Expectation of the squared error with regards to a sub sigma field

Question

I am totally stuck.

Given a probability space $(\Omega, \mathcal F, \mathbf P)$ and a random variable $X$. Let $\mathcal A$ be a sub-$\sigma$-field of $\mathcal F$. Let $Y$ run over all $\mathcal A$-measurable random variables. Show that the squared error $$ E[(X-Y)^2] $$ is minimal when $Y=E[X\mid\mathcal A]$.

Answer 1

Let $\def\P{\mathbf P}\def\A{\mathcal A}\def\E{\mathbb E}$$Y \in L^2(\P|_\A)$, we have, as $Y - \E[X\mid \A]$ is $\A$-measurable \begin{align*} \E\bigl[(X-\E[X\mid \A])(Y - \E[X\mid \A])\bigr] &= \E\Bigl[ \E\bigl[(X-\E[X\mid \A])(Y - \E[X\mid \A])\bigm| \A\bigr]\Bigr]\\ &= \E\Bigl[ \E\bigl[X - \E[X\mid \A] \mid \A\bigr](Y - \E[X\mid A])\Bigr]\\ &= 0. \end{align*} hence \begin{align*} \E[(X-Y)^2] &= \E\bigl[(X-\E[X\mid\A]-Y + \E[X\mid \A])^2\bigr]\\ &= \E[(X-\E[X\mid\A])^2] - 2\E[(X-\E[X\mid A])(Y-\E[X\mid A]) + \E[(Y-\E[X\mid \A])^2]\\ &= \E[(X-\E[X\mid\A])^2] + \E[(Y-\E[X\mid \A])^2]\\ &\ge \E[(X- \E[X\mid \A])^2] \end{align*}

Answer 2

Let $Y={\rm E}[X\mid\mathcal{A}]$ and let $Z$ be any integrable and $\mathcal{A}$-measurable variable. Then we must show that $$ {\rm E}[(X-Y)^2]\leq {\rm E}[(X-Z)^2]. $$ Now try and expand the following expression $$ {\rm E}[(X-Z)^2]={\rm E}[(X-Y+Y-Z)^2] $$ to obtain the inequality.