Financial Risk Management for Pension Plans

Financial Risk Management for Pension Plans Lesaw Gajek Krzysztof M. Ostaszewski ELSEVIER Financial Risk Management...

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Financial Risk Management for Pension Plans

Lesaw Gajek Krzysztof M. Ostaszewski

ELSEVIER


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Lesław Gajek Dean, College of Technical Physics, Computer Science and Applied Mathematics, Technical University of Ło´dz´, Ło´dz´, Poland

Krzysztof M. Ostaszewski Actuarial Program Director, Illinois State University, Normal, Illinois, USA

2004

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v

Abstract This book is devoted to modern financial risk management of pension plans, mostly defined benefit plans. It presents main actuarial funding and valuation methods for defined benefit plans. Modern investment theory is used to discuss ways to value both marketable and nonmarketable assets as well as liabilities. Finally, financial risk management for pension plans is presented in detail, with emphasis on applicable asset –liability management methodologies.


vii

Contents Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

xiii

Part 1. Fundamentals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1

Chapter 1.

Chapter 2.

Retirement plans as a part of economic security system . . . . . . . . . . . . . . . . . . . . . . . . 1.1 Genesis . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Classification of pension plans . . . . . . . . 1.3 The role of the actuary in the management of pension plans . . . . . . . . . . . . . . . . . . .

........... ........... ...........

3 3 5

...........

10

Fundamental concepts of theory of interest . . . . . . . . . . . . . . . 2.1 Accumulation function . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Rate of return . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 Measurement of interest . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4 Cash flows . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5 Interest measurement in a fund . . . . . . . . . . . . . . . . . . . 2.5.1 Discrete-time model . . . . . . . . . . . . . . . . . . . . . 2.5.2 Continuous-time model . . . . . . . . . . . . . . . . . . . 2.5.3 Allocation of interest to accounts . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.6 Interest measurement in a group of funds . . . . . . . . . . . . 2.6.1 Nonrandom continuous-time model . . . . . . . . . . 2.6.2 A discrete-time stochastic model . . . . . . . . . . . . 2.6.3 Deriving the formula for the average rate of return . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.6.4 Martingale-fairness of the average rate of return r . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.6.5 Axiomatic approach to the investment efficiency index . . . . . . . . . . . . . . . . . . . . . . . . . 2.6.6 Merger of funds . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

13 13 13 14 17 19 22 23 23 24 25 26 29 30 32 34 36 38 42 44

viii

Chapter 3.

Contents

Life insurance and annuities . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1 Annuities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Loan amortization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 Bonds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4 Term structure of interest rates . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5 Depreciation of fixed assets . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.6 Capitalized cost . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.7 Random future lifetime . . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.8 Mortality tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.9 Life insurance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.10 Life annuities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.11 Life insurance premiums . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.12 Life insurance reserves . . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.13 Multiple lives models . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.14 Multiple decrements . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

47 47 52 56 59 62 64 65 67 68 72 74 75 80 82 85 90 96 101 108 118 121 125 128 139 141 144 148

Part 2. Valuation of Assets and Liabilities . . . . . . . . . . . . . . . . . . . . . . . .

157

Chapter 4.

159 159 161 162 163 165 165 166 167 172

General principles of pension plan valuation . . . . . . . . . . . . . . 4.1 Objectives and principles of pension funding . . . . . . . . . 4.2 General principles of pension valuation . . . . . . . . . . . . . 4.2.1 Accrued liabilities . . . . . . . . . . . . . . . . . . . . . . . 4.2.2 The normal cost versus accrued liabilities . . . . . 4.2.3 Unfunded actuarial liabilities . . . . . . . . . . . . . . . 4.2.4 Actuarial gain . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.5 Some additional remarks . . . . . . . . . . . . . . . . . . 4.3 Optimal payoff of a liability . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

ix

Contents

Chapter 5.

Chapter 6.

Valuation of pension plan liabilities . . . . . . . . . . . . . . . . . . . . 5.1 Unit credit method . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1.1 Accrued liabilities . . . . . . . . . . . . . . . . . . . . . . . 5.1.2 Normal cost . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1.3 Unfunded actuarial liability . . . . . . . . . . . . . . . . 5.1.4 Actuarial gain . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 Entry Age Normal method of funding . . . . . . . . . . . . . . 5.2.1 Normal cost . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2.2 Accrued liabilities . . . . . . . . . . . . . . . . . . . . . . . 5.2.3 Unfunded actuarial liabilities . . . . . . . . . . . . . . . 5.2.4 Actuarial gain . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3 Individual level premium method of funding . . . . . . . . . 5.3.1 Normal cost . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3.2 Accrued liabilities . . . . . . . . . . . . . . . . . . . . . . . 5.3.3 Unfunded actuarial liabilities . . . . . . . . . . . . . . . 5.3.4 Actuarial gain . . . . . . . . . . . . . . . . . . . . . . . . . . 5.4 The method of Frozen Initial Liability . . . . . . . . . . . . . . 5.4.1 Normal cost . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.4.2 Accrued liabilities . . . . . . . . . . . . . . . . . . . . . . . 5.4.3 Unfunded actuarial liabilities and the actuarial gain . . . . . . . . . . . . . . . . . . . . . . . . . . 5.4.4 A recursive formula for the unit normal cost . . . 5.4.5 Iterative method of normal cost calculation . . . . 5.5 Valuation of plans paying pension benefits . . . . . . . . . . 5.5.1 Accrued liabilities . . . . . . . . . . . . . . . . . . . . . . . 5.5.2 Unfunded actuarial liabilities . . . . . . . . . . . . . . . 5.5.3 Actuarial gain . . . . . . . . . . . . . . . . . . . . . . . . . . 5.6 New plan participants . . . . . . . . . . . . . . . . . . . . . . . . . . 5.6.1 The method of Frozen Initial Liability . . . . . . . . 5.6.2 Unit credit method . . . . . . . . . . . . . . . . . . . . . . 5.6.3 The Entry Age Normal method . . . . . . . . . . . . . 5.6.4 Individual level premium method . . . . . . . . . . . . 5.7 Aggregate pension funding methods . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Random fluctuation of pension liabilities . . . . . . . . . . . . . . . . . 6.1 Formulation of the problem . . . . . . . . . . . . . . . . . . . . . . 6.2 Fluctuation of the present value of a life annuity . . . . . . 6.2.1 The expected value of the random variable a T . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2.2 The variance of the random variable a T . . . . . . . 6.2.3 The coefficient of variation of the random variable a T . . . . . . . . . . . . . . . . . . . . . . . . . . . .

179 179 180 181 182 183 185 186 186 189 190 191 191 192 195 195 196 196 197 199 200 203 204 204 207 207 208 209 212 214 214 215 217 231 231 232 233 233 235

x

Contents

6.2.4 The Gompertz Law of Mortality . . . . . . . . . . . . Confidence intervals for accrued liabilities of a pension plan . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3.1 General case . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3.2 Cohorts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

238

Valuation of pension plan assets . . . . . . . . . . . . . . . . . . . . . . . 7.1 Theories of capital markets . . . . . . . . . . . . . . . . . . . . . . 7.1.1 Optimal consumption decision versus the level of interest rates . . . . . . . . . . . . . . . . . . . . . 7.1.2 Optimization of an investment portfolio . . . . . . . 7.1.3 Capital assets pricing methodologies . . . . . . . . . 7.1.4 The Principle of No Arbitrage . . . . . . . . . . . . . . 7.1.5 The Fundamental Theorem of Asset Pricing . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2 General methodology of arbitrage-free valuation of capital assets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.3 Derivative securities . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.4 A survey of asset valuation methods . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

253 253

6.3

Chapter 7.

242 242 246

253 256 260 263 267 271 273 275 278 280 283

Part 3. Financial Risk Management . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

287

Chapter 8.

General rules of risk classification and measurement . . . . . . . . 8.1 Risk classification . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2 Total risk measurement and capital requirement . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.3 Asset-side risks and investment benchmarks . . . . . . . . . 8.4 Stability of total annual premium . . . . . . . . . . . . . . . . .

289 289 290 295 298 300

Chapter 9.

Interest rate risk—nonrandom approach . . . . . . . . . . . . . . . . . 9.1 The relationship between the price of capital assets and the interest rates . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.1.1 Duration of a security . . . . . . . . . . . . . . . . . . . . 9.1.2 Convexity of a financial asset . . . . . . . . . . . . . . 9.1.3 Nominal interest rates . . . . . . . . . . . . . . . . . . . . 9.1.4 The Babcock Equation . . . . . . . . . . . . . . . . . . . 9.2 Classical immunization . . . . . . . . . . . . . . . . . . . . . . . . . 9.2.1 A critical discussion of classical immunization for a pension plan . . . . . . . . . . . . . . . . . . . . . . .

303 303 303 307 311 313 315 317

xi

Contents

9.2.2 9.2.3

9.3

Multivariate immunization . . . . . . . . . . . . . . . . . Simulation approach to asset – liability management . . . . . . . . . . . . . . . . . . . . . . . . . . . Axiom of Solvency and surplus immunization . . . . . . . . 9.3.1 Monotone shift factors . . . . . . . . . . . . . . . . . . . . 9.3.2 Arbitrary shift factors . . . . . . . . . . . . . . . . . . . . 9.3.3 Conservative asset –liability management . . . . . .

Chapter 10. Stochastic approach to financial risk management . . . . . . . . . . 10.1 One-period analysis of the plan surplus and funded ratio . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.1.1 Risk management by shortfall control . . . . . . . . 10.1.2 Value-at-Risk . . . . . . . . . . . . . . . . . . . . . . . . . . 10.1.3 Funded ratio return . . . . . . . . . . . . . . . . . . . . . . 10.2 Axiom of Solvency and random interest rates . . . . . . . . 10.3 Immunization when the portfolio is not correlated with term structure of interest rates . . . . . . . . . . . . . . . . 10.3.1 Monotone shift factors . . . . . . . . . . . . . . . . . . . . 10.3.2 Arbitrary shift factors . . . . . . . . . . . . . . . . . . . . 10.4 Immunizing arbitrary portfolios . . . . . . . . . . . . . . . . . . . 10.4.1 Monotone shift factors . . . . . . . . . . . . . . . . . . . . 10.4.2 Arbitrary shift factors . . . . . . . . . . . . . . . . . . . . 10.5 Conservative strategy of risk management for pension plans . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.5.1 Conservative choice of TSIR . . . . . . . . . . . . . . . 10.5.2 Portfolio with a minimal immunization risk . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Chapter 11. Conclusion

320 321 323 325 325 326 329 329 332 334 336 337 338 338 340 341 341 343 344 344 346 348

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353

Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

357

Index

361

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xiii

Introduction The key objective of pension plans is the delivery of retirement benefits, typically payable for life or a set period of time, to the specified group of recipients. Such plans are usually organized by employers, although there also exist multi-employer plans, or plans organized by various levels of government. First private retirement plans were created in the United States in the nineteenth century. In both the United States and Canada, private pension plans play a crucial role in the overall financial security system. While other countries initially relied mostly on state pension systems, the last quarter century had brought about new growth in private retirement systems worldwide. Chapter 1 contains more detailed discussion of the role of pension plans in financial security systems. Private pension plans can have a very diverse structure—they may be directed towards a small group, or a group of several thousand employees, or even more in the case of multi-employer plans—in all cases providing for the financial needs of the members of the group. This results in a diversity of forms of pension plans, and creates a need for understanding of a variety of possible funding mechanisms for such funds. These issues are addressed in Chapters 4 and 5. Chapter 4 contains general rules for calculation of the actuarial liabilities of a defined benefit pension plan, its normal cost and possible amortization of initial liability, or unfunded liability. Chapter 5 presents basic methods of actuarial pension plan valuation, discussing both funding and calculation of accrued liabilities. Chapter 6 contains analysis of stability of pension plan liabilities, especially in relation to the plan size. Chapter 7 is devoted to fundamental results of modern investment theory. These results are later used in the analysis of marketable and nonmarketable assets of pension plans, and financial approach to valuing assets and liabilities. Chapter 8 presents general rules of risk classification and measurement for pension plans. In particular, we describe in detail capital requirements to cover the risks. In Chapter 9 we present nonrandom approach to asset –liability management. Chapter 10 is devoted to stochastic approach to financial risk management. Chapter 11 contains a brief presentation of a final conclusion. The whole book is divided into three parts: Part 1 (Chapters 1 –3) contains auxiliary results and information; Part 2 (Chapters 4 –7) is on methods of assets and liabilities valuation; Part 3 (Chapters 8– 11) concerns the methodologies of financial risk management for pension plans. This part is, in our opinion, of particular importance for defined benefit plans, for which we analyze their investment risk, funding ratio risk, and asset – liability management practices.

xiv

Introduction

This book aims to integrate the analysis of assets and liabilities for pension plan. We study the dependence of both assets and liabilities on interest rates, and other random variables, e.g., longevity of plan participants. We present general results concerning asset – liability risk management as well as practical problems and requirements. The intended audience of this work are all decision makers involved in management of assets, liabilities, and integrated portfolios of both sides of the balance sheet, for pension plans. This may include actuaries, investment managers, plan fiduciaries, and government regulators. We assume that our readers are familiar with basic probability theory, as well as mathematical analysis. Chapters 2 and 3 introduce basic terminology and notation of actuarial mathematics for life contingencies, as applied to life insurance and pensions. The authors hope that this work can also become an academic textbook for courses in finance, mathematics, economics, and actuarial science. For that reason we have incorporated plenty of examples and exercises. Early versions of this text have been used in lectures at the Technical University of Ło´dz´, Poland, Summer Actuarial School at the University of Warsaw, Poland, University of Louisville and Illinois State University. These experiences have resulted in numerous improvements of the text. Some of the ideas of this book have appeared in a Polish text Plany emerytalne: Zarza¸dzanie Aktywami i Pasywami, published in 2002 by Wydawnictwo NaukowoTechniczne in Warsaw, Poland, by the same authors. We express our gratitude to the Society of Actuaries and the Casualty Actuarial Society for allowing us the use of past actuarial examination problems as exercises. We dedicate this work to our wives: Elz˙bieta and Patricia, who offered us infinite patience and support and whom we offer love and gratitude. Lesław Gajek Krzysztof M. Ostaszewski

Part 1 Fundamentals


3

Chapter 1 Retirement plans as a part of economic security system

1.1 Genesis Retirement plans constitute one of the key elements of every society’s financial security system. Of course they are of utmost importance for the elderly, but they have great effect on the functioning of the national economy and everyone’s welfare. Let us note that one could divide human life into three main phases: youth, productive years, and retirement. There are many perspectives on this division, but in this work we are interested in the economic one. Youth is therefore the period when a person, through education, acquires human capital, i.e., the ability to earn income throughout one’s productive life. Acquisition of education, i.e., investment in human capital, is among the most important financial issues in a person’s life. Another such issue is, of course, the accumulation of financial capital for retirement, the subject of this work. Human capital acquired mostly during one’s youth is gradually used up during one’s productive life. One may argue that a person should re-acquire education later in life, as the human capital accumulated early in life does become obsolete eventually, as the technology of production changes. Ironically, the blessing of longer life enjoyed by most of the world’s population in the twentieth century has made such a threat of a need for reacquisition of human capital more likely. Nevertheless, eventually one’s physical and mental ability to work is depleted, and somewhere before or at that moment, one needs to replace human capital with financial assets allowing for comparable standard of living. The problem of financial security in the late stages of life has assumed increasing significance as human lifespan expanded. In the prehistoric times, the antiquity, or the Middle Ages, retirement protection was never a social issue, as few people reached the retirement age (defined as the age when productive ability is no longer available). On the other hand, in the twentieth century, providing for retirement became a very important social issue. As that century drew to a close, it also has become a major challenge globally. For example, the life expectancy for a person aged 65, calculated

4

Chapter 1

on life tables1 for 1900 and 1950, respectively, increased by 25.28% in England and Wales and by 29.86% in France. This seems to be a permanent universal tendency; in the United States the life expectancy for an analogous person increased between 1960 and 1999 by 23.40%. The World Bank publication [6] goes as far as to call it the Old Age Crisis. The gravity of the problem results from the fact that many of the world economies appear to be unprepared for increasing longevity of their populations, the resulting ageing of their societies, and increased retirement needs. There are various possible means of support in the old age. Traditionally, the elderly had depended on support from their family members, mostly their children. Churches and charitable organizations had assumed a larger role over time. But major social transformations of the twentieth century had given those two forms of support a secondary role. The simplest solution to the problem of old age is the one used once by the aristocracy: a person can retire as soon as this person has acquired enough wealth to retire on. In short, the wealthy ones can retire. Interestingly enough, this may not be as difficult as it appears, due to the increased longevity we all enjoy. A person earning 4000 a month, saving 10% of income over the span of 40 years, and earning 7% real on the income saved, will accumulate slightly over 1 million monetary units by the end of the period. With a rough approximation of a life annuity factor of 20, this provides enough income to replace one’s wages for the rest of the lifetime. This calculation ignores the effects of inflation, but in a free-market economy both wages and capital market instruments generally grow at rates not only matching inflation but also exceeding it. For example, in the United States, long-term average rate of return of stocks in the twentieth century was roughly 7% above inflation [59]. At the end of the nineteenth century, new state-run retirement schemes were created, beginning with the German social insurance system created by Chancellor Bismarck. New private retirement also started around the same time. American Express created a company pension plan in 1875, and Baltimore and Ohio Railroad Company started one in 1880 [2]. In the United States, private pension plans grew rapidly during and following World War II, when wage increases were limited due to wartime wage and price controls, but collective bargaining for pension benefits was allowed. In 1974, another watershed event for pension plan history in the US occurred—the federal law named Employee Retirement Income Security Act (ERISA) was passed. This law created a requirement for qualified pension plans (i.e., those receiving tax assistance in the form of tax exemption for contributions and tax deferral for investment income) of having a fund appropriate for paying the benefits and making a regular payment of the plan normal cost, as established by a qualified actuary. 1

Source: Human Mortality Database. University of California, Berkeley (USA), and Max Planck Institute for Demographic Research (Germany). Available at www.mortality.org or www.humanmortality.de (data downloaded on April 17, 2004).

Retirement plans as a part of economic security system

5

Most developed economies in the world have gradually evolved towards a system of what is commonly called a three legged stool in the retirement security system. The three legs in this concept refer to: state-sponsored, employer-sponsored, and individual retirement benefits. The first leg is commonly created with a system of social insurance, such as the United States Social Security System (Old Age, Survivors and Disability Insurance or OASDI), or with a privatized mandatory system of individual accounts, the first of which was created in Chile in 1981. In the United States, the second leg is represented by employer-sponsored pension plans of either the defined benefit or the defined contributions variety. The third leg in the US consists of the variety of personal retirement accounts, such as Individual Retirement Accounts (IRA), Roth IRA accounts, and others. In what follows, we will also refer to the three “legs” of the stool as the tiers of the pension system.

1.2 Classification of pension plans In this section we will present a short review of the most important types of pension plans, which may be, for the purpose of our classification, part of any of the three legs of the retirement security stool. We will use various criteria for the classification. The first criterion is the asset base for the liabilities for benefits promised to plan participants. Here we can point out two types of plans: † pension plans without a fund ( pay-as-you-go plans) and † pension plans with a fund ( funded pension plans). A pay-as-you-go plan does not accumulate specific assets designated to create income needed to pay benefits. Instead, they depend on future income generated by the plan sponsor to pay future benefits. In effect, such plans are backed by the credit of the plan sponsor, as opposed to funded plans, where benefits are backed by the portfolio of capital assets. Employer-sponsored pay-as-you-go cannot be qualified for tax purposes in the United States: this is a provision of ERISA. The reasoning behind pay-as-you-go private plans prohibition in ERISA is quite sound. In a pay-as-you-go plan, participants depend on their employer for the security of their benefits. This dramatically increases the risk to plan participants. Such increased risk is present even for participants who are still working. Without legal protection, in the case of a bankruptcy, plan participants’ claims would become a part of all creditors’ claims and may not be safe. It is even possible, that participants’ claims would not even be recognized as employer’s liability. Even if recognized, if, in the course of bankruptcy proceedings, other liabilities were determined to have priority over pension obligations, employees would not have any recourse and any retirement protection. Funded pension plans avoid the potential pitfalls of pay-as-you-go plans. The fund accumulated for the purpose of providing benefits allows for a substantial

6

Chapter 1

increase in the security of participants’ benefits. In the United States, qualified pension plans are funded. However, pay-as-you-go plans are still a dominant form of pension plans among social insurance plans, such as OASDI in the United States. Social insurance is administered and guaranteed by the state, or, in the case of the United States, by the federal government. This makes the credit problems described above practically nonexistent. Insolvency of social insurance would be basically equivalent to insolvency of the state itself. Such an insolvency is not only extremely unlikely, it is also an uninsurable event: one cannot find a realistic protection against it. The second criterion for classification of pension plans is the method of correction of an imbalance between assets and liabilities. From that perspective, we distinguish between: † defined benefit plans and † defined contribution plans. If a defined benefit plan exhibits an imbalance between its assets and liabilities, its normal cost (regular payment required from the contributor to the plan, i.e., the analog of an insurance premium) is adjusted. If a defined contribution plan shows such an imbalance, its future benefits are automatically adjusted. These definitions of defined benefit and defined contribution plans are nonstandard. They were originally proposed by Dr Wojciech Otto of the University of Warsaw in Poland and are adopted by this work. In other works, it is more common to define defined contribution plans as the plans for which only the contributions are prescribed in advance (and benefits are determined by the performance of the assets of the plan), while defined benefit plans are defined as plans for which benefits are prescribed in advance, and asset performance affects contribution levels needed to fund benefits. A pay-as-you-go plan must be, by definition, a defined benefit plan, as it specifies a benefit which will be delivered to a participant without any fund so it shows a permanent imbalance. Private pension plans are also built upon a promise of future benefits, but the promise is backed by a fund, and contributions (normal costs) paid regularly into the fund. The levels of plan liabilities and contributions needed are established by an actuary. There are four basic forms of a promise of future pension benefits that exist in private retirement plans: † † † †

a set dollar amount of monthly (or annual) benefit, a set dollar amount of monthly benefit for each year of employment service, a set percentage of (final or average) wages paid as a monthly pension benefit, a set percentage of (final or average) wages for each year of employment service paid as a monthly pension benefit.


7

There are also combinations of the four above, or variations of them, which are used in practice. For example, both private and social insurance programs set the maximum for the number of years of service counted towards granting of benefits, if such benefit depends on service length. This design serves to encourage workers to retire when they reach a certain age believed to be the maximum age desired for a given professional group. Some professional groups also enjoy certain retirement privileges, e.g., social insurance systems of many countries grant special early retirement benefits to police, military, or miners. Private defined benefit plans are very common in the United States and Canada, but not in other countries. In these two countries, defined benefit plans must be funded. Benefits are typically calculated as the number of years of service (subject to a certain maximum) times a percentage of final year wages, or average wages of the last 5 or 3 years of service. For example, if a plan grants 2% of the average wages of the last 3 years of employment and if the worker’s salaries during the last 3 years of service were: $4000 per month, $4500 per month, and $5000 per month, respectively, and the worker has 40 years of service, this worker’s retirement benefit will be 40 £ 0.02 £ ((4000 þ 4500 þ 5000)/3) ¼ $3600. Private defined contributions plans in the United States and Canada developed rapidly beginning in the late 1970s. The 1970s were marked by high inflation and increased investment risk. These factors may have caused the employers to want to shift the risk to plan participants. In a defined contribution plan, investment risk is absorbed by plan participants, and the only obligation of the employer is the contribution itself. Over the last quarter of the century, a dramatic shift in the relative role of the two types of plans in the United States has occurred: in 1975, defined benefit plans had a near monopoly in the pension provision market, but by the beginning of the twenty-first century, defined contribution plans achieved dominance. The reason for this change has been hotly debated, especially among actuaries. Let us try to understand the positions of this debate. In a defined contribution plan, the employer makes a regular contribution to an individual account of a given employee. Typically, the contribution equals a set percentage of the employee’s wages (although in the United States there are limits due to nondiscrimination rules). Such contributions are subsequently invested based on a choice made by the employee, although the variety of choices available to the employee is generally defined by the employer. In the United States, an employer who wants to fully shift the responsibility for investment choices to the employee must provide the following four investment choices: money market equivalent (very shortterm bonds, considered to be cash equivalent), diversified portfolio of bonds, balanced portfolio (stocks and bonds), and a diversified portfolio of stocks. Employer stock may be an investment choice, but it must be in addition to the above four. If the employer does not provide these four choices, the plan may still be qualified for tax purposes, but the employer will have to accept fiduciary duty and responsibility for investment choices by the employees. Otherwise, such responsibility rests fully with the employee.

8

Chapter 1

This shifting of the investment risk to the employees is often considered to be the main explanation for the seminal shift from defined benefit to defined contribution plans in the United States. Another hypothesis ventured to explain this phenomenon is the relative ease of transfer from one defined contribution plan to another contribution plan. If an employee terminates employment from an employer offering a defined benefit plan, such employee will typically receive pension benefit upon retirement based on a relatively shortened period of service and on the last wages with the said employer. This means that if there are periods of high inflation between termination and retirement, the value of benefits will deteriorate substantially. Defined contribution plans, on the other hand, offer relatively easy transfer opportunities. In the United States, fund balance in the participant’s account can be transferred either to the new employer’s plan or to an IRA. Both defined benefit plans and defined contribution plans can be financed solely by the employer, solely by the employee, or jointly by both parties. Plans which require employee financing are called contributory. Contributory defined contribution plans are fairly common in the United States and Canada alike. On the other hand, contributory defined benefit plans do not exist in the United States (the reason is simple—US tax law does not make employee contribution tax-exempt), while they are quite common in Canada (where they are generally tax-exempt). Pension plan contributions are invested in various capital assets (and sometimes, although generally rarely, in real assets, e.g., real estate or commodities). Regulations usually limit the investment universe. In the United States it is fairly common for defined benefit pension plan to hold a balanced portfolio of approximately 60% stocks and 40% bonds. Investments in employer’s stocks or bonds are generally either prohibited or limited. It is quite common for plan participants in defined contribution plans to choose their own asset allocation. Too often, it seems, many of them choose to invest predominantly in money market instruments, very low-risk asset class, thus limiting their long-term investment opportunities. On the other hand, a typical defined benefit plan portfolio of 60% stocks and 40% bonds may be a good choice for an individual with an intermediate risk tolerance profile. But even the debate concerning the investment portfolio of defined benefit plans is not free from controversy. Some finance theoreticians claim that defined benefit plans, instead of a balanced portfolio of stocks and bonds, should invest only in bonds as such bonds will provide a better match for the liabilities. The problem of construction of an optimal portfolio for a retirement plan is discussed by Leibowitz et al. [42]. Immediately following World War II, and in the 1950s, defined contribution plans constituted virtually the entire second tier of the pension system in the United States. When ERISA was passed, assets of defined benefit plans amounted to 75% of all pension plan assets. As of this writing, they account for less than half of pension plan assets, the rest having been taken over by the defined contribution plans. Over 75% of new contributions flow into defined contribution plans. Even though defined benefit plans have a significantly longer history of existence, more than half the pension


9

benefits are paid by defined contribution plans. And, most importantly, employers no longer start new defined benefit plans. This massive decline in the role of defined benefit plans has received significant attention from researchers, who proposed several explanations for it. We had already mentioned the shifting of risk onto employees as one of them, and the portability of defined contribution plans as another. But there are other theories, some of them contradictory. Some hypothesize it was the ERISA law itself that caused the demise of defined benefit plans, by putting additional reporting and valuation requirements on them, thus causing increasing costs and bureaucratic burdens. Requirement of actuarial valuation for defined benefit plans is not to be ignored here, as it increases costs substantially. Other explanations proposed are: greater ease of explanation and understanding of defined contribution plans (which are quite similar to bank accounts or mutual fund accounts) and high rates of return available to participants’ accounts in the years 1983– 1999 that have encouraged participants to chase those stellar returns [54]. In the face of the shift away from defined benefit plans, their proponents have worked to create new alternative designs, which may help stem the tide of the escape to defined contribution plans. Two main innovations in design are: † target benefit plans and † cash balance plans. Target benefit plans are designed as a hybrid between defined benefit plans and defined contribution plans. They are officially qualified as defined contribution plans, but their funding is based on projected retirement benefits. Normal cost is calculated based on such a projection and estimated by the plan actuary to be at the level which would provide for the expected benefit. However, unlike in a defined benefit plan, the employer does not assume an obligation to deliver that level of benefit. Thus, as in the defined contribution plan, investment risk in a target benefit plan is assumed by the employees. It should be noted, however, that target benefit plans have not been accepted by the retirement markets in the United States. The second new alternative design is the cash balance plan. The first cash balance plan was created by the Bank of America in 1984. In a cash balance plan, a worker has exactly the same benefits and rights to them, as in a defined benefit plan. The key difference lies in the fact that the actuarial value of all benefits, accrued as a result of his/her service up to date, is presented to the worker as an account balance. Each year, this account additionally increases by interest credited according to a set formula and credits due to additional service. The method of crediting due to service is established by the plan actuary, and reported to the Department of Labor. Interest crediting, on the other hand, is set by the employer, although there is a series of Internal Revenue Service (IRS), the US tax authority, regulations concerning

10

Chapter 1

methods of crediting approved by the IRS (such an approval is essential for the plan to be qualified).

