Reflection Principle

Question

Let $(\Omega,\mathcal{F},P)$ be a probability space and $\{W_t ∶ t ≥ 0\}$ be a standard Wiener process. By setting $\tau$ as a stopping time and defining \begin{align} W^*(t)=\Big\{\matrix{W_t\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,,t\leq\tau\cr2 W_{\tau}-W_t\,\,,\,t>\tau} \end{align} Why $W^*(t)$ is standard Wiener process? I want to solve it by Reflection Principle.is it Correct?Please help me

Answer 1

If $\tau$ is finite then from the strong Markov property both the paths $X_t = \{W_{t+\tau} −W_\tau ∶ t\geq 0\}$ and $−X_t = \{−(W_{t+\tau} − W_\tau) ∶ t \geq 0\}$ are standard Wiener processes and independent of $Y_t = \{W_t ∶ 0 \leq t \leq \tau\}$, and hence both $(X_t, Y_t)$ and $(X_t ,−Y_t)$ have the same distribution. Given the two processes defined on $[0, \tau]$ and $[0, \infty)$, respectively, we can paste them together as follows:

\begin{align} (Y,X)\rightarrow\{\,(X_{t-\tau}+W_t)1_{\{t>\tau\}}+Y_t 1_{\{t\leq\tau\}}+:t\geq0\} \end{align} Thus, the process arising from pasting $Y_t$ to $X_t$ has the same distribution ,which is $\{W_t ∶ t \geq 0\}$.In contrast, the process arising from pasting $Y_t$ to $-X_t$ is $\{W_t^* ∶ t ≥ 0\}$.Thus,$\{W_t^* ∶ t ≥ 0\}$ is also a standard Wiener process.

Answer 2

First note that paths are a.s continuous.

Then by strong Markov property and reflection principle, $(W_\tau - W_t)$ is a Brownian motion independant of the before tau part.

Then you can verify that increments are independent and gaussian by decomposing them in before and after tau part.

Or you can décompose the quadratic variation and use Lévy 's characterization.