For every stopping time $\tau$, the martingale case of the <optional sampling theorem for a supermartingale> gives $\mathbb EX_\tau=X_0=0$. Hence
$$
\mathbb E(Z_\tau+X_\tau)=\mathbb EZ_\tau.
$$
Use the optimal stopping time from part c and the pathwise inequality
$$
\max_{0\leq t\leq T}(Z_t+X_t)
\geq Z_{\tau^*}+X_{\tau^*}.
$$
Taking expectations gives
$$
\mathbb E\max_{0\leq t\leq T}(Z_t+X_t)
\geq\mathbb EZ_{\tau^*}=U_0.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 211 / 4 / d / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-211/4/d/solution