# ASYMPTOTICS FOR EXPONENTIAL FUNCTIONALS OF RANDOM WALKS

By Wei Xu111Department of Mathematics, Humboldt-Universität zu Berlin,
Germany. E-mail<EMAIL_ADDRESS>
###### Abstract

This paper provides a detailed description for the asymptotics of exponential
functionals of random walks with light/heavy tails. We give the convergence
rate based on the key observation that the asymptotics depends on the simple
paths with either slowly decreasing local minimum or final value below a low
level. Also, our careful analysis of the interrelationship between the local
minimum and the final value provides the exact expression of the limiting
coefficient in terms of some transformations of the random walk.

MSC2020 subject classifications: Primary 60G50, 60J55; secondary 60F10, 60E05.

Keywords and phrases: Random walk, exponential functional, Spitzer’s
condition, domain of attraction, regularly varying.

##  1 Introduction

Exponential functionals of random walks have been studied deeply in the past
decades because of their wide and important applications in various fields
such as mathematical finance, physics and population evolution. In the general
setting, for a one-dimension random walk $S$ generated by a sequence of
independent copies $X_{1},X_{2},\cdots$ of a random variable $X$, its
exponential functional is usually defined by

$\displaystyle I_{n}:=\sum_{k=1}^{n}e^{-S_{k}},\quad k=1,2,\cdots$ (1.1)

with the convention $I_{0}=0$. It increases to the limit
$I_{\infty}\in(0,\infty]$ almost surely as $n\to\infty$. According to the
Blumenthal zero-one law, the probability that $I_{\infty}$ is finite,
$\mathbf{P}(I_{\infty}<\infty)$, necessarily equals to $0$ or $1$.
Furthermore, the long-term behavior of $S$ shows this probability equals to
$0$ if and only if $S$ does not drift to infinity. In this case, much
attention has been drawn to the speed at which $I_{n}$ increases to infinity,
especially the decay rate of the following expectation

$\displaystyle\mathbf{E}\big{[}F(I_{n})\big{]}=\mathbf{E}\Big{[}F\Big{(}\sum_{k=1}^{n}e^{-S_{k}}\Big{)}\Big{]}$
(1.2)

with $F$ be a positive function on $(0,\infty)$ that vanishes at infinity. The
main aim of this work is to provide a detailed description for the decay rate
of this expectation as $n\to\infty$.

The study of the expectation $\mathbf{E}[F(I_{n})]$ has been initiated to
explore the asymptotic properties of stochastic systems in random environment.
For instance, let $\tau$ be the extinction time of a linear fractional Galton-
Watson process in an i.i.d. random environment. The survival probability of
the population at time $n$ can be written as

$\displaystyle\mathbf{P}(\tau>n)=\mathbf{E}\big{[}(1+I_{n})^{-1}\big{]}$

in which the random walk $S$ is generated by the offspring distribution that
changes randomly as time goes. Under the assumption that $\mathbf{E}[X]=0$ and
$\mathbf{E}[|X|^{2}]<\infty$, Kozlov [24] first showed that
$\mathbf{P}(\tau>n)\sim C\cdot n^{-1/2}$. When $\mathbf{E}[X]<0$, Liu [26]
proved a rough asymptotic result

$\displaystyle\big{(}\mathbf{P}(\tau>n)\big{)}^{1/n}\sim
C\cdot\zeta\quad\mbox{with}\quad\zeta:=\inf_{\lambda\in[0,1]}\mathbf{E}[e^{\lambda
X}].$

The more precise results were provided in [3, 5, 20], i.e,
$\mathbf{P}(\tau>n)\sim C\cdot\zeta^{n}$, $C\cdot n^{-1/2}\zeta^{n}$ or
$C\cdot n^{-3/2}\zeta^{n}$ if $\mathbf{E}[Xe^{X}]<0$, $=0$ or $>0$. As the
second example, we write $\mathcal{M}$ for the maximum of a simple transient
random walk moving in an i.i.d. random environment. The tail probability of
$\mathcal{M}$ can be given by

$\displaystyle\mathbf{P}(\mathcal{M}\geq
n)=\mathbf{E}\big{[}\hat{I}^{\prime}_{\infty}(\hat{I}^{\prime}_{\infty}+I_{n})^{-1}\big{]},\quad
n=1,2,\cdots,$

where $\hat{I}^{\prime}_{\infty}:=1+\sum_{i=-1}^{-\infty}e^{S^{\prime}_{i}}$
and the two random walks $\\{S^{\prime}_{k}:k=-1,-2,\cdot\\}$ and
$\\{S_{k}:k=1,2,\cdot\\}$ are generated by the random environment at negative
and positive integer points respectively. The asymptotics of
$\mathbf{P}(\mathcal{M}\geq n)$ was considered in [1] with $\mathbf{E}[X]<0$,
i.e., it is asymptotically equivalent to $C\cdot\zeta^{n}$, $C\cdot
n^{-1/2}\zeta^{n}$ or $C\cdot n^{-3/2}\zeta^{n}$ if $\mathbf{E}[Xe^{X}]<0$,
$=0$ or $>0$ respectively.

For general function $F$, to our best knowledge the asymptotics of the
expectation (1.2) has only been studied in Hirano [23] under the assumption
that $\mathbf{E}[Xe^{\alpha X}]=0$ and $F$ is completely monotone satisfying
$F(x)\leq Cx^{-\beta}$ for some $\beta>\alpha>0$. His result shows that
$\mathbf{E}[F(I_{n})]\sim C\cdot n^{-3/2}(\mathbf{E}[e^{\alpha X}])^{n}$. In
this work, we provide under natural assumptions, a more complete and accurate
description for the asymptotic behavior of the expectation (1.2) for a general
function $F$ and a general random walk $S$ that is allowed to own a light- or
heavy-tailed distribution. According to the comparison between the decay rates
of function $F$ at infinity and the tail-distribution of $X$, we shall see
that six regimes arise for the convergence rate of the expectation (1.2). The
feature of this work is that we give not only the convergence rate but also
the expression of the limiting coefficient in all regimes. Roughly speaking,
the expectation (1.2) decays at an exponential rate with different regularly
varying modifying factor, i.e.,

$\displaystyle\mathbf{E}[F(I_{n})]\sim\psi_{n}\cdot\varrho^{n}$

with $\varrho:=\inf_{\lambda\in[0,\theta_{F}]}\mathbf{E}[e^{\lambda X}]$ and
$\theta_{F}:=\sup\\{\theta>0:\sup_{x>0}x^{\theta}F(x)<\infty\\}$ being the
optimal polynomial decay rate of $F$. The regularly varying sequence $\psi$
vanishes at infinity and is determined by the fluctuation of the random walk
$S$ under the probability law $\mathbf{P}^{(\Lambda)}$ that is the Esscher
transform of $\mathbf{P}$ associated to the martingale $\\{\varrho^{-n}\cdot
e^{\Lambda S_{n}}:n=0,2,\cdots\\}$ with
$\Lambda:=\operatorname*{arg\,inf}_{\lambda\in[0,\beta_{F}]}\mathbf{E}[e^{\lambda
X}]$. Particularly, when $0\leq\Lambda<\theta_{F}$ the comparison with the
asymptotics of the first entrance time $\tau_{0}^{-}$ of the random walk $S$
in $(-\infty,0)$ shows that for some constant $C$,

$\displaystyle\mathbf{E}[F(I_{n})]\sim C\cdot\mathbf{P}(\tau_{0}^{-}>n).$

Our methodology is based on the fluctuation theory for random walks. We
encounter two major challenges in the asymptotic analysis of the expectation
(1.2). The first one is to find out the simple paths of random walk who makes
the main contribution to the expectation (1.2). Inspired by the asymptotic
analysis in [4, 2, 3, 9, 30] about survival probabilities for Galton-Watson
processes in an i.i.d. random environment, we observe that these simple paths
have varying characteristics in different regimes, i.e., when $\Lambda=0$,
simple paths with local minimum decreasing slowly make the main contribution
to the expectation (1.2); when $\Lambda>0$, the long-term behavior of the
expectation (1.2) is mainly determined by the simple paths with either slowly
decreasing local minimum or final vale below a low level. The second challenge
is to provide an exact expression of the limiting coefficients. In order to
achieve it, we first decompose the simple paths that make the main
contribution to the expectation (1.2) into several divisions at the local
minimum and then approximate the contribution of each division to the limiting
coefficients with some transformations of a finite subdivision.

The second purpose of this work is to offer assistance in our future study of
the asymptotic behavior of exponential functional of a Lévy process
$\\{\xi_{t}:t\geq 0\\}$ defined by

$\displaystyle I_{t}(\xi):=\int_{0}^{t}e^{-\xi_{s}}ds,\quad t\geq 0.$

It is analogous to random walks that $I_{t}(\xi)\to I_{\infty}(\xi)$ a.s. as
$t\to\infty$ with $I_{\infty}(\xi)<\infty$ a.s. if and only if $\xi$ drifts to
infinity. Readers may refer to [10, 15, 28, 29, 31] and references therein for
many interesting results of $I_{t}(\xi)$ and $I_{\infty}(\xi)$. In the case
$I_{\infty}(\xi)=\infty$ a.s., we are usually interested in the decay rate of
the following expectation $\mathbf{E}[F(I_{t}(\xi))]$ defined as in (1.2),
because of its close connection to the long-term properties of random
processes in random environment, e.g. continuous-state branching processes in
Lévy random environment and diffusion in Lévy random environment; see [7, 22,
25, 27]. Under the assumption that $F(x)\leq cx^{\theta_{F}}$ and
$\lambda_{0}:=\sup\\{\lambda\geq
0:\mathbf{E}[e^{\lambda\xi_{1}}]<\infty\\}>\theta_{F}$, four different regimes
for the convergence rate of the expectation (1.2) were provided in [7, 27,
25]. When $\lambda_{0}=0$, Patie and Savov [29] and Xu [32] proved the
polynomial decay rate for $\mathbf{E}[F(I_{t}(\xi))]$ with $\xi$ oscillating
and satisfying the Spitzer’s condition or $\mathbf{E}[\xi_{1}]<0$ and
$\mathbf{P}(\xi_{1}>x)$ regularly varying. For the case
$0<\lambda_{0}\leq\theta_{F}$, to the best of our knowledge, the asymptotics
of $\mathbf{E}[F(I_{t}(\xi))]$ is still an open problem. Notice that $\xi$ can
be well approximated by the random walk
$\xi^{\delta}:=\\{\xi_{n}^{\delta}:n=0,1,\cdots\\}$ with
$\xi_{n}^{\delta}:=\xi_{n\delta}$ for small $\delta>0$. Let $I^{\delta}$ be
the exponential functional of $\xi^{\delta}$, we may conjecture that there
exists a regularly varying function $\Psi$ satisfying $\Psi(n)\sim\psi_{n}$ as
$n\to\infty$ such that

$\displaystyle\mathbf{E}[F(I_{t}(\xi))]\sim\mathbf{E}[F(\delta\cdot
I^{\delta}_{[t/\delta]})]\sim\Psi(t)\cdot\Big{(}\inf_{\lambda\in[0,\theta_{F}]}\mathbf{E}\big{[}e^{\lambda\xi_{1}}\big{]}\Big{)}^{n}.$

The remainder of this paper is organized as follows. In Section 2, we recall
some basic elements of fluctuation theory for random walks and then provide
the accurate asymptotic results for the expectation (1.2). We give in Section
3 some auxiliary asymptotic results for random walks, which will be used in
the proofs for our main results. The asymptotic results for the expectation
(1.2) with $S$ oscillating or drifting to $\infty$ under
$\mathbf{P}^{\Lambda}$ are proved separately in Section 4 and 5.

##  2 Preliminaries and main results

In this section we first introduce some basic notation and elements of
fluctuation theory for random walks. We then provide the main results in this
paper about the asymptotic behavior of exponential functionals of random
walks.

###  2.1 Random walks

Suppose that the random walk $S$ is defined on a complete probability space
$(\Omega,\mathscr{F},\mathbf{P})$. Let $\mathscr{F}^{S}$ and
$(\mathscr{F}^{S}_{n})_{n\geq 0}$ denote the $\sigma$-algebra and the
filtration generated by $S$, i.e.,
$\mathscr{F}^{S}:=\sigma(S_{k}:k=0,1,\cdots)$ and
$\mathscr{F}^{S}_{n}:=\sigma(S_{k}:k=0,1,\cdots,n)$. For any probability
measure $\mu$ on $\mathbb{R}$, we denote by $\mathbf{P}_{\mu}$ and
$\mathbf{E}_{\mu}$ the law and expectation of the random walk $S$ with initial
state $S_{0}$ distributed as $\mu$. When $\mu=\delta_{x}$ is a Dirac measure
at point $x\in\mathbb{R}$, we write $\mathbf{P}_{x}$ for
$\mathbf{P}_{\delta_{x}}$ and $\mathbf{E}_{x}$ for $\mathbf{E}_{\delta_{x}}$.
For simplicity, we also write $\mathbf{P}$ for $\mathbf{P}_{0}$ and
$\mathbf{E}$ for $\mathbf{E}_{0}$. According to its behavior as $n\to\infty$,
the random walk $S$ can be exactly classified into three types: (i) drift to
$\infty$: $S_{n}\to\infty$ a.s.; (ii) drift to $-\infty$: $S_{n}\to-\infty$
a.s.; (iii) oscillation:
$\limsup_{n\to\infty}S_{n}=-\liminf_{n\to\infty}S_{n}=\infty$ a.s.

We write $M:=\\{M_{n}:n=1,2,\cdots\\}$ and $L:=\\{L_{n}:n=1,2,\cdots\\}$ for
the running maximum and minimum processes respectively,

$\displaystyle M_{n}:=\max_{1\leq i\leq n}S_{n}\quad\mbox{and}\quad
L_{n}:=\min_{1\leq i\leq n}S_{n}.$

In addition to $M_{n}$ and $L_{n}$, we will also use the following two random
times:

$\displaystyle\sigma^{+}_{n}:=\inf\big{\\{}0\leq i\leq n:S_{i}=S_{0}\vee
M_{n}\big{\\}}\quad\mbox{and}\quad\sigma^{-}_{n}:=\inf\big{\\{}0\leq i\leq
n:S_{i}=S_{0}\wedge L_{n}\big{\\}}$

the first times up to time $n$ that the maximum and minimum are attained. For
$x\in\mathbb{R}$, denote by $\tau_{x}^{+}$ and $\tau_{x}^{-}$ the first
entrance times of $S$ in $(x,\infty)$ and $(-\infty,x)$ respectively, i.e.,

$\displaystyle\tau_{x}^{+}:=\inf\\{n>0:S_{n}>x\\}\quad\mbox{and}\quad\tau_{x}^{-}:=\inf\\{n>0:S_{n}<x\\}.$

In the sequel, we always denote by $\tilde{S}$ an independent copy of $S$. For
the quantities introduced to $S$, the corresponding ones for $\tilde{S}$ are
denoted by tildes, for instance, $\tilde{I}$ and $\tilde{\tau}_{x}^{-}$.

The renewal function $V$ associated with the strict descending ladder height
process is defined by $V(x)=0$ if $x<0$ and

$\displaystyle
V(x):=1+\sum_{n=1}^{\infty}\mathbf{P}(S_{\gamma_{n}^{-}}\geq-x),\quad x\geq
0,$

where $\\{\gamma_{i}^{-}:i=0,1,\cdots\\}$ are strict descending ladder epochs
of $S$ with $\gamma_{0}^{-}=0$ and

$\displaystyle\gamma_{i}^{-}:=\inf\big{\\{}n>\gamma_{i-1}^{-}:S_{n}<S_{\gamma_{i-1}^{-}}\big{\\}},\quad
i\geq 1.$

It is obvious that $V$ is a non-deceasing right-continuous function with
$V(0)=1$. Using the duality lemma, we also have

$\displaystyle
V(x)=1+\sum_{n=1}^{\infty}\mathbf{P}(S_{n}\geq-x,\sigma_{n}^{-}=n)=1+\sum_{n=1}^{\infty}\mathbf{P}(S_{n}\geq-x,M_{n}<0).$

When the random walk $S$ drifts to $-\infty$, we have
$\mathbf{E}_{x}[\tau_{0}^{-}]<\infty$ and
$V(x)=\mathbf{E}_{x}[\tau_{0}^{-}]/\mathbf{E}[\tau_{0}^{-}]$ for any $x\geq
0$.

The dual process of $S$ is denoted by $\hat{S}$, that is $\hat{S}=-S$ when the
starting point is $0$. Let $\hat{\mathbf{P}}_{x}$ be the law of $\hat{S}$
under $\mathbf{P}_{-x}$. For the quantities introduced to $S$, the
corresponding ones for $\hat{S}$ are denoted by hats, for instance $\hat{I}$,
$\hat{\tau}_{x}^{-}$ and so on. Specially, the renewal process $\hat{V}$
associated with the strict descending ladder height process of $\hat{S}$ is
equal to the renewal process associated with the strict ascending ladder
height process of $S$, i.e.

$\displaystyle\hat{V}(x)=1+\sum_{n=1}^{\infty}\mathbf{P}(S_{n}\leq
x,\sigma_{n}^{+}=n)=1+\sum_{n=1}^{\infty}\mathbf{P}(S_{n}\leq x,L_{n}>0).$

When $S$ drifts to $\infty$, we also have
$\mathbf{E}_{x}[\tau_{0}^{+}]<\infty$ and
$\hat{V}(x)=\mathbf{E}_{x}[\tau_{0}^{+}]/\mathbf{E}[\tau_{0}^{+}]$ for any
$x\geq 0$.

If the random walk $S$ does not drift to $-\infty$, the process
$\\{V(S_{n})\mathbf{1}_{\\{\tau_{0}^{-}>n\\}}:n=0,1,\cdots\\}$ is a
$\mathbf{P}_{x}$-martingale for any $x\geq 0$. In this case, we introduce a
probability $\mathbf{P}^{\uparrow}_{x}$ on
$(\Omega,\mathscr{F},\mathscr{F}^{S})$ defined by

$\displaystyle\mathbf{P}^{\uparrow}_{x}(A):=\int_{A}V(S_{n})\mathbf{1}_{\\{\tau_{0}^{-}>n\\}}\frac{d\mathbf{P}_{x}}{V(x)},\quad
A\in\mathscr{F}^{S}_{n},n\geq 0.$

It is usually well-known as Doob’s $h$-transform of $\mathbf{P}_{x}$.
Particularly, under $\mathbf{P}^{\uparrow}$ the process $S$ turns to be a
homogeneous Markov process on $[0,\infty)$ with transition function

$\displaystyle p^{\uparrow}(x,dy):=\frac{V(y)}{V(x)}\mathbf{P}(x+X\in
dy),\quad x,y\geq 0.$

Similarly, in the case that $\hat{S}$ does not drift to $-\infty$, for any
$x\geq 0$ we can also introduce another Doob’s $h$-transform
$\mathbf{P}^{\downarrow}_{x}$ of $\hat{\mathbf{P}}_{x}$ on
$(\Omega,\mathscr{F},\mathcal{F}^{\hat{S}}_{n})$ by the renewal function
$\hat{V}$, i.e.,

$\displaystyle\mathbf{P}^{\downarrow}_{x}(A):=\int_{A}\hat{V}(\hat{S}_{n})\mathbf{1}_{\\{\hat{\tau}_{0}^{-}>n\\}}\frac{d\hat{\mathbf{P}}_{x}}{\hat{V}(x)},\quad
A\in\mathscr{F}^{\hat{S}}_{n},n\geq 0.$

Under $\mathbf{P}^{\downarrow}$ the process $\hat{S}$ turns to be a
homogeneous Markov process on $[0,\infty)$ with transition function

$\displaystyle
p^{\downarrow}(x,dy):=\frac{\hat{V}(y)}{\hat{V}(x)}\mathbf{P}(x-X\in dy),\quad
x,y\geq 0.$

In order to make our following statements much easier to be understood, we
write $S^{\uparrow}$ and $S^{\downarrow}$ for the two independent Markov
processes with transition functions $p^{\uparrow}$ and $p^{\downarrow}$
respectively. Their exponential functionals are denoted as $I^{\uparrow}$ and
$I^{\downarrow}$. Moreover, we also write $\hat{S}^{\uparrow}$ and
$\hat{S}^{\downarrow}$ for the dual processes of $S^{\uparrow}$ and
$S^{\downarrow}$ respectively and also write $\hat{I}^{\uparrow}$ and
$\hat{I}^{\downarrow}$ for their exponential functionals. Repeating the
preceding argument, we see that $\hat{S}^{\uparrow}$ and
$\hat{S}^{\downarrow}$ are two homogeneous Markov processes with transition
functions

$\displaystyle\hat{p}^{\uparrow}(x,dy):=\frac{\hat{V}(-y)}{\hat{V}(-x)}\mathbf{P}(x+X\in
dy)\quad\mbox{and}\quad\hat{p}^{\downarrow}(x,dy):=\frac{V(-y)}{V(-x)}\mathbf{P}(x-X\in
dy),\quad x,y\leq 0.$

As we have mentioned before, the asymptotics of the tail distribution of
generic step $X$ plays a crucial role in the following classification and
asymptotic analysis of exponential functionals of random walks. Thus we need
the Laplace transform of $X$

$\displaystyle\mathcal{L}_{X}(\lambda):=\mathbf{E}\big{[}e^{\lambda
X}\big{]},\quad\lambda\in\mathbb{R}.$

Certainly, it may happen that $\mathcal{L}_{X}(\lambda)=\infty$ for some
$\lambda\in\mathbb{R}$. Let
$\mathcal{D}_{\it\Phi}:=\\{\lambda\in\mathbb{R}:{\it\Phi}(\lambda)\leq\infty\\}$,
$\mathcal{D}^{+}_{\it\Phi}:=\mathcal{D}_{\it\Phi}\cap[0,\infty)$ and
$\mathcal{D}^{-}_{\it\Phi}:=\mathcal{D}_{\it\Phi}\cap(-\infty,0]$. For each
$\lambda_{0}\in\mathcal{D}_{\it\Phi}$, the process
$\\{\exp\\{\lambda_{0}S_{n}\\}\cdot|\mathcal{L}_{X}(\lambda_{0})|^{-n}:n=0,2,\cdots\\}$
is a $\mathbf{P}$ martingale. Thus we can define a probability measure
$\mathbf{P}^{(\lambda_{0})}$ on $(\Omega,\mathscr{F},\mathscr{F}^{S})$ as
follows

$\displaystyle\mathbf{P}^{(\lambda_{0})}(A):=\int_{A}\exp\\{\lambda_{0}S_{n}\\}\cdot|\mathcal{L}_{X}(\lambda_{0})|^{-n}d\mathbf{P},\quad
A\in\mathscr{F}^{S}_{n},n\geq 0.$

It is known that the process $S$ under $\mathbf{P}^{(\lambda_{0})}$ is still a
random walk and the generic step $X$ has Laplace transform

$\displaystyle\mathbf{E}^{(\lambda_{0})}[e^{\lambda
X}]=\mathcal{L}_{X}(\lambda_{0}+\lambda)/\mathcal{L}_{X}(\lambda_{0}),\quad\lambda\in\mathbb{R}.$

Let $V^{(\lambda_{0})}$ and $\hat{V}^{(\lambda_{0})}$ be the renewal functions
associated with the strict descending ladder height processes of $S$ and
$\hat{S}$ under $\mathbf{P}^{(\lambda_{0})}$.

