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

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

The four-Q0 trimming slab radius is false

Statement refuted

The claim that every point of the correct-scale tail intersection $K_{i_{0}}$ in Adaptive multivariate trimmed mean lies within $4 ε Q_{0}$ of the clean mean is false, even for an absolutely continuous univariate clean law and no actual replacements.

Facts & Assumptions

Given: $X$ uniform on $[- 1, 1]$ , $N = 4096$ , $η = 1 / 10$ , $δ = 1 / 2$ , and no replacements.

[F1]

Adaptive multivariate trimmed mean defines quantiles, slabs of radius $2 ε Q$ and their dyadic tail intersections.

[F2]

Basic finite covariance matrix identities for robust mean bounds supplies the covariance and trace convention; here direct integration gives $μ = 0$ and $Σ = 1 / 3$ .

Counterexample

Given: The stated uniform clean law and parameters.

1.1F1F2givenalgebra

Here $N \geq 2560 \log 4$ because $\log 4 < 3 / 2$ , and $ε = \max {1, 2560 \log 4 / 4096} = 1$ . The proposed scale is $Q_{0} = \max {256 \sqrt{1 / (3 N)}, 16 / \sqrt{3}} = 16 / \sqrt{3}$ . The unique dyadic scale in $[2 Q_{0}, 4 Q_{0})$ is $Q = 32$ , so $i_{0} = 5$ . All parameters belong to the asserted domain, and zero actual replacements is allowed when the known upper fraction is $1 / 10$ .

2.1F1step 1.1algebra∎

Every observed point and directional cutoff lies in $[- 1, 1]$ . For every dyadic $R \geq 32$ , the expanded clipping interval contains $[- 1, 1]$ in each of the two unit directions. Thus $U_{R} (1) = \overset{ˉ}{X}$ , $U_{R} (- 1) = - \overset{ˉ}{X}$ , and $Γ (R) = [\overset{ˉ}{X} - 2 R, \overset{ˉ}{X} + 2 R]$ . These intervals are nested, so $K_{5} = [\overset{ˉ}{X} - 64, \overset{ˉ}{X} + 64]$ . Since $∣ \overset{ˉ}{X} ∣ \leq 1$ , $50 \in K_{5}$ for every realization, but $50 > 64 / \sqrt{3} = 4 ε Q_{0}$ . The proposed high-probability event therefore has probability zero, disproving the radius constant.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

13 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Lugosi and Mendelson, Robust Multivariate Mean Estimation: The Optimality of Trimmed Mean, pp. 6-23 (standard reference, not scraped)