Unbiassed Minimum Variance Estimation in a Class of Discrete Distributions

UNBIASSED MINIMUM VARIANCE ESTIMATION CLASS OF DISCRETE DISTRIBUTIONS JOGABRATA ROY

By

Indian

IN A

and SUJIT KUMAR MITRA

Statistical

Calcutta

Institute,

The problem of estimation of parameters involved in discrete distributions like the and Poisson Binomial and their truncated forms have been considered Binomial, Negative Most of them, e.g., Fisher (1936, 1941), Haldane (1941), Finney (1949, by various authors. 1955) David and Johnson (1952) discuss computational aspects of estimation by maximum likelihood while a few others e.g., Plackett Moore Rider (1953), (1952, 1954), (1953, 1955) of estimation which at times lead to unbiassed give other simpler but inefficient methods estimates.

In this paper, for a wide class of discrete distributions involving one unknown para minimum variance of the parameter the uniformly unbiassed estimate is (UMVU) It is shown that results for derived and the UMVU estimate of its variance is also obtained.

meter

from those for the complete disbribution by are treated as the Poisson distributions and The binomial operation. negative cases. Tables of the UMVU estimate are given for the Poisson distribution truncated

special at

at zero can be obtained

truncated

distributions a difference

zero,

for

up to

size

sample

ESTIMATES IN A CLASS OF DISCRETE DISTRIBUTIONS

1. UMVU Consider

the following prob

where

d>

ten.

0 is an unknown

discrete {X

==

probability x}

=

f(0)

=

x=

a(x)d*lf(d),

0 does not

a(x) >

parameter,

distribution

... ? a(x)0*.

defined

0, 1, 2, ..., involve

by Noak

(1950).

...

(1.1)

0 and

(1.2)

loss of generality, we shall assume a(0) ? 1. The Poisson, occur as special cases of the above. series distributions Logarithmic

Without

Negative

Binomial

and

To derive uniformly minimum variance unbiassed (UMVU) estimate for 6 on the basis = 1, ... n from of size the distribution 2, n) (1.1) we require the sample Xi(i stated below without lemmas proof.

of a random

Lemma

1.1:

If

as

tr (x) is defined tr(x)

?

=

0

a(x?r)

_v?^_

for x < r

for

r

"1 J. ...

^

x >

r

(1.3)

i J

a(x) then E{tr(X)} = Lemma

6r. 1.2:

T =

n 2 X? is a complete sufficient statistic for 6 in the sense of Lehmann

and Scheff? (1950). 371

SANKHY? :THE INDIAN JOURNAL OF STATISTICS

Vol. 18 ] Lemma

The probability

1.3:

prob

=

{T

t)

=

over

summation

denoting

= xx+x2+...+xn

ate),

values

integral

any

positive

of

..., xn

xl9 x2,

to

subject

r,

integer

ur(t) =0 C(t-r9

for

.


r

n)

C(t9 n) and noting that the probability 1.1 that Lemma

distribution

of T is of the

we may

note

that

ur(t)

...

(1.5)

same form as (1.1) it follows

from

=

E{ur(T)}

Incidentally

(1.4)

..

for

Defining

non-negative

...

0, 1, 2, ...,

of {f(6)}n and it may be expressed alternatively

C(t, w)=Sn

S

is given by

= C(t, n) &?f(6)n> t

in the expansion

where C(t, n) is the coefficient of $ as

of T

distribution

[Parts 3 & 4

.

'...

(1.6)

i-i

=

U u^t?i). ...

(1.7)

.=o

of the sufficient statistic T it follows from Rao-Blackwell (1.6) and the completeness 1947 and Rao, 1945) that ur (T) is the UMVU Theorem estimate of 6r. The (Blackwell, variance of ux(T)9 the UMVU estimate of 6 is given by From

V{ux(T)} Hence

UMVU

we

have

the

following:

Theorem

1.1:

For

estimate

because

of the variance

of ux(T)

2. Consider distribution

UMVU

are defined

prob

estimate for d is ux(T) and

the

(1.9)

...

%(%(?)-Wl(T-l)}

(1.10)

by (1.5).

