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Suppose $(\Omega,\mathcal F,\mathbb{P})$ is a probability space and that $\mathcal G$ is a sub-sigma-algebra of $\mathcal F$. If $X$ is an integrable, non-negative random variable with the same distribution as $\mathbb{E}[X|\mathcal G]$, how does one show that $X=\mathbb{E}[X|\mathcal G]$ a.s.?

Thank you.

(I can do the case where $X$ has finite variance, but the method doesn't seem to extend to this...)

Edit

I've just found a hint for this question (but remain stuck):

Show that $f(X)=f(\mathbb{E}[X|\mathcal{G}])$, almost surely, where $f(x)$ equals $1$ if $x>0$, $0$ if $x=0$ and $-1$ otherwise.

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1 Answer 1

By conditional Jensen $ e^{-E(X\vert \mathcal G)} \le E(e^{-X} | \mathcal G) $, by taking expectations equality must hold. I would like to claim here that your result follows by characterization of equality in conditional Jensen, but I'm not sure about that. However, the lhs has the same distribution as $e^{-X}$ , and now you have your original condition with bounded r.v.s

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Thanks. How does this give the original condition with bounded r.v.s? –  Ben Derrett Apr 12 '12 at 9:30

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