Prove or disprove : If $E[Y|X]=X$, $E[X|Y]=Y$, and both $E[X^2]\mbox{ and }E[Y^2]$ are finite, then $P[X=Y]=1$. (Possible Hint: $P[X=Y]=1\mbox{ if }\operatorname{var}[X-Y]=0$)
Basically I need prove that $E[(X-Y)^2] = E[(X-Y)]^2$.
I don't have ideas. Can someone help me please?
$E[(X-Y)^2]= E[X^2] -2 E[XY] + E[Y^2]$ and by the power-property of conditional expectations, ''pulling out known factors'', and the hypothesis we have $E[XY]= E[E[XY|X]] = E[X E[Y|X]] =E[X^2]$. Changing the roles of $X$ and $Y$ we also get $E[XY]=E[Y^2]$. Thus, $E[(X-Y)^2]=0$.
EDIT. The argument can be completed as follows: $Z=(X-Y)^2$ is a non-negative random variable whose expectation is $0$ and this implies $Z=0$ almost surely. Thus, $P[X=Y]=P[Z=0]=1$.