数理解析研究所講究録
第 1672 巻 2010 年 159-176

159

Hitting of a line by two-dimensional symmetricstable L\’evy processes: an approach based on
modified resolvents
Yasuki Isozaki1
Department cf Mathematics, Graduate School of Science,
Osaka University, Toyonaka, Osaka 560-0043, Japan.
Abstractwith $1<\alpha\leq 2$ . We
Let $(X(t), Y(t))$ be a symmetric -stable L\’evy process onannounce a multivariate aasymptotic estimate involving the first hitting time/placeof a half-line. We deduce explicitly the density of the first hitting distribution of aline. The method is based on some modified version of quantities in the celebrated
potential theory. We also discuss properties of quantities arising in our modification.
$\mathbb{R}^{2}$$\alpha$1

Introduction and the result

In [4] and [5], the author stud-ed trivariate asymptotic estimates involving the first hitting
time of the nonnegative-half the first axis. he first hitting place thereon, and the sojoum
time on the first axis up to then, by a random walk and a Brownian motion, respectively.
Note that more precise in-$=ormation$ can be retrieved from this kind of estimates suchas the tail probability $conc_{-}.erning$ both the first hitting time and place, than from the tailprobability of the first hitting time.
Let $1<\alpha\leq 2$ . In this note, we are niainly concerned with the -stable L\’evy process $(X(t), Y(t))$ with rotational symmetry on. Its lawstarting fromand expectation are denoted by, respectively, and are determined byand. Let $L_{Y}(t)$ be the local time at for
for$0_{-}^{-}\wedge$$\uparrow$$\alpha$$(x_{0}, y_{0})\in \mathbb{R}^{2}$$\mathbb{R}^{2}$$E_{(x_{0}.y_{0})}$$P_{(x_{0}.y_{f)})}$$E_{(0,0)}[e^{i\xi_{1}X(t)+i\xi_{2}Y(t)}]=e^{-t(\xi_{1^{+\cdot c2}}^{2}.)^{0/2}}\backslash 2$$Y( \cdot):L_{Y}(t)=\lim_{\epsilonarrow+0^{\frac{1}{2\epsilon}}}\int_{0}^{t}1_{(-\epsilon,\epsilon)}(1^{\nearrow}(s))ds$

For$a\in \mathbb{R}$.

, we set$= \inf\{t\geq 0|Y(t)=0_{5}.X(t)\geq a\}$ .

$\tau(a^{\backslash },$We also set$0$$(\xi_{1}.\xi_{2})\in \mathbb{R}^{2}$$\Phi_{a}(\xi_{1}, \mu_{2})=2\pi/\int_{\mathbb{R}}\frac{(l_{\backslash 2}^{c}}{\mu 2+(\xi_{l}+\backslash 2)^{c1/2}}\dot{\prime}C_{1}((\nu)=\Phi_{a}(1,0)=2\pi/B(\frac{1}{2}, \frac{a-1}{2}),$$\Phi_{a}(0,1)=\alpha\sin\frac{\pi}{\alpha}$$\xi_{1}\in \mathbb{R}$$C_{2}(\alpha)=$, and$I_{\alpha}( \mu_{0}.\mu_{1,L^{l_{2}})}=\int_{-\cdot x}^{\infty}\frac{dt}{2\pi(t^{2}+1)}\log(\mu_{0}+\Phi_{\alpha}(\mu_{1}t_{l}\iota_{2}))$

for

(1.1)

and $\mu_{i}\geq 0(i=0,\cdot 1,2)$ such that$l^{\iota_{0}}+l^{\iota_{1}}+A\iota_{2}>0$

(1.2)

.

lThe author was partly supported by a grant of Japan Society for the Promotion of Science, no.
18740053. E-mail: yasuki@]nath.sc .osaka-u ac$i$$\dot{I}1\supset$

160

To state the main theorem, we introduce a family of holomorphic functions.
and set$\mathbb{C}_{+}=\{z\in \mathbb{C}|\Im z>0\},$(1.3)$\varphi_{a}(z;\mu_{0}, \mu_{2})=\exp(\frac{-1}{2\pi i}\int_{-t\lambda}^{\infty}\frac{z}{t^{2}-z^{2}}\log(\mu_{0}+\Phi_{a}(t, \mu_{2}))dt)$

forand $\mu_{i}\geq 0(i=0,2)$ such that $l^{l_{0}}+\mu_{2}>0$ . We can extend
by continuity. We also set$z\in \mathbb{C}_{+}$$z\in \mathbb{R}$Let$\overline{\mathbb{C}_{+}}=\{z\in \mathbb{C}|\Im z\geq 0\}$$\varphi_{\alpha}(z;0,0)=\frac{1}{\sqrt{C_{1}((y)}}(-iz)^{-(\alpha-1)/2}$

for$z\in\overline{\mathbb{C}_{+}}\backslash \{0\}$$\varphi_{a}(z;\mu_{0}, \mu_{2})$for,where we employ the branch such that $1^{-\{\cap-1)/2}=1$ .
Theorem 1.1 Let $a>0,$ $\mu_{\mathfrak{i}}\geq 0(i=0.1,2)$ , and $\mu_{0}+\mu_{1}+\mu_{2}>0$ .
(i) It holds$E_{(-a,0)}[e^{-\mu oL)(\tau(0))-\mu_{1}X(\tau(0))-\mu 2^{\tau(0)}}]$.

$=e^{\mu_{1}a}- \frac{e^{\prime r_{1}a}}{\varphi_{a}(i\mu_{1};\mu_{0}.\mu_{2})}\int_{-x}^{\infty}d\theta\frac{1-e^{-ia(\theta-i\mu_{1})}}{2\pi i(\theta-i\mu_{1})}\varphi_{\alpha}(\theta;\mu_{0}, \mu_{2})$(ii) As $sarrow+O$ ,$1-E_{(-a,0)}[e^{-\mu_{1)}s^{2(c)- 1)}L_{1}\cdot(\tau(0))-\mu_{1}s^{2}X(\tau(0))-\mu_{2}s^{2\alpha}\tau(0)}]$$\sim$

where$\sim$,$\frac{\exp(I_{o}(l^{l_{0\backslash }}\mu_{1\backslash }l^{\iota_{2}))}-1)/2}{\sqrt{C_{1}((y)}I^{\urcorner}(1+\frac{\cap-1a^{(\alpha}}{2})}s^{o-1}$

means that the ratio of the both sides converges to 1.

, weSince the method of proof applies to symmetric -stable L\’evy processes onrestate the theorem for such processes in the forthcoming paper [6]. Here $(X(t), Y(t))$ issymmetric iff $(X(t)-X(0), Y(t)-Y(0))$ has tlie same law as $(X$ (0) $-X(t), Y(0)-Y(t))$ .
In Section 4, we give a generalization of the theorem for such $(X(t)\dot{\prime}Y(t))$ that $X(t)$ and$Y(t)$ are independent, $X(t)$ is symmetric-stable, and $Y(t)$ is symmetric -stable.
We could not calculate $ex$ ] $)litit.]\}$ t.he dOfinite integral
defined in (1.2)
andbut some marginal values can be evaluated. e.g.,$\mathbb{R}^{2}$$\alpha$$l^{i}$$\alpha$$I_{o}(\mu_{0}, \mu_{1}, \mu_{2})$$c^{3}xp(I_{o}(0, \{\iota_{1},0))=\sqrt{C_{1}(\alpha)}\mu_{1}^{(a-1)/2}$$\exp(I_{\alpha}(0,0, \mu_{2}))=\sqrt{C_{2}(\alpha)}\mu_{2}^{(0}1)/0$

It is elementary to obtain the following corollary by a Tauberian theorem, the strong
Markov property, and Theorem 1.2(i) below.
Corollary 1.1 (i) We have$P_{(-a,0)}[ \tau(0)>A]\sim\frac{\sqrt{C_{2}((v}l)/2}{\sqrt{C_{1}((v)}I^{\backslash }(1+\frac{)a_{1}^{t\prime\}}(\}}{2})\Gamma(1-\frac{n-1}{20})}A^{-\frac{(0-1)}{20}}$

as $Aarrow+\infty$ ,

161

where
(ii) If.
and$sarrow+O$,, we have, as$C_{1}( \alpha)=2\pi/B(\frac{1}{2}, \frac{\alpha-1}{2})$$y_{0}\neq 0$

and$x_{0}\in \mathbb{R}$$C_{2}(()!)= \zeta y\sin\frac{\pi}{\alpha}$$1-E_{(x_{0},yo)}[e^{-\mu 0s^{2(0-1)}L_{Y}(\tau(0))-\mu_{1}s^{2}X(\tau(0))-\mu s^{20}\tau(0)}2]$$\sim$.

