Prove that $ \mathbb{E}[X-Y]^2 \geq \mathbb{E}[Y-\mathbb{E}[Y \mid \mathcal{F}]]^2. $

Question

Let $Y \in L^2(\Omega, \mathcal{M}, P)$, $\mathcal{F}$-$\sigma$-algebra, $\mathcal{F} \subset \mathcal{M}$. Prove that for all $X \in L^2(\Omega, \mathcal{F}, P)$ $$ \mathbb{E}[X-Y]^2 \geq \mathbb{E}[Y-\mathbb{E}[Y \mid \mathcal{F}]]^2. $$

I would appreciate any tips or hints.

Answer 1

This is false. Take $X=Y$. As long as $Y \neq E(Y|\mathcal F)$ the inequality cannot be satisfied. For example take $\mathcal F=\{\emptyset, \Omega\}$ and $Y$ to be any non-constant random variable.

If you assume that $X \in L^{2}(\Omega, \mathcal F, P)$ then the inequality is true: $E(X-Y)^{2} =E (E(X-Y)^{2}|\mathcal F)\geq E (E(X-Y)|\mathcal F)^{2}$ which reduces to RHS since $E(X|\mathcal F)=X$. [I have used conditional Jensen's inequality].

Answer 2

$\newcommand{\E}{\mathbb{E}}$ $\newcommand{\F}{\mathcal{F}}$ This follows from just another version of the theorem of Pythagoras and $\E[Z]=\E[\E[Z|\F]]$ \begin{align} \E[(X-Y)^2]&=\E[((X-\E[Y|\F])-(Y-\E[Y|\F]))^2]\\ &=\E[(X-\E[Y|\F])^2] - 2\E[(X-\E[Y|\F])(Y-\E[Y|\F])]+\E[(Y-\E[Y|\F])^2]\\ &=\E[(X-\E[Y|\F])^2]+\E[(Y-\E[Y|\F])^2] \end{align} because \begin{align} \E[(X-\E[Y|\F])(\E[Y|\F]-Y)]&=\E[\E[(X-\E[Y|\F])(\E[Y|\F]-Y)|\F]]\\ &=\E[(X-\E[Y|\F])\,\E[(\E[Y|\F]-Y)|\F]]=0 \end{align}

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In linear algebra terms, you have and orthogonal projector $P^2=P$, $P=P^T$ and two vectors $X,Y$ with $X=QX$. Then $$ \|X-Y\|^2=\|X-PY\|^2+\|Y-PY\|^2-2\langle P(X-Y), (I-P)Y\rangle $$ and the vectors inside the scalar product are obviously orthogonal.