# Stable Distributions: An introduction

In this series of posts I will introduce the stable distributions and discuss some of its properties. The theorems and proofs presented here are written in a lucid manner, so that it can be accessible to most of the readers. I tried my best to make this discussion self contained as far as possible so that the readers don’t have to look up to different references to follow the proofs. It is based on the presentation that I did in Measure Theory course in MStat 1st year.

## Introduction

In statistics while modelling continuous data we assume that the data are iid observations coming from some ‘nice’ distribution. Then we do statistical inference. The strongest statistical argument we usually use is the Central Limit Theorem, which states that the sum of a large number of iid variables from a finite variance distribution will tend to be normally distributed. But this finite variance assumption is not always true. For example many real life data, such as financial assets exhibit fat tails. Such finite variance assumption does not hold for heavy tails. So there is a need for an alternative model. One such alternative is the Stable distribution. Of course there are other alternatives. But one good reason to use Stable distribution as an alternative is that they are supported by General Central Limit Theorem.

Notations: $U\stackrel{d}{=} V$ means $U$ and $V$ have same distribution. Throughout this section we assume

$\displaystyle X,X_1,X_2,\ldots \stackrel{iid}{\sim} F \ \ \ \mbox{and} \ S_n=\sum_{i=1}^n X_i$

Motivation: Suppose $F= \mbox{N}(0,\sigma^2)$. In that case we know

$S_n\stackrel{d}{=} \sqrt{n}X$

Motivated from the above relation we question ourselves that can we get a distribution such that the above relation holds with some other constants like $n$ or $\log n$ or $n^{1/3}$ instead of $\sqrt{n}$? Hence we generalise the above relation in the following manner.

Definition: The distribution $F$ is stable (in the broad sense) if for each $n$ there exist constants $c_n >0$ and $\gamma_n$ such that

$S_n \stackrel{d}{=} c_nX+\gamma_n$

and $F$ is non degenerate. $F$ is stable in the strict sense if the above relation holds with $\gamma_n=0$.

Examples: $F= \mbox{N}(0,1) \implies c_n=\sqrt{n}$ and $F=\mbox{Cauchy}(0,1) \implies c_n=n$

Observe that $\gamma_n$ is like a location constant. We are interested in the scale constant $c_n$. We want to know what are the possibilities for $c_n$. The following theorem shows that $c_n$ has to be an appropriate power of $n$. So $\log n$ or $e^n$ are ruled out.

Theorem 1: (Feller volume 2) $c_n= n^{1/\alpha}$ where $0<\alpha \le 2$.

Proof: Note that if $F$ is stable, the symmetrized version of $F$, $\bar{F}$ (distribution of $X_1-X_2$) is strictly stable with same constant $c_n$. So without loss of generality assume $F$ is symmetric stable. Hence we have $S_n \stackrel{d}{=} c_nX$.

Step-1: We claim $\displaystyle\sup_{m, n \in \mathbb{N}: m\ge n} \frac{c_n}{c_{m+n}} < \infty$

Note that

$\displaystyle S_{m+n} = S_n+S_{m+n}-S_m \stackrel{d}{=} c_mX_1+c_nX_2$

as $S_n$ and $S_{m+n}-S_m$ are independent. Hence

$\displaystyle c_{m+n}X \stackrel{d}{=} c_mX_1+c_nX_2$

Observe that

\displaystyle \begin{aligned} P(X>t) & = P(c_{m+n}X>c_{m+n}t) \\ & = P(c_mX_1+c_nX_2>c_{m+n}t) \\ & = P\left(\frac{c_m}{c_n}X_1+X_2>\frac{c_{m+n}}{c_n}t\right) \\ & \ge P\left(\frac{c_m}{c_n}X_1\ge 0\right)P\left(X_2 > \frac{c_{m+n}}{c_n}t\right) \\ & \ge \frac12P\left(X_2 > \frac{c_{m+n}}{c_n}t\right) \end{aligned}

Now if our claim is not true, we may get a sequence such that $\displaystyle\frac{c_{n_k}}{c_{m_k+n_k}} \uparrow \infty$ as $k\to \infty$ with $m_k\ge n_k$. Using this sequence in that above inequality we have

$\displaystyle P(X>t) \ge \frac12\lim_{k\to\infty} P\left(X_2 > \frac{c_{m_k+n_k}}{c_{n_k}}t\right) = \frac12P(X_2>0)= \epsilon >0$

The last inequality is strict as $F$ is non degenerate. But then the above holds for all $t>0$ which contradicts tightness. This proves the claim.

