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Aug 27, 2015 at 22:45 vote accept Yan Zhu
Aug 27, 2015 at 20:29 comment added P.Windridge A pedagogical comment: The Smirnov (von Mises - Cramer) statistic uses the integral of the squared empirical process, whereas the Kolmogorov-Smirnov statistic uses the maximum absolute value of the EP.
Aug 27, 2015 at 20:15 answer added P.Windridge timeline score: 1
Aug 27, 2015 at 20:06 comment added P.Windridge Here is an article that summarises the proof I suggested above = projecteuclid.org/euclid.aop/1020107767 (the pdf is open access, projecteuclid.org/download/pdf_1/euclid.aop/1020107767)
Aug 27, 2015 at 19:52 comment added P.Windridge This looks might it may contain hints-> epubs.siam.org/doi/abs/10.1137/1119082 but I don't have access. It makes the useful point that you only need to consider the uniform[0,1] distribution (assuming $F$ is continuous). I guess the easiest proof would proceed by writing the statistic in terms of the empirical process, using the fact that that converges to a Gaussian process, and then using something about the distribution of the latter. But this is just idle speculation :)
Aug 27, 2015 at 19:24 comment added P.Windridge Note the statistic appearing in the Smirnov Law is better known as the Cramer-von Mises statistic (or Cramer-von Mises-Smirnov).
Aug 17, 2015 at 9:42 history edited Davide Giraudo CC BY-SA 3.0
edited tags; edited title
Aug 17, 2015 at 3:26 answer added Zurab Silagadze timeline score: 3
Aug 17, 2015 at 3:01 history edited Yan Zhu CC BY-SA 3.0
some typos
Aug 15, 2015 at 22:24 history asked Yan Zhu CC BY-SA 3.0