# Distribution of first exit time of Brownian motion

Let $$B_t$$ be standard one dimensional Brownian motion and $$\tau = \inf\{s : B_s \notin (a,b) \}$$ where $$a<0 are real numbers.

What is the distribution of $$\tau$$?

I know that for hitting times $$\tau_a = \inf \{s : B_s =a \}$$ the distribution can be calculated with the reflection principle. And clearly $$\tau = \tau_a \wedge \tau_b$$. So how can I continue?

• See this duplicate question. Another approach: The joint law of the running supremum $\sup_{s \leq t} B_s$ and infimum $\inf_{s \leq t} B_s$ is known and this allows you to write down some expression for the density of $\tau$.
– saz
Commented Feb 17, 2020 at 17:24
• @saz Ok, so I could write $P(\tau > t)=P(\sup_{s \leq t} B_s <b, \inf_{s \leq t} B_s >a)$. Is there any standard literature where I can find this joint law? Commented Feb 17, 2020 at 17:32
• E.g. in the book on Brownian motion by Schilling & Partzsch
– saz
Commented Feb 17, 2020 at 19:53

The Laplace transform of the density $$f$$ of $$\tau$$ is given by $$f^*(\lambda)\equiv \mathcal{L}\{f\}(\lambda)=\mathsf{E}e^{-\lambda \tau}=\frac{\cosh((b+a)\sqrt{\lambda/2})}{\cosh((b-a)\sqrt{\lambda/2})}, \quad\lambda>0$$ (see, for example, Exercise 7.5.3 in Durrett's book). Then one needs to calculate the inverse Laplace transform to get $$f$$. For $$t>0$$, $$f(t)=\mathcal{L}^{-1}\{f^*\}(t)=\sum_{k=-\infty}^{\infty}(-1)^k\frac{\varphi_k(a,b)}{\sqrt{2\pi}t^{3/2}}\exp\left\{-\frac{(\varphi_k(a,b))^2}{2t}\right\},$$ where $$\varphi_k(a,b):=\frac{b+a}{2}+\frac{b-a}{2}(1+2k).$$