Explosion Rates for Continuous-State Branching Processes in a Lévy Environment

Abstract

Here, we study the long-term behaviour of the non-explosion probability for continuous-state branching processes in a Lévy environment when the branching mechanism is given by the negative of the Laplace exponent of a subordinator. In order to do so, we study the law of this family of processes in the infinite mean case and provide necessary and sufficient conditions for the process to be conservative, i.e. that the process does not explode in finite time a.s. In addition, we establish precise rates for the non-explosion probabilities in the subcritical and critical regimes, first found by Palau et al. (ALEA Lat Am J Probab Math Stat 13(2):1235–1258, 2016) in the case when the branching mechanism is given by the negative of the Laplace exponent of a stable subordinator.

1 Introduction and Main Results

Let us consider $(\Omega ^{(b)}, {\mathcal {F}}^{(b)}, ({\mathcal {F}}^{(b)}_t)_{t\ge 0}, {\mathbb {P}}^{(b)})$ a filtered probability space satisfying the usual hypothesis. A continuous-state branching process (CSBP for short) $Y=(Y_t, t\ge 0)$ defined on $(\Omega ^{(b)}, {\mathcal {F}}^{(b)}, ({\mathcal {F}}^{(b)}_t)_{t\ge 0}, {\mathbb {P}}^{(b)})$ is a $[0,\infty ]$-valued Markov process with càdlàg paths whose laws satisfy the branching property; that is, the law of Y started from $x+y$ is the same as the law of the sum of two independent copies of Y but issued from x and y, respectively. More precisely, for all $\lambda \ge 0$ and $x,y \ge 0$

$$\begin{aligned} {\mathbb {E}}^{(b)}_{x+y}\left[ e^{-\lambda Y_t}\right] = {\mathbb {E}}_x^{(b)}\left[ e^{-\lambda Y_t}\right] {\mathbb {E}}_y^{(b)}\left[ e^{-\lambda Y_t}\right] . \end{aligned}$$

Moreover, the law of Y is completely characterised by the latter identity (see, for instance, Theorem 12.1 in [17]), i.e. for all $y>0$

$$\begin{aligned} {\mathbb {E}}^{(b)}_y\left[ e^{-\lambda Y_t}\right] = e^{-yu_t(\lambda )}, \end{aligned}$$

where $t\mapsto u_t(\lambda )$ is a differentiable function satisfying the following differential equation

$$\begin{aligned} \frac{\partial }{\partial t} u_t(\lambda ) + \psi (u_t(\lambda )) = 0 \quad \text {and} \quad u_0(\lambda ) = \lambda , \end{aligned}$$

(1)

where the function $\psi$ is called the branching mechanism of the CSBP Y. It is well known that the function $\psi$ is either the negative of the Laplace exponent of a subordinator or the Laplace exponent of a spectrally negative Lévy process (see, for example, Theorem 12.1 in [17]). In this manuscript, we focus on the subordinator case; that is, we assume

$$\begin{aligned} \psi (\lambda )= - \delta \lambda - \int _{(0,\infty )} (1-e^{-\lambda x}) \mu (\textrm{d} x), \qquad \lambda \ge 0, \end{aligned}$$

(2)

where $\delta \ge 0$ and $\mu$ is a measure supported in $(0,\infty )$ satisfying

$$\begin{aligned} \int _{(0,\infty )}(1\wedge x)\mu (\textrm{d} x)<\infty . \end{aligned}$$

(3)

It is important to note that the function $-\psi$ is also known in the literature as a Bernstein function, see, for instance, the monograph of Schilling et al. [23] for further details about this class of functions.

In this manuscript, we are interested in a particular extension of CSBPs by considering a random external force which affects the dynamics of the previous model. More precisely, we consider continuous-state branching processes in a random environment. Roughly speaking, a process in this class is a time-inhomogeneous Markov process taking values in $[0,\infty ]$ with 0 and $\infty$ as absorbing states. Furthermore, such processes satisfy a quenched branching property; that is, conditionally on the environment, the process started from $x+y$ is distributed as the sum of two independent copies of the same process but issued from x and y, respectively.