$s^{\alpha-1} \frac{\exp(I_{a}(\mu_{0},\mu_{1},)}{\sqrt{C_{1}(\alpha)}\Gamma(1+\frac{\mu_{2})\alpha-1}{2})}\int_{-\infty}^{-xo/|yo|}\frac{(1+t^{2\alpha/2}}{B(\frac{1}{2},\frac{(x-1)^{-}}{2})}|x_{0}+|y_{0}|t|^{(\alpha-1)/2}dt$

In the course of the proof of Theorem 1.1. we obtain an explicit formula for the firsthitting distribution of a line.
is determined by the characterisThe law of a L\’evy process $(X(t), Y(t))$ on. If
tic exponent
forsatisfying$(X(t), Y(t))$ is symmetric in the sense that $(X(t)-X(0), Y(t)-Y(O))$ has the same law
as $(X$ (0) $-X(t), Y(0)-Y(t))$ , we haveand hence is real-valued.
$\mathbb{R}^{2}$$E_{(0,0)}[e^{i\xi_{1}X(t)+i\xi_{2}Y(t)}]=e^{-t\Psi(\xi_{1},\xi_{2})}$$\Psi(\xi_{1}, \xi_{2})$$\Psi(\xi_{1}, \xi_{2})=\Psi(-\xi_{1}, -\xi_{2})$$(\xi_{1}, \xi_{2})\in \mathbb{R}^{2}$$\Psi$$:= \inf\{t\geq 0|Y(t)=0\}$ .
Theorem 1.2 Set(i) Let $(X(t), Y(t))$ be an -stable Levy process with rotational symmetry on
andbe a real random variable such that $P[C( \alpha.rot\in dx]=B(\frac{1}{2}, \frac{a-1}{2})^{-1}(1+x^{2})^{-a/2}dx$ .
Then $P_{(x0,yo)}[X(T_{0}^{Y})\in dx]=P[y_{0}C_{\alpha.rot}+x_{0}\in dx]$ .
(ii) More generally, if $(X(t), Y(t))$ is a genuinely two-dimensional symmetric -stable$T_{0}^{Y}$$\mathbb{R}^{2}$$\alpha$$C_{\alpha,rot}$$\alpha$Ldvy process such that

$E_{(0,0)}[e^{r\xi_{1}X(t)+i\xi_{2}Y(t)}]=e^{-t\Psi(\xi_{1}.\xi_{2})}$, set$P[C_{\Psi} \in dx]=\frac{\Psi(1,x)^{-1}dx}{\int_{\mathbb{R}}\Psi(1,t)^{-1}dt}$.

Then $P_{(x_{O},y_{0})}[X(T_{0}^{Y})\in dx]=P[y_{0}C_{\Psi}+x_{0}\in dx]$ .

The proof is given in Section 2 by an approach based on modified resolvents. We
characterize some quantities related t,o modified resolvents in Section 5.
In Section 2, we also study the hitting times of two parallel lines and some formula
concerning the last exit time from a line.
To our knowledge, there are only two papers in the literature concerning explicit hitting distribution of sets by multidimensional stable L\’evy processes. [2] obtained the firsthitting distribution of $\{x\in \mathbb{R}^{d}||x|>7’\}$ aiid $\{x\in \mathbb{R}^{d}||x|<r\}$ , and [7] obtained that of$\{x\in \mathbb{R}^{d}||x|=r\}$ , by an-stable Lcvv pro $(ess^{1}es$ with rotational symmetry. Theorem 1.2
is restricted to the case for dimension 2, but needs not the rotational symmetry. Unfortunately, the author has not succeeded in extending our result to the case for dimension
3 or higher.
It seems interesting to compare Theorem 1.2 with the formula (5.12) in [8], which)
L\’evy process. Let $X(t)$ and $Y(t)$concentrates on the one-dimensional symmetric
L\’evy $1’\in$ with $1<\alpha\leq 2$ and $P[C_{a}\in dx]=$
are independent symmetric. Then it is shown that $P_{(xo,yo)}[X(T_{0}^{Y})\in dx]=P[y_{0}C_{\alpha}+x_{0}\in dx]$ .
Our Theorem 1.2(ii) contains this formula: in this case we have
) $leC_{J}$ is called an
and $P[C_{\Psi}\in dx]=P[C_{\alpha}\in dx]$ . The-Cauchy variable in [8] since$(Y=2$
its law reduces to the Cauchy distribution if.
$\alpha$$cv-\backslash \backslash tal$$]e$$\alpha-\llcorner\backslash t_{\dot{\mathfrak{c}}}\iota I)1e$$\frac{\alpha}{2\pi}\sin(\frac{\pi}{a})(1+|x|^{\alpha})^{-1}dx$$\Psi(\xi_{1}, \xi_{2})=|\xi_{1}|^{\alpha}+|\xi_{2}|^{\alpha}$$varia\dagger$$\alpha$162

Let us also remark that Theorem 12(i) and 8, (5.12) are different stable-analogsof the hitting distribution of a line by a two-dimensional standard Brownian motion,
namely the Cauchy distribution. A two-dimensional standard Brownian motion has theindependent components and is of rotational symmetry. But a two-dimensional symmetric-stable L\’evy process does not have these two properties at the same time. [8] retains
independence of components while Theorem 1.2(i) is based on rotational symmetry. We
may consider
as another -Cauchy variable.
$[$

$]$

$\alpha$$C_{a,rot}$2$\alpha$Modified resolvents and proof of Theorem 1.2

In this section, we introduce the modified resolvents $U(dy;\xi_{1}, \mu)$ and its density $u(y;\xi_{1}, \mu)$and apply them to determine the joint law of the first hitting time and place of a line.
The resolvents $U(dy;\xi_{1}, \mu)$ are modified ones in the sense that they reduce, if $\xi_{1}=0$ , to-resolvents for a one-dimensional L\’evy pro
as in [1, \S I.2].
Let $(X(t), Y(t))$ be a two-dimensional L\’evy process starting from. Its
law and expectation are denoted by, respectively. Letandbe the$\sigma((X(s),Y(s));s\in[0.
t])$. We denote its characteristic exponent by-completion of, i.e. it holds
for.
Assumesatisfies$(e\llcorner ssY(t)$$\mu$$(x_{0}, y_{0})\in \mathbb{R}^{2}$$P_{(x_{0},yo)}$$E_{(x0,y_{0})}$$\mathcal{F}_{t}$$P_{(x_{0},yo)}$$E_{(0,0)}[e^{i\xi_{1}X(t)+i\epsilon_{2}Y(t)}]=e^{-t\Psi(\xi_{1}.\xi_{2})}$$\Psi(\xi_{1}, \xi_{2})$$(\xi_{1}, \xi_{2})\in \mathbb{R}^{2}$$\Psi(\xi_{1}, \xi_{2})$$\int_{\mathbb{R}}|\frac{1}{1+\Psi(0\backslash \xi_{2})}|d\xi_{2}<\infty$

.

(2.1)

Then it is well-known(see [1, Corollary II.20, Theorem V.l, and Proposition V.2]) that
$Y(t)$ admits a local time process)
and $t\mapsto L_{Y}(y, t)$is a.s. continuous.
Note that (2.1) is a bit stronger than the existence of such $L_{Y}(y, t):(2.1)$ implies thatand a single point is regular for itself for $Y(t)$ ; these conditions are
sufficient for the existence ofas above. We assume (2.1) since it facilitates (2.5)
below and the L\’evy processes of ouras symmetric -stable processes with
$1<\alpha\leq 2$ , satisfy (2.1).
One can show that, for any bounded Borel function $f(y)$ on ,$L$$\cdot\cdot(y, t)=\lim_{\epsilonarrow+0^{\frac{1}{2\epsilon}}}\int_{0}^{t}1_{\{|Y(s)-y|<\epsilon\}}ds$$\int_{\mathbb{R}}\Re\frac{1}{1+\Psi(0,\xi_{2})}d\xi_{2}<\infty$$L_{\gamma}\cdot(y.

t)$$intere:.\backslash \uparrow,$$|\backslash \backslash tth$$\alpha$$\mathbb{R}$$\int_{0}^{u}e^{i\xi\iota X(\ell)}f(Y(t))dt=\int_{1R}cfyf(y)\int_{0}^{u}e^{i_{\backslash 1}^{c}X(t)}d_{t}L_{Y}(y, t)$(2.2)

by standard arguments. Set

for$\xi_{1},$$y\in$$U(dy;\xi_{1}.\mu)$$.=$$E_{(0())}[ \int_{0}^{\infty}e’\backslash 11_{\{Y(t)\in dy\}}dtc\chi(\ell)-/t]$$u(y;\xi_{1,l}\iota)$$:=$$E_{(0(1)}[ \int_{0}^{\infty}e^{\prime\xi_{1}\lambda(t)-\mu l}d_{t}L_{Y}(y, t)]$IR and $\mu>0$ .

.

(2.3)
(2.4)

163

Note that these quantities correspond to the following ones in [1] if $\xi_{1}=0$ : (2.2)
reduces to $\int_{0}^{u}f(Y(t))dt=\int_{\mathbb{R}}dyf(y)L_{1}\cdot(y.u)$ in [1, \S V. 1]; $U(dy;0, \mu)$ is the -resolvent$U^{\mu}(O, dy)$ for $Y(t)$ in [1, \S I.2]; and then$\mu$$u(y;0, \mu)=E_{(0_{\partial}0)}[\int_{0}^{\infty}e^{-/4}{}^{t}d_{t}L_{1}\cdot(y, t)]=\frac{1}{2\pi}\int_{\mathbb{R}}\frac{e^{-iy\xi_{2}}}{\mu+\Psi(0,\xi_{2})}d\xi_{2}$is the continuous version of the density for $U^{\mu}(0, dy)$ in [1, \S II.5]. In Section 5, we discuss
their properties from the potential theoretic viewpoint.

