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I want to use the Kolmogorov-Smirnov test to test if a sample is drawn from a Pareto distribution. Unfortunately, the only way to estimate the distribution's parameters is from the sample.

Does anybody know of the existence of tables with modified critical values or any other way to perform the test in this case?

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A Kolmogorov-Smirnov test with estimated parameters is known as Liliiefors' test. The values of the test statistic tend to be smaller than with the KS test.

It is no longer nonparametric; you need to work out the distribution of the test statistic for each distribution type (and it differs again if you estimate a subset of the parameters rather than all of them).

Some packages offer Lilliefors test for the exponential distribution and the normal distribution (which are the cases Lilliefors discusses in his two papers).

Here's a way to use the exponential for the Pareto (I'll use the notation at that link):

Case 1: known $x_m$, unknown $\alpha$:

  • If $x_m$ isn't $1$, divide through by $x_m$, so you have a Pareto with $x_m=1$.

  • take logs, resulting in an exponential with (unknown) scale parameter $\alpha$.

  • apply Lilliefors test for an exponential r.v.

Case 2: unknown $x_m$, unknown $\alpha$. While this case can be done by simulation, we can make use of an exponential Lilliefors test as follows:

  • take logs, yielding a shifted exponential

  • subtract the minimum observation from all observations, and discard the minimum, leaving $n-1$ observations

  • test the reduced sample for exponentiality via the Lilliefors test as before.

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  • $\begingroup$ Thanks for your answer! Do you have a source for Case 2? $\endgroup$
    – Tom
    Commented Aug 12, 2014 at 7:39
  • $\begingroup$ In the meantime, I came across Coefficients of the asymptotic distribution of the kolmogorov-smirnov statistic when parameters are estimated. They propose the following transformation: x_i' = (n - i + 1) * ln[x_i / x_(i-1)] Would this be suitable as well? $\endgroup$
    – Tom
    Commented Aug 12, 2014 at 7:47
  • $\begingroup$ Hi Tom, I can't get the article at the moment, but I believe that the proposed transformation would be for consecutive order statistics (the 'gaps') (in which case your expression is missing a set of parentheses), and also yields an exponential. It should work just fine. Indeed, it may be slightly better (in that it would possibly give better power against more interesting alternatives) $\endgroup$
    – Glen_b
    Commented Aug 14, 2014 at 2:33

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