Looking for a counterexample: Conditioning increases regularity?

Question

Let $p(x,y,z)$ be a joint density (over $\mathbb{R}^3$) under no smoothness or regularity assumptions, besides its existence. I am looking for a (counter)example where $p(y|x)$ is less regular than $p(y|x,z)$, in the sense of differentiability and/or Lipschitz continuity. Thus, conditioning on an extra variable increases regularity.

For example, an example where $p(y|x)$ is at most once differentiable with Lipschitz continuous derivative, but $p(y|x,z)$ is at least twice diferentiable with Lipschitz continuous derivatives. I suppose the weakest counterexample would be the following:

• $p(y|x)$ is Lipschitz continuous, but whose a.e. derivative is not Lipschitz continuous,
• $p(y|x,z)$ has Lipschitz continuous derivative.

Basically, I want to avoid pathologies surrounding everything being non-Lipschitz. For a bonus, give a counterexample over a compact domain such as $[0,1]^3$.

Answer 1

Nice question! Here is an example. Lets set $p(x, y, z)$ independent of $x$ so we can work with just two variables $Y, Z$.

Let $Z > 0$ be a random variable on $[0, 1]$, with distribution to be chosen later, and let $Y$ have conditional distribution

$$f_{Y|Z} (y) = \frac{\mathbb 1_{[0, 1]} (y) g(y, Z)}{\int_0^1 g(s, Z) \, ds}$$

given $Z$, where $g(y, z) := y^3 \sin(\frac{1}{y^{1-z}})$.

The important thing about $g$ is that $g(y, 0)$ is Lipschitz continuous, but its derivative is not.

For $z > 0$, $D_y g(y, z)$ just barely manages to be Lipschitz continuous, with $D_y g(y, z) \sim y^{1+z}.$ It then follows if $Z$ is chosen such that $f_Z (z)$ grows rapidly at $0$, we will have that $D_y p(y)$ is not Lipschitz at $0$, hence is not Lipschitz continuous.