I try to solve it from Cov(X,Y) = E(XY) - E(X)E(Y). However, I get some problems evaluating E(X*E(Y|X)). Any hint would be appreciated.
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4$\begingroup$ I suggest you have a look at the Law of Iterated Expectations. $\endgroup$– Matthias SchmidtblaicherCommented Nov 4, 2016 at 20:06
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5$\begingroup$ Hint: $E[Y]=E[E[Y|X]]$. $\endgroup$– Alex R.Commented Nov 4, 2016 at 20:07
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1$\begingroup$ Please add the tag self-study if it fits (I see you already expect hints towards an answer). $\endgroup$– FirebugCommented Nov 4, 2016 at 20:29
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2 Answers
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We know: \begin{align}Cov(X,Y) = E(XY) - E(X)E(Y)\end{align}
Thus, \begin{align}Cov(X,E[Y|X]) = E[X \cdot E(Y|X)] - E[X]E[E(Y|X)]\end{align}
As such, to solve the problem, we need to show that : \begin{align}E[X\cdot E(Y|X)]= E[XY]\end{align}
as well as: \begin{align}E[E(Y|X)] = E[Y]\end{align}
We want to prove for any function $r: S\rightarrow\mathbb{R}$:
\begin{align} E[r(X) \cdot E(Y|X)] = E[r(X)Y]\end{align}
Proof:
\begin{align} E[r(X)E(Y|X)] &= \int_S r(x)E(Y|X=x)g(x)dx \\ &= \int_S r(x) \left[\int_T y\cdot h(y|x)dy\right]g(x)dx\\ &=\int_S \int_T r(x)\cdot y\cdot h(y|x)\cdot g(x) dydx\\ &= \int_{S\times T} r(x)\cdot y\cdot f(x,y) d(x,y)\\ &= E[r(X)Y] \text{, as wanted.} \end{align}
Additionally, this implies that $E[E(Y|X)] = E[Y]$, by letting $r(x)=1$.
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I believe you could say this:
Given $X$, $X$ is constant and $E[\alpha\cdot Y] = \alpha\cdot E[Y]$ therefore $X\cdot E[Y|X] = E[X\cdot Y|X]$
$\implies E[X\cdot E[Y|X]] = E[E[X\cdot Y|X]] = E[X\cdot Y]$ which will get you to the result you want. I'm sure about the last step though...
