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rmatlabstatisticshypothesis-testkolmogorov-smirnov
rozsasarpi
rozsasarpi

Reputation: 1641

Poorly implemented two-sample Kolmogorov-Smirnov test (kstest2) in Matlab?

Am I missing something obvious or Matlab's kstest2 is giving very poor p-values? Under very poor I mean that I have the suspicion that it is even wrongly implemented.

Help page of kstest2 states that the function calculates asymptotic p-value, though I did not find any reference about which method is used exactly. Anyway, the description further states:

asymptotic p-value becomes very accurate for large sample sizes, and is believed to be reasonably accurate for sample sizes n1 and n2, such that (n1*n2)/(n1 + n2) ≥ 4

Example 1

Let's take Example 6 from Lehman and D'Abrera (1975):

sampleA         = [6.8, 3.1, 5.8, 4.5, 3.3, 4.7, 4.2, 4.9];
sampleB         = [4.4, 2.5, 2.8, 2.1, 6.6, 0.0, 4.8, 2.3];
[h,p,ks2stat]   = kstest2(sampleA, sampleB, 'Tail', 'unequal');

(n1*n2)/(n1 + n2) = 4 in this case so the p-value should be reasonably accurate.

Matlab yields p = 0.0497, while the solution given in the book is 0.0870. To validate the solution I used R, which I trust more than Matlab, especially in statistics.

Using ks.test from stats package and ks.boot from Matching package:

ks.test(sampleA, sampleB, alternative = "two.sided")
ks.boot(sampleA, sampleB, alternative = "two.sided")

Both give p = 0.0870.

Example 2

Lets use kstest2's own example to compare Matlab and R results for larger sample size:

rng(1);     % For reproducibility
x1 = wblrnd(1,1,1,50);
x2 = wblrnd(1.2,2,1,50);
[h,p,ks2stat] = kstest2(x1,x2);

This yields p = 0.0317. Now, using the same x1 and x2 vectors R gives p = 0.03968. About 20% difference when very accurate result is expected (n1*n2)/(n1 + n2) = 25.

Am I missing, messing up something? Is it possible that Matlab's kstest2 performs so poorly as the examples indicate? What approximation, algorithm kstest2 is using? (I can see the implemented code for kstest2, however a reference to book or paper would be much better to understand what is going on.)

I am using Matlab 2016a.

Lehman and D'Abrera (1975). Nonparametrics: Statistical Methods Based on Ranks. 1st edition. Springer.

Upvotes: 5

Views: 1947

Answers (1)

Hack-R
Hack-R

Reputation: 23210

I think that the correct test to compare with R's ks.test in MATLAB or Octave would be kolmogorov_smirnov_test_2:

sampleA         = [6.8, 3.1, 5.8, 4.5, 3.3, 4.7, 4.2, 4.9];
sampleB         = [4.4, 2.5, 2.8, 2.1, 6.6, 0.0, 4.8, 2.3];

kolmogorov_smirnov_test_2(sampleA, sampleB)

pval: 0.0878664

The difference appears to be the use of ks versus lambda, i.e.

ks   = sqrt (n) * d;
pval = 1 - kolmogorov_smirnov_cdf (ks);

versus

lambda =  max((sqrt(n) + 0.12 + 0.11/sqrt(n)) * d , 0);
pval = 1 - kolmogorov_smirnov_cdf (lambda);

I presume the different test statistics arise from the differences in the research papers cited by these 2 functions. If you want a deeper dive into the statistical theory you may want to reach out to CrossValidated.

Upvotes: 4

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