ESTIMATE IN THE CASE OF TRUNCATION ON THE LEFT

the distribution

can then be

(1.1) the UMVU is

ux(T)-u2(T) ...

=

^) r =1,2,

=

(1.8)

in the form:

be put

of (1.1) may

where ur (T) for

E{ux(T)}2 -Q2....

the distribution

v(T) which

=

written {X

=

x}

on the

(1.1) truncated as =

a(x)6xlf8(6),

where f)=L(?)^,

x=

s, s+1,

... a*_=

372

left at x =

s+2,

(2.2)

...,

s?1.

The probability

...

(2.1)

UMVU ESTIMATION IN A CLASS OF DISCRETE DISTRIBUTIONS By an argument exactly = a random sample X{(i for & is ur8(T) where

to the one used

similar

1, it follows

in section

1, 2, ..., n) of size n from the distribution

for

.


over

summation

denoting

The

=

relation

following

we write For

may

CQ(t, n)

=

difference

=

operator

form

t

C8(t,n)

forr =

Complete Negative Negative

{X

=

x]

...

on G(t, x) regarded operator

(2.6)

as a function

of the integer x.

by

y)>

: ...

(2.8)

Illustrations

Binomial

Distribution. distribution:

*" (*+^1)|\K?1):

1, we have

...

Ak defined

^+i)-?WM

Binomial

=

is s ?

(2.7)

A8_1 C8^{t-n(s-l),0}. 3.

prob

(2.5)

:

A?G(t,x)\^

the bivariate

introducing

(2.5) in the alternative

complete

...

n-j),

UA C(t,n-j),

sVl)j(

A* /(?. y) =/(*+*.

form of the

=

established,

at zero, that

of truncation

in the alternative

be expressed

In general,

3.1.

...+xn

xl+x2

C(?, .i).

case of interest

the special

where A is the ordinary

can write

to

subject

I n. {a(s-l)YC8_1{t-j(s-l), )

Cx(t,n)

we

..., xn

x2,

Cs(t, n) can be easily

connecting

Cx(t,n)= which

xv

1, 2, ... n).

C,(t, n) ^(-ly where

(2.4)

i_=l

of

values

integral

II afo),...

?_

XI

0*1(1-0)-*,

373

We

x=

shall, consider

0,1,

2, ...,

the

following

(3.1)

Vol. 18 ]

SANKHY? :THE INDIAN JOURNAL OF STATISTICS

[ Parts 3 & 4

and take k as a known well-known Negative

It is positive integer as in problems of inverse binomial sampling. that the total T of a random sample of size n from this distribution has again the Binomial distribution.

prob

-

{T

.}

=

so that here C(t, n) =

Hence

for the complete Result

from

3.1:

the Negative

The UMVU Binomial

... .

(3.1)

of the variance

estimate

(3.2)

of a random

sample

of size n

is ...

T?(kn+T-1), and the UMVU

...

the following. of 6 on the basis

estimate

distribution

.

0, 1, 2, ...,

(3.3)

^-+t~W

we have

distribution

*=

t\ 0*1(1-0)'^

(^??^| (fen?1)!

(3.4)

is

of this estimate (kn-l)T

'

*

;

(kn+T-~i)2(kn+T-2)

where

T

is the

3.2.

total

Poisson

of

the

sample. truncated on the left at

distribution

zero.

For

the

complete

Poisson

distribution prob

it is well known

{X

=

x}

=

is again of the Poisson

form.

{T

=

t}

for a Poisson

=

(3.6)

of size n is given by

...

(3.7)

7^-dtlene9

=

... "?[.

(3.8)

on the left at zero, with

truncated

distribution

sample

...

Hence, C(i, ?)

Therefore

0, 1, 2, ...,

of the total T of a random

that the distribution prob

which

x=

-L xl dx?ee,

probability

law given.

by prob

we

have

which may

from

(3.3) and

be written

{X

(2.7)

=

x}^~

6xl(ee-l)9

Cx (., n)

in the usual

=

x=

~-\

a;=

1, 2, ...,

Q

form:

Ci(i.?)-^ Using

(2.3) and the results

...

of Theorem

1.1 we

thus have

374

the following: