currently I try to solve this following problem for myself. Let $\tau_a$ be a stopping time such that $\tau_a < \infty \; P\,\text{a.s.}$ (first hitting time of $a$) and consider the following "restarted" Brownian motion i.e. $\tilde{W}_t = W_{t+\tau_a} - W_{\tau_a} = W_{t+\tau_a}-a$. I want to prove that this is indeed again Brownian motion.
What is clear so far
- $\tilde{W}_0 = 0$ is clear
- $\tilde{W}_t - \tilde{W}_s \sim \mathcal{N}(0,t-s)$ is also clear (here one abuses that the theory of semigroups, where one can prove that the restarted Brownian motion shares the same transition group as standard Brownian motion.
What is open? I wanna prove that $\tilde{W}_{t}$ is indeed independent of $\mathcal{F}_{\tau}$. The idea should be that one uses that $W_t - W_s$ is independent of $\mathcal{F}^{W}_s$ and lift this to the more complicated sigma algebra.
My idea: It sufficies to prove that given any bounded Borel-measurable function $g : \mathbb R \to \mathbb R$ one has $$ E[g(\tilde{W_{t}}-\tilde{W_{s}}) Z] = E[g(\tilde{W_{t}}-\tilde{W_{s}})] E[Z] $$ where $Z : \Omega \to \mathbb R$ is a $\mathcal{F}_{\tau}$-measurable random variable that is bounded. This still seems to be hard, to further simplify I would like to give a monotone class argument, hence choose the following sets
- $\mathcal{H}=\{Z : \Omega \to \mathbb{R} : E[g(\tilde{W_{t}}-\tilde{W_{s}}) Z] = E[g(\tilde{W_{t}}-\tilde{W_{s}})] E[Z] \quad Z \in \mathcal{F}_{\tau}]$
But now I am stuck, since I don't know that good monotone class can be choosen such that the claim follows and that generates the sigma algebra $\mathcal{F}_{\tau}$?
Any hints/suggestions are welcome.
Thanks in advance