6
$\begingroup$

Define the class of probability density functions $\mathcal{C}$: $\,p \in \mathcal{C}$ iff $p(x)=p(|x|)$, and $\log p(\!\sqrt{x})$ is convex on $[0,\infty)$.

Conjecture: if $p,q \in \mathcal{C}$, then $p * q \in \mathcal{C}$ (where $*$ denotes convolution).

This would be similar to the closure of the class of log concave densities (P$\mathrm{\acute{o}}$lya frequency functions $\mathrm{PF}_2$) under convolution, but applying to heavy-tailed densities.

Sums of Gaussian random variables are Gaussian, and sums of Gaussian Scale Mixtures (GSMs) are GSMs, specifically, it is easily shown that, $$ \xi\hspace{1pt} Z_1 + \gamma\hspace{1pt} Z_2 \overset{d}{=} (\xi^2+\gamma^2)^{\frac{1}{2}} Z $$ where $Z,Z_1,Z_2$ are standard Normal, and $\xi,\gamma$ are non-negative random variables, all independent. So the class of GSMs, denoted $\mathcal{G}_S$, is closed under convolution.

Gaussian Scale Mixtures, by the Bernstein-Widder Theorem, have densities $p(x)$ such that $p(\!\sqrt{x})$ is completely monotonic, and thus $\log p(\!\sqrt{x})$ is convex on $(0,\infty)$. So $\mathcal{G}_S \subset \mathcal{C}$.

Also, using the defining limit of the exponential function, GSMs are the limit as $n\to \infty$ of the properly normalized $n$-times monotonic in $\sqrt{x}$ classes, denoted $\mathcal{M}_n$, where $p\in\mathcal{M}_n$ iff, $$ p(x) = \int_0^{\infty} (1-t\hspace{1pt}x^2)_+^{n-1} \hspace{1pt} d\mu(t) $$ with $\mu$ non-decreasing and bounded below. It can be shown that the classes $\mathcal{M}_n$ are closed under convolution. And it can be shown that $\mathcal{C} \subset \mathcal{M}_2$.

So $\mathcal{G}_S = \mathcal{M}_{\infty} \subset \mathcal{C} \subset \mathcal{M}_2$, and $\mathcal{G}_S$ and $\mathcal{M}_2$ are closed under convolution.

The modern proof that log concave densities are closed under convolution used the fact that for log concave densities $p(x-t)$ is a totally positive kernel of second order ($\mathrm{TP}_2$), so $\int p(x-t) q(t-y)dt=r(x-y)$ is $\mathrm{TP}_2$. The densities in $\mathcal{C}$ are not sign regular on $(-\infty,\infty)$ so this approach doesn't seem to apply.

It seems that $\mathcal{C}$ cannot be represented as a scale mixture or convolution of simpler functions.

Convex (or convex in $\!\sqrt{x}$) functions can be written as integrals of Heaviside type functions. So $p(x) = \exp(\int f(tx)d\mu(t))$. So $p$ can be written as the limit of a product of functions (that are not densities). But the convolution of a product of functions does not simplify to the product of convolutions e.g.

Can anyone think of a representation of the densities in $\mathcal{C}$ that might me useful?

Can anyone come up with a counterexample (or proof) that $\mathcal{C}$ is closed under convolution.

$\endgroup$
3
  • $\begingroup$ it might make sense to try to see whether the convolution of two extreme points in your class remains in the class. I expect extreme points to be thresholded Gaussians $\endgroup$ Commented Aug 14, 2024 at 18:45
  • $\begingroup$ The class $\mathcal{C}$ can be constructed from products of "truncated continuous" Gaussians. But the general $p$ is not then a convex combination, or a convolution of these ... $\endgroup$ Commented Aug 14, 2024 at 18:51
  • $\begingroup$ hmm yes that's true $\endgroup$ Commented Aug 14, 2024 at 18:57

0

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.