Exponential ergodicity of branching processes

with immigration and competition

Pei-Sen Li1, Zenghu Li2, Jian Wang3 and Xiaowen Zhou4

1School of Mathematics and Statistics, Beijing Institute of Technology,
Beijing

100872, China<EMAIL_ADDRESS>
2Laboratory of Mathematics and Complex Systems, School of Mathematical

Sciences, Beijing Normal University, Beijing 100875, China<EMAIL_ADDRESS>
3School of Mathematics and Statistics & Fujian Key Laboratory of Mathematical

Analysis and Applications & Center for Applied Mathematics of Fujian Province,

Fujian Normal University, Fuzhou 350007, China<EMAIL_ADDRESS>
4Department of Mathematics and Statistics, Concordia University, 1455 De

Maisonneuve Blvd. W., Montreal, Canada<EMAIL_ADDRESS>

Abstract: We study the ergodic property of a continuous-state branching
process with immigration and competition, which is an extension of the models
studied by Lambert (2005), Pardoux (2016) and Berestycki et al. (2018) with an
additional immigration structure. The exponential ergodicity in a weighted
total variation distance is proved under natural assumptions. The main theorem
applies to branching mechanisms in the full range of criticality, including
all stable types. The proof is based on a Markov coupling process and a
nonsymmetric control function for the distance, which are designed to identify
and to take the advantage of the dominating factor from the branching,
immigration and competition mechanisms in different parts of the state space.
The approach provides a way of finding explicitly the ergodicity rate.

Key words: continuous-state branching process; immigration; competition;
exponential ergodicity; stochastic equation; Markov coupling; Lyapunov
function; control function.

MSC (2020) Subject Classification: 60J80, 60J25, 60G51, 60G52

##  1 Introduction

Classical Galton–Watson branching processes are discrete-time Markov processes
taking values of nonnegative integers. They are models for the evolution of
populations where the progenies of individuals are described by i.i.d. random
variables. Standard references on those processes are Athreya and Ney [3] and
Harris [26]. As the quantity of particles can sometimes be expressed by means
other than by counting, it is reasonable to consider branching models with
continuous-state space $\mathbb{R}_{+}:=[0,\infty)$. The study of continuous-
state branching processes (CB-processes) was initiated by Feller [19], who
noticed that a one-dimensional diffusion process may arise as the limit of a
sequence of rescaled Galton–Watson branching processes. The result was
extended by Lamperti [30] to the situation where the limiting process may have
discontinuous sample paths; see also Aliev and Shchurenkov [2] and Grimvall
[24]. Let ${\it\Psi}$ be a function on $[0,\infty)$ with the Lévy–Khintchine
representation:

$\displaystyle{\it\Psi}(\lambda)=b\lambda+c\lambda^{2}+\int_{0}^{\infty}\big{(}\mathrm{e}^{-\lambda
z}-1+\lambda z\mathbf{1}_{\\{z\leq
1\\}}\big{)}\mu(\mathrm{d}z),\quad\lambda\geq 0,$ (1.1)

where $b\in\mathbb{R}$ and $c\geq 0$ are constants and $(1\land
z^{2})\mu(\mathrm{d}z)$ is a finite measure on $(0,\infty)$. The transition
semigroup $(Q_{t})_{t\geq 0}$ of a CB-process with branching mechanism
${\it\Psi}$ is defined by

$\displaystyle\int_{\mathbb{R}_{+}}e^{-\lambda
y}Q_{t}(x,\mathrm{d}y)=\mathrm{e}^{-xv_{t}(\lambda)},\quad x\geq 0,\lambda>0,$
(1.2)

where $t\mapsto v_{t}(\lambda)$ is the unique strictly positive solution to
the differential equation

$\displaystyle\frac{\partial}{\partial
t}v_{t}(\lambda)=-{\it\Psi}(v_{t}(\lambda)),\quad v_{0}(\lambda)=\lambda.$
(1.3)

The CB-process is eventually degenerate in the sense that it tends to either
zero or infinity as $t\to\infty$. In fact, the process may explode at a finite
time. It is conservative if and only if the following condition is satisfied:

$\displaystyle\int_{0+}\frac{\mathrm{d}\lambda}{0\vee[-{\it\Psi}(\lambda)]}=\infty;$
(1.4)

see, e.g., Grey [23]. In order that the integral on the left-hand side of
(1.4) is convergent, we have necessarily that ${\it\Psi}^{\prime}(0)=-\infty$.
Let $v_{t}(0)=\lim_{\lambda\to 0+}v_{t}(\lambda)$ for $t\geq 0$. Under
condition (1.4), we have $v_{t}(0)=0$ for every $t\geq 0$, which is also the
unique solution to (1.3) with $\lambda=0$. The first moments of the transition
probabilities of the CB-process are given by

$\displaystyle\int_{\mathbb{R}_{+}}yQ_{t}(x,\mathrm{d}y)=x\exp\big{\\{}-{\it\Psi}^{\prime}(0+)t\big{\\}},\quad
t,x>0,$ (1.5)

where

$\displaystyle{\it\Psi}^{\prime}(0+)=b-\int_{1}^{\infty}z\mu(\mathrm{d}z).$

The branching mechanism ${\it\Psi}$ is said to be subcritical, critical or
supercritical according as ${\it\Psi}^{\prime}(0+)>0$,
${\it\Psi}^{\prime}(0+)=0$ or ${\it\Psi}^{\prime}(0+)<0$, respectively. Let
$\sigma\geq 0$, $a\in\mathbb{R}$ and $0<\alpha<2$ be constants. The stable
branching mechanism is defined by

$\displaystyle{\it\Psi}(\lambda)=\left\\{\begin{array}[]{ll}a\lambda+c\lambda^{2}+\sigma\lambda^{\alpha},&\quad
1<\alpha<2,\cr
a\lambda+c\lambda^{2}+\sigma\lambda\log\lambda,&\quad\alpha=1,\cr
a\lambda+c\lambda^{2}-\sigma\lambda^{\alpha},&\quad
0<\alpha<1.\end{array}\right.$ (1.9)

Clearly, the conservativeness condition (1.4) fails for the stable branching
mechanism if $\sigma>0$ and $0<\alpha<1$. If $\sigma>0$ and $0<\alpha\leq 1$,
then ${\it\Psi}^{\prime}(0+)=-\infty$ and the CB-process has infinite first
moments by (1.5). The CB-processes defined by (1.2) and (1.3) involve rich
mathematical structures and have been applied to the research in several
important areas. In particular, Bertoin and Le Gall [6] gave a representation
of the genealogical structure of the general CB-processes by Bochner’s
subordination and used the representation to give a precise description of the
connection between the so-called Neveu’s CB-process with branching mechanism
$\lambda\mapsto\lambda\log\lambda$ and the coalescent processes introduced by
Bolthausen and Sznitman [10] in the study of spin glasses. Berestycki et al.
[4] proved Neveu’s CB-process may arise in a limit theorem of some rescaled
particle systems. General stochastic flows associated with coalescent
processes and CB-processes were studied by Bertoin and Le Gall [7, 8, 9].
Constructions of the flows by strong solutions of stochastic equations were
given by Dawson and Li [14].

Continuous-state branching processes with immigration (CBI-processes) were
introduced by Kawazu and Watanabe [28] in their study of scaling limits of
discrete branching processes with immigration; see also Aliev [1]. Let
${\it\Phi}$ be an immigration mechanism, which is a function on $[0,\infty)$
with the Lévy–Khintchine representation:

$\displaystyle{\it\Phi}(\lambda)=\beta\lambda+\int_{0}^{\infty}\big{(}1-\mathrm{e}^{-z\lambda}\big{)}\nu(\mathrm{d}z),\quad\lambda\geq
0,$ (1.10)

where $\beta\geq 0$ and $(1\land z)\nu(\mathrm{d}z)$ is a finite measure on
$(0,\infty)$. The transition semigroup $(P_{t})_{t\geq 0}$ of a CBI-process is
characterized by

$\displaystyle\int_{\mathbb{R}_{+}}e^{-\lambda
y}P_{t}(x,\mathrm{d}y)=\exp\bigg{\\{}-xv_{t}(\lambda)-\int_{0}^{t}{\it\Phi}(v_{s}(\lambda))\mathrm{d}s\bigg{\\}},\quad\lambda>0,$
(1.11)

where $t\mapsto v_{t}(\lambda)$ is defined by (1.3). The CBI-process is a
natural model for the study of ergodicity, where the immigration ensures that
the stationary distribution is not the degenerate distribution at zero. A
sufficient and necessary integrability condition for the ergodicity in weak
convergence of the process was announced in Pinsky [47]; see Li [34] for a
proof of the result. The exponential ergodicity of the CBI-process in the
total variation distance was proved in Li and Ma [37] and the corresponding
results for measure-valued models were established in Li [36], which improve
the earlier results of Stannat [50, 51] for processes with special branching
mechanisms. The exponential ergodicity in a suitably chosen Wasserstein
distance was established by Friesen et al. [20] for affine Markov processes,
which include finite-dimensional CBI-processes as a special case. The key
tools of those explorations are the characterizations of the transition
probabilities by Laplace transforms given as in (1.3) and (1.11). Since
typical critical or supercritical CBI-processes eventually go to infinity, the
studies of their ergodicities have mainly been focused on subcritical
mechanisms. To get the ergodicity in the total variation distance, Li and Ma
[37] also assumed the following Grey’s condition:

$\displaystyle\int^{\infty-}\frac{\mathrm{d}\lambda}{{\it\Psi}(\lambda)}<\infty.$
(1.12)

Those restrictions unfortunately exclude Neveu’s branching mechanism
$\lambda\mapsto\lambda\log\lambda$ and some other very interesting models.

In this work we provide a general framework for exponential ergodicity
applicable to branching mechanisms in the full range of criticality. To do so,
a natural way is to incorporate a competition mechanism into the model. This
consideration has been inspired by the work of Lambert [29], who defined a
logistic growth branching process to model the pairwise competition between
individuals in the population. The process was constructed in [29] by a random
time change from a spectrally positive Ornstein–Uhlenbeck type process. More
general branching population systems with competition were studied in depth by
Berestycki et al. [5] and Pardoux [46]; see also Foucart [21]. Our main
problem is to understand whether and how a strong competition could balance
the branching and the immigration to guarantee the exponential ergodicity.
This is not clear as the process may have infinite means and conditions (1.4)
and (1.12) may not be satisfied. The emphasis here is the interplay among the
branching, immigration and competition mechanisms.

###  1.1 Main results

Suppose that $({\it\Omega},\mathscr{F},\mathscr{F}_{t},\mathbf{P})$ is a
filtered probability space satisfying the usual hypotheses. Let
$\\{L(\mathrm{d}s,\mathrm{d}u)\\}$ be a spectrally one-sided time-space
$(\mathscr{F}_{t})$-Lévy white noise with the Lévy–Itô decomposition:

$\displaystyle L(\mathrm{d}s,\mathrm{d}u)\\!\\!\\!$ $\displaystyle=\\!\\!\\!$
$\displaystyle
W(\mathrm{d}s,\mathrm{d}u)-b\mathrm{d}s\mathrm{d}u+\int_{0}^{1}z\tilde{M}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}u)$
(1.14)
$\displaystyle+\int_{1}^{\infty}zM(\mathrm{d}s,\mathrm{d}z,\mathrm{d}u),$

where $W(\mathrm{d}s,\mathrm{d}u)$ is a Gaussian white noise on
$(0,\infty)^{2}$ based on $2c\mathrm{d}s\mathrm{d}u$ and
$M(\mathrm{d}s,\mathrm{d}z,\mathrm{d}u)$ is a Poisson random measure on
$(0,\infty)^{3}$ with intensity $\mathrm{d}s\mu(\mathrm{d}z)\mathrm{d}u$ and
compensated measure
$\tilde{M}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}u):=M(\mathrm{d}s,\mathrm{d}z,\mathrm{d}u)-\mathrm{d}s\mu(\mathrm{d}z)\mathrm{d}u$.
Here and in the sequel, we make the convention that

$\displaystyle\int_{y}^{x}=-\int_{x}^{y}=\int_{(y,x]}~{}\mbox{and}~{}\int_{x}^{\infty}=\int_{(x,\infty)},\quad
x\geq y\in\mathbb{R}.$

Let $\\{\eta(t):t\geq 0\\}$ be an $(\mathscr{F}_{t})$-subordinator defined by

$\displaystyle\eta(t)=\beta
t+\int_{0}^{t}\int_{0}^{\infty}zN(\mathrm{d}s,\mathrm{d}z),$ (1.15)

where $N(\mathrm{d}s,\mathrm{d}z)$ is a Poisson random measure on
$(0,\infty)^{2}$ with intensity $\mathrm{d}s\nu(\mathrm{d}z)$. Suppose that
$\\{L(\mathrm{d}s,\mathrm{d}u)\\}$ and $\\{\eta(t)\\}$ are independent of each
other. Let $g$ be a competition mechanism, which by definition is a
nondecreasing and continuous function on $[0,\infty)$ satisfying $g(0)=0$. For
any $\mathscr{F}_{0}$-measurable random variable $x(0)\geq 0$, we consider the
stochastic equation:

$\displaystyle x(t)\\!\\!\\!$ $\displaystyle=\\!\\!\\!$ $\displaystyle
x(0)+\int_{0}^{t}\int_{0}^{x(s-)}L(\mathrm{d}s,\mathrm{d}u)-\int_{0}^{t}g(x(s))\mathrm{d}s+\eta(t).$
(1.16)

We say an $(\mathscr{F}_{t})$-adapted càdlàg process $\\{x(t):t\geq 0\\}$
taking values in $\bar{\mathbb{R}}_{+}:=[0,\infty]$ is a solution to (1.16) if
the equation holds almost surely when $t$ is replaced by $t\land\zeta_{n}$ for
each $t\geq 0$ and $n\geq 1$, where $\zeta_{n}=\inf\\{t\geq 0:x(t)\geq n\\}$
with $\inf\emptyset=\infty$. We call $\zeta:=\lim_{n\to\infty}\zeta_{n}$ the
lifetime of $\\{x(t)\\}$ and make the convention that $x(t)=\infty$ for
$t\geq\zeta$. If $\zeta=\infty$ almost surely, we say the process $\\{x(t)\\}$
is conservative. We shall prove that there is a pathwise unique solution to
(1.16); see Theorem 2.1. Then the solution $\\{x(t)\\}$ is a strong Markov
process in $\bar{\mathbb{R}}_{+}$. Let $(P_{t})_{t\geq 0}$ be the transition
semigroup of $\\{x(t)\\}$. We shall call any Markov process with transition
semigroup $(P_{t})_{t\geq 0}$ a continuous-state branching process with
immigration and competition (CBIC-process). Constructions of CB- and CBI-
processes in terms of similar stochastic equations were suggested by Bertoin
and Le Gall [9] and Dawson and Li [13, 14]. The CBIC-process extends the
population models of Berestycki et al. [5] and Pardoux [46] by the additional
immigration structure. A more general continuous-state population model in
random environments was introduced by Palau and Pardo [45] by solving a
stochastic equation driven by Brownian motions and Poisson random measures.

We need to introduce some concepts in order to present our main results. Let
$C^{2}(\mathbb{R}_{+})$ be the linear space of twice continuously
differentiable functions on $\mathbb{R}_{+}$. For any $f\in
C^{2}(\mathbb{R}_{+})$ write

$\displaystyle Lf(x)\\!\\!\\!$ $\displaystyle=\\!\\!\\!$ $\displaystyle
cxf^{\prime\prime}(x)+x\int_{0}^{\infty}\big{[}{\it\Delta}_{z}f(x)-zf^{\prime}(x)\mathbf{1}_{\\{z\leq
1\\}}\big{]}\mu(\mathrm{d}z)$ (1.18) $\displaystyle+\,[\beta-
bx-g(x)]f^{\prime}(x)+\int_{0}^{\infty}{\it\Delta}_{z}f(x)\nu(\mathrm{d}z),\quad
x\geq 0,$

where

$\displaystyle{\it\Delta}_{z}f(x)=f(x+z)-f(x).$

Let $\mathscr{D}(L)$ denote the linear space consisting of functions $f\in
C^{2}(\mathbb{R}_{+})$ such that the two integrals on the right-hand side of
(1.18) are convergent and define continuous functions on $\mathbb{R}_{+}$. Let
$C_{b}^{2}(\mathbb{R}_{+})$ be the space of bounded and continuous functions
on $\mathbb{R}_{+}$ with bounded and continuous derivatives up to the second
order. Then $C_{b}^{2}(\mathbb{R}_{+})\subset\mathscr{D}(L)$. In general, we
allow $f$ and $Lf$ to be unbounded functions. We shall see that
$(L,\mathscr{D}(L))$ is a restriction of the generator of the CBIC-process.

Given a nonnegative Borel function $V$ on $\mathbb{R}_{+}$, we denote by
$\mathscr{P}_{V}(\mathbb{R}_{+})$ the set of Borel probability measures
$\gamma$ on $\mathbb{R}_{+}$ such that

$\displaystyle\int_{\mathbb{R}_{+}}V(x)\gamma(\mathrm{d}x)<\infty.$

Let $W_{V}$ be the $V$-weighted total variation distance on
$\mathscr{P}_{V}(\mathbb{R}_{+})$ defined by

$\displaystyle
W_{V}(\gamma,\eta)=\int_{\mathbb{R}_{+}}[1+V(x)]|\gamma-\eta|(\mathrm{d}x),\quad\gamma,\eta\in\mathscr{P}_{V}(\mathbb{R}_{+}),$
(1.19)

where $|\cdot|$ denotes the total variation measure. We shall see that $W_{V}$
is actually the Wasserstein distance determined by the metric

$\displaystyle d_{V}(x,y)=[2+V(x)+V(y)]\mathbf{1}_{\\{x\neq y\\}},\quad
x,y\in\mathbb{R}_{+}.$ (1.20)

More precisely, we have

$\displaystyle
W_{V}(\gamma,\eta)=\inf_{\pi\in\mathscr{C}(\gamma,\eta)}\int_{\mathbb{R}_{+}^{2}}d_{V}(x,y)\pi(\mathrm{d}x,\mathrm{d}y),$
(1.21)

where $\mathscr{C}(\gamma,\eta)$ is the collection of all probability measures
on $\mathbb{R}_{+}^{2}$ with marginals $\gamma$ and $\eta$; see Lemma 5.3. The
consideration of this distance was inspired by Hairer and Mattingly [25]; see
also [18, 38]. In particular, if $V\equiv 0$, then $W_{V}$ reduces to the
total variation distance. The other two frequently used weight functions are
given by

$\displaystyle V_{1}(x)=x~{}~{}\mbox{and}~{}~{}V_{\log}(x)=\log(1+x),\quad
x\geq 0.$ (1.22)

We say the CBIC-process $\\{x(t):t\geq 0\\}$ or its transition semigroup
$(P_{t})_{t\geq 0}$ is exponentially ergodic in the distance $W_{V}$ with rate
$\lambda>0$ if it possesses a unique stationary distribution $\gamma$ and
there is a nonnegative function $\eta\mapsto C(\eta)$ on
$\mathscr{P}_{V}(\mathbb{R}_{+})$ such that

$\displaystyle W_{V}(\gamma,\eta P_{t})\leq C(\eta)\mathrm{e}^{-\lambda
t},\quad t\geq 0,\eta\in\mathscr{P}_{V}(\mathbb{R}_{+}).$ (1.23)

Clearly, the above property implies that $\\{x(t):t\geq 0\\}$ is conservative
and $\gamma\in\mathscr{P}_{V}(\mathbb{R}_{+})$. The exponential ergodicity
(1.23) follows by standard arguments if there is a constant $K\geq 0$ such
that

$\displaystyle W_{V}(P_{t}(x,\cdot),P_{t}(y,\cdot))\leq K\mathrm{e}^{-\lambda
t}d_{V}(x,y),\quad t\geq 0.$ (1.24)

We call $V\in\mathscr{D}(L)$ a Lyapunov function for the CBIC-process if there
are constants $C_{0}>0$ and $C_{1}>0$ such that

$\displaystyle LV(x)\leq C_{0}-C_{1}V(x),\quad x\geq 0.$ (1.25)

###### Condition 1.1

(Non-triviality condition) There exists a constant $\lambda_{0}>0$ such that
${\it\Psi}(\lambda_{0})>0$ and ${\it\Phi}(\lambda_{0})>0$.

###### Condition 1.2

(Fluctuation condition) We have $c>0$ or there are constants $c_{0}>0$ and
$\kappa_{0}>0$ such that

$\displaystyle\kappa(x):=[\mu\land(\delta_{x}*\mu)](0,\infty)+[\nu\land(\delta_{x}*\nu)](0,\infty)\geq\kappa_{0},\quad|x|\leq
c_{0},$ (1.26)

where “ $*$” denotes the convolution operation.

###### Condition 1.3

(Lyapunov condition) There is a nonnegative Lyapunov function
$V\in\mathscr{D}(L)$ satisfying $V(x)\to\infty$ as $x\to\infty$.

###### Theorem 1.1

Suppose that Conditions 1.1, 1.2 and 1.3 are satisfied. Then the CBIC-process
is conservative and there are constants $K>0$ and $\lambda>0$ such that (1.24)
holds. Consequently, the CBIC-process is exponentially ergodic in the
$V$-weighted total variation distance.

