# Stable Distributions And Limit Theorems

1. Feb 21, 2014

### mikeyork

I'm an oldie and not well-versed in the modern formalism used in stochastic calculus, so please bear with me. I'm aware of Levy's characteristic function for stable distributions, though not well-versed in its practicalities.

I have read that for alpha=2 the stable distribution is Gaussian and also that the Gaussian is the only stable function with finite variance. However, I think I have found a pdf that is a rational function with finite variance and power law tails |x|^-4 and symmetric in x, for which an infinite recurrence of convolutions with itself appears to converge to another rational function with |x|^-4 tails. If so, then does this not imply convergence to a non-Gaussian stable distribution? And since the cumulative distribution will behave as |x|^-3 this implies alpha = 2 does it not?

On a similar note, I have read that the Central Limit Theorem implies that any distribution that is bounded converges to a Gaussian. Does bounded mean that all moments are finite? If we are always dealing with sampled data then of course all moments will be finite. But I have also read statements to the effect that any distribution with finite variance will converge to a Gaussian. But if we have a continuous distribution and make convolutions, as in my case, then it would seem reasonable that the limiting distribution need not be Gaussian if not all moments are finite!

Can anyone clear these issues up for me? Are cases like mine well known and understood? Is there any literature on them?

2. Feb 21, 2014

### Stephen Tashi

It says something about the distribution of the mean of independent samples from a distribution F converging, not that the distribution F itself converges to some other distribution.

3. Feb 21, 2014

### mikeyork

You refer to samples. Any finite number of samples will have all finite moments -- even if F does not. Now suppose we do repeated convolutions of F instead of taking samples. That's the situation I am asking about.

4. Feb 28, 2014

### bpet

You've encountered an example of a two-sided sub-exponential distribution. The one-sided variants are very commonly used in extreme value theory. You may well find a factor of n somewhere in the asymptotic nth convolution (though the two-sidedness makes this more complicated) so the "raw" limit won't exist but with appropriate shifting and scaling the limiting distribution will indeed be a stable Gaussian distribution.

5. Feb 28, 2014

### mikeyork

Thanks a lot for that. I had not encountered the term "sub-exponential" before and I definitely see the relevance.

The convergence I was intending to refer to (sorry I didn't make this clear) was not to the distribution of the sum which comes from a convolution of the distribution, but to the density of the mean which comes from the convolution of the density divided by n. (Since the convolution of the density with itself gives the density of the sum.) The density of the mean has the same variance as the original density (which, unlike the density of the sum, does not have the factor n you referred to). It is this density of the mean that seems to converge to a limiting density (without any further shifting or scaling) and every n I have computed behaves as |x|^-4, the same as the original density. I don't see how, then, it can converge to a Gaussian.

Since I could then, in principle, take that asymptotic density (if it converges) and apply the same convolution, I should surely obtain the same density, because of the convergence, should I not? Is that not the condition that implies a stable distribution?

Perhaps I am confused about the terminology "stable" and "limiting density". What is the proper term to use for the asymptotic (n-->infinity) density of the mean? Is it correct to call the result of asymptotic application of convolution a "stable distribution" if it converges? Can you help out again here with some more explanation of what is going on and/or clarification of terms?

6. Feb 28, 2014

### bpet

It won't (Law of Large Numbers applies). CLT includes a factor of sqrt(n).

HTH

7. Mar 4, 2014

### mikeyork

So am I correct that the density of the mean converges to a stable distribution? It seems to me it must since repeated convolution of asymptotic density divided by n will reproduce itself, won't it?

8. Mar 5, 2014

### bpet

No - remember LLN says the sample mean converges to a constant, so the limiting density doesn't exist. Also check the tail cdf, it should look something like n/|nx|^3.