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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?

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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. –  user10525 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
1  
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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