Studentized Wilcoxon validity under heterogeneous laws

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 ${\mu }_N=(m\mathrm{/}N)P+(n\mathrm{/}N)Q$, its pooled projection $g_N$ is centered by antisymmetry; its second moment ${\sigma }_N^2$ converges to the positive displayed pooled variance because it is a polynomial in $m\mathrm{/}N$ with bounded coefficients.

Conditional on $X_i$, $R_i$ averages $n$ independent variables bounded by one, so $E(R_i-a(X_i){)}^2\le 1\mathrm{/}n$. Consequently $m^{-1}{\sum }_i(R_i-a(X_i){)}^2\to 0$ in probability by the elementary nonnegative bound $P(Z>t)\le EZ\mathrm{/}t$. The corresponding bound for $C_j-b(Y_j)$ is $1\mathrm{/}m$. The averages of $a(X_i)$ and $a(X_i{)}^2$ converge to their expectations by [F5], since their variances are bounded by $1\mathrm{/}m$; 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 $E_Q{b}^2$.

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 $N{d}_i\mathrm{/}(N-1)$ and variance at most $4\mathrm{/}n$. Indeed, sampling k of M bounded numbers without replacement gives variance $(M-k)\mathrm{/}(k(M-1))$ times their population variance when $M>1$, by expanding the indicators with means $k\mathrm{/}M$ and pair probabilities $k(k-1)\mathrm{/}(M(M-1))$; 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 $4\mathrm{/}n+(N-1{)}^{-2}$. For B's column averages the same argument approximates $-d_j$ with error at most $4\mathrm{/}m+(N-1{)}^{-2}$. 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 $4\mathrm{/}\min (m,n)$. 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 $m\mathrm{/}(m-1)$ and $n\mathrm{/}(n-1)$ tend to one.

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 $3N^2$ 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 ${\mu }_N^{\otimes 3}$. Removing the repeated indices changes either expectation by at most $6\mathrm{/}N$, hence $\mathrm{\mid }E{A}_N-{\sigma }_N^2\mathrm{\mid }\le 6\mathrm{/}N$. Two triple terms with disjoint index sets are independent. At most $9N^5$ ordered pairs of triples share an index, and each covariance has absolute value at most two, giving $\mathrm{Var}(A_N)\le 18\mathrm{/}N$. By [F5], $A_N-{\sigma }_N^2\to 0$ 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.

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.

For $T=S$, $-S$ or $\mathrm{\mid }S\mathrm{\mid }$, 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 $\alpha$. For $P=Q$, [F4] instead gives the stated exact finite-sample conclusions directly, whether or not the positive-variance asymptotic hypotheses hold.