Conditional expectation of a product XY given Z with Y independent of Z Let $X,Y$
  and $Z$
  be integrable random variables s.t. $XY$
  is integrable and $Y$
  is independent of $Z$
 .
I was wondering if there are any helpful/common ways of rewriting $\mathbb{E}[XY\mid Z]$ when $Y$ is independent of $Z$
  (and nothing more is known)?
For example, if $X$
  and $Y$
  are conditionally independent given $Z$
 , then (more or less by definition - depending on your point of view)
$\mathbb{E}[XY\mid Z]=\mathbb{E}[X\mid Z]\mathbb{E}[Y\mid Z]$.
But this is quite clearly not true in general.
 A: Not in general.
First, note that if $XY,Z$ are independent:
$$
\mathbb{E}[XY\mid Z] = \mathbb{E}[XY]
$$
which is "just a number."
Now, it's tempting to try and say something in the case that if $XY,Z$ are not independent. But...


*

*If $X=f(Z)$ for some function $f$, then:
$$
\mathbb{E}[XY\mid Z] = \mathbb{E}[f(Z)Y\mid Z] = f(Z)\mathbb{E}[Y\mid Z]
= X\mathbb{E}[Y]
$$
which is not equal to $\mathbb{E}[XY]$ in general. For instance, if $X=Z$ is a Rademacher r.v. (uniform on $\{-1,1\}$) independent of $Y\sim\operatorname{Bern}(1/2)$, then $X\mathbb{E}[Y]=\frac{1}{2}X$ but $\mathbb{E}[XY]=0$.

*If $X,Y$ are independent Rademacher random variables and $Z=XY$, then (easy to check) we do have $X,Y,Z$ pairwise independent, and clearly $XY,Z$ not independent:
$$
\mathbb{E}[XY\mid Z] = \mathbb{E}[Z\mid Z] = Z = XY\neq X\mathbb{E}[Y] = 0.
$$
(and $Z\neq \mathbb{E}[XY] = \mathbb{E}[Z] = 0$ either.)


so you will need more assumptions.
A: In Durret example 4.1.7. there is a nice formula that might be helpful in such a setting

For example it could be that you can write $X=X_{1}X_{2}$ where $X_{1}$ is independent of $Y$ and $X_{2}$ correlates with it and so then you would apply this for $\tilde{X}=X_{1}$ and $\tilde{Y}=X_{2}Y$ and $Z=X_{1}$.
