Page 153 - Elements of Distribution Theory

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

5.3 Exponential Family Models 139

The model function (5.2) is called the canonical form of the model function and the
parameter η is called the natural parameter of the exponential family distribution. Note
that the function k can be obtained from T, h, and Y.For instance, if the distribution is
absolutely continuous, we must have

T
exp{η T (y) − k(η)}h(y) dy = 1,η ∈ H 0
Y
so that

T
k(η) = log exp{η T (y)}h(y) dy.
Y
The set

T
m
H ={η ∈ R : exp{η T (y)}h(y) dy < ∞}
Y
m
is the largest set in R for which (5.2) defines a valid probability density function; it is
called the natural parameter space.A similar analysis, in which integrals are replaced by
sums, holds in the discrete case.
Consider an exponential family of distributions with model function of the form
T
exp{η T (y) − k(η)}h(y), y ∈ Y,
where η ∈ H 0 and H 0 is a subset of the natural parameter space. In order to use this family
of distributions as a statistical model, it is important that the parameter η is identifiable. The
following result shows that this holds provided that the distribution of T (Y) corresponding
to some η 0 ∈ H is nondegenerate; in this case, we say that the rank of the exponential family
is m, the dimension of T .

Lemma 5.1. Consider an m-dimensional exponential family with model function
T
exp{η T (y) − k(η)}h(y), y ∈ Y,
where η ∈ H 0 ⊂ H. The parameter η is identifiable if and only if T (Y) has a nondegenerate
distribution under some η 0 ∈ H.

Proof. We consider the case in which the distribution is absolutely continuous; the argu-
ment for the discrete case is similar. The parameter η is not identifiable if and only if there
exist η 1 ,η 2 ∈ H 0 such that

T T
exp{η T (y) − k(η 1 )}h(y) dy = exp η T (y) − k(η 2 ) h(y) dy
1 2
A A
for all A ⊂ Y. That is, if and only if

T
T
exp{(η 1 − η 0 ) T (y) − [k(η 1 ) − k(η 0 )]} exp η T (y) − k(η 0 ) h(y) dy
0
A

T
T

= exp{(η 2 − η 0 ) T (y) − [k(η 2 ) − k(η 0 )]} exp η T (y) − k(η 0 ) h(y) dy
0
A
where η 0 is an arbitrary element of H.
This is true if and only if
T T
exp η T (Y) − k(η 1 ) = exp η T (Y) − k(η 2 )
1 2