I am having some problems showing the equality below:
Let $Y \in L_2(\Omega, \mathcal{A}, \mathbb{P})$ and $\mathcal{F} \subset \mathcal{A}$. Show that for $\mathcal{G} \subset \mathcal{F}$:
$$ \mathbb{E}((Y - \mathbb{E}(Y \vert \mathcal{G}))^2) + \mathbb{E}((\mathbb{E}(Y\vert \mathcal{G}) - \mathbb{E}(Y\vert \mathcal{F}))^2)= \mathbb{E}((Y - \mathbb{E}(Y \vert \mathcal{F}))^2) $$
I tried the tower property but the whole thing got very messy instantly. So I hope anyone has a good approach on this one.
$\newcommand{\Ex}{\mathbb{E}}$ $\newcommand{\F}{\mathcal{F}}$ $\newcommand{\G}{\mathcal{G}}$ The easiest way is by "adding zero":
$$\begin{align*} &\Ex [(Y - \Ex(Y | \F))^2] \\ &= \Ex [(Y - \Ex (Y | \G) + \Ex (Y|\G) -\Ex(Y | \F))^2] \\ &= \Ex [(Y - \Ex (Y| \G))^2] + \Ex [(\Ex (Y|\G ) - \Ex (Y| \F))^2] - 2 \Ex [(Y - \Ex (Y| \G))(\Ex (Y|\G ) - \Ex (Y| \F))] \end{align*}$$
By using the tower property, and "pulling out what is known" we then get:
$$\begin{align*} \Ex [(Y - \Ex (Y| \G))(\Ex (Y|\G ) - \Ex (Y| \F))] &= \Ex [ \Ex [(Y - \Ex (Y| \G))(\Ex (Y|\G ) - \Ex (Y| \F)) | \F] ] \\ &= (\Ex (Y|\G ) - \Ex (Y| \F)) \Ex [ \Ex [(Y - \Ex (Y| \G))| \F] ] \\ &= (\Ex (Y|\G ) - \Ex (Y| \F)) \underbrace{\Ex [ \Ex (Y | \F) - \Ex (Y|G) ]}_{= \Ex Y - \Ex Y = 0} \\ &=0 \end{align*}$$
Thus, we conclude that $$\Ex [(Y - \Ex(Y | \F))^2] = \Ex [(Y - \Ex (Y| \G))^2] + \Ex [(\Ex (Y|\G ) - \Ex (Y| \F))^2]$$