Robert_V._Hogg,_Joseph_W._McKean,_Allen_T._Craig

7.2. A Sufficient Statistic for a Parameter 421

provided thatx 1 ,x 2 ,...,xnare such that the fixedy 1 =u 1 (x 1 ,x 2 ,...,xn), and
equals zero otherwise. We say thatY 1 =u 1 (X 1 ,X 2 ,...,Xn)isasufficient statistic
forθif and only if this ratio does not depend uponθ. While, with distributions of
the continuous type, we cannot use the same argument, we do, in this case, accept
the fact that if this ratio does not depend uponθ, then the conditional distribution
ofX 1 ,X 2 ,...,Xn,givenY 1 =y 1 , does not depend uponθ. Thus, in both cases, we
use the same definition of a sufficient statistic forθ.

Definition 7.2.1.LetX 1 ,X 2 ,...,Xndenote a random sample of sizenfrom a

distribution that has pdf or pmff(x;θ),θ∈Ω.LetY 1 =u 1 (X 1 ,X 2 ,...,Xn)be a

statistic whose pdf or pmf isfY 1 (y 1 ;θ).ThenY 1 is asufficient statisticforθif

and only if

f(x 1 ;θ)f(x 2 ;θ)···f(xn;θ)

fY 1 [u 1 (x 1 ,x 2 ,...,xn);θ]

=H(x 1 ,x 2 ,...,xn),

whereH(x 1 ,x 2 ,...,xn)does not depend uponθ∈Ω.

Remark 7.2.1.In most cases in this book,X 1 ,X 2 ,...,Xnrepresent the observa-

tions of a random sample; that is, they are iid. It is not necessary, however, in more

general situations, that these random variables be independent; as a matter of fact,

they do not need to be identically distributed. Thus, more generally, the definition

of sufficiency of a statisticY 1 =u 1 (X 1 ,X 2 ,...,Xn) would be extended to read that

f(x 1 ,x 2 ,...,xn;θ)

fY 1 [u 1 (x 1 ,x 2 ,...,xn);θ)]

=H(x 1 ,x 2 ,...,xn)

does not depend uponθ∈Ω, wheref(x 1 ,x 2 ,...,xn;θ) is the joint pdf or pmf of
X 1 ,X 2 ,...,Xn. There are even a few situations in which we need an extension like
this one in this book.

We now give two examples that are illustrative of the definition.

Example 7.2.2.LetX 1 ,X 2 ,...,Xnbe a random sample from a gamma distribu-
tion withα=2andβ=θ>0. Because the mgf associated with this distribution
is given byM(t)=(1−θt)−^2 ,t< 1 /θ,themgfofY 1 =

∑n

i=1Xiis

E[et(X^1 +X^2 +···+Xn)]=E(etX^1 )E(etX^2 )···E(etXn)

=[(1−θt)−^2 ]n=(1−θt)−^2 n.

ThusY 1 has a gamma distribution withα=2nandβ=θ,sothatitspdfis

fY 1 (y 1 ;θ)=

{ 1

Γ(2n)θ^2 ny

2 n− 1

1 e

−y 1 /θ 0 <y 1 <∞

0elsewhere.

Thus we have
[
x^21 −^1 e−x^1 /θ
Γ(2)θ^2

][

x^22 −^1 e−x^2 /θ

Γ(2)θ^2

]

···

[

x^2 n−^1 e−xn/θ

Γ(2)θ^2

]

(x 1 +x 2 +···+xn)^2 n−^1 e−(x^1 +x^2 +···+xn)/θ

Γ(2n)θ^2 n

=

Γ(2n)

[Γ(2)]n

x 1 x 2 ···xn

(x 1 +x 2 +···+xn)^2 n−^1

,