and E2 = [1, ∞) and define pi (θ) = Prθ (Ei ) , i = 1, 2.



283

a) Find the asymptotic distribution of the test statistic ³ ´ Tnφ θb =

2n φ00 (1)

b p, p(θ)), Dφ (b

where θb is the maximum likelihood estimator based on the original data and φ (x) = 12 (x − 1)2 .

b) Prove that the eigenvalue λ³associated with the asymptotic distribu´ φ b tion of the test statistic Tn θ verifies 0 ≤ λ ≤ 1.

6. Solve Exercise 5 using Theorem 6.4.

7. Let X be a random variable with probability mass function Prθ (X = 1) = 0.5 − 2θ, Prθ (X = 2) = 0.5 + θ and Prθ (X = x3 ) = θ, with 0 < θ < 14 . a) Find the minimum power divergence estimator of θ for λ = −2. b) Find the asymptotic distribution of the estimator obtained in part a). c) In 8000 independent trials the events {i} , i = 1, 2, 3, have occurred 2014, 5012 and 974 times respectively. Given the significance level α = 0.05, test the hypothesis that the data are from the population described by the random variable X, using the power-divergence test statistic for λ = −2, −1, −1/2, 0, 2/3 and 1. 8. Let X be a random variable with probability mass function Pr (X Pr (X Pr (X Pr (X

= 1) = = 2) = = 3) = = 4) =

p1 (θ1 , θ2 ) = θ1 θ2 p2 (θ1 , θ2 ) = θ1 (1 − θ2 ) p3 (θ1 , θ2 ) = θ2 (1 − θ1 ) p4 (θ1 , θ2 ) = (1 − θ2 ) (1 − θ1 ) ,

with 0 ≤ θ1 , θ2 ≤ 1. a) Find Fisher information matrix as well as the asymptotic distribution of the minimum φ-divergence estimator. b) Find the minimum φ-divergence estimator with φ (x) = x log x − x + 1 as well as its asymptotic distribution.


284


³ ´ b(0) for φ (x) = c) Find the expression of the test statistic Tnφ θ for testing H0 : p (θ), where

1 2

(x − 1)2

p (θ) = (p11 (θ) , p12 (θ) , p21 (θ) , p22 (θ))T = (θ1 θ2 , θ1 (1 − θ2 ) , θ2 (1 − θ1 ) , (1 − θ2 ) (1 − θ1 ))T . 9. Let X be a random variable with probability density function given by   1 − θ cos x x ∈ [0, 2π) , fθ (x) = 2π  0 otherwise

θ ∈ (−1, 1) and we consider the discretized model obtained on dividing the interval [0, 2π) into M intervals of equal size. a) Find the Fisher information matrix in the discretized model. b) Find the minimum power-divergence estimator for λ = −2. c) Find its asymptotic distribution.

10. Given the model P (θ) =

Ã

1−θ θ 1 0

!

∈ P θ,

and the stationary distribution given by p (θ) =

µ

θ 1 , 1+θ 1+θ

¶T

, θ ∈ Θ = (0, 1) :

a) Find the minimum φ-divergence estimator as well as its asymptotic properties. b) Find the expression of the minimum power-divergence estimator for λ 6= −1.

6.5.


1. In our case we have pb1 = 0.5, pb2 = 0.2, pb3 = 0.15 and pb4 = 0.15, © 2006 by Taylor & Francis Group, LLC


285

then, taking into account Exercise 3 in Chapter 5, we have θbφ2 (y1 , ..., yn ) =

and

(−1+2b p1 −b p2 −b p3 )+((−1+2b p1 −b p2 −b p3 )2 +8b p4 ) 2

1/2

= 0.4,

1 (2 + 0.4) = 0.6 4 1 p2 (θbφ2 ) = (1 − 0.4) = 0.15 4 p3 (θbφ2 ) = 0.15 p4 (θbφ ) = 0.1.

p1 (θbφ2 ) =

2

The expression of the test statistic is given by Ã ! ³ ´ 4 P 2n p b i b, p(θbφ2 ) = 2n pbi log Dφ p φ002 (1) 2 pi (θbφ2 ) i=1 µ 0.2 0.15 0.5 + 0.2 log + 0.15 log = 40 0.5 log 0.6 0.15 0.15 ¶ 0.15 = 1.0878. + 0.15 log 0.1

On the other hand χ24−1−1, 0.05 = 5.991 and we should not reject the null hypothesis. 2. In this case we have b = (b p p1 , pb2 , pb3 )T = (4/20, 6/20, 10/20)T

and

2b p1 + pb3 = 0.45. θbφ2 (y1 , ..., yn ) = 2

Then

Therefore

p1 (θbφ2 ) = 0.2025, p2 (θbφ2 ) = 0.3025, p3 (θbφ2 ) = 0.4950.

!2 Ã ³ ´ 3 P 2n p b i b, p(θbφ2 ) = n −1 pi (θbφ2 ) Dφ p φ001 (1) 1 pi (θbφ2 ) i=1 3 (b P pi − pi (θbφ2 ))2 = 2.0406 × 10−3 . = n pi (θbφ ) i=1 2

On the other hand hypothesis.


χ23−1−1, 0.05

= 3.841 and we should not reject the null

286


3. a) It is necessary to minimize in θ the function 2b p1 2b p2 2b p3 2b p4 + pb2 log + pb3 log + pb4 log θ 1−θ θ (2 − θ) (1 − θ)2 = −b p1 log θ − pb2 log (1 − θ) − pb3 log (2 − θ) − pb3 log θ − pb4 2 log (1 − θ) + c.

g (θ) = pb1 log

Differentiating with respect to θ and equating to zero we have

θ 2 (−1 − pb4 − pb3 ) + θ (4 − pb1 − 2b p2 ) − 2 (b p1 + pb3 ) = 0,

then

θbφ2 =

−(4−b p1 −2b p2 )+((4−b p1 −2b p2 )2 +8(−1−b p4 −b p3 )(b p1 +b p3 )) 2(−1−b p4 −b p3 )

1/2

.

b) We have

∂p1 (θ) 1 ∂p2 (θ) 1 ∂p3 (θ) ∂p4 (θ) = , =− , = 1 − θ and = − (1 − θ) , ∂θ 2 ∂θ 2 ∂θ ∂θ then ³ ´ A (θ) = diag p (θ)−1/2 J (θ) ³√ √ √ ´ ¡ ¢−1/2 = 22 θ−1/2 , − 22 (1 − θ)−1/2 , θ−1/2 2−θ (1 − θ) , − 2 2

and

1 1 A (θ) A (θ) = θ−1 + (1 − θ)−1 + θ−1 2 2 T

µ

2−θ 2

¶−1

(1 − θ)2 + 2.

Therefore, µ ³ ´−1 ¶ √ b L T n(θφ2 (Y1 , ..., Yn ) − θ0 ) −→ N 0, A (θ0 ) A (θ0 ) . n→∞

c) In our case

then

pb1 = 0.447, pb2 = 0.0405, pb3 = 0.5075, pb4 = 0.005, θbφ2 (y1 , ..., yn ) = 0.9129.

We have

p1 (θbφ2 ) = 0.4564, p2 (θbφ2 ) = 0.0435, p3 (θbφ2 ) = 0.4962 © 2006 by Taylor & Francis Group, LLC

287


and p4 (θbφ2 ) = 0.0039,

and the value of the φ-divergence test statistic is !2 Ã ³ ´ 4 P p b 2n i b, p(θbφ2 ) = n −1 pi (θbφ2 ) Dφ p φ001 (1) 1 pi (θbφ2 ) i=1 ³ ´2 pbi − pi (θbφ2 ) 4 P = n pi (θbφ2 ) i=1 = 1.9361.

But χ2M−M0 −1,α = χ22,0.05 = 5.991 and we should not reject the null hypothesis. 4. The Fisher information associated with the original model is 1 =1 σ2

IF (θ) =

and the Fisher information associated with the discretized model is given by I F (θ) = A (θ)T A (θ) where A (θ) = =

Ã

Ã

p1 (θ)−1/2 0 0 p2 (θ)−1/2

−

!µ

1 1 2 2 − √ e−θ /2 , √ e−θ /2 2π 2π !T

p1 (θ)−1/2 −θ2 /2 p2 (θ)−1/2 −θ2 /2 √ e e , √ 2π 2π

then

,

2

I F (θ) = A (θ)T A (θ) = We have M = 2, M0 = 1 and φ (x) =

1 2

e−θ . p1 (θ) p2 (θ) 2π

(1 − x)2 , then

¡ ¢¢2 ¡ n p b − p Y 2n 1 1 L b = D (b p , p( θ)) −→ (1 − λ1 ) χ21 φ φ00 (1) p1 (θ) p2 (θ) n→∞

being λ1 the solution of the equation

det (I F (θ) − λIF (θ)) = det


Ã

! 2 e−θ − λ = 0, p1 (θ) p2 (θ) 2π

¶T

288


i.e., 2

e−θ λ1 = . p1 (θ) p2 (θ) 2π 5. We have Prθ (E1 ) = p1 (θ) = 1 − exp (−1/θ) and Prθ (E2 ) = p2 (θ) = exp (−1/θ) , and the maximum likelihood estimator of θ, based on the original data, is θb = Y .

³ ´ The test statistic Tnφ θb has the expression

Ã ¡ ¢2 ¡ ¢2 ¡ ¢ 2! 2 ³ ´ X (b p (b p (b p − p Y ) − p Y ) − p Y ) i i 1 1 2 2 ¡ ¢ ¡ ¢ ¡ ¢ =n + Tnφ θb = n pi Y p1 Y p2 Y Ãi=1 Ã ¡ ¢2 ¡ ¢ 2! ¡ ¢ ! (b p1 − p1 Y ) (b p1 − p1 Y )2 (b p1 − p1 Y ) ¡ ¢ ¡ ¢ ¡ ¢ ¡ ¢ , =n + =n p1 Y p2 Y p1 Y p2 Y but

¡ ¢ ¡ ¢ P p1 Y p2 Y −→ p1 (θ) p2 (θ) . n→∞ ³ ´ Therefore the asymptotic distribution of Tnφ θb coincides with the asymptotic distribution of the random variable ¡ ¢ ³ ´ (b p1 − p1 Y )2 φ b . Rn θ = n p1 (θ) p2 (θ) We know that

³ 1´ ¡ ¢ ¢ ∂p1 (θ) ¡ p1 Y = p1 (θ) + Y − θ + oP n− 2 , ∂θ

but

∂p1 (θ) 1 = − 2 exp (−1/θ) , ∂θ θ

then we have µ ¶ √ ¡ ¢ ¢ √ n 1 ¡ n(p1 Y − p1 (θ)) = − 2 exp − Y − θ + oP (1) . θ θ

The asymptotic distribution of the random variable √ n(b p1 − p1 (θ) , Y − θ)T



289

is bivariate normal with mean vector (0, 0)T and variance-covariance matrix given by √ 1 V ar [ n(b p1 − p1 (θ))] = nV ar [b p1 ] = n 2 V ar [Ni ] n 1 = n 2 np1 (θ) p2 (θ) = p1 (θ) p2 (θ) n ¢¤ £ ¤ £ ¤ £√ ¡ = nV ar Y − θ = nV ar Y V ar n Y − θ 1 = n V ar [Yi ] = θ2 . n Denoting by T =

n 1 P (I (Yi ) − p1 (θ)), we have n i=1 (0,1)

· ¸ n ¢¤ £√ ¡ 1 P (Yi − θ) Cov n pb1 − p1 (θ) , Y − θ = nCov T, n i=1 ¤ £ = Cov I(0,1) (Yi ) − p1 (θ), Yi − θ ¤ £ = E I(0,1) (Yi ) (Yi − θ) Z 1 (x − θ) 1θ exp (−x/θ) dx = 0

= − exp (−1/θ) .

Therefore ¢T L √ ¡ n pb1 − p1 (θ) , Y − θ −→ N n→∞

ÃÃ

0 0

! Ã ,

p1 (θ) p2 (θ) − exp(- 1θ ) − exp(- 1θ ) θ2

!!

On the other hand

¡ ¢ ¢T √ √ ¡ n pb1 -p1 (θ) , p1 Y -p1 (θ) = n

Ã

1 0 0 - exp(- 1θ )

!Ã

pb1 − p1 (θ) Y −θ

!

,

and denoting by ! Ã ! Ã 1 0 p1 (θ) p2 (θ) − exp (−1/θ) and A = Σ= 0 − exp (−1/θ) − exp (−1/θ) θ2 we have ∗

T

Σ = AΣA =


Ã

p1 (θ) p2 (θ) 1 θ2 exp (−2/θ)

1 θ2 1 θ2

exp (−2/θ) exp (−2/θ)

!

,

.

290


then ¡ ¢ ¢T L √ ¡ n pb1 − p1 (θ) , p1 Y − p1 (θ) −→ N n→∞

and

ÃÃ

0 0

!

∗

,Σ

!

,

¡ ¢¢ L ¡ ¢ √ ¡ n pb1 − p1 Y −→ N 0, p1 (θ) p2 (θ) − θ−2 exp (−2/θ) n→∞

or

¡ ¢¢ √ ¡ n pb1 − p1 Y

L

p −→ N (0, 1) . p1 (θ) p2 (θ) − θ−2 exp (−2/θ) n→∞

Finally

¡ ¢¢2 L ¡ ¢ ¡ −→ p1 (θ) p2 (θ) − θ−2 exp (−2/θ) χ21 n pb1 − p1 Y n→∞

and

being λ = 1 −

¡ ¢¢2 ¡ n pb1 − p1 Y L −→ λχ21 p1 (θ) p2 (θ) n→∞

exp (−2/θ)

θ2 p1 (θ) p2 (θ)

.

Now we are going to see that λ ≥ 0. We know that Σ∗ is nonnegative definite, then µ ¶ exp (−2/θ) exp (−2/θ) ∗ 0 ≤ det (Σ ) = p1 (θ) p2 (θ) − θ2 θ2 µ ¶ exp (−2/θ) p1 (θ) p2 (θ) exp (−2/θ) 1 − , = θ2 θ 2 p1 (θ) p2 (θ) and 0 ≤ λ ≤ 1. 6. The Fisher information associated with the original model is given by 1 , θ2 and the Fisher information associated with the discretized model is IF (θ) =

I F (θ) = A (θ)T A (θ) being A (θ) = =


Ã

p1 (θ)−1/2 0 0 p2 (θ)−1/2

!

¡ 1 −1/θ 1 −1/θ ¢T − θ2 e , θ2 e

´T ³ −1/2 −1/2 − p1 (θ)θ2 e−1/θ , p2 (θ)θ2 e−1/θ ,


then I F (θ) = A (θ)T A (θ) = We have M = 2, M0 = 1 and φ (x) =

1 2

e−2/θ 1 . θ4 p1 (θ) p2 (θ)

(1 − x)2 , then

¡ ¡ ¢¢2 n pb1 − p1 Y L −→ (1 − λ1 )χ21 p1 (θ) p2 (θ) n→∞

being λ1 the root of the equation

det (I F (θ) − λ1 IF (θ)) = 0. It is easy to get λ1 =

exp (−2/θ) θ 2 p1 (θ) p2 (θ)

.

7. a) The power-divergence for λ = −2 has the expression   3 2 X p (θ) 1 j  Dφ(−2) (b p, p (θ)) = − 1 −2 (−2 + 1) pbj j=1 ! Ã 1 (0.5 − 2θ)2 (0.5 + θ)2 θ 2 + + −1 . = 2 pb1 pb2 pb3 Differentiating

g (θ) ≡ Dφ(−2) (b p, p (θ)), with respect to θ and equating to zero we get

then

p1 pb3 + 2b p1 pb2 ) − 2b p2 pb3 + pb1 pb3 = 0, θ(8b p2 pb3 + 2b θbφ(−2) (y1 , ..., yn ) =

b) We have

2b p2 pb3 − pb1 pb3 . 8b p2 pb3 + 2b p1 pb3 + 2b p1 pb2

∂p1 (θ) ∂p2 (θ) = −2, =1 ∂θ ∂θ then

and

´ ³ A (θ) = diag p (θ)−1/2 J (θ)

∂p3 (θ) = 1, ∂θ

= (−2 (0.5 − 2θ)−1/2 , (0.5 + θ)−1/2 , θ−1/2 )T ,


291

292


and

¢ ¡ 2θ + 0.52 A (θ) A (θ) = . (0.5 − 2θ) (0.5 + θ) θ T

Therefore µ ¶ ´ √ ³ (0.5 − 2θ0 ) (0.5 + θ0 ) θ0 L b n θφ(−2) (Y1 , ..., Yn ) − θ0 −→ N 0, . n→∞ (2θ0 + 0.52 ) c) We have

p, p (θ)) = Dφ(λ) (b It is immediate to get θbφ(−2) =

then

and

  3 λ+1 X pj (θ) 1  − 1 . λ λ (λ + 1) p b j j=1

2b p2 pb3 − pb1 pb3 = 0.1057, 8b p2 pb3 + 2b p1 pb3 + 2b p1 pb2

p1 (θbφ(−2) ) = . 2886, p2 (θbφ(−2) ) = 0. 6057 and p3 (θbφ(−2) ) = 0.1057,

φ Tn (λ)

³λ ´ θbφ(−2)

-2

-1

-0.5

0

2/3

1

62.853

63.583

64.017

64.498

65.213

65.603

.

On the other hand χ23−1−1,0.05 = 3.841. Then we should reject the null hypothesis for all the values of λ considered. 8. a) The matrix A (θ) is given by  q

then we have

   A (θ) =     T

I F (θ) = A (θ) A (θ) =


θ2 q θ1 1−θ2 q θ1 θ2 − 1−θ q 1 2 − 1−θ 1−θ1

Ã

q

θ1 q θ2 θ1 − 1−θ 2 q 1−θ1 q θ2 1 − 1−θ 1−θ2



   ;   

0 (θ1 (1 − θ1 ))−1 0 (θ2 (1 − θ2 ))−1

!

.


293

Therefore, ´ ´ ³ √ ³ 1 2 T L n (θbφ2 , θbφ2 ) − (θ10 , θ20 )T −→ N 0, I F (θ0 )−1 . n→∞

b) It is necessary to minimize in θ1 and θ2 the function

pb1 pb2 pb3 + pb2 log + pb3 log θ1 θ2 θ1 (1 − θ2 ) θ2 (1 − θ1 ) pb4 + pb4 log (1 − θ1 ) (1 − θ2 ) = −b p1 log θ1 θ2 − pb2 log θ1 (1 − θ2 ) − pb3 log θ2 (1 − θ1 ) − pb4 log (1 − θ1 ) (1 − θ2 ) + c.

g (θ1 , θ2 ) = pb1 log

Differentiating with respect to θ1 and θ2 and equating to zero we have θb1 = pb1 + pb2 and θb2 = pb1 + pb3 .

The asymptotic distribution has been obtained, in general, for any φ in a). b = (b b = (b p11 , pb12 , pb21 , pb22 )T , we have c) If we denote p p1 , pb2 , pb3 , pb4 )T by p n11 + n12 n1∗ ≡ θb1 = pb1 + pb2 = pb11 + pb12 = pb1∗ = n n n11 + n21 n∗1 θb2 = pb1 + pb3 = pb11 + pb21 = pb∗1 = ≡ . n n

In a similar way

Then we get

n21 + n22 n2∗ ≡ 1 − θb1 = 1 − pb1∗ = pb2∗ = n n n12 + n22 n∗2 1 − θb2 = 1 − pb∗1 = pb∗2 = ≡ . n n

µ 2 X 2 ³ ´ X Tnφ θbφ(0) = n pij (θbφ(0) ) i=1 j=1

pbij pij (θbφ(0) )

¶2

−1

=

2 X 2 X (nn

ij −ni∗ n∗j )

i=1 j=1

ni∗ n∗j

2

.

© ª 2π 9. a) The partition of the interval [0, 2π) is given by Ej = [j − 1 2π M,jM) , j = 1, ..., M. It is clear that µ µ ¶ ¶ 2π 2π 1 − Fθ (j − 1) = + θ cj , pj (θ) = Prθ (Ej ) = Fθ j M M M where cj =

1 2π (sen (j


2π − 1) 2π M − sen j M ).

294


Then, p (θ) = ( and

1 1 + θ c1 , ..., + θ cM )T M M



   A (θ) =    

Then,

A (θ)T A (θ) =

M X j=1

and

c1

¡

cM

¡

¡

+ θ c1 . . .

1 M

+ θ cM

c2j 1 M

+ θ cj

I F (θ) = M

¢−1/2     .    ¢−1/2

1 M

M X j=1

¢ =M

M X j=1

c2j 1 + Mθ cj

c2j . 1 + Mθ cj

b) The power-divergence for λ = −2 is Dφ(−2) (b p, p (θ)) = =

  M 2 X pj (θ) 1  − 1 −2 (−2 + 1) j=1 pbj   M ¢2 1 X 1 ¡ 1 + θcj − 1 . 2 pbj M j=1

Differentiating and equating to zero we have θbφ(−2)

 1 M M 2 X X c cj   1 j = − . M pbj pbj 

j=1

j=1

c) The asymptotic distribution is given by

  M X √ b L n(θφ(−2) (Y1 , ..., Yn ) − θ0 ) → N 0, M n→∞

j=1

−1  c2j  . 1 + Mθ0 cj

10. a) First we are going to obtain the asymptotic properties of the minimum φ-divergence estimator. We have


295


· A (θ0 ) =

1 −1/2

Ã

(1 + θ0 ) Ã ! −1 1 = −1/2 θ0 (1 + θ0 )3/2

!

1 0 −1/2 0 θ0

 1 −  (1 + θ )2  0     1 2 (1 + θ0 ) 

· A (θ0 )T A (θ0 ) = (1 + θ0 )−2 θ0−1 1/2

· ∆T0 = ((1 + θ0 ) θ0 )1/2 (−θ0 , 1) Ã ! 2+θ +1 2 1 θ −θ 0 0 0 . · C −1 0 = (1 + θ0 ) −θ0 2θ0 + 1 Therefore, S0 = Then,

(1−θ0 )θ0 (1+θ0 )3

Ã

1 −1 −1 1

!

and B 0 =

(1−θ0 ) (1+θ0 )2

Ã

1/2

θ0 −θ0 1/2 −θ0 1

!

.

´ ¢ ¡ √ ³b L n θφ2 (Y1 , ..., Yn ) − θ0 −→ N 0, ∆T0 B 0 ∆0 , n→∞

and

∆T0 B 0 ∆0 = θ0 (1 − θ02 ).

The minimum φ-divergence estimator is obtained minimizing, in θ ∈ (0, 1) , the function ¡ ¢¢ 1 ¡ φ2 ((1 + θ) pb1 ) + θφ2 θ−1 pb2 (1 + θ) . g (θ) = Dφ2 (b p, p (θ)) = 1+θ

b) If we consider the power-divergence measure we obtain ³ ´ λ λ −λ λ+1 1 g (θ) = λ(λ+1) pbλ+1 (1 + θ) + (1 + θ) θ p b − 1 ³ 1 ³ ´ 2´ λ λ+1 λ+1 1 −λ −1 . = λ(λ+1) (1 + θ) pb1 + θ pb2

Then,

g 0 (θ) = = =

´ ´ ³ 1 (1 + θ)λ−1 λ pbλ+1 − (1 + θ)λ λ θλ+1 + θ−λ pbλ+1 pbλ+1 1 2 2 Ã ! pbλ+1 pbλ+1 pbλ+1 (1+θ)λ−1 λ λ+1 2 2 2 pb1 + λ − λ+1 − λ λ(λ+1) θ θ θ Ã ! λ+1 pb (1+θ)λ−1 − 2λ pbλ+1 , 1 (λ+1) θ 1 λ(λ+1)


³

296


and therefore g 0 (θ) = 0 implies θ = pb2 /b p1 . Then we can conclude that p1 θbφ(λ) = pb2 /b


if

0 < pb2 /b p1 < 1.

7

Testing Loglinear Models Using Phi-divergence Test Statistics 7.1.

Introduction

One basic and straightforward method for analyzing categorical data is via crosstabulation. For example, a medical researcher may tabulate the frequency of different symptoms by patient’s age and gender; an educational researcher may tabulate the number of high school drop-outs by age, gender, and ethnic background; an economist may tabulate the number of business failures by industry, region, and initial capitalization; a market researcher may tabulate consumer preferences by product, age, and gender, etc. In all of these cases, the most interesting results can be summarized in a multiway frequency table, that is, in a crosstabulation table with two or more factors. Loglinear models provide a more “sophisticated” way of looking at crosstabulation tables. Specifically, one can test the different factors that are used in the crosstabulation (e.g., gender, region, etc.) and their interactions for statistical significance. In this introductory section we present an intuitive approach to loglinear models and in the remaining sections a systematic study of them. Example 7.1 The data in Table 7.1 for 205 married persons, reported initial ly by Galton,


298


give the number of cases in which a tall, medium or short man was married to a tall, medium or short woman.

Husband

Tall 18 20 12 50

Tall Medium Short Totals

Wife Medium 28 51 25 104

Short 14 28 9 51

Totals 60 99 46 205

Table 7.1 Source: Christensen, R. (1997, p. 67).

Are the heights of husband and wife independent ? The answer to the previous question could be given using the classical chisquare test statistic for independence. However, we are going to deal with the problem considering a loglinear model for the data assuming independence. Let X and Y denote two categorical response variables, X and Y having I and J levels, respectively. The responses (X, Y ) of a subject randomly chosen from some population have a probability distribution. Let pij = Pr (X = i, Y = j) , with pij > 0, i = 1, ..., I, j = 1, ..., J. We display this distribution in a rectangular table having I rows for the categories of X and J columns for the categories of Y . The corresponding matrix I × J is called a contingency table. Consider a random sample of size n on (X, Y ) and we denote by nij the observed frequency in the P P (i, j)th-cell for (i, j) ∈ I×J with n = Ii=1 Jj=1 nij and the totals for the ith-row P P and jth-column by ni∗ = Jj=1 nij and Ii=1 nij = n∗j , i = 1, ..., I, j = 1, ..., J, respectively. In the following we assume that nij is the observed value corresponding to a random variable Nij , i = 1, ..., I, j = 1, ..., J, in such a way that the random variable (N11 , ..., NIJ ) is multinomially distributed with parameters n and (p11 , ...., pIJ ). We denote mij = E [Nij ] = npij , i = 1, ..., I, j = 1, ..., J. Under the independence assumption we have H0 : pij = pi∗ p∗j , i = 1, ..., I, j = 1, ..., J ⇔ mij = npi∗ p∗j , where pi∗ =

PJ

j=1 pij


and p∗j =

PI

i=1 pij .

(7.1)

The hypothesis (7.1) can be written as

H0 : log mij = log n + log pi∗ + log p∗j ,

(7.2)

Testing Loglinear Models Using Phi-divergence Test Statistics

299

or equivalently log mij = log n + log pi∗ + log p∗j ¶ ¶ µ I µ J P P = log pi∗ − log ph∗ /I + log p∗j − log p∗h /J h=1 ¶ µ J ¶ µ Ih=1 P P log ph∗ /I + log p∗h /J. + log n + h=1

h=1

If we denote

θ1(i) = log pi∗ −

µ

¶

I P

log ph∗ /I ¶ log p∗h /J θ2(j) = log p∗j − ¶ µ J ¶ µ Ih=1 P P log ph∗ /I + log p∗h /J, u = log n + µh=1 J P h=1

h=1

we have log mij (θ) = u+θ1(i) +θ2(j) , i = 1, ..., I, j = 1, ..., J, where the parameters ª © ª © P P θ1(i) and θ2(j) verify Ii=1 θ1(i) = Jj=1 θ2(j) = 0. Then the hypothesis of independence given in a two-way contingency table can be specified by the model log mij (θ) = u + θ1(i) + θ2(j) , i = 1, ..., I, j = 1, ..., J, where the parameters verify

PI

i=1 θ1(i)

=

PJ

j=1 θ2(j)

= 0.

The number of parameters of the model, initially, is   u 1  I + J − 1, θ1(i) I −1  J −1  θ2(j)

but we are assuming that our data are from a multinomial population with n = PI PJ i=1 j=1 mij (θ) ; therefore actually we have I + J − 2 parameters because u is a function of θ1(i) and θ2(j) . The loglinear model of independence which justifies the data from Example 7.1 is log mij (θ) = u + θ1(i) + θ2(j) , i = 1, ..., 3, j = 1, ..., 3, P P with 3i=1 θ1(i) = 3j=1 θ2(j) = 0. © 2006 by Taylor & Francis Group, LLC

300


Since it may be useful to express the loglinear model in a matrix notation, we obtain the corresponding matrix form for the model of Example 7.1, log m11 (θ) = log m12 (θ) = log m13 (θ) = log m21 (θ) = log m22 (θ) = log m23 (θ) = log m31 (θ) = log m32 (θ) = log m33 (θ) =

u + θ1(1) + θ2(1) u + θ1(1) + θ2(2) u + θ1(1) + θ2(3) u + θ1(2) + θ2(1) u + θ1(2) + θ2(2) u + θ1(2) + θ2(3) u + θ1(3) + θ2(1) u + θ1(3) + θ2(2) u + θ1(3) + θ2(3)

If we denote by



       X =       

= u + θ1(1) − θ2(1) − θ2(2) = = = =

u + θ1(2) − θ2(1) − θ2(2) u − θ1(1) − θ1(2) + θ2(1) u − θ1(1) − θ1(2) + θ2(2) u − θ1(1) − θ1(2) − θ2(1) − θ2(2) .

 1 1 0 1 0 1 1 0 0 1    1 1 0 −1 −1   1 0 1 1 0   1 0 1 0 1  , 1 0 1 −1 −1    1 −1 −1 1 0   1 −1 −1 0 1  1 −1 −1 −1 −1

m (θ) = (m11 (θ) , ..., m13 (θ) , ..., m31 (θ) , ..., m33 (θ))T and ¢T ¡ θ = u, θ1(1) , θ1(2) , θ2(1) , θ2(2) ,

we have log m (θ) = Xθ.

In a two-way contingency table the most general loglinear model is log mij (θ) = u + θ1(i) + θ2(j) + θ12(ij)

(7.3)

where u= θ2(j) =

PJ 1 PI j=1 log mij (θ) IJP i=1 I 1 i=1 log mij (θ) − u I

P θ1(i) = J1 Jj=1 log mij (θ) − u θ12(ij) = log mij (θ) − u − θ1(i) − θ2(j) .

The term θ12(ij) represents the interaction between the two random variables P P X and Y. It is easy to verify Ii=1 θ12(ij) = Jj=1 θ12(ij) = 0. The number of © 2006 by Taylor & Francis Group, LLC


301

parameters in this model is IJ −1. A loglinear model, with multinomial sampling, in which the number of cells minus one is equal to the number of parameters is called saturated model. A saturated model is one which attempts to estimate parameters for all single-variable and all interaction effects. That is, saturated models include all possible terms, including all interaction effects. They provide an exact fit for the observed cell counts. Since observed and expected are the same, there are too many unknown parameters to compute goodness-of-fit test statistics. If we assume I = J = 3, denoting  1 1 0 1 0 1 0 0 0  1 1 0 0 1 0 1 0 0   1 0 −1 −1 −1 −1 0 0  1   1 0 1 1 0 0 0 1 0   X= 1 0 1 0 1 0 0 0 1  1 0 1 1 −1 0 0 −1 −1    1 −1 −1 1 0 −1 0 −1 0   1 −1 −1 0 1 0 −1 0 −1 1 −1 −1 −1 −1 1 1 1 1



       ,       

m (θ) = (m11 (θ) , ..., m13 (θ) , ..., m31 (θ) , ..., m33 (θ))T

and

¢T ¡ θ = u, θ1(1) , θ1(2) , θ2(1) , θ2(2) , θ12(11) , θ12(12) , θ12(21) , θ12(22)

we have that the matrix form of the saturated model is log m (θ) = Xθ. Before introducing the loglinear models in a three-way contingency table we present an example. Example 7.2 Worchester (1971) describes a case-control study of women diagnosed with thromboembolism for the purpose of studying the risks associated with smoking and oral contraceptive use. Their data are summarized in Table 7.2. Contraceptive use? Yes No

Cases Smoker nonsmoker 14 12 7 25

Controls Smoker nonsmoker 2 8 22 84

Table 7.2 Source: Worchester, J. (1971).


302


Loglinear models are useful methods to describe the inter-relationships between these three factors. Let X, Y and Z denote three categorical response variables: X having I levels, Y having J levels and Z having K levels. When subjects are classified based on the three variables, there are IJK possible combinations of classification. The responses (X, Y, Z) of a subject randomly chosen from some population have a probability distribution. Let pijk = Pr (X = i, Y = j, Z = k) , with pijk > 0, i = 1, ..., I, j = 1, ..., J and k = 1, ..., K, and let p =(p111 , ..., pIJK )T be the joint distribution of X, Y and Z. This distribution is displayed in a three-way contingency table. Consider a random sample of size n on (X, Y, Z) and let nijk be the observed frequency in the (i, j, k)th-cell for (i, j, k) ∈ I × J × K with PI PJ PK i=1 j=1 k=1 nijk = n. In the following we assume that nijk is the observed value corresponding to a random variable Nijk , i = 1, ..., I, j = 1, ..., J, k = 1, ..., K, in such a way that the random variable (N111 , ..., NIJK ) is multinomially distributed with parameters n and (p111 , ...., pIJK ). We denote mijk = E [Nijk ] = npijk , i = 1, ..., I, j = 1, ..., J, k = 1, ..., K. We also denote

u θ1(i) θ2(j) θ3(k) θ12(ij) θ13(ik) θ23(jk)

= = = = = = =

J P K I P P

j=1 k=1 K I P P i=1 k=1 J I P P i=1 j=1 K P k=1 J P j=1 I P i=1

log mijk /JK − u

log mijk /IK − u

log mijk /IJ − u

log mijk /K −

log mijk /J −

log mijk /I −

θ123(ijk) = log mijk − +

log mijk /IJK

i=1 j=1 k=1 K J P P

K J P P

K P

k=1

K J P P

j=1 k=1 K J P P

j=1 k=1 K I P P

i=1 k=1


log mijk /JK −

log mijk /IK −

log mijk /K −

log mijk /JK +

j=1 k=1

log mijk /JK −

J P

j=1

K I P P

i=1 k=1

K I P P

log mijk /IK + u

i=1 k=1 J I P P

log mijk /IJ + u

i=1 j=1 J I P P

log mijk /IJ + u

i=1 j=1

log mijk /J −

log mijk /IK +

I P

log mijk /I

i=1 J I P P

i=1 j=1

log mijk /IJ + u.


303

Then we have log mijk (θ) = u + θ1(i) + θ2(j) + θ3(k) + θ12(ij) + θ13(ik) + θ23(jk) + θ123(ijk) . (7.4) It is clear that the parameters verify the following constraints, I P

i=1 I P i=1 I P i=1

θ1(i) =

J P

θ2(j) =

j=1 J P

θ12(ij) =

K P

θ3(k) = 0

k=1

θ12(ij) =

j=1 J P

θ123(ijk) =

I P

θ13(ik) =

i=1

θ123(ijk) =

j=1

K P

θ13(ik) =

k=1 K P

θ123(ijk) = 0.

J P

θ23(jk) =

j=1

K P

θ23(jk) = 0

k=1

k=1

The number of parameters, initially, in this model is u 1 θ1(i) I − 1 θ2(j) J − 1 θ3(k) K − 1 θ12(ij) (I − 1) (J − 1) θ13(ik) (I − 1) (K − 1) θ23(jk) (J − 1) (K − 1) θ123(ijk) (I − 1) (J − 1) (K − 1)

                          

IJK,

but under the assumption of multinomial sampling we have IJK − 1 parameP P P ters because Ii=1 Jj=1 K k=1 mijk (θ) = n. We have the saturated model, in a three-way contingency table, because the number of parameters coincide with the number of cells minus one. A simpler model in a three-way contingency table is the model log mijk (θ) = u + θ1(i) + θ2(j) + θ3(k) .

(7.5)

This is the independence model in a three-way contingency table, i.e., pijk = PJ PK PI PK pi∗∗ p∗j∗ p∗∗k with pi∗∗ = j=1 k=1 pijk , p∗j∗ = i=1 k=1 pijk and p∗∗k = PI PJ i=1 j=1 pijk .

Between the models (7.4) and (7.5) there are different loglinear models that can be considered: i) Variable X is jointly independent of Y and Z. In this case we have log mijk (θ) = u + θ1(i) + θ2(j) + θ3(k) + θ23(jk) ,


304


i.e., pijk = p∗jk pi∗∗ . The number of parameters in this model is (I − 1) + (J − 1) + (K − 1) + (K − 1) (J − 1) . ii) The random variables X and Z are independent given the random variable Y . In this case we have log mijk (θ) = u + θ1(i) + θ2(j) + θ3(k) + θ12(ij) + θ23(jk) , i.e., pijk /p∗j∗ = (pij∗ /p∗j∗ ) (p∗jk /p∗j∗ ) . The number of parameters of this model is (I − 1) + (J − 1) + (K − 1) + (I − 1)(J − 1) + (J − 1)(K − 1). iii) The random variables X, Y and Z are pairwise dependent but the three random variables X, Y and Z are jointly independent. In this case we have log mijk (θ) = u + θ1(i) + θ2(j) + θ3(k) + θ12(ij) + θ13(ik) + θ23(jk) . This model does not admit a representation in terms of probabilities. The number of parameters is given by (I − 1) + (J − 1) + (K − 1) + (I − 1) [(J − 1) + (K − 1)] + (J − 1) (K − 1) . Hierarchical loglinear models require that high order interactions are always accompanied by all of their lower order interactions. All hierarchical models for a three-way contingency table are the following: H1 : log mijk (θ) = u + θ1(i) + θ2(j) + θ3(k) + θ12(ij) + θ13(ik) + θ23(jk) H2 : log mijk (θ) = u + θ1(i) + θ2(j) + θ3(k) + θ13(ik) + θ23(jk) H2∗ : log mijk (θ) = u + θ1(i) + θ2(j) + θ3(k) + θ12(ij) + θ13(ik) H2∗∗ : log mijk (θ) = u + θ1(i) + θ2(j) + θ3(k) + θ12(ij) + θ23(jk) H3 : log mijk (θ) = u + θ1(i) + θ2(j) + θ3(k) + θ23(jk)



305

H3∗ : log mijk (θ) = u + θ1(i) + θ2(j) + θ3(k) + θ12(ij) H3∗∗ : log mijk (θ) = u + θ1(i) + θ2(j) + θ3(k) + θ13(ik) H4 : log mijk (θ) = u + θ1(i) + θ2(j) + θ3(k) H4∗ : log mijk (θ) = u + θ2(j) + θ3(k) + θ23(jk) H4∗∗ : log mijk (θ) = u + θ1(i) + θ2(j) + θ12(ij) H4∗∗∗ : log mijk (θ) = u + θ1(i) + θ3(k) + θ13(ik) H5 : log mijk (θ) = u + θ2(j) + θ3(k) H5∗ : log mijk (θ) = u + θ1(i) + θ2(j) H5∗∗ : log mijk (θ) = u + θ1(i) + θ3(k) H6 : log mijk (θ) = u + θ3(k) H6∗ : log mijk (θ) = u + θ1(i) H6∗∗ : log mijk (θ) = u + θ2(j) H7 : log mijk (θ) = u. It is clear that, from a practical point of view, the problem consists of obtaining the model that presents a better fit to our data. In relation with Example 7.2 the problem is to find the model, among the previous considered models, that is able to explain more clearly the given data. In the procedure that follows in the next Sections it will be necessary to give a method to estimate the parameters of the model and then to choose the best model. In order to do this second step we first choose a nested sequence of loglinear models (two loglinear models are said to be nested when one contains a subset of the parameters in the other) in the way which is explained in Section 7.4. and then we shall give some procedure to choose the best model among the models considered in a nested sequence.

7.2.

Loglinear Models: Definition

Let Y1 , Y2 , ..., Yn be a sample of size n ≥ 1, with independent realizations in the statistical space X = {1, 2, ..., M} , which are identically distributed according to a


306


probability distribution p (θ0 ). For a two-way contingency table we have M = IJ and for a three-way contingency table we have M = IJK. This distribution is assumed to be unknown, but belonging to a known family © ª T = p (θ) = (p1 (θ) , ..., pM (θ))T : θ ∈ Θ , of distributions on X with Θ ⊆ RM0 (M0 < M − 1) . In other words, the true value θ0 of the parameter θ = (θ1 , ..., θM0 )T ∈ Θ ⊆ RM0 is assumed to be fixed b = (b but unknown. We denote p = (p1 , ..., pM )T and p p1 , ..., pbM )T with pbj =

Nj n

and

Nj =

n X i=1

I{j} (Yi ) ; j = 1, ..., M.

(7.6)

The statistic (N1 , ..., NM ) is obviously sufficient for the statistical model under consideration and is multinomially distributed; that is, Pr (N1 = n1 , ..., NM = nM ) =

n! p1 (θ)n1 ...pM (θ)nM , n1 !...nM !

(7.7)

for integers n1 , ..., nM ≥ 0 such that n1 + ... + nM = n. In what follows, we assume that p (θ) belongs to the general class of loglinear models. That is, we assume:

M ¡ ¢ X ¡ ¢ exp wTv θ ; u = 1, ..., M, pu (θ) = exp wTu θ /

(7.8)

v=1

where wTu = (wu1 , ..., wuM0 ) . The M ×M0 matrix W = (w1 , ..., wM )T is assumed to have full column rank M0 (M0 < M − 1) and columns linearly independent of the M × 1 column vector (1, ..., 1)T . This is the model that we shall consider for the theoretical results in the next sections. We restrict ourselves to multinomial random sampling but it is possible to give the results presented in the next sections under the assumptions of either Poisson, multinomial, or product-multinomial sampling jointly. For more details see Cressie and Pardo (2002b).



7.3.

307

Asymptotic Results for Minimum Phi-divergence Estimators in Loglinear Models

In this Section we present some asymptotic results for the minimum φ-divergence estimator under the loglinear model (7.8). These results are obtained particularizing the results obtained in Chapter 5 for a general multinomial model. In the cited Chapter 5 we establish that the Fisher information matrix in a multinomial model has the expression I F (θ) = A (θ)T A (θ) , where A (θ) is the M × M0 matrix defined in (5.7). For a loglinear model we have M X ∂pj (θ) = pj (θ) wjr − pj (θ) wvr pv (θ) . ∂θr v=1

Then

and hence

´ ∂p (θ) ³ = diag (p (θ)) − p (θ) p (θ)T W = Σp(θ) W ∂θ ´ ³ A (θ) = diag p (θ)−1/2 Σp(θ) W .

Then the Fisher information matrix for a loglinear model is given by I F (θ) = W T Σp(θ) W . bφ is the minimum φ-divergence estimator for the loglinear By Theorem 5.2, if θ model given in (7.8), then ³ ¡ ¢−1 ´ √ L bφ − θ0 ) −→ , n(θ N 0, W T Σp(θ 0 ) W n→∞

where

bφ = arg inf Dφ (b p, p (θ)), θ θ∈Θ

(7.9)

and by Theorem 5.1 verifies

¢ ¡ bφ = θ0 + I F (θ0 )−1 W T Σp(θ ) diag p(θ0 )−1 (b b − p(θ0 ) k). p − p(θ0 )) + o(k p θ 0 (7.10) © 2006 by Taylor & Francis Group, LLC

308


Another interesting result, useful later, is the following ´ ´ ³ ¡ ¢−1 T √ ³ L bφ ) − p (θ0 ) −→ n p(θ N 0, Σp(θ 0 ) W W T Σp(θ 0 ) W W Σp(θ 0 ) . n→∞

(7.11)

From a practical point of view in order to find the minimum φ-divergence bφ we must solve the following system of equations estimator θ  p, p (θ))  ∂Dφ (b =0 , ∂θi  i = 1, ..., M0

with the condition that p(θ) verifies (7.8).