Actually the subclass of CSBPs in random environment that we are interested in pertains to instances where the external stochastic perturbation is driven by an independent Lévy process. This family of processes is known as continuous-state branching processes in Lévy environments (or CBLEs for short), and its construction has been given by He et al. [13] and by Palau and Pardo [19], independently, as the unique nonnegative strong solution of a stochastic differential equation which will be specified below.

We now construct the demographic or branching terms of the model. Let $N^{(b)}(\textrm{d} s, \textrm{d} z, \textrm{d} u)$ be a $({\mathcal {F}}_t^{(b)})_{t\ge 0}$-adapted Poisson random measure on ${\mathbb {R}}^3_+$ with intensity $\textrm{d} s \mu (\textrm{d} z)\textrm{d} u$, where $\mu$ is defined as above and satisfies the integral condition in (3). On the other hand, for the environmental term, we consider another filtered probability space $(\Omega ^{(e)}, {\mathcal {F}}^{(e)},({\mathcal {F}}^{(e)}_t)_{t\ge 0}, {\mathbb {P}}^{(e)})$ satisfying the usual hypotheses. Let us consider $\sigma \ge 0$ and $\alpha$ real constants and $\pi$ a measure concentrated on ${\mathbb {R}}\setminus \{0\}$ such that

$$\begin{aligned} \int _{{\mathbb {R}}} (1\wedge z^2)\pi (\textrm{d} z)<\infty . \end{aligned}$$

Suppose that $(B_t^{(e)}, t\ge 0)$ is a $({\mathcal {F}}_t^{(e)})_{t\ge 0}$—adapted standard Brownian motion, $N^{(e)}(\textrm{d} s, \textrm{d} z)$ is a $({\mathcal {F}}_t^{(e)})_{t\ge 0}$—Poisson random measure on ${\mathbb {R}}_+ \times {\mathbb {R}}$ with intensity $\textrm{d} s \pi (\textrm{d} z)$, and ${\widetilde{N}}^{(e)}(\textrm{d} s, \textrm{d} z)$ its compensated version. We denote by $S=(S_t, t\ge 0)$ a Lévy process, that is, a process with stationary and independent increments and càdlàg paths, with the following Lévy–Itô decomposition

$$\begin{aligned} S_t = \alpha t + \sigma B_t^{(e)} + \int _{0}^{t} \int _{(-1,1)} (e^z - 1) {\widetilde{N}}^{(e)}(\textrm{d} s, \textrm{d} z) + \int _{0}^{t} \int _{(-1,1)^c} (e^z - 1) N^{(e)}(\textrm{d} s, \textrm{d} z). \end{aligned}$$

Note that S has no jumps smaller than or equals -1 since the size of the jumps is given by the map $z\mapsto e^z-1$.

In our setting, the population size has no impact on the evolution of the environment and we are considering independent processes for the demography and the environment. More precisely, we work now on the space $(\Omega , {\mathcal {F}}, ({\mathcal {F}}_t)_{t\ge 0}, {\mathbb {P}})$ which is the direct product of the two probability spaces defined above, that is to say, $\Omega := \Omega ^{(e)} \times \Omega ^{(b)}, {\mathcal {F}}:= {\mathcal {F}}^{(e)}\otimes {\mathcal {F}}^{(b)}, {\mathcal {F}}_t:= {\mathcal {F}}^{(e)}_t \otimes {\mathcal {F}}^{(b)}_t$ for $t\ge 0$, and ${\mathbb {P}}:={\mathbb {P}}^{(e)} \otimes {\mathbb {P}}^{(b)}$. Therefore, the continuous-state branching process $Z=(Z_t, t\ge 0)$ in a Lévy environment S is defined on $(\Omega , {\mathcal {F}}, ({\mathcal {F}}_t)_{t\ge 0}, {\mathbb {P}})$ as the unique nonnegative strong solution of the following SDE

$$\begin{aligned} Z_t = Z_0 +\delta \int _{0}^{t} Z_s \textrm{d} s + \int _{0}^{t} \int _{[0,\infty )} \int _{0}^{Z_{s-}}z N^{(b)} (\textrm{d} s, \textrm{d} z, \textrm{d} u) + \int _{0}^{t} Z_{s-} \textrm{d} S_s. \end{aligned}$$

