TITLE: Time changes of the Itō integral
QUESTION [6 upvotes]: 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).

REPLY [4 votes]: 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