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. 



Time Change of a Brownian motion
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.


