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