Past exam of the mathematics course of the University of Cambridge 2008 iii Paper 39 3 ii Solution Created 2026-10-03 Updated 2026-10-14
For a stopping time , is adapted: split according to for and , all events belonging to . It is integrable because . Moreover,The indicator is -measurable, so conditioning and the supermartingale property giveThus . This proves stopping preserves supermartingales in discrete time without assuming itself bounded or integrable.