The advantage of Theorem 1.1 is that it works for general branching mechanisms
without criticality restriction. In particular, it applies to all the stable
branching mechanisms given by (1.9). Furthermore, the proof of the theorem
actually provides a way of finding explicitly the ergodicity rate $\lambda>0$,
which is important in applications; see Remark 5.4. Condition 1.1 is
introduced to avoid some extreme cases of (1.16). If the condition fails, then
either subordinator $\eta(t)$ vanishes or the Lévy field
$L(\mathrm{d}s,\mathrm{d}u)$ only has nonnegative increments. In any of those
cases, the immigration essentially plays no role in the ergodic behavior.
Fluctuation conditions like Condition 1.2 have been considered in the study of
ergodicities of Ornstein-Uhlenbeck type processes with nonlinear drift; see,
e.g., [32, 38, 40, 41, 48]. Our condition (1.26) is actually weaker than the
corresponding assumptions for exponential ergodicities in those previous
papers, where one usually required $\kappa(x)\to\infty$ as $x\to 0$. In
particular, the condition is satisfied if $\mu(\mathrm{d}z)$ or
$\nu(\mathrm{d}z)$ is bounded below by the measure
$r\mathbf{1}_{(u,v)}(z)\mathrm{d}z$ for some $r>0$ and $v>u\geq 0$. The
existence of a suitable Lyapunov function has become a standard condition for
ergodicity following Down et al. [15] and Meyn and Tweedie [42, 43, 44].
Condition 1.3 means roughly the competition mechanism wins its confrontation
against the branching and immigration when the population is large. Typically,
this is equivalent to a simple growth condition for the competition mechanism.

###### Proposition 1.2

Suppose that the measures $\mu$ and $\nu$ satisfy the integrability condition

$\displaystyle\int_{1}^{\infty}z\mu(\mathrm{d}z)+\int_{1}^{\infty}z\nu(\mathrm{d}z)<\infty.$
(1.27)

Then $V_{1}$ is a Lyapunov function for the CBIC-process if and only if

$\displaystyle\liminf_{x\to\infty}\frac{g(x)}{x}+b-\int_{1}^{\infty}z\mu(\mathrm{d}z)>0.$
(1.28)

###### Proposition 1.3

Suppose that the measures $\mu$ and $\nu$ satisfy the integrability condition

$\displaystyle\int_{1}^{\infty}\log(1+z)\mu(\mathrm{d}z)+\int_{1}^{\infty}\log(1+z)\nu(\mathrm{d}z)<\infty.$
(1.29)

Then $V_{\log}$ is a Lyapunov function for the CBIC-process if and only if

$\displaystyle\liminf_{x\to\infty}\bigg{\\{}\frac{g(x)}{x\log x}-\frac{x}{\log
x}\int_{1}^{\infty}\log\Big{(}1+\frac{z}{1+x}\Big{)}\mu(\mathrm{d}z)\bigg{\\}}>0.$
(1.30)

Under the integrability condition (1.27), the CBIC-process has finite first
moments. In this case, condition (1.28) means the competition mechanism should
grow at least linearly with a sufficiently large rate. In particular, if
$g(x)=ax$ for $x\geq 0$, the CBIC-process reduces to a CBI-process with
branching mechanism $\lambda\mapsto a\lambda+{\it\Psi}(\lambda)$ and (1.28)
simply means the branching mechanism should be subcritical. This is consistent
with the subcriticality condition in [37, Theorem 2.5], but here we do not
need to assume Grey’s condition. The conditions of Proposition 1.3 apply to
supercritical branching mechanisms including those with infinite first moments
but finite logarithmic moments.

###  1.2 Stable branching CBIC-processes

The CBIC-processes with stable branching mechanisms are solutions of
stochastic differential equations driven by spectrally one-sided Lévy
processes. Let $m_{\alpha}$ be the $\sigma$-finite measure on $(0,\infty)$
defined by

$\displaystyle
m_{\alpha}(\mathrm{d}z)=\left\\{\begin{array}[]{ll}(\alpha-1){\it\Gamma}(2-\alpha)^{-1}z^{-1-\alpha}\mathrm{d}z,&1<\alpha<2,\cr
z^{-2}\mathrm{d}z,&\alpha=1,\cr{\it\Gamma}(1-\alpha)^{-1}z^{-1-\alpha}\mathrm{d}z,&0<\alpha<1,\end{array}\right.$
(1.34)

where ${\it\Gamma}$ stands for the Gamma function. Let
$\\{M_{\alpha}(\mathrm{d}s,\mathrm{d}z)\\}$ be a time-space
$(\mathscr{F}_{t})$-Poisson random measure on $(0,\infty)^{2}$ with intensity
$\mathrm{d}sm_{\alpha}(\mathrm{d}z)$ and compensated measure
$\tilde{M}_{\alpha}(\mathrm{d}s,\mathrm{d}z)$. Let $\\{z_{\alpha}(t)\\}$ be
the spectrally positive $\alpha$-stable Lévy process defined by

$\displaystyle
z_{\alpha}(t)=\left\\{\begin{array}[]{ll}\displaystyle\int_{0}^{t}\int_{0}^{\infty}z\tilde{M}_{\alpha}(\mathrm{d}s,\mathrm{d}z),&1<\alpha<2,\vskip
3.0pt plus 1.0pt minus
1.0pt\cr\displaystyle\int_{0}^{t}\int_{0}^{1}z\tilde{M}_{\alpha}(\mathrm{d}s,\mathrm{d}z)+\int_{0}^{t}\int_{1}^{\infty}zM_{\alpha}(\mathrm{d}s,\mathrm{d}z),&\alpha=1,\vskip
3.0pt plus 1.0pt minus
1.0pt\cr\displaystyle\int_{0}^{t}\int_{0}^{\infty}zM_{\alpha}(\mathrm{d}s,\mathrm{d}z),&0<\alpha<1.\end{array}\right.$
(1.38)

Let $\\{B(t)\\}$ be a standard $(\mathscr{F}_{t})$-Brownian motion and let
$\\{\eta(t)\\}$ be the $(\mathscr{F}_{t})$-subordinator defined by (1.15).
Suppose that $\\{B(t)\\}$, $\\{z_{\alpha}(t)\\}$ and $\\{\eta(t)\\}$ are
independent of each other. Given any $\mathscr{F}_{0}$-measurable random
variable $x(0)\geq 0$, we shall construct the CBIC-process $\\{x(t):t\geq
0\\}$ with stable branching mechanism defined by (1.9) in terms of the
stochastic differential equations, for $\alpha=1$,

$\displaystyle\mathrm{d}x(t)\\!\\!\\!$ $\displaystyle=\\!\\!\\!$
$\displaystyle\sqrt{2cx(t)}\mathrm{d}B(t)+\sigma
x(t-)\mathrm{d}z_{1}(t)+\sigma x(t)\log(\sigma
x(t))\mathrm{d}t-ax(t)\mathrm{d}t\qquad$ (1.39)
$\displaystyle-\,g(x(t))\mathrm{d}t+\mathrm{d}\eta(t)$

and, for $\alpha\neq 1$,

$\displaystyle\mathrm{d}x(t)=\sqrt{2cx(t)}\mathrm{d}B(t)+\sqrt[\alpha]{\alpha\sigma
x(t-)}\mathrm{d}z_{\alpha}(t)-ax(t)\mathrm{d}t-g(x(t))\mathrm{d}t+\mathrm{d}\eta(t);$
(1.40)

see Theorem 2.4. By applying Theorem 1.1 and Propositions 1.2 and 1.3 to the
stable branching mechanisms, we obtain the following:

###### Corollary 1.4

Suppose that the branching mechanism is given by (1.9) with $1<\alpha<2$. In
addition, assume that

$\displaystyle\int_{1}^{\infty}z\nu(\mathrm{d}z)<\infty~{}~{}\mbox{and}~{}~{}\liminf_{x\to\infty}\frac{g(x)}{x}>-a.$

Then the CBIC-process is exponentially ergodic in the $V_{1}$-weighted total
variation distance.

###### Corollary 1.5

Suppose that the branching mechanism is given by (1.9) with $0<\alpha\leq 1$
and that $a>0$ or $c>0$ for $0<\alpha<1$. In addition, assume that

$\displaystyle\int_{1}^{\infty}\log(1+z)\nu(\mathrm{d}z)<\infty$

and

$\displaystyle\left\\{\begin{array}[]{ll}\displaystyle\liminf_{x\to\infty}\frac{g(x)}{x^{2-\alpha}}>\frac{\sigma\pi}{{\it\Gamma}(1-\alpha)\sin(\alpha\pi)},&\quad
0<\alpha<1,\cr\displaystyle\liminf_{x\to\infty}\frac{g(x)}{x\log
x}>\sigma,&\quad\alpha=1.\end{array}\right.$

Then the CBIC-process is exponentially ergodic in the $V_{\log}$-weighted
total variation distance.

In the situation of Corollary 1.5, the CBIC-process has infinite first moment,
so the result has no counterpart in the setting of CBI-processes, where
exponential ergodicities can only be considered for subcritical branching
mechanisms. The assumption that $a>0$ or $c>0$ for $0<\alpha<1$ in the
corollary comes from Condition 1.1. If this were not satisfied, there would
not be sufficient fluctuations in the CBIC-process as the two remaining noise
terms in (1.40) are both nondecreasing. As we explained before, if $\sigma>0$
and $0<\alpha<1$, then condition (1.4) fails and the corresponding CBI-process
without competition explodes at a finite time.

###  1.3 Examples

The following examples concern further applications of the results presented
above and show the conditions we introduce are sharp in typical situations.

###### Example 1.1

Consider the CBI-process with branching and immigration mechanisms given
respectively by (1.1) and (1.10) with $b>0$, $\beta>0$ and
$c=\nu(0,\infty)=0$. In addition, assume that
$\mu(\mathrm{d}z)=r\mathbf{1}_{(u,v)}(z)\mathrm{d}z$ for some $r>0$ and $0\leq
u<v\leq 1$. By Proposition 1.2, the CBI-process is exponentially ergodic
relative to the $V_{1}$-weighted total variation distance. In this case, the
branching mechanism does not satisfy Grey’s condition (1.12) as

$\displaystyle\lim_{\lambda\to\infty}\lambda^{-1}{\it\Psi}(\lambda)=b+\int_{0}^{1}z\mu(\mathrm{d}z)=b+\frac{r}{2}(v^{2}-u^{2}).$

Then the exponentially ergodicity in the total variation distance of the CBI-
process does not follow from Li and Ma [37, Theorem 2.5]. One can also see
that the conditions of Li and Wang [32, Theorem 1.1] are not satisfied.
$\square$

###### Example 1.2

Consider the CBI-process with branching and immigration mechanisms given by
(1.1) and (1.10), respectively. Suppose that (1.27) holds and the branching
mechanism is critical, that is,

$\displaystyle b_{0}:=b-\int_{1}^{\infty}z\mu(\mathrm{d}z)=0.$

It is known that

$\displaystyle\int_{\mathbb{R}_{+}}yP_{t}(0,\mathrm{d}y)=t\beta+t\int_{0}^{\infty}z\nu(\mathrm{d}z);$

see, e.g., Li [35, p.33]. The right-hand side tends to infinity as
$t\to\infty$, so the CBI-process is not exponentially ergodic in the
$V_{1}$-weighted total variation distance. This shows the conditions of
Corollary 1.4 are sharp. $\square$

###### Example 1.3

Let ${\it\Psi}$ be the stable branching mechanisms given by (1.9). By the
proof of Theorem 2.4, the CBIC-process has generator $L$ defined by (1.18)
with $b=a+\sigma h_{\alpha}$ and $\mu(\mathrm{d}z)=\alpha\sigma
m_{\alpha}(\mathrm{d}z)$, where $m_{\alpha}$ and $h_{\alpha}$ are defined by
(1.34) and (2.12), respectively. Let us consider the case where $\alpha=1$ and
$b=c=\nu(0,\infty)=0$. It is easy to show that

$\displaystyle LV_{\log}(x)=\frac{\sigma
x}{1+x}\big{[}1+\log(1+x)\big{]}-\frac{g(x)}{1+x}+\frac{\beta}{1+x}.$

If $\sigma\beta>0$ and $g(x)=\sigma x\log(1+x)$ for $x\geq 0$, there is a
constant $\lambda>0$ such that $LV_{\log}(x)\geq\lambda$, and so

$\displaystyle\int_{\mathbb{R}_{+}}V_{\log}(y)P_{t}(x,\mathrm{d}y)=V_{\log}(x)+\int^{t}_{0}P_{s}LV_{\log}(x)\mathrm{d}s\geq
V_{\log}(x)+\lambda t.$

Consequently, the CBIC-process is not exponentially ergodic in the
$V_{\log}$-weighted total variation distance. When $0<\alpha<1$, a similar
example can be given by considering the function
$g(x)=[{\it\Gamma}(1-\alpha)\sin(\alpha\pi)]^{-1}\sigma\pi x^{2-\alpha}$. Then
the conditions of Corollary 1.5 are sharp. $\square$

###  1.4 The approach of coupling and distance

The characterizations like (1.3) and (1.11) are not available for the CBIC-
process. Our proof of Theorem 1.1 is based on the approach of coupling and
distance. Those techniques have played important roles in the ergodic theory
of Markov processes. We refer to Chen [11, 12] for systematical treatments of
the techniques. In particular, a number of results on the exponential
ergodicity have been obtained by this method for Ornstein-Uhlenbeck type
processes with nonlinear drifts defined by stochastic differential equations
of the form:

$\displaystyle\mathrm{d}x(t)=\mathrm{d}L(t)-g(x(t))\mathrm{d}t,\quad t\geq 0,$
(1.42)

where $\\{L(t):t\geq 0\\}$ is a Lévy process and the drift coefficient $g$ is
sufficiently regular so that there is a unique solution to the equation. When
the driving noise is a Brownian motion, the exponential ergodicity and related
problems were studied by Eberle and his coauthors [16, 17, 18] and Luo and
Wang [39]. For discontinuous Lévy noises, the ergodicity problem was
investigated by Liang et al. [38], Luo and Wang [40], Schilling and Wang [48]
and Majka [41]. The feature of those processes is that the random noise in
(1.42) is both temporarily and spatially homogeneous.

The main difficulty of the proof of Theorem 1.1 is that the branching
fluctuations become small and rare when the process is close to zero, which is
clear from (1.16) or (1.18). A similar phenomenon has been noticed in
nonlinear branching processes by Li and Wang [32]; see also [31, 33]. The
difficulty was treated in [32] by introducing conditions on the asymptotics at
zero of the branching coefficients. We cannot do that in our setting here
since the branching coefficients here have to be linear. The key of our proof
is the design of a Markov coupling process and a nonsymmetric control
function. More precisely, we construct a two-dimensional Markov coupling
process $\\{(X_{t},Y_{t}):t\geq 0\\}$ on $\mathbb{R}_{+}^{2}$, where both
$\\{X_{t}:t\geq 0\\}$ and $\\{Y_{t}:t\geq 0\\}$ are Markov processes with
transition semigroup $(P_{t})_{t\geq 0}$. The coupling process satisfies
$X_{T+t}=Y_{T+t}$ for every $t\geq 0$, where $T=\inf\\{t\geq 0:X_{t}=Y_{t}\\}$
with $\inf\emptyset=\infty$ by convention is the succeeding or coupling time.
We need to make the coupling succeed as early as possible, which is realized
by reflecting the noises in suitable ways; see Remark 3.6. Then we define a
non-symmetric function $G_{0}$ with the exponential contraction property:

$\displaystyle\tilde{P}_{t}G_{0}(x,y)\leq\mathrm{e}^{-\lambda
t}G_{0}(x,y),\quad t\geq 0,(x,y)\in\mathbb{R}_{+}^{2},$ (1.43)

where $(\tilde{P}_{t})_{t\geq 0}$ is the transition semigroup of the coupling
process. The function $G_{0}$ should control the distance $d_{V}$ in the sense
that there are constants $c_{2}\geq c_{1}>0$ such that, for
$(x,y)\in\mathbb{R}_{+}^{2}$,

$\displaystyle c_{1}G_{0}(x,y)\leq d_{V}(x,y)\leq c_{2}G_{0}(x,y).$ (1.44)

Then we deduce the estimate (1.24) from (1.43) and (1.44). The non-symmetric
control function makes it possible to identify the dominating factor from the
branching, immigration and competition mechanisms in different parts of the
space, which is important in establishing the estimate (1.43). By those
devices we not only overcome the difficulty caused by the degeneracy of the
branching coefficients but also weaken the fluctuation conditions introduced
for the exponential ergodicity of the Ornstein-Uhlenbeck type processes; see,
e.g., [32, 38, 40, 41, 48]. In fact, the assumptions on the drift coefficient
in Corollaries 1.4 and 1.5 are also weaker than those in the previous papers,
where lower bounds of the increment $g(x)-g(y)$ for $x\geq y\geq 0$ were
required. We believe that the method used here could be adapted to more
general classes of processes with Lévy driving noises.

The rest of the paper is organized as follows. In Section 2, the existence and
uniqueness of solutions to the stochastic equations (1.16), (1.39) and (1.40)
are proved. In Section 3, we give the construction of the Markov coupling
process. In Section 4, some necessary estimates for the coupling generator are
established. The proofs of the ergodicity results are given in Section 5.

##  2 Stochastic equations of CBIC-processes

In this section, we prove the existence and pathwise uniqueness of solutions
to the stochastic equations (1.16), (1.39) and (1.40), which give
constructions of the CBIC-processes.

###### Theorem 2.1

For any $\mathscr{F}_{0}$-measurable random variable $x(0)\geq 0$, there is a
pathwise unique solution $\\{x(t):t\geq 0\\}$ to (1.16).

Proof. By Theorem 2.5 of Dawson and Li [14], for each $n\geq 1$ there is a
pathwise unique solution $\\{x_{n}(t):t\geq 0\\}$ to

$\displaystyle x(t)\\!\\!\\!$ $\displaystyle=\\!\\!\\!$ $\displaystyle
x(0)+\int_{0}^{t}\int_{0}^{x(s-)}W(\mathrm{d}s,\mathrm{d}u)+\int_{0}^{t}[\beta-
bx(s)-g(x(s))]\mathrm{d}s$ (2.3)
$\displaystyle+\int_{0}^{t}\int_{0}^{1}\int_{0}^{x(s-)}z\tilde{M}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}u)+\int_{0}^{t}\int_{1}^{\infty}\int_{0}^{x(s-)}(z\land
n)M(\mathrm{d}s,\mathrm{d}z,\mathrm{d}u)$
$\displaystyle+\int_{0}^{t}\int_{0}^{\infty}(z\land
n)N(\mathrm{d}s,\mathrm{d}z).$

Let $\zeta_{n}=\inf\\{t\geq 0:x_{n}(t)\geq n\\}$. From the pathwise uniqueness
for (2.3) it follows that $\zeta_{n}$ is nondecreasing in $n\geq 1$ and
$x_{n+1}(t)=x_{n}(t)$ for $0\leq t<\zeta_{n}$. Clearly, the pathwise unique
solution $\\{x(t):t\geq 0\\}$ to (1.16) is given by $x(t)=x_{n}(t)$ for $0\leq
t<\zeta_{n}$ and $x(t)=\infty$ for $t\geq\zeta:=\lim_{n\to\infty}\zeta_{n}$.
$\square$

The above theorem gives the construction of the CBIC-process. The next result
justifies the fact that the operator $(L,\mathscr{D}(L))$ defined by (1.18) is
a restriction of the generator of the process.

