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Let $B=(B_t)_{t \ge 0}$ be a Brownian motion on $\mathbb{R}^d$ starting from $x \in \mathbb{R}^d$. Let $y \in \mathbb{R}^d$ with $y \neq x$, and $R$ be the mirror reflection with respect to the $(d-1)$-dimensional hyperplane $H:= \{z \in \mathbb{R}^d \mid |z-x|=|z-y|\}$ bisecting $x$ and $y$. We define $W$ by \begin{align*} W_t= \begin{cases} RB_t,\quad &t<\tau,\\ B_t,\quad &t \ge \tau. \end{cases} \end{align*} Here, $\tau=\inf\{t>0 \mid B_t \in H\}$. Because $R$ is an orthogonal matrix, $W_t$ is also a $d$-dimensional Brownian motion starting from $y$. The pair $(B,W)$ is often called the mirror coupling of Brownian moitons.

My question

The mirror coupling $(B,W)$ is a strong Markov process on $\mathbb{R}^d \times \mathbb{R}^d$?

This claim seems to be true. However, I don't quite get it.

Please let me know if you have an simple reason.

sharpe
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  • I don't think one can do simple reasoning with the Strong Markov property. You will have to go through the details, unfortunately. Have you seen a proof of the SMP for Brownian motion? You can see the proof of theorem 0.2 here – Sarvesh Ravichandran Iyer Feb 11 '21 at 03:48
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    @TeresaLisbon Thank you for your comment. I think $(B,W)$ is obtained by connecting two strong Markov processes on $\mathbb{R}^d \times \mathbb{R}^d$. – sharpe Feb 11 '21 at 22:59
  • Yes, agreed. But I don't know how that helps, since the processes are coupled. We will have to think more about this one. By the way, your question is fascinating : where did you get the idea from that this is Strong Markov? Also which text are you consulting? – Sarvesh Ravichandran Iyer Feb 12 '21 at 10:10
  • I had two ideas, apart from going direct : one is that every Feller process is a Strong Markov process, by a result in Partszch and Schilling. The process you have looks like a Feller process to me. The next is to show that the resulting processes' probability distributions satisfy the martingale problem for a specific pair of functions, and then there's a result in Karatzas and Shreve about these probability distributions determining a unique Strong Markov Process. That's more involved, though. – Sarvesh Ravichandran Iyer Feb 12 '21 at 11:53

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