Robert_V._Hogg,_Joseph_W._McKean,_Allen_T._Craig

7.4. Completeness and Uniqueness 433

The statement thatY 1 is a sufficient statistic for a parameterθ, θ∈Ω, and that
the family{fY 1 (y 1 ;θ):θ∈Ω}of probability density functions is complete is lengthy
and somewhat awkward. We shall adopt the less descriptive, but more convenient,
terminology thatY 1 is acomplete sufficient statisticforθ. In the next section,
we study a fairly large class of probability density functions for which a complete
sufficient statisticY 1 forθcan be determined by inspection.

Example 7.4.2(Uniform Distribution).LetX 1 ,X 2 ,...,Xnbe a random sample
from the uniform distribution with pdff(x;θ)=1/θ, 0 <x<θ,θ>0, and
zero elsewhere. As Exercise 7.2.3 shows,Yn=max{X 1 ,X 2 ,...,Xn}is a sufficient
statistic forθ. It is easy to show that the pdf ofYnis

g(yn;θ)=

{

nynn−^1

θn^0 <yn<θ

0elsewhere.

(7.4.1)

To show thatYnis complete, suppose for any functionu(t)andanyθthatE[u(Yn)] =
0; i.e.,

0=

∫θ

0

u(t)

ntn−^1

θn

dt.

Sinceθ>0, this equation is equivalent to

0=

∫θ

0

u(t)tn−^1 dt.

Taking partial derivatives of both sides with respect toθand using the Fundamental
Theorem of Calculus, we have

0=u(θ)θn−^1.

Sinceθ>0,u(θ)=0,forallθ>0. ThusYnis a complete and sufficient statistic
forθ. It is easy to show that

E(Yn)=

∫θ

0

y

nyn−^1

θn

dy=

n

n+1

θ.

Therefore, the MVUE ofθis ((n+1)/n)Yn.

EXERCISES

7.4.1.Ifaz^2 +bz+c= 0 for more than two values ofz,thena=b=c=0. Use

this result to show that the family{b(2,θ):0<θ< 1 }is complete.

7.4.2.Show that each of the following families is not complete by finding at least

one nonzero functionu(x) such thatE[u(X)] = 0, for allθ>0.

(a)

f(x;θ)=

{ 1

2 θ −θ<x<θ, where 0<θ<∞

0elsewhere.