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Let $(X,Y)\in\mathbb{R}^n\times\mathbb{R}^m$ be a pair of random variables with joint density $p(x,y)$. I am interested in the regularity properties of the conditional densities $p(y|x)$ and $p(x|y)$ (in the sense of a regular conditional probability). For example, under what conditions on $p(x,y)$, $p(x)$, and $p(y)$ are the conditional densities differentiable? Continuous everywhere, continuous a.e., etc.?

This question is inspired by this question I asked. The counterexamples suggested there all involve some type of discontinuity on the joint $p(x,y)$, and have not given me much intuition behind this.

I have done some digging but have not found any literature on this question, which seems quite natural. I would be more interested in general literature references on this problem than specific results, but any pointers or suggestions are valuable.

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This question has hardly anything to do with the regular conditional probability or with your previous question (because in the counterexample in the answer to that question no joint density exists).

In your present case, the conditional density $$p(x|y)=\frac{p(x,y)}{p(y)}$$ is defined only for $y$ with $p(y)\ne0$. So, if $p(x,y)$ and $p(y)$ are both continuous on the set $S:=\{y\colon p(y)\ne0\}$, then $p(x|y)$ is continuous on its domain $\mathbb R\times S$; if $p(x,y)$ and $p(y)$ are both differentiable on the set $S$, then $p(x|y)$ is differentiable on its domain; etc.

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  • $\begingroup$ I guess I am not familiar with the "only if" part of your claim: Clearly, if the joint density exists and $p(y)>0$, then the conditional density exists and has the claimed formula. Do you have a reference for the converse? (If you do not wish to invoke regular conditional probabilities, I am not even sure how to formulate this statement.) $\endgroup$
    – user19200
    Commented Jun 3, 2021 at 19:34
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    $\begingroup$ @user19200 : The existence of a joint density was assumed in your own post. In such a case, the display in my answer is the usual definition of the conditional density; see e.g. page 3 in these lecture notes by Pollard: stat.yale.edu/~pollard/Courses/241.fall97/Condit.density.pdf Also, I do not have "only if" in my answer (I have "defined only for"), and there is no such thing as the converse of a definition. $\endgroup$
    – Iosif Pinelis
    Commented Jun 3, 2021 at 20:39

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