The question is all in the title. The only background I can add, is that I am finding a powerlaw to interpolate beautifully between my data points. Only that my data appears in integers only, so I would better use the discrete equivalent of the powerlaw distribution - that is to say, a probability density function equal to $ax^{-b}$ with $a$ and $b$ positive, and $x > 1$ - to use it for a Monte Carlo simulation. From a quick googling it seems like a Zeta distribution might be what I need, but I am unsure.

  • 1
    $\begingroup$ Can you make it more precise what do you mean by powerlaw distribution? $\endgroup$ – Tim Jan 4 '17 at 10:45
  • $\begingroup$ Thanks Tim - just edited the question. Do not know yet how to use Latex in CrossValidated though $\endgroup$ – famargar Jan 4 '17 at 10:49
  • 1
    $\begingroup$ You haven't defined the domain of $x$ though, nor what $ax^{-b}$ represents (the pmf? the survivor function? something else?) so your definition is incomplete (and different authors use different definitions!). On using the subset of LaTeX implemented in MathJax, see here. $\endgroup$ – Glen_b Jan 4 '17 at 11:27
  • $\begingroup$ Thanks Glen, hope my last edit contains all relevant infos. $\endgroup$ – famargar Jan 4 '17 at 11:54

If you want a discrete distribution where $P(X=x)\propto x^{-\alpha}$ for $x=1,2,3,...$ then you would have a zeta distribution.

If you want a discrete distribution where $P(X=x)\propto x^{-\alpha}$ for $x=1,2,3,...,N$ then you would have a truncated zeta distribution, also known as Zipf's law.

On the other hand, if (as at least some authors do) you define a power law up to a slowly varying function -- i.e. where say

$P(X=x) = L(x)\cdot x^{-\alpha}\,,\quad x=1,2,3,...$


$P(X\gt x) = L(x)\cdot x^{-\alpha}\,,\quad x=1,2,3,...$

then neither of these is a zeta unless $L$ is the identity. Indeed, this definition encompasses an infinity of distributions


Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.