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The Central Limit Theorem tells us that for an iid sequence of random variables $(X_n)_{n\geq 0}$ of finite variance $\sigma^2$ and zero mean

$$\lim_{n\to\infty}\frac{S_n}{\sqrt{n}}=^d N(0,\sigma^2)$$

where $S_n=X_1+\cdots+X_n$.

Suppose we have a similar sequence, except now we suppose that $X_n$ has infinite variance. Then is it possible for the sequence $\frac{S_n}{\sqrt{n}}$ to converge in distribution? Is there always some $\alpha$ such that $n^\alpha S_n$ converges to a non-constant distribution?

(It seems to me that the answer to the first question should be no, but I'm having trouble showing this.)

Thank you.

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  • $\begingroup$ There is the Cauchy distribution for instance. But there are many other distributions with infinite variance. You'll need to specify more to have a definite limiting distribution. $\endgroup$ – Raskolnikov Oct 5 '11 at 17:37
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    $\begingroup$ Suppose the random variables have symmetric density $f(x)$ and $f(x)$ is asymptotic to $|x|^{-(1+\alpha)}$ as $|x| \rightarrow \infty$. If $\alpha >2$, then the central limit theorem applies. If $0 < \alpha <2$, then $\frac{S_n}{n^{1/\alpha}}$ converges in distribution to a symmetric $\alpha$-stable distribution. See en.wikipedia.org/wiki/Stable_distribution $\endgroup$ – ShawnD Oct 5 '11 at 18:02
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The generalized central limit theorem states (see this for a summary), that if $X_i$ are i.i.d. such that its density function has left tail power-law asymptotic $\mathbb{P}(X < -x) \sim d x^{-\mu}$ and right tail asymptotic $\mathbb{P}(X > x) \sim 1- c x^{-\mu}$ as $x \to +\infty$, then there exist sequences $a_n$ and $b_n$ such that the random variate $Z_n = ((\sum_{i=1}^n X_i) - a_n )/b_n$ converges in probability to a stable distribution with stability index $\alpha = \min(\mu, 2)$ and asymmetry parameter $\beta = \frac{c-d}{c+d}$.

Details on the constructive choice of sequences $a_n$ and $b_n$ are given in the table found at the link above. Also see page 62 of Zolotarev and Uchaikin on Google books.

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