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I need a goodness of fit test for the exponential distribution. I understand that Kolmogorov-Smirnov is not generally regarded as very powerful and that Anderson-Darling is regarded as superior. However I have two problems.

  • None of the literature I can find specifically talks about goodness of fit for exponential distributions. Anderson-Darling is almost always discussed wrt the normal distribution.
  • Is there an R (or python) package to do an Anderson-Darling goodness of fit for the exponential distribution? I can find them for other distributions.
  • Is there a better test (which exists in R or python)?
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  • $\begingroup$ Is this a completeley specified distribution or are we estimating parameters? $\endgroup$ – Glen_b Aug 15 '14 at 8:08
  • $\begingroup$ @Glen_b In some case I will have to estimate the parameters which I know causes another problem. A method which had a correction factor for that case would be even better of course! $\endgroup$ – felix Aug 15 '14 at 8:28
  • $\begingroup$ When you say A-D's more powerful than K-S, what kind of departures from exponentiality are you thinking of? $\endgroup$ – Scortchi Aug 15 '14 at 9:58
  • $\begingroup$ @Scortchi I am thinking for example if the tail is too long. A test just for that case would be great. $\endgroup$ – felix Aug 15 '14 at 18:17
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    $\begingroup$ I'd guess the LRT for Weibull shape parameter less than one would have good power against long tailed alternatives. $\endgroup$ – Scortchi Aug 18 '14 at 9:54
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The same considerations apply as to the distribution of the Kolmogorov–Smirnov test statistic discussed here. The Anderson–Darling test statistic (for a given sample size) has a distribution that (1) doesn't depend on the null-hypothesis distribution when all parameters are known, & (2) depends only on the functional form of the null-hypothesis distribution when location & scale parameters are estimated. I don't know of an R implementation of the A–D test specifically for the exponential distribution with estimated rate parameter, but you could quickly make a function to calculate the test statistic by adapting the ad.test function from the nortest package: change the distribution function from the best-fit normal, pnorm((x - mean(x))/sd(x)), to the best-fit exponential,pexp(x/mean(x)). Then get critical values for any desired significance level & sample size by simulation.

As to the "best" test, note that different tests are more powerful against different kinds of departure from the null-hypothesis distribution. If you have a quite specific alternative in mind, e.g. a Weibull distribution with shape parameter greater than one, a likelihood ratio test will be more powerful than a general-purpose goodness-of-fit test. For more vaguely specified alternatives it might be helpful to compare the power of various tests against a rogues gallery, following the approach of Stephens (1974), "EDF statistics for goodness of fit and some comparisons", JASA, 69, 347.

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  • $\begingroup$ Thank you very much. I couldn't even find an implementation of the A-D test for the exponential distribution with known rate parameter. I am sorry if this is obvious but are you saying nortest can be used in that case somehow? $\endgroup$ – felix Aug 15 '14 at 18:16
  • $\begingroup$ I'm saying that rather than write a function to calculate the A-D statistic from scratch, you can modify the code used in ad.test. (Type ad.test to see it.) $\endgroup$ – Scortchi Aug 18 '14 at 9:52
  • $\begingroup$ @Scortchi, how did you know to change pnorm((x - mean(x))/sd(x)) for a normal distribution to pexp(x/mean(x)) for an exponential distribution? I am looking to test weibull and lognormal distributions thorough the AD statistic $\endgroup$ – lukeg Oct 11 '15 at 17:36
  • $\begingroup$ @lukeg: I just substituted a standardized exponential variate for a standardized normal variate, & the exponential for the normal distribution function. NB (1) If you're estimating the shape of a Weibull distribution as well as the scale, you need to simulate the distribution of the test statistic under your estimated parameters, & results are approximate; (2) the logarithm of a log-normal random variate has a normal distribution. $\endgroup$ – Scortchi Oct 12 '15 at 9:41

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