# Least-Squares Property of Condtional Expectation

Let $Y$ be a vector-valued random element. Then then for any $\sigma$-field the conditional expectation satisfies the least-squares property $$argmin_{X \in L^2(\mathscr{F})}( \mathbb{E}[(X-Y)^2] )= \mathbb{E}[Y|\mathscr{F}].$$

My question is the following: if $g$ and $f$ are $\sigma(Y)$-measurable then what is the solution to the problem

$$argmin_{X \in L^2(\mathscr{F})}( \mathbb{E}[(g(X)-f(Y))^2])?$$

I expect something of the form $g(\mathbb{E}[f(Y)|\mathscr{F}])$ but I'm not certain, if and when that would be true.

• Your original statement does not seem to be correct to me; the argmin of $E[(X-Y)^2]$ over $L^2(F)$ is the conditional expectation of $Y$. As for your second question, it will depend quite drastically on what $g$ is. $g$ might very well be nonconvex for example... – Ian Oct 4 '16 at 2:41
• It's still wrong, by the way. – Ian Oct 4 '16 at 2:42
• Thanks for pointing that out I missed the crucial exponent in my typo. Ultimately I only need $g\in C^2$ and if necessary (but I would prefer not) I may assume it is convex. – AIM_BLB Oct 4 '16 at 2:52

If we make the added assumption that $g$ is injective and measurable wrt $\mathscr{F}$ then $$Z\triangleq f(Y)$$ then the least squares problem \begin{align} \hat{X}\triangleq \min_{X \in L^2(\mathscr{F})} \mathbb{E}[(X-Z)^2] & = \mathbb{E}[Z|\mathscr{F}] \\ &= \mathbb{E}[f(Y)|\mathscr{F}]. \end{align}
Now if $X$ in te minimization is of the form $g(\tilde{X})$ above then: $$\mathbb{E}[Z|\mathscr{F}] = \hat{X} = g(\tilde{X}),$$ the left-invertibility of $g$ allow us to conclude that $$g^{-1}\left(\mathbb{E}[Z|\mathscr{F}] \right) = \tilde{X} .$$