The assertion is true. The <Dambis-Dubins-Schwarz theorem>, with an independent continuation of the Brownian motion if $f$ is bounded, represents
$$
M_t=W_{[M]_t}=W_{f(t)}.
$$
Because $f$ is deterministic, every finite vector $(M_{t_1},\ldots,M_{t_n})$ is a finite vector of a <Brownian motion> at deterministic times and therefore has a <multivariate normal distribution>. Hence $M$ is a <Gaussian process>. This is the <deterministic quadratic variation characterizes a Gaussian continuous local martingale> result.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 202 / 3 / c / Solution

If a continuous local martingale starts at zero and has deterministic continuous <quadratic variation> $[M]_t=f(t)$, the <Dambis-Dubins-Schwarz theorem> gives $M_t=B_{f(t)}$. It is therefore a centered <Gaussian process>. The converse follows from <independent increments of a Gaussian martingale>.

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