###  2.2 Main results

We now provide the asymptotic results for the expectation
$\mathbf{E}[F(I_{n})]$ in which $F$ is a positive, bounded 222The assumption
on the upper bound for $F$ is not necessary if $\theta_{F}\in\mathcal{D}_{F}$.
function on $(0,\infty)$ and always satisfies the following assumption:

###### Assumption 2.1

The set $\mathcal{D}_{F}:=\\{\theta>0:\sup_{x>0}x^{\theta}F(x)<\infty\\}$ is
not null and the supremum is denoted as $\theta_{F}\in(0,\infty]$.

To simplify the representation of our main results, let us list the following
conditions:

###### Condition 2.2

For each $\delta>0$, there exists a constant $C_{\delta}>0$ such that
$|F(x)-F(y)|\leq C_{\delta}|x-y|$ for any $x,y\geq\delta$.

###### Condition 2.3

There exist a constant $K_{0}>0$ such that $F(x)\sim K_{0}x^{-\theta_{F}}$ as
$x\to\infty$.

It is known that when $S$ does not drift to $\infty$, with probability one it
visits the negative half-line infinite times and hence $I_{\infty}=\infty$
a.s. For converse, when $S$ drifts to $\infty$, Erickson’s theorem in [19]
shows that $\lim_{n\to\infty}S_{n}/n\in(0,\infty]$ a.s. and hence
$I_{\infty}\leq\infty$ a.s. The first asymptotic result for exponential
functionals of random walks is summarized as follows.

###### Lemma 2.4

The following assertions are equivalent:
(i) $I_{\infty}<\infty$ a.s.; (ii) $\mathbf{P}(I_{\infty}<\infty)>0$; (iii)
$S$ drifts to $\infty$, i.e., $S_{n}\to\infty$ a.s.

We in this paper are mainly interested in the case $I_{\infty}=\infty$ a.s.,
that is the random walk $S$ will not drift to $\infty$. As we have mentioned
before, beside of the decay rate of $F$ the long-term behavior of the
expectation $\mathbf{E}[F(I_{n})]$ also heavily depends on the speed at which
the random walk $S$ approaches to $-\infty$. The fluctuation of $S$ is closely
related to the following crucial quantity:

$\displaystyle\varrho:=\inf_{\lambda\in[0,\theta_{F}]}\mathcal{L}_{X}(\lambda).$

It is obvious that $\varrho\leq 1$ and the equality holds if and only if
$\mathcal{D}_{\it\Phi}^{+}=\\{0\\}$. In the sequel of this paper we always
make the following assumption:

###### Assumption 2.5

The infimum of $\mathcal{L}_{X}(\lambda)$ over $[0,\theta_{F}]$ can be
attained, i.e.,

$\displaystyle\Lambda:=\operatorname*{arg\,inf}_{\lambda\in[0,\theta_{F}]}\mathcal{L}_{X}(\lambda)\in\mathcal{D}_{\it\Phi}.$

####  2.2.1 The oscillating cases

We now provide the asymptotic results for the expectation
$\mathbf{E}[F(I_{n})]$ with the random walk $S$ being oscillating under
$\mathbf{P}^{(\Lambda)}$. We first consider the case in which $\Lambda=0$ and
the well-known Spitzer’s condition holds for $S$.

###### Condition 2.6

There exists a constant $\rho\in(0,1)$ such that

$\displaystyle\frac{1}{n}\sum_{n=1}^{\infty}\mathbf{P}(S_{n}>0)\to\rho.$

Spitzer’s condition is of key importance in fluctuation theory of random walks
and it is equivalent to the convergence $\mathbf{P}(S_{n}>0)\to\rho$; see
Theorem 1 in [17]. Specially, all symmetric random walks satisfy Spitzer’s
condition with $\rho=1/2$. To show the exact decay rate of the expectation
$\mathbf{E}[F(I_{n})]$, we need the positive, slowly varying function
$\ell_{1}$ defined by

$\displaystyle\ell_{1}(x):=\frac{1}{\Gamma(\rho)}\exp\Big{\\{}\sum_{k=1}^{\infty}\frac{(1-1/x)^{k}}{k}\big{(}\mathbf{P}(S_{k}\geq
0)-\rho\big{)}\Big{\\}},\quad x\geq 1$

###### Theorem 2.7

If $\theta_{F}>\Lambda=0$, under Condition 2.2 and 2.6, we have as
$n\to\infty$,

$\displaystyle\mathbf{E}[F(I_{n})]\sim
C_{F,1}\cdot\mathbf{P}(\tau_{0}^{-}>n)\sim C_{F,1}\cdot
n^{\rho-1}\ell_{1}(n),$ (2.3)

where the limit coefficient $C_{F,1}$ is given by

$\displaystyle
C_{F,1}=\sum_{k=0}^{\infty}\mathbf{E}\big{[}F\big{(}I_{k}+e^{-S_{k}}\cdot
I^{\uparrow}_{\infty});\sigma_{k}^{-}=k\big{]}\in(0,\infty).$ (2.4)

###### Remark 2.8

The positiveness of $C_{F,1}$ results from the fact that
$I^{\uparrow}_{\infty}<\infty$ a.s., which is a direct consequence of the
limit $\liminf_{n\to\infty}n^{-\rho/2}S^{\uparrow}_{n}>0$ a.s.; see (2.22) in
[4]. Similarly, the other three random variables $I^{\downarrow}_{\infty}$,
$\hat{I}^{\uparrow}_{\infty}$ and $\hat{I}^{\downarrow}_{\infty}$ are also
finite almost surely.

We now turn to consider the asymptotics of the expectation
$\mathbf{E}[F(I_{n})]$ with $0<\Lambda<\theta_{F}$. Using the change of
measure, we have

$\displaystyle\mathbf{E}\big{[}F(I_{n})\big{]}=\varrho^{n}\cdot\mathbf{E}^{(\Lambda)}\big{[}e^{-\Lambda
S_{n}}F(I_{n})\big{]},$ (2.5)

which indicates that different to the previous case, the long-term behavior of
the expectation $\mathbf{E}\big{[}F(I_{n})\big{]}$ depend on not only the
sample paths with slowly decreasing local minimum but also the sample paths
with final value $S_{n}$ below a low level. In order to get an exact
description for the distribution of $S_{n}$, we need the following condition.

###### Condition 2.9

Under $\mathbf{P}^{(\Lambda)}$ the random walk $S$ is in the domain of
attraction of a strictly stable law without centering with index
$\alpha\in(0,2]$ and positivity parameter $\rho\in(0,1)$, we write
$S\in\mathcal{D}_{\alpha,\rho}^{(\Lambda)}$.

This condition is usually stronger than Spitzer’s condition. Indeed, if
$S\in\mathcal{D}_{\alpha,\rho}^{(\Lambda)}$ we have
$\mathbf{P}^{(\Lambda)}(S_{n}>0)\to\rho$. For the converse, Spitzer’s
condition usually does not imply a domain of attraction; readers may refer to
[12, p.380] for more details. In particular,
$S\in\mathcal{D}_{\alpha,\rho}^{(\Lambda)}$ with $\alpha\in(1,2]$ if and only
if Spitzer’s condition holds for $S$ under $\mathbf{P}^{(\Lambda)}$, and in
this case we always have $\rho=1-1/\alpha$; see Proposition 8.9.16 in [12,
p.384]. Moreover, Condition 2.9 holds if and only if there exists a function
$\ell_{2}$ that is slowly varying at $\infty$ such that under
$\mathbf{P}^{(\Lambda)}$,

$\displaystyle\frac{S_{n}}{n^{1/\alpha}\ell_{2}(n)}\to Y_{\alpha}$

in distribution with $Y_{\alpha}$ a strictly stable random variable of
parameter $\alpha$ and positivity parameter $\rho$. To simplify the following
statements, we define

$\displaystyle A_{n}:=n^{-1-1/\alpha}/\ell_{2}(n),\quad n=1,2,\cdots.$

The following quantities are necessary to give an exact decay rate for the
expectation $\mathbf{E}[F(I_{n})]$. We write
$\\{g_{\alpha}(x):x\in\mathbb{R}\\}$ for the probability density function of
$Y_{\alpha}$. The fact that $V^{(\Lambda)}(x)=O(x)$ and
$\hat{V}^{(\Lambda)}(x)=O(x)$ as $x\to\infty$ induces that the following two
integrals are finite:

$\displaystyle\mathcal{L}_{V^{(\Lambda)}}^{(\lambda)}(y):=\int_{0}^{y}e^{-\lambda
z}V^{(\Lambda)}(z)dz\quad\mbox{and}\quad\mathcal{L}_{\hat{V}^{(\Lambda)}}^{(\lambda)}(y):=\int_{0}^{y}e^{-\lambda
z}\hat{V}^{(\Lambda)}(z)dz,\quad\lambda,y>0.$

Let $\mu_{V^{(\Lambda)}}^{(\lambda)}(dx)$ and
$\mu_{\hat{V}^{(\Lambda)}}^{(\lambda)}(dx)$ be two probability measures on
$\mathbb{R}_{+}$ defined by

$\displaystyle\mu_{V^{(\Lambda)}}^{(\lambda)}(dx):=\frac{e^{-\lambda
x}V^{(\Lambda)}(x)}{\mathcal{L}_{V^{(\Lambda)}}^{(\lambda)}(\infty)}dx\quad\mbox{and}\quad\mu_{\hat{V}^{(\Lambda)}}^{(\lambda)}(dx):=\frac{e^{-\lambda
x}\hat{V}^{(\Lambda)}(x)}{\mathcal{L}_{\hat{V}^{(\Lambda)}}^{(\lambda)}(\infty)}dx.$

###### Theorem 2.10

If $\theta_{F}>\Lambda>0$, under Condition 2.2 and 2.9, we have as
$n\to\infty$,

$\displaystyle\mathbf{E}[F(I_{n})]\sim
C_{F,2}\cdot\mathbf{P}(\tau_{0}^{-}>n)\sim C_{F,2}\cdot
g_{\alpha}(0)\mathcal{L}_{\hat{V}^{(\Lambda)}}^{(\Lambda)}(\infty)\cdot\varrho^{n}\cdot
A_{n},$ (2.6)

where the limit coefficient 333The positiveness of $C_{F,2}$ is a direct
consequence of the finiteness of $I_{\infty}^{\uparrow}$ and
$I_{\infty}^{\downarrow}$; see Remark 2.8.  $C_{F,2}\in(0,\infty)$ is given by
444$\mathbf{E}_{(x,y,z)}^{(\Lambda)}$ is the expectation of
$(S,S^{\uparrow},S^{\downarrow})$ under $\mathbf{P}^{(\Lambda)}$ with initial
state $(x,y,z)$.

$\displaystyle C_{F,2}\\!\\!\\!$ $\displaystyle=\\!\\!\\!$
$\displaystyle\sum_{k=0}^{\infty}\Big{[}\int_{0}^{\infty}\mathbf{E}_{(0,0,y)}^{(\Lambda)}\big{[}e^{-\Lambda
S_{k}}F\big{(}I_{k}+e^{-S_{k}}(I^{\uparrow}_{\infty}+e^{-y}+I^{\downarrow}_{\infty})\big{)};\sigma_{k}^{-}=k\big{]}\mu_{\hat{V}^{(\Lambda)}}^{(\Lambda)}(dy)$
$\displaystyle+\frac{\mathcal{L}_{V^{(\Lambda)}}^{(\eta)}(\infty)}{\mathcal{L}_{\hat{V}^{(\Lambda)}}^{(\Lambda)}(\infty)}\int_{0}^{\infty}e^{(\eta+\Lambda)z}\mathbf{E}_{(0,0,z)}^{(\Lambda)}\big{[}e^{-\Lambda
S_{k}}F\big{(}e^{z}(1+I_{k}+I_{\infty}^{\uparrow}+I_{\infty}^{\downarrow})\big{)};\tau_{0}^{-}>k\big{]}\mu_{V^{(\Lambda)}}^{(\eta)}(dz)\Big{]}$

for any $\eta\in(0,\theta_{F}-\Lambda)$ and it is independent of the choice of
$\eta$.

We now continue to consider the asymptotics of the expectation
$\mathbf{E}[F(I_{n})]$ with $\theta_{F}=\Lambda>0$. In this case, we find that
the impact of sample paths with small final value on the expectation on the
right side of (2.5) is more complicated than that in the preceding case. In
precise, if the random walk ends up below a low level at time $n$, we observe
that its local minimum over the time interval $[0,n]$ is not only
approximately equal to the final value but also prefers be attained at the end
of the time interval, and consequently its contribution to the expectation
$\mathbf{E}[F(I_{n})]$ can be roughly represented as

$\displaystyle e^{-\Lambda S_{n}}F(I_{n})=O\big{(}e^{-\Lambda
S_{n}}(I_{n}-I_{\sigma_{n}^{-}-1})^{-\Lambda}\big{)}=O\big{(}(n-\sigma_{n}^{-})^{-\Lambda}\big{)}.$
(2.7)

These make it difficult to provide an exact description for the decay rate of
the expectation $\mathbf{E}[F(I_{n})]$. To keep away from these difficulties,
we consider the aysmptotics of the expectation $\mathbf{E}[F(I_{n})]$ with $F$
satisfying Condition 2.3. Let $\hat{\ell}_{1}^{(\Lambda)}$ be a positive,
slowly varying function defined by

$\displaystyle\hat{\ell}_{1}^{(\Lambda)}(x)\\!\\!\\!$
$\displaystyle=\\!\\!\\!$
$\displaystyle\frac{1}{\Gamma(1-\rho)}\exp\Big{\\{}-\sum_{k=1}^{\infty}\frac{(1-1/x)^{k}}{k}\big{(}\mathbf{P}^{(\Lambda)}(S_{k}>0)-\rho\big{)}\Big{\\}},\quad
x\geq 1$

###### Theorem 2.11

If $\theta_{F}=\Lambda>0$, under Condition 2.3 and 2.9, we have as
$n\to\infty$,

$\displaystyle\mathbf{E}[F(I_{n})]\sim K_{0}\cdot
C_{F,3}\cdot\varrho^{n}\cdot\mathbf{P}^{(\Lambda)}(\tau_{0}^{+}>n)\sim
K_{0}\cdot C_{F,3}\cdot\varrho^{n}\cdot
n^{-\rho}\hat{\ell}_{1}^{(\Lambda)}(n),$ (2.8)

where the coefficient $C_{F,3}\in(0,\infty)$ is given by

$\displaystyle
C_{F,3}=\sum_{k=0}^{\infty}\mathbf{E}^{(\Lambda)}\big{[}\big{(}1+\hat{I}_{k}+e^{-\hat{S}_{k}}\cdot\hat{I}^{\uparrow}_{\infty})^{-\Lambda};\hat{\sigma}_{k}^{-}=k\big{]}\in(0,\infty).$
(2.9)

####  2.2.2 The negative drift cases

We now provide the asymptotic results for the expectation
$\mathbf{E}[F(I_{n})]$ in which under $\mathbf{P}^{(\Lambda)}$ the random walk
$S$ has negative drift, i.e.,
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}a:=-\mathbf{E}^{(\Lambda)}[X]}>0$.
We first consider the case with $\Lambda=\theta_{F}>0$ and $F$ satisfying
Condition 2.3. We notice that the expectation $\mathbf{E}[F(I_{n})]$ is
asymptotically equivalent to $K_{0}\cdot\mathbf{E}[I_{n}^{-\theta_{F}}]$.
Using the change of measure and then the duality lemma, we have

$\displaystyle\mathbf{E}[I_{n}^{-\Lambda}]\sim\mathbf{E}^{(\Lambda)}[(1+\hat{I}_{n-1})^{-\theta_{F}}]\cdot\varrho^{n}.$
(2.10)

By Lemma 2.4, as $n\to\infty$ we have
$\hat{I}_{n-1}\to\hat{I}_{\infty}<\infty$ a.s. under $\mathbf{P}^{(\Lambda)}$
and hence $\mathbf{E}^{(\Lambda)}[(1+\hat{I}_{\infty})^{-\theta_{F}}]>0$.

###### Theorem 2.12

If $\Lambda=\theta_{F}>0$, under Condition 2.3, we have as $n\to\infty$,

$\displaystyle\mathbf{E}[F(I_{n})]\sim
K_{0}\cdot\mathbf{E}^{(\Lambda)}[(1+\hat{I}_{\infty})^{-\theta_{F}}]\cdot
e^{\Phi(\theta_{F})\cdot n}.$ (2.11)

We now provide the asymptotic results for the expectation
$\mathbf{E}[F(I_{n})]$ with $\theta_{F}>\Lambda\geq 0$ and the distribution of
generic step $X$ always satisfying the following two regularly varying
assumptions:

###### Assumption 2.13

For some constant $\beta>1$, the tail-probability
$\mathbf{P}^{(\Lambda)}(X>x)$ is regularly varying with index $-\beta$, i.e.,
there exists a slowly varying function $\ell_{3}(x)$ at $\infty$ such that
$\mathbf{P}^{(\Lambda)}(X>x)\sim x^{-\beta}\ell_{3}(x)$ as $x\to\infty$,

###### Assumption 2.14

For every $\delta>0$, we have
$\mathbf{P}^{(\Lambda)}(X\in(x,x+\delta])\sim\beta
x^{-1-\beta}\ell_{3}(x)\cdot\delta$ as $x\to\infty$.

When $\Lambda=0$, we first observe that the random walk prefers to attain the
local minimum around the arrival of the first large step. Because of the
negative drift, its local minimum will decrease slowly if there is a large
step occurring at the beginning, otherwise it drifts to $-\infty$ very fast
and its contribution to the expectation $\mathbf{E}[F(I_{n})]$ can be
asymptotically ignored. By analyzing the contribution of simple paths before
and after the first large step separately, we give in the next theorem, the
exact decay rate of the expectation $\mathbf{E}[F(I_{n})]$.