Lemma 2.1 Assume (2.1).
(i) The function $y\mapsto u(y;\xi_{1}, \mu)$ is a version of the density for $U(dy;\xi_{1}, \mu)$ .
(ii) Assume
integrable for any fixed . Then we have$\xi_{2}\mapsto 1/|\mu+\Psi(\xi_{1}, \xi_{2})|\iota s$$\xi_{1}$$u(y; \xi_{1}.\mu)=\frac{1}{2\pi}\int_{R}\frac{e^{-iy\xi_{2}}}{l^{l}+\Psi(\xi_{1},\xi_{2})}d\xi_{2}$

(2.5)

.

Proof.

We refer the reader to the forthcoming paper [6] for the proof.
Note that if
is integrable for some $\mu>0$ , then it is integrable for any$\mu>0$ .
Note also thatis integrable if the process is genuinely two-dimensionaland. Indeed,vanishes only at$(0,0)$ , and we have $1/|\mu+\Psi(\xi_{1}, \xi_{2})|\sim|\xi_{2}|^{-}’/|\Psi(\xi_{1}/\xi_{2},1)|\sim|\xi_{2}|^{-a}/|\Psi(0,1)|$ as.
A similar bound holds when.
We next set, for any fixed
and $\mu>0$ ,$\square$$1/|\mu+\Psi(\xi_{1}, \xi_{\wedge}r’)|$$1/|\mu+\Psi(\xi_{1}, \xi_{2})|$$\forall c>0,\forall(\xi_{1}, \xi_{2}),$$\Psi(c\xi_{1}, c\xi_{-},)=c^{\alpha}\Psi(\xi_{1}.\xi_{2})$$\Re\Psi(\xi_{1}, \xi_{2})\geq 0,$$\Psi$$\xi_{2}arrow\infty$$\xi_{2}arrow-\infty$$\xi_{1}\in \mathbb{R}$(2.6)$N(t)=e^{\xi_{1}X(\ell)-\mu t}u(-Y(t);\xi_{1}, \mu)$ .

This process is bounded since $|u(y;\xi_{1,l^{\lambda}})|\leq u(O:0, \mu)$ by (2.4).

Lemma 2.2 Assume (2.1). Then for any starting point(i)
is a $u.i$ . martingale;(ii) $M(t)=e^{(1/u(0,\xi_{1},\mu))L_{Y}(0t)}N(t)$ is a local martingale.

$(x_{0}, y_{0})\in \mathbb{R}^{2}$, under $P_{(x_{0},yo);}$$N(t)+ \int_{0}^{t}e^{i\xi_{1}X(s)-\mu s}d_{s}L_{Y}(0,\cdot s)$

Proof. We refer the reader to the forthcoming paper [6] for the proof.

$\square$Let$L_{Y}^{-1}(t)$Then, under$P_{(x_{0},0)},$$:= \inf\{s\geq 0|L_{Y}(0.s)>t\}$$(\Xi(t), L_{Y}^{-1}(t))$

and$\Xi(t)=X(L_{Y}^{-1}(t))$ .

(2.7)

is a two-dimensional L\’evy $proc\cdot ess$ starting from $(x_{0},0)$ .

Lemma 2.3 Assume (2.1)
the condition in Lemma 2.1(ii).
$(\Xi(t)!L_{Y}^{-1}(t))$
Then the L\’evy process
has the following Fourier-Laplace characteristic$a\gamma_{l}d$exponent:$E_{(0,0)}[e^{i\xi_{1}\Xi(t)-\mu L_{Y}^{-1}(t)}=e^{-t\Phi(\xi_{1},\mu)}u\prime ith$$\Phi(\xi_{1}, \mu)=2\pi/’\int_{R}\frac{1}{l^{l}+\Psi(\xi_{1},\xi_{2})}(l\xi_{2}$for$\xi_{1}\in$

If $(X(t), Y(t))$ is a genuinely two-dimensional symmetricalso valid for
and $\mu=|)$ .
$\xi_{1}\neq 0$$\alpha$IR and$l^{\chi}>0$

.

(2.8)

-stable L\’evy process, (2.8) is164

Proof.

If $\mu>0$ , we stopat$\Lambda f(t)$$L_{Y}^{-.1}(t)$to obtain a bounded martingale. Then

we have $E_{(0,0)}[e^{(1/u(0_{t}\xi_{1},\mu))t}e^{-\mu L_{\}}^{-.1}(\ell)+i\xi_{1}X(L_{)}^{1}(\ell))}u(0;\xi_{1,l}\iota)]=M(0)=u(0;\xi_{1)}\mu)$ , which
implies (2.8) by (2.5).
. If $(X(t), Y(t))$ is a genuinely two-dimensional symmetric -stable L\’evy
Fix
. The
and
as
process, we havecondition in Lemma 2.1(ii) is satisfied, as is seen in the arguments following the proof ofby the domiLemma 2.1. On one hand, we have$\xi_{1}\neq 0$$\alpha$$|\xi_{2}|arrow\infty$$\Psi(\xi_{1}, \xi_{2})\sim|\xi_{2}|^{a}\Psi(0,1)$$\inf_{\xi_{2}\in \mathbb{R}}\Psi(\xi_{1}, \xi_{2})>0$$\lim_{\muarrow+0}\Phi(\xi_{1}, \mu)=2\pi/\int_{\mathbb{R}}\frac{1}{\Psi(\xi_{1r}\xi_{2})}d\xi_{2}$nated convergence. On the other hand,$\exp(-t\lim_{\muarrow+0}\Phi(\xi_{1}, \mu))$.

$E_{(0,0)}[e^{i\xi_{1}\Xi(\ell)}]= \lim_{\muarrow+0}E_{(0,0)}[e^{i\xi_{1}\Xi(t)-\mu L_{Y}^{-1}(t)}]=$$\square$

Proof of Theorem 1.2. Fix$\xi_{1}>0$.

By the saine argument as the proof of Lemma 2.3,

we have$u(y; \xi_{1},0):=\lim_{\muarrow+0}u(y;\xi_{1,l}\iota)=\frac{1}{2\pi}\int_{\mathbb{R}}\frac{e^{-iy\xi_{2}}}{\Psi(\xi_{1},\xi_{2})}d\xi_{2}$, we have $\Psi(\xi_{1}, \xi_{2})>0$ for aiiy
Since
, we have
Stopping $M(t)$ at$\xi_{1}\neq 0$$\xi_{2}\in \mathbb{R}$.

, and then $u(0;\xi_{1},0)\in(0, \infty)$ .

$T_{0}^{Y}$.
$E_{(x_{0},y_{0})}[e^{i\xi_{1}X(T_{0}^{1})-\mu T_{0}^{)}}$$]= \frac{e^{i\xi_{1}x_{0}}u(-y_{0};\xi_{1},\mu)}{u(0;\xi_{1},\mu)}$.

We then let $\muarrow+0$ to obtain$E_{(x_{0},y_{3})}[e^{i\xi_{1}X(T_{0}^{\backslash })}]= \frac{e^{j\xi\iota x_{0}}u(-y_{0}\cdot,\xi_{1},0)}{u(0;\xi_{1},0)}$By substituting$\xi_{2}=\xi_{1}x$

(2.9)

.

, we have$u(y; \xi_{1},0)=\frac{1}{2\pi}\int_{\mathbb{R}}\frac{e^{-iy_{\backslash lJ}^{\zeta}}}{\Psi(\xi_{1\}\xi_{1^{J}}.r)}\xi_{1}dx=\frac{\xi_{1}^{1-a}}{2\pi}\int_{\mathbb{R}}\frac{e^{-iy\xi_{1}x}}{\Psi(1)x)}dx$

since$\Psi(c’\xi_{1}, c\cdot\xi_{2})=c^{\alpha}\Psi(\xi_{1}, \xi_{2})$Putting this into (2.9), we have.

$E_{(x_{0},yo)}[e^{i\xi_{1}X(T_{0}^{Y})}]$

$=$

$=$

$( \int_{1R}\frac{1}{\Psi(J.t)}dt)^{-1}\int_{\mathbb{R}}\frac{e^{i\xi_{1}y_{0}x+i\xi_{1}x_{0}}}{\Psi(1,x)}dx$$\int_{1R}e^{;\epsilon_{1}(yox\dashv x_{0})}(\int_{1R}\frac{1}{\Psi(1.t)}dt)^{-1}\Psi(1, x)^{-1}dx$.

The comlex conjugate of both sides yields the same formula for $\xi_{1}<0$ .
Then the right hand side is equal to $E[\exp(\uparrow\xi_{1}(y_{0}C_{\Psi}+x_{0}))]$ , where $P[C_{\Psi}\in dx]=$$( \int_{\mathbb{R}}\Psi(1, t)^{-1}dt)^{-1}\Psi(1, x)^{-1}dz$

2.1.

$\square$Appendix to Section 2: hitting of a line or two parallel lines

We determine the joint law of the first hitting $time/pla(e$ of a line or lines. We do notneed the content of this subsection in proving Theorem 1.2.

165

For any $a,$$b\in \mathbb{R}$such that $a\neq b$ , set$T_{a}^{y^{r}}$$=$$\inf\{t\geq 0|Y(t)=a\}$ ,$T_{a,b}^{Y}$$=$$\inf\{t\geq 0|Y(t)\in\{a, b\}\}$.