(4)

According to Theorem 1 in [19], the SDE (4) has pathwise uniqueness and strong solution up to explosion and by convention here it is identically equal to $\infty$ after the explosion time. Further, the process Z satisfies the strong Markov property and the quenched branching property. For further details, we refer to He et al. [13] or Palau and Pardo [19] in more general settings.

According to He et al. [13] and Palau and Pardo [19], the long-term behaviour of the process Z is deeply related to the behaviour and fluctuations of the Lévy process $\xi = ( \xi _t, t \ge 0)$, defined as follows

$$\begin{aligned} \xi _t = {\overline{\alpha }} t + \sigma B_t^{(e)} + \int _{0}^{t} \int _{(-1,1)} z {\widetilde{N}}^{(e)}(\textrm{d} s, \textrm{d} z) + \int _{0}^{t} \int _{(-1,1)^c}z N^{(e)}(\textrm{d} s, \textrm{d} z), \end{aligned}$$

(5)

where

$$\begin{aligned} {\overline{\alpha }}:= \alpha +\delta -\frac{\sigma ^2}{2} - \int _{(-1,1)} (e^z -1 -z) \pi (\textrm{d} z). \end{aligned}$$

Note that both processes $(S_t, t\ge 0)$ and $(\xi _t, t\ge 0)$ generate the same filtration. Actually, the process $\xi$ is obtained from S by changing the drift term and the jump sizes as follows

$$\begin{aligned} \alpha \mapsto {\overline{\alpha }}\qquad {\text {and}} \qquad \Delta S_t\mapsto \ln \left( \Delta S_t +1\right) , \end{aligned}$$

where $\Delta S_t=S_t-S_{t-}$, for $t\ge 0$. Furthermore, we point out that the drift term ${\overline{\alpha }}$ involves both branching and environment parameters. The relevance of this process stems from the fact that, in the finite mean case, the process $(Z_te^{-\xi _t}, t\ge 0)$ is a quenched martingale which allows to study the long-term behaviour of the process Z (see Proposition 1.1 in [3]). Moreover, the law of the process $(Z_te^{-\xi _t}, t\ge 0)$ can be characterised via a backward differential equation which is the analogue to (1) when the environment is fixed. In the infinite mean case, it is no longer true that $(Z_te^{-\xi _t}, t\ge 0)$ is a quenched martingale; however, in this paper we show that the law of $(Z_te^{-\xi _t}, t\ge 0)$ can be also characterised via a backward differential equation.

Recall that we say that Z is a conservative process if

$$\begin{aligned} {\mathbb {P}}_z(Z_t<\infty )=1,\quad \quad \text {for all}\quad t>0, \end{aligned}$$

(6)

where ${\mathbb {P}}_z$ denotes the law of Z starting at $z\ge 0$. Thus, when we think about the non-explosion event, the first question that might arise is: Under which conditions the process Z is conservative? When the environment is fixed, Grey in [12] provided a necessary and sufficient condition for the process to be conservative which depends on the integrability at 0 of the associated branching mechanism $\psi$. More precisely, a continuous-state branching process is conservative if and only if

$$\begin{aligned} \int _{0+} \frac{1}{|\psi (\lambda )|} \textrm{d} \lambda = \infty . \end{aligned}$$

