Find a change of time $C_{t}$ (that is, an increasing function of $\left.t\right)$ so that the following processes $M\left(C_{t}\right)$ are Brownian motion.
(c) $M_{t}=e^{B_{t}-t / 2}-1$
My attempt
Firstly, $M(0)=0$, we now find $[M, M]_{t}$ $$ d M_{t}=f^{\prime}\left(B_{t} t\right) d B_{t}+\frac{1}{2} f^{\prime \prime}\left(B_{t}, t\right) d t+\frac{\partial f}{\partial t} d t . $$ Hence
$d M_{t}=e^{B_{t}-\frac{t}{2}} d B_{t}+\frac{1}{2} e^{B_{t} -\frac{t}{2}} dt$ $-\frac{1}{2} e^{B_t - \frac{t}{2}} dt$
$dM_t=e^{B_{t}-t / 2} d B_{t} \Rightarrow$ is a martingale.
So $[M,M]_{t}=\int_0^T e^{2 B_{s}-s} d s$
I know that I need to find the inverse of $[M,M]_{t}^{-1}$ which will give me $C_t$ and then M$(C_t)$ is a brownian motion by the D.D.S theorem. However, I am not sure how to find the inverse of $[M,M]_t$ as there is a brownian motion term. Do I find the M.G.F of $e^{2 B t}$ and then evaluate the integral with respect to t?
Thank you!
For each sample path, $t\mapsto\int_0^t\exp(2B_s-s) ds$ is a continuous strictly increasing map of $[0,\infty)$ onto $[0, D)$, where $D:=[M,M]_\infty$. As such this map is invertible: there is a continuous strictly increasing family $C(t)$, $0\le t<D$ such that $$ t=\int_0^{C(t)}\exp(2B_s-s) ds,\qquad\forall t\in[0,D). $$ You won't be able to find an explicit formula for $C(t)$, but do notice (for what it's worth) that $C$ satisfies the ODE $$ C'(t)=\exp(C(t) - 2B_{C(t)}),\qquad t\ge 0. $$
In addition: (i) each $C(t)$ is a stopping time of the Brownian motion $(B_t)$, and (ii) $P[D<\infty]=1$.