Why for a continous local martingale ,on an enlarged probability space, it possibly holds that $M_t=B_{\langle M \rangle _t}$

Question

Why for a continous local martingale $(M_t)_{t\in \mathbb{R}_+}$ ,on an enlarged probability space, it possibly holds that $M_t=B_{\langle M \rangle _t}$ where $(B_u, u \geq 0)$ is Brownian motion . I read this statement in a Marc Yor paper I am reading for my thesis.Please note I have not studied any semi-martingale thoery. It would be very helpful if someone could explain why this is true?

I know Brownian motion being a martingale with a.s continuous path is a local martingale and if $M_t=B_{\langle M \rangle _t} \implies \langle M \rangle_t=\langle{B_{\langle M \rangle _t}}\rangle =\langle M\rangle_t $ since $\langle B_t \rangle_t=t$

Answer 1

Suppose, for simplicity, that $\langle M\rangle_\infty=\infty$ almost surely. Define $\tau(t):=\inf\{s:\langle M\rangle_s>t\}$. The time-changed process $X_t:=M_{\tau(t)}$ is then a continuous local martingale with $\langle X\rangle_t= \langle M\rangle_{\tau(t)}=t$. By Lévy's characterization theorem, $X$ is Brownian motion.

If $\langle M\rangle_\infty(\omega)<\infty$ then the above construction only determines $X$ up to time $t_0(\omega) = \langle M\rangle_\infty(\omega)$; to get a complete Brownian motion one may need to enlarge the initial (filtered) probability space.