What is the distribution of minimum of Brownian motion on arbitrary interval?

Question

We know that $P(\min_{0 \leq s\leq t} B_t \leq x)=2P(B_t\leq x)$. This can be found in any standard stochastic calculus textbook.

However I am curious about instead of the interval $[0,t]$ if we consider an arbitrary interval $[a,b]$. What will the distribution be then?

By law of total probability we have $P(\min_{a \leq s\leq b} B_t \leq x)=P(\min_{a \leq s\leq b} B_t \leq x|B(a)\geq x)$ then using Bayes' theorem we have that this is $P(\min_{a \leq s\leq b} B_t \leq x,B(a)\geq x)/P(B(a)\geq x)$. Then here I am stuck.

Is there another way?

Answer 1

Let $W_{a,b}:=\min_{a\le s\le b}B_s$. Then \begin{align} \mathsf{P}(W_{a,b}\le w)&=\mathsf{E}[\mathsf{P}(W_{a,b}-B_a\le w-B_a\mid B_a)]\\ &=\int_{\mathbb{R}} \mathsf{P}(W_{0,{b-a}}\le w-x)\phi(x;0,a)\,dx, \end{align} where $\phi(x;\mu,\sigma^2)=(2\pi\sigma^2)^{-1/2}\exp\{-(x-\mu)^2/2\sigma^2\}$.