let Y, YZ $\in$ $\mathcal{L}^1$ and let Z be $\mathcal{G}$ measurable then show E(YZ|$\mathcal{G}$) = ZE(Y|$\mathcal{G}$)

Question

let Y, YZ $\in$ $\mathcal{L}^1$ and let Z be $\mathcal{G}$ measurable then show E(YZ|$\mathcal{G}$) = ZE(Y|$\mathcal{G}$)

I know we want to start with simple functions then move to non negative and integrable functions and for each show $\int_{\Lambda} ZE(Y |\mathcal{G}$)dP = $\int_{\Lambda} YdP$

For the simple case we have Z = $1_A A \in \mathcal{G}$ then

$\int_{\Lambda} ZE(Y |\mathcal{G}$)dP = $\int_{\Lambda \cap A} E(Y |\mathcal{G}$)dP = $\int_{\Lambda \cap A} Y $dP

since $\Lambda \cap A \in \mathcal{G}$

=$\int_{\Lambda} Y1_A $dP = $\int_{\Lambda} YZ $dP

Is this the right path? I am struggling a little bit with the Non negative and integrable functions.

Answer 1

The conditional expectation $E(YZ | \mathcal{G})$ is characterized by the two properties

1. $E(YZ|\mathcal{G})$ is $\mathcal{G}$-measurable, and
2. $\int E(YZ|\mathcal{G}) \cdot \varphi \,d\mathbb{P} = \int YZ\varphi \,d\mathbb{P}$ whenever $\varphi$ is $\mathcal{G}$-measurable.

It's clear that $Z \cdot E(Y|\mathcal{G})$ is $\mathcal{G}$-measurable because both $Z$ and $E(Y|\mathcal{G})$ are. Therefore it's enough to show that $Z E(Y|\mathcal{G})$ also satisfies property $2$ above.

Let $\varphi$ be any $\mathcal{G}$-measurable function. Then $Z\varphi$ is also $\mathcal{G}$-measurable, so $\int E(Y|\mathcal{G}) Z\varphi \,d\mathbb{P} = \int YZ\varphi \,d\mathbb{P}$ by the definition of $E(Y|\mathcal{G})$. Therefore the random variable $Z E(Y|\mathcal{G})$ satisfies both properties 1 and 2, so it must be a version of $E(YZ|\mathcal{G})$.