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Consistency transfers from IID mixture samples to permuted groups

Statement

Assume AC. Let P,Q be probability laws and let mN+nN=N, with pN=mN/N→p∈(0,1). Write PˉN=pNP+(1−pN)Q. For each kN∈{mN,nN}, let TN be a measurable statistic of kN observations and let cN be a deterministic target. If

TN(V1,…,VkN)−cN→0in probability when Vi are IID PˉN,

then TN applied to either group of a uniform random permutation of the pooled independent sample X1,…,XmN∼P, Y1,…,YnN∼Q also converges in probability to cN. The statement does not require pN−p=O(N−1/2) because the IID comparison law uses the exact weight pN.

Facts & Assumptions

Given: Independent IID samples from P and Q, a uniform permutation of their pooled N observations, and an estimator/statistic consistent for cN on IID PˉN samples of the same group size.

[F1]

Hypergeometric and binomial composition counts are contiguous: the hypergeometric label count in a size-kN permuted group is contiguous to the binomial label count in a size-kN IID mixture sample.

[F2]

Identical distribution and IID families: each source group is IID with its stated common law.

[F3]

Independent random elements: the source groups and the random subset selection are independent as specified.

[F4]

Statistics and estimators: a statistic is a measurable function of the ordered observations; an estimator is such a statistic with a specified target.

[A1]

AC is inherited from the likelihood-ratio contiguity result [F1]. The Axiom of Choice

Proof

Given: The assumptions in the Statement. Prove consistency for both permuted group sizes.

1.1F1F2F3

Fix kN equal to either mN or nN. In an IID sample from PˉN, the number BN of observations whose mixture component is P has law Bin⁡(kN,pN). In a uniformly chosen size-kN group from the pooled sample with exactly mN P-observations among N, the corresponding count HN is hypergeometric.

1.2F2F3F4

Conditional on either count being j, the ordered group has the same distribution: its P-positions are uniform among the j-subsets of the kN positions, the values at those positions are IID P, and the remaining values are IID Q. Therefore there is a function fN(j)∈[0,1] giving the conditional probability that ∣TN−cN∣>ε under either sampling scheme.

2.1A1F1step 1.2algebra

The IID mixture consistency assumption gives E[fN(BN)]→0. Markov's inequality implies fN(BN)→0 in probability under the binomial law. By contiguity [F1], fN(HN)→0 in probability under the hypergeometric law. Since 0≤fN≤1, this also gives E[fN(HN)]→0. AC is used through [F1] only.

3.1step 1.1step 1.2step 2.1∎

By step 1.2, E[fN(HN)] is exactly the probability that the statistic differs from cN by more than ε for the randomly permuted size-kN group. This tends to zero. The argument applies separately to kN=mN and kN=nN, proving the assertion.

Depends on

Used by

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