Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This is an exercise of conditional expectations:

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 $|E[Y\mid{\mathcal G}]|\leq c$.

With Jensen's inequality, one immediately has $|E[Y\mid{\mathcal G}]|\leq E(|Y|\mid{\mathcal G})$.

I am trying to show that $E[|Y|\mid{\mathcal G}]\leq |Y|$, which is not necessarily true though. If $Y$ is $\mathcal{G}$-measurable, then $E[|Y|\mid {\mathcal G}]= |Y|$. But I have no idea for the general case.

share|improve this question
    
Hint: Choose $\mathcal G = \{\varnothing, \Omega\}$. –  cardinal Nov 29 '12 at 1:25
    
@cardinal I don't see your point. –  Jack Nov 29 '12 at 2:14
    
Ok. With the given $\mathcal G$, what is $\mathbb E( |Y| \mid \mathcal G)$ equal to? Now, think about what that means and consider a simple case, say, e.g., a discrete nonnegative $Y$. –  cardinal Nov 29 '12 at 2:30
add comment

1 Answer 1

up vote 3 down vote accepted

The following result shows that what you are trying to prove is the exception rather than the rule.

The only case when $\mathbb E(Z\mid\mathcal G)\leqslant Z$ almost surely, for some integrable random variable $Z$ and some sigma-algebra $\mathcal G$, is when $Z$ is measurable with respect to $\mathcal G$. Then $\mathbb E(Z\mid\mathcal G)=Z$ almost surely.

Proof: Call $T=\mathbb E(Z\mid\mathcal G)$, then $\mathbb E(T)=\mathbb E(Z)$. Hence, if $T\leqslant Z$ almost surely, then $T=Z$ almost surely. If this happens, $Z$ is measurable with respect to $\mathcal G$ because $T$ is, by definition of conditional expectation.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.