want to show $E({(X - E(X| \mathcal{F} )}^2) + E({(E(X| \mathcal{F} ) - E(X| \mathcal{G} )}^2) = E({(X - E(X| \mathcal{G})}^2) $

Question

I am new to measure-theory based probablity:

if $\mathcal{G} \subset \mathcal{F}$, and $E(X)^2 < \infty$, then want to show $E({(X - E(X| \mathcal{F} )}^2) + E({(E(X| \mathcal{F} ) - E(X| \mathcal{G} )}^2) = E({(X - E(X| \mathcal{G})}^2) $

I tried to expand the squares and then get stuck in terms of showing how the above...

Answer 1

Let $Y:= X-E[X\mid\mathcal F]$ and $Z:= E[X\mid\mathcal F]-E[X\mid\mathcal G]$. What is wanted is that $E[Y^2]+E[Z^2]=E[(Y+Z)^2]$ hence it is sufficient to show that $YZ$ has expectation zero. This can be done by having a look at $E[YZ\mid \mathcal F]$.