These equations are nonlinear functions of the minimum φ-divergence estibφ . In order to solve these equations numerically the Newton-Raphson mator, θ method is used. We have, µ µ µ ¶ ¶ ¶ M P ∂Dφ (b p, p (θ)) pbl pbl pbl 0 φ = −φ ∂θj pl (θ) pl (θ) pl (θ) l=1 ¶ µ M P wuj pu (θ) , × pl (θ) wlj − pl (θ) u=1

and

∂ ∂θr

µ

∂Dφ (b p, p(θ)) ∂θj

¶

¶ pbl ∂pl (θ) ∂pl (θ) pbl pbl = pl (θ) pl (θ)2 ∂θr ∂θj pl (θ(t) ) l=1 ¶ ¶ ¶ µ µ µ k ∂ 2 p (θ) P pbl pbl pbl l 0 −φ . + φ pl (θ) pl (θ) pl (θ) l=1 ∂θj ∂θr (7.12) M P

φ00

µ

(t+1)

b , in a Newton-Raphson procedure Therefore the (t + 1)th-step estimate, θ (t) b is obtained from θ as  T (t) b p, p(θ ))  b(t) − G(θ b(t) )−1  ∂Dφ (b b(t+1) = θ , θ ∂θj j=1,...,M0

where G(θ(t) ) is the matrix whose elements are given in (7.12).

An interesting simulation study to analyze the behavior of the minimum power-divergence estimator, defined by b(λ) ≡ arg min Dφ (b p, p (θ)), θ (λ) θ∈Θ


(7.13)


309

in a three-way contingency table, has been considered in Pardo, L. and Pardo, b(0) is the maximum likelihood estimator. M. C. (2003). Notice that θ

7.4.

Testing in Loglinear Models

bφ ) the parametric estimator, based on the minimum θ bφ We denote by p(θ 2 2 divergence estimator, of the loglinear model defined in (7.8). For testing if our data are from a loglinear model we can use the family of φ-divergence test statistics ³ ´ ³ ´ bφ = 2n Dφ p bφ ) b Tnφ1 θ (7.14) , p( θ 2 1 2 φ001 (1) which by Theorem 6.1 are asymptotically distributed chi-squared with M −M0 −1 degrees of freedom. In Example 7.1 we have M = 9 and M0 = 4, therefore bφ is the minimum M − M0 − 1 = 4 = (I − 1)(J − 1). In the previous expression θ 2 φ2 -divergence estimator for the parameters of the considered loglinear model. In the rest of the chapter we shall assume the conditions given for the function φ2 in Theorem 5.1 as well as that φi (x) is twice continuously differentiable in a neighborhood of 1 with the second derivative φ00i (1) 6= 0, i = 1, 2.

Based on this result it is possible to select a nested sequence of loglinear models. In order to fix ideas we consider all the possible loglinear models associated with a I × J × K contingency table where the first element of the sequence is given by H1 : log mijk (θ) = u + θ1(i) + θ2(j) + θ3(k) + θ12(ij) + θ13(ik) + θ23(jk) . The second element should be chosen between the models H2 : log mijk (θ) = u + θ1(i) + θ2(j) + θ3(k) + θ13(ik) + θ23(jk) H2∗ : log mijk (θ) = u + θ1(i) + θ2(j) + θ3(k) + θ12(ij) + θ13(ik) H2∗∗ : log mijk (θ) = u + θ1(i) + θ2(j) + θ3(k) + θ12(ij) + θ23(jk) . We consider the three tests of hypotheses HNull : H2 versus HAlt : The saturated model HNull : H2∗ versus HAlt : The saturated model HNull : H2∗∗ versus HAlt : The saturated model


310


and we shall choose the model that better support ³the data. To do the previous ´ φ1 b(l) tests we consider the φ-divergence test statistic Tn θφ2 , given in (7.14), where (l)

b we indicate the minimum φ2 -divergence estimator for the parameters of by θ φ2 the loglinear model l, being l = H2 , H2∗ or H2∗∗ .

If we assume that we have chosen the model H2∗ , the third element in the sequence of nested sequence of loglinear models should be chosen between the models H3∗ : log mijk (θ) = u + θ1(i) + θ2(j) + θ3(k) + θ12(ij) H3∗∗ : log mijk (θ) = u + θ1(i) + θ2(j) + θ3(k) + θ13(ik) . We consider the two tests of hypotheses HN ull : H3∗ versus HAlt : The saturated model HN ull : H3∗∗ versus HAlt : The saturated model and we choose the model that better support the ³ data. ´ To do the previous tests φ1 b(l) we consider the φ-divergence test statistic Tn θφ2 , given in (7.14), where by

b(l) we indicate the minimum φ2 -divergence estimator for the parameters of the θ φ2 loglinear model l, being l = H3∗ or H3∗∗ . We can continue in the same way until getting a convenient nested sequence of loglinear models.

One of the main problems in loglinear models is to test a nested sequence of hypotheses, Hl : p = p (θ) ; θ ∈ Θl ; l = 1, ..., m, m ≤ M0 < M − 1,

(7.15)

where Θm ⊂ Θm−1 ⊂ ... ⊂ Θ1 ⊂ RM0 ; M0 < M −1 and dim(Θl ) = dl ; l = 1, ..., m, with dm < dm−1 < ... < d1 ≤ M0 . (7.16) Our strategy will be to test successively the hypotheses Hl+1 against Hl ;

l = 1, ..., m − 1,

(7.17)

as null and alternative hypotheses respectively. We continue to test as long as the null hypothesis is accepted and choose the loglinear model Hl according to the first l for which Hl+1 is rejected (as a null hypothesis) in favor of Hl (as



311

an alternative hypothesis). This strategy is quite standard for nested models (Read and Cressie, 1988, p. 42). The nesting occurs naturally because of the hierarchical principle, which says that interactions should not be fitted unless the corresponding main effects are present (e.g., Collett, 1994, p. 78). The results presented in this Section were obtained in Cressie, N. and Pardo, L. (2000) and Cressie, N. et al . (2003). Theorem 7.1 Suppose that data (N1 , ..., NM ) are multinomially distributed according to (7.7) and (7.8). Consider the nested sequence of hypotheses given by (7.15) and (7.16). Choose two functions φ1 , φ2 ∈ Φ∗ . Then, for testing hypotheses, HNull : Hl+1 against HAlt : Hl , the asymptotic null distribution of the φ-divergence test statistic, (l) Teφ1 ,φ2 ≡

2n b(l+1) ), p(θ b(l) )), Dφ1 (p(θ φ2 φ2 00 φ1 (1)

(7.18)

(l) is chi-square with dl − dl+1 degrees of freedom; l = 1, ..., m − 1. In (7.18), θbφ2 and (l+1) are the minimum φ2 -divergence estimators under the models Hl and Hl+1 , θb φ2

respectively, where the minimum φ-divergence estimators are defined by (7.9). ³ ´ b(l+1) ), p(θ b(l) ) about Proof. The second-order expansion of Dφ1 ,φ2 ≡ Dφ1 p(θ φ2 φ2 (p (θ0 ) , p (θ0 )) gives Ã ! M ³ ´2 2 D (p, q) X ∂ φ 1 b(l+1) ) − pj (θ0 ) p Dφ1 ,φ2 = 12 ( θ j φ 2 ∂p2j j=1 (p(θ 0 ),p(θ 0 )) Ã ! M ³ ´2 X ∂ 2 Dφ1 (p, q) 1 b(l) ) − pj (θ0 ) p + 2 ( θ j φ2 ∂qj2 j=1 (! p(θ 0 ),p(θ 0 )) Ã M X M ³ ´ X ∂ 2 Dφ1 (p, q) b(l+1) ) − pi (θ0 ) pi (θ + φ2 ∂pi ∂qj i=1 j=1 (p(θ 0 ),p(θ 0 )) ³ ´ b(l) ) − pj (θ0 ) × pj (θ µ° φ2 °2 ° °2 ¶ ° ° b(l) ° ° b(l+1) + o °p(θφ2 ) − p (θ0 )° + °p(θφ2 ) − p (θ0 )° .

We have used in the previous Taylor’s expansion that Dφ1 (p (θ0 ) , p (θ0 )) = 0 (l) and the first order term in Taylor expansion is also zero. Thus, Teφ1 ,φ2 can be


312


written as ´2 1 ³ b(l+1) pj (θφ2 ) − pj (θ0 ) p (θ0 ) j=1 j M ´2 X 1 ³ b(l) n pj (θφ2 ) − pj (θ0 ) p (θ0 ) j=1 j M ´³ ´ X 1 ³ b(l+1) b(l) 2n ) − p (θ ) pj (θφ2 ) − pj (θ0 ) pj (θ j 0 φ2 pj (θ0 ) j=1 µ° °2 ° °2 ¶ ° ° b(l) ° ° b(l+1) n o °p(θφ2 ) − p (θ0 )° + °p(θφ2 ) − p (θ0 )° ³ ´´T ³ ´³ b(l) ) b(l+1) ) − p(θ n1/2 diag p (θ0 )−1/2 p(θ φ2 φ2 ´³ ³ ³ ´´ (l+1) −1/2 b b(l) p(θ n1/2 diag p (θ0 ) φ2 ) − p(θ φ2 ) µ° °2 ° °2 ¶ ° ° b(l) ° ° b(l+1) ) − p (θ ) + θ ) − p (θ ) n o °p(θ . °p( φ2 0 ° 0 ° φ2

(l) Teφ1 ,φ2 = n

+ −

+ = × +

M X

We know from (7.11) that, under the loglinear model (7.8) and the null hypothesis Hl+1 , ´ ´ ³ √ ³ L ∗ b(l+1) ) − p (θ0 ) −→ n p(θ N 0, Σ φ2 (l+1) , n→∞

³ ´−1 W T(l+1) Σp(θ 0 ) , and W (l+1) with Σ∗(l+1) = Σp(θ 0 ) W (l+1) W T(l+1) Σp(θ 0 ) W (l+1) is the loglinear model matrix of explanatory variables under the null hypothesis Hl+1 . °2 ° ¡ −1 ¢ ° ° b(l+1) ) − p (θ ) Then °p(θ , and because it is assumed that θ0 ∈ 0 ° = OP n φ2 ° °2 ¡ ¢ ° b(l) ° ) − p (θ ) Θl+1 ⊂ Θl , we also have that °p(θ ° = OP n−1 . Consequently, 0 φ2 µ° °2 ° °2 ¶ ° b(l+1) ° ° b(l) ° n o °p(θ = oP (1) , ) − p (θ ) + θ ) − p (θ ) °p( φ2 0 ° 0 ° φ2

(l) and hence the asymptotic distribution of the test statistic Teφ1 ,φ2 is the same as the asymptotic distribution of the random variable Z T Z, where

Z≡

´³ ³ ´ √ b(l+1) ) − p(θ b(l) ) . ndiag p (θ0 )−1/2 p(θ φ2 φ2


313


Then, using (7.10), we obtain µ ³ ´−1 (l+1) (l) b b p(θφ2 ) − p(θφ2 ) = Σp(θ 0 ) W (l+1) W T(l+1) Σp(θ 0 ) W (l+1) W T(l+1) Σp(θ 0 ) ¶ ³ ´−1 − Σp(θ 0 ) W (l) W T(l) Σp(θ 0 ) W (l) W T(l) Σp(θ 0 ) ´ ³ p − p (θ0 )) × diag p (θ0 )−1 (b °´ °´ ³° (l) ³° (l+1) ° ° °b °b − o − θ − θ + o °θ θ ° ° 0 0° . φ2 φ2

If we denote

A(i) ≡ ×

´ ³ ´−1 ³ diag p (θ0 )−1/2 Σp(θ 0 ) W (i) W T(i) Σp(θ 0 ) W (i) ´ ³ W T(i) Σp(θ 0 ) diag p (θ0 )−1/2 ; i = l, l + 1,

(7.19)

which is a symmetric matrix, the random vector Z can be written as ´√ ³ ¢ ¡ n (b p − p (θ0 )) Z = A(l+1) − A(l) diag p (θ0 )−1/2 °´ °´ ³° (l+1) ³° (l) °b ° °b ° − θ0 ° − o °θ + o °θ φ2 φ2 − θ 0 ° .

Thus,

´³ ³ ´ √ L b(l+1) ) − p(θ b(l) ) −→ ndiag p (θ0 )−1/2 p(θ N (0, Σ∗ ) , φ2 φ2 n→∞

where

³ ´ ´ ³ ¢ ¡ A(l+1) − A(l) diag p (θ0 )−1/2 Σp(θ 0 ) diag p (θ0 )−1/2 ¢ ¡ × A(l+1) − A(l) µ ³ ´T ¶ ¡ ¡ ¢ ¢ 1/2 1/2 p (θ0 ) = A(l+1) − A(l) I − p (θ0 ) A(l+1) − A(l) ,

Σ∗ =

´T ³ and p (θ0 )1/2 = p1 (θ0 )1/2 , ..., pM (θ0 )1/2 . Then, because ³

can be written as

p (θ0 )1/2

´T

´ ³ diag p (θ0 )−1/2 Σp(θ 0 )

1T diag (p (θ0 )) − 1T p (θ0 ) p (θ0 )T = 0, where 1 = (1, ..., 1)T , we have, using Exercise 10, that ¢¡ ¢ ¡ Σ∗ = A(l+1) − A(l) A(l+1) − A(l) = A(l) − A(l+1) . © 2006 by Taylor & Francis Group, LLC

314


¡ ¢ Therefore the matrix A(l) − A(l+1) is symmetric and idempotent with ¢ ¡ trace A(l) − A(l+1) = dl − dl+1

and Z is asymptotically normal with mean vector 0 and variance-covariance ma¢ ¡ trix A(l) − A(l+1) . Applying Lemma 3, p. 57, in Ferguson (1996) (see Remark 2.6), we have that Z T Z is asymptotically chi-squared distributed with dl − dl+1 degrees of freedom. Finally, (l) Teφ1 ,φ2 ≡

³ ´ (l+1) (l) 2n L b b p( θ ), p( θ ) −→ χ2 . D φ φ2 φ2 n→∞ dl −dl+1 φ001 (1) 1

For the (h, φ)-divergence measures we have the following result: Theorem 7.2 Under the assumptions given in Theorem 7.1, the asymptotic null distribution of the (h, φ)-divergence test statistic, (l) Teφ1 ,φ2 ,h ≡

³ ´´ ³ 2n b(l+1) ), p(θ b(l) ) , p( θ h D φ φ φ 1 2 2 φ001 (1) h0 (0)

is chi-square with dl −dl+1 degrees of freedom; l = 1, ..., m−1, where h is a differentiable function mapping from [0, ∞) onto [0, ∞) , with h (0) = 0 and h0 (0) > 0. Proof. We know ³ ´ 00 ¡ ¢ b(l+1) ), p(θ b(l) ) = φ1 (1) Z T Z + oP n−1 , Dφ1 p(θ φ2 φ2 2n

where Z T Z is asymptotically chi-squared distributed with dl − dl+1 degrees of freedom. Further, because h (x) = h (0) + h0 (0) x + o (x) , we have L (l) Teφ1 ,φ2 ,h = Z T Z + oP (1) −→ χ2dl −dl+1 . n→∞

We can observe that in the two previous theorems the estimated model in the null hypothesis appears in the left argument of the φ-divergence. It is usual to consider the null hypothesis in the right argument of the φ-divergence; for this reason we present the following theorem.



315

Theorem 7.3 Under the assumptions of Theorems 7.1 and 7.2, the asymptotic null distribution of the φ-divergence test statistics, (l)

Tφ1 ,φ2 = and (l)

Tφ1 ,φ2 ,h =

³ ´ (l) (l+1) 2n b b p( θ ), p( θ ) D φ φ2 φ2 φ001 (1) 1

³ ´´ ³ 2n b(l) ), p(θ b(l+1) ) , p( θ h D φ φ2 φ2 1 φ001 (1) h0 (0)

(7.20)

(7.21)

is chi-square with dl −dl+1 degrees of freedom; l = 1, ..., m−1, where h is a differentiable function mapping from [0, ∞) onto [0, ∞) , with h (0) = 0 and h0 (0) > 0. ¡ ¢ Proof. We consider the function ϕ (x) = xφ1 x−1 . It is clear that ϕ (x) ∈ (l) (l) (l) (l) Φ∗ , Teϕ,φ2 = Tφ1 ,φ2 and Teϕ,φ2 ,h = Tφ1 ,φ2 ,h . Then the result follows directly from Theorems 7.1 and 7.2. Remark 7.1 A well known test statistic appears when we choose φ1 (x) = φ2 (x) = x log x − x + 1 (l)

in the test statistic Tφ1 ,φ2 given in (7.20). It is the classical likelihood ratio test statistic (Agresti, 1996, p. 197; Christensen, 1997, p. 322) and it holds the well known result,

(i)

(l)

k X

b L b(l) ) log pj (θ ) −→ G ≡ 2n pj (θ χ2 , (l+1) n→∞ dl −dl+1 b pj (θ ) j=1 2

b is the maximum likelihood estimator of θ under the model Hi (θ ∈ Θi ) ; where θ i = l, l + 1.

Another important test statistic appears when we put φ1 (x) = 12 (1 − x)2 , (l) φ2 (x) = x log x − x + 1 in the test statistic Tφ1 ,φ2 given in (7.20). Then we obtain the chi-square test statistic given in Agresti (1996, p. 197), as well as the result,

X2 ≡ n


´2 ³ b(l) ) − pj (θ b(l+1) ) k pj (θ X j=1

b pj (θ

(l+1)

)

L

−→ χ2dl −dl+1 .

n→∞

316


But this is not the only family of test statistics for testing nested sequences of loglinear models given in (7.17) based on φ-divergence measures. In the following theorem another family of test statistics based on this measure is proposed. Theorem 7.4 Suppose that data (N1 , ..., Nk ) are multinomially distributed according to (7.7) and (7.8). Consider the nested sequence of hypotheses given by (7.15) and (7.16). Choose the two functions φ1 , φ2 ∈ Φ∗ . Then, for testing hypotheses, HNull : Hl+1 against HAlt : Hl ;

l = 1, ..., m − 1,

the φ-divergence test statistics, ³ ´ ´´ (l) 2n ³ ³ (l) b(l+1) b b, p(θ b ) − D ) , Sφ = 00 Dφ p p , p( θ φ φ φ φ (1)

(7.22)

and

(l)

Sφ,h =

´´ ³ ³ ´´´ ³ ³ ³ (l+1) (l) 2n b b b b p , p( θ p , p( θ ) − h D ) h D φ φ φ φ φ00 (1) h0 (0)

(7.23)

are nonnegative and their asymptotic null distribution is chi-square with dl − dl+1 degrees of freedom; l = 1, ..., m − 1, where h is a differentiable increasing function mapping from [0, ∞) onto [0, ∞) , with h (0) = 0 and h0 (0) > 0. (l)

Proof. It is clear that Sφ ≥ 0; l = 1, ..., m − 1, because

³ ³ ´ ´ b(l+1) ) = inf Dφ (b b(l) ) . b, p(θ b, p(θ p, p (θ)) ≥ inf Dφ (b p, p (θ)) = Dφ p Dφ p φ φ θ∈Θl+1

θ∈Θl

(l)

(l)

The proof of the asymptotic distribution of the test statistics Sφ and Sφ,h follows the same steps as the proof given in Theorems 7.1 and 7.2. Remark 7.2 The asymptotic result of Theorem 7.4 can be generalized further to include a b(i) . That is, the statistic φ1 for divergence Dφ , and a φ2 for estimation θ 1

(l)

Sφ1 ,φ2 ≡

under Hl+1 .

φ2

³ ³ ´ ´´ (l+1) (l) 2n ³ L b b b b ) − D ) −→ χ2 D p , p( θ p , p( θ φ1 φ1 φ2 φ2 n→∞ dl −dl+1 φ001 (1)

The special case of φ1 (x) = 12 (1 − x)2 , φ2 (x) = x log x − x + 1 yields a statistic based on the difference of chi-square test statistic with maximum likelihood



317

estimation used to obtain the expected frequencies (e.g., Agresti, 1996, p. 197), namely ³ ´ ´´ ³ ³ b(l+1) ) − D 1 b(l) ) . b, p(θ b n D 1 (1−x)2 p p , p( θ 2 (1−x) 2

2

(l)

However, the nonnegativity of Sφ1 ,φ2 does not hold when φ1 6= φ2 . Thus, for the case above, considered by Agresti, the difference of the chi-square test statistics is not necessarily nonnegative. Since it is common to use maximum (l) likelihood estimation (that is, φ2 (x) = x log x − x + 1), the test statistic Sφ1 ,φ2 ; φ1 6= φ2 is not all that interesting to us. In the following we shall concentrate on (l) the statistics Tφ1 ,φ2 . For testing the nested hypotheses {Hl : l = 1, ..., m} given by (7.15), we test (l) HNull : Hl+1 against HAlt : Hl , using the φ-divergence test statistic Tφ1 ,φ2 given (l)

by (7.20); if it is too large, HNull is rejected. When Tφ1 ,φ2 > c, we reject HN ull in (7.17), where c is specified so that the significance level of the test is α : ´ ³ (l) (7.24) Pr Tφ1 ,φ2 > c | Hl+1 = α; α ∈ (0, 1) . Theorem 7.1 was shown that under (7.7), (7.8) and (7.15), and HNull : Hl+1 , (l) the test statistic Tφ1 ,φ2 converges in distribution to a chi-square distribution with dl − dl+1 degrees of freedom; l = 1, ..., m − 1. Therefore, c could be chosen as the 100 (1 − α) percentile of a chi-square distribution with dl − dl+1 degrees of freedom, c = χ2dl −dl+1 ,α . (7.25) The choice of (7.25) in (7.24) only guarantees an asymptotic size-α test. Here we use (7.25) but ask, in the finite-sample simulations given in Section 5, for (l) what choices of λ in Tφ1 (λ),φ2 (λ) is the relation (7.24) most accurately attained? The asymptotic chi-square approximation, c = χ2dl −dl+1 ,α , is checked for a sequence of loglinear models in the simulation study given in Section 7.5. We give a small illustration of those results now. Figures 7.1 and 7.2 show departures of the exact simulated size from the nominal size of α = 0.05 for one particular choice (specified in Section 7.5) of Hl+1 and Hl , for various choices of λ in φ1 = φ(λ) , and for small to large sample sizes (n = 15, 20, 25, 35, 50, 100, 200). Figure 7.1 represents nonpositive choices of λ, and Figure 7.2 represents nonnegative choices of λ. The positive values of λ perform the best.


l

318


0.1 .08 .06 .04 .02

0

2.5

3

3.5

x

4

4.5

5

02

Figure 7.1. (Exact size − Nominal size of 0.05) as a function of x = log n. Shown are λ = −2 (dashed line), λ = −1 (dotted line), λ = −1/2 (dashdotted line) and λ = 0 (solid line). Source: Cressie, N., Pardo, L. and Pardo, M.C. (2003).

0.1 .08 .06 .04 .02

0

2.5

3

3.5

x

4

4.5

5

02

Figure 7.2. (Exact size − Nominal size of 0.05) as a function of x = log n. Shown are λ = 0 (solid line), λ = 2/3 (dashed line), λ = 1 (dotted line) and λ = 2 (dash-dotted line). Source: Cressie, N., Pardo, L. and Pardo, M.C. (2003).

To test the nested sequence of hypotheses {Hl : l = 1, ..., m} effectively, we



319 (1)

need an asymptotic independence result for the sequence of test statistics Tφ1 ,φ2 , (m∗ )

(2)

Tφ1 ,φ2 , ..., Tφ1 ,φ2 , where m∗ is the integer 1 ≤ m∗ ≤ m for which Hm∗ is true but Hm∗ +1 is not true. This result is given in the theorem below. Notice that our hypothesis-testing strategy is an attempt to find this value of m∗ ; we infer it to be l∗ , the first in the sequence of hypothesis tests for which Hl∗ +1 is rejected as a null hypothesis. Theorem 7.5 Suppose that data (N1 , ..., NM ) are multinomially distributed according to (7.7) and (7.8). Suppose we wish to test first HNull : Hl against HAlt : Hl−1 , followed by HNull : Hl+1 against HAlt : Hl . Then, under the null hypothesis Hl , the φ-divergence test statistics (l−1)

Tφ1 ,φ2

and

(l)

Tφ1 ,φ2

are asymptotically independent and chi-squared distributed with dl−1 − dl and dl − dl+1 degrees of freedom, respectively. Proof. A similar development to the one presented in Theorem 7.1 gives (l)

Tφ1 ,φ2 = and (l−1)

Tφ1 ,φ2 = where

√ √ n (b p − p (θ0 ))T M Tl M l n (b p − p (θ0 )) + oP (1)

√ √ n (b p − p (θ0 ))T M Tl−1 M l−1 n (b p − p (θ0 )) + oP (1) ,

´ ³ ¡ ¢ M i = A(i+1) − A(i) diag p (θ0 )−1/2 ; i = l − 1, l,

A(i) is given in (7.19) and W (i) is the loglinear model matrix of explanatory variables under the ith-loglinear model; i = l − 1, l, l + 1. Now because

¡ ¢ √ L n (b p − p (θ0 )) −→ N 0, Σp(θ 0 ) , n→∞

(l)

(l−1)

from Theorem 4 in Searle (1971, p. 59), Tφ1 ,φ2 and Tφ1 ,φ2 are asymptotically independent if P (l) ≡ M Tl−1 M l−1 Σp(θ 0 ) M Tl M l = 0.


320


We have

µ ¶ q p ¡ ¢ ¢ T ¡ A(l) − A(l−1) I − p (θ0 ) p (θ0 ) A(l+1) − A(l) M l P (l) = ¡ ¢¡ ¢ = M Tl−1 A(l) − A(l−1) A(l+1) − A(l) M l , p since A(i) p (θ0 ) = 0; i = l − 1, l, l + 1. M Tl−1

Applying Exercise 10, we have

A(i) A(i+1) = A(i+1) A(i) = A(i+1) ; A(i) A(i) = A(i) ; i = l − 1, l, l + 1, therefore M Tl−1 M l−1 Σp(θ 0 ) M Tl M l = 0. Similar results to one obtained in this theorem can be obtained for the φ(l) (l) (l) (l) (l) (l) divergence test statistics Teφ1 ,φ2 , Tφ1 ,φ2 ,h , Teφ1 ,φ2 ,h , Sφ , Sφ,h and Sφ1 ,φ2 . (l)

In general, theoretical results for the test statistic Tφ1 ,φ2 under alternative hypotheses are not easy to obtain. An exception to this is when there is a contiguous sequence of alternatives that approaches the null hypothesis Hl+1 at the ¢ ¡ rate of O n−1/2 . Regarding the alternative, Haberman (1974) was the first to study the asymptotic distribution of the chi-square test statistic and likelihood ratio test statistic under contiguous alternative hypotheses, establishing that the asymptotic distribution is noncentral chi-square with dl −dl+1 degrees of freedom. Oler (1985) presented a systematic study of the contiguous alternative hypotheses in multinomial populations, obtaining as a particular case the asymptotic distribution for the loglinear models. Through simulations, she also studied how closely the noncentral chi-square distributions agree with the exact sampling distributions. Fenech and Westfall (1988) presented an interesting analytic study of the noncentrality parameter in the case of loglinear models. Now we generalize their (l) results to tests based on the φ-divergence test statistic Tφ1 ,φ2 given by (7.20). Consider the multinomial probability vector √ pn ≡ p (θ0 ) + d/ n; θ0 ∈ Θl+1 ,

(7.26)

P where d ≡ (d1 , ..., dM )T is a fixed M × 1 vector such that M j=1 dj = 0, and recall that n is the total-count parameter of the multinomial distribution. As n → ∞, the sequence of multinomial probabilities {pn }n∈N converges to a multinomial ¢ ¡ probability in Hl+1 at the rate of O n−1/2 . We call √ Hl+1,n : pn = p (θ0 ) + d/ n; θ0 ∈ Θl+1 (7.27)



321

a sequence of contiguous alternative hypotheses, here contiguous to the null hypothesis Hl+1 . Now consider the problem of testing HNull : Hl+1 against HAlt : Hl+1,n , (l)

using the φ-divergence test statistics Tφ1 ,φ2 given by (7.20). The power of this test is ´ ³ (l) πn(l) ≡ Pr Tφ1 ,φ2 > c |Hl+1,n . (7.28) (l)

In what follows, we show that under the alternative Hl+1,n , and as n → ∞, Tφ1 ,φ2 converges in distribution to a noncentral chi-square random variable with noncentrality parameter δ, where δ is given in Theorem 7.6, and dl − dl+1 degrees of freedom (χ2dl −dl+1 (δ)). Consequently, as n → ∞, ³ ´ πn(l) → Pr χ2dl −dl+1 (δ) > c . (7.29) In (7.10), it was established that the asymptotic expansion of the minimum φ-divergence estimator about θ0 ∈ Θl+1 is given by ´−1 ´ ³ ³ −1 T b(l+1) = θ0 + W T Σp(θ ) W (l+1) W Σ diag p (θ ) θ 0 φ (l+1) (l+1) p(θ 0 ) 0 p − p (θ0 )k) × (b p − p (θ0 )) + o (kb (7.30) where W l+1 is the loglinear-model matrix of explanatory variables under the null hypothesis Hl+1 . Under the hypothesis given in (7.27), we have √ √ n (b p − p (θ0 )) = n (b p − pn ) + d, and hence

¡ ¢ √ L n (b p − p (θ0 )) −→ N d, Σp(θ 0 ) , n→∞

so

³ ´´ ³ ´ ³ o (kb p − p (θ0 )k) = o OP n−1/2 = oP n−1/2 .

(7.31)

Therefore, we have established that under the contiguous alternative hypotheses given in (7.27), and for θ0 ∈ Θl+1 , ´−1 ´ ³ ³ −1 T b(l+1) = θ0 + W T Σp(θ ) W (l+1) W Σ diag p (θ ) θ 0 φ (l+1) (l+1) p(θ 0 ) ¡0 ¢ × (b p − p (θ0 )) + oP n−1/2 . (7.32)


322


This result will be important in the following theorem. Theorem 7.6 Suppose that data (N1 , ..., NM ) are multinomially distributed according to (7.7) and (7.8). The asymptotic distribution of the φ-divergence test statis(l) tic Tφ1 ,φ2 , under the contiguous alternative hypotheses (7.27), is chi-square with dl − dl+1 degrees of freedom and noncentrality parameter δ given by ´¡ ´ ³ ³ ¢ δ = dT diag p (θ0 )−1/2 A(l) − A(l+1) diag p (θ0 )−1/2 d, where d = (d1 , ..., dM )T is defined in (7.26) and satisfies i = l, l + 1, is given in (7.19).

PM

i=1 di

= 0, and A(i) ,

Proof. By Theorem 7.1, we know that µ° °2 ° °2 ¶ ° b(l+1) ° ° b(l) ° (l) T Tφ1 ,φ2 = Z Z + n o °p(θφ2 ) − p (θ0 )° + °p(θφ2 ) − p (θ0 )° , where

Z= But (l+1)

b p(θ φ2

´³ ³ ´ √ b(l+1) ) − p(θ b(l) ) . n diag p (θ0 )−1/2 p(θ φ2 φ2

) − p (θ0 ) =

and (7.32) we have

°´ ³° ∂p (θ0 ) b(l+1) ° b(l+1) ° ) − p (θ ) (θφ2 − θ0 ) + o °p(θ 0 ° , φ2 ∂θ (l+1)

and

b p(θ φ2

³ ´ ) − p (θ0 ) = OP n−1/2

° °2 ¡ ¢ ° b(l+1) ° °p(θφ2 ) − p (θ0 )° = OP n−1 .

In a similar way and taking into account that θ0 ∈ Θl+1 ⊂ Θl , it can be obtained that °2 ° ¡ ¢ ° ° b(l) °p(θ ) − p (θ0 )° = OP n−1 . φ2

Then

(l)

Tφ1 ,φ2 = Z T Z + oP (1) .

From (7.32), we have, under the contiguous alternative hypotheses, that ³ ´ ¢ √ ¡ Z = n A(l+1) − A(l) diag p (θ0 )−1/2 (b p − p (θ0 )) + oP (1) . © 2006 by Taylor & Francis Group, LLC


By (7.31)

323

¡ ¢ √ L n (b p − p (θ0 )) −→ N d, Σp(θ 0 ) , n→∞

and hence

L

Z −→ N (δ, Σ∗ ) , n→∞ ³ ´ ¡ ¢ where δ = A(l+1) − A(l) diag p (θ0 )−1/2 d and

³ ´ ´ ³ ¢ ¡ A(l+1) − A(l) diag p (θ0 )−1/2 Σp(θ 0 ) diag p (θ0 )−1/2 ¢ ¡ × A(l+1) − A(l) µ ¶ q p ¡ ¢ ¢ T ¡ = A(l+1) − A(l) I − p (θ0 ) p (θ0 ) A(l+1) − A(l) .

Σ∗ =

Using the results in the proof of Theorem 7.1, it can be shown that ¢ ¡ Σ∗ = A(l) − A(l+1) ,

and it is an idempotent matrix of rank (dl − dl+1 ).

If we establish that Σ∗ µ = µ, the theorem follows from the Lemma on page 63 in Ferguson (1996), because in this case the noncentrality parameter is given by δ = µT µ. Applying the results obtained in Exercise 10, we have ¡ ¢ Σ∗ µ = A(l) − A(l+1) µ = A(l) µ ³ − A(l+1) µ´ ¡ ¢ ¡ ¢ = A(l) A(l+1) − A(l) diag p (θ0 )−1/2 d − A(l+1) A(l+1) − A(l) ´ ³ × diag p (θ0 )−1/2 d = µ.

Then the noncentrality parameter δ is given by ´¡ ´ ³ ³ ¢ δ = µT µ = dT diag p (θ0 )−1/2 A(l) − A(l+1) diag p (θ0 )−1/2 d. Remark 7.3 Theorem 7.6 can be used to obtain an approximation to the power function of the test (7.17), as follows. Write (l+1)

p(θ(l) ) = p(θ0


´´ 1 ³√ ³ (l+1) )+ √ n p(θ(l) ) − p(θ0 ) , n

324


and define (l+1)

pn (θ(l) ) ≡ p(θ0

1 ) + √ d, n

´ √ ³ (l+1) where d = n p(θ(l) ) − p(θ0 ) . Then substitute d into the definition of δ, and finally δ into the right hand side of (7.29). The asymptotic noncentral chi-square approximation for power is checked for finite samples in the simulation given in Section 7.5. Figures 7.3 and 7.4 show departures of the exact power from the asymptotic power for one particular choice (specified in Section 7.5) of Hl+1 and Hl , for various choices of λ in φ1 = φ(λ) , and for small to large sample sizes (n = 15, 20, 25, 35, 50, 100, 200). Figure 7.3 represents nonpositive choices of λ and Figure 7.4. represents nonnegative choices of λ. These figures need to be interpreted in light of associated exact sizes; see Section 7.5. However, it is immediately apparent that from an asymptoticapproximation point of view, λ = 2/3 seems to perform the best, particularly for small and moderate sample sizes.

0.1 .08 .06 .04 .02

0

2.5

3

3.5

x

4

4.5

5

02

Figure 7.3. (Exact Power − Asymptotic power) as a function of x = log n. Shown are λ = −2 (dashed line), λ = −1 (dotted line), λ = −1/2 (dashdotted line) and λ = 0 (solid line). Source: Cressie, N., Pardo, L. and Pardo, M.C. (2003).



325

0.1 .08 .06 .04 .02

0

2.5

3

3.5

x

4

4.5

5

02

Figure 7.4. (Exact Power − Asymptotic power) as a function of x = log n. Shown are λ = 0 (solid line), λ = 2/3 (dashed line), λ = 1 (dotted line) and λ = 2 (dash-dotted line). Source: Cressie, N., Pardo, L. and Pardo, M.C. (2003).

7.5.

Simulation Study

Section 7.4 presents theoretical results for testing hypotheses in loglinear models. The results give asymptotic distribution theory for the φ-divergence test sta(l) tistic Tφ1 ,φ2 under the null hypothesis and a sequence of contiguous alternative hypotheses. The appropriateness of these results in finite samples is demonstrated in Figures 7.1, 7.2, 7.3 and 7.4. We now describe the finite-sample simulation study from which these figures were obtained, and we give new results that com(l) pare the powers of tests based on Tφ(λ) ,φ(0) for λ = −2, −1, −1/2, 0, 2/3, 1, 2. We consider for the estimation problem the maximum likelihood estimator and for the testing problem the power-divergence test statistics with the values of λ given previously. Consider a 2×2×2 contingency table, so M = 8. We simulate data N1 , ..., NM from a multinomial distribution with sample size n and probability vector p = (p1 , ..., pM )T , where n and p are specified. The motivation for our simulation study comes from a similar one carried out by Oler (1985). The simulation study, presented in this Section, was carried out in Cressie et al . (2003). For


326


(l)

the moment, fix l and consider the power-divergence test statistics Tφ(λ) ,φ(0) for testing HN ull : Hl+1 against HAlt : Hl+1,n , and let p0 ∈ HNull and p1,n ∈ HAlt , where p1,n is a subscript with n because its entries may depend on n. The essence of our simulation study is to obtain the exact probabilities, ´ ³ (l) (l) αn ≡ Pr Tφ(λ) ,φ(0) > c | p0 (l) πn (l)

(l)

≡ Pr

³

(l) Tφ(λ) ,φ(0)

´

(7.33)

> c | p1,n .

In fact, αn and πn are estimated using N = 100000 simulations from the multinomial sampling schemes (n, p0 ) and (n, p1,n ), respectively. For a given p0 (see below), the various choices of n and p1,n represent the design of our simulation study. We choose n = 15, 20, 25, 35, 50, 100, 200 to represent small, moderate and large sample sizes. We simulate multinomial random vectors (N1 , ..., NM ) and compute proba(l) (l) bilities αn for (n, p0 ) and πn for (n, p1,n ). To see what happens for contiguous alternatives, we fix p1 ∈ Hl (see below) and define p∗1,n ≡ p0 + (25/n)1/2 (p1 − p0 ) .

(7.34)

Notice that p∗1,25 = p1 and, as n increases, p∗1,n converges to p0 at the rate ª © n−1/2 ; that is, p∗1,n is a sequence of contiguous alternatives. Our design for the simulation study is to choose (n, p1,n ) as fixed and contiguous alternatives, which we now give. Contiguous alternatives: © ª (n, p∗1,n ) : n = 15, 20, 25, 35, 50, 100, 200 ,

where p∗1,n is given by (7.34) and p1 is specified below. Fixed alternatives:

{(n, p1 ) : n = 15, 20, 25, 35, 50, 100, 200} ,



327

where p1 is specified below. Notice that for n < 25, the contiguous alternatives are further from HN ull than are the fixed alternatives and that the two sequences share the alternative (25, p1 ). These choices allow reasonable coverage of the space of alternatives. In the simulation study, we consider the following nested sequence of loglinear models for the 2 × 2 × 2 table: ª © H1 : pijk (θ) = exp u + θ1(i) + θ2(j) + θ3(k) + θ12(ij) + θ13(ik) + θ23(jk) ª © H2 : pijk (θ) = exp u + θ1(i) + θ2(j) + θ3(k) + θ12(ij) + θ13(ik) ª © H3 : pijk (θ) = exp u + θ1(i) + θ2(j) + θ3(k) + θ12(ij) © ª H4 : pijk (θ) = exp u + θ1(i) + θ2(j) + θ3(k) .

Here, exp (−u) is the normalizing constant and the subscript θ-terms add to zero over each of their indices. Based on Oler’s study, we used a moderate value for each main effect and a small value for the interactions. That is, we used ¡ ¢ ¡ ¢ ¡ ¢ exp θ1(1) = exp θ2(1) = exp θ3(1) = 5/6

¢ ¡ ¢ ¡ ¢ ¡ exp θ12(11) = exp θ13(11) = exp θ23(11) = 9/10. Then the simulation experiment is designed so that p1 ∈ Hl ; l = 1, 2, 3. In Section 7.4, we showed Figures 7.1 and 7.2 (Exact size-Nominal size) for (3) HNull : H4 , using the test statistic Tφ(λ),φ(0) and c = χ21,0.05 as well as Figure 7.3 and 7.4 (Exact power − Asymptotic power) for HNull : p0 ∈ H4 and HAlt : p∗1,n ; (3)

with p1 ∈ H3 , using the test statistic Tφ(λ),φ(0) and c = χ21,0.05 .

In the simulation study, we shall compare members of the power-divergence family of test statistics; our criteria for a good performance are:

i) good exact power and size for small to moderate sample sizes. For this, we consider the following three hypothesis tests with fixed alternatives: HNull : p0 ∈ Hl+1 versus HAlt : (n, p1 ) , where p1 ∈ Hl and n = 15, 20, 25, 35; l = 1, 2, 3.


328


ii) good agreement of exact and asymptotic probabilities for reasonably small and moderate sample sizes. For this we consider the following three hypothesis tests with contiguous alternatives: ¢ ¡ HNull : p0 ∈ Hl+1 versus HAlt : n, p∗1,n , where p∗1,n is given by (7.34), p1 ∈ Hl and n = 15, 20, 25, 35; l = 1, 2, 3.

First of all, we study the closeness of the exact size for HN ull : H2 , HNull : H3 and HNull : H4 to the nominal size α = 0.05. Following Dale (1986), we consider the inequality, ¯ ¯ ¯ ¯ ) − logit(1 − α) ¯logit(1 − α(l) ¯ ≤ e, n

(7.35)

where logit(p) ≡ ln(p/(1 − p)). The two probabilities are considered to be “close” if they satisfy (7.35) with e = 0.35 and “fairly close” if they satisfy (7.35) with (l) e = 0.7. Note that for α = 0.05, e = 0.35 corresponds to αn ∈ [0.0357, 0.0695] , (l) and e = 0.7 corresponds to αn ∈ [0.0254, 0.0959] . From the calculations that yield Figures 7.5-7.7, the test statistics that satisfy (7.35) for e = 0.35 are those corresponding to λ = 2/3, 1, 2. For e = 0.7, only one extra test statistic, that corresponding to λ = 0, is added. Figures 7.8-7.10 give a similar comparison of exact to asymptotic, this time for power under a contiguous alternative. In Figures 7.8 and 7.9, it is clear that the test statistic corresponding to λ = 2/3 has the best behavior. In Figure 7.10, λ = 1, 2 are the best but λ = 2/3 is still competitive. Figures 7.11-7.13 show (Exact power-Exact size) for the three hypothesis tests, H4 versus H3 , H3 versus H2 , and H2 versus H1 . This is a measure of how quickly the power curve increases from its probability of type I error. We see from the figures that the increase in power is a little more for tests based on negative λ than for positive λ. This should be tempered with the fact that for negative λ the exact size is considered not even “fairly close”. This trade-off between size behavior and power behavior is a classical problem in hypothesis testing.



329

1.4 1.2 1 0.8 0.6 0.4 0.2 -2

-1

0 0

1 x

2

Figure 7.5. |logit(1 − Exact size) − logit(1 − Nominal size of .05)| as a function of λ for model H4 . Shown are n = 15 (dashed line), n = 20 (dotted line), n = 25 (dash-dotted line) and n = 35 (solid line). The two horizontal lines correspond to Dale’s bounds of e = .35 and e = .7 in (7.35). Source: Cressie, N., Pardo, L. and Pardo, M.C. (2001).

1.4 1.2 1 0.8 0.6 0.4 0.2 -2

-1

0 0

1 x

2

Figure 7.6. |logit(1 − Exact size) − logit(1 − Nominal size of .0.5)| as a function of λ for model H3 . Shown are n = 15 (dashed line), n = 20 (dotted line), n = 25 (dash-dotted line) and n = 35 (solid line). The two horizontal lines correspond to Dale’s bounds of e = .35 and e = .7 in (7.35). Source: Cressie, N., Pardo, L. and Pardo, M.C. (2001).


