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✓ 13 results · all verified · 10 also independently AI-judged

Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 3 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Studentized permutation tests under weak parameter nulls

1 · Prerequisites

2 · Summary

This page develops two-sample order- $(r, r)$ U-statistics, their first projections, and second-moment control of the higher-order remainder. A finite permutation variance estimate strengthens the exact triangular-mixture coupling and proves the conditional permutation CLT under the weak parameter null, block antisymmetry, mixed second moments and positive pooled projection variance. No assignment-wise centering or sample-fraction convergence rate is required. The all-assignment centered theorems remain useful special cases.

The studentized pivot divides by the estimated standard error $r {{\hat{σ}}_{m}^{2} / m + {\hat{σ}}_{n}^{2} / n}^{1 / 2}$ ; that ratio itself has the standard-normal limit. General joint variance estimates require direct permutation consistency. For estimates depending on their own group alone, IID triangular-mixture consistency is sufficient by the composition-count transfer lemma. The finite-support rank example verifies permutation consistency of its joint estimates directly.

When $P = Q$ , finite-group invariance gives finite-sample exactness for every measurable studentized statistic, with the stated tie conventions. That exactness does not establish validity of unstudentized Wilcoxon statistics under a weak mean or median null.

The explicit finite-enumeration test recomputes both variance estimates on every permuted split and supplies inclusive or tie-randomized p-values. The heterogeneous-law Wilcoxon specialization uses tie-corrected pairwise comparisons, verifies its empirical projection variance estimates under both the true samples and permutation, and tests the probabilistic-index null. It does not replace that null with equality of means or medians.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: AI-adaptedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

Two-sample U-statistic parameter

Definition

Let $(X, A)$ be a measurable space, let $P, Q$ be probability measures on it, and let $r \geq 1$ be a fixed integer. Give $X^{2 r}$ the product sigma-algebra and the product probability measure $P^{\otimes r} \otimes Q^{\otimes r}$ . A two-sample order- $(r, r)$ kernel is a measurable function $h : X^{2 r} \to R$ that is symmetric in its first $r$ arguments and separately symmetric in its last $r$ arguments. If $h$ is integrable under this product law, its two-sample U-statistic parameter is

$θ (P, Q; h) : = \int_{X^{2 r}} h d (P^{\otimes r} \otimes Q^{\otimes r}) .$

The value is finite and real. This definition imposes no null hypothesis and no antisymmetry condition; those are hypotheses of later results.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-luna)Open item page →

Conditional CLT for a random subset of an IID finite population

Statement

Assume the Axiom of Choice. For each $N$ , let $W_{N, 1}, \dots, W_{N, N}$ be IID real variables with mean $0$ , variance $σ_{N}^{2} \to σ^{2} > 0$ , and uniformly integrable squares:

$\lim_{K \to \infty} \sup_{N} E ⁣ [W_{N, 1}^{2} 1_{{∣ W_{N, 1} ∣ > K}}] = 0.$

Let integer sample sizes $m_{N}, n_{N}$ satisfy $m_{N} + n_{N} = N$ and $m_{N} / N \to p \in (0, 1)$ , and put $q = 1 - p$ . It is enough to consider sufficiently large $N$ , for which $N > 1$ and $1 \leq m_{N}, n_{N} \leq N - 1$ ; for each such $N$ , choose a uniformly random $m_{N}$ -element subset $A_{N} \subseteq {1, \dots, N}$ independently of the row. Set

${\overset{ˉ}{W}}_{N} = N^{- 1} \sum_{i = 1}^{N} W_{N, i}, S_{N}^{2} = (N - 1)^{- 1} \sum_{i = 1}^{N} (W_{N, i} - {\overset{ˉ}{W}}_{N})^{2} .$

Define

$T_{N} = \frac{\sum_{i \in A_{N}} (W_{N, i} - {\overset{ˉ}{W}}_{N})}{\sqrt{m_{N} n_{N} S_{N}^{2} / N}},$

and set $T_{N} = 0$ if $S_{N}^{2} = 0$ . Then the conditional distribution of $T_{N}$ given the finite population converges in probability to $N (0, 1)$ , uniformly in its real argument.

Facts & Assumptions

Given: IID rows with mean zero, variances $σ_{N}^{2} \to σ^{2} > 0$ , uniformly integrable squares, and a uniform sample without replacement of size $m_{N}$ with $m_{N} / N \to p \in (0, 1)$ .

[F1]

Martingale central limit theorem: a square-integrable martingale-difference array whose predictable variance clock tends to $1$ in probability and satisfies Lindeberg has a standard-normal terminal limit, assuming AC.

[F2]

Conditional Hoeffding bound for bounded martingale differences: if a conditionally centered increment has conditional range of width at most $c$ , its conditional exponential moment is at most $e^{λ^{2} c^{2} / 8}$ .

[F3]

Square-integrable martingale-difference array and variance clock: for martingale differences $D_{k}$ , the predictable clock is $\sum_{k} E [D_{k}^{2} ∣ F_{k - 1}]$ .

[F4]

Simple random sampling without replacement: every subset of the given size is sampled with equal probability.

[F5]

SRSWOR inclusion covariance and sample-mean variance: for fixed population values with finite-population variance $S_{N}^{2}$ , a size- $m_{N}$ sample sum has variance $m_{N} n_{N} S_{N}^{2} / N$ .

[F6]

Independent random elements: the random subset is independent of the IID row, so conditional on the row its design law is uniform SRSWOR.

[F7]

Tower property of conditional expectation: iterated conditioning preserves conditional-expectation identities.

[A1]

The Axiom of Choice is assumed exactly for the martingale CLT and the conditional-expectation interfaces in [F1]–[F3]. The Axiom of Choice

Proof

Given: The hypotheses in the Statement. Prove conditional normal convergence for the displayed random-subset statistic.

1.1givenalgebra

Since $E W_{N, 1} = 0$ and $σ_{N}^{2}$ is bounded, Chebyshev gives ${\overset{ˉ}{W}}_{N} \to 0$ in probability. To show $N^{- 1} \sum_{i} W_{N, i}^{2} - σ_{N}^{2} \to 0$ , first truncate $W^{2}$ at a fixed $K$ : the truncated variables are bounded, so Chebyshev gives convergence of their row average to its expectation. The expected tail average is at most $\sup_{N} E [W_{N, 1}^{2} 1_{{∣ W_{N, 1} ∣ > \sqrt{K}}}]$ , which tends to zero as $K \to \infty$ by uniform integrability. Hence the untruncated average converges to $σ^{2}$ , and $S_{N}^{2} \to σ^{2}$ .

1.2F3F4F6givenconstruct

Conditional on the row, list its centered values in a uniformly random order $a_{N, π (1)}, \dots, a_{N, π (N)}$ and write $S_{j} = \sum_{i \leq j} a_{N, π (i)}$ . For $j < N$ , $E [a_{N, π (j + 1)} ∣ F_{j}] = - S_{j} / (N - j)$ , since the remaining values sum to $- S_{j}$ . Therefore $M_{j} : = S_{j} / (N - j)$ is a martingale for $j \leq m_{N}$ , and its increments are bounded by $2 \max_{i} ∣ a_{N, i} ∣ / (N - j - 1)$ . For $j \leq m_{N} - 1$ , the denominator is at least $n_{N}$ , which is bounded below by a positive multiple of $N$ .

2.1givenstep 1.1algebra

For every $ε > 0$ , a union bound and $W^{2} \geq ε^{2} N$ on ${∣ W ∣ > ε \sqrt{N}}$ give $P ⁣ (\max_{i \leq N} ∣ W_{N, i} ∣ > ε \sqrt{N}) \leq ε^{- 2} E ⁣ [W_{N, 1}^{2} 1_{{∣ W_{N, 1} ∣ > ε \sqrt{N}}}] ⟶ 0.$ Together with ${\overset{ˉ}{W}}_{N} = O_{P} (1)$ , this implies $\max_{i} ∣ a_{N, i} ∣ / \sqrt{N} \to 0$ for $a_{N, i} = W_{N, i} - {\overset{ˉ}{W}}_{N}$ .

2.2F2F7givenstep 1.1step 1.2algebra

First consider any fixed finite population $c_{1}, \dots, c_{N}$ with $\sum_{i} c_{i} = 0$ and $∣ c_{i} ∣ \leq L$ . For its random order, the martingale $S_{j} / (N - j)$ from step 1.2 has increments of absolute value at most $C_{p} L / N$ through $j \leq m_{N}$ , for all sufficiently large $N$ . Iterating the conditional Hoeffding bound [F2] with [F7] gives, for each fixed $u > 0$ , $P_{π} (∣ S_{j} ∣ / N > u) \leq 2 \exp (- c_{p} u^{2} N / L^{2})$ uniformly in $j \leq m_{N}$ ; here $∣ S_{j} ∣ / N > u$ implies $∣ S_{j} / (N - j) ∣ > u$ because $N - j \leq N$ . A union bound over at most $N$ prefixes makes the maximum divided by $N$ tend to zero. Apply this argument first to the clipped values $W_{N, i}^{(K)} - {\overline{W}}_{N}^{(K)}$ , and then to the centered clipped squares $(W_{N, i}^{(K)} - {\overline{W}}_{N}^{(K)})^{2} - \overline{(W_{N}^{(K)} - {\overline{W}}_{N}^{(K)})^{2}}$ . For fixed $K$ , both are bounded arrays. The average absolute first-power truncation error is at most twice the average of $∣ W_{N, i} ∣ 1_{{∣ W_{N, i} ∣ > K}}$ ; its expectation is at most $2 K^{- 1} \sup_{N} E [W_{N, 1}^{2} 1_{{∣ W_{N, 1} ∣ > K}}]$ . The average absolute difference of the centered squares tends to zero in probability as $K \to \infty$ : Cauchy--Schwarz bounds it by the square root of the average squared truncation error times the square root of the average $(∣ a_{N, i} ∣ + ∣ W_{N, i}^{(K)} - {\overline{W}}_{N}^{(K)} ∣)^{2}$ , whose first factor is small by uniform integrability and whose second is bounded in probability by step 1.1. Thus, taking $N \to \infty$ and then $K \to \infty$ , we obtain both $\max_{j \leq m_{N}} \frac{∣ S_{j} ∣}{N} \to 0, \max_{j \leq m_{N}} \frac{∣ \sum_{i \leq j} (a_{N, π (i)}^{2} - \overline{a_{N}^{2}}) ∣}{N} \to 0$ in probability.

2.3F3F4step 1.2algebra

Let $R_{N, k}$ be the mean of the values remaining before draw $k$ , let $s_{N, k - 1}^{2}$ be their finite-population variance, and let $F_{N, k}$ reveal the first $k$ draws. The Doob martingale for the final sample sum has increments $D_{N, k} = \frac{n_{N}}{N - k} (a_{N, π (k)} - R_{N, k}), 1 \leq k \leq m_{N} .$ Indeed, after draw $k$ the conditional expected contribution of the remaining $m_{N} - k$ draws is $(m_{N} - k)$ times the new remaining mean; subtracting the previous conditional expectation gives the displayed coefficient. The increments are centered and square-integrable.

3.1step 1.1step 2.2algebra

Before draw $j + 1$ , the remaining set has size $N - j \geq n_{N} ≍ N$ . Its mean is $- S_{j} / (N - j)$ , and its raw second moment is $\overline{a_{N}^{2}} - \frac{1}{N - j} \sum_{i \leq j} (a_{N, π (i)}^{2} - \overline{a_{N}^{2}}) .$ Both differ uniformly from $0$ and $\overline{a_{N}^{2}}$ , respectively, by $o_{P} (1)$ by step 2.2. The remaining finite-population variance therefore differs uniformly from $S_{N}^{2} = N \overline{a_{N}^{2}} / (N - 1)$ by $o_{P} (1)$ ; the factors $(N - j) / (N - j - 1)$ tend uniformly to $1$ because $N - j \geq n_{N} ≍ N$ . Hence $\max_{j \leq m_{N}} ∣ s_{N, j}^{2} - S_{N}^{2} ∣ \to 0$ .

