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I would like to show that for any random variable $X$ and $Z$ such that $X$ and $Z$ are independent and for any measurable functions $f$ and $g$,

$$ \mathbb E \left[ f(g(X),Z) | g(X) \right] = \mathbb E \left[ f(g(X),Z) | X \right]. $$

How could I proceed?

Thank you for your help!

EDIT: The $X, Z, g(X), f(g(X),Z)$ are considered square integrable. $f$ and $g$ are continuous.

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My answer, at probably math.SE level:

Note $\nu$ the law of $Z$. Because $X$ and $Z$ are independent, considering the function $f\circ g$ applied to $X$ one has $$ E\left[f(g(X),Z)\vert X\right] = \int f(g(X),z)\nu(dz) \text{ , }$$ and so the left-hand side is $g(X)$-measurable. You then take $g(X)$-conditional-expectations on both sides and conclude because $\sigma(g(X)) \subseteq \sigma(X)$.

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  • $\begingroup$ Than you for your answer. How does the first expression proves that $E[f(g(X),Z)|X]$ is $g(X)$-measurable? $\endgroup$
    – gagaouthu
    Commented Jan 29, 2015 at 8:17
  • $\begingroup$ Because $f$ can be approximated with elementary functions for which the $g(X)$-measurability is trivial. $\endgroup$
    – imateapot
    Commented Jan 29, 2015 at 14:36
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This is correct even if $X$ and $Z$ are dependents. We can use the characterization of the conditional expectations to prove it.

let $\varphi$ a $g(X)$-measurable function, as $\sigma(g(X)) \subset \sigma(X)$ Then we have $$E[f(g(X),Z)\varphi]= E[E(f(g(X),Z)\varphi)|X]]=E[E[f(g(X),Z)|X]\varphi]$$ The last equality comes from the fact that $\varphi$ is $X$-measurable. As this equality is true for all the $g(X)$-measurable functions $\varphi$ we conclude that $$E(f(g(X),Z)|g(X)]=E(f(g(X),Z)|X]$$ At any moment I supposed that $X$ and $Z$ are independents.

$\bf Edit$ The answer is wrong when $X$ and $Z$ are not independents as Jochen pointed below as we have to check that $E(f(g(X),Z)|X]$ is $g(X)$-measurable first.

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  • $\begingroup$ I do not believe in this proof: You show that $Y=E[f(g(X),Z)|X]$ satisfies the Radon-Nikodym equations for the conditional expectation w.r.t $g(X)$. But this is not enough: You also have to show that it is $\sigma(g(X))$-measurable. Without independence, the statement is certainly wrong (e.g. $Z=X$, $g$ constant, $f(g(X),Z)=Z$). $\endgroup$
    – Jochen Wengenroth
    Commented Jan 28, 2015 at 13:22
  • $\begingroup$ You're right, I see that the purpose of the first part of imateapot answer is to justify that this conditional expectation is $g(X)$ measurable when $X$ and $g(X)$ are independent. I'll leave the answer like that so the mistake can be clear to the reader $\endgroup$
    – Hicham
    Commented Jan 28, 2015 at 15:15

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