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Let $X$ and $Y$ be random vectors defined on a common probability space. $X$ takes values in a finite-dimensional space $\mathcal{X} \subset \mathbb{R}^p$, while $Y$ takes values in $\mathbb{R}$. The conditional expectation $E(Y \mid X)$ is then a random variable that is uniquely defined up to null sets.

I am seeking a set of sufficient conditions on the joint distribution of $(X,Y)$ for the following statement to be true:

Given any point $x_0 \in \mathcal{X}$, there exists a function $f \colon \mathcal{X} \to \mathbb{R}$ such that (i) $f(X)$ is a version of $E(Y \mid X)$, and (ii) $f(\cdot)$ is continuous at $x_0$.

Obviously, the continuity part is the non-trivial one. By Lusin's theorem, any measurable function (such as any version of the conditional expectation function) is "nearly continuous", but this is not quite enough for me.

Ideally the sufficient conditions for the above statement would not involve restrictions on the densities or conditional densities of $X$ and $Y$. The problem that motivates this question has a complicated geometry, so it is difficult to characterize densities with respect to fixed dominating measures.

If you require more structure to the problem (but ideally the question would be answered in more generality), you may assume: $Y = g(A)$ for a continuous function $g(\cdot)$, $X = A + B$, and the random vectors $A$ and $B$ are independent. However, $A$ and $B$ may concentrate on different subspaces of $\mathbb{R}^p$, each of which is a complicated manifold.

Thank you in advance for your time!

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1 Answer 1

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I'm not sure, if I'm just answering something that is already obvious to you, but since the question did not receive any answer for so long, I am trying to make an attempt. The most obvious choice for $f$ would be to have the factorization of the conditional expectation, i.e.

$$f(x)=E[Y|X=x],$$

such that $$f(X)=E[Y|X=\cdot\ ]\circ X = E[Y|X].$$

If $X$ is a discrete random variable, the conditional expectation is characterized by

$$P(X=x)E[Y|X=x]=E[Y\cdot\mathbb{I}_{\{X=x\}}]$$

If you are in the fortunate position where there is a constant $C$, such that

$$E[Y|X=x]=E[Y\cdot\mathbb{I}_{\{X=x\}}]/P(X=x)=C$$

for any $P(X=x)>0$, you define $E[Y|X=x]:=C$ and you find

$$P(X=x)E[Y|X=x]=E[Y\cdot\mathbb{I}_{\{X=x\}}]$$

even for $P(X=x)=0$. Otherwise you seem to be out of luck.

If $X$ is a continuous random variable (unfortunately you didn't specify if $A$ is a discrete or continuous random variable), the conditional expectation is characterized by

$$ E[Y|X=x]\int_\mathbb{R}f_{Y,X}(y,x)dy%=E[Y|X=x]f_X(x) =\int_\mathbb{R}yf_{Y,X}(y,x)dy$$

and you should be fine with choosing

$$ E[Y|X=x]=\int_\mathbb{R}y\frac{f_{Y,X}(y,x)}{\int_\mathbb{R}f_{Y,X}(y,x)dy}dy=\int_\mathbb{R}y\frac{f_{Y,X}(y,x)}{f_X(x)}dy$$

This is a parameter-dependent integral and you should be fine, as long as you can find a dominating integrable function $g$, such that $|\max(1,y)f_{X,Y}(y,x)/f_X(x)|\le g(y)$ for every $x\in\mathcal{X}$ and almost all $y$, see

https://encyclopediaofmath.org/wiki/Parameter-dependent_integral

Especially, it does not matter how you define $E[Y|X=x]$ for subsets of $\mathcal{X}$, where $\int_\mathbb{R}f_{Y,X}(y,x)dy=0$. You can just choose any continuation of $E[Y|X=x]$ (again this most likely will only work if there exists the above mentioned dominating function $g$) and the characterizing equality

$$ E[Y|X=x]\int_\mathbb{R}f_{Y,X}(y,x)dy%=E[Y|X=x]f_X(x) =\int_\mathbb{R}yf_{Y,X}(y,x)dy$$

will still hold true.

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