438 SufficiencyThis theorem has useful implications. In a regular case of form (7.5.1), we can
see by inspection that the sufficient statistic isY 1 =
∑n
1 K(Xi). If we can see how
to form a function ofY 1 ,sayφ(Y 1 ), so thatE[φ(Y 1 )] =θ, then the statisticφ(Y 1 )
is unique and is the MVUE ofθ.
Example 7.5.2.LetX 1 ,X 2 ,...,Xndenote a random sample from a normal dis-
tribution that has pdf
f(x;θ)=1
σ√
2 πexp[
−(x−θ)^2
2 σ^2]
, −∞<x<∞, −∞<θ<∞,or
f(x;θ)=exp(
θ
σ^2x−x^2
2 σ^2−log√
2 πσ^2 −θ^2
2 σ^2)
.Hereσ^2 is any fixed positive number. This is a regular case of the exponential class
with
p(θ)=
θ
σ^2,K(x)=x,H(x)=−x^2
2 σ^2−log√
2 πσ^2 ,q(θ)=−θ^2
2 σ^2.Accordingly,Y 1 =X 1 +X 2 +···+Xn=nXis a complete sufficient statistic for
the meanθof a normal distribution for every fixed value of the varianceσ^2 .Since
E(Y 1 )=nθ,thenφ(Y 1 )=Y 1 /n=Xis the only function ofY 1 that is an unbiased
estimator ofθ; and being a function of the sufficient statisticY 1 , it has a minimum
variance. That is,Xis the unique MVUE ofθ. Incidentally, sinceY 1 is a one-to-one
function ofX,Xitself is also a complete sufficient statistic forθ.
Example 7.5.3(Example 7.5.1, Continued).Reconsider the discussion concerning
the Poisson distribution with parameterθfound in Example 7.5.1. Based on this
discussion, the statisticY 1 =
∑n
i=1Xiwas sufficient. It follows from Theorem
7.5.2 that its family of distributions is complete. SinceE(Y 1 )=nθ, it follows that
X=n−^1 Y 1 is the unique MVUE ofθ.EXERCISES
7.5.1.Write the p dff(x;θ)=1
6 θ^4x^3 e−x/θ, 0 <x<∞, 0 <θ<∞,zero elsewhere, in the exponential form. IfX 1 ,X 2 ,...,Xnis a random sample from
this distribution, find a complete sufficient statisticY 1 forθand the unique function
φ(Y 1 ) of this statistic that is the MVUE ofθ.Isφ(Y 1 ) itself a complete sufficient
statistic?
7.5.2.LetX 1 ,X 2 ,...,Xndenote a random sample of sizen>1 from a distribution
with pdff(x;θ)=θe−θx, 0 <x<∞, zero elsewhere, andθ>0. ThenY=
∑n
1 Xi
is a sufficient statistic forθ.Provethat(n−1)/Yis the MVUE ofθ.