###### Theorem 2.15

If $\theta_{F}>\Lambda=0$, assume $F$ is non-increasing 555This theorem still
holds if $F:=F_{+}-F_{-}$ with $F_{+},F_{-}$ being bounded, non-increasing and
satisfying Condition 2.2. and satisfies Condition 2.2, we have as
$n\to\infty$,

$\displaystyle\mathbf{E}[F(I_{n})]\sim\frac{C_{F,4}}{\mathbf{E}[\tau_{0}^{-}]}\cdot\mathbf{P}(\tau_{0}^{-}>n)\sim
C_{F,4}\cdot\mathbf{P}(X\geq an),$ (2.12)

where the coefficient
$C_{F,4}:=\sum_{k=1}^{\infty}\mathbf{E}[C_{F,4}(k)]\in(0,\infty)$ with

$\displaystyle
C_{F,4}(k):=\lim_{n\to\infty}\mathbf{E}\big{[}F(I_{k-1}+e^{-S_{k-1}-X}\tilde{I}_{n})\,\big{|}\,X\geq
an,\mathscr{F}^{S}\big{]}.$ (2.13)

We now start to consider the case $\theta_{F}>\Lambda>0$. As our preceding
asymptotic analysis for the random walk in the previous case, the slow
decreasing of its local minimum also stems from the early arrival of a large
step. Moreover, by (2.5) we see that the expectation $\mathbf{E}[F(I_{n})]$ is
also seriously effected by the final value $S_{n}$. However, we observe that
the random walk would stay above a high level for a long time after the large
step and hence its final value is more likely above a high level, which gives
rise to its meager contribution to the expectation $\mathbf{E}[F(I_{n})]$.
Consequently, the main contribution to the expectation $\mathbf{E}[F(I_{n})]$
is made by the simple paths with not only an early large step but also small
final value. In order to get an exact relationship between the local minimum
and the final value, we need the following additional technic condition.

###### Condition 2.16

Assume that $\beta\neq 2$ and $\mathbf{E}[|X|^{\kappa}]<\infty$ for some
$\kappa\in(1,2)$.

Under this condition, we have $n^{-1/\kappa}(S_{n}+an)$ converges to $0$ in
distribution by the Kolmogorov-Marcinkiewicz-Zygmund law of large numbers; see
Theorem 10.3 in [21, p.311]. These tell that the final value $S_{n}$ falls
into a bounded interval around $0$ if and only if the size of the early large
step is about $an$. In order to simplify the notation, we define the following
quantities:

$\displaystyle B_{n}=\frac{\beta}{an}\mathbf{P}^{(\Lambda)}(X\geq an),\quad
n=1,2,\cdots.$ (2.14)

###### Theorem 2.17

If $\theta_{F}>\Lambda>0$, under Condition 2.2 and 2.16, we have as
$n\to\infty$,

$\displaystyle\mathbf{E}[F(I_{n})]\sim\frac{C_{F,5}}{\mathcal{L}_{\hat{V}^{(\Lambda)}}^{(\Lambda)}(\infty)}\mathbf{P}(\tau_{0}^{-}>n)\sim
C_{F,5}\cdot\varrho^{n}\cdot B_{n},$ (2.15)

where the coefficient $C_{F,5}:=\sum_{k=1}^{\infty}C_{F,5}(k)\in(0,\infty)$
with

$\displaystyle
C_{F,5}(k):=\int_{-\infty}^{\infty}\mathbf{E}^{(\Lambda)}\big{[}e^{-\Lambda
S_{k}-\Lambda
z}F\big{(}I_{k}+e^{-S_{k}-z}(1+\hat{\tilde{I}}_{\infty})\big{)}\big{]}dz.$
(2.16)

##  3 Auxiliary results for random walks

In this section we list as well as generalize some well-known asymptotic
results for random walks under conditions and assumptions introduced in
Section 2 with $\Lambda=0$, e.g., asymptotics of the first passage times,
local probabilities conditioned to stay positive and conditional limit
theorems.

###### Remark 3.1

The right-continuity of $V$ and $\hat{V}$ yields that all the following claims
still hold with $\tau_{0}^{-}>n$ and $\tau_{0}^{+}>n$ replaced by $L_{n}>0$
and $M_{n}<0$ respectively.

We first recall a useful asymptotic result for regularly varying sequences;
readers can find it in [6].

###### Lemma 3.2

Let $\\{b_{n}\\}$ be a regularly varying sequence. Consider two summable
sequences $\\{f_{n}:n=0,1,\cdots\\}$ and $\\{g_{n}:n=0,1,\cdots\\}$ satisfying
that $f_{n}\sim c_{1}\cdot b_{n}$ and $g_{n}\sim c_{2}\cdot b_{n}$ with
$c_{1},c_{2}$ being two nonnegative constants. We have as $n\to\infty$,

$\displaystyle\sum_{k=0}^{n}f_{n-k}g_{k}\sim\Big{(}c_{1}\sum_{k=0}^{\infty}g_{k}+c_{2}\sum_{k=0}^{\infty}f_{k}\Big{)}\cdot
b_{n}.$

###  3.1 Oscillating random walk

The fluctuation theory for oscillating random walks satisfying Spitzer’s
condition has been well developed and abundant results have been gotten. Here
we list a part of them that will be used in our following proofs. Recall the
two slowly varying functions $\ell_{1}$ and $\hat{\ell}_{1}$. A simple
calculation shows that $\ell_{1}(x)\hat{\ell}_{1}(x)$ converges to a positive
constant as $x\to\infty$. The next lemma comes from Theorem 8.9.12 in [12,
p.381] and Lemma 2.1 in [4].

###### Lemma 3.3

Under Condition 2.6, for every $x\geq 0$ we have as $n\to\infty$,

$\displaystyle\mathbf{P}_{x}(\tau_{0}^{-}>n)\sim
V(x)n^{\rho-1}\ell_{1}(n)\quad\mbox{and}\quad\mathbf{P}_{-x}(\tau_{0}^{+}>n)\sim\hat{V}(x)n^{-\rho}\hat{\ell}_{1}(n).$

Moreover, there exists a constant $C>0$ such that for all $x\geq 0$ and $n\geq
0$,

$\displaystyle\mathbf{P}_{x}(\tau_{0}^{-}>n)\leq
CV(x)n^{\rho-1}\ell_{1}(n)\quad\mbox{and}\quad\mathbf{P}_{-x}(\tau_{0}^{+}>n)\leq
C\hat{V}(x)n^{-\rho}\hat{\ell}_{1}(n).$

The following result is a direct consequence of the proof for Lemma 2.2 in [4]
with $u(x)=e^{-\lambda x}$.

###### Lemma 3.4

Under Condition 2.6, we have for every $\lambda>0$, both of the two sequences

$\displaystyle\mathbf{E}[e^{\lambda
S_{n}}:\sigma_{n}^{-}=n]=\mathbf{E}[e^{\lambda
S_{n}}:M_{n}<0]\quad\mbox{and}\quad\mathbf{E}[e^{-\lambda
S_{n}}:\sigma_{n}^{+}=n]=\mathbf{E}[e^{-\lambda S_{n}}:L_{n}>0]$

are summable and can be uniformly bounded by $c_{0}/n$ for some constant
$c_{0}>0$.

The first conditional limit theorem for random walks in the next lemma was
proved by Bertoin and Doney [11] and the second one can be gotten immediately
by using the duality lemma.

###### Lemma 3.5

For $k\geq 1$, let $f$ be a bounded function on $\mathbb{R}^{k+1}$ and
$f(S):=f(S_{0},S_{1},\cdots,S_{k})$. Under Condition 2.6, for any $x\geq 0$ we
have as $n\to\infty$,

$\displaystyle\mathbf{E}_{x}[f(S)\,|\,\tau_{0}^{-}>n]\to\mathbf{E}_{x}[f(S^{\uparrow})]\quad\mbox{and}\quad\mathbf{E}_{-x}[f(S)\,|\,\tau_{0}^{+}>n]\to\mathbf{E}_{-x}[f(\hat{S}^{\uparrow})],$

where
$f(S^{\uparrow}):=f(S^{\uparrow}_{0},S^{\uparrow}_{1},\cdots,S^{\uparrow}_{k})$
and
$f(\hat{S}^{\uparrow}):=f(\hat{S}^{\uparrow}_{0},\hat{S}^{\uparrow}_{1},\cdots,\hat{S}^{\uparrow}_{k})$.

By Remark 2.8, we see that conditioned to stay positive the random walk $S$
drifts to $\infty$ a.s. To meet the needs of the following proofs for our main
theorems, we provide in the following lemma, some large deviation estimates
and uniform upper estimates for the final value of $S$ conditioned to stay
positive. It can be proved by slightly extending and modifying the proofs for
Proposition 2.1 and Corollary 2.4 in [2], who considers the case with
$\alpha\in(1,2]$.

###### Lemma 3.6

Suppose $S\in\mathcal{D}_{\alpha,\rho}$ with $\alpha\in(0,2]$ and
$\rho\in(0,1)$. For any $\lambda>0$, $x\geq 0$ and $y\in[0,\infty]$, we have
as $n\to\infty$,

$\displaystyle\mathbf{E}_{x}[e^{-\lambda S_{n}};S_{n}\leq
y,\tau_{0}^{-}>n]\sim
g_{\alpha}(0)V(x)\mathcal{L}_{\hat{V}}^{(\lambda)}(y)\cdot A_{n},$
$\displaystyle\mathbf{E}_{-x}[e^{\lambda
S_{n}};S_{n}\geq-y,\tau_{0}^{+}>n]\sim
g_{\alpha}(0)\hat{V}(x)\mathcal{L}_{V}^{(\lambda)}(y)\cdot A_{n}.$

Moreover, there exists a constant $C>0$ such that uniformly in $x,y\geq 0$ and
$n\geq 1$,

$\displaystyle\mathbf{E}_{x}[e^{-\lambda S_{n}};S_{n}\leq
y,\tau_{0}^{-}>n]\\!\\!\\!$ $\displaystyle\leq\\!\\!\\!$ $\displaystyle C\cdot
V(x)\mathcal{L}_{\hat{V}}^{(\lambda)}(y)\cdot A_{n},$
$\displaystyle\mathbf{E}_{-x}[e^{\lambda
S_{n}};S_{n}\geq-y,\tau_{0}^{+}>n]\\!\\!\\!$ $\displaystyle\leq\\!\\!\\!$
$\displaystyle C\cdot\hat{V}(x)\mathcal{L}_{V}^{(\lambda)}(y)\cdot A_{n}.$

Let $C_{0}(\mathbb{R}^{2})$ be the space of continuous functions on
$\mathbb{R}^{2}$ vanishing at infinity. The following lemma generalizes the
conditional limit theorem from Lemma 10 in [23].

###### Lemma 3.7

Suppose $S\in\mathcal{D}_{\alpha,\rho}$ with $\alpha\in(0,2]$ and
$\rho\in(0,1)$. Let $G\in C_{0}(\mathbb{R}^{2})$ and $f,g$ be two bounded,
continuous functions on $\mathbb{R}^{k+1}$ and $\mathbb{R}^{m+1}$ respectively
for some $k,m\geq 1$. Let $f(S):=f(S_{0},S_{1},\cdots,S_{k})$ and
$g_{n}(S):=g(S_{n},S_{n-1},\cdots,g_{n-m})$ for $n\geq m$. For any $x\geq 0$
and $\lambda>0$ we have as $n\to\infty$,

$\displaystyle\frac{\mathbf{E}_{x}[G(f((S),g_{n}(S))e^{-\lambda
S_{n}};\tau_{0}^{-}>n]}{\mathbf{E}_{x}[e^{-\lambda
S_{n}};\tau_{0}^{-}>n]}\\!\\!\\!$ $\displaystyle\to\\!\\!\\!$
$\displaystyle\int_{0}^{\infty}\mathbf{E}_{(x,y)}[G(f(S^{\uparrow}),g(S^{\downarrow}))]\mu_{\hat{V}}^{(\lambda)}(dy),$
$\displaystyle\frac{\mathbf{E}_{-x}[G(f((S),g_{n}(S))e^{\lambda
S_{n}};\tau_{0}^{+}>n]}{\mathbf{E}_{-x}[e^{\lambda
S_{n}};\tau_{0}^{+}>n]}\\!\\!\\!$ $\displaystyle\to\\!\\!\\!$
$\displaystyle\int_{0}^{\infty}\mathbf{E}_{(-x,-y)}[G(f(\hat{S}^{\uparrow}),g(\hat{S}^{\downarrow}))]\mu_{V}^{(\lambda)}(dy),$

where $f(S^{\uparrow})$ and $f(\hat{S}^{\uparrow})$ are defined as in Lemma
3.5,
$g(S^{\downarrow}):=g(S^{\downarrow}_{0},S^{\downarrow}_{1},\cdots,S^{\downarrow}_{k})$
and
$g(\hat{S}^{\downarrow}):=g(\hat{S}^{\downarrow}_{0},\hat{S}^{\downarrow}_{1},\cdots,\hat{S}^{\downarrow}_{m})$.

Proof. If $G(f(S),g_{n}(S))=f(S)g_{n}(S)$, the desired results can be gotten
immediately by repeating the proof for Lemma 10 in [23]. The general results
follow by the Stone-Weierstrass theorem. $\Box$

###  3.2  Random walk with negative drift

We now list the asymptotic results for random walks with negative drift
satisfying Assumption 2.13 and 2.14. For every $x\geq 0$, we need the
following important stopping time

$\displaystyle\mathcal{T}^{x}:=\inf\\{k\geq 1:X_{k}>x\\}$

the first arrival of step with size larger than $x$. For each $k\geq 1$, a
simple calculation together with Assumption 2.13 induces that the two events
$\mathcal{T}^{x}=k$ and $X_{k}\geq x$ are asymptotically equivalent as
$x\to\infty$, i.e.,

$\displaystyle\mathbf{P}(\mathcal{T}^{x}=k|X_{k}\geq
x)\sim\mathbf{P}(X_{k}\geq x|\mathcal{T}^{x}=k)\to 1,$ (3.17)

and hence $\mathbf{P}(\mathcal{T}^{x}\leq k)\sim k\mathbf{P}(X\geq x)$. The
following well-known asymptotic results for the tail-probabilities of the
final value $S_{n}$ and the first passage time $\tau_{0}^{-}$ can be found in
many literature, e.g., Theorem 5 in [6] and Theorem 2.2 in [16].

###### Lemma 3.8

For every $x\geq 0$, we have as $n\to\infty$,

$\displaystyle\mathbf{P}(S_{n}\geq x)\sim n\mathbf{P}(X\geq
an)\quad\mbox{and}\quad\mathbf{P}_{x}(\tau_{0}^{-}\geq
n)\sim\mathbf{E}_{x}[\tau_{0}^{-}]\mathbf{P}(X\geq an).$

By this lemma and the Markov property, for each integer $k\geq 0$ we have as
$n\to\infty$,

$\displaystyle\mathbf{P}(\sigma_{n}^{-}=k)=\mathbf{P}(\sigma_{k}^{-}=k)\cdot\mathbf{P}(\tau_{0}^{-}>n-k)\sim\mathbf{P}(\sigma_{k}^{-}=k)\mathbf{E}[\tau_{0}^{-}]\cdot\mathbf{P}(X\geq
an).$

The next lemma comes from Remark 3.7 in [32], which extends Theorem 3.2 in
[18]. It shows that an early large step is necessary to keep the random walk
staying positive for a long time.

###### Lemma 3.9

For every $b\in(0,a]$ and integer $k\geq 1$, we have as $n\to\infty$,

$\displaystyle\mathbf{P}(\mathcal{T}^{bn}>n|\tau_{0}^{-}>n)\to
0\quad\mbox{and}\quad\mathbf{P}(\mathcal{T}^{bn}\leq
k|\tau_{0}^{-}>n)\to\mathbf{E}[\tau_{0}^{-}\wedge
k]/\mathbf{E}[\tau_{0}^{-}].$

Moreover, we also observe that the random walk will stay above a high level
for a long time after the first large step. This implies that the local
minimum of the random walk should be attained near the early large step; see
the next lemma and it can be proved by slightly modifying the proof for Lemma
4.6 in [32].

###### Lemma 3.10

Let $b\in(0,a]$ and integer $k\geq 0$. For every $\epsilon>0$, there exist two
integers $n_{0},t_{0}\geq 1$ such that for any $T\geq t_{0}$ and $n\geq
n_{0}$,

$\displaystyle\mathbf{P}(\mathcal{T}^{bn}\geq
T,\sigma_{n}^{-}=k)\leq\epsilon\cdot\mathbf{P}(X\geq an).$

The next lemma gives several large deviation estimates for the random walk $S$
conditioned to stay positive/negative. It comes from Remark 3.16 in [32] and
generalizes Lemma 6 in [30].

###### Lemma 3.11

For any $x,\lambda>0$ and $y\in(0,\infty]$, we have as $n\to\infty$,

$\displaystyle\mathbf{E}_{x}[e^{-\lambda S_{n}};S_{n}\leq
y,\tau_{0}^{-}>n]\\!\\!\\!$ $\displaystyle\sim\\!\\!\\!$ $\displaystyle
V(x)\int_{0}^{y}e^{-\lambda z}\hat{V}(z)dz\cdot B_{n},$
$\displaystyle\mathbf{E}_{-x}[e^{\lambda
S_{n}};S_{n}\geq-y,\tau_{0}^{+}>n]\\!\\!\\!$ $\displaystyle\sim\\!\\!\\!$
$\displaystyle\hat{V}(x)\int_{0}^{y}e^{-\lambda z}V(z)dz\cdot B_{n}.$

The next lemma provides some asymptotic results for joint local probabilities
of the local maximum/minimum ($M_{n}/L_{n}$) and the final value ($S_{n}$).
Its proof is similar to that of Lemma 3.13 and 3.14 in [32].

###### Lemma 3.12

There exists a constant $\mathcal{C}_{0}>0$ such that for any $x,y\geq 0$, we
have as $n\to\infty$,

$\displaystyle\mathbf{E}\Big{[}e^{-\lambda_{1}M_{n}-\lambda_{2}(M_{n}-S_{n})};M_{n}\leq
x,M_{n}-S_{n}\leq y\Big{]}$
$\displaystyle\qquad\sim\mathcal{C}_{0}\int_{0}^{x}\int_{0}^{y}e^{-\lambda_{1}z_{1}-\lambda_{2}z_{2}}\Big{(}\hat{V}(z_{1})dz_{1}dV(z_{2})+d\hat{V}(z_{1})V(z_{2})dz_{2}\Big{)}\cdot
B_{n},$
$\displaystyle\mathbf{E}\Big{[}e^{\lambda_{1}L_{n}+\lambda_{2}(L_{n}-S_{n})};-L_{n}\leq
x,S_{n}-L_{n}\leq y\Big{]}$
$\displaystyle\qquad\sim\mathcal{C}_{0}\int_{0}^{x}\int_{0}^{y}e^{-\lambda_{1}z_{1}-\lambda_{2}z_{2}}\Big{(}V(z_{1})dz_{1}d\hat{V}(z_{2})+dV(z_{1})\hat{V}(z_{2})dz_{2}\Big{)}\cdot
B_{n}.$

##  4 Proof for Theorem 2.7-2.11

In this section we prove the asymptotic results for the expectation
$\mathbf{E}[F(I_{n})]$ with the random walk $S$ being oscillating under
$\mathbf{P}^{(\Lambda)}$. Although the technic difficulties in the following
proofs vary in different cases, one may find that it is the common key point
to observe that the sample paths with slowly decreasing local infimum make the
main contribution to the expectation $\mathbf{E}[F(I_{n})]$. For simplicity,
we may always assume $\theta_{F}\in\mathcal{D}_{F}$. All the following proofs
still work with $\theta_{F}$ replaced by any
$\theta\in\mathcal{D}_{F}\cap(\Lambda,\infty)$.

###  4.1 Proof for Theorem 2.7

Under Condition 2.6, we first notice that sample paths with slowly decreasing
local infimum would attain the local minimum at the beginning of the time
interval. And then, we prove in the next proposition, that the contribution of
sample paths with the local minimum being late attained to the expectation
$\mathbf{E}[F(I_{n})]$ can be asymptotically ignored.

