# Time changes of the Itō integral

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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You're correct, I did not. I fixed it now, thanks. – user98123 Nov 29 '11 at 3:14