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Suppose $X$ and $Y$ are random variables, $E(Y^2) < \infty$ and $\varepsilon = Y - E(Y|X)$ so that $Y = E(Y|X) + \varepsilon$.

Given that $E(\varepsilon | X) = E(\varepsilon) = 0$, show that $Cov(\varepsilon , E(Y|X)) = 0$.

This question has multiple parts so $E(Y^2) < \infty$ may or may not be applicable in this case.

Here's what I tried so far. I used the fact that $Cov(X,Y) = E(XY) - E(X)E(Y)$ and $Cov(X,Y) = E[(X - E(X))(Y-E(Y))]$ and concluded that $Cov(\varepsilon , E(Y|X)) = E(\varepsilon E(Y|X))$ or in other words, $E(\varepsilon E(Y)) = 0$.

From there, I guess the only thing I have to show is that: $E(\varepsilon E(Y|X)) = 0$, but I'm having trouble doing this.
Am I going in the right track or is this completely the wrong approach to tackling this problem?

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  • $\begingroup$ Substitute for $\varepsilon$ in $E(\varepsilon E(Y|X))$, and use the law of total expectation by conditioning on $X$ $\endgroup$
    – Alecos Papadopoulos
    Nov 3, 2013 at 0:18
  • $\begingroup$ Substituting, I get $E((Y-E(Y|X))E(Y|X)) = E(YE(Y|X) - E(Y|X)^2)$. I don't see how I can use the law of total expectation to get anywhere useful. $\endgroup$
    – Karnage2015
    Nov 3, 2013 at 1:31

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$$E\Big((Y-E(Y\mid X))E(Y\mid X)\Big) = E\Big(YE(Y\mid X) - E(Y\mid X)^2\Big) $$

$$E\Big(YE(Y\mid X)\Big) - E\Big(E(Y\mid X)^2\Big) $$

Now by the law of total (or is it iterated - I always forget) expectation, we have for the first term

$$E\Big(YE(Y\mid X)\Big) = E\Big[E\Big(YE(Y\mid X)\mid X \Big)\Big] = E\Big[E(Y\mid X)E(Y\mid X ) \Big] = E\Big(E(Y\mid X)^2\Big)$$

so the whole expression equals zero.

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  • $\begingroup$ How can you take the expected value of a product of random variables to be the product of expected values? What I'm trying to say is how do we know that $Y$ and $E(Y∣X)$ are independent of each other? $\endgroup$
    – Karnage2015
    Nov 3, 2013 at 3:55
  • $\begingroup$ $Y$ and $E(Y\mid X)$ are not unconditionally independent -but it doesn't matter. $E(Y\mid X)$ is a function of $X$. When we take its expectation conditioning again on $X$, it goes out of the expectations operator, as a matter of basic properties of expected values, irrespective of what other variables may be present. In general, $E(Yh(Z)\mid Z) = h(Z)E(Y|Z)$, irrespective of the unconditional relation between $Y$ and $Z$. $\endgroup$
    – Alecos Papadopoulos
    Nov 3, 2013 at 4:07
  • $\begingroup$ Using that property, it makes solving this question a lot more easier. Thanks! However, why did you tell me to substitute for $\varepsilon$? Wouldn't it be equally correct to do $E(\varepsilon E(Y|X)) = E(E(\varepsilon E(Y|X) | X)) = E(E(\varepsilon | X)E(Y|X))$? $\endgroup$
    – Karnage2015
    Nov 3, 2013 at 13:56
  • $\begingroup$ I took the twisted route, you found the simpler. Bravo! $\endgroup$
    – Alecos Papadopoulos
    Nov 3, 2013 at 14:30

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