Noisy simulator output separates intrinsic noise from input variance

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.