Suppose $X, Y$ are random variables in $L^2$ such that $$X = E(Y | \sigma(X)) $$ $$Y = E(X | \sigma(Y))$$ Then I want to show that $X=Y$ almost everywhere.
What I've done:
By conditional Jensen $$E(X^2|\sigma(Y)) \geq E(X|\sigma(Y))^2 = Y^2 $$ a.e and thus $||X||_2 \geq ||Y||_2 $. Analogously $||Y||_2 \geq ||X||_2 $ and so $||Y||_2 = ||X||_2$. Moreover $$E(XY|\sigma(Y)) = YE(X|\sigma(Y)) = Y^2 $$ This implies that $$\int_{\Omega}(X-Y)^2 dP = 0 $$ and thus $X=Y$ a.e.
Is it correct ?
It is correct. Maybe some details can be added, for example just after the first displayed equation, we could add "and thus, taking the expectation, $\mathbb E\left[X^2\right] \geqslant \mathbb E\left[Y^2\right] $". And there is no need to work with the $\mathbb L^2$-norm as we only consider the square of expectations.