Stable distribution

Remarks

For a stable law there is a unique $\alpha \in (0, 2]$, the index, such that $c^\alpha = a^\alpha + b^\alpha$ always. The characteristic function of a stable law has the form $\varphi(t) = \exp\left(i \mu t - \abs{\gamma t}^\alpha (1 - i \beta \sgn(t) \Phi(t))\right)$, with $\Phi(t) = \tan(\pi\alpha/2)$ for $\alpha \ne 1$ and $\Phi(t) = -\frac{2}{\pi} \log \abs{t}$ for $\alpha = 1$, where $\beta \in [-1, 1]$, $\gamma > 0$ and $\mu \in \R$. Example: let $X$ be normal with mean $0$ and variance $1$. Then $a X_1 + b X_2$ is normal with mean $0$ and variance $a^2 + b^2$. So it has the law of $c X$ with $c = \sqrt{a^2 + b^2}$, and the index is $2$. Example: for the standard Cauchy law, $\varphi(t) = e^{-\abs{t}}$, so $a X_1 + b X_2$ has characteristic function $e^{-(a + b)\abs{t}}$, the law of $(a + b) X$; this is index $1$. Non-example: let $X$ be Poisson with mean $\lambda > 0$. Then $X_1 + X_2$ is Poisson with mean $2\lambda$, and takes all values in $\N$. If it had the law of $cX + d$, comparing the two smallest values of the supports would force $c = 1$ and $d = 0$, but $X$ has mean $\lambda \ne 2\lambda$. For $\alpha < 2$, $E[\abs{X}^p] < \infty$ if and only if $p < \alpha$; so the variance is infinite, and the mean is finite only for $\alpha > 1$. Some authors also call the point masses stable.