Show that $E((E(X\mid \mathcal{G})-E(X\mid \mathcal{F}))^2=E(E(X\mid \mathcal{G}))^{2}-E(E(X\mid \mathcal{F}))^{2}$ 
Let $\mathcal{F} \subseteq \mathcal{G}$.
Show that $$E((E(X\mid \mathcal{G})-E(X\mid \mathcal{F}))^2=E(E(X\mid \mathcal{G}))^{2}-E(E(X\mid \mathcal{F}))^{2}$$

My idea:
$$E((E(X\mid \mathcal{G})-E(X\mid \mathcal{F}))^2=E(E(X\mid \mathcal{G}))^{2}-2E(X\mid \mathcal{F})E(X\mid \mathcal{G})+E(E(X\mid \mathcal{F}))^{2}$$
So I basically need to show that
$$-2E(X\mid \mathcal{F})E(X\mid \mathcal{G})=-2E(E(X\mid \mathcal{F}))^2$$
I am attempting to use the tower property:  $$-2E(X\mid \mathcal{F})E(X\mid \mathcal{G})=-2E(E(X\mid \mathcal{G})\mid \mathcal{F})E(X\mid \mathcal{G})=-2E(E(X\mid \mathcal{F})\mid \mathcal{G})E(X\mid \mathcal{G})$$
And since $E(X\mid \mathcal{G})$ is $\mathcal{G}-$measurable:
$$-2E(E(X\mid \mathcal{F})\mid \mathcal{G})E(X\mid \mathcal{G})=-2E(E(X\mid \mathcal{F})E(X\mid \mathcal{G})\mid \mathcal{G})$$
I do not know how to continue.
 A: 
My idea:
$$E((E(X\mid \mathcal{G})-E(X\mid \mathcal{F}))^2=E(E(X\mid \mathcal{G}))^{2}-2E(X\mid \mathcal{F})E(X\mid \mathcal{G})+E(E(X\mid \mathcal{F}))^{2}$$

There is a mistake (typo?) in there. It should read
$$E((E(X\mid \mathcal{G})-E(X\mid \mathcal{F}))^2=E(E(X\mid \mathcal{G}))^{2}-2\color{red}{E}(E(X\mid \mathcal{F})E(X\mid \mathcal{G}))+E(E(X\mid \mathcal{F}))^{2}$$
(I suppose you use the convention that $EY^2=E(Y^2)$). This means that you actually need to show that
$$-2E(E(X\mid \mathcal{F})E(X\mid \mathcal{G})) = -2E(E(X \mid\mathcal{F})^2),$$
i.e.
$$E(E(X\mid \mathcal{F})E(X\mid \mathcal{G})) = E(E(X\mid\mathcal{F})^2). \tag{1}$$
Because of the tower property, we can write the "outer" expectation on the left-hand side as
$$E(\ldots) = E(E(\ldots \mid \mathcal{F}))$$
Next use that $E(X \mid \mathcal{F})$ is $\mathcal{F}$-measurable and the fact that $\mathcal{F} \subset \mathcal{G}$.
A: $E(X|\mathcal{F})$ is $\mathcal{G}$ measurable as $\mathcal{F}\subset\mathcal{G}$. Thus,
\begin{aligned}
E(X|\mathcal{F})E(X|\mathcal{G})&=E\Big(E(X|\mathcal{F})X|\mathcal{G}\Big)
\end{aligned}
Consequently
\begin{aligned}
E\left(E(X|\mathcal{F})E(X|\mathcal{G})\right)&=E\left(E\Big(E(X|\mathcal{F})X|\mathcal{G}\Big)\right)=E\big(XE(X|\mathcal{F})\big)\\
&=E\Big(E\big(XE(X|\mathcal{F})|\mathcal{F}\big)\Big)=E\Big(\big(E(X|\mathcal{F})\big)^2\Big)
\end{aligned}
A: Let $\mathcal{F} \subseteq \mathcal{G}$ and $X,Y$ two random variable. 
\begin{align}
E\left[(E[X\vert \mathcal{G}]-E[X\vert \mathcal{F}])^2\right] &= E\left[E[X\vert \mathcal{G}]^2-2E[X\vert \mathcal{F}] E[X\vert \mathcal{G}] + E[X\vert \mathcal{F}]^2\right] \\
&= E\left[E[X\vert \mathcal{G}]^2\right]-2E\left[E[X\vert \mathcal{F}]E[X\vert \mathcal{G}]\right] + E\left[E[X\vert \mathcal{F}]^2\right]
\end{align}
but as $E[X\vert \mathcal{F}]$ is $\mathcal{G}-$mesurable.  We have $E\left[E[X\vert \mathcal{F}]E[X\vert \mathcal{G}]\right]=E[E\left[E[X\vert \mathcal{F}]X\vert \mathcal{G}\right]] = E\left[XE[X\vert \mathcal{F}]\right] = E\left[E[E[X\vert \mathcal{F}]X\vert \mathcal{F}\right]]= E\left[(E[X\vert \mathcal{F})^2\right]$.
So \begin{align}
E\left[(E[X\vert \mathcal{G}]-E[X\vert \mathcal{F}])^2\right] &= E\left[E[X\vert \mathcal{G}]^2\right]- E\left[E[X\vert \mathcal{F}]^2\right]
\end{align}
