Up to what order do the moments of the Kolmogorov distribution exist? References would be appreciated.

  • $\begingroup$ The question was answered in Math Overflow. See the following link for more details mathoverflow.net/questions/123875/… $\endgroup$
    – user21692
    Mar 7, 2013 at 19:03

2 Answers 2


Looks like this paper could help you: Evaluating Kolmogorov's Distribution

Check the 3rd header, Limiting Forms, for a mention of how the moments are found.

Not only do all moments exist, but they are all simply expressible in analytic form:

$\left<x^m\right> = \frac{\Gamma(m/2 + 1) \, \eta(m)}{2^{m/2 - 1}} $

You can obtain this formula by using the form of the series definition containing factors $e^{-2 k^2 x^2}$ and integrating term-by-term.

(Sorry to have to add this information via edit, but "protecting" this question from answers by people who have not used the site previously prevented me from putting this in a separate answer.)

  • $\begingroup$ Thanks for your answer. I saw this paper but it seems like this applies only to the first and second moment. I was wondering if other moments exist as well. $\endgroup$
    – Askoli
    Mar 6, 2013 at 23:44
  • $\begingroup$ Couldn't find anything, have you tried finding the third moment? The author of that paper mentions the first two moments being easily-integrable, hopefully the third (or higher-order) one is as well. $\endgroup$
    – RS18
    Mar 6, 2013 at 23:58
  • $\begingroup$ I am thinking that my question is actually trivial. Since the Kolmogorov distribution has support on $(0,1)$, then $\int_0^1 x^n k(x) dx \leq \int_0^1 k(x) dx = 1$. Therefore, all the moments should exist (existence is my main interest). Marsaglia just provides closed expressions for this. Thanks for your help. I will accept this answer since this seems to be a relevant reference. $\endgroup$
    – Askoli
    Mar 7, 2013 at 0:05
  • 2
    $\begingroup$ @Askoli: No, the Kolmogorov distribution does not have support on $(0,1)$. Its support is unbounded. $\endgroup$
    – cardinal
    Mar 7, 2013 at 1:41
  • $\begingroup$ @cardinal Indeed, I realised that later from the definition but I couldn't comment on my question anymore. So, yes, the support is unbounded. Thank you for clarifying this. P.S. I haven't found a good reference. $\endgroup$
    – user21663
    Mar 7, 2013 at 9:58

The Kolmogorov distribution is defined by the distribution of the random variable $K:=\sup_{0\leqslant t\leqslant 1}|B(t)|$, where $B(t)$ is the Brownian Bridge.

The problem of existence of moments for $K$ is actually the same as the study of moments of $K':=\sup_{0\leqslant t\leqslant 1}|W(t)|$, where $W(t)$ is a standard Brownian motion. An application of Doob's (sub)martingale inequality gives that for all $C>0$, $$P(K'\geqslant C)\leqslant \exp\left(-\frac{C^2}2\right).$$

Using the fact that for a non-negative random variable $X$ and $p>1$, we have $$E(X^p)=\int_0^{+\infty}pt^{p-1}P(X\geqslant t)dt,$$ we conclude that Kolmogorov distribution admits moments of any order.

As for $p<0$, we have $K^p\leqslant B(1)^p$, we can say that $K$ admits moments of order $p>-1$ for all $p$.


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