Past exam of the mathematics course of the University of Cambridge 2015 iii Paper 29 5 i Solution Created 2026-10-03 Updated 2026-10-14
For , the first time is a stopping time. On , the bounded increments assumption controls the overshoot:Before that time, . Allowing the possibility , the processis a nonnegative martingale. Indeed, the stopped martingale in discrete time is a martingale, and the integrable -measurable variable is a constant-in-time martingale. If , nonnegativity follows from . Its expectations are constant and finite, so the martingale convergence theorem makes converge to a finite value almost surely.
On this proves finite convergence of . Taking the countable union over positive integer , we conclude that converges finitely on the event . Apply the same reasoning to the martingale to obtain finite convergence on .
Outside the event of finite convergence, both the infimum and the supremum of the sequence must therefore be infinite in the respective directions. Removing any finite initial segment cannot change this, because each of its values is finite. HenceThis is the bounded-increment martingale convergence-or-oscillation dichotomy. The term ensures the proof also covers an unbounded integrable initial value.
Apply part (i) for every positive integer and discard the countable union of its exceptional null sets. A sample path bounded below never reaches for some such , and must therefore converge to a finite limit of a sequence.
Apply the same argument to the martingale . Outside one more null set, every sample path bounded above also converges finitely. Thus a sample path which does not converge finitely must be unbounded in both directions. Each finite initial segment has finite values, so removing any such segment preserves unboundedness. It follows that the limit superior is and the limit inferior is . ConsequentlyThis proves the bounded-increment martingale convergence-or-oscillation dichotomy; in particular one-sided divergence to infinity is impossible with positive probability.