Past exam of the mathematics course of the University of Cambridge 2016 iii Paper 201 1 e Solution Created 2026-10-03 Updated 2026-10-06
Write for the martingale difference sequence. Bounded increments mean that one deterministic satisfies for every , almost surely. Then and, for ,The increments are uncorrelated random variables, uniformly bounded in . Applying the supplied strong law for uniformly L2-bounded uncorrelated random variables gives almost surely. Also , because is finite almost surely. Therefore a martingale with bounded increments has zero linear growth:This is the strong law for martingales with bounded increments.
For the submartingale, let and define its predictable drift and compensator byThis is the Doob decomposition in discrete time. The process is a martingale. Since , we have , so . The result just proved gives almost surely, and . Consequently the strong law for submartingales with bounded increments isThe compensator may grow, so the conclusion is a lower bound rather than a claim of convergence to zero.
Past exam of the mathematics course of the University of Cambridge 2024 iii Paper 211 4 a Solution Created 2026-09-24 Updated 2026-10-03
Definefor , with . The supermartingale property makes , and it is -measurable, so is previsible and nondecreasing. The increments of have conditional mean zero, so is a martingale. Finally,and summation gives the Doob decomposition in discrete time
Past exam of the mathematics course of the University of Cambridge 2024 iii Paper 211 4 e Solution Created 2026-09-24 Updated 2026-09-25
Let be its Doob decomposition in discrete time and choose the martingaleSince and ,For the first optimal stopping time , the complementarity for the Snell envelope compensator implies : all compensator increments before vanish. Since ,Thereforepathwise, and taking expectations proves the asserted equality.
If a submartingale has uniformly bounded increments, then almost surely. In its Doob decomposition in discrete time, the predictable compensator is nondecreasing, while the martingale part has bounded increments. The strong law for martingales with bounded increments makes the latter negligible on the linear scale.