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