330


1.4 1.2 1 0.8 0.6 0.4 0.2 -2

0 0

-1

1 x

2

Figure 7.7. |logit(1 − Exact size) − logit(1 − Nominal size of .05)| as a function of λ for model H2 . Shown are n = 15 (dashed line), n = 20 (dotted line), n = 25 (dash-dotted line) and n = 35 (solid line). The two horizontal lines correspond to Dale’s bounds e = .35 and e = .7 in (7.35). Source: Cressie, N., Pardo, L. and Pardo, M.C. (2001).

0.1 0.08 0.06 0.04 0.02

-2

0 0

-1

1 x

2

-0.02

Figure 7.8. (Exact Power − Asymptotic power) as a function of λ for testing p0 ∈ H4 versus p∗1,n with p1 ∈ H3 . Shown are n = 15 (dashed line), n = 20 (dotted line), n = 25 (dash-dotted line) and n = 35 (solid line). Source: Cressie, N., Pardo, L. and Pardo, M.C. (2001).



0.1 0.08 0.06 0.04 0.02

-2

0 0

-1

1 x

2

-0.02

Figure 7.9. (Exact Power − Asymptotic power) as a function of λ for testing p0 ∈ H3 versus p∗1,n with p1 ∈ H2 . Shown are n = 15 (dashed line), n = 20 (dotted line), n = 25 (dash-dotted line) and n = 35 (solid line). Source: Cressie, N., Pardo, L. and Pardo, M.C. (2001).

0.1 0.08 0.06 0.04 0.02

-2

0 0

-1

1 x

2

-0.02

Figure 7.10. (Exact Power − Asymptotic power) as a function of λ for testing p0 ∈ H2 versus p∗1,n with p1 ∈ H1 . Shown are n = 15 (dashed line), n = 20 (dotted line), n = 25 (dash-dotted) and n = 35 (solid line). Source: Cressie, N., Pardo, L. and Pardo, M.C. (2001).


331

332


0.06 0.05 0.04 0.03 0.02 0.01

-2

0 0

-1

1 x

2

Figure 7.11. (Exact Power − Exact size) as a function of λ for testing H4 versus H3 . Shown are n = 15 (dashed line), n = 20 (dotted line), n = 25 (dash-dotted line) and n = 35 (solid line). Source: Cressie, N., Pardo, L. and Pardo, M.C. (2001).

0.06 0.05 0.04 0.03 0.02 0.01

-2

-1

0 0

1 x

2




333

0.06 0.05 0.04 0.03 0.02 0.01

-2

-1

0 0

1 x

2


In what follows, we consider only the test power-divergence test statistic that satisfy (7.34) with e = 0.7, and to discriminate between them we calculate: ¯ ¯ ¯ ¯ (l) (l) g1 (λ) ≡ ¯APi,n (λ) − SEPi,n (λ)¯

and

³ ´−1 (l) (l) , g2 (λ) ≡ SEPi,n (λ) − ST Si,n (λ) (l)

(l)

where APi,n (λ) is the asymptotic power, SEPi,n (λ) is the simulated exact power, (l)

(l)

and ST Si,n (λ) is the simulated test size of the test statistic Tφ(λ),φ(0) ; l = 1, 2, 3, under the alternative i = F (fixed), C (contiguous) and n = 15, 20, 25, 35. Then, (l) for a given l, we consider a test statistic Tφ(λ1 ),φ(0) to be better than a test statistic (l)

Tφ(λ2 ),φ(0) iff g1 (λ1 ) < g1 (λ2 ) and g2 (λ1 ) < g2 (λ2 ) .

(7.36)

In Figures 7.14-7.16, we plot y = g2 (λ) versus x = g1 (λ) ; from (7.36), we look for values of λ that are as close to (0, 0) as possible in the (x, y) plane.


334


100 80 60 40 20

0 0

0.01

0.02 x

0.03

0.04

(3)

Figure 7.14. y = g2 (λ) versus x = g1 (λ) for Tφ(λ),φ(0) . Shown are λ = 0 (Square), λ = 2/3 (Cross), λ = 1 (Diamond) and λ = 2 (Circle). Source: Cressie, N., Pardo, L. and Pardo, M.C. (2001).

100 80 60 40 20

0 0

0.01

0.02 x

0.03

0.04 (2)

Figure 7.15. y = g2 (λ) versus x = g1 (λ) for Tφ(λ),φ(0) . Shown are λ = 0 (Square), λ = 2/3 (Cross), λ = 1 (Diamond) and λ = 2 (Circle). Source: Cressie, N., Pardo, L. and Pardo, M.C. (2001).

The points (g1 (λ) , g2 (λ)) far away from (0, 0) are those corresponding to smallest sample size n = 15, as expected. For this sample size, Figures 7.14-7.16 show that the (l) (l) exact size of the tests based on Tφ(0),φ(0) is too large in relation to that of Tφ(λ),φ(0) ,



335

λ = 2/3, 1, 2. For n = 15, the points (g1 (2/3) , g2 (2/3)) are closer to (0, 0) than the points (g1 (1) , g2 (1)) and (g1 (2) , g2 (2)) . Thus, according to the criterion (7.36), the (l) test based on Tφ(2/3),φ(0) is the best for n = 15. 100 80 60 40 20

0 0

0.01

0.02 x

0.03

0.04 (1)

Figure 7.16. y = g2 (λ) versus x = g1 (λ) for Tφ(λ),φ(0) . Shown are λ = 0 (Square), λ = 2/3 (Cross), λ = 1 (Diamond) and λ = 2 (Circle). Source: Cressie, N., Pardo, L. and Pardo, M.C. (2001). (l)

(l)

For n = 20, 25, 35, it can be seen that Tφ(2/3),φ(0) is better than Tφ(λ),φ(0) , for λ = (l)

(l)

1, 2, according to (7.36). However, Tφ(2/3),φ(0) is not obviously better than Tφ(0),φ(0) , (l)

(l)

since g1 (2/3) < g1 (0) but g2 (0) < g2 (2/3) . The choice of Tφ(2/3),φ(0) or Tφ(0),φ(0) for n = 20, 25, 35, is going to depend on whether we need to make a very good approximation of the exact power (g1 small) or whether we want to use a test statistic with as much (l) exact power as possible (g2 small). In the first instance, we should choose Tφ(2/3),φ(0) (l)

and in the second instance, Tφ(0),φ(0) . From all the simulation studies we have carried out, our conclusion is that the test (l) based on Tφ(2/3),φ(0) is a very good and often better alternative to the tests based on the (l)

classical statistics Tφ(λ),φ(0) with λ = 0, 1.


336

7.6.


Exercises

1. Find the matrix form of the loglinear models, a) log mij (θ) = u + θi + θj + θij , i, j = 1, 2, 3, with 3 X

θi = 0,

i=1

3 X

θij = 0, j = 1, 2, 3, and θij = θji , i, j = 1, 2, 3.

i=1

This is the symmetry model. b) log mij (θ) = u + θ1(i) + θ2(j) + θ12(ij) , i, j = 1, 2, 3, with 3 X

θ1(i) = 0,

i=1

3 X i=1

3 X

θ2(j) = 0

j=1

θ12(ij) = 0, j = 1, 2, 3 and θ12(ij) = θ12(ji) , i, j = 1, 2, 3, i 6= j.

This is the quasi-symmetry model. 2. Find the matrix form for the loglinear models a) log mij (θ) = u + θ1(i) + θ2(j) + δi cij , i, j = 1, 2, with 2 X

θ1(i) =

i=1

2 X

θ2(j) =

j=1

2 X

δi = 0

i=1

and cij known constants. b) log mij (θ) = u + θ1(i) + θ2(j) + ρj dij , i, j = 1, 2, with 2 X

θ1(i) =

i=1

2 X

θ2(j) =

j=1

2 X

ρj = 0

j=1

and dij known constants. c) log mij (θ) = u + θ1(i) + θ2(j) + λ(lij )2 , i, j = 1, 2, with 2 X i=1

θ1(i) =

2 X

θ2(j) = 0

j=1

and lij known constants. 3. We consider the loglinear model of quasi-symmetry given in Exercise 1. Find the maximum likelihood estimator of the expected values, m (θ) . 4. Find the expression of the minimum power-divergence estimator for the expected values, m (θ) , in the symmetry model.



337

5. We consider the loglinear model, log mij (θ) = u + θ1(i) + θ2(j) + θ12(ij) with u= θ1(i) = θ2(j) =

1 IJ 1 J 1 I

J I P P

log mij

i=1 j=1 J P

j=1 I P

i=1

log mij − u

log mij − u

θ12(ij) = log mij − u − θ1(i) − θ2(j) .

Show that

θ12(ij) =

J I 1 XX mij (θ) mst (θ) . log IJ s=1 t=1 mit (θ) msj (θ)

6. The maximum likelihood estimators for mijk (θ) , i, j, k = 1, 2, in the model log mijk (θ) = u + θ1(i) + θ2(j) + θ3(k) + θ12(ij) + θ13(ik) + θ23(jk) , based on a random sample of size 820, are given in the following table Variable (C) C1

Variable (A)

Variable (B) A1 A2

B1 350.5 25.51

C2 B2 149.5 23.49

B1 59.51 19.49

B2 112.5 79.51

Find the maximum likelihood estimators for the parameters θ1(i) and θ12(ij) . 7. In 1968, 715 blue collar workers, selected from Danish Industry, were asked a number of questions concerning their job satisfaction. Some of these questions were summarized in a measure of job satisfaction. Based on similar questions the job satisfaction of the supervisors was measured. Also included in the investigation there was an external evaluation of the quality of management for each factory. The following table shows 715 workers distributed on the three variables: A: Own job satisfaction, B: Supervisors job satisfaction and C: Quality of management. A Bad C Good

B Low High Low High

Low 103 32 59 78

High 87 42 109 205

Source: Andersen, E. B. (1990, p. 156).

We consider the nested sequence of loglinear models H1 , H2∗∗ , H3 , H4 , H5 and H6 given in Section 7.1.


338


a) Find the maximum likelihood estimators of m (θ) for all the models. b) Using as significance level α = 0.05, test the null hypothesis that data are from model H4 based on the maximum likelihood estimator as well as on the power-divergence test statistic with λ = −2, −1, −1/2, 0, 2/3 and 1.

c) Find the best loglinear model among the models: H1 , H2∗∗ , H3 , H4 , H5 and H6 using the maximum likelihood estimator as well as the power-divergence (l) test statistics Tφ(λ),φ(0) with λ = 0, 1, −1 and taking as significance level α = 0.01. d) The same as in the previous part but with the power-divergence test statistic (l) Sφ(λ),φ(0) .

8. The Swedish traffic authorities investigated in 1961 and 1962 on a trial basis the possible effects of speed limitations. In certain weeks a speed limit of 90 km/hour was enforced, while in other weeks no limits were enforced. The following table shows for two periods of the same length, one in 1961 and one in 1962, the observed number of killed persons in traffic accidents on main roads and on secondary roads.

Speed limit (B)

Type of Roads (C) Main Secondaries 90Km 90Km Free Free (Hour) (Hour)

1961

8

57

42

106

1962

11

45

37

69

Year (A) Source: Andersen, E. B. (1990, p. 158).

We consider the nested sequence of loglinear-models H1 , H2 , H3 , H4 , H5 and H6 given in Section 7.1. a) Find the maximum likelihood estimators of m (θ) for all the models. b) We consider the model H3 . Find the maximum likelihood estimators for the parameters θ1(i) , θ2(j) , θ3(k) , θ23(jk) . c) Using as significance level α = 0.05, test the null hypothesis that data are from model H3 based on the maximum likelihood estimator as well as on the power-divergence test statistic with λ = −2, −1, −1/2, 0, 2/3, 1 and 2. d) Find the best loglinear model among the models: H1 , H2 , H3 , H4 , H5 and H6 using the maximum likelihood estimator as well as the power-divergence (l) test statistics Tφ(λ),φ(0) with λ = −2, −1/2, 0, 2/3, 1 and 2 assuming that for testing HNull : H3 versus HAlt : H2 the null hypothesis was accepted and taking as significance level α = 0.05.



339

e) The same as in the previous part but with the power-divergence test statistic (l) Sφ(λ),φ(0) with λ = 1. 9. In 1973 a retrospective study of cancer of the ovary was carried out. Information was obtained from 299 women, who were operated on for cancer of the ovary 10 years before. The following four dichotomous variables were observed: A: Whether X−ray treatment was received or not. B: Whether the woman had survived the operation by 10 years or not. C: Whether the operation was radical or limited. D: Whether the cancer at the time of operation was in an early or an advanced stage. The observed number of women are shown in the following table: A : X-ray D

C

(Stage)

(Operation)

Early

Radical Limited

Advanced

Radical Limited

B

NO

YES

No YES No YES NO YES NO YES

10 41 1 13 38 6 3 1

17 64 3 9 64 11 13 5

(Survival)

Source: Andersen, E. B. (1998, p. 121).

We consider the following nested sequence of loglinear models: H1 : log mijkl (θ) = u + θ1(i) + θ2(j) + θ3(k) + θ4(l) + θ12(ij) + θ13(ik) + θ14(il) + θ23(jk) + θ24(jl) + θ34(kl) + θ123(ijk) + θ124(ijl) + θ134(ikl) + θ234(jkl) . H2 : log mijkl (θ) = u + θ1(i) + θ2(j) + θ3(k) + θ4(l) + θ12(ij) + θ13(ik) + θ14(il) + θ23(jk) + θ24(jl) + θ34(kl) + θ124(ijl) + θ134(ikl) + θ234(jkl) . H3 : log mijkl (θ) = u + θ1(i) + θ2(j) + θ3(k) + θ4(l) + θ12(ij) + θ13(ik) + θ14(il) + θ23(jk) + θ24(jl) + θ34(kl) + θ134(ikl) + θ234(jkl) . H4 : log mijkl (θ) = u + θ1(i) + θ2(j) + θ3(k) + θ4(l) + θ12(ij) + θ13(ik) + θ14(il) + θ23(jk) + θ24(jl) + θ34(kl) + θ134(ikl) .


340


H5 : log mijkl (θ) = u + θ1(i) + θ2(j) + θ3(k) + θ4(l) + θ13(ik) + θ14(il) + θ23(jk) + θ24(jl) + θ34(kl) + θ134(ikl) . H6 : log mijkl (θ) = u + θ1(i) + θ2(j) + θ3(k) + θ4(l) + θ13(ik) + θ14(il) + θ24(jl) + θ34(kl) + θ134(ikl) . H7 : log mijkl (θ) = u + θ1(i) + θ2(j) + θ3(k) + θ4(l) + θ13(ik) + θ14(il) + θ34(kl) + θ134(ikl) . a) Using as significance level α = 0.05, test the null hypothesis that data are from model H5 based on the maximum likelihood estimator as well as on the power-divergence test statistic with λ = −2, −1, −1/2, 0, 2/3, 1 and 2. b) Find the best loglinear models among the models H1 , H2 , H3 , H4 , H5 , H6 and (l) H7 considered above based on Sφ(0),φ(0) , using as significance level α = 0.05.

10. Show that A(l) A(l+1) = A(l+1) A(l) = A(l+1) and A(l) A(l) = A(l) , where A(i) is defined in (7.19) for i = l, l + 1.

7.7.


1. Under the given restrictions it is easy to    1 2 log m11 (θ)  log m12 (θ)   1 1     log m13 (θ)   1 0     log m (θ)   1 1 21       0  log m22 (θ)  =  1     log m23 (θ)   1 −1     log m31 (θ)   1 0     log m32 (θ)   1 −1 1 −2 log m33 (θ) in the case a) and by    log m11 (θ)  log m12 (θ)       log m13 (θ)       log m (θ)   21        log m22 (θ)  =      log m23 (θ)       log m31 (θ)       log m32 (θ)   log m33 (θ)

in the second case.


verify that the matrix form is given by  0 1 0 0   1 0 1 0   u  −1 −1 −1 0    θ1  1 0 1 0     θ  2 0 0 1  2   θ  0 0 −1 −1   11    θ12  −1 −1 −1 0   θ22 0 0 −1 −1  −2

1

2

1

  1 1 0 1 0 1 0 0 u 1 1 0 0 1 0 1 0    θ1(1)  1 1 0 −1 −1 −1 −1 0   θ 1(2)  1 0 1 1 0 0 1 0     θ2(1) 1 0 1 0 1 0 0 1   θ 1 0 1 −1 −1 0 −1 −1   2(2)   θ12(11)  1 −1 −1 1 0 −1 −1 0    θ12(12)  1 −1 −1 0 1 0 −1 −1 θ12(22) 1 −1 −1 −1 −1 1 2 1

            


341

2. a) The matrix X is given by 

 1 1 1 c11  1 1 −1 c12  . X=  1 −1 1 −c21  1 −1 −1 −c22 b) The matrix X is given by  1 1 1 d11  1 1 −1 −d12  . X =  1 −1 1 d21  1 −1 −1 −d22 

c) The matrix X is given by 

1 1 1  1 1 −1 X=  1 −1 1 1 −1 −1

2  l11 2  l12  2 . l21 2 l22

3. To find the maximum likelihood estimator of m (θ) we must maximize the expresP P sion, Ii=1 Ij=1 nij log mij (θ) . We have I I X X

nij log mij (θ) = nu +

i=1 j=1

+

I I X X

nij θ1(i) +

i=1 j=1 ¶ I X I µ X nij + nji i=1 j=1

2

I I X X

nij θ2(j)

i=1 j=1

θ12(ij) ,

and I X I X

nij log mij (θ) =

i=1 j=1

I X I X i=1 j=1

nij log npij (θ) = n log n +

I X I X i=1 j=1

We denote L (θ) ≡ log Pr θ (N11 = n11 , ..., NII = nII ) ;


nij log pij (θ) .

342


then, n! + L (θ) = log n11 !...n II !

I I X X

nij log pij (θ)

i=1 j=1

= log

n! n11 !...nII !

= log

n! n11 !...nII !

+

I ³ I X X n

− n log n +

I X I X

nij i=1 j=1 I X

− n log n + nu +

ij +nji

2

i=1 j=1

´

log mij (θ)

ni∗ θ1(i) +

i=1

I X j=1

n∗j θ2(j)

θ12(ij)

and Pr θ (N11 = n11 , ..., NII = nII ) = +

n! exp(−n log n) n11 !...n II !

exp (−nu) = +

X

j=1

n∗j θ2(j) +

i=1

i=1 j=1

n! exp(−n log n) n11 !...n exp II !

n11 ,...,nII I X

exp nu +

I X

ni∗ θ1(i)  I I X J ³  ´ X X nij +nji θ . n∗j θ2(j) + 12(ij) 2  j=1

Therefore,

(

I X I ³ X n

ij +nji

2

i=1 j=1

´

( I X i=1 

θ12(ij)

 

ni∗ θ1(i) +

.

Differentiating in the two members of the previous equality with respect to θ12(ij) , we have ( I X X ∂u n! −n exp (−nu) = exp(−n log n) n11 !...n exp ni∗ θ1(i) + II ! ∂θ12(ij) n11 ,...,nII i=1  I I X I ³ ³ ´ ´ X X nij +nji nij +nji θ . n∗j θ2(j) + + 12(ij) 2 2  j=1

i=1 j=1

Then

−n

∂u = ∂θ12(ij) = =


X

³

nij +nji 2

n11h,...,nII i N +N E ij 2 ji mij (θ)+mji (θ) . 2

´

Pr θ (N11 = n11 , ..., NII = nII )


343

On the other hand ∂L (θ) ∂θ12(ij)

∂u nij + nji + ∂θ12(ij) 2 mij (θ) + mji (θ) nij + nji + . = − 2 2 = n

Therefore

∂L (θ) = 0, ∂θ12(ij)

or equivalently 0=−

mij (θ) + mji (θ) nij + nji + . 2 2

Then b + mji (θ) b = nij + nji , i 6= j. mij (θ)

In the same way differentiating with respect to θ1(i) and θ2(j) we have b = ni∗ and m∗j (θ) b = n∗j . mi∗ (θ)

b are obtained as a solution of Finally, the maximum likelihood estimators, mij (θ), the system of equations,  b + mji (θ) b = nij + nji , i 6= j   mij (θ) b = ni∗ . mi∗ (θ)   m (θ) b = n∗j ∗j

The first set of equations jointly with one of the other two is enough to find the b maximum likelihood estimator of mij (θ).

4. To find the minimum power-divergence estimator we must minimize in θ the expression   I X I X p, p (θ)) + µ  pij (θ) − 1 = 0, (7.37) Dφ(λ) (b i=1 j=1

with

p, p (θ)) = Dφ(λ) (b

1 λ(λ+1)

 I I X X  i=1 j=1

=

1 λ(λ+1)

p bλ+1 ij

pij (θ)λ



− 1

I µ λ+1 λ+1 I X X p b +b p ij

ji

2

i=1 j=1

¶ exp (−λ log pij (θ)) − 1

and log pij (θ) = log n + log mij (θ) = log n + u + θi + θj + θij .


(7.38)

344


It is clear that ∂u ∂ log pij (θ) = + 1, ∂θij ∂θij and in a similar way to Exercise 3 we can find that n

∂u = −mij (θ) , ∂θij

then ∂ log pij (θ) = −pij (θ) + 1. ∂θij Also, ∂

I I P P

pij (θ)

i=1 j=1

= pij (θ) (−pij (θ) + 1) .

∂θij

Differentiating in (7.37) with respect to θij , taking into account the expression (7.38), 1 λ (λ + 1)

µ

p bλ+1 +b pλ+1 ij ji pij 2

¶ (θ)−λ (−λ) (−pij (θ) + 1) + µpij (θ) (−pij (θ) + 1) = 0,

i.e., λ+1

1 pbij λ+1

then µ

1 λ+1

Ã 1 ¶ λ+1

and µ

1 λ+1

=

µ

1 λ+1

+ pbλ+1 ji = µpij (θ)λ+1 , 2

+ pbλ+1 pbλ+1 ij ji 2

1 ! λ+1

1

= µ λ+1 pij (θ) ,

1 Ã λ+1 ! λ+1 1 ¶ λ+1 I I X X pbij + pbλ+1 ji

i=1 j=1

2

.

Therefore

b(λ) ) = pij (θ

µ

p bλ+1 pλ+1 ij +b ji 2

I I P P

i=1 j=1

µ

1 ¶ λ+1

p bλ+1 pλ+1 ij +b ji 2

1 ¶ λ+1

1 ¢ λ+1 ¡ λ+1 pbij + pbλ+1 ji = I I . 1 ¢ λ+1 P P ¡ λ+1 pbij + pbλ+1 ji

i=1 j=1

5. We have θ12(ij) = log mij (θ) − u − θ1(i) − θ2(j) ,



345

then θ12(ij) = log mij (θ) − u −

=

1 IJ

I P J P

log

s=1 t=1

J P

1 J

= log mij (θ) −

1 J

j=1

J P

j=1

log mij (θ) + u −

log mij (θ) −

1 I

mij (θ) mst (θ) . mit (θ) msj (θ)

I P

1 I

I P

log mij (θ) + u

i=1

log mij (θ) +

i=1

1 IJ

J I P P

log mij (θ)

i=1 j=1

6. For a three-way contingency table we know J P K I P P

u b=

i=1 j=1 k=1 K J P P

θb1(i) =

and θb12(ij) = −

Substituting we obtain u b=

θb1(1) = θb1(2) =

K P

j=1 k=1

b log mijk (θ)/JK −u b

K J P P b b log mijk (θ)/K − log mijk (θ)/JK

k=1 K I P P

i=1 k=1

1 8

b log mijk (θ)/IJK

j=1 k=1

b log mijk (θ)/IK +u b.

J P K I P P

i=1 j=1 k=1 K J P P

j=1 k=1 K J P P j=1 k=1

b log mijk (θ)

b log m1jk (θ)/JK −u b = 0.74818

b log m2jk (θ)/JK −u b = −0.7418.

Only we have to calculate θb12(11) because

θb12(11) + θb12(12) θb12(11) + θb12(21)

Then we get

= 4.1771

=0 = 0.

θb12(11) = 4.9727 − 4.9189 − 4.0386 + 4.1771 = 0.1923.

7. a) For the models H2∗∗ , H3 , H4 , H5 and H6 , the maximum likelihood estimators are H2∗∗ H3 H4 H5 H6 nij∗ n∗jk n∗jk ni∗∗ ni∗∗ n∗j∗ n∗∗k n∗j∗ n∗∗k n∗∗k . nn∗j∗ n n2 nI IJ


346


This result follows because, for instance, the model H2∗∗ can be characterized by pij∗ p∗jk . p∗j∗

pijk =

For the model H1 we do not have an explicit expression and we must use a statistical package (Statgraphics, SAS, SPSS, etc.). In the following table we present the maximum likelihood estimators for mijk (θ) for the different models. (i, j, k) (1, 1, 1) (1, 1, 2) (1, 2, 1) (1, 2, 2) (2, 1, 1) (2, 1, 2) (2, 2, 1) (2, 2, 2)

nijk 103 59 32 78 87 109 42 205

H2∗∗ 86 76 22.8 87.2 104 92 51.2 195.8

H1 102.3 59.7 32.7 77.3 87.7 108.3 41.3 205.7

H3 72.3 63.9 28.2 107.7 117.7 104.1 45.8 175.3

H4 50.3 85.9 50.1 85.7 81.9 139.9 81.7 139.5

H5 66.1 112.9 65.9 112.6 66.1 112.9 65.9 112.6

H6 66 112.8 66 112.8 . 66 112.8 66 112.8

b) We are going to establish the goodness-of-fit to the data of the model H4 . We consider the power-divergence test statistics 2

(4)

Tφ(λ) φ(0) =

2

2

X X X (nijk )λ+1 2 2n , ¶λ − µ λ (λ + 1) i=1 j=1 λ (λ + 1) (4) k=1 b ) mijk (θ

whose asymptotic distribution is chi-square with 4 degrees of freedom. In the following table the values of the test statistics are presented for some choices of λ. λ (4) Tφ(λ) ,φ(0)

-2 117.738

-1 114.746

-1/2 115.57

0 117.98

2/3 123.85

1 128.057

On other hand we have χ24; 0.05 = 9.49. Then we should reject the null hypothesis. c) For λ = 1, we have

H2∗∗ versus H1

dl − dl+1 1

(l)

Tφ(1) ,φ(0) 19.9

χ2dl −dl+1 ,0.01

.

χ2dl −dl+1 ,0.01

.

χ2dl −dl+1 ,0.01

.

6.635

For λ = 0, we have

H2∗∗ versus H1

dl − dl+1 1

(l)

Tφ(0) ,φ(0) 19.647

6.635

Finally, for λ = −1, H2∗∗ versus H1


dl − dl+1 1

(l)

Tφ(−1) ,φ(0) 19.419

6.635


347

Independently of the value of λ we should choose the model H1 . d) If we consider the test statistic µ ¶ ¶¶ µ µ (l+1) (l) (l) b b b, p(θ b, p(θ ) ) − Dφ(λ) p . Sφ(λ) ,φ(0) = 2n Dφ(λ) p

For l = 1, we have

(l)

Sφ(1) ,φ(0) =

19.884 − 0.0648 = 19. 819

(l) Sφ(0) ,φ(0) = (l) Sφ(−1) ,φ(0) =

19.712 − 0.0646 = 19. 647

19.568 − 0.0851 = 19. 483,

and the conclusion is the same as in c). 8. a) In a similar way to the previous exercise we get (i, j, k) (1, 1, 1) (1, 1, 2) (1, 2, 1) (1, 2, 2) (2, 1, 1) (2, 1, 2) (2, 2, 1) (2, 2, 2)

nijk 8 42 57 106 11 37 45 69

H1 8.8 41.2 56.2 106.8 10.2 37.8 45.8 68.2

H2 10.21 46 54.80 102 8.8 33 47 73

b) The maximum likelihood estimators given by θb1(1) θb2(1) 0.1368 -0.619

H3 10.8 44.9 57.9 99.4 8.2 34.1 44.1 75.6

H4 18 37.7 50.8 106.6 13.7 28.7 38.6 81.1

H5 15.8 33.2 44.7 93.8 15.8 33.2 44.7 93.8

H6 36.3 63.5 30.3 63.5 30.3 63.5 30.3 63.5

for the parameters of the model H3 are θb3(1) -0.4912

θb23(12) . -0.2213

c) We have to check the goodness-of-fit of the data to the model H3 , 2

(3)

Tφ(λ) φ(0) =

2

2

X X X (nijk )λ+1 2 2n . ¶λ − µ λ (λ + 1) i=1 j=1 λ (λ + 1) (3) k=1 b mijk (θ )

In the following table we present the values of the power-divergence test statistic for the different values of λ, λ (3) Tφ(λ) φ(0)

−2 3. 1795

−1/2 3. 1342

−1 3. 1431

0 3. 132

2/3 3. 1391

1 3. 147

2 3. 1882

On the other hand χ23; 0.05 = 7.81. Then we should not reject the null hypothesis.


348


d) First we present the test HNull : H4 versus HAlt : H3 , using the power-divergence test statistic (3)

Tφ(λ) φ(0) We have λ (3) Tφ(λ) φ(0)

  λ (4) b ) 2 mijk (θ b(4) )   − 1 = mijk (θ  . (3) λ (λ + 1) i=1 j=1 b mijk (θ ) k=1 2 X 2 2 X X

-2 10. 069

-0.5 11. 136

-1 10. 714

0 11. 626

2/3 12. 413

1 12. 873

2 14. 579

and χ21; 0.05 = 3.84. Then we should reject the null hypothesis and to choose the model H3 , i.e., H3 : log pijk (θ) = u + θ1(i) + θ2(j) + θ3(k) + θ23(ij) . (l)

e) The results using the test statistics Sφ(λ) φ(0) are given in the following table Model H1 H2 H3 H4 H5 H6

(l)

b )) p, p(θ 2nDφ(0) (b 0.1935 2. 4427 3.142 13. 851 20.809 109.83

(l)

b )) 2nDφ(1) (b p, p(θ 0.1928 2. 4508 3. 147 12. 566 19.638 106.72

(l)

(l)

Sφ(0) φ(0)

Sφ(1) φ(0)

2. 2492 0. 6993 10. 709 6. 958 89. 021

2. 258 0. 6962 9. 419 7. 072 87. 082

On the other hand χ2dl −dl+1 ;α = χ21; 0.05 = 3.84, and we should choose the model H3 . 9. a) The predicted values for the frequencies associated with the model H5 are: (10.69911, 37.45849, 1.781327, 3.034881, 40.30089, 6.541514, 12.21867, 0.0965124, 16.99271, 63.84969, 1.526851, 13.65695, 64.00729, 11.15031, 10.47315, 4.343055) . Based on this vector we have λ (5) Tφ(λ) φ(0)

−2 1. 8629

−1 1. 8986

−0.5 1. 9479

0 2. 0242

2/3 2. 171

1 2. 2667

2 2. 6674

and χ25; 0.05 = 11.1. Therefore we should not reject the null hypothesis.



349

(l)

b) The results obtained with Sφ(0),φ(0) are given in the following table, Model H1 H2 H3 H4 H5 H6 H7

b(l) )) p, p(θ 2nDφ(0) (b 0.60 1.23 1.55 1.93 2.02 4.12 136.73

(l)

Sφ(0) φ(0)

dl − dl+1

0.63 0.32 0.38 0.09 2.10 132.61

1 1 1 1 1 1

.

On the basis of this table we should choose the model H6 because χ21; 0.05 = 3.84. 10. We have ´ ³ ´−1 ³ A(l) A(l) = diag p (θ 0 )−1/2 Σp(θ 0 ) W (l) W T(l) Σp(θ 0 ) W (l) W T(l) ´ ³ ´ ³ × Σp(θ 0 ) diag p (θ0 )−1/2 diag p (θ 0 )−1/2 Σp(θ 0 ) W (l) ´−1 ´ ³ ³ × W T(l) Σp(θ 0 ) W (l) W T(l) Σp(θ 0 ) diag p (θ 0 )−1/2 ´ ³ ´−1 ³ = diag p (θ 0 )−1/2 Σp(θ 0 ) W (l) W T(l) Σp(θ 0 ) W (l) ´ ³ × W T(l) Σp(θ 0 ) diag p (θ0 )−1/2 .

The last equality follows because ´ ³ ´ ³ Σp(θ 0 ) diag p (θ0 )−1/2 diag p (θ 0 )−1/2 Σp(θ 0 ) = Σp(θ 0 ) .

We know that W (l+1) is a submatrix of W (l) ; therefore there exists a matrix B verifying W (l+1) = W (l) B. Therefore, ´ ³ ´−1 ³ A(l) A(l+1) = diag p (θ0 )−1/2 Σp(θ 0 ) W (l) W T(l) Σp(θ 0 ) W (l) W T(l) Σp(θ 0 ) ´ ³ ´ ³ −1/2 diag p (θ0 )−1/2 Σp(θ 0 ) W (l+1) × diag p (θ0 ) ´−1 ´ ³ ³ × W T(l+1) Σp(θ 0 ) W (l+1) W T(l+1) Σp(θ 0 ) diag p (θ0 )−1/2 ´ ´−1 ³ ³ = diag p (θ0 )−1/2 Σp(θ 0 ) W (l) B W T(l+1) Σp(θ 0 ) W (l+1) ´ ³ × W T(l+1) Σp(θ 0 ) diag p (θ0 )−1/2 ³ ´ ³ ´−1 = diag p (θ0 )−1/2 Σp(θ 0 ) W (l+1) W T(l+1) Σp(θ 0 ) W (l+1) ´ ´ ³ ³ −1/2 W T(l+1) Σp(θ 0 ) diag p (θ 0 )−1/2 × W T(l+1) Σp(θ 0 ) diag p (θ0 ) = A(l+1) .

¡ ¢T We know A(l) A(l+1) = AT(l+1) and AT(l+1) AT(l) = AT(l+1) . But A(i) , i = l, l + 1, is a symmetric matrix, therefore A(l+1) A(l) = A(l+1) .


8

Phi-divergence Measures in Contingency Tables 8.1.

Introduction

In this chapter we study the importance of the estimators based on φ-divergences as well as the test statistics based on φ-divergences in some classical problems of contingency tables. First, we study the problems of independence, symmetry, marginal homogeneity and quasi-symmetry in a two-way contingency table and then the classical problem of homogeneity. There are different approaches to these problems but we study them using the result given in Section 5.5 in Chapter 5 regarding the minimum φ-divergence estimator with constraints as well as the result given in Section 6.3.5 in Chapter 6 regarding the testing problem when the null hypothesis can be written in terms of some constraints on the parameter space. Throughout the chapter, the cited results of Chapters 5 and 6 will be very useful. Due to this we reproduce them now. We suppose that we have ν (ν < M0 ) real-valued functions f1 (θ) , ..., fν (θ) that constrain the parameter θ ∈ Θ ⊂ RM0 , fm (θ) = 0, m = 1, ..., ν, such that they verify the conditions i) and ii) given in Section 5.5, and we denote by p (θ0 ) = (p1 (θ0 ) , ..., pM (θ0 ))T , where θ0 is unknown, the probability vector associated with the multinomial model under consideration. If we denote by Θ0 = {θ ∈ Θ : fm (θ) = 0, m = 1, ..., ν} , the restricted minimum φ-divergence estimator of θ0 or the minimum φ-divergence esti-


352


b(r) mator of θ0 in Θ0 , θ φ verifies

´ ³ b(r) = θ0 + H (θ0 ) I F (θ 0 )−1 A (θ 0 )T diag p (θ0 )−1/2 (b p − p (θ 0 )) + o (kb θ p − p (θ0 )k) φ (8.1) where the M0 × M0 matrix H (θ0 ) is defined by ³ ´−1 (8.2) H (θ0 ) = I − I F (θ0 )−1 B (θ0 )T B (θ0 ) I F (θ0 )−1 B (θ0 )T B (θ0 ) ;

I denotes the M0 × M0 identity matrix and also µ ¶ √ (r) L b n θ φ − θ0 → N (0, W (θ0 )) , n→∞

(8.3)

where the M0 × M0 matrix W (θ0 ) is given by µ ¶ ³ ´−1 −1 T −1 T B (θ0 ) I F (θ0 )−1 . I − B (θ0 ) B (θ 0 ) I F (θ 0 ) B (θ0 ) I F (θ0 ) I F (θ0 ) is the Fisher information matrix associated with the multinomial model and B (θ0 ) is the ν × M0 matrix of partial derivatives µ ¶ ∂fm (θ0 ) B (θ0 ) = m=1,...,ν . ∂θk k=1,...,M0 For testing, H0 : p=p (θ0 ) , for some unknown θ 0 ∈ Θ0 ⊂ Θ ⊂ RM0 , we consider the φ-divergence test statistic ¶ µ (r) 2n L b b, p(θ φ2 ) −→ χ2M−M0 +ν−1 , Dφ1 p 00 n→∞ φ1 (1)

(8.4)

(8.5)

b(r) where θ φ2 is the minimum φ2 -divergence estimator of θ 0 in Θ0 .

We shall assume that φ1 , φ2 ∈ Φ∗ are twice continuously differentiable in x > 0 with > 0 and φ002 (1) > 0.

φ001 (1)

8.2.

Independence

One of the most interesting models in a two-way contingency table consists of testing the independence between two random variables X and Y. b = (b We consider a two-way contingency table and let p p11 , ..., pbIJ )T be the nonparametric estimator of the unknown probability vector p = (p11 , ..., pIJ )T , where pij = © 2006 by Taylor & Francis Group, LLC

Phi-divergence Measures in Contingency Tables

353

Pr (X = i, Y = j) , with pij > 0, i = 1, ..., I, j = 1, ..., J. The hypothesis of independence is given by H0 : pij = pi∗ p∗j , i = 1, ..., I, j = 1, ..., J, (8.6) PI PJ where p∗j = i=1 pij and pi∗ = j=1 pij . We consider the parameter space o n T Θ = θ : θ = (pij ; i = 1, ..., I, j = 1, ..., J, (i, j) 6= (I, J))

(8.7)

and we denote by

p (θ) = (p11 , ..., pIJ )T = p, the probability vector characterizing our model with pIJ = 1 −

(8.8) I P

i=1

J P

pij . The hy-

j=1 (i,j)6=(I,J)

pothesis of independence given in (8.6) can be formulated again using the (I − 1) (J − 1) constraints hij (θ) = pij − pi∗ p∗j = 0, i = 1, ..., I − 1, j = 1, ..., J − 1.

(8.9)

Also considering the parameter β = (p1∗ , ..., pI−1∗ , p∗1 , ..., p∗J−1 )T and the set

P PJ−1 B = {(a1 , ..., aI−1 , b1 , ..., bJ−1 )T ∈ RI+J−2 / I−1 i=1 ai < 1, j=1 bj < 1, ai > 0, bj > 0, i = 1, ..., I − 1, j = 1, ..., J − 1},

the hypothesis (8.6) can be expressed for some unknown θ0 ∈ Θ with p (θ0 ) = p0 as H0 : p0 = p (g(β 0 )) ,

β0 ∈B and g(β 0 ) = θ0 ,

(8.10)

where the function g is defined by g = (gij ; i = 1, ..., I, j = 1, ..., J, (i, j) 6= (I, J)) with gij (β) = pi∗ p∗j , and

i = 1, ..., I − 1, j = 1, ..., J − 1

µ

¶ I−1 P 1− pi∗ p∗j , j = 1, ..., J − 1 i=1 Ã ! J−1 P 1− p∗j pi∗ , i = 1, ..., I − 1. giJ (β) =

gIj (β) =

j=1

It is important to remark by (8.8) that p (g(β)) = (gij (β), i = 1, ..., I, j = 1, ..., J)T , where I J X X gij (β). gIJ (β) = 1 − i=1

j=1 (i,j)6=(I,J)

With this approach the hypothesis of independence can be formulated in terms of the results presented in Chapter 6 (see Exercise 8).


354


In this Chapter we consider the approach given in (8.9). We can observe that with this approach the hypothesis of independence can be written as H0 : p=p (θ0 ) , for some unknown θ 0 ∈Θ0 , where Θ0 = {θ ∈Θ : hij (θ) = 0, i = 1, ..., I − 1, j = 1, ..., J − 1} ,

(8.11)

with hij defined in (8.9) and p (θ) defined in (8.8). For I = J = 2, we only have a constraint which is given by h11 (θ) = p11 − p1∗ p∗1 = 0, and for I = J = 3, we have four constraints given by h11 (θ) = h12 (θ) = h21 (θ) = h22 (θ) =

8.2.1.

p11 − p1∗ p∗1 p12 − p1∗ p∗2 p21 − p2∗ p∗1 p22 − p2∗ p∗2

=0 =0 =0 = 0.

Restricted Minimum Phi-divergence Estimator

In the following theorem we present the expression of the restricted minimum φ−divergence estimator for the problem of independence as well as its asymptotic properties. We can observe that this estimator is the minimum φ-divergence estimator under the independence hypothesis. Theorem 8.1 The minimum φ-divergence estimator, ³ ´T bI,φ = pI,φ ; i = 1, ..., I, j = 1, ..., J and (i, j) 6= (I, J) , θ i,j

of θ0 ∈Θ0 (i.e., under the hypothesis of independence) is obtained as a solution of the system of equations  J µ µ ¶ µ ¶¶ X  nij nij 0 nij   φ p∗j φ − − µ = 0, i = 1, ..., I   npi∗ p∗j npi∗ npi∗ p∗j j=1 (8.12) µ ¶ µ ¶¶ I µ  X  nij nij 0 nij   φ pi∗ φ − − µ = 0, j = 1, ..., J,  npi∗ p∗j np∗j npi∗ p∗j i=1

where µ is given by

µ=

µ I X J µ X pi∗ p∗j φ i=1 j=1


nij npi∗ p∗j

¶

−

nij 0 φ n

µ

nij npi∗ p∗j

¶¶

.


355

Its asymptotic distribution is ´ √ ³ I,φ L b − θ0 → n θ N (0, W I (θ0 )) n→∞

where the (IJ − 1) × (IJ − 1) matrix W I (θ0 ) is given by µ ¶ ³ ´−1 T T Σθ 0 I (IJ−1)×(IJ−1) − B I (θ 0 ) B I (θ0 ) Σθ 0 B I (θ0 ) B I (θ0 ) Σθ 0 ,

(8.13)

(8.14)

the (I − 1)(J − 1) × (IJ − 1) matrix B I (θ0 ) is defined by µ ¶ ∂hij (θ 0 ) , B I (θ0 ) = ∂θ i=1,...,I−1,j=1,...,J−1 −1

and Σθ 0 = I F (θ 0 )

is the (IJ −1)×(IJ − 1) matrix verifying Σθ 0 = diag (θ0 )−θ0 θ T0 .

Proof. Instead of getting ³ ´T bI,φ = pI,φ ; i = 1, ..., I, j = 1, ..., J and (i, j) 6= (I, J) θ ij we shall obtain

b p(θ

I,φ

³ ´T ) = pI,φ . ij ; i = 1, ..., I, j = 1, ..., J

The p0ij s minimizing the φ-divergence

I X J X

µ

nij p, p (θ)) = pij φ Dφ (b np ij i=1 j=1

¶

subject to the hypothesis of independence (or the constraints about θ given in (8.9)) may be obtained minimizing the Lagrangian function   µ ¶ I−1 I X I X J J X X J−1 X X nij pij φ λij hij (θ) + µ 1 − pij  (8.15) + np ij i=1 j=1 i=1 j=1 i=1 j=1

where λij and µ are undetermined Lagrangian multipliers. Taking into account the constraints given in (8.9), minimizing the expression (8.15) is equivalent to minimizing the expression   Ã ! µ ¶ I X J I J X X X nij pi∗ p∗j φ p∗j  + µ2 1 − pi∗ . + µ1 1 − npi∗ p∗j i=1 j=1 j=1 i=1 Finally, differentiating with respect to pi∗ and p∗j we get the system of equations given bI,φ . in (8.12), whose solutions provide the minimum φ-divergence estimator, θ The asymptotic distribution follows from (8.3) because I F (θ 0 )−1 = Σθ 0 , under the considered parametric multinomial model.


