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 $\mathbb{E}[\mathbb{E}^2(X|\mathscr{G})]$.
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1$\begingroup$ A useful fact is that for square-integrable $X$, $\mathbb{E}[X \mid \mathcal{G}]$ is the $L^2$ orthogonal projection of $X$ onto the subspace of $\mathcal{G}$-measurable random variables. In particular, $\mathbb{E}[X \mid \mathcal{G}]$ minimizes the $L^2$ distance to that subspace, so if $Y$ is any other $\mathcal{G}$-measurable random variable, then $\mathbb{E}[(X - \mathbb{E}[X \mid \mathcal{G}])^2] \le E[(X-Y)^2]$. Now applying this with $\mathcal{G} = \mathcal{G}_2$ and $Y = \mathbb{E}[X \mid \mathcal{G}_1]$ you should get the result. $\endgroup$– Nate EldredgeCommented Nov 9, 2020 at 15:02
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$\begingroup$ Thanks for your help! $\endgroup$– Haosheng ZhouCommented Nov 10, 2020 at 5:16
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$\newcommand\G{\mathscr G}$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.
answered Nov 9, 2020 at 14:51