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 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 $I_{x,y} \le (C_1\vee C_2)|x-y|^{(\alpha-\epsilon \alpha)/(\alpha+1)}$
 A: 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.
