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The PDF of StableDistribution in terms of FoxH for the case $\alpha\leq 1$ is available at https://reference.wolfram.com/language/ref/FoxH.html namely

enter image description here

StableDistributionPDF[x_, α_, β_, μ_, σ_] := 
 With[{Α = σ (1 + β^2 Tan[(π α)/2]^2)^(1/(2 α)), 
    Β = 2/(π α) Sign[x - μ] ArcTan[β Tan[(π α)/2]]}, 
  1/(α Α)*FoxH[{{{1 - 1/α, 1/α}}, {{(1 - Β)/2, (1 + Β)/
        2}}}, {{{0, 1}}, {{(1 - Β)/2, (1 + Β)/2}}}, 
    Abs[x - μ]/Α]]

How does this vary for the $\alpha> 1$ case?

I checked it numerically that the above formula really gives $S(x;\alpha,\beta,\mu,\sigma)$ for the case $0<\alpha<1$. For the case $1<\alpha<2$, however, the above formula gives completely wrong results.

I found two sources:

[1] Ralf Metzler, Joseph Klafter, The random walk's guide to anomalous diffusion: a fractional dynamics approach, Physics Reports, Volume 339, Issue 1, December 2000, Pages 1-77 https://doi.org/10.1016/S0370-1573(00)00070-3 (relevant part: formulas (C.17) and (C.18) in Appendix C)

[2] A.M. Mathai, Ram Kishore Saxena, and Hans J. Haubold, The H-Function: Theory and Applications (Springer; 2010th edition) (relevant part: formulas (6.168) and (6.169))

The relevant formulas in the above sources seem not to be sufficient to get $S(x;\alpha,\beta,\mu,\sigma)$ in terms of a Fox H-function for the case $1<\alpha<2$.

Could you please help me to find $S(x;\alpha,\beta,\mu,\sigma)$ in terms of a Fox H function for the case $1<\alpha<2$?

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  • $\begingroup$ The StableDistribution allows $0<\alpha\leq 2$. Please describe what happened when you tried a value of $\alpha$ that is between 1 and 2. $\endgroup$
    – JimB
    Jan 17, 2023 at 16:43
  • $\begingroup$ Thanks. StableDistribution works fine for alpha larger than 1. However I need StableDistribution in terms of FoxH function for more complex physics related calculations. $\endgroup$
    – student
    Jan 17, 2023 at 17:38
  • $\begingroup$ I'll repeat: Please describe what happened when you tried a value of α that is between 1 and 2. In other words, was there an error? did it not return a value? did it return a wrong value? $\endgroup$
    – JimB
    Jan 17, 2023 at 18:17
  • $\begingroup$ Sorry for not answering your question. Yes, it returns a wrong value, i.e., not the one which can be obtained by StableDistribution. $\endgroup$
    – student
    Jan 17, 2023 at 19:08
  • $\begingroup$ -1 for not giving specific necessary information despite repeated requests. $\endgroup$
    – JimB
    Jan 17, 2023 at 21:58

1 Answer 1

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Based on

P.N. Rathie, L.C.de S.M. Ozelim, C.E.G. Otiniano, Exact distribution of the product and the quotient of two stable Lévy random variables, Communications in Nonlinear Science and Numerical Simulation, Volume 36, 2016, Pages 204-218, ISSN 1007-5704, https://doi.org/10.1016/j.cnsns.2015.11.012.

the $\alpha$-stable probability density function $S(x;\alpha,\beta,\mu,\sigma)$ in terms of a Fox H-function for the case $1<\alpha<2$ is the following:

$S(x;\alpha,\beta,\mu,\sigma)=\frac{1}{\alpha A}H^{1,1}_{2,2} \left[ \frac{|x-\mu|}{A} \Bigg| \begin{matrix} (1-\frac {1}{\alpha},\frac {1}{\alpha}), & (\frac{1-B(\alpha-2)/\alpha}{2},\frac{1+B(\alpha-2)/\alpha}{2}) \\ (0,1), & (\frac{1-B(\alpha-2)/\alpha}{2},\frac{1+B(\alpha-2)/\alpha}{2}) \end{matrix}\right]$

where

$A=\sigma(1+\beta^2\tan^2(\pi\alpha/2))^\frac{1}{2\alpha}$

and

$B=\frac{2}{\pi(\alpha-2)}{\rm sign}(x-\mu)\arctan(\beta\tan(\pi(\alpha-2)/2)).$

The Wolfram Mathematica code for merging the $0<\alpha<1$ and $1<\alpha<2$ cases is:

enter image description here

Its test, e.g. : enter image description here

enter image description here

with output

enter image description here

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    $\begingroup$ Where is the Mathematica code to show that this works? I'm not doubting that it works but the answer given currently has nothing to do with Mathematica. $\endgroup$
    – JimB
    Jan 22, 2023 at 21:53
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    $\begingroup$ Please include the actual code in your answer, not an image of it., so that interested readers can copy it into a notebook to obtain the outcome themselves. $\endgroup$
    – bbgodfrey
    Jan 22, 2023 at 22:44

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