Let $(M_t)$ be a martingale w.r.t a filtration $(\mathcal F_t)$. The martingale representation theorem implies there exists a Brownian motion $(B_t)$ adapted to $(\mathcal F_t)$ such that $M_t$ is the solution of an SDE of the form
$$dM_t=\sigma(t,M_t)d B_t $$
Suppose that we found a process $(X_t)$ adapted to $(\mathcal F_t)$ such that
$$dM_t=\tilde \sigma(t,M_t)d X_t $$
Does that mean that $(X_t)$ is a Brownian motion? If so, what is the theorem behind that result?
Best Answer
No. If $M_t$ is a martingale which is not a Brownian motion with respect to $(\mathcal{F}_t)$, then you can just pick $dX_t = dM_t$ and $\tilde{\sigma} = 1$.