# Strong Markov property of Brownian motion on $[0,1]$.

Let $$\{B_t\}_{t\geq 0}$$ be standard Brownian motion with respect to filtration $$\{\mathcal{F}_t\}_{t\geq 0}$$. Let $$T$$ be any stopping time. Is it true that $$B_1-B_{T\wedge 1}$$ independent of $$\mathcal{F}_T$$, or is it true that $$\mathbb{E}[|B_1-B_{T\wedge 1}|\,|\,\mathcal{F}_T]=\mathbb{E}[|B_1-B_{T\wedge 1}|]$$?

I am not sure whether $$\{\mathcal{F}_t\}_{t\geq 0}$$ is right continuous or not, but I know $$B_t$$ has continuous path and $$B_{t\wedge 1}$$ is a uniformly integrable martingale. When I search strong Markov property for Brownian motion, textbooks or other notes usually give me $$B_{T+t}-B_T$$ is independent of $$\mathcal{F}_{T+}$$ and they usually assume $$P(T<\infty)>0$$. I am not sure whether these regularity assumption is needed. Can anyone help?

It is not true that $$\mathbb{E}[|B_1-B_{T\wedge 1}|\,|\,\mathcal{F}_T]=\mathbb{E}[|B_1-B_{T\wedge 1}|]$$? If this is true then (since $$(T>1) \in\mathcal F_T$$) we get $$0=\mathbb{E} I_{T>1}[|B_1-B_{T\wedge 1}|=\mathbb{E}[|B_1-B_{T\wedge 1}|]P(T>1)$$ which is clearly false.
• Thanks for your answer @KaviRamaMurthy. It seems we cannot directly remove $\mathcal{F}_T$, then do you think we can get rid of the conditional expectation by the following upper bounding process: $\mathbb{E}[|B_1-B_{T\wedge 1}|\,|\,\mathcal{F}_T]\leq 2\mathbb{E}[\sup_{t\in[0,1]}|B_t|\,|\,\mathcal{F}_T]=2\mathbb{E}[\sup_{t\in[0,1]}|B_t|]$? Feb 15, 2022 at 5:15
• Thanks again @KaviRamaMurthy, I want to do that because I am trying to upper bounding the term $\mathbb{E}[|B_1-B_{T\wedge 1}|\,|\,\mathcal{F}_T]$ by a constant. Currently I know $\mathbb{E}[|B_1-B_{T\wedge 1}|]$ and $\mathbb{E}[\sup_{t\in[0,1]}|B_t|]$ can be bounded above by a constant. If the conditional expectation cannot be avoided, do you know any approach that I could try? Feb 15, 2022 at 5:32