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If $E(X|Y) = E(X)$ then $E(XY)=E(X)E(Y)$.

I was attempting the discrete case. I was trying to show that $E(X|Y)E(Y)=E(XY)$. But I could not proceed much beyond the definition of expectations. Rather I feel that this equality may not always hold. I have failed to find any counter-example for $X$ and $Y$ either.

Soham
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1 Answers1

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What you need is the iterative law of conditional expectation and $E(XY|Y) = YE(X|Y)$: \begin{align*} E(XY) = E(E(XY|Y)) = E(YE(X|Y)) = E(YE(X)) = E(X)E(Y). \end{align*}

Zhanxiong
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  • Why is $E(XY|Y)=YE(X|Y)$? – Soham Nov 14 '20 at 14:00
  • As iterative law, this is an important property of conditional expectation: in general, if $\mathscr{G}$ is a sigma field, and $Y$ is $\mathscr{G}$-measurable, then $E(YX | \mathscr{G}) = Y E(X | \mathscr{G})$ for any random variable $X$. – Zhanxiong Nov 14 '20 at 14:02
  • I think I found my answer in https://math.stackexchange.com/questions/612810/if-ey-mid-x-is-constant-then-x-y-are-uncorrelated/2931211#2931211 and https://math.stackexchange.com/questions/284932/prove-exy-eyexy. It will be best if I close this question. – Soham Nov 14 '20 at 14:03
  • @tatan or because $E(XY\mid Y=y) = E(Xy\mid Y=y) = yE(X\mid Y=y)$ – Henry Nov 14 '20 at 14:03
  • In its general sense, check the section "Basic properties" in this wiki page. I recommend studying its measure-theory based definition (i.e., with respect to sigma field) and properties instead of treating continuous and discrete cases separately. – Zhanxiong Nov 14 '20 at 14:05