Observe that a necessary condition is that $\psi (0)=0$ and a sufficient condition is that $\psi (0)=0$ and $|\psi '(0+)|<\infty$ (see, for instance, Theorem 12.3 in [17]). In contrast, in the particular case when the environment is driven by a Lévy process, Bansaye et al. [3] furnish a necessary condition under which the process Z is conservative. They proved that if the branching mechanism satisfies $|\psi '(0+)|<\infty$, then the associated CBLE is conservative (see Lemma 7 in [3]). It is worth noting that these results remain valid when the branching mechanism is given by the Laplace exponent of a spectrally negative Lévy process. Nevertheless, our focus here is on the case when the branching mechanism is given by (2) and satisfies $\psi '(0+)=-\infty$. So our first aim is twofold: first we characterise the law of the process $(Z_te^{-\xi _t}, t\ge 0)$ in the infinite mean case, which up to our knowledge is unknown, and then provide necessary and sufficient conditions for conservativeness when the branching mechanism is given as in (2) and satisfies $\psi '(0+)=-\infty$.

Our second aim deals with the asymptotic behaviour of the non-explosion probability which, up to our knowledge, has only been studied for the case where the associated branching mechanism is given by the negative of the Laplace exponent of a stable subordinator (see Proposition 2.1 in [20]), that is,

$$\begin{aligned} \psi (\lambda )= C\lambda ^{1+\beta },\quad \text {for all}\quad \lambda \ge 0, \end{aligned}$$

(7)

where $\beta \in (-1,0)$ and C is a negative constant. According to [19], the non-explosion probability for a CBLE with branching mechanism as in (7) (stable CBLE for short) is given by

$$\begin{aligned} {\mathbb {P}}_z(Z_t<\infty ) = {\mathbb {E}}\left[ \exp \left\{ -z\big (\beta C\texttt {I}_{0,t}(\beta \xi )\big )^{-1/\beta }\right\} \right] , \qquad z\ge 0, \end{aligned}$$

(8)

where $\texttt {I}_{0,t}(\beta \xi )$ denotes the exponential functional of the Lévy process $\beta \xi$, i.e.

$$\begin{aligned} \texttt {I}_{0,t}(\beta \xi ): = \int _{0}^{t} e^{-\beta \xi _s} \textrm{d} s. \end{aligned}$$

(9)

Observe from (8) that the process Z explodes with positive probability. Hence, our second aim is to study the rates of the non-explosion probability of CBLEs in a more general setting rather than the stable case.

According to Palau et al. [20], there are three different regimes for the asymptotic behaviour of the non-explosion probability that depends only on the mean of the underlying Lévy process $\xi$. They called these regimes: subcritical, critical and supercritical explosive depending on whether the mean of $\xi$ is negative, zero or positive (see Proposition 2.1 in [20]). Up to our knowledge, this is the only known result in the literature about explosion rates for CBLEs.

In particular, we study the speed of the non-explosion probability of the process Z, when the branching mechanism $\psi$ is given as in (2), in the critical and subcritical explosive regimes. In particular, we show that in the subcritical explosive regime, i.e. when the auxiliary Lévy process $\xi$ drifts to $-\infty$, and under an integrability condition, the limit of the non-explosion probability is positive. In the critical regime, i.e. when $\xi$ oscillates, and in particular when $\xi$ satisfies the so-called Spitzer’s condition plus an integrability condition, the non-explosion probability decays as a regularly varying function at $\infty$. Our arguments use strongly fluctuation theory and the asymptotic behaviour of exponential functionals of Lévy processes.

The supercritical explosive regime, i.e. when $\xi$ drifts to $\infty$, remains unknown, except in the stable case. We may expect, similarly as in the stable case, that under some positive exponential moments of $\xi$, the non-explosion probability decays exponentially as time increases. But it seems that this case requires other techniques than the ones developed in this article.

As we said before, the long-term behaviour of the non-explosion probability of Z is deeply related to the behaviour and fluctuations of $\xi$. Therefore, a few knowledge on fluctuation theory of Lévy process is required in order to state our main results.

