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How can I prove that $\newcommand\cov{\operatorname{Cov}}\cov[X,E(Y|X)]=\cov[X,Y]$?

I tried $\cov[X,E(Y|X)] = E[XE(Y|X)]-E(X)E[E(Y|X)] = E[XE(Y|X)]-E(X)E(Y)$ then I am stuck.

How can $E[XE(Y|X)] = E(XY)$?

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Note that $X$ is $\sigma(X)$-measurable, hence $E[XY\mid X] = X\cdot E[Y\mid X]$, taking expectations, we have $$ E\bigl[X \cdot E[Y\mid X]\bigr] = E\bigl[E[XY\mid X]\bigr] = E[XY]$$ as needed.

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  • $\begingroup$ Thank you for the quick answer. $\endgroup$ – user89855 Dec 10 '13 at 12:20

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