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Are stable distributions smooth enough for each index of stability $\alpha$ between 0 and 2, and skewness parameter $\beta$ between 0 and 1?

Where there any papers that mention this?

kjetil b halvorsen
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rsc05
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  • The methods described in my answer to https://stats.stackexchange.com/questions/41467/consider-the-sum-of-n-uniform-distributions-on-0-1-or-z-n-why-does-the/43075#43075 should help you see what happens and why. – whuber May 10 '17 at 15:40

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Yes, it is this can be observed from the inverse transformation of its characteristic function. Zolotarev was able to compute the probability density function as follow.

(see Lévy Processes and Infinitely Divisible Distributions-Cambridge Studies in Advanced Mathematics- p.88)

for $\alpha >1$% $$ p\left( x\right) =\frac{1}{\pi }\sum_{n=1}^{\infty }\left( -1\right) ^{n-1}% \frac{\Gamma \left( n/\alpha +1\right) }{n!}(\sin \pi n\left( 1+\beta \right) /2)x^{n-1} $$ for $x\in \mathbb{R}$ similarly there are for $\alpha <1$ and $\alpha =1$. Therefore, the pdf of a stable law is a sum series which are all polinomial. Hence, it is differntiable everywhere.

rsc05
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  • This doesn't answer this particular question. – Michael R. Chernick May 14 '17 at 00:36
  • How come you may see it by applying inverse of Fourier transform in the characteristic function – rsc05 May 14 '17 at 00:41
  • This is being automatically flagged as low quality, probably because it is so short. At present it is more of a comment than an answer by our standards. Can you expand on it? You can also turn it into a comment. – gung - Reinstate Monica May 14 '17 at 01:11
  • It is not true that power series necessarily sum to everywhere-differentiable functions. It is invalid to differentiate them term-by-term unless they are absolutely convergent in a neighborhood of zero. (It's not hard to show that $p$ is absolutely convergent for $\alpha \gt 1$.) – whuber May 15 '17 at 00:14
  • Okay @whuber can you show this using this series? – rsc05 May 15 '17 at 00:17
  • Hint: you can bound the absolute values of the coefficients with an obvious constant. – whuber May 15 '17 at 00:22