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I want to prove that for $X,Y$ in $L^2$ and $E[X|Y] = Y$ and $E[Y|X]=X$ almost surely, $X=Y$ almost surely.

I came up with the following solution:

$X = E[Y|X] = E[E[X|Y]|Y] = E[X|Y] = Y$, where all of the equalities hold almost surely and the tower property was used.

Is this proof correct and did I use the fact that $X,Y$ are in $L^2$ implicitly?

max_121
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  • Can you define the space $L$? – jameselmore Feb 08 '22 at 17:29
  • Well for $L^2$, it is the space of all random variables with finite second moment – max_121 Feb 08 '22 at 17:31
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    You didn't use the tower property, you changed a conditioning on $X$ to one on $Y$ (unjustifiedly) and you didn't use that either $X$ or $Y$ are in $\mathscr{L}^2.$ – William M. Feb 08 '22 at 17:40
  • @WilliamM. Isn't $E[E[X|Y] | Y] = E[X|Y]$ the tower property as the sigma algebra generated by the R.V. Y is trivially included in the sigma algebra generated by Y? – max_121 Feb 08 '22 at 17:45
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    but you said $E[Y\mid X] = E[E[X\mid Y]\mid Y]$ – Henry Feb 08 '22 at 17:46
  • $E(E(X\mid Y) \mid Y) = E(X \mid Y)$ can be derived from the fact that if $Z$ is $\mathscr{H}$-measurable, then $E(Z \mid \mathscr{H}) = Z.$ This is not the tower property. – William M. Feb 08 '22 at 17:46
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    Just follow this solution: https://math.stackexchange.com/questions/666843/if-exy-y-almost-surely-and-eyx-x-almost-surely-then-x-y-almost-surel?rq=1 – William M. Feb 08 '22 at 17:47
  • @Henry thanks, I completely overlooked that. – max_121 Feb 08 '22 at 17:47
  • (I am not referring to Did's solution but rather you have to calculate $E((X-Y)^2)$ by using $E(Z) = E(E(Z \mid \mathscr{H}))$ for any sigma algebra $\mathscr{H}.$ You will find that $E((X-Y)^2) = -E((X-Y)^2),$ which can only happen when the expectation is zero.) – William M. Feb 08 '22 at 17:49
  • Intuitively there is no regression to the mean so correlation must be $1$ (having a correlation requires $X$ and $Y$ in $\mathcal L^2$) – Henry Feb 08 '22 at 17:51

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