Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 202 / 1 / b

The Dambis-Dubins-Schwarz theorem states that if $M$ is a continuous local martingale with $M_0=0$ and $[M{]}_{\infty }=\infty$, then, forthe process $W_s=M_{{\tau }_s}$ is a standard Brownian motion and $M_t=W_{[M{]}_t}$. If $[M{]}_{\infty }<\infty$, one obtains the same representation after enlarging the probability space and continuing $W$ independently beyond $[M{]}_{\infty }$.

Set $M_t={\int }_0^t{H}_{s}d{B}_s$. Its quadratic variation is $[M{]}_t={\int }_0^t{H}_s^{2}ds$, which is continuous and tends to infinity almost surely by assumption. The stated stopping time is the inverse clock at level one, so $[M{]}_{\tau }=1$. The Dambis-Dubins-Schwarz theorem gives

The assertion is false: the Brownian motion produced by the Dambis-Dubins-Schwarz theorem need not be independent of its clock. Let $W$ be a standard Brownian motion and setIf the Brownian motion $B$ in $M_t=B_{[M{]}_t}$ were independent of the whole quadratic variation process, then conditioning on $[M]$ would give $\mathbb{E}[M_t[M{]}_t]=\mathbb{E}[B_{[M{]}_t}[M{]}_t]=0$. Instead, the fourth-moment formula for a bivariate normal distribution gives $\mathrm{Cov}(W_t^2,W_s^2)=2s^2$ for $s\le t$, and henceThus $B$ and $[M]$ are dependent.