# Kolmogorov–Smirnov statistic/distance is normally distributed?

Suppose we have two populations X and Y, then compute the KS statistic/distance for samples drawn from them. Is this KS statistic/distance normally distributed? If so, how to prove it? If not, what distribution should it be? I know samples means $$\bar{X}$$ and $$\bar{Y}$$ are normally distributed according to the central limit theory.

• you can't prove this because it's not true Feb 6, 2022 at 18:05
• The Kolmogorov distribution describes the relevant asymptotics when $X$ and $Y$ have identical continuous distributions. It clearly is non-Normal.
– whuber
Feb 6, 2022 at 18:07
• As a fairly general principle, distributions involving maxima won't be normal, nor even asymptotically normal. You typically need something more-or-less "mean-like". Feb 6, 2022 at 23:02

there's no such result regarding K-S statistics applied to different populations. generally, it may not even be possible to calculate K-S stat. consider $$X\sim\mathcal N$$ and $$Y\sim\chi^2$$ - these two have different domains. how do you suppose to obtain K-S stat for them?
• Not sure I understand the point you are making: ys both $\chi^2$ and $\mathcal{N}$ distributions have different supports (unless you just want to think of $P(\chi^2<t) = 0$ for all $t<0$, but the K-S statistic is built around the ordering of observations in each eCDF: in such a case as you describe the K-S statistic can be calculated, and the K-S null would almost certainly be correctly rejected, yes? Or am I misunderstanding? Feb 6, 2022 at 20:12