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.