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Let $X$, $Y$ be r.v. with finite second moments. Suppose $\mathbb{E}(X\mid\sigma (Y))=Y$, and $\mathbb{E}(Y\mid\sigma(X))=X$, show that $\Pr(X=Y)=1$.

So what I have done is this, I first consider $\mathbb{E}((X-Y)^2)$ by conditioning on $X$ and $Y$

$\mathbb{E}((X-Y)^2\mid X)=\mathbb{E}(X^2\mid X)-2\mathbb{E}[XY\mid X]+\mathbb{E}[Y^2\mid X]=X^2-2X^2+\mathbb{E}(Y^2\mid X)=-X^2+\mathbb{E}[Y^2\mid X]$, and similarly for conditioning on $Y$, but I am not sure how to subtract them properly to make use of them. Thanks

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In the end I have $\mathbb{E}((X-Y)^2\mid X)=-X^2+\mathbb{E}[Y^2\mid X]$;

$\mathbb{E}((X-Y)^2\mid Y)=-Y^2+\mathbb{E}[X^2\mid Y]$

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$\operatorname{cov}(X,Y) = \mathbb{E}(XY) – (\mathbb{E}X)(\mathbb{E}Y) = \mathbb{E}(\mathbb{E}(XY \mid X)) – (\mathbb{E}X)(\mathbb{E}Y)$ $= \mathbb{E}(X\mathbb{E}(Y\mid X)) – (\mathbb{E}X)(\mathbb{E}Y)$ $= \mathbb{E}(X^2) – (\mathbb{E}X)(\mathbb{E}Y)$.

Now use the fact that the expectations of $X$ and $Y$ are equal:

$\mathbb{E}(X)= \mathbb{E}(\mathbb{E}(X\mid Y)) = \mathbb{E}(Y)$.

We get $\operatorname{cov}(X,Y) = \mathbb{E}(X^2) – (\mathbb{E}X)^2 = \operatorname{var}(X)$. By the same argument, we get $\operatorname{cov}(X,Y) = \operatorname{var}(Y)$. Hence $\operatorname{cov}(X,Y) = \operatorname{var}(X)=\operatorname{var}(Y)$.

Hence the correlation between $X$ and $Y$ is $1$ (provided neither of them has variance $0$, but proving the result you want when that happens is trivial).

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