356


Corollary 8.1 The minimum power-divergence estimator ³ ´T bI,φ(λ) ) = pI,φ(λ) ; i = 1, ..., I, j = 1, ..., J p(θ ij

of p (θ0 ) (i.e., under the hypothesis of independence (8.9)) is given by I,φ(λ)

pij I,φ(λ)

where pi∗

I,φ(λ)

= pi∗

I,φ(λ)

× p∗j

, i = 1, ..., I, j = 1, ..., J

I,φ

and p∗j (λ) are the solutions of the system of equations   1   λ+1  λ+1  J  X  nij       ³ I,φ ´λ        j=1 (λ)   p b  ∗j I,φ (λ)   pi∗ = i = 1, ..., I   1    λ+1   λ+1 I X J   X nij        ³ ´    λ  I,φ    (λ) i=1 j=1  p b  ∗j     1    λ+1  λ+1 I    X  nij        ³ I,φ ´λ       i=1   pbi∗ (λ)  I,φ(λ)   = p  ∗j   1   λ+1   λ+1 J I   X nij  X     ³ I,φ ´λ       i=1   j=1 pb (λ)

j = 1, ..., J

(8.16)

.

i∗

Proof. In this case the system of the equations (8.12) can be written as   1 λ+1   λ+1    J I J  X  X nλ+1 nλ+1  X  ij ij     =   ³ ´ ³ ´ ³ ´λ  λ+1 λ    I,φ I,φ I,φ  (λ) (λ) (λ) λ+1 λ+1  j=1 n i=1 j=1 n pi∗ p∗j p∗j       1 λ+1   λ+1    I I J   X  X nλ+1 nλ+1 X   ij ij    = .   ³ ´ ³ ´ ³ ´λ   λ+1 λ    I,φ I,φ I,φ (λ) (λ) (λ)  λ+1 λ+1 i=1 n i=1 j=1 n  p p p ∗j i∗ i∗     It is clear that the solution of this system is given by (8.16). For λ = 0, I,φ(0)

pi∗

=


ni∗ n

and

I,φ(0)

p∗j

=

n∗j , i = 1, ..., I, j = 1, ..., J, n


357

hence the maximum likelihood estimator, under the hypothesis of independence, is obtained. If λ → −1 the minimum modified likelihood estimator under independence is obtained as the solution of the system of equations   pI,φ(−1)   J ∗j Y  n  ij       I,φ  (−1)    pI,φ(−1) = j=1 pb∗j  i = 1, ..., I  i∗   pI,φ(−1)   I Y J ∗j  X  n ij       I,φ   (−1)  p i=1 j=1  ∗j    pI,φ(−1)    I Y  nij  i∗       I,φ    pi∗ (−1) I,φ(−1) i=1   p∗j =   pI,φ(−1)    J Y I  X nij  i∗       I,φ  pi∗ (−1) j=1 i=1

j = 1, ..., J

.

We call the resulting estimators for λ = −2, −1/2, 2/3 and 1 the minimum modified chisquare estimator, Freeman-Tukey estimator, Cressie-Read estimator and the minimum chi-square estimator, respectively. Remark 8.1 For I = J = 2, given θ0 = (p∗1 p1∗ , (1 − p∗1 ) p1∗ , p∗1 (1 − p1∗ ))T ∈ Θ0 , it is easy to check that · BI (θ0 ) = (1 − p1∗ − p∗1 , −p∗1 , −p1∗ ) ,

· Σθ 0 = diag (θ0 ) − θ0 θT0 , ´−1 ³ T −1 = (p1∗ p∗1 (1 − p1∗ − p∗1 + p1∗ p∗1 )) , · B I (θ 0 ) Σθ 0 B I (θ 0 ) and it is not difficult to establish that the matrix W I (θ 0 ) is given by 

p∗1  1 − p∗1 −p∗1

 µ ¶µ p1∗ p1∗ (1 − p1∗ ) 0 p∗1  −p1∗ 0 p∗1 (1 − p∗1 ) p1∗ 1 − p1∗

where: · The matrix


µ

0 p1∗ (1 − p1∗ ) 0 p∗1 (1 − p∗1 )

¶

1 − p∗1 −p1∗

−p∗1 1 − p1∗

¶

,

358


is the inverse of the Fisher information matrix corresponding to the parameter β0 = (p1∗ , p∗1 )T , i.e., I F (β 0 )−1 · The matrix

 p1∗ p∗1  1 − p∗1 −p1∗  −p∗1 1 − p1∗ µ ¶T ∂g(β 0 ) is the matrix of partial derivatives . ∂β 

Then we have W I (θ0 ) =

µ

∂g(β 0 ) ∂β

¶T

I F (β0 )−1

µ

∂g(β 0 ) ∂β

¶

.

This relation is true in general. Theorem 8.2 Under the independence model, we have W I (θ0 ) = M Tβ 0 I F (β 0 )−1 M β 0 , where M β0 =

µ

∂g(β0 ) ∂β

¶

(8.17)

,

and g was defined in (8.10). Proof. It is not difficult to establish that T I F (β 0 ) = M β 0 Σ−1 θ0 M β0 .

Then we have ³ ´−1 T M Tβ 0 I F (β0 )−1 M β 0 = M Tβ 0 M β 0 Σ−1 M M β0. β0 θ0

(8.18)

T −1 Multiplying (8.17) by Σ−1 θ 0 M β 0 on the right and by M β 0 Σθ 0 on the left and taking into account (8.18) and the expression of W I (θ 0 ) given in (8.14), we have T T −1 M β 0 Σ−1 θ 0 M β 0 = M β 0 Σθ 0 M β 0

− M β 0 B I (θ0 )


T

³ ´−1 B I (θ 0 ) Σθ B I (θ0 )T B I (θ0 ) M Tβ 0 .


359

But M β 0 B I (θ 0 )T = 0 and B I (θ0 ) M Tβ 0 = 0. Then we have established that AT XA = AT Y A, T

where X = M Tβ 0 I F (β0 )−1 M β 0 , Y = W I (θ0 ) and A = Σ−1 θ 0 M β 0 . But the matrix A has full rank, then X = Y . Remark 8.2 The Fisher information matrix associated with the independence model is given by ³ ´−1 M Tβ 0 I F (β0 )−1 M β 0 , since if we formulate the problem of independence based on (8.10) we have ´ ¢ ¡ √ ³ L b −β n β −→ N 0, I F (β 0 )−1 , 0 n→∞

b is the maximum likelihood estimator of β , and where β 0 ´ ´ ³ √ ³ L b − g(β 0 ) −→ n g(β) N 0, M Tβ 0 I F (β 0 )−1 M β 0 . n→∞

In Theorem 8.1 we have established that ´ √ ³ I,φ L b − θ0 → n θ N(0, W I (θ 0 )). n→∞

Then the restricted minimum φ-divergence estimator will be BAN if and only if W I (θ0 ) = M Tβ 0 I F (β 0 )−1 M β 0 . bI,φ is BAN by Theorem 8.2. Therefore, θ

Now we are going to present a result that establishes sufficient conditions for uniqueness of the minimum φ-divergence estimator in the independence problem. Theorem 8.3 bI,φ = The minimum φ-divergence estimator under the hypothesis of independence, θ ³ ´T pI,φ , is unique if ij ; i = 1, ..., I, j = 1, ..., J and (i, j) 6= (I, J) k2 (i, j) > −k1 (i, j) > 0, i = 1, ..., I; j = 1, ..., J

where

µ

¶ µ ¶ nij nij nij 0 k1 (i, j) = φ φ − , i = 1, ..., I; j = 1, ..., J npi∗ p∗j µ npi∗ p∗j¶ npi∗ p∗j 2 nij nij k2 (i, j) = 2 2 2 2 φ00 , i = 1, ..., I; j = 1, ..., J. n pi∗ p∗j npi∗ p∗j


360


Proof. It is necessary to establish that the function µ

I X J X

nij pi∗ p∗j φ f(p1∗ , ..., pI∗ , p∗1 , ..., p∗J ) = np i∗ p∗j i=1 j=1 is strictly convex. Consider the functions fij (p1∗ , ..., pI∗ , p∗1 , ..., p∗J ) = pi∗ p∗j φ

µ

nij npi∗ p∗j

¶

¶

, i = 1, ..., I, j = 1, ..., J.

We will use the fact that if fij is strictly convex for all i and j, then strictly convex.

PI

i=1

PJ

j=1 fij

is

For the sake of brevity, we forget indexes and write f (x, y) = xyφ(

k ), xy

where k is a positive constant. The determinant of the Hessian matrix of the function f is µ ¶2 µ µ ¶ µ ¶ µ ¶¶2 k4 00 k k k 0 k k2 00 k φ φ − φ − + 2 2φ x4 y4 xy xy xy xy x y xy which can be written as follows µ µ ¶ µ ¶¶ µ µ ¶ µ ¶ µ ¶¶ k k 0 k k k 0 k k2 00 k φ φ − φ − φ − +2 2 2φ . xy xy xy xy xy xy x y xy Hence the result holds. It is not difficult to establish that in the case of the power-divergence family the result is verified for λ > −1/2.

8.2.2.

Test of Independence

Based on the φ-divergence test statistic given in (8.5) we should reject the hypothesis of independence if ³ I,φ2 ´ ³ ´ 2n b bI,φ2 ) > c, b, p(θ ≡ 00 Dφ1 p Inφ1 θ φ1 (1) where c is a positive constant. In ³the following theorem we establish the asymptotic ´ I,φ2 φ1 b . distribution of the test statistic In θ

Theorem 8.4 The asymptotic distribution of the φ-divergence test statistics ³ I,φ2 ´ ³ ´ 2n b bI,φ2 ) b, p(θ Inφ1 θ ≡ 00 Dφ1 p φ1 (1) © 2006 by Taylor & Francis Group, LLC

(8.19)

361


and

³ I,φ2 ´ b ≡ Inφ1 ,h θ

³ ³ ´´ 2n bI,φ2 ) , b p , p( θ h D φ 1 h0 (0) φ001 (1)

(8.20)

for testing the hypothesis of independence, is chi-square with (I − 1) (J − 1) degrees of freedom. Proof. In our case we have IJ cells and by (8.7) IJ − 1 parameters that are necessary to estimate. Using relation (8.9) the number of constraints is (I − 1) × (J − 1) . Then by (8.5) the asymptotic distribution of the family of test statistics (8.19) is chi-square with IJ |{z} Cells

−

IJ − 1 | {z } Estimated Parameters

+ (I − 1) (J − 1) −1 = (I − 1) (J − 1) {z } | {z } | Constraints Degrees of . freedom

In relation with the family of (h, φ)-divergence test statistics given in (8.20), we have h (x) = h (0) + h0 (0) x + o (x) , then

³ ³ ³ ´´ ´ bI,φ2 ) = h0 (0) Dφ p bI,φ2 ) + oP (1) b, p(θ b , p( θ h Dφ1 p 1

and we get that the asymptotic distribution of the family of test statistics given in (8.20) is also chi-square with (I − 1) (J − 1) degrees of freedom. Remark 8.3 ³ I,φ2 ´ b For φ1 (x) = φ2 (x) = x log x − x + 1 = φ(0) (x) we get that Inφ1 θ coincides with the classical likelihood ratio test statistic for testing independence given by J I X ³ I,φ(0) ´ X pbij φ b G2 ≡ In (0) θ = 2n pbij log , p b pb∗j i∗ i=1 j=1

and for φ2 (x) = x log x − x + 1 = φ(0) (x) and φ1 (x) = chi-square test statistic given by

1 2

(x − 1)2 = φ(1) (x) the classical

I X J 2 ³ I,φ(0) ´ X (b pij − pbi∗ pb∗j ) φ b X 2 ≡ In (1) θ =n . pbi∗ pb∗j i=1 j=1

It is important to note that for φ2 (x) = x log x − x + 1, we get the family of test statistics studied by Zografos (1993) and Pardo, L. et al . (1993a), which is ³ ´ ³ I,φ ´ b ≡ I φ1 θ b (0) = 2n Dφ (b b p, p(θ)), (8.21) Inφ1 θ n φ001 (1) 1

b ≡θ bI,φ(0) is the maximum likelihood estimator, where θ ³ ´T b = ni∗ × n∗j , i = 1, ..., I, j = 1, .., J, (i, j) 6= (I, J) . θ n n © 2006 by Taylor & Francis Group, LLC

362


Based on the previous test we should reject the hypothesis of independence with significance level α if ³ I,φ2 ´ ³ ³ I,φ2 ´´ b b Inφ1 θ or Inφ1 ,h θ ≥ χ2(I−1)(J−1),α . (8.22)

Power of the Test Let q = (q11 , ..., qIJ )T be a point at the alternative hypothesis, i.e., there exist at least two indexes i and j for which qij 6= qi∗ × q∗j . We denote by θφa 2 the value of θ verifying θφa 2 =arg min Dφ2 (q, p (θ)), θ ∈Θ0

where Θ0 was defined in (8.11). It is clear that θφa 2 = (fij (q); i = 1, ..., I, j = 1, ..., J, (i, j) 6= (I, J) )T and p(θφa 2 ) = (fij (q); i = 1, ..., I, j = 1, ..., J )T ≡ f(q), with fIJ (q) = 1 −

I X i=1

J X

fij (q).

j=1 (i,j)6=(I,J)

The notation fij (q) indicates that the elements of the vector θφa 2 depend on q. For instance, for φ2 (x) = x log x − x + 1, p(θφa 2 ) = (qi∗ × q∗j ; i = 1, ..., I, j = 1, ..., J )T , and fij (q) = qi∗ × q∗j . We also denote

and then

³ ´T bI,φ2 = pI,φ2 ; i = 1, ..., I, j = 1, ..., J, (i, j) 6= (I, J) θ ij b p(θ

I,φ2

³ ´T 2 ) = pI,φ ≡ f (b p), ij ; i = 1, ..., I, j = 1, ..., J

T b where f = (fij ; i = 1, ..., I, j = 1, ..., J) . If the alternative q is true we have that p I,φ2 φ2 b tends to q and p(θ ) to p(θ ) in probability. a



363

If we define the function Ψφ1 (q) = Dφ1 (q, f(q)), we have p) = Ψφ1 (q) + Ψφ1 (b

I X J X ∂Dφ1 (q, f (q)) (b pij − qij ) + o (kb p − qk) . ∂qij i=1 j=1

Hence the random variables √ n (Dφ1 (b p, f (b p)) − Dφ1 (q, f (q))) and

I X J √ X ∂Dφ1 (q, f(q)) n (b pij − qij ) ∂qij i=1 j=1

have the same asymptotic distribution. If we denote by lij =

∂Dφ1 (q, f(q)) ∂qij

(8.23)

and by l = (lij ; i = 1, ..., I, j = 1, ..., J )T we have √ where

³ ´ L n (Dφ1 (b p, f(b p)) − Dφ1 (q, f(q))) −→ N 0, lT Σq l , n→∞

¡ ¡ ¢¢ Σq = qi1 j1 δ(i1 j1 )(i2 j2 ) − qi2 j2 i

1 ,i2 =1,...,I j1 ,j2 =1,...,J

(8.24)

.

In the following theorem we present the asymptotic distribution using the maximum likelihood estimator. In this case we have: Theorem 8.5 b = (b Let p(θ) pi∗ × pb∗j , i = 1, ..., I, j = 1, ..., J)T the maximum likelihood estimator of p (θ) = (p11 , ..., pIJ )T , under the independence hypothesis, and q a point at the alternative hypothesis. Then, ¡ ¢´ L ¡ ¢ √ ³ b − Dφ q, q n Dφ1 (b p, p(θ)) −→ N 0, σφ2 1 (q) 1 I×J n→∞

where

σφ2 1 (q) =

I X J X i=1 j=1

lij

 2 I X J X 2 qij lij − qij lij  , i=1 j=1

µ µ ¶ µ ¶¶ I P qrj qrj 0 qrj = φ1 qr∗ φ1 − r=1 µ µ qr∗ q∗j ¶ q∗j µ qr∗ q∗j¶¶ µ ¶ J P qis qis 0 qis qij + φ1 q∗s φ1 − + φ01 qi∗ q∗s qi∗ qi∗ q∗s qi∗ q∗j s=1


364


and qI×J = (qi∗ × q∗j , i = 1, ..., I, j = 1, ..., J)T whenever σφ2 1 (q) > 0. Proof. We have to calculate the elements lij , given in (8.23) taking into account that in this case fij (q) = qi∗ × q∗j . We can write the function Ψφ1 (q) as Ψφ1 (q) = + +

µ ¶ µ ¶ qij qij qi∗ q∗j φ1 = qi∗ q∗j φ1 qi∗ q∗j q q i=1 j=1 µ ¶ µ i∗ ∗j ¶ J I P P qis qrj qi∗ q∗s φ1 qr∗ q∗j φ1 + qi∗ q∗s qr∗ q∗j s=1 r=1 s6=j r6=i µ ¶ I P J P qrs qr∗ q∗s φ1 = G1 + G2 + G3 + G4 . qr∗ q∗s r=1 s=1 J I P P

r6=i s6=j

Then we have µ ¶ µ ¶ ∂G1 qij qij qi∗ q∗j − qij (q∗j + qi∗ ) = φ1 (q∗j + qi∗ ) + φ01 ∂qij q q q q qi∗ q∗j ¶ i∗ ∗j i∗ ∗j µ ¶ µ ¶µ qij qij qij qij 0 = φ1 − (q∗j + qi∗ ) + φ1 1− qi∗ q∗j qi∗ q∗j qi∗ q∗j ½ µ ¶ µ ¶ ¾ 1 qis qis qis (−1) 2 q∗s φ1 + qi∗ q∗s φ01 qi∗ q∗s qi∗ q∗s q∗s qi∗ s=1 s6=j µ ¶ µ ¶ J J P P qis qis qis = q∗s φ1 φ01 − qi∗ q∗s qi∗ q∗s qi∗ s=1 s=1

∂G2 = ∂qij

J P

s6=j

∂G3 = ∂qij

I P

s6=j

qr∗ φ1

r=1 r6=i

µ

qrj qr∗ q∗j

¶

I q P rj 0 φ1 − q r=1 ∗j r6=i

µ

qrj qr∗ q∗j

¶

∂G4 = 0. ∂qij Therefore,

lij =

∂Ψφ1 (q) ∂qij


µ µ ¶ µ ¶¶ qrj qrj 0 qrj φ1 qr∗ φ1 − r=1 µ µ qr∗ q∗j ¶ q∗j µ qr∗ q∗j¶¶ J P qis qis 0 qis + φ1 q∗s φ1 − q q q q q∗s i∗ ∗s i∗ i∗ s=1µ ¶ qij 0 + φ1 qi∗ q∗j =

I P


365

and based on (8.24) we have σφ2 1 (q) = lT Σq l =

I X J X i=1 j=1

 2 I X J X 2 qij lij − qij lij  . i=1 j=1

Using this result it is possible to get the power of the test of independence. This is given by ¶¶ µ √ µ 00 n φ1 (1) 2 χ βn,φ1 (q) = 1 − Φn − Dφ1 (q, qI×J ) , σφ1 (q) 2n (I−1)(J−1),α where Φn (x) is a sequence of distribution functions leading uniformly to the standard normal distribution Φ (x) . Remark 8.4 For the Kullback-Leibler divergence we have  2 I X J X q q ij ij  σ2 (q) = qij log2 − qij log . q q q q∗j i∗ ∗j i∗ i=1 j=1 i=1 j=1 I X J X

³ I,φ2 ´ ³ I,φ2 ´ b b or Inφ1 ,h θ , In general, theoretical results for the test statistics Inφ1 θ under alternative hypotheses, with φ2 (x) 6= x log x − x + 1, are not easy to obtain. An exception to this fact is when there is a contiguous sequence of alternatives that approaches the null hypothesis H0 : p = p (θ0 ) , for some unknown θ 0 ∈ Θ0 , at the rate ¡ ¢ O n−1/2 . Consider the multinomial probability vector dij pn,ij = pij (θ0 ) + √ , i = 1, ..., I, j = 1, ..., J, n

where d = (d11 , ..., dIJ )

T

is a fixed IJ × 1 vector such that

I X J X

dij = 0; recall that

i=1 j=1

n is the total count parameter of the multinomial distribution and θ0 is unknown but belonging to Θ0 . As n → ∞, the sequence of multinomial probabilities {pn }n∈N with T

pn = (pn,ij , i = 1, ..., I, j = 1, ..., J)

¢ ¡ converges to a multinomial probability in H0 at the rate of O n−1/2 . Let d H1,n : pn = p (θ 0 ) + √ , n


θ 0 ∈Θ0

(8.25)

366


be a sequence of contiguous alternative hypotheses, here contiguous to the null hypothesis H0 : p=p (θ 0 ) , for some unknown in studying the asymptotic ³ I,φ2 θ ´0³∈Θ0 . We³areI,φinterested ´´ 2 φ1 b φ1 ,h b θ or I θ under contiguous alternative hybehavior of the power of I n

n

potheses given in (8.25).

The power of this test is ³ I,φ2 ´ ³ ³ I,φ2 ´´ ´ ³ b b πn ≡ Pr Inφ1 θ or Inφ1 ,h θ > χ2(I−1)(J−1),α /H1,n .

In what follows, we prove that under the alternative H1,n , and as n → ∞, ³ I,φ2 ´ ³ ³ I,φ2 ´´ b b or I φ1 ,h θ I φ1 θ n

n

converge in distribution to a noncentral chi-square random variable with³noncentrality pa´ rameter δ given in Theorem 8.6, and (I − 1) (J − 1) degrees of freedom χ2(I−1)(J−1) (δ) . Consequently, as n → ∞ ´ ³ πn → Pr χ2(I−1)(J−1) (δ) > χ2(I−1)(J−1),α .

Theorem 8.6 Under H1,n : pn = p (θ0 ) + √dn , θ0 some unknown value in Θ0 , the φ-divergence test statistics ³ I,φ2 ´ ³ ³ I,φ2 ´´ b b Inφ1 θ or Inφ1 ,h θ (8.26)

are asymptotically noncentrally chi-squared distributed with (I − 1) (J − 1) degrees of freedom and noncentrality parameter δ=

I X I J J X X d2ij d2i∗ X d2∗j − − . p p p p i=1 j=1 i∗ ∗j i=1 i∗ j=1 ∗j

Proof. Pardo, J. A. et al . (2002) established that for testing H0 : p = p (θ0 ) , for some unknown θ0 ∈Θ0 ⊂ Θ ⊂ RM0 with Θ0 = {θ ∈ Θ : fm (θ) = 0, m = 1, ..., ν} ,

√ versus H1,n : p = p (θ 0 ) + d/ n, the asymptotic distribution of the test statistic given in (8.26), under H1,n , is noncentral chi-square with M − M0 + ν − 1 degrees of freedom and noncentrality parameter given by δ = µT µ, where ´ ³ −1/2 (I − L (θ 0 )) d, µ = diag p (θ 0 ) ´ ´ ³ ³ L (θ 0 ) = diag p (θ 0 )−1/2 A (θ 0 ) H (θ 0 ) Σθ 0 A (θ 0 )T diag p (θ0 )−1/2 ,



367

and H (θ 0 ) was given in (8.2). In our case it is not difficult to establish that ´ n³ ³ µT µ = dT diag p (θ0 )−1/2 I − A (θ0 ) Σθ 0 A (θ0 )T ¶¾ ³ ´−1 + A (θ0 ) Σθ 0 B I (θ0 )T B I (θ0 ) Σθ 0 BI (θ0 )T B I (θ0 ) Σθ 0 A (θ0 )T ´ ³ −1/2 d. × diag p (θ0 ) Applying (8.17) we have ³ ´−1 Σθ 0 − Σθ 0 B I (θ0 )T B I (θ0 ) Σθ 0 B I (θ0 )T B I (θ 0 ) Σθ 0 = M Tβ 0 I F (β0 )−1 M β 0 .

Multiplying, in the previous expression, on the right by A (θ 0 ) and on the left by A (θ0 )T we have ´³ ´ ³ µT µ = dT diag p (θ0 )−1/2 I − A (θ0 ) M Tβ 0 I F (β0 )−1 M β 0 A (θ0 )T ³ ´ × diag p (θ 0 )−1/2 d and

p (θ0 ) = p (g(β0 )) , A (θ0 ) M β 0 = A (g(β0 )) , then

´ ¢¢ ³ ¡ ¡ µT µ = dT diag p g(β 0 )−1 I − A (g(β 0 )) I F (g(β 0 ))−1 A (g(β 0 ))T ´ ³ ×diag p (g(β 0 ))−1/2 d. ´ ³ It is not difficult to establish that the matrix dT diag p (g(β0 ))−1/2 is given by µ ¶ d11 d12 d1J dI1 dI2 dIJ ,√ , ..., √ , ......, √ ,√ , ..., √ √ p1∗ p∗1 p1∗ p∗2 p1∗ p∗J pI∗ p∗1 pI∗ p∗2 pI∗ p∗J and the matrix I F (β 0 )−1 by

µ

¶ 0 , ΣpJ where ΣpI = diag (pI ) − pI pTI and ΣpJ = diag (pJ ) − pJ pTJ , being ΣpI 0

pI = (p1∗ , ..., pI−1∗ )T and pJ = (p∗1 , ..., p∗J−1 )T .

With these expressions one gets δ = µT µ =

J I X I J X X d2ij d2i∗ X d2∗j − − . p p p p i=1 j=1 i∗ ∗j i=1 i∗ j=1 ∗j

For more details about the problem of independence in a two-way contingency table see Menéndez et al. (2005b).


368

8.2.3.


Multiway Contingency Tables

The parameter space associated with a three-way contingency table is o n Θ = θ : θ = (pijk ; i = 1, ..., I, j = 1, ..., J, k = 1, ..., K, (i, j, k) 6= (I, J, K))T ,

where pijk = Pr (X = i, Y = j, Z = k) . The hypothesis of independence,

H0 : pijk = pi∗∗ p∗j∗ p∗∗k , i = 1, ..., I, j = 1, ..., J, k = 1, ..., K, can be formulated by using the IJK − I − J − K + 2 constraints, hijk (θ) = pijk − pi∗∗ p∗j∗ p∗∗k = 0, where i, j, k ∈ D, being D the set, {(i, j, k)/i = 1, ..., I, j = 1, ..., J, k = 1, ..., K, (i, j, k) 6= (i1 , J, K), i1 = 1, ..., I,

(i, j, k) 6= (I, j1 , K), j1 = 1, ..., J − 1, (i, j, k) 6= (I, J, k1 ), k1 = 1, ..., K − 1}

In this situation similar results to those of the previous subsections can be obtained (see Pardo, J. A. (2004)).

8.3.

Symmetry

An interesting problem in a two-way contingency table is to investigate whether there are symmetric patterns in the data: Cell probabilities on one side of the main diagonal are a mirror image of those on the other side. This problem was first discussed by Bowker (1948) who gave the maximum likelihood estimator as well as a large sample chi-square type test for the null hypothesis of symmetry. In Ireland et al . (1969) it was proposed the minimum discrimination information estimator and in Quade and Salama (1975) the minimum chi-squared estimator. Based on the maximum likelihood estimator and on the family of φ-divergence measures, in Menéndez et al. (2001e) a new family of test statistics was introduced. This family contains as a particular case the test statistic given by Bowker (1948) as well as the likelihood ratio test. The state-of-the art in relation with the symmetry problem can be seen in Bishop et al . (1975), Agresti (2002), Andersen (1998) and references therein. b = (b We consider a two-way contingency table with I = J and let p p11 , ..., pbII )T be the nonparametric estimator of the unknown probability vector p with components pij = Pr(X = i, Y = j), with pij > 0 and i, j = 1, ..., I. The hypothesis of symmetry is H0 : pij = pji , , i, j = 1, ..., I.


(8.27)


We consider the parameter space o n T Θ = θ : θ = (pij ; i, j = 1, ..., I, (i, j) 6= (I, I)) ,

369

(8.28)

and we denote by

p (θ) = (p11 , ..., pII )T = p,

(8.29) PI

PI

the probability vector characterizing our model, with pII = 1 − i=1 j=1 pij with (i, j) 6= (I, I). The hypothesis of symmetry introduced in (8.27) can be formulated using the following I (I − 1) /2 constraints hij (θ) = pij − pji = 0, i < j, i = 1, ..., I − 1, j = 1, ..., I.

(8.30)

Also considering the parameter β = (p11 , ..., p1I , p22 , ..., p2I , ..., pI−1I−1 , pI−1I )T and the set I(I+1) B = {(a11 , ..., a1I , a22, ..., a2I , ..., aI−1I−1 , aI−1I )T ∈ R 2 −1 : P i≤j aij < 1, 0 < aij , i, j = 1, .., I}, the hypothesis (8.27) can be expressed for some unknown θ0 ∈ Θ with p (θ 0 ) = p0 as H0 : p0 = p (g(β0 )) ,

β 0 ∈B and g(β 0 ) = θ 0 ,

(8.31)

where the function g is defined by g = (gij ; i, j = 1, ..., I, (i, j) 6= (I, I)) with gij (β) =

½

pij pji

i≤j , i>j

i, j = 1, ..., I − 1,

and gIj (β) = pjI , j = 1, ..., I − 1 giI (β) = piI , i = 1, ..., I − 1. Note that p (g(β)) = (gij (β); i, j = 1, ..., I)T , where gII (β) = 1 −

I X

gij (β).

i,j=1 (i,j)6=(I,I)

This approach can be formulated in terms of the results presented in Chapter 6. In this Chapter we consider the approach given in (8.30). We can observe that with this approach the hypothesis of symmetry can be written as H0 : p=p (θ0 ) , for some unknown θ0 ∈Θ0 , with Θ0 = {θ ∈Θ : hij (θ) = 0, i < j, i = 1, ..., I − 1, j = 1, ..., I} , p (θ) defined in (8.29) and hij in (8.30).


(8.32)

370


It is clear that the

¡ ¢ × I 2 − 1 matrix B S (θ0 ) given by µ ¶ ∂hij (θ0 ) B S (θ 0 ) = I(I−1) ∂θij ×(I 2 −1)

I(I−1) 2

(8.33)

2

has rank I (I − 1) /2, because the matrix of partial derivatives B S (θ0 ) has the same rank as the matrix ³ ´ I I(I−1) × I(I−1) , −I I(I−1) × I(I−1) , O I(I−1) ×I 2

2

2

2

2

where I I(I−1) × I(I−1) is the identity matrix of order I(I −1)/2 and O I(I−1) ×I is the matrix 2 2 2 with I(I − 1)/2 rows and I columns whose elements are all zero. For I = 2, we only have a constraint which is given by h12 (θ) = p12 − p21 = 0,

and for I = 3, we have three constraints given by h12 (θ) = p12 − p21 = 0 h13 (θ) = p13 − p31 = 0 h23 (θ) = p23 − p32 = 0. In the following we obtain the expression of the minimum φ-divergence estimator of θ0 ∈ Θ0 , i.e., the minimum φ-divergence estimator under the hypothesis of symmetry (8.30). Theorem 8.7 The minimum φ-divergence estimator, ³ ´T bS,φ = pS,φ ; i, j = 1, ..., I and (i, j) 6= (I, I) , θ ij

of θ0 in Θ0 (i.e., under the hypothesis of symmetry) is obtained as a solution of the system of equations µ µ ¶ µ ¶ µ ¶ µ ¶¶ pbij pbji pbij 0 pbij pbji 0 pbji 1 φ φ φ + φ − − − µ = 0, i, j = 1, ..., I, 2 pij pij pij pij pij pij (8.34) where µ has the expression · µ ¶ µ ¶¸ µ ¶ µ ¶¾ I J ½ 1 XX pbij pbji pbij pbji 0 0 µ= pij φ +φ − pbij φ − pbji φ . 2 i=1 j=1 pij pij pij pij

Its asymptotic distribution is given by ´ √ ³ S,φ L b − θ0 → n θ N (0, W S (θ0 )), n→∞


(8.35)


371

¡ ¢ ¡ ¢ where the I 2 − 1 × I 2 − 1 matrix W S (θ 0 ) is ³ ´−1 W S (θ0 ) = Σθ 0 − Σθ 0 B S (θ0 )T B S (θ0 ) Σθ 0 B S (θ0 )T B S (θ 0 ) Σθ 0 , and B S (θ 0 ) is the matrix defined in (8.33).

³ ´T bS,φ = pS,φ ; i, j = 1, ..., I and (i, j) 6= (I, I) Instead of getting θ we shall ij ³ ´T S,φ b ) = pS,φ ; i, j = 1, ..., I . The p0 s, i, j = 1, ..., I, which minimize the obtain p(θ ij ij φ-divergence p, p(θ)) Dφ (b

Proof.

subject to the null hypothesis of symmetry (or the constraints about θ given in (8.30)), may be obtained by minimizing   µ ¶ X I X I X I X I I I X X pbij pij φ λij (pij − pji ) + µ 1 − pij  (8.36) + p ij i=1 j=1 i=1 j=1 i=1 j=1 i<j

with respect to the pij , where µ and λij are undetermined Lagrangian multipliers. Minimizing the expression (8.36) is equivalent to minimizing the expression   µ µ ¶ µ ¶¶ I I X I I X 1 XX pbij pbji pij φ pij  . +φ + µ 1 − 2 i=1 j=1 pij pij i=1 j=1 Differentiating with respect to pij , i, j = 1, ..., I we get the system of equations given in bS,φ . (8.34), whose solutions provide the minimum φ-divergence estimator θ

The asymptotic distribution follows from (8.3) because in our model I F (θ 0 )−1 = Σθ 0 . Corollary 8.2

³ ´T bS,φ(λ) ) = pS,φ(λ) ; i, j = 1, ..., I of The minimum power-divergence estimator, p(θ ij p (θ 0 ) , under the hypothesis of symmetry (8.27), is given by S,φ pij (λ)

=

1 ¢ λ+1 ¡ λ+1 pbij + pbλ+1 ji

I X I X 1 ¡ λ+1 ¢ λ+1 pbij + pbλ+1 ji

,

i, j = 1, ..., I.

i=1 j=1

Proof. In this case the system of equations (8.34) can be written as λ+1  1 µ λ+1 λ+1 ¶ I µ λ+1 λ+1 ¶ λ+1 I X X p b +b p p b +b p ij ji ij ji 1  , i, j = 1, ..., I = 2 2 pλ+1 ij


i=1 j=1

(8.37)

372


and it is clear that the solution of this system of equations is given by (8.37). For λ = 0, pbij + pbji , i, j = 1, ..., I, 2 so we obtain the maximum likelihood estimator for symmetry introduced by Bowker (1948). S,φ(0)

pij

=

For λ → −1, we obtain as a limit case, S,φ(−1)

pij

=

(b pij pbji )1/2 , i, j = 1, ..., I I X I X 1/2 (b pij pbji ) i=1 j=1

i.e., the minimum modified likelihood estimator for symmetry introduced and studied in Ireland et al . (1969). For λ = 1, S,φ pij (1)

=

¢1/2 ¡ 2 pbij + pb2ji

I X I X ¡ 2 ¢1/2 pbij + pb2ji i=1 j=1

we get the minimum chi-square estimator for symmetry introduced in Quade and Salama (1975). Other interesting estimators for symmetry may be: For λ = −2 the minimum modified chi-square estimator, for λ = −1/2 the Freeman-Tukey estimator and finally for λ = 2/3 the Cressie-Read estimator. Remark 8.5 For I = 2, given θ0 = (p11 , p12 , p21 )T ∈ Θ0 , it is easy to check that · B S (θ0 ) = (0, 1, −1)

· Σθ 0 = diag (θ0 ) − θ 0 θT0 , ´−1 ³ ´−1 ³ · B S (θ0 ) Σθ 0 B S (θ0 )T = − (p12 − p21 )2 + p21 + p12 , and it is not difficult to establish that the matrix W S (θ0 ) has the expression 

 −p11 p12 p11 (1 − p11 ) −p11 p12  −p11 p12 − 12 p12 (2p12 − 1) − 12 p12 (2p12 − 1)  . −p11 p12 − 12 p12 (2p12 − 1) − 12 p12 (2p12 − 1)

This matrix can be written as



373



 µ ¶µ ¶ 1 0 1 0 0 11 p12  0 1  p11 (1 − p11 ) −p , 1 −p11 p12 0 1 1 2 p12 (1 − 2p12 ) 0 1 where: · The matrix

µ

p11 (1 − p11 ) −p11 p12 1 −p11 p12 2 p12 (1 − 2p12 )

¶

is the inverse of the Fisher information matrix corresponding to the parameter T β0 = (p11 , p12 ) , i.e., I F (β 0 )−1 . · The matrix



 1 0  0 1  0 1 ³ ´T 0) is the matrix of partial derivatives ∂g(β . ∂β

Then we have W S (θ0 ) =

µ

∂g(β0 ) ∂β

¶T

I F (β 0 )

−1

µ

∂g(β0 ) ∂β

¶

.

This result can be proved in general in the same way as the one corresponding to the problem of independence in Section 8.2 . Theorem 8.8 Under the hypothesis of symmetry, we have W S (θ0 ) = M Tβ 0 I F (β0 )M β 0 .

Proof. The proof follows the same steps as the proof of Theorem 8.2. bS,φ2 is BAN. From Theorem 8.8, it can be seen that θ

8.3.1.

Test of Symmetry

Based on the test statistic given in (8.5) we should reject the hypothesis of symmetry if

³ S,φ2 ´ b ≡ Snφ1 θ


2n

Dφ φ001 (1) 1

³ ´ bS,φ2 ) > c, b, p(θ p

374


where c is a positive constant. In ³ the following theorem we establish the asymptotic ´ S,φ2 φ1 b . distribution of the test statistic Sn θ

Theorem 8.9 The asymptotic distribution of the φ-divergence test statistics ³ S,φ2 ´ ³ ´ 2n b bS,φ2 ) b, p(θ = 00 Dφ1 p Snφ1 θ φ1 (1) and

³ S,φ2 ´ b ≡ Snφ1 ,h θ

(8.38)

³ ³ ´´ 2n bS,φ2 ) , b p , p( θ h D φ 1 h0 (0) φ001 (1)

for testing the hypothesis of symmetry, given in (8.27), is chi-square with of freedom.

(8.39) I(I−1) 2

degrees

Proof. In our case we have I 2 cells and by (8.28) I 2 − 1 parameters that are necessary to estimate. Using the relation (8.30) the number of constraints is I (I − 1) /2. Then by (8.5) the asymptotic distribution of the family of φ-divergence test statistics (8.38) is chi-square with I2 |{z} Cells

−

I2 − 1 | {z } Estimated Parameters

+

I (I − 1) /2 | {z } Constraints

−1 =

I (I − 1) /2 | {z } Degrees of . freedom

In relation with the family of test statistics given in (8.39), we have h (x) = h (0) + h0 (0) x + o (x) , then

³ ³ ³ ´´ ´ bS,φ2 ) = h0 (0) Dφ p bS,φ2 ) + oP (1) b, p(θ b h Dφ1 p , p( θ 1

and we get that the asymptotic distribution of the family of test statistics given in (8.39) is also chi-square with I (I − 1) /2 degrees of freedom. Based on this result we should reject the hypothesis of symmetry if ³ S,φ2 ´ ³ ³ S,φ2 ´´ b b > χ2I(I−1)/2,α or Snφ1 ,h θ > χ2I(I−1)/2,α . Snφ1 θ

(8.40)

Remark 8.6 ³ S,φ2 ´ b For φ1 (x) = φ2 (x) = φ(0) (x) = x log x − x + 1 we get that Snφ1 θ coincides with the classical likelihood ratio test statistic for symmetry I ³ S,φ(0) ´ X φ b =2 nij log G2 ≡ Sn (0) θ i,j=1 i6=j


2nij . nij + nji


375

³ S,φ2 ´ b For φ2 (x) = φ(0) (x) = x log x−x+1 and φ1 (x) = φ(1) (x) = 12 (x − 1)2 , Snφ1 θ coincides with the classical chi-square test statistic of Bowker (1948) and is given by I ³ S,φ(0) ´ X (nij − nji )2 φ b = X 2 ≡ Sn (1) θ . nij + nji i,j=1 i<j

It is also important to note that for φ1 (x) = φ2 (x) = φ(1) (x) = 12 (x − 1)2 we get the test statistic given by Quade and Salama (1975) and for φ2 (x) = x log x − x + 1 and whatever φ1 (x) we get the family of test statistics introduced and studied in Menéndez et al . (2001c).

Power of the Test Let q =(q11 , ..., qII )T be a point at the alternative hypothesis, i.e., there exist at least two indexes i and j for which qij 6= qji . We denote by θφa 2 the point on Θ verifying θ φa 2 =arg min Dφ2 (q, p (θ)), θ∈Θ0

where Θ0 was defined in (8.32). It is clear that θφa 2 = (fij (q); i, j = 1, ..., I, (i, j) 6= (I, I))T and p(θφa 2 ) = (fij (q); i, j = 1, ..., I, )T ≡ f(q), with fII (q) = 1 −

I X

fij (q).

i,j=1 (i,j)6=(I,I)

The notation fij (q) indicates that the elements of the vector θ φa 2 depend on q. For instance, for the power-divergence family φ(λ) (x) we have fij (q) =

¢ 1 ¡ λ+1 λ+1 λ+1 qij + qji

I X I X ¡ λ+1 ¢ 1 λ+1 λ+1 qij + qji

,

i, j = 1, ..., I.

i=1 j=1

We also denote ³ ´T bS,φ2 = pS,φ2 ; i, j = 1, ..., I, (i, j) 6= (I, I) θ ij © 2006 by Taylor & Francis Group, LLC

376


and then b p(θ

S,φ2

³ ´T 2 ) = pS,φ ; i, j = 1, ..., I ≡ f (b p), ij

b tends to q where f = (fij ; i, j = 1, ..., I)T . If the alternative q is true we have that p ³ ´ S,φ2 φ b and p(θ ) to p θa 2 in probability. If we define the function

Ψφ1 (q) = Dφ1 (q, f (q)), we have Ψφ1 (b p) = Ψφ1 (q) +

J I X X ∂Dφ1 (q, f(q)) (b pij − qij ) + o (kb p − qk) . ∂qij i=1 j=1

Then the random variables √ n (Dφ1 (b p, f (b p)) − Dφ1 (q, f (q))) and

I X J √ X ∂Dφ1 (q, f (q)) n (b pij − qij ) ∂qij i=1 j=1

have the same asymptotic distribution. If we define lij =

∂Dφ1 (q, f(q)) ∂qij

(8.41)

and l = (lij ; i, j = 1, ..., I)T , we have ´ ³ ´ √ ³ L T b − Dφ (q, f (q)) −→ n Dφ1 (b p, f (p)) N 0, l Σ l q , 1 n→∞

(8.42)

where Σq = diag (q) −qqT .

In the following theorem we present the asymptotic distribution using the maximum likelihood estimator. In this case we have: Theorem 8.10 ³ ´T b = pbij +bpji , i, j = 1, ..., I Let p(θ) the maximum likelihood estimator, under the 2

T symmetry hypothesis, of p (θ 0 ) = (p11 , ..., pII ) and let q be a point at the alternative hypothesis (q 6= p(θ0 )). Then, ´ ¡ ¢ √ ³ L b − Dφ (q, f (q)) −→ n Dφ1 (b p, p(θ)) N 0, σφ2 1 (q) , 1 n→∞

where

σφ2 1 (q) =

I X I X i=1 j=1 i6=j




I X I X

 qij m2ij −  

i=1 j=1 i6=j

2

 qij mij   ,


377

and mij =

1 2 φ1

³

2qij qij +qji

´

+ 12 φ1

³

2qji qij +qji

´

+

qij qij +qji

³ ³ ´ ³ ´´ ij ji φ01 qij2q+q − φ01 qij2q+q . ji ji

Proof. The result follows from (8.41) and (8.42). In the same way as in the problem of independence it is not easy to get the expression of the power function of the test statistics given in (8.40) when φ2 (x) 6= x log x − x + 1; however we can consider a contiguous sequence of alternative hypotheses that approach ¢ ¡ the null hypothesis H0 : pij = pji at the rate O n−1/2 . Consider the multinomial probability vector dij pn,ij = pij (θ 0 ) + √ , i, j = 1, ..., I, n P where d = (d11 , ..., dII )T is a fixed vector such that Ii,j=1 dij = 0 and θ0 is unknown but belonging to Θ0 . As n → ∞, the sequence of multinomial probabilities {pn }n∈N with ¶T µ dij pn = pn,ij + √ , i, j = 1, ..., I n

¢ ¡ converges to a multinomial probability in H0 at the rate of O n−1/2 . We name d H1,n : pn = p (θ) + √ , θ ∈Θ0 . n

Then, we have the following result: Theorem 8.11 Under H1,n : pn = p (θ0 ) + √dn , θ0 some unknown value in Θ0 , the test statistics ³ S,φ2 ´ b Snφ1 θ

and

³ S,φ2 ´ b Snφ1 ,h θ

are asymptotically noncentrally chi-squared distributed with I (I − 1) /2 degrees of freedom and noncentrality parameter δ=

I I X dij dji 1 X d2ij − . 2 i,j=1 pij i,j=1 pij i6=j

i<j

An interesting analysis of the problem of symmetry as well as a simulation study can be seen in Menéndez et al. (2005a).


