Robert_V._Hogg,_Joseph_W._McKean,_Allen_T._Craig

7.5. The Exponential Class of Distributions 437

3. Var(Y 1 )=np′(^1 θ) 3 {p′′(θ)q′(θ)−q′′(θ)p′(θ)}.

Example 7.5.1. LetXhave a Poisson distribution with parameterθ∈(0,∞).
Then the support ofXis the setS={ 0 , 1 , 2 ,...}, which does not depend onθ.
Further, the pmf ofXon its support is

f(x, θ)=e−θ

θx

x!

=exp{(logθ)x+log(1/x!) + (−θ)}.

Hence the Poisson distribution is a member of the regular exponential class, with
p(θ) = log(θ),q(θ)=−θ,andK(x)=x. Therefore, ifX 1 ,X 2 ,...,Xndenotes
a random sample onX, then the statisticY 1 =

∑n
i=1Xiis sufficient. But since
p′(θ)=1/θandq′(θ)=−1, Theorem 7.5.1 verifies that the mean ofY 1 isnθ.It
is easy to verify that the variance ofY 1 isnθalso. Finally, we can show that the
functionR(y 1 ) in Theorem 7.5.1 is given byR(y 1 )=ny^1 (1/y 1 !).

For the regular case of the exponential class, we have shown that the statistic
Y 1 =

∑n
1 K(Xi) is sufficient forθ. We now use the form of the pdf ofY^1 given in
Theorem 7.5.1 to establish the completeness ofY 1.

Theorem 7.5.2.Letf(x;θ),γ<θ<δ, be a pdf or pmf of a random variableX
whose distribution is a regular case of the exponential class. Then ifX 1 ,X 2 ,...,Xn
(wherenis a fixed positive integer) is a random sample from the distribution ofX,
the statisticY 1 =

∑n
1 K(Xi)is a sufficient statistic forθand the family{fY 1 (y^1 ;θ):
γ<θ<δ}of probability density functions ofY 1 is complete. That is,Y 1 is a
complete sufficient statistic forθ.

Proof: We have shown above thatY 1 is sufficient. For completeness, suppose that
E[u(Y 1 )] = 0. Expression (7.5.2) of Theorem 7.5.1 gives the pdf ofY 1. Hence we
have the equation
∫

SY 1

u(y 1 )R(y 1 )exp{p(θ)y 1 +nq(θ)}dy 1 =0

or equivalently since exp{nq(θ)}
=0,
∫

SY 1

u(y 1 )R(y 1 )exp{p(θ)y 1 }dy 1 =0

for allθ. However,p(θ) is a nontrivial continuous function ofθ, and thus this
integral is essentially a type of Laplace transform ofu(y 1 )R(y 1 ). The only function
ofy 1 transforming to the 0 function is the zero function (except for a set of points
with probability zero in our context). That is,

u(y 1 )R(y 1 )≡ 0.

However,R(y 1 ) =0forally 1 ∈SY 1 because it is a factor in the pdf ofY 1. Hence
u(y 1 )≡0 (except for a set of points with probability zero). Therefore,Y 1 is a
complete sufficient statistic forθ.