LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge 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.

Second-moment order of the degenerate remainder

Statement

In the setup of Two-sample Hoeffding decomposition, fix $r$ and suppose $m / N \to p \in (0, 1)$ , $n / N \to 1 - p$ , with $N = m + n$ . At row $N$ , let the two groups be independent IID samples from probability measures $P_{N}, Q_{N}$ and use the same block-symmetric kernel $h$ . If $B : = \sup_{N} \max_{η \in {P_{N}, Q_{N}}^{2 r}} E_{η} [h^{2}] < \infty,$ where $E_{η}$ integrates independent arguments with the assigned marginal laws. Write $E_{N}$ for expectation under the row- $N$ sample law. Then the remainder in the first-projection decomposition satisfies

$E_{N} [R_{m, n}^{2}] = O (N^{- 2}), \sqrt{N} R_{m, n} \to 0 in L^{2} .$

Facts & Assumptions

Given: Fixed order $r$ , proportionally growing $m, n$ , independent IID groups with row laws $P_{N}, Q_{N}$ , and the canonical Hoeffding components from the preceding lemma. Fixed $P, Q$ is the special case $P_{N} = P$ , $Q_{N} = Q$ .

[F1]

Two-sample Hoeffding decomposition: $R_{m, n}$ is the finite sum of the terms $(\binom{r}{a}) (\binom{r}{b}) U_{a, b} (h_{a, b})$ with $a + b \geq 2$ ; every selected-coordinate kernel is canonical.

[F2]

Moments, variance, and covariance on a probability space: variance is the second centered moment and covariance is the expectation of the product of centered variables.

[F3]

Cauchy-Schwarz inequality for $L^{2}$ : $∣ E [A B] ∣ \leq (E A^{2})^{1 / 2} (E B^{2})^{1 / 2}$ .

Proof

Given: The hypotheses in the Statement. Prove the remainder is $O_{L^{2}} (N^{- 1})$ .

1.1F1F2algebra

Fix $(a, b)$ with $a + b \geq 1$ . Expand $U_{a, b} (h_{a, b})$ as the average over its $(\binom{m}{a}) (\binom{n}{b})$ pairs of index subsets. Two summands with different X-index sets or different Y-index sets have covariance zero: choose an observation appearing in one selected-coordinate set but not the other, condition on all other observations, and integrate its canonical coordinate to zero by [F1]. The identical-index terms each have second moment at most $E_{N} [h_{a, b}^{2}]$ , where $E_{N}$ is expectation under the row- $N$ sample law. Therefore $E_{N} [U_{a, b} (h_{a, b})^{2}] \leq \frac{E_{N} [h_{a, b}^{2}]}{(\binom{m}{a}) (\binom{n}{b})},$ with the convention $(\binom{m}{0}) = (\binom{n}{0}) = 1$ .

1.2F1F3algebra

The coordinate operators in the decomposition have $L^{2}$ norm at most $1$ for $P_{u}$ by Cauchy–Schwarz on a probability section. The pointwise inequality $∣ f - P_{u} f ∣^{2} \leq 2 ∣ f ∣^{2} + 2 ∣ P_{u} f ∣^{2}$ then bounds the $L^{2}$ norm of $I - P_{u}$ by $2$ . Hence each $E_{N} [h_{a, b}^{2}] \leq 4^{2 r} E_{N} [h^{2}] \leq 4^{2 r} B$ , uniformly over $N$ and the finitely many pairs $(a, b)$ . Since $m, n ≍ N$ and $a, b$ are fixed, $(\binom{m}{a}) (\binom{n}{b}) ≍ N^{a + b}$ .

2.1F1step 1.1step 1.2algebra∎

Distinct Hoeffding components in [F1] are orthogonal by the same canonical-coordinate argument used in step 1.1. Summing their variances with the fixed coefficients ${(\binom{r}{a})}^{2} {(\binom{r}{b})}^{2}$ gives $E_{N} [R_{m, n}^{2}] = \sum_{a + b \geq 2} {(\binom{r}{a})}^{2} {(\binom{r}{b})}^{2} E_{N} [U_{a, b} (h_{a, b})^{2}] \leq C_{r, p} B N^{- 2},$ because there are finitely many pairs and every denominator has order at least $N^{2}$ . Multiplying by $N$ proves $E_{N} [(\sqrt{N} R_{m, n})^{2}] \to 0$ .

Depends on

Used by

Positive kernel second moment does not imply nondegenerate first projections Counterexample
Permutation CLT for a two-sample U-statistic under a weak null Theorem
Projection CLT for a two-sample U-statistic Theorem

Dependency tree · two levels

9 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

E. Y. Chung and J. P. Romano, Asymptotically Valid and Exact Permutation Tests Based on Two-Sample U-Statistics, Stanford Technical Report 2011-09, Appendix A, Lemma A.2 (standard reference, not scraped)
Péter Major, Some Results About U-statistics, in On the Estimation of Multiple Random Integrals and U-Statistics, Lecture Notes in Mathematics 2079, Theorem 9.1 (standard reference, not scraped)