4.1F3F5step 1.1step 3.1algebra

A uniform draw from the $N - k + 1$ remaining values has variance $((N - k) / (N - k + 1)) s_{N, k - 1}^{2}$ . On ${S_{N}^{2} > 0}$ , the predictable clock of $\sum_{k \leq m_{N}} D_{N, k} / \sqrt{m_{N} n_{N} S_{N}^{2} / N}$ is $\frac{1}{m_{N} n_{N} S_{N}^{2} / N} \sum_{k = 1}^{m_{N}} \frac{n_{N}^{2} s_{N, k - 1}^{2}}{(N - k) (N - k + 1)} .$ Define the normalized increments and their clock to be zero on ${S_{N}^{2} = 0}$ . By step 1.1, $S_{N}^{2} \to_{P} σ^{2} > 0$ , so this zero-variance event has probability tending to zero. On its complement, step 3.1 permits replacing each $s_{N, k - 1}^{2}$ uniformly by $S_{N}^{2}$ with relative $o_{P} (1)$ error. The remaining deterministic sum telescopes: $n_{N}^{2} \sum_{k = 1}^{m_{N}} \frac{1}{(N - k) (N - k + 1)} = n_{N}^{2} (\frac{1}{n_{N}} - \frac{1}{N}) = \frac{m_{N} n_{N}}{N} .$ Thus the defined normalized clock tends to $1$ in probability, in agreement with the finite-population variance in [F5]. The normalized terminal sum equals $T_{N}$ also on ${S_{N}^{2} = 0}$ by the statement's convention.

5.1givenstep 1.1step 2.1step 4.1algebra

On ${S_{N}^{2} > 0}$ , since $n_{N} / (N - k) \leq 1$ and $∣ R_{N, k} ∣ \leq \max_{i} ∣ a_{N, i} ∣$ , step 2.1 gives $\max_{k \leq m_{N}} \frac{∣ D_{N, k} ∣}{\sqrt{m_{N} n_{N} S_{N}^{2} / N}} \leq \frac{2 \max_{i} ∣ a_{N, i} ∣}{\sqrt{m_{N} n_{N} S_{N}^{2} / N}} ⟶ 0$ in probability. Under the zero-variance convention of step 4.1, the normalized increments are zero on ${S_{N}^{2} = 0}$ as well; since $S_{N}^{2} \to_{P} σ^{2} > 0$ , the maximum normalized increment tends to zero in probability. Therefore, for each fixed $η > 0$ , every term in the normalized martingale Lindeberg sum is zero with probability tending to one.

6.1A1F1step 1.1step 2.1step 4.1step 5.1algebra

Suppose conditional convergence failed along a subsequence. From steps 1.1 and 2.1, choose a further subsequence on which $S_{N}^{2} \to σ^{2} > 0$ and $\max_{i} ∣ a_{N, i} ∣ / \sqrt{N} \to 0$ almost surely. For each positive rational $η$ , the probability that the normalized clock differs from $1$ by more than $η$ tends to zero by step 4.1. Its conditional probability given the row therefore tends to zero in probability by Markov's inequality. A diagonal further subsequence makes all those conditional probabilities tend to zero almost surely. For almost every resulting sequence of rows, the conditional clock converges in probability to $1$ ; the normalized increments have a deterministic uniform bound tending to zero by step 5.1, so for every fixed $η > 0$ the Lindeberg sum on that row's permutation space is exactly zero for all sufficiently large rows. The conditional permutation spaces may vary with $N$ , whereas the array definition [F3] is stated on one fixed probability space. To justify using the martingale CLT in this setting, repeat the proof of [F1] with the row-specific conditional laws $P_{N}^{π}$ and expectations $E_{N}^{π}$ . Every conditional expectation and variance estimate is formed within one row, and no joint event involving different rows is used. In particular, the only bounded-convergence step in [F1]'s proof can be read rowwise: if $Γ_{N} \to 1$ in probability under laws $P_{N}$ , then for bounded continuous $f$ and any $δ > 0$ , $∣ E_{N} f (Γ_{N}) - f (1) ∣ \leq \sup_{∣ x - 1 ∣ \leq δ} ∣ f (x) - f (1) ∣ + 2 ∥ f ∥_{\infty} P_{N} (∣ Γ_{N} - 1 ∣ > δ),$ which tends to zero after $N \to \infty$ and then $δ ↓ 0$ . All other steps of the proof of [F1], including the Lindeberg estimate and the predictable-clock truncation, are already rowwise; hence that characteristic-function proof yields the martingale CLT for these finite permutation rows. Pad each row with zero increments. The conditional CDF converges to $Φ$ at every $t$ , contradicting the assumed failure. Thus it converges in probability at each $t$ .

7.1step 1.1step 6.1algebra∎

Finally, choose $T$ so that $Φ (- T)$ and $1 - Φ (T)$ are small, and partition $[- T, T]$ into finitely many points where adjacent $Φ$ values differ by at most $ε$ . Monotonicity of each conditional CDF bounds its deviation between grid points by the endpoint deviations plus $ε$ . Pointwise convergence at the finite grid and the tail bounds yield uniform convergence in probability. The convention when $S_{N}^{2} = 0$ affects an event whose probability tends to zero by step 1.1 and $σ^{2} > 0$ .

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passOpen item page →

Hypergeometric and binomial composition counts are contiguous

Statement

Assume AC. For each $N$ , let $M_{N}$ of $N$ objects be marked, with $p_{N} = M_{N} / N \to p \in (0, 1)$ , and let $k_{N} / N \to s \in (0, 1)$ . Let $B_{N} \sim Bin (k_{N}, p_{N})$ and let $H_{N}$ count marked objects in a uniform size- $k_{N}$ sample without replacement from the $N$ objects. Then the law of $H_{N}$ is contiguous to the law of $B_{N}$ : for every sequence of sets $A_{N}$ , if $P (B_{N} \in A_{N}) \to 0$ , then $P (H_{N} \in A_{N}) \to 0$ .

More precisely, the likelihood ratio $L_{N} (j) = P (H_{N} = j) / P (B_{N} = j)$ satisfies

$L_{N} (B_{N}) \Rightarrow (1 - s)^{- 1 / 2} \exp ⁣ (- \frac{s}{2 (1 - s)} Z^{2}), Z \sim N (0, 1),$

and this nonnegative limit has mean $1$ .

Facts & Assumptions

Given: AC, marked-object proportions $p_{N} \to p \in (0, 1)$ , sampling proportions $k_{N} / N \to s \in (0, 1)$ , a binomial count, and a uniform sample without replacement.

[F1]

Lindeberg-Feller central limit theorem: sufficiency gives a standard-normal limit for a centered independent triangular array under the Lindeberg condition, assuming AC.

[F2]

Stirling's formula for factorials states $j! \sim \sqrt{2 π j} (j / e)^{j}$ as positive integers $j \to \infty$ .

[F3]

Identical distribution and IID families defines IID as mutual independence together with a common law.

[F4]

Moments, variance, and covariance on a probability space defines variance as the centered second moment.

[F5]

For sigma-finite factors, the product measure exists, has the rectangle formula, is sigma-finite, and is unique supplies the finite product probability measure used for that Bernoulli row.

[F6]

Expectations factor over finite products of independent random variables factors expectations of products of integrable functions of independent variables.

[A1]

AC is used exactly through [F1]. The Axiom of Choice

Proof

Given: The hypotheses in the Statement. Prove the likelihood-ratio limit and the stated direction of contiguity.

1.1givenalgebra

The exact mass ratio, with $q_{N} = 1 - p_{N}$ , is $L_{N} (j) = \frac{(\binom{M_{N}}{j}) (\binom{N - M_{N}}{k_{N} - j})}{(\binom{N}{k_{N}}) (\binom{k_{N}}{j}) p_{N}^{j} q_{N}^{k_{N} - j}}$ on the hypergeometric support, and is zero off that support.

1.2F3F5givenconstructalgebra

By [F5], form the $k_{N}$ -fold product of the law assigning masses $p_{N}$ and $q_{N}$ to $1$ and $0$ . The rectangle formula makes its coordinate indicators $I_{N, i}$ independent with that common Bernoulli law, hence IID in the sense of [F3]. Each binary vector with $j$ ones has mass $p_{N}^{j} q_{N}^{k_{N} - j}$ , and there are $(\binom{k_{N}}{j})$ such vectors. Their sum therefore has the specified binomial law, so we may realize $B_{N} = \sum_{i = 1}^{k_{N}} I_{N, i}$ on this space.

2.1F4F6step 1.2algebra

Each indicator has mean $p_{N}$ and variance $p_{N} q_{N}$ by [F4]. Factorization [F6] gives $E [I_{N, i} I_{N, j}] = p_{N}^{2}$ for $i \neq j$ , so the row sum has mean $k_{N} p_{N}$ and variance $k_{N} p_{N} q_{N}$ .

3.1A1F1step 2.1givenalgebra

Each centered summand has magnitude at most $1$ , while the row standard deviation is $\sqrt{k_{N} p_{N} q_{N}} ≍ \sqrt{N}$ . Thus for each fixed $ε > 0$ all Lindeberg tail terms are zero for large $N$ ; under AC [A1], [F1] gives $Z_{N} : = \frac{B_{N} - k_{N} p_{N}}{\sqrt{k_{N} p_{N} q_{N}}} \Rightarrow Z \sim N (0, 1) .$

4.1F2step 1.1step 3.1algebra

Put $s_{N} = k_{N} / N$ and $δ_{N} = B_{N} - k_{N} p_{N}$ . By [step 3.1], $δ_{N} = O_{P} (\sqrt{N})$ , so the event $∣ δ_{N} ∣ \leq N^{2 / 3}$ has probability tending to one. On this event, all factorial arguments in [step 1.1] are bounded below by a positive multiple of $N$ , since $p_{N} \to p$ and $s_{N} \to s$ with $p, s \in (0, 1)$ . Let $H (a, x) = a \log a - x \log x - (a - x) \log (a - x)$ and $H (t) = H (1, t)$ . Stirling's formula [F2], uniformly for these arguments, expands each log-binomial coefficient as $H (a, x) - \frac{1}{2} \log (2 π x (a - x) / a) + o (1)$ . The order- $N$ terms at $δ = 0$ cancel because $H (p_{N} N, s_{N} p_{N} N) = p_{N} N H (s_{N})$ , $H (q_{N} N, s_{N} q_{N} N) = q_{N} N H (s_{N})$ , and $H (s_{N} N, s_{N} p_{N} N) = s_{N} N H (p_{N})$ . For $F_{N} (δ) = H (p_{N} N, s_{N} p_{N} N + δ) + H (q_{N} N, s_{N} q_{N} N - δ) - H (N, s_{N} N) - H (s_{N} N, s_{N} p_{N} N + δ) - (s_{N} p_{N} N + δ) \log p_{N} - (s_{N} q_{N} N - δ) \log q_{N}$ , use $H_{a}^{'} (x) = \log ((a - x) / x)$ and $H_{a}^{''} (x) = - 1 / x - 1 / (a - x)$ ; direct substitution gives $F_{N} (0) = 0$ , $F_{N}^{'} (0) = 0$ and $F_{N}^{''} (0) = - \frac{1}{p_{N} N s_{N} (1 - s_{N})} - \frac{1}{q_{N} N s_{N} (1 - s_{N})} + \frac{1}{s_{N} N p_{N} q_{N}} = - \frac{1}{p_{N} q_{N} N (1 - s_{N})} .$ The third derivative is bounded by $C / N^{2}$ throughout this event because every argument remains at least $c N$ . The square-root terms at $δ = 0$ sum to $- \frac{1}{2} \log \frac{(2 π p_{N} N s_{N} (1 - s_{N})) (2 π q_{N} N s_{N} (1 - s_{N}))}{(2 π N s_{N} (1 - s_{N})) (2 π s_{N} N p_{N} q_{N})} = - \frac{1}{2} \log (1 - s_{N}),$ and their change is $O (∣ δ_{N} ∣ / N) = o_{P} (1)$ . Taylor's formula therefore gives $\log L_{N} (B_{N}) = - \frac{1}{2} \log (1 - s_{N}) - \frac{δ_{N}^{2}}{2 p_{N} q_{N} N (1 - s_{N})} + O_{P} (∣ δ_{N} ∣^{3} / N^{2}) + o_{P} (1) = - \frac{1}{2} \log (1 - s_{N}) - \frac{s_{N}}{2 (1 - s_{N})} Z_{N}^{2} + o_{P} (1) .$ The Taylor remainder is $o_{P} (1)$ because $δ_{N} = O_{P} (\sqrt{N})$ , and the omitted event has probability tending to zero by [step 3.1].

5.1step 3.1step 4.1algebra

Since $s_{N} \to s$ , [step 3.1] and [step 4.1] imply the displayed likelihood-ratio limit. Its mean is $(1 - s)^{- 1 / 2} E \exp ⁣ (- \frac{s}{2 (1 - s)} Z^{2}) = (1 - s)^{- 1 / 2} \sqrt{1 - s} = 1,$ by direct integration against the standard-normal density.

