# Coupling times of subordinate Brownian motions

This is a question about coupling times of subordinate Brownian motions.

We fix $$y \in \mathbb{R}^d$$ with $$y \neq x$$ and define a map $$R_{x,y} \colon \mathbb{R}^d \to \mathbb{R}^d$$ by \begin{align*} R_{x,y}(z)=z-2 (z-(x+y)/2,x-y)\frac{x-y}{|x-y|^2},\quad z \in \mathbb{R}^d. \end{align*} We note that $$R_{x,y}$$ is the reflection with respect to the hyper plane $$H_{x,y}$$ such that the vector $$x-y$$ is normal with respect to $$H_{x,y}$$ and such that $$(x+y)/2 \in H_{x,y}$$. We write $$B^x=(\{B_t^{x}\}_{t \ge 0}$$ for the $$d$$-dimensional Brownian motion starting at $$x \in \mathbb{R}^d$$. We define $$W^y=(\{W_t^y\}_{t \ge 0})$$ by

\begin{align} W_t^y= \begin{cases} R_{x,y}(B_t^x),&\quad t0 \mid B_s^x \in H_{x,y}\}, \\ B_t^x,&\quad t \ge T_{x,y}. \end{cases} \end{align} The couple $$(B^x,W^y)$$ is called the mirror coupling of Brownian motions.

Let $$\{S_t\}_{t \ge 0}$$ be a subordinator, which is an increasing pure-jump Lévy process starting at zero independent of $$(B^x,W^y)$$. If we set $$X_t^x=B_{S_t}^x$$ and $$Y_t^y=W_{S_t}^y$$, $$t \ge 0$$. Then, $$(X^x,Y^y)$$ becomes a coupling of subordinate Brownian motions. Then, we denote by $$U_{x,y}$$ the coupling time of $$(X^x,Y^y)$$. By the result of this paper BSW, Theorem 2.1, we obtain that \begin{align*} U_{x,y}=\inf\{ t \ge 0 \mid S_t \ge T_{x,y}\}. \end{align*}

We denote by $$P_{x,y}$$ the law of $$(X^x,Y^y)$$ and $$f$$ the corresponding Bernstein function. By using the identity $$U_{x,y}=\inf\{ t \ge 0 \mid S_t \ge T_{x,y}\}$$, we obtain that \begin{align} (1)\quad P_{x,y}(U_{x,y} \ge t)\le \frac{|x-y|}{2\sqrt{2} \pi} \int_{0}^{\infty}\frac{e^{-tf(r)}}{\sqrt{r}}\,dr. \end{align}

See the proof of BSW, Theorem 2.1 for details. In particular, if $$X^x$$ is a symmetric $$\alpha$$-stable process, \begin{align*} (2)\quad P_{x,y}(U_{x,y} \ge t) \le C|x-y|/t^{1/\alpha},\quad t>0 \end{align*} Here, $$C$$ is a explicit constant.

My question

I think the equation $$U_{x,y}=\inf\{ t \ge 0 \mid S_t \ge T_{x,y}\}$$ (or (1)) is very useful, but I don't think it shows some geometric information such as where and how $$X^x$$ and $$Y^y$$ couple.

For some reasons, I study an asymptotic behavior of the probability $$I_{x,y}:=P_{x,y}(U_{x,y} \ge \tau_{B(x,|x-y|^{\varepsilon})}^X)$$ as $$x \to y$$ when $$X^x$$ is a symmetric $$\alpha$$-stable process. Here, $$\varepsilon<1$$ is a small number and $$\tau_{B(x,r)}^X=\inf\{t>0 \mid |X_t^{x}-x|>r\}$$, $$r>0$$.

We can easily deduce from the equation (1) that $$I_{x,y} \lesssim |x-y|^{(\alpha-\epsilon \alpha)/(1+\alpha)}$$ as $$x \to y$$. Just using (1), however, we do not know whether the index $$(\alpha-\epsilon \alpha)/(1+\alpha)$$ is optimal. Because $$I_{x,y}$$ should be a potential theoretical quantity, I also think that it should be possible to use another suitable method for a more precise estimate of $$I_{x,y}$$. Is there such a method?

ADD: By using (2), we have for any $$t>0$$, \begin{align*} I_{x,y} \le P_{x,y}(U_{x,y} >t)+P_{x,y}(\tau_{B(x,r)}^X \le t)\le C_1|x-y|/t^{1/\alpha}+C_2 tr^{-\alpha}. \end{align*} Here, $$C_1, C_2$$ are positive constant. If we set $$t=|x-y|^{\eta}$$, $$r=|x-y|^{\varepsilon}$$, we arrive at $$I_{x,y} \le (C_1\vee C_2)(|x-y|^{1-\eta/\alpha}+|x-y|^{\eta-\varepsilon \alpha})$$. Thus, if we take $$\eta>0$$ such that $$1-\eta/\alpha=\eta-\varepsilon \alpha$$, we have $$I_{x,y} \le (C_1\vee C_2)|x-y|^{(\alpha-\epsilon \alpha)/(\alpha+1)}$$

