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.


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}


\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))<M

Then due to symmetricity

\displaystyle \begin{aligned} E(X^2)& \le 2+2\sum_{k=1}^{\infty} EX^21_{2^{k-1}<X\le 2^k} \\ & \le 2+2\sum_{k=1}^{\infty} 2^{2k}P(2^{k-1}<X<2^k) \\ & \le 2+2\sum_{k=1}^{\infty} 2^{2k}P(X>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.


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


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