1.3 The role of the actuary in the management of pension plans Pension funding mathematics exhibits many similarities with the mathematics of life insurance. In both cases the actuary must rationally value products, which provide the customer with future income stream, in return for a premium stream paid to the provider. There are, however, some fundamental differences. First, most pension plans have fewer participants than the typical number of customers of an insurance company. Pension plans vary in their member count from as few as one or several participants to, rarely, as many as hundreds of thousands, but a typical customer base of a life insurance company is at least in tens of thousands, if not hundreds of thousands. This smaller number of participants means that random fluctuations of assets and liabilities of pension plans may have a more profound effect on plan funded status than in the case of a life insurance company. This increased uncertainty must be taken into account by the actuary when establishing pension plan liabilities and normal cost. The second key difference lies in the timing of benefit payments. Pension plan participants may withdraw from the plan early, due to termination of employment. They also generally have great latitude in choosing their retirement age, within the bounds set by the early retirement age and the latest age allowed by the plan. The actual amount of the benefit will be directly influenced by the date chosen and, additionally, indirectly, the date will affect the final salary (or final salary average), again influencing the benefit amount. This makes the work of the plan actuary more challenging, especially if one wants to achieve stable normal cost, a common desire among employers. Pension plan management requires substantial involvement of the plan actuary. In the case of defined contribution plans, the actuary must assure that all applicable regulations are followed, and that existing plan assets provide appropriate level of projected benefits for plan participants. In the case of defined benefit plans the role of the plan actuary is especially pronounced: it is the actuary who values plan benefits granted and calculates the normal cost. In general, a pension plan actuary has the following responsibilities: † † † † † † † †

to know generally the accepted pension valuation and funding methods, to know which methods are applicable to the plan under consideration, to establish appropriate assumptions for valuation, to estimate the effect of plan size on the stability of its funding, to value benefits other than retirement benefits, if granted (e.g., disability benefits), to model future cash flows of the plan, to value plan assets appropriately and to model sensitivity of the plan to changing parameters such as interest rates, mortality, or general economic variables.


11

The above list illustrates that the plan actuary must possess vast knowledge and experience in order to meet such a variety of responsibilities. In the following chapters we will lay the foundation for such a knowledge. We hope that this book can serve as a, naturally limited, substitute for experience. We will present methods of valuation of plan liabilities, assets and normal cost calculation. We will next study the methods of financial risk management applicable to pension plans, with emphasis on asset – liability management.


13

Chapter 2 Fundamental concepts of theory of interest

2.1 Accumulation function If a unit amount is placed in an account, its accumulated value at time t is denoted by aðtÞ and called the accumulation function. It must be equal to 1 (amount invested) plus the interest earned over the period ½0; t: We assume generally that aðtÞ is an increasing continuous function. If the amount invested is different than 1, we typically denote the accumulated value in the account at time t by AðtÞ and call it the amount function. The two functions are related via AðtÞ ¼ Að0Þ · aðtÞ;

ð2:1:1Þ

which effectively states that every monetary unit invested earns interest at the same rate.

2.2 Rate of return Assume at first that rate of return (interest rate) i is constant in the period of time considered. We will use year as a unit of time. If the interest is compounded (i.e., added to the principal invested) once a year (i.e., annually), the rate of return i is called the effective rate of return (over a year). If the compounding frequency is different than once a year, the annual rate is called a nominal rate of return and denoted as iðmÞ ; where m is the compounding frequency per year. The relationship between these two measurements of interest is !m iðmÞ 1þi¼ 1þ : ð2:2:1Þ m If the interest rate is not constant, we define the effective annual interest rate as the interest earned over a given year divided by the amount invested at the year’s

14

Chapter 2

beginning. Thus the effective annual rate it during the tth year, i.e., in the period ½t 2 1; t; is defined by it ¼

aðtÞ 2 aðt 2 1Þ AðtÞ 2 Aðt 2 1Þ ¼ ; aðt 2 1Þ Aðt 2 1Þ

ð2:2:2Þ

assuming that the only principal invested is either 1 for the function aðtÞ or Að0Þ for the function AðtÞ: We typically assume that the period of time over which the interest rate is defined is a year, but this is not in any way required, it is merely a convention. We should, however, note that the unit of time over which the interest rate is defined should be the same as the one used for actual counting of time.

2.3 Measurement of interest If the effective rate of interest i is constant every year and the interest earned is reinvested in the account, we call the resulting accumulation function the compound interest accumulation function. Its form is aðtÞ ¼ ð1 þ iÞt ;

ð2:3:1Þ

i.e., it is exponential. If, however, the interest earned every year does not earn additional interest, the resulting form of the accumulation function aðtÞ ¼ 1 þ ti

ð2:3:2Þ

is termed the simple interest accumulation function. The two functions are equal for t ¼ 0 or t ¼ 1; and the compound interest function increases much more rapidly for t . 1: However, it should be noted that for 0 , t , 1 simple interest actually exceeds compound interest. Under compound interest, the expression 1 þ i is called the accumulation factor, and equals the accumulated value of a monetary unit after 1 year. Its reciprocal, v¼

1 ; 1þi

ð2:3:3Þ

is called the discount factor, and is the present value of a monetary unit paid a year from now. Note also that vt ; the t-year discount factor, is the present value of a monetary unit paid t years from now (t need not be an integer). The effective annual rate of discount in the year t, dt, is defined as the interest earned over a year divided by the accumulated amount at the year-end. In the case of the compound interest accumulation function, the effective rate of discount is constant and written as d. We have then d¼

að1Þ 2 að0Þ aðtÞ 2 aðt 2 1Þ i ¼ ¼ : að1Þ aðtÞ 1þi

ð2:3:4Þ

15

Fundamental concepts of theory of interest

Note that vþd ¼

1 i 1 þ ¼ 1; v ¼ 1 2 d; d ¼ 1 2 v; and vi ¼ i ¼ d: 1þi 1þi 1þi

It is also interesting to observe that if i ¼ 1=n for an integer n, then d ¼ 1=ðn þ 1Þ: The accumulation function produced by a constant rate of discount is aðtÞ ¼ ð1 2 dÞ2t ¼

1 : ð1 2 dÞt

ð2:3:5Þ

Two measurements of interest are said to be equivalent if for any amount of principal invested for any length of time they yield equal accumulated values at any time. It can be shown easily that two measurements of interest are equivalent if, and only if, they produce the same accumulation function. The following identities hold for equivalent measurements of interest: !m 1 1 iðmÞ 1þi¼ ¼ ¼ 1þ : ð2:3:6Þ v 12d m For the nominal annual interest rate iðmÞ compounded m times a year, the number m need not be an integer, as the formula works equally well for an annual rate compounded, e.g., every 8 months, with m ¼ 1:5 in such a case. Nominal rates of discount can also be considered. Constant nominal annual rate of discount compounded m times a year is denoted by d (m), and the accumulation function produced by it is !2mt dðmÞ ð2:3:7Þ að t Þ ¼ 1 2 m so that the following holds for equivalent measurements of interest: !m !2r 1 1 iðmÞ dðrÞ ¼ 1þ ¼ 12 1þi¼ ¼ v 12d m r

ð2:3:8Þ

(we are using a different notation for the compounding frequency of nominal interest and nominal discount to indicate that the two need not be equal). We also have ðmÞ

12

d ¼ v1=m ; m

iðmÞ d m ¼ ; m iðmÞ 1þ m ðmÞ

v1=m ·

iðmÞ d ðmÞ ¼ : m m

ð2:3:9Þ

As the compounding frequency for a deposit earning a fixed annual nominal interest rate i (m) increases, the amount of interest earned in a year increases, but only to a

16

Chapter 2

certain upper bound. The accumulated value at the end of 1 year will be: lim

m!1

iðmÞ 1þ m

!m ¼ lim

m!1

1þ

1 m=iðmÞ

m=iðmÞ !iðmÞ

ðmÞ

¼ ei :

ð2:3:10Þ

We obtain the same kind of limit result when working with compound discount. The limiting nominal annual interest rate is said to be compounded continuously. In actuarial literature, it is commonly called the force of interest and denoted by d. The accumulation function given by it is aðtÞ ¼ edt :

ð2:3:11Þ

We have the following identities for equivalent measurements of interest: !m !2r 1 1 iðmÞ dðrÞ ¼ 1þ ¼ 12 ¼ ed : ð2:3:12Þ 1þi¼ ¼ v 12d m r It should be noted that for equivalent measurements of interest we have i . ið2Þ . ið3Þ . · · · . d . · · · . dð3Þ . dð2Þ . d:

ð2:3:13Þ

In general, the force of interest can vary over time and in such general case it is defined as

dt ¼

a0 ðtÞ d A0 ðtÞ ¼ lnðaðtÞÞ ¼ : aðtÞ dt AðtÞ

ð2:3:14Þ

Note that Aðt þ DtÞ 2 AðtÞ < AðtÞ · dt · Dt:

ð2:3:15Þ

Under varying force of interest aðtÞ ¼ exp

ðt 0

ds ds :

In particular, if

dt ¼ c

f 0 ðtÞ f ðtÞ

for some function f ðtÞ then aðtÞ ¼

f ðtÞ f ð0Þ

c :

ð2:3:16Þ


17

Also, if we move money forward in time (accumulate) from time t1 to time t2, with t2 . t1 ; we multiply it by ðt 2 aðt1 Þ ¼ exp 2 ds ds : aðt2 Þ t1 Sometimes it is useful to know the rate of change, i.e., the derivative, of one measurement of interest with respect to another. This is particularly important when considering the derivatives of financial assets’ prices with respect to the interest rate or the force of interest. We have, e.g., dd d 1 lnð1 þ iÞ ¼ ¼ v ¼ 1 2 d; ¼ di 1þi di

ð2:3:17Þ

dv d 1 1 ¼ ¼2 ¼ 2v2 ; di di 1 þ i ð1 þ iÞ2

ð2:3:18Þ

dvt ¼ 2tvtþ1 ; di

ð2:3:19Þ

dd dð1 2 vÞ ¼ ¼ v2 ¼ ð1 2 dÞ2 : di di

ð2:3:20Þ

and

Exercises Exercise 2.3.1 May 2003 Society of Actuaries (SOA)/Casualty Actuarial Society (CAS) Course 2 Examination, Problem No. 1. Bruce deposits 100 into a bank account. His account is credited interest at a nominal rate of interest i convertible semiannually. At the same time, Peter deposits 100 into a separate account. Peter’s account is credited interest at a force of interest of d. After 7.25 years, the value of each account is 200. Calculate ði 2 dÞ: Solution Bruce’s account after 7.25 years is worth: 100ð1 þ i=2Þ2£7:25 ¼ 200: Therefore, 1 þ i=2 ¼ 21=14:5 so that i ¼ 9:79285%: Peter’s account after 7.25 years is worth: 100e7:25d ¼ 200: Thus, 7:25d ¼ ln 2 and d ¼ 9:56065%: We conclude that ði 2 dÞ ¼ 0:2322%: A Exercise 2.3.2 May 2003 SOA/CAS Course 2 Examination, Problem No. 12. Eric deposits X into a savings account at time 0, which pays interest at a nominal rate of i, compounded semiannually. Mike deposits 2X into a different savings account at time 0, which pays simple interest at an annual rate of i. Eric and Mike earn the same amount of interest during the last 6 months of the 8th year. Calculate i.

18

Chapter 2

Solution Eric’s interest during the last 6 months of the 8th year is: i 15 i X 1þ · : 2 2 Mike’s interest in the last 6 months of the 8th year is 2X · ði=2Þ: Therefore i 15 i i i 15 X 1þ · ¼ 2X · ) 1 þ ¼2 2 2 2 2 so that i ¼ 9.46%.

A

Exercise 2.3.3 May 2003 SOA/CAS Course 2 Examination, Problem No. 50. Jeff deposits 10 into a fund today and 20 fifteen years later. Interest is credited at a nominal discount rate of d compounded quarterly for the first 10 years, and at a nominal interest rate of 6% compounded semiannually thereafter. The accumulated balance in the fund at the end of 30 years is 100. Calculate d. Solution Writing the equation of value we have d 240 10 1 2 ð1:03Þ40 þ 20ð1:03Þ30 ¼ 100: 4 This implies d 240 d 10 1 2 ¼ 15:77; 1 2 ¼ 0:98867052 4 4 and therefore d ¼ 0:0453:

A

Exercise 2.3.4 November 2001 SOA/CAS Course 2 Examination, Problem No. 1. Ernie makes deposits of 100 at time 0, and X at time 3. The fund grows at a force of interest

dt ¼

t2 ; t . 0: 100

The amount of interest earned from time 3 to time 6 is X. Calculate X. Solution The accumulation function is ðt 2 t 3 3 aðtÞ ¼ exp s =100 ds ¼ es =300l0 ¼ et =300 : 0

19


The amount in the account at time 3 is 3

100að3Þ ¼ 100 e3 =300 ¼ 100 e0:09 ¼ 109:41743: Thus between times 3 and 6 the accumulation process can be summarized by the equation 3

3

ð100 e0:09 þ XÞe6 =30023 =300 2 ð100 e0:09 þ XÞ ¼ X: This is a linear equation that solves to 96.025894 ¼ 0.1223894X, and then X ¼ 784.59. A

2.4 Cash flows In practice, deposits are made more frequently than just at the beginning of the period considered. Withdrawals can be made, as well. Consider an account in which an amount of C0 is initially invested, and at the end of the kth year, additional amount of gk ; k ¼ 1; …; n is invested (the amounts could, in general, be positive or negative). Such a combination of payments will be termed a cash flow C0, g1 ; …; gn of this account. Calculation of the combined value of such cash flows is a common practical financial problem. In general, when we value a cash flow or a set of cash flows, its value today is called the present value, while its value at some future time is termed the accumulated value. If the value is calculated at a time somewhere in between now and the future time horizon, it is called the current value. Let Ck be the account balance at the end of the kth year. Then Ck ¼ Ck21 þ iCk21 þ gk ;

k ¼ 1; …; n:

After a simple rearrangement of the above formula we obtain Cn ¼ ð1 þ iÞn C0 þ

n X

ð1 þ iÞn2k gk :

ð2:4:1Þ

k¼1

The formula (2.4.1) basically tells us that the balance in the account at the end of the nth year will be the accumulated value of the initial deposit plus the accumulated value of all additional deposits made up to the time n. Example 2.4.1 Suppose that we invest $1000 a year at the beginning of every year over a span of 3 years. Assume also that the effective annual interest rate is 10% throughout this period of time. Interest earned is compounded annually, i.e., it is added to the account at the end of each year. The amount accumulated after

20

Chapter 2

3 years will be C3 ¼ ð1:10Þ3 · 1000 þ ð1:1Þ2 · 1000 þ ð1:1Þ · 1000 ¼ 3641:00:

A

Using the discount factor v, we can also rewrite (2.4.1) as vn Cn ¼ C0 þ

n X

vk gk :

ð2:4:2Þ

k¼1

Example 2.4.2 The present value at time 0 of the cash flow described in Example 2.4.1 is: 1000 þ 1000ð1:1Þ21 þ 1000ð1:1Þ22 ¼ 2735:54:

A

Now assume that the payments made into an account are made continuously at a rate of gðtÞ; a function of time t. This means that the payment amount in the infinitesimal period of time ðt; t þ dtÞ is equal gðtÞdt: Let CðtÞ be the account value at time t. Assume also that the interest is compounded continuously with the force of interest equal to dt ¼ dðtÞ: This means that if at the time t the account balance is C, then after the infinitesimal period of time ðt; t þ dtÞ the account value will increase by C dðtÞdt (assuming dðtÞ is positive; if dðtÞ is negative, this is a decrease). The total increase in the account value in the time period ðt; t þ dtÞ is therefore equal to dCðtÞ ¼ CðtÞdðtÞdt þ gðtÞdt:

ð2:4:3Þ

This gives us a differential equation C0 ðtÞ ¼ CðtÞdðtÞ þ gðtÞ:

ð2:4:4Þ

Ð By multiplying both sides of (2.4.4) by exp 2 t0 dðsÞ ds we arrive at the following ðt ðt C0 ðtÞexp 2 dðsÞds 2 CðtÞdðtÞexp 2 dðsÞds 0

0

ðt ¼ gðtÞexp 2 dðsÞds :

ð2:4:5Þ

0

We now note left-hand side of (2.4.5) is the derivative of the function Ð that the CðtÞexp 2 t0 dðsÞds and by integrating both sides of (2.4.5) from 0 to t; and moving Cð0Þ to the right-hand side, we conclude that ðt ðt ðt CðtÞexp 2 dðsÞds ¼ Cð0Þ þ exp 2 dðsÞds gðtÞdt: 0

0

0

ð2:4:6Þ

21


Observe that (2.4.6) is a continuous version of (2.4.2). Thus the present value of an amount Ð t CðtÞ at time 0 can be obtained by multiplying it by the discount factor exp 2 0 dðsÞds ; and that present value is equal to the sum of Cð0Þ and the integral of the discounted value of the continuous payment stream gðtÞ: We also note that by accumulating (2.4.6) we obtain ðt ðt ðt dðsÞds þ gðtÞexp dðsÞds dt: ð2:4:7Þ CðtÞ ¼ Cð0Þexp 0

0

t

In particular, the accumulated value at time t of a single payment C0 made at time t, with t . t; is: ðt dðsÞds : ð2:4:8Þ CðtÞ ¼ C0 exp t

If dðsÞ ¼ d is constant and the effective annual interest rate is i, (2.4.8) leads to

d ¼ lnð1 þ iÞ

ð2:4:9Þ

Example 2.4.3 Assume that in a span of 2 years, initial deposit in an account has increased by 21%. The effective annual interest rate i is then calculated from the equation ð1 þ iÞ2 ¼ 1:21; and we conclude that i ¼ 0:10: Using (2.4.9) we can also calculate the force of interest equivalent to i ¼ 0:10 as d ¼ 0:09531: A One practical problem in dealing with cash flows is the question of an interest rate that makes certain cash flows equivalent. For example, suppose that we have 30 monetary units paid every year at the end of the year for 4 years in return for 100 paid today. What is the interest rate that makes these two equal? It is called the internal rate of return (IRR) or the dollar-weighted rate of return. In order to find that interest rate i, we set up the equation 100 ¼

30 30 30 30 þ þ þ : 1þi ð1 þ iÞ2 ð1 þ iÞ3 ð1 þ iÞ4

ð2:4:10Þ

If the payments are level, most financial calculators can find the interest rate. The problem can be also solved using the Solver function in Microsoft Excel software. The solution is i < 7:71%: The IRR is also called the yield rate for the cash flow analyzed. It is important to note that finding the yield rate involves solving a polynomial equation, and such an equation may have multiple solutions, no solution, or solutions, which are complex numbers. There are situations when a unique real solution exists: if all outflows occur before all inflows or if the cash outflows can be treated as a ‘deposit’ and outflows as ‘withdrawals’, with the resulting account balance remaining positive throughout the period under consideration [40].

22

Chapter 2

There is also a problem with the use of the IRR as a tool for evaluating a project in which the outflows are invested and from which the inflows are returns—using IRR implies that cash flows received from the project are reinvested at that IRR, which may or may not (usually is not) be possible as one has to generally reinvest at the prevailing market rates. Note that if two cash flows are equivalent at one point in time, they are equivalent at any other time. Also, a value of a linear combination of cash flows is a linear combination of values of individual cash flows. Finally, if a value VðtÞ is established at time t then the value VðsÞ at any other time may be calculated as: VðsÞ ¼ vt2s VðtÞ: One important consequence is also that you can compare cash flows or calculate their values, at various points in time, and it is often convenient to pick a different reference date than just today or a future horizon. Exercises Exercise 2.4.1 November 2001 SOA/CAS Course 2 Examination, Problem No. 24. David can receive one of the following two payment streams: (i) 100 at time 0, 200 at time n, and 300 at time 2n, (ii) 600 at time 10. At an annual effective interest rate of i, the present values of the two streams are equal. Given that vn ¼ 0:75941; determine i. Solution As we know, the two streams are equal, therefore 100 þ 200vn þ 300v2n ¼ 600v10 : We are given that vn ¼ 0:75941; and using this, we obtain 100 þ 151:882 þ 173:10 ¼ 600v10 : This solves to v10 ¼ 0:708; so that i < 3:50%:

A

Exercise 2.4.2 November 2001 SOA/CAS Course 2 Examination, Problem No. 28. Payments are made to an account at a continuous rate of 8k þ tk; where 0 # t # 10. Interest is credited at a force of interest

dt ¼

1 : 8þt

After 10 years, the account is worth 20 000. Calculate k.

23


Solution Note that ð 10 exp n

1 dt 8þt

10

¼ elnð8þtÞln ¼

18 : 8þn

Therefore, by (2.4.7), 20 000 ¼

ð 10

kð8 þ tÞ

0

18 dt ¼ 18ktlt¼10 t¼0 ¼ 180k: 8þt

This implies that k¼

20 000 : 180

A

2.5 Interest measurement in a fund 2.5.1 Discrete-time model In the practice of insurance companies a problem of interest measurement in a fund over a year arises naturally. First of all, it is used to reflect the assets’ increase by showing the investment efficiency. But it is also an important indicator for the participants of defined contribution pension plans for the increase of their share in the plan assets is usually defined via the rate of return of the participation (accounting) unit. Let us introduce the following notation At ¼ fund balance at time t [ ½0; 1; I ¼ interest earned during the year, between t ¼ 0 and t ¼ 1; Ct ¼ principal contributed (if positive) or withdrawn (if negative) at time t, t [ ½0; 1; P C ¼ t Ct ¼ total net contributions to the fund (if negative, this is an indication of net withdrawals), i ¼ effective interest rate between t ¼ 0 and t ¼ 1: We have A1 ¼ A0 þ C þ I;

and

I ¼ iA0 þ

X

Ct ðð1 þ iÞ12t 2 1Þ:

0#t#1

In order to find i we would have to solve the second equation for it, and the first equation would actually be only the source of the value of I. This is not an easy problem, and in practice one often uses a simple interest approximation ð1 þ iÞ12t 2 1 < 1 þ ð1 2 tÞi 2 1 ¼ ð1 2 tÞi;

24

Chapter 2

which allows us to obtain a solution i
> > < x e sðxÞ ¼ 1 2 ; 1 # x , 4:5; > 100 > > : 0; 4:5 # x: Calculate mð4Þ: Solution Recall the basic definition of the force of mortality: e4 0 2s ð4Þ e4 100 ¼ ¼ mð4Þ ¼ ¼ 1:202553422: 4 sð4Þ 100 2 e4 e 12 100

A

Exercise 3.7.3 November 2001 SOA/CAS Course 3 Examination, Problem No. 1. You are given ( 0:04; 0 , x , 40; mðxÞ ¼ 0:05; x . 40: Calculate e 25: 25 : Solution Because of the peculiar form of the force of mortality formula, it is best to break the calculation at the age where the force changes. Thus: ð 15 ð 10 e 25: 25 ¼ e 25: 15 þ 15 p25 · e 40: 10 ¼ p dt þ p · t 25 t p40 dt 15 25 0 0 ð 10 ð 15 e20:04t dt þ e20:04£15 · e20:05t dt ¼ a 15 d¼0:04 þ e20:60 · a 10 ¼ 0

20:60

0 20:50

d¼0:05

12e 12e þ e20:6 ¼ 25 2 25e20:60 þ 20e20:60 2 20e21:1 0:04 0:05 ¼ 25 2 5e20:60 2 20e21:1 ¼ 15:5985: ¼

A

Exercise 3.7.4 November 2000 SOA/CAS Course 3 Examination, Problem No. 36. Given (i) mðxÞ ¼ F þ e2x ; x $ 0 (ii) 0:4 p0 ¼ 0:50: Calculate F:

82

Chapter 3

Solution ð 0:4 ð 0:4 2x 0:50 ¼ 0:4 p0 ¼ exp 2 mðxÞ dx ¼ exp 2 ðF þ e Þ dx 0

0

e0:8 2 1 ¼ exp 2 0:4F þ : 2 Taking natural logarithms of both the sides and rearranging we obtain F ¼ 0:2009: Exercise 3.7.5 May 2000 SOA/CAS Course 3 Examination, Problem No. 1. Given (i) e 0 ¼ 25; (ii) lx ¼ v 2 x; 0 # x # v; (iii) TðxÞ is the future lifetime random variable. Calculate VarðTð10ÞÞ: Solution Since we are working with De Moivre’s Law, v is twice the life expectancy of a newborn. Hence v ¼ 50: For the 10 year old, the distribution of the remaining lifetime is uniform on ½0; 40; and its variance is 402 400 ¼ 133:333… ¼ 3 12

A

3.8 Mortality tables In the early models of life insurance, it was common to work with a deterministic cohort, i.e., a group of survivors from a group of initial l0 newborns. The size of such a group at age x is lx ; and the number of people who die between ages x and x þ 1 is denoted by dx : The collection of data describing the size of the cohort and its mortality is called a mortality table. Mortality tables are constructed from data collected by governments and insurance companies. In the United States, national census is performed every 10 years, and the data so collected is used to create tables adopted by the insurance regulators. The tables are, in turn, the source for estimation of survival and death probabilities. We have, for example: lxþ1 : lx

ð3:8:1Þ

lxþk ¼ px · pxþ1 · pxþ2 · · ·pxþk21 : lx

ð3:8:2Þ

px ¼ Similarly, k px

¼

83

Life insurance and annuities

Valuation of pension plans often requires the use of commutation functions. We will now define two functions that will be used often later in this chapter: D x ¼ vx · l x ; Nx ¼

þ1 X

ð3:8:3Þ ð3:8:4Þ

Dxþk :

k¼0

Additional discussion of commutation functions is given by Aitken [1] and Bowers et al. [14]. In the insurance practice, a group underwritten for coverage usually has different mortality than the general population. A cohort for which some additional information about mortality is obtained is called a select cohort. If the effect of underwriting wears off after some period of time, we say that the group ultimately becomes like the general population, and the table so obtained is called select and ultimate. We use the following notation for a select table: t p½x ; t q½x ; where ½x represents select mortality, and q½xþr ¼ q½x2jþrþj ¼ qxþr ;

ð3:8:5Þ

where r is the select period. Mortality tables created in practice show data only for the integer ages. This creates some degree of difficulty for insurance firms and pension plans, as they have to work with the insured people and plan participants who may die, retire, or terminate contracts at noninteger ages. In order to better model those situations, we introduce the random variable: S ¼ T 2 K:

ð3:8:6Þ

We have 0 # S , 1 and S is the fraction of a year lived in the year of death. Because the standard assumption is that T has an absolutely continuous distribution, so does S: The CDF of the random variable S conditional on K ¼ k is given by the formula PrðS # slK ¼ kÞ ¼

s qxþk

qxþk

:

ð3:8:7Þ

Assume now that the random variables S and K are independent. Then PrðS # slK ¼ kÞ does not depend on k and for a certain function FS ðsÞ; for each s [ ½0; 1; s qxþk

¼ FS ðsÞ · qxþk ;

ð3:8:8Þ

for every k ¼ 0; 1; …: Of course, FS in (3.8.8) is the CDF of S: If we assume that S is uniformly distributed, then (3.8.10) becomes an equality. Furthermore, the independence of K and S implies in this case (uniform distribution of S and we will return to this special case in detail later): VarðTÞ ¼ VarðKÞ þ

1 : 12

84

Chapter 3

Also, the expected value of the conditional distribution of S given that K ¼ k will be termed the fractional life expectancy. In what follows, we will use the following discrete version of the random variable S: For an m [ : (the set of positive integers) we define a new random variable 1 bmS þ 1c: m

SðmÞ ¼

ð3:8:9Þ

This variable is obtained by rounding S down to the smallest multiple of the fraction 1=m that already exceeds S: The random variable SðmÞ is discrete, with a jump in its CDF on the atoms 1=m; 2=m; …; 1: If K and S are independent, then so are K and SðmÞ : If S is uniformly distributed, then SðmÞ has a uniform distribution on the set of its atoms. While we do not generally know the exact probability distribution of S; several approximation assumptions are commonly used. They are called the fractional age assumptions. The first such assumption is called the Uniform Distribution of Deaths (UDD) assumption, which states that S has the uniform distribution on the unit interval ½0; 1: Let us note that De Moivre’s Law also assumes uniform distribution, but UDD is different, as it gives a different uniform distribution in each year of age. Under UDD, K and S are independent, and EðTÞ ¼ EðKÞ þ UDD

1 ; 2

VarðTÞ ¼ VarðKÞ þ UDD

1 : 12

ð3:8:10Þ

ð3:8:11Þ

We also have the following UDD identities for 0 # s þ t # 1; s [ ½0; 1; t [ ½0; 1; t qx

¼ t · qx ;

ð3:8:12Þ

UDD

mðx þ tÞ ¼

UDD

qx ; 1 2 tqx

qx ¼ fT ðtÞ ¼ t px mðx þ tÞ ¼ ð1 2 t qx Þmðx þ tÞ ¼ ð1 2 tqx Þmðx þ tÞ; UDD

UDD

lxþt ¼ lx 2 tdx ; UDD

ð3:8:13Þ ð3:8:14Þ ð3:8:15Þ

and s qxþt

¼

UDD

sqx : 1 2 tqx

ð3:8:16Þ

85


The second partial age assumption assumes constant force (CF) of mortality in the year of death. We have then: smxþ1=2 ¼ 2 ln s px ;

ð3:8:17Þ

¼ e2smxþ1=2 ¼ psx :

ð3:8:18Þ

CF

as well as s px

Furthermore, the conditional distribution of S; given that K ¼ k; is PrðS # slK ¼ kÞ ¼

1 2 psxþk ; 1 2 pxþk

0 # s # 1:

ð3:8:19Þ

This is the truncated exponential distribution. Note that in this case the random variables S and K are not, in general, independent, but in the special case when pxþk ¼ px (e.g., when the force of mortality is constant throughout all the ages) is independent of k; K and S are independent. The last commonly used partial age assumption is the one of Balducci, denoted as BAL. Under this model, the probability of death by the end of the year for a person who has survived a fraction s of the year between ages x and x þ 1 is a linear function of the form: 12s qxþs

¼ ð1 2 sÞqx :

