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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!

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  • $\begingroup$ What restrictions do you have on $F$? If not, then take $F$ to be a $\sigma$-algebra generated by $X+Y$. $\endgroup$
    – Boby
    Commented Feb 10, 2020 at 18:05
  • $\begingroup$ @Boby Sorry but I don't know how to continue. The sigma algebra generated by $X+Y$ looks very foreign to me... $\endgroup$
    – No One
    Commented Feb 10, 2020 at 18:07

1 Answer 1

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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} $

On the other hand, we have

$\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 $1/2$.

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