# Limit of sums of iid random variables which are not square-integrable

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.

#### Solutions Collecting From Web of "Limit of sums of iid random variables which are not square-integrable"

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.