ANOVA variance identity for nestedly scrambled point sets

For a contrast label $r$, its normalized child vector is $v_r=(1,\dots ,1,-r,0,\dots ,0)\mathrm{/}\sqrt{r(r+1)}$. These vectors have coordinate sum zero and $v_r\cdot v_s=\mathbf{1}\{r=s\}$: if $r<s$, the $r+1$ nonzero entries of $v_r$ meet entries equal to $1$ in $v_s$ and sum to $r-r=0$; if $r=s$, the squared norm is $(r+r^2)\mathrm{/}(r(r+1))=1$. In a fixed coordinate $q$, consider $h_{k,t,r}(X_{i,q})h_{\mathrm{\ell },t^{\mathrm{\prime }},s}(X_{j,q})$. If $k<\mathrm{\ell }$, condition on all permutations used by these two factors except the one at the length-$\mathrm{\ell }$ input prefix of $a_{j,q}$. The coarse factor and both output parent locations are then fixed, while the finer output child is uniform, so the conditional product expectation is zero. The case $\mathrm{\ell }<k$ is symmetric. At equal levels with distinct input parents, condition on the preceding prefix permutations; the two independent next-node permutations give mean-zero child factors, hence zero expected product. With a common input parent, both points have the same output parent, so distinct output translation indices $t\mathrm{\ne }{t}^{\mathrm{\prime }}$ give zero product. For a common parent at level $k$, its output parent is uniform among the $b^k$ parents. If the two points use the same next child, the expectation of the product is $b^{-k}(b^{k+1}\mathrm{/}b)(v_r\cdot v_s)=\mathbf{1}\{r=s\}$. If they use distinct next children, their ordered output children are uniform among the $b(b-1)$ distinct pairs; since ${\sum }_c{v}_r(c)={\sum }_c{v}_s(c)=0$, $\frac{b^{k+1}}{b^{k}b(b-1)}{\sum }_{c\mathrm{\ne }{c}^{\mathrm{\prime }}}{v}_r(c)v_s(c^{\mathrm{\prime }})=-\frac{\mathbf{1}\{r=s\}}{b-1}\mathrm{.}$ Thus, for a common parent at equal levels, the coordinate covariance is $\mathbf{1}\{r=s\}{\mathrm{{\rm Y}}}_{ij,q,k}$; the zero cases above cover all other level or parent pairs.

Scrambling is independent across coordinates. Therefore products from distinct coordinate subsets have zero cross expectation, since a coordinate appearing in only one product contributes a mean-zero contrast independent of the other coordinates; [F8] factors that expectation across coordinates. For a common subset and scale, step 1.1 leaves only identical parent and contrast indices; multiplying the coordinate expectations gives ${\prod }_{q\in u}{\mathrm{{\rm Y}}}_{ij,q,k_q}$. Summing squared coefficients over the parent and contrast indices gives ${\sigma }_{u,\kappa }^2(f)$ by [F3].

For each $i$, the singleton $(a_i)$ is a $(1,0,0,d)$-net by [F9]. Using the same node permutations as in the full array gives exactly $X_i$ as its scrambled point by [F1], so [F4] gives $X_i\sim \mathrm{Unif}([0,1{)}^d)$ without a net assumption on the full array. For any finite ANOVA sum $g$ with constant coefficient $I(f)$, each nonconstant basis product consequently has mean zero under each $X_i$ by [F2] and [F6]. Expanding its centered variance and using [F7] and step 2.1 yields the corresponding finite sum $n^{-2}{\sum }_{u,\kappa }{M}_{u,\kappa }{\sigma }_{u,\kappa }^2$. By [F3], take increasing finite sums of whole ANOVA components $g_m\to f$ in $L^2$. For real $y_1,\dots ,y_n$, $0\le {\sum }_{i<j}(y_i-y_j{)}^2=n{\sum }_i{y}_i^2-{\left({\sum }_i{y}_i\right)}^2,$ so pointwise $\mathrm{\mid }{n}^{-1}{\sum }_i{y}_i{\mathrm{\mid }}^2\le n^{-1}{\sum }_i{y}_i^2$. Apply this to $y_i=g_m(X_i)-f(X_i)$ and integrate; uniformity and [F6] give $\mathbb{E}\mathrm{\mid }{\widehat{I}}_n(g_m)-{\widehat{I}}_n(f){\mathrm{\mid }}^2\le \frac{1}{n}{\sum }_i\mathbb{E}\mathrm{\mid }{g}_m(X_i)-f(X_i){\mathrm{\mid }}^2=\mathrm{\parallel }{g}_m-f{\mathrm{\parallel }}_2^2⟶0.$ Thus the estimators converge in $L^2$. Uniformity and [F6] give the limit mean $I(f)$ and finite second moment, and each finite estimator has the same mean because all its nonconstant terms are centered, using [F7]. The centered estimators therefore converge in $L^2$, so their variances converge by [F5]. Finally, $\mathrm{\mid }{\mathrm{\Gamma }}_{u,\kappa }\mathrm{\mid }\le n$ because every $\mathrm{\mid }{\mathrm{{\rm Y}}}_{ij,q,k}\mathrm{\mid }\le 1$; [F3] makes ${\sum }_{u,\kappa }{\sigma }_{u,\kappa }^2$ finite, so the series converges absolutely and the finite-sum identity passes to the limit.

Substituting ${\mathrm{\Gamma }}_{u,\kappa }=M_{u,\kappa }\mathrm{/}n$ into the identity of step 3.1 gives both displayed forms.