# Stable Distributions: Properties

This is a continuation of the previous post that can be found here. We will discuss some of the properties of the Stable distributions in this post.

Theorem 2. $Y$ is the limit of $\displaystyle\frac{X_1+X_2+\cdots+X_n-b_n}{a_n}$ for some iid sequence $X_i$ and sequence $a_n >0$ and $b_n$ if and only if $Y$ has a stable law.

Remark: This kind of explains why Stable laws are called Stable!

Proof: The if part follows by taking $X_i$ to be stable itself. We focus on the only if part.

Let $\displaystyle Z_n=\frac{X_1+X_2+\cdots+X_n-b_n}{a_n}$. Fix a $k\in \mathbb{N}$. Let

$\displaystyle S_n^{j}= \sum_{i=1}^n X_{(j-1)n+i}= X_{(j-1)n+1}+X_{(j-1)n+2}+\cdots+X_{jn} \ \ \mbox{for} \ (1\le j \le k)$

Now consider $Z_{nk}$. Observe that

$\displaystyle Z_{nk} = \frac{S_n^1+S_n^2+\cdots+S_n^k-b_{nk}}{a_{nk}}$

$\displaystyle \implies \frac{a_{nk}Z_{nk}}{a_n} = \frac{S_n^1-b_n}{a_n}+\frac{S_n^2-b_n}{a_n}+\cdots+\frac{S_n^k-b_n}{a_n}+\frac{kb_n-b_{nk}}{a_n}$

As $n\to \infty$,

$\displaystyle \frac{S_n^1-b_n}{a_n}+\frac{S_n^2-b_n}{a_n}+\cdots+\frac{S_n^k-b_n}{a_n} \stackrel{d}{\to} Y_1+Y_2+\cdots+Y_k$

where $Y_i$‘s are iid copies of $Y$. Let $Z_{nk}=W_n$ and let

\displaystyle \begin{aligned} W_n' & :=\frac{a_{nk}Z_{nk}}{a_n}-\frac{kb_n-b_{nk}}{a_n} \\ & = \alpha_nW_n+\beta_n \end{aligned}

where $\displaystyle\alpha_n:=\frac{a_{nk}}{a_n}$ and $\displaystyle\beta_n:=-\frac{kb_n-b_{nk}}{a_n}$.

Now $W_n \stackrel{d}{\to} Y$. and $W_n'\stackrel{d}{\to} Y_1+Y_2+\cdots+Y_k$. If we can show that $\alpha_n \to \alpha$ and $\beta_n \to \beta$, then this will ensure

$\alpha Y+\beta \stackrel{d}{=} Y_1+Y_2+\cdots+Y_k$

Note that $\alpha$ and $\beta$ depends only on $k$. Hence the law of $Y$ is stable.

$\square$

The fact that $\alpha_n \to \alpha$ and $\beta_n \to \beta$ follows from the well known convergence of types theorem which we state below.

Theorem 3. (Convergence of types) If $W_n \stackrel{d}{\to} W$ and there are constants $\alpha_n>0$ and $\beta_n$ so that $W_n'=\alpha_nW_n+\beta_n \stackrel{d}{\to} W'$. If $W,W'$ are non degenerate distributions, then there are constants $\alpha$ and $\beta$ so that $\alpha_n \to \alpha$ and $\beta_n \to \beta$.

Proof: The proof involves analysis arguments using characteristic functions. We will skip it. Interested readers may look into Durret[1] for proof.

$\square$

So far we have introduced stable distributions in the abstract sense. Except for $\alpha=1$ (Cauchy) and $\alpha=2$ (Normal) we dont know if Stable distributions exist at all for other $\alpha$’s. Now we will show the existence through characteristic functions. We will now focus only on the Symmetric stable distributions. Let us assume

$\displaystyle X_1+X_2+\cdots+X_k=k^{1/\alpha}X \ \ \forall \ k\in\mathbb{N} \ \ \ \ \ \ (*)$

where $X, X_1,X_2, \ldots$ iid $F$ where $F$ is symmetric.

Theorem 4. The characteristic function of $F$ is given by $\phi(t)=e^{-c|t|^{\alpha}}$.

Comment: I didn’t find a simple proof of this fact in books. I asked few professors in our institute, they were also not aware any simple proof. The following proof is my own, largely inspired by Avi levy‘s idea. Well, I do not claim that we are first to discover it!

