Reflection Principle
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
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:
[Math Processing Error]$$ \begin{align} (Y,X)\rightarrow{,(X_{t-\tau}+W_t)1_{{t>\tau}}+Y_t 1_{{t\leq\tau}}+:t\geq0} \end{align} $$ 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.