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I'm asked to show that if $X, Y$ are random variables (discrete or continuous) and $E(X|Y) = Y$, $E(Y|X) = X$, then $X = Y$ almost surely.

I'd like to argue as follows: Suppose not. Then

$$\{X\neq Y\} = \bigcup _{r\in\mathbb{Q}} \left(\{Xr\} \cap \{r>Y\}\right)$$

has positive measure, and hence without loss of generality there's some $r\in \mathbb{Q}$ such that $\{X has positive measure. Let's call this set $C$. Then we get:

$$\int_CX<\mu(C)\cdot r<\int_CY=\int_CE(Y|X)=\int_CX$$

where the second-last equality comes from the definition of conditional expectation, and the last equality comes from one of the given equations. But the above is a contradiction, so that does it.

The problem here is that I've only used one of the given assumptions, so have I implicitly used the other one somewhere? Is there some measurability-related concern I'm overlooking?

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    Here's the flaw: the equation $\int_C Y=\int_C E(Y|X)$ is only valid for sets $C\in \sigma(X)$.2012-01-29
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    See also http://math.stackexchange.com/questions/34101/conditional-expectation-eab-b-and-eba-a-implies-a-b2012-01-29
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    Alright, so for $C\in\sigma(X)$, $\int_CY = \int_CE(Y|X) = \int_CX$ (since $E(Y|X) = X$). Similarly, for $C\in\sigma(Y)$ using the other given. So $\int_C(Y-X)=0$ for all $C\in\sigma(X)\cup\sigma(Y)$. Does this get me anywhere?2012-01-29
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    I don't think so. Please look at the argument in the MSE question that I linked to above.2012-01-29
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    Thanks! I wasn't familiar with the definition Didier was using for conditional expectation, but I get it now.2012-01-29

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