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Let $(B_t)_{t\geq0}$ be a standard Brownian motion in $\mathbb{R}$ and let $T$ be a stopping time. Then the Strong Markov Property gives that $(B_{T+t}-B_T)_{t\geq0}$ is a standard Brownian motion independent of $\mathscr{F}_T$.

I am wondering if the same also holds for negative times: suppose that $T\geq1$ is a stopping time always greater or equal than $1$ (e.g. $T=\inf\{t\geq 1\,|\,B_t=0\}$), is it true that $(B_T-B_{T-t})_{t\in[0,1]}$ is a Brownian motion? The usual proof does not work since we must now work in a new filtration.

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No. Note that the stopping time $T$ can depend on all the values of $B_t$ with $t\le T$. So you could take something like $$T=\inf\{t\ge 1: B_t=0, B_{t-1/2}>0\}.$$ In this case $(B_T-B_{T-t})_{t\in [0,1]}$ cannot be a Brownian motion, since its value at $t=1/2$ is always negative.

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  • $\begingroup$ Great answer. Do you know if in the particular case of $T=\inf\{t\geq1\,|\,B_t=0\}$ we still get a Brownian motion? $\endgroup$
    – No-one
    Commented Nov 6, 2022 at 18:24
  • $\begingroup$ I now believe that the answer to my question is no because of Blumenthal's $0-1$ law: $X=(B_T-B_{T-t})_{t\in [0,1]}$ cannot be a Brownian motion because for each $\omega\in \{T>1\}$ (which is a set of positive measure), there exists an $\epsilon>0$ (depending on $\omega$) such that $X_t$ is not $0$ for all $t\in[0,\epsilon]$, while using Blumenthal's law it is easy to prove that a standard BM changes sign infinitely many times near the origin. $\endgroup$
    – No-one
    Commented Nov 6, 2022 at 23:09

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