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Robust Means under Heavy Tails — Examples

1 · Prerequisites

2 · Summary

These examples use a continuous polynomial-tail law with finite variance, an exact five-block calculation showing rejection of one arbitrarily displaced block, and indistinguishable point-mass samples at a replacement fraction of one half. They keep the clean finite-variance guarantee, sample-replacement model, and iid Huber-mixture model distinct.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: AI-generatedVerification: AI-generatedjudge pass (gpt-6-luna)Open item page →

One corrupted block is rejected by the geometric median

Example

Take five real block means 0,0,0,0,M with M>0. Their geometric median is exactly 0, regardless of how large M is.

Facts & Assumptions

Given: Countable Choice and the five block means 0,0,0,0,M with M>0. Countable Choice is a prerequisite of the minimum-norm selector used by Geometric median of block means, through Measurable minimum-norm geometric-median selection.

[F1]

A geometric median minimizes the sum of distances to the listed block means (Geometric median of block means).

[F2]

With inner product ⟨x,y⟩=xy, R is a real inner-product space: ordered-field arithmetic gives bilinearity, symmetry, and x2≥0 with equality only for x=0 (The reals form a totally ordered field, Real and complex inner-product spaces and their induced length). Its induced norm is x2=∣x∣ (Square roots exist: a unique a≥0 with (a)2=a; the positives are {x2:x≠0}, Basic properties of the absolute value), so completeness of the real line makes it a Hilbert space (The reals are complete, Hilbert space). The embedded rationals are countable and dense, so this Hilbert space is separable (Q is countably infinite, The rationals embed densely in the reals, Separability: the existence of an at most countable dense subset).

Verification

technique · direct

Given: M>0 and the objective F(z)=4∣z∣+∣z−M∣.

1.1F1F2algebra

If z≤0, then F(z)=M−5z≥M, with equality only at z=0. If 0≤z≤M, then F(z)=M+3z≥M, again with equality only at 0.

2.1F1step 1.1algebra∎

If z≥M, then F(z)=5z−M≥4M>M=F(0). Therefore 0 is the unique minimizer, and the minimum-norm selection rule also returns 0. Four of five centers are exact while the fifth can be arbitrarily far away.

CounterexampleConstruction: AI-generatedVerification: AI-generatedjudge pass (gpt-6-luna)Open item page →

Half-sample replacement prevents covariance-only mean recovery

Statement refuted

For sample-replacement fractions 1/2≤η≤1, no deterministic Borel estimator T:R2N→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 2N and a replacement fraction 1/2≤η≤1.

[F1]

Under sample replacement, an adversary may change at most ⌊2ηN⌋ 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+ηQ, 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 P0=δ0 and PM=δM on R.

1.1F1givenconstruct

Under P0, the clean sample of size 2N is all zero. Replace N coordinates by M to obtain the observed vector (0,…,0,M,…,M) with N entries of each value. This uses at most ⌊2ηN⌋ replacements by [F1].

2.1F1givenstep 1.1

Under PM, the clean sample is all M. Replacing N coordinates by 0 produces exactly the same observed vector and also uses at most ⌊2ηN⌋ 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≥1/2. The Huber mixture model in [F2] is not being used in this construction.

ExampleConstruction: AI-generatedVerification: AI-generatedjudge pass (gpt-6-luna)Open item page →

A polynomial-tail law with finite variance

Example

Let X have density f(x)=32(1+∣x∣)4 on R. This law has mean 0 and variance 1, but it is not sub-Gaussian in the moment-generating-function sense. The geometric-median-of-means theorem therefore applies even though no finite sub-Gaussian proxy exists.

Facts & Assumptions

Given: Countable Choice, the density f(x)=3/(2(1+∣x∣)4), and the definitions of sub-Gaussianity and variance. Fix 0<δ<1 and a positive integer n for which k=⌊log⁡(1/δ)/ψ(7/18;1/10)⌋+1≤n/2, with ψ as defined in Finite-variance Hilbert mean bound for geometric medians of means. Let X1,…,Xn be iid Borel real-valued observations with this density, and let μ^ be their geometric-median-of-block-means output using k blocks and the minimum-norm selection rule specified by that theorem. Countable Choice is an explicit hypothesis of Finite-variance Hilbert mean bound for geometric medians of means, whose selector uses Measurable minimum-norm geometric-median selection.

[F1]

A centered real random variable is sub-Gaussian with proxy s2 only if EeλX≤eλ2s2/2 for every real λ (Sub-Gaussian variables and vectors, and sub-exponential variables).

[F2]

For the density displayed in the Example, symmetry and finite second moment identify the mean and variance; the geometric-median theorem requires a separable Hilbert-valued iid sample with finite second moment (Finite-variance Hilbert mean bound for geometric medians of means).

[F3]

With inner product ⟨x,y⟩=xy, R is a real inner-product space: ordered-field arithmetic gives bilinearity, symmetry, and x2≥0 with equality only for x=0 (The reals form a totally ordered field, Real and complex inner-product spaces and their induced length). Its induced norm is x2=∣x∣ (Square roots exist: a unique a≥0 with (a)2=a; the positives are {x2:x≠0}, Basic properties of the absolute value), so completeness of the real line makes it a Hilbert space (The reals are complete, Hilbert space). The embedded rationals are countable and dense, so this Hilbert space is separable (Q is countably infinite, The rationals embed densely in the reals, Separability: the existence of an at most countable dense subset).

Verification

technique · direct

Given: Countable Choice, the displayed density, the iid sample X1,…,Xn, the admissible confidence parameter δ and block count k, and the estimator μ^ specified in Facts & Assumptions.

1.1F2algebra

The density is normalized because ∫Rf(x) dx=3∫0∞(1+x)−4 dx=1. Its second moment is 3∫0∞x2(1+x)−4 dx=3∫1∞(u−2−2u−3+u−4) du=1. Since ∣x∣≤1+x2, the first absolute moment is finite, and symmetry then gives EX=0.

1.2F1algebra

For every λ>0, EeλX≥32∫1∞eλx(1+x)−4 dx=∞, since λx−4log⁡(1+x)→∞ and the integrand is eventually at least 1. Thus the defining MGF inequality in [F1] fails for every finite proxy.

2.1F2F3givenalgebra∎

The law is Borel and takes values in the separable Hilbert space H=R by [F3], with finite second moment 1. For the iid sample and geometric-median-of-means estimator specified in the Given, the admissible n, δ, and k therefore satisfy all hypotheses of Finite-variance Hilbert mean bound for geometric medians of means, which gives P(∣μ^∣≥11log⁡(7/(5δ))/n)≤δ. The conclusion uses only the finite variance just computed, not a sub-Gaussian assumption.

Sources