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Basu’s independence theorem

Statement

Let (X,A) be standard Borel, and let T and U be statistics from this sample space into standard Borel spaces for a family (Pθ). If T is complete and sufficient and U is ancillary, then T and U are independent under every Pθ.

Facts & Assumptions

Given: the family on a standard Borel sample space, complete sufficient T, ancillary U, and their standard Borel target spaces.

[F1]

Sufficient statistics: one kernel K gives the conditional law of the data given T for every parameter.

[F2]

Complete statistics: an integrable function of T with zero expectation for every parameter vanishes almost surely under every law.

[F3]

Ancillary statistics: the law of U is parameter-independent.

[F4]

Measure kernel and probability kernel: y↦K(y,A) is measurable and K(y,⋅) is a probability measure.

Proof

1.1F1F4givenconstruct

Fix a Borel set C in the target of U. The function qC(y)=K(y,U−1(C)) is Borel by [F4]. For every θ, the conditional-law identity [F1] shows qC(T) is a version of Pθ(U∈C∣T).

2.1F2F3step 1.1givenalgebra

By ancillarity [F3], pC=Pθ(U∈C) is constant in θ. Hence the bounded function gC(y)=qC(y)−pC satisfies EθgC(T)=0 for every θ. Completeness [F2] implies qC(T)=pC Pθ-almost surely for each θ.

3.1F1F3step 2.1givenalgebra∎

For any Borel D in the target of T, Pθ(T∈D,U∈C)=Eθ[1{T∈D}qC(T)]=pCPθ(T∈D)=Pθ(U∈C)Pθ(T∈D). This holds for every pair of Borel sets D,C, which is the definition of independence.

Depends on

Used by

Dependency tree · two levels

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Sources