###### Proposition 4.1

For every $\epsilon>0$, there exist two integers $k_{0},n_{0}\geq 1$ such that
for any $K\geq k_{0}$ and $n\geq n_{0}$,

$\displaystyle\mathbf{E}\big{[}F(I_{n});\sigma_{n}^{-}\geq
K\big{]}\leq\epsilon\cdot\mathbf{P}(\tau_{0}^{-}>n).$

Moreover, there exists a constant $C>0$ such that for any $n\geq 0$,

$\displaystyle\mathbf{E}\big{[}F(I_{n})\big{]}\leq
C\cdot\mathbf{P}(\tau_{0}^{-}>n).$

Proof. By Assumption 2.1, we first have $F(I_{n})\leq
C\cdot\exp\\{\theta_{F}L_{n}\\}$. Using the Markov property of $S$, we have

$\displaystyle\mathbf{E}\big{[}F(I_{n});\sigma_{n}^{-}\geq K\big{]}\\!\\!\\!$
$\displaystyle\leq\\!\\!\\!$ $\displaystyle
C\sum_{k=K}^{n}\mathbf{E}\big{[}e^{\theta_{F}S_{k}};\sigma_{k}^{-}=k\big{]}\cdot\mathbf{P}(\tau_{0}^{-}>n-k).$

Applying Lemma 3.2 together with Lemma 3.3 and 3.4 to the foregoing partial
sum, we have for large $n$,

$\displaystyle\mathbf{E}\big{[}F(I_{n});\sigma_{n}^{-}\geq K\big{]}\\!\\!\\!$
$\displaystyle\leq\\!\\!\\!$ $\displaystyle
C\sum_{k=K}^{\infty}\mathbf{E}\big{[}e^{\theta_{F}S_{k}};\sigma_{k}^{-}=k\big{]}\cdot\mathbf{P}(\tau_{0}^{-}>n).$

The desired two claims follow directly from the summability of the sequence
$\mathbf{E}[e^{\theta_{F}S_{k}};\sigma_{k}^{-}=k]$; see Lemma 3.4, and the
fact that $\mathbf{E}\big{[}F(I_{n});\sigma_{n}^{-}=0\big{]}\leq
C\mathbf{P}(\sigma_{n}^{-}=0)\sim C\mathbf{P}(\tau_{0}^{-}>n)$. $\Box$

We now consider the contribution of sample paths with the local minimum
attained at the beginning of the time interval, i.e.,
$\mathbf{E}[F(I_{n});\sigma_{n}^{-}=k]$ for each fixed $k\geq 0$. Using the
Markov property of $S$ and then the duality lemma, we have

$\displaystyle\mathbf{E}\big{[}F(I_{n});\sigma_{n}^{-}=k\big{]}\\!\\!\\!$
$\displaystyle=\\!\\!\\!$
$\displaystyle\mathbf{E}\big{[}\mathbf{E}\big{[}F\big{(}I_{k}+e^{-S_{k}}\cdot\tilde{I}_{n-k}\big{)};\tilde{\tau}_{0}^{-}>n-k\,\big{|}\,\mathscr{F}^{S}\big{]};\sigma_{k}^{-}=k\big{]}.$
(4.18)

Thus it is a crucial step to analyze the asymptotics of the foregoing
conditional expectation.

###### Proposition 4.2

Let $H$ be a bounded, Lipschitz continuous function on $(0,\infty)$. For every
$\epsilon>0$, there exist two integers $k_{0},n_{0}\geq 1$ such that for any
$K\geq k_{0}$ and $n\geq n_{0}$,

$\displaystyle\mathbf{E}\big{[}|H(I_{n})-H(I_{K})|;\tau_{0}^{-}>n\big{]}\leq\epsilon\cdot\mathbf{P}(\tau_{0}^{-}>n).$

Proof. By Remark 3.1, the boundedness of $H$ induces that
$\mathbf{E}\big{[}|H(I_{n})-H(I_{K})|;L_{n}=0\big{]}=o(\mathbf{P}(\tau_{0}^{-}>n))$
uniformly in $K$ as $n\to\infty$ and hence it suffices to prove

$\displaystyle\mathbf{E}\big{[}|H(I_{n})-H(I_{K})|\,;L_{n}>0\big{]}\leq\epsilon\cdot\mathbf{P}(\tau_{0}^{-}>n).$

By the Lipschitz continuity of $H$, there exists a constant $C>0$ such that

$\displaystyle\mathbf{E}\big{[}|H(I_{n})-H(I_{K})|\,;L_{n}>0\big{]}\\!\\!\\!$
$\displaystyle\leq\\!\\!\\!$ $\displaystyle
C\sum_{k=K+1}^{n}\mathbf{E}\big{[}e^{-S_{k}};L_{n}>0\big{]}.$ (4.19)

Applying the Markov property and then the second claim in Lemma 3.3 to
$\mathbf{E}[e^{-S_{k}};L_{n}>0]$, we see that it can be bounded by

$\displaystyle\mathbf{E}\big{[}e^{-S_{k}}\mathbf{P}_{S_{k}}(\tilde{L}_{n-k}>0)\,;L_{k}>0\big{]}\\!\\!\\!$
$\displaystyle\leq\\!\\!\\!$ $\displaystyle
C\mathbf{E}\big{[}e^{-S_{k}}V(S_{k})\,;L_{k}>0\big{]}\cdot\mathbf{P}(L_{n-k}>0).$
(4.20)

The fact that $V(x)=O(x)$ as $x\to\infty$ shows $V(x)e^{-x}\leq Ce^{-x/2}$ for
all $x\geq 0$. From this and Lemma 3.4, we have for large $k$,

$\displaystyle\mathbf{E}\big{[}e^{-S_{k}}V(S_{k})\,;L_{k}>0\big{]}=o\big{(}\mathbf{P}(\tau_{0}^{-}>k)\big{)}.$

Taking this and (4.20) back into (4.19) and then using Lemma 3.2, we have for
large $n$,

$\displaystyle\mathbf{E}\big{[}\left|H(I_{n})-H(I_{K})\right|;L_{n}>0\big{]}\\!\\!\\!$
$\displaystyle\leq\\!\\!\\!$ $\displaystyle
C\sum_{k=K}^{\infty}\mathbf{E}\big{[}e^{-\theta_{F}S_{k}/2};\sigma_{k}^{-}=k\big{]}\cdot\mathbf{P}(\tau_{0}^{-}>n),$

Here the summation above vanishes as $K\to\infty$; see Lemma 3.4, and the
desired result follows immediately. $\Box$

###### Proposition 4.3

Let $H$ be a bounded function on $(0,\infty)$. For $K\geq 1$, we have

$\displaystyle\lim_{n\to\infty}\mathbf{E}\big{[}H(I_{K})\,|\,\tau_{0}^{-}>n\big{]}=\mathbf{E}\big{[}H(I_{K}^{\uparrow})\big{]}>0.$

Moreover, the sequence $\mathbf{E}[H(I_{K}^{\uparrow})]$ converges to
$\mathbf{E}[H(I_{\infty}^{\uparrow})]>0$ as $K\to\infty$.

Proof. The first claim follows from Lemma 3.5. The second one follows from the
dominated convergence theorem and the fact that $I_{K}^{\uparrow}$ increases
to $I_{\infty}^{\uparrow}<\infty$ a.s. as $K\to\infty$; see Remark 2.8. $\Box$

Proof for Theorem 2.7. We first prove this theorem with $F$ being globally
Lipschitz continuous on $(0,\infty)$. By Proposition 4.1,

$\displaystyle\lim_{n\to\infty}\frac{\mathbf{E}[F(I_{n})]}{\mathbf{P}(\tau_{0}^{-}\geq
n)}\\!\\!\\!$ $\displaystyle=\\!\\!\\!$
$\displaystyle\lim_{K\to\infty}\lim_{n\to\infty}\frac{\mathbf{E}[F(I_{n});\sigma_{n}^{-}\leq
K]}{\mathbf{P}(\tau_{0}^{-}\geq
n)}=\sum_{k=0}^{\infty}\lim_{n\to\infty}\frac{\mathbf{E}[F(I_{n});\sigma_{n}^{-}=k]}{\mathbf{P}(\tau_{0}^{-}\geq
n)}.\quad$ (4.21)

Applying Proposition 4.2 and 4.3 with $H(x)=F(I_{k}+e^{-S_{k}}\cdot x)$ to the
conditional expectation on the right side of (4.18), for large $n$ we see that
it is asymptotically equivalent to

$\displaystyle\mathbf{E}\big{[}F\big{(}I_{k}+e^{-S_{k}}\cdot
I^{\uparrow}_{\infty}\big{)}\,\big{|}\,\mathscr{F}^{S}\big{]}\cdot\mathbf{P}(\tilde{\tau}_{0}^{-}>n-k)\sim\mathbf{E}\big{[}F\big{(}I_{k}+e^{-S_{k}}\cdot
I^{\uparrow}_{\infty}\big{)}\,\big{|}\,\mathscr{F}^{S}\big{]}\cdot\mathbf{P}(\tau_{0}^{-}>n).$

Taking this back into (4.18) and then (4.21), we have

$\displaystyle\lim_{n\to\infty}\frac{\mathbf{E}[F(I_{n});\sigma_{n}^{-}=k]}{\mathbf{P}(\tau_{0}^{-}>n)}=\mathbf{E}\big{[}F\big{(}I_{k}+e^{-S_{k}}\cdot
I^{\uparrow}_{\infty}\big{)};\sigma_{k}^{-}=k\big{]}>0.$

Hence the desired asymptotic equivalences in (2.3) hold and the limit
coefficient $C_{F,1}$ is finite because of Proposition 4.1. For general $F$
satisfying Condition 2.2 and $\delta>0$, we define
$F_{\delta}(x)=F(x\vee\delta)$, which is globally Lipschitz continuous. By
Chebyshev’s inequality,

$\displaystyle\mathbf{E}\big{[}|F(I_{n})-F_{\delta}(I_{n})|\big{]}\leq
2C\cdot\mathbf{E}[I_{n}^{-\theta_{F}};I_{n}\leq\delta]\leq
2C\delta\cdot\mathbf{E}[I_{n}^{-\theta_{F}-1}].$

Like the proof of Proposition 4.1, we also can prove
$\mathbf{E}[I_{n}^{-\theta_{F}-1}]\leq C\cdot\mathbf{P}(\tau_{0}^{-}\geq n)$
for any $n\geq 1$ and hence

$\displaystyle\lim_{n\to\infty}\frac{\mathbf{E}[F(I_{n})]}{\mathbf{P}(\tau_{0}^{-}\geq
n)}=\lim_{\delta\to
0}\lim_{n\to\infty}\frac{\mathbf{E}[F_{\delta}(I_{n})]}{\mathbf{P}(\tau_{0}^{-}\geq
n)}.$

The preceding result shows that
$\mathbf{E}[F_{\delta}(I_{n})]/\mathbf{P}(\tau_{0}^{-}\geq n)\to
C_{F_{\delta},1}\in(0,\infty)$ as $n\to\infty$, where $C_{F_{\delta},1}$ can
be represented as the summation in (2.4) with $F$ replaced by $F_{\delta}$.
Specially, when $F(x)=Cx^{-\theta_{F}}$ for any $C>0$, the monotone
convergence theorem induces that $C_{F_{\delta},1}\to C_{F,1}$ as $\delta\to
0+$. This can be extended to the general $F$ satisfying Condition 2.1 by the
dominated convergence theorem. $\Box$

###  4.2 Proof for Theorem 2.10

By (2.5), we observe that the asymptotics of the expectation
$\mathbf{E}^{(\Lambda)}[F(I_{n})e^{-\Lambda S_{n}}]$ is mainly determined by
sample paths with either the local minimum attained at the beginning of the
time interval or the final value below a low level. Moreover, we also observe
that if the random walk ends up at time $n$ below a low level, it will tend to
attain the local minimum at the end of the time interval $[0,n]$. Precisely,
the next proposition proves that the contribution of sample paths with the
local minimum attained at neither the beginning nor the end of the time
interval $[0,n]$ to the expectation $\mathbf{E}[F(I_{n})]$ can be
asymptotically ignored.

###### Proposition 4.4

For every $\epsilon>0$, there exist two integers $k_{0},n_{0}\geq 1$ such that
for any $K\geq k_{0}$ and $n\geq n_{0}$,

$\displaystyle\mathbf{E}\big{[}F(I_{n});\sigma_{n}^{-}\in[K,n-K]\big{]}\leq\epsilon\cdot\mathbf{P}(\tau_{0}^{-}>n).$

Moreover, there exists a constant $C>0$ such that for any $n\geq 1$,

$\displaystyle\mathbf{E}\big{[}F(I_{n})\big{]}\leq
C\cdot\mathbf{P}(\tau_{0}^{-}>n).$

Proof. By the change of measure, it suffices to prove

$\displaystyle\mathbf{E}^{(\Lambda)}\big{[}F(I_{n})e^{-\Lambda
S_{n}};\sigma_{n}^{-}\in[K,n-K]\big{]}\leq\epsilon\cdot\mathbf{E}^{(\Lambda)}\big{[}e^{-\Lambda
S_{n}};\tau_{0}^{-}>n\big{]}.$

By Assumption 2.1 we have $\mathbf{E}^{(\Lambda)}[F(I_{n})e^{-\Lambda
S_{n}};\sigma_{n}^{-}=k]\leq
C\cdot\mathbf{E}^{(\Lambda)}[e^{\beta_{F}S_{k}-\Lambda
S_{n}};\sigma_{n}^{-}=k]$ for any $k\geq 1$. By the Markov property and the
independent increments of $S$,

$\displaystyle\mathbf{E}^{(\Lambda)}[e^{\theta_{F}S_{k}-\Lambda
S_{n}};\sigma_{n}^{-}=k]=\mathbf{E}^{(\Lambda)}[e^{(\theta_{F}-\Lambda)S_{k}}:\sigma_{k}^{-}=k]\cdot\mathbf{E}^{(\Lambda)}[e^{-\Lambda
S_{n-k}};\tau_{0}^{-}>n-k].$

Using Lemma 3.2 together with Lemma 3.6, we have for large $n$,

$\displaystyle\mathbf{E}^{(\Lambda)}\big{[}F(I_{n})e^{-\Lambda
S_{n}};\sigma_{n}^{-}\in[K,n-K]\big{]}$ $\displaystyle\leq\\!\\!\\!$
$\displaystyle
C\sum_{k=K}^{n-K}\mathbf{E}^{(\Lambda)}[e^{(\theta_{F}-\Lambda)S_{k}}:\sigma_{k}^{-}=k]\cdot\mathbf{E}^{(\Lambda)}[e^{-\Lambda
S_{n-k}};\tau_{0}^{-}>n-k]$ $\displaystyle\sim\\!\\!\\!$ $\displaystyle
C\sum_{k=K}^{\infty}\Big{(}\mathbf{E}^{(\Lambda)}[e^{(\theta_{F}-\Lambda)S_{k}}:\sigma_{k}^{-}=k]+\mathbf{E}^{(\Lambda)}[e^{-\Lambda
S_{k}};\tau_{0}^{-}>k]\Big{)}\cdot A_{n}$

Using Lemma 3.6 again, the last summation vanishes as $K\to\infty$ and the
first claim follows. The second claim follows from the first one and the fact
that $\mathbf{E}^{(\Lambda)}[F(I_{n})e^{-\Lambda S_{n}};\sigma_{n}^{-}=0]\leq
C\mathbf{E}^{(\Lambda)}[e^{-\Lambda S_{n}};\tau_{0}^{-}>n]$. $\Box$

We now start to consider the contribution of sample paths with the local
minimum attained early to the expectation $\mathbf{E}[F(I_{n})]$, i.e.
$\mathbf{E}[F(I_{n});\sigma_{n}^{-}=k]$ for $k\geq 0$. Like the proof for
Theorem 2.7, we first consider the conditional expectation on the right side
of (4.18). Different to Proposition 4.2, the next proposition shows that
conditioned on $\tau_{0}^{-}>n$ for large $n$, the exponential functional
$I_{n}$ can be well approximated by the sum of $I_{J}$ and
$I_{n-J,n}:=I_{n}-I_{n-J}$ for large $J<n/2$.

###### Proposition 4.5

Let $H$ be a bounded, Lipschitz continuous function. For every $\epsilon>0$,
there exists two integers $j_{0},n_{0}\geq 0$ such that for any $J\geq j_{0}$
and $n\geq n_{0}$,

$\displaystyle\mathbf{E}\big{[}|H(I_{n})-H(I_{J}+I_{n-J,n})|;\tau_{0}^{-}>n\big{]}\leq\epsilon\cdot\mathbf{P}(\tau_{0}^{-}>n).$

Proof. By the Lipschitz continuity of $F$, we first have

$\displaystyle\mathbf{E}\big{[}|H(I_{n})-H(I_{J}+I_{n-J,n})|;\tau_{0}^{-}>n\big{]}\\!\\!\\!$
$\displaystyle\leq\\!\\!\\!$
$\displaystyle\sum_{j=J+1}^{n-J}\mathbf{E}\big{[}e^{-S_{j}};\tau_{0}^{-}>n\big{]}.$
(4.22)

By the change of measure, we have

$\displaystyle\mathbf{E}\big{[}e^{-S_{j}}\,\big{|}\,\tau_{0}^{-}>n\big{]}\\!\\!\\!$
$\displaystyle=\\!\\!\\!$
$\displaystyle\frac{\mathbf{E}[e^{-S_{j}};\tau_{0}^{-}>n]}{\mathbf{P}(\tau_{0}^{-}>n)}=\frac{\mathbf{E}^{(\Lambda)}[e^{-S_{j}-\Lambda
S_{n}};\tau_{0}^{-}>n]}{\mathbf{E}[e^{-\Lambda S_{n}};\tau_{0}^{-}>n]}.$
(4.23)

Using the Markov property and the independent increments of $S$ to the
numerator of the last fraction, we see that it is equal to

$\displaystyle\mathbf{E}^{(\Lambda)}\big{[}e^{-(1+\Lambda)S_{j}}\mathbf{E}^{(\Lambda)}_{S_{j}}\big{[}e^{-\Lambda\tilde{S}_{n-j}};\tilde{\tau}_{0}^{-}>n-j\big{]};\tau_{0}^{-}>j\big{]}.$

Together with the fact that $V(x)=O(x)$ as $x\to\infty$, the second claim in
Lemma 3.6 induces that there exists a constant $C>0$ such that

$\displaystyle\mathbf{E}^{(\Lambda)}_{S_{j}}\big{[}e^{-\Lambda\tilde{S}_{n-j}};\tilde{\tau}_{0}^{-}>n-j\big{]}\\!\\!\\!$
$\displaystyle\leq\\!\\!\\!$ $\displaystyle C\cdot S_{j}\cdot A_{n-j}$

and hence by the fact that $xe^{-(1+\Lambda)x}\leq Ce^{-(1+\Lambda)x/2}$ for
any $x\geq 0$,

$\displaystyle\mathbf{E}^{(\Lambda)}[e^{-S_{j}-\Lambda
S_{n}};\tau_{0}^{-}>n]\\!\\!\\!$ $\displaystyle\leq\\!\\!\\!$ $\displaystyle
C\mathbf{E}^{(\Lambda)}\big{[}e^{-(1+\Lambda)S_{j}/2};\tau_{0}^{-}>j\big{]}\cdot
A_{n-j}\leq C\cdot A_{j}A_{n-j}.$

Taking this back into (4.23) and then using the first claim in Lemma 3.6 with
$y=\infty$, we have for large $n$,

$\displaystyle\mathbf{E}\big{[}e^{-S_{j}}\,\big{|}\,\tau_{0}^{-}>n\big{]}\\!\\!\\!$
$\displaystyle\leq\\!\\!\\!$ $\displaystyle\frac{C\cdot A_{j}A_{n-j}}{A_{n}}.$

Taking this back into (4.22) and then using Lemma 3.2, we have for large $n$,

$\displaystyle\mathbf{E}\big{[}|H(I_{n})-H(I_{J}+I_{n-J,n})|\,\big{|}\,\tau_{0}^{-}>n\big{]}\leq
C\sum_{j=J+1}^{\infty}A_{j},$

which goes to $0$ as $J\to\infty$ and the desired result follows. $\Box$

###### Proposition 4.6

Let $H$ be a bounded, continuous function on $(0,\infty)$ vanishing at
infinity. For each $J\geq 1$, we have as $n\to\infty$,

$\displaystyle\mathbf{E}\big{[}H(I_{J}+I_{n-J,n})\,\big{|}\,\tau_{0}^{-}>n\big{]}\\!\\!\\!$
$\displaystyle\to\\!\\!\\!$
$\displaystyle\int_{0}^{\infty}\mathbf{E}_{(0,y)}^{(\Lambda)}\big{[}H(I^{\uparrow}_{J}+e^{-y}+I^{\downarrow}_{J-1})\big{]}\mu_{\hat{V}^{(\Lambda)}}^{(\Lambda)}(dy),$

Moreover, the limit coefficient converges as $J\to\infty$ to a finite limit
given by

$\displaystyle\int_{0}^{\infty}\mathbf{E}_{(0,y)}^{(\Lambda)}\big{[}H(I^{\uparrow}_{\infty}+e^{-y}+I^{\downarrow}_{\infty})\big{]}\mu_{\hat{V}^{(\Lambda)}}^{(\Lambda)}(dy).$

Proof. By the change of measure

$\displaystyle\mathbf{E}\big{[}H(I_{J}+I_{n-J,n})\,\big{|}\,\tau_{0}^{-}>n\big{]}\\!\\!\\!$
$\displaystyle=\\!\\!\\!$
$\displaystyle\frac{\mathbf{E}^{(\Lambda)}\big{[}H(I_{J}+I_{n-J,n})e^{-\Lambda
S_{n}};\tau_{0}^{-}>n\big{]}}{\mathbf{E}^{(\Lambda)}\big{[}e^{-\Lambda
S_{n}};\tau_{0}^{-}>n\big{]}}.$