• There's something fishy going on here: by scale-invariance, if $|z| = 1$, then $I_{x,y} = P_{0,z}(U_{0,z} \ge \tau_{B(0, |x-y|^{\epsilon-1})}) \to 1$ as $|x - y| \to 0$. Am I missing something? Jul 14, 2020 at 9:39
• @MateuszKwaśnicki Thank you for your reply. I'm sorry. I assume that $\varepsilon<1$. In this case, $\tau_{B(0,|x-y|^{\varepsilon-1})}$ should go to infinity as $x \to y$. So, $I_{x,y}$ should go to $0$ as $x \to y$ because $(X^x,Y^y)$ is known to be successful. Jul 14, 2020 at 9:47
• Ah, sorry, I was seeing the inequality in the opposite direction... Jul 14, 2020 at 9:54
• @MateuszKwaśnicki I see. But using the scale-invariance makes $I_{x,y}$ easier to see. Thank you. Jul 14, 2020 at 10:00
• @MateuszKwaśnicki Sorry, I missed to calculate the index. So, I modified. Jul 14, 2020 at 11:04

Here is how I would approach the problem.

The coupling time $$U_{x,y}$$ is not greater than the first exit time from $$H_{x,y}^+$$, the half-space bounded by $$H_{x,y}$$ and containing $$x$$, by the process $$X_t^x$$. Thus, $$I_{x,y} \le \mathbb{P}^x(\tau_{H_{x,y}^+} \ge \tau_{B(x, r)}) ,$$ where $$r = |x-y|^\epsilon$$. By scale-invariance, we find that $$I_{x,y} \le \mathbb{P}^z(\tau_{H^+} \ge \tau_{B(z, 2)}) ,$$ where $$H^+ = \{x : x_1 > 0\}$$, $$z = (z_1, 0, \ldots, 0)$$ and $$z_1 = \tfrac{2}{r} |x - y| = 2 |x - y|^{1 - \epsilon}$$. It follows that as long as $$z_1 < 1$$, $$I_{x,y} \le \mathbb{P}^z(\tau_{H^+} \ge \tau_{B(0, 1)}) .$$ The right-hand side decays as $$z_1^{\alpha/2}$$ when $$z \to 0$$ (it is a positive $$\alpha$$-harmonic function of $$z$$ in $$B(0, 1) \cap H^+$$). Therefore, $$I_{x,y} \le C z_1^{\alpha/2} = C' |x - y|^{(1 - \epsilon) \alpha/2} .$$

Remarks:

• This bound should be optimal, I believe.
• The above remark seems to be in conflict with your claim with exponent $$(1 - \epsilon) \alpha / (1 + \alpha)$$, so I may have made an error in the above calculation.

Edit: OK, now I think both bounds are sub-optimal, but the minimum of the two could be sharp. Here is a possible approach.

First of all, the problem is essentially one-dimensional: the probability of leaving a ball before coupling time should be comparable with the probability of leaving a strip.

In dimension one, consider the process $$X_t$$ (started at some $$x > 0$$) killed at the coupling time (with another process started at $$y = -x$$). This is a decent "stable-like" process in $$(0, \infty)$$, with intensity of jumps of the form $$c (|y - x|^{-1-\alpha} - |y + x|^{-1 - \alpha}),$$ and additionally killed with intensity $$c' x^{-\alpha}.$$ Locally this process behaves as the $$\alpha$$-stable one, but globally the intensity of jumps decays as $$|y - x|^{-2 - \alpha}$$.

I bet this process has been studied before, and estimates for the probability of hitting $$(r, \infty)$$ prior to death are known. And even if not, tools are readily available: this is a positive self-similar Markov process, and one can use the Lamperti–Kiu transformation together with fluctuation theory for Lévy processes to study these kind of problems.

• I am grateful for your very kind reply. I'll take a closer look at your reply from now on, but as you say, my bound is different from yours (that's why I added the proof of my bound above). Maybe my bound is wrong... Jul 14, 2020 at 14:01
• Your wrote "$I_{x,y} \le P^z(\tau_{H^+} \ge \tau_{B(0,1)})$ provided that $z_1<1/2$". However, this should be "$I_{x,y} \le P^z(\tau_{H^+} \ge \tau_{B(0,1/2)})$ provided that $z_1<1/2$", right (of course, this is a minor issue)?. I think your argument is correct and boundary Harnack principle is suitable to derive the bound. However, my bound seems optimal when $\alpha<1$... On the other hand, your bound seems optimal when $\alpha \ge 1$. Jul 14, 2020 at 16:14
• @sharpe: Right, I got the inequality wrong the second time today. :-) I added some more thoughts on the problem, but unfortunately I do not have time to further think about it now. Jul 14, 2020 at 16:51
• Thank you very much for your very thoughtful comments. Unfortunately, I don't have the ability to carry out your ideas right now, but I would love to study. Jul 14, 2020 at 17:07