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Suppose that X and Y are random variables on a common probability space such that $E(X^2+Y^2) \lt \infty$, $E(X|Y)=Y$, $E(Y|X)=X$. Prove that $$P(X=Y)=1$$

My work: $E(X)=E(E(X|Y))=E(Y)$ and $E(Y)=E(E(Y|X))=E(X)$

But I don't know what to do next and I'm sure how to use the condition $E(X^2+Y^2) \lt \infty$.

Thanks in advance.

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We have the following,

\begin{align*} E( (X-Y)^2 )&= E(X^{2}+Y^{2}-2XY ) \\ &= E(X^{2}+Y^{2})-E(XY )-E(XY ) \\ &= E(X^{2} )+E(Y^{2} )-E(E(XY |X ))-E(E(XY |Y )) \\ &= E(X^{2} )+E(Y^{2} )-E(XE(Y |X ))-E(YE(X |Y )) \\ &= E(X^{2} )+E(Y^{2} )-E(X^2 )-E(Y^2 ) \\ &=0 \end{align*} Therefore, $X=Y$ almost surely.

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  • $\begingroup$ Intuitively it seems like we ought to be able to prove this with only EX and EY finite without worrying about the second moments. Is there a coutnerexample? $\endgroup$ – C Monsour Jun 8 '18 at 2:20
  • $\begingroup$ @CMonsour The same holds when $X,Y \in L^{1}$. $\endgroup$ – clark Jun 8 '18 at 2:22
  • $\begingroup$ $NYRAHHH it is hard to answer this in a comment since the proof is slightly longer. Why don't you ask a separate question? $\endgroup$ – Kavi Rama Murthy Jun 8 '18 at 6:07
  • $\begingroup$ For the proof without the $L_2$ inegrability, see e.g. this. math.stackexchange.com/questions/666843/… $\endgroup$ – Clement C. Jun 8 '18 at 13:30

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