Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 39 / 3

Relative to a filtration $({\mathcal{F}}_t)$, a supermartingale is an adapted process $U_t$ with $\mathbb{E}|U_t|<\infty$ for every $t$ andBy the tower property this also gives $\mathbb{E}[U_u\mid {\mathcal{F}}_t]\le U_t$ for every $u\ge t$.

For a stopping time $\tau$, $U_{t\wedge \tau }$ is adapted: split according to $\{\tau =j\}$ for $j<t$ and $\{\tau \ge t\}$, all events belonging to ${\mathcal{F}}_t$. It is integrable because $|U_{t\wedge \tau }|\le {\sum }_{j=0}^t|U_j|$. Moreover,The indicator is ${\mathcal{F}}_t$-measurable, so conditioning and the supermartingale property giveThus $\boxed{(U_{t\wedge \tau }{)}_t\text{ is a supermartingale}}$. This proves stopping preserves supermartingales in discrete time without assuming $\tau$ itself bounded or integrable.

Use the tree's natural filtration. The stopping times take values in the process horizon $\{0,1,2\}$. Construct a dominating supermartingale $Y$ by backward comparison, the finite-horizon Snell envelope. At time two put $Y_2={\xi }_2$. At time one the continuation values areCompare these with immediate rewards $11$ and $7$ to obtain $Y_1=12$ on the upper branch and $Y_1=7$ on the lower branch. Finally setThese definitions explicitly give $Y_t\ge {\xi }_t$, $Y_0=\mathbb{E}{Y}_1$, and $Y_1\ge \mathbb{E}[Y_2\mid {\mathcal{F}}_1]$. Thus $Y$ is a supermartingale. By part (ii), its stopped process is a supermartingale, so for every allowed $\tau$,The reward process itself need not be a supermartingale; the constructed majorant is what proves the bound.

Continue at time zero. At time one stop on the lower branch and continue on the upper branch:This is a stopping time since the branch decision is measurable at time one. Its expected reward isIt therefore attains the upper bound in part (iii).