 Solution

Because $A$ is nondecreasing and $A_{n+1}$ is ${\mathcal{F}}_n$-measurable,The convention $A_{N+1}=\infty$ ensures ${\tau }^{\ast }\le N$, so ${\tau }^{\ast }$ is a bounded stopping time.

By minimality, $A_{{\tau }^{\ast }}=0$ and $A_{{\tau }^{\ast }+1}-A_{{\tau }^{\ast }}>0$. Part (d) then givesMoreover $M_{{\tau }^{\ast }}=V_{{\tau }^{\ast }}+A_{{\tau }^{\ast }}=Z_{{\tau }^{\ast }}$. Optional sampling for the martingale $M$ now yieldsTogether with part (a), this proves that ${\tau }^{\ast }$ is optimal and $V_0$ is the optimal stopping value.