Applying Lemma 3.7 to the foregoing expectations with
$G(x_{1},x_{2})=H(x_{1}+x_{2})$, $f(S)=I_{J}$ and $g_{n}(S)=I_{n-J,n}$, we can
get the desired convergence immediately. The second claim follows by the
dominated convergence theorem and the fact that
$I^{\uparrow}_{\infty},I^{\downarrow}_{\infty}<\infty$ a.s. under
$\mathbf{P}^{(\Lambda)}$; see Remark 2.8. $\Box$

###### Corollary 4.7

For each $k\geq 0$, we have as $n\to\infty$,

$\displaystyle\frac{\mathbf{E}[F(I_{n});\sigma_{n}^{-}=k]}{\mathbf{P}(\tau_{0}^{-}>n)}\\!\\!\\!$
$\displaystyle\to\\!\\!\\!$
$\displaystyle\int_{0}^{\infty}\mathbf{E}_{(0,0,y)}^{(\Lambda)}\big{[}e^{-\Lambda
S_{k}}F\big{(}I_{k}+e^{-S_{k}}(I^{\uparrow}_{\infty}+e^{-y}+I^{\downarrow}_{\infty})\big{)};\sigma_{k}^{-}=k\big{]}\mu_{\hat{V}^{(\Lambda)}}^{(\Lambda)}(dy).$

Proof. Applying Proposition 4.5 and 4.6 with $H(x)=F(I_{k}+e^{-S_{k}}x)$ to
the conditional expectation on the right side of (4.18), we have

$\displaystyle\lim_{n\to\infty}\frac{\mathbf{E}\big{[}F(I_{k}+e^{-S_{k}}\tilde{I}_{n-k});\tilde{\tau}_{0}^{-}>n-k\,\big{|}\,\mathscr{F}^{S}\big{]}}{\mathbf{P}(\tau_{0}^{-}>n)}$
$\displaystyle=\\!\\!\\!$
$\displaystyle\lim_{n\to\infty}\frac{\mathbf{P}(\tau_{0}^{-}>n-k)}{\mathbf{P}(\tau_{0}^{-}>n)}\cdot\lim_{n\to\infty}\mathbf{E}\big{[}F(I_{k}+e^{-S_{k}}\tilde{I}_{n-k})\,\big{|}\,\tilde{\tau}_{0}^{-}>n-k,\mathscr{F}^{S}\big{]}$
$\displaystyle=\\!\\!\\!$ $\displaystyle e^{-{\it\Phi}(\Lambda)\cdot
k}\cdot\lim_{J\to\infty}\int_{0}^{\infty}\mathbf{E}_{(0,y)}^{(\Lambda)}\big{[}F\big{(}I_{k}+e^{-S_{k}}(I^{\uparrow}_{J}+e^{-y}+I^{\downarrow}_{J-1})\big{)}\,\big{|}\,\mathscr{F}^{S}\big{]}\mu_{\hat{V}^{(\Lambda)}}^{(\Lambda)}(dy)$
$\displaystyle=\\!\\!\\!$ $\displaystyle e^{-{\it\Phi}(\Lambda)\cdot
k}\int_{0}^{\infty}\mathbf{E}_{(0,y)}^{(\Lambda)}\big{[}F\big{(}I_{k}+e^{-S_{k}}(I^{\uparrow}_{\infty}+e^{-y}+I^{\downarrow}_{\infty})\big{)}\,\big{|}\,\mathscr{F}^{S}\big{]}\mu_{\hat{V}^{(\Lambda)}}^{(\Lambda)}(dy).$

Here the second equality comes from the first claim in Lemma 3.6 with
$y=\infty$. Taking this back into expectation on the right side of (4.18)
yields the desired result. $\Box$

We now start to consider the impact of sample paths with the local minimum
attained at the end of the time interval on the expectation
$\mathbf{E}[F(I_{n})]$. Applying the Markov property to
$\mathbf{E}[F(I_{n});\sigma^{-}_{n}=n-k]$ for $n>k\geq 0$, we see it equals to

$\displaystyle\mathbf{E}\big{[}\mathbf{E}\big{[}F(I_{n-k}+e^{-S_{n-k}}\tilde{I}_{k});\sigma^{-}_{n-k}=n-k\,\big{|}\,\mathscr{F}^{\tilde{S}}\big{]};\tilde{\tau}_{0}^{-}>k\big{]}.$
(4.24)

Like the previous argument, we first need to consider the aysmptotics of the
conditional expectation. In the next proposition, we show that it is rarely
contributed by sample paths with final value below a very low level.

###### Proposition 4.8

For every $x\geq 0$ and $\epsilon>0$, there exist two integers
$y_{0},n_{0}\geq 1$ such that for any $y\geq y_{0}$ and $n\geq n_{0}$,

$\displaystyle\mathbf{E}\big{[}F\big{(}I_{n}+xe^{-S_{n}}\big{)};S_{n}<-y,\sigma^{-}_{n}=n\big{]}\leq\epsilon\cdot\mathbf{P}\big{(}\tau_{0}^{-}>n\big{)}.$

Proof. By the change of measure, we see that the desired inequality holds if
and only if

$\displaystyle\mathbf{E}^{(\Lambda)}\big{[}e^{-\Lambda
S_{n}}F\big{(}I_{n}+xe^{-S_{n}}\big{)};S_{n}<-y,\sigma^{-}_{n}=n\big{]}\leq\epsilon\cdot\mathbf{E}^{(\Lambda)}\big{[}e^{-\Lambda
S_{n}};\tau_{0}^{-}>n\big{]}.$

Applying the duality lemma to the expectation on the left side of this
inequality, we see it equals to

$\displaystyle\mathbf{E}^{(\Lambda)}\big{[}e^{-\Lambda
S_{n}}F\big{(}e^{-S_{n}}(1+\hat{I}_{n-1}+x)\big{)};S_{n}<-y,M_{n}<0\big{]},$

which can be bounded by
$C\mathbf{E}^{(\Lambda)}\big{[}e^{(\theta_{F}-\Lambda)S_{n}};S_{n}<-y,M_{n}<0\big{]}$
because of the assumption that $F(x)\leq Cx^{-\theta_{F}}$. From Lemma 3.6,
there exists a constant $C>0$ such for any $n\geq 1$,

$\displaystyle\lim_{n\to\infty}\frac{\mathbf{E}^{(\Lambda)}[e^{(\theta_{F}-\Lambda)S_{n}};S_{n}<-y,M_{n}<0]}{\mathbf{E}^{(\Lambda)}[e^{-\Lambda
S_{n}};\tau_{0}^{-}>n]}\leq
C\int_{y}^{\infty}e^{-(\theta_{F}-\Lambda)z}V(z)dz,$

which goes to $0$ as $y\to\infty$ and hence the desired result follows. $\Box$

###### Proposition 4.9

Suppose $F$ is globally Lipschitz continuous on $(0,\infty)$. For each
$x,y\geq 0$ and $\epsilon>0$, there exist two integers $j_{0},n_{0}>1$ such
that for any $J\geq j_{0}$ and $n\geq n_{0}$,

$\displaystyle\mathbf{E}\big{[}\big{|}F(I_{n}+xe^{-S_{n}})-F(I_{J}+I_{n-J,n}+xe^{-S_{n}})\big{|};S_{n}\geq-y,\sigma_{n}^{-}=n\big{]}\leq\epsilon\mathbf{P}(\tau_{0}^{-}>n).$
(4.25)

Proof. By the Lipschitz continuity of $F$, the expectation on the left side of
(4.25) can be bounded by

$\displaystyle
C\mathbf{E}\big{[}I_{J+1,n-J};S_{n}\geq-y,\sigma_{n}^{-}=n\big{]}\\!\\!\\!$
$\displaystyle=\\!\\!\\!$ $\displaystyle
C\sum_{j=J+1}^{n-J}\mathbf{E}\big{[}e^{-S_{j}};S_{n}\geq-y,\sigma_{n}^{-}=n\big{]}.$
(4.26)

Applying the duality lemma and then the change of measure to
$\mathbf{E}\big{[}e^{-S_{j}};S_{n}\geq-y,\tau_{0}^{+}>n\big{]}$, we see that
it equals to

$\displaystyle\mathbf{E}\big{[}e^{S_{n-j}-S_{n}};S_{n}\geq-y,M_{n}<0\big{]}\\!\\!\\!$
$\displaystyle=\\!\\!\\!$
$\displaystyle\mathbf{E}^{(\Lambda)}\big{[}e^{S_{n-j}-(1+\Lambda)S_{n}};S_{n}\geq-y,M_{n}<0\big{]},$
(4.27)

which can be bounded by
$e^{(\kappa+1+\Lambda)y}\mathbf{E}^{(\Lambda)}[e^{S_{n-j}+\kappa
S_{n}};M_{n}<0]$ for any $\kappa>0$. By the Markov property and the second
claim in Lemma 3.6, there exists a constant $C>0$ such that

$\displaystyle\mathbf{E}^{(\Lambda)}\big{[}e^{S_{n-j}+\kappa
S_{n}};M_{n}<0\big{]}\\!\\!\\!$ $\displaystyle=\\!\\!\\!$
$\displaystyle\mathbf{E}^{(\Lambda)}\big{[}e^{S_{n-j}}\mathbf{E}^{(\Lambda)}_{S_{n-j}}[e^{\kappa\tilde{S}_{j}};\tilde{M}_{j}<0];M_{n-j}<0\big{]}$
$\displaystyle\leq\\!\\!\\!$ $\displaystyle
C\mathbf{E}^{(\Lambda)}\big{[}e^{S_{n-j}}S_{n-j};M_{n-j}<0\big{]}\cdot
A_{j}\leq CA_{j}A_{n-j}.$

Taking this and (4.27) back into (4.26) and then using Lemma 3.2, we have for
large $n$,

$\displaystyle\frac{\mathbf{E}\big{[}I_{J+1,n-J};S_{n}\geq-y,\sigma_{n}^{-}=n\big{]}}{\mathbf{P}(\tau_{0}^{-}>n)}\\!\\!\\!$
$\displaystyle\leq\\!\\!\\!$
$\displaystyle\frac{C}{A_{n}}\sum_{j=J+1}^{n-J}A_{j}A_{n-j}\sim
2C\sum_{j=J}^{\infty}A_{j},$

which goes to $0$ as $J\to\infty$ and the desired result follows. $\Box$

###### Proposition 4.10

For any $x,y\geq 0$, $\eta\in(0,\theta_{F}-\Lambda)$ and $J\geq 1$, we have as
$n\to\infty$,

$\displaystyle\frac{\mathbf{E}[F(I_{J}+I_{n-J,n}+xe^{-S_{n}});S_{n}\geq-y,\sigma_{n}^{-}=n]}{\mathbf{P}(\tau_{0}^{-}>n)}$
$\displaystyle\to\\!\\!\\!$
$\displaystyle\frac{\mathcal{L}_{V^{(\Lambda)}}^{(\eta)}(\infty)}{\mathcal{L}_{\hat{V}^{(\Lambda)}}^{(\Lambda)}(\infty)}\int_{0}^{y}e^{(\eta+\Lambda)z}\mathbf{E}_{(0,z)}^{(\Lambda)}\big{[}F\big{(}e^{z}\big{(}1+x+I_{J-1}^{\uparrow}+I_{J-1}^{\downarrow}\big{)}\big{)}\big{]}\mu_{V^{(\Lambda)}}^{(\eta)}(dz).$

Moreover, the limit coefficient converges as $J\to\infty$ and then
$y\to\infty$ to

$\displaystyle\frac{\mathcal{L}_{V^{(\Lambda)}}^{(\eta)}(\infty)}{\mathcal{L}_{\hat{V}^{(\Lambda)}}^{(\Lambda)}(\infty)}\int_{0}^{\infty}e^{(\eta+\Lambda)z}\mathbf{E}_{(0,z)}^{(\Lambda)}\big{[}F\big{(}e^{z}\big{(}1+x+I_{\infty}^{\uparrow}+I_{\infty}^{\downarrow}\big{)}\big{)}\big{]}\mu_{V^{(\Lambda)}}^{(\eta)}(dz).$

Proof. By the duality lemma, we first have

$\displaystyle\mathbf{E}^{(\Lambda)}\Big{[}F\big{(}I_{J}+I_{n-J,n}+xe^{-S_{n}}\big{)}e^{-\Lambda
S_{n}};S_{n}\geq-y,\sigma_{n}^{-}=n\Big{]}$ $\displaystyle=\\!\\!\\!$
$\displaystyle\mathbf{E}^{(\Lambda)}\Big{[}F\Big{(}e^{-S_{n}}\Big{(}1+x+\sum_{j=1}^{J-1}e^{S_{j}}+\sum_{i=n-J}^{n-1}e^{S_{i}}\Big{)}\Big{)}e^{-\Lambda
S_{n}};S_{n}\geq-y,M_{n}<0\Big{]}$ $\displaystyle=\\!\\!\\!$
$\displaystyle\mathbf{E}^{(\Lambda)}\Big{[}e^{-(\eta+\Lambda)S_{n}}F\Big{(}e^{-S_{n}}\Big{(}1+x+\sum_{j=1}^{J-1}e^{S_{j}}+\sum_{i=n-J}^{n-1}e^{S_{i}}\Big{)}\Big{)}\mathbf{1}_{\\{S_{n}\geq-y\\}}\cdot
e^{\eta S_{n}},M_{n}<0\Big{]}.$

By Lemma 3.7, we have as $n\to\infty$,

$\displaystyle\frac{\mathbf{E}[F(I_{J}+I_{n-J,n}+xe^{-S_{n}});S_{n}\geq-y,\sigma_{n}^{-}=n]}{\mathbf{P}(\tau_{0}^{-}>n)}$
$\displaystyle\sim\\!\\!\\!$
$\displaystyle\frac{\mathbf{E}^{(\Lambda)}[F(I_{J}+I_{n-J,n}+xe^{-S_{n}})e^{-\Lambda
S_{n}};S_{n}\geq-y,\sigma_{n}^{-}=n]}{\mathbf{E}^{(\Lambda)}[e^{\eta
S_{n}},M_{n}<0]}$ $\displaystyle\to\\!\\!\\!$
$\displaystyle\int_{0}^{y}\mathbf{E}_{(0,-z)}^{(\Lambda)}\Big{[}e^{(\eta+\Lambda)z}F\Big{(}e^{z}\Big{(}1+x+\sum_{j=1}^{J-1}e^{\hat{S}^{\uparrow}}+\sum_{i=1}^{J}e^{\hat{S}^{\downarrow}}\Big{)}\Big{)}\Big{]}\mu_{V^{(\Lambda)}}^{(\eta)}(dz)$
$\displaystyle=\\!\\!\\!$
$\displaystyle\int_{0}^{y}e^{(\eta+\Lambda)z}\mathbf{E}_{(0,z)}^{(\Lambda)}\Big{[}F\Big{(}e^{z}\big{(}1+x+I_{J-1}^{\uparrow}+I_{J-1}^{\downarrow}\big{)}\Big{)}\Big{]}\mu_{V^{(\Lambda)}}^{(\eta)}(dz).$

Here the last equality follows from the fact that
$\hat{S}^{\uparrow}=-S^{\uparrow}$ and $\hat{S}^{\downarrow}=-S^{\downarrow}$.
The first claim follows directly from Lemma 3.6. The second one can be proved
by using the dominated convergence theorem when $J\to\infty$ and then the
monotone convergence theorem when $y\to\infty$. $\Box$

Applying the preceding proposition to (4.24), we can get the following
corollary immediately.

###### Corollary 4.11

For each $k\geq 0$, we have as $n\to\infty$

$\displaystyle\frac{\mathbf{E}[F(I_{n});\sigma_{n}^{-}=n-k]}{\mathbf{P}(\tau_{0}^{-}>n)}$
$\displaystyle\to\\!\\!\\!$
$\displaystyle\frac{\mathcal{L}_{V^{(\Lambda)}}^{(\eta)}(\infty)}{\mathcal{L}_{\hat{V}^{(\Lambda)}}^{(\Lambda)}(\infty)}\int_{0}^{\infty}e^{(\eta+\Lambda)z}\mathbf{E}_{(0,0,z)}^{(\Lambda)}\big{[}e^{-\Lambda
S_{k}}F\big{(}e^{z}(1+I_{k}+I_{\infty}^{\uparrow}+I_{\infty}^{\downarrow})\big{)};\tau_{0}^{-}>k\big{]}\mu_{V^{(\Lambda)}}^{(\eta)}(dz).$

Proof for Theorem 2.10. Here we just prove this theorem with $F$ being
globally Lipschitz continuous. Proceeding as in the proof of Theorem 2.7, we
can prove this theorem for general $F$. With the help of the preceding
propositions, we have

$\displaystyle\lim_{n\to\infty}\frac{\mathbf{E}[F(I_{n})]}{\mathbf{P}(\tau_{0}^{-}>n)}\\!\\!\\!$
$\displaystyle=\\!\\!\\!$
$\displaystyle\sum_{k=0}^{\infty}\lim_{n\to\infty}\frac{\mathbf{E}[F(I_{n});\sigma_{n}^{-}=k]}{\mathbf{P}(\tau_{0}^{-}>n)}+\sum_{k=0}^{\infty}\lim_{n\to\infty}\frac{\mathbf{E}[F(I_{n});\sigma_{n}^{-}=n-k]}{\mathbf{P}(\tau_{0}^{-}>n)}.$

and the desired convergence follows directly from Corollary 4.7 and 4.11. The
finiteness of the limit coefficient $C_{F,2}$ can be gotten from Proposition
4.4. $\Box$

###  4.3 Proof for Theorem 2.11

Condition 2.3 tells that for every $\epsilon>0$, there exits a constant
$x_{0}>0$ such that $|F(x)-K_{0}\cdot x^{\Lambda}|<\epsilon\cdot x^{-\Lambda}$
for any $x\geq x_{0}$. Thus we need to consider the asymptotic behavior of
$\mathbf{E}[I_{n}^{-\Lambda}]$ at first. Using the change of measure and then
duality lemma, we have

$\displaystyle\mathbf{E}[I_{n}^{-\Lambda}]=\varrho^{n}\cdot\mathbf{E}^{(\Lambda)}\big{[}I_{n}^{-\Lambda}e^{-\Lambda
S_{n}}\big{]}=\varrho^{n}\cdot\mathbf{E}^{(\Lambda)}\big{[}(1+\hat{I}_{n-1})^{-\Lambda}\big{]}.$

According to our previous argument, the dual process $\hat{S}$ satisfies
Spitzer’s condition with positivity parameter $1-\rho$. Using Theorem 2.7 with
$F(x)=(1+x)^{-\Lambda}$, we have as $n\to\infty$,

$\displaystyle\mathbf{E}^{(\Lambda)}\big{[}(1+\hat{I}_{n-1})^{-\Lambda}\big{]}\\!\\!\\!$
$\displaystyle\sim\\!\\!\\!$ $\displaystyle
C_{F,3}\cdot\mathbf{P}^{(\Lambda)}(\hat{\tau}_{0}^{-}>n)=C_{F,3}\cdot\mathbf{P}^{(\Lambda)}(\tau_{0}^{+}>n).$

It remains to prove that
$\mathbf{E}^{(\Lambda)}\big{[}|F(I_{n})-K_{0}I_{n}^{-\Lambda}|e^{-\Lambda
S_{n}}\big{]}=o\big{(}\mathbf{P}^{(\Lambda)}(\tau_{0}^{+}>n)\big{)}$ as
$n\to\infty$. For any $x\geq x_{0}$, from the previous result we first have
for large $n$,

$\displaystyle\mathbf{E}^{(\Lambda)}\big{[}|F(I_{n})-K_{0}I_{n}^{-\Lambda}|e^{-\Lambda
S_{n}};I_{n}>x\big{]}\leq\epsilon\cdot\mathbf{E}^{(\Lambda)}\big{[}I_{n}^{-\Lambda}e^{-\Lambda
S_{n}}\big{]}\leq\epsilon\cdot C\cdot\mathbf{P}^{(\Lambda)}(\tau_{0}^{+}>n).$

On the other hand, by Chebyshev’s inequality we also have

$\displaystyle\mathbf{E}^{(\Lambda)}\big{[}|F(I_{n})-K_{0}I_{n}^{-\Lambda}|e^{-\Lambda
S_{n}};I_{n}\leq x\big{]}\\!\\!\\!$ $\displaystyle\leq\\!\\!\\!$
$\displaystyle x\mathbf{E}^{(\Lambda)}\big{[}I_{n}^{-\Lambda-1}e^{-\Lambda
S_{n}}\big{]}.$

For each $k\geq 0$, by the Markov property we have

$\displaystyle\mathbf{E}^{(\Lambda)}\big{[}I_{n}^{-\Lambda-1}e^{-\Lambda
S_{n}};\sigma_{n}^{-}=k\big{]}\\!\\!\\!$ $\displaystyle\leq\\!\\!\\!$
$\displaystyle\mathbf{E}^{(\Lambda)}\big{[}e^{(\Lambda+1)S_{k}-\Lambda
S_{n}};\sigma_{n}^{-}=k\big{]}$ $\displaystyle=\\!\\!\\!$
$\displaystyle\mathbf{E}^{(\Lambda)}\big{[}e^{S_{k}};\sigma_{k}^{-}=k\big{]}\mathbf{E}^{(\Lambda)}\big{[}e^{-\Lambda
S_{n-k}};\tau_{0}^{-}>n-k\big{]}.$

Applying Lemma 3.2 together with Lemma 3.6, we have

$\displaystyle\mathbf{E}^{(\Lambda)}\big{[}I_{n}^{-\Lambda-1}e^{-\Lambda
S_{n}}\big{]}\\!\\!\\!$ $\displaystyle=\\!\\!\\!$
$\displaystyle\sum_{k=0}^{n}\mathbf{E}^{(\Lambda)}\big{[}I_{n}^{-\Lambda-1}e^{-\Lambda
S_{n}};\sigma_{n}^{-}=k\big{]}$ $\displaystyle\leq\\!\\!\\!$
$\displaystyle\sum_{k=0}^{n}\mathbf{E}^{(\Lambda)}\big{[}e^{S_{k}};\sigma_{k}^{-}=k\big{]}\mathbf{E}^{(\Lambda)}\big{[}e^{-\Lambda
S_{n-k}};\tau_{0}^{-}>n-k\big{]},$

which can be bounded by $C\cdot
A_{n}=o(\mathbf{P}^{(\Lambda)}(\tau_{0}^{+}>n))$. Putting all estimates above
together, we can get the desired result immediately.

