# Tail behavior of a stable distribution according its moments

I'm studying about stable distributions and I would like to understand a statement that relates the moments to the behavior of their tails. More specifically, the characteristic function of a $$\alpha-$$stable distribution is given by: $$$$\varphi(t; \alpha, \beta, c, \mu) = \exp \left ( i t \mu - |c t|^\alpha \left ( 1 - i \beta sgn(t) \Phi \right ) \right )$$$$ where $$sgn(t)$$ is just the sign function and $$\Phi$$ is $$\tan \left (\frac{\pi \alpha}{2} \right)$$ if $$\neq 1$$ and $$\frac{-2}{\pi}\log|t|$$ if $$\alpha = 1$$

Now, it is simple to show that if $$\alpha >1$$ then $$E|X|< \infty$$. This is not the case for $$\alpha<1$$, where the derivative at zero does not exist. However, regarding the other moments, we have: If $$X$$ is stable with $$0 < \alpha < 2$$, then for any $$p > 0$$,

$$$$E |X|^p < \infty \iff 0 < p < \alpha.$$$$

This property of the moments $$\underline{suggests}$$ that the tails of a stable law behave as $$x^{-\alpha}$$.

I honestly have no idea how this suggestion is so obvious. Are there any intuition or fact that help to understand this "suggestion"?

• In fact, the tail probabilities are related to the behavior of the characteristic function at $0$. Commented Jun 1, 2022 at 10:23
• Do you have some equation?
– PSE
Commented Jun 1, 2022 at 17:04
• Loosely, for $\alpha<1$, $\log \phi_X(t) \sim -c|t|^\alpha$, as $t\to 0$, or for $\alpha>1$, $\log \phi_X(t) \sim it\mu -c|t|^\alpha$, is equivalent to $P(|X|\ge x) \propto x^{-\alpha}$, $x\to+\infty$ (with $E X = \mu$ is the latter case). Actually, even a stronger equivalence holds. Should be in any textbook on stable distributions. Commented Jun 1, 2022 at 18:11

I think I got an insight. I will use the following formula for the the $$p-$$moment, for any $$p>0$$, $$E|X|^{p}=\int_{0}^{\infty}px^{p-1}P(|X|>x)dx< \infty \iff 0< p <\alpha.$$
Since $$\int_{0}^{\infty} x^{-q}dx< \infty$$, for $$q > 1$$, this suggests that
$$P(|X|>x) = x^r,\quad r + (p-1) = -q$$ with $$q>1$$. Thus $$r= 1 - p - q$$.
If $$r = -\alpha$$, we have $$\alpha = q+p-1$$. Equivalently, $$q = \alpha-p+1$$. Since $$\alpha -p > 0$$, we have $$q>1$$. Thus, all this suggests that
$$P(|X|>x) = x^{-\alpha}.$$