Conditional Expectation: Orthogonal Projection in $L^2$

Question

Let $X \in L^2(\Omega, \mathcal{A}, P)$ and $\mathcal{F}\subseteq \mathcal{A}$ a sub-sigma-Algebra.

Then $E[X \mid \mathcal{F}]$ is the orthogonal projection of $X$ onto $L^2(\Omega, \mathcal{F}, P)$.

I have some questions regarding this result.

1. We have restated this result as $E[(X-Z')^2] \geq E[(X-E[X \mid \mathcal{F})^2]$ for all $Z' \in L^2(\Omega, \mathcal{F}, P)$. Why is this the same?
2. What does it mean to have scalar product 0 in $L^2$?
3. General question concerning $X \in L^2(\Omega, \mathcal{A}, P)$. So this means that $X$ is a random variable with $E^P[|X|^2]^{1/2} = (\int_{\Omega} X(\omega)^2 dP(\omega))^{1/2} < \infty$ (which is Independent of the sigma-algebra $\mathcal{A}$, but depends on $P$) and is $\mathcal{A}$-measurable. Is this true?

Thanks a lot.

Answer 1

The scalar product on $L^2$ is $$\langle X, Y \rangle = E(XY) = \int_\Omega XY\, dP.$$

Note that if $H$ is a Hilbert space and $M$ is a closed subspace, the projection of $x$ onto $M$, $P_M(x)$, is the closest point in $M$ to $X$ in the $L^2$ norm. This is a very basic property of Hilbert spaces.

The random variable $E(X|\mathcal{A})$ is $\mathcal{A}$ measurable and for all $Q\in\mathcal{A}$, $$ \int_{Q} X\, dP = \int_Q E(X|\mathcal{A})\, dP.$$