In the lecture we learned the following strong Markov property of the Brownian motions
Theorem. Let $T$ be a stopping time, and let $Y$ be a bounded random variable. Then we have $$E[Y\circ \theta^T|\mathcal{F}_T]=E_{B_T}[Y] \text{ on } {T<\infty}.$$
Here $\theta$ denotes the shift operator. It is the version that is stated in the book of Durrett. I saw another version of the Markov Strong Property. Can we prove that they are equivalent?
Theorem. Let $T$ be a stopping time that is almost surely finite. Then the process $X$ defined by $$X_t=B_{T+t}-B_{T}$$ is a standard Brownian motion independent of $\mathcal{F}_T$.
Thanks for any comment!