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6. Sufficiency, Completeness, and Ancillarity 323

the statistic R (X ), j = 1, ..., k. Then, the (jointly) minimal suffi-
j i
cient statistic T = (T , ..., T ) for θ is complete.
1 k
Theorem 6.6.2 will sometimes provide an easy route to claim
the completeness of a sufficient statistic for θ as long as f(x; θ)
is given by (6.6.9) and the parameter space Θ(⊆ ℜ )
k
includes a k-dimensional rectangle.
Example 6.6.7 Let X , ..., X be iid N(µ, σ ) with (µ, σ) ∈ Θ = ℜ×ℜ +
2
1
n
where µ and σ are both assumed unknown. The pdf of N(µ, σ ) has the form
2
2
given in (6.6.9) where k = 2, θ = (θ , θ ) with θ = µ/σ , θ = 1/σ , R (x) = x,
2
1
1
2
1
2
2
R (x) = x , so that the minimal sufficient statistic is T = T(X) = (T (X), T (X))
1
2
2
where and . In view of the Theorem 6.6.2,
the statistic T(X) is complete. !
Example 6.6.8 Let X , ..., X be iid Gamma(α, β) so that f(x; α, β) =
α 1 n
-1
+
+
+
{β Γ(α)} exp(–x/β)x with (α, β) ∈ ℜ × ℜ , x ∈ ℜ where the parameters
α-1
α and β are both assumed unknown. This pdf has the form given in (6.6.9)
where k = 2, θ = (θ , θ ) with θ = 1/β, θ = α, R (x) = –x, R (x) = log(x), so
2
1
2
1
1
2
that the sufficient statistic is T(X) = (T (X), T (X)) where
1
2
and . In view of the Theorem 6.6.2,
the statistic T is complete. !
Example 6.6.9 In the Example 6.6.7, if µ is assumed unknown, but σ is
known, then or equivalently is a complete sufficient sta-
tistic for µ. In that same example, if µ is known, but σ is unknown, then
is a complete sufficient statistic for σ. In the Ex-
ample 6.6.8, if α is unknown, but β is known, then is a
complete sufficient statistic for α. In this case if α is known, but β is un-
known, then or equivalently is a complete sufficient sta-
tistic for β. These results follow immediately from the Theorem 6.6.2. !
Example 6.6.10 (Example 6.6.4 Continued) Suppose that we have X , ...,
1
X iid Poisson(λ) where λ(> 0) is the unknown parameter. The common pmf
n
–λ x
f(x; λ) = e λ /x! with χ = {0, 1, 2, ...}, θ = log(λ) has the same representa-
–θ
tion given in (6.6.9) where k = 1, p(x) = (x!) , q(λ) = exp{e } and R(x) = x.
-1
Hence, in view of the Theorem 6.6.2, the sufficient statistic is
then complete. !
Theorem 6.6.2 covers a lot of ground by helping to prove
the completeness property of sufficient statistics. But it fails to
reach out to non-exponential family members. The case in
point will become clear from the Example 6.6.11.