These are respectively the first hitting times of a line and two parallel lines.
The hitting time
can be decomposed at the last exit time from the line $\{Y=Y(0)\}$ :$T_{a}^{Y}$$G_{a}^{Y}= \inf\{t\leq T_{a}^{Y}|Y(t)=Y(0)\}$is independent of.
In the following lemma, (i) is an extension of a well-known fact, see $e.g$ . Corollary II. 18
in [1]. Moreover, (ii), (iii) and (iv) are extensions of Proposition 5.4, 5.5, and Theorem
5.8 in [8], respectively.
$T_{a}^{Y}-G_{a}^{Y}$

Lemma 2.4 Assume (2.1) and let(i) it holds(ii) it holds$\xi_{1}\in \mathbb{R},$$l^{\iota}>0,$$a\neq 0,$ $b\neq 0$, and $a\neq b$ . Then;

$E_{(0)0)}[e^{i\xi_{1}X(T_{a}^{Y})-\mu T_{a}^{y/}}]= \frac{u(a;\xi_{1},\mu)}{u(0;\xi_{1},\mu)}$$E_{(0,0)}[e^{i\xi_{1}X(T_{a,b}^{Y})-\mu T_{a,b}^{Y}}]$$=$;$\frac{(u(0;\xi_{1},\mu\grave{)}-u(b-a;\xi_{1\}}\mu,))u(a;\xi_{1\backslash l}x)+.(u(0;\xi_{1},\mu)-u(a-b;\xi_{1},\mu))u(b;\xi_{1},\mu)}{u(0;\xi_{1},\mu)^{2}-u(a-b)\xi_{1},\mu)u(b-a;\xi_{1},\mu)}$

if $(X(t), -Y(t))^{1aw}=(X(t), Y(t))$ then$E_{(0_{t}0)}[e^{i\xi_{1}X(T_{ab}^{Y})-\mu T_{o\dagger)}^{)}}.]= \frac{u(a;\xi_{1},\mu)+u(b;\xi_{1},\mu)}{\uparrow\iota(0;\xi_{1},\mu)+u(b-a;\xi_{1},\mu)}$;(iii) it holds,$E_{(0,0)}[e^{i\xi_{1}X(T_{b}^{Y})-\mu T_{b}^{Y}};T_{b}^{Y}<T_{a}^{\iota}.]= \frac{-u(b-a;\xi_{1\backslash }\mu)u(a;\xi_{1},\mu)+u(0)\xi_{1},\mu,)u(b;\xi_{1},\mu)}{u(0;\xi_{1\backslash }\mu)^{2}-u(a-b;\xi_{1},\mu)u(b-a;\xi_{1},\mu)}$(iv) it holds, with$h^{(\alpha)}(a)= \frac{|0|^{o- 1}}{2\Gamma(0)\sin\frac{(c)- 1)\pi}{2}}$,$E_{(0,0)}[e^{\xi_{1}X(G_{\alpha}^{1})1}-l^{l(}7rJ]$$=$$\frac{u(0,\xi_{1},\mu)^{2}-u(a_{\backslash }\xi_{1\backslash }\mu)u(-a;\xi_{1,l}\iota)}{2h^{(\circ)}(a)\Psi(0,1)^{-1}u(0;\xi_{1},\mu)}$$E_{(0,0)}[e^{i\xi_{1}(X(T_{\alpha}^{Y})-X(G_{a}^{\}}))-\mu(T_{a}^{)}-c_{o}^{Y})}]$$=$$\frac{2h^{((\})}(a)\Psi(0,1)^{-1}u(a;\xi_{1)}\mu)}{u(0,\xi_{1},\mu)^{2}-u(a\cdot\xi_{1},\mu,)u(-a;\xi_{1},\mu)}$

Proof. Let our process start from$(0$

. $-0)$ .

We stop$1\mathfrak{h}[(t)$at$E_{(0,-a)}[e^{\prime\xi_{1}X(T_{\{)}^{\}})-\mu T_{0}^{\}}}.]= \frac{u(a)\xi_{1},’\mu)}{u(0,\xi_{\iota l}\iota)}$.

$T_{0}^{Y}$

.

,

.

Since $L_{Y}(0, T_{0}^{Y})=0$ ,166

By the translation invariance, we have the statement of (i).
(ii) Letbe such thatand$c_{a}$$c_{b}$

1$=$$c_{a}u(0;\xi_{\iota l}\iota)+c_{b}u(b-a;\xi_{1}, \mu)$

,

1

$=$

$c_{a}u(a-b;\xi_{1}, \mu)+c_{b}u(0;\xi_{1}, \mu)$.

As a corollary to (i) we have
which ensures that the solution$|u(y)\xi_{1},$$\mu)|<|u(0;\xi_{1}, \mu)|$$(c_{a}.c_{b})$for any $y\neq 0,$$\xi_{1}\in \mathbb{R}$, and $\mu>0$ ,exist:$c_{a}$$=$$\frac{u(0,\xi_{1l}\iota)-u(a-b;\xi_{1},\mu)}{u(0;\xi_{1},\mu)^{2}-u(a-b;\xi_{1},\mu)u(b-a;\xi_{1},\mu)}$,$c_{b}$$=$$\frac{u(0;\xi_{1},\mu)-u(b-a;\xi_{1},\mu)}{u(0;\xi_{1},\mu)^{2}-u(a-b_{1}\cdot\xi_{1},\mu)u(b-a;\xi_{1},\mu)}$.

We define$M_{a,b}(t)=e^{i\zeta_{1}X(t)-\mu t}(c_{a}u(a-Y(t);\xi_{\iota\cdot l}\iota)+c_{b}u(b-Y(t);\xi_{1}, \mu))$Then$M_{a_{1}b}(t\wedge T_{a,b}^{Y})$.

is a bounded martingale. Now the statement in (ii) is equivalent to.

$E_{(0,0)}[e^{i\epsilon_{1}x(T_{a,b}^{Y})-\mu T_{\alpha,b}^{Y}}]=E_{(0,0)}[1lf(T_{ob}^{Y})]=M_{a_{y}b}(0)=c_{a}u(a;\xi_{1}, \mu)+c_{b}u(b;\xi_{1}, \mu)$If we put the symmetry assumption in (ii), $u(y;\xi_{1}, \mu)=u(-y;\xi_{1}, \mu)$ and hence $c_{a}=$$c_{b}=1/(u(0;\xi_{1}, \mu)+u(b-a;\xi_{1}, \mu))$ .
(iii) Letand be such that$c_{a}$$c_{b}$

,

$0$

$=$

$c_{a}u(0:\xi_{1}.\mu)+c_{b}u(b-a;\xi_{1}, \mu)$1$=$$c_{a}u(a-b:\xi_{1}.\mu)+c_{b}u(0;\xi_{1}, \mu)$ .

Then$c_{a}$$=$$\frac{-u(b-a;\xi_{1},\mu)}{u(0_{\backslash }\xi_{1\backslash l}\iota)^{2}-u(a-b\backslash \xi_{1},\mu)u(b-a;\xi_{1},\mu)}$,$c_{b}$$=$$\frac{v(0;\xi_{1},\mu)}{u(0;\xi_{1 l}/)^{2}-u(a-b,\cdot\xi_{1\backslash }\mu)u(b-a;\xi_{1},\mu)}$.

We define$N_{a,b}(t)=e^{i\xi_{1}X(t)-\mu t}(c_{a}u(0-Y(t);\xi_{1}, \mu)+c_{b}u(b-Y(t);\xi_{1}, \mu))$so that$N_{a_{1}b}(t\wedge T_{a,b}^{Y})$is another bounded martingale. Finally,$E_{(0,0)}[e^{j\xi_{1}\lambda(T_{b}^{\}})}\mu T_{:T_{b}^{)}}^{\backslash },\cdot\cdot<T_{a}^{)}.\cdot]$. $Y(T_{o,b}^{Y})=b]$$=$$E_{(0())}[c^{l_{\backslash I}^{t}\lambda(T_{o}^{\backslash },,)}0^{\cdot}l^{rT_{\cap}^{1}}.|)$$=$$E_{(0.0)}[N(T_{a.b}^{)})]$$=$$N_{ab}(0)=c_{o}v(a:\xi_{1}.\mu)+c_{b}u(b;\xi_{1}, \mu)$ .

167(iv) Recall that we normalize the local time of$Y(\cdot)$

at

by

$0$

$E_{(0,0)}[ \int_{0}^{\infty}e^{i\xi_{1}X(t)-\mu t}d_{t}L_{Y}(0.t)]=u(0, \xi_{1\cdot l^{l}})=\frac{1}{2\pi}\int_{\mathbb{R}}\frac{1}{\mu+\Psi(\xi_{1},\xi_{2})}d\xi_{2}$.

