Robert_V._Hogg,_Joseph_W._McKean,_Allen_T._Craig

424 Sufficiency

Thus the joint pdf ofX 1 ,X 2 ,...,Xnmay be written

(

1

σ

√

2 π

)n

exp

[

−

∑n

i=1

(xi−θ)^2 / 2 σ^2

]

={exp[−n(x−θ)^2 / 2 σ^2 ]}

⎧

⎪⎪

⎪⎪

⎨

⎪⎪

⎪⎪

⎩

exp

[

−

∑n

i=1

(xi−x)^2 / 2 σ^2

]

(σ

√

2 π)n

⎫

⎪⎪

⎪⎪

⎬

⎪⎪

⎪⎪

⎭

.

Because the first factor of the right-hand member of this equation depends upon

x 1 ,x 2 ,...,xnonly throughx, and the second factor does not depend uponθ,the

factorization theorem implies that the meanXof the sample is, for any particular

value ofσ^2 , a sufficient statistic forθ, the mean of the normal distribution.

We could have used the definition in the preceding example because we know
thatX isN(θ, σ^2 /n). Let us now consider an example in which the use of the
definition is inappropriate.

Example 7.2.5.LetX 1 ,X 2 ,...,Xndenote a random sample from a distribution

with pdf

f(x;θ)=

{

θxθ−^10 <x< 1

0elsewhere,

where 0<θ. The joint pdf ofX 1 ,X 2 ,...,Xnis

θn

(n

∏

i=1

xi

)θ− 1

=

⎡

⎣θn

(n

∏

i=1

xi

)θ⎤

⎦

(

1

∏n

i=1xi

)

,

where 0<xi< 1 ,i=1, 2 ,...,n. In the factorization theorem, let

k 1 [u 1 (x 1 ,x 2 ,...,xn);θ]=θn

(n

∏

i=1

xi

)θ

and

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

1

∏n

i=1xi

.

Sincek 2 (x 1 ,x 2 ,...,xn) does not depend uponθ, the product

∏n

i=1Xiis a sufficient

statistic forθ.

There is a tendency for some readers to apply incorrectly the factorization theo-
rem in those instances in which the domain of positive probability density depends
upon the parameterθ. This is due to the fact that they do not give proper consid-
eration to the domain of the functionk 2 (x 1 ,x 2 ,...,xn). This is illustrated in the
next example.