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.
Best Answer
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.