# CLT and Stable Distributions

One of the questions I posed at the end of the previous post about the Central Limit Theorem was this: what is special about the normal distribution?

More precisely, for a large class of variables (those with finite variance) the limit in distribution of $S_n$ after a natural rescaling is distributed as N(0,1). As a starting point for investigating similar results for a more general class of underlying distributions, it is worth considering what properties we might require of a distribution if it is to appear as a limit in distribution of sums of IID RVs, rescaled if necessary.

The property required is that the distribution is stable. In the rest of the post I am going to give an informal precis of the content of the relevant chapter of Feller.

Throughout, we assume a collection of IID RVs, $X,X_1,X_2,\ldots$, with the initial sums $S_n:=X_1+\ldots+X_n$. Then we say $X$ is stable in the broad sense if

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

for some deterministic parameters $c_n,\gamma_n$ for every n. If in fact $\gamma_n=0$ then we say $X$ is stable in the strict sense. I’m not sure if this division into strict and broad is still widely drawn, but anyway. One interpretation might be that a collection of distributions is stable if they form a non-trivial subspace of the vector space of random variables and also form a subgroup under the operation of adding independent RVs. I’m not sure that this is hugely useful either though. One observation is that if $\mathbb{E}X$ exists and is 0, then so are all the $\gamma_n$s.

The key result to be shown is that

$c_n=n^{1/\alpha}$ for some $0<\alpha\leq 2$.

Relevant though the observation about means is, a more useful one is this. The stability property is retained if we replace the distribution of $X$ with the distribution of $X_1-X-2$ (independent copies naturally!). The behaviour of $c_n$ is also preserved. Now we can work with an underlying distribution that is symmetric about 0, rather than merely centred. The deduction that $\gamma_n=0$ still holds now, whether or not X has a mean.

Now we proceed with the proof. All equalities are taken to be in distribution unless otherwise specified. By splitting into two smaller sums, we deduce that

$c_{m+n}X=S_{m+n}=c_mX_1+c_nX_2.$

Extending this idea, we have

$c_{kr}X=S_{kr}=S_k^{(1)}+\ldots+S_k^{(r)}=c_kX_1+\ldots+c_kX_r=c_kS_r=c_kc_rX.$

Note that it is not even obvious yet that the $c_n$s are increasing. To get a bit more control, we proceed as follows. Set $v=m+n$, and express

$X=\frac{c_m}{c_v}X_1+\frac{c_n}{c_v}X_2,$

from which we can make the deduction

$\mathbb{P}(X>t)\geq \mathbb{P}(X_1>0,X_2>t\frac{c_v}{c_n})=\frac12\mathbb{P}(X_2>t\frac{c_v}{c_n}).$ (*)

So most importantly, by taking $t>>0$ in the above, and using that X is symmetric, we can obtain an upper bound

$\mathbb{P}(X_2>t\frac{c_v}{c_n})\leq \delta<\frac12,$

in fact for any $\delta<\frac12$ if we take $t$ large enough. But since

$\mathbb{P}(X_2>0)=\frac12(1-\mathbb{P}(X_2=0)),$

(which should in most cases be $\frac12$), this implies that $\frac{c_v}{c_n}$ cannot be very close to 0. In other words, $\frac{c_n}{c_v}$ is bounded above. This is in fact regularity enough to deduce that $c_n=n^{1/\alpha}$ from the Cauchy-type functional equation (*).

It remains to check that $\alpha\leq 2$. Note that this equality case $\alpha=2$ corresponds exactly to the $\frac{1}{\sqrt{n}}$ scaling we saw for the normal distribution, in the context of the CLT. This motivates the proof. If $\alpha>2$, we will show that the variance of X is finite, so CLT applies. This gives some control over $c_n$ in an $n\rightarrow\infty$ limit, which is plenty to ensure a contradiction.

To show the variance is finite, we use the definition of stable to check that there is a value of t such that

$\mathbb{P}(S_n>tc_n)<\frac14\,\forall n.$

Now consider the event that the maximum of the $X_i$s is $>tc_n$ and that the sum of the rest is non-negative. This has, by independence, exactly half the probability of the event demanding just that the maximum be bounded below, and furthermore is contained within the event with probability $<\frac14$ shown above. So if we set

$z(n)=n\mathbb{P}(X>tc_n)$

we then have

$\frac14>\mathbb{P}(S_n>tc_n)\geq\frac12\mathbb{P}(\max X_i>tc_n)=\frac12[1-(1-\frac{z}{n})^n]$

$\iff 1-e^{-z(n)}\leq \frac12\text{ for large }n.$

So, $z(n)=n(1-F(tc_n))$ is bounded as $n$ varies. Rescaling suitably, this gives that

$x^\alpha(1-R(x))

This is exactly what we need to control the variance, as:

$\mathbb{E}X^2=\int_0^\infty \mathbb{P}(X^2>t)dt=\int_0^\infty \mathbb{P}(X^2>u^2)2udu$

$=\int_0^\infty 4u\mathbb{P}(X>u)du\leq \int_0^\infty 1\wedge\frac{4M}{u^{-(\alpha-1)}}du<\infty,$

using that X is symmetric and that $\alpha>2$ for the final equalities. But we know from CLT that if the variance is finite, we must have $\alpha=2$.

All that remains is to mention how stable distributions fit into the context of limits in distribution of RVs. This is little more than a definition.

We say F is in the domain of attraction of a broadly stable distribution R if

$\exists a_n>0,b_n,\quad\text{s.t.}\quad \frac{S_n-b_n}{a_n}\stackrel{d}{\rightarrow}R.$

The role of $b_n$ is not hugely important, as a broadly stable distribution is in the domain of attraction of the corresponding strictly stable distribution.

The natural question to ask is: do the domains of attraction of stable distributions (for $0<\alpha\leq 2$) partition the space of probability distributions, or is some extra condition required?

Next time I will talk about stable distributions in a more analytic context, and in particular how a discussion of their properties is motivated by the construction of Levy processes.