Exact equality null versus weak parameter null

In both cases, $E_{P}X=E_{Q}Y=0$, so [F1] gives $\theta =0$. The kernel is antisymmetric, and for either assignment of P and Q to its two arguments its mean is $0-0=0$. All mixed second moments are finite and positive because the two laws have finite positive variances.

In the first case $P=Q$, so the joint law of the pooled IID sample is invariant under coordinate permutations; [F2] gives finite-sample superuniformity and exactness after tie randomization. In the second case $P\mathrm{\ne }Q$ (for example, $P(X=0)=0$ and $Q(Y=0)=1\mathrm{/}2$), so swapping coordinates across groups does not preserve the sampling law. The parameter null alone therefore does not supply the finite-sample exchangeability premise of [F2].

Under AC [A1], in each case the two population variances are positive, and the pooled-mixture variance is positive. For any bounded law $R$, [F5] applied to $W$ and $W^2$ shows that their sample averages converge to $E_{R}W$ and $E_R{W}^2$; the identity $\stackrel{}{\mathrm{Var}}(W)=\frac{k}{k-1}\{\overline{W^2}-{\bar{W}}^2\}$ then proves sample-variance consistency. For the triangular pooled laws ${\bar{P}}_N$, the same Chebyshev bounds are uniform because the support stays in $\{-2,-1,0,1,2\}$, and their variances converge to $1$ in the first case and $\frac{3}{2}$ in the second. Thus the estimators in [F4] satisfy the conditions of [F3]. In the second case, studentization has the asymptotic standard-normal calibration even though finite-sample group exchangeability is absent.