Deriving the limiting distribution of a sum of Pareto distributed variables For a series of independent and identical Pareto distributed variables $X_i$ with $\alpha > 2$, their sum $S_n = \sum_{i=1}^{n} X_i$ has a normal distribution as limiting distribution for $n\to \infty$
$$\frac{1}{\sqrt{n}\sigma_X} S_n -\mu_X \quad \sim \quad N(0,1)$$
But what is the situation when we have a shape parameter $\alpha \leq 2$? Are there some series of constants $a_n$ and $b_n$ such that the following scaled and translated sum approaches a distribution?
$$a_n S_n + b_n \quad \sim \quad ?$$

Currently I am thinking about trying to derive that it must be a stable distribution by using the characteristic function for $a_n S_n + b_n$ (for simplicity I set the scale parameter $x_m =1$).
$$\begin{array}{}
\varphi_{a_n S_n + b_n}(t) & =& e^{it\,b_n} \alpha^n (-it \, a_n)^{n\alpha} \Gamma(-\alpha, -it \, a_n)^n
\end{array}$$
For the $\alpha > 2$ case we would scale by $a_n = \sigma_X \,n^{-0.5}$, and for the $\alpha \leq 2$ case we will, I guess/suspect, need something like $a_n \propto n^{-1/\alpha}$.

Partial results

*

*Below is a trial by comparing the sum of Pareto variables (with $\alpha = 0.5$) with a Levy distribution. The shifting and scaling are done based on the median and interquartile range. The convergence is not very fast but it does seem to work.


### function to get a scaled sample mean 
require(actuar)
sum_sample = function(n,s) {
  a_n = n^(-1/s)
  b_n = 1*(0.5)^(-1/s)
  sum( rpareto1(n,s,1)-b_n ) * a_n
}


### get a sample
set.seed(1)
x <- replicate(10^4,sum_sample(10^5,0.5))

### scale the sample according to median and interquartile range
x = x-median(x)
iqr = diff(quantile(x, probs = c(0.25,0.75)))
x = x*(rmutil::qlevy(0.75)-rmutil::qlevy(0.25))/iqr

### histogram
hist(x, freq = 0, breaks = c(seq(-5,5, 0.1),max(x)),
     xlim = c(-4,4), main = "histogram of shifted and scaled sample means \n compared with levy distribution")
### levy curve
xs = seq(0.001,10,0.01)
lines(xs-rmutil::qlevy(0.5),rmutil::dlevy(xs),col=2)