BAL

ð3:8:20Þ

As px ¼ ð12s pxþs Þs px (3.8.20) implies that s px

¼

BAL

1 2 qx : 1 2 ð1 2 sÞqx

ð3:8:21Þ

Recalling the definition of the force of mortality, we obtain

mðx þ sÞ ¼

BAL

qx ; 1 2 ð1 2 sÞqx

ð3:8:22Þ

and PrðS # slK ¼ kÞ ¼

1 2s pxþk s ¼ : 1 2 ð1 2 sÞqxþk 1 2 pxþk

ð3:8:23Þ

Note that (3.8.23) implies that the random variables S and K are not independent. The Balducci assumption is also commonly called the hyperbolic assumption, because the graph of lxþs for 0 # s # 1 is a hyperbola in this case. Exercises Exercise 3.8.1 Using the constant force assumption, calculate the fractional life expectancy of S given that K ¼ k; under the constant force assumption for women,

86

Chapter 3

and men. You have the following data: qxþk ¼ 0:00576 for women and qxþk ¼ 0:01652 for men (this data comes from the mortality table for Poland, age x ¼ 50; k ¼ 5). Solution We have EðSlK ¼ kÞ ¼

ð1

ð1

s d PrðS # slK ¼ kÞ ¼ ð1 2 PrðS # slK ¼ kÞÞds 0 ð1 1 1 qxþk s ¼12 1 2 pxþk ds ¼ 1 2 1þ : qxþk qxþk ln pxþk 0 0

For the data given, we obtain: EðSlK ¼ 5Þ ¼ 0:499952 for women and EðSlK ¼ 5Þ ¼ 0:49861 for men. Let us note that these fractional life expectancies differ very little from the approximation of 0.50 given by the UDD assumption. A Exercise 3.8.2 November 2003 SOA/CAS Course 3 Examination, Problem No. 16. For a space probe to Mars: (i) The probe has three radios, whose future lifetimes are independent, each with mortality following k lq0

¼ 0:1ðk þ 1Þ;

k ¼ 0; 1; 2; 3;

where time 0 is the moment the probe lands on Mars. (ii) The failure time of each radio follows the hyperbolic assumption within each year. (iii) The probe will transmit until all three radios have failed. Calculate the probability that the probe is no longer transmitting 2.25 years after landing on Mars. Solution We have 0 lq0 ¼ 0:1; 1 lq0 ¼ 0:2; 2 lq0 ¼ 0:3; 3 lq0 ¼ 0:4; and therefore, q0 ¼ 1=10; q1 ¼ 2=9, q2 ¼ 3=7; q4 ¼ 1: We want to find the probability 3 7 p2 · ð2:5 q0 Þ ¼ ð1 2 2:25 p0 Þ ¼ ð1 2 2 p0 · 0:25 p2 Þ ¼ 1 2 BAL 10 1 2 ð1 2 0:25Þq2 3

3

3

0

13 4 3 7 39 B C 7 · ¼ @1 2 ¼ 0:06918676: A¼ 3 3 10 95 12 £ 4 7

A

87


Exercise 3.8.3 November 2001 SOA/CAS Course 3 Examination, Problem No. 2. For a select-and-ultimate mortality table with a 3-year ultimate period: q½60 q½61 q½62 q½63 q½64

¼ 0:09; ¼ 0:10; ¼ 0:11; ¼ 0:12; ¼ 0:13;

q½60þ1 q½61þ1 q½62þ1 q½63þ1 q½64þ1

¼ 0:11; ¼ 0:12; ¼ 0:13; ¼ 0:14; ¼ 0:15;

q½60þ2 q½61þ2 q½62þ2 q½63þ2 q½64þ2

¼ 0:13; ¼ 0:14; ¼ 0:15; ¼ 0:16; ¼ 0:17;

q½60þ3 q63 q½61þ3 q64 q½62þ3 q65 q½63þ3 q66 q½64þ3 q67

¼ 0:15; ¼ 0:16; ¼ 0:17; ¼ 0:18; ¼ 0:19;

(i) White was a newly selected life on 01/01/2000. (ii) White’s age on January 1, 2001 is 61. (iii) P is the probability on January 1, 2001 that White will be alive on January 1, 2006. Calculate P: Solution The probability sought is 5 p½60þ1

¼ p½60þ1 · p½60þ2 · p63 · p64 · p65 ¼ ð1 2 0:11Þð1 2 0:13Þð1 2 0:15Þ ð1 2 0:16Þð1 2 0:17Þ ¼ 0:4589:

A

Exercise 3.8.4 May 2001 SOA/CAS Course 3 Examination, Problem No. 1. For a given life age 30, it is estimated that an impact of a medical breakthrough will be an increase of 4 years in e 30 ; the complete expectation of life. Prior to the medical breakthrough, sðxÞ followed De Moivre’s Law with v ¼ 100 as the limiting age. Assuming De Moivre’s Law still applies after the medical breakthrough, calculate the new limiting age. Solution With De Moivre’s Law, TðxÞ is uniformly distributed on ½0; v 2 x; and e x ¼

ðv 2 xÞ : 2

Prior to the medical breakthrough, v ¼ 100 and e 30 ¼

70 ¼ 35: 2

After the breakthrough the limiting age value increases by 4 and therefore: e 30 ¼ 39 ¼ 35 þ 4 ¼ Hence, the new limiting age value is 108.

ðv 2 30Þ : 2

88

Chapter 3

Exercise 3.8.5 May 2001 SOA/CAS Course 3 Examination, Problem No. 13. Mr. Ucci has only three hairs left on his head and he will not be growing any more. (i) The future mortality of each hair is k lqx

¼ 0:1ðk þ 1Þ;

where k ¼ 0; 1; 2; 3; and x is Mr. Uccis age. (ii) Hair loss follows the hyperbolic assumption at fractional ages. (iii) The future lifetimes of the three hairs are independent. Calculate the probability that Mr. Ucci is bald (has no hair left) at age x þ 2:5: Solution Since three independent hairs must ‘die’ by age x þ 2:5; the probability sought is ð2:5 qx Þ3 : We are given that qx ¼ 0:1; 1lqx ¼ 0:2; and 2lqx ¼ 0:3: From the first two of the probabilities, it follows that 2 qx ¼ 0:1 þ 0:2 ¼ 0:3 and hence 2 px ¼ 0:7: From this, combined with 2lqx ¼ 0:3; it follows that 0:3 ¼2lqx ¼ 2 px · qxþ2 ¼ 0:7qxþ2 so that qxþ2 ¼ 3=7: With Balducci’s Law t pxþ2

¼

pxþ2 : pxþ2 þ t · qxþ2

Therefore

0:5 qxþ2

¼ 1 2 0:5 pxþ2 ¼ 1 2

4 7 4 3 þ 0:5 · 7 7

¼

1:5 3 ¼ : 5:5 11

Finally, 2:5 qx

¼ 2 qx þ 2 px ·

0:5 qxþ2

¼ 0:3 þ 0:7 ·

3 ¼ 0:490909; 11

and ð2:5 qx Þ3 ¼ 0:1183:

A

Exercise 3.8.6 November 2000 SOA/CAS Course 3 Examination, Problem No. 4. Mortality for Audra, age 25, follows De Moivre’s Law with v ¼ 100: If she takes up hot air ballooning for the coming year, her assumed mortality will be adjusted so that for the coming year only, she will have a constant force of mortality of 0.1. Calculate the decrease in the 11-year temporary complete life expectancy for Audra if she takes up hot ballooning.

89


Solution From De Moivre’s Law, t p25

¼12

t : 75

Before the mortality adjustment ð 11 ð 11 t 112 e 25: 11 ¼ ¼ 10:1933: p dt ¼ 1 2 dt ¼ 11 2 t 25 75 2 · 75 0 0 To analyze the effect of 1-year mortality adjustment on the temporary life expectancy, we use the recursive relation: e 25: 11 ¼ e 25:1 þ p25 · e 26: 10 : Only e 25: 1 and p25 are affected by the change in mortality. With the change, we have p25 ¼ e20:10 and e 25:1 ¼

ð1

e20:10t dt ¼

0

1 2 e20:10 ¼ 0:951626: 0:10

Also e 26: 10 ¼

ð 10 0

t p26

dt ¼

ð 10 t 102 ¼ 9:324324: 12 dt ¼ 10 2 74 2 · 74 0

Substituting into the recursion relation we get the adjusted life expectancy: e 25: 11 ¼ e 25:1 þ p25 · e 26: 10 ¼ 0:951626 þ e20:10 · 9:324324 ¼ 9:388624: The decrease is 10:1933 2 9:3886 ¼ 0:8047:

A

Exercise 3.8.7 May 2000 SOA/CAS Course 3 Examination, Problem No. 12. For a certain mortality table, you are given: (i) mð80:5Þ ¼ 0:0202; (ii) mð81:5Þ ¼ 0:0408; (iii) mð82:5Þ ¼ 0:0619; (iv) deaths are uniformly distributed between integral ages. Calculate the probability that a person aged 80.5 will die within 2 years. Solution Under UDD 0:0408 ¼ mð81:5Þ ¼

UDD

q81 ; 1 2 0:5q81

90

Chapter 3

and therefore q81 ¼ 0:04: Similarly, q80 ¼ 0:02 and q82 ¼ 0:06: Hence 2 q80:5

¼ 0:5 q80:5 þ 0:5 p80:5 · ðq81 þ p81 · ¼

0:5 q82 Þ

0:01 0:98 þ · ð0:04 þ 0:96 · 0:03Þ ¼ 0:0782: 0:99 0:99

A

3.9 Life insurance We will now consider insurance contracts, which make a one-time payment of a specified amount of insurance in the case of death of the insured, such payment made to a named policy beneficiary. In order to calculate the present value of the benefit, one should use the actual future interest rates between now and the death of the insured. But these interest rates are not known ex ante, and in practice actuaries choose an appropriate valuation rate for valuation (calculation of reserve) purposes, or for premium calculation. Similarly, for pension plans, a valuation rate is used for establishing actuarial liabilities and normal cost of a given plan. Such an interest rate chosen by the actuary will be denoted by i; and the corresponding discount factor will be written as v: As in the previous sections, the age of a person under consideration (i.e., person entering into the life insurance contract) will be denoted by x: The actual benefit amount and the time of payment of the benefit for the life insurance policy will depend on the random variable TðxÞ; the future lifetime of a life age ðxÞ; starting from the policy effective date. The present value of the said benefit will be denoted by Z; and Z is, just as TðxÞ; a random variable. The expected value of Z; EðZÞ; is called the actuarial present value of the benefit, or the single benefit premium, or net single premium, of the policy. The first form of a life insurance contract we will consider is traditionally called the whole life insurance. In this contract, the insurance amount (which, for simplification, we will assume to be a monetary unit) is paid upon death of the insured. The payment can be made either immediately (the policy is then termed continuous) or at the end of the year of death (the policy is then termed discrete). Using the notation introduced earlier in this chapter, we assume that the insured was aged when the policy was issued, T is the future lifetime of the insured, and K ¼ bTc; the number of whole years lived by the insured. The present value of the benefit payment random variable is denoted by Z: For the discrete policy, the death benefit is assumed to be paid at the end of the year of death, and we have Z ¼ vKþ1 :

ð3:9:1Þ

Furthermore EðZÞ ¼

1 X k¼0

vkþ1 PrðK ¼ kÞ ¼

1 X k¼0

vkþ1 · k px · qxþk :

ð3:9:2Þ

91


Recall that this expected value is also termed single benefit premium for the policy. We will denote it for the discrete whole life insurance by Ax : Therefore Ax ¼

1 X


ð3:9:3Þ

k¼0

Note that Ax ; Ax ðvÞ is a function of the discount factor v: In all practical applications v is a fraction, i.e., v [ ð0; 1Þ: Let us also observe that Ax ð·Þ is the probability generating function of the random variable K þ 1 as expressed by the function GKþ1 ðvÞ ¼

1 X

vkþ1 PrðK þ 1 ¼ k þ 1Þ:

ð3:9:4Þ

kþ1¼1

It is not hard to see that Ax ð·Þ is an increasing function of v [ ð0; 1Þ: In fact, if EðKÞ , 1; then 1 X d Ax ðvÞ ¼ ðk þ 1Þvk PrðK ¼ kÞ; dv k¼0

v [ ð0; 1Þ:

ð3:9:5Þ

Using (3.9.5) and the definition of curtate life expectancy, we infer that d A ð1Þ ¼ EðK þ 1Þ ¼ ex þ 1; dv x

v [ ð0; 1Þ:

ð3:9:6Þ

If EðK 2 Þ , 1 then the second derivative of the function Ax ð·Þ equals 1 X d2 A ðvÞ ¼ ðk þ 1Þkvk21 PrðK ¼ kÞ . 0; x dv2 k¼0

v [ ð0; 1Þ:

ð3:9:7Þ

It is easy to see, based on the above, that for any v [ ð0; 1Þ d2 Ax ðvÞ . 0; dv2

ð3:9:8Þ

as long as PrðK ¼ 0Þ , 1: Thus, if the random variable K is not degenerate at zero (i.e., there is at least one person alive at age x who survives for another year), then Ax ð·Þ is strictly convex. From (3.9.7) we also conclude that d2 Ax ð1Þ ¼ EðK 2 Þ þ EðKÞ: dv2

ð3:9:9Þ

Therefore, by combining the above with (3.9.6) and (3.9.8) we get the formula for the variance s 2 of the random variable K d2 d d 2 A ð1Þ A ð1Þ 2 1 : s ¼ 2 Ax ð1Þ 2 ð3:9:10Þ dv x dv x dv

92

Chapter 3

The properties of the function Ax ð·Þ help us to calculate all higher moments of the random variable Z: For example, the second moment is EðZ 2 Þ ¼

1 X

v2ðkþ1Þ · k px · qxþk ¼ Ax ðv2 Þ:

ð3:9:11Þ

vnðkþ1Þ · k px · qxþk ¼ Ax ðvn Þ ¼ Ax ðe2nd Þ:

ð3:9:12Þ

k¼0

In fact, it is quite easy to see that EðZ n Þ ¼

1 X k¼0

The formula (3.9.12) assumes such a simple form for this type of life insurance, because the benefit amount is 1, and any power of 1 is equal to 1 itself. There is only one other number that can replace 1 and still allow for this property, that number is 0. This gives rise to the popular Rule of Moments for the calculation of the moments of the random variable Z: as long as the amount of insurance benefit can be only 0 or 1, the nth moment of Z is calculated by replacing the force of interest d; in the formula for EðZÞ; by its multiple nd: In particular, the single benefit premium calculated with the force of interest multiplied by n is denoted by a superscript of n placed in front of the regular notation for the single benefit notation, e.g., 2 Ax for discrete whole life. From (3.9.12) we also conclude that VarðZÞ ¼ Ax ðv2 Þ 2 ðAx ðvÞÞ2 :

ð3:9:13Þ

The above considerations were concerned with the case of a discrete whole life insurance, with benefit payable at the end of the year of death. Of course, in the insurance practice, benefit is paid very soon after the death of the insured. To model the reality of the insurance marketplace, it is far more natural to use the random variable T; the future lifetime of life age x: The random variable Z; the present value of the insurance benefit for the continuous whole life insurance, is then Z ¼ vT ;

ð3:9:14Þ

and the single benefit premium for this policy is ð1 ð1 vt fT ðtÞ dt ¼ vt t · px · qxþt · dt: A x ¼ EðZÞ ¼ 0

ð3:9:15Þ

0

Recall that T ¼ K þ S; where K ¼ bTc: Therefore A x ¼ EðZÞ ¼ EðvKþ1 vS21 Þ:

ð3:9:16Þ

If we assume UDD in the year of death, then the random variables K þ 1 and 1 2 S are independent, and 1 2 S has uniform distribution on ð0; 1Þ: Therefore, ð1 Ax ¼ EðZÞ ¼ EðvKþ1 Þ · EðvS21 Þ ¼ Ax e2dðs21Þ ds ¼ i Ax : ð3:9:17Þ UDD d 0

93


Let us now assume that a year is divided into m equal parts (e.g., 12 months, with m ¼ 12; or four quarters, with m ¼ 4) and the benefit of the policy is paid at the end of the part of the year in which death occurs. The single benefit premium for such a plan ðmÞ of insurance will be denoted by Ax : We have ðmÞ

¼ EðvKþS Þ; AðmÞ x

ð3:9:18Þ

where SðmÞ is defined by (3.8.9). Using the UDD assumption, in a manner analogous to the one for A x ; we can derive AðmÞ ¼ x

i

UDD iðmÞ

ð3:9:19Þ

Ax :

One more important observation we should make is that the value of the single benefit premium for the unit whole life insurance can be calculated quite easily under the simple assumptions of the constant force of mortality or De Moivre’s Law. Indeed, if m is the constant force, then A x ¼

ð1

CF

e2td e2tm · m · dt ¼ 2

0

m 2tðmþdÞ t!1 m e ¼ : mþd m þd t¼0

ð3:9:20Þ

For the discrete whole life policy, if we write p ¼ px ¼ e2d ; and q ¼ 1 2 p; then we have Ax ¼

CF

1 X

vkþ1 · k px · qxþk ¼

k¼0

1 X

vkþ1 · pk · ð1 2 pÞ ¼ vð1 2 pÞ

k¼0

¼ vð1 2 pÞ

1 X

v k · pk

k¼0

1 q q ¼ ¼ : 1 2 vp ð1 þ iÞ 2 p qþi

ð3:9:21Þ

We can see that (3.9.20) and (3.9.21) are direct analogues of each other. On the other hand, under De Moivre’s Law, we have ð v2x a 1 · dt ¼ v2x ; ð3:9:22Þ A x ¼ vt · DML 0 v2x v2x and Ax ¼

vX 2x21

vkþ1 · k px · qxþk ¼

DML

k¼0

¼

vX 2x21 k¼0

vkþ1 ·

vX 2x21 k¼0

1 a ¼ v2x : v2x v2x

vkþ1 ·

v2x2k 1 · v2x v2x2k ð3:9:23Þ

94

Chapter 3

Term life insurance is a simple modification of the whole life insurance policy. Unit term life contract calls for payment of death benefit of one monetary unit if the death of the insured occurs within a prescribed term, n years, from the policy effective date. If the insured survives n years, the policy ends without a benefit payment. First, let us consider the discrete model in which the payment is made at the end of the year of death. Then the present value of the death benefit equals ( Kþ1 v ; K # n 2 1; Z¼ ð3:9:24Þ 0; K $ n: The single benefit premium for this contract is denoted by A1x: n : By definition A1x: n ¼

n21 X


k¼0

n21 X


ð3:9:25Þ

k¼0

Just as for Ax ; A1x: n ; A1x: n ðvÞ is also a function of v: The second moment of the present value of benefit random variable Z is given by the formula: EðZ 2 Þ ¼

n21 X k¼0

v2ðkþ1Þ PrðK ¼ kÞ ¼

n21 X

v2ðkþ1Þ · k px · qxþk ¼ A1x: n ðv2 Þ:

ð3:9:26Þ

k¼0

The Rule of Moments applies to this unit term life insurance. Also, VarðZÞ ¼ A1x: n ðv2 Þ 2 ðA1x: n ðvÞÞ2 :

ð3:9:27Þ

Unit term life insurance is also considered in the continuous case, for which ( T v ; T # n; Z¼ ð3:9:28Þ 0; T . n: For the continuous case, the single benefit premium is denoted by A 1x: n : The Rule of Moments also applies. A pure endowment insurance unit contract pays a monetary unit after a prescribed number of years, e.g., n; if and only if the insured survives that number of years. While in the definition there is no mention of the length of future lifetime of the insured, the present value of the benefit does depend on it. For this policy, there is no difference between the discrete or continuous model, and present value of benefit random variable is ( ( 0; K # n 2 1; 0; T , n; Z¼ ¼ ð3:9:29Þ n v ; K $ n; vn ; T $ n:

95


Therefore, Z is the product of vn and a zero-one Bernoulli Trial random variable with the probability of success n px : The single benefit premium for this pure endowment policy is denoted by A1x: n or by n Ex ; and it equals EðZÞ ¼ A1x: n ¼ vn · n px :

ð3:9:30Þ

The variance of Z is given by the formula VarðZÞ ¼ v2n · n px · n qx :

ð3:9:31Þ

The Rule of Moments applies to the unit pure endowment. A similar type of life insurance policy is an endowment, which pays a monetary unit at death of the insured if such death occurs before the n years of policy term and also pays a monetary unit if the insured survives those n years. Effectively, this insurance is a sum of a unit pure endowment and a unit year term insurance. The present value of benefit random variable for the discrete version of this policy is ( Kþ1 v ; K # n 2 1; Z¼ ð3:9:32Þ vn ; K $ n: Of course, the present value of benefit for this plan of insurance is the sum of the present value of benefit for a pure n-year term insurance (let us denote that random variable by Z1 ) and the present value of benefit for pure n-year endowment (let us denote that random variable by Z2 ). Using that observation, we establish the formula for the single benefit premium Ax: n for endowment insurance: Ax: n ¼ EðZ1 Þ þ EðZ2 Þ ¼ A1x: n þ Ax: n 1 :

ð3:9:33Þ

This insurance also has a continuous version. Since the product of the benefit payment for Z1 and Z2 is always zero, CovðZ1 ; Z2 Þ ¼ EðZ1 · Z2 Þ 2 A1x: n · Ax: n 1 ¼ 2A1x: n · Ax: n 1 ;

ð3:9:34Þ

and therefore VarðZÞ ¼ VarðZ1 Þ þ VarðZ2 Þ 2 2 · A1x: n · Ax: n 1 :

ð3:9:35Þ

The above equation implies that the risk (as measured by the variance) of an endowment policy is less than the sum of risks of its two building blocks: term insurance and pure endowment insurance. The endowment insurance can also be considered for the continuous model.

96

Chapter 3

The last type of life insurance we will consider is a deferred whole life insurance, which does not pay any benefit for a beginning period, m years, and then becomes a regular whole life insurance. The present value of benefit random variable for the discrete model is ( Z¼

K # m 2 1;

0; v

Kþ1

; K $ m:

ð3:9:36Þ

The single benefit premium for this plan of insurance is denoted by m lAx and it equals

m lAx

¼

1 X


k¼m

1 X


ð3:9:37Þ

k¼m

By using a substitution k ¼ i þ m in (3.9.36) we also obtain:

m lAx

¼

1 X

viþmþ1 ·

iþm px

· qxþiþm ¼ vm · m px ·

i¼0

1 X

viþ1 · i pxþm · qxþmþi

i¼0

¼ vm · m px · Axþm :

ð3:9:38Þ

Exercises Exercise 3.9.1 November 2003 SOA/CAS Course 3 Examination, Problem No. 2. For a whole life insurance of 1000 on ðxÞ with benefits payable at the moment of death: ( ðiÞ

dt ¼

( ðiiÞ

mðx þ tÞ ¼

0:04; 0 , t # 10; 0:05;

t . 10:

0:06;

0 , t # 10;

0:07;

t . 10:

Calculate the single benefit premium for this insurance.

97


Solution The single benefit premium is ðt ðt ð þ1 1000 exp 2 ds ds exp 2 mx ðsÞ ds mx ðtÞ dt 0 0 0 ðt ðt ð 10 exp 2 ds ds exp 2 mx ðsÞ ds mx ðtÞ dt ¼ 1000 0 0 0 ðt ðt ð þ1 exp 2 ds ds exp 2 mx ðsÞ ds mx ðtÞ dt þ 1000 10 0 0 ð þ1 ðt ð 10 20:04t 20:06t e e 0:06 dt þ 1000 exp 20:4 2 0:05 ds ¼ 1000 0

10

10

20:1t 10 ð þ1 e þ70 e21 £ exp 2 0:6 þ 0:07 ds 0:07 dt ¼ 60 e20:12ðt210Þ dt 2 0:1 0 10 10 ð þ1 1 e21 0:2 2 ¼ 60 e20:12t dt þ 70 e 0:1 0:1 10 21:2 0 70 21 21 0:2 e e ¼ 593:87: 2 ¼ 600 2 600 e þ 70 e ¼ 600 2 600 e21 þ 0:12 0:12 0:12 ðt

A Exercise 3.9.2 November 2003 SOA/CAS Course 3 Examination, Problem No. 22. For a whole life insurance of 1 on (41) with death benefit payable at the end of year of death, you are given: (i) i ¼ 0:05; (ii) p40 ¼ 0:9972; (iii) A41 2 A40 ¼ 0:00822; (iv) 2 A41 2 2 A40 ¼ 0:00433; (v) Z is the present-value random variable for this insurance. Calculate VarðZÞ: Solution Note that A40 ¼ vq40 þ vp40 · A41 ; and this implies that 0:9972 0:0028 A41 2 A40 ¼ 2 : 1:05 1:05 Combining this with A41 2 A40 ¼ 0:00822;

98

Chapter 3

we get 0:9972 0:0028 ; 12 A41 ¼ 0:00822 þ 1:05 1:05 and this means that A41 ¼ 0:21649621: Similarly 2

A40 ¼ v2 q40 þ v2 p40 · 2 A41 ;

and we infer that 0:9972 2 0:0028 A41 2 2 A40 ¼ 2 : 2 1:05 1:052 Combining this with 2

A41 2 2 A40 ¼ 0:00433;

we get 0:9972 0:0028 ; 12 · 2 A41 ¼ 0:00433 þ 2 1:05 1:052 and 2

A41 ¼ 0:07192616:

Finally VarðZÞ ¼ 2 A41 2 A241 ¼ 0:07192616 2 0:216496212 ¼ 0:02505555:

A

Exercise 3.9.3 November 2001 SOA/CAS Course 3 Examination, Problem No. 8. Each of 100 independent lives purchases a single premium 5-year deferred whole life insurance of 10 payable at the moment of death. You are given: (i) m ¼ 0:004: (ii) d ¼ 0:006: (iii) F is the aggregate amount the insurer receives from the 100 lives. Using the normal approximation, calculate F such that the probability the insurer has sufficient funds to pay all claims is 0.95. You are given that the 95th percentile of the standard normal distribution is 1.645.

