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$$\mathbb{E}[g(X)h(Y)]=\mathbb{E}[h(Y)\,\mathbb{E}[g(X)|Y]]$$

I am reading the book "An Introduction to Stochastic Modeling". This equation appears a lot but I can not see why.

Can anyone please provide some proofs and examples? Thanks a lot.

Yilin Wang
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    This is only true under the assumption that $X$ and $Y$ are conditionally independent given the sigma-algebra of $Y$; the equation you wrote in fact could be taken as the definition of this. As to why these particular RVs are conditionally independent in the context given, you would have to supply us with more information about where specifically in the book this is occurring. – Chill2Macht Apr 30 '16 at 22:25
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    @William I think you might be misreading the equation. It is valid as long as $g(X)$ and $h(Y)$ are integrable, and follows from basic properties of conditional expectation. –  Apr 30 '16 at 22:34
  • Also, this is not really related to stochastic processes per se... – Clement C. Apr 30 '16 at 23:06
  • @ByronSchmuland $X$ and $Y$ are conditionally independent if and only if $g(X)$ and $h(Y)$ are conditionally independent for any measurable $g,h$. But yeah it looks like I misread the braces -- that is not how I would write that equation. – Chill2Macht May 01 '16 at 00:30

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For any random variables $Z$ and $Y$ with $Z$ integrable we have $E(Z) = E[ E(Z\mid Y) ] $. Apply this to $$Z:=g(X)h(Y)$$ and note that $$E(g(X)h(Y)\mid Y ) = h(Y) E(g(X)\mid Y), $$ since $h(Y)$ is measurable with respect to $\sigma(Y)$.

grand_chat
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