I have some data which looks like this when I plot a normalized histogram.

enter image description here

The full data set is available here and here (the second link is pastebin). It is 20,000 lines long.

My guess is that it is a sample from a (generalized) gamma distribution but I have failed to show this.

I attempted in python to fit a generalized gamma distribution using


but it returns

(12.406504335648364, 0.89229023296508592, 9.0627571605857646, 0.51700107895010139)

and I am not sure what to make of it.

Overall, how can I work out what distribution my data is likely to be and how could I test it in R or preferably python?

Assuming my coding/understanding is not broken, it now seems unlikely this data is from a generalised gamma distribution.

  • I simulated 100 samples of size 20,000 using the parameters (12.406504335648364, 0.89229023296508592, 9.0627571605857646, 0.51700107895010139) given above.
  • I computed the Kolmogorov-Smirnov statistic for each using stats.kstest(simul_data, 'gengamma', args = (a,c,loc,scale)).statistic .
  • I found that the Kolmogorov-Smirnov statistic for the data is larger than all 100 from the simulation.
  • $\begingroup$ Any chance you could post the data without me having to download that linked file? $\endgroup$
    – epp
    Nov 3, 2016 at 12:20
  • $\begingroup$ @StatsPlease There are 20,000 lines so I am not sure how to post it otherwise. It's just a plain text file that I have linked to. If you know a better way to get it to you please let me know. $\endgroup$
    – graffe
    Nov 3, 2016 at 12:43
  • $\begingroup$ Can you use GitHub? $\endgroup$ Nov 3, 2016 at 12:56
  • $\begingroup$ @AntoniParellada I never have but I could try to learn. Does just downloading the text file linked not work for you? $\endgroup$
    – graffe
    Nov 3, 2016 at 13:19
  • $\begingroup$ It was just a suggestion. I personally don't like downloading material from websites I'm not familiar with. $\endgroup$ Nov 3, 2016 at 13:20

2 Answers 2


Some other graphical methods, maybe more helpful than the simple histogram. I will illustrate with graphics produced with the help of R. First a plot visualizing the data together with some possible theoretical distributions in a skewness-kurtosis space:

library(fitdistrplus)  # on CRAN 
dat  <-  read.table("gamma2.test", header=FALSE) # poster's data
gamma2  <-  dat$V1
descdist(gamma2, boot=1000)  # Cullen and Frey-plot

Example Cullen and Frey graph with poster's data

The orange values around the blue (data) point are based on bootstrapping. We can see that kurtosis are somewhat larger than what can be accommodated by gamma, weibull or lognormal distributions, while skewness is smaller.

We can make some more detailed comparisons:

gammafit  <-  fitdistrplus::fitdist(gamma2, "gamma")
weibullfit  <-  fitdistrplus::fitdist(gamma2, "weibull")
lnormfit  <-  fitdistrplus::fitdist(gamma2, "lnorm")  # not supported?

library(flexsurv) # on CRAN

gengammafit  <-  fitdistrplus::fitdist(gamma2, "gengamma",
                                       start=function(d) list(mu=mean(d),

then we can compare the fits with qqplots:

qqcomp(list(gammafit, weibullfit, lnormfit, gengammafit),
       legendtext=c("gamma", "lnorm", "weibull", "gengamma") )

example QQ plots with posters data

and we can see that indeed the generalized gamma seems to fit better, especially in the left (lower) tail. None of the distributions fit very well in the right (upper) tail, but the generalized gamma is best. For general help on qqplots, see How to interpret a QQ plot .

Another way of doing the comparison is a relative density plot, let us use the best fitting generalized gamma distribution as reference distribution. Then if the data exactly follows the reference distribution, the plot will be a uniform density. See What are good data visualization techniques to compare distributions? for details.

library(reldist) # on CRAN
N  <-  length(gamma2)
q_rel  <-  qgengamma(ppoints(N), mu=gengammafit$estimate["mu"],
reldist(gamma2, q_rel, main="Relative distribution comp to generalized gamma")

Example relative distribution plot with posters data

(please note the vertical (density) scale, all relative density values are quite close to 1, so the theoretical reference model is quite good)


Using MATLAB, I fit a few distributions to the data using MLE.

Basing fit on the Bayesian Information Criterion (BIC), the better fits I could find were:

$\text{Gamma}(\alpha=42.0827,\beta=0.4229)$, $\text{BIC}=96,829$

$\text{GEV}(k=-0.0614,\sigma=2.3768,\mu=16.5714)$, $\text{BIC}=96,137$

Obviously, you could fit other distributions to compare fits. Essentially, MLE is one way to guage the fit of a particular distribution to the data you have. You can compare the fits of the different distributions using Aikaike Information Criterion (AIC) or BIC (above). AIC penalizes fits by the number of fitted parameters whereas BIC's penalty is also related to the size of the sample.

The smaller the BIC (or AIC), the better.

For example, I'll provide the output from a normal distribution fit, which from the above histogram, is very unlikely to be the underlying distribution. This will highlight a poor fit, and the BIC that goes along with it.

$N(\mu=17.798,\sigma=2.8195)$, $\text{BIC}=98,238$

Other common parametric fits:

$\text{Weibull}(A=19.022,B=5.8626)$, $\text{BIC}=102,356$

$LN(\mu=2.8672,\sigma=0.1531)$, $\text{BIC}=96,399$

  • $\begingroup$ Thank you but what can we conclude from these numbers? $\endgroup$
    – graffe
    Nov 3, 2016 at 13:28
  • $\begingroup$ @Lembik That, based on your sample, the data conforms to a GEV better than a Gamma (using MLE). And that, given whatever distribution you select, the parameters of that distribution which maximum the likelihood function are those provided above. $\endgroup$
    – epp
    Nov 3, 2016 at 13:30
  • $\begingroup$ @Lembik So clearly, if you were to fit a parametric distribution to the data, the Gamma distribution wouldn't be a bad idea. Particularly if there are other reasons to believe the underlying distribution is Gamma. $\endgroup$
    – epp
    Nov 3, 2016 at 13:45
  • 1
    $\begingroup$ @Lembik Perhaps if you could clarify exactly what your problem is and what you are trying to achieve? $\endgroup$
    – epp
    Nov 3, 2016 at 13:53
  • 1
    $\begingroup$ This seems related. $\endgroup$ Nov 3, 2016 at 14:54

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