##  5 Proof for Theorem 2.15-2.12

In this section we prove the asymptotic results for the expectation
$\mathbf{E}[F(I_{n})]$ with $S$ drifting to $-\infty$ under
$\mathbf{P}^{(\Lambda)}$. Different to the oscillating cases, we observe that
the slow decreasing of the local minimum of $S$ usually results from an early
large step. For simplicity, we again assume $\theta_{F}\in\mathcal{D}_{F}$.

###  5.1 Proof for Theorem 2.12

It suffices to prove $\mathbf{E}^{(\Lambda)}[e^{-\Lambda S_{n}}F(I_{n})]\sim
K_{0}\mathbf{E}^{(\Lambda)}[e^{-\Lambda S_{n}}I_{n}^{-\Lambda}]$ as
$n\to\infty$. By Condition 2.3, for any $\epsilon>0$ there exists a constant
$x_{0}>0$ such that for any $n\geq 1$ and $x\geq x_{0}$

$\displaystyle\mathbf{E}^{(\Lambda)}[e^{-\Lambda
S_{n}}|F(I_{n})-K_{0}I_{n}^{-\Lambda}|;I_{n}\geq x]\\!\\!\\!$
$\displaystyle\leq\\!\\!\\!$
$\displaystyle\epsilon\cdot\mathbf{E}^{(\Lambda)}[e^{-\Lambda
S_{n}}I_{n}^{-\Lambda}]$ $\displaystyle=\\!\\!\\!$
$\displaystyle\epsilon\cdot\mathbf{E}^{(\Lambda)}[(1+\hat{I}_{n-1})^{-\Lambda}]\leq\epsilon.$

On the other hand, by Assumption 2.1, for any $K>0$ we have

$\displaystyle\mathbf{E}^{(\Lambda)}[e^{-\Lambda
S_{n}}|F(I_{n})-K_{0}I_{n}^{-\Lambda}|;I_{n}<x]\\!\\!\\!$
$\displaystyle\leq\\!\\!\\!$ $\displaystyle
C\mathbf{E}^{(\Lambda)}[e^{-\Lambda
S_{n}}I_{n}^{-\Lambda};S_{n}\geq-K,I_{n}<x]$
$\displaystyle+C\mathbf{E}^{(\Lambda)}[e^{-\Lambda
S_{n}}I_{n}^{-\Lambda};S_{n}<-K,I_{n}<x].$

By the duality lemma, the first expectation on the right side of the above
inequality can be bounded by
$\mathbf{E}^{(\Lambda)}[(1+\hat{I}_{n-1})^{-\Lambda};S_{n}\geq-K]\leq\mathbf{P}^{(\Lambda)}(S_{n}\geq-K)\to
0$ as $n\to\infty$. For the second expectation, by Chebyshev’s inequality, we
see that uniformly in $n\geq 1$ it can be bounded by

$\displaystyle x^{\kappa}\mathbf{E}^{(\Lambda)}[e^{-\Lambda
S_{n}}I_{n}^{-\Lambda-1};S_{n}<-K]\\!\\!\\!$ $\displaystyle\leq\\!\\!\\!$
$\displaystyle
xe^{-K}\mathbf{E}^{(\Lambda)}[e^{-(\Lambda+1)S_{n}}I_{n}^{-\Lambda-1}]$
$\displaystyle=\\!\\!\\!$ $\displaystyle
xe^{-K}\mathbf{E}^{(\Lambda)}[(1+\hat{I}_{n-1})^{-\Lambda-1}]\leq xe^{-K},$

which vanishes as $K\to\infty$. We can get the desired result and finish the
proof by putting all estimates above together.

###  5.2 Proof for Theorem 2.15

As we have mentioned before, the slow decreasing of the local minimum of $S$
usually results from an early large step, thus the contribution of sample
paths with late arrival of the first large step,
$\mathbf{E}[F(I_{n});\mathcal{T}^{an}>T]$ for large $T$, can be asymptotically
ignored. By Lemma 3.10, we have the local minimum is usually not far from the
first large step. Thus for any $\epsilon>0$ and $K\geq 1$, there exist two
integers $t_{0},n_{0}>0$ such that for any $T>t_{0}$ and $n\geq n_{0}$,

$\displaystyle\mathbf{E}\big{[}F(I_{n});\mathcal{T}^{an}>T,\sigma_{n}^{-}\leq
K\big{]}\leq C\cdot\mathbf{P}(\mathcal{T}^{an}>T,\sigma_{n}^{-}\leq
K)\leq\epsilon\cdot\mathbf{P}(\tau_{0}^{-}>n).$

On the other hand, we prove in the next proposition, that the contribution of
simple paths with the local minimum attained late,
$\mathbf{E}[F(I_{n});\mathcal{T}^{an}>T,\sigma_{n}^{-}>K]\leq\mathbf{E}[F(I_{n});\sigma_{n}^{-}>K]$
for large $K$, also can be asymptotically ignored.

###### Proposition 5.1

For every $\epsilon>0$, there exist two integers $k_{0},n_{0}\geq 1$ such that
for any $K\geq k_{0}$ and $n\geq n_{0}$,

$\displaystyle\mathbf{E}\big{[}F(I_{n});\sigma^{-}_{n}\geq K\big{]}\\!\\!\\!$
$\displaystyle\leq\\!\\!\\!$
$\displaystyle\epsilon\cdot\mathbf{P}(\tau_{0}^{-}>n).$

Moreover, there exists a constant $C>0$ such that for any $n\geq 1$,

$\displaystyle\mathbf{E}\big{[}F(I_{n})\big{]}\leq
C\cdot\mathbf{P}(\tau_{0}^{-}>n).$

Proof. By Assumption 2.1, we first have $F(I_{n})\leq
C\exp\\{\theta_{F}L_{n}\\}$ and hence

$\displaystyle\mathbf{E}[F(I_{n});\sigma_{n}^{-}\geq K]\\!\\!\\!$
$\displaystyle\leq\\!\\!\\!$ $\displaystyle
C\sum_{k=K}^{n}\mathbf{E}[e^{\theta_{F}S_{k}};\sigma_{n}^{-}=k]=C\sum_{k=K}^{n}\mathbf{E}[e^{\theta_{F}S_{k}};\sigma_{k}^{-}=k]\mathbf{P}(\tau_{0}^{-}>n-k).$

Here the equality above follows from the Markov property of $S$ and the
duality lemma. Applying Lemma 3.2 together with Lemma 3.11 with $y=\infty$ to
the last sum, we have as $n\to\infty$ it is asymptotically equivalent to

$\displaystyle\mathbf{P}\\{\tau_{0}^{-}>n\\}\cdot\sum_{k=K}^{\infty}\mathbf{E}[e^{\theta_{F}S_{k}};\sigma_{k}^{-}=k]\leq\mathbf{P}(\tau_{0}^{-}>n)\cdot\sum_{k=K}^{\infty}\frac{1}{k}\mathbf{P}(X\geq
ak).$

Since the sequence $k^{-1}\mathbf{P}(X\geq ak)$ is $(-\alpha-1)$-regularly
varying and summable, the two desired inequalities follow directly. $\Box$

We now start to consider the contribution of sample paths with early large
step, i.e. $\mathbf{E}[F(I_{n});\mathcal{T}^{an}=k]$ for each $k\geq 1$. By
(3.17), we have for large $n$,

$\displaystyle\frac{\mathbf{E}[F(I_{n});\mathcal{T}^{an}=k]}{\mathbf{P}(\tau_{0}^{-}>n)}\sim\frac{\mathbf{E}[F(I_{n});X_{k}\geq
an]}{\mathbf{P}(\tau_{0}^{-}>n)}.$

Applying the strong Markov property to $\mathbf{E}[F(I_{n});X_{k}\geq an]$, we
see it is equal to

$\displaystyle\mathbf{E}\big{[}\mathbf{E}[F(I_{k-1}+e^{-S_{k-1}-X}\tilde{I}_{n-k})|X\geq
an,\mathcal{F}^{S}]\big{]}\cdot\mathbf{P}(X\geq an).$ (5.28)

It is obvious that we need to consider the asymptotics of the conditional
expectation at first.

###### Proposition 5.2

Let $H$ be a bounded, Lipschitz continuous function on $(0,\infty)$. For every
$\epsilon>0$, there exist two integers $k_{0},n_{0}\geq 1$ such that for any
$K\geq k_{0}$ and $n\geq n_{0}$,

$\displaystyle\mathbf{E}\big{[}|H(e^{-X}I_{n})-H(e^{-X}I_{K})|;X\geq
an\big{]}\leq\epsilon\cdot\mathbf{P}(X\geq an).$

Proof. Let $I_{K,n}:=I_{n}-I_{K}$ for $0\leq K\leq n$. The boundedness and
Lipschitz continuity of $H$ induces that there exists a constant $C>0$ such
that

$\displaystyle\mathbf{E}\big{[}|H(e^{-X}I_{n})-H(e^{-X}I_{K})|;X\geq
an\big{]}\\!\\!\\!$ $\displaystyle\leq\\!\\!\\!$ $\displaystyle
C\mathbf{E}\big{[}1\wedge\big{(}e^{-X}I_{K,n}\big{)};X\geq an\big{]}.$

For any $b>a$, it is obvious that the foregoing quantities can bounded by the
sum of

$\displaystyle\varepsilon(b,n):=C\mathbf{P}(X\in[an,bn))\quad\mbox{and}\quad\varepsilon(b,K,n):=C\mathbf{E}\big{[}1\wedge\big{(}e^{-X}I_{K,n}\big{)};X\geq
bn\big{]}.$

Condition 2.13 induces that $\varepsilon(b,n)/\mathbf{P}(X\geq an)\to 0$ as
$n\to\infty$ and then $b\to a+$. Let $S_{n}^{b}:=S_{n}+nb$, which drifts to
$\infty$ as $n\to\infty$. By the independence between $X$ and $I_{K,n}(S)$,

$\displaystyle\varepsilon(b,K,n)\leq
C\mathbf{E}\big{[}1\wedge\big{(}e^{-bn}I_{K,n}\big{)}\big{]}\cdot\mathbf{P}\\{X\geq
bn\\}\leq C\mathbf{E}\big{[}1\wedge I_{K,n}^{b}\big{]}\cdot\mathbf{P}\\{X\geq
bn\\},$

where $I_{K,n}^{b}:=I_{n}^{b}-I_{K}^{b}$ and $I^{b}$ is the exponential
functional of $S^{b}$. By Lemma 2.4, we have $I_{\infty}^{b}<\infty$ a.s. and
hence $I_{K,n}^{b}\leq I_{K,\infty}^{b}\to 0$ a.s. as $K\to\infty$. Applying
the dominated convergence theorem, we immediately get
$\varepsilon(b,K,n)/\mathbf{P}(X\geq an)\to 0$ as $n\to\infty$ and then
$K\to\infty$. The desired result following by putting all preceding estimates
together. $\Box$

###### Proposition 5.3

Let $H$ be a bounded, non-increasing function on $(0,\infty)$. For any $K>0$,
we have as $n\to\infty$,

$\displaystyle\mathbf{E}\big{[}H(e^{-X}I_{K})\,\big{|}\,X\geq an\big{]}\to
C_{H,K}>0.$

Moreover, the sequence $C_{H,K}$ decreases to a limit $C_{H}>0$ as
$K\to\infty$ and

$\displaystyle\lim_{n\to\infty}\mathbf{E}\big{[}H(e^{-X}I_{n})\,\big{|}\,X\geq
an\big{]}=C_{H}.$

Proof. It is easy to see that the first result follows directly from our
observation that the monotonicity of $H$ results in
$\mathbf{E}\big{[}H(e^{-X}I_{K})\,\big{|}\,X\geq an\big{]}$ being non-
decreasing in $n$, i.e., for any $z>0$ a simple calculation show that the
expectation $\mathbf{E}\big{[}H(ze^{-X})\,\big{|}\,X\geq an\big{]}$ can
written as the sum of $\mathbf{E}\big{[}H(ze^{-X})\,\big{|}\,X\geq
a(n+1)\big{]}$ and

$\displaystyle\frac{\mathbf{P}\\{X\geq[an,a(n+1))\\}}{\mathbf{P}\\{X\geq
an\\}}\Big{(}\mathbf{E}\big{[}H(ze^{-X})\,\big{|}\,X\geq[an,a(n+1))\big{]}-\mathbf{E}\big{[}H(ze^{-X})\,\big{|}\,X\geq
a(n+1)\big{]}\Big{)}.$

It is obvious that the foregoing quantity is non-positive and hence
$\mathbf{E}\big{[}H(e^{-X}I_{K})\,\big{|}\,X\geq
an\big{]}\geq\mathbf{E}\big{[}H(e^{-X}I_{K})\,\big{|}\,X\geq a(n+1)\big{]}$.
We now turn to prove the second claim. Because of the monotonicity of $H$ and
$I_{K}$, the sequence $C_{H,K}$ is non-increasing and converges to a limit
$C_{H}\in[0,\infty)$ as $K\to\infty$. The second desired convergence follows
directly from Proposition 5.2. It remains to prove $C_{H}>0$. Recall the
modified random walk $S^{b}_{n}:=S_{n}+bn$ for $b>a$. We have

$\displaystyle\mathbf{E}\big{[}H(e^{-X}I_{n})\,\big{|}\,X\geq
an\big{]}\geq\frac{\mathbf{E}\big{[}H(e^{-X}I_{n});X\geq
bn\big{]}}{\mathbf{P}\\{X\geq
an\\}}\geq\mathbf{E}\big{[}H(I_{n}^{b})\big{]}\frac{\mathbf{P}\\{X\geq
bn\\}}{\mathbf{P}\\{X\geq an\\}}.$

From Lemma 2.4 and the dominated convergence theorem, we have
$\mathbf{E}\big{[}H(I_{n}^{b})\big{]}\to\mathbf{E}\big{[}H(I_{\infty}^{b})\big{]}>0$
as $n\to\infty$ and hence
$C_{H}=\liminf_{n\to\infty}\mathbf{E}\big{[}H(e^{-X}I_{n})\,\big{|}\,X\geq
an\big{]}>0$. $\Box$

###### Corollary 5.4

For each $k\geq 1$, we have $\mathbf{E}[F(I_{n});X_{k}\geq
an]/\mathbf{P}(X\geq an)\to\mathbf{E}[C_{F,4}(k)]>0$ as $n\to\infty$.

Proof. Applying Proposition 5.2 and 5.3 with
$H(x):=\mathbf{E}[F(I_{k-1}+xe^{-S_{k-1}}|\mathcal{F}^{S}]$ to the conditional
expectation in (5.28), we have as $n\to\infty$ it converges to the limit
$C_{F,4}(k)>0$ a.s. Taking this back into (5.28) we can get the desired result
directly. $\Box$

Proof for Theorem 2.15. Based on our previous analysis, we have

$\displaystyle\lim_{n\to\infty}\frac{\mathbf{E}[F(I_{n})]}{\mathbf{P}(X\geq
an)}\\!\\!\\!$ $\displaystyle=\\!\\!\\!$
$\displaystyle\sum_{k=1}^{\infty}\lim_{n\to\infty}\frac{\mathbf{E}[F(I_{n});X_{k}\geq
an]}{\mathbf{P}(X\geq an)}$

and the asymptotic results (2.12) follows directly from Corollary 5.4. The
finiteness of the coefficient $C_{F,4}$ follows from Proposition 5.1. $\Box$

###  5.3 Proof for Theorem 2.17

By (2.5) we first show in the next proposition that the impact of sample paths
with fast decreasing local minimum on the expectation
$\mathbf{E}^{(\Lambda)}\big{[}e^{-\Lambda S_{n}}F(I_{n})\big{]}$ can be
asymptotically ignored.

###### Proposition 5.5

For every $\epsilon>0$, there exist two integers $y_{0},n_{0}\geq 1$ such that
for any $Y\geq y_{0}$ and $n\geq n_{0}$,

$\displaystyle\mathbf{E}^{(\Lambda)}[F(I_{n})e^{-\Lambda
S_{n}};L_{n}\leq-Y]\leq\epsilon\cdot B_{n}.$

Proof. Assumption 2.1 induces that $F(I_{n})\leq Ce^{\Lambda L_{n}}$. By Lemma
3.12, for large $n$ we have

$\displaystyle\frac{1}{B_{n}}\mathbf{E}^{(\Lambda)}[F(I_{n})e^{-\Lambda
S_{n}};L_{n}\leq-Y]\\!\\!\\!$ $\displaystyle\leq\\!\\!\\!$ $\displaystyle
e^{-\theta_{F}Y/2}\cdot\frac{C}{B_{n}}\mathbf{E}^{(\Lambda)}\big{[}e^{\theta_{F}L_{n}/2-\Lambda
S_{n}}\big{]}\leq Ce^{-\theta_{F}Y/2},$

which goes to $0$ as $Y\to\infty$ and hence the desired result follows. $\Box$

###### Proposition 5.6

For every $\epsilon>0$, there exist two integers $k_{0},n_{0}\geq 1$ such that
for any $K\geq k_{0}$ and $n\geq n_{0}$,

$\displaystyle\mathbf{E}^{(\Lambda)}\big{[}e^{-\Lambda
S_{n}}F(I_{n});\sigma_{n}^{-}\in[K,n-K]\big{]}\leq\epsilon\cdot B_{n}.$

Moreover, there exists a constant $C>0$ such that for any $n\geq 1$,

$\displaystyle\mathbf{E}^{(\Lambda)}\big{[}e^{-\Lambda
S_{n}}F(I_{n})\big{]}\leq C\cdot B_{n}$

Proof. By Assumption 2.1, we first have $F(I_{n})\leq
C\exp\\{\theta_{F}L_{n}\\}$. By the independent increments of $S$, we have for
any $k\geq 1$,

$\displaystyle\mathbf{E}^{(\Lambda)}\big{[}e^{-\Lambda
S_{n}}F(I_{n});\sigma_{n}^{-}=k\big{]}\\!\\!\\!$ $\displaystyle\leq\\!\\!\\!$
$\displaystyle C\mathbf{E}^{(\Lambda)}\big{[}e^{\theta_{F}S_{k}-\Lambda
S_{n}};\sigma_{n}^{-}=k\big{]}$ $\displaystyle=\\!\\!\\!$
$\displaystyle\mathbf{E}^{(\Lambda)}\big{[}e^{(\theta_{F}-\Lambda)S_{k}};\sigma_{k}^{-}=k\big{]}\cdot\mathbf{E}^{(\Lambda)}\big{[}e^{-\Lambda
S_{n-k}};\tau_{0}^{-}>n-k\big{]}.$

Using Lemma 3.2 together with Lemma 3.11, for large $n$ we have

$\displaystyle\mathbf{E}^{(\Lambda)}\big{[}e^{-\Lambda
S_{n}}F(I_{n});\sigma_{n}^{-}\in[K,n-K]\big{]}$ $\displaystyle\leq\\!\\!\\!$
$\displaystyle
C\sum_{k=K}^{n-K}\mathbf{E}^{(\Lambda)}[e^{(\theta_{F}-\Lambda)S_{k}};\sigma_{n}^{-}=k]\mathbf{E}^{(\Lambda)}[e^{-\Lambda
S_{n-k}};\tau_{0}^{-}>n-k]$ $\displaystyle\sim\\!\\!\\!$ $\displaystyle
C\sum_{k=K}^{\infty}\Big{(}\mathbf{E}^{(\Lambda)}[e^{(\theta_{F}-\Lambda)S_{k}};\sigma_{n}^{-}=k]+\mathbf{E}^{(\Lambda)}[e^{-\Lambda
S_{k}};\tau_{0}^{-}>k]\Big{)}\cdot B_{n}.$

The two desired inequalities follows from
$\mathbf{E}^{(\Lambda)}\big{[}e^{-\Lambda
S_{n}}F(I_{n});\sigma_{n}^{-}=0\big{]}\leq C\cdot B_{n}$ and the fact that the
two sequences
$\mathbf{E}^{(\Lambda)}[e^{(\theta_{F}-\Lambda)S_{k}};\sigma_{n}^{-}=k]$ and
$\mathbf{E}^{(\Lambda)}[e^{-\Lambda S_{k}};\tau_{0}^{-}>k]$ are summable; see
Lemma 3.11. $\Box$

As we mentioned before, the contribution of simple paths with large final
value on the expectation $\mathbf{E}[F(I_{n})]$ also can be ignored; see the
following two propositions.

