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.

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    $\begingroup$ 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... $\endgroup$ – Ian Oct 4 '16 at 2:41
  • $\begingroup$ It's still wrong, by the way. $\endgroup$ – Ian Oct 4 '16 at 2:42
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    $\begingroup$ 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. $\endgroup$ – AIM_BLB Oct 4 '16 at 2:52

(Partial Answer)

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 $$ g^{-1}\left(\mathbb{E}[Z|\mathscr{F}] \right) = \tilde{X} . $$

Note: If someone knows the solution dropping the assumption of injectiveness I would be happy to hear that solution also.


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