Let $W(t)$ be a Standard Brownian motion. Let $\tau$ be the first time that $W(t)$ hits either level "$a$" or level "$-b$". What is the most straightforward way to compute $\mathbb{E}[\tau]$?
I am able to show the probability that $W(t)$ hits level "$a$" before "$-b$" and vice versa, but I am unable to easily compute the expectation of the stopping time $\tau$.
To show the probability that $W(t)$ hits "$a$" before "$b$", I assume $\mathbb{E[\tau]}\leq \infty$, so that by Doob's optional stopping theorem, $\mathbb{E}[W_{\tau}]=W(0)=0$ (i.e. stopped process is a martingale). Then:
$$ 0=\mathbb{E}[W_{\tau}]=a*\mathbb{P}(W_{\tau}=a)+ (-b)\mathbb{P}(W_{\tau}=-b) $$
By definition of $\tau$, we have that $\mathbb{P}(W_{\tau}=a)+\mathbb{P}(W_{\tau}=-b)=1$, so that:
$$ a*\mathbb{P}(W_{\tau}=a)+ (-b)\mathbb{P}(W_{\tau}=-b) = \\ = a*\mathbb{P}(W_{\tau}=a)-b(1-\mathbb{P}(W_{\tau}=a)$$
Solving for $\mathbb{P}(W_{\tau}=a)$ gives: $\mathbb{P}(W_{\tau}=a)=\frac{b}{a+b}$
Question 1: how could I easily show that $\mathbb{E}[\tau]\leq \infty$, so that I can verify that I can indeed use Doob's optional stopping theorem?
Question 2: how can I compute $\mathbb{E}[\tau]$ in the simpliest possible way?
Best Answer
You probably know (and if not, could easily check) that the process $X_{t}=B_{t}^{2}-t$ is a martingale.
Now consider, for $n \in \mathbb{N}$, the (bounded) stopping times $$T_{n}=T \wedge n$$
Apply the Optional Stopping Theorem on $T_{n}$ noting that $B_{T_{n}} \le \max(a,b)$ and $T_{n} \le n$
Use the monotone convergence theorem to get $$E[T]=\lim_{n\rightarrow \infty}E[T_{n}]= \lim_{n\rightarrow \infty} E[B_{T_{n}}^{2}] \le \max(a^2,b^2)< \infty$$
Now use dominated convergence to conclude $$\lim_{n\rightarrow \infty} E[B_{T_{n}}^{2}] = E[B_{T}^{2}] = a^2 P(B_{T}=a) + b^2 P(B_{T}=b)$$
which you know already.
This gives you $E[T]$.