Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let's first state the theorem

$\forall M$ continuous local martingale, there exists a unique increasing continuous process $\langle M\rangle $ zero at $t=0$ and such that $M^2-\langle M \rangle $ is again a continuous local martingale. Further for all stopping times $\tau$, we have $\langle M^\tau\rangle = \langle M \rangle^\tau$.

The last statement is not clear for me. I know that $L:=M^2-\langle M \rangle $ is a local martingale. Then we look at $L^\tau=(M^\tau)^2-\langle M \rangle^\tau $, and we apply the stopping theorem. Why do we can apply the stopping theorem? For that, $L$ should be uniformly integrable, or $\tau$ must be finite?

thanks for your help

share|cite|improve this question
up vote 3 down vote accepted

All you need to use is that the set of local martingales are stable under stopping, i.e. $M^\tau$ is local martingale if $M$ is a local martingale and $\tau$ is any stopping time. Then $$ (M^{\tau})^2-\langle M\rangle^\tau=(M^2-\langle M\rangle)^{\tau} $$ yields the result.

share|cite|improve this answer
Just a small question after a while. Why are local martingales stable under stopping? Thank you for your help – user20869 Jul 19 '12 at 13:34
Simply because martingales are stable under stopping. – Stefan Hansen Jul 22 '12 at 11:38
But this is just, as far as I know, true if $\tau$ is bounded or the martingale is uniformly integrable? – user20869 Jul 22 '12 at 12:00
It is true for a general $\tau$, but it's true that it uses the optional sampling theorem on the bounded stopping time $\tau\wedge t$. – Stefan Hansen Jul 22 '12 at 16:58

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.