Page 315 - Probability and Statistical Inference

292 6. Sufficiency, Completeness, and Ancillarity

h(x , ..., x ) = for all (x , ..., x ) ∈ ℜ . In other words, T = T(X , ...,
n
1 n 1 n 1
X ) = is jointly sufficient for (µ, σ ). !
2
n
If T is a sufficient statistic for θθ θθ θ, then any statistic T’ which
is a one-to-one function of T is also sufficient for θθ θθ θ.

Example 6.2.11 (Example 6.2.10 Continued) We have ,
, and so it is clear that the transfor-
mation from to T’ = ( , S ) is one-to-one. Hence, in
2
2
the Example 6.2.10, we can also claim that ( , S ) is jointly sufficient for θθ θθ θ =
2
(µ, σ ). !
Let us emphasize another point. Let T be a sufficient statistic for
θ θ θ θ θ. Consider another statistic T’, an arbitrary function of T. Then,
the statistic T’ itself is not necessarily sufficient for θθ θθ θ.
Look at the earlier Example 6.2.7.

An arbitrary function of a sufficient statistic T need not be sufficient for θθ θθ θ.
Suppose that X is distributed as N(θ, 1) where –∞ < θ < ∞ is the unknown
parameter. Obviously, T = X is sufficient for θ. One should check that the
statistic T’ = | X |, a function of T, is not sufficient for θ.
Remark 6.2.2 In a two-parameter situation, suppose that the Neyman
factorization (6.2.9) leads to a statistic T = (T , T ) which is jointly sufficient
1
2
for θθ θθ θ = (θ , θ ). But, the joint sufficiency of T should not be misunderstood to
1
2
imply that T is sufficient for θ or T is sufficient for θ .
1 1 2 2
From the joint sufficiency of the statistic T = (T , ..., T )
1
p
for θθ θθ θ = (θ , ..., θ ), one should not be tempted to claim that the
p
1
statistic T is componentwise sufficient for θ , i = 1, ..., p.
i
i
Look at the Example 6.2.12. In some cases, the statistic T
and θθ θθ θ may not even have the same number of components!
Example 6.2.12 (Example 6.2.11 Continued) In the N(µ, σ ) case when
2
both the parameters are unknown, recall from the Example 6.2.11 that ( ,
S ) is jointly sufficient for (µ, σ ). This is very different from trying to answer
2
2
a question like this: Is sufficient for µ or is S sufficient for σ ? We can
2
2
legitimately talk only about the joint sufficiency of the statistic ( , S ) for θθ θθ θ
2
= (µ, σ ).
2