6.1step 5.1algebra

Each $L_{N} (B_{N}) \geq 0$ and $E [L_{N} (B_{N})] = 1$ by the likelihood-ratio definition. Nonnegative convergence in distribution to a variable with the same mean implies uniform integrability: for fixed $K$ , $E [L_{N} \land K] \to E [L \land K]$ , so ${lim sup}_{N} E [(L_{N} - K)_{+}] = 1 - E [L \land K] \to 0$ as $K \to \infty$ . This also controls $E [L_{N} 1_{{L_{N} > 2 K}}]$ .

7.1step 6.1algebra∎

If $P (B_{N} \in A_{N}) \to 0$ , then for every $K > 0$ , $P (H_{N} \in A_{N}) = E [L_{N} (B_{N}) 1_{{B_{N} \in A_{N}}}] \leq K P (B_{N} \in A_{N}) + E [L_{N} (B_{N}) 1_{{L_{N} (B_{N}) > K}}] .$ The first term tends to zero for fixed $K$ and the second is uniformly small for large $K$ by step 6.1. Therefore $P (H_{N} \in A_{N}) \to 0$ . This proves the required direction of contiguity; the reverse direction is not needed here.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-luna)Open item page →

Finite-group rank validity on a measurable sample space

Statement

Let a finite group $G$ act on a measurable space $(S, S)$ by measurable maps, and let $P$ be an invariant probability law: $P (g^{- 1} A) = P (A)$ for every $g \in G$ and $A \in S$ . For a measurable statistic $T : S \to R$ , define

$p_{0} (x) = \frac{# {g \in G : T (g \cdot x) \geq T (x)}}{∣ G ∣}, p_{U} (x, u) = \frac{# {g : T (g \cdot x) > T (x)} + u # {g : T (g \cdot x) = T (x)}}{∣ G ∣} .$

If $X \sim P$ and $U$ is independent uniform on $[0, 1]$ , then $p_{0} (X)$ is superuniform and $p_{U} (X, U)$ is uniform on $[0, 1]$ . The sample space need not be finite; only the acting group is finite.

Facts & Assumptions

Given: A finite group action by measurable maps, an invariant probability law, a measurable real statistic, and an independent uniform tie randomizer.

[F1]

Group and abelian group and Left group actions, transitive actions, and faithful actions: $G$ is a group acting with $e \cdot x = x$ and $(g h) \cdot x = g \cdot (h \cdot x)$ ; hence right multiplication by a fixed element is a bijection of $G$ .

[F2]

A measurable function between measurable spaces: the action maps and $T$ are measurable. The product sigma-algebra and its finite iterates makes the coordinate projections on $S \times [0, 1]$ measurable, and Arithmetic and lattice operations preserve measurability whenever they are defined preserves measurability under the finite sums and products in the rank formulas.

[F3]

Tied upper ranks on a finite multiset: for a finite list of real numbers, an independent uniform index gives a superuniform inclusive upper rank and a uniform tie-randomized upper rank.

[F4]

Independent random elements: the uniform group element and the tie randomizer are independent of the data and of each other.

Proof

Given: The hypotheses in the Statement. Prove both rank conclusions.

1.1F1F3construct

Fix $x \in S$ and form the finite list $(T (g \cdot x))_{g \in G}$ . For any $h \in G$ , the list $(T (g \cdot (h \cdot x)))_{g \in G}$ has exactly the same multiset of values, because right multiplication $g \mapsto g h$ is a bijection of $G$ by [F1]. Therefore, when $H$ is uniform on $G$ , $p_{0} (H \cdot x)$ and $p_{U} (H \cdot x, U)$ are exactly the inclusive and tie-randomized ranks of a uniform index in this list.

1.2F2givenalgebra

For each $g$ , $x \mapsto T (g \cdot x)$ is measurable by [F2]. Its difference from $T (x)$ is measurable, so the indicators of the strict and equal comparisons are measurable; finite sums and products with the second coordinate $u$ remain measurable by [F2]. Thus $p_{0}$ and the joint map $(x, u) \mapsto p_{U} (x, u)$ are measurable.

1.3F4givenalgebra

Let $H$ be uniform on $G$ , independent of $(X, U)$ . For every measurable $A$ , $P (H \cdot X \in A) = ∣ G ∣^{- 1} \sum_{h \in G} P (X \in h^{- 1} A) = P (A)$ by invariance. Since $H$ is independent of $U$ , the pair $(H \cdot X, U)$ has the same law as $(X, U)$ .

2.1F3step 1.1step 1.2step 1.3algebra∎

For each fixed $x$ , steps 1.1–1.2 and [F3] give $P_{H} (p_{0} (H \cdot x) \leq t) \leq t$ for $0 \leq t \leq 1$ , and $p_{U} (H \cdot x, U)$ is uniform. Integrating these statements over $X$ , then using equality in law from step 1.3, proves the two conclusions for $p_{0} (X)$ and $p_{U} (X, U)$ . No choice principle is used: all group averages and ranks are finite.

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A successive-revelation variance bound for a uniform permutation

Statement

Let $π$ be uniform on the permutations of ${1, \dots, N}$ , where $N \geq 1$ , and let $f$ be any real function on this finite set. Write $π (k, l)$ for the permutation obtained by exchanging its entries in positions $k, l$ , with $(k, k)$ the identity. Then

$Var (f (π)) \leq \frac{1}{2} \sum_{k = 1}^{N} \frac{1}{N - k + 1} \sum_{l = k}^{N} E [(f (π) - f (π (k, l)))^{2}] .$

In particular, if every transposition difference has second moment at most $b$ , then $Var (f (π)) \leq N b / 2$ . These are finite-sum statements and require no choice axiom.

Facts & Assumptions

Given: A positive integer $N$ , a uniform permutation, and a real-valued function on its finite state space.

[F1]

The uniform probability space on a nonempty finite set assigns equal probability to every permutation.

[F2]

Moments, variance, and covariance on a probability space defines variance and covariance by centered products.

[F3]

The Cauchy-Schwarz inequality for finite sums implies that the square of an average of finitely many real numbers is at most the average of their squares.

Proof

Given: The hypotheses in the Statement. Prove the variance estimate by revealing successive entries.

1.1F1F2givenalgebra

Let $F_{k}$ reveal the first $k$ entries of $π$ , and let $M_{k}$ be the average of $f$ over permutations agreeing with those entries. Thus $M_{0} = E f$ , $M_{N} = f$ , and the average of $M_{k}$ conditional on $F_{k - 1}$ is $M_{k - 1}$ . For $j < k$ , averaging $(M_{j} - M_{j - 1}) (M_{k} - M_{k - 1})$ first over completions of the first $k - 1$ entries gives zero. Expanding the square of $f - E f = \sum_{k} (M_{k} - M_{k - 1})$ therefore gives $Var (f) = \sum_{k} E [(M_{k} - M_{k - 1})^{2}]$ . All these conditional averages are finite sums.

2.1F1F2F3step 1.1algebra

Fix a prefix of length $k - 1$ and put $q = N - k + 1$ . For each remaining image $a$ , let $u_{a}$ be the average of $f$ over completions whose kth entry is $a$ . The conditional variance of $M_{k}$ is $\frac{1}{2 q^{2}} \sum_{a, b} (u_{a} - u_{b})^{2}$ . Choose a uniform completion $π$ of the prefix and independently choose $l$ uniformly from ${k, \dots, N}$ . Given $π (k) = a$ , the image $b = π (l)$ is uniform among the $q$ remaining images. For fixed $a, b$ , swapping position $k$ with the location of $b$ is a bijection from completions with kth image $a$ onto completions with kth image $b$ . Hence $E [f (π) - f (π (k, l)) ∣ a, b, F_{k - 1}] = u_{a} - u_{b}$ . By [F3], the conditional variance just displayed is at most $\frac{1}{2} E [(f (π) - f (π (k, l)))^{2} ∣ F_{k - 1}]$ .

3.1step 1.1step 2.1algebra∎

Average step 2.1 over prefixes and sum over $k$ using step 1.1. The independent uniform choice of $l$ gives the displayed estimate. If every squared difference has expectation at most $b$ , each inner average is at most $b$ , so the sum is at most $N b / 2$ . For $N = 1$ both sides of the displayed estimate are zero.

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Two-sample order-r U-statistic and first projections

Definition

Fix $m, n \geq r$ and independent samples $X_{1}, \dots, X_{m} \sim P$ and $Y_{1}, \dots, Y_{n} \sim Q$ , each IID within its group. For a kernel from Two-sample U-statistic parameter, define

$U_{m, n} : = \frac{1}{(\binom{m}{r}) (\binom{n}{r})} \sum_{I \in (\binom{{1, \dots, m}}{r})} \sum_{J \in (\binom{{1, \dots, n}}{r})} h (X_{I}; Y_{J}) .$

Here $I, J$ are unordered index subsets, and $h (X_{I}; Y_{J})$ means that the selected observations are listed in increasing index order in their respective blocks. Block symmetry makes the value independent of that ordering. Under $E ∣ h ∣^{2} < \infty$ for the $P^{\otimes r} \otimes Q^{\otimes r}$ law, write

$g_{X} (X_{1}) : = E [h (X_{1}, X_{2}, \dots, X_{r}; Y_{1}, \dots, Y_{r}) ∣ X_{1}] - θ (P, Q; h),$ $g_{Y} (Y_{1}) : = E [h (X_{1}, \dots, X_{r}; Y_{1}, Y_{2}, \dots, Y_{r}) ∣ Y_{1}] - θ (P, Q; h) .$

These are the centered first projections under $P$ and $Q$ ; they are square-integrable equivalence classes, so changes on their respective null sets do not change any later claim.

Remarks

The displayed conditional-expectation notation is shorthand for the completed section-integral versions under the product laws. The corresponding functions are measurable and square-integrable by First projections are measurable and square-integrable ↗.

The finite sum defining $U_{m, n}$ is a measurable real statistic: the finite sample-vector coordinate maps are measurable by Coordinate maps and permutations are measurable on finite products, composition with $h$ is measurable, and its finite sum and constant scaling remain measurable by Arithmetic and lattice operations preserve measurability whenever they are defined. This is a statistic in the sense of Statistics and estimators.

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Consistency transfers from IID mixture samples to permuted groups

Statement

Assume AC. Let $P, Q$ be probability laws and let $m_{N} + n_{N} = N$ , with $p_{N} = m_{N} / N \to p \in (0, 1)$ . Write ${\overset{ˉ}{P}}_{N} = p_{N} P + (1 - p_{N}) Q$ . For each $k_{N} \in {m_{N}, n_{N}}$ , let $T_{N}$ be a measurable statistic of $k_{N}$ observations and let $c_{N}$ be a deterministic target. If

$T_{N} (V_{1}, \dots, V_{k_{N}}) - c_{N} \to 0 in probability when V_{i} are IID {\overset{ˉ}{P}}_{N},$

then $T_{N}$ applied to either group of a uniform random permutation of the pooled independent sample $X_{1}, \dots, X_{m_{N}} \sim P$ , $Y_{1}, \dots, Y_{n_{N}} \sim Q$ also converges in probability to $c_{N}$ . The statement does not require $p_{N} - p = O (N^{- 1 / 2})$ because the IID comparison law uses the exact weight $p_{N}$ .

Facts & Assumptions

Given: Independent IID samples from $P$ and $Q$ , a uniform permutation of their pooled $N$ observations, and an estimator/statistic consistent for $c_{N}$ on IID ${\overset{ˉ}{P}}_{N}$ samples of the same group size.

[F1]

Hypergeometric and binomial composition counts are contiguous: the hypergeometric label count in a size- $k_{N}$ permuted group is contiguous to the binomial label count in a size- $k_{N}$ IID mixture sample.

[F2]

Identical distribution and IID families: each source group is IID with its stated common law.

[F3]

Independent random elements: the source groups and the random subset selection are independent as specified.

[F4]

Statistics and estimators: a statistic is a measurable function of the ordered observations; an estimator is such a statistic with a specified target.

[A1]

AC is inherited from the likelihood-ratio contiguity result [F1]. The Axiom of Choice

Proof

Given: The assumptions in the Statement. Prove consistency for both permuted group sizes.

1.1F1F2F3

Fix $k_{N}$ equal to either $m_{N}$ or $n_{N}$ . In an IID sample from ${\overset{ˉ}{P}}_{N}$ , the number $B_{N}$ of observations whose mixture component is $P$ has law $Bin (k_{N}, p_{N})$ . In a uniformly chosen size- $k_{N}$ group from the pooled sample with exactly $m_{N}$ $P$ -observations among $N$ , the corresponding count $H_{N}$ is hypergeometric.

