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Working through the book Brownian Motion & Stochastic Calculus by Revuz and Yor, I got confronted with the following problem:

Let $X$ be a random variable s.t. $ E[X^+] < \infty $.

Now assume that using the integrability of $\max(X,-n), \ \forall n \in \mathbb{N} $ we were able to show that $\max(X,-n) \leq E[\max(X,-n) \mid \mathcal{F} \ ], \ \forall n \in \mathbb{N} $.

Can we somehow deduce from this that also $\ X \leq E[X | \mathcal{F} ] $ ?

Or in other words, does $E[\max(X,-n) \ | \ \mathcal{F} \ ] \ $ tend to $E[X \mid \mathcal{F} ] \ $ as $ n \rightarrow + \infty $ ? (which would of course imply the above inequality)

Can we maybe somehow apply monotone convergence or dominated convergence for conditional expectations?

Thanks a lot for any help towards a solution!

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up vote 5 down vote accepted

If $Y$ is an integrable random variable such that $\mathrm E(Y\mid \mathcal F)\geqslant Y$ almost surely, then $Z=\mathrm E(Y\mid \mathcal F)-Y$ is an almost surely nonnegative random variable such that $\mathrm E(Z)=0$, hence $Z=0$ almost surely, and $\mathrm E(Y\mid \mathcal F)=Y$ almost surely.

In your setting, this proves that $\mathrm E(\max(X,-n)\mid \mathcal F)=\max(X,-n)$ almost surely hence $\max(X,-n)$ is $\mathcal F$-measurable for every $n$. Since $\max(X,-n)\to X$ pointwise when $n\to+\infty$, this proves that $X$ is $\mathcal F$-measurable. In particular, $\mathrm E(X\mid \mathcal F)=X$ almost surely.

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He Didier! Thanks a lot for your help! Regards, Si – Mad Si Nov 17 '11 at 14:52

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