Conditional expectation with respect to product of two random variables

Question

If $X$ and $Y$ are independent discrete random variable, how to compute $E(X|XY)$? Thanks.

Answer 1

\begin{align*} \mathbb E[X\mid XY](\omega ')&=\mathbb E[X\mid XY=X(\omega ')Y(\omega ')]\\ &=\sum_{\omega \in \Omega }X(\omega )\mathbb P\{X=X(\omega )\mid XY=X(\omega ')Y(\omega ')\}\\ &=\sum_{\omega \in \Omega }X(\omega )\frac{\mathbb P\{X=X(\omega ),XY=X(\omega ')Y(\omega ')\}}{\mathbb P\{XY=X(\omega ')Y(\omega ')\}}. \end{align*}

Without further information, it's not possible to do better.

Answer 2

If $P(A)>0$ $$E(X|A)=\frac{E(X1_A)}{P(A)}$$

Conditional_expectation_with_respect_to_an_event

So $$E(X|XY=t)=\frac{E(X1_{\{XY=t\}})}{P(XY=t)}$$

$$=\frac{\sum_{(x,y)} x1_{\{xy=t\}} P(X=x,Y=y)}{P(XY=t)}$$

$$=\frac{\sum_{\{(x,y)| xy=t\}} x P(X=x,Y=y)}{P(XY=t)}$$

$$=\frac{\sum_{\{(x,y)| xy=t\}} x P(X=x)P(Y=y)}{P(XY=t)}$$