1.2F2F3F4

Conditional on either count being $j$ , the ordered group has the same distribution: its $P$ -positions are uniform among the $j$ -subsets of the $k_{N}$ positions, the values at those positions are IID $P$ , and the remaining values are IID $Q$ . Therefore there is a function $f_{N} (j) \in [0, 1]$ giving the conditional probability that $∣ T_{N} - c_{N} ∣ > ε$ under either sampling scheme.

2.1A1F1step 1.2algebra

The IID mixture consistency assumption gives $E [f_{N} (B_{N})] \to 0$ . Markov's inequality implies $f_{N} (B_{N}) \to 0$ in probability under the binomial law. By contiguity [F1], $f_{N} (H_{N}) \to 0$ in probability under the hypergeometric law. Since $0 \leq f_{N} \leq 1$ , this also gives $E [f_{N} (H_{N})] \to 0$ . AC is used through [F1] only.

3.1step 1.1step 1.2step 2.1∎

By step 1.2, $E [f_{N} (H_{N})]$ is exactly the probability that the statistic differs from $c_{N}$ by more than $ε$ for the randomly permuted size- $k_{N}$ group. This tends to zero. The argument applies separately to $k_{N} = m_{N}$ and $k_{N} = n_{N}$ , proving the assertion.

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Mixture coupling for permuted two-sample U-statistics

Statement

Use the setup of Two-sample order-r U-statistic and first projections with fixed $r$ , $m + n = N$ , and $m / N \to p \in (0, 1)$ . Assume $h$ is symmetric within each block and antisymmetric under exchange of its two blocks. For every assignment of independent P- and Q-distributed arguments to its $2 r$ positions, assume $E h^{2} < \infty$ .

Let ${\overset{ˉ}{P}}_{N} = (m / N) P + (n / N) Q$ . Couple an IID row from ${\overset{ˉ}{P}}_{N}$ to the original labelled sample by drawing mixture labels and matching observations until one labelled group is exhausted, then drawing fresh observations for the remaining labels. Reorder the coupled row to agree with the original $P^{m} Q^{n}$ ordering except at $D$ unmatched positions. For a uniform permutation $π$ independent of this coupling,

$\sqrt{m} (U_{m, n} (Z^{π}) - U_{m, n} ({\overset{ˉ}{Z}}^{π π_{0}})) ⟶ 0 in L^{2} .$

Neither the weak null nor assignment-wise kernel centering is needed for this coupling bound. No rate for $m / N - p$ is assumed.

Facts & Assumptions

Given: Fixed order, proportional sample sizes, a block-symmetric and block-antisymmetric kernel, all mixed second moments finite, and the exact triangular mixture.

[F1]

Two-sample order-r U-statistic and first projections defines the kernel average $U_{m, n}$ over $M = (\binom{m}{r}) (\binom{n}{r})$ pairs of subsets.

[F2]

Identical distribution and IID families defines the IID convention used for each source group and the constructed mixture row.

[F3]

Moments, variance, and covariance on a probability space defines variance and covariance.

[F4]

Cauchy-Schwarz inequality for $L^{2}$ bounds expected absolute deviations and yields the triangle inequality for $L^{2}$ norms.

[F5]

Independent random elements defines independence of random elements. The source groups are independent by hypothesis; independence of the auxiliary draws and permutation is imposed by the product construction [F6].

[F6]

For sigma-finite factors, the product measure exists, has the rectangle formula, is sigma-finite, and is unique constructs the finite product of these laws.

[F7]

The uniform probability space on a nonempty finite set gives uniform permutations and finite conditional averaging.

[F8]

A successive-revelation variance bound for a uniform permutation bounds a permutation function's variance by half the sum of its successive-revelation averaged squared transposition differences.

[A1]

$B = \max_{η \in {P, Q}^{2 r}} E_{η} h^{2} < \infty$ by the finite family of moment assumptions.

Proof

Given: The hypotheses in the Statement. Prove the $L^{2}$ coupling bound.

1.1F2F5F6F7givenconstruct

Enlarge the source sample space by the finite product [F6] with $N$ Bernoulli $(p_{N})$ label factors, $N$ fresh P-draw factors, $N$ fresh Q-draw factors, and one uniform permutation factor [F7], where $p_{N} = m / N$ . Its rectangle formula makes all auxiliary factors mutually independent and independent of the source samples, in the sense of [F5]. Match each P label to the next unused original X and each Q label to the next unused original Y, using distinct fresh draws of the requested law after exhaustion. Conditional on the labels the constructed coordinates are independent with their respective label laws, so their unconditioned row $\overset{ˉ}{Z}$ is IID ${\overset{ˉ}{P}}_{N}$ . Align every shared observation with its original position, putting the fresh observations in the remaining positions by a fixed index order; call the resulting array $\tilde{Z} = {\overset{ˉ}{Z}}^{π_{0}}$ . With $B_{N}$ the number of P labels, its mismatch set $J$ has size $D = ∣ B_{N} - m ∣$ . Conditional on the labels and alignment, the pairs $(Z_{i}, {\tilde{Z}}_{i})$ are independent across $i$ ; at a shared position both entries coincide, and at a mismatch the original and fresh entries are independent.

1.2F1F7givenalgebra

For any deterministic array $a$ of length $N$ , the permutation average of $U_{m, n} (a^{π})$ is zero: the arguments of a fixed selected kernel term form a uniform ordered distinct $2 r$ -tuple, and exchanging its two r-blocks preserves that distribution and negates the kernel. Therefore $f (π) : = U_{m, n} (Z^{π}) - U_{m, n} ({\tilde{Z}}^{π})$ has permutation mean zero for every realized pair of arrays. This is conditional centering of the full statistic, not centering of each mixed-assignment kernel.

2.1F1F3F4step 1.1algebra

Since $B_{N}$ is binomial with parameters $N, p_{N}$ , $E B_{N} = m$ and $E D^{2} = N p_{N} (1 - p_{N}) \leq N / 4$ . Thus $E D \leq \sqrt{E D^{2}} = O (\sqrt{N})$ by [F4]. Also $m, n ≍ N$ and $M ≍ N^{2 r}$ . Constants below depend only on $r$ and bounds keeping $p_{N}$ away from $0, 1$ , not on the labels or values.

3.1A1F1F4step 1.1step 2.1algebra

Fix two positions $k, l$ and a permutation $π$ , conditioning on labels and alignment. A selected term in $U (a^{π}) - U (a^{π (k, l)})$ vanishes unless its subset pair meets ${k, l}$ ; there are at most $C N^{2 r - 1}$ such pairs. Each term in $f (π) - f (π (k, l))$ is a signed sum of four kernel evaluations, each on distinct independent population coordinates and hence with conditional second moment at most $B$ by [A1]. The $L^{2}$ triangle inequality [F4], followed by division by $M$ , bounds the norm of this difference by $C \sqrt{B} / N$ . If neither $π (k)$ nor $π (l)$ is in $J$ , a contribution also vanishes unless its selected positions meet $π^{- 1} J$ outside ${k, l}$ . There are at most $C D N^{2 r - 2}$ such pairs, giving the sharper norm bound $C \sqrt{B} D / N^{2}$ . Overlap among evaluations causes no independence requirement for this triangle inequality.

4.1F7F8step 1.2step 3.1algebra

Averaging the squared bounds in step 3.1 over uniform $π$ , the probability that either swapped entry is a mismatch is at most $2 D / N$ . Thus, for every fixed position transposition, $E_{π, v a l u e s} [(f (π) - f (π (k, l)))^{2} ∣ l a b e l s] \leq C B (D / N^{3} + D^{2} / N^{4})$ . First apply [F8] for fixed realized arrays, where its expectations and variance are only over $π$ , and then average over values conditional on labels. Its sum of inner-average coefficients is $N / 2$ . By step 1.2 the conditional permutation mean is identically zero, so $E_{π, v a l u e s} [f (π)^{2} ∣ l a b e l s] \leq C B (D / N^{2} + D^{2} / N^{3})$ .

5.1step 2.1step 4.1algebra∎

Averaging step 4.1 over labels and using $m ≍ N$ and step 2.1 yields $E [m f (π)^{2}] \leq C B (E D / N + E D^{2} / N^{2}) = O (N^{- 1 / 2}) \to 0$ . Since ${\tilde{Z}}^{π} = {\overset{ˉ}{Z}}^{π π_{0}}$ with the alignment convention in the Statement, this is the required $L^{2}$ convergence.

Remarks

The 2011 report's Lemma A.1 uses assignment-wise zero means and a fixed-mixture fraction rate. The local proof instead centers the complete statistic by antisymmetry, controls its permutation variance by transpositions, and uses the exact triangular mixture. It does not assert those extra hypotheses follow from the weak null.

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First projections are measurable and square-integrable

Statement

For the setup of Two-sample order-r U-statistic and first projections, suppose $h \in L^{2} (P^{\otimes r} \otimes Q^{\otimes r})$ . Then the section integrals defining $g_{X}$ and $g_{Y}$ are finite and measurable almost everywhere, can be assigned measurable values on exceptional null sets, and define square-integrable centered projections. In particular, $E_{P} [g_{X} (X_{1})] = E_{Q} [g_{Y} (Y_{1})] = 0,$ $E_{P} [g_{X} (X_{1})^{2}] \leq E_{P^{\otimes r} \otimes Q^{\otimes r}} [h^{2}], E_{Q} [g_{Y} (Y_{1})^{2}] \leq E_{P^{\otimes r} \otimes Q^{\otimes r}} [h^{2}] .$

Facts & Assumptions

Given: Probability measures $P, Q$ , a product-measurable kernel $h$ with finite second moment under $P^{\otimes r} \otimes Q^{\otimes r}$ , and the projection definitions.

[F1]

Two-sample order-r U-statistic and first projections defines $g_{X}$ and $g_{Y}$ as the corresponding conditional section integrals minus $θ (P, Q; h)$ .

[F2]

Fubini's theorem for L^1 functions on a sigma-finite product gives integrable sections almost everywhere, measurable section-integral versions, and equality of iterated and product integrals.

[F3]

Cauchy-Schwarz inequality for $L^{2}$ gives $∣ \int a b ∣ \leq ∥ a ∥_{2} ∥ b ∥_{2}$ .

[F4]

The product sigma-algebra and its finite iterates supplies the finite product sigma-algebras on which the section maps and product integrals are defined.

Proof

Given: $h \in L^{2} (P^{\otimes r} \otimes Q^{\otimes r})$ and the definitions in [F1]. Prove both projections are measurable, centered, and square-integrable with the displayed bounds.

1.1F1F3

The product measure is a probability measure, so applying [F3] to $h$ and the constant function $1$ gives $E ∣ h ∣ \leq (E h^{2})^{1 / 2} < \infty$ . Thus $h$ is integrable and $θ (P, Q; h) = E h$ is finite.

1.2F1F2F4

By [F4], the product decomposition uses the stated finite product sigma-algebra. Apply [F2] with first coordinate $P$ -distributed and the remaining $2 r - 1$ coordinates distributed according to $P$ or $Q$ . For $P$ -almost every $x$ , the section integral $f_{X} (x) = \int h (x, x_{2}, \dots, x_{r}; y_{1}, \dots, y_{r}) d P^{r - 1} d Q^{r}$ is finite; [F2] supplies a measurable version after setting it to zero on the exceptional null set. Reordering the coordinates and using the $Y_{1}$ section gives a finite measurable $f_{Y}$ for $Q$ -almost every $y$ .

2.1F2F3step 1.2

For each good $x$ , Cauchy–Schwarz [F3] on the remaining probability product measure gives $∣ f_{X} (x) ∣^{2} \leq \int h (x, x_{2}, \dots, x_{r}; y_{1}, \dots, y_{r})^{2} d P^{r - 1} d Q^{r}$ . Integrating in $P$ and using [F2] yields $E_{P} [f_{X} (X_{1})^{2}] \leq E [h^{2}]$ . The same argument for the $Y_{1}$ section gives $E_{Q} [f_{Y} (Y_{1})^{2}] \leq E [h^{2}]$ .

3.1F1F2step 1.1step 2.1∎

By [F2], $E_{P} f_{X} = E h = θ (P, Q; h)$ and $E_{Q} f_{Y} = E h = θ (P, Q; h)$ . Subtracting this finite mean gives the centered projections in [F1]; step 2.1 and finiteness of $θ$ show both remain square-integrable.