99


Solution We have ( Z¼

T , 5;

0; T

10 · v ;

T $ 5;

and ( 2

Z ¼

T , 5;

0; 2T

100 · v ;

T $ 5; 0:04 ¼ 2:426123; 0:04 þ 0:06 0:04 ¼ 11:233224; · e25£0:04 Þ 0:04þ0:12

EðZÞ ¼ v5 · 5 px · 10 · Axþ5 ¼ 10ðe25ð0:06þ0:04Þ Þ EðZ 2 Þ ¼ v10 · 5 px · 100 · 2 Axþ5 ¼ 100ðe210

· 0:06

and hence VarðZÞ ¼ EðZ 2 Þ2ðEðZÞÞ2 ¼ 5:347153: The aggregate present value for the 100 lives is ZAGG ¼ Z1 þ···þZ100 : The 95th percentile given by the normal approximation is pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 100 EðZÞ þ 1:645 100 VarðZÞ ¼ 100ð2:426123Þ þ 1:645 100ð5:347153Þ ¼ 280:65: A Exercise 3.9.4 May 2000 SOA/CAS Course 3 Examination, Problem No. 13. An investment fund is established to provide benefits on 400 independent lives age x: (i) On January 1, 2001, each life is issued a 10-year deferred whole life insurance of 1000, payable at the moment of death. (ii) Each life is subject to a constant force of mortality of 0.05. (iii) The force of interest is 0.07. Calculate the amount needed in the investment fund on January 1, 2001, so that the probability, as determined by the normal approximation, is 0.95 that the fund will be sufficient to provide these benefits. You are given that the 95th percentile of the standard normal distribution is 1.645. Solution We have m 210ðmþdÞ 5 21:2 EðZÞ ¼ 1000 e ¼ 1000 ¼ 125:5; e mþd 12

100

Chapter 3

and VarðZÞ ¼ 10002

m e210ðmþ2dÞ 2 m þ 2d

5 12

2

e22:4

¼ 23 610:16:

As S¼

400 X

Z;

i¼1

we have EðSÞ ¼ 400 EðZÞ ¼ 50 200; and VarðSÞ ¼ 400 VarðZÞ ¼ 9 444 064: The amount needed is the 95th percentile of the distribution of S pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1:645 · 9 444 064 þ 50 200 ¼ 55 255:

A

Exercise 3.9.5 May 2000 SOA/CAS Course 3 Examination, Problem No. 17. The future lifetimes of a certain population can be modeled as follows: (i) Each individual’s future lifetime is exponentially distributed with constant hazard rate u: (ii) Over the population, u is uniformly distributed over ð1; 11Þ: Calculate the probability of surviving to time 0.5 for an individual randomly selected at time 0. Solution We use the Fubini Theorem for the appropriate double integral: ð1 ð 1 ð 11 PrðT . 0:5Þ ¼ fT ðtÞ dt ¼ fT ðtlQ ¼ uÞ · fQ ðuÞ du dt 0:5 0:5 u¼1 ð 11 ð 1 ð 1 ð 11 1 · du dt ¼ u e2ut · u e2ut dt · 0:10 du ¼ 10 0:5 t¼0:5 u¼1 u¼1 0 1 ð 11 25:5 20:5 2e Be C A ¼ ðe20:5u Þ · 0:10 du ¼ 0:10 · @ A ¼ 0:1205: 1 u¼1 2 2 Exercies 3.9.6 May 2000 SOA/CAS Course 3 Examination, Problem No. 21. A risky investment with a constant rate of default will pay principal and accumulated interest at 16% compounded annually at the end of 20 years if it does not

101


default, and zero if it defaults. A risk-free investment will pay principal and accumulated interest at 10% compounded annually at the end of 20 years. The principal amounts of the two investments are equal. The actuarial present values of the two investments are equal at time zero. Calculate the median time until default or maturity of the risky investment. Solution Let p be the constant annual rate of survival from default for the risky investment. Then its probability of survival over 20 years is p ¼ p20 : Therefore, the actuarial present value of a unit invested in the risky investment is

p · 1:1620 · 1:10220 þ ð1 2 pÞ · 0 ¼

1:16 1:10

20

p20 :

The actuarial present value of a unit investment in the risk-free investment is 1 and the two quantities are equal; therefore p ¼ 1:10=1:16: The problem effectively assumes constant force of default, i.e., exponential distribution of time to default for times between 0 and 20 and then the residual probability of paying the maturity amount at time 20. What is the median of that distribution? If the cumulative probability of default is more than 0.50, it is somewhere between 0 and 20, otherwise it is at 20. The constant force of default is m ¼ lnð1:16=1:10Þ and the median m; if it falls below 20, satisfies 1:10 m 1 e2mm ¼ ¼ : 1:16 2 Luckily, this equation has a nice solution: m¼

ln 2 ¼ 13:0512043: ln 1:16 2 ln 1:10

A

3.10 Life annuities We will consider series of payments lasting till the end of life of the insured or in some form dependent on the survival of the insured. Such streams of payments are termed life annuities. Annuities may be paid not only over the physical life of the insured, but also as long as the insured is a participant in an insured group of some type (e.g., a pension plan). For a unit annuity, the amount of payment is a monetary unit. They play a double role in the study of insurance. First, annuity policies pay benefits in this form. Secondly, life insurance premiums are paid by the insured in a form of such an annuity. The actuarial present value or single benefit premium (also called net premium) of an annuity is the expected present value of the stream of payments. If an annuity is payable as long as the insured

102

Chapter 3

lives, but no longer than a prescribed term (e.g., n years) we term it a temporary life annuity. Let T; as always, denote the future lifetime of an insured aged x (this future lifetime could be his/her length of future employment service towards pension plan benefits accrual), K ¼ bTc; and let Y be the present value of a unit life annuity payable as long as the insured survives. This could be a continuous, or discrete, annuity. For the discrete model, we will usually assume that this is an annuity due. For such a model we have Y ¼ 1 þ v þ · · · þ vK ¼ a€ Kþ1 :

ð3:10:1Þ

Recall that K is a random variable with probability distribution: PrðK ¼ kÞ ¼ k px · qxþk ;

k ¼ 0; 1; 2; …

ð3:10:2Þ

The actuarial present value of this life annuity due is denoted by a€ x and it equals: a€ x ¼ EðYÞ ¼

þ1 X

a€ kþ1 · k px · qxþk :

ð3:10:3Þ

vk I{k#K } ;

ð3:10:4Þ

k¼0

Since Y¼

1 X k¼0

where ( I{k#K } ¼

1;

k # K;

0;

k . K;

ð3:10:5Þ

then a€ x ¼ EðYÞ ¼

þ1 X

vk · PrðK $ kÞ ¼

k¼0

þ1 X

v k · k px :

ð3:10:6Þ

k¼0

Recalling the definition of the commutation functions (3.8.3) and (3.8.4) we arrive at the following formula: þ1 X

a€ x ¼

þ1 X k¼0

v

k lxþk

lx

þ1 xþk X v lxþk ¼ ¼ vx lx k¼0

þ1 X

vxþk lxþk

k¼0 x

v lx

¼

Dxþk

k¼0

Dx

¼

Nx : Dx

ð3:10:7Þ

Example 3.10.1 Consider a life annuity due payable once a year for a Polish man aged x ¼ 65 in the amount of 12 000 per annum. Let the valuation rate be i ¼ 0:05:

103


Mortality table for Poland gives Nx ¼ 24 896:14 and Dx ¼ 2676:52 for male aged 65 (see [68]). Calculate the single benefit premium for this annuity for the Polish man. From Eq. (3.10.7) we get the following net single premium for the annuity for the man: 12 000

Nx ¼ 111 620:20: Dx

Assume also that for a Polish woman aged x ¼ 60; with the same valuation interest rate, Nx ¼ 60 742:84 and Dx ¼ 4774:29 based on the same table. The actuarial present value of a life annuity, in the same amount as the one for the man, for the woman is

12 000

Nx ¼ 152 674:86: Dx

A

Recall that 1 2 vKþ1 : d

a€ Kþ1 ¼ Therefore a€ x ¼

1 2 Ax : d

ð3:10:8Þ

Furthermore, we also have the following variance relationship:

Varð€a Kþ1

1 2 vKþ1 Þ ¼ Var d

2

¼

Ax 2 A2x : d2

ð3:10:9Þ

Similar variance relationships between life annuities and the corresponding life insurances also hold for continuous and modal (i.e., paid m times a year) annuities. We also have the following recursion formula a€ xþ1 ¼ To prove (3.10.10) note that a€ xþ1 ¼

1 X

kþ1 px

vk k pxþ1 ¼

k¼0

¼

1 ð€a 2 p Þ: vpx x 0 x

a€ x 2 1 : vpx

ð3:10:10Þ

¼ px · k pxþ1 and 1 1 1 X 1 X vk kþ1 px ¼ vi p px k¼0 vpx i¼1 i x

ð3:10:11Þ

104

Chapter 3

Consider now a life annuity due payable m times a year, in the amount of 1=m: The present value of the such a stream of payments is written as Y ðmÞ ¼

1 v1=m v2=m vKþL=m þ þ þ ··· þ ; m m m m

ð3:10:12Þ

where K is the number of whole years lived, and L the number of mth of the K þ 1st year in which a payment is made. Therefore Y ðmÞ ¼

1 2 ðv1=m ÞKmþLþ1 : mð1 2 v1=m Þ

ð3:10:13Þ

Since mSðmÞ ¼ L þ 1;

ð3:10:14Þ

d ðmÞ ¼ mð1 2 v1=m Þ;

ð3:10:15Þ

and

then ðmÞ

Y ðmÞ ¼ ðmÞ

Thus the single benefit premium a€ x

1 2 vKþS d ðmÞ

for this life annuity equals ðmÞ

a€ ðmÞ ¼ EðY ðmÞ Þ ¼ x

ð3:10:16Þ

:

ðmÞ

1 2 EðvKþS Þ 1 2 Ax ¼ dðmÞ d ðmÞ

;

ð3:10:17Þ

where the last step uses (3.9.18). Now let us turn our attention to the UDD assumption about the year of death. We ðmÞ will use it to find the relationship between a€ x and a€ x : From (3.10.17) it follows that ðmÞ

a€ ðmÞ x

1 2 Ax ¼ dðmÞ ¼ a€ x

i 1 i ¼ 1 2 ðmÞ Ax ¼ ðmÞ 1 2 ðmÞ ð1 2 da€ x Þ UDD d ðmÞ i d i 1

id i 2 iðmÞ 2 : iðmÞ dðmÞ iðmÞ d ðmÞ

ð3:10:18Þ

Let us denote the coefficient of a€ x by aðmÞ in (3.10.18) and the free term by 2bðmÞ: It can be shown the Taylor series expansions of the coefficients a and b with respect to d are of the following form:

aðmÞ ¼ 1 þ

m2 2 1 2 d þ ··· 12m2

ð3:10:19Þ

105


and

bðmÞ ¼

m21 m2 2 1 þ d þ ··· 2m 6m2

ð3:10:20Þ

From the above and (3.10.18) we get the following approximation a€ ðmÞ < a€ x 2 x

m21 : 2m

ð3:10:21Þ

We will now turn our attention to the continuous model. Let us now consider a series of payments made continuously and uniformly, with a total of one monetary unit paid over Ð the whole year. The nominal value of all payments made from time 0 till time t is t0 dt ¼ t; and the present value, denoted by a t ; equals ðt ðt a t ¼ exp 2 dðsÞ ds dt; ð3:10:22Þ 0

0

where dðsÞ is the force of interest equivalent to the discount factor used. Let T be the random future lifetime of a life age ðyÞ: Our standard assumption is that T has an absolutely continuous distribution with density fT ðtÞ: The expected present value of a life annuity paid continuously to this person is denoted by a y and it equals: ðt ð1 ð 1 ð t a y ; Eða T Þ ¼ a t fT ðtÞ dt þ exp 2 dðsÞ ds dt fT ðtÞ dt: ð3:10:23Þ 0

0

0

0

If the force of interest dð·Þ is a constant function of time and it equals d; then ð1 ð1 1 1 a y ¼ ð1 2 e2dt Þf ðtÞ dt ¼ ð1 2 e2dt Þt py · mðy þ tÞ dt; ð3:10:24Þ 0 d 0 d where the last step uses (3.7.9). Recalling the formula for the single benefit premium for the continuous whole life insurance (3.9.15) we obtain ð ð 1 1 1 2dt 1 1 1 t a y ¼ 2 e t py · mðy þ tÞ dt ¼ 2 v p · mðy þ tÞ dt d d 0 d d 0 t y ¼

1 1 2 A: d d y

ð3:10:25Þ

Let us now assume that the annuity payment is made continuously with intensity hðxÞ; such intensity being dependent on the age of the insured. The present value of such a life annuity, calculated at the moment when the person reaches retirement age y; will be denoted by a hy : We then have ðt ð 1 ð t h a y ¼ exp 2 dðsÞ ds hðy þ tÞ dt fT ðtÞ dt; ð3:10:26Þ 0

0

0

where fT ð·Þ is the density of the random variable T; future lifetime of a random person aged y: By integrating the right-hand side of (3.10.26) by parts, we get the following

106

Chapter 3

equivalent formula: ðt !1 ðt h a y ¼ 2 exp 2 dðsÞ ds hðy þ tÞ dt · ð1 2 FT ðtÞÞ 0 0 0 ðt ð1 þ exp 2 dðsÞ ds hðy þ tÞð1 2 FðtÞÞ dt 0 0 ðt ð1 ¼ exp 2 dðsÞ ds hðy þ tÞð1 2 FðtÞÞ dt: 0

ð3:10:27Þ

0

We should note that the formulas (3.10.26) and (3.10.27) are valid only under the assumption that the integral on the right-hand side of (3.10.26) is finite. Since 1 2 FT ðtÞ ¼

lyþt ; ly

(3.10.27) can be written in the following form: ðt ð1 lyþt h a y ¼ exp 2 dðsÞ ds hðy þ tÞ dt: ly 0 0

ð3:10:28Þ

ð3:10:29Þ

If we denote by w the age at which a given person joins a pension plan and by sðxÞ ¼x2w pw the survival function within the pension plan, then ðt ð1 sðy þ tÞ dt: ð3:10:30Þ a hy ¼ exp 2 dðsÞ ds hðy þ tÞ sðyÞ 0 0 In the case when the force of interest is constant and equal to d we get: ð1 ð1 sðy þ tÞ sðxÞ h 2dt dt ¼ e2dðx2yÞ hðxÞ dx; a y ¼ e hðy þ tÞ sðyÞ sðyÞ 0 y where x ¼ y þ t is the age of the person t years after retiring. A temporary life annuity is an annuity, which is payable only for a specified contract term, as long as the insured is still alive. The frequency of the payment, just as with life annuities, may be annual, m times a year (monthly, quarterly, etc.), or continuous. Temporary life annuities can be treated as regular life annuities with a modified mortality table—in the modified table the probability of dying would be equal to one at the end of the term of the annuity. For the purposes of this work, however, it is more appropriate to consider separately the ending of plan participation due to death, or withdrawal, or disability, or some other reason. Thus, without modifying the mortality tables in any form, let us denote by a€ x: n the single benefit premium for an n-year temporary unit life annuity due for a life age ðxÞ: The random present value of this annuity’s benefit payments, Y; equals: ( a€ Kþ1 ; for K # n 2 1 Y¼ ð3:10:31Þ a€ n ; for K $ n:

107


Therefore a€ x: n ¼ EðYÞ ¼

n21 X

a€ kþ1 k px · qxþk þ a€ n > n px :

ð3:10:32Þ

k¼0

By introducing a new random variable K 0 ¼ KI{K,n} þ ðn 2 1ÞI{K,n} ;

ð3:10:33Þ

we obtain Y¼

1 X

vk I{k#K 0 } :

ð3:10:34Þ

k¼0

From the above and from (3.10.32) we infer that a€ x: n ¼

1 X

vk PrðK 0 $ kÞ ¼

k¼0

n21 X

vk PrðK $ kÞ ¼

k¼0

n21 X

vk k px :

ð3:10:35Þ

k¼0

Recalling the definitions of the commutation functions (3.8.3) and (3.8.4), we can also write: a€ x: n ¼

1 X

v k k px 2

k¼0

¼

1 X

v k k px ¼

k¼n

1 1 X X vxþk lxþk vxþk lxþk 2 x v lx vx l x k¼0 k¼n

Nx 2 Nxþn : Dx

ð3:10:36Þ

Before we proceed to the exercises, let us note the values of the annuities single benefit premium in the two simple mortality cases: constant force and De Moivre’s Law. If the force of mortality is constant and equal to m (and the force of interest is constant and equal to d) then 1 2 A x a x ¼ ¼ d

12

m 1 mþd ¼ : mþd d

ð3:10:37Þ

For the discrete life annuity due, we have 1 2 Ax ¼ a€ x ¼ d

12

q 1þi qþi : ¼ qþi d

ð3:10:38Þ

When De Moivre’s Law holds, we have a 1 2 v2x 1 2 A x v 2 x ¼ ðv 2 xÞ 2 a v2x ¼ ðDaÞ v2x ; a x ¼ ¼ d d dðv 2 xÞ v2x

ð3:10:39Þ

108

Chapter 3

and in the discrete case: a 1 2 v2x 1 2 Ax v 2 x ¼ ðv 2 xÞ 2 a v2x ¼ ðD€aÞ v2x : ¼ a€ x ¼ d d dðv 2 xÞ v2x

ð3:10:40Þ

Exercises Exercise 3.10.1 November 2003 SOA/CAS Course 3 Examination, Problem No. 9. For an annuity payable semi-annually, you are given: (i) Deaths are uniformly distributed over each year of age. (ii) q69 ¼ 0:03: (iii) i ¼ 0:06: (iv) 1000A 70 ¼ 530: ð2Þ

Calculate a€ 69 : Solution We know that ð2Þ

a€ 69 ¼ að2Þ€a69 2 bð2Þ; UDD

where i i id 1 þ i pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 1:0002163; að2Þ ¼ ð2Þ ð2Þ ¼ pffiffiffiffiffiffiffiffiffiffi i d 2ð 1 þ i 2 1Þ · 2ð1 2 1=ð1 þ iÞÞ and pffiffiffiffiffiffiffiffiffiffi i 2 ið2Þ i 2 2ð 1 þ i 2 1Þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 0:2573991: bð2Þ ¼ ð2Þ ð2Þ ¼ pffiffiffiffiffiffiffiffiffiffi i d 2ð 1 þ i 2 1Þ · 2ð1 2 1=ð1 þ iÞÞ We also know that a€ 69 ¼

1 2 A69 : d

Observe that A69 ¼ vq69 þ vp69 A70 ¼

q69 p 0:03 0:97 þ A : þ 69 A70 ¼ 1:06 1:06 70 1þi 1þi

Furthermore A70 ¼

UDD

d ln 1:06 0:53 ¼ 0:5147087: A70 ¼ 0:06 i

109


This gives A69 ¼ 0:4993089; and a€ 69 ¼

1 2 0:4993089 ¼ 8:8455075: 0:06 1:06

Finally ð2Þ

a€ 69 ¼ að2Þ€a69 2 bð2Þ ¼ 1:0002163 · 8:8455075 2 0:2573991 < 8:59: UDD

A

Exercise 3.10.2 November 2003 SOA/CAS Course 3 Examination, Problem No. 15. For a group of individuals all aged x; of which 30% are smokers and 70% are nonsmokers, you are given (i) d ¼ 0:10: (ii) A Smoker ¼ 0:444: x (iii) A Nonsmoker ¼ 0:286: x (iv) T is the future lifetime of ðxÞ: (v) VarðaSmoker Þ ¼ 8:818: T (vi) VarðaNonsmoker Þ ¼ 8:503: T Calculate Varða T Þ for an individual chosen at random from this group. Solution Let T be the future lifetime of an individual randomly chosen from this population. We have Ax ¼ EðvT Þ ¼ 0:30A Smoker þ 0:70A Nonsmoker ¼ 0:30 · 0:444 þ 0:70 · 0:286 ¼ 0:3334: x x Also VarðaSmoker Þ ¼ 8:818 ¼ T

2 Smoker Ax 2 ðA Smoker Þ x 2

d

and hence 2 Smoker Ax

¼ 0:285316:

¼

2 Smoker Ax 2 0:4442 ; 0:102

110

Chapter 3

Furthermore, VarðaNonsmoker Þ ¼ 8:503 ¼ T

2 Nonsmoker 2 ðA Nonsmoker Þ2 Ax x 2

d

¼

2 Nonsmoker 2 0:2862 Ax ; 0:102

and 2 Nonsmoker Ax

¼ 0:166826:

Hence 2

Ax ¼ 0:30 · 0:285316 þ 0:70 · 0:166826 ¼ 0:202373:

Finally Varða T Þ ¼

Ax 2 ðA x Þ2 ¼ 9:121744: d2

2

A

Exercise 3.10.3 November 2003 SOA/CAS Course 3 Examination, Problem No. 32. Your company currently offers a whole life annuity product that pays the annuitant 12 000 at the beginning of each year. A member of your product development team suggests enhancing the product by adding a death benefit that will be paid at the end of the year of death. Using a discount rate, d; of 8%, calculate the death benefit that minimizes the variance of the present value random variable of the new product. Solution Let B be the death benefit amount. Then the random present value of the total payment made by this product is

12 000

1 2 vKþ1 þ BvKþ1 ¼ 150 000 þ ðB 2 150 000ÞvKþ1 : 0:08

Thus, the variance of the present value random variable is ðB 2 150 000Þ2 VarðvKþ1 Þ: This is minimized (it is zero) when B ¼ 150 000: Why do we get this strange answer of zero variance? Because 150 000 is the present value of a perpetuity of 12 000 at d ¼ 8%; and this new product becomes effectively such a perpetuity, but paid till death, with perpetuity value being paid upon death to the heirs. In a sense, the product is no longer random, it becomes deterministic. A

111


Exercise 3.10.4 November 2002 SOA/CAS Course 3 Examination, Problem No. 4. You are given that (i) mðx þ tÞ ¼ 0:01; 0 # t , 5; (ii) mðx þ tÞ ¼ 0:02; t $ 5; (iii) d ¼ 0:06: Calculate a x : Solution We have a x ¼

ð1 0

e2dt t px dt:

Thus we need to use the information about the force to find t px : We have 8 20:01t ; 0#t,5 ðt > <e ðt p ¼ exp 2 m ðsÞ ds ¼ t x x > 0 : e20:05 · exp 2 0:02 ds ; t $ 5 5 8 20:01t <e ; 0#t,5 ¼ : e20:05 · e20:02ðt25Þ ; t$5 8 < e20:01t ; 0#t,5 ¼ : e0:05 · e20:02t ; t$5 and a x ¼

ð1 0

e2dt t px dt ¼

¼

2

ð5

e20:06t · e20:01t dt þ

ð1

0

1 20:35 1 e þ 0:07 0:07

e20:06t · e0:05 · e20:02t dt

5

þ e0:05 20 þ

1 20:40 e 0:08

¼ 13:0273426969309: A

Exercise 3.10.5 November 2002 SOA/CAS Course 3 Examination, Problem No. 31. For a 20-year deferred whole life annuity due of 1 per year on (45), you are given: (i) Mortality follows De Moivre’s Law with v ¼ 105: (ii) i ¼ 0%: Calculate the probability that the sum of the annuity payments actually made will exceed the actuarial present value at issue of the annuity.

112

Chapter 3

Solution Note that v 2 45 2 1 ¼ 59; and vk ¼ 1 for any k: Therefore a45 20 l€

¼

59 X k¼20

k px

¼

59 59 59 X X 60 2 k 1 X 41 ¼ ¼ 13:666… 12 k¼ 60 60 3 k¼20 k¼20 k¼20

The sum of the annuity payments will exceed the actuarial present value at issue of the annuity if there are 14 payments, i.e., if the (45) insured survives the 20-year deferral plus 13 more years (as the annuity payments are made at the beginning of each year). What is the probability of that? It is 60 2 33 ¼ 0:45: A 33 p45 ¼ 60 Exercise 3.10.6 November 2001 SOA/CAS Course 3 Examination, Problem No. 3. For a continuous whole life annuity of 1 on ðxÞ: (i) TðxÞ is the future lifetime random variable for ðxÞ: (ii) The force of interest and force of mortality are constant and equal. (iii) a x ¼ 12:50: Calculate the standard deviation of a TðxÞ . Solution Since m ¼ d and 12:50 ¼ a x ¼

1 ; mþd

then m ¼ d ¼ 0:04: From the standard annuity variance formula: 2 2 2 1 1 m m 2 2 Varða TðxÞ Þ ¼ ð Ax 2 ðAx Þ Þ ¼ 2 m þ 2d mþd d d 2 1 0:04 0:04 2 2 ¼ ¼ 52:08: 0:04 0:12 0:08 pffiffiffiffiffiffiffiffiffiffiffi Hence, the standard deviation is 52:08 ¼ 7:22:

A

Exercise 3.10.7 November 2001 SOA/CAS Course 3 Examination, Problem No. 17. For a group of individuals all aged x; you are given: (i) 30% are smokers and 70% are nonsmokers. (ii) The constant force of mortality for smokers in 0.06. (iii) The constant force of mortality for nonsmokers is 0.03. (iv) d ¼ 0:08:

113


Calculate Varða TðxÞ Þ for an individual chosen at random from this group. Solution Let us use the superscript n for nonsmokers and s for smokers. Because of the mixed population and constant force of mortality, moments for the related whole life insurance random variable Z ¼ vT can be calculated as follows: EðZ k Þ ¼ k A x ¼ 0:30 ·

ms mn þ 0:70 · n ; m þ kd m þ kd s

where ms ¼ 0:06; mn ¼ 0:03; and d ¼ 0:08: For k ¼ 1; 2 this results in A x ¼ 0:319481; and 2 A x ¼ 0:192344: From a standard annuity variance formula Varða TðxÞ Þ ¼

1 2 1 ð Ax 2ðA x Þ2 Þ ¼ · ð0:19234420:3194812 Þ ¼ 14:11: 2 0:082 d

A

Exercise 3.10.8 November 2001 SOA/CAS Course 3 Examination, Problem No. 26. You are given: Ax ¼ 0:28; Axþ20 ¼ 0:40; A1x: 20 ¼ 0:25; and i ¼ 0:05: Calculate ax: 20 : Solution We have A1x: 20 ¼ Ax 220lAx ¼ Ax 2 A1x: 20 · Axþ20 ¼ 0:28 2 0:25 · 0:40 ¼ 0:18: and a€ x: 20 ¼

1 2 Ax: 20 ¼ d

1 2 A1x: 20 þ A1x: 20 i 1þi

¼

1:05 · ð1 2 0:18 2 0:25Þ 0:05

¼ 21 · 0:57 ¼ 11:97: Finally, ax: 20 ¼ a€ x: 20 þ A1x: 20 2 1 ¼ 11:97 þ 0:25 2 1 ¼ 11:22:

A

Exercise 3.10.9 May 2001 SOA/CAS Course 3 Examination, Problem No. 24. For a disability insurance claim (i) The claimant will receive payments at the rate of 20 000 per year, payable continuously as long as she remains disabled. (ii) The length of the payment period in years is a random variable with the gamma distribution with parameters a ¼ 2 and u ¼ 1: (iii) Payments begin immediately. (iv) d ¼ 0:05:

114

Chapter 3

Calculate the actuarial present value of the disability payments at the time of disability. Solution The actuarial present value of this disability annuity is ð1 ð1 ð 1 2 e20:05t t e2t 20000 1 20000 a t fT ðtÞ dt ¼ 20000 ðt e2t 2 t e21:05t Þ dt dt ¼ 0:05 0 0:05 Gð2Þ 0 0 ð 20000 1 20000 1:05t þ 1 21:05t 1 2t 21:05t 2t 2ðt þ 1Þe þ ¼ ðt e 2 t e Þdt ¼ e 0:05 0 0:05 1:052 0 20000 1 12 ¼ ¼ 37188: A 0:05 1:052 Exercise 3.10.10 November 2000 SOA/CAS Course 3 Examination, Problem No. 3. A person aged 40 wins 10 000 in the actuarial lottery. Rather than receiving the money at once, the winner is offered the actuarially equivalent option of receiving an annual payment of K (at the beginning of each year) guaranteed for 10 years and continuing thereafter for life. You are also given that i ¼ 0:04; A40 ¼ 0:30; A50 ¼ 0:35; and A140: 10 ¼ 0:09: Calculate K: Solution Since the actuarial present value (APV) of an annuity due to ðxÞ of one per year with 10 payments guaranteed and the rest contingent on survival is a€ 10 þ 10 l€ax ; K is determined from the APV relation We have

10 000 ¼ Kð€a 10 þ a€ 10 ¼

Furthermore,

ax Þ: 10 l€

ð1 2 1:04210 Þ ¼ 8:4353: 0:04 1:04

0:21 ¼ 0:30 2 0:09 ¼ A40 2 A140: 10 ¼

10 lA40

¼ v10 · 10 p40 · A50 ¼ v10 ·

Therefore v10 ·

10 p40

¼

0:21 : 0:35

Since a€ 50 ¼

1 2 A50 1 2 0:35 ¼ ¼ 16:90; 0:04 d 1:04

10 p40

· 0:35:

115


it follows that a40 10 l€

¼ v10 ·

10 p40

· a€ 50 ¼

0:21 · 16:90 ¼ 10:14: 0:35

Substituting these results into the first equation gives 10 000 ¼ Kð€a 10 þ10 l€ax Þ ¼ Kð8:4353 þ 10:14Þ;

K ¼ 538:35:

A

Exercise 3.10.11 November 2000 SOA/CAS Course 3 Examination, Problem No. 20. Y is the present value random variable for a special 3-year temporary life annuity due on ðxÞ: You are given that t px ¼ 0:9t ; K is the curtate-future lifetime of ðxÞ; and that 8 1:00; K ¼ 0; > > < Y ¼ 1:87; K ¼ 1; > > : 2:72; K ¼ 2; 3; … Calculate VarðYÞ: Solution We have PrðK ¼ 0Þ ¼ 1 2 px ¼ 0:10; PrðK ¼ 1Þ ¼ 1 px 2 2 px ¼ 0:90 2 0:81 ¼ 0:09; PrðK . 1Þ ¼ 2 px ¼ 0:81; EðYÞ ¼ 0:10 · 1 þ 0:09 · 1:87 þ 0:81 · 2:72 ¼ 2:4715; EðY 2 Þ ¼ 0:10 · 12 þ 0:09 · 1:872 þ 0:81 · 2:722 ¼ 6:407; VarðYÞ ¼ 6:407 2 2:47152 ¼ 0:299:

A

Exercise 3.10.12 November 2000 SOA/CAS Course 3 Examination, Problem No. 26. A fund is established to pay annuities to 100 independent lives age x: Each annuitant will receive 10 000 per year continuously until death. You are given d ¼ 0:06; A x ¼ 0:40; 2 A x ¼ 0:25: Calculate the amount needed in the fund so that the probability, using the normal approximation, is 0.90 that the fund will be sufficient to provide the payments.

