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6. Sufficiency, Completeness, and Ancillarity 322
U, and hence by the Theorem 6.3.2, T itself is a minimal sufficient statistic for
n
θ. But recall that E (X ) = (1 + n )θ and E [(n 1) ∑ (X -X )] = θ for all
-1
1
θ
θ
i=1
i
n:1
n:1
θ > 0. Let us denote h(t) = (1 + n ) x (n 1) for all t
1
1 1
n:1
+
1 1
∈ T = (θ, ∞) × ℜ , so that E [h(T)] = E [(1 + n ) X (n 1) 1
θ θ n:1
= 0 for all θ > 0. But obviously h(t) is not identically zero
for all t ∈ T ≡ (θ, ∞) × ℜ . Hence, the minimal sufficient statistic T can not be
+
complete. !
In the Examples 6.6.5-6.6.6, one can easily find other nontrivial real val-
ued functions h(t) such that E [h(T)] ≡ 0 for all θ, but h(t) is not identically
θ
zero. We leave these as exercises.
Theorem 6.6.1 Suppose that a statistic T = (T , ..., T ) is complete. Let U
1
k
= (U , ..., U ) be another statistic with U = f(T) where g: T → U is one-to one.
1
k
Then, U is complete.
Proof For simplicity, let us pretend that T, U have continuous distributions
and that T, U and θ are all real valued. Let the pdf of T be denoted by g(t; θ).
Since f(.) is one-to-one, we can write:
Now, assume that U is not complete. Then, there is a function a(U) such that
E [a(U)] ≡ 0 but P {u : a(u) ≠ 0} > 0 for all θ ∈ Θ. From (6.6.7) we can then
θ
θ
claim that E [h(T)] ≡ 0 for all θ ∈ Θ where h ≡ a f stands for the composition
θ
o
mapping. But, this function h(.) is non-zero with positive probability which
contradicts the assumed completeness property of T. !
Now, we state a remarkably general result (Theorem 6.6.2) in the case of
the exponential family of distributions. One may refer to Lehmann (1986, pp.
142-143) for a proof of this result.
Theorem 6.6.2 (Completeness of the Minimal Sufficient Statistic in
the Exponential Family) Suppose that X , ..., X are iid with the common
n
1
pmf or the pdf belonging to the k-parameter exponential family defined by
with some appropriate forms for p(x) ≥ 0, q(θ) ≥ 0, θ and R (x), i = 1, ...,
i
i
k. Suppose that the regulatory conditions stated in (3.8.5) hold. Denote