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Pooled-mixture projection and variance

Statement

Use the setup of Two-sample order-r U-statistic and first projections and suppose $h$ is antisymmetric under exchange of its two blocks and has finite second moment under every independent P/Q assignment to its $2 r$ positions. For $t \in (0, 1)$ put ${\overset{ˉ}{P}}_{t} = t P + (1 - t) Q$ and define

$g_{t} (z) = \int h (z, z_{2}, \dots, z_{r}; w_{1}, \dots, w_{r}) d {\overset{ˉ}{P}}_{t}^{\otimes (2 r - 1)} - θ ({\overset{ˉ}{P}}_{t}, {\overset{ˉ}{P}}_{t}; h) .$

Then $θ ({\overset{ˉ}{P}}_{t}, {\overset{ˉ}{P}}_{t}; h) = 0$ , the first projection in the second block is $- g_{t}$ , and $σ_{t}^{2} : = E_{{\overset{ˉ}{P}}_{t}} [g_{t}^{2}]$ is finite. The map $t \mapsto σ_{t}^{2}$ is continuous on $(0, 1)$ . In particular, with $N = m + n$ and $p_{N} = m / N \to p \in (0, 1)$ ,

$r^{2} (1 + \frac{m}{n}) σ_{p_{N}}^{2} ⟶ \frac{r^{2}}{1 - p} σ_{p}^{2} .$

Facts & Assumptions

Given: A block-symmetric kernel, its antisymmetry under exchange of blocks, and finite second moments under all mixed P/Q assignments.

[F1]

Two-sample order-r U-statistic and first projections defines the block kernel and its first projections.

[F2]

First projections are measurable and square-integrable gives the section-integral versions and their L2 bounds.

[F3]

Moments, variance, and covariance on a probability space defines the second-moment variance $E [g_{t}^{2}]$ when $E g_{t} = 0$ .

[F4]

Cauchy-Schwarz inequality for $L^{2}$ makes cross products of the finitely many square-integrable mixed-assignment section functions integrable.

[F5]

Fubini's theorem for L^1 functions on a sigma-finite product supplies measurable integrable section functions and equality of iterated integrals on the mixed product probability spaces.

Proof

Given: The hypotheses in the Statement. Prove the pooled mean, sign relation, finiteness, and continuity claims.

1.1F1givenalgebra

Under ${\overset{ˉ}{P}}_{t}^{\otimes 2 r}$ , swapping the first and second blocks preserves the product law and changes $h$ to $- h$ . Therefore $θ ({\overset{ˉ}{P}}_{t}, {\overset{ˉ}{P}}_{t}; h) = - θ ({\overset{ˉ}{P}}_{t}, {\overset{ˉ}{P}}_{t}; h)$ , so it is zero.

1.2F2F3F4F5givenalgebra

Expand each factor ${\overset{ˉ}{P}}_{t} = t P + (1 - t) Q$ in the section integral. This writes $g_{t}$ as a finite sum $\sum_{η} c_{η} (t) f_{η} (z)$ , where $η$ ranges over assignments of P and Q to the remaining $2 r - 1$ coordinates, $c_{η} (t)$ is a polynomial, and $f_{η}$ is the corresponding section integral. For either first-coordinate law $R \in {P, Q}$ , write $E_{R, η}$ for product expectation with independent coordinates having law $R$ first and the laws specified by $η$ in the remaining positions. The mixed-moment assumption gives both $E_{P, η} [h^{2}] < \infty$ and $E_{Q, η} [h^{2}] < \infty$ . Cauchy--Schwarz [F4] also makes $h$ integrable with first-coordinate law $(P + Q) / 2$ . Fubini [F5] on that product space supplies one measurable version of $f_{η}$ , valid $(P + Q)$ -almost everywhere after setting it to zero on the common exceptional set. Cauchy--Schwarz on the remaining probability product then gives $\int ∣ f_{η} ∣^{2} d R \leq E_{R, η} [h^{2}] < \infty$ for each $R \in {P, Q}$ , as in the section-integral argument of [F2]. Consequently every cross integral $\int f_{η} f_{η^{'}} d R$ is finite by [F4]. Expanding $E_{{\overset{ˉ}{P}}_{t}} g_{t}^{2} = t E_{P} [g_{t}^{2}] + (1 - t) E_{Q} [g_{t}^{2}]$ now gives a finite sum of polynomial coefficients times these finite cross integrals, so it is finite and continuous in $t$ .

2.1F1F2step 1.1algebra

Holding the first coordinate of $h$ fixed and integrating the other $2 r - 1$ coordinates with law ${\overset{ˉ}{P}}_{t}$ defines $g_{t}$ , since the kernel mean in [F1] is zero by step 1.1. For the projection in the second block, exchange the blocks inside the integral. Antisymmetry changes the sign and the common product law is unchanged, so the second-block projection is $- g_{t}$ .

3.1step 1.2algebra∎

Since $m / n = p_{N} / (1 - p_{N})$ and $p_{N} \to p$ , step 1.2 gives $σ_{p_{N}}^{2} \to σ_{p}^{2}$ , while $1 + m / n = 1 / (1 - p_{N}) \to 1 / (1 - p)$ . Multiplication by $r^{2}$ proves the displayed limit.

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Two-sample Hoeffding decomposition

Statement

In the setup of Two-sample order-r U-statistic and first projections, assume $h \in L^{2} (P^{\otimes r} \otimes Q^{\otimes r})$ . For each $X$ -coordinate $u$ let $P_{u}$ integrate that coordinate under $P$ ; for each $Y$ -coordinate let $P_{u}$ integrate under $Q$ , and put $Q_{u} = I - P_{u}$ . These coordinate operators commute on $h$ by Fubini. For $0 \leq a, b \leq r$ , choose any $a$ of the $X$ -coordinates and any $b$ of the $Y$ -coordinates, apply $Q_{u}$ in those coordinates and $P_{u}$ in all others, and call the resulting (reindexed) kernel $h_{a, b}$ . Separate block symmetry makes it independent of the chosen coordinate subsets. It is canonical in each of its $a + b$ arguments: integrating it in any such argument gives zero. Let $U_{a, b} (h_{a, b})$ be its average over all $a$ -subsets of the $X$ sample and all $b$ -subsets of the $Y$ sample, with the zero-order convention $U_{0, 0} = θ (P, Q; h)$ . Then

$U_{m, n} - θ (P, Q; h) = \sum_{\begin{array}{c} 0 \leq a, b \leq r \\ a + b \geq 1 \end{array}} (\binom{r}{a}) (\binom{r}{b}) U_{a, b} (h_{a, b}) .$

In particular,

$U_{m, n} - θ (P, Q; h) = \frac{r}{m} \sum_{i = 1}^{m} g_{X} (X_{i}) + \frac{r}{n} \sum_{j = 1}^{n} g_{Y} (Y_{j}) + R_{m, n},$

where $R_{m, n}$ is the sum of the terms with $a + b \geq 2$ .

Facts & Assumptions

Given: Independent IID samples from $P$ and $Q$ , a symmetric order- $(r, r)$ kernel with finite second moment, and the projection definitions.

[F1]

Two-sample order-r U-statistic and first projections defines $U_{m, n}$ , $g_{X}$ , and $g_{Y}$ .

[F2]

First projections are measurable and square-integrable supplies measurable square-integrable coordinate integrals and the Fubini equalities needed for each projection operator.

[F3]

Fubini's theorem for L^1 functions on a sigma-finite product permits the coordinate-integration operators on different coordinates to commute.

[F4]

Cauchy-Schwarz inequality for $L^{2}$ gives the $L^{2}$ contraction bound for integration over a probability coordinate.

Proof

Given: The hypotheses in the Statement. Prove the finite decomposition and identify its order-one terms.

1.1F2F3F4algebra

For each coordinate $u$ , Cauchy–Schwarz [F4] on its probability marginal and Fubini [F3] give $∥ P_{u} h ∥_{2} \leq ∥ h ∥_{2}$ . The pointwise inequality $∣ h - P_{u} h ∣^{2} \leq 2 ∣ h ∣^{2} + 2 ∣ P_{u} h ∣^{2}$ then gives $∥ Q_{u} h ∥_{2} \leq 2 ∥ h ∥_{2}$ . Thus every finite product of the coordinate operators is a well-defined $L^{2}$ class. The operator $P_{u}$ is idempotent because integrating an already coordinate-independent function changes nothing. For distinct coordinates $u, v$ , Fubini [F3] gives $P_{u} P_{v} = P_{v} P_{u}$ on each integrable section, so $P_{u}, Q_{u}$ commute across distinct coordinates.

2.1step 1.1algebra

Since $I = P_{u} + Q_{u}$ for every one of the $2 r$ coordinates, multiplying these finitely many identities and using commutativity expands $h$ as the sum over all subsets of coordinates of the corresponding products of $P_{u}$ and $Q_{u}$ . This is a finite algebraic expansion; its empty selected subset is $P_{1} \dots P_{2 r} h = θ (P, Q; h)$ .

3.1step 1.1step 2.1algebra

If coordinate $u$ is selected, applying $P_{u}$ to that component gives zero because $P_{u} Q_{u} = P_{u} - P_{u}^{2} = 0$ ; the other commuting factors do not alter this identity. Thus each component is canonical in each selected coordinate. Block symmetry identifies the $(\binom{r}{a}) (\binom{r}{b})$ components with $a$ selected $X$ -coordinates and $b$ selected $Y$ -coordinates as reindexed copies of $h_{a, b}$ .

4.1F1step 2.1step 3.1algebra

Apply the expansion in step 2.1 to each summand $h (X_{I}; Y_{J})$ in $U_{m, n}$ , summing over sample subsets and selected kernel coordinates together. Each fixed $a$ -subset of the $X$ sample is contained in exactly $(\binom{m - a}{r - a})$ sample $r$ -subsets; in each, exactly one choice of $a$ kernel coordinates selects it. Thus its coefficient after averaging is $(\binom{m - a}{r - a}) / (\binom{m}{r}) = (\binom{r}{a}) / (\binom{m}{a})$ . Similarly, each fixed $b$ -subset of the $Y$ sample has coefficient $(\binom{r}{b}) / (\binom{n}{b})$ . These identities also hold when $a = 0$ or $b = 0$ . Block symmetry identifies the resulting functions with $h_{a, b}$ , so the combined contribution is $(\binom{r}{a}) (\binom{r}{b}) U_{a, b} (h_{a, b})$ . Removing the constant component gives the first displayed decomposition.

5.1F1step 4.1algebra∎

For $(a, b) = (1, 0)$ , the integrated kernel is exactly $g_{X}$ from [F1], and its subset average is $m^{- 1} \sum_{i} g_{X} (X_{i})$ ; its coefficient is $(\binom{r}{1}) = r$ . For $(a, b) = (0, 1)$ the average is $n^{- 1} \sum_{j} g_{Y} (Y_{j})$ with the same coefficient $r$ . All remaining pairs have $a + b \geq 2$ and are, by definition, the remainder $R_{m, n}$ .

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

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Studentized two-sample U-statistic and permutation distribution

Definition

For the two-sample statistic in Two-sample order-r U-statistic and first projections, let ${\hat{σ}}_{m}^{2}$ and ${\hat{σ}}_{n}^{2}$ be nonnegative measurable statistics of the two samples intended to estimate the variances of $g_{X} (X_{1})$ and $g_{Y} (Y_{1})$ . Each may depend on both groups; separate-group dependence is an additional property, not part of this definition. Define the estimated standard error and studentized statistic by

$V_{m, n}^{S E} : = r \sqrt{\frac{{\hat{σ}}_{m}^{2}}{m} + \frac{{\hat{σ}}_{n}^{2}}{n}}, S_{m, n} : = {\begin{cases} U_{m, n} / V_{m, n}^{S E}, & V_{m, n}^{S E} > 0, \\ 0, & V_{m, n}^{S E} = 0. \end{cases}$

For a uniform random permutation $π$ of the pooled $N = m + n$ observations, recompute both group statistics and both variance estimates on the permuted groups; the permutation distribution of $S_{m, n}$ is the conditional law of $S_{m, n}^{π}$ given the pooled observations.

