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 $(B_t^1, \ldots, B_t^d)$ be a standard $d$-dimensional Brownian motion, and $H_t^j$, $j=1, \ldots d$ be continuous processes adapted to the filtration $\{\mathcal{F}_t\}$. Let $$Z_t = \sum_{j=1}^d \int_0^t H_s^j dB_s^j$$ $$\langle Z \rangle_t = \sum_{j=1}^d \int_0^t (H_s^j)^2 ds$$ where the above integrals are the Itō integral against Brownian motion. Suppose that with probability $1$, we have $$\lim_{t \to \infty} \langle Z \rangle_t = \infty$$ and define stopping times $\tau_r$ by $$\tau_r = \inf\{t : \langle Z \rangle_t = r\}$$ Then I want to show that $W_r = Z_{\tau_r}$ is a standard Brownian motion with respect to the filtration $\mathcal{F}_{\tau_r}$.

The way to do this is to let $y \in \mathbb{R}$ and apply Itō's formula to $Y_t := \exp(iyZ_t + y^2 \langle Z\rangle_t/2)$ in order to show that $Y_t$ is a local martingale. I do not see how to do this, can anyone help? What are we integrating to apply Itō's fomula? (Please explain all the steps as thoroughly as possible, I am very new to stochastic calculus and cannot fill in gaps to arguments yet).

share|cite|improve this question
You're correct, I did not. I fixed it now, thanks. – user98123 Nov 29 '11 at 3:14
up vote 3 down vote accepted

I think the best way to show that a continuous process is actually a Brownian motion is to use Paul Lévy's characterisation. That is, to show that the quadratic variation of this process is equal to r.

By the way this is a particular case of a slightly more general result known as Dambis-Dubins-Schwarz theorem. You can find its proof for example in the book of Karatzas and Shreve "Brownian Motion and Stochastic Calculus"

Best Regards

share|cite|improve this answer
+1. But Lester E. Dubins and Gideon Schwarz, see here. – Did Nov 29 '11 at 7:45
@Didier : thank's for the spelling check, I edited my post accordingly. Best regards – TheBridge Nov 29 '11 at 9:49
@Didier : I added Dambis as he also has credit for the result (cf. Karatzas and Shreve's book). – TheBridge Nov 29 '11 at 10:35
So it seems. – Did Nov 29 '11 at 16:32

Your Answer


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

Not the answer you're looking for? Browse other questions tagged or ask your own question.