Does this equality hold? Is it true that 
$E[XY|Z]=E[X|Z]E[Y|Z]$ if $X$ and $Y$ are independent each other, but $X$, $Y$ are not independent with $Z$? Can anyone prove this? Thank you.
 A: This is not true in general. Take $X$ and $Y$ to be independent Bernoulli random variables with success probability $p = 1/2$ and $Z = X + Y$. When $Z = 1$ the left hand side is zero while the right hand side is not.
A: The equation you have specified does not hold in general, but it would hold in the case where $X$ and $Y$ are conditionally independent given $Z$, and in some broader cases.  We can decompose the conditional expectation of the product by applying the law of iterated expectation to get:
$$\begin{equation} \begin{aligned}
\mathbb{E}(XY|Z) 
&= \mathbb{E} \Big( \mathbb{E}(XY|Y,Z) \Big|Z \Big) \\[6pt]
&= \mathbb{E} \Big( \mathbb{E}(X|Y,Z) \cdot Y \Big|Z \Big) \\[6pt]
&= \mathbb{E} (\mathbb{E} (X|Y,Z) |Z) \cdot \mathbb{E}(Y|Z). \\[8pt]
\end{aligned} \end{equation}$$
So you can see that the equation $\mathbb{E}(XY|Z) = \mathbb{E}(X|Z) \mathbb{E}(Y|Z)$ holds only in the special case where $\mathbb{E}(X|Z) = \mathbb{E} ( \mathbb{E}(X|Y,Z) |Z)$.  This occurs in the case where $X$ and $Y$ are conditionally independent given $Z$ and can also occur in some broader cases.