116

Chapter 3

Solution For each person Y ¼ 10 000a TðxÞ is the random present value of that person’s annuity. If Y1 ; Y2 ; …; Y100 are independent and identically distributed as Y; then YAGG ¼ Y1 þ · · · þ Y100 ; is the aggregate present value. Assuming that YAGG is approximately normal in distribution, we can compute the fund needed as the 90th percentile as pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi EðYAGG Þ þ 1:282 VarðYAGG Þ: We have EðYAGG Þ ¼ 100 EðYÞ; VarðYAGG Þ ¼ 100 VarðYÞ; 1 2 0:40 ¼ 100 000; EðYÞ ¼ 10 000ax ¼ 10 000 · 0:06 2 0:25 2 0:16 A x 2 A 2x VarðYÞ ¼ 10 0002 ¼ 25 · 108 : ¼ 108 · 2 0:062 d The fund required is EðYAGG Þ þ 1:282

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi VarðYAGG Þ ¼ 100 · 100 000 þ 1:282 100 · 25 · 108

¼ 10 000 000 þ 1:282 · 500 000 ¼ 10 641 000:

A

Exercise 3.10.13 November 2000 SOA/CAS Course 3 Examination, Problem No. 35. For a special 30-year deferred annual whole life annuity due of 1 on (35): (i) If death occurs during the deferral period, the single benefit premium is refunded without interest at the end of the year of death, (ii) a€ 65 ¼ 9:90; (iii) A35: 30 ¼ 0:21; (iv) A135: 30 ¼ 0:07: Calculate the single benefit premium for this special deferred annuity. Solution If P is the single benefit premium, then the actuarial present value of the refund of P without interest if death occurs during the deferral period is PA135: 30 : Hence P is the solution of the equation: P ¼ P · A135: 30 þ v30 ·

30 p35

· a€ 65 :

117


From the information given, v30 · 30 p35 ¼ A135: 30 ¼ A35: 30 2 A135: 30 ¼ 0:21 2 0:07 ¼ 0:14: Hence P¼

v30 · 30 p35 · a€ 65 0:14 · 9:90 ¼ 1:49: ¼ 1 2 0:07 1 2 A135: 30

A

Exercise 3.10.14 May 2000 SOA/CAS Course 3 Examination, Problem No. 29. For a whole life annuity due of 1 on ðxÞ; payable annually qx ¼ 0:01; qxþ1 ¼ 0:05; i ¼ 0:05; and a€ xþ1 ¼ 6:951: Calculate the change in the actuarial present value of this annuity due if pxþ1 is increased by 0.03. Solution We have 6:951 ¼ a€ xþ1 ¼ 1 þ v · pxþ1 · a€ xþ2 ¼ 1 þ

0:95 a€ ; 1:05 xþ2

and therefore a€ xþ2 ¼ 6:57742: Using this we calculate a€ x ¼ a€ x: 2 þ v2 · px · pxþ1 · a€ xþ2 ; ðnewÞ

¼ a€ x: 2 þ v2 · px · pxþ1 · a€ xþ2 : a€ ðnewÞ x The difference is equal to v2 · px · 0:03 · a€ xþ2 ¼

1 1:05

2 · 0:99 · 0:03 · 6:57742 ¼ 0:1772:

A

Exercise 3.10.15 May 2000 SOA/CAS Course 3 Examination, Problem No. 39. For a continuous whole life annuity of 1 on ðxÞ: (i) TðxÞ; the future lifetime of ðxÞ; has a constant force of mortality 0.06. (ii) The force of interest is 0.04. Calculate Prða TðxÞ . a x Þ:

118

Chapter 3

Solution We have 1 2 vT 1 2 vT 1 2 A x Prða T . a x Þ ¼ Pr . a x ¼ Pr . ¼ Prð2vT . 2A x Þ d d d m 21 m ln ¼ Pr e2dT , ¼ Pr T . ¼ t 0 px ; mþd dþm d where 21 m t0 ¼ ln : dþm d Therefore m=d 6=4 m 6 2 mt 0 ¼ ¼ ¼ 0:4648: A t0 px ¼ e mþd 10 Exercise 3.10.16 November 2002 SOA/CAS Course 3 Examination, Problem No. 39. For a whole life insurance of 1 on ðxÞ; you are given: (i) mðx þ tÞ is the force of mortality. (ii) The benefit is payable at the moment of death. (iii) d ¼ 0:06: (iv) A x ¼ 0:60: Calculate the revised actuarial present value of the benefit of this insurance if mðx þ tÞ is increased by 0.03 for all t and d is decreased by 0.03. Solution ðt ð1 2tð0:0620:03Þ ¼ e exp 2 ð m ðsÞþ0:03Þ ds ðmx ðtÞþ0:03Þ dt A New x x 0

¼

0

ð1 e 0

2t · 0:06

ð t exp 0

ð1 mx ðsÞ ds mx ðtÞ dtþ0:03 e2t

¼ A Old aOld x þ0:03 x ¼0:60þ0:03

0

120:6 ¼0:80: 0:06

· 0:06

ðt exp 2 mx ðsÞ ds dt 0

A

3.11 Life insurance premiums Life insurance and annuities are typically not purchased with one lump sum, but rather with a series of premiums. These premiums are paid only as long as the insured is alive, and therefore are life annuities. What principles can be used for setting

119


premiums? The following three principles have been put forth as possible premium calculation principles [14]: † Percentile premiums. Charge premium large enough to ensure that the company only suffers financial loss with sufficiently low probability. † Benefit premiums. Base the premium level on the condition of the insurer’s loss having the expected value of zero. This is also referred to as the Equivalence Principle (EP). † Exponential premiums. Premium is set as the number P such that using the utility of wealth function uðxÞ ¼ 2e20:1x ; the insurer is indifferent between accepting and not accepting the risk. The most important random variable studied in derivation of the premium is the loss-at-issue of the insurer. The random present value of the excess of outflows of the insurer over the inflows of the insurer for a unit policy, if the premium is paid as an annuity, is L ¼ Z 2 PY; where P is the premium charged by the insurer, Z the unit benefit random present value, and Y the unit annuity random present value. The premium P is set by the insurer, and, as explained above, there may be various principles for setting it. The simplest and the most fundamental principle is the EP: P is chosen so that EðLÞ ¼ 0: This means, of course, that P¼

EðZÞ : EðYÞ

The models for premium combine the mode of death benefit payment (continuous, discrete, or modal) and the mode of premium payment (continuous, discrete, modal). Also note that it is possible to pay the premium for a period shorter than the one for which the coverage is offered. For a fully continuous model, death benefit is paid at the moment of death and premium is paid continuously. For a fully discrete model, death benefit is paid at the end of the year of death and the premium is paid as a life annuity due. For a semi-continuous model, death benefit is paid at the moment of death and premiums are paid as a life annuity due. Premiums are generally assumed to be paid as a level annuity. For a fully continuous model, whole life insurance, we have the following form of the loss at issue function: T

aT ¼ L ¼ v 2 P

P P 1þ vT 2 ; d d

ð3:11:1Þ

where P is the annual level of the premium paid as a continuous life annuity. We have ax ; EðLÞ ¼ A x 2 P

ð3:11:2Þ

120

Chapter 3

and VarðLÞ ¼

P 2 2 1þ ð Ax 2 A 2x Þ: d

ð3:11:3Þ

For this continuous model, the EP premium is A x Þ ¼ Ax ¼ 1 2 d ¼ dAx : Pð a x a x 1 2 A x

ð3:11:4Þ

Furthermore, the variance of the loss at issue under the EP premium is VarðLÞ ¼

2 Ax 2 A 2x Ax 2 A 2x ¼ : 2 ðda x Þ ð1 2 A x Þ2

2

ð3:11:5Þ

We have analogous formulas for the fully discrete whole life insurance. The loss at issue random variable is P Kþ1 P ð3:11:6Þ L ¼ vKþ1 2 P€a Kþ1 ¼ 1 þ 2 : v d d The expected value of the loss is EðLÞ ¼ Ax 2 P€ax ; and the variance of the loss at issue is P 2 2 ð Ax 2 A2x Þ: VarðLÞ ¼ 1 þ d

ð3:11:7Þ

ð3:11:8Þ

The EP premium (i.e., benefit premium) is Px ¼

Ax 1 dAx ¼ 2d ¼ ; a€ x a€ x 1 2 Ax

ð3:11:9Þ

and the variance of the loss at issue under the EP premium is 2

VarðLÞ ¼

2 Ax 2 A2x Ax 2 A2x ¼ : ðda€ x Þ2 ð1 2 Ax Þ2

ð3:11:10Þ

The constant force model has some interesting implications for the life insurance premium calculation. We have

m x A m þd A x Þ ¼ ¼ ¼ m; Pð 1 a x CF mþd

ð3:11:11Þ

121


and q Ax qþi ¼ vq ¼ A1x: 1 : Px ¼ ¼ a€ x CF 1 þ i qþi

ð3:11:12Þ

Also, under the constant force assumption, with the EP premiums, the variance of the loss at issue function has a very simple form. For the fully continuous model it is 2 m m m ð m þ dÞ 2 2 2 m2 2 2 Ax 2 A x m þ 2d mþd m þ 2d ¼ ¼ VarðLÞ ¼ 2 d2 ð1 2 A x Þ2 CF m 12 mþd m þ 2d md2 m3 þ 2m2 d þ md2 m2 2 m2 þ 2 m2 m þ 2d m þ 2d m þ 2d ¼ ¼ d2 d2 m ¼ ¼ 2 A x ; ð3:11:13Þ m þ 2d and for the fully discrete model it is VarðLÞ ¼ p · 2Ax : CF

ð3:11:14Þ

Exercises Exercise 3.11.1 November 2003 SOA/CAS Course 3 Examination, Problem No. 31. For a block of fully discrete whole life insurances of one on independent lives age x; you are given i ¼ 0:06; Ax ¼ 0:24905; 2 Ax ¼ 0:09476; and p ¼ 0:025; where p is the contract premium for each policy. You are also given that losses are based on the contract premium. Using the normal approximation, calculate the minimum number of policies the insurer must issue so that the probability of a positive total loss on the policies issued is less than or equal to 0.05. Solution The individual loss at issue is L ¼ vKþ1 2 pa€ Kþ1 ¼

173 Kþ1 53 v : 2 120 120

Therefore VarðLÞ ¼

173 120

2 ·ð

2

Ax 2 A2x Þ ¼

173 120

2

ð0:09476 2 0:279052 Þ < 0:06803464;

122

Chapter 3

and

sL < 0:26083451; where sL is the standard deviation of the individual loss at issue. The total loss on n policies is

L ¼ L1 þ ·· · þ Ln ; with EðLÞ ¼ 20:0826196n; and pffiffiffi VarðLÞ ¼ 0:26083451 n: Using the normal approximation, we estimate the probability of a positive total loss as

L 2 EðLÞ 0 þ 0:0826196n pffiffiffi . sL 0:26083451 n ¼ 0:05:

PrðL . 0Þ ¼ Pr

> < 10 000v2 2 p€a 2 ¼ 3178:80; 0L ¼ > > : 10 000v3 2 p€a 3 ¼ 283:52;

for K ¼ 0; for K ¼ 1; for K $ 2:

Furthermore PrðK ¼ 0Þ ¼ q50 ¼ 0:00832; PrðK ¼ 1Þ ¼ p50 · q51 ¼ 0:99168 · 0:00911 ¼ 0:0090342; PrðK $ 2Þ ¼ 1 2 PrðK ¼ 0Þ 2 PrðK ¼ 1Þ ¼ 0:98265;

137


and Eð0 LÞ ¼ 0; since p is the benefit premium. Therefore, we can calculate the variance of 0 L as Varð0 LÞ ¼ Eð0 L2 Þ ¼ 0:00832 · 65392 þ 0:00903 · 3178:802 þ 0:98265 · ð283:52Þ2 ¼ 453 895:

A

Exercise 3.12.12 May 2000 SOA/CAS Course 3 Examination, Problem No. 3. For a fully discrete 2-year term insurance of 400 on ðxÞ : i ¼ 10%; 400P1x: 2 ¼ 74:33; 400 · 1 Vx:1 2 ¼ 16:58: The contract premium equals the benefit premium. Calculate the variance of the loss at issue. Solution Let us start by finding qx and qxþ1 : We have 400 · 1 Vx:1 2 ¼ 16:58 ¼ 400vqxþ1 2 400P1x: 2 ¼

400 q 2 74:33: 1:10 xþ1

Therefore, qxþ1 ¼ 0:25: Also, 400 · 0 Vx:1 2 ¼ 0 ¼ 400vqx 2 400P1x: 2 þ 400vpx · 1 Vx:1 2 : Hence qx ¼

400P1x: 2 ð1 þ iÞ 2 400 · 1Vx:1 2 ¼ 0:17: 400 2 400 · 1Vx:1 2

The loss random variable is described as follows: 8 400 > > 2 74:33 ¼ 289:30; K ¼ 0; > > 1:10 > > > > < 400 1 2 74:33 1 þ ¼ 188:68; K ¼ 1; L ¼ 2 0 1:10 1:10 > > > > > > > > 274:33 1 þ 1 ¼ 2141:90 K $ 2: : 1:10 Since the contract premium is the benefit premium, the premium is set by the EP, and the variance of the loss at issue is the second moment of the loss. It equals: Varð0 LÞ ¼ ð289:30Þ2 · 0:17 þ ð188:68Þ2 · 0:25 · ð1 2 0:17Þ þ ð141:90Þ2 · ð1 2 0:25Þ · ð1 2 0:17Þ ¼ 31 450:

A

138

Chapter 3

Exercise 3.12.13 May 2000 SOA/CAS Course 3 Examination, Problem No. 24. For a fully discrete whole life insurance with nonlevel benefits on (70): (i) The level benefit premium for this insurance is equal to P50 : (ii) q70þk ¼ q50þk þ 0:01; k ¼ 0; 1; …; 19: (iii) q60 ¼ 0:01368: (iv) k V ¼ k V50 ; k ¼ 0; 1; …; 19: (v) 11 V50 ¼ 0:16637: Calculate b11 ; the death benefit in year 11. Solution For the fully discrete unit whole life insurance issued to (50) the relation between the 10th and 11th reserves is 10 V50

þ P50 ¼ v · q60 · 1 þ v · p60 · 11V50 :

For the variable benefit plan on (70) in this problem, the relation between the 10th and 11th reserve is 10 V

þ P50 ¼ v · q80 · b11 þ v · p80 · 11V;

where b11 is the unknown death benefit in the 11th year. Since 10 V50 ¼ 10 V and 11 V50 ¼ 11 V; we see from the above equations that the left-hand sides are equal. Setting the right-hand sides equal and substituting q60 ¼ 0:01368; q80 ¼ q60 þ 0:01 and 11 V50 ¼ 11 V ¼ 0:16637 results in a linear equation in the unknown b11 : The solution is b11 ¼ 0:648: A Exercise 3.12.14 May 2000 SOA/CAS Course 3 Examination, Problem No. 26 For a fully discrete 3-year endowment insurance of 1000 on ðxÞ : qx ¼ qxþ1 ¼ 0:20; i ¼ 0:06; and 1000Px: 3 ¼ 373:63: Calculate 1000ð2 Vx: 3 2 1Vx: 3 Þ: Solution ð10000 Vx: 3 þ 1000Px: 3 Þð1 þ iÞ ¼ px · 1000 · 1Vx: 3 þ qx · 1000; and ð10001 Vx: 3 þ 1000Px: 3 Þð1 þ iÞ ¼ px · 1000 · 2Vx: 3 þ qx · 1000: The first formula gives us ð1000 · 0 þ 373:63Þ · 1:06 ¼ 0:80 · 1000 · 1Vx: 3 þ 200; so that 10001 Vx: 3 ¼ 245:06:

139


For the second year we have ð245:06 þ 373:63Þ · 1:06 ¼ 0:80 · 1000 · 2Vx: 3 þ 200; and therefore 10002 Vx: 3 ¼ 569:76: The difference is 569:76 2 245:06 ¼ 324:70:

A

3.13 Multiple lives models In the United States, pensions and pension rights are generally considered a common property if one of the spouses of a married couple is a pension plan participant. Pension benefit payment default is a joint-life-and-survivor annuity: such an annuity is paid as long as at least one of the spouses is alive, and pays at least 50% of the original benefit to the surviving spouse if the participant dies. Because of these requirements, it is sometimes important to model annuities and life insurance related to the survival of multiple lives. These models generally consider a new artificial entity, called a status, which is alive as long as one, or both, or all, of several real entities are alive. One artificial entity that is also quite useful, which we have considered already is n : this status is alive for exactly n years, and dies instantly at the end of the n-year period. If x is a person aged x; and y is a person aged y; then x : y denotes the status alive as long as both x and y are alive, while x : y is the status, which is alive as long as at least one of the two lives x or y is alive. In relation to these statuses, we also consider their future lifetime random variables, e.g., 4

Tðx : yÞ ¼ minðTðxÞ; TðyÞÞ;

Tðx : yÞ ¼ maxðTðxÞ; TðyÞÞ:

Let us also note that the semicolon symbol is often dropped in the notation for joint life statuses, as long as the meaning is clear. The most commonly used property of multiple lives models relates to an important literary work of Western Civilization, which we will quote now:

Then he dropped two (fir-cones) in (the river) at once, and leant over the bridge to see which one of them would come out first; and one of them did; but as they were both the same size, he didn’t know if it was the one which he wanted to win, or the other one. So the next time he dropped one big one and one little one, and the big one came out first, which was what he said it would do, and the little one came out last, which was what he had said it would do, so he had won twice … (Milne, [51]).

140

Chapter 3

This led to the creation of the Poohsticks game, and for that reason, we will term the following the Poohsticks Formulas: TðxyÞ þ TðxyÞ ¼ TðxÞ þ TðyÞ;

ð3:13:1Þ

TðxyÞ · TðxyÞ ¼ TðxÞ · TðyÞ;

ð3:13:2Þ

aTðxyÞ þ aTðxyÞ ¼ aTðxÞ þ aTðyÞ ;

ð3:13:3Þ

for any parameter a; t pxy

þ t pxy ¼ t px þ t py ;

ð3:13:4Þ

t qxy

þ t qxy ¼ t qx þ tqy ;

ð3:13:5Þ

and fTðxyÞ ðtÞ þ fTðxyÞ ðtÞ ¼ fTðxÞ ðtÞ þ fTðyÞ ðtÞ:

ð3:13:6Þ

The life expectancies for joint lives are defined as follows: EðTðxyÞÞ ¼ e xy ;

ð3:13:7Þ

EðTðxyÞÞ ¼ e xy ;

ð3:13:8Þ

and for KðxyÞ ¼ bTðxyÞc; KðxyÞ ¼ bTðxyÞc; EðKðxyÞÞ ¼ exy ; EðKðxyÞÞ ¼ exy : We have the following Poohsticks formulas for life expectancies:

ð3:13:9Þ ð3:13:10Þ

e xy þ e xy ¼ e x þ e y ;

ð3:13:11Þ

exy þ exy ¼ ex þ ey :

ð3:13:12Þ

It should be noted that Poohsticks can only be used when the relationship between TðxÞ and TðyÞ is not relevant. For example, the covariance of TðxÞ and TðyÞ is an expression of their statistical dependence, and the Poohsticks approach cannot be used for it CovðTðxyÞ; TðxyÞÞ " #" # ¼ CovðTðxÞ; TðyÞÞ þ EðTðxÞÞ 2 EðTðxyÞÞ EðTðyÞÞ 2 EðTðxyÞÞ ; ð3:13:13Þ and CovðvTðxyÞ ; vTðxyÞ Þ ¼ ðA x 2 A xy ÞðA y 2 A xy Þ:

ð3:13:14Þ

Furthermore, if TðxÞ and TðyÞ are uncorrelated then CovðTðxyÞ; TðxyÞÞ ¼ ðex 2 e xy Þðey þ e xy Þ:

ð3:13:15Þ

141


Exercises Exercise 3.13.1 November 2003 SOA/CAS Course 3 Examination, Problem No. 39. You are given: (i) Mortality follows De Moivre’s Law with v ¼ 105: (ii) (45) and (65) have independent future lifetimes. Calculate e 45:65 : Solution The appropriate Poohsticks formula says e 45:65 ¼ e 45 þ e 65 2 e 45:65 : We have e 45 ¼

105 2 45 ¼ 30; 2

e 65 ¼

105 2 65 ¼ 20; 2

and

as well as e 45:65 ¼

ð 40 0

ð 40 t t 5t t2 þ 12 12 · 12 dt ¼ dt 60 40 120 2400 0

40 5t2 t3 ¼ 40 2 5 · 1600 þ 64 000 ; þ ¼ t2 240 7200 240 7200 0 ¼ 40 2

100 80 360 2 300 þ 80 140 þ ¼ ¼ : 3 9 9 9

Thus e 45:65 ¼ 30 þ 20 2

140 < 34:44: 9

A

Exercise 3.13.2 November 2002 SOA/CAS Course 3 Examination, Problem No. 33. XYZ Co. has just purchased two new tools with independent future lifetimes. Each tool has its own distinct De Moivre survival pattern. One tool has a 10-year maximum lifetime and the other one has a 7-year maximum lifetime. Calculate the expected time until both the tools have failed.

142

Chapter 3

Solution We have, because of what we know about De Moivre’s Law, e x ¼ 5; e y ¼ 3:5: Also ð ð1 ð7 7 2 t 10 2 t 1 7 · dt ¼ e xy ¼ ð70 2 17t þ t2 Þ dt ¼ 2:683: t px · t py dt ¼ 7 10 70 0 0 0 Therefore e xy ¼ e x þ e y 2 e xy ¼ 5 þ 3:5 2 2:683 < 5:81:

A

Exercise 3.13.3 May 2001 SOA/CAS Course 3 Examination, Problem No. 9. ðxÞ and ðyÞ are two lives with identical expected mortality. You are given (i) Px ¼ Py ¼ 0:1: (ii) Pxy ¼ 0:06; where Pxy is the annual benefit premium for a fully discrete insurance of one on ðxyÞ: (iii) d ¼ 0:06: Calculate the premium Pxy ; the annual benefit premium for a fully discrete insurance of one on ðxyÞ: Solution We use the relation 1 ¼ d þ Px ; a€ x and its analogues 1 ¼ d þ Pxy ; a€ xy and 1 ¼ d þ Pxy : a€ xy First 1 ¼ d þ Px ¼ 0:06 þ 0:10 ¼ 0:16; a€ x so a€ x ¼ a€ y ¼

1 : 0:16

Finally a€ xy is determined from a€ xy ¼ a€ x þ a€ y 2 a€ xy ¼ 4:167:

143


Hence Pxy ¼

1 1 2 0:06 ¼ 0:18: 2d ¼ a€ xy 4:167

A

Exercise 3.13.4 May 2001 SOA/CAS Course 3 Examination, Problem No. 23. A continuous two-life annuity pays: † 100 while both (30) and (40) are alive, † 70 while (30) is alive but (40) is dead, and † 50 while (40) is alive but (30) is dead. The actuarial present value of this annuity is 1180. Continuous single life annuities paying 100 per year are available for (30) and (40) with actuarial present values of 1200 and 1000, respectively. Calculate the actuarial present value of a two-life continuous annuity that pays 100 while at least one of them is alive. Solution We are given that 1180 ¼ 70a30 þ 50a40 2 20a30:40 ¼ 70 · 12 þ 50 · 10 2 20a30:40 : Therefore, a 30:40 ¼ 8: Now we can use the Poohsticks formula to obtain a 30:40 ¼ a 30 þ a 40 2 a 30:40 ¼ 12 þ 10 2 8 ¼ 14: We conclude that 100a30:40 ¼ 1400:

A

Exercise 3.13.5 November 2000 SOA/CAS Course 3 Examination, Problem No. 1. For independent lives ðxÞ and ðyÞ; qx ¼ 0:05; qy ¼ 0:10: Deaths are uniformly distributed over each year of age. Calculate 0:75 qxy : Solution Thanks to independence, 0:75 qxy

¼ 1 20:75 pxy ¼ 1 20:75 px ·

0:75 py

¼ 1 2 ð1 20:75 qx Þ · ð1 20:75 qy Þ ¼ 1 2 ð1 2 0:75qx Þ · ð1 2 0:75qy Þ UDD

¼ 1 2 0:9625 · 0:925 ¼ 0:10969:

A

144

Chapter 3

Exercise 3.13.6 November 2000 SOA/CAS Course 3 Examination, Problem No. 30. For independent lives (50) and (60):

mðxÞ ¼

1 ; 100 2 x

0 # x , 100:

Calculate e 50:60 : Solution To calculate the last survivor life expectancy we use the Poohsticks identity: e x:y þ e x:y ¼ e x þ e y : Under de Moivre’s Law with v ¼ 100; TðxÞ is uniformly distributed on ½0; v and e x ¼

v2x 100 2 x ¼ : 2 2

Hence e 50 ¼ 25; e 60 ¼ 20; and e 50:60 ¼ ¼

ð1 0

t p50 t p60

dt ¼

ð 40 0

ð 40 t t 12 12 dt ¼ 50 40

ð2000 2 90t þ t2 Þdt

0

2000

29 333:33 ¼ 14:667: 2000

Finally e 50:60 ¼ 25 þ 20 2 14:667 ¼ 30:333:

A

3.14 Multiple decrements In practical actuarial work on pension plans, one must consider various modes of participants leaving the plan, such as retirement, death, disability, or withdrawal. We do that by considering multiple decrements models for survival, with decrements being those modes of leaving the group. We have, as before, T being the future lifetime of ðxÞ; a continuous random variable. As always, it has a discrete cousin: K ¼ bTc: We introduce a new random variable J; describing the mode of decrement. This is a discrete random variable. Effectively, we are now studying the joint distribution of T and J (or K and J). We use the upper right superscript to denote the mode of decrement, and t denotes the effect of all modes of decrement combined. We have the following notation for the multiple decrement tables: ð tÞ

† lx is the (expected) number of people in the group (cohort) at age x: ð jÞ † dx is the (expected) number of people departing the group (cohort) between ages x and x þ 1 due to cause j:

145


ð jÞ

ð jÞ

ð tÞ

† qx ¼ dx =lx is the probability that ðxÞ departs the group in the next year due to cause j; with other decrements also after ðxÞ (i.e., with competition from other causes).P ðtÞ ð jÞ † qx ¼ m j¼1 qx is the probability of departure, regardless of cause, when all causes compete. We also have fT;J ðt; jÞ ¼ t pðxtÞ · mð jÞ ðx þ tÞ;

ð3:14:1Þ

fT ðtÞ ¼ t pðxtÞ mðtÞ ðx þ tÞ;

ð3:14:2Þ ð jÞ

fK;J ðk; jÞ ¼ PrðK ¼ k; J ¼ jÞ ¼ k lqðx jÞ ¼ k pðxtÞ · qxþk ; ðt ð jÞ ðtÞ ð jÞ q ¼ PrðT # t; J ¼ jÞ ¼ s px · m ðx þ sÞ ds; t x 0

FT ðtÞ ¼ t qðxtÞ ¼ fJ ð jÞ ¼

ð1 0

m X j¼1

ð tÞ t px

· mð jÞ ðx þ tÞ dt ¼

ð jÞ t qx ; 1 X k

ð3:14:3Þ ð3:14:4Þ ð3:14:5Þ

lqxð jÞ ¼

ð jÞ 1 qx ;

ð3:14:6Þ

k¼0

and fJ ð jlT ¼ tÞ ¼

ðtÞ fT;J ðt; jÞ px · mð jÞ ðx þ tÞ mð jÞ ðx þ tÞ ¼ ðtÞ ¼ t ð tÞ : ðtÞ fT ðtÞ m ðx þ tÞ t px · m ðx þ tÞ

ð3:14:7Þ

An important consideration in multiple decrement models is the interaction between them. Models of that nature are commonly termed competing risks in statistics. If we want to understand the nature of the individual decrements, i.e., individual risks, we are interested in modeling a population, which is subjected only to one of the risks. This single-risk population is termed the associated single decrement table, and its actuarial functions are denoted with a prime. In order to study these models, we consider these alternative random variables: Tj is the waiting time until decrement ð jÞ removes ðxÞ from the population, if other decrements do not compete with decrement ð jÞ: We assume that the force of mortality (hazard rate) for individual decrements is the same in this model as in the multiple decrement model. We use the following notation: d sTj ðtÞ ; mð jÞ ðx þ tÞ ¼ 2 dt sTj ðtÞ ðt 0 ð jÞ ð jÞ sTj ðtÞ ¼ t px ¼ exp 2 m ðx þ sÞ ds ;

ð3:14:8Þ ð3:14:9Þ

0

0 ð jÞ t qx

¼ 1 2 t p0x ð jÞ ¼ FTj ðtÞ;

fTj ðtÞ ¼ t p0x ð jÞ mð jÞ ðx þ tÞ:

ð3:14:10Þ ð3:14:11Þ

146

Chapter 3

We also have the following relationships between the multiple decrement model and the associated single decrement tables (all these follow from simple probability considerations):

mðtÞ ðx þ tÞ ¼

m X

mð jÞ ðx þ tÞ;

ð3:14:12Þ

j¼1

ð tÞ t px

fT ðtÞ ¼ t pxðtÞ mðtÞ ðx þ tÞ; ðt m Y 0 ð tÞ ¼ PrðT . tÞ ¼ m ðx þ sÞ ds ; t px ð jÞ ¼ exp 2

ð3:14:13Þ ð3:14:14Þ

0

j¼1 ð tÞ t qx

¼

m X

ð jÞ t qx ;

ð3:14:15Þ

j¼1

and 12

m X

ð jÞ ðtÞ ð tÞ t qx ¼ 1 2 t q x ¼ t px ¼

j¼1

m Y

ð1 2 t q0x ð jÞÞ:

ð3:14:16Þ

j¼1

In the simplest case of only two decrements ð1Þ t qx

ð tÞ 0 ð1Þ þ t qð2Þ þ t q0x ð2Þ 2 t q0x ð1Þ t q0x ð2Þ : x ¼ t qx ¼ t qx

ð3:14:17Þ

If we assume the constant force over each year of age in each decrement mð jÞ ðx þ tÞ ¼ ð jÞ mx then ð jÞ

q0x ð jÞ ¼ 1 2 p0x ð jÞ ¼ 1 2 ð pðxtÞ Þqx

ðt Þ

=qx

;

ð jÞ

p0x ð jÞ ¼ ð pðxtÞ Þqx

ðtÞ

=qx

;

ð3:14:18Þ

and ð jÞ

ln p0x ð jÞ ¼

qx

ð tÞ

qx

ð jÞ

· ln pðxtÞ ;

qx

ðtÞ

ð jÞ

¼

qx

mx

ð tÞ

:

mx

To show this, observe that ð1 ð1 ðt Þ ðtÞ ð jÞ e2mx mxð jÞ dt ð jÞ ð jÞ t px mx ðtÞ dt qx mx 0 0 ¼ ¼ ¼ : ð ð 1 1 ðtÞ ð tÞ ðtÞ qx mx ðtÞ ðtÞ 2 mx ð tÞ p m ðtÞ dt e m dt x x t x 0

ð3:14:19Þ

ð3:14:20Þ

0

If we make the assumption of the uniform distribution of each decrement ð jÞ ð jÞ within each year of age in the multiple decrement table (i.e., t qx ¼ t · qx ; ð jÞ ð jÞ t px ¼ 1 2 t · qx ) then we have ð jÞ

q0x ð jÞ ¼ 1 2 p0x ð jÞ ¼ 1 2 ðpðxtÞ Þqx

ðtÞ

=qx

;

ð3:14:21Þ

147


and this turns out to be the same as (3.14.18). To show it in this case, note that ðt ð jÞ tqx ¼ s pðxtÞ mð jÞ ðx þ sÞ ds: 0

Differentiate this with respect to t and obtain qxð jÞ ¼ t pðxtÞ mð jÞ ðx þ tÞ: Therefore 2

ð jÞ

qx

ð jÞ

m ðx þ tÞ ¼

ðtÞ

1 2 t qx

¼

d ð tÞ ð1 2 tqx Þ qð jÞ x dt : ð tÞ ðtÞ qx 1 2 tqx

By integrating both sides we get ðt ð jÞ ð jÞ qx qx 2 mð jÞ ðx þ sÞ ds ¼ ðtÞ lnð1 2 tqðxtÞ Þ ¼ ðtÞ lnðt pðxtÞ Þ: 0 qx qx

ð3:14:22Þ

ð3:14:23Þ

We also know that lnð t p0x ð jÞ Þ ¼ 2

ðt

mð jÞ ðx þ sÞ ds:

0

Therefore 0 ð jÞ t px

ð jÞ

¼ ð t pðxtÞ Þqx

ðt Þ

=qx

:

