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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 ?

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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 $P(X = Y) = 1$.


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,

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

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    $\begingroup$ You mean to say $E(X|Y)$ is a constant random variable !? Also I would like to point out that $E(X|X)=X$ ... $\endgroup$ – user521337 Dec 20 '18 at 2:40
  • $\begingroup$ @user521337 Yes, since $X$ is independent of $Y$. $\endgroup$ – Moreblue Dec 20 '18 at 2:42
  • $\begingroup$ Also, $Z$ is $N(0 , \sigma^2)$ ... $\endgroup$ – user521337 Dec 20 '18 at 2:49
  • $\begingroup$ @user521337 No, $Z \sim N(0, 2 \sigma^2)$. For the variance, check here $\endgroup$ – Moreblue Dec 20 '18 at 2:50
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    $\begingroup$ again ... I would like to point out that $E(X|X)=X$ and not $E(X)$ ... $\endgroup$ – user521337 Dec 20 '18 at 2:52

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