Expectation of Exit Time of Brownian Motion from Interval I am trying to solve the following:
Let $W_t$ be a brownian motion and $a,b >0.$  Define $\tau$ to be the exit time of $W_t$ from $[-a,b]$, that is $$\tau = \inf\{t \ge 0\ :\ W_{t}(\omega) \notin [-a,b]\}.$$  Show that $\tau$ is integrable and compute $\mathbb{E}(\tau)$.
I have seen this stated without proof multiple places that $\mathbb{E}(\tau)=ab.$  What is the logical argument that backs up this statement?  How are we guaranteed $\tau$ is integrable?
 A: Truncation is always a good idea to start with. Applying the optional stopping theorem to the bounded stopping time $\tau \wedge n$, we have
$$ \Bbb{E}[W_{\tau \wedge n}] = 0 \qquad \text{and} \qquad \Bbb{E}[W_{\tau \wedge n}^2 - (\tau \wedge n)] = 0. \tag{*} $$


*

*Even at this point we can infer that $\tau$ is integrable. Indeed, let $c = \max\{a,b\}$ and note that $|W_{\tau \wedge n}| \leq c$. Then applying the monotone convergence theorem to the inequality $\Bbb{E}[\tau\wedge n]\leq c^2$ shows that $\tau$ is integrable.

*Taking $n\to\infty$ to $\text{(*)}$ and applying both the bounded convergence theorem and the monotone convergence theorem gives
$$ \Bbb{E}[W_{\tau}] = 0, \qquad \Bbb{E}[W_{\tau}^2] = \Bbb{E}[\tau]. $$
Now the remaining computation is straightforward: Solving the system of equations
$$ \Bbb{P}(W_{\tau} = -a) + \Bbb{P}(W_{\tau} = b) = 1, \qquad (-a)\cdot\Bbb{P}(W_{\tau} = -a) + b\cdot\Bbb{P}(W_{\tau} = b) = 0 $$
gives $\Bbb{P}(W_{\tau} = -a) = \frac{b}{a+b}$ and $\Bbb{P}(W_{\tau} = b) = \frac{a}{a+b}$, and plugging this to $\Bbb{E}[W_{\tau}^2] = \Bbb{E}[\tau]$ gives
$$ \Bbb{E}[\tau] = a^2 \cdot \Bbb{P}(W_{\tau} = -a) + b^2 \cdot \Bbb{P}(W_{\tau} = b) = ab. $$
