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Let $X$ and $Y$ be square integrable random variables s.t. $E[X|Y]=Y$ and $E[Y|X]=X$. Prove $P(X=Y)=1$.

Furthermore, when the condition changes to $X$ and $Y$ are integrable, show that the conclusion is still true.

So far, I only know $E[X]=E[Y]$, which means $E[X-Y]=0$. So, that means I need to show $var(X-Y)=0$, right? Any hints would be appreciated.

hil316
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2 Answers2

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From $\mathbb E[X\mid Y]=Y$ it follows that:

$$\mathbb{E}\left[X^{2}-2XY+Y^{2}\mid Y\right]=\mathbb{E}\left[X^{2}\mid Y\right]-2Y\mathbb{E}\left[X\mid Y\right]+Y^{2}=\mathbb{E}\left[X^{2}\mid Y\right]-Y^{2}$$

Taking the expectation on both sides we find: $$\mathbb{E}\left(X-Y\right)^{2}=\mathbb{E}X^{2}-\mathbb{E}Y^{2}$$

Likewise on base of $\mathbb E[Y\mid X]=X$ it can be shown that:$$\mathbb{E}\left(X-Y\right)^{2}=\mathbb{E}Y^{2}-\mathbb{E}X^{2}$$

This implies $\mathbb{E}\left(X-Y\right)^{2}=0$ and consequently $\Pr(X=Y)=1$.

drhab
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  • Same spirit, slightly faster: $E(XY|X)=X^2$ gives $E(XY)=E(E(XY|X))=E(X^2)$ and similarly $E(XY)=E(Y^2)$. So $E[(X-Y)^2]=E(X^2+Y^2 - 2XY)=0$. (probably what JD had in mind). – H. H. Rugh Jul 29 '16 at 21:25
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    @H.H.Rugh How to show the same conclusion when $X$ and $Y$ are integrable? – hil316 Jul 30 '16 at 03:05
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The $L^1$ case seems more difficult. The following should be a proof:

Let $X=X_+ - X_-$, with $X_+=\max\{0,X\}$ and $X_-=\min\{0,X\}$. Then since $X_+$ and $X_-$ are positive random variables: $$ Y=E(X|Y) = E(X_+|Y) - E(X_-|Y) , \ \ \ |Y| \leq E(X_+|Y) + E(X_-|Y) = E(|X|\; | \; Y).$$
Taking expectation in the latter inequality we get $E|Y| \leq E(E(|X|\; | \; Y))=E|X|$. But by the symmetry in $X$ and $Y$ we must have $E|X|=E|Y|$ and then also (a.s.) equality in the previous equation, i.e. $$ \left| E(X_+|Y) - E(X_-|Y) \right| = E(X_+|Y) + E(X_-|Y) \ \ (a.s).$$ Now, this implies that the two random variables have disjoint support: When one is non-zero the other is zero (a.s). Set $\Omega_{\pm} = \{ E (X_\pm | Y) > 0 \}$. Then e.g. $\Omega_+$ is $Y$-measurable, and on this set we have $E(X_-|Y)=0$ (a.s.), so $Y = E(X_+|Y)>0$ (a.s.) on $\Omega_+$. We get: $$ \int_{\Omega_+} Y = \int_{\Omega_+} X_+ \leq \int X_+=E (X_+). $$ We have similar results for $\Omega_-$ and then: $$ E (|Y|) = \int_{\Omega_+} Y + \int_{\Omega_-} (-Y)=\int_{\Omega_+}X_+ + \int_{\Omega_-} X_- \leq \int X_+ + \int X_-= E(|X|)=E(|Y|).$$ Therefore, again equality and thus $X_+=0$ on $\Omega_-$ (a.s.) and $X_-=0$ on $\Omega_+$ (a.s.). We deduce that $$ \{ X>0\} \subset \{Y>0\} \subset \{X\geq 0\} \ \ {\rm (a.s.)} $$ Replacing $X$ and $Y$ by $X-a$ and $Y-a$ we deduce similarly for every $a\in {\Bbb R}$: $$ \{ X>a\} \subset \{Y>a\} \subset \{X\geq a\} \ \ {\rm (a.s.)} $$ It follows that $\{Y>a\} \subset \cup_{n\geq 1} \{X\geq a+1/n\} =\{X>a\} \subset \{Y>a\} $ (a.s.) and by looking at rationals $X=Y$ (a.s.). [Later edit: Just noticed that the question was marked as duplicate. Too bad, but fun to solve]

H. H. Rugh
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