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 $a_n \propto n^{-1/\alpha}$.

  • $\begingroup$ In order to get a more intuitive feel for this, I started to look for a derivation of that characteristic function myselve. But it seems not so easy. I have posted a question about it on math.stackexchange.com/questions/4309671 $\endgroup$ Nov 18, 2021 at 14:16

1 Answer 1


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.

computation compared with Levy distribution

### function to get a scaled sample mean 
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
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)

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