Page 160 - Elements of Distribution Theory

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146 Parametric Families of Distributions

n

provided that t = y j . This probability simplifies to
j=1
1
n

t
n
for all y 1 ,..., y n taking values in the set {0, 1} such that j=1 y j = t.
That is, given that n Y j = t, each possible arrangement of 1s and 0s summing to t
j=1
is equally likely.
Example 5.15 (Exponential random variables). Let Y 1 ,..., Y n denote independent, iden-
tically distributed random variables each distributed according to the absolutely continuous
distribution with density function

θ exp{−θy}, y > 0
where θ> 0. Then Y = (Y 1 ,..., Y n ) has model function
n

n
n
θ exp −θ y j , y = (y 1 ,..., y n ) ∈ (0, ∞) ;
j=1
hence, this is a one-parameter exponential family of distributions with natural parameter
n
η =−θ, H = (−∞, 0), and T (y) = y j .
j=1
n
Let a 1 ,..., a n denote real-valued, nonzero constants and let Z = a j log Y j . Then
j=1
Z has moment-generating function
n n

a j t

M Z (t; θ) = E exp t a j log(Y j ) ; θ = E Y j ; θ
j=1 j=1
n n
(a j t + 1) j=1 (a j t + 1)
= a j t+1 = n , |t| < 1/ max(|a 1 |,..., |a n |).
θ n+t j=1 a j
j=1 θ
n
It follows that the distribution of Z does not depend on θ if and only if j=1 a j = 0.
n n
Hence, since H 0 = H,by Theorem 5.4, j=1 a j log(Y j ) and j=1 Y j are independent

n
if and only if j=1 a j = 0.
In applying the second part of Theorem 5.4 it is important that H 1 contains an open
m
subset of R . Otherwise, the condition that the distribution of Z does not depend on θ ∈
is not strong enough to ensure that Z and T (Y) are independent. The following example
illustrates this possibility.
Example 5.16. Let Y 1 and Y 2 denote independent Poisson random variables such that Y 1 has
mean θ and Y 2 has mean 1 − θ, where 0 <θ < 1. The model function for the distribution
of Y = (Y 1 , Y 2 ) can then be written
1
exp{log θy 1 + log(1 − θ)y 2 } , y 1 = 0, 1,... ; y 2 = 0, 1,....
y 1 !y 2 !
Hence, c(θ) = (log θ, log(1 − θ)) and T (y) = (y 1 , y 2 ).
Let Z = Y 1 + Y 2 . Then, by Example 4.15, Z has a Poisson distribution with mean
θ + (1 − θ) = 1so that the distribution of Z does not depend on θ.However, Z and (Y 1 , Y 2 )
are clearly not independent; for instance, Cov(Z, Y 1 ; θ) = θ.