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

  • $\begingroup$ Have you tried to use the cumulative distribution function? $\endgroup$ – Davide Giraudo Mar 7 '13 at 16:37
  • $\begingroup$ @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. $\endgroup$ – Askoli Mar 7 '13 at 17:40
  • $\begingroup$ 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. $\endgroup$ – Askoli Mar 7 '13 at 17:45

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.

| cite | improve this answer | |
  • $\begingroup$ Many thanks. Very nice answer. Before accepting, can we say something about the existence of moments for $p<0$ using the same argument? $\endgroup$ – Askoli Mar 7 '13 at 18:34
  • $\begingroup$ We have for $p<0$ that $K^p\leqslant |B(1)|^p$ hence there are moments of order $p$ for $0-1<p<0$. $\endgroup$ – Davide Giraudo Mar 7 '13 at 18:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.