Difference between running maximum and reflected Brownian motion

Question

Let $(B_t)_{t > 0}$ be a Brownian motion and let $M_t = \sup_{0 \leq s \leq t} B_s$. I was wondering whether $M_t - |B_t|$ was a Brownian motion as well ? If so, why ?

From this question, I would naively expect $M_t - |B_t|$ to behave like $B_t$.

Answer 1

For $x>0$, we have the disjoint union $$\{B_t<-x\}=\{M_t-|B_t|<-x\} \cup \{-x> B_t \ge-x-M_t\}$$ so $$P[B_t<-x]>P[M_t-|B_t|<-x] \,.$$

Addendum: Note that $$\{-x> B_t \ge-x-M_t\} \supset \{2>B_{t/2}>1\}\cap \{-x>B_{t}>-x-1\}$$ and using the Markov property and the monotonicity of the Gaussian density, $$P[ 2>B_{t/2}>1 \; \text{and} \; -x>B_{t}>-x-1]$$ $$ \ge P[2>B_{t/2}>1] \cdot \inf_{h\in [1,2]} P[-h-x>B_{t/2}>-h-1-x] $$ $$ = P[2>B_{t/2}>1] \cdot P[-2-x>B_{t/2}>-3-x] >0\,.$$