Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 2 / 30K (Stochastic Financial Models)

The optional sampling theorem for a supermartingale at a bounded stopping time giveswhere $V_0$ is constant because ${\mathcal{F}}_0$ is trivial. Since $Z_{\tau }\le V_{\tau }$ almost surely,

For $0\le k<N$, putThis is ${\mathcal{F}}_k$-measurable, and it is nonnegative because $V$ is a supermartingale. Hence $A_n$, being the sum of increments known by time $n-1$, is ${\mathcal{F}}_{n-1}$-measurable for $n\ge 1$, so $A$ is previsible; and its nonnegative increments make it nondecreasing.

The process $M$ is adapted and integrable. Since $A_{n+1}$ is ${\mathcal{F}}_n$-measurable,Thus $M$ is a martingale. This is the finite-horizon Doob decomposition $V=M-A$.

For $n<N$,The recursionshows that if $V_n>Z_n$, then $V_n=\mathbb{E}[V_{n+1}\mid {\mathcal{F}}_n]$ and the increment of $A$ is zero. Conversely, if that increment is positive, the maximum must be attained by $Z_n$, so $V_n=Z_n$. The two nonnegative quantities therefore satisfyFor $n=N$, this also holds because $V_N=Z_N$, regardless of the convention $A_{N+1}=\infty$.

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.