An extra square-root sample-size factor destroys standard-error studentization

The two population means are zero and the two variances are one, so $\theta (P,Q;h)=0$ by [F1]. The kernel is antisymmetric, and both P/Q assignments have mean zero. For each assignment, $h^2$ equals $0$ on the two equal-sign pairs and $4$ on the two opposite-sign pairs, each of probability $1\mathrm{/}4$, so $E[h^2]=2$. The first projections are $g_X(x)=E[x-Y]-\theta =x$ and $g_Y(y)=E[X-y]-\theta =-y$; each has variance $1$.

Under AC [A1], the sample variances are measurable finite-sample statistics. Here $W_i^2=1$ exactly and, by [F4], the sample mean $\bar{W}$ converges in probability to $0$; the identity $\stackrel{}{\mathrm{Var}}(W)=\frac{k}{k-1}(1-{\bar{W}}^2)$ then gives consistency to $1$. The pooled mixture is the same Rademacher law, so this consistency holds in both permuted groups and the positive pooled projection variance is also $1$. Thus [F6] applies and gives the standard-normal limits for $S_{m,n}$ itself and its conditional permutation distribution.

For every fixed $M>0$ and every $\epsilon >0$, eventually $M\mathrm{/}\sqrt{m}<\epsilon$, hence $P(\mathrm{\mid }\sqrt{m}{S}_{m,n}\mathrm{\mid }\le M)\le P(\mathrm{\mid }{S}_{m,n}\mathrm{\mid }\le \epsilon )\mathrm{.}$ By the true-sampling convergence in step 1.2, the limsup of the right side is at most $\mathrm{\Phi }(\epsilon )-\mathrm{\Phi }(-\epsilon )$. Letting $\epsilon \downarrow 0$ proves $P(\mathrm{\mid }\sqrt{m}{S}_{m,n}\mathrm{\mid }\le M)\to 0$; thus $\mathrm{\mid }\sqrt{m}{S}_{m,n}\mathrm{\mid }\to \mathrm{\infty }$ in probability.

Under AC [A1], let $F_N^{\pi }$ be the conditional permutation CDF of $S_{m,n}^{\pi }$. For fixed $\epsilon >0$, eventually $M\mathrm{/}\sqrt{m}<\epsilon$, so $P_{\pi }(\mathrm{\mid }\sqrt{m}{S}_{m,n}^{\pi }\mathrm{\mid }\le M\mid Z)\le F_N^{\pi }(\epsilon )-F_N^{\pi }(-\epsilon -),$ where $F_N^{\pi }(-\epsilon -)$ denotes the left limit of the conditional CDF. By [F6], the right side converges in probability to $\mathrm{\Phi }(\epsilon )-\mathrm{\Phi }(-\epsilon )$. Letting $\epsilon \downarrow 0$ shows that the conditional probability of any fixed bounded interval tends to zero in probability. The standard-normal pivot is $S_{m,n}$; the extra $\sqrt{m}$ factor is erroneous when the denominator is already a standard error.