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Let $Y$ be a real-valued random variable, $X, Z$ be random elements, such that $X,Y$ are independent of $Z$. Then is it true that $E[Y|X, f(X,Z)]=E[Y|X]$ for any measurable function $f$?

NXWang
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  • Maybe this answer helps. – jro Mar 03 '23 at 19:14
  • @jro I think it's different here and my question is trickier because $Z$ is mixed with $X$ along with $f$. i.e. I cannot really rearrange $E[Y|X, f(X,Z)]$ into $E[Y|h(X), g(Z)]$ – NXWang Mar 03 '23 at 19:34

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If $σ(X,Y)$ is independent of $σ(Z)$, this follows with the tower property of conditional expectation from the more general fact that in this case, $$ \mathbb E[Y | σ(X,Z)] = E[Y|X]. \tag{$*$}$$

To see this, note that both $ω ↦ X(ω)$ and $ω ↦ Z(ω)$ are $σ(X,Z)$-measurable, hence $φ: ω ↦ (X(ω),Z(ω))$ is, and this extends to the composition $ f \circ φ = f(X,Z) $. Hence we have the inclusions $$ σ(X) ⊆ σ(X, f(X,Z)) ⊆ σ(X,Z).$$ Conditioning both sides of $(*)$ on $σ(X, f(X,Z))$ and using the tower property then yields the claim.

jro
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