Remarks

The report's scale is $V_{m, n}^{C R} = r \sqrt{{\hat{σ}}_{m}^{2} + (m / n) {\hat{σ}}_{n}^{2}}$ and its statistic is $U_{m, n} / V_{m, n}^{C R}$ , with a $\sqrt{m}$ multiplier. Since $V_{m, n}^{S E} = V_{m, n}^{C R} / \sqrt{m},$ the usual standard-error studentization is $S_{m, n} = \sqrt{m} U_{m, n} / V_{m, n}^{C R} .$ The standard-normal limits for studentized statistics use $S_{m, n}$ itself as the pivot, under the null, nondegeneracy, and variance-consistency hypotheses of the corresponding limit theorem. An additional $\sqrt{m}$ multiplying this $S_{m, n}$ is not part of the studentized pivot.

Consistency and permutation consistency are hypotheses of asymptotic validity results, not consequences of this definition. For an estimate depending on both groups, consistency on an IID pooled-mixture row alone does not imply permutation consistency; that condition must be proved for the actual estimate.

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Projection CLT for a two-sample U-statistic

Statement

Assume AC and the setup of Two-sample order-r U-statistic and first projections, $r$ is fixed, $m / N \to p \in (0, 1)$ , $θ (P, Q; h) = 0$ , and $0 < E_{P} [g_{X}^{2}] = σ_{X}^{2} < \infty$ and $0 < E_{Q} [g_{Y}^{2}] = σ_{Y}^{2} < \infty$ . If the fixed-order kernel has finite second moments under all mixed assignments, then

$\sqrt{m} U_{m, n} \Rightarrow N ⁣ (0, r^{2} (σ_{X}^{2} + \frac{p}{1 - p} σ_{Y}^{2})) .$

Facts & Assumptions

Given: Independent IID samples from $P$ and $Q$ , fixed $r$ , $m / N \to p \in (0, 1)$ , a zero parameter, and positive finite first-projection variances.

[F1]

Two-sample order-r U-statistic and first projections defines $U_{m, n}$ and its centered projections $g_{X}, g_{Y}$ .

[F2]

Second-moment order of the degenerate remainder gives $\sqrt{N} R_{m, n} \to 0$ in $L^{2}$ under the mixed second-moment premises.

[F3]

Lindeberg-Levy iid central limit theorem: under AC, each IID centered real sample mean with positive finite variance has its normalized normal limit.

[F4]

Characteristic functions under affine maps and independent sums, Characteristic function of a normal law, and Characteristic function criterion for weak convergence give the independent-sum normal limit.

[F5]

Slutsky's theorem for real random variables transfers a weak limit after adding a term converging to zero in probability.

[A1]

AC is assumed exactly for the CLT and characteristic-function suppliers in [F3]–[F4]. The Axiom of Choice

Proof

Given: The assumptions in the Statement. Prove the centered normal limit at scale $\sqrt{m}$ .

1.1F1F2givenalgebra

By [F1] and [F2], $\sqrt{m} U_{m, n} = r \frac{1}{\sqrt{m}} \sum_{i = 1}^{m} g_{X} (X_{i}) + r \sqrt{\frac{m}{n}} (\frac{1}{\sqrt{n}} \sum_{j = 1}^{n} g_{Y} (Y_{j})) + \sqrt{m} R_{m, n},$ and the final term tends to zero in probability because $m / N$ is bounded and $E [(\sqrt{N} R_{m, n})^{2}] \to 0$ .

1.2F1F3A1given

The summands $g_{X} (X_{i})$ are IID, centered, and have variance $σ_{X}^{2}$ ; [F3] gives $m^{- 1 / 2} \sum_{i} g_{X} (X_{i}) \Rightarrow N (0, σ_{X}^{2})$ . The $g_{Y} (Y_{j})$ are IID, centered, and have variance $σ_{Y}^{2}$ , so the same result gives $n^{- 1 / 2} \sum_{j} g_{Y} (Y_{j}) \Rightarrow N (0, σ_{Y}^{2})$ . The two sums are independent because the two samples are independent.

2.1F1F4step 1.1step 1.2algebra

Since $\sqrt{m / n} \to \sqrt{p / (1 - p)}$ , the characteristic function of the sum of the first two terms in step 1.1 is the product of the two independent characteristic functions. By [F4] this product converges to the characteristic function of a centered normal with variance $r^{2} (σ_{X}^{2} + (p / (1 - p)) σ_{Y}^{2})$ ; the positivity assumptions make that variance nonzero.

3.1F5step 1.1step 2.1∎

The first two terms in step 1.1 therefore have the stated weak limit. The last term converges to zero in probability by step 1.1, so [F5] gives the claimed limit for $\sqrt{m} U_{m, n}$ . All divisions are by positive finite variances, and no claim is made when either first-projection variance is zero.

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Finite-sample exactness under equal distributions

Statement

For the studentized statistic in Studentized two-sample U-statistic and permutation distribution, assume $P = Q$ , $m, n \geq r$ , and use the convention there when a permuted denominator is zero. For every such finite $m, n$ , the inclusive permutation p-value is superuniform under this null. If an independent uniform $[0, 1]$ variable randomizes within the tied boundary rank, the resulting permutation p-value is exactly uniform.

Facts & Assumptions

Given: Independent IID samples of sizes $m, n$ from a common probability law, the measurable coordinate-permutation action on their pooled vector, and the studentized statistic with its zero-denominator convention.

[F1]

Identical distribution and IID families: the pooled observations are IID when $P = Q$ .

[F2]

For sigma-finite factors, the product measure exists, has the rectangle formula, is sigma-finite, and is unique: the finite product law is determined by its rectangle values.

[F3]

Coordinate maps and permutations are measurable on finite products: each coordinate permutation is measurable.

[F4]

Studentized two-sample U-statistic and permutation distribution: the statistic is a measurable real function, including on zero-denominator inputs.

[F5]

Exact finite-group permutation p-values and tests: under an invariant law, every measurable statistic on a finite measurable group action has a superuniform inclusive orbit-rank p-value and an exactly uniform tie-randomized p-value.

Proof

Given: $P = Q$ , independent IID samples with $m, n \geq r$ , all coordinate permutations of the pooled sample, and the rank conventions in [F5].

1.1F1F2givenalgebra

By [F1], the pooled vector has product law $P^{\otimes N}$ . On every measurable rectangle $A_{1} \times \dots \times A_{N}$ , its probability is $\prod_{i = 1}^{N} P (A_{i})$ ; permuting coordinates merely reorders these factors. By uniqueness in [F2], the product law is invariant under every coordinate permutation.

2.1F3F4F5step 1.1

The symmetric group on $N$ coordinates is finite and acts by measurable maps by [F3]. The statistic is measurable on the entire pooled sample space by [F4], including any zero-denominator permutation. Thus the hypotheses of [F5] hold for this action and invariant law.

3.1F5step 2.1∎

Apply [F5] to conclude that the inclusive p-value is superuniform and the independent-uniform tie-randomized p-value is uniform on $[0, 1]$ . The result is finite-sample and does not require a nondegenerate first projection.

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Permutation CLT for a two-sample U-statistic under a weak null

Statement

Assume AC. Let $X_{1}, \dots, X_{m}$ be IID $P$ and independently let $Y_{1}, \dots, Y_{n}$ be IID $Q$ , with $N = m + n$ , $m / N \to p \in (0, 1)$ , and $θ (P, Q; h) = 0$ . Fix $r$ and a measurable kernel $h$ that is symmetric in each block and antisymmetric under block exchange. Assume for every assignment of independent P- and Q-distributed arguments to the $2 r$ positions that $0 < E [h^{2}] < \infty$ . Let $σ_{p}^{2} = E_{\overset{ˉ}{P}} [g_{\overset{ˉ}{P}}^{2}] > 0$ for $\overset{ˉ}{P} = p P + (1 - p) Q$ , where $g_{\overset{ˉ}{P}}$ is its first projection. Then the conditional permutation CDF of $\sqrt{m} U_{m, n}$ satisfies

$\sup_{t \in R} ∣ {\hat{R}}_{m, n}^{\sqrt{m} U} (t) - Φ ⁣ (\frac{t}{τ_{p}}) ∣ \to_{P} 0, τ_{p}^{2} = \frac{r^{2}}{1 - p} σ_{p}^{2} .$

Facts & Assumptions

Given: The sample and kernel hypotheses in the Statement, including finite positive second moments under every mixed argument assignment.

[F1]

Two-sample order-r U-statistic and first projections defines $U_{m, n}$ and its two first projections.

[F2]

Two-sample Hoeffding decomposition gives the first-order projection plus the finite sum of canonical higher-order terms.

[F3]

Second-moment order of the degenerate remainder makes the scaled higher-order remainder negligible.

[F4]

Pooled-mixture projection and variance identifies the pooled projection $g_{\overset{ˉ}{P}}$ and its variance scale $τ_{p}^{2}$ .

[F5]

Conditional CLT for a random subset of an IID finite population gives conditional normal convergence for a uniform sample without replacement from an IID pooled row.

[F6]

Mixture coupling for permuted two-sample U-statistics couples the original permuted U-statistic to the IID-mixture permuted U-statistic under antisymmetry and all mixed second moments.

[A1]

AC is used through [F5]. The Axiom of Choice

Proof

Given: The weak-null and mixed-moment kernel hypotheses and $m / N \to p \in (0, 1)$ . Prove the conditional permutation CDF limit for $\sqrt{m} U_{m, n}$ .

1.1F1F2F3F4given

Let $p_{N} = m / N$ and ${\overset{ˉ}{P}}_{N} = p_{N} P + (1 - p_{N}) Q$ . Under an IID ${\overset{ˉ}{P}}_{N}$ row, antisymmetry gives pooled parameter zero and first projections $g_{N}$ and $- g_{N}$ by [F4]. Expanding the $2 r$ mixture factors shows that $E_{{\overset{ˉ}{P}}_{N}^{\otimes 2 r}} h^{2}$ is a finite convex combination of the mixed-assignment second moments in the hypotheses, hence is uniformly bounded in $N$ . The same bound applies to every mixed assignment when both row laws are ${\overset{ˉ}{P}}_{N}$ , so the triangular version of [F3] applies to the two groups formed by a uniform permutation and makes $\sqrt{m}$ times the degenerate remainder tend to zero in joint probability.

2.1F1F4F5givenalgebra

The linear projection for a uniform size- $m$ subset $A$ of the pooled row is $\sqrt{m} L_{N}^{π} = \frac{r N}{n \sqrt{m}} (\sum_{i \in A} g_{N} ({\overset{ˉ}{Z}}_{i}) - \frac{m}{N} \sum_{i = 1}^{N} g_{N} ({\overset{ˉ}{Z}}_{i})) .$ Expanding each factor of ${\overset{ˉ}{P}}_{N}$ shows that $g_{N}$ is a finite linear combination of the section-integral functions for the finitely many mixed P/Q assignments. The assignment-wise second-moment bounds make these functions square-integrable under P and Q; their coefficients are bounded as $p_{N} \to p$ . If $H$ is the sum of their squared absolute values, then $H$ is integrable under both P and Q and $g_{N}^{2} \leq C H$ for one constant $C$ . Hence, for $K > 0$ , $E_{{\overset{ˉ}{P}}_{N}} [g_{N}^{2} 1_{{g_{N}^{2} > K}}] \leq C (E_{P} [H 1_{{H > K / C}}] + E_{Q} [H 1_{{H > K / C}}]) ⟶ 0,$ so the row $g_{N} ({\overset{ˉ}{Z}}_{i})$ has uniformly integrable squares. Its variance is $σ_{N}^{2} \to σ_{p}^{2}$ . Let $S_{g, N}^{2} = (N - 1)^{- 1} \sum_{i = 1}^{N} (g_{N} ({\overset{ˉ}{Z}}_{i}) - {\overset{ˉ}{g}}_{N})^{2}$ be its empirical finite-population variance. Apply step 1.1 of [F5] to this IID row: the uniformly integrable squares give $N^{- 1} \sum_{i} g_{N} ({\overset{ˉ}{Z}}_{i})^{2} - σ_{N}^{2} \to_{P} 0$ , while $E {\overset{ˉ}{g}}_{N} = 0$ and $Var ({\overset{ˉ}{g}}_{N}) = σ_{N}^{2} / N \to 0$ . Therefore $S_{g, N}^{2} - σ_{N}^{2} \to_{P} 0$ and $S_{g, N}^{2} \to_{P} σ_{p}^{2}$ .

