Let X be a random variable defined in a probability space ($\Omega$, $A$,$P$) such that $E(X) < \infty$. Being $F_1$ and $F_2$ two $\sigma$-algebra and $F_1 \subset F_2 \subset A$ , how to prove that:
$E(X - E(X|F_2))^2\leq E(X - E(X|F_1))^2$
And being $F$ $\subset A$ $\sigma$-algebra and $F \subset A$, if we define the conditional variance as:
$Var(X|F)=E((X - E(X|F)^2|F)$
How to prove that:
$Var(X) = E[Var(X|F)]+Var[E(X|F)]$
Let $U=X-E(X|\mathcal F_2)$, $V=X-E(X|\mathcal F_1)$. Claim: $E(U(V-U))=0$. To see this note that $Z\equiv V-U$ is measurable with respect to $\mathcal F_2$. Hence $EZU=EX-EZ(E(X|\mathcal F_2))$ and since $Z$ can be taken inside the conditional expectation in the last term we get $ZU=0$. Now $EV^{2}=E(U+(V-U))^{2}=EU^{2}+2EU(V-U)+E(V-u)^{2}=EU^{2}+E(V-u)^{2}$ by teh claim. Hence $EV^{2} \geq EU^{2}$ which proves the first part. For the second part let $S=X-E(X|\mathcal F)$ and $T=E(X|\mathcal F)$. Then $E(ST)=0$ (this is proved exactly as before). Hence $EX^{2}=E(S+T)^{2}=ES^{2}+ET^{2}$. Just subtracting $(EX)^{2}$ from both sides gives the second part.