Minimizing MMSE over positive random variables

Question

Let X be a random variable with a finite second moment. We know that

argmin E(X-Y)^2 = E(X|g),

Where the minimum is taken over all g-measurable random variables Y. How can I find

argmin E(X-Y)^2

Over all positive g-measurable random variables Y?

Answer 1

I have a heuristic, very intuitive idea, so this is not a proof, just some thinking direction. Short: we possibly can take the conditional expectation over all the subsets of subalgebra $\sigma(C) \subset \mathcal{C}$, where $\mathbb{E}[X \mathbb{1}_{C}]\geq 0, \ C \in \mathcal{C}$ and set $Y=0$ elsewhere.

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At first, it is obvious that if $X$ takes only negative values then the closest approximation is $0$.

Suppose that $A=\{\omega: X(\omega)\geq 0\}$ and $B=\{\omega: X(\omega)<0\}$, $\Omega = A\cup B, \ A\cap B = \emptyset$. Both $A$ and $B$ belong to $\mathcal{F}$ as $X$ is $\mathcal{F}$-measurable, and let $\mathcal{C}$ be a subalgebra of $\mathcal{F}$ and $Y$ be $\mathcal{C}$-measurable conditional expectation. Then we can see that positive values of $Y$ on $B$ make our mean square distance bigger, so for the negative values on $A$:

$$E[(X-Y)^2] = E[(X-Y)^2\mathbb{1}_{A}] + E[(X-Y)^2\mathbb{1}_{B}] $$

(for example, in discrete case $E = \sum_{i=1}^n(X(\omega_i)-Y(\omega_i))^2p(\omega_i)$)

So, $Y$ should better have the same sign as $X$ on $A$ and $B$. Lets focus more carefully on the different subsets.

At first, let $X$ have all the values positive on some subset $U \in A$ that also belongs to $\mathcal{C}$. It is known, and possible to prove that $Y$ has the same sign, because of $\mathbb{E}[X\mathbb{1}_U]=\mathbb{E}[Y\mathbb{1}_U]$ and so on for all the subsets of $U$ that belong to $\mathcal{C}$. So, we can include subsets from $\sigma(A) \cap \mathcal{C}$.

Suppose we have zeroed $Y$ on all "pure" subsets of $B$ that belong to $\mathcal{C}$ (which affects the second addend above): $$\forall U \in \sigma(B)\cap \mathcal{C}: \ \ Y(\omega)=0, \omega \in U$$

Lets check the "mixed" subsets. Lets take one of the minimal possible ones, a part of which belongs to $A$, and a part - to $B$, and the subset cannot be split further. $Y$ has some one value for all the $\omega$ from this subset (otherwise it can be split, and recall that $\mathcal{F}$ is bigger algebra, so the parts belong to it too, so we can choose smaller that is "mixed"). Lets take the set $W$ of such subsets, where $X$ (and, consequently, $Y$) has some positive expectation. Then $Y$ has some positive value on all the $\omega$ from this set and

$$E[(X-Y)^2] = E[(X-Y)^2\mathbb{1}_{A\setminus W}] +E[(X-Y)^2\mathbb{1}_{W}]+ E[(X-Y)^2\mathbb{1}_{B\setminus W}] $$

The result about conditional expectation still tells us that the overall expectation is minimal, and we have zeroed $Y$ on some sets inside the last addend, making it smaller(for positive $Y$)*. So we can include the subsets, where $X$ has positive expectation and that cannot be split further in $\mathcal{C}$ - and this will not deform the expectation on the subsets higher in the hierarchy.

Now lets take the subset where the expectation is negative, so for the values of $Y$. Such values make the first addend $\mathbb{E}[(X-Y)\mathbb{1}_A]$ bigger (if few of such $Y$s trap into it). So zeroing them will make mean squared distance smaller.

Conclusion: we have the mean squared minimal for positive $Y$ by taking the conditional expectation over all the subsets from $\mathcal{C}$, where $\mathbb{E}[X \mathbb{1}_{C}]\geq 0, \ C \in \mathcal{C}$ and setting $Y=0$ on everything from $B\setminus W$.

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(*) I don't see in this split any possibility of a "balance", where the bigger negative values of $X-Y$ somewhere on $B$ are compensated by lesser positive values somewhere on $A$. But may be this needs more careful research.