###### Theorem 2.2

Let $\\{x(t):t\geq 0\\}$ be the pathwise unique solution to (1.16) and let
$\zeta_{n}=\inf\\{t\geq 0:x(t)\geq n\\}$. Then for any $n\geq 1$ and
$f\in\mathscr{D}(L)$ we have

$\displaystyle
f(x(t\land\zeta_{n}))=f(x(0))+\int_{0}^{t\land\zeta_{n}}Lf(x(s))\mathrm{d}s+M_{n}(t),$
(2.4)

where $\\{M_{n}(t):t\geq 0\\}$ is a martingale defined by

$\displaystyle M_{n}(t)\\!\\!\\!$ $\displaystyle=\\!\\!\\!$
$\displaystyle\int_{0}^{t\land\zeta_{n}}\int_{0}^{x(s-)}f^{\prime}(x(s-))W(\mathrm{d}s,\mathrm{d}u)+\int_{0}^{t\land\zeta_{n}}\int_{0}^{\infty}{\it\Delta}_{z}f(x(s-))\tilde{N}(\mathrm{d}s,\mathrm{d}z)$
(2.6)
$\displaystyle+\int_{0}^{t\land\zeta_{n}}\int_{0}^{\infty}\int_{0}^{x(s-)}{\it\Delta}_{z}f(x(s-))\tilde{M}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}u).$

Proof. By (1.16) the process $\\{x(t):t\geq 0\\}$ is a semimartingale. For any
$f\in\mathscr{D}(L)$, we can use Itô’s formula to see that

$\displaystyle f(x(t\land\zeta_{n}))\\!\\!\\!$ $\displaystyle=\\!\\!\\!$
$\displaystyle
f(x(0))-\int_{0}^{t\land\zeta_{n}}f^{\prime}(x(s))g(x(s))\mathrm{d}s+\int_{0}^{t\land\zeta_{n}}\int_{0}^{x(s-)}f^{\prime}(x(s-))L(\mathrm{d}s,\mathrm{d}u)$
$\displaystyle+\int_{0}^{t\land\zeta_{n}}f^{\prime}(x(s-))\eta(\mathrm{d}s)+c\int_{0}^{t\land\zeta_{n}}f^{\prime\prime}(x(s))x(s)\mathrm{d}s$
$\displaystyle+\int_{0}^{t\land\zeta_{n}}\int_{0}^{\infty}\int_{0}^{x(s-)}\big{[}{\it\Delta}_{z}f(x(s-))-zf^{\prime}(x(s-))\big{]}M(\mathrm{d}s,\mathrm{d}z,\mathrm{d}u)$
$\displaystyle+\int_{0}^{t\land\zeta_{n}}\int_{0}^{\infty}\big{[}{\it\Delta}_{z}f(x(s-))-zf^{\prime}(x(s-))\big{]}N(\mathrm{d}s,\mathrm{d}z)$
$\displaystyle=\\!\\!\\!$ $\displaystyle
f(x(0))-\int_{0}^{t\land\zeta_{n}}f^{\prime}(x(s))g(x(s))\mathrm{d}s+\int_{0}^{t\land\zeta_{n}}\int_{0}^{x(s-)}f^{\prime}(x(s-))W(\mathrm{d}s,\mathrm{d}u)$
$\displaystyle+\int_{0}^{t\land\zeta_{n}}\big{[}\beta-
bx(s)\big{]}f^{\prime}(x(s))\mathrm{d}s+c\int_{0}^{t\land\zeta_{n}}x(s)f^{\prime\prime}(x(s))\mathrm{d}s$
$\displaystyle+\int_{0}^{t\land\zeta_{n}}\int_{0}^{\infty}\int_{0}^{x(s-)}{\it\Delta}_{z}f(x(s-))\tilde{M}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}u)$
$\displaystyle+\int_{0}^{t\land\zeta_{n}}x(s)\mathrm{d}s\int_{0}^{\infty}\big{[}{\it\Delta}_{z}f(x(s))-zf^{\prime}(x(s))\mathbf{1}_{\\{z\leq
1\\}}\big{]}\mu(\mathrm{d}z)$
$\displaystyle+\int_{0}^{t\land\zeta_{n}}\int_{0}^{\infty}{\it\Delta}_{z}f(x(s-))\tilde{N}(\mathrm{d}s,\mathrm{d}z)+\int_{0}^{t\land\zeta_{n}}\mathrm{d}s\int_{0}^{\infty}{\it\Delta}_{z}f(x(s))\nu(\mathrm{d}z).$

Then (2.4) holds with $M_{n}(t)$ defined by (2.6). Since $x(s-)\leq n$ for
$0<s\leq\zeta_{n}$, it is easy to show that $\\{M_{n}(t):t\geq 0\\}$ is a
martingale. $\square$

###### Proposition 2.3

Suppose that Condition 1.3 is satisfied. Then the solution $\\{x(t):t\geq
0\\}$ to (1.16) is conservative and

$\displaystyle\mathbf{E}[V(x(t))]\leq\mathbf{E}[V(x(0))]\mathrm{e}^{-C_{1}t}+C_{0}C_{1}^{-1}\big{(}1-\mathrm{e}^{-C_{1}t}\big{)},\quad
t\geq 0.$ (2.7)

Proof. There is no loss of generality to assume $\mathbf{E}[V(x(0))]<\infty$.
By applying (2.4) to the function $V$ and using integration by parts we have

$\displaystyle\mathrm{e}^{C_{1}t}V(x(t))=V(x(0))+\int_{0}^{t}\mathrm{e}^{C_{1}s}\big{[}C_{1}V(x(s))+LV(x(s))\big{]}\mathrm{d}s+M(t),$
(2.8)

where

$\displaystyle M(t)\\!\\!\\!$ $\displaystyle=\\!\\!\\!$
$\displaystyle\int_{0}^{t}\int_{0}^{x(s)}\mathrm{e}^{C_{1}s}f^{\prime}(x(s))W(\mathrm{d}s,\mathrm{d}u)+\int_{0}^{t}\int_{0}^{\infty}\mathrm{e}^{C_{1}s}{\it\Delta}_{z}f(x(s-))\tilde{N}(\mathrm{d}s,\mathrm{d}z)$
$\displaystyle+\int_{0}^{t}\int_{0}^{\infty}\int_{0}^{x(s-)}\mathrm{e}^{C_{1}s}{\it\Delta}_{z}f(x(s-))\tilde{M}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}u).$

Recall that $\zeta_{n}=\inf\\{t\geq 0:x(t)\geq n\\}$ and
$\zeta=\lim_{n\to\infty}\zeta_{n}$. In view of (1.25), we can take the
expectations in both sides of (2.8) at time $t\land\zeta_{n}$ to obtain

$\displaystyle\mathbf{E}[\mathrm{e}^{C_{1}(t\land\zeta_{n})}V(x(t\land\zeta_{n}))]\leq\mathbf{E}[V(x(0))]+C_{0}C_{1}^{-1}\big{(}\mathrm{e}^{C_{1}t}-1\big{)}.$

By Fatou’s lemma it follows that

$\displaystyle\mathbf{E}[\mathrm{e}^{C_{1}(t\land\zeta)}V(x(t\land\zeta))]\leq\mathbf{E}[V(x(0))]+C_{0}C_{1}^{-1}\big{(}\mathrm{e}^{C_{1}t}-1\big{)}$

with the convention $V(\infty)=\infty$. Then $\mathbf{P}(\zeta\leq t)=0$ and
(2.7) follows. Since $t\geq 0$ was arbitrary, we have $\zeta=\infty$ almost
surely. $\square$

###### Theorem 2.4

For any $\mathscr{F}_{0}$-measurable random variable $x(0)\geq 0$, there are
pathwise unique solutions to (1.39) and (1.40), and the solutions are CBIC-
processes with competition mechanism $g$, stable branching mechanisms
${\it\Psi}$ given by (1.9) and immigration mechanism ${\it\Phi}$ given by
(1.10).

Proof. Let $\\{x(t)\\}$ be the solution to (1.16) with
$\\{M(\mathrm{d}s,\mathrm{d}z,\mathrm{d}u)\\}$ being a Poisson random measure
on $(0,\infty)^{3}$ with intensity
$\alpha\sigma\mathrm{d}sm_{\alpha}(\mathrm{d}z)\mathrm{d}u$, where
$m_{\alpha}(\mathrm{d}z)$ is given by (1.34). Moreover, we take $b=a+\sigma
h_{\alpha}$, where

$\displaystyle
h_{\alpha}=\left\\{\begin{array}[]{ll}-\alpha{\it\Gamma}(2-\alpha)^{-1},&\quad
1<\alpha<2,\cr
0,&\quad\alpha=1,\cr\alpha(1-\alpha)^{-1}{\it\Gamma}(1-\alpha)^{-1},&\quad
0<\alpha<1.\end{array}\right.$ (2.12)

If $c>0$, one can see that

$\displaystyle
B(t)=\int_{0}^{t}\int_{0}^{x(s)}\frac{1}{\sqrt{2cx(s)}}\mathbf{1}_{\\{x(s)>0\\}}W(\mathrm{d}s,\mathrm{d}u)+\frac{1}{\sqrt{2c}}\int_{0}^{t}\int_{0}^{1}\mathbf{1}_{\\{x(s)=0\\}}W(\mathrm{d}s,\mathrm{d}u)$

defines a standard Brownian motion and

$\displaystyle\int_{0}^{t}\sqrt{2cx(s)}\mathrm{d}B(s)=\int_{0}^{t}\int_{0}^{x(s)}W(\mathrm{d}s,\mathrm{d}u).$

In the sequel, we assume $\sigma>0$, for otherwise the proof is simpler. Let
$\\{M_{\alpha}(\mathrm{d}s,\mathrm{d}z)\\}$ be the random measure on
$(0,\infty)^{2}$ defined by, for $B\in\mathscr{B}(0,\infty)$,

$\displaystyle M_{\alpha}((0,t]\times B)\\!\\!\\!$ $\displaystyle=\\!\\!\\!$
$\displaystyle\int_{0}^{t}\int_{0}^{\infty}\int_{0}^{x(s-)}\mathbf{1}_{\\{x(s-)>0\\}}\mathbf{1}_{B}\Big{(}\frac{z}{\sqrt[\alpha]{\alpha\sigma
x(s-)}}\Big{)}M(\mathrm{d}s,\mathrm{d}z,\mathrm{d}u)$
$\displaystyle+\int_{0}^{t}\int_{0}^{\infty}\int_{0}^{1/(\alpha\sigma)}\mathbf{1}_{\\{x(s-)=0\\}}\mathbf{1}_{B}(z)M(\mathrm{d}s,\mathrm{d}z,\mathrm{d}u).$

One can show as in Li [34, p.230] that
$\\{M_{\alpha}(\mathrm{d}s,\mathrm{d}z)\\}$ is a Poisson random measure with
intensity $\mathrm{d}sm_{\alpha}(\mathrm{d}z)$. Let $\\{z_{\alpha}(t)\\}$ be
the $\alpha$-stable Lévy processes defined by (1.38). It is easy to see that

$\displaystyle\int_{0}^{t}\sqrt[\alpha]{\alpha\sigma
x(s-)}\mathrm{d}z_{\alpha}(s)\\!\\!\\!$ $\displaystyle=\\!\\!\\!$
$\displaystyle\int_{0}^{t}\int_{0}^{1}\int_{0}^{x(s-)}z\tilde{M}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}u)+\sigma
h_{\alpha}\int_{0}^{t}x(s)\mathrm{d}s$
$\displaystyle+\int_{0}^{t}\int_{1}^{\infty}\int_{0}^{x(s-)}zM(\mathrm{d}s,\mathrm{d}z,\mathrm{d}u).$

Then (1.39) and (1.40) hold. By Fu and Li [22, Theorem 5.1], for any $n\geq 1$
there is a pathwise unique solution to the stochastic equation:

$\displaystyle x(t)\\!\\!\\!$ $\displaystyle=\\!\\!\\!$ $\displaystyle
x(0)+\int_{0}^{t}\sqrt{2cx(s)}\mathrm{d}B(s)+\int_{0}^{t}\int_{0}^{1}\sigma
x(s-)z\tilde{M}_{1}(\mathrm{d}s,\mathrm{d}z)$
$\displaystyle+\int_{0}^{t}\int_{1}^{\infty}\sigma x(s-)(z\land
n)M_{1}(\mathrm{d}s,\mathrm{d}z)+\sigma\int_{0}^{t}x(s)\log(\sigma
x(s))\mathrm{d}s$
$\displaystyle-\int_{0}^{t}[ax(s)+g(x(s))]\mathrm{d}s+\eta(t).$

Since $n\geq 1$ is arbitrary, the pathwise uniqueness also holds for (1.39). A
similar argument gives the pathwise uniqueness of (1.40). These give the
results of the theorem; see, e.g., Situ [49, p.76 and p.104]. $\square$

##  3 Construction of the coupling process

In this section, we give the construction of the Markov coupling of the CBIC-
process. Given $x\in\mathbb{R}$ and a general $\sigma$-finite measure $m$ on
$(0,\infty)$, we define another $\sigma$-finite measure on $(0,\infty)$ by

$\displaystyle
m_{x}(\mathrm{d}z)=\frac{1}{2}\big{[}m\land(\delta_{x}*m)\big{]}(\mathrm{d}z).$
(3.1)

It is easy to see that

$\displaystyle m_{x}=\delta_{x}*m_{-x}~{}~{}\mbox{and}~{}~{}m_{x}(0,0\vee
x]=0.$ (3.2)

Moreover, we have

$\displaystyle m_{x}(0,\infty)=m_{-x}(0,\infty)\leq\frac{1}{2}m(|x|,\infty).$
(3.3)

For any $x,y\geq 0$ write $m_{x,y}=m_{y-x}+m_{x-y}$. One can show that both
$x\mapsto m_{x}(\mathrm{d}z)$ and $(x,y)\mapsto m_{x,y}(\mathrm{d}z)$ are
kernels. For notational convenience, we also write those kernels as
$m(x,\mathrm{d}z)$ and $m(x,y,\mathrm{d}z)$, respectively. We shall use those
notations and facts to the Lévy measures $\mu$ and $\nu$ of the CBIC-process.
For $x\in\mathbb{R}$ and $z>0$ we set

$\displaystyle\rho_{1}(x,z)=\frac{\mu(x,\mathrm{d}z)}{\mu(\mathrm{d}z)},~{}~{}r_{1}(x,z)=\frac{\nu(x,\mathrm{d}z)}{\nu(\mathrm{d}z)}.$
(3.4)

By (3.1), the above Radon-Nikodym derivatives exist and satisfy

$\displaystyle\sup_{x\in\mathbb{R}}\sup_{z>0}\rho_{1}(x,z)\leq\frac{1}{2},\quad\sup_{x\in\mathbb{R}}\sup_{z>0}r_{1}(x,z)\leq\frac{1}{2}.$

In view of (3.2), we have

$\displaystyle\rho_{1}(x,z)=r_{1}(x,z)=0,\quad 0<z\leq 0\vee x.$ (3.5)

For $x,y\geq 0$ and $z>0$ let

$\displaystyle\rho_{2}(x,y,z)=\rho_{1}(x-y,z)+\rho_{1}(y-x,z)$

and

$\displaystyle r_{2}(x,y,z)=r_{1}(x-y,z)+r_{1}(y-x,z).$

Suppose that $({\it\Omega},\mathscr{F},\mathscr{F}_{t},\mathbf{P})$ is a
filtered probability space satisfying the usual hypotheses. Let
$\\{W_{0}(\mathrm{d}s,\mathrm{d}u)\\}$ be an $(\mathscr{F}_{t})$-Gaussian
white noise on $(0,\infty)^{2}$ based on $2c\mathrm{d}s\mathrm{d}u$. Let
$\\{M_{0}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}u,\mathrm{d}v)\\}$ be a time-space
$(\mathscr{F}_{t})$-Poisson random measure on $(0,\infty)^{3}\times(0,1]$ with
intensity $\mathrm{d}s\mu(\mathrm{d}z)\mathrm{d}u\mathrm{d}v$ and compensated
measure
$\\{\tilde{M}_{0}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}u,\mathrm{d}v)\\}$. Let
$\\{N_{0}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}v)\\}$ be a time-space
$(\mathscr{F}_{t})$-Poisson random measure on $(0,\infty)^{2}\times(0,1]$ with
intensity $\mathrm{d}s\nu(\mathrm{d}z)\mathrm{d}v$. We assume those random
noises are independent of each other. Here the extra component $(0,1]$ of the
Poisson random measures is introduced for the convenience of disassembling the
jumps in the construction of the coupling process. Let $\\{\eta_{t}\\}$ be the
$(\mathscr{F}_{t})$-subordinator defined by

$\displaystyle\eta_{t}=\beta
t+\int_{0}^{t}\int_{0}^{\infty}\int_{0}^{1}zN_{0}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}v).$

Let $\\{L_{0}(\mathrm{d}s,\mathrm{d}u)\\}$ and
$\\{L_{0}^{*}(\mathrm{d}s,\mathrm{d}u)\\}$ be two accompanying spectrally one-
sided time-space $(\mathscr{F}_{t})$-Lévy white noises defined by

$\displaystyle L_{0}(\mathrm{d}s,\mathrm{d}u)\\!\\!\\!$
$\displaystyle=\\!\\!\\!$ $\displaystyle
W_{0}(\mathrm{d}s,\mathrm{d}u)-b\mathrm{d}s\mathrm{d}u+\int_{\\{0<z\leq
1\\}}\int_{\\{0<v\leq
1\\}}z\tilde{M}_{0}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}u,\mathrm{d}v)$ (3.7)
$\displaystyle+\int_{\\{1<z<\infty\\}}\int_{\\{0<v\leq
1\\}}zM_{0}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}u,\mathrm{d}v)$

and

$\displaystyle L_{0}^{*}(\mathrm{d}s,\mathrm{d}u)\\!\\!\\!$
$\displaystyle=\\!\\!\\!$
$\displaystyle-\,W_{0}(\mathrm{d}s,\mathrm{d}u)-b\mathrm{d}s\mathrm{d}u+\int_{\\{0<z\leq
1\\}}\int_{\\{0<v\leq
1\\}}z\tilde{M}_{0}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}u,\mathrm{d}v)$ (3.9)
$\displaystyle+\int_{\\{1<z<\infty\\}}\int_{\\{0<v\leq
1\\}}zM_{0}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}u,\mathrm{d}v).$

By Theorem 2.1, for any $\mathscr{F}_{0}$-measurable random variable
$X_{0}\geq 0$ we can construct a CBIC-process $\\{X_{t}:t\geq 0\\}$ by the
pathwise unique solution to the stochastic integral equation:

$\displaystyle
X_{t}=X_{0}+\int_{0}^{t}\int_{0}^{X_{s-}}L_{0}(\mathrm{d}s,\mathrm{d}u)-\int_{0}^{t}g(X_{s})\mathrm{d}s+\eta_{t}.$
(3.10)

###### Theorem 3.1

Suppose that the CBIC-process $\\{X_{t}:t\geq 0\\}$ defined by (3.10) is
conservative. Let $Y_{0}$ be an $\mathscr{F}_{0}$-measurable random variable
satisfying $X_{0}\geq Y_{0}\geq 0$. Then there is a pathwise unique solution
$\\{(Y_{t},\xi_{t}):t\geq 0\\}$ to the system of stochastic equations:

$\displaystyle Y_{t}\\!\\!\\!$ $\displaystyle=\\!\\!\\!$ $\displaystyle
Y_{0}+\int_{t\land
T}^{t}\int_{0}^{Y_{s-}}L_{0}(\mathrm{d}s,\mathrm{d}u)-\int_{0}^{t}g(Y_{s})\mathrm{d}s+\eta_{t}\quad$
(3.12) $\displaystyle+\int_{0}^{t\land
T}\int_{0}^{Y_{s-}}L_{0}^{*}(\mathrm{d}s,\mathrm{d}u)+\xi_{t}$

and

$\displaystyle\xi_{t}\\!\\!\\!$ $\displaystyle=\\!\\!\\!$
$\displaystyle\int_{0}^{t\land
T}\int_{0}^{\infty}\int_{0}^{Y_{s-}}\int_{0}^{\rho_{1}(Y_{s-}-X_{s-},z)}(X_{s-}-Y_{s-})M_{0}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}u,\mathrm{d}v)$
(3.16) $\displaystyle+\int_{0}^{t\land
T}\int_{0}^{\infty}\int_{0}^{r_{1}(Y_{s-}-X_{s-},z)}(X_{s-}-Y_{s-})N_{0}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}v)$
$\displaystyle+\int_{0}^{t\land
T}\int_{0}^{\infty}\int_{0}^{Y_{s-}}\int_{\rho_{1}(Y_{s-}-X_{s-},z)}^{\rho_{2}(X_{s-},Y_{s-},z)}(Y_{s-}-X_{s-})M_{0}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}u,\mathrm{d}v)$
$\displaystyle+\int_{0}^{t\land
T}\int_{0}^{\infty}\int_{r_{1}(Y_{s-}-X_{s-},z)}^{r_{2}(X_{s-},Y_{s-},z)}(Y_{s-}-X_{s-})N_{0}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}v),$

where

$\displaystyle T=\inf\\{t\geq 0:X_{t}\leq Y_{t}\\}=\inf\\{t\geq
0:X_{t}=Y_{t}~{}\mbox{or}~{}X_{t-}=Y_{t-}\\}.$ (3.17)

Moreover, we have $X_{T+t}=Y_{T+t}$ for every $t\geq 0$ if $T<\infty$.