Let us introduce Ito’s excursion measure (see standard texbooks; we adopt the notations in [8, \S 3] . Let ID $=D([0, \infty);\mathbb{R}^{2})$ be the space of -valued c\’adl\’ag paths equippedwith the Skorohod topology. We define the random set$\mathbb{R}^{2}$$)$$D:=\{l>0|L_{Y}^{-1}(l)>L_{1’}^{-.1}(l-)\}$

and a point function $p(l)\in D$ on $D$ by$p(l)(t):=\{\begin{array}{ll}(X (t+L_{Y}^{-1}(l-)), Y(t+L_{Y}^{-1}(l-))), if t\in[0, L_{Y}^{-1}(l)-L_{Y}^{-1}(l-)),(X (L_{Y}^{-1}(1)), Y(L_{Y}^{-1}(l))), otherwise.\end{array}$Remark. $Y(L_{Y}^{-1}(l))=0$ but $X(L_{1}^{-,1}(l))$ needs not to be since $X$ is running freely.’
Then it is well-known thatis a Poisson point process. Ito’s excursion measure isdefined as follows: if$:=\{1\in D|p(l)\in U\}$ , set
and
$($

$0$

$p(\cdot)$$D_{U}$$U\in \mathcal{B}(D)$$n^{\Psi}[U]:=E_{(0.0)}[\#(D_{U}\cap(0,1])]$ .

The formula$E_{(0,0)}[e^{i_{\backslash 1}^{c}X(L_{)}^{-.1}(t))L^{-.1}(t)}-/l,]=e^{-t/u(0,\xi_{1},\mu)}$as in Lemma 2.3 implies that$n^{\Psi}[1-\circ xp(i\xi_{1}v_{1}(\zeta)-l^{\iota\zeta)}]=1/u(0;\xi_{1}, \mu,)$where$\zeta$(2.10),

is the lifetime of a generic excursion $u(\cdot)=(u_{1}(\cdot), u_{2}(\cdot))\in$ IID:$(= \zeta(u):=\sup\{t\geq 0|u_{2}(t)=0\}$ .

Note thatLet$T_{a}(u_{2})$($\mathbb{R}^{2}$needs not to end at .
be the first hitting time of$u_{1}(t)$$0$$a\in \mathbb{R}$by the second component$u_{2}(\cdot)$of a generic-valued) excursion $u\in D$ .

Set $U_{a}:=\{u\in D|T_{a}(u_{2})<\zeta(u)\}$ and recal that$:=\{l\in D|p(l)\in U\}$ . Then it
is well-known thatand
, are independent. Moreover, $n^{\Psi}[U_{a}]=n^{\Psi}[T_{a}(u_{2})<$$((u)]<\infty$ and then the first $ext^{-}ursion$ of, determines the hitting place; moreprecisely, if we set$D_{\zeta t}$$p|_{D_{U_{\alpha}}}$$p|_{D_{t_{tJ}^{r}}}$$X(T_{a}^{Y})$$p|_{D_{t_{rr}}}$$\kappa_{a}$$:= \inf\{l>0|p(l)\in U_{o}\}=\inf D_{U_{a}}$ ,we have.

.

$T_{a}^{Y}-G_{o}^{)}$$=$$T_{o}(p(\wedge\cdot O))$

$X(T_{a}^{Y})-X(G_{rx}^{)’}.)$$=$$p(\bigwedge_{(1})_{1}(T_{a}(p(\kappa_{a})))$

,

168

.
$=$

$G_{a}^{)}$$(0. \wedge CJ)\cap D_{\iota;_{o}}\sum_{\ulcorner}((p(l))$

,

$IE$

$X(G_{o}^{)}.)$$=$$l \in(0_{\tau}\kappa_{a})\cap D_{t\prime^{\Gamma}}\sum_{a}p(\kappa_{a})_{1}(\zeta(p(l)))$.

. is the first component of the first excursion.
inNote that
By the standard argument concerning the Poisson point processes, we can deduce thatare independent, so that $(T_{a}^{)’}-G_{a}^{Y}, X(T_{a}^{Y})-X(G_{a}^{Y}))$ and. Henceisare independent. The law of$p(\kappa_{a})_{1}$$\{p|_{D_{U_{a}^{C}}},$$\kappa_{a},$$($$p(\kappa_{a})$$)$$p|_{D_{U_{a}}}$$(G_{a}^{Y}, X(G_{a}^{Y}))$$p(\kappa_{a})\}$$n^{\Psi}[\cdot :$p(\kappa_{a})$U_{o}]/n^{\Psi}[U_{a}]$$E_{(0_{I}0)}[e^{c\epsilon_{1}(X(T_{a}^{Y})-X(G_{a}^{\gamma}))-\mu(T_{a}^{\backslash }-c_{a}^{Y})}]= \frac{n^{\Psi}[\exp(i\xi_{1}u_{1}(T_{a}(u_{2}))-\mu T_{a}(u_{2}));U_{a}]}{n^{\Psi}[U_{a}]}$.

(2.11)

By the independence described above,$E_{(0,0)}[e^{i\xi_{1}(X(T_{\alpha}^{Y})-X(G_{\cap}^{)}))-\mu(T_{a}^{)}-c_{\alpha}^{1})}]$

$=$

Since

.

$E_{(0,0)}[e^{i\xi_{1}X(G_{\alpha}^{Y})-\mu G_{\alpha}^{Y}}]$.

$E_{(0_{1}0)}[e^{i\xi_{1}X(T_{n}^{\}})-\mu T_{a}^{)}}]= \frac{u(a_{7}\cdot\xi_{1},\mu)}{u(0_{\backslash }\cdot\xi_{1\backslash }l\iota)}$$\{p(l);l\in(0, \kappa_{a})\cap L_{U_{a}^{t}}^{1}\}$(2.12)

is a Poisson point process stopped at an independent

exponential variable, we have$E_{(0,0)}[e^{i\xi_{1}X(G_{a}^{Y})-\mu G_{a}^{Y’}}]$$=$$\int_{0}^{\infty}dln^{\Psi}[U_{a}]e^{-ln^{\Psi}|\iota l_{a}|}\exp(-ln^{\Psi}[1-\exp(i\xi_{1}u_{1}(\zeta(u))-\mu\zeta(u)))U_{a}^{c}])$$=$$\frac{n^{\Psi}[U_{a}]}{n^{\Psi}[U_{a}]+n^{\Psi}[1-\exp(i\xi_{1}u_{1}(((u))-l^{\iota((u));U_{o}^{c}]}}$

By the strong Markov property of$n^{\phi}$(2.13),$n^{\Psi}[\exp(i\xi_{1}u_{1}(\zeta(u)^{\backslash }, -l^{l}\zeta(u));U_{o}]$$=$$n^{\Psi}[\exp(i\xi_{1}u_{1}(\tau_{\mathfrak{a}}(u_{2}))-l^{\iota T,(u_{2})):U_{o}]\cdot E_{(0,a)}}|[e^{i\xi_{1}X(T_{0}^{Y})-\mu T_{0}^{\gamma}}]$$=$$n^{\Psi}[\exp(i\xi_{1}u_{1}(\tau_{a}(u_{2}))-l^{\iota T,(v_{2}))_{\}\cdot U_{o}]\cdot\frac{u(-a;\xi_{1},\mu)}{u(0;\xi_{1},\mu)}}\cdot$

An elementary manipulation of these equalities yields$n^{\Psi}[\exp(i\xi_{1}u_{1}(\tau_{a}(u_{2}))-l^{\iota I_{l}^{1}((\iota_{2})):U_{\cap}]}=\frac{u(a;\xi_{1},\mu)}{14(0:\xi_{1l}\iota)^{2}-u(a;\xi_{1}.\mu)u(-a;\xi_{1},\mu)}$among others. Then$n^{\Psi}[U_{a}]$$=$$\lim$$\backslash c_{1}-,r)_{1’}-\cdot\dashv t)$$n^{\Psi}[t^{Y}X|)(i\xi_{1}?\iota_{1}$$(T.(u_{2}))-l^{\iota T_{a}(u_{2}));U_{a}]}$(2.14)

169$=$$\lim_{\muarrow+0}\frac{u(a)0,\mu)}{u(0;0\backslash \mu)^{2}-u(a)0,\mu)u(-a,0,\mu)}$.

This quantity is concerned with the one-dimensional symmetric -stable L\’evy process$Y(t)$ . Although we omit the further detail,can be evaluated by the same method$\alpha$$n^{\Psi}[U_{a}]$as Lemma 4.1 in [8]:.

