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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

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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    $E(X^2\mid X)=X^2$ and not $E(X^2)$.2011-10-21
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    Does $\sigma$ mean standard deviation or are you saying that the equalities hold for all (measurable) functions $\sigma$?2011-10-21
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    $\sigma(X)$ is the $\sigma$ field generated by X2011-10-21
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    @Dilip: I took $\sigma(Y)$ to mean the sigma-algebra generated by the random variable $Y$, i.e. the coarsest sigma-algbra that makes $Y$ measurable.2011-10-21
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    @Glenn: certainly $\mathbb{E}(X^2\mid X) = X^2$.2011-10-21
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    Yes, I have corrected that. Thanks I am stuck on how to use those two equation in the end to show P(X=Y)=12011-10-21
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    Is there anyway to simplify $\mathbb{E}[Y^2|X]$?2011-10-21
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    I've posted an answer using a different approach. In the mean time, "anyway" is a perfectly good word---an adverb---that does not mean the same thing as "any way".2011-10-21
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    And now you are ready for the real stuff, which is to prove that the same conclusion holds without the hypothesis that $X$ and $Y$ are square integrable but with the minimal hypothesis required for the exercise to make sense, namely, that $X$ and $Y$ are integrable.2011-10-21

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