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Let $\{B_t, \mathcal{F}_t; t\ge 0\}$ be a standard, one-dimensional Brownian motion. Can we construct a random time $S$ such that $P[0\le S < \infty] = 1$ and $W_t = B_{S+t} - B_S$ is not a Brownian motion?

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    $\begingroup$ Of course we can! For example, set $S$ to be the last time $t$ such that $B_t = -t$. Then $W_t > -t$ for every $t$, and hence $W_t$ is not a Brownian motion. $\endgroup$ Feb 20 '20 at 21:57
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    $\begingroup$ Or, perhaps even simpler: let $S$ be the first time $t$ such that $B_{t+1} = B_t$. Then $W_1 = 0$ with probability one. $\endgroup$ Feb 20 '20 at 21:58
  • $\begingroup$ I don't understand. Is not $P[S < \infty] = 0$? $\endgroup$ Feb 20 '20 at 22:43
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    $\begingroup$ @DieterKadelka: You mean in Mateusz's first example, where $S$ is the last time that $B_t = -t$? No, this is finite a.s., using for instance the strong law of large numbers. $\endgroup$ Feb 21 '20 at 2:29
  • $\begingroup$ Thank you, guys! Your comments are really helpful! $\endgroup$
    – Jacob Lu
    Feb 21 '20 at 5:07

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