# Is this process a Brownian motion? A martingale problem

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?

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 $$\tilde{\sigma} = 1$$.
• Makes sense. But in the case where $M_t$ and $X_t$ are distinct, would the answer be different ? May 18, 2021 at 16:08
• By scaling, the answer would still be no. More interestingly, if $dM_t = s(t,M_t) r(t, M_t) dB_t$ with $r, s \in L^2$, then you could always have $dX_t = r(t, M_t) dB_t$. May 18, 2021 at 16:13