Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 35 / 2 / c / Solution

Use $A_t={\int }_0^t{H}_s^{2}ds$ and its inverse ${\tau }_u$ as above. Strict increase and continuity make $u\mapsto {\tau }_u$ continuous, with $A_{{\tau }_u}=u$ and ${\tau }_{A_t}=t$. Thus $W_u=M_{{\tau }_u}$ is continuous and adapted to ${\mathcal{G}}_u={\mathcal{F}}_{{\tau }_u}$, with $W_0=0$.

The essential remaining point is independence of increments, which does not follow merely from the Gaussian marginal in part (a). Fix $0\le u<v$ and $\lambda \in \mathbb{R}$. The complex exponentialis bounded by $e^{{\lambda }^{2}v/2}$ in modulus, hence is a uniformly integrable martingale. The optional sampling theorem at the finite, possibly unbounded, times ${\tau }_u\le {\tau }_v$ givesDivide by the nonzero right-hand exponential. ThenThis deterministic conditional characteristic function says that $W_v-W_u$ is $N(0,v-u)$ and independent of ${\mathcal{G}}_u$. Successively conditioning proves independence of every finite family of increments. Together with continuity and the zero initial value, this proves that $W$ is Brownian motion. The inverse-clock identity gives $W_{A_t}=M_{{\tau }_{A_t}}=M_t$, completing the requested special-case proof of the Dubins-Schwarz theorem.