kstest for uniform distribution

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John Smith
John Smith on 12 Oct 2021
Commented: John Smith on 14 Oct 2021
Hi,
I have been testing the use of kstest for the detection of a discrete uniform distribution. However, I believe that I am encountering an error, or using the function incorrectly. For example, if I define a variable array
x = randi([1 4],1900,1);
where tabluate gives the following very uniform distribution
tablate(x)
Value Count Percent
1 449 23.63%
2 482 25.37%
3 482 25.37%
4 487 25.63%
and I then run the test
kstest(x, 'CDF', [x unidcdf(x,4)])
I get a result of h = 1, i.e. rejection of the hypothesis that x is discrete uniform, which is clearly not the case (at least in my eyes). Would someone with more experience with this test potentially be able to helpfully provide an explanation as to why I'm getting this result? And whether I'm doing something wrong?
Many thanks.

Accepted Answer

Jeff Miller
Jeff Miller on 12 Oct 2021
you can test for the fit of a discrete distribution, including uniform, with chi2gof. One of the examples (about 1/3 of the way down the page) shows how to test for a Poisson distribution. With the uniform discrete, your expected counts expCounts are just the total number of observations divided by the number of possible discrete values.
  3 Comments
Jeff Miller
Jeff Miller on 14 Oct 2021
Just a minor correction of the terminology at the end: the null hypothesis is that X is a discrete uniform, and h=0 means that this null hypothesis should not be rejected based on the observed X values.
John Smith
John Smith on 14 Oct 2021
Thank you, I appreciate the correction, and have edited the reply to avoid anyone reading this making the same mistake.

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More Answers (1)

the cyclist
the cyclist on 12 Oct 2021
From the documentation: "The one-sample Kolmogorov-Smirnov test is only valid for continuous cumulative distribution functions." (Emphasis added.)
  1 Comment
John Smith
John Smith on 13 Oct 2021
Thank you, I missed that part on the first read. I suspect Jeff Miller's answer is the way to go then, as chi square will allow for a test of discrete distributions.

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