generall question for conditional expectation

Question

Let $X,Y\geq 0$ be random variables and $\mathcal{F}$ be a arbitrarily $\sigma$-algebra.

Under which conditions does the conditional expectation holds:

$$\mathbb{E}[X\cdot Y|\mathcal{F}]=\mathbb{E}[X|\mathcal{F}]\cdot\mathbb{E}[Y|\mathcal{F}]$$

Answer 1

These facts might help. X and Y are assumed to be $L^1$ on ${\cal A}$ (and associated given sample space/probability measure), and ${\cal F}\subseteq {\cal A}$.

1. If Y is independent of both X and ${\cal F}$, that is, independent of $\sigma(X,{\cal F})$, then $\mathbb{E}[XY|{\cal F}]=\mathbb{E}[X|{\cal F}]\mathbb{E}[Y]$. (Note that, under the assumed conditions, $\mathbb{E}[Y|{\cal F}]=\mathbb{E}[Y]$.)

2. If Y is ${\cal F}$-measurable and bounded, then $\mathbb{E}[XY|{\cal F}]=\mathbb{E}[X|{\cal F}]Y$. (Note that, under the assumed conditions, $\mathbb{E}[Y|{\cal F}]=Y$.)

This http://www.swansonsite.com/W/instructional/condexp.pdf is a good reference on conditional expectation that might help (classic texts might help too) to take your investigation further.

Answer 2

The statement holds if $X$ and $Y$ are independent conditionally on $\mathcal{F}$ which means that $\mathbb{P}(X \in A, Y \in B \vert \mathcal{F}) = \mathbb{P}(X \in A\vert \mathcal{F})\mathbb{P}(Y \in B \vert \mathcal{F})$ a.s. for all Borel sets $A$ and $B$.