Page 321 - Probability and Statistical Inference

298 6. Sufficiency, Completeness, and Ancillarity

consists of the union of possibilities such as {X = 0 ∩ X = 0 ∩ X = 0}, {X 1
2
3
1
= 0 ∩ X = 1 ∩ X = 0}, and {X = 1 ∩ X = 0 ∩ X = 0}. Hence, if the event
1
2
3
2
3
{U = 0} is observed, we know then that either T = 0 or T = 1 must be
observed. But, the point is that we cannot be sure about a unique observed
value of T. Thus, T can not be a function of U and so there is a contradiction.
Thus, U can not be sufficient for p. !
Example 6.3.3 (Example 6.2.10 Continued) Suppose that X , ..., X are iid
1
n
N(µ, σ ), where θθ θθ θ = (µ, σ) and both µ, σ are unknown, –∞ < µ < ∞, 0 < σ <
2
∞. Here, χ = ℜ and Θ = ℜ × ℜ . We wish to find a minimal sufficient statistic
+
for θθ θθ θ. Now, for two arbitrary data points x = (x , ..., x ) and y = (y , ..., y ),
1
n
1
n
both from χ, we have:

From (6.3.3), it becomes clear that the last expression would not involve the
unknown parameter θθ θθ θ = (µ, σ) if and only if as well as
, that is, if and only if = and
. Hence, by the theorem of Lehmann-Scheffé, we claim
that is minimal sufficient for (µ, σ). !
Theorem 6.3.2 Any statistic which is a one-to-one function of a minimal
sufficient statistic is itself minimal sufficient.
Proof Suppose that a statistic S is minimal sufficient for θθ θθ θ. Let us consider
another statistic T = h(S) where h(.) is one-to-one. In Section 6.2.2, we
mentioned that a one-to-one function of a (jointly) sufficient statistic is (jointly)
sufficient and so T is sufficient. Let U be any other sufficient statistic for θθ θθ θ.
Since, S is minimal sufficient, we must have S = g(U) for some g(.). Then,
we obviously have T = h(S) = h(g(U)) = h g(U) which verifies the minimality
of the statistic T. !
Example 6.3.4 (Example 6.3.3 Continued) Suppose that X , ..., X are iid
1
n
N(µ, σ ), where θθ θθ θ = (µ, σ) and both µ, σ are unknown, –∞ < µ < ∞, 0 < σ <
2
∞. We know that is minimal sufficient for (µ, σ).
Now, ( , S ) being a one-to-one function of T, we can claim that ( , S ) is
2
2
also minimal sufficient. !
Example 6.3.5 (Example 6.3.3 Continued) Suppose that X , X , X are
3
1
2
iid N(µ, σ ) where µ is unknown, but σ is assumed known, –∞ < µ < ∞,
2