Page 157 - Elements of Distribution Theory

P1: JZP
052184472Xc05 CUNY148/Severini May 24, 2005 17:53

5.3 Exponential Family Models 143

Theorem 5.3. Let Y denote a random variable with model function of the form
T
exp{c(θ) T (y) − A(θ)}h(y), y ∈ Y,
where θ ∈ . Then the conditional distribution of Y given T (Y) does not depend on θ.

Proof. Let η = c(θ), H denote the natural parameter space of the model, and
H 0 ={η ∈ H: η = c(θ),θ ∈ }.

Then the model function for this model can be written
T
exp{η T (y) − k(η)}h(y), y ∈ Y,
where η ∈ H 0 . Hence, it suffices to show that the conditional distribution of Y given T (Y),
based on this model, with the parameter space enlarged to H, does not depend on η.
We prove this result by showing that for any bounded, real-valued function g on Y,
E[g(Y)|T ; η] does not depend on η.
Fix η 0 ∈ H. The idea of the proof is that the random variable
Z = E[g(Y)|T ; η 0 ]

satisfies
E[Zh(T ); η] = E[g(Y)h(T ); η]
for any η ∈ H, for all bounded functions h of T . Hence, by Theorem 2.6,

Z = E[g(Y)|T ; η].
That is, for all η 0 ,η ∈ H,
E[g(Y)|T ; η] = E[g(Y)|T ; η 0 ],

which proves the result.
Wenowconsiderthedetailsoftheargument.Leth denoteabounded,real-valuedfunction
on the range of T . Then, since Z and g(Y) are bounded,
E[|Zh(T )|; η] < ∞ and E[|g(Y)h(T )|; η] < ∞;

by Lemma 5.2,
T
E[Zh(T ); η] = exp{k(η) − k(η 0 )}E[Zh(T )exp{(η − η 0 ) T }; η 0 ]
and

T
E[g(Y)h(T ); η] = exp{k(η) − k(η 0 )}E[g(Y)h(T )exp{(η − η 0 ) T }; η 0 ].
Let
T
h 0 (T ) = h(T )exp{(η − η 0 ) T }.
Note that
E[|h 0 (T )|; η 0 ] = exp{k(η 0 ) − k(η)}E[|h(T )|; η] < ∞.

It follows that
E[Zh 0 (T ); η 0 ] = E[g(Y)h 0 (T ); η 0 ]