Vrftsjtryk

Assume that $X$ and $Y$ are two random variables such that $Y=E[X|Y]$ almost surely and $X= E[Y|X]$ almost surely. Prove that $X=Y$ almost surely.

The hint I was given is to evaluate:
$$E[X-Y;X>a,Yleq a] + E[X-Y;Xleq a,Yleq a]$$

which I can write as: $$int_A(X-Y)dP +int_B(X-Y)dP$$ where $A={X>a, Yleq a}$ and $B={Xleq a,Yleq a}$.

But I need some more hints.

Simply follow the hint... First note that, since $E(Xmid Y)=Y$ almost surely, for every $c$, $$E(X-Y;Yleqslant c)=E(E(Xmid Y)-Y;Yleqslant c)=0,$$ and that, decomposing the event $[Yleqslant c]$ into the disjoint union of the events $[X>c,Yleqslant c]$ and $[Xleqslant c,Yleqslant c]$, one has $$E(X-Y;Yleqslant c)=U_c+E(X-Y;Xleqslant c,Yleqslant c),$$ with $$U_c=E(X-Y;X>c,Yleqslant c).$$ Since $U_cgeqslant0$, this shows that $$E(X-Y;Xleqslant c,Yleqslant c)leqslant 0.$$ Exchanging $X$ and $Y$ and following the same steps, using the hypothesis that $E(Ymid X)=X$ almost surely instead of $E(Xmid Y)=Y$ almost surely, one gets $$E(Y-X;Xleqslant c,Yleqslant c)leqslant0,$$ that is $$E(Y-X;Xleqslant c,Yleqslant c)=0,$$ which, coming back to the first decomposition of an expectation above, yields $U_c=0$, that is, $$E(X-Y;X>cgeqslant Y)=0.$$ This is the expectation of a nonnegative random variable hence $(X-Y)mathbf 1_{X>cgeqslant Y}=0$ almost surely, which can only happen if the event $[X>cgeqslant Y]$ has probability zero. Now, $$[X>Y]=bigcup_{cinmathbb Q}[X>cgeqslant Y],$$ hence all this proves that $P(X>Y)=0$. By symmetry, $P(Y>X)=0$ and we are done.

If $X,Y$ are square-integrable we can give a quick proof.

Consider the random variable $(X-Y)^2 = X^2 - 2XY + Y^2$. Let's compute its expectation by conditioning on $X$. We have
$$begin{align*} E[(X-Y)^2] &= E[E[(X-Y)^2 mid X]] \
&= E[E[X^2 - 2XY + Y^2 mid X]] \
&= E[X^2 - 2 X E[Y mid X] + E[Y^2 mid X]] \
&= E[X^2 - 2 X^2 + E[Y^2 mid X]] \
&= E[-X^2 + Y^2]end{align*}$$
If we condition on $Y$ instead we get $E[(X-Y)^2] = E[X^2 - Y^2]$. Comparing these, we see that we have $E[(X-Y)^2] = -E[(X-Y)^2]$, i.e. $E[(X-Y)^2]=0$. This means $X=Y$ almost surely.

Unfortunately I don't quite see a way to handle the case where $X,Y$ are merely integrable, since in that case some of the expectations used above may be undefined.

Acknowledgement of priority. After writing this I found (thanks to Did) that essentially the same proof was given by Michael Hardy in Conditional expectation and almost sure equality.

Let $h:{mathbb R}to{mathbb R}$ be bounded and strictly increasing. (For example, $h(x)=1/(1+e^{-x})$.) Since $X$ and $Y$ are integrable and $h$ is bounded, the random variable $Z:=(X-Y)(h(X)-h(Y))$ is integrable, with expectation
$$
E[Xh(X) -Yh(X)-Xh(Y) +Yh(Y)].tag1
$$
But $E[Yh(X)]=Eleft[E(Yh(X)mid X)right]=E[h(X)E(Ymid X)]=E[Xh(X)]$ and similarly $E[Xh(Y)]=E[Yh(Y)]$. Plugging into (1), we find the expectation of $Z$ is zero. But $Z$ is non-negative, since $h$ is increasing. It follows that $Z$ is zero almost surely. To finish the proof, the fact that $h$ is one-to-one implies the set inclusion
$$left{(X-Y)(h(X)-h(Y))=0right}subsetleft{X=Yright}.
$$