I'm trying to change the time variable in the following SDE: $$ dF(t) = A(t) dB(t) $$ I'm interested to find the SDE for the process $\overline{F}(t) = F\left(a t\right)$ where $a$ is a non-negative scalar. I'm looking at the book by Øksendal, Theorem 8.5.7. However, I'm not understanding correctly I think. The book says explicitly
We now use this to prove that a time change of an Ito integral is again an Ito integral, but driven by a different Brownian motion.
Then, the Theorem defines this new Brownian motion in the following manner $$W(s) = \int_0^{\alpha(u)} \sqrt{c(u,\omega)}dB_u$$ where $B_s$ was our original Brownian motion. Then, $$\int_0^{\alpha(t)} v(s, \omega) dB_s = \int_0^t v(\alpha(r),\omega) \sqrt{\alpha'(r)} dW(r).$$
So, if I am not mistaken, $$ d\overline{F}(s) = A(s) \sqrt{a} dW(s).$$
However, I am a little confused when I have to compute the increment for the Brownian motion $dW(s)$. I have a number of these processes that are correlated and therefore I need to compute expressions such as $dW_i(s)dW_j(s)$. However, what exactly is the increment of $dW(s)$ in this case? Judging from $$W(s) = \int_0^{\alpha(s)} \sqrt{c(u,\omega)}dB_u$$ it could either be $$dW(s) = \sqrt(c(\alpha(s)))dB_{\alpha(s)}$$ or $$dW(s) = \sqrt(c(s))dB_{s}.$$