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

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    $\begingroup$ 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 $\tau$. $\endgroup$
    – saz
    Commented Feb 17, 2020 at 17:24
  • $\begingroup$ @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? $\endgroup$
    – nabla
    Commented Feb 17, 2020 at 17:32
  • $\begingroup$ E.g. in the book on Brownian motion by Schilling & Partzsch $\endgroup$
    – saz
    Commented Feb 17, 2020 at 19:53

1 Answer 1

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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 $$ \varphi_k(a,b):=\frac{b+a}{2}+\frac{b-a}{2}(1+2k). $$

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  • $\begingroup$ Isn't the inverse Laplace transform usually a complex integral? (Mellin's inverse formula) How do you get the relatively neat looking sum? $\endgroup$ Commented Nov 23, 2020 at 15:10
  • $\begingroup$ @ThomasAhle I found this result on page 282 (18) in this book. $\endgroup$
    – user140541
    Commented Nov 23, 2020 at 21:19

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