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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?

Dilip Sarwate
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Acuzio Leigh
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  • Maybe 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? – DanielV Jan 11 '17 at 08:55
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    @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. – spaceisdarkgreen Jan 11 '17 at 09:04
  • Ah, I guess ${\rm E}(X | Y) = \lambda y. {\rm E}(X | Y = y)$. – DanielV Jan 11 '17 at 09:24
  • Yes, 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 $E[(X-Y)^2]$. – Michael Jan 11 '17 at 09:42
  • https://math.stackexchange.com/questions/103472/ex-mid-y-y-and-ey-mid-x-x-implies-x-y-almost-surely, https://math.stackexchange.com/questions/34101/conditional-expectation-ea-mid-b-b-and-eb-mid-a-a-implies-a-b?noredirect=1&lq=1. – StubbornAtom May 04 '19 at 11:43

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