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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.

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  • $\begingroup$ 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. $\endgroup$
    – charlus
    Commented Apr 1, 2021 at 2:53
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    $\begingroup$ 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. $\endgroup$
    – statfan
    Commented Apr 1, 2021 at 3:32
  • $\begingroup$ 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. $\endgroup$
    – charlus
    Commented Apr 1, 2021 at 14:18

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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 $P[\tau=\tau'<1]=1$.

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