In the proof of the above theorem, we shall see that the pure jump process
$\\{\xi_{t}:t\geq 0\\}$ has at most a finite number of jumps. In order to give
the proof, we make some preparations. By Theorem 2.1, there is a CBIC-process
$\\{Z_{0}(t):t\geq 0\\}$ defined by the pathwise unique solution to

$\displaystyle
Z_{0}(t)=Y_{0}+\int_{0}^{t}\int_{0}^{Z_{0}(s-)}L_{0}^{*}(\mathrm{d}s,\mathrm{d}u)-\int_{0}^{t}g(Z_{0}(s))\mathrm{d}s+\eta_{t}.$
(3.18)

###### Lemma 3.2

Let $U_{1}=\inf\\{t\geq 0:X_{t}\leq Z_{0}(t)\\}$. Under the assumptions of
Theorem 3.1, we have

$\displaystyle U_{1}=\inf\\{t\geq 0:X_{t}=Z_{0}(t)\\}=\inf\\{t\geq
0:X_{t-}=Z_{0}(t-)\\}.$ (3.19)

Proof. From (3.10) and (3.18) it is easy to see that $X_{t-}=Z_{0}(t-)$
implies $X_{t}=Z_{0}(t)$. It follows that

$\displaystyle U_{1}\leq\inf\\{t\geq 0:X_{t}=Z_{0}(t)\\}\leq\inf\\{t\geq
0:X_{t-}=Z_{0}(t-)\\}.$

Then (3.19) holds if $U_{1}=\infty$. In the case of $U_{1}<\infty$, we clearly
have $X_{U_{1}-}\geq Z_{0}(U_{1}-)$ and $X_{t}>Z_{0}(t)$ for $0\leq t<U_{1}$.
By (3.10) and (3.18),

$\displaystyle{\it\Delta}X_{U_{1}}\\!\\!\\!$ $\displaystyle=\\!\\!\\!$
$\displaystyle\int_{\\{U_{1}\\}}\int_{0}^{\infty}\int_{0}^{X_{U_{1}-}}\int_{0}^{1}zM_{0}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}u,\mathrm{d}v)$
$\displaystyle+\int_{\\{U_{1}\\}}\int_{0}^{\infty}\int_{0}^{1}zN_{0}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}v)$
$\displaystyle\geq\\!\\!\\!$
$\displaystyle\int_{\\{U_{1}\\}}\int_{0}^{\infty}\int_{0}^{Z_{0}(U_{1}-)}\int_{0}^{1}zM_{0}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}u,\mathrm{d}v)$
$\displaystyle+\int_{\\{U_{1}\\}}\int_{0}^{\infty}\int_{0}^{1}zN_{0}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}v)={\it\Delta}Z_{0}(U_{1}).$

On the other hand, by the right continuity of the processes,

$\displaystyle X_{U_{1}-}+{\it\Delta}X_{U_{1}}=X_{U_{1}}\leq
Z_{0}(U_{1})=Z_{0}(U_{1}-)+{\it\Delta}Z_{0}(U_{1}-).$

Then we must have $X_{U_{1}-}=Z_{0}(U_{1}-)$ and
${\it\Delta}X_{U_{1}}={\it\Delta}Z_{0}(U_{1})$, implying
$X_{U_{1}}=Z_{0}(U_{1})$. Those yield (3.19). $\square$

By (3.19) and the definition of $U_{1}$ we have $X_{s-}>Z_{0}(s-)\geq 0$ for
$0<s<U_{1}$. Let us consider the pure jump process $\\{\xi_{0}(t):t\geq 0\\}$
given by

$\displaystyle\xi_{0}(t)\\!\\!\\!$ $\displaystyle=\\!\\!\\!$
$\displaystyle\int_{0}^{t\land
U_{1}}\int_{0}^{\infty}\int_{0}^{Z_{0}(s-)}\int_{0}^{\rho_{1}(Z_{0}(s-)-X_{s-},z)}[X_{s-}-Z_{0}(s-)]M_{0}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}u,\mathrm{d}v)$
(3.23) $\displaystyle+\int_{0}^{t\land
U_{1}}\int_{0}^{\infty}\int_{0}^{r_{1}(Z_{0}(s-)-X_{s-},z)}[X_{s-}-Z_{0}(s-)]N_{0}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}v)$
$\displaystyle+\int_{0}^{t\land
U_{1}}\int_{0}^{\infty}\int_{0}^{Z_{0}(s-)}\int_{\rho_{1}(Z_{0}(s-)-X_{s-},z)}^{\rho_{2}(X_{s-},Z_{0}(s-),z)}[Z_{0}(s-)-X_{s-}]M_{0}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}u,\mathrm{d}v)$
$\displaystyle+\int_{0}^{t\land
U_{1}}\int_{0}^{\infty}\int_{r_{1}(Z_{0}(s-)-X_{s-},z)}^{r_{2}(X_{s-},Z_{0}(s-),z)}[Z_{0}(s-)-X_{s-}]N_{0}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}v).$

The process makes its first jump at time $\sigma_{1}\land\tau_{1}$, where

$\displaystyle\sigma_{1}=\inf\\{t\geq
0:{\it\Delta}\xi_{0}(t)=X_{t-}-Z_{0}(t-)\\}$

and

$\displaystyle\tau_{1}=\inf\\{t\geq
0:{\it\Delta}\xi_{0}(t)=Z_{0}(t-)-X_{t-}\\}.$

In view of (3.3) and (3.4), for any $x>y\geq 0$,

$\displaystyle\rho_{1}(x,y):=\int_{0}^{\infty}\rho_{1}(x-y,z)\mu(\mathrm{d}z)=\int_{0}^{\infty}\rho_{1}(y-x,z)\mu(\mathrm{d}z)\leq\frac{1}{2}\mu(x-y,\infty)<\infty$

and

$\displaystyle
r_{1}(x,y):=\int_{0}^{\infty}r_{1}(x-y,z)\nu(\mathrm{d}z)=\int_{0}^{\infty}r_{1}(y-x,z)\nu(\mathrm{d}z)\leq\frac{1}{2}\nu(x-y,\infty)<\infty.$

Then the stopping times $\sigma_{1}$ and $\tau_{1}$ occur at the same rate

$\displaystyle\big{[}Z_{0}(s-)\rho_{1}(X_{s-},Z_{0}(s-))+r_{1}(X_{s-},Z_{0}(s-))\big{]}\mathrm{d}s.$

It follows that

$\displaystyle\mathbf{P}\big{(}\sigma_{1}<\tau_{1}\big{|}\sigma_{1}\land\tau_{1}<\infty\big{)}=\mathbf{P}\big{(}\tau_{1}<\sigma_{1}\big{|}\sigma_{1}\land\tau_{1}<\infty\big{)}=\frac{1}{2}.$
(3.24)

For $t\geq 0$ let

$\displaystyle Y_{0}(t)=Z_{0}(t\land
U_{1}\land\sigma_{1}\land\tau_{1})+\xi_{0}(t\land
U_{1}\land\sigma_{1}\land\tau_{1}).$ (3.25)

###### Lemma 3.3

Under the assumptions of Theorem 3.1, we have
$X_{\sigma_{1}}=Y_{0}(\sigma_{1})>0$ on the event
$\\{\sigma_{1}<\tau_{1}\leq\infty\\}$ and $X_{\tau_{1}}>Y_{0}(\tau_{1})>0$ on
the event $\\{\tau_{1}<\sigma_{1}\leq\infty\\}$.

Proof. (1) On the event $\\{\sigma_{1}<\tau_{1}\leq\infty\\}$ we have
$\sigma_{1}<U_{1}$ and so $\sigma_{1}<U_{1}\land\tau_{1}$. By (3.23) and the
definition of $\sigma_{1}$ we have

$\displaystyle{\it\Delta}\xi_{0}(\sigma_{1})=X_{\sigma_{1}-}-Z_{0}(\sigma_{1}-)>0$
(3.26)

and

$\displaystyle\int_{\\{\sigma_{1}\\}}\int_{0}^{\infty}\int_{0}^{Z_{0}(\sigma_{1}-)}\int_{0}^{\rho_{1}(Z_{0}(\sigma_{1}-)-X_{\sigma_{1}-},z)}M_{0}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}u,\mathrm{d}v)$
$\displaystyle\qquad\qquad+\int_{\\{\sigma_{1}\\}}\int_{0}^{\infty}\int_{0}^{r_{1}(Z_{0}(\sigma_{1}-)-X_{\sigma_{1}-},z)}N_{0}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}v)=1.$

By the temporarily homogeneous nature of the Poisson random measures, we can
enlarge the integration intervals of $u$ and $v$ without changing the equality
above. In particular, we have

$\displaystyle\int_{\\{\sigma_{1}\\}}\int_{0}^{\infty}\int_{0}^{Z_{0}(\sigma_{1}-)}\int_{0}^{1}M_{0}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}u,\mathrm{d}v)+\int_{\\{\sigma_{1}\\}}\int_{0}^{\infty}\int_{0}^{1}N_{0}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}u)$
$\displaystyle\qquad=\int_{\\{\sigma_{1}\\}}\int_{0}^{\infty}\int_{0}^{X_{\sigma_{1}-}}\int_{0}^{1}M_{0}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}u,\mathrm{d}v)+\int_{\\{\sigma_{1}\\}}\int_{0}^{\infty}\int_{0}^{1}N_{0}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}u)=1.$

It follows that

$\displaystyle z(\sigma_{1})\\!\\!\\!$ $\displaystyle:=\\!\\!\\!$
$\displaystyle\int_{\\{\sigma_{1}\\}}\int_{0}^{\infty}\int_{0}^{Z_{0}(\sigma_{1}-)}\int_{0}^{1}zM_{0}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}u,\mathrm{d}v)+\int_{\\{\sigma_{1}\\}}\int_{0}^{\infty}\int_{0}^{1}zN_{0}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}u)$
$\displaystyle=\\!\\!\\!$
$\displaystyle\int_{\\{\sigma_{1}\\}}\int_{0}^{\infty}\int_{0}^{X_{\sigma_{1}-}}\int_{0}^{1}zM_{0}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}u,\mathrm{d}v)+\int_{\\{\sigma_{1}\\}}\int_{0}^{\infty}\int_{0}^{1}zN_{0}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}u),$

which means
$z(\sigma_{1})={\it\Delta}Z_{0}(\sigma_{1})={\it\Delta}X_{\sigma_{1}}$ by
(3.10) and (3.18). This together with (3.26) yields

$\displaystyle
Z_{0}(\sigma_{1})+\xi_{0}(\sigma_{1})=Z_{0}(\sigma_{1}-)+{\it\Delta}Z_{0}(\sigma_{1})+{\it\Delta}\xi_{0}(\sigma_{1})=X_{\sigma_{1}-}+{\it\Delta}X_{\sigma_{1}}=X_{\sigma_{1}}>0.$

By (3.25) we have $Y_{0}(\sigma_{1})=X_{\sigma_{1}}>0$.

(2) On the event $\\{\tau_{1}<\sigma_{1}\leq\infty\\}$ we have
$\tau_{1}<U_{1}$ and so $\tau_{1}<U_{1}\land\sigma_{1}$. By (3.23) and the
definition of $\tau_{1}$ we have

$\displaystyle{\it\Delta}\xi_{0}(\tau_{1})=Z_{0}(\tau_{1}-)-X_{\tau_{1}-}<0$
(3.27)

and

$\displaystyle\int_{\\{\tau_{1}\\}}\int_{0}^{\infty}\int_{0}^{Z_{0}(\tau_{1}-)}\int_{\rho_{1}(Z_{0}(\tau_{1}-)-X_{\tau_{1}-},z)}^{\rho_{2}(X_{\tau_{1}-},Z_{0}(\tau_{1}-),z)}M_{0}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}u,\mathrm{d}v)$
$\displaystyle\qquad\qquad\qquad+\int_{\\{\tau_{1}\\}}\int_{0}^{\infty}\int_{r_{1}(Z_{0}(\tau_{1}-)-X_{\tau_{1}-},z)}^{r_{2}(X_{\tau_{1}-},Z_{0}(\tau_{1}-),z)}N_{0}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}v)=1.$

In view of (3.5) and (3.27), for $0<z\leq-{\it\Delta}\xi_{0}(\tau_{1})$ we
have

$\displaystyle\rho_{2}(X_{\tau_{1}-},Z_{0}(\tau_{1}-),z)-\rho_{1}(Z_{0}(\tau_{1}-)-X_{\tau_{1}-},z)=\rho_{1}(-{\it\Delta}\xi_{0}(\tau_{1}),z)=0$

and

$\displaystyle
r_{2}(X_{\tau_{1}-},Z_{0}(\tau_{1}-),z)-r_{1}(Z_{0}(\tau_{1}-)-X_{\tau_{1}-},z)=r_{1}(-{\it\Delta}\xi_{0}(\tau_{1}),z)=0.$

It follows that

$\displaystyle z(\tau_{1})\\!\\!\\!$ $\displaystyle:=\\!\\!\\!$
$\displaystyle\int_{\\{\tau_{1}\\}}\int_{0}^{\infty}\int_{0}^{Z_{0}(\tau_{1}-)}\int_{\rho_{1}(Z_{0}(\tau_{1}-)-X_{\tau_{1}-},z)}^{\rho_{2}(X_{\tau_{1}-},Z_{0}(\tau_{1}-),z)}zM_{0}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}u,\mathrm{d}v)$
(3.29)
$\displaystyle+\int_{\\{\tau_{1}\\}}\int_{0}^{\infty}\int_{r_{1}(Z_{0}(\tau_{1}-)-X_{\tau_{1}-},z)}^{r_{2}(X_{\tau_{1}-},Z_{0}(\tau_{1}-),z)}zN_{0}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}u)>-{\it\Delta}\xi_{0}(\tau_{1})>0.$

As in Part (1) of this the proof, one sees
$z(\tau_{1})={\it\Delta}Z_{0}(\tau_{1})={\it\Delta}X_{\tau_{1}}$, and so

$\displaystyle 0\leq
Z_{0}(\tau_{1}-)<Z_{0}(\tau_{1}-)+{\it\Delta}Z_{0}(\tau_{1})+{\it\Delta}\xi_{0}(\tau_{1})<X_{\tau_{1}-}+{\it\Delta}X_{\tau_{1}}=X_{\tau_{1}}.$

Then we have $0<Y_{0}(\tau_{1})<X_{\tau_{1}}$ by (3.25). $\square$

Proof of Theorem 3.1. We give an inductive construction of the process
$\\{(Y_{t},\xi_{t}):t\geq 0\\}$ in a sequence of steps. The construction also
yields the pathwise uniqueness of the solution of the equation system
(3.12)-(3.16).

Step 1.  We start with the process $\\{(X_{t},Z_{0}(t)):t\geq 0\\}$ defined by
(3.10) and (3.18). Consider separately the cases $\sigma_{1}=\tau_{1}=\infty$,
$\sigma_{1}<\tau_{1}$ and $\tau_{1}<\sigma_{1}$. In the case of
$\sigma_{1}=\tau_{1}=\infty$, we have $\xi_{0}(t)=0$ for all $t\geq 0$. Then
the process $\\{(Y_{t},\xi_{t}):t\geq 0\\}$ is defined by $T=U_{1}$ and

$\displaystyle\xi_{t}=0,~{}Y_{t}=Z_{0}(t\land T)+X_{t}-X_{t\land T}.$

In the case of $\sigma_{1}<\tau_{1}\leq\infty$, we have
$X_{\sigma_{1}}=Y_{0}(\sigma_{1})>0$ by Lemma 3.3. Then the process
$\\{(Y_{t},\xi_{t}):t\geq 0\\}$ is given by $T=\sigma_{1}$ and

$\displaystyle\xi_{t}=\xi_{0}(t\land T),~{}Y_{t}=Y_{0}(t\land
T)+X_{t}-X_{t\land T}.$

In case of $\tau_{1}<\sigma_{1}\leq\infty$, we clearly have $Y_{t}=Z_{0}(t)$
and $\xi_{t}=0$ for $0\leq t<\tau_{1}$. The continuing construction of
$\\{(Y_{\tau_{1}+t},\xi_{\tau_{1}+t}):t\geq 0\\}$ is given in the next step.

Step 2.  Suppose that $\tau_{1}<\sigma_{1}\leq\infty$. In this case, we have
$X_{\tau_{1}}>Y_{0}(\tau_{1})>0$ by Lemma 3.3. Let $X_{1}(t)=X_{\tau_{1}+t}$
for $t\geq 0$. Then $\\{X_{1}(t):t\geq 0\\}$ is also a CBIC-process. From
(3.10) it follows that

$\displaystyle
X_{1}(t)=X_{\tau_{1}}+\int_{0}^{t}\int_{0}^{X_{1}(s-)}L_{1}(\mathrm{d}s,\mathrm{d}u)-\int_{0}^{t}g(X_{1}(s))\mathrm{d}s+\eta_{\tau_{1}+t},$
(3.30)

where
$L_{1}(\mathrm{d}s,\mathrm{d}u)=L_{0}(\tau_{1}+\mathrm{d}s,\mathrm{d}u)$. By
Theorem 2.1 we can construct another CBIC-process $\\{Z_{1}(t):t\geq 0\\}$ by
the pathwise unique solution to

$\displaystyle
Z_{1}(t)=Y_{0}(\tau_{1})+\int_{0}^{t}\int_{0}^{Z_{1}(s-)}L_{1}^{*}(\mathrm{d}s,\mathrm{d}u)-\int_{0}^{t}g(Z_{1}(s))\mathrm{d}s+\eta_{\tau_{1}+t},$
(3.31)

where
$L_{1}^{*}(\mathrm{d}s,\mathrm{d}u)=L_{0}^{*}(\tau_{1}+\mathrm{d}s,\mathrm{d}u)$.
Then we repeat the procedure of in the first step for the process
$\\{(X_{1}(t),Z_{1}(t)):t\geq 0\\}$.

In view of (3.24), we only need a finite number of sequential steps to
complete the construction of $\\{(Y_{t},\xi_{t}):t\geq 0\\}$, which means that
the process $\\{\xi_{t}:t\geq 0\\}$ has at most a finite number of jumps. By
the construction, we have (3.17) and $X_{T+t}=Y_{T+t}$ for every $t\geq 0$.
$\square$

###### Remark 3.4

By Theorem 3.1, the process $\\{(X_{t},Y_{t}):t\geq 0\\}$ defined by (3.10)
and (3.12) with $X_{0}\geq Y_{0}\geq 0$ actually lives in the space
$D:=\\{(x,y):x\geq y\geq 0\\}\subset\mathbb{R}_{+}^{2}$. Here we think of the
coupling time $T$ as a part of the solution. We may also say that
$\\{(X_{t},Y_{t},\xi_{t}):t\geq 0\\}$ is a pathwise unique solution to the
system of equations (3.10), (3.12) and (3.16).

###### Theorem 3.5

The solution $\\{Y_{t}:t\geq 0\\}$ to (3.12) is a CBIC-process.

Proof. Let $\\{N(\mathrm{d}s,\mathrm{d}z)\\}$ be the optional random measure
defined by, for $t\geq 0$ and $A\in\mathscr{B}(0,\infty)$,

$\displaystyle N((0,t]\times A)\\!\\!\\!$ $\displaystyle=\\!\\!\\!$
$\displaystyle\int_{0}^{t}\int_{0}^{\infty}\int_{0}^{r_{1}(Y_{s-}-X_{s-},z)}\mathbf{1}_{A}(z+X_{s-}-Y_{s-})N_{0}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}v)$
$\displaystyle+\int_{0}^{t}\int_{0}^{\infty}\int_{r_{1}(Y_{s-}-X_{s-},z)}^{r_{2}(X_{s-},Y_{s-},z)}\mathbf{1}_{A}(z+Y_{s-}-X_{s-})N_{0}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}v)$
$\displaystyle+\int_{0}^{t}\int_{0}^{\infty}\int_{r_{2}(X_{s-},Y_{s-},z)}^{1}\mathbf{1}_{A}(z)N_{0}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}v).$

Then $N(\mathrm{d}s,\mathrm{d}z)$ has predictable compensator
$\hat{N}(\mathrm{d}s,\mathrm{d}z)$ determined by

$\displaystyle\hat{N}((0,t]\times A)\\!\\!\\!$ $\displaystyle=\\!\\!\\!$
$\displaystyle\int_{0}^{t}\mathrm{d}s\int_{0}^{\infty}\mathbf{1}_{A}(z+X_{s-}-Y_{s-})r_{1}(Y_{s-}-X_{s-},z)\nu(\mathrm{d}z)$
$\displaystyle+\int_{0}^{t}\mathrm{d}s\int_{0}^{\infty}\mathbf{1}_{A}(z+Y_{s-}-X_{s-})r_{1}(X_{s-}-Y_{s-},z)\nu(\mathrm{d}z)$
$\displaystyle+\int_{0}^{t}\mathrm{d}s\int_{0}^{\infty}\mathbf{1}_{A}(z)[1-r_{2}(X_{s-},Y_{s-},z)]\nu(\mathrm{d}z)$
$\displaystyle=\\!\\!\\!$
$\displaystyle\int_{0}^{t}\mathrm{d}s\int_{0}^{\infty}\mathbf{1}_{A}(z+X_{s-}-Y_{s-})\nu(Y_{s-}-X_{s-},\mathrm{d}z)$
$\displaystyle+\int_{0}^{t}\mathrm{d}s\int_{0}^{\infty}\mathbf{1}_{A}(z+Y_{s-}-X_{s-})\nu(X_{s-}-Y_{s-},\mathrm{d}z)$
$\displaystyle+\int_{0}^{t}\mathrm{d}s\int_{0}^{\infty}\mathbf{1}_{A}(z)[\nu(\mathrm{d}z)-\nu(X_{s-},Y_{s-},\mathrm{d}z)].$

In view of (3.2), for any $x\geq y\geq 0$ we have

$\displaystyle\int_{0}^{\infty}\mathbf{1}_{A}(z+x-y)\nu(y-x,\mathrm{d}z)+\int_{0}^{\infty}\mathbf{1}_{A}(z+y-x)\nu(x-y,\mathrm{d}z)$
$\displaystyle\qquad=\int_{0}^{\infty}\mathbf{1}_{A}(z)\nu(x-y,\mathrm{d}z)+\int_{0}^{\infty}\mathbf{1}_{A}(z)\nu(y-x,\mathrm{d}z)$
$\displaystyle\qquad=\int_{0}^{\infty}\mathbf{1}_{A}(z)\nu(x,y,\mathrm{d}z).$

Then $\hat{N}((0,t]\times A)=t\nu(A)$, and so $N(\mathrm{d}s,\mathrm{d}z)$ is
a Poisson random measure on $(0,\infty)^{2}$ with intensity
$\mathrm{d}s\nu(\mathrm{d}z)$; see, e.g., Theorem III.7.4 in Ikeda and
Watanabe [27, p.93]. Let $\\{M(\mathrm{d}s,\mathrm{d}z,\mathrm{d}u)\\}$ be the
optional random measure defined by, for $t\geq 0$ and
$A,B\in\mathscr{B}(0,\infty)$,

$\displaystyle M((0,t]\times A\times B)\\!\\!\\!$ $\displaystyle=\\!\\!\\!$
$\displaystyle\int_{0}^{t}\int_{0}^{\infty}\int_{B}\int_{0}^{\rho_{1}(Y_{s-}-X_{s-},z)}\mathbf{1}_{A}(z+X_{s-}-Y_{s-})M_{0}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}u,\mathrm{d}v)$
$\displaystyle+\int_{0}^{t}\int_{0}^{\infty}\int_{B}\int_{\rho_{1}(Y_{s-}-X_{s-},z)}^{\rho_{2}(X_{s-},Y_{s-},z)}\mathbf{1}_{A}(z+Y_{s-}-X_{s-})M_{0}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}u,\mathrm{d}v)$
$\displaystyle+\int_{0}^{t}\int_{0}^{\infty}\int_{B}\int_{\rho_{2}(X_{s-},Y_{s-},z)}^{1}\mathbf{1}_{A}(z)M_{0}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}u,\mathrm{d}v).$

