Given $ \mathbb{E}X^2<\infty $, how can I show that if two $\sigma$-algebras $\mathscr{G}\_1\subset \mathscr{G}\_2$, then $\mathbb{E}[Var(X|\mathscr{G}\_2)]\leq \mathbb{E}[Var(X|\mathscr{G}\_1)]$ ?
I have noticed that $\mathbb{E}[Var(X|\mathscr{G})] = \mathbb{E}X^2-\mathbb{E}[\mathbb{E}^2(X|\mathscr{G})]$ and was thinking of the tower property of conditional expectation in order to use the size relationship between $\sigma$-fields, but I have currently got stuck, and I am wondering how to generate the size relationship between $\mathbb{E}[\mathbb{E}^2(X|\mathscr{G})]$.



Why does the dispersion of X about its conditional mean decreases as the σ−algebra grows?
$\newcommand\G{\mathscr G}$For $j=1,2$, let $E\_j$ and $V\_j$ denote, respectively, the conditional expectation and the conditional variance given $\G\_j$. Given that $\G\_1\subset\G\_2$, you want to show that $E V\_2 X\le E V\_1 X$. You have already established that $E V\_j X=EX^2-EY\_j^2$ for $j=1,2$, where $Y\_j:=E\_j X$. So, it is enough to show that $EY\_2^2\ge EY\_1^2$.


We have $E\_1Y\_2^2=E^2\_1Y\_2+V\_1 Y\_2\ge E^2\_1Y\_2=Y\_1^2$, since $\G\_1\subset\G\_2$. So, $EY\_2^2=EE\_1Y\_2^2\ge EY\_1^2$, as desired.