378


In some real problems (i.e., medicine, psychology, sociology, etc.) the categorical response variables (X, Y ) represent the measure after or before a treatment. In such situations our interest is to determine the treatment effect, i.e., if X ≥ Y (we assume that X represents the measure after the treatment and Y before the treatment). In the following we understand that X is preferred or indifferent to Y , according to joint likelihood ratio ordering, if and only if pij ≥ pji ∀i ≥ j. In this situation the alternative hypothesis is H1 : pij ≥ pji , for all i ≥ j. This problem was first considered by Barmi & Kochar (1995) who presented the likelihood ratio test statistic for the problem of testing H0 : pij = pji against H1 : pij ≥ pji , ∀i ≥ j,

(8.43)

and considered the application of it to a real life problem: He tested if the vision of both the eyes, for 7477 women, is the same against the alternative that the right eye has better vision than the left eye. Menéndez et al . (2003c) considered the three hypotheses, H0 and H1 given in (8.43) and H2 no restriction over the pij ’s and studied some φ-divergence test statistics for testing H0 against H1 and H1 against H2 .

8.3.2.

Symmetry in a Three-way Contingence Table

If we consider a three-way contingency table, the parameter space, given in (8.7), is o n T Θ = θ : θ = (pijk ; i, j, k = 1, ..., I, (i, j, k) 6= (I, I, I)) ,

where pijk = Pr (X = i, Y = j, Z = k) , i, j, k = 1, ....., I, (i, j, k) 6= (I, I, I) and pIII = 1 −

I X

pijk .

i,j,k=1 (i,j,k)6=(I,I,I)

The hypothesis of symmetry is given by H0 : pijk = pi0 j 0 k0

∀(i, j, k) and ∀ permutations (i0 , j 0 , k0 ) of (i, j, k).

(8.44)

We denote by p (θ) = (p111 , ..., pIII )T the probability vector characterizing our model. The problem of symmetry formulated in (8.44) can be formulated using the following I (I − 1) (5I + 2)/6 constraints about θ hijk (θ) = pijk − pi0 j 0 k0 = 0, ∀ permutations (i0 , j 0 , k0 ) of (i, j, k). We denote Θ0 = {θ ∈ Θ/hijk (θ) = 0, ∀ permutations (i0 , j 0 , k0 ) of (i, j, k)} .


(8.45)

379


The matrix B ∗S (θ0 ) =

µ

∂hijk (θ0 ) ∂θijk

¶

(8.46) I(I−1)(5I+2) ×I 3 6

has rank I (I − 1) (5I +2)/6, because the matrix B ∗S (θ0 ) has the same rank as the matrix µ ¶ −I I(I−1)(5I+2) × I(I−1)(5I+2) , C I(I−1)(5I+2) × I(I−1)(I+4) , 0 I(I−1)(5I+2) , 6

6

6

6

×I

6

where I I(I−1)(5I+2) is the identity matrix of order I(I − 1)(5I + 2)/6, 0 I(I−1)(5I+2) 6

6

×I

is

the matrix with I(I − 1)(5I + 2)/6 rows and I columns whose elements are all zero and rows and I(I−1)(I+4) columns C I(I−1)(5I+2) × I(I−1)(I+4) is the matrix with I(I−1)(5I+2) 6 6 6 6 ´T ³ whose rank is the same as the rank of the matrix I I(I−1)(I+4) , I I(I−1)(I+4) . 6

6

Now we present in the following theorem the expression of the minimum φ-divergence estimator of θ0 in Θ0 . Theorem 8.12 The minimum φ-divergence estimator, ³ ´T bS,φ = pS,φ ; i, j, k = 1, ..., I, (i, j, k) 6= (I, I, I) , θ ijk

of θ0 in Θ0 is obtained as a solution of the system of equations ´ ³ ´ ³ ´ ³ ´ ³ ´ ³ ´´  ³ ³ p bijk p bikj p bjik p bjki p bkij p bkji  6 φ + φ + φ + φ + φ + φ −  p p p p p p  ijk ijk ijk³ ijk ³ ´ ijk ³ ´ ³ijk ´´   p bijk 0 p bijk p bikj 0 p bikj p bjik 0 p bjik  − pijk φ pijk + pijk φ pijk + pijk φ pijk ³ ³ ´ ³ ´ ³ ´´ p b bjki p bkij 0 p bkij p bkji 0 p bkji  jki 0 p  − + + − 6µ = 0 φ φ φ  p p p p p p  ijk ijk ijk ijk ijk ijk   i, j, k = 1, ..., I (8.47) where µ has the expression µ µ µ ¶ µ ¶ µ ¶ µ ¶ µ ¶ µ ¶¶ I X pbijk pbikj pbjik pbjki pbkij pbkji pijk φ +φ +φ +φ +φ +φ pijk pijk pijk pijk pijk pijk i,j,k=1 µ ¶ µ ¶ µ ¶¶ µ 1 pbijk pbikj pbjik − + pbikj φ0 + pbjik φ0 pbijk φ0 6 pijk pijk pijk µ ¶ µ ¶ µ ¶¶¶ µ 1 p b p b pbkji jki kij − + pbkij φ0 + pbkji φ0 pbjki φ0 6 pijk pijk pijk and its asymptotic distribution is given by √ L bS,φ − θ0 ) → n(θ N (0, W S (θ0 )) n→∞

where the matrix W ∗S (θ 0 ) is ¶ µ ³ ´−1 B S (θ0 ) Σθ0 , Σθ0 I (I 3 −1)×(I 3 −1) − B ∗S (θ0 )T B ∗S (θ0 ) Σθ0 B ∗S (θ0 )T


(8.48)

380


and B ∗S (θ0 ) is the matrix given in (8.46). Proof. Instead of getting ³ ´T bS,φ = pS,φ ; i, j, k = 1, ..., I, (i, j, k) 6= (I, I, I) θ ijk

³ ´T bS,φ ) = pS,φ ; i, j, k = 1, ..., I . The pijk , i, j, k = 1, ..., I, which we shall obtain p(θ ijk p, p(θ)) subject to the null hypothesis of symmetry may minimize the φ-divergence Dφ (b be obtained minimizing   µ ¶ I I I X X X pbijk pijk φ λijk (pijk − pi0 j 0 k0 ) + µ 1 − pijk  (8.49) + pijk i,j,k=1

i,j,k=1

i,j,k=1

with respect to pijk , where µ and λijk are undetermined Lagrangian multipliers. Minimizing the expression (8.49) is equivalent to minimizing the expression I X

i,j,k=1

µ µ ¶ µ ¶ µ ¶ µ ¶ µ ¶ pbijk pbikj pbjik pbjki pbkij pijk φ +φ +φ +φ +φ pijk pijk pijk pijk pijk

µ

pbkji +φ pijk

¶

µ

pbkji +φ pijk

¶¶



+ µ 1 −

I X

i,j,k=1



pijk  .

Differentiating with respect to pijk , i, j, k = 1, ..., I, we get the system of equations given bS,φ . in (8.47) whose solutions provide the minimum φ-divergence estimator θ

The asymptotic distribution follows from (8.3) because in our model I F (θ0 )−1 = Σθ 0 . Corollary 8.3 The minimum power-divergence estimator ³ ´ bS,φ(λ) ) = pS,φ ; i, j, k = 1, ..., I , p(θ ijk

under the hypothesis of symmetry, is given by ´ 1 ³ λ+1 λ+1 λ+1 λ+1 λ+1 λ+1 λ+1 + p b + p b + p b + p b + p b p b ijk ikj jik jki kij kji S,φ pijk (λ) = I , i, j, k = 1, ..., I. ´ 1 X ³ λ+1 λ+1 λ+1 λ+1 λ+1 λ+1 λ+1 pbijk + pbikj + pbjik + pbjki + pbkij + pbkji i,j,k=1

(8.50)

For λ = 0, S,φ

pijk (0) =

pbijk + pbikj + pbjik + pbjki + pbkij + pbkji , i, j, k = 1, ..., I, 6



381

hence, we obtain the maximum likelihood estimator. This estimator was introduced and studied by Bowker (1948). For λ → −1, we have S,φ

pijk (−1) =

(b pijk × pbikj × pbjik × pbjki × pbkij × pbkji )1/6 , i, j, k = 1, ..., I, I X 1/6 (b pijk × pbikj × pbjik × pbjki × pbkij × pbkji )

i,j,k=1

the minimum modified likelihood estimator and finally for λ = 1, ³ ´1/2 2 2 2 2 2 2 p b + p b + p b + p b + p b + p b ijk ikj jik jki kij kji S,φ , pijk (1) = I ³ ´1/2 X 2 2 2 2 2 2 pbijk + pbikj + pbjik + pbjki + pbkij + pbkji i,j,k=1

we get the minimum chi-square estimator for symmetry.

Based on the test statistic given in (8.5) we should reject the hypothesis of symmetry if

³ S,φ2 ´ b Snφ1 θ ≡

where c is a positive constant.

³ ´ 2n bS,φ2 ) > c, b, p(θ D φ1 p 00 φ1 (1)

In the following theorem we establish the asymptotic distribution of the φ-divergence ³ S,φ ´ 2 φ1 b test statistic Sn θ . Theorem 8.13 The asymptotic distribution of the φ-divergence test statistics ³ S,φ2 ´ ³ ´ 2n b bS,φ2 ) b, p(θ Snφ1 θ = 00 Dφ1 p φ1 (1)

and

³ S,φ2 ´ b ≡ Snφ1 ,h θ

³ ³ ´´ 2n bS,φ2 ) , b p , p( θ h D φ 1 h0 (0) φ001 (1)

(8.51)

(8.52)

for testing the hypothesis of symmetry, given in (8.44), is chi-square with I(I − 1)(5I + 2)/6 degrees of freedom. Proof. In our case we have I 3 cells and by (8.28) I 3 −1 parameters that are necessary to estimate. Using the relation (8.45) the number of constraints is I(I − 1)(5I + 2)/6. Then by (8.5) the asymptotic distribution of the family of test statistics (8.51) is a chi-square distribution with I3 |{z} Cells

−

− 1} I| 3{z Estimated Parameters


+ I(I − 1)(5I + 2)/6 −1 = I(I − 1)(5I + 2)/6 {z } | {z } | . Constraints Degrees of freedom

382


In relation with the family of test statistics given in (8.52), we have h (x) = h (0) + h0 (0) x + o (x) , then

³ ³ ³ ´´ ´ bS,φ2 ) = h0 (0) Dφ p bS,φ2 ) + oP (1) b, p(θ b , p( θ h Dφ1 p 1

and we get that the asymptotic distribution of the family of test statistics given in (8.52) is also chi-square with I(I − 1)(5I + 2)/6 degrees of freedom. A study of this problem with maximum likelihood estimator and φ-divergence test statistics can be seen in Menéndez et al . (2004b).

8.4.

Marginal Homogeneity

In case the pattern is not completely symmetric, one likes to check whether, at least, the two sets of marginal totals have the same distribution: marginal symmetry (or marginal homogeneity) or the local odd ratios are symmetric: quasi-symmetry. The problem of marginal homogeneity was first discussed by Stuart (1955), who defined a test statistic which is a quadratic form in the differences of the corresponding marginal values, whose matrix is the inverse of a consistent estimate of the covariance matrix of the differences under the null hypothesis and its asymptotic distribution is chi-square with I − 1 degrees of freedom under the null hypothesis of marginal homogeneity. This hypothesis has been discussed by several authors (e.g., Bhapkar 1966, 1979, Ireland et al . 1969, Bishop et al. 1975, Agresti 1983, Bhapkar and Darroch 1990, Kullback 1971). b = (b We consider a two-way contingency table with I = J and let p p11 , ...., pbII )T be the nonparametric estimator of the unknown probability vector p = (p11 , ..., pII )T , where pij = Pr (X = i, Y = j) , with pij > 0, and i, j = 1, ..., I. The hypothesis of marginal homogeneity is given by H0 :

I X i=1

pji =

I X i=1

pij,

j = 1, ..., I − 1,

and the parameter space is o n Θ = θ : θ = (pij ; i, j = 1, ..., I, (i, j) 6= (I, I))T .

(8.53)

(8.54)

We denote by

p (θ) = (p11 , ..., pII )T = p

(8.55)

the probability vector characterizing the marginal homogeneity model with pII = 1 − PI PI i=1 j=1 pij with (i, j) 6= (I, I). © 2006 by Taylor & Francis Group, LLC


383

The hypothesis of marginal homogeneity formulated in (8.53) can be formulated using the following I − 1 constraints about the parameter θ, hj (θ) =

I X i=1

pji −

I X

pij = 0,

j = 1, ..., I − 1,

i=1

(8.56)

or considering the parameter β = (p11 , p21 , .., p2I , ..., pI1 , .., pII−1 )T and the set B=

(

(a11 , a21 , .., a2I , ..., aI1 , .., aII−1 ) : aij > 0,

P

)

aij < 1

(i,j)∈L

where L = {(i, j) : i, j = 1, ..., I, (i, j) 6= (I, I) , (i, j) 6= (1, j) , j = 2, ..., I} . The hypothesis (8.53) can be expressed for some unknown θ0 ∈ Θ with p(θ 0 ) = p0 , by H0 : p0 =p (g(β0 )) , β 0 ∈B and g(β 0 ) = θ0 ,

(8.57)

where the function g is defined by gij (β) = −

I X

i6=1,j

gij (β) = pij g11 (β) = p11 .

pij +

I X

pjl

i = 1, j = 2, ..., I

l6=j

i 6= 1, j = 1, ..., I, (i, j) 6= (I, I)

We recall p (g(β)) = (gij (β), i, j = 1, ..., I)T , where gII (β) = 1 −

I X

gij (β).

i,j=1 (i,j)6=(I,I)

This approach can be formulated and solved in terms of the results presented in Chapter 6. In this Chapter we consider the approach given in (8.56). We can observe that with this approach the problem of marginal homogeneity can be written as H0 : p = p (θ0 ) , for some unknown θ0 ∈Θ0 , with Θ0 = {θ ∈Θ: hj (θ) = 0, j = 1, ..., I − 1} ,


384


p (θ) defined in (8.55) and hj , j = 1, ..., I − 1, in (8.56). It is clear that the matrix B MH (θ 0 ) =

µ

∂hj (θ 0 ) ∂θij

¶

(8.58)

(I−1)×(I 2 −1)

has rank I − 1 because the matrix I (I−1)×(I−1) is a submatrix of it. For I = 2, we only have a constraint and it is given by h1 (θ) = p12 − p21 = 0, so that the problem of marginal homogeneity coincides with the problem of symmetry. For I = 3, we have the two following constraints, h1 (θ) = p12 + p13 − p21 − p31 = 0 h2 (θ) = p21 + p23 − p12 − p32 = 0. In the following we present the expression of the minimum φ-divergence estimator of θ0 under the constraints given in (8.56). Theorem 8.14 The minimum φ-divergence estimator, ³ ´T bMH,φ = pMH,φ ; i, j = 1, ..., I, and (i, j) 6= (I, I) , θ ij

of θ0 in Θ0 (i.e., under the hypothesis of marginal homogeneity) is obtained as a solution of the system of equations µ ¶  µ ¶ pbij pbij 0 pbij     φ pij − pij φ pij − µ + λi − λj = 0, i, j = 1, ..., I, i 6= j I I (8.59) X X   p − p = 0, j = 1, ..., I  ji ij  i=1

i=1

where µ has the expression µ=

µ µ ¶ ¶ µ ¶¶ J µ I X X pbij pbij pij φ + λi − λj − pbij φ0 . pij pij i=1 j=1

Its asymptotic distribution is √ bMH,φ L n(θ − θ 0 ) → N (0, W MH (θ0 )), n→∞

where the (I 2 − 1) × (I 2 − 1) matrix W MH (θ0 ) is given by ³ ´−1 T T B MH (θ0 ) Σθ 0 . Σθ 0 − Σθ0 B MH (θ 0 ) B MH (θ0 ) Σθ 0 B MH (θ0 ) © 2006 by Taylor & Francis Group, LLC

(8.60)


385

³ ´T bMH,φ = pMH,φ ; i, j = 1, ..., I, and (i, j) 6= (I, I) we shall Proof. Instead of getting θ ij ³ MH,φ ´ ³ ´T MH,φ b obtain p θ = pij ; i, j = 1, ..., I . The p0ij s, i, j = 1, ..., I, which minimize the φ-divergence p, p(θ)) Dφ (b subject to the null hypothesis of marginal homogeneity (or the constraints about θ given in (8.56)), may be obtained minimizing     µ ¶ X I X I I I I X I I X X X X pbij pij φ λi  pij − pji  + µ 1 − pij  (8.61) + p ij i=1 j=1 i=1 j=1 j=1 i=1 j=1

with respect to the pij , where µ and λi are undetermined Lagrangian multipliers. Minimizing the expression (8.61), is equivalent to solving the system of equations given in (8.59). The asymptotic distribution is obtained from (8.3) because in our case I F (θ0 )−1 = Σθ 0 . Corollary 8.4

³ ´T bMH,φ ) = pMH,φ ; i, j = 1, ..., I of The minimum power-divergence estimator, p(θ ij θ 0 under the hypothesis of marginal homogeneity (8.56), is given by MH,φ(λ)

pij

pbij

=

i, j = 1, ..., I

1

((λ + 1) (−µ + λi − λj )) λ+1

where µ and λi , i = 1, ..., I, must satisfy the system of equations  I I XX  p bij   =1 1   ((λ+1)(−µ+λi −λj )) λ+1     

i=1 j=1 I X j=1

p bij

1

=

((λ+1)(−µ+λi −λj )) λ+1

I X l=1

p bli

. 1

, i = 1, ..., I

((λ+1)(−µ+λl −λi )) λ+1

The maximum likelihood estimator, obtained for λ = 0, is MH,φ(0)

pij where the λi holds I X j=1

=

pbij , 1 + λi − λj

i, j = 1, ..., I

I

X pbij pbji = , i = 1, ..., I. 1 + λi − λj 1 + λj − λi j=1

The minimum modified likelihood estimator, obtained for λ → −1, is a

MH,φ(−1) pij

= PI

i=1


pbij aji PI

j=1

a

pbij aji

,

i, j = 1, ..., I,

386


where the ai holds ai

I X pbji j=1

I 1 X = pbil al , i = 1, ..., I. aj ai l=1

Remark 8.7 T For I = 3, given θ0 = (p11 , p21 + p23 − p32 , p31 + p32 − p23 , p21 , p22 , p23 , p31 , p32 ) ∈ Θ0 , it is easy to check that

· B MH (θ 0 ) =

µ

0 1 1 −1 0 0 −1 0 0 −1 0 1 0 1 0 −1

¶

,

· Σθ 0 = diag (θ0 ) − θ 0 θT0 , ³ ´−1 · The matrix B MH (θ 0 ) Σθ 0 B TMH (θ0 ) is 1 4p21 p31 +4p23 p31 +4p21 p32 −(p23 −p32 )2

µ

2 (p23 + p21 ) 2p21 + p23 − p32

2p21 + p23 − p32 2 (p31 + p21 )

¶

and it is not difficult to establish that the matrix W MH (θ0 ) is W MH (θ0 ) = where

µ

∂g(β0 ) ∂β 

   µ ¶T   ∂g(β0 )  =  ∂β    

1 0 0 0 0 0 0 0

¶T 0 1 0 1 0 0 0 0

−1

I F (β 0 )

µ

∂g(β0 ) ∂β

¶

 0 0 0 0 0 1 0 −1   0 −1 1 1    0 0 0 0   1 0 0 0   0 1 0 0   0 0 1 0  0 0 0 1

and I F (β 0 )−1 is the inverse of the Fisher information matrix for the parameter β 0 = (p11 , p21 , p22 , p23 , p31 , p32 )T . This result can be proved in general in the same way as the results in Sections 2 and 3 corresponding to hypotheses of independence and symmetry, respectively. From this bMH,φ2 is also BAN. result, θ Theorem 8.15 Under the marginal homogeneity model, we have

W MH (θ 0 ) = M Tβ 0 I F (β0 )−1 M β 0 .



387

Proof. The proof follows the same steps as the proof of Theorem 8.2 In the following theorem we present two families of test statistics for testing marginal homogeneity based on the minimum φ-divergence estimator. Theorem 8.16 The asymptotic distribution of the φ-divergence test statistics ³ MH,φ2 ´ ³ ´ 2n b bMH,φ2 ) b, p(θ M Hnφ1 θ ≡ 00 Dφ1 p φ1 (1) and

³ MH,φ2 ´ b ≡ M Hnφ1 ,h θ

(8.62)

³ ³ ´´ 2n bMH,φ2 ) b p , p( θ h D φ 1 h0 (0) φ001 (1)

(8.63)

for testing the hypothesis of marginal homogeneity given in (8.53) is chi-square with I −1 degrees of freedom. Proof. In our case we have I 2 cells and by (8.54) I 2 −1 parameters that are necessary to estimate. Using the relation (8.56) the number of constraints is I − 1. Then by (8.5), the asymptotic distribution of the family of φ-divergence test statistics (8.62) is chi-square with I2 |{z} Cells

−

− 1} I| 2{z Estimated Parameters

+

(I − 1) | {z } Constraints

−1 =

(I − 1) | {z } Degrees of . freedom

In relation with the family of test statistics given in (8.63), we have h (x) = h (0) + h0 (0) x + o (x) , therefore

³ ³ ³ ´´ ´ bMH,φ ) = h0 (0) Dφ p bMH,φ ) + oP (1) b, p(θ b h Dφ1 p , p( θ 1

and we get that the asymptotic distribution of the family of test statistics given in (8.63) is also chi-square with I − 1 degrees of freedom.

³ MH,φ2 ´ ³ ³ MH,φ2 ´´ b b MHnφ1 ,h θ for testing the If we use the test statistics MHnφ1 θ marginal homogeneity we should reject ³ MH,φ ´ ³the null hypothesis, ³ MH,φ2 ´í.e., the hypothesis of mar2 φ1 b φ1 ,h b MHn θ is too large. Based on the ginal homogeneity if MHn θ previous theorem we should reject the hypothesis of marginal homogeneity, with significance level α, if ³ MH,φ2 ´ ³ MH,φ2 ´ b b > χ2I−1,α or MHnφ1 ,h θ > χ2I−1,α . MHnφ1 θ © 2006 by Taylor & Francis Group, LLC

388

8.5.


Quasi-symmetry

The symmetry concept is stronger than that of marginal symmetry in the sense that the latter is implied by the former. The hypothesis of quasi-symmetry was introduced by Caussinus (1965) who introduced the maximum likelihood estimator for quasi-symmetry as well as a chi-square type statistic for testing this hypothesis. For additional discussion of quasi-symmetry, see Darroch (1981, 1986), McCullagh (1982), Darroch and McCullagh (1986) and Agresti (2002). Recently, Matthews and Growther (1997) have studied quasisymmetry and independence for cross-classified data in a two-way contingency table. These models are expressed in terms of the cross-product ratios and a maximum likelihood estimation procedure is proposed for estimating the expected frequencies subject to the constraints imposed on the frequencies through the cross-product ratios. b = (b We consider a two-way contingency table with I = J and let p p11 , ..., pbII )T be the nonparametric estimator of the unknown probability vector p = (p11 , ..., pII )T , where pij = Pr (X = i, Y = j) , with pij > 0 and i, j = 1, ..., I. The hypothesis of quasi-symmetry is given by H0 : pij pjk pki = pik pkj pji ,

(8.64)

for all i, j, k = 1, ..., I. We consider the parameter space o n Θ = θ : θ = (pij ; i, j = 1, ..., I, (i, j) 6= (I, I))T ,

(8.65)

and we denote by

p (θ) = (p11 , ..., pII )T = p the probability vector characterizing our model with pII = 1− (I, I).

(8.66) PI

i=1

PI

j=1 pij

and (i, j) 6=

There are other two equivalent ways to characterize the quasi-symmetry showed by Caussinus (1965). The hypothesis (8.64) is equivalent to H0 : sij = sji , i 6= j with sij =

pij pII , i, j = 1, ..., I − 1, piI pIj

(8.67)

as well as to H0 : cij = cji , i 6= j, cij > 0 with pij = ai cij , ai > 0.


(8.68)


389

The hypothesis of quasi-symmetry in the way given by (8.67) could be formulated using the following (I − 1)(I − 2)/2 constraints about θ, hij (θ) = pij pjI pIi − piI pIj pji = 0,

(8.69)

for all i, j = 1, ..., I − 1, i < j. Also considering the parameter T

β = (p11 , p1I , p21 , p22 , p2I , ..., pI1 , pI2 , ..., pII−1 ) and the set

B = {(a11 , a1I , a21 , a22 , a2I , ..., aI1 , aI2 , ..., aII−1 )T ∈ R P aij < 1, aij > 0, i, j = 1, ..., I},

(I−1)(I+4) 2

:

i,j

the hypothesis (8.69) can be expressed as

H0 : p=p (g(β0 )) , β0 ∈B and g(β0 ) = θ0 , where the function g is defined by piI pIj pji pjI pI1 gij (β) = pij giI (β) = piI gij (β) =

i, j = 1, ..., I − 1, i < j i = 1, ..., I, j = 1, ..., I − 1, i ≥ j i = 1, ..., I − 1.

We observe that p (g(β)) = (gij (β), i, j = 1, ..., I)T , where gII (β) = 1 −

I X

gij (β).

i,j=1 (i,j)6=(I,I)

This approach can be formulated and solved in terms of the results presented in Chapter 6. In this Section we consider the approach given in (8.67). We can observe that with this approach the problem of quasi-symmetry can be written as H0 : p = p (θ0 ) ,

for some unknown θ0 ∈ Θ0 ,

with Θ0 = {θ ∈ Θ : hij (θ) = 0, i, j = 1, ..., I − 1, i < j} , p (θ) defined in (8.66) and hij (θ) in (8.69). In this case the matrix of partial derivatives µ ¶ ∂hij (θ0 ) B QS (θ0 ) = ∂θij (I−1)(I−2)/2×(I 2 −1)


(8.70)

390


has rank (I −1)(I − 2)/2. For more details about quasi-symmetry hypothesis see Bhapkar (1979). For I = 3, we have a constraint given by h12 (θ) = p12 p23 p31 − p13 p32 p21 = 0. Next theorem presents the expression of the minimum φ-divergence estimator of θ0 P under the constraints given in (8.69). We shall denote pi∗ = Ij=1 pij . Theorem 8.17 The minimum φ-divergence estimator, ³ ´T bQS,φ = pQS , i, j = 1, ..., I, (i, j) 6= (I, I) , θ ij

of θ0 ∈ Θ0 (i.e., under the hypothesis of quasi-symmetry) is obtained as a solution of the system of equations ³ ´ ³ ´ ³ ´ ³ ´  p b p b p b p b  pij φ pij + pji φ pji − pbij φ0 pij − pbji φ0 pji = µ (pij + pji ) , i, j = 1, ..., I  ij ji ij ji    I X

    

pij φ

j=1

with

³

p bij pij

´

−

I X j=1

pbij φ0

³

p bij pij

´

= µpi∗ , i = 1, ..., I

(8.71)

µ ¶ µ ¶¶ I X I µ X pbij pbij 0 µ= pij φ − pbij φ . p p ij ij i=1 j=1

Its asymptotic distribution is

√ bQS,φ L n(θ − θ 0 ) → N (0, W QS (θ0 )), n→∞

(8.72)

where the (I 2 − 1) × (I 2 − 1) matrix W QS (θ0 ) has the expression ³ ´−1 Σθ0 − Σθ0 B QS (θ0 )T B QS (θ0 ) Σθ0 B QS (θ 0 )T B QS (θ 0 ) Σθ0 , and the matrix B QS (θ0 ) is defined in (8.70).

³ ´T bQS,φ = pQS,φ ; i, j = 1, ..., I, and (i, j) 6= (I, I) Proof. Instead of getting θ we shall ij obtain ³ ´T bQS,φ ) = pQS,φ ; i, j = 1, ..., I . p(θ ij

The

p0ij s

which minimize the φ-divergence

p, p(θ)) Dφ (b



391

subject to the null hypothesis of quasi-symmetry may be obtained minimizing   µ ¶ I X I X I I X X X pbij pij φ λij (pij pjI pIi − piI pIj pji ) + µ 1 − pij  + pij i=1 j=1 i=1 j=1 (i,j)∈C

where C = {(i, j) : i, j = 1, ..., I − 1, i < j} . By using the characterization given in (8.68), we must obtain the minimum of the function   µ ¶ I X I X I I X X X pbij ai cij φ λij (cij − cji ) + µ 1 − ai cij  (8.73) + ai cij i=1 j=1 i=1 j=1 (i,j)∈C

with respect to ai and cij , where µ and λij are undetermined Lagrangian multipliers. Minimizing the expression (8.73) is equivalent to minimizing the expression   µ µ ¶ ¶¶ I µ I I I X X 1 XX pbij pbji 1 (ai + aj ) cij  ai cij φ + aj cij φ + µ 1 − 2 i=1 j=1 ai cij aj cij 2 i=1 j=1 and the values of ai and cij are obtained as a solution of the system of equations given in (8.71). The asymptotic distribution is obtained from (8.3) because in this case I F (θ 0 )−1 = Σθ 0 . Corollary 8.5

´T ³ QS,φ(λ) ´ ³ QS,φ b = pij (λ) ; i, j = 1, ..., I The minimum power-divergence estimator, p θ of p (θ 0 ) under the hypothesis of quasi-symmetry (8.64) is given by the solution of the system of equations  µ ¶ p bλ+1 p bλ+1  ij ji 1  − + p − p = µ (pij + pji ) , i, j = 1, ..., I, i 6= j  ij ji  pij λ pji λ  λ+1   I X p bλ+1  ij 1    = µpi∗ , i = 1, ..., I, pi∗ −   λ+1 pij λ  j=1

where µ is

  I I X X pbλ+1 1  ij  1− µ= . λ λ+1 p i=1 j=1 ij

It is clear that for λ → 0, we obtain the maximum likelihood estimator introduced by Caussinus (1965), i.e., ½ pij + pji = pbij + pbji . pi∗ = pbi∗ © 2006 by Taylor & Francis Group, LLC

392


Remark 8.8 For I = 3, given µ ¶T p13 p32 p31 , p13 , p21 , p22 , p23 , p31 , p32 ∈ Θ0 θ0 = p11 , p21 p31 it is easy to check that · B QS (θ 0 ) =

¡

0, p23 p31 , −p32 p21 , −p13 p32 , 0, p12 p31 , p12 p23 , −p13 p21

· Σθ 0 = diag (θ0 ) − θ 0 θT0 , ´−1 ³ · The matrix B QS (θ0 ) Σθ 0 B QS (θ 0 )T is given by

¢

p23 p31 p13 p32 p21 (p23 p31 (p23 p31 + p21 p32 + p13 p32 + p13 p21 ) + p13 p32 p21 (p23 + p31 ))

and it is not difficult to establish that the matrix W QS (θ0 ) is W QS (θ 0 ) =

µ

∂g(β 0 ) ∂β

¶T

I F (β0 )−1

µ

∂g(β 0 ) ∂β

¶

,

where 

1  0    ¶T  0 µ  0 ∂g(β 0 ) =  0 ∂β    0   0 0

0

0

p21 p32 p23 p31

p13 p32 p23 p31

1 0 0 0 0 0

0 1 0 0 0 0

0 0 p32 p21 0 − p13 p223 p31 0 0 0 0 1 0 0 1 0 0 0 0

0

0

p32 p21 − p13 p23 p231 0 0 0 0 1 0

p13 p21 p23 p31

0 0 0 0 0 1

             

and I F (β 0 )−1 is the inverse of the Fisher information matrix for the parameter β 0 = (p11 , p13 , p21 , p22 , p23 , p31 , p32 )T . This result can be obtained in general in the same way bQS,φ2 is BAN. as in the previous problems. Therefore, θ Now we shall establish a theorem to get the asymptotic distribution of two new families of test statistics introduced for testing the hypothesis of quasi-symmetry given in (8.64). Theorem 8.18 The asymptotic distribution of the φ-divergence test statistics ³ QS,φ2 ´ ³ ´ 2n b bQS,φ2 ) b, p(θ QSnφ1 θ ≡ 00 Dφ1 p φ1 (1) © 2006 by Taylor & Francis Group, LLC

(8.74)

,


and

³ QS,φ2 ,h ´ b ≡ QSnφ1 θ

³ ³ ´´ 2n bQS,φ ) b p , p( θ h D φ 1 h0 (0) φ001 (1)

393

(8.75)

for testing the hypothesis of quasi-symmetry is chi-square with (I − 1) (I − 2) /2 degrees of freedom. Proof. In our case we have I 2 cells and I 2 − 1 parameters that are necessary to estimate. Using the relation (8.69) the number of constraints is (I − 1) (I − 2) /2. Then the asymptotic distribution of the family of statistics (8.74) is chi-square with I2 |{z} Cells

−

I2 − 1 | {z } Estimated Parameters

+ (I − 1) (I − 2) /2 −1 = (I − 1) (I − 2) /2 | {z } {z } | . Constraints Degrees of freedom

In relation with the family of tests statistics given in (8.75), we have h (x) = h (0) + h0 (0) x + o (x) , then

³ ³ ³ ´´ ´ bQS,φ2 ) = h0 (0) Dφ p bQS,φ2 ) + oP (1) b, p(θ b, p(θ h Dφ1 p 1

and we get that the asymptotic distribution of the family of tests statistics given in (8.75) is also chi-square with (I − 1) (I − 2) /2 degrees of freedom. Caussinus (1965) showed that symmetry is equivalent to quasi-symmetry and marginal homogeneity simultaneously holding; then we have Quasi-Symmetry + Marginal homogeneity = Symmetry.

(8.76)

From this idea we consider the two following families of φ-divergences test statistics ³ ³ ³ ´ ´´ 2n bS,φ2 ) − Dφ p bQS,φ2 ) (8.77) b b D p , p( θ , p( θ = WφMH 00 φ 1 1 φ (1) 1 ,φ2 1

and

= SφMH 1 ,φ2

2n

φ00 2 (1)

³ ´ bQS,φ2 ), p(θ bS,φ2 ) . Dφ1 p(θ

(8.78)

In the following theorem, its proof can be seen in Menéndez et al. (2005c), we present their asymptotic distribution. Theorem 8.19 For testing hypotheses, H0 : Symmetry versus H1 : Quasi-Symmetry, the asymptotic null distribution of the test statistics WφMH and SφMH given in (8.77) 1 ,φ2 1 ,φ2 and (8.78) respectively is chi-square with I − 1 degrees of freedom.


394

8.6.


Homogeneity

Suppose we have ν independent random samples and we are interested in testing the null hypothesis that the samples are homogeneous, i.e., are from the same un´T ³ (1) (1) ,..., X (ν) = known population. We denote the samples by X (1) = X1 , ..., Xn1 ³ ´T (ν) (ν) X1 , ..., Xnν , of sizes n1 , ..., nν respectively, and we are interested in deciding if the samples X (1) , ..., X (ν) are all derived from the same distribution function F (x) = Q (X ≤ x) , x ∈ R, where Q is a probability measure on the real line. In this direction let P = {Ei }i=1,...,M be a partition of ´the real line into M mutually exclusive and exhaus³ (i)

tive intervals, where Pr Xk ∈ Ej = pij for i = 1, ..., ν, j = 1, ..., M and k = 1, ..., ni . PM T We denote θi = (pi1 , ..., piM−1 ) , where 0 < pij < 1, j=1 pij = 1, ∀i = 1, ..., ν, and p (θi ) = (pi1 , ..., piM )T . If X (1) , ..., X (ν) are all drawn, at random, from the same distribution function F, then it is expected that pij = Q (Ej ) , for every i = 1, ..., ν and j = 1, ..., M and therefore the problem is now reduced to a problem of testing homogeneity in multinomial populations, i.e., the null hypothesis, H0 : p1j = ... = pνj = Q (Ej ) = p∗j , j = 1, ..., M, where 0 < p∗j < 1 and

PM

∗ j=1 pj

(8.79)

= 1.

We can observe that our parameter space is given by n o Θ = θ : θ = (p11 , ..., p1M−1 , ..., pν1 , ..., pνM−1 )T ,

´T ³ that is, θ = θ T1 , ..., θTν and its dimension is (M − 1) ν.

(8.80)

T

The usual test statistics for testing (8.79), if p∗ = (p∗1 , ..., p∗M ) is completely unknown, are the chi-square test statistic and the likelihood ratio test given, respectively, by ³ ni∗ n∗j ´2 M ν X nij − X n X2 = (8.81) ni∗ n∗j i=1 j=1 n and µ ¶ ν X M X n∗j nij G2 = 2 nij log − log . (8.82) ni∗ n i=1 j=1 Their asymptotic distribution under the null hypothesis H0 given in (8.79) is chi-square with (M − 1)(ν − 1) degrees of freedom. In the expressions (8.81) and (8.82), nij is the observed number of the components of X (i) , i = 1, .., ν, belonging to the interPν PM val Ej , j = 1, ..., M, n∗j = i=1 nij , j = 1, ..., M, ni∗ = j=1 nij , i = 1, ..., ν and © 2006 by Taylor & Francis Group, LLC


395

P P PM Pν n = νi=1 M j=1 nij = j=1 n∗j = i=1 ni∗ . In the two previous test statistics the unknown parameters have been estimated using the maximum likelihood estimator in all the parameter space (nij /ni∗ ) and the maximum likelihood estimator under the null hypothesis (n∗j /n). The test statistics given in (8.81) and (8.82) were extended by Pardo, L. et al . (1999) by considering test statistics based on φ-divergence measures and using the maximum likelihood estimator. An extension of that result when the probability distribution p∗ = (p∗1 , ..., p∗M )T depends on some unknown parameters can be seen in Pardo L. et al . (2001). In the following we shall denote by bi ) = p(θ

µ

ni1 niM , ..., ni∗ ni∗

¶T

,

the maximum likelihood estimator of p (θi ) , i.e., the probabilities corresponding to the ith-population. We also consider the two following probability vectors ³n ´T ³ n n1M nν1 nνM ´T 1∗ 11 b1 )T , ..., nν∗ p(θ bν )T b= p p(θ , ..., , ..., , ..., = n n n n n n and

´T n1∗ nν∗ nν∗ p1M , ..., pν1 , ..., pνM n n ´T n ³ nn nν∗ 1∗ T T p (θ1 ) , ..., p (θν ) = . n n

p∗ (θ) =

³n

1∗

p11 , ...,

b and p∗ (θ) is given by The φ-divergence between the vectors p ∗

Dφ (b p, p (θ)) =

µ

ν X M X ni∗ i=1 j=1

nij pij φ n ni∗ pij

¶

.

It is clear that Hnφ

³ ´ b ≡ θ

ν

M

2n 2n X X ni∗ n∗j ∗ b D φ (b p , p ( θ)) = φ φ00 (1) φ00 (1) i=1 j=1 n n

µ

nij n ni∗ n∗j

¶

b is the coincides with the test statistic X 2 , given in (8.81), for φ (x) = 12 (x − 1)2 , where θ maximum likelihood estimator of θ ∈ Θ, under the null hypothesis, given by ³ n∗M−1 ´T (ν n∗1 b = n∗1 , ..., n∗M−1 , ..., θ , ..., , (8.83) n n n n b is obtained from p∗ (θ) replacing θ by θ. b The likelihood ratio test statistic G2 , and p∗ (θ) given in (8.82), is obtained for φ (x) = x log x − x + 1. Hnφ

Pardo, ³ ´ L. et al . (1999) established that the asymptotic distribution of the test statistic b is chi-square with (M − 1)(ν − 1) degrees of freedom. In this section instead of θ


396


considering the maximum likelihood estimator we consider the minimum φ-divergence estimator. The hypothesis of homogeneity given in (8.79) can be formulated using the (ν − 1) (M − 1) constraints, hij (θ) = pij − pνj = 0, i = 1, ..., ν − 1, j = 1, ..., M − 1.

(8.84)

In this case Θ0 = {θ ∈Θ : hij (θ) = 0} , ´T ³ (ν therefore if θ0 ∈ Θ0 , θ 0 is given by θ0 = p1 , ..., pM−1 , ..., p1 , ..., pM−1 . We denote

T θ01 = (p1 , ..., pM−1 ) and

¡ ¢ I λ (θ0 ) = diag (λ) ⊗ IF θ01 ,

where λi = lim ni∗ /n, i = 1, ..., ν, λ = (λ1 , ..., λν )T and n→∞

¡ ¢ IF θ01 =

(

1 pi

+ 1 pM

1 pM

if i = j i, j = 1, ..., M − 1. if i = 6 j

¡ ¢ By ⊗ we are denoting the Kronecker product of the matrices diag (λ) and IF θ01 .

In the following theorem, its proof can be seen in Menéndez et al. (2003b), we present the expression of the minimum φ−divergence estimator of θ under the constraints given in (8.84). Theorem 8.20 ³ ´T bH,φ = pH,φ , j = 1, ..., M − 1 , of θ 0 under The minimum φ-divergence estimator, θ j the hypothesis of homogeneity, is obtained as a solution of the system of the equations ! Ã !! Ã Ã ν X nij 0 nij ni∗ nij φ − − µ = 0, j = 1, ..., M, φ (8.85) n ni∗ p∗j ni∗ p∗j ni∗ p∗j i=1 where µ=

ν M X X j=1 i=1

ni∗ p∗j n

Ã Ã φ

nij ni∗ p∗j

!

nij 0 φ − ni∗ p∗j

Ã

nij ni∗ p∗j

Its asymptotic distribution is ´ √ ³bH,φ L n θ − θ0 → N (0, W H (θ0 )) , n→∞

!!

.

(8.86)

where the matrix W H (θ0 ) has the expression µ ¶ ³ ´−1 T T −1 −1 I −1 B (θ ) I − B (θ ) (θ ) I (θ ) B (θ ) B (θ ) I (θ ) , 0 H 0 H 0 0 H 0 H 0 0 λ λ λ



Ãµ

B H (θ0 )(ν−1)(M−1)×ν(M−1) =

∂hij (θ) ∂θst

¶

θ =θ 0

!

397

,

i=1,...,ν−1;j=1,...,M−1 s=1,...,ν; t=1,...,M−1

= (I, M H ) I denotes the (M − 1)(ν − 1) × (M − 1)(ν − 1) identity matrix, M H is the matrix given by ´T ³ ... , −I (M−1)×(M−1) M H = −I (M−1)×(M−1) , (ν−1 and the rank of B H (θ0 ) is (M − 1) × (ν − 1) .

