CounterexampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)

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Half-sample replacement prevents covariance-only mean recovery

Statement refuted

For sample-replacement fractions $1 / 2 \leq η \leq 1$ , no deterministic Borel estimator $T : R^{2 N} \to R$ can satisfy $T = μ$ almost surely for every clean law and every admissible replacement coupling: two different zero-covariance means can generate the same observed sample within the replacement budget. In particular, no covariance-only error bound that forces zero error whenever the clean covariance is zero can hold uniformly.

Facts & Assumptions

Given: A sample size $2 N$ and a replacement fraction $1 / 2 \leq η \leq 1$ .

[F1]

Under sample replacement, an adversary may change at most $⌊ 2 η N ⌋$ coordinates of a clean iid sample; the observed vector need not be iid (Adversarial sample replacement).

[F2]

Huber contamination instead makes each observed coordinate iid from $(1 - η) P + η Q$ , which is a different model from a data-dependent joint replacement coupling (Huber contamination model).

Counterexample

technique · direct

Given: Fix any $M > 0$ . Consider the two clean laws $P_{0} = δ_{0}$ and $P_{M} = δ_{M}$ on $R$ .

1.1F1givenconstruct

Under $P_{0}$ , the clean sample of size $2 N$ is all zero. Replace $N$ coordinates by $M$ to obtain the observed vector $(0, \dots, 0, M, \dots, M)$ with $N$ entries of each value. This uses at most $⌊ 2 η N ⌋$ replacements by [F1].

2.1F1givenstep 1.1

Under $P_{M}$ , the clean sample is all $M$ . Replacing $N$ coordinates by $0$ produces exactly the same observed vector and also uses at most $⌊ 2 η N ⌋$ replacements. Both clean laws have variance zero, while their means are $0$ and $M$ .

3.1F2step 1.1step 2.1algebra∎

The two deterministic replacement couplings put probability one on the same observed vector. Hence the same Borel map $T$ has the same output value in both experiments. Exact recovery would require that value to equal $0$ under the first coupling and $M$ under the second, which is impossible because $M > 0$ . Both clean variances are zero, so any bound depending only on covariance and forcing zero error there would fail uniformly at $η \geq 1 / 2$ . The Huber mixture model in [F2] is not being used in this construction.

Depends on

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Dependency tree · two levels

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Sources

Gábor Lugosi and Shahar Mendelson, Robust Multivariate Mean Estimation: The Optimality of Trimmed Mean, lower-bound discussion before Theorem 2 (standard reference, not scraped)