Show that $\mathbb{E}[\tau] = b^2$, where $\tau = \inf[t>0: |W_t| >b]$.

Question

Let $\tau = \inf[t > 0 : |W_t| > b]$. The author says that $\mathbb{E}[\tau] = b^2$, and I want to verify it. I am aware of $T_b$ having the same distribution as $b^2/Y$, where $T_b = \inf [t > 0: W_t >b]$, and $Y$ is a standard normal. My attempt is something like: $\mathbb{E}[\tau] = \int_0^\infty \mathbb{P}(\tau > t) dt$. Since $\tau$ is also equal to $\inf[ t > 0 : M_t > b]$, where $M_t = \max_{0 \le s \le t} |W_s|$, $\mathbb{P}(\tau >t) = \mathbb{P}(M_t < b)$. I hope $\mathbb{P}(M_t < b) = b^2 e^{-t}$, but not sure about this.

Answer 1

Here is a way you can proof it:
You want to apply Doob's stopping theorem to the martingale $X_t= W_t^2-t$. (You can easily check that it is a martingale).
Doob's gives us: $E[W_{\tau}^2-\tau]=E[W_0^2-0]=W_0^2-0=0$
Hence: \begin{align} &E[W_{\tau}^2]-E[\tau]=0 \\ \Leftrightarrow &E[W_{\tau}^2]=E[\tau] \end{align} Since $W_{\tau}^2=b^2$ we are almost there.
Unfortunately our martingale $X_t$ does not satisfy any of the 3 requirements of Doob's theorem. So we can not apply it here.
So you need to take the stopped martingale $X_{t\wedge \tau}$. Which again satisfies: \begin{align} E[W_{t\wedge \tau}^2- t \wedge \tau]=0 \end{align} Now we let $t\to \infty$ and we get : \begin{align} W_{t\wedge \tau}^2 \to W_{\tau}^2 \text{ a.s. } &\Rightarrow E[W_{t\wedge \tau}^2]\to E[W_{\tau}^2] \\ t\wedge \tau \to \tau \text{ a.s. } &\Rightarrow E[t\wedge \tau] \to E[\tau] \end{align} (The first implication holds, because of dominated convergence. The second, because of monoton convergence.)
Using this you get: \begin{align} 0=E[W_{t\wedge \tau}^2- t \wedge \tau]\to E[W_{\tau}^2-\tau] \end{align} And furthermore: \begin{align} E[W_{\tau}^2-\tau]&=0 \\ E[W_{\tau}^2]&=E[\tau] \\ b^2&=E[\tau] \end{align} Hope this wasn't too lenghty.