Corollary 8.6 bH,φ(λ) ) = The minimum power-divergence estimator, p(θ p (θ 0 ) under the hypothesis of homogeneity (8.84) is given by

H,φ(λ)

pj

³ ´T pH,φ , j = 1, ..., M , of j

!1/(λ+1) ν nλ+1 P ij λ i=1 ni∗ = Ã !1/(λ+1) , j = 1, ..., M. M ν nλ+1 P P ij λ j=1 i=1 ni∗ Ã

(8.87)

The proof follows from (8.85) and taking into account the expression of φ(λ) . It is interesting to observe that for λ = 0, H,φ(0)

pj

=

n∗j , j = 1, ..., M, n

i.e., we obtain the classical maximum likelihood estimator under homogeneity and for λ = 1, !1/2 Ã ν n2 P ij i=1 ni∗ H,φ pj (1) = Ã !1/2 , j = 1, ..., M M ν n2 P P ij j=1 i=1 ni∗

the minimum chi-square estimator under homogeneity. This estimator was obtained, for the first time, in a different way from the one presented here, by Quade and Salama (1975). Other interesting estimators for homogeneity are: For λ = −2 the minimum modified chi-square estimator; for λ → −1, the minimum modified likelihood estimator; for λ = −1/2 Freeman-Tukey estimator and finally for λ = 2/3 Cressie-Read estimator. The asymptotic distribution of the φ-divergence test statistic for testing homogeneity, based on the restricted minimum φ-divergence estimator, is given in the following theorem. Its proof can be seen in Menéndez et al . (2003b).


398


Theorem 8.21 The asymptotic distribution of the φ-divergence test statistic ³ H,φ2 ´ ³ ´ 2n b bH,φ2 ) b, p∗ (θ Hnφ1 θ ≡ 00 Dφ1 p φ1 (1)

for testing the hypothesis of homogeneity is chi-square with (ν − 1) (M − 1) degrees of freedom. Based on this theorem we should reject the null hypothesis of homogeneity given in (8.79), with significance level α, iff ³ H,φ2 ´ b Hnφ1 θ > χ2(ν−1)(M−1),α . (8.88)

Remark 8.9 ³ H,φ2 ´ b coincides with the For φ1 (x) = φ2 (x) = x log x − x + 1 we get that Hnφ1 θ classical likelihood ratio test statistic for homogeneity and for φ2 (x) = x log x − x + 1 and φ1 (x) = 12 (x − 1)2 coincides with the chi-square test statistic.



8.7.

399

Exercises

1. Find the power function of the test statistic given in (8.88) using the maximum likelihood estimator and the alternative hypothesis p∗ = where

³n

1∗

n

(p∗1 )T , ...,

nυ∗ ∗ T ´T (pν ) , n

p∗i = (p∗i1 , ..., p∗iM )T , 0 < p∗ij < 1, for i = 1, ..., ν and j = 1, ..., M and there exists an index j with 1 ≤ j ≤ M and two indexes i and k, 1 ≤ i, k ≤ ν and i 6= k such that p∗ij 6= p∗kj . 2. The following data represent the blood types and the predisposition to suffer from diabetes (A ≡ Low, B ≡ Average and C ≡ High) O A B AB

Low 137 42 19 14 212

Average 86 23 17 7 133

High 35 11 7 2 55

Total 258 76 43 23 400

Is there evidence to conclude that blood type is independent of predisposition to suffer from diabetes? Use the power-divergence family with λ = −1 − 1/2, 0, 2/3 and 1 for testing and λ = 0 for estimation, taking as significance level α = 0.05. 3. Fifteen 3-year-old boys and fifteen 4-year-old girls were observed during 30 minutes play sessions, and each child’s play during these two periods was scored as follows for incidence and degree of aggression: Boys Girls

96, 65, 74, 82, 121, 68, 79, 111, 48, 53, 92, 81, 31, 48 12, 47, 32, 59, 83, 14, 32, 15, 17, 82, 21, 34, 9, 15, 50

Test the hypothesis that there were not sex differences in the amount of aggression shown, using the power-divergence test statistic with λ = −1, −1/2, 0, 2/3 and 1, taking as significance level α = 0.05. P P 4. Let Sn = n1/2 Ii=1 Jj=1 lij (b pij − qij ) be the first term in the Taylor expansion ¢ ¡ b of Dφ1 (b p, p(θ)) around q, qI×J in Theorem 8.5. Prove that Sn = 0 ∀n if and only if σφ2 1 (q) = 0.

5. If qij = qi∗ × q∗j then σφ2 1 (q) = 0 in Theorem 8.5 because lij = 0 ∀i = 1, ..., I, j = 1, ..., J. Find a counter example, using the Kullback-Leibler divergence, in which σ2 (q) = 0 but qij 6= qi∗ × q∗j .


400


6. Find the expression of the test statistics: Likelihood ratio test, chi-square test statistic, modified likelihood ratio test statistic, Freeman-Tukey and modified chisquare test statistic for the problem of symmetry. 7. Find the expression of the power for Pearson test statistic as well as for the likelihood ratio test in the problem of symmetry using the maximum likelihood estimator. 8. The data in the following table report the relative heights of 205 married couples

Tall 18 20 12

Tall Medium Short

Men

Women Medium 28 51 25

Short 14 28 9

Source: Christensen, R. (1997, p. 67).

Find the minimum power-divergence estimators, under the hypothesis of symmetry, for λ = −2, −1, −0.5, 0, and 1. 9. Find the expression of σφ2 (q) in Theorem 8.5 for the φ-divergences defined by: √ 2 φ1 (x) = 12 (x − 1)2 ; φ2 (x) = (1 − x) ; φ3 (x) = (s − 1)−1 (xs − x) ; φ4 (x) = ¡ ¢ (λ (λ + 1))−1 xλ+1 − x and φ5 (x) = (1−a)(1−x) a+(1−a)x .

8.8.


1. We denote qs∗ =

ν X nl∗ l=1

n

∗ T p∗ls , s = 1, ..., M, q = (q1∗ , ..., qM )

and q∗ =

³n

1∗

n

qT , ...,

Under the alternative hypothesis

nν∗ T ´T q . n

P

b −→ p∗ p n→∞

and

P b −→ q∗ . p∗ (θ) n→∞

It is clear that in this case we have:

 nij  ν M X X n n i∗ ∗j b = 2n φ  ni∗nn∗j  p, p∗ (θ)) · 2nDφ1 (b n j=1 n l=1 n n



∗

∗

· 2nDφ1 (p , q ) = 2n and

M ν X X ni∗ l=1 j=1

n

qj∗ φ

Ã

p∗ij qj∗

!

´ √ ³ b − 2nDφ (p∗ , q∗ ) n 2nDφ1 (b p, p∗ (θ)) 1

takes the expression   Ã ! ν X M X ¢ ¡ √ ∂L p − p∗ k) , n pbij − p∗ij  + o (kb ∗ ∂p ij i=1 j=1

where

L=2

ν X M X

ni∗ qj∗ φ

l=1 j=1

Ã

p∗ij qj∗

!

.

It is not difficult to establish that µ ∗ ¶ µ ∗ ¶ µ ∗ ¶¶ µ υ X ph1 phs p ∗ ∗ ∗ nh∗ q1 φ L= 2 + ..... + qs φ + ... + qM φ hM ∗ q1∗ qs∗ qM h=1 h6=i Ã ! p∗ij ∗ + k, +2ni∗ qj φ qj∗ where k is independent of p∗ij . Taking into account that ∂qs∗ = ∂p∗ij

( ni∗ n 0

s=j s 6= j

,

we have Ã Ã ! Ã !! Ã ! υ p∗hj p∗hj 0 p∗hj p∗ij ∂L ni∗ X 0 = 2 n φ φ φ − + 2n . h∗ i∗ ∂p∗ij n qj∗ qj∗ qj∗ qj∗ h=1

Then, if we denote Aφij =

∂L , ∂p∗ij

we have that the random variables   Ã ! υ X M X ¢ ¡ ¢T √ ¡ √ ∂L n n (b p − p∗ ) , pbij − p∗ij  = Aφ ∗ ∂p ij i=1 j=1

´T ³ being Aφ = Aφ11 , ..., Aφ1M , ..., Aφν1 , ..., AφυM , and

´ √ ³ b − 2nDφ (p∗ , q∗ ) n 2nDφ1 (b p, p∗ (θ)) 1


401

402


have the same asymptotic distribution. But √ L n (b p − p∗ ) −→ N (0,Σ∗ ) , n→∞

¢ ¡ where Σ∗ = diag λ−1 , Σp∗i = diag (p∗i ) − p∗i (p∗i )T , i = 1, ..., ν , ..., λ−1 υ Σp∗ 1 Σp∗ 1 ν ni∗ and λi = limn→∞ . n Therefore, ´ ¡ ¢ √ ³ L b − 2nDφ (p∗ , q∗ ) → n 2nDφ1 (b p, p∗ (θ)) N 0, σ2φ1 (p∗ ) , 1 n→∞

where

σφ2 1 (p∗ ) =

υ X i=1

  2  M ³ M ´2 X X   Aφij p∗ij −  λ−1 Aφij p∗ij   . i  j=1

j=1

Based on this result, we have that the power of the test statistic is given by !! Ã √ Ã 2 χ(ν−1)(M−1),α n ∗ ∗ ∗ − Dφ1 (p , q ) , βn,φ1 (p ) = 1 − Φn σφ1 (p∗ ) 2n where Φn (x) is a sequence of distribution functions leading uniformly to the standard normal distribution Φ (x) . 2. If we consider the maximum likelihood estimator, the power-divergence test staφ bI,φ(0) ) has the expression tistic, In (λ) (θ   J I X λ+1 X n 2n φ(λ) bI,φ(0) ij nλ−1 )= − 1 . In (θ λ nλ λ (λ + 1) n i∗ ∗j i=1 j=1 In our case,

n1∗ = 258, n2∗ = 76, n3∗ = 43 and n4∗ = 23 n∗1 = 212, n∗2 = 133 and n∗3 = 55. Then we have λ φ bI,φ(0) ) In (λ) (θ

-1

-1/2

0

2/3

1

2.543

2.499

2.462

2.422

2.405

.

On the other hand χ2(I−1)(J−1),α = χ26, 0.05 = 12.59, then we should not reject the hypothesis that the two variables are independent. 3. If we consider the maximum likelihood estimator, the power-divergence test staφ bH,φ(0) ) has the expression tistic Hn (λ) (θ   ¡ nij ¢λ+1 k m X X 2 φ(λ) bH,φ(0)  )= ni∗ ¡nn ¢λ − n . Hn (θ ∗j λ (λ + 1) i=1 j=1 n



If we consider the following partition of the interval [8, 122] , E1 = [8, 43) , E2 = [43, 78) and E3 = [78, 122] , we have E1 1 10 11

Boys Girls n∗j Then, φ bH,φ(0) ) Hn (λ) (θ

2 λ(λ+1)

=

+

n2∗

E2 6 3 9

E3 8 2 10

ni∗ 15 . 15

½ µµ ¶ µ ¶ µ ¶¶ 1 λ+1 6 λ+1 8 λ+1 ( 15 ( 15 ( 15 ) ) ) n1∗ + + λ λ 9 λ ( 11 ( 30 ) ( 10 30 ) 30 )

µµ

λ+1 ( 10 15 ) λ ( 11 30 )

¶

+

µ

3 λ+1 ( 15 ) 9 λ ( 30 )

¶

+

µ

2 λ+1 ( 15 ) λ ( 10 30 )

¶¶

¾ −n

and λ φ bH,φ(0) ) Hn (λ) (θ

-1

-1/2

0

2/3

1

17.69

15.051

13.42

12.24

11.96

.

On the other hand χ22,0.05 = 5.991, then we should reject the null hypothesis. 4. If Sn = 0 a.s. ∀n, we have V ar (Sn ) = 0 ∀n and then lim V ar (Sn ) = σφ2 1 (q) = 0.

n→∞

Suppose that σφ2 1 (q) = 0. We have Sn

=

J I P √ P n lij (b pij − qij ) , i=1 j=1

then E [Sn ]

=

J I P √ P n lij E [b pij − qij ] = 0 i=1 j=1

"

V ar [Sn ] = E n =

I P

J P

li1 j1 li2 j2 (b pi1 j1 − qi1 j1 )(b pi2 j2 i1 ,i2 =1 j1 ,j2 =1 i h J I P P n li21 j1 E (b pi1 j1 − qi1 j1 )2 i1 =1 j1 =1 I J P P

+ n

i1 ,i2 =1 j1 ,j2 =1 i1 6=i2 j1 6=j2

− qi2 j2 )

#

li1 j1 li2 j2 E [(b pi1 j1 − qi1 j1 ) (b pi2 j2 − qi2 j2 )] .

But,

E [(b qi1 j1

i h qi j (1 − pi1 j1 ) = V [b pi1 j1 ] = 1 1 E (b pi1 j1 − qi1 j1 )2 n q q i j i j − qi1 j1 ) (b pi2 j2 − qi2 j2 )] = Cov [b pi1 j1 , pbi2 j2 ] = − 1 1 2 2 , n


403

404


therefore V ar [Sn ] =

J I P P

li21 j1 qi1 j1 (1 − qi1 j1 ) −

J I P P

li21 j1 qi1 j1 −

i1 =1 j1 =1

=

i1 =1 j1 =1

Then Sn = 0.

Ã

J I P P

i1 =1 j1 =1

I P

J P

li1 j1 l i2 j2 qi1 j1 qi2 j2

i1 ,i2 =1 j1 ,j2 =1 i1 6=i2 j1 6=j2 !2

li1 j1 qi1 j1

= σφ2 1 (q) = 0.

5. We consider the bivariate random variable (X, Y ) with probability distribution q11 = Pr (X = x1 , Y = y1 ) = 1/2 q21 = Pr (X = x2 , Y = y1 ) = 0

q12 = Pr (X = x1 , Y = y2 ) = 0 q22 = Pr (X = x2 , Y = y2 ) = 1/2.

We have l11 = l22 = − log 2, l12 = l21 = 0 with σ 2 (q) =

2 X 2 X i=1 j=1

On the other hand

 2 2 X 2 X 2 qij lij − qij lij  = 0. i=1 j=1

q1∗ = 1/2, q2∗ = 1/2, q∗1 = 1/2 and q∗2 = 1/2. Then qij 6= qi∗ × q∗j and σ 2 (q) = 0. 6. It is immediate to get for the power-divergence family and for the maximum likelihood estimator that ! Ãµ ¶λ X 2 2nij φ(λ) bS,φ(0) Sn (θ )= nij −1 . λ (λ + 1) i,j nij + nji i6=j

Then for λ → 0 (likelihood ratio test statistic), we have X 2nij φ bS,φ(0) ) ≡ G2 = 2 Sn(0) (θ nij log , n ij + nji i,j i6=j

for λ → −1 (modified likelihood ratio test statistic), X nij + nji φ bS,φ(0) ) = (nij + nji ) log , Sn (−1) (θ 2nij i,j i6=j

for λ = 1 (chi-square test statistic), φ bS,φ(0) ) ≡ X 2 = Sn (1) (θ


X (nij − nji )2 i,j i<j

nij + nji

,


405

for λ = −1/2 (Freeman-Tukey test statistic), Ã ¶1/2 ! µ X nij + nji φ(−1/2) bS,φ(0) (θ )=8 nij 1 − Sn 2nij i,j i6=j

and finally for λ = −2 (modified chi-square test statistic) φ bS,φ(0) ) Sn (−2) (θ

=

X

nij

i,j i6=j

Ãµ

nij + nji 2nij

¶2

!

−1 .

7. The expression of the m0ij s given in Theorem 8.10 for the power-divergence family is as follows ! Ãµ ¶λ+1 µ ¶λ+1 1 2qji 2qij (λ) mij = + −2 2λ (λ + 1) q + qji qij + qji Ãµ ij ! ¶λ µ ¶λ 2qji 1 qji 2qij − + . λ qij + qji qij + qji qij + qji Then for λ → 0 and λ = 1 we get (0)

mij = log

2 2 − 3qji + 2qij qji qij 2qji (1) , mij = . qij + qji 2 (qij + qji )2

Then we have X µ σφ2 (0) (q) = qij log

2qij qij + qji

¶2

2 2 − 3qji + 2qij qji qij

!2

i,j i6=j

and

σφ2 (1) (q) =

X i,j i6=j

qij

Ã

2

2 (qij + qji )

X − qij log  i,j i6=j

2qij   qij + qji 

2



2 2 X qij − 3qji + 2qij qji   . − q ij 2   2 (q + q ) ij ji i,j

Ã

!! Ã 2 √ χI(I−1)/2,α n − Dφ(λ) (q, f(q)) , σφ(λ) (q) 2n

then it is easy to obtain βn,φ(0) (q) and βn,φ(1) (q).


2

i6=j

According to Theorem 8.10 βn,φ(λ) (q) = 1 − Φn



406


8. In (8.37) it was established that 1 ¢ λ+1 ¡ λ+1 pbij + pbλ+1 S,φ(λ) ji = I I , pij XX¡ ¢ 1 λ+1 λ+1 λ+1 pbij + pbji

i, j = 1, ..., I.

i=1 j=1

In the following table we present the minimum power-divergence estimators S,φ

λ = −2 λ = −1 λ = −1/2 λ=0 λ=1

S,φ

p11 (λ) 0.0885 0.0881 0.0879 0.0878 0.0874

S,φ

p12 (λ) 0. 1147 0.1159 0. 1165 0. 1171 0. 1182

S,φ

p13 (λ) 0.0635 0.0634 0.0634 0.0634 0.0633

p22 (λ) 0. 2508 0. 2498 0. 2493 0. 2487 0. 2477

S,φ

S,φ

p23 (λ) 0. 1299 0.1295 0. 1294 0. 1293 0. 1290

p33 (λ) 0.0442 0.0440 0.0439 0.0439 0.0437

9. Taking into account the expression of σ2 (q) in Theorem 8.5, we get the following table Divergence φ1 (x)

φ2 (x)

φ3 (x)

φ4 (x)

φ5 (x)

lij qij qi∗ q∗j

3− −

I P

1 2

r=1

r=1

1/2

−

1/2

q∗j s qrj

s−1 s qr∗ q∗j

µ I P −

r=1

(

(1 − a) Ã J P + s=1

2 qrj 2 qr∗ q∗j

1/2 qrj qr∗

r=1

I P

1 λ+1

−

I P

I P

−

J P

s=1

J P

s=1

λ+1 qrj λ q λ+1 qr∗ ∗j

r=1

Ã

−

1 2

J P

s=1

1/2 1/2

1/2

1/2 qis q∗s 1/2

qi∗

s qis s−1 s q∗s qi∗

−

2 qis 2 qi∗ q∗s

J P

s=1

+

−

λ+1 qis λ q λ+1 q∗s i∗

³ q q∗j a+(1−a) qr∗rj q

∗j

1 − q 2 (a+(1−a) q∗qij∗s )

¾

qis i∗ q∗s

.

1/2

qij

1 1−s

qrj

qis ³ q∗j a+(1−a) q

qi∗ q∗j

´2

+

µ ¶ s−1 qij 1 − s qs−1 qs−1 i∗

¶ +2 +

1 λ

∗j

µ

λ qij λ qλ qi∗ ∗j

¶ −1

qr∗ ³ ´2 + q (1−a) a+(1−a) qr∗rj q

q∗s ³ ´ q (1−a) a+(1−a) q rj q∗s i∗

!

∗j

´

!

The previous φ-divergences correspond to Pearson, Matusita (a=1/2), RathieKanappan, Cressie-Read and Rukhin. The expressions of Rathie-Kanappan and Rukhin φ-divergences presented in Chapter 1 are obtained from here with φ3 (x) − φ03 (1)(x − 1) and φ5 (x) − φ05 (1)(x − 1), respectively.


9

Testing in General Populations 9.1.

Introduction

The domains of application of φ-divergence test statistics go far beyond that of multinomial testing presented in previous chapters. Thus in this chapter the φ-divergence test statistics are introduced and studied in general populations. Let (X , βX , Pθ )θ∈Θ be the statistical space associated with the random variable X, where βX is the σ-field of Borel subsets A ⊂ X and {Pθ }θ ∈Θ is a family of probability distributions defined on the measurable space (X , βX ) where Θ is an open subset of RM0 , with M0 ≥ 1. Probability measures Pθ are assumed to be described by densities fθ (x) = dPθ /dµ (x) , where µ is a σ-finite measure on (X , βX ). Let Y1 , . . . , Yn be a random sample from a population described by the random variable X. For testing H0 : θ = θ0 against H1 : θ = θ 1 (simple null hypothesis against simple alternative), the uniformly most powerful test is given by the Neyman-Pearson criterion. If µ is the Lebesgue measure, i.e., X is a continuous random variable, the criterion establishes: reject H0 if hn = hn (Y1 , . . . , Yn ) = L(θ 1 ; Y1 , . . . , Yn )/L(θ0 ; Y1 , . . . , Yn ) ≥ kα , with L(θ; Y1 , . . . , Yn ) =

n Y

fθ (Yi ) ,

i=1

and accept H0 otherwise, where kα = kα (n, α, θ0 , θ1 ) is determined by Prθ 0 (hn ≥ kα ) = α, where α (0 < α < 1) is the significance level.


408


More generally, for testing H0 : θ = θ0 against H1 : θ 6= θ 0 , the most powerful or uniformly most powerful test does not exist and we have to rely on other criteria for the choice of an appropriate test statistic. In such situations the classical solutions are Wald test statistic, Rao test statistic, likelihood ratio test statistic and more recently the test statistics based on φ-divergence measures: φ-divergence test statistics. The same problem appears with composite null hypotheses of the type H0 : θ ∈ Θ0 ⊂ Θ and again the previous test statistics provide good solutions. In this chapter we study the properties of φ-divergence test statistics for testing simple and composite null hypotheses. We assume that the statistical model (X , βX , Pθ )θ∈Θ satisfies the standard regularity assumptions considered in parametric asymptotic statistics, i.e., conditions i)-v) in Section 2 of Chapter 2, as well as the following assumptions in relation to the functions φ involved in the definition of general φ-divergence test statistics: (Φ1) The function φ ∈ Φ∗ is twice continuously differentiable, with φ00 (1) > 0 ; (Φ2) For each θ 0 ∈ Θ there exists an open neighborhood N (θ0 ) such that for all θ ∈ N (θ0 ) and 1 ≤ i, j ≤ M0 it holds: µ µ µ ¶ ¶¶ Z Z fθ (x) fθ (x) ∂ ∂ fθ 0 (x)φ fθ 0 (x)φ dµ (x) = dµ (x) , ∂θi X fθ 0 (x) fθ 0 (x) X ∂θi µ µ µ ¶ ¶¶ Z Z fθ (x) fθ (x) ∂2 ∂2 fθ (x)φ fθ 0 (x)φ dµ (x) = dµ (x) , ∂θi ∂θj X 0 fθ 0 (x) fθ 0 (x) X ∂θi ∂θj and these expressions are continuous on N (θ0 ) .

9.2.

Simple Null Hypotheses: Wald, Rao, Wilks and Phi-divergence Test Statistics

The classical test statistics for testing H0 : θ = θ0 against H1 : θ 6= θ0 (simple null hypothesis against composite alternative hypothesis) are the following: · Wald test statistic b − θ0 )T IF (θ)( b θ b − θ0 ), Wn0 = n(θ

(9.1)

b is the maximum likelihood estimator of θ in Θ obtained from the sample where θ Y1 , ..., Yn and IF (θ) is the Fisher information matrix for the original model.

· Likelihood ratio test statistic

³ ´ b − λn (θ0 ) , L0n = 2n λn (θ) © 2006 by Taylor & Francis Group, LLC

(9.2)

Testing in General Populations

where

409

n

λn (θ) = · Rao test statistic

R0n =

being U n (θ 0 ) =

1X log fθ (Yi ) . n i=1

1 U n (θ0 )T IF (θ 0 )−1 U n (θ 0 ) n

Ã n X ∂ log fθ (Yi ) i=1

∂θ1

, ...,

n X ∂ log fθ (Yi ) i=1

∂θM0

(9.3) !T

.

θ =θ 0

Kupperman (1957, 1958) established that the test statistic based on the KullbackLeibler divergence measure b θ0 ) ≡ 2nDKull (θ, b θ0 ) = 2n TnKull (θ,

Z

X

fbθ (x) log

fθb (x) dµ (x) fθ 0 (x)

is asymptotically chi-squared distributed with M0 degrees of freedom. This result allows to test the null hypothesis H0 : θ = θ 0 by means of the Kullback-Leibler divergence measure. More recently, in the line of Kupperman, Salicr´ u et al . (1994) introduced the φdivergence test statistic and studied its properties including its asymptotic behavior. · φ-divergence test statistic b θ0) = Tnφ (θ,

2n φ00 (1)

b θ0 ). Dφ (θ,

(9.4)

In the following theorem we present the asymptotic distribution of the test statistics given in (9.1), (9.2), (9.3) and (9.4). Theorem 9.1 Let the model (X , βX , Pθ )θ∈Θ . Suppose φ satisfy the assumptions i)-v) considered in Section 2 of Chapter 2 and (Φ1)-(Φ2) respectively. Under the null hypothesis H0 : θ = θ 0

(9.5)

the asymptotic distribution of the test statistics given in (9.1), (9.2), (9.3) and (9.4) is chi-square with M0 degrees of freedom. Proof. We start with Wald test statistic. Since under H0 : θ = θ 0 , P b −→ θ θ0, n→∞


410


and by assumption v), the elements of IF (θ) are continuous in θ, it holds P

b −→ IF (θ0 ) . IF (θ) n→∞

Then, the asymptotic distribution of the quadratic form b θ b − θ0) b − θ0 )T IF (θ)( n(θ

coincides with that of the quadratic form

where X =

√

b − θ 0 )T IF (θ 0 ) (θ b − θ0 ), X T X ≡ n(θ

1/2

nIF (θ 0 )

b − θ 0 ). (θ

We also know that under H0 ,

´ ³ √ L b − θ0 ) −→ n(θ N 0, IF (θ0 )−1 n→∞

which implies

L

X −→ N (0, I M0 ×M0 ) . n→∞

Therefore, L b θ b − θ 0 ) −→ b − θ 0 )T IF (θ)( χ2M0 , n(θ n→∞

and the Wald test statistic should reject the null hypothesis whenever b θ b − θ 0 ) > χ2M ,α . b − θ 0 )T IF (θ)( n(θ 0

As to Rao efficient score test statistic by assumption iv), we know that · ¸ ∂ log fθ (X) Eθ = 0, i = 1, ..., M0 . ∂θi Further, by the Central Limit Theorem, under H0 the random vector Ã n !T n 1 X ∂ log fθ 0 (Yi ) 1 X ∂ log fθ 0 (Yi ) 1 U n (θ 0 ) = , ..., n n i=1 ∂θ1 n i=1 ∂θM0 satisfies

√ 1 L n U n (θ0 ) −→ N (0, Σ) , n→∞ n where Σ is the variance-covariance matrix of the random vector ¶T µ ∂ log fθ 0 (X) ∂ log fθ 0 (X) , ..., , ∂θ1 ∂θM0 which is obviously IF (θ0 ) . Therefore 1 L √ U n (θ0 ) −→ N (0, IF (θ 0 )) n→∞ n



411

and consequently Rn =

1 L U n (θ0 )T IF (θ0 )−1 U n (θ0 ) −→ χ2M0 . n→∞ n

Now let us establish the asymptotic distribution of the φ-divergence test statistic, b θ0 ) = Tnφ (θ,

2n b θ 0 ). Dφ (θ, φ00 (1)

b gives A second order Taylor expansion of Dφ (θ, θ0 ) around θ = θ0 at θ =θ µ ¶ M0 b θ 0 ) = Dφ (θ0 , θ 0 ) + P ∂Dφ (θ, θ0 ) (θbi − θ0i ) Dφ (θ, ∂θ i i=1 θ=θ 0 Ã ! µ° °2 ¶ M0 M0 ∂Dφ2 (θ, θ 0 ) °b ° 1 P P − θ0° , + 2 (θbi − θ0i )(θbj − θ0j ) + o °θ ∂θi ∂θj i=1 j=1 θ=θ 0

where by (1.6), it is clear that

Dφ (θ 0 , θ0 ) = φ (1) = 0. Now we prove that the second term in the previous expansion is zero. Indeed, ∂Dφ (θ, θ 0 ) = ∂θi

Z

φ0

X

µ

fθ (x) fθ 0 (x)

¶

∂fθ (x) dµ (x) , ∂θi

and for θ = θ0 µ ¶ Z Z ∂ ∂fθ (x) ∂Dφ (θ, θ0 ) 0 0 = φ (1) dµ (x) = φ (1) fθ (x) dµ (x) = 0. ∂θi ∂θi ∂θi θ=θ 0 X

X

This means that the random variables b θ0 ) 2nDφ (θ,

and n

¶ M0 µ 2 M0 X X ∂ Dφ (θ, θ 0 ) ∂θi ∂θj

i=1 j=1

θ =θ 0

(θbi − θ0i )(θbj − θ0j )

µ° °2 ¶ ¡ ¢ °b ° have the same asymptotic distribution, since under H0 o °θ − θ 0 ° = oP n−1 . The second order derivatives are µ ¶ Z ∂ 2 Dφ (θ, θ 0 ) fθ (x) 1 ∂fθ (x) ∂fθ (x) = φ00 dµ (x) ∂θi ∂θj fθ 0 (x) fθ 0 (x) ∂θi ∂θj ¶ ZX µ fθ (x) ∂ 2 fθ (x) 0 + φ dµ (x) , fθ 0 (x) ∂θi ∂θj X


412


and then µ

∂ 2 Dφ (θ, θ0 ) ∂θi ∂θj

¶

00

= φ (1) θ=θ 0

Z

1 ∂fθ 0 (x) ∂fθ 0 (x) dµ (x) fθ 0 (x) ∂θi ∂θj

X

ij = φ00 (1) IF (θ 0 ) ,

ij (θ0 ) is the (i, j)th-element of the Fisher information matrix. Therefore, where IF

b θ0 ) = Tnφ (θ,

2n φ00 (1)

and finally, this implies

b θ 0 ) = n(θ b − θ0 )T IF (θ0 ) (θ b − θ 0 ) + oP (1) , Dφ (θ,

b θ0) = Tnφ (θ,

2n φ00 (1)

L b θ0 ) −→ Dφ (θ, χ2M0 . n→∞

In order to obtain the asymptotic distribution of L0n under H0 , we consider the following Taylor expansion µ ¶ M P0 ∂λn (θ) b λn (θ) = λn (θ0 ) + (θbi − θ0i ) ∂θi i=1 θ=θ 0 µ° µ ¶ °2 ¶ M P0 M P0 ∂ 2 λn (θ) °b ° − θ0 ° . + 12 (θbi − θ0i )(θbj − θ0j ) + o °θ ∂θi ∂θj θ=θ 0 i=1 j=1 But

µ

∂λn (θ) ∂θi

¶

n

= θ=θ 0

1 X ∂ log fθ 0 (Yl ) . n ∂θi l=1

Then, applying Khintchine Law of Large Numbers we get ¶ µ Z ∂ log fθ 0 (x) ∂λn (θ) a.s. −→ fθ 0 (x) dµ (x) = 0. n→∞ ∂θi ∂θi θ=θ 0 X

On the other hand, the second derivatives are µ 2 ¶ n n 1X ∂ λn (θ) 1 ∂fθ 0 (Yl ) ∂fθ 0 (Yl ) 1 X ∂ 2 fθ 0 (Yl ) 1 . =− + 2 ∂θi ∂θj θ=θ 0 n ∂θi ∂θj n ∂θi ∂θj fθ 0 (Yl ) fθ 0 (Yl ) l=1

l=1

But by Khintchine Law of Large Numbers, n

− and

1X 1 ∂fθ 0 (Yl ) ∂fθ 0 (Yl ) a.s. ij −→ I (θ 0 ) 2 n→∞ F n ∂θi ∂θj fθ 0 (Yl )

n

l=1

1 X ∂ 2 fθ 0 (Yl ) 1 a.s. −→ n→∞ n ∂θi ∂θj fθ 0 (Yl ) l=1


Z

X

∂ 2 fθ 0 (x) dµ (x) = 0. ∂θi ∂θj


413

Therefore, ³ ´ b − λn (θ0 ) = √n(θ b − θ 0 )T IF (θ 0 ) √n(θ b − θ0 ) + oP (1) , 2n λn (θ)

and this implies

L

L0n −→ χ2M0 . n→∞

Remark 9.1 The (h, φ)-divergence test statistics associated with the (h, φ)-divergence measures satisfy 2n L b θ0) ≡ b θ0 ) −→ Tnφ,h (θ, Dh (θ, χ2 . n→∞ M0 h0 (0) φ00 (1) φ This result is immediate from the relation

³ ´ b θ0 ) = h0 (0) Dφ (θ, b θ0 ) + o Dφ (θ, b θ0 ) . Dφh (θ,

Based on Theorem 9.1 and Remark 9.1, we should reject the null hypothesis given φ,h b 2 b θ 0 ) > χ2 in (9.5), with significance level α, if Tnφ (θ, M0 ,α (or if Tn (θ, θ 0 ) > χM0 ,α ). The 0 0 0 rejection rule is analogous for Wn , Ln and Rn . In most cases, the power function of this testing procedure can not be calculated explicitly. In the following theorem we present a useful asymptotic result for approximating the power function. Theorem 9.2 Let the model and φ ∈ Φ∗ satisfy the assumptions i)-v) considered in Section 2 of Chapter 2 and (Φ1)-(Φ2) respectively. Let θ∗ be the true parameter, with θ∗ 6= θ 0 . Then it holds ´ ¡ ¢ √ ³ L b θ0 ) − Dφ (θ∗ , θ 0 ) −→ n Dφ (θ, N 0, σφ2 (θ∗ ) , n→∞ ´ ³ −1 ∂Dφ (θ,θ 0 ) T , j = where σφ2 (θ∗ ) = T T IF (θ∗ ) T , T = (t1 , ..., tM ) and tj = ∂θj θ =θ ∗ 1, ..., M. Proof. A first order Taylor expansion gives °´ ³° b θ0 ) = Dφ (θ∗ , θ0 ) + T T (θ b − θ∗ ) + o ° b − θ∗ ° Dφ (θ, ° . °θ

By the Central Limit Theorem,

´ ³ √ b L −1 n(θ − θ∗ ) −→ N 0, IF (θ∗ ) n→∞

and


°´ ³° √ ° °b n o °θ − θ ∗ ° = oP (1) .

414


Then, it is clear that the random variables, ´ √ ³ b θ 0 ) − Dφ (θ∗ , θ0 ) n Dφ (θ, have the same asymptotic distribution.

√ b − θ∗ ), and T T n(θ

Remark 9.2 Through Theorem 9.2, a first approximation to the power function, at θ∗ 6= θ0 , is given by Ã √ !! Ã φ00 (1) χ2M0 ,α n ∗ ∗ 1 βn,φ (θ ) = 1 − Φ − Dφ (θ , θ0 ) , (9.6) σφ (θ∗ ) 2n where Φ (x) is the standard normal distribution function. If some alternative θ ∗ 6= θ0 ³is the´true parameter, then the probability of rejecting θ0 with the rejection rule b θ 0 > χ2 Tnφ θ, M0 ,α , for fixed significance level α, tends to one as n → ∞. The test is consistent in Fraser’s sense. In order to produce some less trivial asymptotic powers that are not all equal to 1 we can use a Pitman-type local analysis, as developed by LeCam (1960), confining the attention to n−1/2 -neighborhoods of the true parameter values. More specifically, we consider the power at contiguous alternative hypotheses described by H1,n : θn = θ0 + n−1/2 d, where d is a fixed vector in RM0 such that θn ∈ Θ ⊂ RM0 . A fundamental tool to get the asymptotic distribution of the φ-divergence test stab θ0 ) under the contiguous alternative hypotheses is LeCam’s third lemma, as tistic Tnφ (θ, presented in H´ ajek and Sidàk (1967). Instead, in the following theorem we present a simpler proof. Theorem 9.3 Let the model and φ ∈ Φ∗ satisfy the assumptions i)—v) considered in Section 2 of Chapter 2 and (Φ1)—(Φ2) respectively. Under the contiguous alternative hypotheses H1,n : θn = θ0 + n−1/2 d, where d is a fixed vector in RM0 such that θn ∈ Θ ⊂ RM0 , the asymptotic distribution b θ 0 ) is noncentral chi-square, with M0 degrees of of the φ-divergence test statistic Tnφ (θ, freedom and noncentrality parameter δ = dT IF (θ0 ) d. Proof. We can write √ b − θ0 ) = √n(θ b − θ n ) + √n (θn − θ0 ) = √n(θ b − θ n ) + d. n(θ

Under H1,n , it holds ´ ³ √ b L −1 n(θ − θn ) −→ N 0, IF (θ0 ) n→∞


and

´ ³ √ L −1 b − θ0 ) −→ . n(θ N d, IF (θ0 ) n→∞


415

By applying the delta method, it is not difficult to establish that µ° °2 ¶ °b ° φ b T b b Tn (θ, θ0 ) = n(θ − θ 0 )IF (θ 0 ) (θ − θ0 ) + n o °θ − θ0 ° .

b θ0 ) has the same asymptotic distribution as the quadratic form Then, Tnφ (θ, ³ ´T ³ ´ √ √ b − θ0 ) b − θ0 ) . X T X ≡ IF (θ)1/2 n(θ IF (θ)1/2 n(θ

On the other hand

X ≡IF (θ0 )1/2 Therefore

´ ³ √ L 1/2 b − θ 0 ) −→ n(θ N IF (θ0 ) d, I M0 ×M0 . n→∞

³ ´T ³ ´ √ √ L b − θ0 ) b − θ0 ) −→ IF (θ0 )1/2 n(θ IF (θ0 )1/2 n(θ χ2M0 (δ) , n→∞

being δ = dT IF (θ0 ) d.

Remark 9.3 Using Theorem 9.3, we get a second approximation to the power function, at θn = θ 0 + n−1/2 d, by means of ¡ ¢ βn2 (θn ) = 1 − Gχ2 (δ) χ2M0 ,α , M0

where Gχ2

M0

(δ)

is the distribution function of a noncentral chi-square random variable

with M0 degrees of freedom and noncentrality parameter δ = dT IF (θ 0 ) d. If we want to approximate the power at the alternative θ 6= θ0 , then we can take d = d(n, θ, θ0 ) = √ n(θ − θ 0 ). We can observe that this approximation is independent of φ. The same result can be obtained for the test statistics Wn0 , L0n and Rn0 (see, e.g., Sen and Singer (1993) or Serfling (1980)). Example 9.1 Let Y1 , ..., Yn be a random sample from a normal population with mean µ and variance σ 2 . Consider the parameter θ = (µ, σ) . Find the expression of the test statistics given in (9.1), (9.2) and (9.3), as well as the expression of Rényi test statistic, for testing H0 : θ = θ 0 = (µ0 , σ0 )

versus

H1 : θ 6= θ0 .

It is well known that the maximum likelihood estimators of µ and σ are given by v u n u1 X¡ ¢2 Yi − Y . and σ b=t µ b=Y n i=1 © 2006 by Taylor & Francis Group, LLC

416


The likelihood ratio test statistic becomes ! Ã Ã Ã µ ¶!! 1 1 n nσ e2 0 Ln = 2 log , ¡√ ¢n exp(− ) − log ¡√ ¢n exp − 2 2 σ02 σ bn 2π σ0n 2π

where

v u n u1 X (Yi − µ0 )2 , σ e=t n i=1

i.e.,

L0n = 2n log The Fisher information matrix is

µ 2 ¶ σ e σ0 +n − 1 . σ b σ02

IF (µ, σ) =

µ

1 σ2

0

0

2 σ2

¶

.

Therefore, the Wald test statistic has the expression ¶µ ¶ µ 1 ¢ ¡ 0 Y − µ0 2 0 σ b b − σ0 Wn = n Y − µ0 , σ 2 0 σ b − σ0 σ b2 ¡ ¢2 2 = n Y − µ0 /b σ + 2n (b σ − σ0 )2 /b σ2.

Concerning the Rao test statistic, the score vector is Ã n !T n X ∂ log fµ,σ (Yi ) X ∂ log fµ,σ (Yi ) U n (µ0 , σ0 ) = , ∂µ ∂σ i=1 i=1 (µ,σ)=(µ0 ,σ0 ) µ µ ¶ µ 2 ¶¶ T Y − µ0 σ e − σ02 = n ,n σ02 σ03 and then

Rn0

¢2 ¡ ¶2 µ 2 Y − µ0 n σ e − σ02 =n + . σ02 2 σ02

Finally, we obtain the expression of Rényi test statistic.

The Rényi divergence may be regarded as a (h, φ)-divergence; that is, ³ ´ b θ0 ) = h Dφ (θ, b θ0 ) , Dr1 (θ,

with

h (x) =

1 log (r (r − 1) x + 1) r (r − 1)

(9.7)

φ (x) =

1 (xr − r (x − 1) − 1) . r (r − 1)

(9.8)

and



417

Rényi test statistic has the expression b θ0 ) = Tnr (θ,

2n b θ0 ). D1 (θ, h0 (0) φ00 (1) r

It is easy to see that h0 (0) = 1 and φ00 (1) = 1. Therefore, b θ0) Tnr (θ,

2n log = r (r − 1)

Z

X

fθb (x)r fθ 0 (x)1−r dµ (x) , r 6= 1, 0.

In Chapter 1 it was established that Rényi’s divergence measure between two normal populations is Dr1 ((µ, σ) , (µ0 , σ0 )) =

2 rσ02 + (1 − r) σ2 1 1 (µ − µ0 ) log − r . 2 rσ02 + (1 − r) σ 2 2r (r − 1) (σ 2 )1−r (σ02 )

(9.9)

Using this formula, we get

b θ0) Tnr (θ,

=n

Ã

! ¢2 ¡ Y − µ0 rσ02 + (1 − r) σ 1 b2 log − , r rσ02 + (1 − r) σ b2 r (r − 1) (b σ2 )1−r (σ02 )

b = (Y , σ b2 ) and θ 0 = (µ0 , σ0 ) . where θ

When r → 1, we get the test statistic based on Kullback-Leibler divergence, b θ0 ) = 2nDKull (θ, b θ0 ) = 2n log lim Tnr (θ,

r→1

σ0 +n σ b

µ

¶ σ b2 b θ0 ). − 1 ≡ TnKull (θ, σ02

b θ 0 ) = T Kull (θ, b θ0 ) coincides with L0 . The following Observe that in this case Tn1 (θ, n n Remark provides a sufficient condition for the equality of these test statistics. Remark 9.4 When the density fθ (x) belongs to the exponential family, that is,

fθ (x) = q (θ) t (x) exp

ÃM 0 X i=1

!

Si (θ) ti (x)

x ∈ X,

likelihood ratio test statistic coincides with the test statistic based on Kullback-Leibler divergence measure.


418


For the exponential family, likelihood ratio test statistic is given by ³ ´ b − λn (θ0 ) = 2 log fθb (Y1 , ..., Yn ) L0n = 2n λn (θ) Ã fθ 0 (Y1 , ..., Yn ) ! n M n Q P P0 b i (Yj ) bn t (Yi ) exp Si (θ)t q(θ) i=1

= 2 log

q (θ0 )

n

n Q

t (Yi ) exp

i=1

Ã

j=1 i=1 n M P P0

!

Si (θ 0 ) ti (Yj )

j=1 i=1

! ´ n M bn P P0 ³ b q(θ) Si (θ) − Si (θ 0 ) ti (Yj ) = 2 log n + q (θ0 ) j=1 i=1 !! Ã Ã ´ 1 P M n b P0 ³ b q(θ) Si (θ) − Si (θ 0 ) + ti (Yj ) , = 2n log q (θ 0 ) i=1 n j=1 Ã

while the test statistic based on Kullback-Leibler divergence takes the form

´ M0 ³ b b θ0 ) = 2n log q(θ) + 2n P Si (θ) b − Si (θ0 ) Eb [ti (X)] . TnKull (θ, θ q(θ0 ) i=1

It is clear that the two statistics coincide asymptotically since n 1 P ti (Yj ) n j=1

and

Ebθ [ti (X)]

a.s.

−→

Eθ 0 [ti (X)]

n→∞

a.s.

−→

n→∞

Eθ 0 [ti (X)] .

