7

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?

user56834
  • 2,157
  • 13
  • 35
  • 5
    Relevant [wikipedia article](https://en.wikipedia.org/wiki/Stable_distribution#A_generalized_central_limit_theorem). –  Feb 10 '18 at 16:21
  • @smith, very interesting. – user56834 Feb 10 '18 at 16:43
  • 3
    Closely related answers: https://stats.stackexchange.com/questions/29497 and https://stats.stackexchange.com/questions/8515. – whuber Feb 10 '18 at 17:07
  • 2
    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. – Dilip Sarwate Feb 11 '18 at 15:01
  • @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? – user56834 Feb 11 '18 at 15:11
  • I have no idea what you think; I can only go by what you write. A precise formulation _is_ indeed relevant to the question, regardless of what you think; (I know what you have written about this issue). – Dilip Sarwate Feb 11 '18 at 15:20

1 Answers1

2

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 $k>0$

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

Sextus Empiricus
  • 43,080
  • 1
  • 72
  • 161