What can I say about $E(X \hat{X})$ where $\hat{X}$ is a version of $E(X|\mathcal{G})$, where $X \in \mathcal{L}^2(\Omega,\mathcal{F},\mathbb{P})$ and $\hat{X} \in \mathcal{L}^2(\Omega,\mathcal{G},\mathbb{P})$ and $\mathcal{G}$ is a sub-$\sigma$-algebra of $\mathcal{F}$?

I know I have to use the tower property of expectations to make this collapse, but two things confuse me - the power (is then inside the conditional or outside the conditional expectation) - and independence - nothing is stated about $\sigma(X)$ being independent of $\mathcal{G}$, so I'm not sure if I can collapse them then.

Any suggestions?


$E(X\hat {X} |\mathcal G)=\hat {X} E(X|\mathcal G)=\hat {X}^{2}$ because $\hat {X}$ is already measurable with respect to $\mathcal G$. Take expectation to get $EX\hat {X}=E\hat {X}^{2}$.

  • $\begingroup$ Thanks, that makes sense. I just miss one step inbetween - is it always justified to say that $E(X \hat{X}) = E(X\hat {X} |\mathcal G)$? $\endgroup$ – Daniel Oct 3 '18 at 10:20
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    $\begingroup$ No, the left side in your equation is a number and the right side is a random variable. You have to take expectation again on the right side to make this equation correct. $\endgroup$ – Kavi Rama Murthy Oct 3 '18 at 10:23
  • $\begingroup$ Got it! Now it all makes sense. Thanks a lot $\endgroup$ – Daniel Oct 3 '18 at 10:29

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