Continuity gives $X_{\tau_a}=a$ on $\{\tau_a<\infty\}$. The stopped process $X^{\tau_a}$ is bounded by $a$ and is therefore a true martingale. Hence
$$
1=\mathbb E X_{t\wedge\tau_a}
=a\mathbb P(\tau_a\leq t)+\mathbb E[X_t\mathbf1_{\{\tau_a>t\}}].
$$
The second term tends to zero by bounded convergence because $X_t\to0$ and it is bounded by $a$. Thus
$$
\boxed{\mathbb P(\tau_a<\infty)=\mathbb P(\sup_{t\geq0}X_t>a)=\frac1a}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 202 / 4 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-202/4/a/solution