# Time scaling of Brownian motion

Let $(B_t)_{t\geq 0}$ be a standard Brownian motion and $A_t$ be an increasing continuous process adapted to the filtration generated by the Brownian Motion and $A_0 = 0$. I am trying to prove that $(B_{A_t})_{t\geq 0}$ is a local martingale.

My Attempt : If $A_t$ happens to be quadratic variation of some local martingale $M$, then it is clear that $(B_{A_t})_{t\geq 0}$ will be $(M_t)_{t \geq 0}$ (at least in distribution sense) by Dubin's Theorem. However, I am not able to proceed much in the general case. Any help is highly appreciated. Thanks

• At least if $(A_t)_t$ is absolutely continuous, you can use a similar argumentation by writing $$A_t = \int_0^t m_s \, ds$$ for some positive (measurable) process $(m_t)_{t \geq 0}$. – saz Dec 21 '13 at 17:27