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Reflecting Brownian Motion After a Stopping Time

Article Quant Q&A · Author: user16731
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Summary

The document asks why a process formed by keeping Brownian motion unchanged up to a stopping time and reflecting its later path around its value at that time is itself standard Brownian motion. The response uses the strong Markov property: after a finite stopping time, the Brownian increments form a Brownian motion independent of the path before the stopping time. Negating those future increments preserves their Brownian distribution.

The process can thus be viewed as joining the original pre-stopping path to either the future increments or their negatives. Both joined processes have the same law, so the reflected process is standard Brownian motion. A second response sketches alternate verification through independent Gaussian increments or Lévy’s characterization using quadratic variation. The argument assumes the stopping time is finite and relies on standard Brownian continuity and strong Markov properties. It concerns a probability result, not a trading strategy, and does not provide empirical evidence or broader applications.

Key ideas

  • After a finite stopping time, Brownian increments are independent of the path observed before that time.
  • Negating the post-stopping Brownian increments leaves their distribution unchanged.
  • Joining the unchanged past with the reflected future therefore yields a standard Brownian motion.
  • Independent Gaussian increments or Lévy’s characterization can also be used to verify the result.

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# Reflection Principle


# 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

## Answer by user16651 (score 1, accepted)

https://quant.stackexchange.com/a/18600

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 by AFK (score 2)

https://quant.stackexchange.com/a/18597

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.

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

This summary was written by Stratmill's research agent from the original; it is not a copy of the source.