# "co-relatedness" of conditional expectation of two independent random variables

Let $$\mathcal{F}$$ be a $$\sigma$$-algebra and $$X,Y$$ be two independent (not just uncorrelated) random variables, I wonder if the following statement true

$$\mathbb E(XY|\mathcal{F})=\mathbb E(X|\mathcal{F})\mathbb E(Y|\mathcal{F}).$$

I have a feeling that it is false but I can't come up with a counterexample. Thanks in advance!

• What restrictions do you have on $F$? If not, then take $F$ to be a $\sigma$-algebra generated by $X+Y$.
– Boby
Commented Feb 10, 2020 at 18:05
• @Boby Sorry but I don't know how to continue. The sigma algebra generated by $X+Y$ looks very foreign to me... Commented Feb 10, 2020 at 18:07

Building on Boby's hint, take $$X,Y \overset{iid}{\sim} \mbox{Ber}_{\pm}(1/2)$$ and $$\mathcal{F} = \sigma(X+Y)$$. Then it is easy to compute the conditional expectations:
$$\mathbb{E}[XY| X+Y] = \begin{cases} 1, &X+Y = -2\\ -1,& X+Y = 0 \\ 1, &X+Y = 2 \end{cases}$$
$$\mathbb{E}[X|X+Y] = \mathbb{E}[Y|X+Y]= \begin{cases} -1 &X+Y = -2\\ 0,& X+Y = 0 \\ 1, &X+Y = 2 \end{cases}$$
Thus we see that a.s. $$\mathbb{E}[XY|X+Y] \ne 0$$, whereas $$\mathbb{E}[X|\mathcal{F}]\cdot \mathbb{E}[Y| \mathcal{F}] = 0$$ with probability $$1/2$$.