1.1 Preliminaries on Lévy Processes

For simplicity, we denote by ${\mathbb {P}}^{(e)}_x$ the law of the process $\xi$ starting from $x\in {\mathbb {R}}$, and when $x=0$, we use the notation ${\mathbb {P}}^{(e)}$ for ${\mathbb {P}}^{(e)}_0$ (resp. ${\mathbb {E}}^{(e)}$ for ${\mathbb {E}}_0^{(e)}$). Let us denote by ${\widehat{\xi }}=-\xi$ the dual process. For every $x\in {\mathbb {R}}$, let $\widehat{{\mathbb {P}}}_x^{(e)}$ be the law of $x+\xi$ under $\widehat{{\mathbb {P}}}^{(e)}$, that is, the law of ${\widehat{\xi }}$ under ${\mathbb {P}}_{-x}^{(e)}$. In the sequel, we assume that $\xi$ is not a compound Poisson process since it is possible that in this case the process visits the same maxima or minima at distinct times which can make our analysis more involved.

Let us introduce the running infimum and supremum of $\xi$, by ${\underline{\xi }}=({\underline{\xi }}_t, t\ge 0)$ and ${\overline{\xi }}= ({\overline{\xi }}_t, t\ge 0)$, with

$$\begin{aligned} {\underline{\xi }}_t = \inf _{0\le s\le t} \xi _s \qquad {\text { and }} \qquad {\overline{\xi }}_t = \sup _{0 \le s \le t} \xi _s, \qquad t \ge 0. \end{aligned}$$

(10)

It is well known that the reflected process $\xi -{\underline{\xi }}$ (resp. ${\overline{\xi }}-\xi$) is a Markov process with respect to the filtration $({\mathcal {F}}^{(e)}_t)_{t\ge 0}$, see, for instance, Proposition VI.1 in [5]. We denote by $L=(L_t, t \ge 0 )$ and ${\widehat{L}}=({\widehat{L}}_t, t \ge 0 )$ the local times of ${\overline{\xi }}-\xi$ and $\xi -{\underline{\xi }}$ at 0, respectively, in the sense of Chapter IV in [5]. Next, define

$$\begin{aligned} H_t={\overline{\xi }}_{L_t^{-1}} \qquad \text {and}\qquad {\widehat{H}}_t=-{\underline{\xi }}_{{\widehat{L}}_t^{-1}} \qquad t\ge 0, \end{aligned}$$

(11)

where $L^{-1}$ and ${\widehat{L}}^{-1}$ are the right continuous inverse of the local times L and ${\widehat{L}}$, respectively. The range of the inverse local times, $L^{-1}$ (resp. ${\widehat{L}}^{-1}$), corresponds to the set of times at which new maxima (resp. new minima) occur. Hence, the range of the process H (resp. ${\widehat{H}}$) corresponds to the set of new maxima (resp. new minima). The pairs $(L^{-1}, H)$ and $({\widehat{L}}^{-1}, {\widehat{H}})$ are bivariate subordinators known as the ascending and descending ladder processes, respectively. The Laplace transform of $(L^{-1}, H)$ is such that for $\theta ,\lambda \ge 0$,

$$\begin{aligned} {\mathbb {E}}^{(e)}\left[ \exp \left\{ -\theta L^{-1}_t-\lambda H_t \right\} \right] =\exp \left\{ -t \kappa (\theta , \lambda )\right\} ,\qquad t\ge 0, \end{aligned}$$

(12)

where $\kappa (\cdot ,\cdot )$ denotes its bivariate Laplace exponent (resp. ${\widehat{\kappa }}(\cdot , \cdot )$ for the descending ladder process). Similarly to the absorption rates studied in [3, 8, 9], the asymptotic analysis of the event of explosion and the role of the initial condition involve the renewal functions U and ${\widehat{U}}$, associated with the supremum and infimum respectively, which are defined, as follows

$$\begin{aligned} U(x):= {\mathbb {E}}^{(e)}\left[ \int _{[0,\infty )} {\textbf{1}}_{\left\{ H_t\le x\right\} } \textrm{d} t\right] \quad {\text {and}}\quad {\widehat{U}}(x):= {\mathbb {E}}^{(e)}\left[ \int _{[0,\infty )} {\textbf{1}}_{\left\{ {\widehat{H}}_t\le x\right\} } \textrm{d} t\right] , \qquad x>0. \nonumber \\ \end{aligned}$$

