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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 $\alpha <1$?

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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.

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  • $\begingroup$ Could you explain a bit about how positive definiteness of the kernel relates to infinite divisibility? $\endgroup$
    – Noah Stein
    Commented May 15, 2014 at 17:03
  • $\begingroup$ The following paper explorers this connection: arxiv.org/pdf/1403.7304 $\endgroup$
    – Suvrit
    Commented May 15, 2014 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
    Commented May 15, 2014 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
    Commented May 16, 2014 at 0:23

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