Page 320 - Probability and Statistical Inference

6. Sufficiency, Completeness, and Ancillarity 297

for some appropriate g (.; θθ θθ θ) and h (.). Here, h (.) does not depend upon θθ θθ θ.
0
0
0
Now, for any two sample points x = (x , ..., x ), y = (y , ..., y ) from χ such
n
n
1
n
1
that U(x) = U(y), we obtain

Thus, h(x, y; θθ θθ θ) is free from θθ θθ θ. Now, by invoking the “only if” part of the
statement in (6.3.1), we claim that T(x) = T(y). That is, T is a function of U.
Now, the proof is complete. !
Example 6.3.1 (Example 6.2.8 Continued) Suppose that X , ..., X are iid
1
n
Bernoulli(p), where p is unknown, 0 < p < 1. Here, χ = {0, 1}, θ = p, and Θ
= (0, 1). Then, for two arbitrary data points x = (x , ..., x ) and y = (y , ..., y ),
1
n
1
n
both from χ, we have:

From (6.3.2), it is clear that would become free
from the unknown parameter p if and only if , that is, if
and only if Hence, by the theorem of Lehmann-Scheffé,
we claim that is minimal sufficient for p. !

We had shown non-sufficiency of a statistic U in the Example 6.2.5.
One can arrive at the same conclusion by contrasting U with
a minimal sufficient statistic. Look at the Examples 6.3.2 and 6.3.5.

Example 6.3.2 (Example 6.2.5 Continued) Suppose that X , X , X are
2
1
3
iid Bernoulli(p), where p is unknown, 0 < p < 1. Here, χ = {0, 1}, θ = p,
and Θ = (0, 1). We know that the statistic is minimal suffi-
cient for p. Let U = X X + X , as in the Example 6.2.5, and the question is
2
3
1
whether U is a sufficient statistic for p. Here, we prove again that the
statistic U can not be sufficient for p. Assume that U is sufficient for p,
and then must be a function of U, by the Definition 6.3.1 of
minimal sufficiency. That is, knowing an observed value of U, we must be
able to come up with a unique observed value of T. Now, the event {U = 0}