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

Definefor $t\ge 1$, with $A_0=M_0=0$. The supermartingale property makes $\Delta A_t\ge 0$, and it is ${\mathcal{F}}_{t-1}$-measurable, so $A$ is previsible and nondecreasing. The increments of $M$ have conditional mean zero, so $M$ is a martingale. Finally,and summation gives the Doob decomposition in discrete time

The recursion makes $U$ a supermartingale and gives $U_t\ge Z_t$ at every date. The optional sampling theorem for a supermartingale for the bounded stopping time $\tau$ therefore gives

Take the first entry into the stopping region:This set is nonempty because $U_T=Z_T$. Before ${\tau }^{\ast }$, the recursion hasso the stopped process $U_{t\wedge {\tau }^{\ast }}$ is a martingale. Optional sampling and $U_{{\tau }^{\ast }}=Z_{{\tau }^{\ast }}$ giveThus ${\tau }^{\ast }$ is an optimal stopping time.

For every stopping time $\tau$, the martingale case of the optional sampling theorem for a supermartingale gives $\mathbb{E}{X}_{\tau }=X_0=0$. HenceUse the optimal stopping time from part c and the pathwise inequalityTaking expectations gives

Let $U=U_0+M-A$ be its Doob decomposition in discrete time and choose the martingaleSince $Z_t\le U_t$ and $A_t\ge 0$,For the first optimal stopping time ${\tau }^{\ast }$, the complementarity for the Snell envelope compensator implies $A_{{\tau }^{\ast }}=0$: all compensator increments before ${\tau }^{\ast }$ vanish. Since $Z_{{\tau }^{\ast }}=U_{{\tau }^{\ast }}$,Thereforepathwise, and taking expectations proves the asserted equality.