Brownian Motion with Optional Stopping Theorem (OST) Let $(B_t)_{t \geq 0}$ be a standard Brownian Motion and let $T:=\inf\{t \geq 0: B_t=at-b\}$ for some positive constant $a,b>0$. Calculate $\mathbb{E}[T]$.
How do i begin it?
 A: Set $X_t := B_t-at$, then
$$\tau:= \inf\{t \geq 0; B_t=at-b\} = \inf\{t \geq 0; X_t = -b\}.$$
If we already knew that $\mathbb{E}\tau<\infty$, then Wald's identity would imply
$$-b=\mathbb{E}(X_{\tau}) = \underbrace{\mathbb{E}(B_{\tau})}_{0}-a \mathbb{E}\tau,$$
i.e. $\mathbb{E}(\tau)=\frac{b}{a}$. Unfortunately, we do not know whether $\mathbb{E}\tau<\infty$, and therefore we have to use a different approach.

Define a sequence of stopping times by
$$\tau_n := \inf\{t \geq 0; X_t \notin [-b,n]\}.$$
Obviously, by definition, $$\tau_n \uparrow \tau \qquad \qquad |X_{t \wedge \tau_n}| \leq \max\{b,n\}. \tag{1}$$ Using Itô's formula, it is not difficult to see that $$M_t := \exp\big(2aX_t \big)$$ is a (local) martingale. By the optional stopping theorem, $(M_{t \wedge \tau_n})_{t \geq 0}$ is a martingale; hence, in particular
$$\mathbb{E}M_{t \wedge \tau_n} = \mathbb{E}\exp\big(2a X_{t \wedge \tau_n}\big) = 1.$$
From $(1)$ we see that we may apply the dominated convergence theorem and get
$$\mathbb{E}\exp\big(2aX_{\tau_n}\big)=1. \tag{2}$$
On the other hand,
$$\mathbb{E}\exp\big(2a X_{\tau_n}\big) = e^{-2ab} \cdot \mathbb{P}(X_{\tau_n}=-b)+e^{2an} \cdot \mathbb{P}(X_{\tau_n}=n). \tag{3}$$
Combing $(2)$ and $(3)$ we obtain
$$p_n := \mathbb{P}(X_{\tau_n}=-b) = 1- \mathbb{P}(X_{\tau_n}=n)= \frac{e^{-2an}-1}{e^{-2a(b+n)}-1}. \tag{4}$$
Recall that, by the optional stopping theorem, $(B_{t \wedge \tau_n})_{t \geq0}$ is a martingale; thus, $\mathbb{E}(B_{t \wedge \tau_n})=0$. This means that
$$\mathbb{E}(X_{t \wedge \tau_n}) = \mathbb{E}(B_{t \wedge \tau_n})-a \mathbb{E}(t \wedge \tau_n)= -a \cdot \mathbb{E}(t \wedge \tau_n).$$
Using the monotone convergence theorem and dominated convergence theorem, cf. $(1)$, yields
$$\mathbb{E}\tau_n = \sup_{t \geq 0} \mathbb{E}(t \wedge \tau_n) = - \frac{1}{a} \mathbb{E}(X_{\tau_n}) \stackrel{(4)}{=} - \frac{1}{a} \bigg(-b \cdot p_n + n \cdot (1-p_n)\bigg).$$
Now as $n (1-p_n) \to 0$, $p_n \to 1$ as $n \to \infty$ and $\tau_n \uparrow \tau$ we finally conclude
$$\mathbb{E}\tau = \sup_{n \in \mathbb{N}} \mathbb{E}\tau_n = \frac{b}{a}.$$