ð3:14:24Þ

The formula (3.14.21) is the special case of (3.14.24) when t ¼ 1: The uniform distribution of each decrement within each year of age can be also assumed in each single decrement table t q0x ðiÞ ¼ tq0x ðiÞ ; and this assumption is significantly different from uniform distribution of each decrement in the observed combined table. In this new case, we have ð1 ðtÞ ðiÞ qðiÞ ¼ t px m ðx þ tÞ dt x 0 1 0 ! ð1 Y B C ð1 2 t · q0x ð jÞ Þ · @t p0x ðiÞ mðiÞ ðx þ tÞ Adt ¼ |fflfflfflfflfflfflfflfflfflffl ffl {zfflfflfflfflfflfflfflfflfflffl ffl } 0 j–i ! PDF of Ti ¼q0x ðiÞ ð1 Y ¼ q0x ðiÞ ð1 2 t · q0x ð jÞ Þ dt: ð3:14:25Þ 0

j–i

For example, when there are only two decrements, under uniform distribution in the associated decrement table: 1 0 ð2Þ ð1Þ 0 ð1Þ 1 2 qx ; ð3:14:26Þ qx ¼ qx 2

148

Chapter 3

and with three decrements:

qð1Þ x

¼

q0x ð1Þ

1 0 ð2Þ 1 0 ð2Þ 0 ð3Þ 0 ð3Þ 1 2 ðqx þ qx Þ þ ðqx qx Þ ; 2 3

0 ðiÞ qðiÞ x ¼ qx

ð1 0

Y

ð3:14:27Þ

! 0 ð jÞ t px

fTi ðtlTi # 1Þ dt:

ð3:14:28Þ

j–i

In pension and insurance practice, it is common to have benefits depending on the mode of decrement. An important and simple principle for working with such plans is that their single benefit premium can be calculated by adding the single benefit premiums for all decrements that can be calculated separately first, i.e.,

t

EðBðt; jÞ · v Þ ¼

m ð1 X j¼1

Bðt; jÞ · vt · t pðxtÞ · mxð jÞ ðtÞ dt : |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

ð3:14:29Þ

0

APV of benefit payable for decrement j

Exercises Exercise 3.14.1 November 2003 SOA/CAS Course 3 Examination, Problem No. 7. A whole life policy provides that upon accidental death as a passenger on an airplane a benefit of 1 000 000 will be paid. If death occurs from other accidental causes, a death benefit of 500 000 will be paid. If death occurs from a cause other than an accident, a death benefit of 250 000 will be paid. You are given (i) Death benefits are payable at the moment of death. (ii) mð1Þ ¼ 1=2 000 000 where (1) indicates accidental death as a passenger on an airplane. (iii) mð2Þ ¼ 1=250 000 where (2) indicates death from other accidental causes. (iv) mð3Þ ¼ 1=10 000 where (3) indicates nonaccidental death. (v) d ¼ 0:06: Calculate the single benefit premium for this insurance. Solution Note that

mðtÞ ¼

1 1 1 þ þ ¼ 0:0001405: 2 000 000 250 000 10 000

149


Therefore, the single benefit premium is ð þ1 ð þ1 1 20:06t 20:0001405t dtþ 1000000 e e · 2000000 0 0 ð þ1 1 1 dt þ dt 500000 e20:06t e20:0001405t · 250000 e20:06t e20:0001405t · 250000 10000 0 ¼

1 1 1 1 27:5 þ2 · þ 25 · ¼ ¼457:262577: 2 0:0601405 0:0601405 0:0601405 0:0601405 A

Exercise 3.14.2 November 2003 SOA/CAS Course 3 Examination, Problem No. 27. For a double decrement table, you are given ð1Þ

ðtÞ

(i) mx ðtÞ ¼ 0:2mx ðtÞ; t . 0; ðtÞ (ii) mx ðtÞ ¼ kt2 ; t . 0; (iii) q0x ð1Þ ¼ 0:04: ð2Þ

Calculate 2 qx : Solution We know that ðtÞ mð1Þ x ðtÞ ¼ 0:2mx ðtÞ ¼

k 2 t : 5

But ðt 3 1 0:04 ¼ q0x ð1Þ ¼ 1 2 p0x ð1Þ ¼ 1 2 exp 2 ðk þ 5Þt2 dt ¼ 1 2 e2ðk=5Þðð1=3Þt Þl0 0

¼ 1 2 e2ðk=15Þ : Therefore e2ðk=15Þ ¼ 0:96;

k ¼ ð215Þ · ln 0:96 ¼ lnð0:96215 Þ:

Also

ðtÞ t px

mðxtÞ ðtÞ ¼ t2 lnð0:96215 Þ; t . 0; ðt 3 3 2 215 ¼ exp 2 r lnð0:96 Þ dr ¼ e15 ln 0:96ð1=3Þt ¼ 0:965t : 0

Finally ð tÞ 2 212 mð2Þ Þ: x ðtÞ ¼ 0:8mx ðtÞ ¼ t lnð0:96

150

Chapter 3

Therefore ð2Þ 2 qx ¼

ð2 0

ðtÞ ð2Þ t px mx ðtÞ

1 ¼ lnð0:96212 Þ 3 ¼ 24 ln 0:96

dt ¼

ð2

0:965t

3

0

ð8

& & & & u ¼ t3 t ¼ 0 ) u ¼ 0 & & · t2 lnð0:96212 Þ dt ¼ & 2 & & t dt ¼ 1 du t ¼ 2 ) u ¼ 8 & 3

0:965u du ¼ lnð0:9624 Þ

0

ð8

e5ðln

0:96Þu

du

0

1 4 40 ð0:965u Þlu¼8 u¼0 ¼ 2 ð0:96 2 1Þ < 0:64370708: 5 ln 0:96 5

A

Exercise 3.14.3 November 2002 SOA/CAS Course 3 Examination, Problem No. 2. For a triple decrement table, you are given ð1Þ

(i) mx ðtÞ ¼ 0:3; t . 0; ð2Þ (ii) mx ðtÞ ¼ 0:5; t . 0; ð3Þ (iii) mx ðtÞ ¼ 0:7; t . 0; ð2Þ

Calculate qx : Solution Note that ð2Þ ð3Þ mðxtÞ ðtÞ ¼ mð1Þ x ðtÞ þ mx ðtÞ þ mx ðtÞ ¼ 0:3 þ 0:5 þ 0:7 ¼ 1:5:

Under individual constant forces, we have ð tÞ qðiÞ x ¼ qx

lnðp0x ðiÞ Þ ðtÞ

lnðpx Þ

¼ qxðtÞ

ln e2m ln e

ðiÞ

2mðtÞ

¼ qðxtÞ

mðiÞ 0:5 ¼ 0:258956613: ¼ ð1 2 e21:5 Þ 1:5 mðtÞ A

Exercise 3.14.4 November 2002 SOA/CAS Course 3 Examination, Problem No. 13. For a double-decrement table where cause 1 is death and cause 2 is withdrawal, you are given (i) Deaths are uniformly distributed over each year of age in the singledecrement table. (ii) Withdrawals occur only at the end of each year of age. ð tÞ (iii) lx ¼ 1000: ð2Þ (iv) qx ¼ 0:40: ð1Þ ð2Þ (v) dx ¼ 0:45dx : Calculate p0x ð2Þ :

151


Solution The key observation is that since withdrawals occur at the end of the year, while deaths have the continuous uniform distribution, the probability of withdrawal does not affect the probability of death, only the probability of death affects the probability of withdrawal. In other words, withdrawals at the end of the year do not change the exposure to death during the year (they change it at one point only and this has no affect on probabilities in a continuous distribution). ð1Þ Thus p0x ð1Þ ¼ px : We can see quickly that ð2Þ qð1Þ x ¼ 0:45qx ¼ 0:18; ð2Þ pðxtÞ ¼ 1 2 qð1Þ x 2 qx ¼ 1 2 0:18 2 0:40 ¼ 0:42; ð tÞ

p0x ð2Þ ¼

ð tÞ

px px 0:42 ¼ 0:512195: ¼ ð1Þ ¼ 0 ð1Þ 0:82 px px

A

Exercise 3.14.5 November 2002 SOA/CAS Course 3 Examination, Problem No. 14. You intend to hire 200 employees for a new management training program. To predict the number of people who will complete the program, you build a multiple decrement table. You decide that the following associated single decrement assumptions are appropriate: (i) Of 40 hires, the number who fail to make adequate progress in each of the first three years is 10, 6, and 8, respectively. (ii) Of 30 hires, the number who resign from the company in each of the first three years is 6, 8, and 2, respectively. (iii) Of 20 hires, the number who leave the program for other reasons in each of the first three years is 2, 2, and 4, respectively. (iv) You use the uniform distribution of decrement assumption in each year in the multiple decrement table. Calculate the expected number of those employees who fail to make adequate progress in the third year. Solution From the data given we get the following: 10 1 ¼ ; 40 4

q01 ð1Þ ¼

q00 ð2Þ ¼ q00 ð3Þ

6 1 ¼ ; 30 5 2 1 ¼ ; ¼ 20 10

q00 ð1Þ ¼

6 1 ¼ ; 30 5

q02 ð1Þ ¼

q01 ð2Þ ¼

q02 ð2Þ ¼

q01 ð3Þ

8 1 ¼ ; 24 3 2 1 ¼ ; ¼ 18 9

q02 ð3Þ

8 1 ¼ ; 24 3

2 1 ¼ ; 16 8 4 1 ¼ : ¼ 16 4

152

Chapter 3

Therefore ðtÞ p0

ðtÞ

1 1 1 ¼ 12 12 12 ¼ 0:54; 4 5 10

p1 ¼

1 1 1 12 12 12 ¼ 0:47407; 5 3 9

and ðtÞ p2

1 1 1 ¼ 12 12 12 ¼ 0:4375: 3 8 4

It follows that ð tÞ

ð tÞ

ðtÞ

ð tÞ

l2 ¼ l0 · p0 · p1 ¼ 200 · 0:54 · 0:47407 ¼ 51:20; ð1Þ

q2 ¼

ln p02 ð1Þ ln

ð tÞ p2

ð1Þ

ð tÞ

· q2 ¼ 0:257590;

ð1Þ

ð tÞ

d2 ¼ q2 · l2 ¼ 14:

A

Exercise 3.14.6 November 2002 SOA/CAS Course 3 Examination, Problem No. 34. XYZ Paper Mill purchases a 5-year special insurance paying a benefit in the event its machine breaks down. If the cause is ‘minor’ (1), only a repair is needed. If the cause is ‘major’ (2), the machine must be replaced. Given (i) The benefit for cause (1) is 2000 payable at the moment of breakdown. (ii) The benefit for cause (2) is 500 000 payable at the moment of breakdown. (iii) Once a benefit is paid, the insurance contract is terminated. ð1Þ ð2Þ (iv) mx ðtÞ ¼ 0:100 and mx ðtÞ ¼ 0:004 for t . 0: (v) d ¼ 0:04: Calculate the actuarial present value of this insurance. Solution We have

mðxtÞ ðtÞ ¼ 0:104; for all t; so that ð tÞ t px

¼ e20:104t :

153


Therefore, the cost of the first benefit is ð5 2000 e20:04t e20:104t · 0:1 dt ¼ 200 0

5 1 e20:144t ð20:144Þ 0

¼ 1388:8889 · ð1 2 e20:72 Þ ¼ 712:84409: The value of the second benefit is ð5 500000

e

20:04t

e

20:104t

0

5 1 20:144t e · 0:004 dt ¼ 2000 ð20:144Þ 0 ¼ 13888:889 · ð1 2 e20:72 Þ ¼ 7128:4409:

The total of the two is 7841.28.

A

Exercise 3.14.7 November 2001 SOA/CAS Course 3 Examination, Problem No. 20. Don, age 50, is an actuarial science professor. His survival in his career is subject to two decrements (i) Decrement 1 is mortality. The associated single decrement table follows De Moivre’s Law with v ¼ 100: (ii) Decrement 2 is leaving academic employment with mð2Þ ð50 þ tÞ ¼ 0:05; t $ 0: Calculate the probability that Don remains an actuarial science professor for at least five but less than 10 years. Solution From (i) and De Moivre’s Law, it follows that 0 ð1Þ t p50

¼12

t 50

for t # 50:

From (ii), t p050 ð2Þ ¼ e20:05t for t . 0: Hence t 20:05t ðtÞ p ¼ 1 2 e t 50 50

for t # 50:

The probability sought is ðtÞ

ð tÞ

Prð5 # T , 10Þ ¼ sT ð5Þ 2 sT ð10Þ ¼ 5 p50 2 10 p50 45 20:25 40 20:5 e e ¼ 2 ¼ 0:216: 50 50

A

154

Chapter 3

Exercise 3.14.8 May 2001 SOA/CAS Course 3 Examination, Problem No. 10. For students entering a college, you are given the following from a multiple decrement model: (i) 1000 students enter the college at t ¼ 0: (ii) Students leave the college for failure (1) or all other reasons (2). (iii) ( ð1Þ m ðtÞ ¼ m ¼ constant; 0 # t # 4;

mð2Þ ðtÞ ¼ 0:04;

0 # t # 4:

(iv) 48 students are expected to leave the college during their first year due to all causes. Calculate the expected number of students who will leave because of failure during their fourth year. ð tÞ

Solution Since mðtÞ ðtÞ ¼ m þ 0:04 is constant, we have t p0 ¼ e2ðmþ0:04Þt : The expected decrement of 48 in the first year is the same as ðtÞ

ð tÞ

q0 · 1000 ¼ ð1 2 p0 Þ · 1000 ¼ ð1 2 e2ðmþ0:04Þ1 Þ · 1000: Hence 0:048 ¼ 1 2 e2ðmþ0:04Þ1 ; and m ¼ 2ln 0:952 2 0:04: The expected number leaving due to failure (cause 1) in the fourth year is ð1Þ

1000 · 3 lq0 ¼ 1000 · ¼ 1000 ·

ð4 3

ð4

ð tÞ

ð1Þ t p0 m ðtÞ dt ¼ 1000 ·

e2ð2ln

0:95220:04þ0:04Þt

ð4

e2ðmþ0:04Þt ð2ln 0:952 2 0:04Þ dt

3

ð2ln 0:952 2 0:04Þ dt

3

¼ 1000 ·

ð4

0:952t ð2ln 0:952 2 0:04Þ dt

3

¼ 1000 · ð2ln 0:952 2 0:04Þ ·

0:9524 2 0:9523 ¼ 7:74: ln 0:952

A

Exercise 3.14.9 November 2000 SOA/CAS Course 3 Examination, Problem No. 7. For a multiple decrement table, you are given that decrement (1) is death, decrement (2) is disability, and decrement (3) is withdrawal. You also have q060 ð1Þ ¼ 0:010; q060 ð2Þ ¼ 0:050; and q060 ð3Þ ¼ 0:100: Withdrawals occur only at the end of

155


the year. Mortality and disability are uniformly distributed over each year of age in ð3Þ the associated single decrement tables. Calculate q60 : Solution Only decrements (1) and (2) occur during the year, and therefore the probability of survival to the moment just before end the year is 0 ð1Þ 1 p60

· 1 p060 ð2Þ ¼ ð1 2 0:01Þð1 2 0:05Þ ¼ 0:9405:

Probability of survival through the entire year is 0 ð1Þ 1 p60

· 1 p060 ð2Þ · 1 p060 ð3Þ ¼ 0:9405ð1 2 0:10Þ ¼ 0:84645:

We have, therefore ð3Þ

q60 ¼ Probability of dying of cause 3 when other causes operate ¼ Probability of surviving to the moment just before the end of the year BUT not surviving the whole year ¼ 0:9405 2 0:84645 ¼ 0:09405: A Exercise 3.14.10 November 2000 SOA/CAS Course 3 Examination, Problem No. 9. For a special whole life insurance of 100 000 on ðxÞ; you are given that (d ¼ 0:06 and the death benefit is payable at the moment of death. If death occurs by accident during the first 30 years, the death benefit is doubled. Furthermore, mðtÞ ðx þ tÞ ¼ 0:008; t $ 0; and

mð1Þ ðx þ tÞ ¼ 0:001;

t $ 0;

where mð1Þ is the force of decrement due to death by accident. Calculate the single benefit premium for this insurance. Solution The benefit here depends on both the time and mode of decrement. It can be viewed as the combination of 100 000 at death plus an extra 100 000 at death if death is accidental and within 30 years. Thus the single benefit premium is given by ðt Þ

100 000 · mmðtÞ þd |fflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflffl}

Constant force ðmain insuranceÞ

þ

EðZÞ |ffl{zffl}

;

Accidental Death

where ( T

Z ¼ BðT; JÞv ¼

100 000 e20:06T ; if J ¼ 1 and T # 30; 0;

otherwise:

156

Chapter 3

where Z is a function of both T and J; which have the joint density ( 0:001 e20:008t ; J ¼ 1; fT;J ðt; jÞ ¼ t pðxtÞ · mð jÞ ðx þ tÞ ¼ 0:007 e20:008t ; J ¼ 2: So EðZÞ ¼

2 ð1 X j¼1

0

t

Bðt; jÞv · fT;J ðt; jÞ dt ¼

ð 30

100 000 e20:06t · e20:008t · 0:001 dt

0

1 2 e20:068ð30Þ ¼ 1279:37: 0:068 The total single benefit premium is ¼ 100 ·

100 000 ·

mðtÞ 0:008 þ 1279:37 ¼ 13 044:08: þ EðZÞ ¼ 100 000 · 0:068 mðtÞ þ d

A

Exercise 3.14.11 May 2000 SOA/CAS Course 3 Examination, Problem No. 15. ð tÞ ð1Þ In a double decrement table: l30 ¼ 1000; q030 ð1Þ ¼ 0:100; q030 ð2Þ ¼ 0:300; 1 lq30 ¼ ðtÞ 0:075; l32 ¼ 472: ð2Þ Calculate q31 : Solution We have ðtÞ

p30 ¼ p030 ð1Þ · p030 ð2Þ ¼ 0:9 £ 0:7 ¼ 0:63; and ð tÞ

ðtÞ

ðtÞ

l31 ¼ p30 · l30 ¼ 630: Also ð1Þ

ð1Þ

d31 ¼

1

lq30

ð tÞ

· l30 ¼ 75:

Furthermore, ðtÞ

ðtÞ

ðtÞ

d31 ¼ l31 2 l32 ¼ 630 2 472 ¼ 158; and ð2Þ

ðtÞ

ð1Þ

d31 ¼ d31 2 d32 ¼ 158 2 75 ¼ 83: Therefore ð2Þ

q31 ¼

ð2Þ

d31

ð tÞ l31

¼

83 ¼ 0:1317: 630

A

Part 2 Valuation of Assets and Liabilities


159

Chapter 4 General principles of pension plan valuation

4.1 Objectives and principles of pension funding The financing of future liabilities by an individual has the following two risks: the risk that the person at some point may not have sufficient funds to continue providing resources for the liabilities (due to deterioration of financial condition), and the risk of fluctuation in the value of liabilities (apart from any deterioration of provider’s financial condition). An individual enters into the insurance arrangements precisely in order to decrease or, at last, control them. Of course, the other side of the transaction, the insurer or the plan provider, increases the exposure to such risks with every new customer. Why do then insurers accept such increasing risk by taking on new customers? Because a large portfolio of risks provides a pooling diversification benefit, resulting from the Central Limit Theorem or the Law of Large Numbers. Sufficiently diversified liabilities may in fact have their random nature dramatically reduced and become nearly deterministic. This is especially true of retirement benefit liabilities for a large group. Creation of a fund designated for provision of retirement benefit payments affords additional advantages to plan participants: † increase the financial security of future benefit payments; † by creating an asset structure which appropriately matches the liabilities, make the benefits payments less exposed to market risk and to macro-economic factors. The same rewards derived from a fund accumulation are also present when an employer or another pension plan provider accumulates a fund in order to provide pension benefits. In view of these considerations, we must view accumulation of capital assets in a fund to be one of the central functions of financial intermediaries engaged in the provision of pension benefits, such as insurance firms and pension plans. In order to purchase those capital assets, such intermediaries collect premium payments from their customers. In Section 4.2 we will discuss general principles of

160

Chapter 4

calculation of a premium for a given set of liabilities. In Chapter 5 we will discuss the actuarial methods of calculation of an appropriate premium for pension plans, and present details of four basic methods of valuation. However, the existing insurance literature does not provide an answer to a very natural question—what form of accumulation of funds for a given set of future liabilities is optimal, from the point of view of the insurance firm, pension plan, or the customer? The method of accumulation of premium may turn out to be a controversial issue in practice. Actuaries are concerned with solvency of the insurance enterprise or the pension plan under consideration and generally prefer early accumulation of assets, as this provides greater financial security. But other divisions of an insurance firm or the plan may be more interested in using the financial resources provided by the premiums or contributions for physical plant investment, marketing, or other forms of growing the business. In the case of pension plans, the employer usually feels that the resources could be better used when applied towards growing the core business of the firm, not pension plan assets. Such a redirection would move the actual payment to plan participants into the future, but may be beneficial to plan participants if indeed the employer does deliver extraordinary growth in the meantime. In fact, this financing redirection has been used for quite a while in Germany, following World War II. However, in the United States funding a pension plan through internal growth is not allowed on the basis of the 1974 Employee Retirement Income Security Act (ERISA). In the case of an insurance firm, one can look at the question of proper funding as a question of maximization of profit. Chalke [17] discusses a conflict between actuarial and marketing divisions of an insurer, so common in the United States. He postulates that the conflict can be resolved by optimization of premium level, and focusing the firm on maximization of profits, with constraints set by solvency margins. Of course, maximization of profits is a standard objective in pricing of products other than insurance, but Chalke’s work was the first one to point out that this standard approach can be applied to insurance. General theory of pricing is presented by Stigler [71]. One can, of course, argue that in the case of a pension plan, maximization of profits is never a correct objective and the true goal is to provide secure delivery of future benefits. The issue of solvency of the insurance firm, and setting appropriate levels of premiums and reserves, are of course at the very core of the work of that firm’s actuary. But the problem of finding an optimal scheme of premium payments has not received any attention in actuarial literature. Level premium has been traditionally utilized in both life insurance and other forms of insurance. There has been some move away from the traditional approach with the introduction of universal life policies. But with the sole exception of the recent work of these authors [32], the existing literature on pension plans concerned with their methods of funding (see, for example Aitken [1], McGill [47]) does not even consider the question of optimality of a funding method.

General principles of pension plan valuation

161

In this book, we propose the following general approach to the problem. We will divide the premium paid for the pension fund cost into the following two parts: † planned premium, covering the planned growth of liabilities to the plan participants; and † additional premium, used to pay for amortization of the difference between the planned and actual growth of unfunded liabilities. The planned premium is called the normal cost of the plan. It is paid at the beginning of the fiscal period and established by the actuary. A pension actuary has more flexibility in the choice of valuation assumptions than a typical life actuary. Actual methods of establishing the normal cost are shown in Chapter 5. But the flexibility of the pension plan actuary is not absolute: pension valuation is subject to government regulation, and in the United States the actuary has the fiduciary duty to plan participants, as prescribed by the 1974 ERISA Law. The additional premium is an unexpected cost, typically termed supplemental cost, to the plan provider and, as such, a source of additional risk (see Section 8.4). Minimization of risk would most likely call for spreading this additional cost equally over time. This issue will be discussed in more detail in Section 4.3.

4.2 General principles of pension valuation The basis for valuation of pension liabilities is the same actuarial mathematics developed for life insurance in Chapter 2. It is not difficult to note similarities of various concepts used in the areas of pensions and life insurance, for example, ‘accrued liabilities’ is the pension analog of ‘benefit reserve’. Nevertheless, the mathematics of pension plans is treated as a separate area of actuarial science. There are several reasons for that. First, the number of plan participants is often significantly smaller than a typical number of customers of a life insurance company. Random fluctuations of mortality play a greater role in the planning of pension plan funding than in life insurance funding. We will address this issue of random fluctuations in Chapter 6. Second, pension plan valuation interest rate is not set as strictly by law as life insurance valuation rate, and this creates additional uncertainty in the pension case. Also, pension benefits come in a much greater variety than life insurance benefits, because of many ancillary benefits available for plan participants (such as survivor benefits, disability benefits, etc.). This also adds uncertainty to the cost of plan funding. And, to make things even more complicated, pension benefits are based on salary history, but there is a great variety in benefit formulas, from straight career average to various forms of final average. Third, when the plan funding experience differs from actuarial assumptions, the consequences are not a direct analog of the life insurance situation. Careful conservative valuation basis in life insurance can only bring

162

Chapter 4

eventually higher profits to the firm and those can be returned to policyholders, especially for mutual companies. Excessive normal cost level for pension funding will show a higher pension expense for the plan provider, and such higher cost will cause a private provider to show smaller profitability, while a government provider will have to increase tax burden to fund pensions. If the normal cost is too low, this is also undesirable, as it creates a possible threat to plan solvency. In the context of the issues raised above, we will now present the fundamental concepts and the methodology of pension valuation and the calculation of normal cost and plan surplus. The key concepts of pension valuation are: † † † †

accrued liabilities, normal cost, unfunded liabilities, and actuarial gain.

We will discuss the meaning of these concepts and then present a general rule of pension valuation dynamics. We will then introduce basic pension funding methods, which will be explained in detail in Chapter 5. 4.2.1 Accrued liabilities Assume that a plan participant earns a right to retire (with a life or temporary annuity) at age y: Let us also assume that this participant will receive the retirement benefit as a monthly annuity, with the total annual benefit amount BðyÞ: A properly funded pension plan should accumulate financial resources sufficient to pay these benefits. If the annuity is a life annuity, then the present value of benefits at retirement is BðyÞ€að12Þ y ; ð12Þ

where a€ y is the actuarial present value (i.e., expected present value, see (3.10.17)) of a monthly life annuity due payable beginning at age y; with monthly installments equal to 1/12 of the nominal annual amount. If upon retirement at age y the beneficiary will instead receive a temporary annuity for a term of n years, then the present value of these benefits is ð12Þ

BðyÞ€ay: n ; ð12Þ

where BðyÞ is the annual benefit amount, and a€ y: n ; given by the formula (3.10.36), is the actuarial present value of an n year temporary annuity, payable monthly, with monthly installments equal to 1/12. In what follows, we will concentrate on the case of life annuities, but of course the case of temporary annuities will require just a minor adjustment. If we replace in all ð12Þ ð12Þ formulas a€ y by a€ y: n ; we obtain the temporary annuity analogs of life annuity

163


formulas. One can also use the same formulas as for life annuities, and merely modify the laws of mortality used. For temporary n year life annuities we would simply need to assume that all plan beneficiaries do not survive beyond age y þ n; in other words, replace the single life status ðyÞ by the joint life status y : n : Typically, a pension plan participant earns benefit rights with the passage of years of service. There are two ð12Þ approaches in determining the value at retirement BðyÞ€ay of the benefit earned by a participant. The first approach used, for example, in the unit credit method (see Section 5.1) determines the accrued benefit BðxÞ directly, as a function of age x; or service years, of a participant. If the participant starts service at age w; and at current time t is aged x; with w , x , y; then 0 , BðxÞ , BðyÞ; as it is usually assumed that BðwÞ ¼ 0: Denote by ALjt the accrued liability with respect to the jth plan participant at time t; defined here as ALjt ¼ Bðxj Þ€að12Þ yj

Dyj Dxj

;

where the index j means that the value is assigned to the jth participant, and Dyj =Dxj is the actuarial discount factor used for discounting to time t; accounting for mortality (or, more generally, probability of any form of decrement, i.e., departure from the group of plan participants). Total accrued liability for the entire plan, denoted by ALt ; is then equal to X j ALt ; ALt ¼ j[At

where At is the group of active participants at time t: The method described above is especially useful when the rules of the plan clearly define the function BðxÞ: Otherwise, it is more convenient to use a second approach, which, although less natural, leads to fewer problems of the kind described in Example 5.1.1. This second approach first determines the normal cost, and then, using it, establishes accrued liabilities. This will be discussed in Section 4.2.2. 4.2.2 The normal cost versus accrued liabilities The normal cost, NCt ; is defined as the premium which needs to be paid into the plan at the beginning of every year (or, more generally, every fiscal period) in order to pay that year increment of all liabilities, assuming that all actuarial assumptions concerning mortality, salaries, interest rates, and expenses, turn out to be correct. Thus the normal cost is effectively the annual cost of the plan, if the actuarial assumptions turn out to be exact predictions. In reality, assumptions are not perfect, and differences between assumptions and reality (termed actuarial gains or

164

Chapter 4

actuarial losses) need to be regularly amortized. The total premium paid must include this amortization, and such premium is termed the plan cost. The Fundamental Principle of Pension Plan Actuarial Valuation is:

The normal costs of a pension plan must be spread over time in such a way that at the time of retirement of a given participant all normal costs to be paid on that participant’s behalf have the same accumulated actuarial value as the actuarial present value of the benefits to be paid to this participant.

The above rule is, of course, a version of the Equivalence Principle used widely in actuarial mathematics in life insurance (see Section 3.11 and e.g., Bowers et al. [14]). The following illustrates the application of this principle to valuation of a pension plan, by relating the normal cost and the accrued liabilities.

The accrued liabilities with respect to the jth participant at every year are equal to the accumulated actuarial value of the normal costs paid on behalf of this participant.

The above is the retrospective formula for the accrued liabilities, in terms of the normal costs paid up to date. By using the Equivalence Principle, we can also derive this prospective formula for accrued liabilities:

At each year the accrued liabilities with respect to the jth plan participant are equal to the actuarial present value of future benefits ðPVFBjt Þ minus the actuarial present value of future normal costs ðPVFNCjt Þ: ALjt ¼ PVFBjt 2 PVFNCjt : It is easy to see that the retrospective formula follows from the prospective formula combined with the Equivalence Principle. In fact, the two formulas are equivalent, given the Equivalence Principle.

