$\mathbb{E}[\;X \;\mathbb{E}[X| \mathcal{A}]\;]=\mathbb{E}[\;\mathbb{E}[X\;| \mathcal{A}]^2\;]$?

Question

Let $X \in \mathcal{L}^2(\Omega,\mathcal{F},\mathbb{P})$ and $\mathcal{A} \subset \mathcal{F}$ be a sub-$\sigma$-algebra. From the definition of the conditional expectation it follows that $\mathbb{E}[X]=\mathbb{E}[\mathbb{E}[X\;| \mathcal{A}]]$. But why do we obtain that even $$\mathbb{E}[\;X \;\mathbb{E}[X| \mathcal{A}]\;]=\mathbb{E}[\;\mathbb{E}[X\;| \mathcal{A}]^2\;]$$ holds?

Answer 1

$\mathbb{E}[\;X \;\mathbb{E}[X| \mathcal{A}]\;]=\mathbb{E}[ \xi]$ where $\xi = X \mathbb{E}[X| \mathcal{A}]$. Put $\eta = \mathbb{E}[X| \mathcal{A}]$. Then $\xi = \eta X$ and $E(\xi|A) = E( X \eta |A) = \eta E(X|A) = \eta \eta = \eta^2$, so $E(\xi|A) = \eta^2$ and $E (E(\xi|A)) = E\eta^2$. Thus $E \xi = E (E(\xi|A)) = E\eta^2$.

Answer 2

$$\newcommand\E{\mathbb E}\newcommand{\A}{\mathcal A} \E[\E[X\mid \A]^2]=\E[\color{blue}{\E[X\mid \A]}\cdot \E[X\mid \A]]=\E[\E[\color{blue}{\E[X\mid \A]}\cdot X\mid \A]]=\E[\E[X\mid A]\cdot X] $$ Step $2$ uses the fact that $\E[ZX\mid A]=Z\cdot E[X\mid A]$ whenever $Z\in \A$. Step $3$ uses the fact that $\E[\E[Y\mid \A]]=\E[Y]$.