# Time Change of a Brownian motion

We know that for if $X$ is a stochastic integral of the form below -

$X_t = \int_0^t v(s,\omega) db(s,\omega)$. then we can use time change formula to claim that $X_t = W_{\alpha(t)}$ where $W$ is a different brownian motion with a changed clock.

This holds true, under mild conditions, even when $b$ is a d-dimensional brownian motion.

I was wondering if I can use the same result to say that if $X$ is a stochastic integral of the form - $X_t = \int_0^t v(s,X_s) db(s,\omega)$.

Intuitively I don't see why it might be an issue as $v(s,X_s)$ is essentially a function of $\omega$ only. So I wanted to know if I can write $X_t = W_{\alpha(t)}$ for some other brownian motion.

• Do you mean that $X$ is a solution to a stochastic differential equation? Sep 6, 2015 at 18:54
• Sorry if I wasn't clear. I meant that $X_t$ is a solution to an SDE $dX_t = v(s,X_t) db_t$, where $b_t$ is an $d-$dimensional brownian motion. $X_t$ is a scalar. The way Oksendal would probably write this SDE is- $dX_t = \sum_i v_i(t,X_t) db_{it}$. I want to say that since $X_t$ is a martingale $X_t = W_{\alpha_t}$ where $\alpha$ is the new clock. Thanks. Sep 6, 2015 at 21:44

I think you're right and the new clock $\alpha_t$ is proportional to $\int_0^t |v(s,X_s)|^2\ ds$, if the $b_i$ are i.i.d.. It needs a proof, but to my intuition the result is clear enough: a stochastic integral $\int_0^t \lambda_s\cdot db_s$, where the vector-valued process $\lambda_t$ is adapted (continuous, etc) and $|\lambda_t|^2\equiv 1$, should be a continuous martingale with independent stationary increments, i.e. brownian.