Tough Conditional Expectation Problem

[meiji1]

Homework Statement

Suppose that $Y$ is a random variable, $\mathcal{G}$ a $\sigma$-algebra, $E|Y| < \infty$. Show that $Y = E(Y|\mathcal{G})$ a.s. (a.s. = almost surely).

Homework Equations

We're given $Y$ integrable.

The Attempt at a Solution

It's recommended as a hint to prove sgn($E(Y|\mathcal{G})$) = sgn($Y$), and then apply it to $X = Y - c$ somehow. Here's an attempt at the first part which seemed promising, but which I now realize is wrong:

Let $Y^+$ denote the positive part of $Y$ and $Y^-$ its negative part, and let $A = \{ x : E(Y|\mathcal{G})(x) > 0\}$. Then $A \in \mathcal{G}$ since $E(Y|\mathcal{G})$ is measurable, and so by definition of conditional expectation,

\[
\int_A Y dP = \int_A E(Y|\mathcal{G}) dP
= \int_A E(Y|\mathcal{G})^+ dP = = \int_A E(Y^+|\mathcal{G}) dP
\]

the last line derived by supposing $E(Y^+|\mathcal{G}) = E(Y|\mathcal{G})^+$, the very thing I'm trying to prove. From this point I went on to conclude that $Y^- = 0$ a.s., and similarly, that $E(Y|\mathcal{G}) < 0$ implies $Y^+ = 0$ a.s., and $E(Y|\mathcal{G}) = 0$ implies $Y^+ = Y^- = 0$ a.s., which alone is not enough to get sgn($E(Y|\mathcal{G})$) = sgn($Y$) a.s.

I've tried to get $Y = E(Y|\mathbb{G})$ a.s. by showing

\[
0 = \int |E(Y|\mathcal{G}) - Y| dP = \int ||E(Y|\mathcal{G})| - |Y|| dP
\]

to no avail. I have no idea how else you might show it, or how sgn($E(Y|\mathcal{G})$) = sgn($Y$) a.s. may point to the desired conclusion.

Sorry that I can't seem to get the board to render my tex, much of which is wrong anyhow, I'm sure.

 

[mighty2000]

Thank you for your question. Let me try to help you with your problem. First, let's recall the definition of conditional expectation: $E(Y|\mathcal{G})$ is a random variable that is $\mathcal{G}$-measurable and satisfies $\int_A E(Y|\mathcal{G}) dP = \int_A Y dP$ for all $A \in \mathcal{G}$.

Now, for the first part, you are on the right track by considering the positive and negative parts of $Y$. However, instead of trying to show that $E(Y^+|\mathcal{G}) = E(Y|\mathcal{G})^+$, try to show that $E(|Y|\,|\mathcal{G}) = |E(Y|\mathcal{G})|$ a.s. This can be done by breaking up the integral into positive and negative parts, and then using the definition of conditional expectation to simplify. From there, you should be able to show that $E(Y|\mathcal{G}) = Y$ a.s. for $Y$ non-negative, and similarly for $Y$ non-positive.

For the second part, try to use the fact that $E(Y|\mathcal{G})$ is $\mathcal{G}$-measurable to show that $Y - E(Y|\mathcal{G})$ is also $\mathcal{G}$-measurable. From there, you can use the definition of conditional expectation to show that $\int_A (Y - E(Y|\mathcal{G}))dP = 0$ for all $A \in \mathcal{G}$, which implies that $Y = E(Y|\mathcal{G})$ a.s.

I hope this helps. Let me know if you need any further clarification. Good luck with your problem!