(13)

The renewal functions U and ${\widehat{U}}$ are finite, subadditive, continuous and increasing. Moreover, they are identically 0 on $(-\infty , 0]$, strictly positive on $(0,\infty )$ and satisfy

$$\begin{aligned} U(x)\le C_1 x \qquad {\text {and}}\qquad {\widehat{U}}(x)\le C_2 x \quad \text { for any } \quad x\ge 0, \end{aligned}$$

(14)

where $C_1, C_2$ are finite constants (see, for instance, Lemma 6.4 and Section 8.2 in the monograph of Doney [10]). Moreover, $U(0)=0$ if 0 is regular upwards and $U(0)=1$ otherwise, similarly ${\widehat{U}}(0)=0$ if 0 is regular upwards and ${\widehat{U}}(0)=1$ otherwise.

Roughly speaking, the renewal function U(x) (resp. ${\widehat{U}}(x)$) “measures” the amount of time that the ascending (resp. descending) ladder height process spends on the interval [0, x] and in particular induces a measure on $[0,\infty )$ which is known as the renewal measure. The latter implies

$$\begin{aligned} \int _{[0,\infty )} e^{-\theta x} U( \textrm{d}x) = \frac{1}{\kappa (0,\theta )}, \qquad \theta >0. \end{aligned}$$

(15)

A similar expression holds true for the Laplace transform of the measure ${\widehat{U}}(\textrm{d}x)$ in terms of ${\widehat{\kappa }}$. For a more in-depth account of fluctuation theory, we refer the reader to the monographs of Bertoin [5], Doney [10] and Kyprianou [17].

1.2 Main Results

Our first main result determines the law of the reweighted process $(Z_te^{-\xi _t}, t\ge 0)$ via a backward differential equation in terms of $\psi _0(\lambda ):=\psi (\lambda )+\lambda \delta$ and the process $\xi$. The function $\psi _0$ plays now the role of the branching mechanism for the CBLE Z since the demographic term $\delta$ can be added to the environment S without changing the structure of S and the definition of Z in (4). In fact, we observe that the resulting process $(\delta t+S_t, t\ge 0)$ is still a Lévy process. To avoid any confusion, we call $\psi _0$ as the pure branching mechanism of the CBLE Z.

We denote by ${\mathbb {P}}_{(z,x)}$ the law of the process $(Z, \xi )$ starting at (z, x).

Theorem 1.1

For every $z > 0$, $x\in {\mathbb {R}}$, $\lambda \ge 0$ and $t\ge s\ge 0$, we have

$$\begin{aligned} {\mathbb {E}}_{(z,x)}\Big [\exp \{-\lambda Z_te^{-\xi _t}\}\ \Big |\Big . \ \xi , {\mathcal {F}}^{(b)}_s\Big ] = \exp \big \{- Z_se^{-\xi _s}v_t(s,\lambda ,\xi )\big \}, \end{aligned}$$

where for any $\lambda , t \ge 0$, the function $(v_t(s,\lambda , \xi ), s\in [0,t])$ is an a.s. solution of the backward differential equation

$$\begin{aligned} \frac{\partial }{\partial s} v_t(s,\lambda ,\xi ) =e^{\xi _s} \psi _0\big (v_t(s,\lambda ,\xi )e^{-\xi _s}\big ), \quad \text{ a.e. }\quad s\in [0,t] \end{aligned}$$

