I am using scipy's ks_samp in order to apply the Kolmogorov-Smirnov-test.

The data I use is twofold:

  1. I have a dataset d1 which is an evaluation-metric applied on the forecast of a machine-learning model m1 (namely the MASE - Mean Average Scaled Error). These are around 6.000 data points meaning the MASE-result of 6.000 forecasts using m1.
  2. My second dataset d2 is analogous to d1 with the difference that I used a second model m2, which slightly differs from m1.

The distribution of both datasets looks like:


As can be seen, the distribution looks pretty much alike. I wanted to underline this fact with a Kolmogorov-Smirnov test. However, the results I get applying k2_samp indicate the contrary:

from scipy.stats import ks_2samp

k2_samp(d1, d2)

# Ks_2sampResult(statistic=0.04779414731236298, pvalue=3.8802872942682265e-10)

As I understand, such a pvalue indicates that the distribution is not alike (rejection of H0). But as can be seen on the images it definitely should.


  1. Am I misunderstand the usage of Kolmogorov-Smirnov and this test is not applicable for the use-case/kind of distribution?
  2. If first can be answered with yes, what alternative do I have?


Below is the overlay-graph. Concluding from your answers and comments I assume that the divergence in the "middle" might be the cause since KS is sensitive there.

  • 1
    $\begingroup$ With a sample size of 6000, it's not surprising that you're picking up really small differences between the empirical cdfs. $\endgroup$ – stats134711 Aug 25 '19 at 0:10
  • $\begingroup$ @stats134711: Thanks for the hint with the sample size! Any suggestions how to avoid this (e.g. another test)? $\endgroup$ – Markus Aug 25 '19 at 1:23
  • 1
    $\begingroup$ The K-S test is usually more sensitive to differences in the center of the distributions and less sensitive to differences at the tails. The test of comparing two distributions depends on what your goals are. You could consider using a different statistic/distance (e.g. Hellinger, Anderson-Darling, etc.). $\endgroup$ – stats134711 Aug 25 '19 at 1:31

A P-value below 0.05 would indicate that the two samples are from different distributions. Your P-value is smaller than 0.05, so you would reject the null hypothesis that the two samples are from the same distribution.

A difficulty with the Kolmogorov-Smirnov test, used with large sample sizes, is that small, unimportant differences between two samples are sometimes detected as 'significantly different'.

Here are two large samples of size $n = 4000$ generated from the same distribution in R:

x1 = rnorm(4000, 100, 15);  x2 = rnorm(4000, 100, 15)

A K-S test in R (correctly) does not find a difference between them:

ks.test(x1, x2)

        Two-sample Kolmogorov-Smirnov test

data:  x1 and x2
D = 0.0165, p-value = 0.6476
alternative hypothesis: two-sided

By contrast, here are two large samples from slightly different distributions, for which the K-S test (correctly) rejects the null hypothesis with a P-value smaller than 0.05.

y1 = rnorm(4000, 99, 15);  y2 = rnorm(4000, 100, 15)

        Two-sample Kolmogorov-Smirnov test

data:  y1 and y2
D = 0.03625, p-value = 0.01043
alternative hypothesis: two-sided

Addendum, as per Comment: Two ECDF plots separately look the same; with overlay, the slight difference may be visible.

 plot(ecdf(y1), col="blue"); plot(ecdf(y2), col="orange")
 plot(ecdf(y1), col="blue", main="Overlay")
  lines(ecdf(y2), col="orange")

enter image description here

enter image description here

In this specific example, the difference is is population means. Because data are normal, a two-sample t test 'finds' this difference:

t.test(y1, y2)

        Welch Two Sample t-test

data:  y1 and y2
t = -3.4689, df = 7997.9, p-value = 0.0005253
alternative hypothesis: 
  true difference in means is not equal to 0
95 percent confidence interval:
 -1.8404842 -0.5114389
sample estimates:
mean of x mean of y 
 98.39611  99.57208 
  • 1
    $\begingroup$ Thank you for your clarifications on difficulties with the sample size (+1). Anyhow, my p-value equals 0.00000000038803 which definetly is much smaller than 0.05 (which is my underlying "problem"). $\endgroup$ – Markus Aug 25 '19 at 1:22
  • 1
    $\begingroup$ OK, right. Misread that. Slightly rewritten. // Have you tried plotting empirical CDFs on top of each other, possibly to reveal slight mis-matches? See addendum. To make ECDF plot from scratch: sort data, jump of $1/n$ at each sorted value; approximates theoretical CDF. $\endgroup$ – BruceET Aug 25 '19 at 2:46
  • 1
    $\begingroup$ Thanks for the hint and further explanation! I added the overlay-graph to my question. Right in the middle is some divergence which might be the root cause. $\endgroup$ – Markus Aug 25 '19 at 11:37
  • $\begingroup$ Yes, that would do it. BTW, when showing graphs (these and others) it is always a good idea to label (or explain) the axes. $\endgroup$ – BruceET Aug 25 '19 at 17:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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