Pick-freeze covariances recover first and total Sobol numerators

Fix $i$. For fixed $x_i$, integrate $f(x_i,x_{-i})f(x_i,x_{-i}^{\mathrm{\prime }})$ over two independent copies of the other coordinates. Fubini [F5] factors this integral as $(P_{\{i\}}f(x_i){)}^2$. Averaging over $x_i$ gives $\mathbb{E}[f(X)f(X^{[i]})]=\mathrm{\parallel }{P}_{\{i\}}f{\mathrm{\parallel }}_2^2$. For a nonconstant ANOVA component with support $u≠\{i\}$, choose $j\in u\setminus \{i\}$ if $i\in u$, or any $j\in u$ otherwise. Dependence gives $P_u{f}_u=f_u$. Hence [F7] gives $P_{\{i\}}{f}_u=P_{\{i\}}{P}_u{f}_u=P_{\{i\}\cap u}{f}_u$ and $P_{\{i\}}{P}_{u\setminus \{j\}}{f}_u=P_{\{i\}\cap (u\setminus \{j\})}{f}_u$. The two index sets are equal by the choice of $j$, so these averages agree; the latter is zero by the coordinate-centering property [F1]. Thus only the constant and singleton components remain, and $P_{\{i\}}f={\mu }_f+f_{\{i\}}$. Orthogonality [F1] gives $\mathrm{\parallel }{P}_{\{i\}}f{\mathrm{\parallel }}_2^2={\mu }_f^2+V_{\{i\}}$, the first-order numerator by [F2]. Subtracting ${\mu }_f^2$ proves the first identity.

Fix $x_{-i}$. The two values $f(X_i,x_{-i})$ and $f(X_i^{\mathrm{\prime }},x_{-i})$ are independent with the same conditional product integral over coordinate $i$. Expanding their squared difference gives half its conditional expectation equal to the variance of that one-coordinate section. Integrating over $x_{-i}$ by Fubini yields $\frac{1}{2}\mathbb{E}[(f(X)-f({\tilde{X}}^{[i]}){)}^2]=\mathrm{\parallel }f-P_{-i}f{\mathrm{\parallel }}_2^2$.

If $i\mathrm{\notin }u$, then $P_{-i}{f}_u=f_u$ because $f_u$ depends only on $X_u$. If $i\in u$, dependence gives $P_u{f}_u=f_u$, and [F7] gives $P_{-i}{f}_u=P_{-i}{P}_u{f}_u=P_{u\setminus \{i\}}{f}_u=0$ by [F1]. Thus $f-P_{-i}f={\sum }_{u\ni i}{f}_u$. Orthogonality [F1] makes its squared norm ${\sum }_{u\ni i}{V}_u$, which is the total-effect numerator by [F2] and proves the second identity.