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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).

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  • $\begingroup$ You're correct, I did not. I fixed it now, thanks. $\endgroup$
    – user98123
    Nov 29, 2011 at 3:14

1 Answer 1

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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

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  • $\begingroup$ +1. But Lester E. Dubins and Gideon Schwarz, see here. $\endgroup$
    – Did
    Nov 29, 2011 at 7:45
  • $\begingroup$ @Didier : thank's for the spelling check, I edited my post accordingly. Best regards $\endgroup$
    – TheBridge
    Nov 29, 2011 at 9:49
  • $\begingroup$ @Didier : I added Dambis as he also has credit for the result (cf. Karatzas and Shreve's book). $\endgroup$
    – TheBridge
    Nov 29, 2011 at 10:35
  • $\begingroup$ So it seems. $\endgroup$
    – Did
    Nov 29, 2011 at 16:32

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