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 set
$$
M_t=\int_0^tW_s\,dW_s=\frac12(W_t^2-t),
\qquad
[M]_t=\int_0^tW_s^2ds.
$$
If 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 $\operatorname{Cov}(W_t^2,W_s^2)=2s^2$ for $s\leq t$, and hence
$$
\mathbb E[M_t[M]_t]=\frac12\int_0^t\operatorname{Cov}(W_t^2,W_s^2)ds=\int_0^ts^2ds=\frac{t^3}{3}>0.
$$
Thus $B$ and $[M]$ are dependent.

Solution

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

{c}

The Brownian motion $B$ in $M_t=B_{[M]_t}$ is generally constructed from $M$, so it need not be independent of the <quadratic variation> $[M]$. For example, $M_t=(W_t^2-t)/2=\int_0^tW_s\,dW_s$ has clock $\int_0^tW_s^2ds$, and their dependence can be detected by a nonzero mixed moment.

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