Past exam of the mathematics course of the University of Cambridge 2014 iii Paper 26 3 1 Solution Created 2026-10-03 Updated 2026-10-06
First prove the strong law for Brownian motion, with probability one. For each , the Gaussian tail bound givesBy stationary increments and the Brownian reflection principle, followed by the same tail bound,Both bounds are summable in . The First Borel-Cantelli lemma therefore implies that, eventually, both quantities inside these probability events are at most . For this gives . Intersect over a sequence of positive rational tending to zero to obtain the asserted continuous-time limit.
On this one almost sure event,so for and to for . For each real , the path is eventually strictly on the corresponding side of . HenceThe same event works simultaneously for all levels , giving the requested transience of Brownian motion with drift.
Past exam of the mathematics course of the University of Cambridge 2014 iii Paper 26 4 1 Solution Created 2026-10-03 Updated 2026-10-06
Use the Kolmogorov continuity theorem in its one-parameter form: if on a compact interval a process satisfies for some , it has a modification whose paths are Hölder continuous of every order on that interval.
For Brownian motion, normal increments give, for every ,Every such normal moment is finite. Taking yields , hence any order below . For a prescribed , choose with .
The continuous modification and the given continuous Brownian motion agree at all rational times on one almost sure event; continuity makes them agree everywhere on the interval. To obtain all exponents and all compact intervals simultaneously, apply the theorem to integer intervals and a countable sequence of positive exponents increasing to , then intersect these almost sure events. A bound at exponent implies a bound at on a compact interval. Consequently the Brownian Hölder regularity conclusion iswith finite random constants on one common event of probability one. This uses the usual positive-exponent meaning of Hölder continuity.