Up to what order do the moments of the Kolmogorov distribution exist? References would be appreciated.
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$\begingroup$ The question was answered in Math Overflow. See the following link for more details mathoverflow.net/questions/123875/… $\endgroup$– user21692Commented Mar 7, 2013 at 19:03
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2 Answers
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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.)
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$\begingroup$ 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. $\endgroup$– AskoliCommented Mar 6, 2013 at 23:44
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$\begingroup$ 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. $\endgroup$– RS18Commented Mar 6, 2013 at 23:58
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$\begingroup$ 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. $\endgroup$– AskoliCommented Mar 7, 2013 at 0:05
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2$\begingroup$ @Askoli: No, the Kolmogorov distribution does not have support on $(0,1)$. Its support is unbounded. $\endgroup$– cardinalCommented Mar 7, 2013 at 1:41
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$\begingroup$ @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. $\endgroup$– user21663Commented Mar 7, 2013 at 9:58
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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 $p$.
answered Mar 7, 2013 at 19:17
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$\begingroup$ The first inequallity is also studied in The Tight Constant in the Dvoretzky-Kiefer-Wolfowitz Inequality. $\endgroup$– MacallanCommented Apr 12, 2013 at 10:44