Random variables $X,Y$ such that $E(X|Y)=E(Y|X)$ a.s.

Question

Let $X$ and $Y$ be random variables such that $E(|X|), E(|Y|)<\infty$ and $E(X|Y)=E(Y|X)$ a.s. Then is it true that $X=Y$ a.s. ? If this is not true in general, what happens if we also assume $X,Y$ are identically distributed ?

Answer 1

Not true at all.

Let $X, Y$ be indepedently, identically distributed.

$$ \mathbb{E}(X ) = \mathbb{E}(X | Y ) = \mathbb{E}(Y|X ) = \mathbb{E}(Y ) $$

but this does not mean $P(X = Y) = 1$.

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Suppose $X, Y \overset{iid}{\sim} Normal(\mu, \sigma^2)$.

Note that $P(Z = 0) = 0$, where $Z := X -Y$, because $Z \sim N(0, 2\sigma^2)$ is a continuous R.V.

So the probability measure of the event $X = Y$ is $0$, not $1$ (as in almost sure case).

But since $X, Y$ are independent,

$$ \mu = \mathbb{E}(X ) = \mathbb{E}(X | Y ) = \mathbb{E}(Y|X ) = \mathbb{E}(Y ) $$