Proof: (Sayan and Avi Levy) Let $\psi(t)$ be the characteristic function. $\psi(t)$ is real and even function. Observe that from $(*)$ relation we have

$\displaystyle E(e^{it(X_1+X_2+\cdots+X_k)})=E(e^{itk^{1/\alpha}X}) \ \ \forall \ k\in \mathbb{N} \ \ \mbox{and} \ \ \forall \ t$
$\displaystyle \implies (\psi(t))^k= \psi(tk^{1/\alpha}) \ \ \forall \ k\in \mathbb{N} \ \ \mbox{and} \ \ \forall \ t \ \ \ \ \ \ \ (1)$

Note that if we put $\displaystyle t=\frac{x}{2^{1/\alpha}}$ and $k=2$ in $(1)$, then

$\displaystyle \phi(x)=\phi\left(\frac{x}{2^{1/\alpha}}\cdot 2^{1/\alpha}\right)=\phi\left(\frac{x}{2^{1/\alpha}}\right)^2 \ge 0$

Thus $\phi$ is non negative. Hence $\phi$ has a unique $k$-th root for all positive integers $k$. We will now use it.

Observe that for all $m,n \in \mathbb{N}$ we have

\displaystyle \begin{aligned} \psi\left(\left(\frac{m}{n}\right)^{1/\alpha}\right) & = \psi\left(\frac{1}{n^{1/\alpha}}\cdot m^{1/\alpha}\right) \\ & = \left(\psi\left(\frac1{n^{1/\alpha}}\right)\right)^m \ \ \mbox{(Putting} \ t=\dfrac{1}{n^{1/\alpha}} \ \mbox{and} \ k=m \ \mbox{in} \ (1)) \\ & = (\psi(1))^{m/n} \ \ \mbox{(Putting} \ t=\frac{1}{n^{1/\alpha}} \ \mbox{and} \ k=n \ \mbox{in} \ (1)) \end{aligned}

Suppose $\psi(1)= e^{-c}$ (to be justified later). As $\psi(t)=\psi(-t)$. We have

$\displaystyle\psi(t)=e^{-c|t|^\alpha} \ \ \ \ \ \ \ (2)$

whenver $|t|^\alpha$ is rational. Now since $\psi$ is continuous and $(2)$ holds on a dense set of $\mathbb{R}$. We have

$\displaystyle \psi(t)=e^{-c|t|^\alpha} \ \ \forall t \in \mathbb{R}$

Now we only have to justify the existence of constant $c \ge 0$. Note that $\psi(1)=E(\cos X) \le 1$ and $\psi(1) \ge 0$. If $\psi(1)=0$, then for all $n$, $\displaystyle \psi\left(\frac{1}{n^{1/\alpha}}\right)=0$ which forces $\psi(0)=0$ by continuity of $\psi$ which is a contradiction. Thus $0 < \psi(1) \le 1$ and $c$ can be taken as $- \log \psi(1)$.

$\square$

Finally we show that symmetric stable distributions actually exists by showing that $e^{-c|t|^\alpha}$ is a characteristic function of some random variable for $latex 0<\alpha\le 2$.

Theorem 6: $e^{-c|t|^{\alpha}}$ where $0<\alpha\le 2$ is a characteristic function.

Proof: The case $\alpha=2$ is settled by normal distribution. So let us assume $0<\alpha <2$. Note that for any $\beta$ and $latex|x|\le 1$ we have

$\displaystyle (1-x)^\beta = \sum_{n=0}^\infty \binom{\beta}{n}(-x)^n$

where $\displaystyle \binom{\beta}{n}=\frac{\beta(\beta-1)\cdots (\beta-n+1)}{n!}$.

Let $\displaystyle\psi(t)=1-(1-\cos t)^{\alpha/2}=\sum_{n=1}^{\infty} c_n(\cos t)^n$ where

\displaystyle \begin{aligned} c_n & = \binom{\alpha/2}{n}(-1)^{n+1} \\ & = \frac{\frac{\alpha}2\left(1-\frac{\alpha}{2}\right)\left(2-\frac{\alpha}{2}\right)\cdots\left(n-1-\frac{\alpha}{2}\right)}{n!} \end{aligned}

Since $\alpha<2$ we have $c_n \ge 0$ and $\displaystyle \psi(0)=1 \implies \sum_{n=1}^{\infty} c_n =1$. Note that $\cos(t)$ is a characteristic function, and hence $(\cos(t))^n$ (for all $n$) is a characteristic function. Therefpre, $\psi(t)$ is characteristic function as it is a convex combination of characteristic functions. Thus

$\displaystyle \psi_n(t)=[\psi(t\cdot 2^{1/2}\cdot n^{-1/\alpha})]^n \ \ \mbox{ is a characteristic function.}$

For $t>0$ we have

\displaystyle \begin{aligned} \lim_{n\to\infty} \psi_n(t) & = \lim_{n\to\infty} \left[1-\left(1-\cos \frac{t\sqrt{2}}{n^{1/\alpha}}\right)^{\alpha/2}\right]^n = e^{-t^\alpha} \end{aligned}

