Past exam of the mathematics course of the University of Cambridge 2004 iii Paper 30 5 a Solution Created 2026-10-03 Updated 2026-10-14
Denote the density process by . The given identities for all implyThus is a strictly positive true martingale, not merely a local martingale. The Itô formula gives , and henceFirst work with the zero-starting corrected increments . Integration by parts givesThe finite-variation bracket has zero quadratic covariation with , so no other term appears. Thus is a P-local martingale.
To justify the change of measure, stop at , with optional further localization if needed. The product remains a P-local martingale: before the stopping time the cancellation above applies, and afterwards only its bounded constant factor multiplies . Its absolute value is bounded by . On a finite horizon the stopped values of are uniformly integrable, so Q1(c) upgrades this product to a true martingale. The Bayes formula for conditional expectation now givesFinite-horizon equivalence of the measures transfers the localizing sequence to Q. Therefore is a continuous Q-local martingale. If , adding its constant-in-time process provesThis proves the Girsanov theorem assertion under the usual zero or deterministic initial-value convention. More generally the normalization makes , so P and Q agree on and the required initial integrability is automatic.
There is a literal initial-integrability qualification if neither convention is imposed. On positive integers let , , with , and let every filtration equal the full sigma-algebra. Constant processes and are P-martingales, and is precisely the prescribed density. But and , so the asserted Q-local martingale cannot have an integrable initial value. This is why initial integrability under a density change is part of the precise formulation.