###### Proposition 5.7

For every $\epsilon>0$ and integer $k\geq 0$, there exist two integers
$n_{0},y_{0}\geq 1$ such that for any $n\geq n_{0}$ and $Y\geq y_{0}$,

$\displaystyle\mathbf{E}^{(\Lambda)}[e^{-\Lambda S_{n}}F(I_{n});S_{n}\geq
Y;\sigma_{n}^{-}=k]\leq\epsilon\cdot B_{n}.$ (5.29)

Proof. When $k=0$, by Lemma 3.11 and the duality lemma we have for large $n$,

$\displaystyle\frac{1}{B_{n}}\mathbf{E}^{(\Lambda)}[e^{-\Lambda
S_{n}}F(I_{n});S_{n}\geq Y;\sigma_{n}^{-}=0]\\!\\!\\!$
$\displaystyle\leq\\!\\!\\!$
$\displaystyle\frac{C}{B_{n}}\mathbf{E}^{(\Lambda)}[e^{-\Lambda
S_{n}};S_{n}\geq Y;\tau_{0}^{-}>n]$ $\displaystyle\leq\\!\\!\\!$
$\displaystyle C\int_{Y}^{\infty}e^{-\lambda y}\hat{V}(z)dz,$

which vanishes as $Y\to\infty$. For $k\geq 1$, by Assumption 2.1 we have

$\displaystyle\mathbf{E}^{(\Lambda)}[e^{-\Lambda S_{n}}F(I_{n});S_{n}\geq
Y;\sigma_{n}^{-}=k]\\!\\!\\!$ $\displaystyle\leq\\!\\!\\!$ $\displaystyle
C\mathbf{E}^{(\Lambda)}[e^{\theta_{F}S_{k}-\Lambda S_{n}};S_{n}\geq
Y;\sigma_{n}^{-}=k].$

Conditioned on $\sigma_{n}^{-}=k$, we notice that $\\{S_{n}\geq
Y\\}\subset\\{S_{n}-S_{k}\geq Y\\}$ and hence

$\displaystyle\mathbf{E}^{(\Lambda)}[e^{-\Lambda S_{n}}F(I_{n});S_{n}\geq
Y;\sigma_{n}^{-}=k]\\!\\!\\!$ $\displaystyle\leq\\!\\!\\!$ $\displaystyle
C\mathbf{E}^{(\Lambda)}[e^{-\Lambda S_{n}+\theta_{F}S_{k}};S_{n}-S_{k}\geq
Y;\sigma_{n}^{-}=k].$

The independent increments of $S$ induces that the expectation on the right
side of this inequality equals to

$\displaystyle\mathbf{E}^{(\Lambda)}[e^{(\theta_{F}-\Lambda)S_{k}};\sigma_{k}^{-}=k]\cdot\mathbf{E}^{(\Lambda)}[e^{-\Lambda
S_{n-k}};S_{n-k}\geq Y;\tau_{0}^{-}>n-k].$

Using Lemma 3.11, there exists a constant $C>0$ such that for large $n$,

$\displaystyle\frac{1}{B_{n}}\mathbf{E}^{(\Lambda)}[e^{-\Lambda
S_{n}}F(I_{n});S_{n}\geq Y;\sigma_{n}^{-}=k]$ $\displaystyle\leq\\!\\!\\!$
$\displaystyle\frac{C}{B_{n}}\mathbf{E}^{(\Lambda)}[e^{-\Lambda
S_{n-k}};S_{n-k}\geq Y;\tau_{0}^{-}>n-k]\leq C\int_{Y}^{\infty}e^{-\lambda
y}\hat{V}(z)dz,$

which vanishes as $Y\to\infty$ and the desired result follows. $\Box$

###### Proposition 5.8

For every $\epsilon>0$ and integer $k\geq 0$, there exist two integers
$n_{0},y_{0}\geq 1$ such that for any $n\geq n_{0}$ and $Y\geq y_{0}$,

$\displaystyle\mathbf{E}^{(\Lambda)}[e^{-\Lambda S_{n}}F(I_{n});S_{n}\geq
Y;\sigma_{n}^{-}=n-k]\leq\epsilon\cdot B_{n}.$ (5.30)

Proof. An argument similar to that in the proof for Proposition 5.7 shows that
the expectation in the above inequality can be bounded by

$\displaystyle
C\mathbf{E}^{(\Lambda)}[e^{(\theta_{F}-\Lambda)S_{n-k}};\sigma_{n-k}^{-}=n-k]\cdot\mathbf{E}^{(\Lambda)}[e^{-\Lambda
S_{k}};S_{k}\geq Y,\tau_{0}^{-}>k].$ (5.31)

Dividing it by $B_{n}$ and the using Lemma 3.12, for large $n$ we see that it
can be bounded by $C\cdot\mathbf{E}^{(\Lambda)}[e^{-\Lambda S_{k}};S_{k}\geq
N,\tau_{0}^{-}>k]$, which vanishes as $N\to\infty$. $\Box$

We now turn to consider the impact of simple paths with late arrival of the
first large step, slowly decreasing local minimum and the final value below
some fixed level, i.e. for $b\in(0,a]$ and $Y,T\geq 1$,

$\displaystyle\mathbf{E}^{(\Lambda)}[e^{-\Lambda S_{n}}F(I_{n});-Y\leq
L_{n}\leq S_{n}\leq Y,\mathcal{T}^{bn}\geq T].$ (5.32)

The next two propositions show that the late arrival of the first large jump
will cause the random walk to drift to a low level. It extends Lemma 9 in [8],
who considered the case $\beta>2$.

###### Proposition 5.9

Let $r>\frac{\beta+1}{\beta-1}$, $b\in(0,a/r)$ and $Y\geq 0$. For any
$\epsilon>0$, there exists an integer $n_{0}\geq 1$ such that for any $n\geq
n_{0}$,

$\displaystyle\mathbf{P}^{(\Lambda)}(S_{n}\geq-Y,\mathcal{T}^{bn}\geq
n)\leq\epsilon\cdot B_{n}.$

Proof. Here we just prove this proposition for $\beta\in(1,2)$. For $\beta>2$,
it can be proved in the same way with the help of Theorem 4.1.2(i) in [13,
p.183]. We first assume that there exists a constant $C>0$ such that

$\displaystyle\mathbf{P}^{(\Lambda)}(X\leq-x)\leq
C\cdot\mathbf{P}^{(\Lambda)}(X\geq x),\quad x\geq 0.$ (5.33)

By Lemma 3.1 in [14], there exists a constant $C>0$ such that for large $n$,

$\displaystyle\mathbf{P}^{(\Lambda)}(S_{n}+an>rbn,\mathcal{T}^{bn-a}>n)\leq
C\big{|}n\mathbf{P}^{(\Lambda)}(X\geq bn-a)\big{|}^{r}\leq
Cn^{r(1-\beta)}(\ell_{3}(n))^{r}$

From this and the assumption on $br<a$ and $r(\beta-1)>\beta+1$, we have for
large $n$,

$\displaystyle\mathbf{P}^{(\Lambda)}(S_{n}\geq-Y,\mathcal{T}^{bn}>n)\leq
Cn^{r(1-\beta)}(\ell_{3}(n))^{r}=o(B_{n})$

We consider the general case with $X$ satisfying Assumption 2.13. For some
$K<0$ we define $X^{K}:=X\vee K$ and $X^{K}_{i}=X_{i}\vee K$ such that
$a_{K}:=-\mathbf{E}^{(\Lambda)}[X^{K}]\in(rb,a)$. It is obvious that the
random variable $X^{K}$ satisfies the inequality (5.33). Let $S^{K}_{n}$ be
the random walk generated by the sequence $\\{X_{i}^{K}:i=1,2,\cdots\\}$. It
is obvious that $\mathbf{P}^{(\Lambda)}(S_{n}\geq-Y,\mathcal{T}^{bn}\geq
n)\leq\mathbf{P}^{(\Lambda)}(S_{n}^{K}\geq-Y,\mathcal{T}^{bn}\geq n)$ and our
previous result induces that it is $o(B_{n})$. $\Box$

###### Proposition 5.10

Let $b>0$ and $Y\geq 0$. For any $\epsilon>0$, there exist two integers
$t_{0},n_{0}>1$ such that for any $T\geq t_{0}$ and $n\geq n_{0}$,

$\displaystyle\mathbf{P}^{(\Lambda)}(-Y\leq L_{n}\leq S_{n}\leq
Y,T\leq\mathcal{T}^{bn}\leq n)\leq\epsilon\cdot B_{n}.$

Proof. For any $2\leq k\leq n$, we have $L_{k-1}\geq L_{n}$ and by the Markov
property

$\displaystyle\mathbf{P}^{(\Lambda)}(-Y\leq L_{n}\leq S_{n}\leq Y,X_{k}\geq
bn)$ (5.34) $\displaystyle\leq\\!\\!\\!$
$\displaystyle\int_{-Y}^{\infty}\mathbf{P}^{(\Lambda)}(L_{k-1}\geq-Y,S_{k-1}\in
dx)\cdot\int_{bn}^{\infty}\mathbf{P}^{(\Lambda)}(x+y+\tilde{S}_{n-k}\in[-Y,Y])\mathbf{P}^{(\Lambda)}(X_{k}\in
dy)$ (5.35) $\displaystyle\leq\\!\\!\\!$
$\displaystyle\mathbf{P}^{(\Lambda)}(L_{k-1}\geq-Y)\cdot\sup_{x\geq-Y}\int_{bn}^{\infty}\mathbf{P}^{(\Lambda)}(x+y+\tilde{S}_{n-k}\in[-Y,Y])\mathbf{P}^{(\Lambda)}(X_{k}\in
dy).$ (5.36)

By Assumption 2.14, we have $\mathbf{P}^{(\Lambda)}(X_{k}\in[bn+i,bn+i+1))\leq
2B_{n}$ for any $i\geq 1$ and large $n$. Thus the last integral in (5.34) can
be bounded by

$\displaystyle
2B_{n}\sum_{i=0}^{\infty}\mathbf{P}^{(\Lambda)}(x+bn+i+\tilde{S}_{n-k}\in[-Y-1,Y+1])$
$\displaystyle\leq\\!\\!\\!$ $\displaystyle
2B_{n}\int_{bn}^{\infty}\mathbf{P}^{(\Lambda)}(x+y+\tilde{S}_{n-k}\in[-Y-2,Y+2])dy.$

Applying the integration by parts to this integral, we have it can be bounded
by

$\displaystyle\int_{-\infty}^{\infty}\int_{-x-y-Y-2}^{-x-y+Y+2}\mathbf{P}^{(\Lambda)}(\tilde{S}_{n-k}\in
dz)dy=\int_{-\infty}^{\infty}\mathbf{P}^{(\Lambda)}(\tilde{S}_{n-k}\in
dy)\int_{-x-y-Y-2}^{-x-y+Y+2}dz\leq 2Y+4.$

Taking these back into (5.34), we have

$\displaystyle\mathbf{P}^{(\Lambda)}(-Y\leq L_{n}\leq S_{n}\leq Y,X_{k}\geq
bn)\\!\\!\\!$ $\displaystyle\leq\\!\\!\\!$ $\displaystyle
4(Y+2)\mathbf{P}^{(\Lambda)}(L_{k-1}\geq-Y)\cdot B_{n}.$

Hence by Lemma 3.8, we have for large $n$,

$\displaystyle\frac{1}{B_{n}}\mathbf{P}^{(\Lambda)}(-Y\leq L_{n}\leq S_{n}\leq
Y,T\leq\mathcal{T}^{bn}\leq n)\\!\\!\\!$ $\displaystyle\leq\\!\\!\\!$
$\displaystyle C\sum_{k=T}^{n}\mathbf{P}^{(\Lambda)}(L_{k-1}\geq-Y)$
$\displaystyle\leq\\!\\!\\!$ $\displaystyle
C\sum_{k=T}^{\infty}\mathbf{P}^{(\Lambda)}(X\geq a(k-1)),$

which vanishes as $T\to\infty$ and the desired result follows. $\Box$

Putting all preceding estimates together, we see that the main contribution to
the expectation $\mathbf{E}^{(\Lambda)}[F(I_{n})e^{-\Lambda S_{n}}]$ is made
by the simple paths with an early large step, i.e. as $n\to\infty$,

$\displaystyle\mathbf{E}^{(\Lambda)}[F(I_{n})e^{-\Lambda
S_{n}}]\sim\sum_{k=1}^{\infty}\mathbf{E}^{(\Lambda)}[F(I_{n})e^{-\Lambda
S_{n}};\mathcal{T}^{bn}=k],$ (5.37)

where $b\in(0,a/r)$ and $r>\frac{\beta+1}{\beta-1}$. By (3.17), for each
$k\geq 1$ we have

$\displaystyle\mathbf{E}^{(\Lambda)}[F(I_{n})e^{-\Lambda
S_{n}};\mathcal{T}^{bn}=k]\sim\mathbf{E}^{(\Lambda)}[F(I_{n})e^{-\Lambda
S_{n}};X_{k}\geq bn].$ (5.38)

Using Proposition 5.5 again, for large $Y$ we see that it is also
asymptotically equivalent to

$\displaystyle\mathbf{E}^{(\Lambda)}[F(I_{n})e^{-\Lambda
S_{n}};S_{k-1}\geq-Y,X_{k}\geq bn].$ (5.39)

Moreover, as we have mentioned before, the local minimum is usually attained
around the early large step and hence the impact of simple paths that stay
above a hight level before the early large step on the expectation
$\mathbf{E}^{(\Lambda)}[F(I_{n})e^{-\Lambda S_{n}}]$ can be ignored; see the
following proposition.

###### Proposition 5.11

Let $b\in(0,a]$ and integer $k\geq 1$. For each $\epsilon>0$, there exist two
integers $n_{0},y_{0}\geq 1$ such that for any $Y\geq y_{0}$ and $n\geq
n_{0}$,

$\displaystyle\mathbf{E}^{(\Lambda)}[F(I_{n})e^{-\Lambda S_{n}};S_{k-1}\geq
Y,X_{k}\geq bn]\leq\epsilon\cdot B_{n}.$

Proof. Let $\gamma\in(1/\beta,1)$. By Assumption 2.1, we see the foregoing
expectation can be bounded by

$\displaystyle C\mathbf{E}^{(\Lambda)}\big{[}e^{\theta_{F}L_{n}-\Lambda
S_{n}};S_{k-1}\geq n^{\gamma},X_{k}\geq
bn\big{]}+C\mathbf{E}^{(\Lambda)}\big{[}e^{\theta_{F}L_{n}-\Lambda
S_{n}};N\leq S_{k-1}\leq n^{\gamma},X_{k}\geq bn\big{]}.$ (5.40)

Applying Hölder’s inequality to the first expectation, for large $n$ we have
it can be bounded by

$\displaystyle\big{(}\mathbf{E}^{(\Lambda)}[e^{p\theta_{F}L_{n}-p\Lambda
S_{n}}]\big{)}^{1/2}\big{(}\mathbf{P}^{(\Lambda)}(S_{k-1}\geq
n^{\gamma})\mathbf{P}^{(\Lambda)}(X_{k}\geq bn)\big{)}^{1/2}.$

Using Lemma 3.12 and then Lemma 3.8, we see that the forgoing quantity can be
bounded by $C\cdot B_{n}\big{(}n\cdot\mathbf{P}^{(\Lambda)}(S_{k-1}\geq
n^{\gamma})\big{)}^{1/2}=o(B_{n})$. On the other hand, by the Markov property
the second expectation in (5.40) equals to

$\displaystyle\int_{Y}^{n^{\gamma}}\mathbf{P}^{(\Lambda)}(S_{k-1}\in
dy)\int_{bn}^{\infty}\mathbf{E}^{(\Lambda)}_{y+x}[e^{\theta_{F}\tilde{L}_{n-k}-\Lambda\tilde{S}_{n-k}}]\mathbf{P}^{(\Lambda)}(X_{k}\in
dx)$ (5.41) $\displaystyle=\\!\\!\\!$
$\displaystyle\int_{Y}^{n^{\gamma}}\mathbf{P}^{(\Lambda)}(S_{k-1}\in
dy)\int_{bn+y}^{\infty}\mathbf{E}^{(\Lambda)}_{x}[e^{\theta_{F}\tilde{L}_{n-k}-\Lambda\tilde{S}_{n-k}}]\mathbf{P}^{(\Lambda)}(X_{k}+y\in
dx).$ (5.42)

By Assumption 2.13 and 2.14, there exists a constant $C>0$ such that for large
$n$,

$\displaystyle\mathbf{P}^{(\Lambda)}(X_{k}+y\in[bn+i,bn+i+1))\leq
C\cdot\mathbf{P}^{(\Lambda)}(X\in[bn+i,bn+i+1))$

uniformly in $i\geq 0$ and $y\in[N,n^{\gamma}]$. Thus the inner integral in
the double integral on the right side of (5.41) can be bounded by

$\displaystyle\sum_{i=0}^{\infty}\mathbf{E}^{(\Lambda)}_{bn+i}[e^{\theta_{F}(\tilde{L}_{n-k}+1)-\Lambda(\tilde{S}_{n-k}-1)}]\cdot\mathbf{P}^{(\Lambda)}(X_{k}+y\in[bn+i,bn+i+1))$
$\displaystyle\leq\\!\\!\\!$ $\displaystyle
C\sum_{i=0}^{\infty}\mathbf{E}^{(\Lambda)}_{bn+i}[e^{\theta_{F}\tilde{L}_{n-k}-\Lambda\tilde{S}_{n-k}}]\cdot\mathbf{P}^{(\Lambda)}(X\in[bn+i,bn+i+1))$
$\displaystyle\leq\\!\\!\\!$ $\displaystyle
C\int_{bn}^{\infty}\mathbf{E}^{(\Lambda)}_{x}[e^{\theta_{F}\tilde{L}_{n-k}-\Lambda\tilde{S}_{n-k}}]\cdot\mathbf{P}^{(\Lambda)}(X\in
dx)$ $\displaystyle=\\!\\!\\!$ $\displaystyle
C\cdot\mathbf{E}^{(\Lambda)}\big{[}e^{\theta_{F}\tilde{L}_{n-k+1}-\Lambda\tilde{S}_{n-k+1}};\tilde{X}_{1}\geq
bn\big{]}\leq
C\mathbf{E}^{(\Lambda)}\big{[}e^{\theta_{F}\tilde{L}_{n-k+1}-\Lambda\tilde{S}_{n-k+1}}\big{]}\leq
C\cdot B_{n}.$

Here the last inequality follows from Lemma 3.12. Taking this back into
(5.41), we have for large $n$ and $Y$,

$\displaystyle\mathbf{E}^{(\Lambda)}\big{[}e^{\theta_{F}L_{n}-\Lambda
S_{n}};Y\leq S_{k-1}\leq n^{\gamma},X_{k}\geq bn\big{]}\leq C\cdot
B_{n}\cdot\mathbf{P}^{(\Lambda)}(S_{k-1}\geq Y)=o(B_{n}).$

The desired result follows by putting all estimates above together. $\Box$

Taking the estimates in (5.39) and Proposition 5.11 back into (5.38), we have
for large $n$ and $Y$,

$\displaystyle\mathbf{E}^{(\Lambda)}[F(I_{n})e^{-\Lambda
S_{n}};\mathcal{T}^{bn}=k]\sim\mathbf{E}^{(\Lambda)}[e^{-\Lambda S_{k-1}}\cdot
F(I_{n})\cdot e^{-\Lambda(S_{n}-S_{k-1})};|S_{k-1}|\leq Y,X_{k}\geq bn].$
(5.43)

The next proposition shows that for large $Z>0$, the contribution of simple
path with $|S_{n}-S_{k-1}|>Z$ to the forgoing expectations can be
asymptotically ignored.

