Global sensitivity anova and sobol estimation — Examples

Symmetry gives $\mathbb{E}Z=0$ and $\mathbb{E}{Z}^2=1$, so $V=1$. Since $Y$ is measurable with respect to either $X_1$ or $X_2$, [F5] fixes its conditional expectation for either singleton, giving $v(\mathrm{\varnothing })=0$, $v(\{1\})=v(\{2\})=1$, and $v(\{1,2\})=1$.

For permutation $(1,2)$, the increments are $1-0=1$ for input 1 and $1-1=0$ for input 2. For permutation $(2,1)$, they are $1$ for input 2 and $0$ for input 1. There are exactly these two permutations, so [F3] returns ${\varphi }_1={\varphi }_2=(1+0)\mathrm{/}(2V)=1\mathrm{/}2$.

Both effects are nonnegative and ${\varphi }_1+{\varphi }_2=1$, in agreement with [F2]. The allocation splits variance that either input fully reveals; it does not assert an independent-input Sobol decomposition.

The inputs are independent by assumption [F4], so Fubini [F5] factors products of coordinate functions. Symmetry gives $\mathbb{E}\sin X_1=0$, $\mathbb{E}{\sin }^2{X}_1=1\mathrm{/}2$, and $\mathbb{E}{\sin }^4{X}_1=3\mathrm{/}8$. Direct integration gives $\mathbb{E}{X}_3^4={\pi }^4\mathrm{/}5$ and $\mathbb{E}{X}_3^8={\pi }^8\mathrm{/}9$. Also $\mathbb{E}{\sin }^2{X}_2=1\mathrm{/}2$ and $\mathrm{Var}({\sin }^2{X}_2)=3\mathrm{/}8-(1\mathrm{/}2{)}^2=1\mathrm{/}8$ by [F6].

Set $m_4={\pi }^4\mathrm{/}5$. Integrating the defining function over complementary independent coordinates gives $f_{\{1\}}=(1+b{m}_4)\sin X_1$, $f_{\{2\}}=a({\sin }^2{X}_2-1\mathrm{/}2)$, and $f_{\{1,3\}}=b\sin X_1(X_3^4-m_4)$. Integrating any other nonempty subset's alternating projection sum gives zero, since the function is a sum of these three terms and each displayed factor is centered in every coordinate in its support. This is exactly the component convention in [F1].

The component variances are $V_1=(1+b{m}_4{)}^2\mathrm{/}2$, $V_2=a^2\mathrm{/}8$, and $V_{13}=\frac{b^2}{2}({\pi }^8\mathrm{/}9-{\pi }^8\mathrm{/}25)=8b^2{\pi }^8\mathrm{/}225$ by [F6]. They are mutually orthogonal by [F2]. Adding them yields $V=1\mathrm{/}2+a^2\mathrm{/}8+b{\pi }^4\mathrm{/}5+b^2{\pi }^8\mathrm{/}18$. This variance is positive: if $b\mathrm{\ne }0$ then $V_{13}>0$, and if $b=0$ then $V_1=1\mathrm{/}2$; in either case $V_2\ge 0$.

By [F3], $S_1=V_1\mathrm{/}V$, $S_2=V_2\mathrm{/}V$, and $S_3=0$, while $S_{T_1}=(V_1+V_{13})\mathrm{/}V$, $S_{T_2}=V_2\mathrm{/}V$, and $S_{T_3}=V_{13}\mathrm{/}V$. Thus the interaction contributes to the total effects of inputs 1 and 3 but not to either first-order effect.

Each coordinate has mean zero and second moment one, so its degree-one monomial Gram matrix is the identity. The construction [F2] therefore gives $1$ and $x_j$, whose products in [F3] are the four displayed functions. On the four equiprobable outcomes $(1,1),(1,-1),(-1,1),(-1,-1)$, their value vectors are $(1,1,1,1)$, $(1,1,-1,-1)$, $(1,-1,1,-1)$, and $(1,-1,-1,1)$. These vectors are pairwise orthogonal and each has squared length four, so under weights $1\mathrm{/}4$ the functions are orthonormal. Four independent functions on a four-point support form a basis of its four-dimensional $L^2$ space.

The functions in step 1.1 are linearly independent, so the exact expansion has unique coefficients: $c_{00}=2$, $c_{10}=3$, $c_{01}=0$, and $c_{11}=4$. The centered output is $3X_1+4X_1{X}_2$. Orthogonality gives $V=\mathbb{E}[(3X_1+4X_1{X}_2{)}^2]=9+16=25$ by [F7]. The exact displayed finite expansion needs no infinite Parseval assertion.

Product averaging [F4] and $\mathbb{E}{X}_j=0$ give $P_{\mathrm{\varnothing }}f=2$, $P_{\{1\}}f=2+3X_1$, and $P_{\{2\}}f=2$. The component definition in [F5] yields $f_{\{1\}}=3X_1$, $f_{\{2\}}=0$, and $f_{\{1,2\}}=4X_1{X}_2$. Their variance contributions are $9,0,16$ because $\mathbb{E}{X}_j^2=1$, matching the coefficient sums grouped by support.

