# Construct a random time such that the strong Markov property of Brownian motion fails

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?

• 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. – Mateusz Kwaśnicki Feb 20 '20 at 21:57
• 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. – Mateusz Kwaśnicki Feb 20 '20 at 21:58
• I don't understand. Is not $P[S < \infty] = 0$? – Dieter Kadelka Feb 20 '20 at 22:43
• @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. – Nate Eldredge Feb 21 '20 at 2:29
• Thank you, guys! Your comments are really helpful! – Jacob Lu Feb 21 '20 at 5:07