It is well known that The Kolmogorov distribution is the distribution of the random variable $$ {\displaystyle K=\sup _{t\in [0,1]}|B(t)|} $$ where B(t) is the Brownian bridge:

$$ B(t) = (W_t|W_1=0) $$

My question is the following - can we say anything about distribution of $K$ if $B(t)$ is defined as:

$$ B(t) = (W_t|W_1=\textbf{a}) $$


The most relevant thing I could find is the paper An explicit expression for the distribution of the supremum of brownian motion with a change point by Benzion Boukai. The relevant result is in Lemma 2 in Section 2.

(If you can't access that paper, then the technical report has the same result in Theorem 1.) I reproduce the result here.

Let $B_{t_0}^{(x,y)}(t) $ be the Brownian bridge with $B^{(x,y)}_{t_0}(0) = x$ and $B^{(x,y)}_{t_0}(t_0) = y$. Then

$$\Pr\left(\sup_{0 < t\leq t_0}B_{t_0}^{(x,y)}(t) > z \right) = \begin{cases} e^{-2(z-x)(z-y)/t_0} & z > \max(x,y) \\ 1 & \text{otherwise}\end{cases}\,.$$

In your case, since $x = y = a$,

$$\Pr\left(\sup_{0 < t\leq t_0}B_{t_0}^{(a,a)}(t) > z \right) = \begin{cases} e^{-2(z-a)^2/t_0} & z > a \\ 1 & \text{otherwise}\end{cases}\,.$$


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.