# Finding $\alpha$-stable distributions by (probably) Levy's continuity theorem

Let $$X_1,...,X_n$$ be iid random variables uniformly distributed on $$[-1,1]$$. Now let's put $$Y_n = \frac{sgn X_n}{|X_n|^{1/\alpha}}, \, n= 1,2,...,$$ with a set value $$\alpha \in (0,2)$$.

The goal is to:

$$\textbf{(a)}$$ Show that $$Z_n := \frac{Y_1 + ... + Y_n}{n^{1/\alpha}}$$ converges by distribution.

$$\textbf{(b)}$$ Find the characteristic function of the limit.

And that would mean that the limiting distribution is stable as it has its own domain of attraction consisting of $$Z_n$$.

My best guess is to do this by showing that the characteristic function of $$Z_n$$ converges to a certain function that's continuous at $$t=0$$ and then using the Levy's continuity theorem (https://en.wikipedia.org/wiki/L%C3%A9vy's_continuity_theorem), but I was sadly unable to do so.

Do You perhaps have other ideas that would answer this problem or maybe it's indeed solvable by Levy's theorem? If so, how?

Edit:

I've calculated that the characteristic function of $$Z_n$$ should be: $$\phi_{Z_n}(u) = \left[ \frac{1}{2} \int_0^1 exp(iu \frac{-1}{x^{1/\alpha}} n^{-\frac{1}{\alpha}}) dx + \frac{1}{2} \int_0^1 exp(iu \frac{1}{x^{1/\alpha}} n^{-\frac{1}{\alpha}}) dx \right]^n,$$ which is $$\left[ \int_0^1 cos \left( \frac{ u}{(xn)^{1/\alpha}} \right) dx \right]^n$$

Edit 2:

I've found an answered question that shows what the limit should be, but still doesn't have what I need. Symmetric alpha stable distributions with $X_1+X_2+\cdots+X_n \stackrel{d}{=} n^{1/\alpha}X$ as definition

• Were you able to get an expression for the characteristic function of $Y_i$? Jan 20, 2021 at 16:52
• @TeresaLisbon No :/ The sgn and |*| functions are giving me a really hard time in finding the characteristic function rigoristically. Jan 20, 2021 at 16:59
• It was asked as an exercise under the chapter about Levy's theorem in my uni's lecture notes. Jan 20, 2021 at 17:21
• @TeresaLisbon I've placed an edit with the corrected characteristic Jan 20, 2021 at 17:54
• @TeresaLisbon I've added further calculations. Sorry that my previous characteristic was wrong, now it should be proper. Jan 20, 2021 at 18:18

Problem Let $$X_1,X_2,\cdots$$ be a sequence of iid random variables uniformly distributed on $$[-1,1]$$. Now let's put $$\begin{equation*} Y_n=\frac{\mathrm{sgn}X_n}{|X_n|^{1/\alpha}}, \qquad n\ge 1, \end{equation*}$$ with a set value $$\alpha\in(0,2)$$.
(a) Show that $$Z_n=\frac{Y_1+\cdots+Y_n}{n^{1/\alpha}}$$ converges by distribution.
And that would mean that the limiting distribution is stable as it has its own domain of attribution consisting of $$Z_n$$
Answer Let $$\phi_n(t)=\mathsf{E}\Big[\exp\Big(\frac{itY_1}{n^{1/\alpha}}\Big)\Big] =\int_{0}^{1}\cos\Big(\frac{t}{(xn)^{1/\alpha}}\Big)dx.$$ Then $$\phi_{Z_n}(t)=\mathsf{E}[\exp(itZ_n)]=(\phi_n(t))^n.$$ Meanwhile, \begin{align*} n[1-\phi_n(t)]&=n\int_0^1\Big[1-\cos\Big(\frac{t}{(nx)^{1/\alpha}}\Big)\Big]\,dx\\ &=\alpha |t|^\alpha\int_{|t|/n^{1/\alpha}}^{\infty}\frac{1-\cos(z)}{z^{\alpha +1}}\,dz\\ &\to C(\alpha) |t|^\alpha, \qquad \text{as}\quad n\to\infty. \end{align*} where(cf. Sato, Lévy Processes and Infinitively Divisible Distributions, Cambridge University Press, 1999, Lemma 14.1 p.84, or Y. S. Chow & H. Teicher, Probability Theory, 3rd ed., Springer Verlag, 1997, p.469--.) \begin{align*} C(\alpha)&=\alpha\int_{0}^{\infty}\frac{1-\cos(z)}{z^{\alpha +1}}\,dz\\ &=\begin{cases} \cos\Big(\dfrac{\alpha\pi}{2}\Big)\Gamma(1-\alpha),& \alpha \in (0,1)\cup(1,2)\\ \quad\dfrac{\pi}{2}, &\alpha=1. \end{cases} \end{align*} Then $$\lim_{n\to\infty}\phi_{Z_n}(t)=\lim_{n\to\infty}[1-(1-\phi_n(t))]^n=\exp[-C(\alpha) |t|^\alpha].$$