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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

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

If i perform the same calculation for a normal distribution i have:

$$ \lim_{x\rightarrow +\infty}\frac{P_N(X>x_0)}{f_N(x,\alpha,\beta)}\sim \frac{1}{2x} $$

Why there is such a big difference in the tails? namely if i can think a gauss distribution as an $\alpha$-stable distribution with $\alpha=2,\beta=0$ why they are so different?

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1 Answer 1

The tails of $\alpha$-stable distributions for $\alpha\lt2$ and for $\alpha=2$ are indeed quite different:

  • When $\alpha\lt2$, the density $f_\alpha$ is such that $f_\alpha(x)\sim c/x^{1+\alpha}$ when $x\to+\infty$. When $\alpha=2$, the density $f_2$ is gaussian hence $f_2(x)\sim c\mathrm e^{-c'x^2+c''x}$ when $x\to+\infty$. Thus, for every $\alpha\lt2$, $\lim\limits_{x\to+\infty}\dfrac{\log f_\alpha(x)}{\log x}=-(1+\alpha)$ is finite while $\lim\limits_{x\to+\infty}\dfrac{\log f_2(x)}{\log x}=-\infty$.
    In particular $\dfrac{\log f_2(x)}{\log x}$ does not converge to $-1-\alpha=-3$ for $\alpha=2$.

  • Another way to express this discontinuity is to note that, for every $\alpha\lt2$, $\lim\limits_{x\to+\infty}\dfrac{\log f_\alpha(x)}{x}=0$ while $\lim\limits_{x\to+\infty}\dfrac{\log f_2(x)}{x}=-\infty$.

  • Still another way is that, for every $\alpha\lt2$, $\lim\limits_{x\to+\infty}\dfrac{\mathrm P(X_\alpha\geqslant x+c)}{\mathrm P(X_\alpha\geqslant x)}=1$ for every $c\gt0$ while $\lim\limits_{x\to+\infty}\dfrac{\mathrm P(X_2\geqslant x+c)}{\mathrm P(X_2\geqslant x)}=0$ for every $c\gt0$.

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