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

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.