Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 3 / 29K / b / iii / Solution

The two processes with the same distribution areIndeed, $(M_t{)}_{t\ge 0}$ has almost surely nondecreasing paths, whereas $|B_t|$ and $M_t-B_t$ do not, so $M$ cannot be one member of the equal pair. The stated conclusion then follows from the premise that two of the three processes have the same law. More directly, it is the first coordinate of the Lévy identity for Brownian motion; both $|B|$ and $M-B$ are Reflected Brownian motion.