Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 202 / 3 / d / Solution

Put $A_t=[M{]}_t$ and use the Dambis-Dubins-Schwarz theorem to write $M_t=B_{A_t}$, enlarging the space if necessary after the terminal clock value. In the time-changed filtration, $A_t$ is a stopping time. For $b<0$, let $S_b=\inf \{s:B_s-s=b\}$. Applying part c with $R=A_t$ givesOn $\{S_b\le A_t\}$, the first integrand is at most $e^b{e}^{A_t/2}$. The assumed Novikov condition therefore impliesas $b\to -\infty$, uniformly for $t\le T$. Meanwhile $mathbf{1}_{\{A_t<S_b\}}\uparrow 1$, so the monotone convergence theorem gives $\mathbb{E}\mathcal{E}(M{)}_t=1$. A nonnegative local martingale with constant expectation is a martingale. Thus $\mathcal{E}(M^T)$ is a martingale, proving the Novikov condition.