Expected value of first hitting time

Question

Let $(W_t)_{t\geq0}$ be a standard Wiener process. For each $a>0$ and $b>0$, let $\tau_{a,b}:=\inf\{t\geq0:W_t \notin (-a,b)\}$, and $\inf\emptyset:=\infty.$ We would like to calculate $\mathbb{E}\left(\text{e}^{-\lambda\tau_{a,b}}\right)$, where $\lambda>0$, but unfortunately we got stuck. :(

Answer 1

You may use the hint given in Exercise 7.5.3 in Durrett's book (page 379). The answer is $$ \mathsf{E}e^{-\lambda \tau_{a,b}}=\frac{\cosh((b-a)\sqrt{\lambda/2})}{\cosh((b+a)\sqrt{\lambda/2})}. $$

Answer 2

We know that $(W_{t\wedge\tau_{a,b}})_{t\geq0}$ and $(e^{\theta W_{t\wedge\tau_{a,b}}-\frac{\theta^2}2t\wedge\tau_{a,b}})_{t\geq0}$ are bounded martingales.

Therefore $$\mathbb E[e^{\theta W_{\tau_{a,b}}-\frac{\theta^2}2\tau_{a,b}}] = 1.$$ Therefore $$1 = \mathbb E[e^{\theta W_{\tau_{a,b}}-\frac{\theta^2}2\tau_{a,b}}] = e^{-a\theta}\mathbb E[\chi_{\{W_{\tau_{a,b}} = -a\}}e^{-\frac{\theta^2}2\tau_{a,b}}]+e^{b\theta}\mathbb E[\chi_{\{W_{\tau_{a,b}} = b\}}e^{-\frac{\theta^2}2\tau_{a,b}}] $$ and ($\theta \mapsto -\theta$) $$1 = \mathbb E[e^{-\theta W_{\tau_{a,b}}-\frac{\theta^2}2\tau_{a,b}}] = e^{a\theta}\mathbb E[\chi_{\{W_{\tau_{a,b}} = -a\}}e^{-\frac{\theta^2}2\tau_{a,b}}]+e^{-b\theta}\mathbb E[\chi_{\{W_{\tau_{a,b}} = b\}}e^{-\frac{\theta^2}2\tau_{a,b}}] $$ Thus $$\mathbb E[\chi_{\{W_{\tau_{a,b}} = -a\}}e^{-\frac{\theta^2}2\tau_{a,b}}] = \frac{e^{(2b+a)\theta}-e^{a\theta}}{e^{2(b+a)\theta}-1}$$ and $$\mathbb E[\chi_{\{W_{\tau_{a,b}} = b\}}e^{-\frac{\theta^2}2\tau_{a,b}}] = \frac{e^{(2a+b)\theta}-e^{b\theta}}{e^{2(b+a)\theta}-1}.$$ Adding, we obtain that $$ \mathbb E[e^{-\frac{\theta^2}2\tau_{a,b}}]= \frac{(e^{a\theta}+e^{b\theta})(e^{(a+b)\theta}-1)}{e^{2(a+b)\theta}-1} = \frac{e^{a\theta}+e^{b\theta}}{e^{(a+b)\theta}+1}$$