Let $B$ be a standard Brownian motion in $1$ dimension. Define \begin{equation} \tau = \inf \bigg\{ t \geq 0 : B_t = \max_{0 \leq s \leq 1} B_s \bigg\}. \end{equation} We want to show that $(B_{t+ \tau} - B_{\tau} )_{t \geq 0}$ is not a Brownian motion.

My ideas:

Let $\tilde{B}_t= B_{t+ \tau} - B_{\tau} $. Suppose on the contrary that $\{\tilde{B}_t\}$ is a Brownian motion.

I can only deduce by the definition that $$\mathbb{P} \bigg( \bigcup_{q \in \mathbb{Q}} \bigcap_{t \in [0,q]} \{\tilde{B}_t \leq 0 \} \bigg) =1.$$ Any ideas of how that leads to a contradiction?

(Should I somehow use the fact that $\tau <1$ a.s.?)

  • $\begingroup$ Yes, you need to use the fact that $\tau<1$ a.s. Once you know this, the process $\tilde{B}_t$ can't be Brownian because $\tilde{B}_{t} \le 0$ for $t = 1-\tau$, contradicting the fact that $\tilde{B}_{t}$ is normally distributed with mean $0$ and variance $t$. $\endgroup$ Dec 21, 2014 at 4:50
  • $\begingroup$ @LukasGeyer The fact that $\tilde B_t\leqslant0$ at a random time $t$ does not contradict that $\tilde B$ is a Brownian motion. $\endgroup$
    – Did
    Dec 22, 2014 at 6:05
  • $\begingroup$ @Did, of course you are correct. $\endgroup$ Dec 22, 2014 at 15:42
  • $\begingroup$ Got something from the answer? $\endgroup$
    – Did
    Jan 4, 2015 at 15:22

1 Answer 1


Every Brownian motion starting from $0$ enters $(0,+\infty)$ instantaneously, thus, $$\mathbb{P} \bigg( \bigcup_{q \in \mathbb{Q}} \bigcap_{t \in [0,q]} \{B_t \leqslant 0 \} \bigg) =0.$$


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