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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?

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    $\begingroup$ you mean a sum of a sequence of independent RVs? (see also en.wikipedia.org/wiki/… ) $\endgroup$
    – shabbychef
    Commented May 30, 2012 at 17:17
  • $\begingroup$ 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. $\endgroup$
    – jbowman
    Commented May 30, 2012 at 18:23
  • $\begingroup$ Well, how much can we relax "identically distributed"? $\endgroup$
    – Frank
    Commented May 30, 2012 at 18:31
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    $\begingroup$ 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). $\endgroup$
    – cardinal
    Commented May 30, 2012 at 20:03
  • $\begingroup$ @cardinal I see, good point. Can you point to papers? $\endgroup$
    – Frank
    Commented May 30, 2012 at 20:26

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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.

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    $\begingroup$ 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. $\endgroup$
    – cardinal
    Commented May 30, 2012 at 19:58
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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 $t→∞$.

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    $\begingroup$ Cauchy IS stable. en.wikipedia.org/wiki/Cauchy_distribution $\endgroup$
    – PeterR
    Commented May 30, 2012 at 17:27
  • $\begingroup$ Which is it? Should we fix Wikipedia? (Wouldn't be the first time Wikipedia is wrong :-) $\endgroup$
    – Frank
    Commented May 30, 2012 at 17:28
  • $\begingroup$ 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. $\endgroup$
    – Michael R. Chernick
    Commented May 30, 2012 at 17:35
  • $\begingroup$ 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. $\endgroup$
    – Michael R. Chernick
    Commented May 30, 2012 at 17:43
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    $\begingroup$ 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. $\endgroup$
    – PeterR
    Commented May 30, 2012 at 19:36
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As far as max likelihood, start with this: the Gaussian distribution is stable, so certainly it is possible in some circumstances.

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