The last one is true because as $x\to 0$ we have $1-\cos x \sim x^2/2$. Hence

$\displaystyle \lim_{n\to\infty} n\left(1-\cos \frac{t\sqrt{2}}{n^{1/\alpha}}\right)^{\alpha/2}=\lim_{n\to\infty} n\left(\frac{2t^2}{n^{2/\alpha}}\cdot\frac12\right)^{\alpha/2} =t^\alpha$

Since $\psi_n(t)=\psi_n(-t)$, we have $\psi_n(t) \to e^{-|t|^\alpha}$. Since $e^{-|t|^\alpha}$ is continuous at $t=0$. Using Levy’s continuity theorem we get that $e^{-|t|^\alpha}$ is a characteristic function. Hence $e^{-c|t|^\alpha}$ is a characteristic function.

$\square$

We will now use the notation $S\alpha S(c)$ to denote a symmetric stable random variable with characteristic function $\phi(t)=e^{c|t|^{\alpha}}$ We will drop the parameter $c$ from the notation when it is of no interest. From now on we assume $0<\alpha <2$. So we do not consider the normal case any more.

Theorem 7: $X\sim S\alpha S$, then $P(|X|>\lambda) \sim k_{\alpha}\lambda^{-\alpha}$ as $\lambda \to \infty$ where $k_\alpha > 0$.

Proof: The proof is out of scope of our discussion. I will probably add it (if required) later. Readers may look into Samorodnitsky and Taqqu[5] for proof.

Theorem 8: $X\sim S\alpha S$, then $E|X|^p <\infty$ for all $0 and $E|X|^\alpha =\infty$.

Proof: By Theorem 7 we infer that there exist a constant $M$ such that $x^\alpha(P(|X|>x)) < M \ \ \forall \ x>0$. Let $p<\alpha$.

\displaystyle \begin{aligned} E(|X|^p) & \le 1+\sum_{k=0}^\infty E|X|^p\mathbb{I}_{2^{k-1}<|X|\le 2^k} \\ & \le 1+\sum_{k=0}^\infty 2^{kp}P(2^{k-1}<|X|\le 2^k) \\ & \le 1+\sum_{k=0}^\infty 2^{kp}P(|X|>2^{k-1}) \\ & \le 1+\sum_{k=0}^\infty 2^{kp}\cdot\frac{M}{2^{(k-1)\alpha}} = 1+2^{\alpha}M\sum_{k=0}^{\infty} 2^{k(p-\alpha)} < \infty \end{aligned}

\displaystyle \begin{aligned} E(|X|^\alpha) & \ge \sum_{k=0}^\infty E|X|^\alpha\mathbb{I}_{2^{k-1}<|X|\le 2^k} \\ & \ge 1+\sum_{k=0}^\infty 2^{(k-1)\alpha}P(2^{k-1}<|X|\le 2^k) \\ & = 1+\sum_{k=0}^\infty 2^{(k-1)\alpha}[P(|X|>2^{k-1})-P(|X|>2^{k})] \end{aligned}

Using the tail asymptotics of $S\alpha S$ random variable we have

$\lim_{k\to \infty} 2^{(k-1)\alpha}[P(|X|>2^{k-1})-P(|X|>2^{k})] = k_{\alpha}(1-2^{-\alpha}) > 0$

Hence the last sum diverges implying $E|X|^\alpha = \infty$.

$\square$

We end this section with an application of Theorem 8.

Application: $Y_1,Y_2,\ldots$ are iid random variables with $E\sqrt{|Y_1|} <\infty$. It is well known via Marcinkiewics-Zygmund law that

$\displaystyle \frac{Y_1+Y_2+\cdots+Y_n}{n^2} \stackrel{a.s.}{\to} 0$

We wish to know whether there exist a constant $\delta >0$ small enough such that

$\displaystyle \frac{Y_1+Y_2+\cdots+Y_n}{n^{2-\delta}} \stackrel{a.s.}{\to} 0 \ ?$

Assume there exist such a $\delta >0$. Without loss of generality assume $\delta<1$. Set $\alpha=\dfrac{1}{2-\delta}$. Observe that $\frac12 <\alpha <1$. If we consider $S\alpha S$ distribution as the law of $Y_i$‘s. Then $E\sqrt{|Y_1|}<\infty$ by Theorem 8. But by definition of $S\alpha S$ distribution

$\displaystyle \frac{Y_1+Y_2+\cdots+Y_n}{n^{2-\delta}} =\frac{Y_1+Y_2+\cdots+Y_n}{n^{1/\alpha}} \stackrel{d}{=} Y_1 \not\equiv 0$

Hence we get a contradiction. Thus there exists no such $\delta$.

$\square$

In the next post we will look at The Generalised CLT which makes stable distributions so famous.