3.1A1F5step 1.1step 2.1algebra

Put $T_{N}^{π} = \frac{\sum_{i \in A} (g_{N} ({\overset{ˉ}{Z}}_{i}) - {\overset{ˉ}{g}}_{N})}{\sqrt{m n S_{g, N}^{2} / N}}$ when $S_{g, N}^{2} > 0$ , and set it to zero otherwise. Apply [F5] to this random subset. The exact identity on ${S_{g, N}^{2} > 0}$ is $\sqrt{m} L_{N}^{π} = c_{N} T_{N}^{π}$ , where $c_{N} = r \sqrt{N / n} S_{g, N}$ . Since $S_{g, N}^{2} \to_{P} σ_{p}^{2} > 0$ , $c_{N} \to_{P} r σ_{p} / \sqrt{1 - p} = τ_{p}$ and the event $S_{g, N}^{2} = 0$ has probability tending to zero. The conditional CDF of $T_{N}^{π}$ converges uniformly in probability to $Φ$ by [F5]; thus, for each fixed $t$ , the conditional CDF of $c_{N} T_{N}^{π}$ converges in probability to $Φ (t / τ_{p})$ . Indeed, on $c_{N} > 0$ its conditional CDF is $F_{N, \overset{ˉ}{Z}} (t / c_{N})$ , whose difference from $Φ (t / c_{N})$ is bounded by $\sup_{u} ∣ F_{N, \overset{ˉ}{Z}} (u) - Φ (u) ∣$ , and $Φ (t / c_{N}) \to_{P} Φ (t / τ_{p})$ . The zero-variance event is negligible. The conditional probability of the remainder exceeding any fixed threshold tends to zero in probability, since its expectation tends to zero by step 1.1 and Markov's inequality. Another threshold squeeze and continuity of the normal CDF give the same pointwise conditional limit for $\sqrt{m} U_{m, n} ({\overset{ˉ}{Z}}^{π})$ .

4.1F6step 3.1algebra

Let $π, π^{'}$ be conditionally independent uniform permutations. By step 3.1, the two mixture-row statistics have asymptotically independent $N (0, τ_{p}^{2})$ limits: conditionally on $\overset{ˉ}{Z}$ , their joint CDF is the product of the two conditional CDFs, and each factor converges in probability to a constant. Apply the coupling [F6] to both permutations; the union bound and its $L^{2}$ conclusion preserve this joint limit for the original sample.

5.1step 3.1step 4.1algebra∎

Write $R_{N} (t)$ for the original conditional permutation CDF. The one-permutation convergence gives $E R_{N} (t) \to Φ (t / τ_{p})$ ; the two independent permutations in step 4.1 give $E [R_{N} (t)^{2}] \to Φ (t / τ_{p})^{2}$ . Therefore $Var (R_{N} (t)) \to 0$ , so $R_{N} (t) \to_{P} Φ (t / τ_{p})$ at every $t$ . Since the limit CDF is continuous, choose finite tail cutoffs and a finite grid with arbitrarily small $Φ$ increments; monotonicity bounds the supremum between grid values. This proves the stated uniform convergence.

Remarks

The proof uses the exact triangular mixture and the locally strengthened coupling estimate. It requires neither zero kernel means for other assignments nor a convergence rate for $m / N - p$ . The weak null is needed for the accompanying true-sampling limit; antisymmetry alone centers the permutation distribution.

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Permutation CLT for a two-sample U-statistic under mixed-assignment centering

Statement

Assume AC. Let $X_{1}, \dots, X_{m}$ be IID $P$ and independently let $Y_{1}, \dots, Y_{n}$ be IID $Q$ , with $N = m + n$ , $m / N \to p \in (0, 1)$ , and $θ (P, Q; h) = 0$ . Fix $r$ and a measurable kernel $h$ that is symmetric in each block and antisymmetric under block exchange. Assume for every assignment of independent P- and Q-distributed arguments to the $2 r$ positions that $E [h] = 0$ and $0 < E [h^{2}] < \infty$ . Let $σ_{p}^{2} = E_{\overset{ˉ}{P}} [g_{\overset{ˉ}{P}}^{2}] > 0$ for $\overset{ˉ}{P} = p P + (1 - p) Q$ , where $g_{\overset{ˉ}{P}}$ is its first projection. Then the conditional permutation CDF of $\sqrt{m} U_{m, n}$ satisfies

$\sup_{t \in R} ∣ {\hat{R}}_{m, n}^{\sqrt{m} U} (t) - Φ ⁣ (\frac{t}{τ_{p}}) ∣ \to_{P} 0, τ_{p}^{2} = \frac{r^{2}}{1 - p} σ_{p}^{2} .$

Facts & Assumptions

Given: The sample, all-assignment centering and moment assumptions, and positive pooled projection variance in the Statement.

[F1]

Permutation CLT for a two-sample U-statistic under a weak null gives the conditional permutation limit under the weak null, block antisymmetry, finite positive mixed second moments and positive pooled projection variance.

[A1]

AC is inherited through [F1]. The Axiom of Choice

Proof

Given: The hypotheses in the Statement. Apply the general weak-null result.

1.1F1given

The stated all-assignment assumptions include the weak null and every finite positive mixed second moment required by [F1]. Block symmetry, antisymmetry, proportional growth and positive pooled projection variance are the same hypotheses as in [F1].

2.1A1F1step 1.1algebra∎

Under AC [A1], apply [F1] to obtain the stated uniform conditional CDF limit and the identical scale $τ_{p}^{2} = r^{2} σ_{p}^{2} / (1 - p)$ . The additional assignment-wise centering assumptions cause no change in that scale.

Remarks

This is the all-assignment-centered specialization of the general weak-null theorem. Its extra centering assumptions are sufficient but are not needed by the locally proved permutation coupling.

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Studentized permutation validity under a weak two-sample parameter null

Statement

Assume AC. Let independent IID samples $X_{1}, \dots, X_{m} \sim P$ and $Y_{1}, \dots, Y_{n} \sim Q$ satisfy $N = m + n$ , $m / N \to p \in (0, 1)$ , and the weak null $θ (P, Q; h) = 0$ . Fix $r$ and a measurable kernel symmetric within each block and antisymmetric under block exchange. Assume $0 < E_{η} h^{2} < \infty$ for every independent P/Q assignment $η$ to the $2 r$ positions. Suppose the direct first-projection variances $σ_{X}^{2}, σ_{Y}^{2}$ and limiting pooled-mixture first-projection variance $σ_{p}^{2}$ are positive and finite. Put $p_{N} = m / N$ , ${\overset{ˉ}{P}}_{N} = p_{N} P + (1 - p_{N}) Q$ , and $σ_{N}^{2} = E_{{\overset{ˉ}{P}}_{N}} g_{N}^{2} \to σ_{p}^{2}$ .

Use the nonnegative measurable variance estimates and zero-denominator convention in Studentized two-sample U-statistic and permutation distribution. Assume ${\hat{σ}}_{m}^{2} \to_{P} σ_{X}^{2}$ , ${\hat{σ}}_{n}^{2} \to_{P} σ_{Y}^{2}$ under the original samples. For general estimates depending on both groups, assume also permutation consistency: for each $ε > 0$ and $k \in {m, n}$ ,

$P_{π} (∣ {\hat{σ}}_{k, π}^{2} - σ_{N}^{2} ∣ > ε ∣ Z) \to_{P} 0.$

A sufficient condition for this last hypothesis is that each estimate is a statistic of its own group alone, and is consistent for $σ_{N}^{2}$ on an IID ${\overset{ˉ}{P}}_{N}$ sample of that group's size. Joint estimates may instead establish permutation consistency directly.

Then the conditional permutation CDF of $S_{m, n} = U_{m, n} / V_{m, n}^{S E}$ converges uniformly in probability to $Φ$ , and $S_{m, n} \Rightarrow N (0, 1)$ under the true sampling law, where

$V_{m, n}^{S E} = r \sqrt{{\hat{σ}}_{m}^{2} / m + {\hat{σ}}_{n}^{2} / n} .$

Facts & Assumptions

Given: The weak-null, moment, positive-projection and estimator-consistency hypotheses in the Statement.

[F1]

Permutation CLT for a two-sample U-statistic under a weak null gives the conditional permutation limit of $\sqrt{m} U$ with variance $τ_{p}^{2} = r^{2} σ_{p}^{2} / (1 - p)$ . Its proof also shows $σ_{N}^{2} \to σ_{p}^{2}$ from the finite family of mixed second moments.

[F2]

Consistency transfers from IID mixture samples to permuted groups transfers consistency of a statistic of one IID triangular-mixture group to that permuted group.

[F3]

Projection CLT for a two-sample U-statistic gives the true-sampling CLT with variance $a^{2} = r^{2} (σ_{X}^{2} + (p / (1 - p)) σ_{Y}^{2}) > 0$ .

[F4]

Studentized two-sample U-statistic and permutation distribution defines $V^{S E}$ and sets the ratio to zero if that standard error is zero.

[F5]

Slutsky's theorem for real random variables divides a weak limit by a positive probability limit.

[A1]

AC is inherited through [F1]–[F3]. The Axiom of Choice

Proof

Given: The hypotheses in the Statement. Prove both standard-normal limits and the sufficient consistency criterion.

1.1A1F2givenalgebra

If an estimate depends on its own group alone and has the stated IID triangular-mixture consistency, [F2] gives its unconditional convergence under a random permutation. The expectation of its conditional failure probability given $Z$ is that unconditional failure probability, which tends to zero. Markov's inequality makes the conditional failure probability tend to zero in probability, proving the sufficient condition. For general joint estimates, this conditional convergence is assumed directly. A union bound treats the two estimates simultaneously.

2.1F4step 1.1givenalgebra

Thus $\sqrt{m} V^{S E, π} = r \sqrt{{\hat{σ}}_{m, π}^{2} + (m / n) {\hat{σ}}_{n, π}^{2}}$ converges in conditional probability, in probability over $Z$ , to $τ_{p} = r \sqrt{σ_{p}^{2} / (1 - p)} > 0$ . In particular, the conditional probability of a zero denominator tends to zero.

3.1A1F1step 2.1algebra

For fixed $t$ and $0 < δ < τ_{p} / 2$ , restrict to $∣ \sqrt{m} V^{S E, π} - τ_{p} ∣ < δ$ . There the event $S^{π} \leq t$ is bounded between the numerator events with thresholds $t (τ_{p} - δ)$ and $t (τ_{p} + δ)$ , reversing their order for $t < 0$ . The excluded event has conditional probability tending to zero by step 2.1. Apply [F1] at those thresholds and then let $δ ↓ 0$ to obtain conditional CDF convergence to $Φ (t)$ . A finite grid on a bounded interval with small normal tail probabilities and small adjacent normal CDF increments, together with monotonicity, makes this convergence uniform in $t$ .

4.1A1F3F4F5givenalgebra∎

Under the original sampling law, estimator consistency and [F4] give $\sqrt{m} V^{S E} \to_{P} a > 0$ , while [F3] gives $\sqrt{m} U \Rightarrow N (0, a^{2})$ . By [F5], their ratio converges to $N (0, 1)$ . The zero-denominator convention changes an event of probability tending to zero.

Remarks

The standard-error ratio $S$ is the pivot. The report's normalization is $V^{C R} = \sqrt{m} V^{S E}$ , so its $\sqrt{m} (U / V^{C R})$ equals $S$ . An extra factor $\sqrt{m}$ on the present $S$ would be erroneous.

Assignment-wise centering is not required by the locally proved coupling and permutation CLT. The report's stronger coupling assumptions are not inferred from the weak null, and no unavailable 2013 supplementary proof is used. IID-mixture consistency alone transfers by [F2] only for separate-group statistics; general joint estimates require the explicit permutation condition above.

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Studentized permutation validity under mixed-assignment centering

Statement

Assume AC and the sampling and kernel conditions of Permutation CLT for a two-sample U-statistic under mixed-assignment centering. Let the direct first-projection variances $σ_{X}^{2}, σ_{Y}^{2}$ and pooled variance $σ_{p}^{2}$ be positive and finite, and put $σ_{N}^{2} = E_{{\overset{ˉ}{P}}_{N}} g_{N}^{2} \to σ_{p}^{2}$ . Use the studentized statistic in Studentized two-sample U-statistic and permutation distribution. Suppose both variance estimates are consistent for their direct projection variances under the original samples and are permutation-consistent for $σ_{N}^{2}$ in the conditional sense of Studentized permutation validity under a weak two-sample parameter null. A sufficient condition for the latter is that each estimate depends on its own group alone and is consistent for $σ_{N}^{2}$ on an IID ${\overset{ˉ}{P}}_{N}$ sample of that size.

Then the conditional permutation CDF of $S_{m, n} = U_{m, n} / V_{m, n}^{S E}$ converges uniformly in probability to $Φ$ , and under the true sampling law $S_{m, n} \Rightarrow N (0, 1)$ .