The following question then arises: Is it possible to find assumptions under which the two exact tests coincide? Thus, we need to find the conditions under which n

The likelihood function is

Eθb [ti (X)] =

L (θ,Y1 , ..., Yn ) = q (θ)

n Y n

i=1

and the loglikelihood function is

1X ti (Yj ) . n j=1

  M0 n X X t (Yi ) exp  Si (θ) ti (Yj ) ,

log L (θ,Y1 , ..., Yn ) = n log q (θ) + log

i=1

n Y

i=1

t (Yi ) +

j=1

M0 X i=1

Si (θ)

n X

ti (Yj ) .

j=1

b is a solution of the system of equations Therefore, the maximum likelihood estimator, θ,  M n b P b 1 , ..., Yn ) b P0 ∂Si (θ) n ∂q(θ)  ∂ log L(θ,Y = + ti (Yk ) = 0, ∂θj q (θ) ∂θj i=1 ∂θj k=1  j = 1, ..., M0 . © 2006 by Taylor & Francis Group, LLC


419

On the other hand, we know that Z

q (θ) t (x) exp

X

ÃM 0 X

!

Si (θ) ti (x) dµ (x) = 1.

i=1

Taking derivatives with respect to θj in both sides we get µM ¶ P0 ∂q (θ) t (x) exp Si (θ) ti (x) dµ (x) ∂θj i=1 XZ µM ¶M P0 P0 ∂Si (θ) + q (θ) t (x) exp Si (θ) ti (x) ti (x) dµ (x) = 0, ∂θj i=1 i=1

Z

X

which is equal to  

M P0 ∂Si (θ) n ∂q (θ) +n Eθ [ti (X)] = 0, q (θ) ∂θj ∂θj i=1  j = 1, ..., M0 .

Thus, for the exponential family it holds

 M n b 1 P  P0 ∂Si (θ) ti (Yk ) = ∂θj n k=1  i=1 j = 1, ..., M0 ,

b ∂Si (θ) Eθb [ti (X)] , i=1 ∂θj M P0

and this can be written as a homogeneous system  M  P0 aij zi = 0 i=1  j = 1, ..., M0 with

n

aij =

b ∂Si (θ) 1X and zi = ti (Yk ) − Eθb [ti (X)] . ∂θj n k=1

This system has the unique solution zi = 0, i = 1, ..., M0 , if and only if the determinant of the matrix A = (aij )i,j=1,...,M0 is different from zero. Therefore, the exact likelihood ratio test statistic coincides with the exact test based on the Kullback-Leibler divergence measure if and only if the determinant of the matrix Ã ! b ∂Si (θ) ∂θj i,j=1,...,M0


420


is different from zero. This condition means that the equality is obtained when the exponential family is not overparametrized. Buse (1982), by means of simple diagrams, gave an intuitive meaning of the likelihood ratio test statistic, Rao test statistic and Wald test statistic. Interesting survey articles about these test procedures were given by Breusch and Pagan (1980) and Engle (1981).

9.2.1.

Confidence Regions

We can easily construct families of confidence regions with a prescribed asymptotic confidence coefficient 1 − α for θ ∈ Θ ⊂ RM0 . We consider the problem of testing H0 : b θ0 ). Let Aφ (θ0 ) , θ 0 ∈ Θ ⊂ RM0 , θ = θ0 based on the φ-divergence test statistic Tnφ (θ, the region of acceptance associated with the test statistic “reject the null hypothesis H0 , b θ 0 ) > χ2 with significance level α, if Tnφ (θ, M0 ,α ”, i.e., o n b θ0 ) < χ2 ,α . Aφ (θ0 ) = (y1 , ..., yn ) : Tnφ (θ, M0

For each observation (y1 , ..., yn ) let Sφ (y1 , ..., yn ) denote the set

Sφ (y1 , ..., yn ) = {θ : (y1 , ..., yn ) ∈ Aφ (θ) , θ ∈ Θ} . Then Sφ (y1 , ..., yn ) is a family of confidence regions for θ ∈ Θ ⊂ RM0 with asymptotic confidence coefficient 1 − α.

9.3.

Composite Null Hypothesis

We assume that the statistical model (X , βX , Pθ )θ ∈Θ , satisfies the standard regularity assumptions i)-v) given in Section 2 of Chapter 2 (see also Serfling (1980)). As to the composite null hypothesis H0 : θ ∈ Θ0 ⊂ Θ, (9.10) we assume the following: (H1) Θ0 is a subset of RM0 , and there exist 1 ≤ d0 ≤ M0 , an open subset B ⊂ RM0 −d0 and mappings g : Θ −→ Rd0 and h : B −→ Θ such that Θ0 = {h(β) : β ∈ B} and g(θ) = 0 on Θ0 . (H2) The M0 × d0 matrix B(θ) =

µ

∂gj (θ) ∂θs

¶

j=1,...,d0 s=1,...,M0

exists and is of rank d0 for all θ ∈ Θ0 , with all elements continuous on Θ0 .



(H3) The M0 × (M0 − d0 ) matrix Mβ =

µ

∂hj (θ) ∂βs

421

¶

j=1,...,M0 s=1,...,M0 −d0

exists and is of rank M0 − d0 for all β ∈ B, with all elements continuous on B. (H4) The statistical submodel ³ n o ´ (X , βX ) , pβ = fh(β ) : β ∈ B , µ satisfies regularity conditions i)-v).

In order to solve the testing problem

H0 : θ ∈ Θ0 ⊂ Θ versus H1 : θ ∈ Θ − Θ0 we shall consider the test statistics given in (9.1), (9.2), (9.3) and (9.4), but adapted to current context: · Wald test statistic

³ ´−1 b B(θ)I b F (θ) b −1 B T (θ) b b T, Wn = ng(θ) g(θ)

(9.11)

b is the maximum likelihood estimator of θ in Θ, IF (θ) b −1 denotes the where θ inverse of the Fisher information matrix, g (θ) , the mapping defined in (H1) and B(θ) the matrix in (H2). · Likelihood ratio test statistic

³ ´ b − λn (θ) e Ln = 2n λn (θ)

(9.12)

e = h(β) b is the maximum likelihood estimator restricted to the null hywhere θ pothesis Θ0 , and n 1X λn (θ) = log fθ (Yi ) . n i=1 · Rao test statistic where e = U n (θ) © 2006 by Taylor & Francis Group, LLC

Rn =

1 e F (θ) e −1 U n (θ) e T, U n (θ)I n

Ã n X ∂ log fθ (Yi ) i=1

∂θ1

, ...,

n X ∂ log fθ (Yi ) i=1

∂θM

(9.13) !

e θ=θ

.

422


For testing the special composite null hypotheses H ≡ Θ0 = Θ1 × {θ20 } and H ∗ ≡ Θ∗0 = {θ10 } × Θ2 in models with Θ = Θ1 × Θ2 , Salicr´ u et al. (1994) proposed the φ-divergence test statistics ³ ³ ´ ´ 2n b1 , θ b1 , θ b2 ), (θ b1 , θ20 ) b2 ), (θ b1 , θ20 ) , ( θ D = Tnφ (θ φ φ00 (1) ³ ³ ´ ´ 2n b1 , θ b1 , θ b2 ), (θ 10 , θ b2 ) b2 ), (θ 10 , θ b2 ) ( θ D = Tnφ (θ φ φ00 (1) b2 ). b = (θ b1 , θ using the maximum likelihood estimator θ

Later Morales et al. (1997), using the φ-divergence test statistics, studied the problem under any hypothesis H with the properties (H1) — (H4). They dealt with the following family of test statistics: · φ-divergence test statistic b θ) e ≡ Tnφ (θ,

2n φ00 (1)

b θ), e Dφ (θ,

(9.14)

b is the maximum likelihood estimator of θ in Θ and θ e is the maximum where θ likelihood estimator in the null hypothesis Θ0 . Theorem 9.4 Let the model and φ satisfy the assumptions i) — v) considered in Section 2 of Chapter 2 and (Φ1) , (Φ2) . Then, under any hypothesis H with properties (H1) — (H4), the asymptotic distribution of the test statistics given in (9.11), (9.12), (9.13) and (9.14) is chi-square with d0 degrees of freedom. The proof of this theorem for the φ-divergence test statistic given in (9.14) was established by Morales et al . (1997). For the rest of test statistics, see Serfling (1980) or Sen and Singer (1993). For the test statistic based on the (h, φ)-divergence measures, b θ), e we have a similar result. Tnφ,h (θ, For composite null hypothesis a result analogous to Remark 9.4 can be established.

Consider the null hypothesis H0 : θ ∈ Θ0 ⊂ Θ and assume that conditions (H1) — (H4) b θ) e ≥ χ2 . The hold. By Theorem 9.4, the null hypothesis should be rejected if Tnφ (θ, d0 ,α following theorem can be used to approximate the power function. Assume that θ∗ ∈ / Θ0 a.s. ∗ b is the true value of the parameter so that θ −→ θ and that there exists θ0 ∈ Θ0 such n→∞ a.s. e −→ that the restricted maximum likelihood estimator satisfies θ θ 0 . Then, it holds n→∞

´ √ ³b e L n (θ, θ) − (θ ∗ , θ0 ) −→ N n→∞


Ãµ

0 0

¶ Ã −1 IF (θ∗ ) , A21

A12 A22

!!

,


423

where A12 = A12 (θ∗ , θ0 ) , A21 = AT12 and A22 = A22 (θ∗ , θ 0 ) are d0 × d0 matrices. We have the following result. Theorem 9.5 Let the model φ ∈ Φ∗ and Θ0 satisfy the assumptions i)—v), (φ1)—(φ2) and (H1)—(H4) respectively. Then ´ ¡ ¢ √ ³ L b θ) e − Dφ (θ ∗ , θ0 ) −→ n Dφ (θ, N 0, σφ2 (θ ∗ ) , n→∞

where

−1

σφ2 (θ ∗ ) = T T IF (θ∗ ) T + T T A12 S + S T A21 T + S T A22 S, ´T ´T ³ ³ ∂Dφ (θ 1 ,θ 0 ) ∂Dφ (θ ∗ ,θ 2 ) and S = . T = ∂θ 1 ∂θ 2 ∗ θ 1 =θ

θ 2 =θ 0

Proof. The result follows straightforward by making a first order Taylor expansion of b θ) e Dφ (θ, ° ° °´ ³° °e ° b θ) e = Dφ (θ∗ , θ0 ) + T T (θ b − θ ∗ ) + S T (θ e − θ0) + o ° b − θ∗° − θ0 ° . Dφ (θ, ° + °θ °θ Remark 9.5 On the basis of this ³ theorem we can ´ get an approximation of the power function at b θ) e > χ2 θ ∗ , βn,φ (θ∗ ) = Prθ ∗ Tnφ (θ, d0 ,α , in the following way: βn,φ (θ∗ ) = 1 − Φ

µ

1 σφ (θ∗ )

µ

√ φ00 (1) 2 √ χd0 ,α − nDφ (θ ∗ , θ0 ) 2 n

¶¶

,

where Φ (x) is the standard normal distribution function. b θ) e at an alternative close to We may also find an approximation of the power of Tnφ (θ, the null hypothesis. Let θn ∈ Θ−Θ0 be a given alternative and let θ0 be the element in Θ0 closest to θn in the Euclidean distance sense. A first possibility to introduce contiguous alternative hypotheses is to consider a fixed d ∈ RM0 and to permit θn moving towards θ 0 as n increases in the following way H1,n : θ n = θ0 + n−1/2 d. A second approach is to relax the condition g (θ) = 0 defining Θ0 . Let δ ∈ RM0 and consider the following sequence, θ n , of parameters moving towards θ 0 according to ∗ H1,n : g (θn ) = n−1/2 δ.

Note that a Taylor series expansion of g (θn ) around θ0 yields g (θn ) = g (θ 0 ) + B (θ0 )T (θn − θ0 ) + o (kθ n − θ0 k) .


(9.15)

424


By substituting θn = θ0 + n−1/2 d in (9.15) and taking into account that g (θ0 ) = 0, we get g (θ n ) = n−1/2 B (θ0 )T d + o (kθn − θ0 k) , T

so that the equivalence in the limit is obtained for δ = B (θ0 ) d. In Morales and Pardo (2001), the following result was established: Theorem 9.6 b θ) e under H ∗ , Under some regularity conditions, the asymptotic distribution of Tnφ (θ, 1,n is noncentral chi-square with d0 degrees of freedom and noncentrality parameter δ = dT IF (θ 0 )−1 d and noncentral chi-square with d0 degrees of freedom and noncentrality ³ ´−1 −1 d under H1,n . parameter δ ∗ = dT B (θ 0 ) IF (θ0 ) B T (θ0 ) Davidson and Lever (1970) obtained the same result for the test statistic Ln .

Example 9.2 (Morales et al. 1997) Let Y1 , ..., Yn be a random sample from a normal population with mean µ and variance σ2 , θ = (µ, σ) ∈ Θ = (−∞, ∞) × (0, ∞) . We shall test composite hypotheses H0 : σ = µ/3

versus

H1 : σ 6= µ/3.

The maximum likelihood estimators of µ and σ are µ b = Y = 1 Pn 2 1/2 ( n i=1 (Yi − Y ) ) , respectively. In this case,

1 n

Pn

i=1 Yi

Θ0 = {(µ, σ) ∈ Θ : σ = µ/3}.

and σ b =

e of θ in Θ0 . The Now we are going to calculate the maximum likelihood estimator, θ, density function is ½ ½ ³ ´2 ¾ 1 ³ ´2 ¾ 3 1 x−µ 3 9 x fN(µ,σ=µ/3) (x) = µ(2π) exp − exp − − 1 = (2π)1/2 . 1/2 2 µ/3 2 µ µ Then, the likelihood function is L(µ; Y1 , ..., Yn ) =

µ

3 (2π)1/2

¶n

( !) Ã n n 9 X Yi2 2X 1 exp − +n− Yi µn 2 i=1 µ2 µ i=1

and the loglikelihood function µ ¶ 3 log L(µ; Y1 , ..., Yn ) = n − log µ + log √ − 2π Therefore 1 ∂ log L(µ; Y1 , ..., Yn ) = ∂µ µ


Ã

9 2

µ

n Y2 P i +n− 2 i=1 µ

n n 9 X 2 9X Yi − Yi −n + 2 µ i=1 µ i=1

!

2 µ

n P

i=1

=0

¶

Yi .


and µ e is a solution of the equation 2

−nµ + 9

This solution is

n X

n X

425

n

1X 2 − 9µ Yi = 0 ⇔ µ + 9Y µ − 9 Y = 0. n i=1 i i=1

Yi2

i=1

2

v  u n X u 1 1 −9Y ± t(9Y )2 + 4 × 9 Y 2 µ= 2 n i=1 i 

e = (e and the maximum likelihood estimator of θ in Θ0 is θ µ, µ e/3), where q ¶ µ 3 2 µ e= σ2 . −3Y + 13Y + 4b 2

Now, taking into account the expression of Rényi divergence for two normal populations given in (9.9), we get the expression of the Rényi test statistic b θ) e = Tnr (θ,

=

2n φ00 (1) h0 (0) n

µ

1 r(r−1)

Dr1 ((b µ, σ b), (e µ, µe3 ))

log

(σb 2 )

1−r

(e µ/3)2r 2 r(e µ/3) +(1−r)b σ2

+

(b µ−e µ)2 (r(eµ/3)2 +(1−r)bσ2 )

¶

.

The likelihood ratio test statistic is given by b θ) e = lim Dr1 (θ, b θ) e =n Ln = TnKull (θ, r→1

and for r → 0, we have b θ) e Tn0 (θ,

=

lim

r→0

b θ) e Tnr (θ,

=n

µ

µ

(b µ−µ e)2 σ b2 µ e2 9 + 9 2 − 1 + log 2 2 µ e µ e 9b σ

(b µ−µ e)2 µ e2 1 9b σ2 + − 1 + log σ b2 9 σ b2 µ e2

¶

¶

.

Now we consider the test statistics Wn and Rn . For the Wald test statistic Wn , we need the Fisher information matrix, the function g and the matrix B (θ) . The estimated Fisher information matrix is µ −2 ¶ σ b 0 IF (b µ, σ b) = , 0 2b σ −2 b = g(b b = (1, −3), the function g and the matrix B (θ) are g(θ) µ, σ b) = µ b − 3b σ and B(θ) respectively. Then, the Wald test statistic is

³ ´−1 µ − 3b σ )2 b B(θ)I b F (θ) b −1 B T (θ) b b T = n 2(b Wn = ng(θ) g(θ) . 11b σ2

Regarding Rao test statistic, it is necessary to calculate


426


Ã n ! n X ∂ log fµ,σ (Yi ) X ∂ log f (Y ) µ,σ i e = , U n (θ) ∂µ ∂σ i=1 i=1

.

(µ,σ)=(e µ,e µ/3)

It holds

! Ã n X ∂ log fµ,σ (Yi ) ∂µ

i=1

and

¶ n µ X ∂ log fµ,σ (Yi ) ∂σ

i=1

Therefore

Rn

=

=n

(µ, e µ/3) e

(µ, e µ/3) e

Ã

9 (Y − µ e), µ e2

! n 27 1 X 3 =n − + 3 (Yi − µ e)2 . µ e µ e n i=1

1 eT e −1 e n U n (θ) IF (θ) U n (θ)

= n

Ã

µ e2 9 (Y − µ e)2 + 2 µ e 18

¶2 ! µ n 27 1 P 3 2 (Yi − µ e) . − + 3 µ e µ e n i=1

In the following table we summarize the final expressions of the different test statistics presented here Statistic

Expression 2(b µ − 3b σ )2 n 2 11b σ

Wn

Rn

Ln ≡

n

TnKull

9 µ e2 2 (Y − µ e ) + µ e2 18 n

Tn0

Tnr

Ã

n

n

Ã

µ

µ

¶2 ! µ n 27 1 P 3 (Yi − µ e)2 − + 3 µ e µ e n i=1

(b µ−µ e)2 σ b2 µ e2 9 + 9 − 1 + log µ e2 µ e2 9b σ2

(b µ−µ e)2 µ e2 1 9b σ2 + − 1 + log σ b2 9 σ b2 µ e2

¶

¶

! ¡ 2 ¢1−r σ b (e µ/3)2r 1 (b µ−µ e)2 log + . 2 2 r(r − 1) r (e µ/3) + (1 − r)b σ2 r (e µ/3) + (1 − r)b σ2

µ/3)2 + (1 − r)b σ2 > 0. When The formula for Tnr holds for r 6= 0, r 6= 1 when r(e 2 2 r r(e µ/3) + (1 − r)b σ < 0, then Tn = ∞. The null hypothesis, Θ0 , is given by the line σ = µ/3. Now we consider the perpendicular line to the null hypothesis across the point (0, c). This line is given by σ = −3µ + c.



427

The intersection of both lines is a point at the null hypothesis P5 = (3c/10, c/10). We consider 10 points on the line σ = −3µ + c, four of them under P5 and six over P5 , namely µ ¶ c 101 17 c , (−1, 3), j = 1, 2, ..., 11. Pj = +j 16 20 20 320 For c = 1/2 we have the following 11 points:

P1 = (.1562, .0312) , P2 = (.1547, .0359) , P3 = (.1531, .0400) , P4 = (.1516, .0453) , P5 = (.1500, .0500) , P6 = (.1484, .0547) , P7 = (.1469, .0594) , P8 = (.1453, .0641) , P9 = (.1437, .0687), P10 = (.1422, .0734), P11 = (.1406, .0781). Given, for instance, P2 = (.1547, .0359) , we calculate the power of the test statistics simulating random normal variables with mean 0.1547 and standard deviation 0.0359. For any statistic T ∈ S = {Wn , Rn , Tnr , r = −1, −.6, −.3, 0, .3, .5, .7, 1, 1.3, 1.6, 2} and any point Pj , the power is βT (Pj ) = Pr(T > χ21,0.05 /Pj ) = Pr(T > 3.84/Pj ). This power will be obtained as follows: • For each Pj , j = 1, ..., 11, repeat N = 1000 times: - Generate n = 50 normal random variables with parameter Pj and obtain the maximum likelihood estimators. - Evaluate Tj,i (value of T for the random sample i (i = 1, ..., 1000) of parameter Pj (j = 1, ..., 11) with c = 1/2). • Estimate

N umber of Tj,i > 3.84 (i = 1, ..., 1000) . βbT (j) = 1000 In Table 9.1 we present the powers corresponding to the test statistics Wn , Rn and Tnr , r = −1, −.6, −.3, 0, .3, .5, .7, 1, 1.3, 1.6, 2. For r = 1 we have the likelihood ratio test statistic. Let us denote S = {Wn , Rn , Tnr , r = −1, −.6, −.3, 0, .3, .5, .7, 1, 1.3, 1.6, 2} the set of test statistics considered here. We define βMAX (Pj ) = supT ∈S βT (Pj ), for j = 1, ..., 11, the maximum power achieved in the class S for each alternative hypothesis Pj . Then iT (Pj ) = βMAX (Pj ) − βT (Pj )


T

428


is the inefficiency on the set of alternative hypotheses for each T ∈ S. Thus, the quantity η (T ) = maxj6=5 {iT (Pj )} can be regarded as a measure of inefficiency of the test statistic T. The minimax criterion chooses the test statistic T ∈ S minimizing η (.) . Powers Pj 1 2 3 4 5 6 7 8 9 10 11

Wn 1.00 .965 .675 .252 .068 .083 .301 .544 .811 .910 .972

Rn .997 .853 .431 .099 .044 .194 .477 .715 .907 .962 .989

r = −1 1.00 .979 .742 .312 .095 .067 .222 .459 .754 .872 .964

r = −.6 1.00 .975 .710 .280 .078 .069 .264 .504 .786 .895 .967

r = −.3 1.00 .970 .689 .260 .074 .082 .294 .536 .806 .907 .969

Tnr r=0 1.00 .960 .664 .238 .067 .095 .326 .563 .825 .915 .974

r = .3 1.00 .951 .640 .215 .059 .109 .349 .596 .841 .931 .977

r = .5 1.00 .946 .615 .205 .056 .119 .363 .620 .850 .933 .979

r = .7 1.00 .936 .602 .191 .054 .133 .377 .633 .861 .943 .980

Table 9.1 Powers Tnr Pj 1 2 3 4 5 6 7 8 9 10 11

r=1 1.00 .927 .573 .176 .055 .147 .404 .659 .874 .950 .981

r = 1.3 .999 .913 .540 .159 .052 .162 .437 .677 .888 .958 .986

r = 1.6 .997 .894 .516 .147 .049 .181 .463 .701 .897 .960 .989

r=2 .997 .880 .477 .126 .050 .214 .496 .730 .915 .966 .992

βMAX (j) 1.00 .979 .742 .312 .050 .214 .496 .730 .915 .966 .992

Table 9.1 (Continuation) In Table 9.2 we present the inefficiencies η(T ) of the test statistics Tnr .



Pj 1 2 3 4 5 6 7 8 9 10 11

Wn .000 .014 .067 .060 -.018 .131 .195 .186 .104 .056 .020

Rn .003 .126 .311 .213 .006 .020 .019 .015 .008 .004 .003

r = −1 .000 .000 .000 .000 -.045 .147 .274 .271 .161 .094 .028

Inefficiencies Tnr r = −.6 r = −.3 r = 0 .000 .000 .000 .004 .009 .019 .032 .053 .078 .032 .052 .074 -.028 .024 -.017 .145 .132 .119 .232 .202 .170 .226 .194 .167 .129 .109 .090 .071 .059 .051 .025 .023 .018

.195

.311

.274

.232

.202

.170

429

r = .3 .000 .028 .102 .097 -.009 .105 .147 .134 .074 .035 .015

r = .5 .000 .033 .127 .107 -.006 .095 .133 .110 .065 .033 .013

r = .7 .000 .043 .140 .121 -.004 .081 .119 .097 .054 .023 .012

.147

.133

.140

Table 9.2

j 1 2 3 4 5 6 7 8 9 10 11

r=1 .003 .052 .169 .136 -.005 .067 .092 .071 .041 .016 .011

Inefficiencies Tnr r = 1.3 r = 1.6 .001 .003 .066 .085 .202 .226 .153 .165 -.002 .001 .052 .033 .059 .033 .053 .029 .027 .018 .008 .006 .006 .003

r=2 .003 .099 .265 .186 .000 .000 .000 .000 .000 .000 .000

.169

.202

.265

.226

Table 9.2 (Continuation)

We can see from the last line in Table 9.2 that the test statistic Tn0.5 presents the minimum relative inefficiency. The relative inefficiency of the test statistic using, e.g., the Rao test statistic, is more than 100% higher.


430


Observe that Tn−0.5 = −8n log(1 − TnF re /8n), where TnF re is the Freeman-Tukey test statistic given by Z ³q q ´2 TnF re = 4n fθb (x) − fθe (x) dx. ³ Furthermore, Tn2 = n log 1 +

2 Xn n

Xn2

´

R

where

=n

Z ¡ R

is the chi-square test statistic.

9.4.

¢2 fθb (x) − fθe (x) dx fθe (x)

Multi-sample Problem

This section deals with testing a composite null hypothesis H0 about parameters from s populations whose distributional structure differs just in the value of a parameter. From each population i, a sample of size ni is drawn at random, i = 1, ..., s. Let b1 , . . . , θ bs denote the maximum likelihood estimators and θ e1 , . . . , θ es the maximum likeθ lihood estimators under H0 . When n1 = . . . = ns , Morales et al . (1997) developed a testing procedure based on the φ-divergence test statistic ³ ´ b1 , . . . , θ bs ), (θ e1 , . . . , θ es ) , Tnφ (θ (9.16) which is obtained by calculating a φ—divergence between the joint densities s Y

i=1

fθb i (xi )

and

s Y

i=1

fθe i (xi ) .

When the sample sizes are different, the φ-divergence test statistics given in (9.16) cannot be used unless they were generalized in some sense. In the literature of Statistical Information Theory, problems related to s samples have been treated by using families of divergences between s populations (see Menéndez et al . (1997e), Zografos (1998a), Morales et al . (1998)). This is a nice possibility, but it is not the natural extension of the likelihood ratio test statistic, and in some situations the asymptotic distribution of these test statistics are based on a linear combination of chi-square distributions instead of on a chi-square distribution. The likelihood ratio test statistic uses the ratio between ni s Y Y

i=1 j=1


fθb i (Yij )

and

ni s Y Y

i=1 j=1

fθe i (Yij )

(9.17)


431

for introducing the decision rule. As a parallel approach, in this section we consider a divergence between the two estimated likelihood functions appearing in (9.17) and provide a decision rule on the basis of the resulting statistic. Let (X1 , βX1 , P1,θ 1 )θ 1 ∈Θ1 , . . . , (Xs , βXs , Ps,θ s )θ s ∈Θs be statistical spaces associated with independent populations. For i = 1, · · · , s, Xi ⊂ Rni is the sample space, βXi is the Borel σ-field of subsets of Xi , Θi ⊂ Rki is an open set and fi,θ i is the probability density function of Pi,θ i with respect to a σ-finite measure µi , i = 1, ..., s. In this section, we deal with the product statistical space (X1 × · · · × Xs , βX1 × · · · × βXs , P1,θ 1 × · · · × Ps,θ s ) , with (θ1 , . . . , θs ) ∈ Θ1 × · · · × Θs . Let µ , µn1 1 × · · · × µns s be the product measure and X , X1n1 × · · · × Xsns be the product sample space. Consider s independent random samples, Yi1 , Yi2 , . . . , Yini , i = 1, ..., s, from independent and identically distributed random variables with common probability density function fi,θ i , θi ∈ Θi , i = 1, . . . , s. Assume P that the sample sizes ni tend to infinity with the same rate, that is, if n = si=1 ni , then ni −→ λi ∈ (0, 1) , i = 1, ..., s n ni →∞

(9.18)

Ps where i=1 λi = 1. Further, assume that the parameters θi = (θi,1 , . . . , θi,ki ), i = 1, . . . , s, have the same k first components, that is, θ1,1 θ1,2 .. .

= θ2,1 = θ2,2 .. .

= ... = = ... =

θ1,k

= θ2,k

= . . . = θs,k ,

θs,1 θs,2 .. .

where k ≤ min{k1 , · · · , ks }. Let us consider the joint sample Y = (Y11 , . . . , Y1n1 ; Y21 , . . . , Y2n2 ; . . . ; Ys1 , . . . , Ysns ), and the joint parameter γ = (θ1,1 , . . . , θ1,k1 ; θ2,k+1 , . . . , θ2,k2 ; . . . ; θs,k+1 , . . . , θs,ks ), P with γ ∈ Γ, where Γ is an open subset of RM and M = si=1 ki − (s − 1)k. Let fi,θ i (Y i ) =

ni Y

fi,θ i (Yij ) and L(θi ; Yi1 , ..., Yini ) =

j=1

ni X

log fi,θ i (Yij )

j=1

be the likelihood and the log-likelihood function of θi based on the i-th sample. The likelihood and the log-likelihood function of γ = (γ1 , . . . , γM ) ∈ Γ based on the joint sample are fγ (Y ) =

s Y

i=1


fi,θ i (Yi ) and

l(γ) =

s X i=1

L(θi ; Yi1 , ..., Yini )

432


respectively. The likelihood equations are s

∂l(γ) X ∂L(θi ; Yi1 , ..., Yini ) = = 0, ∂γp ∂γp i=1

p = 1, . . . , M ,

where, for each i, n

i ∂L(θi ; Yi1 , ..., Yini ) X ∂ = log fi,θ i (Yij ). ∂γp ∂γ p j=1

Let I i (θi ) be the Fisher information matrix of the parameter θi associated with the i ith-population with density function fi,θ i and let Ip,q (θi ) the (p, q)th-element of I i (θ i ). i We split I (θi ) into blocks as follows Ã ! i i I (θ ) I (θ ) i i k,k k,ki I i (θi ) = , Iki i ,k (θ i ) Iki i ,ki (θ i ) ki ×ki

i i where Ik,k (θ i ), Ik,k (θ i ), Iki i ,k (θi ) and Iki i ,ki (θ i ) are the submatrices whose lower-right i i i corner elements are respectively Ik,k (θi ), Ik,k (θi ), Iki i ,k (θi ) and Iki i ,ki (θ i ), and whose i sizes are k × k, k × (ki − k), (ki − k) × k and (ki − k) × (ki − k), respectively.

The following M × M matrix, denoted by V (γ), plays the fundamental role of Fisher information matrix,  Ps  i 1 s λ1 Ik,k (θ1 ) · · · λs Ik,k (θ s ) i=1 λi Ik,k (θ i ) 1 s  λ I 1 (θ )  λ1 Ik11 ,k1 (θ1 ) ... 0   1 k1 ,k 1  . V (γ) =  .. .. .. ..  .   . . . s s λs Iks ,k (θs ) 0 ... λs Iks ,ks (θs )

The regularity assumption (vi) is needed to derive the asymptotic normality of the maximum likelihood estimators. (vi) The matrix V (γ) is positive definite. A composite null hypothesis H0 can be usually described by a subset Γ0 of the parametric space and, consequently, the alternative hypothesis is associated with Γ1 = Γ − Γ0 . Suppose that Γ0 can be expressed as Γ0 = {γ ∈ Γ : γi = hi (β), i = 1, . . . , M} = {γ ∈ Γ : h (β) = γ} , where β = (β1 , . . . , βM0 )T ∈ B, h = (h1 , ..., hM ) and B ⊂ RM0 is an open subset. Suppose that Γ0 can be also described by the M − M0 restrictions gi (γ) = 0,


i = 1, . . . , M − M0 ,


433

where the functions hi and gj have continuous first order partial derivatives and the ranks of the matrices µ µ ¶ ¶ ∂gi (γ) ∂hi (β) Tγ = , Mβ = ∂γj i=1,...,M−M0 ∂βj i=1,...,M j=1,...,M0

j=1,...,M

are M − M0 and M0 respectively. Further, suppose that the s submodels (X1 , βX1 , P1,θ 1 )θ 1 ∈Θ1 , . . . , (Xs , βXs , Ps,θ s )θ s ∈Θs restricted to null hypothesis (i.e., with γ ∈ Γ0 ) satisfy i)—v), with derivatives taken with respect to the new parameter β. For testing the null hypothesis H0 : γ = h(β), we shall use the family of φ-divergence test statistics b ≡ 2 Dφ (b b Tnφ (b γ , h(β)) γ , h(β)), φ00 (1) b is the φ-divergence between fγb (x) and f b (x). γ , h(β)) where Dφ (b h(β )

The following theorem was proved by Morales et al . (2001) for k = 0 and by Hobza et al. (2001) for k > 1. Theorem 9.7 Let H0 : γ = h(β) be true. For each i = 1, . . . , s, let Yi1 , . . . , Yini be independent and identically distributed random variables with common probability density function fi,θ i (x) satisfying the regularity assumptions (9.18) and i)-v). Suppose also that the necessary conditions for differentiating inside the integrals hold. Then b = Tnφ (b γ , h(β))

2 φ00 (1)

L b −→ Dφ (b γ , h(β)) χ2M−M0 . n→∞

In the multi-sample case with exponential models, the Kullback-Leibler and the likelihood ratio test statistics coincide for k = 0; however this result does not hold for k > 1. In the case of the (h, φ)-divergence measure the analogous result holds, i.e., b = Tnφ,h (b γ , h(β))

2 L b −→ Dh (b γ , h(β)) χ2M−M0 . n→∞ φ00 (1) h0 (0) φ

For instance, in the special case of Rényi’s divergence, the functions h and φ were given in (9.7) and (9.8) and Rényi test statistic is Z 1 L b ≡ 2D1r (b b = log γ , h(β)) γ , h(β)) fγbr (x)f 1−rb (x)dµ(x) −→ χ2M−M0 . Tnr (b h (β ) n→∞ r(r − 1) X © 2006 by Taylor & Francis Group, LLC

434


The example below, due to Morales et al . (2001), shows the behaviour of Rényi test statistic with normal populations. Example 9.3 Let (Y11 , ..., Y1n1 ) , ..., (Ys1 , ..., Ysns ) be s independent random samples from normal populations with unknown parameters (µ1 , σ1 ) , ..., (µs , σs ) respectively. We are interested in testing H0 : σ12 = ... = σs2 versus H1 : ∃ i 6= j with σi2 6= σj2 . In this case, the joint parameter space is © ª Γ = (µ1 , ..., µs , σ12 , ..., σs2 ) / µi ∈ R, σi2 > 0, i = 1, ..., s ,

and its restriction to H0 is

© ª Γ0 = (µ1 , ..., µs , σ12 , ..., σs2 ) ∈ Γ / σ12 = ... = σs2 > 0 .

Using the functions hi and gi given previously, the null hypothesis Γ0 can be written in the following alternative forms: i) Consider the set B defined by © ª B = β = (µ1 , ..., µs , σ, ...σ) ∈ R2s : µi ∈ R and σ ∈ R+

and the functions,

hi (β) =

½

µi σ2

i = 1, ..., s . i = s + 1, ..., 2s

Then Γ0 = {γ ∈ Γ : γi = hi (β), i = 1, ..., 2s} . ii) Consider the function gj : Θ → Rs−1 , j = 1, ..., s − 1, defined by gj (µ1 , ..., µs , σ1 , ..., σs ) = σ1 − σj , j = 1, ..., s − 1. Obviously gj (µ1 , ..., µs , σ1 , ..., σs ) = 0 for (µ1 , ..., µs , σ1 , ..., σs ) ∈ Γ0 . Therefore, Γ0 = {γ ∈ Γ : gj (γ) =0, j = 1, ..., s − 1} . ¡ ¢ b = Y 1∗ , . . . , Y s∗ , σ The maximum likelihood estimator is γ b12 , . . . , σ bs2 and the maxi¢ ¡ b = Y 1∗ , . . . , Y s∗ , σ b2 , . . . , σ b2 mum likelihood estimator under the null hypothesis is h(β) P where σ b2 = n1 si=1 σ bi2 . The Rényi test statistic for testing H0 : σ12 = . . . = σs2 is given by 2 b = b = 2Dr1 ((b b 1 ), (e e 2 )), D1 (b γ , h(β)) γ , h(β)) µ, Σ µ, Σ Tnr (b φ00 (1)h0 (0) r © 2006 by Taylor & Francis Group, LLC


with

´T ³ b Ts b= µ b T1 , ..., µ , µ ´ ³ b 1 = diag Σ b ∗1 , . . . , Σ b ∗s , Σ ´T ³ e= µ e T1 , ..., µ e Ts µ , ´ ³ ∗ e 2 = diag Σ e 1, . . . , Σ e ∗s , Σ

435

¢T ¡ b Ti = Y i∗ , . . . , Y i∗ , i = 1, . . . , s, µ ¢ ¡ 2 b ∗ = diag σ Σ bi , . . . , σ bi2 , i = 1, . . . , s, i ¢T ¡ e Ti = Y i∗ , . . . , Y i∗ , i = 1, . . . , s, µ ¢ ¡ 2 e ∗ = diag σ Σ b ,...,σ b2 , i = 1, . . . , s. i

After straightforward algebra, we get

 s P (rb σ 2 +(1−r)b σi2 ) 1  ni log  2(1−r) r(1−r) 2r  σ b σ bi  i=1   P s 2 σ b r b = ni log σbi 2 Tn (b γ , h(β))  i=1  ³ 2 ´  s 2 P    ni σb 2 − 1 + log σσbb i2 σbi

i=1

if

r 6= 0, r 6= 1

if

r=1

if

r = 0.

b is chi-square with s − 1 γ , h(β)) By Theorem 9.7, the asymptotic distribution of Tnr (b degrees of freedom. Therefore, an asymptotically test, with significance level α, for the b > χ2 problem of testing the equality of variances should reject H0 when Tnr (b γ , h(β)) s−1,α . In Morales et al . (2001), a power simulation study was carried out for comparing several members of the family of Rényi test statistics.

As a result of this Monte Carlo simulation experiment, they recommended any Rényi divergence test statistic with r ∈ [5/4, 3/2]. They also emphasize that r = 5/4 emerges as a good alternative since it presents the best numerical results. Under the assumption σ12 = ... = σs2 , it is not difficult to establish that Rényi test statistic for testing H0 : µ1 = .... = µs versus H1 : ∃ i 6= j with µi 6= µj rejects the null hypothesis if 1 s

s X i=1

¡ ¢2 ni Y i∗ − Y

ni s X X ¡ ¢2 1 Yij − Y i∗ n−s

> Fs−1,n−s,α ,

i=1 j=1

where Fs−1,n−s,α is the 100 (1 − α) percentile of the F distribution with s − 1 and n − s degrees of freedom.


436

9.5.


Some Topics in Multivariate Analysis

A considerable part of this book is devoted to the use of the φ-divergence concept, developing methods of estimation and testing for the analysis of categorical data in several contexts like analysis of cross-classified data, log-linear models, etc. From this point of view, the φ−divergence is exploited here for the analysis of discrete multivariate data. The use of divergences in order to meet and study problems of Multivariate Analysis is not new. It starts in the early 1950, when Kullback (1959), in his pioneer book, dedicated five chapters to the use of the minimum information discrimination principle for the study of several problems in multivariate analysis. The bridge which links Statistical Information Theory and Multivariate Analysis is founded on the fact that multivariate analysis methods are mainly created on the notion of the distance between observations or distance among their respective distributions, while on the other hand, Statistical Information Theory is mainly concerned with the definition of statistical distances or divergences between distributions and on the development of metric geometries based mainly on the Fisher information matrix. The operational link between these two statistical areas has received great attention over the last four decades. Since Kullback’s pioneer work, there is a vast amount of contributions based on the information theoretic formulation of multivariate statistical topics, like distribution theory, statistical inference using multivariate continuous, categorical or mixed data, concepts of multivariate dependence, discrimination and classification etc. An indicatory, nonexhaustive literature of the subject is the following: i) Construction of Multivariate Distributions (Kapur (1989), Cuadras (1992a), Cuadras et al . (1997a), Zografos (1999)), ii) Statistical Inference (Krzanowski (1983), Bar-Hen and Daudin (1995), Morales et al . (1997, 1998), Zografos (1998a), Garren (2000)), iii) Measures of Multivariate Association (Kent (1983), Inaba and Shirahata (1986), Joe (1989), Zografos (1998b, 2000)), iv) Discriminant Analysis (Matusita (1966, 1973), Krzanowski (1982), Cacoullos and Koutras (1985), Cuadras (1992b), Koutras (1992), Bar-Hen, A. (1996), Cuadras et al. (1997b), Menéndez et al . (2004, 2005d)). In Sy and Gupta (2004) it can be seen that Information Theory is a useful tool for data mining.



9.6.

437

Exercises

b θ0 ), where θ b is the maxi1. Find the asymptotic distribution of the statistic Dφh (θ, mum likelihood estimator obtained from a population with parameter θ 6= θ0 . b θ0) 2. Find the asymptotic distribution of the (h, φ)-divergence test statistic Tnφ,h (θ, under the contiguous alternative hypotheses H1,n : θn = θ 0 + n−1/2 d,

where d is a fixed vector in RM0 such that θn ∈ Θ ⊂ RM0 . 3. We consider the divergence measure of order r and degree s. b θ0 ) under the hypotha) Find the asymptotic distribution of the statistic Drs (θ, esis that the observations are from a population with θ 6= θ0 . b) We consider the random sample

0.0789 0.1887 0.0828 0.0086 0.0572 0.0041 0.3551 0.0783 0.0732 0.1839 0.1439 0.1681 0.0115 0.1155 0.0566 from a exponential population of unknown parameter θ. Using the (h, φ)divergence test statistic based on the divergence measure of order r and degree s (s = 2 and r = 0.5) test if θ = 10 versus θ 6= 10 using as significance level α = 0.05. b θ0 ) for testing H0 : θ = θ0 . Ob4. We consider the φ-divergence test statistic Tnφ (θ, tain the approximate size n, guaranteeing a power β at a given alternative θ 6= θ0 .

5. We consider two random samples from two populations with parameters θ1 and θ2 of sizes n and m respectively, and the corresponding maximum likelihood estimab1 = (θb11 , ..., θb1M )T and θ b2 = (θb21 , ..., θb2M )T , associated with them. Find tors, θ 0 0 b1 , θ b2 ) under the two following the asymptotic distribution of the test statistic Dφ (θ assumptions: a) θ1 6= θ2 and b) θ 1 = θ2 .

6. Use the result obtained in Exercise 5 for testing H0 : θ1 = θ 2 versus H1 : θ 1 6= θ2 . 7. Let Y1 , ..., Yn be a random sample from a exponential distribution of parameter θ.

b θ0 ), for λ = 2 a) Find the expression of Cressie-Read test statistic, Dφ(λ) (θ, where θb is the maximum likelihood estimator of θ. b) Test H0 : θ = 3/2 versus H1 : θ 6= 3/2, on the basis of the exact test based b θ0 ) and using as significance level α. on Dφ(λ) (θ,

8. Let X be a M -variate normal population with unknown mean vector µ and known variance-covariance matrix Σ0 .

a) Find a test statistic for testing µ = µ0 versus µ 6= µ0 using Kullback-Leibler divergence measure as well as the (1 − α) confidence region in R2 for µ.


438


b) The measurements on the first and second adult sons in a sample of 25 families are the following: First son (X1 )

191 208 179 Second son (X2 ) 179 192 186

195 189 183 201 190 174

181 197 174 185 189 185

183 188 190 188 197 195

176 192 188 177 187 187

163 175 197 161 165 200

195 192 190 183 185 187

186 181 174 176 173 182 178 176

.

Source: Mardia, K. V., Kent, J. T. and Bibby, J. M. (1979, p. 121).

We assume that X1 and X2 are independent and that each one is normally distributed with variance 100, i.e., µ ¶ 100 0 Σ = Σ0 = . 0 100 Test using as significance level α, H0 : µ = µ0 = (µ01 , µ02 ) = (182, 182)T against H1 : µ 6= µ0 and get a 95% confidence region for the mean of X1 and X2 . 9. Let X be a M -variate normal population with known vector mean µ0 and unknown variance-covariance matrix Σ. a) Find a test statistic for testing Σ = Σ0 versus Σ 6= Σ0 using the KullbackLeibler divergence measure as well as the 100 (1 − α) confidence region for Σ and M = 2. b) Test using as significance level α = 0.05 for the data given in Exercise 8 H0 : Σ = Σ0 =

µ

100 0 0 100

¶

assuming that µ = (182, 182)T .

