Let $(\Omega,\mathcal{F},P)$ be a probability space. Let $G\subset \mathcal F$ be a sub-sigma algebra.
prove that for z r.v s.t $z\in G$ then E[zX|G]=zE[X|G]
Our defenition of Y=E[X|G] is a r.v with the properties:
- $Y\in G$
- $\forall A\in G: E[Y*1_A|G]=E[X*1_A|G]$
The solution we've been told to do is starting with z to be indicator and then to build up sum of indicator and then positive function.
But I feel like my approach is much simpler so I would like to know what I'm missing.
We need to prove $\forall A\in G: E[zE[X|G]*A]=E[z*X*A]$ we'll notice we can define $g=z*A\in G$ and so $E[E[X|G]*g]=E[X*g]$. But this is the defenition of E[X|G].
I also havn't manage to prove it for indicators using a method that is diffrent from this approach.
Thank you very much,
The shortest demonstration I know of is this: $$\mathbb E[zX\mid G]=\int zxdP_{X/G}=z\mathbb E[X\mid G]$$