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

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.