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given a population of discrete values (not a sample), is there any way to check whether the distribution is a power-law, and with what parameters? Of course I can just plot the distribution and see its shape, but that does not allow me extract the power-law parameters.

Any help on how to do that is highly appreciated.

leco
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  • Not a duplicate. The post you pointed out refers to sampled data, not the whole population data. – leco Oct 08 '14 at 18:11
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    If you have the whole distribution, why would you need to estimate the parameters? – abaumann Oct 08 '14 at 18:34
  • Because I want to prove that the distribution adheres to a power-law (a graph is not enough), as power-laws have specific properties that are important in my context; stated otherwise, I want to assert that a power-law is a plausible model for the population data. – leco Oct 09 '14 at 01:12
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    But why not just use the empirical distribution? Do you want to assert that the model might hold in other domains (in which case you still only have a sample from the "superpopulation")? – abaumann Oct 09 '14 at 14:47
  • Ok, I partially agree. I could still claim that the population snapshot is a sample from this continuously evolving population (this is in tune with what you said). However, how would I interpret the resulting p-value, as there is no randomization in the sample process? This is the point where I get lost :( – leco Oct 09 '14 at 22:59
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    The so-called superpopulation inference approach springs to mind, where a starting point might be this blog post by Gelman j.mp/cZ1WSI and this CV thread http://stats.stackexchange.com/questions/2628/statistical-inference-when-the-sample-is-the-population. However, if you read the paper by Clauset et al., you'll see that they apply their method to cases where they have "whole population" too, in the case of waiting times between murders, if I'm not mistaken :-) – abaumann Oct 10 '14 at 09:20
  • Thanks, that is very insightful, specially the link you have provided. You can put is as an answer; I will vote for it :) – leco Oct 10 '14 at 20:49
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    Can't do that, since the question is marked as a duplicate. But glad it helped! :-) – abaumann Oct 11 '14 at 10:19

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