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Let $X$ and $Y$ be r.v. such that $E[X^2] < \infty$, $E[Y^2] < \infty$. Suppose that $E[X|Y] = Y$ and $E[Y|X] = X$. Prove that $Y=X$ with probability 1.

My professor gave us a hint: Compute $E[X-Y]^2$ using conditioning on X and Y.

My attempt:

$E[X-Y]^2 = E[X^2 - 2XY +Y^2] = E[X^2] - 2E[XY] + E[Y^2]$ $= E[E[Y|X]^2] - 2E[E[X|Y] E[Y|X]] + E[E[X|Y]^2]$

I am stuck here. I don't understand why computation of $E[X-Y]^2$ will help me prove that $X=Y$. I am looking for a hint here.

Reveillark
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Kerry
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  • The point of the calculation is that $E[Z^2]=0$ if and only if $Z=0$ with probability $1$. – Ian Sep 27 '15 at 15:40
  • @Ian Okay that makes sense. Am I in the right direction with the computation? I am not sure what to do next? I think I should use the law of iterated expectations? – Kerry Sep 27 '15 at 15:43
  • Hint: $E[XY|Y]=YE[X|Y]=Y^2$. Similarly $E[XY|X]=XE[Y|X]=X^2$. – Ian Sep 27 '15 at 15:48
  • @Ian - I was able to solve the solution using your hint. Could you please explain why $E[XY|Y] = YE[X|Y]$? – Kerry Sep 27 '15 at 16:03
  • This is sometimes called "factoring out what is known": if $Z$ is a $Y$-measurable random variable then $E[XZ|Y]=ZE[X|Y]$. This is one of the fundamental properties of conditional expectation. – Ian Sep 27 '15 at 16:04
  • @Ian - Of course. Thanks! Also, I've always had difficulty understanding conditional expectations? Which resources would you recommend to better understand these concepts? – Kerry Sep 27 '15 at 16:05

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