Page 319 - Probability and Statistical Inference

296 6. Sufficiency, Completeness, and Ancillarity

between a statistic and so called partitions it induces on the sample space.
Let us look at the original data X = (X , ..., X ) where x = (x , ..., x ) ∈ χ .
n
n
1
1
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Consider a statistic T ≡ T(X , ..., X ), that is T is a mapping from χ onto some
n
n
1
space say. For t ∈ , let χ = {x : x ∈ χ such that T(x) = t} which are
n
t
disjoint subsets of χ and also χ = ? t∈ χ . In other words, the collection of
n
n
t
subsets {χ : t ∈ } forms a partition of the space χ . Often, {χ : t ∈ } is
n
t
t
also called the partition of χ induced by the statistic T.
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Theorem 6.3.1 (Minimal Sufficient Statistics) Let us consider the func-
tion , the ratio of the likelihood func-
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tions from (6.2.9) at x and y, where θθ θθ θ is the unknown parameter and x, y ∈ χ .
Suppose that we have a statistic T ≡ T(X , ..., X ) = (T , ..., T ) such that the
k
1
n
1
following conditions hold:
Then, the statistic T is minimal sufficient for the parameter θθ θθ θ.
Proof We first show that T is a sufficient statistic for θθ θθ θ and then we verify
that T is also minimal. For simplicity, let us assume that f(x; θθ θθ θ) is positive for
all x ∈ χ and θθ θθ θ.
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Sufficiency part: Start with {χ : t ∈ } which is the partition of χ in-
n
t
duced by the statistic T. In the subset χ , let us select and fix an element x . If
t
t
we look at an arbitrary element x ∈ χ , then this element x belongs to χ for
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t
some unique t so that both x and x belong to the same set χ . In other words,
t
t
one has T(x) = T(x ). Thus, by invoking the “if part” of the statement in
t
(6.3.1), we can claim that h(x, x ; θθ θθ θ) is free from θθ θθ θ. Let us then denote h(x)
t
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≡ h(x, x ; θθ θθ θ), x ∈ χ . Hence, we write
t
where x = (x , ..., x ). In view of the Neyman Factorization Theorem, the
t t1 tn
statistic T(x) is thus sufficient for θθ θθ θ.¿
Minimal part: Suppose U = U(X) is another sufficient statistic for θθ θθ θ. Then,
by the Neyman Factorization Theorem, we can write