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I am trying to estimate the alpha parameter of a supposed $\alpha$-stable distributed set of data. I have tried from the Hill estimator to more advanced fitting method, but they are or too approximated or too slow for my power of calculation. So after a lot of thinking i have found this way.

I know that in a $\alpha$-stable distribution we have:

$$ \lim_{x\rightarrow +\infty}f(x,\alpha,\beta)\sim -\alpha \gamma^\alpha \frac{\Gamma(\alpha)}{\pi}sin(\frac{\pi \alpha}{2})(1+\beta)x^{-(\alpha+1)} $$

and

$$ \lim_{x\rightarrow +\infty}P(X>x_0)\sim \gamma^\alpha \frac{\Gamma(\alpha)}{\pi}sin(\frac{\pi \alpha}{2})(1+\beta)x^{-\alpha} $$

so plotting $P/f$ we must have straight line at x>>1 such that

$$ \lim_{x\rightarrow+\infty}\frac{P(X>x_0)}{f(x,\alpha,\beta)}\sim -\frac{x}{\alpha} $$

and indeed i found a straight line at the tail of every data sample.

Now i have a question:

-because i found a straight line in the tail of every data sample, is this a general property of distributions?

kjetil b halvorsen
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emanuele
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  • thanx, very good, but now i have bigger doubt :( . As i know a normal distribution can be thought as an alpha stable distribution with $\alpha=2$. Why there is such a big difference? – emanuele Jun 21 '12 at 12:09
  • here: www.webalice.it/marcojl/papers/Thesis.pdf – emanuele Jun 21 '12 at 13:17
  • It is explicitly stated in the thesis that $\alpha<2$, then there is no contradiction. –  Jun 21 '12 at 13:19
  • yes you right, but what i means is why there is not a smooth transition from one behaviour to another? – emanuele Jun 21 '12 at 13:42
  • I guess it is related to the parameterisation. Consider the reparameterisation $\alpha^{\star}=\dfrac{1}{2-\alpha}$. Then $\alpha=2$ corresponds to $\alpha^{\star}=\infty$ which is the "smooth transition" you are looking for. –  Jun 21 '12 at 13:48
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    Good :). If you put it in form of answer i will give you a +1. – emanuele Jun 21 '12 at 14:04

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