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

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    $\begingroup$ Use the fact that $\widetilde{W}_t := W_{t \vee \tau} - W_\tau$ is a Brownian Motion independent of $W_\tau$ by the strong markov property, and that subsequently so is $-\widetilde{W}_t$. $\endgroup$
    – quasi
    Commented Jun 30, 2015 at 20:00

2 Answers 2


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.

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    $\begingroup$ I like the patching idea but there are typos in your patching map. And actually proving that the law of the patched process only depends on the initial laws takes a few lines. $\endgroup$
    – AFK
    Commented Jul 1, 2015 at 6:27
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    $\begingroup$ @AFK it was edited. $\endgroup$
    – user16651
    Commented Jul 1, 2015 at 10:43
  • $\begingroup$ Stil some typos: $W_t$ should be $Y_\tau$ and one inequality should be strict. $\endgroup$
    – AFK
    Commented Jul 1, 2015 at 20:36

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.


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