The function $u(x)=\epsilon/|x|$ is <harmonic function>[harmonic] on $\{|x|>\epsilon\}$, equals one on the target sphere, and tends to zero at infinity. Optional stopping at the first hit of radius $\epsilon$ and the first exit from a ball of radius $R$ gives
$$
\mathbb P_x(\tau_\epsilon<\tau_R)
=\frac{|x|^{-1}-R^{-1}}{\epsilon^{-1}-R^{-1}}.
$$
Letting $R\to\infty$ and taking $|x|=1$ yields $\mathbb P_x(\tau_\epsilon<\infty)=\epsilon$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 201 / 5 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-201/5/a/solution