Let $B_t$ be a Brownian motion and $M_t := \max_{0\leq s \leq t} B_t$. I know that for all $t\geq 0$, $$M_t - B_t =^d |B_t|$$, i.e., $M_t -B_t$ and $|B_t|$ have the same distribution.
Does it follow directly that the zeros of the process $M_t -B_t$ behave the same as the zeros of the $B_t$? My guess is that for this to be true, we need the finite-dimensional distributions (joint distribution on any finite set of times $t_i$) to be equal.
Any help is appreciated.