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

Question

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.

Answer 1

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 $\mathcal{F} \subset \mathcal{G}$.

Answer 2

$E(X|\mathcal{F})$ 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}

Answer 3

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}