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

$\mathbb{E}((X-Y)^2|X)=\mathbb{E}(X^2|X)-2\mathbb{E}[XY|X]+\mathbb{E}[Y^2|X]=X^2-2X^2+\mathbb{E}(Y^2|X)=-X^2+\mathbb{E}[Y^2|X]$, 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 $\mathbb{E}((X-Y)^2|X)=-X^2+\mathbb{E}[Y^2|X]$;
$\mathbb{E}((X-Y)^2|Y)=-Y^2+\mathbb{E}[X^2|Y]$

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$E(X^2\mid X)=X^2$ and not $E(X^2)$. – Did Oct 21 '11 at 21:41
Does $\sigma$ mean standard deviation or are you saying that the equalities hold for all (measurable) functions $\sigma$? – Dilip Sarwate Oct 21 '11 at 21:44
$\sigma(X)$ is the $\sigma$ field generated by X – Glenn M Oct 21 '11 at 21:57
@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. – Michael Hardy Oct 21 '11 at 21:58
@Glenn: certainly $\mathbb{E}(X^2\mid X) = X^2$. – Michael Hardy Oct 21 '11 at 22:02
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1 Answer

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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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Here's another point of view on this: The hypotheses amount to saying there is no regression towards the mean. Therefore the two random variables must be perfectly correlated. – Michael Hardy Oct 22 '11 at 1:31
how does correlation of 1 imply a.s. convergence? thanks – Glenn M Oct 22 '11 at 2:55
@GlennM: The Cauchy–Schwarz inequality says that $|\operatorname{cov}(X,Y)| \le \sqrt{\operatorname{var}(X)\operatorname{var}(Y)}$ with equality if and only if $\Pr(X=aY+b)=1$ for some constants $a$ and $b$. In the latter case, we have $\mathbb{E}(X \mid Y) = aY+b$. And the hypotheses you started with tell you that $a=1$ and $b=0$. – Michael Hardy Oct 22 '11 at 3:16

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