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This is an exercise mentioned in this question:

Let $Y$ be an integrable random variable on the space $(\Omega,{\mathcal A},{\bf P})$ and $\mathcal{G}$ be a sub $\sigma$-algebra of $\mathcal{A}$. Show that $|Y|\leq c$ implies $|E[Y\mid{\mathcal G}]|\leq c$.

It has been shown that using $E[|y|\mid{\mathcal G}]$ to estimate $|E[Y \mid{\mathcal G}]|$ is not a good idea due to did's answer to the previous question.

One more step I can go so far is $E[Y]=E[E[Y\mid{\mathcal G}]]$. I don't think this would work since expectation only gives the average.

So how can I get $|E[Y\mid{\mathcal G}]|\leq c$ using $|Y|\leq c$?

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3 Answers 3

up vote 2 down vote accepted

First $$ \left|E[Y\mid\mathcal G]\right|\leq E[|Y|\mid \mathcal G] $$ by Jensen's inequality. Now, let $Z=E[|Y|\mid \mathcal G]$, then for every $G\in\mathcal G$ we have $$ \int_G Z=\int_G|Y|\,\mathrm dP\leq \int_G c\;\mathrm dP, $$ and since both $Z$ and $c$ are non-negative and $\mathcal{G}$-measurable, we have that $Z\leq c$ almost surely.


To see that the last statement is true, we consider the non-negative random variable $(Z-c)1_{\{Z>c\}}$ which has mean $0$ as $$ 0\leq E\left[(Z-c)1_{\{Z>c\}}\right]= \int_{\{Z>c\}} Z\,\mathrm dP-\int_{\{Z>c\}} c\,\mathrm dP\leq 0 $$ since ${\{Z>c\}}\in\mathcal{G}$. This implies $(Z-c)1_{\{Z>c\}}=0$ a.s. and hence $Z=c$ a.s.

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thank you for your answer. how do you come up with the last sentence? –  Jack Nov 29 '12 at 21:27
    
@Jack: See the edit. –  Stefan Hansen Nov 30 '12 at 6:24
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Let $F_n:=\{E[Y\mid\cal G]\geqslant c+n^{-1}\}\in\cal G$. Then $$c\mathbf P(F_n)\geqslant (c+n^{-1})\mathbf P(F_n),$$ so $\mathbf P(F_n)=0$ for all $n$, hence $E[Y\mid\cal G]\leqslant c$ a.s. Doing the same for $-Y$, we get the wanted result.

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If $P(Z \leq X)=1$ then $E[Z \vert \mathcal{G}]\leq E[X\vert \mathcal{G}])$ a.s.

Use $Z=|Y|$ and that $c$ is a $\mathcal{G}$-measurable stochastic variable. Then your own observation that $|E[Y\vert \mathcal{G}] | \leq E[|Y| \vert \mathcal{G}]$ pretty much seals the deal.

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