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


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_ns.

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


Extending this idea, we have


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


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


(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_is 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


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))<M\,\forall x,\,\text{for some }M<\infty.

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.


3 thoughts on “CLT and Stable Distributions

  1. Pingback: Gaussian tail bounds and a word of caution about CLT | Eventually Almost Everywhere

  2. Pingback: The Levy-Khintchine Formula | Eventually Almost Everywhere

  3. Pingback: Analytic vs Probabilistic Arguments for Supercritical BP | Eventually Almost Everywhere

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