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The central limit theorems state roughly that under a certain set of properties of a sampling process, the distribution of a statistic from that sample will converge in distribution to the normal distribution.

As the canonical example, let me take the basic central limit theorem: If we take i.i.d. samples $X_1,X_2...$ from a distribution $f$, then the sample average will converfe in diatribution to the normal distribution if

  1. The mean of $f$ exists.

  2. The variance of $f$ exists.

My question is:

Assume that $f$ does not have a variance, and that the sample average of an i.i.d.sample from $f$ does not converge in distribution to the normal distribution.

Are there situations where this sample average nevertheless converges in distribution to some other (non-normal) distribution?

In other words, is there a "central limit theorem" for distributions that don't converge to the nornal distribution?

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    $\begingroup$ Relevant wikipedia article. $\endgroup$
    – user179309
    Commented Feb 10, 2018 at 16:21
  • $\begingroup$ @smith, very interesting. $\endgroup$
    – user56834
    Commented Feb 10, 2018 at 16:43
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    $\begingroup$ Closely related answers: stats.stackexchange.com/questions/29497 and stats.stackexchange.com/questions/8515. $\endgroup$
    – whuber
    Commented Feb 10, 2018 at 17:07
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    $\begingroup$ Your statement that the sample average will converge to "the normal distribution" is false if by "the normal distribution" you mean the standard normal distribution. The statement is continues to be false even if you say that by the normal distribution you mean a normal distribution with mean equal to $\mu$, the mean of $f$, and variance equal to the variance of $f$. The sample mean converges to the constant $\mu$, that is, a normal distribution with mean $\mu$ and variance $0$, a far cry from what you seem to think happens. $\endgroup$ Commented Feb 11, 2018 at 15:01
  • $\begingroup$ @DilipSarwate, do you think that that is what I think, or that I was describing the CLT informally, since a precise formulation is not relevant to the question anyway? $\endgroup$
    – user56834
    Commented Feb 11, 2018 at 15:11

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An answer is already found in the wikipedia link from the first comment by Smith.

Q: Are there situations where this sample average nevertheless converges in distribution to some other (non-normal) distribution?

A: Yes, iff a distribution is a stable distribution then it is a limit to sums of the type:

$$\zeta_n = \frac{\xi_1 + \xi_2 + \dots + \xi_n}{B_n} - A_n $$

with the $\xi$ independent and identically distributed random variables, $B_n>0$ and $\vert A_n\vert<\infty$


The type of distribution laws for $\xi$ that let the above sum converge to a stable distribution (the domain of attraction for that distribution) are described by a theorem in 'Limit distributions for sums of independent random variables' by Gnedenko and Kolmogorov (page 175 in the translated version 1954, link via google)

Theorem 2.* In order that the distribution function F(x) belong to the domain of attraction of a stable law with the characteristic exponent $\alpha$ ($0 \leq \alpha \leq 2$) it is necessary and sufficient that

1) $$\frac{F(-x)}{1-F(x)} \to \frac{c_1}{c_2} \qquad \text{as } k \to \infty$$

2) for every constant $k>0$

$$\frac{1 - F(x) + F(-x)}{1-F(xk) + F(-kx)} \to k^\alpha \qquad \text{as } k \to \infty$$

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