By similar calculations as the above one can see
$\\{M(\mathrm{d}s,\mathrm{d}z,\mathrm{d}u)\\}$ is a Poisson random measure on
$(0,\infty)^{3}$ with intensity $\mathrm{d}s\mu(\mathrm{d}z)\mathrm{d}u$. Let
$\\{\eta(t)\\}$ be the $(\mathscr{F}_{t})$-subordinator defined by (1.15). Let
$\\{L(\mathrm{d}s,\mathrm{d}u)\\}$ be the time-space $(\mathscr{F}_{t})$-Lévy
noise defined by

$\displaystyle L(\mathrm{d}s,\mathrm{d}u)\\!\\!\\!$ $\displaystyle=\\!\\!\\!$
$\displaystyle-\,\mathbf{1}_{\\{s\leq
T\\}}W_{0}(\mathrm{d}s,\mathrm{d}u)-b\mathrm{d}s\mathrm{d}u+\int_{\\{0<z\leq
1\\}}z\tilde{M}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}u)$
$\displaystyle+\,\mathbf{1}_{\\{s>T\\}}W_{0}(\mathrm{d}s,\mathrm{d}u)+\int_{\\{1<z<\infty\\}}zM(\mathrm{d}s,\mathrm{d}z,\mathrm{d}u)$

From (3.12) and (3.16) we obtain

$\displaystyle
Y_{t}=Y_{0}+\int_{0}^{t}\int_{0}^{Y_{s-}}L(\mathrm{d}s,\mathrm{d}u)-\int_{0}^{t}g(Y_{s})\mathrm{d}s+\eta(t).$

Then $\\{Y_{t}:t\geq 0\\}$ is a CBIC-process by the uniqueness of the solution
to the equation. $\square$

The pathwise uniqueness of the solutions to (3.10) and (3.12) implies that
$\\{(X_{t},Y_{t}):t\geq 0\\}$ is a Markov process. Then it is a Markov
coupling of the CBIC-process. Since zero is a trap for $\\{X_{t}-Y_{t}:t\geq
0\\}$, we have

$\displaystyle X_{t}-Y_{t}\\!\\!\\!$ $\displaystyle=\\!\\!\\!$ $\displaystyle
X_{0}-Y_{0}+\int_{0}^{t}\int_{Y_{s}}^{X_{s}}W_{0}(\mathrm{d}s,\mathrm{d}u)+2\int_{0}^{t\wedge
T}\int_{0}^{Y_{s}}W_{0}(\mathrm{d}s,\mathrm{d}u)$ (3.34)
$\displaystyle-\int_{0}^{t}\big{[}g(X_{s})-g(Y_{s})\big{]}\mathrm{d}s+\int_{0}^{t}\int_{0}^{1}\int_{Y_{s-}}^{X_{s-}}\int_{0}^{1}z\tilde{M}_{0}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}u,\mathrm{d}v)$
$\displaystyle+\int_{0}^{t}\int_{1}^{\infty}\int_{Y_{s-}}^{X_{s-}}\int_{0}^{1}zM_{0}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}u,\mathrm{d}v)-\xi_{t}.$

###### Remark 3.6

Concerning the coupling process, what is important to us is its movement
before the succeeding time. To establish (1.24) and (1.43), we need to make
the coupling succeed as early as possible. The Gaussian and Poissonian
integrals on the right-hand side of (3.34) are important in bringing the
process to the success. Clearly, the second Gaussian integral comes from the
reflection of the Gaussian component in (3.9). The Poisson noises cannot be
reflected directly as the CBIC-processes have no negative jumps. We call
$\\{\xi_{t}:t\geq 0\\}$ the lassoing process, which provides reflections of
the Poisson noises in a proper way. In view of (3.24), a possible jump of this
pure jump process at time $s>0$ would take the values $X_{s-}-Y_{s-}>0$ and
$Y_{s-}-X_{s-}<0$ with equal probability $1/2$. The introduction of the
lassoing process was inspired by the construction of a coupling process of Luo
and Wang [40]; see also [32, 38]. In fact, if Grey’s condition (1.12) is not
satisfied, the success of the coupling process can only come with the first
positive jump of $\\{\xi_{t}:t\geq 0\\}$.

##  4 The coupling generator

In this section, we give a characterization for the generator of the Markov
coupling process $\\{(X_{t},Y_{t}):t\geq 0\\}$ defined by (3.10) and (3.12) in
terms of a martingale problem. We also establish some estimates for the
generator, which provide the basis for the proof of the exponential
ergodicity. Recall that $\\{(X_{t},Y_{t}):t\geq 0\\}$ has state space
$D:=\\{(x,y):x\geq y\geq 0\\}$. Let ${\it\Delta}=\\{(z,z):z\geq 0\\}\subset D$
and ${\it\Delta}^{c}=D\setminus{\it\Delta}$. Given a function $F$ on $D$ twice
continuously differentiable on ${\it\Delta}^{c}$, we write

$\displaystyle\tilde{L}F(x,y)=\tilde{L}_{0}F(x,y)+\tilde{L}_{1}F(x,y),\quad(x,y)\in{\it\Delta}^{c},$
(4.1)

where

$\displaystyle\tilde{L}_{0}F(x,y)\\!\\!\\!$ $\displaystyle=\\!\\!\\!$
$\displaystyle\big{[}\beta-bx-g(x)\big{]}F^{\prime}_{x}(x,y)+\big{[}\beta-
by-g(y)\big{]}F^{\prime}_{y}(x,y)+cxF^{\prime\prime}_{xx}(x,y)$ (4.4)
$\displaystyle+\,cyF^{\prime\prime}_{yy}(x,y)-2cyF^{\prime\prime}_{xy}(x,y)+\int_{0}^{\infty}\big{[}F(x+z,y+z)-F(x,y)\big{]}\nu(\mathrm{d}z)$
$\displaystyle+\,(x-y)\int_{0}^{\infty}\big{[}F(x+z,y)-F(x,y)-F_{x}^{\prime}(x,y)z\mathbf{1}_{\\{z\leq
1\\}}\big{]}\mu(\mathrm{d}z)$
$\displaystyle+\,y\int_{0}^{\infty}\big{[}F(x+z,y+z)-F(x,y)-(F_{x}^{\prime}+F_{y}^{\prime})(x,y)z\mathbf{1}_{\\{z\leq
1\\}}\big{]}\mu(\mathrm{d}z)$

and

$\displaystyle\tilde{L}_{1}F(x,y)\\!\\!\\!$ $\displaystyle=\\!\\!\\!$
$\displaystyle
y\int_{0}^{\infty}\big{[}F(x+z,2y+z-x)-F(x+z,y+z)\big{]}\mu_{x-y}(\mathrm{d}z)$
(4.8)
$\displaystyle+\,y\int_{0}^{\infty}\big{[}F(x+z,x+z)-F(x+z,y+z)\big{]}\mu_{y-x}(\mathrm{d}z)$
$\displaystyle+\int_{0}^{\infty}\big{[}F(x+z,2y+z-x)-F(x+z,y+z)\big{]}\nu_{x-y}(\mathrm{d}z)$
$\displaystyle+\int_{0}^{\infty}\big{[}F(x+z,x+z)-F(x+z,y+z)\big{]}\nu_{y-x}(\mathrm{d}z).$

Let $\mathscr{D}(\tilde{L})$ denote the linear space consisting of the
functions $F$ such that the integrals in (4.4) and (4.8) are convergent and
define functions locally bounded on compact subsets of ${\it\Delta}^{c}$. The
operator $\tilde{L}$ determines the movement of the coupling process before
its succeeding time. In particular, the component $\tilde{L}_{1}$ is induced
by the lassoing process. We call $(\tilde{L},\mathscr{D}(\tilde{L}))$ the
coupling generator of the CBIC-process. The precise meaning of this
terminology is made clear by the martingale problem given in the following:

###### Theorem 4.1

Let $\\{(X_{t},Y_{t}):t\geq 0\\}$ be the Markov coupling process defined by
(3.10) and (3.12) with $X_{0}>Y_{0}\geq 0$ and let $\zeta_{n}^{*}=\inf\\{t\geq
0:X_{t}\geq n$ or $X_{t}-Y_{t}\leq 1/n\\}$. Then for any $n\geq 1$ and
$F\in\mathscr{D}(\tilde{L})$ we have

$\displaystyle
F(X_{t\land\zeta_{n}^{*}},Y_{t\land\zeta_{n}^{*}})=F(X_{0},Y_{0})+\int_{0}^{t\land\zeta_{n}^{*}}\tilde{L}F(X_{s},Y_{s})\mathrm{d}s+M_{n}(t),$
(4.9)

where $\\{M_{n}(t):t\geq 0\\}$ is a martingale.

Proof. By (3.10), (3.12) and (3.16), the process
$\\{(X_{t\land\zeta_{n}^{*}},Y_{t\land\zeta_{n}^{*}}):t\geq 0\\}$ is a
semimartingale taking values in ${\it\Delta}\cup D_{n}$, where
$D_{n}=\\{(x,y)\in D:x-y\geq 1/n\\}$. In view of the three equations, a
possible jump $({\it\Delta}X_{s},{\it\Delta}Y_{s})$ of the process at time
$s>0$ is brought about by a point $(s,z,u,v)\in\mathrm{supp}(M_{0})$ with
$0<u\leq X_{s-}$ or by a point $(s,z,v)\in\mathrm{supp}(N_{0})$. The details
of the jump in the two cases are given, respectively, by

$\displaystyle\quad({\it\Delta}X_{s},{\it\Delta}Y_{s})=\left\\{\begin{array}[]{ll}(z,z+X_{s-}-Y_{s-}),&u\in(0,Y_{s-}],v\in(0,\hat{\rho}_{1}(s,z)],\vskip
3.0pt plus 1.0pt minus
1.0pt\cr(z,z+Y_{s-}-X_{s-}),&u\in(0,Y_{s-}],v\in(\hat{\rho}_{1}(s,z),\hat{\rho}_{2}(s,z)],\vskip
3.0pt plus 1.0pt minus
1.0pt\cr(z,z),&u\in(0,Y_{s-}],v\in(\hat{\rho}_{2}(s,z),1],\vskip 3.0pt plus
1.0pt minus 1.0pt\cr(z,0),&u\in(Y_{s-},X_{s-}],v\in(0,1],\end{array}\right.$

where $\hat{\rho}_{1}(s,z)=\rho_{1}(X_{s-}-Y_{s-},z)$ and
$\hat{\rho}_{2}(s,z)=\rho_{2}(X_{s-},Y_{s-},z)$, and

$\displaystyle~{}~{}({\it\Delta}X_{s},{\it\Delta}Y_{s})=\left\\{\begin{array}[]{ll}(z,z+X_{s-}-Y_{s-}),&v\in(0,\hat{r}_{1}(s,z)],\vskip
3.0pt plus 1.0pt minus
1.0pt\cr(z,z+Y_{s-}-X_{s-}),&v\in(\hat{r}_{1}(s,z),\hat{r}_{2}(s,z)],\vskip
3.0pt plus 1.0pt minus
1.0pt\cr(z,z),&v\in(\hat{r}_{2}(s,z),1].\end{array}\right.$

where $\hat{r}_{1}(s,z)=r_{1}(X_{s-}-Y_{s-},z)$ and
$\hat{r}_{2}(s,z)=r_{2}(X_{s-},Y_{s-},z)$. Then the coupling process may have
totally seven different types of jumps. Observe also that $(X_{s-},Y_{s-})\in
D_{n}\subset{\it\Delta}^{c}$ when $0<s\leq\zeta_{n}^{*}$. For any
$(x,y)\in{\it\Delta}^{c}$ write

$\displaystyle\tilde{L}_{2}F(x,y)=cxF^{\prime\prime}_{xx}(x,y)+cyF^{\prime\prime}_{yy}(x,y)-2cyF^{\prime\prime}_{xy}(x,y)-g(x)F^{\prime}_{x}(x,y)-g(y)F^{\prime}_{y}(x,y).$

We can use Itô’s formula to see

$\displaystyle F(X_{t\land\zeta_{n}^{*}},Y_{t\land\zeta_{n}^{*}})\\!\\!\\!$
$\displaystyle=\\!\\!\\!$ $\displaystyle
F(X_{0},Y_{0})+\int_{0}^{t\land\zeta_{n}^{*}}\tilde{L}_{2}F(X_{s},Y_{s})\mathrm{d}s+\int_{0}^{t\land\zeta_{n}^{*}}\int_{0}^{X_{s}}F^{\prime}_{x}(X_{s-},Y_{s-})L_{0}(\mathrm{d}s,\mathrm{d}u)$
$\displaystyle+\int_{0}^{t\land\zeta_{n}^{*}}\int_{0}^{Y_{s}}F^{\prime}_{y}(X_{s-},Y_{s-})L^{*}_{0}(\mathrm{d}s,\mathrm{d}u)+\int_{0}^{t\land\zeta_{n}^{*}}(F^{\prime}_{x}+F^{\prime}_{y})(X_{s-},Y_{s-})\mathrm{d}\eta_{s}$
$\displaystyle+\int_{0}^{t\land\zeta_{n}^{*}}\int_{0}^{\infty}\int_{0}^{Y_{s-}}\int_{0}^{\hat{\rho}_{1}(s,z)}F^{\prime}_{y}(X_{s-},Y_{s-})(X_{s-}-Y_{s-})M_{0}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}u,\mathrm{d}v)$
$\displaystyle+\int_{0}^{t\land\zeta_{n}^{*}}\int_{0}^{\infty}\int_{0}^{\hat{r}_{1}(s,z)}F^{\prime}_{y}(X_{s-},Y_{s-})(X_{s-}-Y_{s-})N_{0}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}v)$
$\displaystyle+\int_{0}^{t\land\zeta_{n}^{*}}\int_{0}^{\infty}\int_{0}^{Y_{s-}}\int_{\hat{\rho}_{1}(s,z)}^{\hat{\rho}_{2}(s,z)}F^{\prime}_{y}(X_{s-},Y_{s-})(Y_{s-}-X_{s-})M_{0}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}u,\mathrm{d}v)$
$\displaystyle+\int_{0}^{t\land\zeta_{n}^{*}}\int_{0}^{\infty}\int_{\hat{r}_{1}(s,z)}^{\hat{r}_{2}(s,z)}F^{\prime}_{y}(X_{s-},Y_{s-})(Y_{s-}-X_{s-})N_{0}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}v)$
$\displaystyle+\int_{0}^{t\land\zeta_{n}^{*}}\int_{0}^{\infty}\int_{0}^{Y_{s-}}\int_{0}^{\hat{\rho}_{1}(s,z)}\big{[}F(X_{s-}+z,X_{s-}+z)-F(X_{s-},Y_{s-})$
$\displaystyle\qquad-\,zF^{\prime}_{x}(X_{s-},Y_{s-})-(z+X_{s-}-Y_{s-})F^{\prime}_{y}(X_{s-},Y_{s-})\big{]}M_{0}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}u,\mathrm{d}v)$
$\displaystyle+\int_{0}^{t\land\zeta_{n}^{*}}\int_{0}^{\infty}\int_{0}^{Y_{s-}}\int_{\hat{\rho}_{1}(s,z)}^{\hat{\rho}_{2}(s,z)}\big{[}F(X_{s-}+z,2Y_{s-}+z-X_{s-})-F(X_{s-},Y_{s-})$
$\displaystyle\qquad-\,zF^{\prime}_{x}(X_{s-},Y_{s-})-(z+Y_{s-}-X_{s-})F^{\prime}_{y}(X_{s-},Y_{s-})\big{]}M_{0}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}u,\mathrm{d}v)$
$\displaystyle+\int_{0}^{t\land\zeta_{n}^{*}}\int_{0}^{\infty}\int_{0}^{Y_{s-}}\int_{\hat{\rho}_{2}(s,z)}^{1}\big{[}F(X_{s-}+z,Y_{s-}+z)-F(X_{s-},Y_{s-})$
$\displaystyle\qquad\qquad\qquad\qquad-\,z(F^{\prime}_{x}+F^{\prime}_{y})(X_{s-},Y_{s-})\big{]}M_{0}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}u,\mathrm{d}v)$
$\displaystyle+\int_{0}^{t\land\zeta_{n}^{*}}\int_{0}^{\infty}\int_{Y_{s-}}^{X_{s-}}\int_{0}^{1}\big{[}F(X_{s-}+z,Y_{s-})-F(X_{s-},Y_{s-})$
$\displaystyle\qquad\qquad\qquad\qquad-\,zF^{\prime}_{x}(X_{s-},Y_{s-})\big{]}M_{0}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}u,\mathrm{d}v)$
$\displaystyle+\int_{0}^{t\land\zeta_{n}^{*}}\int_{0}^{\infty}\int_{0}^{\hat{r}_{1}(s,z)}\big{[}F(X_{s-}+z,X_{s-}+z)-F(X_{s-},Y_{s-})$
$\displaystyle\qquad-\,zF^{\prime}_{x}(X_{s-},Y_{s-})-(z+X_{s-}-Y_{s-})F^{\prime}_{y}(X_{s-},Y_{s-})\big{]}N_{0}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}v)$
$\displaystyle+\int_{0}^{t\land\zeta_{n}^{*}}\int_{0}^{\infty}\int_{\hat{r}_{1}(s,z)}^{\hat{r}_{2}(s,z)}\big{[}F(X_{s-}+z,2Y_{s-}+z-X_{s-})-F(X_{s-},Y_{s-})$
$\displaystyle\qquad-\,zF^{\prime}_{x}(X_{s-},Y_{s-})-(z+Y_{s-}-X_{s-})F^{\prime}_{y}(X_{s-},Y_{s-})\big{]}N_{0}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}v)$
$\displaystyle+\int_{0}^{t\land\zeta_{n}^{*}}\int_{0}^{\infty}\int_{\hat{r}_{2}(s,z)}^{1}\big{[}F(X_{s-}+z,Y_{s-}+z)-F(X_{s-},Y_{s-})$
$\displaystyle\qquad\qquad\qquad\qquad-\,z(F^{\prime}_{x}+F^{\prime}_{y})(X_{s-},Y_{s-})\big{]}N_{0}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}v)$
$\displaystyle=\\!\\!\\!$ $\displaystyle
F(X_{0},Y_{0})+\int_{0}^{t\land\zeta_{n}^{*}}\tilde{L}_{2}F(X_{s},Y_{s})\mathrm{d}s+\int_{0}^{t\land\zeta_{n}^{*}}\int_{0}^{X_{s}}F^{\prime}_{x}(X_{s-},Y_{s-})L_{0}(\mathrm{d}s,\mathrm{d}u)$
$\displaystyle+\int_{0}^{t\land\zeta_{n}^{*}}\int_{0}^{Y_{s}}F^{\prime}_{y}(X_{s-},Y_{s-})L^{*}_{0}(\mathrm{d}s,\mathrm{d}u)+\int_{0}^{t\land\zeta_{n}^{*}}(F^{\prime}_{x}+F^{\prime}_{y})(X_{s-},Y_{s-})\mathrm{d}\eta_{s}$
$\displaystyle+\int_{0}^{t\land\zeta_{n}^{*}}\int_{0}^{\infty}\int_{0}^{Y_{s-}}\int_{0}^{\hat{\rho}_{1}(s,z)}\big{[}F(X_{s-}+z,X_{s-}+z)-F(X_{s-},Y_{s-})$
$\displaystyle\qquad\qquad\qquad\qquad-\,z(F^{\prime}_{x}+F^{\prime}_{y})(X_{s-},Y_{s-})\big{]}M_{0}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}u,\mathrm{d}v)$
$\displaystyle+\int_{0}^{t\land\zeta_{n}^{*}}\int_{0}^{\infty}\int_{0}^{Y_{s-}}\int_{\hat{\rho}_{1}(s,z)}^{\hat{\rho}_{2}(s,z)}\big{[}F(X_{s-}+z,2Y_{s-}+z-X_{s-})$
$\displaystyle\qquad\qquad-\,F(X_{s-},Y_{s-})-z(F^{\prime}_{x}+F^{\prime}_{y})(X_{s-},Y_{s-})\big{]}M_{0}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}u,\mathrm{d}v)$
$\displaystyle+\int_{0}^{t\land\zeta_{n}^{*}}\int_{0}^{\infty}\int_{0}^{Y_{s-}}\int_{\hat{\rho}_{2}(s,z)}^{1}\big{[}F(X_{s-}+z,Y_{s-}+z)-F(X_{s-},Y_{s-})$
$\displaystyle\qquad\qquad\qquad\qquad-\,z(F^{\prime}_{x}+F^{\prime}_{y})(X_{s-},Y_{s-})\big{]}M_{0}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}u,\mathrm{d}v)$
$\displaystyle+\int_{0}^{t\land\zeta_{n}^{*}}\int_{0}^{\infty}\int_{Y_{s-}}^{X_{s-}}\int_{0}^{1}\big{[}F(X_{s-}+z,Y_{s-})-F(X_{s-},Y_{s-})$
$\displaystyle\qquad\qquad\qquad\qquad-\,zF^{\prime}_{x}(X_{s-},Y_{s-})\big{]}M_{0}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}u,\mathrm{d}v)$
$\displaystyle+\int_{0}^{t\land\zeta_{n}^{*}}\int_{0}^{\infty}\int_{0}^{\hat{r}_{1}(s,z)}\big{[}F(X_{s-}+z,X_{s-}+z)-F(X_{s-},Y_{s-})$
$\displaystyle\qquad\qquad\qquad\qquad-\,z(F^{\prime}_{x}+F^{\prime}_{y})(X_{s-},Y_{s-})\big{]}N_{0}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}v)$
$\displaystyle+\int_{0}^{t\land\zeta_{n}^{*}}\int_{0}^{\infty}\int_{\hat{r}_{1}(s,z)}^{\hat{r}_{2}(s,z)}\big{[}F(X_{s-}+z,2Y_{s-}+z-X_{s-})$
$\displaystyle\qquad\qquad-\,F(X_{s-},Y_{s-})-z(F^{\prime}_{x}+F^{\prime}_{y})(X_{s-},Y_{s-})\big{]}N_{0}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}v)$
$\displaystyle+\int_{0}^{t\land\zeta_{n}^{*}}\int_{0}^{\infty}\int_{\hat{r}_{2}(s,z)}^{1}\big{[}F(X_{s-}+z,Y_{s-}+z)-F(X_{s-},Y_{s-})$
$\displaystyle\qquad\qquad\qquad\qquad-\,z(F^{\prime}_{x}+F^{\prime}_{y})(X_{s-},Y_{s-})\big{]}N_{0}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}v).$

