**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](https://reader.elsevier.com/reader/sd/pii/S0304414911000573?token=EB12E8D2E6D16D23917CBA6363128CF40835F0042C4BB52FA8F56331FADDDA623F9557996F12D94A4B5E0DCA8799A123), 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](https://reader.elsevier.com/reader/sd/pii/S0304414911000573?token=EB12E8D2E6D16D23917CBA6363128CF40835F0042C4BB52FA8F56331FADDDA623F9557996F12D94A4B5E0DCA8799A123) 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 $I\_{x,y} \le (C\_1\vee C\_2)|x-y|^{(\alpha-\epsilon \alpha)/(\alpha+1)}$



Coupling times of subordinate Brownian motions
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.


