I always thought, that Pareto distriubtion is continuous.

I found a paper, that states that $$P(X=c)=\frac{1}{c^{\alpha}} - \frac{1}{{c+1}^{\alpha}}$$

for $c=1,2,3,...$, where $X$ is a Pareto random variable.

Is that right?

Maybe they meant an other distribution?


1 Answer 1


This seems like a natural discretization of the continuous Pareto distribution. I agree it's not what people typically mean. $P(X=c)$ in that paper is simply the probability under the true Pareto distribution that $c\le X\le c+1$. The key characteristic of the Pareto distribution in many cases is that $P(X>c)=O(1/c^\alpha)$, which holds here.

  • $\begingroup$ Thank you very much. Is there any way to compute the convolution of n $X_i$, which are i.i.d,meaning: $P(X_1+...+X_n=j)$. Can I make any statement on the common distribution? $\endgroup$
    – Tim
    Oct 29, 2020 at 16:46
  • $\begingroup$ I don't think there's a closed form for the sum of Paretos. However, the Pareto distribution is subexponential, so you'll get that $P(X_1+X_2>j)\sim 2P(X_1>j)$, i.e. the ratio of those two objects is 1 as $j \rightarrow \infty$. $\endgroup$
    – k_moreno
    Oct 29, 2020 at 17:20
  • $\begingroup$ Thank you again $\endgroup$
    – Tim
    Oct 29, 2020 at 17:42

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