Some proof involving expectation

Question

Let $X$ and $Y$ be two random variables with finite second moments and $E(X \mid Y) = Y$ , $E(Y \mid X)= X$. Show that $P(X = Y)= 1$.

I tried using conditional expectation, so $E(E(X \mid Y))=E(Y) \Rightarrow E(X)=E(Y)$, but then I'm totally stuck on how to continue. I don't even know how to relate it to probability or using the fact of finite second moment.

Any help is appreciated. Thank you.

Answer 1

Probably not the fastest or most elegant, but this is the first thing that came to my mind:

Using $E(XY)=E(XE(Y\vert X))=E(X^2)$ and your results, we conclude that $Cov(X,Y)=V(X)=V(Y)$. So $\rho_{XY}=1$ which implies that $Y=aX+b$ for some $a>0$ and $b$.

Since $V(Y)=a^2V(X)=V(X)$ we deduce $a=1$ and from the equality of expectations we deduce $b=0$, so $Y=X$ w.p. $1$.

Answer 2

Because of the symmetry, it holds that $X$ is $\sigma(Y)$-measurable, and $Y$ is $\sigma(X)$-measurable. But, one of the main characteristics of the conditional expectation is that it minimize $E[(X-Y)^2]$. But, because $Y$ is $X$-measurable, this is necessary $0$, because also $X$ lies in the space of $X$-measurable random variables. Therefore, it holds that $E[(X-Y)^2]=0$, from which your inquiry follows.

Answer 3

Slightly different approach to showing $E[(X-Y)^2]=0$ than Albert's way (I'm slightly confused on how to conclude $X$ is $\sigma(Y)$-measurable), but we can directly compute $$E[X(X-Y)] = E[XE[(X-Y)|X]] = E[X(X-E[Y|X])] = 0$$ and similarly $E[Y(X-Y)] = 0$ so $$E[(X-Y)^2] = E[X(X-Y)] - E[Y(X-Y)] = 0.$$