$n^{\Psi}[U_{a}]=\frac{\Psi(0,1)}{2h(a)(a)}=\frac{\Gamma(a)\sin\frac{(\alpha-1)\pi}{a1^{\alpha^{2}-1}}\Psi(0,1)}{1}$$\square$

Proof of Theorem 1.1

3.
Let $\tau(a)=\inf\{t\geq 0|Y(t)=|), X(t)\geq a\}$ and $\sigma(a)=\inf\{t\geq 0|\Xi(t)\geq a\}$ for
Then $\sigma(a)=L_{Y}(\tau(a)),$ $\Xi(\sigma(a))=X(\tau(a))$ , aiid $L_{Y}^{-1}(\sigma(a))=\tau(a)$ . Hence the firstand its companions.
, can be studied via
hitting time of interest,. The coincidence of two definitions can be
We now redefine the function$\mathbb{C}_{+}=\{z\in\mathbb{C}|\Imz>0\},$and setchecked. Let$a\in \mathbb{R}$$\sigma(a)$$\tau(a)$$\varphi_{\alpha}(z, \mu_{0}, \mu_{2})$$\overline{\mathbb{C}_{+}}=\{z\in \mathbb{C}|\Im z\geq 0\}$(3.1)$\varphi_{\alpha}(z;\mu_{0}, \mu_{2})=\sqrt{\mu_{0}+\Phi_{\mathfrak{a}}(0_{l}\iota_{2})}\int_{0}^{\infty}dtE_{(0,0)}[e^{-\mu t+iz-(t)-\mu_{2}L_{Y}^{-1}(t)}0=-]$for$z\in\overline{\mathbb{C}_{+}}$

and $\mu_{i}\geq 0(i=0,2)$ such that$l^{\iota_{0}}+\mu_{2}>0$$\varphi_{a}(z;0,0)=\frac{}{\backslash \frac{\Phi^{1}1,0}{a()}}(-iz)^{-(\mathfrak{a}-1)/2}$

For $\mu_{0}=\mu_{2}=0$ , we set.

for$z\in\overline{\mathbb{C}_{+}}\backslash \{0\}$

(3.2)

,

where we employ the branch such that $1^{-(a-1)/2}=1$ . For $\mu_{i}\geq 0(i=0,1,2)$ such that$\mu_{0}+\mu_{1}+\mu_{2}>0$ , we define
(3.3)

,

$I_{a}( \mu_{0}, \mu_{1)}\mu_{2})=\int_{-\infty}^{\infty}\frac{1}{2\pi(1+t^{2})}\log(\mu_{0}+\Phi_{\mathfrak{a}}(\mu_{1}t, \mu_{2}))dt$

convergence of which is verified using$0\leq\Phi_{\mathfrak{a}}(\xi_{1}, \mu_{2})=|\xi_{1}|^{a-1}\Phi_{o}(1.

as$|\xi_{1}|^{\alpha}/\mu_{2}arrow+\infty$|\xi_{1}|^{-0_{l}}\iota_{2})\sim\Phi_{\alpha}(1,0)|\xi_{1}|^{a-1}$

(3.4)

,

.

If $\mu_{0}=\mu_{2}=0$ , it is elementary to verify
Proof of Theorem 1.1. We use the following in an crucial way:$I_{(\}}(0.\ell\iota_{1}.0)=\log(\sqrt{\Phi_{\mathfrak{a}}(1,0)}\mu_{1}^{((1-1)/2})$$\bullet$

We use Theorem 1 in [3]: for any$z\in\overline{\mathbb{C}_{+}}$

and any$\theta\in \mathbb{R}$.

, it holds$\sqrt{\mu_{0}+\Phi_{\alpha}(0,\mu_{2})}\varphi_{t\}}(z;\mu\eta.l^{\iota_{2})}$$= \exp(l^{\infty}\frac{e^{-l^{4}o^{t}}dt}{t}E[(e^{i_{-}^{-}\Xi(t)}-1)e^{-\mu_{2}L_{\}}^{-.1}(t)};\Xi(t)>0])$$|\varphi_{\alpha}(\theta;\mu_{0},$,$\mu_{2^{1}}.|^{2}=\varphi_{Y}(\theta;l^{\iota_{0}.\mu_{2})\varphi_{f}(-\theta;\mu_{0},\mu_{2})}=\frac{1}{\mu_{0}+\Phi_{\mathfrak{a}}(\theta,\mu_{2})}\cdot$

(3.5)
(3.6)

170

$\bullet$On the real line, we have$\varphi_{a}(\theta;\mu_{c0}, \mu_{2})\sim\frac{\exp((sgn\theta)\frac{\pi}{41}(\alpha-1)i)}{\sqrt{\Phi_{o}(10)}\theta|^{(0-1)/2}}$$\bullet$as$|\theta|arrow\infty$On the positive imaginary axis, we have$\varphi_{a}(j\mu_{1;l^{\iota_{0}}\cdot l^{l_{2}})=\exp(-I_{\alpha}(t^{\iota_{0},\mu_{1},\mu_{2}))}},$$\bullet$.

.

For any $a>0$ and $\mu_{\mathfrak{i}}\geq 0(i=0.1.2)$ such that $\ell\iota_{0}+\mu_{1}+\mu_{2}>0$ ,$1-E_{(0,0)}[e^{-\mu 0\sigma(a)-\mu_{1}\Xi(\sigma(a))-\mu_{2}L_{\}}^{-.1}(\sigma(a))}]$$=$$\frac{1}{\varphi_{\alpha}(i\mu_{1};\mu_{0\backslash }\{\iota_{2})}\int_{-x}^{\infty}(l\theta\frac{1-e^{-ia(\theta-i\mu_{1})}}{2\pi i(\theta-i\mu_{1})}\varphi_{\alpha}(\theta;\mu_{0}, \mu_{2})$.

We refer the reader to the forthcoming paper [6] for the detail of the proof.

$\square$Remark 1 In the terminology of
Vl in [1], $\Xi(\sigma(a))-a$ is the overshoot for a$(\alpha-1)sta1_{J}1e$
L\’evv process
one-dimensional symmetric. Adopting Exercise VI.1 andLemma VIII.1 in [1], we have the following double Laplace transform:$(^{\tau}ha])\uparrow.er$$\Xi(t)$$\int_{0}^{\infty}dae^{-qa}(1-E_{(0.0)}[e^{-/r\Xi(\sigma(0))}])=\frac{\mu^{(a-1)/2}}{q(q+l^{l})^{(\alpha-1)/2}}$.

On the other hand, we set $\mu_{0}=\mu_{=}0$ and take the Laplace transform of the both sides ofTheorem 1.1 (i) to obtain$\int_{0}^{\infty}dae^{-qa}\mu^{(\prime J-1)/2}\int_{-x}^{\infty}\backslash \prime l\theta\frac{1-e^{-i(\theta-\prime l^{J})a}}{2\pi i(\theta-i\mu)}\frac{1}{(-i\theta)^{(a-1)/2}}$$=$$\mu^{(\alpha-1)/2}J_{-\infty}^{\infty}(l\theta\frac{\frac{1}{q}-\frac{1}{q+\mu+i\theta-j_{l^{\iota}}}}{2\pi i(\theta)}\frac{1}{(-i\theta)^{(0-1)/2}}$$=$$\frac{\mu^{(\alpha-1)/2}}{q}\oint_{-u}^{\infty}d\theta\frac{1}{2\pi i(\theta-i(q+l^{l}))}\frac{1}{(-i\theta)^{(\alpha-1)/2}}$

The coincidence of theqe is verified by a $si_{I11])}1ea$ ]4$)])1i_{t\dot{c}}\iota tion$.

of the residue theorem.

The case for independent symmetric stable L\’evy
processes with different indices

Let $1<\alpha\leq 2,0<\beta\leq 2$ , and $(X (t). Y(t))$ be such that $X(t)$ and $Y(t)$ are independent,$X(t)$ is symmetric -stable, and $Y(t)$ is symmetric-stable. In terms of the characteristic
exponent,. When $(X (t).Y(t))$ is started from, its$\beta$$\Psi(\xi_{1}, \xi_{2})=|\xi_{1}|^{\beta}+|\xi_{2}|’\}$$($$(x_{0}, y_{0})\in \mathbb{R}^{2}$

171

law and expectation are denoted by, respectively. Let $L_{Y}(t)$ be theandlocal time at for.
$\tau(a)=\inf\{t\geq
0|Y(t)=0_{\dot{\mathfrak{l}}}X(t)\geq
a\}$
,
.
For
we setWe define, for, and $\mu_{\eta}\geq 0(i=0,1,2)$ such that $\mu_{0}+\mu_{2}>0$ ,$P_{(coy_{()})}$$E_{(x_{0},yo)}$$Y( \cdot):L_{Y}(t)=\lim_{\epsilonarrow+0^{\frac{1}{2\epsilon}}}/0f1_{(-\epsilon,\epsilon)}(Y(s))ds$$0$$a\in \mathbb{R}$$z\in \mathbb{C}_{+},$$\xi_{1}\in \mathbb{R}$,$\Phi_{\alpha,\beta}(\xi_{1}, \mu_{2})$$=$$2 \pi/\int_{\mathbb{R}}\frac{(l\xi_{2}}{l^{l_{2}}+|\xi_{1}|^{\beta}+|\xi_{2}|^{\alpha}}$$I_{\alpha,\beta}(\mu_{0}, \mu_{1}, \mu_{2})$$=$$\int_{-\infty}^{\infty}\frac{dt}{2\pi(t^{2}+1)}\log(\mu_{0}+\Phi_{a,\beta}(\mu_{1}t, \mu_{2}))$$\varphi_{a,\beta}(z;\mu_{0}, \mu_{2})$$=$$\exp(\frac{-1}{2\pi\uparrow}\int_{-x}^{\infty}\frac{z}{t^{2}-z^{2}}\log(\mu_{0}+\Phi_{\alpha,\beta}(t, \mu_{2}))dt)$

For $\mu_{0}=\mu_{2}=0$ , we define$\sqrt{C_{2}(a)}^{1}(-iz)^{-\beta(\alpha-1)/(2a)}$,$I_{\alpha,\beta}(0_{\{}\iota_{1},0)=\log(\sqrt{C_{2}(\alpha)}\mu_{1}^{\beta(\mathfrak{a}-1)/(2\alpha)}’)$

and.

