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Given three random variables $X_1,X_2,Y$ such that $X_2,Y$ are independent, how does one prove that

$$ E[g_1(X_1)g_2(X_2)|Y] = E[g_1(X_1)|Y]E[g_2(X_2)] $$

Attempt:

\begin{equation} \begin{aligned} E[g_1(X_1)g_2(X_2)|Y=y] =& \int \int_{\mathbb{R}^2} g_1(x_1)g_2(x_2)f_{X_1,X_2|Y}(x_1,x_2|y) dx_1dx_2\\ =& \int \int_{\mathbb{R}^2} g_1(x_1)g_2(x_2)\frac{f_{X_1,X_2,Y}(x_1,x_2,y)}{f_Y(y)} dx_1dx_2\\ =& \int \int_{\mathbb{R}^2} g_1(x_1)g_2(x_2)\frac{f_{X_1|X_2,Y}(x_1|x_2,y)f_{X_2,Y}(x_2,y)}{f_Y(y)} dx_1dx_2\\ =& \int \int_{\mathbb{R}^2} g_1(x_1)g_2(x_2)\frac{f_{X_1|X_2,Y}(x_1|x_2,y)f_{X_2}(x_2)f_Y(y)}{f_Y(y)} dx_1dx_2\\ \end{aligned} \end{equation}

I'm stuck here.

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Is this claim true? Let $X_2,Y$ be independent bernoulli and $X_1$ the sum. $E[X_1X_2\mid Y=1]=2\cdot1/2 + 0\cdot1/2=1$ whereas $E[X_1\mid Y=1]E[X_2]=(2\cdot 1/2 + 1\cdot 1/2)\cdot 1/2=3/4.$

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  • $\begingroup$ I'm investigating into correctness of the said result, I may have copied it wrongly in class. $\endgroup$ – Anant Joshi Nov 3 '18 at 6:19
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The claim is incorrect. The result requires conditional independence of $X_1$ and $X_2$ given $Y$.

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