Does this statement hold and how to prove it correctly? $$ \mathbb{E}(\mathbb{E}(X\mid \mathbb{F})^2) = \mathbb{E}(X\mathbb{E}(X\mid \mathbb{F})) $$
Any help? Thanks.
$\mathbb{F}$ is a sigma algebra. $X$ may or may not be $\mathbb{F}$-measurable.
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1Did you intend the left-hand side to be $\mathbb{E}(\mathbb{E}(X\mid\mathbb{F})^2)$ or what you wrote? – Stefan Hansen Jan 16 '14 at 08:00
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Hint: Use that $$ \mathbb{E}(Z)=\mathbb{E}(\mathbb{E}(Z\mid\mathbb{F})) $$ with $Z=X\mathbb{E}(X\mid\mathbb{F})$.
Stefan Hansen
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That makes one side ${\mathbb E}(({\mathbb E}(X|{\mathbb F})^2)$, which is not the same as $({\mathbb E}({\mathbb E}(X|{\mathbb F}))^2$. It's not clear which one was intended in the original question. – Robert Israel Jan 16 '14 at 07:41
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We have to prove that if $X\in\mathbb L^2$, then $$\mathbb E[(\mathbb E(X\mid \mathcal F))^2]=\mathbb E(X\cdot\mathbb E(X\mid \mathcal F)).$$
Using the definition of conditional expectation, we have for each $\mathcal F$-measurable random variable $Y$ that $$\mathbb E(XY)=\mathbb E(\mathbb E(X\mid\mathcal F)Y).$$ Indeed, this is true if $Y=\chi_F$ for some $F\in\mathcal F$ and this can be extended by an approximation argument.
Davide Giraudo
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