$\varphi_{\alpha,\beta}(z;0,0)=$.

We obtain the following theorem by the same method as in \S 3. We refer the reader tothe forthcoming paper [6] for the detail.
$\square$

Theorem 4.1 (i) Let $a>0,$ $/h\geq 0(i=0,1,2)$ , and $\mu_{0}+\mu_{1}+\mu_{2}>0$ .
(i) It holds$E_{(-a,0)}[e^{-\mu oL_{Y}(\tau(0))-\mu_{1}X(\tau(0))-l^{l}27(0)}]$$=$$e^{\mu_{1}a}-e^{\mu_{1}a} \exp(I_{\alpha,\beta}(\mu_{0}, \mu_{1}.\mu_{2}))\int_{-\infty}^{\infty}d\theta\frac{1-e^{-ia(\theta-i\mu_{1})}}{2\pi i(\theta-i\mu_{1})}\varphi_{\alpha,\beta}(\theta;\mu_{0}, \mu_{2})$.

(ii) As $sarrow+O$ it holds$1-E_{(-a_{\gamma}0)}[e^{-l^{l}t)s^{2((\prime-1)}L)(\tau(0))-\mu_{1}s^{20/\beta}X(\tau(0))-\mu_{2}s^{2a}\tau(0)}]$$\sim$$where\sim$5$\frac{\exp(I_{\alpha,\beta}(l^{l_{f)}},l^{\iota_{1\backslash }\mu_{2}))}1)/(2a)}{\sqrt{C_{2}(\alpha)}\Gamma(J+\frac{13(\mathfrak{a}-1)a^{l3(\mathfrak{a}-}}{2\mathfrak{a}})}s^{\mathfrak{a}-1}$,

means that the ratio of the both sides converges to 1.

Some properties of modified resolvents

We modified the resolvents for $Y(t)$ in Section 2 and presented minimal(except Subsection
2.1) arguments for our application. The aim of this section is to characterize the modified
resolvents in terms of the modified (itar} nieasnre for $Y(t)$ . Since the polarity of sets
is determined solely by the process $Y(t)$ . there is no addition to the (.lassification resultsin our modification. We focus on the modified identities between some quantites in thepotential theory for $Y(t)$ . We do not need symmetry or (2.1) but state the results interms ofandthe law aiid the expectation of the dual process, respectively.
$ap(\downarrow\langle$$\hat{P}_{(x_{0},yo)}$$\hat{E}_{(x_{0},y_{0}))}$

172

In this section, we as sume $(X(t).Y(t))]s$ alowing notations for resolvents:

process on$L’\supset\lrcorner vy$$\mathbb{R}^{2}$and employ the fol-,

$P_{t}^{\xi}f(y)$$:=$$E_{(0.y)}[e^{;c}\backslash ^{Y(\ell)}f(Y(t))]$$U^{\xi,\mu}f(y)$$:=$$E_{(0.y)}[ \int_{0}^{\infty}e^{i\xi X\langle\ell)-\mu t}f(Y(t))dt]$. So we have $U^{\xi_{l^{l}}},f(y)= \int_{R}f(y+z)U(dz;\xi, \mu)$ , where $U(dy;\xi, \mu)$foris defined by (2.3) in Section 2. These quantit,ies reduce, if $\xi=0$ , to $P_{t}f(y)$ and $U^{\mu}f(y)$in [1, p.19,22], which employes for . Our resolvents obey the same resolvent equation
as the case $\xi=0$ :$f(y)\in \mathcal{L}^{\infty}(\mathbb{R})\cup \mathcal{L}^{1}(\mathbb{R})$$q$

Lemma 5.1 Let $C_{0}:=$ {
maps
(i)
into
$P_{t}^{\xi}$satisfies$C_{0}$

is continuous and goes toforms a semigroup if$\mathbb{R}arrow \mathbb{R}|f$$0$$P_{0}^{\xi}=$$C_{0};(\prime P_{t}^{\xi})_{t\geq 0}$$\Vert P_{t}^{\xi}f\Vert\leq\Vert f\Vert$(ii) For any

:

$f$

$\mu$

; for each $f\in C_{0},$$f(y)\in \mathcal{L}^{\infty}(\mathbb{R})\cup \mathcal{L}^{1}(\mathbb{R}),$at infinity.}.
Id; not Markovian but

uniformly as $tarrow+O$ .
, and $\lambda>0$ , we have$P_{t}^{c}\backslash farrow f$$l^{\iota}>0$$U^{\xi,\lambda}f(y)-U^{\xi.\mu}f(y)+(\lambda-\mu)U^{\xi_{J}\lambda}U^{\xi,\mu}f(y)=0$(iii) The range of$f$

uniformly as$U^{\xi,\mu}$

.

(5.1)

does not depend on $l^{l}>0$ ; we denote the range by
is a bijection.
is a dense subspace;$\mathcal{D};\mu U^{\xi,\mu}farrow$$U^{\xi_{2}\mu}:C_{0}arrow \mathcal{D}$$\muarrow\infty;D\subset c_{0}^{\neg}$(i) is a modified version of Proposition I.5 in [1, p. 19]. (ii) can be checked bya standard argument. (iii) is shown by the same argument as in [1, p.23].

Proof.

$\square$Obviously,(5.2)$( \mu+I\int(\xi, 0))\int_{R}U^{c}\backslash \cdot/\prime f(y)dy=\int_{\mathbb{R}}f(y)dy$. Set $T_{B}^{Y}= \inf\{t\geq 0|\}’(t)\in B\}$ and define the semigroup/resolventforwith the killing upon entrance of $B$ :$f(y)\in \mathcal{L}^{1}(\mathbb{R})$,$P_{t}^{B,\xi}f(y)$$:=$$E_{(0,y)}[e^{j}\xi X(t)f(Y(t)):t<T_{B}^{)}]$$U_{B}^{\xi,\mu}f(y)$$:=$$\int_{0}^{\infty}e^{-\mu t}P_{\ell}^{B.\xi}f(y)dt=E_{(0,y)}[\int_{0}^{T_{B}^{)}}e^{i\xi.Y(t)-\mu t}f(Y(t))dt]$

in [1, p.47], which employes for
andThese quantities reduce, if $\xi=0$ , to
. Theorem II.5 in [1, p.47] is called Hunt switching identity. We also have an analog
for the modifiOd semigroup and resolvent.
$P_{f}^{B}f(y)$$U_{B}^{l^{l}}f(y)$$q$$\cdot$

$($

$s$

$\mu$

Theorem 5.1 (modified Hunt’s switching identity) Let the modified dual semi}
and
group
and theclual
the same way as, respectively, with $(X(t)-X(0). Y(t)-Y(0))$ replaced by $(X$ (0) $-X(t), Y(O)-Y(t))$ ,i. e., the process travels in the opposite way.
$\hat{P}_{t}^{B,\xi}$$U_{B}^{\xi,\mu}$$mod\iota fi\not\in$ $d$)$\prime^{\backslash }r.solm^{B}r$$f\hat{U}_{B}^{c}\backslash \cdot/;\})^{}(lefine(l/n$$P_{t}^{B.\xi}$173

If either$f\in \mathcal{L}^{\infty}(\mathbb{R}),$or$g\in \mathcal{L}^{1}(\mathbb{R})$$g\in \mathcal{L}^{\infty}(\mathbb{R}),$$\int_{\mathbb{R}}dyg(y)P_{t}^{B,\xi}f(y)$’$\int_{R}.t$$f\in \mathcal{L}^{1}(\mathbb{R})$, we have$=$$\int_{R}dzf(z)\hat{P}_{t}^{B,-\xi}g(z)$$=$$\int_{\mathbb{R}}dzf(z)\hat{U}_{B}^{-\xi,\mu}g(z)$,.

To prove this theorem, we need two Lemmas. The first is a straightforward extensionof Prop.II. 1 in [1, p.44].
Lemma 5.2 The following eqvality for two measures on $\mathbb{R}^{3}=\{(x, y, z)\}$ holds.
(5.3)$dyP_{(0_{1}y)}[X(t)\in dx, Y(t)\in dz]=dz\hat{P}_{(0,z)}[-X(t)\in dx, Y(t)\in dy]$

Proof.

Let $f,g,$
two sides of (5.3) coincides.

$h\in \mathcal{B}_{+}(\mathbb{R})$$\int_{\mathbb{R}})$.

We prove that the integrations of $g(y)f(z)h(x)$ by the$=$$\int_{1R}dyg(y)E_{(0,0)}[h(X(t))f(y+Y(t))]$$=$$E_{(0.0)}[h(X(t)) \int_{R}dyg(y)f(y+Y(t))]$$=$$E_{(0.0)}[h(X(t)) \int_{\mathbb{R}}dzg(z-Y(t))f(z)]$$=$$\int_{R}dzf(z)E_{(0,0)}[h(X(t))g(z-Y(t))]$$=$$\int_{1\mathbb{R}}dzf(z)\hat{E}_{(0,0)}[h(-X(t))g(z+Y(t))]$$=$$\int_{R}dzf(z)\hat{E}_{(0,z)}[h(-X(t))g(Y(t))]$.

