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 $\langle B_t \rangle_t=t$


1 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.

  • $\begingroup$ thank you could you please elaborate a little bit more on the last two line, i would be grateful $\endgroup$ Oct 18, 2015 at 13:18
  • $\begingroup$ I think I partly understand your explanation, how does enlarging the probability space help ? When we stop the process at $\tau(t)$ and $\langle M \rangle_{\infty}(\omega) < \infty$,i.e the quadratic variation is finite for some $\omega$ then $ \exists t>0 ,\text{s.th} \tau(t)(\omega)=\inf(\phi)= \infty$ because $\langle M \rangle_{\infty}(\omega) < \infty$ and therefore for all $t'$ greater that this we will have $X_t(\omega)=M_{\infty}(\omega)$ for that particular path. So you are saying that we need to add some more $\omega$ to $\Omega$ to take care of such cases right? How do we do it? $\endgroup$ Oct 18, 2015 at 14:10
  • $\begingroup$ Am I right or wrong John? $\endgroup$ Oct 19, 2015 at 12:12
  • $\begingroup$ Dawins I understand most of your argument but I do not see why enlarging the filtration determines the entire brownian motion. By enlarging the filtration means adding more sets to the increasing sequence of sub-sigma algebras. I mean if we have that $\sup \limits_{\omega \in \Omega} \langle M \rangle_{\infty}=1 $, then how would be extend the B.M from times greater than 1 just by enlarging the filtration? $\endgroup$ Nov 26, 2015 at 10:05

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