Using the above principles, one can first choose an appropriate method of determining normal costs, and then calculate accrued liabilities. This approach is used in several methods, for example the Entry Age Normal cost method. In this method, one assumes that all participants join the plan at approximately the same age w and

165


later retire at approximately the same age y: Assume that the normal cost for the jth participant is level and equal to NCj : Then from the Fundamental Principle of Pension Plan Actuarial Valuation, we infer that NCj a€ w: yw ¼ Bj ðyÞ€að12Þ y

Dy ; Dw

from which we can calculate NCj : Using the retrospective formula we calculate accrued liabilities ALjt with respect to this participant in the year t; in which this participant was aged x; arriving at ALjt ¼ Bj ðyÞ€að12Þ y

Nw 2 Nx Dy ; Nw 2 Ny Dx

where Nx is the commutation function given by the formula (3.8.4). This calculation will be elaborated further in Chapter 5. We can also obtain its analog using the prospective approach: ALjt ¼ Bj ðyÞ€að12Þ y

Dy Nx 2 Ny 2 NCj ; Dx Dx

where the first expression is the actuarial present value of future benefits, and the second one is the actuarial present value of future normal costs. Of course, both formulas are equivalent. A similar approach is used in the Individual Level Premium method (see Section 5.3), but in that case the assumption of approximately equal age of each new participant is no longer used. 4.2.3 Unfunded actuarial liabilities Let Ft denote the assets of the plan at time t: If the plan is in ideal balance then its assets Ft equal accrued liabilities ALt : In reality there is usually a nonzero difference UALt ¼ ALt 2 Ft ; which will be termed the actuarial deficit or the unfunded accrued liability. Of course, one must establish values of ALt and Ft to calculate the unfunded actuarial liability. Chapter 5 covers valuation of liabilities, while Chapter 7 is devoted to questions related to valuation of assets. Unfunded actuarial liabilities should be amortized. This is discussed in Chapter 5. If UALt , 0; then we will term its opposite (its absolute value) the actuarial surplus (not to be confused with the actuarial gain). A large actuarial surplus should also be amortized. 4.2.4 Actuarial gain Let C be the contribution paid by the plan provider in a given year t: Let I be the amount by which the assets grow in a given year, as a result of investment portfolio

166

Chapter 4

performance, and let P be the amount paid for purchases of annuities for retiring participants. Then the assets at time t þ 1 are Ftþ1 ¼ Ft þ I þ C 2 P: Let us also denote by IC the return earned by the contribution C paid in the year t; and we will assume that the contribution is paid at the beginning of the year. If the actuarial assumptions about the interest rate turn out to be correct, and the plan provider pays normal cost and the entire unfunded actuarial liability from the previous year, then we have: ðUALt þ NCt Þð1 þ iÞ 2 ðC þ IC Þ ¼ 0:

ð4:2:1Þ

The left-hand side of (4.2.1) is called the expected unfunded actuarial liability. In the situation described by (4.2.1), this expected unfunded liability has been reduced to zero. If the actuarial assumptions turn out to be correct, then the contribution C together with return on it IC will reduce to zero the unfunded actuarial liability UALtþ1 in the year t þ 1: In reality, of course, this usually does not happen, and thus Ga ; ðUALt þ NCt Þð1 þ iÞ 2 ðC þ IC Þ 2 UALtþ1 is nonzero. Since the above expression Ga is the difference between the expected and actual unfunded actuarial liability, it is called the actuarial gain. The actuarial gain is positive if the actuarial unfunded liability is less than the expected unfunded actuarial liability. As we had pointed out already, achieving an actuarial gain is not an objective of a pension plan. Thus, for example, in the method of frozen initial liabilities, the valuation of ALt and NCt is set in such a way that Ga ; 0 by definition (see Section 5.4). 4.2.5 Some additional remarks There are two additional concepts used in the actuarial literature in relation to the accrued actuarial liabilities. The two represent methods of establishing liabilities of a pension system. † Accumulated Benefit Obligation (ABO) calculated as those liabilities, which have been already earned by the plan participants, based on their service up to date, but not taking into consideration expected future service (if we were to include future expected service, we would also have to include future expected normal costs paid). This is an analog of the retrospective definition of accrued liabilities. † Projected Benefit Obligation (PBO) calculated as the liabilities with respect to current participants, which have been earned based on their service up to date, and on their expected future service, after subtracting expected future normal costs to be paid on their behalf. This is an analog of the prospective definition of accrued liabilities.


167

The ABO method is similar to approaches used in accounting and in the economic literatures, while the PBO method is similar to the typical approach used in life insurance reserving. If the Fundamental Principle of Pension Plan Actuarial Valuation holds, then both the methods, if used at exactly the same time, should produce exactly the same result. Otherwise, there can be dramatic differences between values produced by them. For example, in social insurance (such as the Old Age, Survivors and Disability Insurance, known popularly as Social Security), the ABO method is almost never used. It is common instead to use the PBO method, and often assume an infinite or nearly infinite time horizon. Because social insurance is assumed to pay out exactly the same total amount as they receive in premiums paid, their PBO must equal zero, unless the premium is not calculated correctly. Of course, this effectively means that premiums for the benefits of current workers will be eventually paid by future workers. At the same time the ABO liabilities of the system can be quite substantial. The duty of the actuary, be it in life insurance or in pension plans, is an appropriate design of the system so that liabilities are paid when due. The unfunded liabilities of a pension plan can be quite large, or in the case of social insurance, even larger than just large. The actuary must pay careful attention to the solvency of the entire insurance company, and to economic value as well as liquidity of the assets held. But, surprisingly, the actuarial and insurance literature has largely ignored the question of what is the optimal method of paying off an existing liability. This problem appeared naturally during the process of pension reform in Poland, when the old system of special pension privilege liabilities to the coal miners had to be discharged separately from the new pension system. The first of the authors of this work had then proposed the solution presented in Section 4.3, and had performed the calculations for the system, together with K. Makarski and M. Weretka. The authors of the present book had later presented the theoretical solution in their work [32], and proved that the solution is optimal with respect to stability of the premium flow. Section 4.3 will present that result in detail.

4.3 Optimal payoff of a liability In this section we address the problem of finding an optimal funding method for a given set of liability cash flows. In the context of a pension plan, we can view this as an optimal approach of paying off existing unfunded or any other actuarial liabilities. We will address this problem for deterministic liabilities cash flows. The discount function used to produce present values is described by the force of interest {dt : t $ 0}: Assume that all liabilities cash flows are contained in the time interval ½0; T; and that those cash flows are Ct paid at times t ¼ 1; 2; …; T (some of the cashÐ flows Ct t may be equal to zero). Their present values at time t ¼ 0 are: Lt ¼ Ct expð2 P 0 ds dsÞ; t ¼ 1; 2; …; T: Define the sum of all present values up to time t; as ALt ¼ ti¼1 Li : Let AL0 ¼ 0:

168

Chapter 4

The liabilities will be paid with a stream of premiums Pt ; also indexed by t ¼ 1; 2; …; T (and we also allow some of the Pt values to equal zero). We will additionally assume that the premiums are paid at the beginning of a period corresponding to each one of them so that the premium PÐt is paid at time t 2 1: The present value of Pt at time t ¼ 0 is PVðPt Þ ¼ Pt expð2 t21 The sum of all 0 ds dsÞ:P present values of premiums up to time t will be denoted by APt ¼ ti¼1 PVðPi Þ: Our objective is to find an optimal stream of premium payments. But first we must ask ourselves: what patterns of premium payments are acceptable? One possible solution is to prepay all liabilities completely in advance by a premium P1 paid at time t ¼ 0: Another alternative is to pay level premiums throughout the period ½0; T (or premiums with level present values). Our discussion in Section 4.1 implies that acceptable funding methods should satisfy the following conditions: Constraint 4.3.1 PVðPt Þ is a nonincreasing function of time, i.e., PVðPt Þ $ PVðPtþ1 Þ

for t ¼ 1; …; T 2 1:

This condition means that delaying payments into the future is severely restrained. But a full prepayment, or level premium do meet this condition. Constraint 4.3.2 for t ¼ 1; 2; …; T 2 1;

APt $ ALt

ð4:3:1Þ

and APT ¼ ALT : Using actuarial terminology this condition means that the reserve is never negative (in fact it is equal to at least one premium), and this in practice means that there never is a need to borrow funds to pay the liability cash flows. In Chapters 9 and 10 we call (4.3.1) Axiom of Solvency and investigate its influence on the immunization of terminal surplus APT 2 ALT against the interest rate perturbations. Under the above two constraints, we seek to minimize simultaneously all of the following quantities: APt 2 ALt

for t ¼ 1; 2; …; T:

ð4:3:2Þ

It should be noted that we intend to minimize all of these quantities simultaneously, seeking the minimum among all cash flows of premiums {P1 ; …; PT } satisfying Constraints 4.3.1 and 4.3.2. Because the premium Pt ; is paid at the beginning of the year, and if APt ¼ ALt ; t ¼ 1; 2; …; T; the reserve would be minimal and equal to one premium. The problem of simultaneous minimization of the expressions (4.3.2), under Constraints 4.3.1 and 4.3.2, when applied to pension plans, means that seeking such a method of amortization of unfunded actuarial liability, which assures liquidity

169


(Constraint 4.3.2), and does not allow for payments to be inappropriately postponed (Constraint 4.3.1), at the same time puts the least financial burden of cost on the plan provider. As we will show later, the optimal funding method, which solves this problem also gives the premium flow, which is the most stable in time in terms of premiums present values. Suppose, in contrast to the normal cost, which was discussed in Section 4.2 and will be presented in detail in Chapter 5, that this premium Pt is determined for amortization of unfunded actuarial liability that appeared as a result of a difference between expectations of the growth of assets and liabilities, and the actual growth of them. This is, in fact, an unexpected cost for the provider, creating an additional risk (see also Section 8.4). Stable distribution of the premium paying for this cost lowers the risk to the provider. It should be noted that the inequality in Constraint 4.3.1 cannot be replaced by an equality, as this could result in a contradiction with the Constraint 4.3.2. Let F be a continuous, real-valued, piecewise linear function defined on ½0; T; such that Fð0Þ ¼ 0: The derivative f of this function exists everywhere but at the bentpoints of the graph of F: The function f (where the derivative exists) is piecewise constant, and can be extended to the entire interval ½0; T; by letting it have the value of 0 at 0 and making it continuous from the left-hand side on ½0; T: After performing such an extension, this function f ; which we will term the continuous from the left derivative of the function F; is uniquely defined and for each t [ ½0; T FðtÞ ¼

ðt

f ðsÞds:

0

Now consider the function FðtÞ ¼ ALt ; which is defined at each integer value t [ ½0; T: Extend it to the entire interval ½0; T by making it piecewise linear and continuous, with the smallest possible number of bent-points. Let F p be the smallest concave majorant of F; i.e., the smallest function F p on ½0; T; such that F p is concave and F p ðtÞ $ FðtÞ for t [ ½0; T: Let f p be the continuous from the left derivative of the function F p (note that f p ð0Þ ¼ 0 by definition). Theorem 4.3.1 Premiums Ppt minimize APt 2 ALt for t ¼ 1; 2; …; T among all cash flows of premiums satisfying Constraints 4.3.1 and 4.3.2 if, and only if PVðPpt Þ ¼ f p ðtÞ

for t ¼ 1; …; T;

where f p is the continuous from the left derivative of the smallest concave majorant of the function giving the accumulated through time t present values of liabilities. Proof To prove this contention, consider an arbitrary cash flow of premiums {P1 ; …; PT }; satisfying Constraints 4.3.1 and 4.3.2 and minimizing all APt 2 ALt for t ¼ 1; 2; …; T:

170

Chapter 4

Because PVðPt Þ is a nonincreasing function of t; the piecewise linear function F p ; defined by linear interpolation of the relationship t 7 ! APt for t [ ½0; T; is concave. Based on Constraint 4.3.2 we conclude that F p ðtÞ $ FðtÞ; where FðtÞ ¼ ALt for t ¼ 1; 2; …; T: Since APt 2 ALt is minimized for every t ¼ 1; 2; …; T; F p is the smallest concave majorant. On the other hand, if F p is the smallest concave majorant of the function FðtÞ ¼ ALt ; defined first for all integers t; and then extended to all t [ ½0; T; then F p satisfies the Constraint 4.3.2, because it is a majorant, and the Constraint 4.3.1, because it is concave. Additionally F p minimizes APt 2 ALt for every t ¼ 1; 2; …; T; as it is the smallest concave majorant. A The cash flow of premiums defined above has two additional optimality properties. The present values Qpt ; PVðPpt Þ minimize their ‘square error’ P distance from a premium cash flow with level present values, Q ¼ T 21 Tt¼1 Lt ; assuming Constraint 4.3.2. More precisely, the following theorem holds: Theorem 4.3.2 ðQp1 ; …; QpT Þ minimizes EðQ1 ; …; QT Þ ;

T X

ðQt 2 QÞ2 ;

t¼1

under Constraints 4.3.1 and 4.3.2. Proof Let D be the set of all present values of premium flows satisfying the Constraints 4.3.1 and 4.3.2. It is easy to see that D is a convex set. Furthermore, E attains its minimum on D for a set of present values of premiums ðQp1 ; …; QpT Þ; if for every other set of present values of premiums ðQ1 ; …; QT Þ [ D : EðQ1 ; …; QT Þ ¼

T X

ðQpt 2 QÞ2 þ

t¼1

$

T X

T X

ðQt 2 Qpt Þ2 þ 2

T X

t¼1

ðQpt 2 QÞðQt 2 Qpt Þ

t¼1

ðQpt 2 QÞ2 :

t¼1

The above inequality is satisfied if the following inequality holds: T X

ðQpt 2 QÞðQt 2 Qpt Þ $ 0:

t¼1

Let us note an identity T X t¼1

at bt ¼ aT

T X t¼1

bt 2

T 21 X t¼1

ðatþ1 2 at Þ

t X s¼1

bs ;

ð4:3:3Þ

171


true for any ða1 ; …; aT Þ and ðb1 ; …; bT Þ: Denote at ¼ Qpt 2 Q and bt ¼ Qt 2 Qpt : By using the above identity we obtain T X

ðQpt 2 QÞðQt 2 Qpt Þ

t¼1

¼

ðQpT

2 QÞ

T X

Qt 2

t¼1

¼2

T21 X t¼1

ðQptþ1 2 Qpt Þ

T X

! Qpt

t¼1 t X

2

T21 X

ðQptþ1 2 Qpt Þ

t¼1

Qs 2

s¼1

t X

t X

ðQs 2 Qps Þ

s¼1

! Qps

$ 0;

s¼1

where the last inequality is true because Qpt is a nonincreasing function of t; and Pt p; accumulated s¼1 Qs t ¼ 1; …; T; creates the smallest concave majorant for Pthe t present values of liabilities cash flows, which implies that the sum Q s¼1 s cannot be P less than ts¼1 Qps : A Let us note that the property proved above means also that the premium flow given by ðPp1 ; …; PpT Þ gives the smallest value of max PVðPt Þ;

t¼1;…;T

while satisfying Constraints 4.3.1 and 4.3.2, and even only under Constraint 4.3.2. The optimal premium flow given by the derivative of the smallest concave majorant of the function determined by present values of liabilities cash flows accumulated to time t; depends on the force of interest dt : Could we use stochastic interest rates in this procedure? The Fundamental Theorem of Asset Pricing (see Section 7.2, Ross [62], and Dybvig and Ross [20]) states that in an efficient market, lack of arbitrage is equivalent to the price of a financial instrument being equal to the expected present value (with respect to an appropriate probability measure, the so-called risk neutral measure, see Sections 7.1 and 7.2) of its future cash flows. In practice this means that we can at least estimate the price of nonmarketable instruments, e.g., insurance liabilities, by calculating the average of present values of future cash flows for a large sample of future interest rate scenarios. Seemingly, our method of optimizing premiums cannot be applied to such a situation, as the optimal premium flow is established for each scenario separately, and there is no global solution. However, if we use a premium flow which, on the average, is the smallest concave majorant of the stream of the expectations of liabilities, we will minimize EðAPt Þ 2 EðALt Þ, for all t ¼ 1; 2; …; T; as well as some immunization risk measure Re (see Section 10.5.2 for details).

172

Chapter 4

The connection between existing unfunded actuarial liabilities and the stream Ct , t ¼ 1; 2; …; T; is quite often obvious. If, for example, the plan has no assets, then one way liabilities can be settled is by buying life annuities for each participant upon retirement. If the fund has unfunded liabilities that need to be amortized over time, then the amortization payments can be treated as a stream of liabilities to be settled. If the group of plan participants is diversified enough, these liabilities do become nearly deterministic. Most importantly, the method of smallest concave majorant is optimal in combining the need for liquidity of the pension plan with the postulate of minimal funding cost to the plan provider.

Exercises Exercise 4.4.1 A pension plan provides an annual benefit equal to 2% of final year compensation for each year of service at normal retirement age of 65. A participant has accumulated 25 years of service at normal retirement age and has the final salary of 40 000. What is the annual pension benefit earned by this participant, and what is the single benefit premium for the pension annuity if D65 ¼ 210; and N65 ¼ 2201? Solution The annual benefit earned by this participant is: 0:02 · 40 000 · 25 ¼ 20 000: The benefit premium for the pension annuity is (see (3.10.7)) 20 000 · a€ 65 ¼ 20 000 ·

N65 2201 ¼ 209 619:05: ¼ 20 000 · 210 D65

A

Exercise 4.4.2 At the beginning of the year 2003, accrued liabilities of a pension plan for the employees of The Honorable Life Insurance Company were 100 000. The assets of the plan at the same time were 150 000. On January 1 2003, a contribution of 10 000 is made to the plan. At the same time, 50 000 is used for purchases of life annuities fully discharging the liabilities for retiring participants, although based on the actuarial assumptions the cost of those annuities should have been 60 000. During the year 2003, assets of the plan earn 45%, while the actuarial valuation rate is 8%. The normal cost of the plan for 2003 is established by the plan actuary as 20 000 (as of January 1, 2003). There are no deaths or terminations other than retirement in 2003. Calculate the actuarial gain for this plan for the calendar year 2003. Solution Initial unfunded actuarial liability is UAL2003 ¼ AL2003 2 F2003 ¼ 100 000 2 150 000 ¼ 250 000:


173

Assets at the end of the year are F2004 ¼ F2003 þ C 2 P þ I ¼ 150 000 2 50 000 þ 10 000 þ ð150 000 2 50 000 þ 10 000Þ · 0:45 ¼ 159 500: Accrued liabilities during 2003 increase with the valuation rate, but are decreasing by the discharge of liabilities of 60 000, and increased by the normal cost of 20 000. Therefore, the accrued liabilities as of January 1, 2004 are ð100 000 2 60 000 þ 20 000Þ · 1:08 ¼ 64 800: The unfunded actuarial liabilities on January 1, 2004, are therefore: 64 800 2 159 500 ¼ 294 700: The expected unfunded actuarial liability on January 1, 2004, on the other hand, is ðUAL2003 þ NC2004 Þð1 þ iÞ 2 ðC þ IC Þ ¼ ð250 000 þ 20 000Þ · 1:08 2 10 000 · 1:08 ¼ 240 000 · 1:08 ¼ 243 200: The actuarial gain is Ga ¼ ð243 200Þ 2 ð294 700Þ ¼ 51 500:

A

Exercise 4.4.3 Mr. Romanov, who is 25 years old, is planning for retirement, and in order to provide for it, he estimates that he will need 2 200 000 monetary units at the age of 65, adjusted for inflation, i.e., in terms of today’s purchasing power. He does not have any current savings. His daughter, Anastasia, was just born, and Mr. Romanov also plans to provide for her college education, which will cost 180 000 monetary units when she turns 18, also in terms of today’s purchasing power. Mr. Romanov invests conservatively in securities with rates of return matching the rate of inflation. Calculate the optimal annual savings amount that Mr. Romanov should invest every year at the beginning of the year, to fund his liabilities in an optimal fashion. Solution Effectively, Mr. Romanov will earn 0% real rate of return on his savings, and for all funds, present value equals future value (in purchasing power). We can use the smallest concave majorant funding with the following future liabilities: – 2 200 000 in 40 years, and – 180 000 in 18 years.

174

Chapter 4

The linear function connecting the point ð0; 0Þ and the point ð18; 180 000Þ has the slope 180 000 2 0 ¼ 10 000: 18 2 0 The linear function connecting the point ð18; 180 000Þ and the point ð40; 2 380 000Þ; where 2 380 000 ¼ 180 000 þ 2 200 000; has the slope 2 380 000 2 180 000 2 200 000 ¼ ¼ 100 000: 40 2 18 22 Since the second slope is higher, the smallest concave majorant will require the level funding of the terminal amount of 2 380 000 over 40 years, i.e., 59 500 per year or about 4958.33 per month. A Exercise 4.4.4 Assume now that Mr. Romanov of Exercise 4.4.3 has decided instead to invest in stocks, earning an annual real (i.e., after inflation) rate of return of 7%. How will this change his optimal funding structure? Solution We can use the smallest concave majorant methodology again, but in this case we should work with all liabilities and funding amounts being expressed as present values at 7%. Mr. Romanov’s retirement funding requires 2 200 000 < 146 916:84; 1:0740 in terms of present value at time 0, while Anastasia Romanov’s college education funding requires 180 000 < 53 255:50; 1:0718 in today’s purchasing power. The slope of the line connecting the point ð0; 0Þ and ð18; 53 255:50Þ is 53 255:50 2 0 < 2958:64; 18 2 0 while the slope of the line connecting the point ð18; 53 255:50Þ and the point ð40; 200 172:34Þ; where 200 172:34 ¼ 53 255:50 þ 146 916:84; is 146 916:84 < 6678:04; 22 therefore, level funding of the amount of 200 172.34 over 40 years, i.e., 5004.31 a year, in terms of today’s present value, is called for. This means that the actual optimal funding is


175

† Year 1: 5004.31, † Year 2: 5004:31 · 1:07 < 5354:61; † Year 3: 5004:31 · 1:072 < 5729:43; etc., with Year 40 funding equal to: 5004:31 · 1:0739 < 70 034:40: It should be noted that the total savings at the end of Year 18, when Anastasia heads for college, would be 18 · 5004:31 · 1:0718 < 304 456:03; significantly in excess of the college funding needs, indicating the simultaneous funding of retirement achieved while saving for college. A Exercise 4.4.5 Now assume now that Mr. Romanov of Exercise 4.4.3 is willing to take a very high level of risk and, as a result, will earn substantially higher rate of return. How high will that rate of return have to be so that Mr. Romanov does not save for retirement while saving for the cost of college education of Anastasia Romanov? Solution The situation desired will require that the smallest concave majorant should be linear, so that these three points: 180 000 180 000 2 200 000 ð0; 0Þ; 18; þ ; 40; ; ð1 þ iÞ18 ð1 þ iÞ18 ð1 þ iÞ40 lie on a straight line. This implies that 180 000 180 000 2 200 000 180 000 20 þ 2 18 18 40 ð1 þ iÞ ð1 þ iÞ ð1 þ iÞ ð1 þ iÞ18 ; ¼ 40 2 18 18 2 0 i.e., 10 000 100 000 ¼ ; 18 ð1 þ iÞ ð1 þ iÞ40 and ð1 þ iÞ22 ¼ 10; i < 11:03%:

A

Exercise 4.4.6 Mr. Louis Bourbon is a king of a Western European country plagued by high levels of poverty among its elderly population. Mr. Bourbon proposes to introduce a flat benefit, paid in gold, to the elderly aged 65 and above, paid commencing on the 65th birthday of each citizen, until death of the citizen. The benefit will be funded by general revenue budget, i.e., by all taxes levied by the kingdom, not by special taxes allocated to this program. Because of recent expenditures of support of a war of independence for major competing power’s overseas colonies, Mr. Bourbon is concerned about the cost of the program and would

176

Chapter 4

like to minimize the expenditures, while not being forced to borrow to finance the program. All payments will be made in gold, which earns 0% real rate of return, i.e., it exactly retains its purchasing power. Mr. Bourbon estimates that the total benefit payment cost will be 100 000 gold coins the first year, 200 000 gold coins the second year, 300 000 gold coins the third year, then 500 000 gold coins for seven years, and after that 400 000 every year forever. Calculate the optimal funding for this retirement scheme designed by Mr. Louis Bourbon. Solution Since all payments are in gold, and gold is assumed to neither appreciate nor depreciate in purchasing power, we can take all amounts as present values. The optimal funding is determined by the smallest concave majorant. The slope of the line connecting the two initial points: ð0; 0Þ and ð1; 100 000Þ is 100 000, while the slope of the line connecting ð0; 0Þ and ð2; 300 000Þ is 150 000, thus Year 2 payment will need to be pre-funded. The slope of the line connecting ð0; 0Þ and ð3; 600 000Þ is 200 000, so Year 3 payment will also need to be pre-funded. The slope of the line connecting the points ð0; 0Þ and ð4; 1 100 000Þ is 275 000, and as this is greater than 200 000, indicating further pre-funding need. Since the funding needs following Year 4 are 500 000 every year, they are always bigger than the funding needs in the first 3 years, and we should find the slope of the line connecting point ð0; 0Þ and ð10; 4 100 000Þ; which is 410 000, and this indicates the optimal funding level for the first 10 years. The optimal funding for the years beyond the 10th year is the constant 400 000 liability payment the same year. A Exercise 4.4.7 November 2002 SOA Course 5 Examination, Problem No. 10, Section C. John is 30 years old and has been working 5 years for Company A, which did not have a pension plan. His company is going to adopt a pension plan. He is now also considering a new position with two other companies. Company A provides a salary with 2% annual increase. The pension benefit is 2% of the employee average salary during the last 5 years of service times the number of years of service. They will recognize in the pension formula the 5 years previously worked. There is no penalty if an employee retires before age 60. Company B offers a starting salary 20% higher than Company A with 2% annual increase. The pension benefit is 2% of the employee average salary during the last 3 years of service times the number of years of service. They offer no recognition of past work. There is no penalty if an employee retires before age 60. Company C offers a starting salary 45% higher than Company A with 2% annual increase. The pension benefit is 2% of the employee final salary times the number of years of service. They offer no recognition of past work. Furthermore, there is a 2% penalty per year if an employee retires before age 60. Compare each offer based on the pension benefit John would receive if he retires at age 55.


177

Solution Let us assume for simplicity that the current salary of John in Company A is 100 000 annually. We will compare the annual benefit rates for the three choices. Choice 1: Company A. Since John started with Company A at age 25, upon retirement at age 55, John will have 30 years of service. All of the service will be recognized in the pension formula, and there will be no reduction due to retirement before age 60. John’s salary in the last 5 years of work, counting backwards, will be 100 100 100 100 100

000 · 1.0224 < 160 000 · 1.0223 < 157 000 · 1.0223 < 154 000 · 1.0222 < 151 000 · 1.0221 < 148

843.72, 689.93, 597.97, 566.63, 594.74.

The average of these quantities is 154 658.60. The annual benefit rate will be 30 (years of service) times 2% (per service year) of 154 658.60, i.e., approximately 90 795.16. Choice 2: Company B. In this company, John’s starting salary will be 120 000. During the last 3 years of service, John’s salary will be: 120 000 · 1.0224 < 193 012.47, 100 000 · 1.0223 < 189 227.91, 100 000 · 1.0223 < 185 517.56. The average of these quantities is 189 252.65. The annual benefit rate will be 25 (years of service) times 2% (per service year) of 189 252.65, i.e., approximately 94 626.32. Choice 3: Company C. In this company, John’s starting salary will be 145 000. His final annual salary will be 145 000 · 1:0224 < 233 223:40: The benefit upon retirement at age 55 will be 2% times 25 (years of service) times 233 223.40 times 0.90 (a total of 10% benefit reduction for early retirement), i.e., 104 950.53. As we see, benefit is the highest with the Company C choice, second highest with Company B, and the lowest with Company A. A Exercise 4.4.8 November 2001 SOA Course 5 Examination, Problem No. 4. You are given the following retirement plan information for an individual. Current age is 40, entry age was 30, and retirement age is 65. The social insurance (such as Social Security in the United States) annual benefit is 11 700. Current salary is 80 000. Annual salary growth is 3%. Personal savings accumulation rate is 7%. Final salary (at age 64) will be 162 624. The annuity conversion factor at age 65 is 8.1958. Retirement benefit is 1% of final 5-year average salary times the number of years of service.

