Conditional expected value (using measure theory)

[Barioth]

Hi, I'm trying to show that
Givien a probability triplet $$(\theta,F,P)$$
with $$G\in F$$ a sub sigma algebra
$$E(E(X|G))=E(X)$$

Now I want to use $$E(I_hE(X|G))=E(I_hX)$$
for every $$h\in G $$

since that's pretty much all I've for the definition of conditional expected value.

I know this property should use the definition of expected value, but I can't get it to work.

Thanks

 

[Siron]

Let $\mathcal{G} \subset \mathcal{F}$ be a sub $\sigma$-algebra of $\mathcal{F}$ then we have to prove:
$$\mathbb{E}[\mathbb{E}[X|\mathcal{G}]] = \mathbb{E}[X]$$

Set $Z = \mathbb{E}[X|\mathcal{G}]$ then by definition $Z$ is $\mathcal{G}$-measurable and $\forall G \in \mathcal{G}: \mathbb{E}[ZI_G] = \mathbb{E}[XI_G]$. Since $\mathcal{G}$ is a sub $\sigma$-algebra it has to contain $\Omega$ thus in particular $\mathbb{E}[ZI_{\Omega}] = \mathbb{E}[XI_{\Omega}]$ which means $\mathbb{E}[Z] = \mathbb{E}[X]$. Hence $\mathbb{E}[Z] = \mathbb{E}[\mathbb{E}[X|\mathcal{G}]] = \mathbb{E}[X]$.

 

[Barioth]

Thanks, very clean!