Dividing these components by $V=25$ according to [F6] gives $S_1=9\mathrm{/}25$, $S_2=0$, $S_{T_1}=(9+16)\mathrm{/}25=1$, and $S_{T_2}=16\mathrm{/}25$. The family is complete here by the finite four-point argument in step 1.1. Finite cutoff and positive moment Gram hypotheses alone do not imply completeness for a general input law.

For every $J\subseteq [d]$, the coordinate average [F3] gives $P_{J}f=c$. The finite alternating sum in [F3] for a nonempty component therefore has coefficient sum $(1-1{)}^{\mathrm{\mid }u\mathrm{\mid }}=0$, so $f_u=0$ for $u\mathrm{\ne }\mathrm{\varnothing }$.

The variance is $\mathbb{E}[(c-c{)}^2]=0$, and all nonempty component variances vanish. Since the denominator condition in [F2] fails, neither $S_i$ nor $S_{T_i}$ is defined.

Put $\mathcal{G}=\sigma (X)$. For each $A\in \mathcal{G}$, independence [F1]–[F2] factors $\mathbb{E}[{\mathbf{1}}_Aϵ]=\mathbb{P}(A)\mathbb{E}ϵ=0$. Thus the $\mathcal{G}$-measurable integrable variable $f(X)$ satisfies $\mathbb{E}[{\mathbf{1}}_{A}Y]=\mathbb{E}[{\mathbf{1}}_{A}f(X)]$ for every $A\in \mathcal{G}$, so [F7] identifies $\mathbb{E}[Y\mid X]=f(X)$. Also [F8] makes the cross moment integrable, and the same independence factorization gives $\mathrm{Cov}(f(X),ϵ)=0$. Expanding variance by [F3] and using linearity [F9] yields $\mathrm{Var}(Y)=V_f+{\tau }^2$, where $V_f=\mathrm{Var}(f(X))$.

In a first-order pick-freeze pair with fresh independent noise at both evaluations, the covariance of the two outputs is $\mathrm{Cov}(f(X),f(X^{[i]}))$: all cross covariances with noise vanish by the product factorization, and the two noise copies have zero covariance. The required products are integrable by [F8]. By [F5], this numerator is $V_{\{i\}}(f)$. Only when $\mathrm{Var}(Y)=V_f+{\tau }^2>0$ does division yield the normalized noisy ratio $V_{\{i\}}(f)\mathrm{/}(V_f+{\tau }^2)$; if $V_f={\tau }^2=0$, that ratio is undefined. The conditional-mean index is $V_{\{i\}}(f)\mathrm{/}{V}_f$ only if $V_f>0$ and is undefined if $V_f=0$.

In a total-effect pair, expand $Y-\tilde{Y}=(f(X)-f({\tilde{X}}^{[i]}))+(ϵ-{ϵ}^{\mathrm{\prime }})$. The cross term has zero expectation because the noises are centered and independent of inputs; [F8] ensures its integrability. Linearity [F9] and independence give $\frac{1}{2}\mathbb{E}[(ϵ-{ϵ}^{\mathrm{\prime }}{)}^2]={\tau }^2$. Thus half the expected squared difference is the input total-effect numerator from [F5] plus ${\tau }^2$. Therefore raw noisy total-effect estimation includes intrinsic noise unless the target is the conditional mean or the noise term is separately accounted for.

Since $\mathbb{E}X=1\mathrm{/}2$ and $\mathrm{Var}(X)=1\mathrm{/}4$, the only ANOVA component is $f_{\{1\}}=X-1\mathrm{/}2$, so $V_{\{1\}}=V=1\mathrm{/}4$ and [F3] gives the true index $S_1=1$.

In each row $r$, the first-order hybrid equals $X_r$, so $N_r=X_r-U_r{V}_r$ and $\mathbb{E}{N}_r=1\mathrm{/}4$. Also $D_r=1\mathrm{/}2$ exactly when $W_r\mathrm{\ne }{W}_r^{\mathrm{\prime }}$, an event of probability $1\mathrm{/}2$, and is zero otherwise. All numerator draws are independent of all denominator draws. If $K$ is the number of positive denominators in the two rows, then $K\sim \mathrm{Binomial}(2,1\mathrm{/}2)$ and is independent of $N_1+N_2$.

On the return event $K>0$, $\bar{D}=K\mathrm{/}4$ and $\bar{N}=(N_1+N_2)\mathrm{/}2$. Independence and $\mathbb{E}[N_1+N_2]=1\mathrm{/}2$ give $\mathbb{E}[\bar{N}\mathrm{/}\bar{D}\mid K]=1\mathrm{/}K$. Since $\mathbb{P}(K=1\mid K>0)=2\mathrm{/}3$ and $\mathbb{P}(K=2\mid K>0)=1\mathrm{/}3$, the conditional reported-ratio expectation is $2\mathrm{/}3+(1\mathrm{/}3)(1\mathrm{/}2)=5\mathrm{/}6$, different from the population index $1$ in step 1.1. Each $N_r$ and $D_r$ remains unbiased for $1\mathrm{/}4$, but their ratio is biased when returned.