178

Chapter 4

This individual begins saving for retirement at his current age. Calculate the level percentage of salary that should be allocated to personal savings each year to provide this individual with a 70% replacement ratio at retirement. Solution The 70% replacement ratio at retirement means that the initial total retirement benefit must be equal to 70% of the final salary, i.e., 0:70 · 162 624 ¼ 113 836:80: This person will receive retirement benefits from three sources: social insurance (this is known to be 11 700), personal savings and the employer-sponsored plan. Therefore, personal savings, and the pension plan must provide together the annual benefit of 113 836:80 2 11 700 ¼ 102 136:80 Let us now calculate the pension benefit that this person will receive. Because the salary is assumed to increase 3% annually, the final average salary (calculated over the last 5 years of work) is 162 624 þ

162 624 162 624 162 624 162 624 þ þ þ 1:03 1:032 1:033 1:034 < 153 422:68: 5

The benefit amount will be 1% times this final average salary times the number of years of service (equal to 65 minus 30 for this person), i.e., 0:01 · 153 422:68 · 35 < 53 697:94: Thus the annual benefit provided by the personal savings must be 102 136:80 2 53 697:94 ¼ 48 438:86: Since the annuity conversion factor at age 65 is 8.1958, the amount needed to fund such a lifetime annuity with personal savings is 48 438:86 · 8:1958 ¼ 396 995:21: This amount is to be accumulated by contributions amounting to a level percentage of salary, where we know that savings earn 7%, while salary increases 3% every year. Current salary is 80 000, and contributions will be made over 25 years. If we denote by x the fraction of salary that must be contributed to this personal savings plan, then we must have 396 995:21 ¼ x · 80 000 · ð1:0725 þ 1:0724 · 1:03 þ · · · þ 1:0324 · 1:07Þ 1:03 25 12 1:07 ¼ x · 80 000 · 1:0725 · : 1:03 12 1:07 This solves to x < 0:0556: Therefore, 5.56% of salary must be saved every year in order to achieve 70% replacement ratio upon retirement. A

179

Chapter 5 Valuation of pension plan liabilities

5.1 Unit credit method We will for now assume that retirement benefit rights are earned continuously during participant’s years of service. At the time of hire, at assumed age w; participant’s benefit earned is 0; at retirement age y; benefit earned is BðyÞ; and at any age x between w and y; it equals an intermediate value BðxÞ; where 0 , BðxÞ , BðyÞ: The quantity BðxÞ will be termed the accrued benefit at age x: If BðyÞ equals the actual annual salary, the method of valuation will be called Traditional Unit Credit, but if the benefit rate is based on salary projected to the retirement age – we will call it Projected Unit Credit. The present value of accrued benefit for the jth participant at age x is D yj Bj ðxj Þ€að12Þ ; yj D xj where Dx ¼ lx vx ; lx denotes the cohort size at age x; surviving from the original newborn cohort size l0 ; v is the discount factor, i.e., v ¼ 1=ð1 þ iÞ; i the valuation rate (assumed long-term rate of return on plan assets). Note that the ratio Dyj Dxj

¼

l yj l xj

ð1 þ iÞxj 2yj ¼ ð12yj 2xj qxj Þð1 þ iÞxj 2yj

is calculated based on the tables of pension plan participation (analogue of ordinary mortality table for survival as a pension plan participants group), with the understanding that the current value at time t (corresponding to age xj of the jth plan participant) of a monetary unit, which will be paid at age yj ; is equal to ð1 þ iÞxj 2yj : Denote by At the group of all plan participants at time t; but only including active, not retired participants. If the plan assets at time t are X j[At

Bj ðxj Þ€að12Þ yj

D yj D xj

;

180

Chapter 5

then they are sufficient, independently of the age distribution among participants, to pay the cost of purchasing life annuities for all participants at retirement age y: In what follows we will assume that a participant attaining retirement age y purchases a life annuity with annual benefit BðyÞ and thus is no longer included in the balance sheet of assets and liabilities of the plan. In practice, of course, the participant may not receive the benefit in the form of such a benefit buyout, but instead receive a life annuity from the plan. We will, however, treat such a situation as essentially equivalent to a buyout, with such a participant considered to be a part of a separate plan, devoted only to payment of benefits. 5.1.1 Accrued liabilities The unit credit valuation method treats accrued liabilities of the plan as discounted accrued benefits of all active plan participants. Therefore, accrued liabilities of the plan at time t; denoted by ALt ; are equal to ALt ¼

X

Bj ðxj Þ€að12Þ yj

j[At

D yj D xj

ð5:1:1Þ

:

An ideally balanced plan has assets equal to ALt at every time t: The accrued liabilities change over time, because the group of active plan participants changes, and the participants who remain in the plan earn additional time of service. We will assume, for simplicity, that no new participants join the plan (we can assign them to a new, separate plan). This way, the group of active participants can only decline in size over time. Let T be the set of all participants who leave the plan in the time period between t and t þ 1; without attaining any benefit rights. Let R be the set of plan participants who retire during the same time period and start receiving benefits at age y: Then Atþ1 ¼ At w ðT < RÞ: We will now consider the change in accrued liabilities during the same year. We have, by definition ALtþ1 ¼

¼

X

Bj ðxj þ 1Þ€að12Þ yj

Dyj

j[Atþ1

Dxj þ1

X

Dyj

j[At


Dxj þ1

2

X j[TA X

Btj a€ ð12Þ y

t

tþ1

Therefore (5.4.7) becomes

PVFBtþ1 ¼ PVFBt ð1 þ iÞ þ

X

j Btþ1 a€ ð12Þ y

j[N

2

X

g j 2 PVFB tþ1

j[T

þ

X j[At >Atþ1

X

g j qx PVFB tþ1

j[At

DBtj a€ ð12Þ y

Dy Dxþ1

Dy : Dxþ1

! 2

X j[R

g j PVFB tþ1

210

Chapter 5

Subsequently, (5.4.8) becomes

ALtþ1 ¼ ALt ð1 þ iÞ þ

X


j[N

2

X

g j 2 PVFB tþ1

j[T

þ

X

Dy Dxþ1 !

g j qx PVFB tþ1

2

DBtj a€ ð12Þ y

j[At >Atþ1

g j PVFB tþ1

j[R

j[At

X

X

Dy 1 X Nxþ1 2 Ny 2 NCtþ1 ntþ1 j[A Dxþ1 Dxþ1 tþ1

1 X Nx 2 Ny þ ð1 þ iÞNCt : nt j[A Dx t

The formula (5.4.10), which gives us the actuarial gain, becomes

Ga ¼ ðI 2 iFt 2 IC þ IP Þ 2

X


j[N

þ

X

g j 2 PVFB tþ1

j[T

2

X

X

Dy Dxþ1 !

g j qx PVFB tþ1

j[At

DBtj a€ ð12Þ y

j[At >Atþ1

Dy 1 X Nxþ1 2 Ny þ NCtþ1 ntþ1 j[A Dxþ1 Dxþ1 tþ1

þ ð1 þ iÞNCt 1 2

1 X Nx 2 Ny nt j[A Dx

! :

t

(5.4.12) is transformed into X Nxþ1 2 Ny X Nx 2 Ny X Nxþ1 2 Ny ¼ 2 1 ð1 þ iÞ þ Dxþ1 Dx Dxþ1 N A A tþ1

t

2

X Nxþ1 2 Ny X Nxþ1 2 Ny 2 qx Dxþ1 Dxþ1 T A t

! ;

211

Valuation of pension plan liabilities

and (5.4.13) becomes ( "

1

Utþ1 ¼ Ut 2 X Nxþ1 2 Ny Dxþ1 A

Ut

X Nx 2 Ny

2 1 ð1 þ iÞ

Dx

At

tþ1

X Nxþ1 2 Ny þ 2 Dxþ1 N 2PVFBt ð1 þ iÞ 2

X Nxþ1 2 Ny X Nxþ1 2 Ny 2 qx Dxþ1 Dxþ1 T A t

X


N

X

þ

g j 2 PVFB tþ1

T

þ

X

!#

X

Dy Dxþ1 !

g j qx PVFB tþ1

At

g j 2 PVFB tþ1

X

DBj a€ ð12Þ y

At >Atþ1

R

) Dy þ ALtþ1 : Dxþ1

Furthermore, the formula (5.4.15) must be adjusted accordingly

1 Utþ1 ¼ Ut 2 X Nxþ1 2 Ny Dxþ1 A tþ1

( £

" ðPVFBt 2 ALt 2 NCt Þð1 þ iÞ þ

X g j PVFB

tþ1 2 Ut

T

# X Nxþ1 2 Ny j g qx PVFBtþ1 2 Ut 2 Dxþ1 At X Nxþ1 2 Ny j ð12Þ Dy 2 Btþ1 a€ y 2 Ut Dxþ1 Dxþ1 N 2

X At >Atþ1

þ

X R

DBj a€ ð12Þ y

Dy Dxþ1

) j g PVFBtþ1 2 PVFBt ð1 þ iÞ þ ALtþ1 :

Nxþ1 2 Ny Dxþ1

212

Chapter 5

And we finally arrive at the following analogue of the formula (5.4.16) 1

Utþ1 ¼ Ut 2 X Nxþ1 2 Ny Dxþ1 A

(

" I 2 iFt 2 IC þ IP þ

X

g j 2 Ut PVFB tþ1

T

Nxþ1 2 Ny Dxþ1

tþ1

# X Nxþ1 2 Ny j g 2 qx PVFB tþ1 2 Ut Dxþ1 At X j ð12Þ Dy X Nxþ1 2 Ny Dy 2 Btþ1 a€ y 2 Ut DBj a€ ð12Þ 2 y D D D xþ1 xþ1 xþ1 N A >A t

þ

X

!

tþ1

)

g j 2 P 2 IP 2 Ga : PVFB tþ1

ð5:6:2Þ

R

If we substitute in the above formula Ga ¼ 0; we obtain an analog of the formula (5.4.17). From (5.6.2) we infer that if based on the unit normal cost Ut, calculated at time t, expected actuarial present value at time t þ 1 of the future normal cost for the group N, Ut

X Nxþ1 2 Ny N

Dxþ1

;

is less than the actuarial present value (still at time t þ 1) of the future benefits for that group, i.e., less than X j Dy Btþ1 a€ ð12Þ ; y D xþ1 N then the new unit normal cost at time t þ 1; Utþ1 ; must be more than the unit normal cost without consideration for new plan participants. Even if the unit normal cost is less as a result of new participants joining the plan (if they are relatively young, the expected present value of their future service years is relatively large), the normal cost of the entire plan is bound to increase somewhat because of increased burden for the plan coming from the increased future benefits obligation (compare the formula (5.4.3)).

5.6.2 Unit credit method In the standard analysis of the unit credit method (Section 5.1) we assume that at the time of entering, the plan has no earned benefit rights. In practice, this is not always true, especially if plan participants can transfer pension rights when changing jobs. When an employee terminates employment before actual retirement,

213


such employee may have the following options for the pension rights earned up to date: † receive an equivalent lump sum payment, † transfer pension rights to a new employer, together with an appropriate plan assets transfer, † remain a participant in the current plan, if the new employer does not offer a pension plan or a transfer of pension rights is not possible. In the last case, because of the lack of future service years, pension benefits will be appropriately adjusted downward. If the current plan allows employee contributions (or, more generally, contributions from plan participants, even if they are not active employees)—such plans are termed contributory plans—then the employee may contribute plan participation by making appropriate contributions voluntarily. If, after some period of time, the employee under consideration returns to active employment with the plan sponsor, accrued liabilities for this employee may exceed plan assets set aside with respect to the said employee. Similarly, an employee entering a retirement plan with a new employer may acquire pension rights in the new plan, and receive asset transfer, which does not appropriately correspond to the pension rights acquired. In all such cases, we must modify the formulas expressing the accrued liabilities and the actuarial gain for the unit credit method. Thus, the formula (5.1.2), and the two expressions for ALtþ1 preceding it, must be corrected by adding to the right-hand side X


j[N

D yj Dxj þ1

:

The formula (5.1.2) then becomes ALtþ1 ¼

ALt þ

X

DBj a€ ð12Þ yj

j[At

2

X

2

j[R


! ð1 þ iÞ þ

Dxj


j[T

X

Dyj

X


j[N

D yj Dxj þ1

Dyj Dxj þ1

2

X

qxj Bj ðxj þ 1Þ€að12Þ yj

j[At

D yj Dxj þ1 ! Dyj

Dxj þ1

:

The formula (5.1.3), defining the normal cost, is unchanged. But the formula (5.1.5) becomes Ftþ1 ¼ Ft þ I þ C þ Tr 2 P;

214

Chapter 5

where Tr is the amount of the assets transferred, possibly zero, which is associated with the new participants entering the plan. The unfunded actuarial liability is therefore adjusted by the following quantity X


j[N

D yj Dxj þ1

ð5:6:3Þ

2 Tr 2 ITr ;

where ITr is the dollar return from the investment of the transferred assets Tr, calculated from the date of the transfer till the end of the year t þ 1: In particular, the formula (5.1.6) becomes UALtþ1 ¼ UALt ð1 þ iÞ 2 ðI 2 iFt 2 IC 2 ITr þ IP Þ 2 ½C þ IC 2 NCt ð1 þ iÞ ! X Dyj j ð12Þ 2 Tr þ ITr 2 B ðxj þ 1Þ€ayj Dxj þ1 j[N 2

X

j

B ðxj þ

1Þ€að12Þ yj

T

2

X R

j

B ðxj þ

1Þ€að12Þ yj

D yj Dxj þ1 D yj Dxj þ1

2

X

j

qxj B ðxj þ

At

1Þ€að12Þ yj

Dyj

!

Dxj þ1

! 2 P 2 IP :

Because the unfunded actuarial liability changed by the amount (5.6.3), the actuarial gain Ga is reduced by exactly the same amount, but the formula (5.1.7) remains correct. 5.6.3 The Entry Age Normal method In the Entry Age Normal method all participants entering the plan at time t þ 1 are of the same age x ¼ w: From the formula (5.2.5) we conclude that the accrued liability with respect to these new entrants is equal to zero. Thus the arrival of the new participants does not change the accrued liability of the plan, and its unfunded accrued liability, if we assume that the new participants do not bring any assets with them. The only modification needed in the formulas in Section 5.2 is that we should assume that DBj ¼ 0 on the set N. Then the formulas (5.2.11), (5.2.13) and (5.2.14) are unchanged, but the sum over the set Atþ1 includes the elements of the set N also. An alternative modification would be to sum over the set At > Atþ1 instead of the set Atþ1. 5.6.4 Individual level premium method In this method the accrued liability with respect to the new plan participant is, by definition, equal to zero, in a manner analogous to that for the Entry Age Normal

215


method. As a result of this situation, the actuarial gain is unaffected by new plan entrants. In the subsequent year t þ 1; the plan normal cost changes, as the normal cost associated with the new plan participants must be added.

5.7 Aggregate pension funding methods The pension funding methods we have studied up to this point all started by calculating the normal cost and the actuarial liability for each plan participant and then assessing those quantities for the entire plan by adding individual participants’ quantities. There is an alternative to this approach, although less frequently used. Instead of assigning liabilities to individual participants, one can assign assets of the plan to such individuals and then assess the funding by the need to pay the liabilities not covered by the assets in an amortization process over an appropriate time interval. All methods using this approach are termed aggregate methods. There is, however, a variety of aggregate approaches. The simplest is termed individual aggregate. In this method, one starts by assigning a share of the entire fund of assets held by the plan to each individual participant. There are various approaches used in such assignment. Assets can be assigned on the pro-rata basis in proportion to the present value of accrued benefits or by actuarial liabilities as established by another funding method (e.g., Entry Age Normal or individual level premium) or by some other reasonable approach. Once the assets are assigned to each individual, that individuals’ normal cost is established as NCj ¼

APVðBj Þ 2 F j a€ n

where NCj is this jth participant’s normal cost, APVðBj Þ that participant’s actuarial present value of accrued benefits, Fi that participant’s fund share, and a€ n an appropriate annuity (assumed here payable at the beginning of each year for a period of amortization n, typically until this individual’s normal retirement age). Such individual normal costs can then be summed up for the entire plan’s normal cost. One very important property of the individual aggregate method is that it never produces any unfunded actuarial liability. All funding needs are amortized into the normal cost. Thus the normal cost is likely to be larger than the one produced by methods developing unfunded liabilities. Some employers may desire lower normal cost, if funding of the normal cost is required statutorily, while supplementary cost may be delayed. On the other hand, the aggregate methodology produces realistic results. The second aggregate method is usually termed just aggregate, but sometimes, in contrast with the individual aggregate, it is also called aggregate aggregate.

216

Chapter 5

Its approach is very similar to the individual aggregate method, with one key difference: that the normal cost pays the difference between total assets and total plan liabilities in a form of so-called average annuity. Let N be the number of plan participants and F the total assets of the plan. Thus, the normal cost, total for the entire plan, under the aggregate method is given by the formula N X

NC ¼

APVðBj Þ 2 F

j¼1

a€

where a€ is the average annuity, a concept specific to aggregate methods. In case of funding cost defined as a dollar amount (as opposed to percentage of salaries), if xj is the jth participant’s age and rj that participant’s retirement age, such average annuity is defined as N X

a€ ¼

a€ xj : rj 2xj

j¼1

N

:

This concept is also used for plans that define normal cost as a percentage of payroll (i.e., combined wages of all employees). In that case, the average annuity is N X

a€ ¼

APVðAll Future Salaries j Þ

j¼1 N X

: j

APVðCurrent Salary Þ

j¼1

In practice, it is quite common to express the salary-based calculations with the salary-based commutation functions. A salary function Sx is a function giving the salary at age x, or an appropriate multiple or a fraction of salary. Most commonly, it is assumed that the salary increases by a constant percentage every year, so that we multiply current year’s salary by a constant factor 1 þ s: Another approach is to specify the salary function with the use of another function sx such that Sx =sx ¼ Sy =sy for any two ages x; y: While the aggregate method has a desirable property of producing no unfunded actuarial liability, it produces a relatively high normal cost. Because of this, alternative forms of aggregate approaches have been developed, which produce a separate initial unfunded liability, which is amortized and paid separately, with the payment termed supplemental cost. The first such method is called Frozen Initial Liability (Entry Age Normal), also abbreviated as FIL (EAN). In this method, the initial actuarial liability at the pension plan inception is calculated for each plan participant using the Entry Age Normal method and then summed over all plan participants. This sum is treated as a separate

217


Frozen Initial Liability, denoted by FIL. Then the normal cost is calculated as N X

NC ¼

APVðBj Þ 2 F 2 FIL

j¼1

a€

:

Another alternative is to calculate the Frozen Initial Liability using the unit credit method, and the resulting aggregate method is then termed Frozen Initial Liability (Attained Age Normal) or FIL (AAN). The normal cost under FIL (AAN) is higher than under FIL (EAN), because the initial unfunded actuarial liability is smaller, while the supplementary cost is smaller than under FIL (EAN).

Exercises Exercise 5.8.1 November 2000 SOA Course 5 Examination, Problem No. 3. You are given the following pension plan information. Normal retirement benefit is 1% final 1-year salary for each year of service. The valuation interest rate is 6% per year. Salary increase is 3% (used for Projected Unit Credit Cost method only). There are no pre-retirement deaths and terminations. Retirement age is 65. The life annuity factor at age 65 is 10. There is only one participant as of January 1, 2001, hired at age 30, and age 45 on January 1, 2001. The year 2001 annual salary of the participant is 50 000, while the year 2002 annual salary is 52 500. Under each of the Traditional Unit Credit Cost method and the Projected United Credit Cost method, calculate the following: the actual liability as of January 1, 2001, the expected liability as of January 1, 2002, the actual liability as of January 1, 2002; and the liability-side actuarial gain or loss at January 1, 2002. Compare and explain the difference between the liability gain or loss under the Traditional Unit Credit Cost method and the Projected Unit Credit Cost method. Solution Traditional Unit Credit. Under Traditional Unit Credit, annual benefit accrued at age x is Bx ¼ 0:01 · Sx · ðx 2 eÞ; where Sx is the salary at age x and e the entry age. In the case of this one participant, as of January 1, 2001, accrued liability is ð12Þ

B45 · v20 · 20 p45 · a€ 65 ¼ ð0:01 · 50 000 · 15Þ · 1:06220 · 1 · 10 ¼ 23 385:35: The expected liability as of January 1, 2002 is based on the 2001 annual salary rate of 50 000 (the 3% projected increase is only used for the Projected Unit Credit method), and it equals · v19 · BExpected 46

19 p45

ð12Þ

· a€ 65 ¼ ð0:01 · 50 000 · 16Þ · 1:06219 · 1 · 10 ¼ 26 441:04:

218

Chapter 5

However, the actual liability as of January 1, 2002 is based on the actual annual salary in 2002, i.e., 52 500. Therefore, it equals B46 · v19 ·

19 p45

ð12Þ

· a€ 65 ¼ ð0:01 · 52 500 · 16Þ · 1:06219 · 1 · 10 ¼ 27 763:09:

The actuarial gain due to the liability side of the balance sheet as of January 1, 2002 is the excess of the expected liability on January 1, 2002 over the actual liability on January 1, 2002, i.e., 26 441:04 2 27 763:09 ¼ 21; 322:05: This, of course, is an actuarial loss, as it is negative. Projected Unit Credit. This funding method uses the same basic methodology as the Traditional Unit Credit method, but the benefit rate is based on salary projected to the retirement age. This method is actually very commonly used in the United States, because it agrees with the standard generally accepted accounting principles (GAAP) concerning pension funding and valuation (as opposed to the statutory valuation performed by the actuary, which is the subject of this chapter). Salary increases are assumed to be 3% annually for this funding method. At age 45 (after 15 years of service), the annual benefit rate accrued to the participant is B45 ¼ 0:01 · 50 000 · 1:0320 · 15 < 13 545:83: This represents the annual benefit rate payable in equal monthly installments accrued as of January 1, 2001. The corresponding accrued liability as of the valuation date on January 1, 2001 is B45 ·

20 E45

ð12Þ

· a€ 65 < 13 545:83 ·

1 · 10 < 42 236:55: 1:0620

The expected liability as of January 1, 2002 is based on the 2001 annual salary rate of 50 000 increased by 3%, and it equals BExpected · 46

19 E46

ð12Þ

· a€ 65 ¼ ð0:01 · 50 000 · 1:03 · 1:0319 · 16Þ · 1:06219 · 1 · 10 < 47 755:46:

The actual liability as of January 1, 2002 is based on the actual salary in 2002, i.e., 52 500. Therefore, it equals ð12Þ

B46 · v19 · 19 p45 · a€ 65 ¼ ð0:01 · 52 500 · 1:0319 · 16Þ · 1:06219 · 1 · 10 < 48 682:75: The actuarial gain due to the liability side of the balance sheet as of January 1, 2002 is the excess of the expected liability on January 1, 2002 over the actual liability on January 1, 2002, i.e., 47 755:46 2 48 682:75 ¼ 2927:29: Again, this is an actuarial loss, as it is negative.

219


Both methods produce an actuarial loss, but the loss is smaller under Projected Unit Credit. This is caused by the fact that Traditional Unit Credit does not assume any salary increases, and while Projected Unit Credit assumes salary increases, the 3% assumed here turns out to be less than the actual salary increase (actual increase from 50 000 to 52 500 represents 5%). A Exercise 5.8.2 November 2000 SOA Course 5 Examination, Problem No. 18. For a newly established plan, you are given the following data. The normal retirement benefit is 10 per month per year of service. Normal retirement age is 65. Valuation interest rate is 6%. There are no pre-retirement terminations other than death. There are no pre-retirement death benefits. You are given this selected annuity ð12Þ value: a€ 65 ¼ 10: The method for amortization of Frozen Initial Liability is: level dollar amount over 20 years. Plan effective date is January 1, 2001. Plan assets as of January 1, 2001 are 0. The following selected commutation factors are given: D32 ¼ 1468;

N32 ¼ 23 018;

D33 ¼ 1382;

N33 ¼ 21 550;

D63 ¼ 215;

N63 ¼ 2287;

D65 ¼ 171;

N65 ¼ 1689:

There is only one participant, Mr Jones, who was hired on January 1, 1970 and born on January 1, 1938. For Mr Jones, calculate the normal cost under the following methods as of January 1, 2001: Frozen Initial Liability (Attained Age Normal), Frozen Initial Liability (Entry Age Normal), and Aggregate. Solution Frozen Initial Liability (Attained Age Normal). In this method, the normal cost is given by the following formula NC ¼

APVðBenefitsÞ 2 Assets 2 UAL ; a€ 63: 2

where the UAL (unfunded actuarial liability) is calculated initially using the Traditional Unit Credit funding method, and a€ 63: 2 is the average future service annuity (the notation is caused by the assumption of age 63 now, and 65 being the normal retirement age). The projected annual benefit at retirement is: 120 times 33 years of service, i.e., 3960. Its portion accrued through January 1, 2001 is 31 · 3960 ¼ 3720: 33 Accrued liability under the Traditional Unit Credit method is ð12Þ

3720 · a€ 65 ·

D65 171 < 29 586:98: ¼ 3720 · 10 · 215 D63

220

Chapter 5

There are no assets in the plan on January 1, 2001, and the above is therefore the unfunded actuarial liability on the valuation date. The actuarial present value of projected benefits is ð12Þ

3960 · a€ 65 ·

D65 171 < 31 495:81: ¼ 3960 · 10 · 215 D63

Combining this with the information about the unfunded actuarial liability we obtain the normal cost under the Frozen Initial Liability (Attained Age Normal) method NC ¼

APVðBenefitsÞ 2 Assets 2 UAL 31495:81 2 0 2 29 586:98 < 686:29: ¼ N63 2 N65 2287 2 1689 D63 215

Recall that under this method, the unfunded actuarial liability of 29 586.98 is amortized and paid separately, in level amortization over 20 years, giving the annual supplemental cost of 29586:98 < 2433:52: a€ 20 Frozen Initial Liability (Entry Age Normal). In this method, the calculation of normal cost is the same as in the Frozen Initial Liability (Entry Age Normal), but the initial unfunded actuarial liability is based on the Entry Age Normal accrued liability calculation. The Entry Age Normal method normal cost is (note that the entry age was 32) ð12Þ D65

APV32 ðBenefitsÞ ¼ a€ 32: 33

120 · 33 · a€ 65

N32 2 N65 D32

D32

171 1468 < 317:50: 23 018 2 1689 1468

120 · 33 · 10 · ¼

The Entry Age Normal accrued liabilities at age 63 (current attained age of the participant) can now be calculated as the excess of the actuarial present value of future benefits over the actuarial present value of future normal costs, i.e., as ð12Þ

120· 33· a€ 65 ·

D65 N 2 N65 2 317:50 · 63 < 31495:81 2 883:09 < 30612:72: D63 D63

Since there are no assets, the unfunded actuarial liability is 30 612.72. The normal cost is then calculated as NC ¼

APVðBenefitsÞ 2 Assets 2 UAL 31495:81 2 0 2 30 612:72 < 317:50: ¼ N63 2 N65 2287 2 1689 D63 215

221


Again, the Frozen Initial Liability of 30 612.72 is amortized and paid off separately, and the annual supplemental cost is 30612:72 < 2668:96: a€ 20 Aggregate method. In this method, the normal cost is calculated as NC ¼

APV63 ðBenefitsÞ 2 Assets 31 495:81 2 0 < 11323:74: ¼ N63 2 N65 2287 2 1689 D63 215

There is no supplemental cost funding in this method.

A

Exercise 5.8.3 November 2002 SOA Course 5 Examination, Problem No. 15, Section C. You are given the following for a pension plan covering one individual. The pension formula is: 2.0% of salary times the number of years of participation in plan up to 25 years of service, plus 1.5% times the number of years of participation in plan in excess of 25 years of service. The pension benefit is payable as a life annuity. The valuation date is January 1, 2002. The valuation interest rate is 8.0%. The normal retirement age is 65. There are no terminations or deaths assumed prior to normal retirement age. The date of birth of the sole participant is January 1, 1952, the date of hire is January 1, 1977, and the effective date of participation is January 1, 1977. The 2001 annual salary of the participant is 100 000, and the future salary is expected to stay the same. You are also given that the actuarial present value at age 65 of 1 payable for life in the manner provided by this plan’s pension (assumed to be monthly), starting at age 65, is 10. The assets of the plan on January 1, 2002 are 150 000, while on December 31, 2002 they are 180 000. Calculate the normal cost and the unfunded actuarial liability as of January 1, 2002 under the following cost methods: Traditional Unit Credit, Projected Unit Credit, Entry Age Normal (level percentage of pay), and Aggregate (level percentage of pay). Solution Traditional Unit Credit. As of January 1, 2002, the participant has 25 years of service. Under this method, the benefit accrued up to date is B50 ¼ 0:02 · 25 · 100 000 ¼ 50 000: The accrued liability is AL50 ¼ B50 ·

15 E50

ð12Þ

· a€ 65 ¼ 50 000 · 1:08215 · 10 < 157 620:85:

The unfunded actuarial liability is therefore UAL ¼ AL 2 F ¼ 157 620:85 2 150 000 ¼ 7620:85:

222

Chapter 5

The normal cost is the actuarial present value of retirement benefit accrued in this year of service (note that only 1.5% of salary accrues this year, as this is the 26th year of service): NC50 ¼ 0:015 · 100 000 · 10 · 1:08215 < 4728:63: Projected Unit Credit. The only (and key) difference between this method and the Traditional Unit Credit is the assumption of salary increases in the Projected Unit Credit. In this problem, no salary increases are assumed, thus the results are the same for the Projected Unit Credit as for the Traditional Unit Credit. Entry Age Normal (level percentage of pay). Note again that no future salary increases are assumed. This also means that no past salary increases should be assumed, and a level percentage of pay is a fixed nominal monetary amount. The entry age for the participant was 25, and at retirement, the participant will have 40 years of service. The normal cost will be paid as an annuity over the entire service period, so that NC25 a€ 25: 40 ¼ APV25 ðBenefitsÞ: The actuarial present value of benefits at the entry age is APV25 ðBenefitsÞ ¼ B65 ·

40 E25

ð12Þ

· a€ 65

¼ ð0:02 · 25 þ 0:015 · 15Þ · 100 000 · 1:08240 · 10 < 33 372:43: Therefore NC25 ¼

APV25 ðBenefitsÞ 33 372:43 33 372:43 < 2591:31: < < a€ 25: 40 a€ 40 0:08 12:879

This represents approximately 2.59% of salary. Furthermore, the accrued liabilities as of age 50 are AL50 ¼ APV50 ðBenefitsÞ 2 NC25 · a€ 50: 15 < ð0:02 · 25 þ 0:015 · 15Þ · 100 000 · 10 · 1:08215 2 2591:31 · a€ 15 < 204 595:56 2 23 954:68 ¼ 228 550:24: Unfunded actuarial liability is UAL50 ¼ 228 550:24 2 150 000 ¼ 78 550:24:


223

Aggregate (level percentage of pay). Under this method, the normal cost calculation is performed as follows: NC50 ¼

APV50 ðBenefitsÞ 2 F a€ 50: 15

¼

ð0:02 · 25 þ 0:015 · 15Þ · 100 000 · 10 · 1:08215 2 150 000 a€ 15

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