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

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The claimed univariate clean-quantile count exponent is too strong

Statement refuted

The auxiliary bound used in the source proof,

P ⁣ (B_{N} > \frac{3}{4} ε N) \leq e^{- ε N / 12}, B_{N} \sim Bin (N, ε / 2),

is false for admissible theorem parameters: with $η = 0$ , $ε = 1 / 10$ , $N$ sufficiently large and divisible by $40$ , and $δ = 4 e^{- N / 120}$ , one has

P ⁣ (B_{N} > \frac{3 N}{40}) > e^{- N / 120} = e^{- ε N / 12} .

For absolutely continuous clean data, An absolutely continuous upper quantile has its specified tail mass makes this binomial variable exactly the source's upper-quantile count. This refutes that proof step, not the univariate estimator theorem.

Facts & Assumptions

Given: An absolutely continuous real law, iid copies $Z_{1}, \dots, Z_{N}$ , and $N$ a sufficiently large multiple of $40$ .

[A1]

Countable Choice enters only through the published singleton-null result used by the exact-tail lemma (The Axiom of Countable Choice ( ${A C}_{ω}$ ), An absolutely continuous upper quantile has its specified tail mass).

[F1]

For $0 < u < 1$ , the upper quantile has exact tail mass $P (Z_{i} > Q_{u} (Z)) = 1 - u$ (An absolutely continuous upper quantile has its specified tail mass).

[F2]

The iid indicators of a common event are independent Bernoulli variables, and their sum has the binomial law (Identical distribution and IID families, Bernoulli random variables and binomial random variables as sums of independent Bernoulli trials).

[F3]

For $x > 0$ , $\log x = \int_{1}^{x} d t / t$ (The natural logarithm has derivative 1/x and equals the integral from 1 to x of 1/t).

[F4]

The logarithm is continuous and satisfies its product and quotient laws, and it is inverse to the exponential (Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm, The natural logarithm as the inverse of the exponential function).

[F5]

For $w > 0$ , the exponential power series has nonnegative terms and $e^{w} \geq 1 + w + w^{2} / 2$ (The real exponential function and the number $e$ by a power series).

Counterexample

technique · direct

Given: The absolutely continuous law, iid copies, and the stated $N$ .

1.1F4F5givenalgebra

Put $η = 0$ and $δ = 4 e^{- N / 120}$ . For sufficiently large $N$ , $0 < δ < 1$ and $δ \geq e^{- N / 4}$ because $4 e^{29 N / 120} > 1$ . The theorem's definition then gives $ε = 8 η + 12 \log (4 / δ) / N = 1 / 10$ . Hence $p = ε / 2 = 1 / 20$ , $q = 3 ε / 4 = 3 / 40$ , and the asserted failure exponent is $ε / 12 = 1 / 120$ .

2.1A1F1F2step 1.1algebra

By [F1], each indicator $1 {Z_{i} > Q_{19 / 20} (Z)}$ has success probability $1 / 20$ ; iid sampling and [F2] give $B_{N} \sim Bin (N, p)$ with $p = 1 / 20$ . At $q = 3 / 40$ , its relative-entropy rate is $D (q ∥ p) = q \log (q / p) + (1 - q) \log ((1 - q) / (1 - p)) = \frac{3}{40} \log \frac{3}{2} + \frac{37}{40} \log \frac{37}{38}$ .

3.1F3step 2.1algebra

To bound this rate rigorously, put $x = 1 / 5$ and $y = 1 / 38$ . From [F3] and $s = (t - 1) / (t + 1)$ , $\log (3 / 2) = 2 \int_{0}^{x} (1 - s^{2})^{- 1} d s$ . On $[0, x]$ , $(1 - s^{2})^{- 1} = 1 + s^{2} + s^{4} + s^{6} + s^{8} / (1 - s^{2})$ and $s^{8} / (1 - s^{2}) \leq (25 / 24) s^{8}$ , so $\log (3 / 2) < 23947783 / 59062500 < 405466 / 10^{6}$ . Also $\log (37 / 38) = - \int_{0}^{y} (1 - s)^{- 1} d s \leq - y - y^{2} / 2 - y^{3} / 3 = - 2195 / 82308 < - 26668 / 10^{6}$ , since $(1 - s)^{- 1} \geq 1 + s + s^{2}$ . Therefore $D (q ∥ p) < (3 \cdot 405466 - 37 \cdot 26668) / (40 \cdot 10^{6}) = 229682 / 40000000 < 23 / 4000 < 1 / 120$ .

4.1F2F4F5step 1.1step 2.1step 3.1algebra∎

Let $m = q N + 1$ , an integer because $40 ∣ N$ . For $p_{m} = m / N$ , the binomial point-mass ratio $P_{p_{m}} (B_{N} = j + 1) / P_{p_{m}} (B_{N} = j) = ((N - j) / (j + 1)) (p_{m} / (1 - p_{m}))$ is greater than $1$ at $j = m - 1$ and less than $1$ at $j = m$ , and decreases with $j$ . Thus $m$ is a mode of $Bin (N, p_{m})$ , whose $N + 1$ masses sum to one; consequently $P_{p_{m}} (B_{N} = m) \geq 1 / (N + 1)$ . The binomial formula and [F4] give $P_{p} (B_{N} = m) = P_{p_{m}} (B_{N} = m) e^{- N D (p_{m} ∥ p)} \geq e^{- N D (p_{m} ∥ p)} / (N + 1)$ . By continuity of $D (r ∥ p)$ in $r$ , [F4] and step 3.1 give $D (p_{m} ∥ p) \to D (q ∥ p) < 1 / 120$ . Choose $c > 0$ with $D (q ∥ p) + c < 1 / 120$ . For all sufficiently large such $N$ , $D (p_{m} ∥ p) < 1 / 120 - c$ and [F5] gives $e^{c N} \geq 1 + c N + (c N)^{2} / 2 > N + 1$ . Therefore $P_{p} (B_{N} = m) > e^{- N / 120}$ . Since $m > 3 N / 40$ , this mass lies in ${B_{N} > 3 N / 40}$ , disproving the displayed source bound at the admissible parameters from step 1.1.

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Sources

Gábor Lugosi and Shahar Mendelson, Robust Multivariate Mean Estimation: The Optimality of Trimmed Mean, Theorem 1 and §2.1 (standard reference, not scraped)