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Let $X,Y$ be independent random variables, $f$ a bounded borel measurable function. Prove that $E(f(X,Y)|Y)=Ef(X,t)_{|t=Y}$. My approach:

I understand the RHS as "We evaluate $Ef(X,t)$ only with respect to $X$ and then we plug $Y$ in place of $t$" Is it correct? I started with LHS as I had troubles introducing $Y$ in RHS.

LHS expands as $E(f(X,Y)|Y) = \sum_{i \in I} E(f(X,Y) \mid B_i)\mathbf{1}_{B_i}$ we can fix some $B_i$ and call it $B$ and it suffices to show that $E(f(X,Y)|B) = \mathbf{1}_{B}Ef(X,t)_{|t=Y}$.

$E(f(X,Y)|B) = \frac{1}{P(\Omega \times B)}\int_{\Omega \times B} f(X, Y) \, dP = \int_{\mathbb{R} \times B'} f(x, y)g_X(x)g_Y(y)dxdy = \int_{B'} Ef(X, y)g_Y(y)dy$. I split densities because $X,Y$ are independent and $B' = \{x: Y^{-1}(x) \in B\}$.

Here I am stuck and I dont know how to proceed. How should I continue. Any help appreciated

Addy137
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    Your attempt does not make much sense. Do you know the general definition of conditional expectations (not involving densities, for r.v.'s which are not necessarily discrete). – Kavi Rama Murthy Aug 13 '24 at 09:58
  • Do you mean either $E(X|A) = \frac{1}{P(A)} \int_{A} XdP$ or $E(X|A) = \int_{\Omega} XdP(\cdot |A)$?. Either way I studied the basic definitions and theorems of conditional expectation – Addy137 Aug 13 '24 at 13:05
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    This question is a duplicate of Conditional Expectation of function of two RVs, one measurable, one independent. – Kurt G. Aug 13 '24 at 13:51
  • I see but the solution in your link introduces third variable which is a method I have never seen. I would like to know if there is simpler solution. – Addy137 Aug 13 '24 at 14:51
  • The introduction of the third variable $Y$ is warranted directly by the very definition of the conditional expectation. Calling that a "complication" is unbelievable. If you want the simplest proof of all consider the special case $f(X,Z)=a(X)b(Z),.$ Then think of ways to generalize this to all functions $f(X,Z),.$ Hint: monotone class theorem. – Kurt G. Aug 13 '24 at 16:11
  • Yes it seems that the linked post answers the OP question. My question is why it suffices to show that for every bounded and A -measurable Y , one has E(UY)=E(f(X,Z)Y). Can you explain this to me Kurt? – somerndguy Aug 13 '24 at 19:59
  • As a healthy exercise take the definition of conditional expectation and prove it. For the non trivial direction use monotone or dominated convergence. – Kurt G. Aug 14 '24 at 01:15

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