Suppose $X$, $Y$ are random variables and $E(X|Y)=Y, E(Y|X)=X$. Is this sufficient to show that $X=Y$? Or at least $X = Y$ almost surely?
Is it true that if $E(X\mid Y)=Y$ and $E(Y\mid X)=X$, then $X$ equals $Y$?
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probability
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0Maybe I'm just not used to the notation, but what does $E(X|Y) = Y$ mean? Expectation inputs a distribution over a domain and outputs a single value from that domain. How can a distribution equal a domain value? – 2017-01-11
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1@DanielV Conditional expectation conditioned on a random variable is a random variable. For instance the say we have $X$ and $Y$ two N(0,1)'s with correlation $\rho$. Then the expectation of $X$ conditional on $Y$ is going to depend on what $Y$ is: $E(X|Y) = \rho Y$ in this case. Y is a random variable. – 2017-01-11
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0Ah, I guess ${\rm E}(X | Y) = \lambda y. {\rm E}(X | Y = y)$. – 2017-01-11
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0Yes, this is sufficient for $X=Y$ almost surely. This has been asked before but I cannot find the link right now. The easiest proof is if you assume $X$ and $Y$ have finite second moments and then compute $E[(X-Y)^2]$. – 2017-01-11