Question about Conditional Expectation

Question

I have a seemingly trivial question regarding conditional expectation. Consider $x$ and $y$ be two integrable random variables on Probability space (X, $\Sigma$, $P$) such that $$E(X|Y) =_{a.s} Y$$ and $$E(Y|X) =_{a.s} X$$ Show that $$X=_{a.s} Y$$

The thing looks like a simple proof result, but I don't know how to move on after showing that$$\int_{B} Y = \int_{B} X$$ for all $B \in \sigma(x)$ and $B \in \sigma(y)$

Now any hint on how to proceed? Thank you so much in advance

Answer 1

Hints:

• For every real number $c$, consider $u(c)=E[X-Y;Y\leqslant c\lt X]$. Use the hypothesis that $E[Y\mid X]=X$ almost surely to show that $u(c)=E[Y-X;c\lt X,c\lt Y]$.
• Deduce that, if furthermore $E[X\mid Y]=Y$ almost surely, then $u(c)=-E[Y-X;X\leqslant c\lt Y]$.
• Studying the sign of the LHS and the RHS, deduce from the above that $u(c)=0$.
• Let $v(c)=P[Y\leqslant c\lt X]$. Deduce from the above that $v(c)=0$ for every real number $c$.
• Show that $\int\limits_\mathbb Rv(c)\mathrm dc=E[(X-Y)^+]$.
• Deduce that $X\leqslant Y$ almost surely.
• Conclude.