# Hitting time of the maximum of a Brownian motion

Let B be a standard Brownian motion in 1 dimension. Define $$\tau = \inf\{t\in[0,1]:B_t=\underset{s\in[0,1]}{\max}B_s\},$$ and $$\tau' = \sup\{t\in[0,1]:B_t=\underset{s\in[0,1]}{\max}B_s\}.$$ Show that $$\tau$$ and $$\tau'$$ are not stopping times. My idea was to first assume that $$\tau$$ is a stopping time, then by strong Markov property $$\{B_{\tau+t}-B_{\tau}:t \geq 0\}$$ is a standard Brownian motion, and so the process fluctuates above and below $$B_\tau$$ at $$\tau$$, which contradicts the fact that $$B_\tau$$ is the maximum. I am wondering if I am heading towards the right direction.

• Your guess is exactly right. If $\tau$ were a stopping time, by the strong Markov property $\beta_t:=B_{T+t}-B_T$ would be a standard Brownian motion, but then, a.s. $\beta_t\leq 0$, a contradiction. You can reach contradictions for $\tau'$ in the same spirit. Apr 1, 2021 at 2:53
• But one has to show that $\tau < 1$ and $\tau' < 1$ a.s. in order to use the strong Markov property. It's easy for the inf since if $\tau =1$, then the maximum must be attained at 1, which is not possible. However, for sup, I am not sure how to rule out the possibility that there exists a sequence approaching 1 where the max is obtained on this sequence. Apr 1, 2021 at 3:32
• That's a nice point. Maybe you can get out of it using some sort of time-reversal or time-inversion but I'm not sure. Regardless, I think you only need to verify $P(\tau'<1)>0$ to apply (a version of) the strong Markov property. Apr 1, 2021 at 14:18

As noted, if $$\tau$$ were a stopping time, then $$\beta_t:=B_{t+\tau}-B_\tau$$, $$t\ge 0$$, would be a standard Brownian motion. On the event $$\{\tau<1\}$$ you would then have $$\beta_t\le 0$$ for all sufficiently small $$t$$, which is inconsistent with the Brownian character of $$\beta$$. Therefore $$P[\tau<1]$$ would be $$0$$. But $${1\over 2} =P[B_1<0]\le P[\tau<1],$$ because $$\max_{0\le s\le 1}B_s\ge 0$$. Therefore $$\tau$$ cannot be a stopping time.
Likewise for $$\tau'$$. In fact, one can show that the time at which Brownian motion on the time inerval $$[0,1]$$ attains its maximum is unique; that is $$P[\tau=\tau'<1]=1$$.