# Strong Markov Property for Brownian Motion in reverse time

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.

## 1 Answer

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.

• Great answer. Do you know if in the particular case of $T=\inf\{t\geq1\,|\,B_t=0\}$ we still get a Brownian motion? Commented Nov 6, 2022 at 18:24
• 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. Commented Nov 6, 2022 at 23:09