Let $B(t)$ be a standard Brownian motion and let $a<0<b$. Let $\tau=\inf\{t\geq0:B(t)\notin[a,b]\}$. I want to calculate its distribution.
I tried to use pde mothod to calculate its characteristic function and then its distribution:
Let $h(t,B(t))=E[e^{iu(\tau-t)}|B(t)]$, then by Ito's formula we get: $$ d(e^{iut}h(t,B(t)))=e^{iut}(h_tdt+h_xdB(t)+\frac{1}{2}h_{xx}dt+iuhdt) $$ Since $e^{iut}h(t,B(t))$ is a martingale and $h_t=0$, we can get a pde: $$ \begin{cases} \frac{1}{2}h_{xx}+iuh=0\\ h(t,a)=h(t,b)=1\\ \end{cases} $$ Then we get: $$ \varphi_\tau(u)=E[e^{iu\tau}]=h(0)=\frac{1+pq}{p+q} $$ where $$ p=e^{\sqrt{u}(1+i)a}, q=e^{\sqrt{u}(1+i)b} $$ then: $$ F_\tau(\beta)-F_\tau(\alpha)=\frac{1}{2\pi}\lim_{T\to+\infty}\int_{-T}^{T}\frac{e^{-iu\alpha}-e^{-iu\beta}}{iu}\frac{1+e^{\sqrt{u}(1+i)(a+b)}}{e^{\sqrt{u}(1+i)a}+e^{\sqrt{u}(1+i)b}}du $$ I am not familiar with complex variables and don't know how to calculate this integral. Thank you for any help!
And I also thought of calculating $E[\tau^n]$ by setting $g(t,x)=E[\tau^n|B(t)=x]$, but I have difficulty solving this pde: $$ \begin{cases} g_t+\frac{1}{2}g_{xx}=0\\ g(t,a)=g(t,b)=t^n\\ \end{cases} $$
And I also notice that we can calculate $E[\tau^n]$ by constructing martingales like $B^2(t)-t$ and $B^3(t)-3tB(t)$. How do we construct this kind of martingales then?
Thank you!