Take the first entry into the stopping region:
$$
\tau^*=\inf\{t\leq T:U_t=Z_t\}.
$$
This set is nonempty because $U_T=Z_T$. Before $\tau^*$, the recursion has
$$
U_t=\mathbb E[U_{t+1}\mid\mathcal F_t],
$$
so the stopped process $U_{t\wedge\tau^*}$ is a martingale. Optional sampling and $U_{\tau^*}=Z_{\tau^*}$ give
$$
U_0=\mathbb E U_{\tau^*}=\mathbb E Z_{\tau^*}.
$$
Thus $\tau^*$ is an <optimal stopping time>.

Solution

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

An optimal stopping time attains the supremum of the expected reward over an admissible class of <stopping times>. For a finite-horizon <Snell envelope>, the first time at which its value equals the current reward is optimal.

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