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Let X and Y be bounded random variables on a probability space $(\Omega, \mathcal{F}, \mathbb{P})$. Consider two independent sub-$\sigma$ fields $\mathcal{G}$ and $\mathcal{H}$ of $\mathcal{F}$. We assume X is $\mathcal{G}$-measurable and Y is $\mathcal{H}$-measurable. Prove the following equality holds: $$\mathbb{E} (XY\,\, |\, \sigma (\mathcal{H},\mathcal{G})) = \mathbb{E} (X\,\, |\mathcal{H}) \mathbb{E} (Y\,\, |\mathcal{G})$$

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up vote 4 down vote accepted

It doesn't.

If $X$ is $\mathcal G$-measurable and $Y$ is $\mathcal H$-measurable then $XY$ is $\sigma(\mathcal G, \mathcal H)$-measurable. Therefore $\mathbb E(XY|\sigma(\mathcal G, \mathcal H))=XY$.

But if $\mathcal G$ and $\mathcal H$ are independent then $\mathbb E(X|\mathcal H)= \mathbb E(X)$ and $\mathbb E(Y|\mathcal G)= \mathbb E(Y)$.

So you claim that $XY = \mathbb E(X)\mathbb E(Y)$, which holds only if $X$ and $Y$ are almost surely constant.

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