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Let $X$, $Y$, and $Z$ be three random variables in $L^{1}(\Omega, \mathcal{F}, P)$. Suppose that we have $E(X|Y)=Z$, $E(Y|Z)=X$, and $E(Z|X)=Y$. Show that $X=Y=Z$ a.s.

I am able to show this in $L^{2}$; in fact, there, it's a very easy proof. I'm having trouble proving it for $L^{1}$, though, because in $L^{1}$, conditional expectations are defined in terms of sub $\sigma$-algebras of $\mathcal{F}$, and that's not the way they're expressed here. I suppose my main problem is expressing the conditional expectations given in the problem as ones I can use results for $L^{1}$ on. Can somebody please help me do that?

Did
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1 Answers1

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$E(X\mid Y) = Z$ implies that $Z$ is $\sigma(Y)$-measurable and in the same way $E(Z \mid X) = Y$ implies that $Y$ is $\sigma(X)$-measurable. By transitivity, this proves that $Z$ is actually $\sigma(X)$ measurable, so that $E(Z\mid X)=Z = Y$.

Et caetera.

Siméon
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  • WOW! Thank you so very much!! There's not much more to show, right? B/c $X$ is certainly $\sigma(X)$-measurable... I like your avatar, by the way! –  Dec 13 '13 at 20:07
  • @GGG: Indeed. You can also finish with $X = E(Y\mid Z) = Y$ (because $Y=Z$). – Siméon Dec 13 '13 at 20:09