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.
Best Answer
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.