CLT and stable distributions

Question

I have a few questions about generalizations of the CLT and stable distributions. I'm trying to correct my understanding and make it precise. Please forgive my naivete, I am not a professional statistician :-)

If I take the sum of a large enough sequence of independent R.V.'s, do they always converge to a stable distribution? (I've heard about generalizations of the CLT, but I'm looking for more precision).

When working with real data, what would be a hint that I need to model with a stable distribution? Is it possible to perform max likelihood with stable distributions?

you mean a sum of a sequence of independent RVs? (see also http://en.wikipedia.org/wiki/Stable_distribution#A_generalized_central_limit_theorem ) — shabbychef, May 30 '12 at 17:17
I assume you also mean independent and identically distributed; otherwise, we can, for example, change the distribution for each element of the sequence in such a way that the sum doesn't converge to any distribution. — jbowman, May 30 '12 at 18:23
One practical complication with maximum likelihood for stable distributions is that a closed-form expression of the pdf is only known in a select handful of cases (normal, Cauchy and [Levy](http://en.wikipedia.org/wiki/L%C3%A9vy_distribution)). — cardinal, May 30 '12 at 20:03
In R you can calculate/approximate the MLE using the command `stableFit` from the package [fbasics](http://127.0.0.1:19048/library/fBasics/html/dist-DistributionFits.html) using McCulloch's quantile method or direct maximisation. [John Nolan](http://academic2.american.edu/~jpnolan) has also done a lot of work on this. — , May 30 '12 at 20:37
Stable distributions have an important property which is [infinite divisibility](http://en.wikipedia.org/wiki/Infinite_divisibility_%28probability%29). They are used, in particular, for modelling data (such as log-returns) using [Lévy Processes](http://en.wikipedia.org/wiki/L%C3%A9vy_process) which require infinite divisibility of the increments. Due to this, the stable family is an attractive option because it contains the normal distribution as a particular case as well as skewed and heavy tailed members. — , May 30 '12 at 20:50

score 1 · Answer 1 · edited May 30 '12 at 22:07

1

No there are distributions that do not satisfy the conditions to be in the domain of attraction of a stable law. Theorem 2(a) Feller "An Introduction to Probability Theory and Its Applications Volume II page 577: In order that a distribution F belongs to some domain of attraction it is necessary that the truncated moment function μ(x) varies regularly with an exponent $2-α$, $(0<α<=2)$. $μ(x) =∫y^2 dF(y)$ where the limits of integration are from $-x$ to $x$. $μ(x)$ varies regularly means $μ(x) ~ x^{2-α} L(x)$ where $L(x)$ is a slowly varying function and slowly varying means $L(tx)/L(t) → 1$ as $t→∞$.

edited May 30 '12 at 22:07

Macro

40,561
8
143
148

answered May 30 '12 at 17:18

Michael R. Chernick

39,640
28
74
143

1

Cauchy IS stable. http://en.wikipedia.org/wiki/Cauchy_distribution – PeterR May 30 '12 at 17:27
Which is it? Should we fix Wikipedia? (Wouldn't be the first time Wikipedia is wrong :-) – Frank May 30 '12 at 17:28
Sorry. I didn't really check that. I will fix the answer. My point is that the there are distributions that don't satisfy the tail conditions for the stasble lw. – Michael R. Chernick May 30 '12 at 17:35
Cauchys sum to Cauchy's. So they do converge to a limit distribution. I assume they satisfy the tail condition. I will look and try to find an example that is not stable and get back on this later. – Michael R. Chernick May 30 '12 at 17:43
@Michael I thought "stable" was defined in terms of linear combinations. I've read something about power laws for the tails, but I don't know if it is a condition for stable distributions. It's not immediately obvious that linear combinations implies power law in the tails. – Frank May 30 '12 at 17:44
2

Is the sum of two independently Bernoulli distributed random variables a Bernoulli random variable? Nope. So the original question needs to be rephrased as something about independent and suitably scaled random variables. – PeterR May 30 '12 at 19:36
That is right for most rvs satisfying the stable law including the special case of the normal distribution you must take normalized sums. – Michael R. Chernick May 30 '12 at 19:56
@Frank. I was not talking about the definition of a stable distribution but rather the tail condition for a distributionto be in the domain of attraction of a stable distribution with index 0< alpha<=2. – Michael R. Chernick May 30 '12 at 20:15
I tried to fix the formatting, Michael. Have a look to make sure I haven't changed anything incorrectly. – Macro May 30 '12 at 22:07

score 1 · Answer 2 · answered May 30 '12 at 18:22

1

If I take the sum of a sequence of independent R.V.'s, do they always converge to a stable distribution? (I've heard about generalizations of the CLT, but I'm looking for more precision).

I think we need more restrictions on this statement to say anything useful. You could have a sequence of independent random uniform RVs on the interval [0, i]. The convolution of any two RVs from this sequence certainly does not follow the same distribution and the asymptotic distribution of the sample mean converges to a non-stable distribution.

answered May 30 '12 at 18:22

AdamO

52,330
5
104
209

1

If we're *really* interested in (unnormalized) *sums* of independent random variables (on the same probability space), we can say something quite a bit stronger. It is a theorem that $\sum_{i=1}^\infty X_i$ converges in distribution *if and only if* it converges almost surely. Now, if they are iid, then the Kolmogorov three-series theorem implies that $X_i = 0$ almost surely and so the limit distribution is completely characterized. – cardinal May 30 '12 at 19:58

score 0 · Answer 3 · answered May 30 '12 at 19:37

0

As far as max likelihood, start with this: the Gaussian distribution is stable, so certainly it is possible in some circumstances.

answered May 30 '12 at 19:37

PeterR

1,712
1
16
13

CLT and stable distributions

3 Answers3

Linked