 By some further cancellations of the terms,

$\displaystyle F(X_{t\land\zeta_{n}^{*}},Y_{t\land\zeta_{n}^{*}})\\!\\!\\!$
$\displaystyle=\\!\\!\\!$ $\displaystyle
F(X_{0},Y_{0})+\int_{0}^{t\land\zeta_{n}^{*}}\tilde{L}_{2}F(X_{s},Y_{s})\mathrm{d}s+\int_{0}^{t\land\zeta_{n}^{*}}\int_{0}^{X_{s}}F^{\prime}_{x}(X_{s},Y_{s})W_{0}(\mathrm{d}s,\mathrm{d}u)$
(4.17)
$\displaystyle+\int_{0}^{t\land\zeta_{n}^{*}}\int_{0}^{X_{s-}}\int_{0}^{1}\int_{0}^{1}F^{\prime}_{x}(X_{s-},Y_{s-})z\tilde{M}_{0}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}u,\mathrm{d}v)$
$\displaystyle+\int_{0}^{t\land\zeta_{n}^{*}}(\beta-
bX_{s})F^{\prime}_{x}(X_{s},Y_{s})\mathrm{d}s+\int_{0}^{t\land\zeta_{n}^{*}}(\beta-
bY_{s})F^{\prime}_{y}(X_{s},Y_{s})\mathrm{d}s$
$\displaystyle-\int_{0}^{t\land\zeta_{n}^{*}}\int_{0}^{Y_{s}}F^{\prime}_{y}(X_{s},Y_{s})W_{0}(\mathrm{d}s,\mathrm{d}u)$
$\displaystyle+\int_{0}^{t\land\zeta_{n}^{*}}\int_{0}^{Y_{s-}}\int_{0}^{1}\int_{0}^{1}F^{\prime}_{y}(X_{s-},Y_{s-})z\tilde{M}_{0}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}u,\mathrm{d}v)$
$\displaystyle+\int_{0}^{t\land\zeta_{n}^{*}}\int_{0}^{\infty}\int_{0}^{Y_{s-}}\int_{0}^{\hat{\rho}_{1}(s,z)}\big{[}F(X_{s-}+z,X_{s-}+z)-F(X_{s-},Y_{s-})$
$\displaystyle\qquad\qquad\qquad\qquad-\,\mathbf{1}_{\\{z\leq
1\\}}z(F^{\prime}_{x}+F^{\prime}_{y})(X_{s-},Y_{s-})\big{]}\tilde{M}_{0}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}u,\mathrm{d}v)$
$\displaystyle+\int_{0}^{t\land\zeta_{n}^{*}}\int_{0}^{\infty}\int_{0}^{Y_{s-}}\int_{\hat{\rho}_{1}(s,z)}^{\hat{\rho}_{2}(s,z)}\big{[}F(X_{s-}+z,2Y_{s-}+z-X_{s-})$
$\displaystyle\qquad-\,F(X_{s-},Y_{s-})-\mathbf{1}_{\\{z\leq
1\\}}z(F^{\prime}_{x}+F^{\prime}_{y})(X_{s-},Y_{s-})\big{]}\tilde{M}_{0}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}u,\mathrm{d}v)$
$\displaystyle+\int_{0}^{t\land\zeta_{n}^{*}}\int_{0}^{\infty}\int_{0}^{Y_{s-}}\int_{\hat{\rho}_{2}(s,z)}^{1}\big{[}F(X_{s-}+z,Y_{s-}+z)-F(X_{s-},Y_{s-})$
$\displaystyle\qquad\qquad\qquad\qquad-\,\mathbf{1}_{\\{z\leq
1\\}}z(F^{\prime}_{x}+F^{\prime}_{y})(X_{s-},Y_{s-})\big{)}\big{]}\tilde{M}_{0}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}u,\mathrm{d}v)$
$\displaystyle+\int_{0}^{t\land\zeta_{n}^{*}}\int_{0}^{\infty}\int_{Y_{s-}}^{X_{s-}}\int_{0}^{1}\big{[}F(X_{s-}+z,Y_{s-})-F(X_{s-},Y_{s-})$
$\displaystyle\qquad\qquad\qquad\qquad-\,\mathbf{1}_{\\{z\leq
1\\}}zF^{\prime}_{x}(X_{s-},Y_{s-})\big{]}\tilde{M}_{0}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}u,\mathrm{d}v)$
$\displaystyle+\int_{0}^{t\land\zeta_{n}^{*}}\int_{0}^{\infty}\int_{0}^{\hat{r}_{1}(s,z)}\big{[}F(X_{s-}+z,X_{s-}+z)$
$\displaystyle\qquad\qquad\qquad\qquad-\,F(X_{s-},Y_{s-})\big{]}\tilde{N}_{0}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}v)$
$\displaystyle+\int_{0}^{t\land\zeta_{n}^{*}}\int_{0}^{\infty}\int_{\hat{r}_{1}(s,z)}^{\hat{r}_{2}(s,z)}\big{[}F(X_{s-}+z,2Y_{s-}+z-X_{s-})$
$\displaystyle\qquad\qquad\qquad\qquad-\,F(X_{s-},Y_{s-})\big{]}\tilde{N}_{0}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}v)$
$\displaystyle+\int_{0}^{t\land\zeta_{n}^{*}}\int_{0}^{\infty}\int_{\hat{r}_{2}(s,z)}^{1}\big{[}F(X_{s-}+z,Y_{s-}+z)$
$\displaystyle\qquad\qquad\qquad\qquad-\,F(X_{s-},Y_{s-})\big{]}\tilde{N}_{0}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}v)$
$\displaystyle+\int_{0}^{t\land\zeta_{n}}\tilde{L}_{3}F(X_{s},Y_{s})\mathrm{d}s,$

 where

$\displaystyle\tilde{L}_{3}F(x,y)\\!\\!\\!$ $\displaystyle=\\!\\!\\!$
$\displaystyle
y\int_{0}^{\infty}\big{[}F(x+z,y+z)-F(x,y)-(F_{x}^{\prime}+F_{y}^{\prime})(x,y)z\mathbf{1}_{\\{z\leq
1\\}}\big{]}\mu(\mathrm{d}z)$
$\displaystyle+\,y\int_{0}^{\infty}\big{[}F(x+z,x+z)-F(x+z,y+z)\big{]}\mu_{y-x}(\mathrm{d}z)$
$\displaystyle+\,y\int_{0}^{\infty}\big{[}F(x+z,2y+z-x)-F(x+z,y+z)\big{]}\mu_{x-y}(\mathrm{d}z)$
$\displaystyle+\,(x-y)\int_{0}^{\infty}\big{[}F(x+z,y)-F(x,y)-F_{x}^{\prime}(x,y)z\mathbf{1}_{\\{z\leq
1\\}}\big{]}\mu(\mathrm{d}z)$
$\displaystyle+\int_{0}^{\infty}\big{[}F(x+z,y+z)-F(x,y)\big{]}\nu(\mathrm{d}z)$
$\displaystyle+\int_{0}^{\infty}\big{[}F(x+z,x+z)-F(x+z,y+z)\big{]}\nu_{y-x}(\mathrm{d}z)$
$\displaystyle+\int_{0}^{\infty}\big{[}F(x+z,2y+z-x)-F(x+z,y+z)\big{]}\nu_{x-y}(\mathrm{d}z).$

Clearly, we have

$\displaystyle\tilde{L}F(x,y)=(\beta-bx)F^{\prime}_{x}(x,y)+(\beta-
by)F^{\prime}_{y}(x,y)+\tilde{L}_{2}F(x,y)+\tilde{L}_{3}F(x,y).$

Since $Y_{s-}\leq X_{s-}\leq n$ for $0<s\leq\zeta_{n}^{*}$, the stochastic
integrals on the right-hand side of (4.17) define a martingale
$\\{M_{n}(t):t\geq 0\\}$. Then the desired result follows. $\square$

We next consider a special form of the function $F\in\mathscr{D}(\tilde{L})$.
Recall that $C_{b}^{2}(\mathbb{R}_{+})$ denotes the space of bounded and
continuous functions on $\mathbb{R}_{+}$ with bounded and continuous
derivatives up to the second order.

###### Lemma 4.2

Suppose that $f\in C_{b}^{2}(\mathbb{R}_{+})$ is a nonnegative and concave
nondecreasing function and $\phi\in C_{b}^{2}(\mathbb{R}_{+})$ is a
nonnegative nonincreasing function. Let

$\displaystyle F(x,y)=\phi(x)f(x-y)\mathbf{1}_{\\{x\neq y\\}},\quad(x,y)\in
D.$ (4.18)

Then $F\in\mathscr{D}(\tilde{L})$ and, for $(x,y)\in{\it\Delta}^{c}$,

$\displaystyle\tilde{L}F(x,y)\\!\\!\\!$ $\displaystyle\leq\\!\\!\\!$
$\displaystyle cy\phi(x)f^{\prime\prime}(x-y)+\big{[}\beta-
bx-g(x)\big{]}\phi^{\prime}(x)f(x-y)$ (4.19)
$\displaystyle+\,y\phi(x)\big{[}f(2(x-y))-2f(x-y)\big{]}\mu_{x-y}(0,\infty)$
$\displaystyle+\,\phi(x)\big{[}f(2(x-y))-2f(x-y)\big{]}\nu_{x-y}(0,\infty)$
$\displaystyle+\,\tilde{A}_{1}F(x,y)+\tilde{A}_{2}F(x,y),$

where

$\displaystyle\tilde{A}_{1}F(x,y)\\!\\!\\!$ $\displaystyle=\\!\\!\\!$
$\displaystyle
f(x-y)\bigg{[}cx\phi^{\prime\prime}(x)-y\int_{0}^{\infty}\big{[}\phi(x+z)-\phi(x)\big{]}\mu_{y-x}(\mathrm{d}z)$
(4.22)
$\displaystyle\qquad\qquad+\,x\int_{0}^{\infty}\big{[}\phi(x+z)-\phi(x)-\phi^{\prime}(x)z\mathbf{1}_{\\{z\leq
1\\}}\big{]}\mu(\mathrm{d}z)$
$\displaystyle\qquad\qquad+\,\int_{0}^{\infty}\big{[}\phi(x+z)-\phi(x)\big{]}(\nu-\nu_{y-x})(\mathrm{d}z)\bigg{]}$

and

$\displaystyle\tilde{A}_{2}F(x,y)\\!\\!\\!$ $\displaystyle=\\!\\!\\!$
$\displaystyle(x-y)\phi(x)\bigg{[}cf^{\prime\prime}(x-y)-bf^{\prime}(x-y)+\int_{0}^{\infty}\big{[}f(x-y+z)$
(4.24)
$\displaystyle\qquad\qquad-\,f(x-y)-f^{\prime}(x-y)z\mathbf{1}_{\\{z\leq
1\\}}\big{]}\mu(\mathrm{d}z)\bigg{]}.$

Proof. According to the definition of $F$, we have $F(z,z)=0$ for $z\geq 0$.
By (4.4) and (4.8) one can see that $F\in\mathscr{D}(\tilde{L})$ and, for
$(x,y)\in{\it\Delta}^{c}$,

$\displaystyle\tilde{L}_{0}F(x,y)\\!\\!\\!$ $\displaystyle=\\!\\!\\!$
$\displaystyle\big{[}\beta-bx-g(x)\big{]}\phi^{\prime}(x)f(x-y)-\big{[}bx-
by+g(x)-g(y)\big{]}\phi(x)f^{\prime}(x-y)$
$\displaystyle+\,c(x+3y)\phi(x)f^{\prime\prime}(x-y)+2c(x+y)\phi^{\prime}(x)f^{\prime}(x-y)+cx\phi^{\prime\prime}(x)f(x-y)$
$\displaystyle+\,(x-y)\int_{0}^{\infty}\big{[}\phi(x+z)f(x-y+z)-\phi(x)f(x-y)$
$\displaystyle\qquad\qquad-\,\phi(x)f^{\prime}(x-y)z\mathbf{1}_{\\{z\leq
1\\}}-\phi^{\prime}(x)f(x-y)z\mathbf{1}_{\\{z\leq
1\\}}\big{]}\mu(\mathrm{d}z)$
$\displaystyle+\,yf(x-y)\int_{0}^{\infty}\big{[}\phi(x+z)-\phi(x)-\phi^{\prime}(x)z\mathbf{1}_{\\{z\leq
1\\}}\big{]}\mu(\mathrm{d}z)$
$\displaystyle+\,f(x-y)\int_{0}^{\infty}\big{[}\phi(x+z)-\phi(x)\big{]}\nu(\mathrm{d}z)$

and

$\displaystyle\tilde{L}_{1}F(x,y)\\!\\!\\!$ $\displaystyle=\\!\\!\\!$
$\displaystyle
y\big{[}f(2(x-y))-f(x-y)\big{]}\int_{0}^{\infty}\phi(x+z)\mu_{x-y}(\mathrm{d}z)$
$\displaystyle-\,yf(x-y)\int_{0}^{\infty}\phi(x+z)\mu_{y-x}(\mathrm{d}z)-f(x-y)\int_{0}^{\infty}\phi(x+z)\nu_{y-x}(\mathrm{d}z)$
$\displaystyle+\,\big{[}f(2(x-y))-f(x-y)\big{]}\int_{0}^{\infty}\phi(x+z)\nu_{x-y}(\mathrm{d}z).$

Recall that $f$ and $g$ are nondecreasing and $\phi$ is nonincreasing. Then we
have

$\displaystyle\tilde{L}_{0}F(x,y)\\!\\!\\!$ $\displaystyle\leq\\!\\!\\!$
$\displaystyle\big{[}\beta-
bx-g(x)\big{]}\phi^{\prime}(x)f(x-y)-b(x-y)\phi(x)f^{\prime}(x-y)+cx\phi(x)f^{\prime\prime}(x-y)$
$\displaystyle+\,cx\phi^{\prime\prime}(x)f(x-y)+(x-y)\int_{0}^{\infty}\big{[}\phi(x+z)f(x-y+z)$
$\displaystyle\qquad\qquad\qquad\qquad-\,\phi(x)f(x-y+z)-\phi^{\prime}(x)f(x-y)z\mathbf{1}_{\\{z\leq
1\\}}\big{]}\mu(\mathrm{d}z)$
$\displaystyle+\,(x-y)\phi(x)\int_{0}^{\infty}\big{[}f(x-y+z)-f(x-y)-f^{\prime}(x-y)z\mathbf{1}_{\\{z\leq
1\\}}\big{]}\mu(\mathrm{d}z)$
$\displaystyle+\,yf(x-y)\int_{0}^{\infty}\big{[}\phi(x+z)-\phi(x)-\phi^{\prime}(x)z\mathbf{1}_{\\{z\leq
1\\}}\big{]}\mu(\mathrm{d}z)$
$\displaystyle+\,f(x-y)\int_{0}^{\infty}\big{[}\phi(x+z)-\phi(x)\big{]}\nu(\mathrm{d}z)$
$\displaystyle\leq\\!\\!\\!$ $\displaystyle\big{[}\beta-
bx-g(x)\big{]}\phi^{\prime}(x)f(x-y)-b(x-y)\phi(x)f^{\prime}(x-y)+cx\phi(x)f^{\prime\prime}(x-y)$
$\displaystyle+\,cx\phi^{\prime\prime}(x)f(x-y)+\,xf(x-y)\int_{0}^{\infty}\big{[}\phi(x+z)-\phi(x)-\phi^{\prime}(x)z\mathbf{1}_{\\{z\leq
1\\}}\big{]}\mu(\mathrm{d}z)$
$\displaystyle+\,(x-y)\phi(x)\int_{0}^{\infty}\big{[}f(x-y+z)-f(x-y)-f^{\prime}(x-y)z\mathbf{1}_{\\{z\leq
1\\}}\big{]}\mu(\mathrm{d}z)$
$\displaystyle+\,f(x-y)\int_{0}^{\infty}\big{[}\phi(x+z)-\phi(x)\big{]}\nu(\mathrm{d}z)$

and

$\displaystyle\tilde{L}_{1}F(x,y)\\!\\!\\!$ $\displaystyle=\\!\\!\\!$
$\displaystyle y\phi(x)\big{[}f(2(x-y))-2f(x-y)\big{]}\mu_{x-y}(0,\infty)$
$\displaystyle+\,y\big{[}f(2(x-y))-f(x-y)\big{]}\int_{0}^{\infty}\big{[}\phi(x+z)-\phi(x)\big{]}\mu_{x-y}(\mathrm{d}z)$
$\displaystyle-\,yf(x-y)\int_{0}^{\infty}\big{[}\phi(x+z)-\phi(x)\big{]}\mu_{y-x}(\mathrm{d}z)$
$\displaystyle+\,\phi(x)\big{[}f(2(x-y))-2f(x-y)\big{]}\nu_{x-y}(0,\infty)$
$\displaystyle+\,\big{[}f(2(x-y))-f(x-y)\big{]}\int_{0}^{\infty}\big{[}\phi(x+z)-\phi(x)\big{]}\nu_{x-y}(\mathrm{d}z)$
$\displaystyle-\,f(x-y)\int_{0}^{\infty}\big{[}\phi(x+z)-\phi(x)\big{]}\nu_{y-x}(\mathrm{d}z)$
$\displaystyle\leq\\!\\!\\!$ $\displaystyle
y\phi(x)\big{[}f(2(x-y))-2f(x-y)\big{]}\mu_{x-y}(0,\infty)$
$\displaystyle-\,yf(x-y)\int_{0}^{\infty}\big{[}\phi(x+z)-\phi(x)\big{]}\mu_{y-x}(\mathrm{d}z)$
$\displaystyle+\,\phi(x)\big{[}f(2(x-y))-2f(x-y)\big{]}\nu_{x-y}(0,\infty)$
$\displaystyle-f(x-y)\int_{0}^{\infty}\big{[}\phi(x+z)-\phi(x)\big{]}\nu_{y-x}(\mathrm{d}z),$

where we have used the relation $\mu_{x-y}(0,\infty)=\mu_{y-x}(0,\infty)$.
Returning to (4.1) and reorganizing the terms we get (4.19). $\square$

Now let us make some special choices of the functions $f\in
C_{b}^{2}(\mathbb{R}_{+})$ and $\phi\in C_{b}^{2}(\mathbb{R}_{+})$ in (4.18).
In the sequel, we assume Conditions 1.1 is satisfied. By the non-triviality
assumption for the branching mechanism ${\it\Psi}$, we can fix a constant
$\lambda_{0}>0$ such that ${\it\Psi}(\lambda_{0})>0$ and let

$\displaystyle\psi(x)=1-e^{-\lambda_{0}x},\quad x\geq 0.$ (4.25)

For constants $0<x_{0}<1$ and $\theta\geq 4$ to be specified later, we define

$\displaystyle\phi(x)=\bigg{\\{}\begin{array}[]{ll}\theta+(1-x/x_{0})^{3},&\quad
0\leq x<x_{0},\vskip 3.0pt plus 1.0pt minus 1.0pt\cr\theta,&\quad x\geq
x_{0}.\end{array}$ (4.28)

Then define the function

$\displaystyle F_{0}(x,y)=\phi(x)\big{[}1+\psi(x-y)\big{]}\mathbf{1}_{\\{x\neq
y\\}},\quad(x,y)\in D.$ (4.29)

By Lemma 4.2 we have $F_{0}\in\mathscr{D}(\tilde{L})$. It is easy to see that

$\displaystyle\theta\leq F_{0}(x,y)\leq
2(1+\theta),\quad(x,y)\in{\it\Delta}^{c}.$ (4.30)

###### Lemma 4.3

Suppose that Conditions 1.1 is satisfied. For any $l\geq 1$ there is a
constant $\lambda_{1}>0$ such that, for $(x,y)\in{\it\Delta}^{c}$,

$\displaystyle\tilde{L}F_{0}(x,y)\\!\\!\\!$ $\displaystyle\leq\\!\\!\\!$
$\displaystyle-c\lambda_{0}^{2}\theta ye^{-\lambda_{0}(x-y)}-\theta
y\mu_{x-y}(0,\infty)-\theta\nu_{x-y}(0,\infty)-\lambda_{1}\theta\psi(x-y)\mathbf{1}_{\\{x-y\leq
l\\}}$ (4.31)
$\displaystyle+\,\big{[}y\mu_{x-y}(0,\infty)+I(x)+J(x)\big{]}\big{[}1+\psi(x-y)\big{]}\mathbf{1}_{\\{x\leq
x_{0}\\}}$
$\displaystyle+\,\big{[}1+\psi(x-y)\big{]}\int_{0}^{\infty}\big{[}\phi(x+z)-\phi(x)\big{]}(\nu-\nu_{y-x})(\mathrm{d}z)\mathbf{1}_{\\{x\leq
x_{0}\\}},$

where

$\displaystyle I(x)=[\beta-bx-g(x)]\phi^{\prime}(x)$ (4.32)

and

$\displaystyle
J(x)=\frac{3x}{x_{0}^{2}}\bigg{(}2c+\int_{0}^{1}z^{2}\mu(\mathrm{d}z)\bigg{)}.$
(4.33)

