# Kolmogorov distribution as the sup of a Brownian bridge

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:

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

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