# About independence in Strong Markov Property of Brownian Motion

Let $$T$$ be a finite stopping time and $$B_t$$ be a Brownian motion (with continuous paths).

The strong Markov Property states that $$B^T_t = B_{T+t} - B_{T}$$ is a Brownian motion and independent of $$\mathscr{F}_T$$.

In the proof, it claims if the following condition holds, then $$B^T$$ is independent of $$\mathscr{F}_T$$:

$$\mathbb{E}[1_{A}F(B^T_{t_1},\cdots,B^T_{t_p})] = \mathbb{P}(A) \mathbb{E}[F(B_{t_1},\cdots,B_{t_p})]$$

for every $$A \in\mathscr{F}_T$$ and $$0 \le t_1 < \cdots < t_p$$, and every bounded nonnegative continuous function $$F$$.

However, I think the above condition means $$\mathbb{E}[F(B^T_{t_1},\cdots,B^T_{t_p}) \mid \mathscr{F}_T]=\mathbb{E}[F(B_{t_1},\cdots,B_{t_p})]$$?

(I am reading Measure Theory, Probability, and Stochastic Processes by Jean-François Le Gall and the strong Markov property shows up in page 367.)

• I just found out that I missed something. First, it is shown that $B^T_{t}$ is a Brownian motion. Since the law of the Brownian motion is uniquely defined, $\mathbb{E}[F(B_{t_1},\cdots,B_{t_p})] = \mathbb{E}[F(B^T_{t_1},\cdots,B^T_{t_p})]$ Feb 2, 2023 at 7:40

A rv $$X$$ is independent from a $$\sigma$$-algebra $$\mathscr{G}$$ (i.e. $$\sigma(X)$$ is independent of $$\mathscr{G}$$) iff $$E[\mathbf{1}_Gf(X)]=P(G)E[f(X)],\,\forall G\in \mathscr{G},\,\forall f$$ nonnegative continuous and bounded.
Indeed: $$(\Rightarrow)$$ since $$\mathbf{1}_G,f(X)$$ are nonnegative and respectively are $$\mathscr{G}$$- and $$\sigma(X)$$-measurable, the statement follows from the result $$E[Zf(X)]=E[Z]E[f(X)]$$ for all $$Z$$ nonnegative s.t. $$\sigma(Z)\subseteq \mathscr{G}$$ and $$f:\mathbb{R}\to [0,\infty)$$ Borel (recall continuous implies Borel measurable). $$(\Leftarrow)$$: let $$y$$ be arbitrary, and choose a sequence $$f_n$$ continuous nonnegative bounded by $$1$$ s.t. pointwise $$f_n\downarrow \mathbf{1}_{(-\infty,y]}$$. We get by DCT $$E[\mathbf{1}_G\mathbf{1}_{(-\infty,y]}(X)]=\lim_{n \to \infty}E[\mathbf{1}_Gf_n(X)]=\lim_{n \to \infty}E[\mathbf{1}_G]E[f_n(X)]=P(G)E[\mathbf{1}_{(-\infty,y]}(X)]$$ This means $$P(G\cap\{X\leq y\})=P(G)P(X\leq y)$$ for all $$y \in \mathbb{R}$$. Since the family $$\{(-\infty,y],y \in \mathbb{R}\}$$ is $$\cap$$-stable and generates the Borel sets, it follows that $$\sigma(X)$$ and $$\mathscr{G}$$ are independent.