Proof. Since $\psi(0)=0$ and $\psi$ is concave, we have $\psi(2z)\leq
2\psi(z)$ for $z\geq 0$. Then from (4.19) it follows that, for
$(x,y)\in{\it\Delta}^{c}$,

$\displaystyle\tilde{L}F_{0}(x,y)\\!\\!\\!$ $\displaystyle\leq\\!\\!\\!$
$\displaystyle cy\phi(x)\psi^{\prime\prime}(x-y)+[\beta-
bx-g(x)]\phi^{\prime}(x)\big{[}1+\psi(x-y)\big{]}$
$\displaystyle+\,y\phi(x)\big{[}\psi(2(x-y))-2\psi(x-y)-1\big{]}\mu_{x-y}(0,\infty)$
$\displaystyle+\,\phi(x)\big{[}\psi(2(x-y))-2\psi(x-y)-1\big{]}\nu_{x-y}(0,\infty)$
$\displaystyle+\,\tilde{A}_{1}F_{0}(x,y)+\tilde{A}_{2}F_{0}(x,y)$
$\displaystyle\leq\\!\\!\\!$ $\displaystyle-c\lambda_{0}^{2}\theta
ye^{-\lambda_{0}(x-y)}+I(x)\big{[}1+\psi(x-y)\big{]}-\theta
y\mu_{x-y}(0,\infty)$
$\displaystyle-\,\theta\nu_{x-y}(0,\infty)+\tilde{A}_{1}F_{0}(x,y)+\tilde{A}_{2}F_{0}(x,y),$

where we have also used the fact that $\phi(x)\geq\theta$. By the definition
of $\phi$, we know that for all $x,z\geq 0$,

$\displaystyle 0\leq\phi(x)-\phi(x+z)\leq\mathbf{1}_{\\{x\leq x_{0}\\}},\quad
0\leq-\phi^{\prime}(x)z\leq\frac{3z}{x_{0}}\mathbf{1}_{\\{x\leq x_{0}\\}},$
$\displaystyle\phi^{\prime\prime}(x)\leq\frac{6}{x_{0}^{2}}\mathbf{1}_{\\{x\leq
x_{0}\\}},\quad\phi(x+z)-\phi(x)-\phi^{\prime}(x)z\leq\frac{3z^{2}}{x_{0}^{2}}\mathbf{1}_{\\{x\leq
x_{0}\\}}.$

From (4.22) it follows that

$\displaystyle\tilde{A}_{1}F_{0}(x,y)\\!\\!\\!$ $\displaystyle=\\!\\!\\!$
$\displaystyle\big{[}1+\psi(x-y)\big{]}\bigg{[}cx\phi^{\prime\prime}(x)-y\int_{0}^{\infty}\big{[}\phi(x+z)-\phi(x)\big{]}\mu_{y-x}(\mathrm{d}z)$
$\displaystyle\qquad\qquad\qquad+\,x\int_{0}^{1}\big{[}\phi(x+z)-\phi(x)-\phi^{\prime}(x)z\big{]}\mu(\mathrm{d}z)$
$\displaystyle\qquad\qquad\qquad+\int_{0}^{\infty}\big{[}\phi(x+z)-\phi(x)\big{]}(\nu-\nu_{x-y})(\mathrm{d}z)\bigg{]}$
$\displaystyle\leq\\!\\!\\!$
$\displaystyle\big{[}1+\psi(x-y)\big{]}\bigg{[}\frac{6cx}{x_{0}^{2}}+y\mu_{x-y}(0,\infty)+x\int_{0}^{1}\frac{3z^{2}}{x_{0}^{2}}\mu(\mathrm{d}z)\bigg{]}\mathbf{1}_{\\{x\leq
x_{0}\\}}$
$\displaystyle+\,\big{[}1+\psi(x-y)\big{]}\int_{0}^{\infty}\big{[}\phi(x+z)-\phi(x)\big{]}(\nu-\nu_{y-x})(\mathrm{d}z)\mathbf{1}_{\\{x\leq
x_{0}\\}}$ $\displaystyle=\\!\\!\\!$
$\displaystyle\big{[}1+\psi(x-y)\big{]}\big{[}y\mu_{x-y}(0,\infty)+J(x)\big{]}\mathbf{1}_{\\{x\leq
x_{0}\\}}$
$\displaystyle+\,\big{[}1+\psi(x-y)\big{]}\int_{0}^{\infty}\big{[}\phi(x+z)-\phi(x)\big{]}(\nu-\nu_{y-x})(\mathrm{d}z)\mathbf{1}_{\\{x\leq
x_{0}\\}}.$

In view of (4.24), we have

$\displaystyle\tilde{A}_{2}F_{0}(x,y)\\!\\!\\!$ $\displaystyle=\\!\\!\\!$
$\displaystyle(x-y)\phi(x)\bigg{[}c\psi^{\prime\prime}(x-y)-b\psi^{\prime}(x-y)+\int_{0}^{\infty}\big{[}\psi(x+z-y)$
$\displaystyle\qquad\qquad\qquad-\,\psi(x-y)-\psi^{\prime}(x-y)z\mathbf{1}_{\\{z\leq
1\\}}\big{]}\mu(\mathrm{d}z)\bigg{]}$ $\displaystyle=\\!\\!\\!$
$\displaystyle-(x-y)\phi(x)e^{-\lambda_{0}(x-y)}{\it\Psi}(\lambda_{0})$
$\displaystyle\leq\\!\\!\\!$
$\displaystyle-\,\lambda_{0}^{-1}(1-e^{-\lambda_{0}(x-y)})\theta
e^{-\lambda_{0}l}{\it\Psi}(\lambda_{0})\mathbf{1}_{\\{x-y\leq l\\}}.$

By putting together the above estimates we get (4.31) with
$\lambda_{1}=\lambda_{0}^{-1}e^{-\lambda_{0}l}{\it\Psi}(\lambda_{0})>0$.
$\square$

###### Proposition 4.4

Suppose that Conditions 1.1 and 1.2 are satisfied. Then for any $l\geq 1$
there is a constant $\lambda_{2}>0$ such that

$\displaystyle\tilde{L}F_{0}(x,y)\leq-\lambda_{2}F_{0}(x,y)\mathbf{1}_{\\{x-y\leq
l\\}},\quad(x,y)\in{\it\Delta}^{c}.$ (4.34)

Proof. The idea of the proof is to identify and take the advantage of the
dominating factor among the branching, immigration and competition mechanisms
in different parts of the space ${\it\Delta}^{c}$. Under Condition 1.2, we can
choose constants $\kappa>0$ and $x_{0}\in(0,c_{0})$ such that

$\displaystyle
c\lambda_{0}^{2}e^{-\lambda_{0}x}+\mu_{x}(0,\infty)+\nu_{x}(0,\infty)\geq
2\kappa,\quad 0\leq x\leq x_{0}.$ (4.35)

Note that $I(x)=0$ for all $x>x_{0}$. By (4.31) we have, for $x>x_{0}$,

$\displaystyle\tilde{L}F_{0}(x,y)\\!\\!\\!$ $\displaystyle\leq\\!\\!\\!$
$\displaystyle-\theta
y\big{[}c\lambda_{0}^{2}e^{-\lambda_{0}(x-y)}+\mu_{x-y}(0,\infty)\big{]}-\theta\nu_{x-y}(0,\infty)$
(4.36) $\displaystyle-\,\lambda_{1}\theta\psi(x-y)\mathbf{1}_{\\{x-y\leq
l\\}}.$

Since $x_{0}\leq 1\leq l$, using (4.31) again we see, for $0\leq x\leq x_{0}$,

$\displaystyle\tilde{L}F_{0}(x,y)\\!\\!\\!$ $\displaystyle\leq\\!\\!\\!$
$\displaystyle-c\lambda_{0}^{2}\theta
ye^{-\lambda_{0}(x-y)}-y\mu_{x-y}(0,\infty)\big{[}\theta-1-\psi(x-y)\big{]}-\theta\nu_{x-y}(0,\infty)$
(4.37)
$\displaystyle-\lambda_{1}\theta\psi(x-y)+\big{[}I(x)+J(x)\big{]}\big{[}1+\psi(x-y)\big{]}$
$\displaystyle+\,\big{[}1+\psi(x-y)\big{]}\int_{0}^{\infty}\big{[}\phi(x+z)-\phi(x)\big{]}(\nu-\nu_{y-x})(\mathrm{d}z).$

With these estimates at hand, we prove the desired assertion by considering
the following three cases.

(i) We first consider the case of $x>x_{0}$. We will apply (4.36) and consider
separately three subcases. From (4.36) it is easy to see
$\tilde{L}F_{0}(x,y)\leq 0$ when $x-y>l$. When $x_{0}/2<x-y\leq l$, noticing
that $\psi$ is nondecreasing and using the facts $\phi\leq 1+\theta$ and
$\psi\leq 1$, we have

$\displaystyle\tilde{L}F_{0}(x,y)\\!\\!\\!$ $\displaystyle\leq\\!\\!\\!$
$\displaystyle-\lambda_{1}\theta\psi(x-y)\leq-\frac{\lambda_{1}\theta}{2}[\psi(x-y)+\psi(x_{0}/2)]$
$\displaystyle\leq\\!\\!\\!$
$\displaystyle-\frac{\lambda_{1}\theta}{2}\psi(x_{0}/2)\big{[}1+\psi(x-y)\big{]}\leq-\frac{\lambda_{1}\theta\psi(x_{0}/2)}{2(1+\theta)}F_{0}(x,y).$

When $x-y\leq x_{0}/2$, we have $y\geq x_{0}/2$ and so, by (4.35) and (4.36),

$\displaystyle\tilde{L}F_{0}(x,y)\\!\\!\\!$ $\displaystyle\leq\\!\\!\\!$
$\displaystyle-\frac{\theta
x_{0}}{2}\big{[}c\lambda_{0}^{2}e^{-\lambda_{0}(x-y)}+\mu_{x-y}(0,\infty)+\nu_{x-y}(0,\infty)\big{]}-\lambda_{1}\theta\psi(x-y)$
$\displaystyle\leq\\!\\!\\!$ $\displaystyle-\theta
x_{0}\kappa-\lambda_{1}\theta\psi(x-y)\leq-(x_{0}\kappa\land\lambda_{1})\theta\big{[}1+\psi(x-y)\big{]}$
$\displaystyle\leq\\!\\!\\!$
$\displaystyle-(x_{0}\kappa\land\lambda_{1})\frac{\theta}{1+\theta}F_{0}(x,y).$

(ii) Let us consider the case of $x\in(0,rx_{0}]$, where $r\in(0,1/2]$ will be
specified later. Recall that $\theta\geq 4$ and $\psi\leq 1$. Then, according
to (4.37), for $x\in[0,rx_{0}]$,

$\displaystyle\tilde{L}F_{0}(x,y)\\!\\!\\!$ $\displaystyle\leq\\!\\!\\!$
$\displaystyle\big{[}I(x)+J(x)\big{]}\big{[}1+\psi(x-y)\big{]}-\theta\nu_{x-y}(0,\infty)$
$\displaystyle+\,\big{[}1+\psi(x-y)\big{]}\int_{0}^{\infty}\big{[}\phi(x+z)-\phi(x)\big{]}(\nu-\nu_{y-x})(\mathrm{d}z)\mathbf{1}_{\\{x\leq
x_{0}\\}}$ $\displaystyle\leq\\!\\!\\!$
$\displaystyle\big{[}1+\psi(x-y)\big{]}\bigg{[}I(x)+J(x)-\nu_{x-y}(0,\infty)$
$\displaystyle+\int_{0}^{\infty}\big{[}\phi(x+z)-\phi(x)\big{]}(\nu-\nu_{y-x})(\mathrm{d}z)\bigg{]}.$

Since $0<x_{0}<1$, we can choose $r_{*}\in(0,1/2]$ and $q>0$ such that, for
$0\leq x\leq r_{*}x_{0}$,

$\displaystyle
I(x)-\nu_{x-y}(0,\infty)+\int_{0}^{\infty}\big{[}\phi(x+z)-\phi(x)\big{]}(\nu-\nu_{y-x})(\mathrm{d}z)$
(4.38) $\displaystyle\qquad\leq
I(x)-\nu_{x-y}(0,\infty)-\int_{x_{0}-x}^{\infty}\Big{(}1-\frac{x}{x_{0}}\Big{)}^{3}(\nu-\nu_{y-x})(\mathrm{d}z)$
(4.39)
$\displaystyle\qquad\qquad-\int^{x_{0}-x}_{0}\Big{[}\Big{(}1-\frac{x}{x_{0}}\Big{)}^{3}-\Big{(}1-\frac{x+z}{x_{0}}\Big{)}^{3}\Big{]}(\nu-\nu_{y-x})(\mathrm{d}z)$
(4.40)
$\displaystyle\qquad\leq\frac{3}{x_{0}}\big{[}|b|x+g(x)-\beta\big{]}\Big{(}1-\frac{x}{x_{0}}\Big{)}^{2}-\nu_{x-y}(0,\infty)-\frac{1}{8}\int_{x_{0}-x}^{\infty}(\nu-\nu_{y-x})(\mathrm{d}z)$
(4.41)
$\displaystyle\qquad\qquad-\frac{1}{x_{0}^{3}}\int^{x_{0}-x}_{0}z^{3}(\nu-\nu_{y-x})(\mathrm{d}z)$
(4.42)
$\displaystyle\qquad\leq\frac{3}{x_{0}}\big{[}|b|x+g(x)\big{]}-\frac{3\beta}{4x_{0}}-\frac{1}{8}\int_{0}^{\infty}(1\land
z^{3})\nu(\mathrm{d}z)\leq-q,$ (4.43)

where for the last inequality we have used the non-triviality assumption for
the immigration mechanism ${\it\Phi}$ in Condition 1.1 and the fact that $g$
is a continuous function with $g(0)=0$. Now we take

$\displaystyle
r=r_{*}\land\Big{[}\frac{x_{0}q}{6}\Big{(}2c+\int_{0}^{1}z^{2}\mu(\mathrm{d}z)\Big{)}^{-1}\Big{]}.$
(4.44)

From (4.33) it follows that, for $x\in[0,rx_{0}]$,

$\displaystyle
J(x)\leq\frac{3r}{x_{0}}\Big{(}2c+\int_{0}^{1}z^{2}\mu(\mathrm{d}z)\Big{)}\leq\frac{q}{2}.$

Recall that $\theta\geq 4$ and $\psi\leq 1$. Then, for $x\in[0,rx_{0}]$,

$\displaystyle\tilde{L}F_{0}(x,y)\\!\\!\\!$ $\displaystyle\leq\\!\\!\\!$
$\displaystyle-\big{[}q-J(x)\big{]}\big{[}1+\psi(x-y)\big{]}$
$\displaystyle\leq\\!\\!\\!$
$\displaystyle-\frac{q}{2}\big{[}1+\psi(x-y)\big{]}\leq-\frac{q}{2(1+\theta)}F_{0}(x,y).$

(iii) We finally consider the case of $x\in(rx_{0},x_{0}]$. In this case, we
have

$\displaystyle I(x)\leq\frac{3}{x_{0}}\big{[}|b|x_{0}+g(x_{0})\big{]},\quad
J(x)\leq\frac{3}{x_{0}}\bigg{(}2c+\int_{0}^{1}z^{2}\mu(\mathrm{d}z)\bigg{)}.$

Now we take

$\displaystyle
H=\frac{3}{x_{0}}\bigg{[}2c+|b|x_{0}+g(x_{0})+\int_{0}^{1}z^{2}\mu(\mathrm{d}z)\bigg{]}$
(4.45)

and

$\displaystyle\theta=\max\Big{\\{}4,\frac{2H}{\lambda_{1}},\frac{4H}{r\kappa
x_{0}},\frac{8H}{\lambda_{1}\psi(rx_{0}/2)}\Big{\\}}.$ (4.46)

From (4.37) it follows that

$\displaystyle\tilde{L}F_{0}(x,y)\\!\\!\\!$ $\displaystyle\leq\\!\\!\\!$
$\displaystyle-c\lambda_{0}^{2}\theta
ye^{-\lambda_{0}(x-y)}-y\mu_{x-y}(0,\infty)\big{[}\theta-1-\psi(x-y)\big{]}-\theta\nu_{x-y}(0,\infty)$
(4.47)
$\displaystyle+\,\psi(x-y)\big{[}I(x)+J(x)-\lambda_{1}\theta\big{]}+I(x)+J(x)$
$\displaystyle\leq\\!\\!\\!$ $\displaystyle-\,c\lambda_{0}^{2}\theta
ye^{-\lambda_{0}(x-y)}-y\mu_{x-y}(0,\infty)\big{[}\theta-1-\psi(x_{0})\big{]}-\theta\nu_{x-y}(0,\infty)$
$\displaystyle+\,\psi(x-y)(H-\lambda_{1}\theta)+H$
$\displaystyle\leq\\!\\!\\!$ $\displaystyle-\,c\lambda_{0}^{2}\theta
ye^{-\lambda_{0}(x-y)}-\frac{\theta}{2}y\mu_{x-y}(0,\infty)-\theta\nu_{x-y}(0,\infty)$
$\displaystyle-\,\frac{\lambda_{1}\theta}{2}\psi(x-y)+H.$

When $y\geq rx_{0}/2$, we have $x-y\leq rx_{0}/2$ and, by (4.35),

$\displaystyle-c\lambda_{0}^{2}\theta
ye^{-\lambda_{0}(x-y)}-\frac{\theta}{2}y\mu_{x-y}(0,\infty)-\theta\nu_{x-y}(0,\infty)\leq-\frac{\theta
r\kappa x_{0}}{2}.$

Returning to (4.47), we obtain

$\displaystyle\tilde{L}F_{0}(x,y)\\!\\!\\!$ $\displaystyle\leq\\!\\!\\!$
$\displaystyle-\frac{\theta r\kappa
x_{0}}{2}-\frac{\lambda_{1}\theta}{2}\psi(x-y)+H$ $\displaystyle\leq\\!\\!\\!$
$\displaystyle-\frac{\theta r\kappa
x_{0}}{4}-\frac{\lambda_{1}\theta}{2}\psi(x-y)$ $\displaystyle\leq\\!\\!\\!$
$\displaystyle-\Big{(}\frac{r\kappa
x_{0}}{2}\land\lambda_{1}\Big{)}\frac{\theta}{2}\big{[}1+\psi(x-y)\big{]}$
$\displaystyle\leq\\!\\!\\!$ $\displaystyle-\Big{(}\frac{r\kappa
x_{0}}{2}\land\lambda_{1}\Big{)}\frac{\theta}{2(1+\theta)}F_{0}(x,y).$

When $y\in[0,rx_{0}/2)$, we have $x-y\geq rx_{0}/2$ and hence

$\displaystyle\tilde{L}F_{0}(x,y)\\!\\!\\!$ $\displaystyle\leq\\!\\!\\!$
$\displaystyle-\frac{\lambda_{1}\theta}{2}\psi(x-y)+H$
$\displaystyle\leq\\!\\!\\!$
$\displaystyle-\frac{\lambda_{1}\theta}{4}\psi(x-y)-\frac{\lambda_{1}\theta}{4}\psi(rx_{0}/2)+H$
$\displaystyle\leq\\!\\!\\!$
$\displaystyle-\frac{\lambda_{1}\theta}{8}\psi(x-y)-\frac{\lambda_{1}\theta}{8}\psi(rx_{0}/2)$
$\displaystyle\leq\\!\\!\\!$
$\displaystyle-\frac{\lambda_{1}\theta\psi(rx_{0}/2)}{8}\big{[}1+\psi(x-y)\big{]}$
$\displaystyle\leq\\!\\!\\!$
$\displaystyle-\frac{\lambda_{1}\theta\psi(rx_{0}/2)}{8(1+\theta)}F_{0}(x,y).$

Combining all the estimates in the three cases, we obtain (4.34). $\square$

##  5 The exponential ergodicity

In this section, we prove the exponential contraction property (1.43) for a
suitable control function. From this property we derive the exponential
ergodicity of the CBIC-process. Let $V$ be the Lyapunov function given as in
Condition 1.3. Let $\varepsilon=4C_{0}/(\lambda_{2}\theta)$, where $C_{0}>0$,
$\lambda_{2}>0$ and $\theta\geq 4$ are given as in (1.25), (4.34) and (4.46),
respectively. Let $F_{0}\in\mathscr{D}(\tilde{L})$ be given by (4.29) and let

$\displaystyle V_{2}(x,y)=V(x)+V(y),\quad
V_{0}(x,y)=V_{2}(x,y)\mathbf{1}_{\\{x\neq y\\}}$ (5.1)

and

$\displaystyle G_{0}(x,y)=\varepsilon F_{0}(x,y)+V_{0}(x,y),\quad(x,y)\in D.$
(5.2)

###### Proposition 5.1

Suppose that Conditions 1.1, 1.2 and 1.3 are satisfied. Then
$G_{0}\in\mathscr{D}(\tilde{L})$ and there is a constant $\lambda>0$ such that

$\displaystyle\tilde{L}G_{0}(x,y)\leq-\lambda
G_{0}(x,y),\quad(x,y)\in{\it\Delta}^{c}.$ (5.3)

