Up to what order do the moments of the Kolmogorov distribution exist? References would be appreciated.
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$\begingroup$ Have you tried to use the cumulative distribution function? $\endgroup$– Davide GiraudoMar 7, 2013 at 16:37
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$\begingroup$ @Davide Thanks. I know that I could try by integrating the density or by using $E[X^n]=\int n x^{n-1}P(X>x)$. However, since both approaches involve the integration of an alternating series, I was looking for a credible reference that justifies the steps. My google-foo has not lead me to a helpful reference yet. $\endgroup$– AskoliMar 7, 2013 at 17:40
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$\begingroup$ In addition, by wildly interchanging the series and the integration process, I came up to the conclusion that all the moments exist. But I would like to confirm this. $\endgroup$– AskoliMar 7, 2013 at 17:45
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1 Answer
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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.
answered Mar 7, 2013 at 18:31
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$\begingroup$ Many thanks. Very nice answer. Before accepting, can we say something about the existence of moments for $p<0$ using the same argument? $\endgroup$– AskoliMar 7, 2013 at 18:34
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$\begingroup$ We have for $p<0$ that $K^p\leqslant |B(1)|^p$ hence there are moments of order $p$ for $0-1<p<0$. $\endgroup$ Mar 7, 2013 at 18:56