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 t<T_{x,y}:=\inf\{s>0 \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 $I_{x,y} \le (C_1\vee C_2)|x-y|^{(\alpha-\epsilon \alpha)/(\alpha+1)}$

  • $\begingroup$ 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? $\endgroup$ Jul 14, 2020 at 9:39
  • $\begingroup$ @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. $\endgroup$
    – sharpe
    Jul 14, 2020 at 9:47
  • $\begingroup$ Ah, sorry, I was seeing the inequality in the opposite direction... $\endgroup$ Jul 14, 2020 at 9:54
  • $\begingroup$ @MateuszKwaśnicki I see. But using the scale-invariance makes $I_{x,y}$ easier to see. Thank you. $\endgroup$
    – sharpe
    Jul 14, 2020 at 10:00
  • $\begingroup$ @MateuszKwaśnicki Sorry, I missed to calculate the index. So, I modified. $\endgroup$
    – sharpe
    Jul 14, 2020 at 11:04

1 Answer 1


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, $$ I_{x,y} \le C z_1^{\alpha/2} = C' |x - y|^{(1 - \epsilon) \alpha/2} .$$


  • 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.

  • $\begingroup$ 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... $\endgroup$
    – sharpe
    Jul 14, 2020 at 14:01
  • $\begingroup$ 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 $\alpha \ge 1$. $\endgroup$
    – sharpe
    Jul 14, 2020 at 16:14
  • $\begingroup$ @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. $\endgroup$ Jul 14, 2020 at 16:51
  • $\begingroup$ 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. $\endgroup$
    – sharpe
    Jul 14, 2020 at 17:07

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