10. Let X be a M -variate normal population with unknown mean vector µ and unknown variance-covariance matrix Σ. Find a test statistic for testing H0 : µ = µ0

versus H1 : µ 6= µ0 ,

based on the Kullback-Leibler divergence measure. 11. Let X be a M -variate normal population with unknown mean vector µ and unknown variance-covariance matrix Σ. Find a test statistic for testing H0 : Σ = Σ0

versus H1 : Σ 6= Σ0 ,

based on the Kullback-Leibler divergence measure.



439

12. Given the independent random samples: 0.4068 1.7698 1.7830 1.0186 1.5880 1.9616 1.1334 0.1288 0.8306 and 1.3863 2.5470 0.9480 0.0420 0.1449 0.7971 1.2858 2.9358 1.4829 0.7971 1.2858 2.5335 1.4829 from uniform distributions in the intervals (0, θ1 ) and (0, θ2 ) respectively, using as significance level α = 0.05, find the exact test based on Kullback-Leibler divergence measure for testing θ1 = θ2 versus θ1 6= θ2 (θ1 ≤ θ2 ). 13. Let X1 , ..., Xn and Y1 , ..., Ym be two independent random samples from the distributions fθi (x) = exp(−(x − θi ))

θi < x < ∞,

i = 1, 2.

Find the test statistic based on Kullback-Leibler divergence measure for testing H0 : θ1 = θ2

versus

H1 : θ1 > θ2 .

14. We consider a population X with probability density function fθ (x) =

θ2θ xθ+1

x ≥ 2, θ > 0.

Find the expression of the test statistics of Wald, Rao, Likelihood ratio and Rényi for testing H0 : θ = θ0

versus

H1 : θ 6= θ0 ,

based on a sample of size n. 15. We consider a population with exponential distribution with parameter θ and we wish to test H0 : θ = 1 versus H1 : θ 6= 1 using as significance level α = 0.05. a) Find Rényi test statistic for r = 1/4, 3/4, 1, 5/4, 7/4 and 2. b) Study the accuracy of powers approximations given for n = 20, 40, 80 and 200, r = 1/4, 3/4, 1, 5/4, 7/4, 2 and θ = 0.5, 0.6, 0.65, 0.70, 0.75, 0.80, 0.85, 0.90, 0.95, 1.10, 1.20, 1.30, 1.40, 1.50, 1.60, 1.70, 1.80, 1.90, 2, 2.10, 2.20, 2.30, 2.40 and 2.50. c) Obtain the sample size to get a power of β = 0.8 for θ = 0.5, 0.6, 0.65, 0.70, 0.75, 0.80, 0.85, 0.90, 0.95, 1.10, 1.20, 1.30, 1.40, 1.50, 1.60, 1.70, 1.80, 1.90, 2, 2.10, 2.20, 2.30, 2.40 and 2.50.


440

9.7.



1. We have, Dφh (θ1 , θ2 ) = h (Dφ (θ1 , θ2 )) , where h is a differentiable increasing function mapping from [0, ∞) onto [0, ∞) , with h (0) = 0 and h0 (0) > 0. We know that h (x) = h0 (x0 ) (x − x0 ) + o (x − x0 ) , then ´ h√ ³ í √ ³ h b b θ0 ) − Dφ (θ, θ0 ) n Dφ (θ, θ0 ) − Dφh (θ, θ 0 ) = h0 (Dφ (θ, θ0 )) n Dφ (θ, + oP (1) ,

and

´ ¡ ¢ √ ³ h L 2 b θ 0 ) − Dφh (θ, θ0 ) −→ n Dφ (θ, N 0, σh,φ (θ) , n→∞

2 σh,φ (θ)

0

where = (h (Dφ (θ, θ 0 ))) Theorem 9.2.

2

σφ2 (θ)

and σφ2 (θ) = T T IF (θ)−1 T is given in

2. In a similar way to the previous Exercise we have, under H1,n , that the statistics b θ 0 ) and h0 (0) Dφ (θ, b θ0 ) Dφh (θ,

have the same asymptotic distribution. Therefore, 2n L b θ0 ) = b θ0 ) −→ Dh (θ, χ2 (δ) , Tnφ,h (θ, n→∞ M0 φ00 (1) h0 (0) φ being δ = dT IF (θ0 ) d.

If we want to approximate the power at some alternative θ, then d = d(n, θ, θ0 ) = √ n(θ − θ0 ) should be used in the formula of the noncentrality parameter. 3. The divergence of order r and degree s is given by    s−1 r−1 Z 1   Drs (θ, θ0 ) = fθ (x)r fθ0 (x)1−r dµ (x) − 1 ,  s−1 X

and it can be considered as a (h, φ)-divergence with ´ s−1 1 ³ (1 + r (r − 1) x) r−1 − 1 ; s, r 6= 1 h (x) = s−1 and xr − r (x − 1) − 1 φ (x) = ; r 6= 0, 1. r (r − 1) a) From Exercise 1 it is only necessary to get the elements of the vector T . These are given by   s−r r−1 Z r  r 1−r ti = fθ (x) fθ 0 (x) dµ (x) r−1 X Z ∂fθ (x) × fθ 0 (x)1−r fθ (x)r−1 dµ (x) , i = 1, ..., M. ∂θi X



441

b) The divergence measure of order r and degree s between two exponential distributions with parameters λ and µ is given for λr + µ (1 − r) > 0 by ! Ãµ s−1 ¶ r−1 λr µ1−r −1 s −1 , Dr (λ, µ) = (s − 1) λr + µ (1 − r) and for λr + µ (1 − r) ≤ 0, Drs (λ, µ) = +∞. The maximum likelihood estib = Y −1 (Y is the sample mean) and for mator in the exponential model is λ b = 9.3379. the given random sample takes on the value λ Now it is necessary to evaluate the test statistic b 10) ≡ Tnr,s (λ,

2n b 10). Ds (λ, h0 (0) φ00 (1) r

In our case h0 (0) = r and φ00 (1) = 1, then we have

b 10) = 2n D2 (λ, b 10) = 0.0704. Tnr,s (λ, 0.5 0.5

On the other hand χ21, 0.05 = 3.841, and we should not reject the null hypothesis. 4. If we consider the expression of the power given in (9.6) the problem will be solved if we consider the sample size n∗ obtained as a solution of the equation Ã √ Ã !! φ00 (1) χ2M0 ,α n 1 − Dφ (θ, θ0 ) , βn,φ (θ) = 1 − Φ σφ (θ) 2n i.e.,

p A + B + A(A + 2B) , (9.19) n = 2Dφ (θ, θ0 )2 ³ ´2 1 where A = σφ2 (θ) Φ−1 (1 − βn,φ (θ)) and B = φ00 (1) χ2M0 ,α Dφ (θ, θ0 ). The required sample size is n = [n∗ ] + 1, where [·] is used to denote “integer part of”. ∗

5. First we consider part a). a) We have

then q

nm n+m

M P0

∂Dφ (θ1 , θ 2 ) b (θ1i − θ1i ) ∂θ1i °´ ³° M P0 ∂Dφ (θ 1 , θ2 ) b ° °b (θ2i − θ2i ) + o °θ + 1 − θ1° ∂θ 2i°´ i=1 ³° °b ° + o °θ 2 − θ2 ° ,

b1 , θ b2 ) = Dφ (θ 1 , θ2 ) + Dφ (θ

i=1

³ ´ b1 , θ b2 ) − Dφ (θ 1 , θ2 ) b 1 − θ 1 ) + S T (θ b2 − θ 2 ) Dφ (θ = T T (θ °´ °´ ³° ³° °b ° ° °b + o θ + o °θ − θ − θ ° ° 1 1 2 2° ,


442


where T = (t1 , ..., tM0 )T and S = (s1 , ..., sM0 )T with ti =

Z

0

φ

X

µ

fθ 1 (x) fθ 2 (x)

¶

∂fθ 1 (x) dµ (x) ∂θ1i

and si =

Z µ

X

µ µ ¶ ¶ ¶ fθ 1 (x) ∂fθ 2 (x) fθ 1 (x) fθ 1 (x) 0 φ −φ dµ (x) . ∂θ2i fθ 2 (x) fθ 2 (x) fθ 2 (x)

On the other hand √ T b nT (θ 1 − θ 1 ) √ b2 − θ 2 ) mS T (θ

L

−→

n→∞ L

−→

n→∞

³ ´ −1 N 0, T T IF (θ1 ) T ³ ´ −1 N 0, S T IF (θ 2 ) S .

Therefore the test statistic r ´ nm ³ b1 , θ b2 ) − Dφ (θ1 , θ 2 ) Dφ (θ n+m

is asymptotically distributed as a normal distribution with mean zero and variance λT T IF (θ1 )−1 T + (1 − λ) S T IF (θ 2 )−1 S,

where

m , m+n °´ °´ ³° ³° ¡ −1/2 ¢ ¡ −1/2 ¢ ° ° °b °b and o °θ . because o °θ 1 − θ 1 ° = oP n 2 − θ 2 ° = oP m λ=

lim

n,m→∞

b) In this case we have

¶ ∂ 2 Dφ (θ1 , θ2 ) (θb1i − θ1i )(θb1j − θ1j ) ∂θ1i ∂θ1j i,j=1 θ =θ µ ¶ 1 2 M P0 ∂ 2 Dφ (θ 1 , θ2 ) + (θb1i − θ1i )(θb2j − θ2j ) ∂θ1i ∂θ2j i,j=1 θ¶1 =θ 2 µ M P0 ∂ 2 Dφ (θ1 , θ2 ) (θb2i − θ2i )(θb2j − θ2j ) + 12 ∂θ ∂θ 2i 2j i,j=1 µ° µ° θ 1 =θ 2 °2 ¶ °2 ¶ °b ° °b ° + o °θ − θ + o °θ 2 − θ 2 ° . 1 1°

b1 , θ b2 ) = Dφ (θ

We have

∂Dφ (θ1 , θ 2 ) = ∂θ1i

1 2

Z

X


M P0

0

φ

µ

µ

fθ 1 (x) fθ 2 (x)

¶

∂fθ 1 (x) dµ (x) ∂θ1i

i = 1, ..., M0


then µ µ µ

∂ 2 Dφ (θ1 , θ 2 ) ∂θ1i ∂θ1j

¶

∂ 2 Dφ (θ1 , θ 2 ) ∂θ1i ∂θ2j

¶

= −

∂ 2 Dφ (θ1 , θ 2 ) ∂θ2i ∂θ2j

¶

µ

= φ00 (1)

θ 1 =θ 2

θ 1 =θ 2

Z

X

=

θ 1 =θ 2

µ

443

1 ∂fθ 1 (x) ∂fθ 1 (x) dµ (x) fθ 1 (x) ∂θ1i ∂θ1j

∂ 2 Dφ (θ 1 , θ2 ) ∂θ1i ∂θ1j

∂ 2 Dφ (θ1 , θ 2 ) ∂θ1i ∂θ1j

¶

θ 1 =θ 2

¶

.

θ 1 =θ 2

Therefore, 2 b1 , θ b 2 ) = (θ b1 − θ1 )T IF (θ1 ) (θ b1 − θ 1 ) Dφ (θ φ00 (1) b2 − θ 2 ) b1 − θ1 )T IF (θ1 ) (θ − 2(θ

On the other hand

then

q

q

µ° °2 ¶ °b ° T b b + (θ2 − θ1 ) IF (θ1 ) (θ2 − θ1 ) + o °θ1 − θ1 ° µ° ¶ °2 °b ° + o °θ 2 − θ2 ° µ° °2 ¶ °b ° T b b b b = (θ1 − θ2 ) IF (θ1 ) (θ1 − θ2 ) + o °θ1 − θ1 ° µ° ¶ °2 °b ° + o °θ . 2 − θ2 °

√ b n(θ 1 − θ 1 ) √ b m(θ 2 − θ 2 ) mn b m+n (θ 1 mn b m+n (θ 2

− θ1 ) − θ2 )

´ ³ N 0, IF (θ 1 )−1 ´ ³ N 0, IF (θ 2 )−1

L

−→

n→∞ L

−→

n→∞

L

−→

n,m→∞ L

−→

n,m→∞

´ ³ N 0, λIF (θ1 )−1 ´ ³ N 0, (1 − λ) IF (θ2 )−1 .

Under the hypothesis θ 1 = θ 2 , we have q L mn b b −→ m+n (θ 1 − θ 2 )

n,m→∞

therefore

´ ³ N 0, IF (θ1 )−1

2 mn L b1 , θ b2 ) −→ Dφ (θ χ2 , n,m→∞ M0 φ00 (1) m + n µ° µ° °2 ¶ °2 ¶ ¡ −1 ¢ ¡ ¢ °b °b ° ° θ n and o − θ − θ because o °θ = o = oP m−1 . ° 2 1 1° P 2°


444


6. We should reject the null hypothesis if b1 , θ b2 ) > c, Dφ (θ

where c is a positive constant. Now it is possible to find two situations: b1 , θ b2 ) is known; then independently i) The distribution of the test statistic Dφ (θ of the regularity conditions, we choose c in such a way that the significance level of the test is α, ³ ´ b1 , θ b2 ) > c = α. Pr Dφ (θ

b1 , θ b2 ) is unknown. In this ii) The exact distribution of the test statistic Dφ (θ case we have to use the asymptotic distribution given in Exercise 5 and we should reject the null hypothesis if 2 mn b1 , θ b2 ) > χ2 ,α . Dφ (θ M0 (1) m + n

φ00

7. a) The divergence of Cressie and Read for λ = 2 between two exponential distributions is given by ! ÃZ 3 ∞ f (x) θ Dφ(2) (θ, θ0 ) = 16 2 dx − 1 ¶ µ 0Z f∞θ0 (x) 3 = 16 θθ2 exp (− (3θ − 2θ0 ) x) dx − 1 ³ 30 0 ´ 1 = 16 θθ2 3θ−2θ − 1 0 0

for θ > 23 θ0 . If θ ≤ 23 θ0 we have Dφ(2) (θ, θ0 ) = +∞.

b) We should reject the null hypothesis, H0 : θ = θ0 , if b θ0 ) ≡ 2nDφ (θ, b θ0 ) > c. Tn2 (θ, (2)

−1 The maximum likelihood estimator for θ is θb = Y . Therefore, we can write !  Ã  4  n if Y < 1 ¢ −1 2¡ 3 b θ0 ) = Tn2 (θ, . 27Y 1 − Y   +∞ if Y ≥ 1

We denote

¶ 4 − 1 if y < 1 g (y) = 27y2 (1 − y)  +∞ if y ≥ 1  

n 3

µ

and it is clear that if y → 0 or 1 then g tends to +∞ and also the minimum is obtained for y = 2/3. Therefore, g(y) > k



445

if and only if y < k1 or y > k2 . The values k1 and k2 are obtained as a solution of the equation system ½ g (k1 ) = g (k2 ) ¡ ¢ Prθ=3/2 Y ∈ (k1 , k2 ) = α.

It is well known that if Y has an exponential distribution with parameter θ0 , P then ni=1 Yi has a gamma distribution with parameters a = θ0 and p = n. Also Z = 2nθ0 Y is chi-square with n degrees of freedom. One procedure to calculate the values k1 and k2 determining the critical region in an easier way is the following: Reject H0 : θ = 3/2 when Y < k1 or Y > k2 is equivalent to reject H0 when 3nY < c1 or 3nY > c2 , where c1 and c2 are obtained in such a way that ½ 0.95 = Fχ2n (c2 ) − Fχ2n (c1 ) ¡ c1 ¢ ¡ c2 ¢ . g 3n = g 3n Then c1 and c2 are the solutions of the equation system Z c2  1 1 n  0.95 = ¡ n ¢ e− 2 x x 2 −1 dx n/2 Γ 2 . c1 2  2 c2 (3n − c2 ) = c21 (3n − c1 )

8. a) Given two M -variate normal distributions N (µ1 , Σ1 ) and N (µ2 , Σ2 ) , KullbackLeibler divergence (see Chapter 1) between them is ³ ¡ −1 ¢´ T DKull (θ1 , θ 2 ) = 12 (µ1 − µ2 ) Σ−1 2 (µ1 − µ2 ) + trace Σ2 Σ1 − I |Σ2 | , + 12 log |Σ1 | with θ1 = (µ1 , Σ1 ) and θ2 = (µ2 , Σ2 ) . In our case θ 1 = (µ, Σ0 ) and θ 2 = (µ0 , Σ0 ) with Σ0 known. Given the random sample Y 1 = (Y11 , ..., Y1M )T , Y 2 = (Y21 , ..., Y2M )T , ..., Y n = (Yn1 , ..., YnM )T from the population distributed N (µ, Σ0 ), the maximum likelihood estimator of ¢T ¡ µ is Y = Y 1 , Y 2 , ..., Y M , where n

Yi =

1X Yli , i = 1, ..., M. n l=1

Then the test statistic for testing µ = µ0 ¡ ¢ ¡ ¢ ¡ ¢T ¡ ¢ TnKull Y , µ0 ≡ 2nDKull Y , µ0 = n Y − µ0 Σ−1 Y − µ0 0 © 2006 by Taylor & Francis Group, LLC

446


is asymptotically chi-squared distributed with M degrees of freedom. Then we should reject the null hypothesis if ¡ ¢ ¡ ¢T ¡ ¢ Y − µ0 > χ2M,α . TnKull Y , µ0 = n Y − µ0 Σ−1 0

In this case the exact distribution of the random variable ¢T ¡ ¢ ¡ Y − µ0 n Y − µ0 Σ−1 0

is also chi-square with M degrees of freedom. Then the exact and asymptotic tests coincide. A (1 − α) 100% confidence region in R2 for the mean of X1 and X2 is given by the values µ = (µ1 , µ2 ) ∈ R2 verifying ¡ ¡ ¢T ¢ n Y − µ Σ−1 Y − µ < χ22,α . 0

b) Applying a) it is necessary to evaluate

¢T ¡ ¢ ¡ A = n Y − µ0 Σ−1 Y − µ0 . 0

Hence

A = 25 (3.74, 1.84)

µ

100 0 0 100

¶−1 µ

3.72 1.84

¶

= 4.343,

and χ22,0.05 = 5.99. Then the null hypothesis should not be rejected. The 95% confidence region for (µ1 , µ2 ) is given by n o (µ1 , µ2 ) ∈ R2 : (185.72 − µ1 )2 + (183.84 − µ2 )2 < 23.96 . 9. a) In this case θ1 = (µ0 , Σ) and θ2 = (µ0 , Σ0 ). Given the random sample T

T

T

Y 1 = (Y11 , ..., Y1M ) , Y 2 = (Y21 , ..., Y2M ) , ..., Y n = (Yn1 , ..., YnM )

the maximum likelihood estimator of Σ, with µ = µ0 known (see Mardia et al. 1979, p. 104), is given by b = S + ddT Σ where d = Y − µ0 and S is the sample variance covariance matrix given by ! Ã n ¢¡ ¢ 1 X¡ Yil − Y i Yjl − Y j . S = (sij )i,j=1,...,M = n l=1

i,j=1,...,M

The expression of the test statistic based on Kullback-Leibler divergence is   ³ ´ ³ ´ ³ ´ |Σ | 0 b Σ0 = 2nDKull Σ, b Σ0 = n trace Σ−1 Σ b − I + log ¯ ¯  . TnKull Σ, 0 ¯b¯ ¯Σ¯ © 2006 by Taylor & Francis Group, LLC

447


b Σ0 ) > χ2 2 Therefore we should reject the null hypothesis if 2nDKull (Σ, . M +M ,α

2

The confidence region for σ = (σ11 , σ22 , σ12 ) is given by       ³ ´ b − I + log ¯|Σ|¯  < 7.82 . (σ11 , σ22 , σ12 ) : n trace Σ−1 Σ ¯b¯   ¯Σ¯ b) With the data given in Exercise 8, we have b = Σ

i.e.,

µ

93.48 66.87 66.87 96.77

¶

+

µ

13.83 6.84 6.84 3.38

¶

=

µ

107.31 73.71 73.71 100.11

¶

,

¯ ¯ ³ ´ ¯b¯ b ¯Σ¯ = 5308.96, |Σ0 | = 10000 and trace Σ−1 0 Σ − I = 0.074.

Then

³ ´ b Σ0 = 17.679. TnKull Σ,

On the other hand χ23,0.05 = 7.82 and we should reject the null hypothesis. 10. In this case the unknown parameter is θ = (µ, Σ) . Given the random sample T T T Y 1 = (Y11 , ..., Y1M ) , Y 2 = (Y21 , ..., Y2M ) , ..., Y n = (Yn1 , ..., YnM ) ,

from the normal population N (µ, Σ), the maximum likelihood estimator of θ, under the null hypothesis, is given by ³ ´ e = µ , S + ddT , θ 0

where d and S were given in Exercise 9. The maximum likelihood estimator of θ in all the parameter space is given by ¡ ¢ b = Y ,S . θ

Therefore b θ) e = DKull (θ,

+

´−1 ¡ ¡ ¢T ³ ¢ S + ddT Y − µ0 Y − µ0 ¯ ¯ µ µ³ ¶ T¯¶ ¯ ´−1 ¯ S +dd ¯ T 1 S + dd S − I M×M + log , 2 trace |S | 1 2

and we must reject the null hypothesis if b θ) e ≡ 2nDKull (θ, b θ) e > c. TnKull (θ, © 2006 by Taylor & Francis Group, LLC

448


We are going to find the exact test statistic. We have M = trace(I M×M ) µ³ ´−1 ³ ´¶ T T S + dd = trace S + dd µ³ µ³ ¶ ´−1 ¶ ´−1 T T T S + trace S + dd dd = trace S + dd µ³ µ ¶ ´−1 ¶ ³ ´−1 ¡ ¡ ¢T ¢ T T S + dd = trace S + dd S + trace Y − µ0 Y − µ0 µ³ ´−1 ¶ ¡ ³ ´ −1 ¢T ¡ ¢ S + ddT S + Y − µ0 Y − µ0 , = trace S + ddT

´−1 ¡ ¢T ³ ¢ ¡ S + ddT Y − µ0 is a scalar. because Y − µ0 Then we have

Taking into account that

b θ) e = n log TnKull (θ,

¯ ¯ ¯ ¯ ¯S + ddT ¯ |S|

.

¯ ¯ ¯ |Ap×p + B p×n C n×p | = |Ap×p | ¯I n×n + C n×p A−1 p×p B p×n ,

we have for Ap×p = S, B p×n = d and C n×p = dT that ´ ³ ³ ´ |S| 1 + dT S −1 d b θ) e = n log TnKull (θ, = n log 1 + dT S −1 d . |S| b θ) e > c is equivalent to Hence TnKull (θ,

(n − 1) dT S −1 d > T 2M, n−1, α ,

(9.20)

2 2 where TM, n−1 is a Hotelling T distribution with parameters M and n − 1. For more details about this distribution see Mardia et al . (1979). Finally, the inequality given in (9.20) is equivalent to

n − M T −1 d S d > FM, n−M, α . n When we use the asymptotic distribution we have L

b θ) e −→ χ2 . TnKull (θ, M(M−1) −(M+ M(M−1) )=M n→∞ 2M+ 2 2

11. In this case, θ = (µ, Σ) , then

¡ ¢ e = Y , Σ0 , θ © 2006 by Taylor & Francis Group, LLC

and

¡ ¢ b = Y ,S . θ


Therefore,

449

µ ¶ ¡ −1 ¢ |Σ0 | b e 2nDKull (θ, θ) = n trace Σ0 S − I + log |S|

whose asymptotic distribution is chi-square with degrees of freedom given by dim (Θ) − dim (Θ0 ) = M + M +

M2 + M 1 (M (M − 1)) − M = . 2 2

12. In this case we are interested in a exact test because the regularity assumptions are not verified and then it is not possible to use the asymptotic results. It is necessary to get the exact distribution of the test statistic DKull (θb1 , θb2 ).

For θ1 > θ2 we have DKull (θ1 , θ2 ) =

Z

0

and

θ1

1/θ1 1 log dx + θ1 1/θ2

Z

θ2

0 log

θ1

0 θ2 dx = log , 1/θ2 θ1

θb2 DKull (θb1 , θb2 ) = log , θb1

with θb1 = max (X1 , ..., Xn ) , θb2 = max (Y1 , ..., Ym ) , where X1 , ..., Xn is a random sample from the population U (0, θ1 ) and Y1 , ..., Ym is a random sample from the population U (0, θ2 ) . We should reject the null hypothesis if log(θb2 /θb1 ) > c, where c is obtained under the assumption that the test has a nominal size α. This is equivalent to reject H0 if and only if θb2 /θb1 > k. The probability density functions of θb1 and θb2 are given by fθb1 (t1 ) = ntn−1 /θ1n 1 m−1 m fθb2 (t2 ) = mt2 /θ2

t1 ∈ (0, θ1 ) t2 ∈ (0, θ2 ) ,

and the joint probability density function of the random variable (θb1 , θb2 ) by fθb1 θb2 (t1 , t2 ) =

mn n−1 m−1 t t2 θ1n θ2m 1

t1 ∈ (0, θ1 ) , t2 ∈ (0, θ2 ) .

Now we are going to get the distribution of the random variable θb2 /θb1 under the null hypothesis θ1 = θ2 . It is immediate to establish that the distribution of the random variable (W, T ) where W =


θb2 θb1

and

T = θb1

450


is given by g(W,T ) (w, t) = where

nm n+m m−1 t w θ1n+m

(w, t) ∈ A

© ª A = (w, t) ∈ R2 : wt < θ2 , 0 < t < θ1 and w > 0 .

Therefore the probability density function of the random variable W is  nm m−1  w ∈ (0, 1]  n+m w . gW (w) =   nm w−n−1 w ∈ (1, ∞) n+m Finally, if we assume that α < m/(m + n), from the equation Z ∞ nm −n−1 w dw α = Pr (W > k/ H0 ) = k n+m

we have k=

³α

´−1/n (n + m) .

m Then, the null hypothesis should be rejected if

´−1/n θb2 ³ α (n + m) > . m θb1

In our case we have for α = 0.05,

θb1 = 1.9616, θb2 = 2.9358, θb2 /θb1 = 1.49 and

hence we should reject the null hypothesis.

³α

m

´−1/n (n + m) = 1.31;

13. It is immediate to establish that Z ∞ DKull (θ1 , θ2 ) = exp (− (x − θ1 )) (θ1 − θ2 ) dx = θ1 − θ2 . θ1

Therefore DKull (θb1 , θb2 ) = θb1 − θb2

with θb1 = min (X1 , ..., Xn ) and θb2 = min (Y1 , ..., Ym ) and the null hypothesis should be rejected if θb1 − θb2 > c where c is obtained in such a way the test has size α. We know that θ1 ≤ t1 < ∞ fθb1 (t1 ) = n exp (−n (t1 − θ1 )) . fθb2 (t2 ) = m exp (−m (t2 − θ2 )) θ2 ≤ t2 < ∞ © 2006 by Taylor & Francis Group, LLC


451

Then the probability density function of the random variable (θb1 , θb2 ) is given by where

fθb

b 1 ,θ2

(t1 , t2 ) = nm exp (−n (t1 − θ1 )) exp (−m (t2 − θ2 )) θ1 ≤ t1 < ∞, θ2 ≤ t2 < ∞.

Under θ1 = θ2 , the distribution of the bivariate random variable (W, T ), with W = θb1 − θb2

is given by

T = θb2 ,

and

fW,T (w, t) = mn exp (−n (w + t − θ1 ) exp (−m (t − θ1 )))

(w, t) ∈ A

where the domain A is © ª A = (w, t) ∈ R2 /t > θ2 , w > θ1 − t .

The distribution of W = θb1 − θb2 under θ1 = θ2 is  nm  exp (mw) if w ∈ (−∞, 0)   n+m . fθb −θb (w) = 1 2    nm exp (−nw) if w ∈ (0, ∞) n+m

Therefore c is obtained, assuming that α ≤ m/(m + n), by solving the equation Z ∞ nm α= exp (−nw) dw. n +m c

It is immediate to get 1 c = log n

µ

m (m + n) α

¶

and we should reject the null hypothesis if µ ¶ m 1 θb1 − θb2 > log . n (m + n) α In our case

1 log n

µ

m (m + n) α

¶

= 0.2349

and θb1 − θb2 = 0.002; then we should not reject the null hypothesis.

14. We have

θn 2nθ L (θ; Y1 , ..., Yn ) = Q n Yiθ+1 i=1


452


then

n X n ∂ log L (θ; Y1 , ..., Yn ) = + n log 2 − log Yi ∂θ θ i=1

and the maximum likelihood estimator is given by n θb = P . n log Y2i i=1

We have L0n











b n 2nθb   n nθ0    (θ)  − log  θ0 2   = 2 log n n  Q   Q b θ0 +1 θ+1 Yi Yi i=1

= 2n

Ã

i=1

! Ã ! θ0 − θb θb . + 2n log θ0 θb

In order to obtain the expression of the test statistics Wn and Rn it is necessary to obtain the Fisher information. We have 1 ∂ log fθ (x) = + log 2 − log x ∂θ θ then

· µ 2 ¶¸ ∂ log fθ (x) 1 IF (θ) = E − = 2, 2 ∂θ θ

and Wn = n

Ã

On the other hand, Un (θ0 ) =

µ

θ0 − θb θb

∂ log L (θ; Y1 , ..., Yn ) ∂θ

and then

¶

!2

L

−→ χ21 .

n→∞

=n

θ=θ0

Ã

n

1 1X + log 2 − log Yi θ0 n i=1

Ã

n θ0 X Yi Rn = n −1 − log n i=1 2

!2

.

Now Rényi’s divergence is given by Dr1 (θ1 , θ2 )

=

=


1 log r (r − 1)

Z

∞ µ θ1 2

2 θ1 xθ1 +1

¶r µ

θ1r θ21−r 1 log r (r − 1) rθ1 + (1 − r) θ2

2θ2 θ2 xθ2 +1

¶1−r

dx

!


453

if rθ1 + (1 − r) θ2 > 0. Therefore Rényi test statistic b θ0 ) = Tnr (θ,

and for r → 1 we have lim

r→1

b θ0 ) Tnr (θ,

θbr θ01−r 2n log , r (r − 1) rθb + (1 − r) θ0 Ã

!

+ 2n log

fθ (x) = θ exp (−θx) I(0,∞) (x) ,

θ > 0,

=

b θ0 ) TnKull (θ,

= 2n

15. For the exponential model with

θ0 − θb θb

Ã

θb θ0

!

= L0n .

we consider the problem of testing H0 : θ = 1 versus H1 : θ 6= 1 b θ0 ). First, we study the accuracy of power based on Rényi test statistic Tnr (θ, approximations of Remarks 9.1—9.3. Rényi divergence between two exponential distributions is  θ1r θ21−r 1  ln if θ1 r + θ2 (1 − r) > 0 1 Dr (θ1 , θ2 ) = r (r − 1) θ1 r + θ2 (1 − r)  ∞ if θ1 r + θ2 (1 − r) ≤ 0,

when r 6= 0, 1. Limiting cases are obtained for r = 1, DKull (θ1 , θ2 ) = lim Dr1 (θ1 , θ2 ) = r→1

θ2 θ1 − 1 + ln θ1 θ2

(Kullback-Leibler)

and for r = 0, D01 (θ1 , θ2 ) = DKull (θ2 , θ1 ) =

θ1 θ2 − 1 + ln θ2 θ1

the minimum discrimination information. b 1) > c, where θb = n/ Pn Yi = Y −1 is the a) H0 should be rejected if Tnr (θ, i=1 b 1) is given by maximum likelihood estimator of θ. In this case Tnr (θ,  ´ ³ 1  b +1−r >0 br (θr b + 1 − r)−1 if r 6= 0, 1 and θr 2n ln θ   r(r−1)   b +1−r ≤0 ∞ if r 6= 0, 1 and θr −1  b b  2n(θ − 1 + ln θ) if r = 1    2n(θb − 1 − ln θ) b if r = 0, b 1) > c is equivalent to but Tnr (θ,  r r−1  > c1  (1 − r)Y + rY Y − ln Y > c2   1 + ln Y > c 3 Y


if if if

b +1−r >0 θr r=1 r = 0.

454


If we define the function  r r−1   (1 − r)y + ry y − ln y gr (y) =   1 + ln y y we have

gr0 (y)

=

½

if r 6= 0, 1 if r = 1 if r = 0,

>0 ⇔ y c ⇐⇒ Zn < cr,1 or Zn > cr,2 , and the rejection rule is Tnr (θ, 2 where Zn = 2nY ∼ χ2n under H0 . Constants cr,1 and cr,2 are obtained by solving the equations ( 1 − α = Pr(cr,1 < Zn < cr,2 ) = Fχ22n (cr,2 ) − Fχ22n (cr,1 ), ¡ cr,1 ¢ ¡ cr,2 ¢ gr 2n = gr 2n

where

 r r−1  (1 − r)c + rc gr (c) = c − ln c  1 c + ln c

if r 6= 0, 1 if r = 1 if r = 0.

The values of cr,1 and cr,2 , for n = 20, 40, 80, 200 and r = 1/4, 3/4, 1, 5/4, 7/4, 2, are presented in Table 9.3. r 1/4 3/4 1 5/4 7/4 2

n c1/4,1 c1/4,2 c3/4,1 c3/4,2 c1,1 c1,2 c5/4,1 c5/4,2 c7/4,1 c7/4,2 c2,1 c2,2

n = 20 25.855 64.033 25.253 61.323 24.879 60.275 24.449 59.371 23.379 57.898 22.700 57.300

n = 40 58.861 110.541 58.100 108.388 57.659 107.479 57.173 106.658 56.039 105.239 55.374 104.626

n = 80 128.804 200.373 127.914 198.531 127.421 197.709 126.892 196.944 125.714 195.565 125.057 194.943

n = 200 348.637 460.411 347.618 458.802 347.074 458.051 346.505 457.334 345.290 455.990 344.639 455.361

Table 9.3. Constants cr,1 and cr,2 .

b) Exact powers of tests: We know that under an alternative hypothesis θ 6= 1, the random variable 2θY is chi-squared distributed with 2n degrees of freedom.



455

Therefore the power function of Rényi test statistic, βn,r (θ) , is given by βn,r (θ) = 1 − Pr θ (cr,1 < Zn < cr,2 ) ¡ ¢ = 1 − Pr θ θcr,1 < 2θY < θcr,2 ¡ ¢ = 1 − Pr θ θcr,1 < χ22n < θcr,2 . For n = 20, r = 0.25, 0.75, 1, 1.25, 1.75, 2 and different values of θ in the interval [0.5, 2.5], these exact powers are presented in Table 9.4. θ 0.50 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95 1.10 1.20 1.30 1.40 1.50 1.60 1.70 1.80 1.90 2.00 2.10 2.20 2.30 2.40 2.50

β20,1/4 (θ) 0.8117 0.5417 0.4005 0.2778 0.1821 0.1149 0.0729 0.0511 0.0446 0.0879 0.1557 0.2481 0.3578 0.4750 0.5892 0.6922 0.7788 0.8473 0.8984 0.9348 0.9595 0.9756 0.9857 0.9919

β20,3/4 (θ) 0.8560 0.6155 0.4768 0.3478 0.2399 0.1581 0.1021 0.0683 0.0522 0.0763 0.1344 0.2180 0.3208 0.4339 0.5476 0.6532 0.7447 0.8192 0.8766 0.9187 0.9481 0.9679 0.9807 0.9887

β20,1 (θ) 0.8714 0.6437 0.5074 0.3771 0.2652 0.1779 0.1160 0.0768 0.0561 0.0699 0.1221 0.2002 0.2983 0.4084 0.5211 0.6278 0.7219 0.8000 0.8613 0.9071 0.9397 0.9621 0.9768 0.9862

β20,5/4 (θ) 0.8838 0.6676 0.5339 0.4032 0.2883 0.1964 0.1294 0.0852 0.0600 0.0631 0.1089 0.1806 0.2731 0.3792 0.4902 0.5975 0.6943 0.7763 0.8420 0.8921 0.9287 0.9542 0.9714 0.9827

β20,7/4 (θ) 0.9025 0.7057 0.5774 0.4473 0.3284 0.2295 0.1539 0.1010 0.0676 0.0490 0.0803 0.1366 0.2143 0.3085 0.4124 0.5184 0.6192 0.7092 0.7854 0.8466 0.8937 0.9285 0.9532 0.9701

β20,2 (θ) 0.9094 0.7207 0.5950 0.4656 0.3455 0.2440 0.1649 0.1083 0.0712 0.0418 0.0652 0.1123 0.1803 0.2657 0.3633 0.4662 0.5675 0.6611 0.7429 0.8110 0.8652 0.9065 0.9369 0.9585

Table 9.4. Powers for n = 20

c) We know that Rényi divergence is a (h, φ)-divergence with h and φ given in (9.7) and (9.8). Therefore using Exercise 1 we have ´ √ ³ 1 b L 2 n Dr (θ, 1) − Dr1 (θ, 1) −→ N(0, σh,φ (θ)), n→∞

2 (θ) = (h0 (D (θ, 1)))2 σ 2 (θ) and σ 2 (θ) = T T I (θ)−1 T. In where σh,φ φ F φ φ


456


our case, IF (θ) = θ−2 ,   

d T = Dr1 (θ, 1) =  dθ 

and

Dr1 (θ, 1) =

  

1 r(r−1)

1 r−1 θ−1 θ2 θ−1 θ

³

0.6088 0.4320 0.3463 0.2677 0.1989 0.1412 0.0944 0.0571 0.0268 0.0571 0.1412 0.2677 0.4320 0.6088 0.7653 0.8791 0.9471 0.9805 0.9940 0.9985 0.9997 0.9999 1.0000 1.0000

1 θ

−

θr θr+(1−r)

θ − 1 − ln θ   ln θ − 1 + θ−1

2 β20 (θ)

θ .50 .60 .65 .70 .75 .80 .85 .90 .95 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 2.1 2.2 2.3 2.4 2.5

³

r=1/4 0.8083 0.5900 0.4543 0.3105 0.1728 0.0637 0.0086 0.0000 0.0000 0.0000 0.0324 0.1692 0.3293 0.4666 0.5754 0.6600 0.7255 0.7766 0.8165 0.8480 0.8731 0.8931 0.9092 0.9223

r=3/4 0.8165 0.6342 0.5116 0.3714 0.2236 0.0920 0.0147 0.0001 0.0000 0.0000 0.0202 0.1259 0.2697 0.4067 0.5238 0.6199 0.6973 0.7592 0.8083 0.8473 0.8781 0.9025 0.9218 0.9371

1 θr+1−r

´

if r = 1 if r = 0

´ − 1 if r 6= 0, 1 if r = 0 if r = 1

1 β20,r (θ) r=1 r=5/4 0.8271 0.8436 0.6559 0.6787 0.5385 0.5651 0.4006 0.4291 0.2495 0.2756 0.1079 0.1249 0.0188 0.0235 0.0001 0.0002 0.0000 0.0000 0.0000 0.0000 0.0155 0.0116 0.1059 0.0872 0.2392 0.2087 0.3742 0.3399 0.4951 0.4637 0.5979 0.5734 0.6830 0.6669 0.7521 0.7443 0.8075 0.8069 0.8514 0.8563 0.8859 0.8947 0.9128 0.9239 0.9336 0.9457 0.9497 0.9618

Table 9.5. Powers for n = 20.


if r 6= 0, 1

r=7/4 0.8804 0.7310 0.6205 0.4860 0.3282 0.1616 0.0351 0.0004 0.0000 0.0000 0.0060 0.0551 0.1492 0.2667 0.3917 0.5136 0.6254 0.7222 0.8019 0.8642 0.9106 0.9435 0.9657 0.9800

.

r=2 1.0000 0.7588 0.6505 0.5155 0.3552 0.1812 0.0421 0.0006 0.0000 0.0000 0.0042 0.0420 0.1215 0.2288 0.3510 0.4772 0.5980 0.7058 0.7956 0.8655 0.9163 0.9509 0.9728 0.9858


457

θ r = 1/4 r = 3/4 r = 1 r = 5/4 r = 7/4 r = 2 (1) 0.50 20 19 18 17 13 0.60 36 35 34 32 27 25 0.65 50 48 47 45 41 38 0.70 72 70 69 67 62 59 0.75 109 107 106 104 99 95 0.80 180 178 176 174 168 165 0.85 338 334 332 329 323 319 0.90 801 795 792 788 780 775 0.95 3370 3358 3352 3345 3331 3323 1.10 972 979 981 984 989 991 1.20 266 269 270 272 273 274 1.30 129 131 132 132 133 134 1.40 78 80 81 81 82 82 1.50 54 56 56 56 57 57 1.60 41 42 42 42 43 43 1.70 32 33 33 34 34 34 1.80 26 27 27 28 28 28 1.90 22 23 23 23 24 24 2.00 19 20 20 20 20 21 2.10 17 18 18 18 18 18 2.20 15 16 16 16 16 16 2.30 14 14 14 14 15 15 2.40 13 13 13 13 13 14 2.50 12 12 12 12 12 12 ∗ ∗ Table 9.6. Sample sizes n = [n ] + 1, where n is the root of the equation 0.8 = βn1 ∗ ,r (θ) . (1) No value of n is obtained because approximate power is 1. 2 (θ) = |θ−1| . Therefore σh,φ θr 1 (θ), is given by First power approximation, βn,r ´´ ³√ r ³ nθ 3.84145 1 θr if r 6= 0, 1 · 1 − Φ |θ−1| − ln 2n θr+1−r r(r−1) ³ √ ´ ¡ ¢ n · 1 − Φ |θ−1| θ 3.84145 − 1θ + 1 − ln θ if r = 1 2n

and

· 1−Φ

³

√ n |θ−1|

¡ 3.84145 2n

− θ + 1 + ln θ

¢´

if r = 0,

where Φ(·) is the c.d.f. of the standard normal random variable. Second power approximation (see Exercise 2) is ¡ ¢ βn2 (θ) = 1 − Gχ21 (δ) χ21,α , © 2006 by Taylor & Francis Group, LLC

458


where Gχ21 (δ) is the distribution function of a noncentral chi-square random variable with 1 degree of freedom and noncentrality parameter δ = n(θ − 1)2 IF (1) = n(θ − 1)2 . It is interesting to note that this approximation does not depend on the functions h and φ considered in the test statistic Tnr . 1 (θ) and We are interested in the approximation of βn,r (θ) by βn,r In Table 9.5 we present their values for n = 20, r = 0.25, 0.75, 1 (θ) 1, 1.25, 1.75, 2 and several values of θ 6= 1. We conclude that β20,r 2 (θ) are good approximations for β and β20 20,r (θ) in the present model. Approximations improve as n increases. For the sake of brevity, we do not present tables for n > 20. It is also interesting to note that for values of θ near to 1, and more concretely if θ ∈ (0.8, 1.3), the approx2 (θ) is better than the approximation β 1 (θ). Otherwise, imation β20 20,r 1 (θ) is better than β 2 (θ). We conclude that the approximation β20,r 20 both approximations can be recommended for practical applications.

βn2 (θ).

For a value of the power function equal to 0.8, r = 0.25, 0.75, 1, 1.25, 1.75, 2 and several values of θ 6= 1, in Table 9.6 we present the approximate sample size n = [n∗ ] + 1, where n∗ is the positive root of the equation β = βn1∗ ,r (θ). As expected, larger sample sizes are obtained in the neighborhood of θ = 1 (e.g., θ ∈ (0.8, 1.3)). Observe that n∗ can be obtained from β = βn1∗ ,r (θ) in explicit form (cf. (9.19)); however, this is not the case for n∗∗ such that β = βn∗∗ ,r (θ). In the present 1 (θ) and n∗ are good and easy computable apnumerical example βn,r proximations for βn,r (θ) and n∗∗ respectively.


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Statistical inference based on divergence measures

Statistical Inference

Statistical Inference