Facts & Assumptions

Given: All-assignment centering and moments, positive direct and pooled projection variances, direct consistency, and either direct permutation consistency or the sufficient separate-group triangular-mixture condition.

[F1]

Studentized permutation validity under a weak two-sample parameter null proves both studentized limits and the sufficient separate-group consistency condition under the weak null and mixed moments.

[F2]

Permutation CLT for a two-sample U-statistic under mixed-assignment centering states the sampling and all-assignment centered kernel conditions inherited here.

[A1]

AC is inherited through [F1]. The Axiom of Choice

Proof

Given: The hypotheses in the Statement. Apply the general studentized theorem.

1.1F1F2given

The all-assignment centered conditions in [F2] include the weak null, block antisymmetry, fixed order, proportional sample growth and all mixed moments needed by [F1]. The remaining projection and estimator conditions here are exactly the corresponding conditions of [F1].

2.1A1F1step 1.1∎

Under AC [A1], apply [F1] to obtain both standard-normal limits and, when used, its separate-group sufficient consistency criterion.

Remarks

This specialization retains the original all-assignment centered branch. Joint two-group variance estimates satisfy the theorem when their permutation consistency is established directly; the separate-group transfer lemma does not by itself prove consistency of arbitrary joint estimates.

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Studentized two-sample U-statistic permutation test

Definition

Inputs: Two samples of sizes $m, n \geq r$ , a measurable order- $(r, r)$ kernel symmetric within each group, two supplied nonnegative measurable variance-estimation routines, and a significance level $α \in (0, 1)$ . Use the U-statistic in Two-sample order-r U-statistic and first projections and the standard-error pivot $S$ in Studentized two-sample U-statistic and permutation distribution, including its value zero when the estimated standard error is zero. Choose the upper alternative statistic $T = S$ , lower alternative statistic $T = - S$ , or two-sided statistic $T = ∣ S ∣$ . Optionally supply an independent uniform random variable $U \in [0, 1]$ for ties.

Pool the observations into $z = (X_{1}, \dots, X_{m}, Y_{1}, \dots, Y_{n})$ and compute $T (z)$ .
Enumerate all $N!$ coordinate permutations $π$ of $N = m + n$ . For each permutation, treat its first $m$ observations as the first group and its last $n$ as the second. Recompute the U-statistic and both variance estimates from those groups, apply the same zero-denominator rule, and compute $T (z^{π})$ .
Return the inclusive p-value $p_{i n c} = N!^{- 1} # {π : T (z^{π}) \geq T (z)}$ and reject when $p_{i n c} \leq α$ . If tie randomization was requested, instead return $p_{U} = N!^{- 1} {# {π : T (z^{π}) > T (z)} + U # {π : T (z^{π}) = T (z)}}$ and reject when $p_{U} \leq α$ .

This is complete finite enumeration, with permutations counted with their multiplicities even when observations repeat. It does not define a Monte Carlo approximation using a subset of permutations. No generic weak-null validity is asserted by the procedure alone; that requires the applicable studentization and estimator-consistency theorem.

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Studentized Wilcoxon validity under heterogeneous laws

Statement

Assume AC. Let independent IID real samples $X_{1}, \dots, X_{m} \sim P$ and $Y_{1}, \dots, Y_{n} \sim Q$ satisfy $m, n \geq 2$ and $m / (m + n) \to p \in (0, 1)$ . Define the tie-corrected Wilcoxon kernel and weak null by $h (x, y) = 1 {y > x} - 1 {y < x}, θ = E h (X, Y) = 0.$ Equivalently, $P (X < Y) + \frac{1}{2} P (X = Y) = 1 / 2$ for independent $X \sim P, Y \sim Q$ . Put $a (x) = E_{Q} h (x, Y)$ , $b (y) = E_{P} h (X, y)$ and $g_{p} (z) = E_{p P + (1 - p) Q} h (z, V)$ . Assume $E_{P} a^{2} > 0$ , $E_{Q} b^{2} > 0$ and $E_{p P + (1 - p) Q} g_{p}^{2} > 0$ .

Set $U = (m n)^{- 1} \sum_{i, j} h (X_{i}, Y_{j})$ , $R_{i} = n^{- 1} \sum_{j} h (X_{i}, Y_{j})$ and $C_{j} = m^{- 1} \sum_{i} h (X_{i}, Y_{j})$ . Use the measurable nonnegative projection variance estimates ${\hat{σ}}_{m}^{2} = \frac{1}{m - 1} \sum_{i} (R_{i} - U)^{2}, {\hat{σ}}_{n}^{2} = \frac{1}{n - 1} \sum_{j} (C_{j} - U)^{2} .$ Recompute them in every permutation. With $S = U / \sqrt{{\hat{σ}}_{m}^{2} / m + {\hat{σ}}_{n}^{2} / n}$ and the zero-denominator convention of Studentized two-sample U-statistic and permutation distribution, both $S \Rightarrow N (0, 1)$ and its conditional permutation CDF converges uniformly in probability to $Φ$ . Consequently, each of the upper, lower and two-sided tests in Studentized two-sample U-statistic permutation test has rejection probability tending to $α$ under this weak null. If $P = Q$ , inclusive p-values are finite-sample superuniform and tie-randomized p-values are exactly uniform, without a nondegeneracy requirement for this finite-sample assertion.

The laws may be different and may have atoms. This null is a probabilistic-index null, not an assertion that equality of means or medians suffices for Wilcoxon validity.

Facts & Assumptions

Given: AC and the laws, positive projection variances, samples and estimators in the Statement.

[F1]

For a fixed antisymmetric two-sample U-kernel with positive direct and pooled first-projection variances, finite positive mixed second moments, sampling variance consistency and direct conditional permutation consistency of the estimates imply both standard-normal limits. Studentized permutation validity under a weak two-sample parameter null

[F2]

The standard-error pivot uses $r = 1$ here and has value zero when its standard error vanishes. Studentized two-sample U-statistic and permutation distribution

[F3]

The test recomputes both variances for every permutation and uses inclusive or randomized orbit ranks, for $S$ , $- S$ or $∣ S ∣$ . Studentized two-sample U-statistic permutation test

[F4]

When $P = Q$ , the pooled product law is invariant under coordinate permutations, as proved by the exactness corollary. For any measurable score on that orbit, inclusive ranks are superuniform and independent-uniform tie ranks are uniform, including its zero-denominator rule. Finite-sample exactness under equal distributions Finite-group rank validity on a measurable sample space

[F5]

For a square-integrable variable $Z$ , $P (∣ Z - E Z ∣ > t) \leq Var (Z) / t^{2}$ . Chebyshev's inequality for random variables

Proof

1.1F1givenalgebra

The bounded measurable kernel is antisymmetric and $h (x, x) = 0$ . All mixed second moments are finite. Positivity of $E_{P} a^{2}$ excludes a point-mass $P$ because $a$ is constant then; likewise positivity of $E_{Q} b^{2}$ excludes a point-mass $Q$ . A non-point-mass law has positive probability that two independent draws differ: otherwise every event would have probability zero or one, and on the real line its distribution function would identify a point mass. Thus the P/P and Q/Q second moments of $h$ are positive. The P/Q second moment is positive because its vanishing would give $a = 0$ P-almost surely; the Q/P case follows by antisymmetry. The first projections under the null are exactly $a, b$ . For $μ_{N} = (m / N) P + (n / N) Q$ , its pooled projection $g_{N}$ is centered by antisymmetry; its second moment $σ_{N}^{2}$ converges to the positive displayed pooled variance because it is a polynomial in $m / N$ with bounded coefficients.

2.1F5step 1.1algebra

Conditional on $X_{i}$ , $R_{i}$ averages $n$ independent variables bounded by one, so $E (R_{i} - a (X_{i}))^{2} \leq 1 / n$ . Consequently $m^{- 1} \sum_{i} (R_{i} - a (X_{i}))^{2} \to 0$ in probability by the elementary nonnegative bound $P (Z > t) \leq E Z / t$ . The corresponding bound for $C_{j} - b (Y_{j})$ is $1 / m$ . The averages of $a (X_{i})$ and $a (X_{i})^{2}$ converge to their expectations by [F5], since their variances are bounded by $1 / m$ ; similarly for $b$ . For bounded lists in $[- 1, 1]$ , replacing a list by one with mean squared error $e$ changes its mean by at most $\sqrt{e}$ and its mean square by at most $2 \sqrt{e}$ , by the finite-sum Cauchy–Schwarz inequality. Since the two row/column means equal $U$ , the stated sample variances are therefore consistent for $E_{P} a^{2}$ and $E_{Q} b^{2}$ .

3.1F5step 2.1algebra

Fix a deterministic pooled list $z_{1}, \dots, z_{N}$ . Put $d_{i} = N^{- 1} \sum_{k} h (z_{i}, z_{k})$ and $A_{N} = N^{- 1} \sum_{i} d_{i}^{2}$ . Antisymmetry gives $\sum_{i} d_{i} = 0$ . For a uniform split into groups A and B of sizes m,n, condition on $i \in A$ . Then B is a uniform n-subset of the other N−1 indices, so its average of $h (z_{i}, z_{j})$ has mean $N d_{i} / (N - 1)$ and variance at most $4 / n$ . Indeed, sampling k of M bounded numbers without replacement gives variance $(M - k) / (k (M - 1))$ times their population variance when $M > 1$ , by expanding the indicators with means $k / M$ and pair probabilities $k (k - 1) / (M (M - 1))$ ; their population variance is at most one. The loose bound four covers every permitted size. Thus the expected within-A average of $(R_{i} - d_{i})^{2}$ is at most $4 / n + (N - 1)^{- 2}$ . For B's column averages the same argument approximates $- d_{j}$ with error at most $4 / m + (N - 1)^{- 2}$ . Finally, the sample means of $d_{i}$ and $d_{i}^{2}$ in either split group converge to the full-list means in conditional probability, with variance at most $4 / \min (m, n)$ . The finite-list bound from step 2.1 shows both recomputed variance estimates converge to $A_{N}$ in conditional permutation probability, uniformly over every pooled list. The factors $m / (m - 1)$ and $n / (n - 1)$ tend to one.

4.1F5step 3.1algebra

Under the original independent pooled observations, $A_{N} = N^{- 3} \sum_{i, j, k} h (Z_{i}, Z_{j}) h (Z_{i}, Z_{k})$ is a bounded ordered-triple average. Terms with repeated indices comprise at most $3 N^{2}$ terms. For distinct indices, expectations factor according to their fixed P/Q labels; allowing all index triples in this deterministic label average is exactly integration of $h (z, v) h (z, w)$ under $μ_{N}^{\otimes 3}$ . Removing the repeated indices changes either expectation by at most $6 / N$ , hence $∣ E A_{N} - σ_{N}^{2} ∣ \leq 6 / N$ . Two triple terms with disjoint index sets are independent. At most $9 N^{5}$ ordered pairs of triples share an index, and each covariance has absolute value at most two, giving $Var (A_{N}) \leq 18 / N$ . By [F5], $A_{N} - σ_{N}^{2} \to 0$ in probability. Combining with the uniform conditional bound of step 3.1 proves the exact direct joint-estimator permutation consistency required by [F1], without asserting that joint-estimator consistency follows automatically from IID-mixture consistency.

5.1F1F2step 1.1step 2.1step 3.1step 4.1algebra

Steps 1.1, 2.1, 3.1, 4.1 verify every premise of [F1] for $r = 1$ and the actual empirical estimates. Its two limits apply to [F2], including its vanishing-probability zero-denominator event. This proves the two asserted standard-normal limits for heterogeneous laws, with or without ties.

6.1F3F4step 5.1algebra∎

For $T = S$ , $- S$ or $∣ S ∣$ , the two limits in step 5.1 imply convergence of the sampling CDF and the conditional permutation CDF to the same continuous limit $H$ , respectively the standard normal or absolute-standard-normal law. Uniform convergence for the absolute value follows by subtracting the two normal CDF tails; the limiting CDF is continuous. The maximal conditional atom is at most twice the sup-norm CDF error from $H$ , and therefore vanishes in probability. Both inclusive and randomized ranks in [F3] consequently differ from $1 - H (T)$ by a quantity tending to zero in probability. Since $T$ converges to the law with continuous CDF $H$ , $1 - H (T)$ converges to uniform: at each u in (0,1) this follows by the continuous strictly increasing quantile of H on its support. Thus rejection probability tends to $α$ . For $P = Q$ , [F4] instead gives the stated exact finite-sample conclusions directly, whether or not the positive-variance asymptotic hypotheses hold.

5 · Examples, counterexamples and false statements

None yet.