Past exam of the mathematics course of the University of Cambridge 2003 iii Paper 29 2 Solution Created 2026-10-03 Updated 2026-10-14
A convenient precise version is the almost sure submartingale convergence theorem: if is a real submartingale and , then converges almost surely to a finite integrable random variable. In particular the martingale convergence theorem applies to any martingale with . It asserts neither convergence in L1 nor preservation of the initial expected value without additional uniform integrability.
Here is a proof including the crossing estimate. Fix real , and let count completed upcrossings by time . Use a predictable indicator which holds one unit during each crossing attempt: enter when , exit when . Witheach completed trade gains at least , while the possible unfinished trade loses no more than . Hence . The submartingale property and give : the complementary predictable gains have nonnegative expected value. ThereforeThe right side is uniformly bounded, since . By the monotone convergence theorem, every rational interval has finitely many upcrossings almost surely. A real sequence whose lower and upper limits differ crosses some rational interval infinitely often, so the countable intersection of these events gives an extended-real limit. Furthermoreso . The Fatou lemma makes the limit finite and integrable, completing the proof. Applying this theorem to gives the almost sure supermartingale convergence theorem for nonnegative supermartingales.
For the given adapted multiplicative drift, putThe next denominator is -measurable, and makes integrable. ThusThe nonnegative supermartingale has a finite almost sure convergence limit . On the probability-one event where the specified series is finite,Consequently and the multiplicative correction for summable adapted drift givesThe argument does not require a finite expected value for the infinite product or for the total drift.