I would like to test if they are distributed according to an exponential distribution. I can apply the Kolmogorov-Smirnov in python using scipy.stats.kstest(data, 'expon') .

However, I assume I have to normalize my data somehow first. Otherwise it is comparing it with an exponential distribution with unknown rate $\lambda$. Is this right and how should I do that?


3 Answers 3


You can standardize the exponential distribution easily enough multiplying variates by the rate parameter (it's a reciprocal scale parameter). But if you're estimating the rate parameter from the data, the Kolmogorov–Smirnov statistic doesn't have the same distribution as when the exponential distribution is completely specified.

See Lilliefors (1969), "On the Kolmogorov–Smirnov tests for the exponential distribution with mean parameters", JASA, 64, 325. And https://stats.stackexchange.com/a/392686/17230 for an intuitive explanation of the phenomenon in general.

You can compare the observed value of the KS test statistic calculated from the data to the tabulated critical values given in the reference. Or simulate the distribution of the statistic yourself as @Glen_b & @soakley have suggested. Note that Lilliefors points out its distribution doesn't depend on the true values of the parameters—generally true for scale & location parameters—, so for a given sample size you can do this once simulating from the standard exponential distribution, & keep the results for future reference; you don't need to repeat the simulation for each new data-set of the same sample size. And there's therefore no approximation involved (except that coming from simulation error). The difference made to the distribution of the KS statistic $D$ by estimating rather than pre-specifying the parameters is not trivial: Kernel-smoothed density estimate of D's distribution under null (n=100)

Lilliefors does give some asymptotic results (worked out rather crudely, but good enough for government work). Stephens has tabulated quantiles for the modified statistic

$$T(n) = \left(D - \frac{0.2}{n}\right)\left(\sqrt{n} + 0.26 + \frac{0.5}{\sqrt{n}}\right)$$

where $D$ is the KS test statistic & $n$ the sample size. According to Durbin (1975), "Kolmogorov–Smirnov tests when parameters are estimated with applications to tests of exponentiality and tests on spacings", Biometrika, 62, 1, these are very close to the exact values for larger sample sizes. They can be found in Pearson & Hartley (1972), Biometrika Tables for Statisticians, CUP, or in Stephens (1974), "EDF Statistics for goodness of fit and some comparisons", JASA, 69, 347. I'm not aware of any published correction to the p-value of the ordinary KS test to approximate that of the Lilliefors test; a power-law relationship seems like it might be useful: KS test p-values (n-100)

  • $\begingroup$ Thank you. As I commented to the other question, what should one do in my case? $\endgroup$
    – Simd
    Commented Aug 1, 2014 at 16:45
  • 1
    $\begingroup$ felix - one should (plainly) use a Lilliefors test; essentially, you use the KS test statistic, but its distribution is different. Some packages will give Lilliefors when you ask for a KS test with some distribution but don't specify a parameter value. Some have a separate Lilliefors function (R has nortest::lillie.test for example). In other cases you must simulate the distribution yourself (which is straightforward) to find the p-value. $\endgroup$
    – Glen_b
    Commented Aug 2, 2014 at 1:27
  • $\begingroup$ Thank you again. Reading around it seems that the Anderson-Darling test doesn't suffer from this problem. Would it be better just to use that? $\endgroup$
    – Simd
    Commented Aug 2, 2014 at 8:23
  • 1
    $\begingroup$ @Scortchi Thank you. Do you have a feeling for how far wrong the p-value for the exponential distribution will be assuming large sample sizes? I have no feeling for whether you could just get something completely wrong or if its a small correction term we are talking about. $\endgroup$
    – Simd
    Commented Aug 2, 2014 at 9:04
  • 2
    $\begingroup$ Not wrong at all if you use the correct procedure. If you were to use the KS test assuming known parameter values, the critical values of the test statistic tend to a constant multiple of the correct ones as sample size increases, e.g. for a 5% significance level it's 78% of what it should be. So doing it wrong is very conservative. (Lilliefors discusses this in the paper.) $\endgroup$ Commented Aug 2, 2014 at 9:32

You don't need to normalize, but can get the p-value for a goodness-of-fit test by simulation. Here is some sample R code, taken from Greg Snow's answer to a similar question (KS test - R, Minitab (and Systat)):

data <- c(7.2,10.5,10.67,0.15,3.92,3.28,0.89,2.29,13.82,0.43)

simp <- replicate(100000, {x <- rexp(length(data),rate=1/mean(data));
     ks.test(x,"pexp",rate=1/mean(x))$p.value} )

mean(simp <= ks.test(data,"pexp",1/mean(data))$p.value)

The method is described by Clauset et. al in a SIAM paper "Power-Law Distributions in Empirical Data."

  • $\begingroup$ Thank you. I am always grateful for code, even if I have to translate it now. $\endgroup$
    – Simd
    Commented Aug 2, 2014 at 9:05
  • 1
    $\begingroup$ (+1) Note you can use any rate for the simulated distributions so may as well skip rate=1/mean(data) in the 2nd line (not in the 3rd!). & can save simp for re-use with any data set having the same sample size. $\endgroup$ Commented Aug 2, 2014 at 11:19
  • $\begingroup$ Interesting point - is this true for other distributions besides the exponential? $\endgroup$
    – soakley
    Commented Aug 4, 2014 at 13:28
  • 1
    $\begingroup$ Any distribution where it's location and/or scale parameters that are being estimated (with some very general conditions on the estimation method). David & Johnson (1948), "The probability integral transformation when parameters are estimated from the sample", Biometrika, 35, pp182-190. So Lilliefors test for the normal distribution relies on this too; but for, say, a gamma distribution with estimated shape parameter you'd need to simulate from the best-fit gamma to get an approximate distribution for the KS statistic. $\endgroup$ Commented Aug 4, 2014 at 14:17

No, you don't need to normalise your data since the KS statistic is defined in terms of the raw data (actually in terms of the empirical distribution of these data):


I don't know Python, but in R you can conduct this test as follows:

x = rexp(100,1)

For this purpose, and by construction, you need to know the parameters of the distribution. You should not plug estimators in it, this breaks the convergence of the statistic and you have to use a different test (see the wikipedia article).

If you want to estimate the parameters and check whether the fitted model is good, then what you actually need is a goodness of fit test, for which you have a variety of options:


  • $\begingroup$ Thanks. I do exactly want to check whether the fitted model is good. I didn't know the KS test wasn't suitable for this. What should one do instead in my case where I am comparing to the exponential distribution? The wiki page is helpful but maybe there is a specific answer in the exponential case? $\endgroup$
    – Simd
    Commented Aug 1, 2014 at 16:54
  • 1
    $\begingroup$ (+1) But your last sentence seems to imply the KS test isn't a goodness-of-fit test, which it very much is. And of course it's still an option if you get the distribution of the test-statistic right when parameters are estimated. $\endgroup$ Commented Aug 2, 2014 at 9:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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