If we define $$ \psi(t|\alpha, \beta, \gamma, \mu) = -it\mu+|\gamma t|^\alpha(1-i\beta \mathrm{sgn}(t) \Phi) $$ with $$\Phi = \tan \frac{\pi \alpha}{2} \mathbf{1}_{\{ \alpha \neq 1 \} } - \frac{2}{\pi} \log(t) \mathbf{1}_{\{ \alpha = 1 \} },$$ then $$ \mathbb{E}[e^{itX}] = \exp(-\psi(t)) $$ is a characteristic function of $X$ with a well-known Stable distribution. On the other hand the value $$ \mathbb{E}[e^{itY}] = \frac{1}{1+\psi(t)} $$ is a characteristic function of $Y$ - random variable with Geometric Stable distribution. There is a relation between $X$ and $Y$: if $X \sim \mathrm{S}_\alpha(\beta, 1, 0)$, then $$ Y = \mu W + \gamma W^{1/\alpha} \Big(X + \frac{2}{\pi} \beta \log(\gamma W)\mathbf{1}_{\{ \alpha = 1 \}} \Big) \sim \mathrm{GS}_\alpha(\beta, \gamma, \mu), $$ where $W$ has a standard exponential distribution.

About Stable distribution it is known that for $\alpha=2$ it corresponds to Normal distribution, for $\alpha=1$ with $\beta=0$ to Cauchy distribution and for $\alpha = \frac{1}{2}$ with $|\beta|=1$ to ±Lévy distribution. And these are the only distributions, for which the density of $X$ has an analytical expression. Therefore, I assume that the same can be said about Geometric Stable distribution, e.g. for $\alpha=2$ $Y$ has an (Asymmetric) Laplace distribution.

I was able to derive the density of $Y$ when $X$ is Lévy distributed, but only for $\frac{\mu}{\sigma} \leq \frac{1}{2}.$ (my C++ implementation), however I failed in the other cases. I have an intuition that one should use complex analysis to solve the task, but it's only a guess.

So, to finalize, my question is: are there any published articles about the density of Geometric Stable distributions, and when/how can it be expressed analytically?

  • $\begingroup$ Wikipedia says that the probability density function has no closed form; why do you expect otherwise? $\endgroup$ – Carlo Beenakker Nov 3 '17 at 17:58
  • $\begingroup$ It does have a close form, but only for certain parameters, namely those I‘ve stated above. $\endgroup$ – Aleksandr Samarin Nov 3 '17 at 19:40

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