Proof. Clearly, for any $(x,y)\in{\it\Delta}^{c}$ the expression (4.4) of
$\tilde{L}_{0}F(x,y)$ only depends on the restriction of $F$ to
${\it\Delta}^{c}$. Then it is easy to show that

$\displaystyle\tilde{L}_{0}V_{0}(x,y)=\tilde{L}_{0}V_{2}(x,y)=LV(x)+LV(y).$
(5.4)

Since $0=V_{0}(z,z)\leq V_{2}(z,z)$ for $z\geq 0$, by (4.8) we have
$\tilde{L}_{1}V_{0}(x,y)\leq\tilde{L}_{1}V_{2}(x,y)$. But, by (4.8), (5.1) and
the first equality in (3.2),

$\displaystyle\tilde{L}_{1}V_{2}(x,y)\\!\\!\\!$ $\displaystyle=\\!\\!\\!$
$\displaystyle
y\int_{0}^{\infty}\big{[}V(2y+z-x)-V(y+z)\big{]}\mu_{x-y}(\mathrm{d}z)$
$\displaystyle+\,y\int_{0}^{\infty}\big{[}V(x+z)-V(y+z)\big{]}\mu_{y-x}(\mathrm{d}z)$
$\displaystyle+\int_{0}^{\infty}\big{[}V(2y+z-x)-V(y+z)\big{]}\nu_{x-y}(\mathrm{d}z)$
$\displaystyle+\int_{0}^{\infty}\big{[}V(x+z)-V(y+z)\big{]}\nu_{y-x}(\mathrm{d}z)=0.$

It follows that

$\displaystyle\tilde{L}G_{0}(x,y)\leq\varepsilon\tilde{L}F_{0}(x,y)+LV(x)+LV(y).$

Then, by (1.25) and Proposition 4.4,

$\displaystyle\tilde{L}G_{0}(x,y)\leq-\varepsilon\lambda_{2}F_{0}(x,y)\mathbf{1}_{\\{x-y\leq
l\\}}+2C_{0}-C_{1}\big{[}V(x)+V(y)\big{]}.$ (5.5)

Let $l\geq 1$ be sufficiently large such that $V(z)>12C_{0}/C_{1}$ for $z>l$.
Recall that $\varepsilon=4C_{0}/(\lambda_{2}\theta)$. When $x\geq l$, we can
use (4.30) and (5.5) to see

$\displaystyle\tilde{L}G_{0}(x,y)\\!\\!\\!$ $\displaystyle\leq\\!\\!\\!$
$\displaystyle 2C_{0}-C_{1}\big{[}V(x)+V(y)\big{]}$
$\displaystyle\leq\\!\\!\\!$
$\displaystyle-4C_{0}-\frac{C_{1}}{2}\big{[}V(x)+V(y)\big{]}$
$\displaystyle\leq\\!\\!\\!$
$\displaystyle-\frac{\lambda_{2}\theta}{2(1+\theta)}\varepsilon
F_{0}(x,y)-\frac{C_{1}}{2}\big{[}V(x)+V(y)\big{]}$
$\displaystyle\leq\\!\\!\\!$
$\displaystyle-\frac{1}{2}\Big{(}\frac{\lambda_{2}\theta}{1+\theta}\land
C_{1}\Big{)}G_{0}(x,y).$

When $x\leq l$, using (4.30) and (5.5) again we have

$\displaystyle\tilde{L}G_{0}(x,y)\\!\\!\\!$ $\displaystyle\leq\\!\\!\\!$
$\displaystyle-\varepsilon\lambda_{2}F_{0}(x,y)+2C_{0}-C_{1}\big{[}V(x)+V(y)\big{]}$
$\displaystyle\leq\\!\\!\\!$
$\displaystyle-\frac{\varepsilon\lambda_{2}}{2}F_{0}(x,y)-\frac{\varepsilon\lambda_{2}\theta}{2}+2C_{0}-C_{1}\big{[}V(x)+V(y)\big{]}$
$\displaystyle\leq\\!\\!\\!$
$\displaystyle-\frac{\varepsilon\lambda_{2}}{2}F_{0}(x,y)-C_{1}\big{[}V(x)+V(y)\big{]}$
$\displaystyle\leq\\!\\!\\!$ $\displaystyle-\Big{(}\frac{\lambda_{2}}{2}\land
C_{1}\Big{)}G_{0}(x,y).$

Then (5.3) holds with $\lambda=(C_{1}\land\lambda_{2})/2>0$. $\square$

###### Theorem 5.2

Suppose that Conditions 1.1, 1.2 and 1.3 are satisfied. Let
$G_{0}\in\mathscr{D}(\tilde{L})$ be defined by (5.2) and let $\lambda>0$ be
given by Proposition 5.1. Then we have (1.43) for every $(x,y)\in D$.

Proof. It suffices to consider $(x,y)\in{\it\Delta}^{c}$. Let
$\\{(X_{t},Y_{t}):t\geq 0\\}$ be the Markov coupling defined by (3.10) and
(3.12) with $(X_{0},Y_{0})=(x,y)$. Recall that $\zeta_{n}^{*}=\inf\\{t\geq
0:X_{t}\geq n$ or $X_{t}-Y_{t}\leq 1/n\\}$. By Proposition 2.3 and Theorem
3.1, the process $\\{(X_{t},Y_{t}):t\geq 0\\}$ is conservative. Then (3.17)
implies $\lim_{n\to\infty}\zeta_{n}^{*}=\lim_{n\to\infty}T_{n}=T$, where
$T_{n}=\inf\\{t\geq 0:X_{t}-Y_{t}\leq 1/n\\}$. By Theorem 4.1 and integration
by parts, for any $n\geq 1$ we have

$\displaystyle\mathrm{e}^{\lambda(t\land\zeta_{n}^{*})}G_{0}(X_{t\land\zeta_{n}^{*}},Y_{t\land\zeta_{n}^{*}})=G_{0}(x,y)+\int_{0}^{t\land\zeta_{n}^{*}}\mathrm{e}^{\lambda
s}(\tilde{L}+\lambda)G_{0}(X_{s},Y_{s})\mathrm{d}s+M_{n}(t),$ (5.6)

where $\\{M_{n}(t):t\geq 0\\}$ is a martingale. From (5.3) and (5.6) it
follows that

$\displaystyle\mathbf{E}\big{[}\mathrm{e}^{\lambda(t\land\zeta_{n}^{*})}G_{0}(X_{t\land\zeta_{n}^{*}},Y_{t\land\zeta_{n}^{*}})\big{]}\leq
G_{0}(x,y),$

and so

$\displaystyle\mathbf{E}\big{[}\mathrm{e}^{\lambda(t\land\zeta_{n}^{*})}G_{0}(X_{t\land\zeta_{n}^{*}},Y_{t\land\zeta_{n}^{*}})\mathbf{1}_{\\{t<T\\}}\big{]}\leq
G_{0}(x,y).$

Since $G_{0}(x,x)=0$ for $x\geq 0$ and $X_{T+t}=Y_{T+t}$ for $t\geq 0$, we can
let $n\to\infty$ and use Fatou’s lemma to get

$\displaystyle\mathbf{E}\big{[}\mathrm{e}^{\lambda
t}G_{0}(X_{t},Y_{t})\big{]}=\mathbf{E}\big{[}\mathrm{e}^{\lambda
t}G_{0}(X_{t},Y_{t})\mathbf{1}_{\\{t<T\\}}\big{]}\leq G_{0}(x,y),$

which clearly implies (1.43). $\square$

The result of the next lemma should be already known, but we could not find a
reference. For the convenience of the reader, we give a simple proof of the
result here.

###### Lemma 5.3

The expressions (1.19) and (1.21) for the $V$-weighted total variation
distance $W_{V}$ are equivalent.

Proof. For $\gamma,\eta\in\mathscr{P}_{V}(\mathbb{R}_{+})$ let
$U=\mathrm{supp}((\gamma-\eta)^{+})$ and
$U^{c}=\mathrm{supp}((\gamma-\eta)^{-})$. If $\pi\in\mathscr{C}(\gamma,\eta)$,
then

$\displaystyle\int_{\mathbb{R}_{+}^{2}}d_{V}(x,y)\pi(\mathrm{d}x,\mathrm{d}y)\\!\\!\\!$
$\displaystyle=\\!\\!\\!$
$\displaystyle\int_{\mathbb{R}_{+}^{2}}[2+V(x)+V(y)]\mathbf{1}_{\\{x\neq
y\\}}\pi(\mathrm{d}x,\mathrm{d}y)$ $\displaystyle\geq\\!\\!\\!$
$\displaystyle\int_{\mathbb{R}_{+}}[1+V(x)]\gamma(\mathrm{d}x)+\int_{\mathbb{R}_{+}}[1+V(y)]\eta(\mathrm{d}y)$
$\displaystyle-\,2\int_{U\times U}[1+V(x)]\pi(\mathrm{d}x,\mathrm{d}y)$
$\displaystyle-\,2\int_{U^{c}\times
U^{c}}[1+V(y)]\pi(\mathrm{d}x,\mathrm{d}y)$ $\displaystyle\geq\\!\\!\\!$
$\displaystyle\int_{\mathbb{R}_{+}}[1+V(x)]\gamma(\mathrm{d}x)+\int_{\mathbb{R}_{+}}[1+V(y)]\eta(\mathrm{d}y)$
$\displaystyle-2\int_{U}[1+V(x)]\gamma(\mathrm{d}x)-2\int_{U^{c}}[1+V(y)]\eta(\mathrm{d}y)$
$\displaystyle=\\!\\!\\!$
$\displaystyle\int_{\mathbb{R}_{+}}[1+V(x)]\gamma(\mathrm{d}x)+\int_{\mathbb{R}_{+}}[1+V(y)]\eta(\mathrm{d}y)$
$\displaystyle-2\int_{U}[1+V(x)](\gamma\land\eta)(\mathrm{d}x)$
$\displaystyle=\\!\\!\\!$
$\displaystyle\int_{\mathbb{R}_{+}}[1+V(x)]|\gamma-\eta|(\mathrm{d}x).$

One the other hand, let $\pi_{*}\in\mathscr{C}(\gamma,\eta)$ be defined by

$\displaystyle\pi_{*}(\mathrm{d}x,\mathrm{d}y)=\frac{(\gamma\land\eta)_{*}(\mathrm{d}x,\mathrm{d}x)}{(\gamma\land\eta)(\mathbb{R}_{+})}+\frac{(\gamma-\eta)^{+}(\mathrm{d}x)(\gamma-\eta)^{-}(\mathrm{d}y)}{(\gamma-\eta)^{+}(\mathbb{R}_{+})}.$

where $(\gamma\land\eta)_{*}$ is the image of $\gamma\land\eta$ under the
mapping $x\mapsto(x,x)$ from $\mathbb{R}_{+}$ to $\mathbb{R}_{+}^{2}$. It is
easy to see that

$\displaystyle\int_{\mathbb{R}_{+}^{2}}d_{V}(x,y)\pi_{*}(\mathrm{d}x,\mathrm{d}y)\\!\\!\\!$
$\displaystyle=\\!\\!\\!$
$\displaystyle\frac{1}{(\gamma-\eta)^{+}(\mathbb{R}_{+})}\int_{U\times
U^{c}}[1+V(x)](\gamma-\eta)^{+}(\mathrm{d}x)(\gamma-\eta)^{-}(\mathrm{d}y)$
$\displaystyle+\frac{1}{(\gamma-\eta)^{+}(\mathbb{R}_{+})}\int_{U\times
U^{c}}[1+V(y)](\gamma-\eta)^{+}(\mathrm{d}x)(\gamma-\eta)^{-}(\mathrm{d}y)$
$\displaystyle=\\!\\!\\!$
$\displaystyle\int_{U}[1+V(x)](\gamma-\eta)^{+}(\mathrm{d}x)+\int_{U^{c}}[1+V(y)](\gamma-\eta)^{-}(\mathrm{d}y)$
$\displaystyle=\\!\\!\\!$
$\displaystyle\int_{\mathbb{R}_{+}}[1+V(x)]|\gamma-\eta|(\mathrm{d}x).$

Those clearly imply the desired result. $\square$

Proof of Theorem 1.1.  By Theorem 5.2, we have (1.43) for $(x,y)\in D$. It is
easy to see that $\tilde{P}_{t}((x,y),\cdot)$ is a coupling of
$P_{t}(x,\cdot)$ and $P_{t}(y,\cdot)$. Then (1.24) follows from (1.43) and
(1.44) with $K=c_{2}/c_{1}$. By the convexity of the Wasserstein distance we
have, for any $\gamma,\zeta\in\mathscr{P}_{V}(\mathbb{R}_{+})$ and
$\pi\in\mathscr{C}(\gamma,\zeta)$,

$\displaystyle W_{V}(\gamma P_{t},\zeta P_{t})\\!\\!\\!$
$\displaystyle\leq\\!\\!\\!$ $\displaystyle
W_{V}\bigg{(}\int_{\mathbb{R}_{+}^{2}}P_{t}(x,\cdot)\pi(\mathrm{d}x,\mathrm{d}y),\int_{\mathbb{R}_{+}^{2}}P_{t}(y,\cdot)\pi(\mathrm{d}x,\mathrm{d}y)\bigg{)}$
(5.7) $\displaystyle\leq\\!\\!\\!$
$\displaystyle\int_{\mathbb{R}_{+}^{2}}W_{V}(P_{t}(x,\cdot),P_{t}(y,\cdot))\pi(\mathrm{d}x,\mathrm{d}y)\leq
Ke^{-\lambda t}W_{V}(\gamma,\zeta);$ (5.8)

see, e.g., Villani [52, Theorem 4.8]. Then for sufficiently large $r>0$, the
operator $P_{r}$ on $\mathscr{P}_{V}(\mathbb{R}_{+})$ is a contraction. By the
Banach fixed point theorem and the completeness of
$\mathscr{P}_{V}(\mathbb{R}_{+})$, there exists a unique
$\gamma_{r}\in\mathscr{P}_{V}(\mathbb{R}_{+})$ such that
$\gamma_{r}P_{r}=\gamma_{r}$. Now we fix such an $r>0$ and define
$\gamma=r^{-1}\int_{0}^{r}\gamma_{r}P_{s}\mathrm{d}s$. By the Chapman-
Kolmogorov equation, for $0\leq t<r$ we have

$\displaystyle\gamma P_{t}\\!\\!\\!$ $\displaystyle=\\!\\!\\!$
$\displaystyle\frac{1}{r}\int_{t}^{r}\gamma_{r}P_{s}\mathrm{d}s+\frac{1}{r}\int_{r}^{r+t}\gamma_{r}P_{s}\mathrm{d}s$
$\displaystyle=\\!\\!\\!$
$\displaystyle\frac{1}{r}\int_{t}^{r}\gamma_{r}P_{s}\mathrm{d}s+\frac{1}{r}\int_{0}^{t}\gamma_{r}P_{r+s}\mathrm{d}s$
$\displaystyle=\\!\\!\\!$
$\displaystyle\frac{1}{r}\int_{t}^{r}\gamma_{r}P_{s}\mathrm{d}s+\frac{1}{r}\int_{0}^{t}\gamma_{r}P_{s}\mathrm{d}s=\gamma.$

More generally, for any $t\geq 0$ there is a unique integer $k\geq 0$ such
that $0\leq t-kr<r$. Then $\gamma P_{t}=\gamma P_{kr}P_{t-kr}=\gamma
P_{t-kr}=\gamma$. By applying (5.7) again we obtain (1.23) with
$C(\zeta)=KW_{V}(\gamma,\zeta)$. $\square$

Proof of Propositions 1.2 and 1.3.  Under the integrability condition (1.27)
we have $V_{1}\in\mathscr{D}(L)$. By (1.18) it is easy to see that

$\displaystyle LV_{1}(x)=[\beta-
bx-g(x)]+x\int_{1}^{\infty}z\mu(\mathrm{d}z)+\int_{0}^{\infty}z\nu(\mathrm{d}z).$

Then (1.25) is equivalent to (1.28). The integrability condition (1.29)
implies $V_{\log}\in\mathscr{D}(L)$. By (1.18) we have

$\displaystyle LV_{\log}(x)\\!\\!\\!$ $\displaystyle=\\!\\!\\!$
$\displaystyle-\frac{c}{(1+x)^{2}}+x\int_{0}^{\infty}\Big{[}\log\Big{(}1+\frac{z}{1+x}\Big{)}-\frac{z}{1+x}\mathbf{1}_{\\{z\leq
1\\}}\Big{]}\mu(\mathrm{d}z)$ $\displaystyle+\,\frac{\beta-
bx-g(x)}{1+x}+\int_{0}^{\infty}\log\Big{(}1+\frac{z}{1+x}\Big{)}\nu(\mathrm{d}z).$

By Taylor’s expansion, we have

$\displaystyle\frac{z}{1+x}-\log\Big{(}1+\frac{z}{1+x}\Big{)}=z^{2}\int_{0}^{1}\frac{(1-u)\mathrm{d}u}{(1+x+uz)^{2}}$

By the dominated convergence theorem,

$\displaystyle\lim_{x\to\infty}x^{2}\int_{0}^{1}\Big{[}\frac{z}{1+x}-\log\Big{(}1+\frac{z}{1+x}\Big{)}\Big{]}\mu(\mathrm{d}z)=\frac{1}{2}\int_{0}^{1}z^{2}\mu(\mathrm{d}z).$

It follows that

$\displaystyle\lim_{x\to\infty}\frac{x}{\log
x}\int_{0}^{1}\Big{[}\log\Big{(}1+\frac{z}{1+x}\Big{)}-\frac{z}{1+x}\Big{]}\mu(\mathrm{d}z)=0.$
(5.9)

Then (1.25) is equivalent to (1.30). $\square$

Proof of Corollaries 1.4 and 1.5.  Corollary 1.4 follows easily by Proposition
1.2. Then it remains to prove Corollary 1.5. By the proof of Theorem 2.4, the
CBIC-process with stable branching mechanism has generator $L$ defined by
(1.18) with $b=a+\sigma h_{\alpha}$ and $\mu(\mathrm{d}z)=\alpha\sigma
m_{\alpha}(\mathrm{d}z)$, where $m_{\alpha}$ and $h_{\alpha}$ are defined by
(1.34) and (2.12), respectively. For $\alpha=1$, by elementary calculus we
have

$\displaystyle\int_{1}^{\infty}\log\Big{(}1+\frac{z}{1+x}\Big{)}\frac{1}{z^{2}}\mathrm{d}z=\Big{(}\frac{1}{1+x}\Big{)}\log(2+x)+\log\Big{(}\frac{2+x}{1+x}\Big{)}.$

For $0<\alpha<1$, by Zwillinger [53, 3.194.4, p.318] we have

$\displaystyle\int_{0}^{\infty}\log\Big{(}1+\frac{z}{1+x}\Big{)}\frac{1}{z^{1+\alpha}}\mathrm{d}z=\frac{\pi}{\alpha\sin(\alpha\pi)(1+x)^{\alpha}}.$

It follows that

$\displaystyle\int_{1}^{\infty}\log\Big{(}1+\frac{z}{1+x}\Big{)}\frac{1}{z^{1+\alpha}}\mathrm{d}z\\!\\!\\!$
$\displaystyle=\\!\\!\\!$
$\displaystyle\frac{\pi}{\alpha\sin(\alpha\pi)(1+x)^{\alpha}}-\int_{0}^{1}\log\Big{(}1+\frac{z}{1+x}\Big{)}\frac{1}{z^{1+\alpha}}\mathrm{d}z$
$\displaystyle=\\!\\!\\!$
$\displaystyle\int_{0}^{1}\Big{[}\frac{z}{1+x}-\log\Big{(}1+\frac{z}{1+x}\Big{)}\Big{]}\frac{1}{z^{1+\alpha}}\mathrm{d}z$
$\displaystyle+\frac{\pi}{\alpha\sin(\alpha\pi)(1+x)^{\alpha}}-\frac{1}{(1-\alpha)(1+x)}.$

Then Corollary 1.5 follows by Proposition 1.3 and (5.9). $\square$

###### Remark 5.4

Recall that the constants $\lambda_{0}>0$, $c_{0}>0$, $C_{0}>0$ and $C_{1}>0$
are given by Conditions 1.1, 1.2 and 1.3. Our approach provides a way of
finding (a lower bound of) the ergodicity rate $\lambda>0$. The procedure is
given in the following:

  1. (1)

determine $\kappa=\kappa(\lambda_{0},c_{0})$ and
$x_{0}=x_{0}(\lambda_{0},c_{0})$ by (4.35);

  2. (2)

choose $l=l(C_{0},C_{1})$ as in the proof of Proposition 5.1;

  3. (3)

let $\lambda_{1}=\lambda_{0}^{-1}e^{-\lambda_{0}l}{\it\Psi}(\lambda_{0})$ as
in the proof of Lemma 4.3;

  4. (4)

determine $q=q(x_{0})$, $r_{*}=r_{*}(x_{0})$ and $r=r(x_{0})$ by (4.38) and
(4.44);

  5. (5)

define $H=H(x_{0})$ and $\theta=\theta(\lambda_{1},x_{0},\kappa,r)$ by (4.45)
and (4.46);

  6. (6)

choose $\lambda_{2}=\lambda_{2}(\lambda_{1},x_{0},\kappa,r,\theta,q)$ as
indicated in the proof of Proposition 4.4;

  7. (7)

let $\lambda=(C_{1}\land\lambda_{2})/2$ as in the proof of Proposition 5.1.

Acknowledgements.  This research is supported by the National Key R&D Program
of China (No. 2020YFA0712900), the National Natural Science Foundation of
China (Nos. 11731012, 11831014, 11901570 and 12071076) and the Education and
Research Support Program for Fujian Provincial Agencies.

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