The weak null does not center every mixed kernel assignment

The kernel is symmetric in $(x_1,x_2)$ and in $(y_1,y_2)$, and exchanging the two blocks changes its sign. By independence within each product block, [F1] gives $\theta =(E_{P}X{)}^2-(E_{Q}Y{)}^2=(1\mathrm{/}2{)}^2-(-1\mathrm{/}2{)}^2=0$.

Under the mixed assignment $(X_1,X_2)\sim P^{\otimes 2}$ and $(Y_1,Y_2)\sim P\otimes Q$, independence gives $Eh=(E_{P}X{)}^2-(E_{P}X)(E_{Q}Y)=1\mathrm{/}4-(-1\mathrm{/}4)=1\mathrm{/}2\mathrm{\ne }0$. Thus the weak null does not force all mixed-assignment means to vanish.

Every law in $\{P,Q\}$ is bounded and assigns positive mass both to zero and to a nonzero value. Therefore every product of two independently assigned variables has positive probability at zero and positive probability at a nonzero value, so it has positive finite variance. For any two assignments, $h$ is the difference of two independent such products; its second moment is finite and strictly positive.

The direct projections are $g_X(x)=x\mathrm{/}2-1\mathrm{/}4$ and $g_Y(y)=1\mathrm{/}4+y\mathrm{/}2$. Each has mean zero and variance $3\mathrm{/}16$ under its respective law, so both direct first-projection variances are positive. At mixture weight $p=1\mathrm{/}4$, $\bar{P}=pP+(1-p)Q$ has mean $-1\mathrm{/}4$ and positive variance; its pooled projection is $g_{\bar{P}}(z)=(-1\mathrm{/}4)(z+1\mathrm{/}4)$, which also has positive variance.

The example satisfies block symmetry and antisymmetry, the weak parameter null, all mixed second-moment requirements, and positive direct and pooled first-projection variances. Step 1.2 nevertheless gives a mixed assignment with nonzero mean, refuting the Statement. This is a gap in the cited coupling route, not a counterexample to the broader asymptotic theorem by itself.