Let $B_t$ be a linear Brownian motion. Heuristically, let $X_t$ be $B_t$ with a time change such that the clock runs twice faster when $B_t > 0$ and runs normally when $B_t <=0$. Formally define $s(t) = \int_0^t 0.5 * 1_{[B_u>0]} + 1_{[B_u \le 0]}du$ and define the process $\{X_t\}$ by setting $X_{s(t)} = B_t$.
My question is how I should understand the process $X_t$. For example what is its generator? What SDE it satisfies?
My guess is that it satisfies the SDE $dX_t = a(X_t) dB_t$ where $a(x) = \begin{cases} 2, x>0 \\1, x \le 0 \end{cases}$, but how do I prove this? Thanks!