In a set of lecture notes with properties on conditional variance, I found this inequality:
E[Var(y|x)] ≥ E[Var(y|x,z)]
The intuition is clear: as you add more information, the expected variance is smaller. However, I cannot find the rigorous derivation of the inequality. I don't think it is a hard one, but I cannot get the trick which will lead me to the result.
Hint: By definition of conditional variance, it suffices to show: $$ E[ E(Y\mid \mathcal F)^2] \ge E[ E(Y\mid\mathcal G)^2]\qquad\text{whenever ${\mathcal G}\subset{\mathcal F}.$} $$ This in turn follows from setting $U:=E(Y\mid \mathcal G)$ and $V:=E(Y\mid\mathcal F)-E(Y\mid \mathcal G)$ in the following identity (which you should prove):
Claim: If $U$ and $V$ are square integrable and $U$ is $\mathcal G$-measurable and $E(V\mid\mathcal G)=0$, then $$ E[(U+V)^2] = E[U^2] + E[V^2]. $$