 Solution

For real $\lambda$, the normal moment-generating formula gives $\mathbb{E}{e}^{\lambda (B_t-B_s)}=e^{{\lambda }^2(t-s)/2}$. Using independent increments and conditioning on the Brownian filtration,All terms are integrable by the same normal formula. Thus the displayed process is an Exponential martingale for Brownian motion.

For the interval exit time, we first show both almost-sure finiteness and finite expectation. On $\{T>n\}$, the position at time $n$ lies in $(-b,a)$. If $X_{n+1}-X_n\ge a+b$, the continuous path must exit by time $n+1$. This increment is independent of ${\mathcal{F}}_n$ and has law $N(\mu ,1)$, so the conditional probability of this event is a fixed $c>0$. ThereforeThis geometric tail bound from a uniform escape probability implies $T<\infty$ almost surely and $\mathbb{E}T<\infty$.

Take $\lambda =-2\mu$. Then $e^{-2\mu X_t}=e^{-2\mu B_t-2{\mu }^{2}t}$ is the exponential martingale just proved. Its stopped values lie between $e^{-2\mu a}$ and $e^{2\mu b}$. The optional stopping theorem at $T\wedge t$, followed by bounded convergence, gives $\mathbb{E}{e}^{-2\mu X_T}=1$. With $v=\mathbb{P}(X_T=a)$ and the two possible boundary values,Solving yieldsNext apply bounded-time optional stopping to $B_t$. Since $B_{T\wedge t}=X_{T\wedge t}-\mu (T\wedge t)$, we obtain $\mathbb{E}{X}_{T\wedge t}=\mu \mathbb{E}(T\wedge t)$. The stopped position is bounded, and the stopped time increases to an integrable $T$. Passing to the limit gives $\mu \mathbb{E}T=av-b(1-v)$. ThereforeAs a check, letting $\mu \downarrow 0$ gives the driftless formulas $v\to b/(a+b)$ and $\mathbb{E}T\to ab$. The expectation calculation uses a justified stopping limit, rather than assuming that optional stopping holds at every unbounded time.