2
$\begingroup$

We consider the probability density, often called a generalized Gaussian density, $$p_{\alpha}(t) \propto \exp (- |t|^\alpha),$$ with parameter $0<\alpha<\infty$. For $p = 2$, we recognize a Gaussian distribution and for $p = 1$ a Laplace one. These distributions are known to be infinitely divisible.

For which $\alpha$ is the probability distribution $p_{\alpha}$ infinitely divisible?

Partial answer: The only infinitely divisible distributions that decay faster than $\exp(−O(\lvert t \rvert \log \lvert t \rvert))$ are the Gaussians (see the Theorem 7 of this paper). Hence, for $\alpha>1$ and $\alpha\neq 2$, $p_\alpha$ is not infinitely divisible. I am interested to know what happens in the case $\alpha <1$?

$\endgroup$
1
$\begingroup$

It is known that the kernel $\phi(x,y) = \exp(-|x-y|^p)$ for $x, y \in R$ and $0< p < \infty$ is positive definite if and only if $p \le 2$. Thus, the infinite divisibility of your generalized Gaussian should hold for the $0 < p \le 2$ case.

$\endgroup$
4
  • $\begingroup$ Could you explain a bit about how positive definiteness of the kernel relates to infinite divisibility? $\endgroup$
    – Noah Stein
    May 15 '14 at 17:03
  • $\begingroup$ The following paper explorers this connection: arxiv.org/pdf/1403.7304 $\endgroup$
    – Suvrit
    May 15 '14 at 17:53
  • $\begingroup$ If I understand correctly the paper you suggest, the authors show that if a continuous and symmetric density is infinitely divisible, then the kernel you defined is positive-definite. In particular it shows that if $p>2$, the probability density that I defined is not infinitely-divisible. But, if I am not mistaken, I cannot say anything for $p<2$. Is there any kind of converse result to the one of the paper you gave? $\endgroup$
    – Goulifet
    May 15 '14 at 20:49
  • $\begingroup$ Since the density function actually generates an infinitely divisible kernel---$\phi(x,y)^\alpha = \exp(-\alpha|x-y|^p)$ is ID---I believe we should be able to obtain a converse to conclude ID of the generalized Gaussian distribution; I hope you manage to work out the details (I gotta run). $\endgroup$
    – Suvrit
    May 16 '14 at 0:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.