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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: \begin{equation} \varphi(t; \alpha, \beta, c, \mu) = \exp \left ( i t \mu - |c t|^\alpha \left ( 1 - i \beta sgn(t) \Phi \right ) \right ) \end{equation} 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 $p > 0$,

\begin{equation} E |X|^p < \infty \iff 0 < p < \alpha. \end{equation}

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"?

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    $\begingroup$ In fact, the tail probabilities are related to the behavior of the characteristic function at $0$. $\endgroup$
    – zhoraster
    Commented Jun 1, 2022 at 10:23
  • $\begingroup$ Do you have some equation? $\endgroup$
    – PSE
    Commented Jun 1, 2022 at 17:04
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    $\begingroup$ 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. $\endgroup$
    – zhoraster
    Commented Jun 1, 2022 at 18:11

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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
$$P(|X|>x) = x^{-\alpha}.$$

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