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Countable factorization criterion for sufficiency

Statement

Let X be a nonempty countable set with its full σ-field, let Θ≠∅, let T:X→Y be a statistic into a standard Borel space, and write pθ(x)=Pθ({x}). Then T is sufficient if and only if there are nonnegative functions h:X→[0,∞) and Borel gθ:Y→[0,∞) such that

pθ(x)=gθ(Tx)h(x)(θ∈Θ, x∈X).

The pointwise convention is gθ(t)=0 whenever M(t)=0, where M(t)=sup⁡θPθ(T=t) on the countable image T(X); outside that image set gθ=0.

Facts & Assumptions

Given: the countable experiment, the statistic T, and the displayed factorization or a common kernel, according to the implication being proved.

[F1]

Sufficient statistics: sufficiency means one probability kernel K satisfying the conditional-law identity for every model law.

[F2]

Measure kernel and probability kernel: for each target event, a probability kernel is measurable in its conditioning argument and is a probability measure in its event argument.

[F3]

Standard Borel spaces: singletons in a standard Borel space are Borel, and countable subsets are Borel.

Proof

1.1F1F2F3givenconstruct

Suppose T is sufficient and fix a common kernel K from [F1]. For t∈T(X) set mθ(t)=Pθ(T=t) and M(t)=sup⁡θmθ(t). The image is countable by countability of X, hence Borel by [F3]; every function on this image extended by zero is Borel. Define h(x)=M(Tx)K(Tx,{x}) and define gθ(t)=mθ(t)/M(t) when M(t)>0, and zero when M(t)=0. Then gθ is Borel and nonnegative.

2.1F1step 1.1givenalgebra

If M(Tx)>0, choose θ0 with mθ0(Tx)>0. The conditional identity in [F1] gives K(Tx,T−1{Tx})=1 and pθ(x)=mθ(Tx)K(Tx,{x}) for every θ; if mθ(Tx)=0, both sides vanish by the same identity. Thus pθ(x)=gθ(Tx)h(x). If M(Tx)=0, every law gives that fiber probability zero, so pθ(x)=h(x)=gθ(Tx)=0. This proves the forward implication pointwise.

2.2F2F3givenchooseconstructstep 1.1

Conversely suppose the factorization holds. For each t∈T(X) let Ft=T−1{t} and H(t)=∑x∈Fth(x). If some mθ(t)>0, then gθ(t)>0 and 0<H(t)=mθ(t)/gθ(t)<∞, since the sum of the probabilities on the fiber is at most one. In that case put K(t,{x})=h(x)/H(t) on Ft and zero off Ft. If all mθ(t)=0, choose the first point of Ft in a fixed enumeration and let K(t,⋅) be its point mass. For t∉T(X) choose one fixed x0∈X and set K(t,⋅)=δx0. The countable image and enumeration make every coordinate t↦K(t,{x}) Borel. For each A⊆X, K(t,A)=∑x∈AK(t,{x}) is a countable sum of Borel functions, so it is Borel; the point-mass assignments are countably additive probability measures.

3.1F1step 2.2givenalgebra∎

On an active fiber, for every A⊆X, Pθ(A∩Ft)=gθ(t)∑x∈A∩Fth(x)=mθ(t)K(t,A). On an inactive fiber both sides are zero. Summing over the countably many t∈C∩T(X) gives the conditional-law identity for every Borel C⊆Y. Thus [F1] holds and T is sufficient.

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