$\square$

The second lemma is an extension of page 48, line 7 in [1].
Lemma 5.3 If$B\subset \mathbb{R}$is either open or closed,$P_{(0_{t}y)arrow(x,\approx)}[t<T_{B}^{1}.]=\hat{P}_{(0.z)arrow(-x.y)}[t<T_{B}^{Y}]$Proof.

(5.4).

By the saine method as Corollary II.3 in [1, p.45], we can prove$((X_{(t-s)-}, Y_{(\ell-s)-;S\in}[0, t]),$ $\lrcorner D(0_{2}y)arrow(\tau_{\sim}^{-}))=((x+X_{s}.Y_{s}:_{1}s\in[0, t]),\hat{P}_{(0,z)arrow(-x_{7}y)})$

(5.5)

.

If $B$ is open, it is clear that (see page 48, line 3 in [1]).
$\{t<T_{B}^{)_{\sigma}}\}=’$(5.6)$\{t<T_{B}^{\iota_{(1- s)-}}.\}$and hence$P_{(0_{7}y)arrow(x,z)}[t<T_{B}^{Y}](5.6)=P_{(0y)arrow(\tau_{\sim}^{-})}[t<T_{B}^{1_{\dot{(}l-s)-}}](55)=\hat{P}_{(0,z)arrow(-x,y)}[t<T_{B}^{Y}]$

.

174

If $B$ is closed, we take a sequence of open setssuch that. Then wehave
and 1 $\{t<T_{B_{n}}^{Y}\}\nearrow 1\{t<T_{B}^{1!}\}$ by Corollary I.8 in [1, p.22]. It is thenelementary to observe$\bigcap_{n}\overline{B_{n}}=B$$B_{n}\backslash B$$T_{B_{n}}^{Y}\nearrow T_{B}^{Y}$$P_{(0_{J}y)arrow(x,z)}[t<T_{B_{1}}^{Y},]$$=$$\hat{P}_{(0,z)arrow(-x_{t}y)}[t<T_{B_{n}}^{Y}]$$\downarrow$$\downarrow$$P_{(0_{1}y)arrow(x,z1}[t<T_{B}^{Y}]$$\hat{P}_{(0,z)arrow(-x,y)}[t<T_{B}^{Y}]$$\square$

Proof of Theorem 5.1. We start with$f,$ $g\in \mathcal{L}^{\infty}(\mathbb{R})\cap \mathcal{L}^{1}(\mathbb{R})$.

By Lemma 5.3, the following functions are equal to each other.
$g(y)f(z)e^{i\xi x}P_{(0,y)arrow(x,z)}[t<T_{B}^{Y}]=g(y)f(z)e^{i\xi x}\hat{P}_{(0,z)arrow(-x_{1}y)}[t<T_{B}^{Y}]$We then integrate the both sides by the measures in the both sides of Lemma 5.2, respectively.
$\int_{R}dyg(y)E_{(0,y)}[e^{i\xi X\langle t)}f(Y(t));t<T_{B}^{)}.]$$=$$\Vert$$\int_{R}dzf(z)\hat{E}_{(0,z)}[e^{-i\xi X(\ell)}g(Y(t));t<T_{B}^{Y}]$$\Vert$$\int_{R}dyg(y)P_{\ell}^{B,\xi}f(y)$$\int_{\mathbb{R}}dzf(z)\hat{P}_{t}^{B,-\xi}g(z)$To loosen the condition $f,$
. we first set $\xi=0$ to verify the bothsides is absolutely convergent by using Fubini’s theorem; next use truncation and thebounded convergence for any.
$g\in \mathcal{L}^{\infty}(\mathbb{R})\cap \mathcal{L}^{1}(\mathbb{R})$$\xi\in \mathbb{R}$$\square$

The capacitary measure is defined in [1], p.49. We define the modified capacitary
measure forwhich is either open or closed:$B\subset \mathbb{R}$$\mu_{\grave{B}}^{c_{l}}\mu(dz):=(\mu+\Psi(\xi.0))\int_{R}E_{(0.y)}[e^{i\xi.Y(T_{B}^{\}^{J}})-\mu T_{B}^{)’}};Y(T_{B}^{Y})\in dz]dy$

Lemma 5.4 For$f(y)\in \mathcal{L}^{1}(\mathbb{R})$.

,$\int_{\mathbb{R}}f(y)dy=(\mu+\Psi(\xi, 0))\int_{\mathbb{R}}U_{B}^{\xi,\mu}f(y)dy+\int_{R}U^{\xi,\mu}f(y)\mu_{B}^{\xi,\mu}(dy)$Proof. Use the strong Markov property at the instantthe equation (1) in [1, p.51].
The next theorem is a modified
the capacitary measure.

$T_{B}^{Y}$.

. The version for $\xi=0$ is$\square$

Theorem 5.2 Define the measureLet$\mu>0$ and suppose that$\xi\in \mathbb{R},$$ver\iota\backslash \backslash ion$of Theorem II.7 in [1, p.50], which characterizebyis either open or closed. Then$\mu_{B}^{\xi,\mu}U^{c_{\mu}}\backslash$$B$$\int_{R}f(z)\mu_{B}^{\xi.\mu}U^{\zeta,\mu}(dz)=\int_{1R}U^{\epsilon_{\mu}},f(y)\mu_{B}^{\xi,\mu}(dy)$$\mu_{B}^{\xi,\mu}U^{\xi.\mu}(dz)=\hat{E}_{(0.z)}[e\prime^{\zeta\backslash }\backslash t(T_{B}^{)}.)-\mu T_{B}^{Y}]dz$

Moreover,$\mu_{B}^{\xi,\mu}$is the unique$\mathbb{C}$.

.

-valued Radon measure on IR satisfying the above equation.

175

Uniqueness follows from the denseness ofLemma 5.4, we have

Proof.

$U^{\xi,\mu}f$in$C_{0}$, see Lemma 5.1. By$\int_{\mathbb{R}}f(z)\mu_{B}^{\xi,\mu}U^{\xi_{J}\mu}(dz)=\int_{\mathbb{R}}f(y)dy-(l^{\iota}+\Psi(\xi, 0))\int_{\mathbb{R}}U_{B}^{\xi,\mu}f(y)dy$

We setright side.

$g\equiv 1\in \mathcal{L}^{\infty}(\mathbb{R})$and$f \frac{\sim}{\sim}\mathcal{L}^{1}(\mathbb{R})$.

in Theorem 5.2 to obtain the second term in the$( \mu+\Psi(\xi,0))\int_{R}U_{P}^{\xi,\mu}f(y)dy$$=$$( \mu+\Psi(\xi, 0))\int_{\mathbb{R}}dzf(z)\hat{U}_{B}^{-\xi,\mu}1_{R}(z)$$=$$( \mu+\Psi(\xi, 0))\int_{\mathbb{R}}dzf(z)\int_{0}^{\infty}e^{-\mu t}\hat{P}_{t}^{B,-\xi}1_{R}(z)dt$$=$$( \mu+\Psi(\xi, 0))\int_{R}dzf(z)\int_{0}^{\infty}e^{-\mu}\hat{E}_{(0,z)}[e^{-i\xi X(\ell)};t<T_{B}^{Y}]dt$.

The first term in the right side is handled with (5.2) for the dual resolvent:$\int_{\mathbb{R}}f(y)dy$$=$$( \mu+\Psi(-(-\xi)’.0))\int_{IR}\hat{U}^{-\xi,\mu}f(y)dy$$=$$( \mu+\Psi(\xi, 0))\int_{\mathbb{R}}dzf(z)\int_{0}^{\infty}e^{-\mu t}\hat{E}_{(0_{2}z)}[e^{-i\xi X(\ell)}]dt$.

Putting these together, we have$\int_{\mathbb{R}}f(z)\mu_{B}^{\xi,\mu}U^{\xi,\mu}(dz)$$=$$+( \mu+\Psi(\xi, 0))\int_{\mathbb{R}}dzf(z)\int_{0}^{\infty}e^{-l^{4}}{}^{t}\hat{E}_{(0.z)}[e^{-i\xi,Y(\ell)}]dt$$-( \mu+\Psi(\xi, 0))\int_{\mathbb{R}}dzf(z)\int_{0}^{\infty}e^{-\mu t}\hat{E}_{(0.z)}[e^{-i\xi.Y(t)},$

Since

$f$

$t<T_{B}^{Y}]dt$$t\geq T_{B}^{Y}]dt$$=$$( \mu+\Psi(\xi, 0))\int_{R}dzf(z)\int_{0}^{\infty}e^{-\mu t}\hat{E}_{(0.z)}[e^{-i_{\backslash }^{c}.Y(t)},$$=$$( \mu+\Psi(\xi, 0))\int_{\mathbb{R}}dzf(z)\hat{E}_{(0_{\vee})}[e^{-i\xi.Y\langle T_{B}^{)})-\mu T_{B]}^{\}’}}\int_{0}^{\infty}e^{-\mu t}\hat{E}_{(0,0)}[e^{-i\xi X(t)}]dt$$=$$\int_{\mathbb{R}}dzf(z)\hat{E}_{(0_{i}z)}[e^{-\prime^{\zeta}X(T_{B}^{)})-\mu T_{B}^{)}}\backslash \cdot\cdot]$.

is arbitrary integrable function, lhe proof is complete. 口References[1] J. Bertoin, L\’evy processes, Cambridge University Press, Cambridge, 1996.

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