Page 159 - Elements of Distribution Theory

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052184472Xc05 CUNY148/Severini May 24, 2005 17:53

5.3 Exponential Family Models 145

a bounded function on the range of Z; then there exists a δ 1 > 0 such that exp[tg(Z)] is
bounded for |t| <δ 1 . Hence, by Lemma 5.2, for any η ∈ H,
T
E{exp[tg(Z)]; η}= exp{k(η 0 ) − k(η)}E[exp[tg(Z)] exp{(η − η 0 ) T (Y)}; η 0 ].
Using the fact that
T
E[exp{(η − η 0 ) T (Y)}; η 0 ] = exp{k(η) − k(η 0 )},
it follows that, for all η ∈ H and all |t| <δ 1 ,
T
T
E{exp[tg(Z) + (η − η 0 ) T (Y)]; η 0 }= E{exp[tg(Z)]; η}E[exp{(η − η 0 ) T (Y)}; η 0 ].
For δ> 0, let

H(δ) ={η ∈ H: ||η − η 0 || <δ}
m
and let δ 2 > 0be such that H(δ 2 ) ⊂ H 1 ; since H 1 is an open subset of R and η 0 ∈ H 1 ,
such a δ 2 must exist. Then, since the distribution of g(Z) does not depend on η for η ∈ H 0 ,
for η ∈ H(δ 2 ) and |t| <δ 1 ,
E{exp[tg(Z)]; η}= E{exp[tg(Z)]; η 0 }.

It follows that, for all η ∈ H(δ 2 ) and all |t| <δ 1 ,
T
T
E{exp[tg(Z) + (η − η 0 ) T (Y)]; η 0 }= E{exp[tg(Z)]; η 0 }E[exp{(η − η 0 ) T (Y)}; η 0 ].
That is, the joint moment-generating function of g(Z) and T (Y) can be factored into the
product of the two marginal moment-generating functions. Hence, by Corollary 4.2 g(Z)
and T (Y) are independent and by part (ii) of Theorem 2.1, Z and T (Y) are independent,
proving part (ii) of the theorem.

Example 5.14 (Bernoulli random variables). Let Y = (Y 1 ,..., Y n ) where Y 1 ,..., Y n are
independent random variables such that
Pr(Y j = 1; θ) = 1 − Pr(Y j = 0; θ) = θ, j = 0,..., n,

where 0 <θ < 1. Then, for all y 1 ,..., y n in the set {0, 1},
n n
Pr{Y = (y 1 ,..., y n ); θ}= θ j=1 y j (1 − θ) n− j=1 y j .
It follows that the model function of Y can be written

n
θ
exp log y j + n log(1 − θ) ,
1 − θ
j=1
and, hence, this is a one-parameter exponential family of distributions, with natural param-
n
eter η = log θ − log(1 − θ) and T (y) = y j .
j=1
We have seen that the distribution of T (Y)isa binomial distribution with parameters n
and θ. Hence,
n n
θ j=1 y j (1 − θ) n− j=1 y j
Pr{Y = (y 1 ,..., y n )|T (Y) = t; θ}= ,
n t n−t

t θ (1 − θ)