Conditional expectations of two variables that point to each other

Question

Let $X,Y\in L^{1}$. If $E[X|Y]=Y$ and $E[Y|X]=X$, show that $Y=X$ a.s.

Any tips for this would be much appreciated! I think one may have to start by showing it for the case $X,Y\in L^{2}$ and then carrying that over to $L^{1}$ by the monotone convergence theorem.

Answer 1

Here is a hint for the $L^2$-case : can you prove that $\mathbf{E}\left[(X-Y)^2 \right] = 0$ ?

As John wrote, you can use the properties of conditional expectations to show that $\mathbf{E}[X^2] = \mathbf{E}[Y^2]$. To compute $\mathbf{E}[(X-Y)^2]$, start to expand the expression : $$\mathbf{E}[(X-Y)^2] = \mathbf{E}[X^2] - 2 \mathbf{E}[XY] + \mathbf{E}[Y^2]$$ But using again the same trick as in John's comment (i.e., conditionning on $Y$), the term in the middle is also equal to $\mathbf{E}[Y^2]$. Therefore, $\mathbf{E}[(X-Y)^2] = \mathbf{E}[X^2] - \mathbf{E}[Y^2] = 0$. But as $(X-Y)^2$ is positive and its integral is zero, it must be zero a.e., which means that $X=Y$ a.e.

Answer 2

For the general case an approach similar to that suggested by Tlön Uqbar Orbis Tertius is available. Fix a bounded continuous strictly increasing function $h:\Bbb R\to\Bbb R$ and argue that $(X-Y)(h(X)-h(Y))\ge 0$ (and $=0$ only on $\{X=Y\}$) but that $\Bbb E[(X-Y)(h(X)-h(Y))]=0$.