Step-2: We will show $c_n= n^{1/\alpha}$. Note that

\displaystyle \begin{aligned} c_{rk}X & \stackrel{d}{=} S_{rk}=S_k+(S_{2k}-S_k)+(S_{3k}-S_{2k})+\cdots+(S_{rk}-S_{(r-1)k}) \\ & \stackrel{d}{=} c_kX_1+c_kX_2+c_kX_3+\cdots+c_kX_r \\ & \stackrel{d}{=} c_kc_rX \end{aligned}

Thus $c_{rk}=c_rc_k \ \ \forall \ r,k$.

Thus $c_1=c_1\cdot c_1 \implies c_1=1$ and if for some some $k>1$, we have $c_k=1$, then $X_1+X_2+\cdots+X_k \stackrel{d}{=} X$. Taking characteristic functions we have

$\displaystyle (\psi(t))^{k} = \psi(t) \implies (\psi(t))^{k-1}=1$

As $F$ is symmetric, $\psi$ is real. Hence $\psi(t)=1$ or $-1$. But since $\psi$ is continuous and $1$ at $t=0$, we have $\psi\equiv 1$ which forces $F$ to be degenerate.

Also if for some $k>1$, we have $c_k<1$. Then $c_{k^n}=c_kc_k\cdots c_k =(c_k)^n \to 0$. But then it contradicts Step 1. Thus we must have $c_k>1$ for all $k>1$. Finally for each $k$, we choose an $\alpha$ (depending on $k$) such that $c_k=k^{1/\alpha}$. We will show the choice of $\alpha$ is independent of $k$. This will prove Step 2.

Therefore we wish to show

$\displaystyle c_{r}=r^{1/\alpha} \ \ \mbox{and} \ \ c_s=s^{1/\beta} \implies \alpha=\beta$

Note that $c_{r^j}=r^{j/\alpha}$ and $c_{s^k}=s^{k/\beta}$. For each $s^k$, there exists $r^j$ such that $s^k < r^j \le s^{k+1}$. Then

$\displaystyle c_{r^j}=r^{j/\alpha}\le s^{(k+1)/\alpha}=s^{1/\alpha}(c_{s^k})^{\beta/\alpha}$

Thus

$\displaystyle \frac{c_{s^k}}{c_{r^j}} \ge \frac{(c_{s^k})^{(\alpha-\beta)/\alpha}}{s^{1/\alpha}}$

But as $k\to \infty$, $c_{s^k} \to \infty$. But by Step 1, the left hand side of the above inequality always remains bounded. Hence $\alpha \le \beta$. Similarly $\beta \le \alpha$ by changing the roles of $r$ and $s$. Hence $\alpha=\beta$.

Step-3: We will show $\alpha \le 2$. Note the if $F$ has finite second moment, taking variances we have

$\displaystyle n\mbox{Var}(X_1)=n^{2/\alpha}Var(X_1) \implies \alpha=2$

We will show that for $\alpha >2$, $F$ will have finite second moment which will force $\alpha=2$ which is a contradiction. For this, we need the following lemma

Lemma 1: $X_1,X_2,\cdots \stackrel{iid}{\sim} F$. $F$ is symmetric. $S_n=X_1+X_2+\cdots+X_n$. Then

$\displaystyle P(|S_n|\ge t) \ge \frac12(1-e^{-n(1-F(t))})$

We skip the proof of this lemma. One can find a proof of the lemma in Feller Volume 2. Since $\left\{\frac{S_n}{n^{1/\alpha}}\right\}$ is tight, get $t$ such that $P(|S_n|>tn^{1/\alpha}) <\frac14$. Clearly $n(1-F(tn^{1/\alpha}))$ must be bounded. Otherwise it will contradict the lemma. Thus $x^{\alpha}(1-F(x))$ is bounded for all $x>t$. So this implies $x^\alpha(1-F(x))$ is also bounded for all $x>0$ So suppose

$\displaystyle x^\alpha(1-F(x))

Then due to symmetricity

\displaystyle \begin{aligned} E(X^2)& \le 2+2\sum_{k=1}^{\infty} EX^21_{2^{k-1}2^{k-1}) \\ & \le 2+2\sum_{k=1}^{\infty} 2^{2k}\frac{M}{2^{(k-1)\alpha}} = 2+2^{1+\alpha}M\sum_{k=1}^{\infty} 2^{k(2-\alpha)} < \infty \end{aligned}

This completes the proof.

$\square$

We look into some standard properties of stable distributions in next post.