# The maximum process of a Brownian motion

Let $$B=(B_t)_{t \in[0,1]}$$ be a standard Brownian motion. The maximum process $$M$$ of Brownian motion is defined by $$M_s=\sup _{t \in[0, s]} B_t$$, for all $$s \in[0,1]$$. Let $$T=\inf \left\{t \in[0,1] \mid B_t=M_1\right\}$$ be the first time where $$B$$ achieves its maximum on $$[0,1]$$. Prove that for all $$s \in[0,1]$$, $$\mathbb{P}(T \leq s)=\mathbb{P}\left(M_s^1>M_{1-s}^2\right)$$ where $$M^1_s$$ is the maximum process of the Brownian motion $$\left(B_t^1\right)_{t \in[0, s]}$$, where $$B_t^1=B_{s-t}-B_t$$, and $$M^2_s$$ is the maximum process of the independent Brownian motion $$\left(B_t^2\right)_{t \in[0,1-s]}$$, where $$B_t^2=B_{s+t}-B_t$$.

My Attempt: I rewrite $$M_s^1$$ and $$M_s^2$$ as the follows: $$M_s^1 = \sup_{t_1\in[0,s/2]}|B_{s-t_1}-B_{t_1}| \sim \sup_{t_1\in[0,s/2]} |N(0,s-2t_1)|$$ and $$M_s^2 \sim N(0,s)$$ independent of $$t$$ (maybe? I am not sure because t varies with $$s$$). And now I have no idea how to proceed.

The formulae you give are not correct if you want $$B^1$$ and $$B^2$$ to be independant Brownian motions. We use equalities between events. $$[T \le s] = \big[\max_{t \in [0,s]} B_t \ge \max_{t \in [s,1]} B_t\big].$$ $$[T \le s] = \big[\max_{t \in [0,s]} B_t-B_s \ge \max_{t \in [s,1]} B_t-B_s\big].$$ $$[T \le s] = \big[\max_{r \in [0,s]} B_{s-r}-B_s \ge \max_{r \in [0,1-s]} B_{s+r}-B_s \big].$$ Then note that $$(B_{s-r}-B_s)_{r \in [0,s]}$$ and $$(B_{s+r}-B_s)_{r \in [0,1-s]}$$ are independant Brownian motions.

• Wow, this is so simple, Thank you very much! Commented Nov 15, 2022 at 17:05
• Wait, but we define $B_t^1 = B_{s-t}-B_t$, how do you get from $\max_{r \in [0,s]} B_{s-r}-B_s$ to $\max_{r \in [0,s]} B_{s-r}-B_r$? Commented Nov 15, 2022 at 17:12
• @Tab1e I do not deal with $(B_{s-r}-B_r)_{r \in [0,s]}$, this is NOT a Brownian motion. Commented Nov 15, 2022 at 17:57
• Oh, sorry I didn't follow your argument, so you are saying the question itself is written wrong? Commented Nov 15, 2022 at 19:14
• @Tab1e Yes. You must change the definitions of $B^1$ and $B^2$ to get Brownian motions (namely subtrack $B_s$ instead of $B_t$ from $B_{s-t}$ and from $B_{s+t}$). Commented Nov 15, 2022 at 21:55