I am trying to better understand the connection between the power law distribution and Zipf's distribution (law). There is a neat explanation in [1].

The article suggests that as we can derivate the power law function from Pareto's law, combined with the relationship between Pareto's law and Zipf's law, the power law parameter alpha is 1 + 1/b. From my understand, this would mean that we can directly determine Zipf's law's b parameter by simply having the power law alpha parameter. So e.g., an alpha of 2 would lead to a b of 1.

Is this true? So can I calculate alpha by e.g., using the methods by Clauset et al. [2] on my data and then directly determine the Zipf parameter b by the definition? This would allow me to use the exact methods by Clauset instead of the non-exact methods like fitting a straight line on a log-log plot. So I would also overcome the necessity of producing rankings etc.

[1] http://www.hpl.hp.com/research/idl/papers/ranking/ranking.html

[2] http://arxiv.org/abs/0706.1062


Zipf's law is generally understood to simply be a power-law distribution with integer values, that is, a probability distribution with the form

$p(x) \propto x^{-\alpha}$ for $x\geq x_{\min}>0$, $\alpha>1$ and $x\in \mathbb{N}_{>0}$

where $x_{\min}$ is the smallest value for which the power law holds, and is generally 1 for Zipf's Law (although not always; there is some ambiguity in the literature as to whether the term Zipf's Law is reserved for the $x_{\min}=1$ case or whether it can be used for $x_{\min}>1$).

But, power-law distributions have the special property that the complementary cumulative distribution function (ccdf) is also a power law form, $P(x) \propto x^{-\beta}$ but now where $\beta>0$ (and $\beta=\alpha-1$). This presents some ambiguity in interpreting what exactly people mean when they state that the estimated such-and-such a parameter for Zipf's Law. Do they mean $\alpha$ or $\beta$? It's important to be clear about which one you are stating. So long as you say whether the parameter you estimate is the pdf or cdf parameter, you should be fine.

Another small point: when people talk about Pareto distributions and data, they often talk about "rank-frequency" plots. These are the same thing as the ccdf (a point we discuss a little more in our SIAM Review paper that you link to), just with the axes reversed. That means you can easily transform an exponent someone has estimated from a rank-frequency plot (what Lada Adamic calls the Pareto form) to a regular pdf exponent by taking the reciprocal. But, people don't really distinguish between Zipf and Pareto laws like that. Both are just power-law distributions, so it's better to just talk about $\alpha$.

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  • $\begingroup$ Thanks Aaron, this makes the connections clearer. So do I get this right: I can describe both Zipf and Pareto distributions by first doing power law fitting and then just keep talking about $\alpha$? This should hold at least for Zipf as it's basically the same as you say, but maybe its not so easy for Pareto as we would need to fit it on the rank frequency plots? $\endgroup$ – fsociety Sep 7 '14 at 8:58
  • $\begingroup$ BTW, Newman states in "Power laws, Pareto distributions and Zipf’s law": " Since power-law cumulative distributions imply a powerlaw form for p(x), “Zipf’s law” and “Pareto distribution” are effectively synonymous with “power-law distribution”. (Zipf’s law and the Pareto distribution differ from one another in the way the cumulative distribution is plotted—Zipf made his plots with x on the horizontal axis and P(x) on the vertical one; Pareto did it the other way around. This causes much confusion in the literature, but the data depicted in the plots are of course identical.)" $\endgroup$ – fsociety Sep 7 '14 at 9:12
  • $\begingroup$ How you discuss it depends a little on your audience. For instance, the Pareto distribution is actually a whole family of distributions, and their parameterizations are slightly different from a regular power-law distribution. But, so long as you are clear what form you are talking about, you should be just fine. $\endgroup$ – aaronclauset Sep 8 '14 at 4:02

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