###### Proposition 5.12

Let $Y\geq 0$ and integer $k\geq 1$. For any $\epsilon>0$, there exist two
integers $z_{0},n_{0}\geq 1$ such that for any $Z\geq z_{0}$ and $n\geq
n_{0}$,

$\displaystyle\mathbf{E}^{(\Lambda)}[F(I_{n})e^{-\Lambda S_{n}};|S_{k-1}|\leq
Y,|S_{n}-S_{k-1}|\geq Z]\leq\epsilon\cdot B_{n}.$ (5.44)

Proof. By Assumption 2.1, it suffices to prove this equality with $F(I_{n})$
replaced by $e^{\theta_{F}L_{n}}$. We first notice that $L_{n}\leq
S_{k-1}+L_{k,n}$ with $L_{k,n}:=\min\\{S_{i}-S_{k-1};i=k-1,\cdots,n\\}$. By
the Markov property,

$\displaystyle\mathbf{E}^{(\Lambda)}\Big{[}e^{\theta_{F}L_{n}-\Lambda
S_{n}};|S_{k-1}|\leq Y,|S_{n}-S_{k-1}|\geq Z\Big{]}$
$\displaystyle\leq\\!\\!\\!$
$\displaystyle\mathbf{E}^{(\Lambda)}\Big{[}e^{\theta_{F}(S_{k-1}+L_{k,n})-\Lambda
S_{n}};|S_{k-1}|\leq Y,|S_{n}-S_{k-1}|\geq Z\Big{]}$
$\displaystyle\leq\\!\\!\\!$
$\displaystyle\mathbf{E}^{(\Lambda)}\Big{[}e^{(\theta_{F}-\Lambda)S_{k-1}};|S_{k-1}|\leq
Y\Big{]}\cdot\mathbf{E}^{(\Lambda)}\Big{[}e^{\theta_{F}L_{n-k+1}-\Lambda
S_{n-k+1}};|S_{n-k+1}|\geq Z\Big{]}.$

It is obvious that the first expectation on the right side of the last
inequality can be bounded by $e^{(\theta_{F}-\Lambda)Y}$. For the second
expectation, we see it can be bounded by

$\displaystyle\mathbf{E}^{(\Lambda)}\Big{[}e^{\theta_{F}L_{n-k+1}-\Lambda
S_{n-k+1}};L_{n-k+1}\leq-Z\Big{]}+\mathbf{E}^{(\Lambda)}\Big{[}e^{\theta_{F}L_{n-k+1}-\Lambda
S_{n-k+1}};L_{n-k+1}-S_{n-k+1}\leq Z\Big{]}.$

By Lemma 3.12, there exists a constant $C>0$ such that for large $n$, these
two expectations can be bounded by

$\displaystyle
C\Big{(}\int_{Z}^{\infty}e^{-(\theta_{F}-\Lambda)y}\big{(}V(dy)+V(y)dy\big{)}+\int_{Z}^{\infty}e^{-\Lambda
y}\big{(}\hat{V}(dy)+\hat{V}(y)dy\big{)}\Big{)}\cdot B_{n}.$

It is obvious that these two integrals vanishes as $Z\to\infty$ and the
desired result follows. $\Box$

Let $r>\frac{\beta+1}{\beta-1}$ and $b\in(0,a/r)$. Combining all estimates
above together, we have for large $Y$, $Z$ and $n$,

$\displaystyle\mathbf{E}^{(\Lambda)}[e^{-\Lambda S_{n}}F(I_{n})]\\!\\!\\!$
$\displaystyle\sim\\!\\!\\!$
$\displaystyle\sum_{k=1}^{\infty}\mathbf{E}^{(\Lambda)}[e^{-\Lambda
S_{n}}F(I_{n});|S_{k-1}|\leq Y,|S_{n}-S_{k-1}|\leq Z,X_{k}\geq bn].$

By the Markov property, the summand can be written into

$\displaystyle\mathbf{E}^{(\Lambda)}\big{[}e^{-\Lambda
S_{k-1}}F_{\mathscr{F}^{S}}^{b,Z}(k,n);|S_{k-1}|\leq Y\big{]}$ (5.45)

with

$\displaystyle
F_{\mathscr{F}^{S}}^{b,Z}(k,n):=\mathbf{E}^{(\Lambda)}\big{[}e^{-\Lambda\tilde{S}_{n-k+1}}F\big{(}I_{k-1}+e^{-S_{k-1}}\tilde{I}_{n-k+1}\big{)};|\tilde{S}_{n-k+1}|\leq
Z,\tilde{X}_{1}\geq bn\big{|}\mathscr{F}^{S}\big{]}.$ (5.46)

It is obvious that $F_{\mathscr{F}^{S}}^{b,Z}(k,n)$ plays the main role in the
asymptotic analysis of the expectation $\mathbf{E}[F(I_{n})]$. The next
proposition that $\\{|\tilde{S}_{n-k+1}|\leq Z,\tilde{X}_{1}\geq bn\\}$ is
asymptotically equivalent to $\\{|\tilde{S}_{n-k+1}|\leq
Z,|\tilde{X}_{1}-an|\geq n^{1/\kappa}\\}$.

###### Proposition 5.13

For any $b\in(0,a)$, $Z\geq 1$ and $\epsilon>0$, there exists an integer
$n_{0}$ such that for any $n\geq n_{0}$,

$\displaystyle\mathbf{P}^{(\Lambda)}\big{(}X_{1}\in[bn,an-n^{1/\kappa}]\cup[an+n^{1/\kappa},\infty);|S_{n}|\leq
Z\big{)}\leq\epsilon\cdot B_{n}.$ (5.47)

Proof. An argument similar to that in the proof of Proposition 5.11 shows that
for large $n$,

$\displaystyle\mathbf{P}^{(\Lambda)}(X_{1}\in[bn,an-n^{1/\kappa}];|S_{n}|\leq
Z)\\!\\!\\!$ $\displaystyle\leq\\!\\!\\!$ $\displaystyle C\cdot
B_{n}\cdot\int_{bn}^{an-n^{1/\kappa}}\mathbf{P}^{(\Lambda)}(|x+S_{n-1}|\leq
Z+2)dx$ $\displaystyle\leq\\!\\!\\!$ $\displaystyle C\cdot
B_{n}\cdot\int_{bn}^{an-n^{1/\kappa}}\int_{-Z-2-x}^{Z+2-x}\mathbf{P}^{(\Lambda)}(S_{n-1}\in
dz)dx$ $\displaystyle\leq\\!\\!\\!$ $\displaystyle C(Z+2)\cdot
B_{n}\cdot\int_{-Z-2-an+n^{1/\kappa}}^{Z+2-bn}\mathbf{P}^{(\Lambda)}(S_{n-1}\in
dz)$ $\displaystyle\leq\\!\\!\\!$ $\displaystyle C(Z+2)\cdot
B_{n}\cdot\mathbf{P}^{(\Lambda)}(S_{n-1}\geq-Z-2-an+n^{1/\kappa})$
$\displaystyle\leq\\!\\!\\!$ $\displaystyle 4NC\cdot
B_{n}\cdot\mathbf{P}^{(\Lambda)}\Big{(}\frac{S_{n-1}+a(n-1)}{n^{1/\kappa}}\geq
1/2\Big{)}.$

The statements after Condition 2.16 shows that the last probability goes to
$0$ as $n\to\infty$. Similarly, we also have for large $n$,

$\displaystyle\mathbf{P}^{(\Lambda)}\big{(}X_{1}>an+n^{1/\kappa};|S_{n}|\leq
N\big{)}\\!\\!\\!$ $\displaystyle\leq\\!\\!\\!$ $\displaystyle C\cdot
B_{n}\cdot\int_{an+n^{1/\kappa}}^{\infty}\mathbf{P}^{(\Lambda)}(|x+S_{n-1}|\leq
2N)dx$ $\displaystyle\leq\\!\\!\\!$ $\displaystyle C\cdot
B_{n}\cdot\mathbf{P}^{(\Lambda)}(S_{n-1}\leq 2N-an-n^{1/\kappa})$
$\displaystyle\leq\\!\\!\\!$ $\displaystyle C\cdot
B_{n}\cdot\mathbf{P}^{(\Lambda)}\Big{(}\frac{S_{n-1}+a(n-1)}{n^{1/\kappa}}\leq-1/2\Big{)}$

and the last probability also goes to $0$ as $n\to\infty$. $\Box$

By this proposition, we see that $F_{\mathscr{F}^{S}}^{b,Z}(k,n)$ can be well
approximated as $n\to\infty$ by

$\displaystyle\mathbf{E}^{(\Lambda)}\big{[}e^{-\Lambda\tilde{S}_{n-k+1}}F\big{(}I_{k-1}+e^{-S_{k-1}}\tilde{I}_{n-k+1}\big{)};|\tilde{S}_{n-k+1}|\leq
Z,|\tilde{X}_{1}-an|\leq n^{1/\kappa}\big{|}\mathscr{F}^{S}\big{]}$ (5.48)

Conditioned on $|\tilde{S}_{n-k+1}|\leq Z$ we in the next proposition prove
that the contribution of $\tilde{I}_{n-k+1}$ to this conditional expectation
can be well approximated by that of $\tilde{I}_{n-k+1-W,n-k+1}$ for large $W$.

###### Proposition 5.14

Let $H$ be a bounded, Lipschitz continuous function on $(0,\infty)$. For every
$Z,\epsilon>0$, there exist two integers $w_{0},n_{0}\leq 1$ such that for any
$W\geq w_{0}$ and $n\geq n_{0}$,

$\displaystyle\mathbf{E}^{(\Lambda)}\big{[}e^{-\Lambda
S_{n}}\big{|}H(I_{n})-H(I_{n-W,n})\big{|};|S_{n}|\leq Z,|X_{1}-an|\leq
n^{1/\kappa}\big{]}\leq\epsilon\cdot B_{n}.$ (5.49)

Proof. Using the Lipschitz continuity of $H$ and then the duality lemma, the
foregoing expectation can be bounded by

$\displaystyle Ce^{\Lambda Z}\mathbf{E}^{(\Lambda)}\big{[}\big{(}I_{n-W}\wedge
1\big{)};|S_{n}|\leq Z,|X_{1}-an|\leq n^{1/\kappa}\big{]}$
$\displaystyle=\\!\\!\\!$ $\displaystyle Ce^{\Lambda
Z}\mathbf{E}^{(\Lambda)}\big{[}\big{(}\big{(}e^{-S_{n}}\hat{I}_{W-1,n-1}\big{)}\wedge
1\big{)};|S_{n}|\leq Z,|X_{n}-an|\leq n^{1/\kappa}\big{]}.$

As argument similar to that in the proof of Proposition 5.10 shows that there
exists a constant $C>0$ such that for large $n$, the foregoing quantities can
be bounded by

$\displaystyle Ce^{\Lambda
Z}B_{n}\cdot\int_{an-n^{1/\kappa}}^{an+n^{1/\kappa}}\mathbf{E}^{(\Lambda)}\big{[}\big{(}(e^{Z}\hat{I}_{W-1,n-1})\wedge
1\big{)};|S_{n-1}+x_{n}|\leq Z\big{]}dx_{n}.$

Changing the variables it turns to be

$\displaystyle
CB_{n}\cdot\int_{-Z}^{Z}\mathbf{E}^{(\Lambda)}\big{[}\big{(}(e^{Z}\hat{I}_{W-1,n-1})\wedge
1\big{)};|S_{n-1}+an-z|\leq n^{1/\kappa}\big{]}dz,$

which can be bounded by $2CZe^{\Lambda Z}\cdot
B_{n}\cdot\mathbf{E}^{(\Lambda)}[((e^{Z}\hat{I}_{W-1,n-1})\wedge 1)]$. The
desired result follows by the dominated convergence theorem and the fact that
$\hat{I}_{\infty}<\infty$ a.s. under $\mathbf{P}^{(\Lambda)}$. $\Box$

###### Proposition 5.15

Let $H$ be a nonnegative, bounded and continuous function on $(0,\infty)$. For
any two integers $W,Z>0$, we have as $n\to\infty$,

$\displaystyle B_{n}^{-1}\cdot\mathbf{E}^{(\Lambda)}\Big{[}e^{-\Lambda
S_{n}}H(I_{n-W,n});|S_{n}|\leq Z,|X_{1}-an|\leq n^{1/\kappa}\Big{]}\\!\\!\\!$
$\displaystyle\to\\!\\!\\!$
$\displaystyle\int_{-Z}^{Z}\mathbf{E}^{(\Lambda)}\big{[}e^{-\Lambda
z}H(e^{-z}(1+\hat{I}_{W-1}))\big{]}dz.$

Moreover, the limit coefficient converges as $W\to\infty$ and then
$Z\to\infty$ to

$\displaystyle\int_{-\infty}^{\infty}\mathbf{E}^{(\Lambda)}\big{[}e^{-\Lambda
z}H(e^{-z}(1+\hat{I}_{\infty}))\big{]}dz.$

Proof. By the duality lemma, we first have

$\displaystyle\mathbf{E}^{(\Lambda)}\Big{[}e^{-\Lambda
S_{n}}H(I_{n-W,n});|S_{n}|\leq Z,|X_{1}-an|\leq n^{1/\kappa}\Big{]}$
$\displaystyle=\\!\\!\\!$
$\displaystyle\mathbf{E}^{(\Lambda)}\Big{[}e^{-\Lambda
S_{n}}H(e^{-S_{n}}(1+\hat{I}_{W-1}));|S_{n}|\leq Z,|X_{n}-an|\leq
n^{1/\kappa}\Big{]}$ $\displaystyle=\\!\\!\\!$
$\displaystyle\int_{an-n^{1/\kappa}}^{an+n^{1/\kappa}}\mathbf{E}^{(\Lambda)}\Big{[}e^{-\Lambda(S_{n-1}+x)}H(e^{-S_{n-1}-x}(1+\hat{I}_{W-1}));|S_{n-1}+x|\leq
Z\Big{]}\mathbf{P}^{(\Lambda)}(X\in dx).$

By Assumption 2.13 and 2.14, for any $\delta,\epsilon>0$ there exists an
integer $n_{0}>1$ such that for any $n\geq n_{0}$ and $|y|\leq n^{1/\kappa}$,

$\displaystyle(1-\epsilon)B_{n}\cdot\delta\leq\mathbf{P}^{(\Lambda)}(X\in[an+y,an+y+\delta])\leq(1+\epsilon)B_{n}\cdot\delta.$

For $i\in\mathbb{Z}$, let $H_{n,\delta}(i)$ be the supremum of
$H(e^{-S_{n-1}-an-x}(1+\hat{I}_{W-1}))$ on the set $i\delta\leq
x<(i+1)\delta$. For any $y\in\mathbb{R}$, let $H_{\delta}(y)$ be the supremum
of $H(e^{-x}(1+\hat{I}_{W-1}))$ on the set $y-\delta\leq x<y+\delta$. A simple
calculation shows that

$\displaystyle B_{n}^{-1}\mathbf{E}^{(\Lambda)}\Big{[}e^{-\Lambda
S_{n}}H(I_{n-K,n});|S_{n}|\leq Z,|X_{1}-an|\leq n^{1/\kappa}\Big{]}$
$\displaystyle\leq\\!\\!\\!$
$\displaystyle(1+\epsilon)\sum_{i=-[n^{1/\kappa}/\delta]-1}^{[n^{1/\kappa}/\delta]+1}\mathbf{E}^{(\Lambda)}\Big{[}e^{-\Lambda(S_{n-1}+an+i\delta)}H_{n,\delta}(i);|S_{n-1}+an+i\delta|\leq
Z+\delta\Big{]}\cdot\delta$ $\displaystyle\leq\\!\\!\\!$
$\displaystyle(1+\epsilon)e^{\Lambda\delta}\int_{-n^{1/\kappa}-\delta}^{n^{1/\kappa}+\delta}\mathbf{E}^{(\Lambda)}\Big{[}e^{-\Lambda(S_{n-1}+an+y)}H_{\delta}(S_{n-1}+an+y);|S_{n-1}+an+y|\leq
Z+2\delta\Big{]}dy$ $\displaystyle=\\!\\!\\!$
$\displaystyle(1+\epsilon)e^{\Lambda\delta}\int_{-Z-2\delta}^{Z+2\delta}\mathbf{E}^{(\Lambda)}\Big{[}e^{-\Lambda
z}H_{\delta}(z);|S_{n-1}+an-z|\leq n^{1/\kappa}+\delta\Big{]}dz.$

Here the last equality following by changing the variables. The statements
below Condition 2.16 together with the arbitrariness of $\epsilon$ yields that
as $n\to\infty$ the last integral converges to

$\displaystyle
e^{\Lambda\delta}\int_{-Z-2\delta}^{Z+2\delta}\mathbf{E}^{(\Lambda)}\Big{[}e^{-\Lambda
z}H_{\delta}(z)\Big{]}dz.$

Letting $\delta\to 0+$, we have $H_{\delta}(z)\to H(e^{-z}(1+\hat{I}_{W-1}))$
for any $z\in\mathbb{R}$ and then by the dominated convergence theorem,

$\displaystyle\limsup_{n\to\infty}B_{n}^{-1}\mathbf{E}^{(\Lambda)}\Big{[}e^{-\Lambda
S_{n}}H(I_{n-K,n});|S_{n}|\leq Z,|X_{1}-an|\leq n^{1/\kappa}\Big{]}$
$\displaystyle\qquad\leq\int_{-Z}^{Z}\mathbf{E}^{(\Lambda)}\Big{[}e^{-\Lambda
z}H(e^{-z}(1+\hat{I}_{W-1}))\Big{]}dz.$

On the other hand, a similar argument also can yields that

$\displaystyle\liminf_{n\to\infty}B_{n}^{-1}\mathbf{E}^{(\Lambda)}\Big{[}e^{-\Lambda
S_{n}}H(I_{n-K,n});|S_{n}|\leq Z,|X_{1}-an|\leq n^{1/\kappa}\Big{]}$
$\displaystyle\qquad\geq\int_{-Z}^{Z}\mathbf{E}^{(\Lambda)}\Big{[}e^{-\Lambda
z}H(e^{-z}(1+\hat{I}_{W-1}))\Big{]}dz.$

Thus the first claim follows. The second one can be gotten by using dominated
convergence theorem as $W\to\infty$ and then the monotone convergence theorem
as $Z\to\infty$. $\Box$

Applying these two propositions to $F_{\mathscr{F}^{S}}^{b,Z}(k,n)$, we can
get the following corollary immediately.

###### Corollary 5.16

For every $b\in(0,1)$ and $k,Z\geq 1$, we have as $n\to\infty$

$\displaystyle
B_{n}^{-1}F_{\mathscr{F}^{S}}^{b,Z}(k,n)\to\int_{-Z}^{Z}\mathbf{E}^{(\Lambda)}\big{[}e^{-\Lambda
z}F\big{(}I_{k-1}+e^{-S_{k-1}-z}(1+\tilde{\hat{I}}_{\infty})\big{)}\big{|}\mathscr{F}^{S}\big{]}dz.$

Proof for Theorem 2.17. By all arguments and proposition above, we have

$\displaystyle\lim_{n\to\infty}B_{n}^{-1}\mathbf{E}^{(\Lambda)}[e^{-\Lambda
S_{n}}F(I_{n})]=\sum_{k=1}^{\infty}\lim_{Y\to\infty}\lim_{Z\to\infty}\lim_{n\to\infty}B_{n}^{-1}\mathbf{E}^{(\Lambda)}\big{[}e^{-\Lambda
S_{k-1}}F_{\mathscr{F}^{S}}^{b,Z}(k,n);|S_{k-1}|\leq Y\big{]}.$

Using Proposition 5.14 and 5.14, we have
$B_{n}^{-1}\mathbf{E}^{(\Lambda)}\big{[}e^{-\Lambda
S_{k-1}}F_{\mathscr{F}^{S}}^{b,Z}(k,n);|S_{k-1}|\leq Y\big{]}$ converges as
$n\to\infty$ to

$\displaystyle\int_{-Z}^{Z}\mathbf{E}^{(\Lambda)}\big{[}e^{-\Lambda
S_{k-1}-\Lambda
z}F\big{(}I_{k-1}+e^{-S_{k-1}-z}(1+\tilde{\hat{I}}_{\infty})\big{)};|S_{k-1}|\leq
Y\big{]}dz,$

which converges as $Z\to\infty$ and then $Y\to\infty$ to $C_{F,5}(k)$ by the
monotone convergence theorem and the dominated convergence theorem. $\Box$

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