Continuous square integrable martingale with non-random quadratic variation.

Question

In a comment on this question @NateEldredge postulates that if a martingale (continuous, square integrable and starting at $0$) $M=(M_t)$ is such that $$E(M_t^2)=\langle M\rangle_t$$ i.e. the quadratic variation is non-random then it "ought to follow from Levy's characterization that such a process $M$ must be a deterministic time change of Brownian motion."

I think this is good enough to be a question itself hence if someone could explain me this in more detail it would be nice.

Thanks in advance.

Answer 1

It actually follows from the Dambis, Dubins-Schwarz theorem, i.e. if $\lim_{t\to\infty}\langle M\rangle_t=\infty$ a.s., then $M_t=B_{\langle M\rangle_t}$, where $B$ is a Brownian motion. So when the quadratic variation of $M$ is deterministic, it is a deterministic time change of a BM.