(16)

and with terminal condition $v_t(t,\lambda , \xi )=\lambda$. In particular, for every $z, \lambda , t > 0$ and $x\in {\mathbb {R}}$, we have

$$\begin{aligned} {\mathbb {E}}_{(z,x)}\Big [\exp \big \{-\lambda Z_t e^{-\xi _t}\big \} \Big |\Big . \ \xi \Big ] = \exp \left\{ -z v_t(0,\lambda e^{-x},\xi -x)\right\} . \end{aligned}$$

(17)

Our second main result furnishes a necessary and sufficient condition for a CBLE to be conservative, in our setting. The result is an extension of the original characterisation given by Grey [12] in the classical case for continuous-state branching process with constant environment.

Proposition 1.2

A continuous-state branching process in a Lévy environment with pure branching mechanism $\psi _0$ is conservative if and only if

$$\begin{aligned} \int _{0+} \frac{1}{|\psi _0(\lambda )|} \textrm{d} \lambda = \infty . \end{aligned}$$

(18)

In what follows, we assume that the pure branching mechanism $\psi _0$ satisfies

$$\begin{aligned} \int _{0+} \frac{1}{|\psi _0(\lambda )|} \textrm{d} \lambda < \infty , \end{aligned}$$

(19)

or in other words that Z may explode in finite time with positive probability. Under such assumption, we are interested in the asymptotic behaviour of the non-explosion probability in the following regimes: subcritical and critical explosive.

First we focus on the subcritical explosive regime, i.e. when the Lévy process drifts to $-\infty$. We recall that ${\overline{\alpha }}$ and $\pi$ are the drift term and the Lévy measure of $\xi$, respectively.

We introduce the following real function

$$\begin{aligned} A_{\xi }(x):= -{\overline{\alpha }} + {\bar{\pi }}^{(-)}(-1) + \int _{-x}^{-1}{\bar{\pi }}^{(-)}(y)\textrm{d} y, \qquad \text {for}\quad x>0, \end{aligned}$$

where ${\bar{\pi }}^{(-)}(-x)=\pi (-\infty , -x)$. We also introduce the function

$$\begin{aligned} \Phi _\lambda (u):= \int _{0}^{\infty } \exp \{- \lambda e^{u} y\} {\bar{\mu }}(y)\textrm{d} y, \qquad \text {for}\quad \lambda > 0, \end{aligned}$$

(20)

where ${\bar{\mu }}(x):= \mu (x,\infty )$. Further, let us denote by $\texttt {E}_1$ the exponential integral, i.e.

$$\begin{aligned} \texttt {E}_1(w)= \int _{1}^{\infty } \frac{e^{-wy}}{y}\textrm{d} y, \qquad w>0. \end{aligned}$$

(21)

We can then formulate the following theorem.

Theorem 1.3

(Subcritical explosive regime) Suppose that (19) holds, $\xi$ drifts to $-\infty$, ${\mathbb {P}}^{(e)}$ -a.s., and that there exists $\lambda >0$ such that

$$\begin{aligned} \int _{(a,\infty )} \frac{y}{A_{\xi }(y)}|\textrm{d}\Phi _\lambda (y)|<\infty , \qquad \text {for some}\quad a>0. \end{aligned}$$

(22)

Then, for any $z>0$,

$$\begin{aligned} \lim \limits _{t\rightarrow \infty } {\mathbb {P}}_{z}(Z_t < \infty ) >0. \end{aligned}$$

In particular, if  ${\mathbb {E}}^{(e)}\big [\xi _1\big ]\in (-\infty , 0)$, then the integral condition (22) is equivalent to

$$\begin{aligned} \int _{0}^{\infty } {\texttt {E}_1}(\lambda y) {\bar{\mu }}(y)\textrm{d} y<\infty . \end{aligned}$$

(23)

Note that, for $y> 0$, the following inequality for the exponential integral holds

$$\begin{aligned} {\texttt {E}_1}(\lambda y) \le e^{-\lambda y}\log \left( 1+\frac{1}{\lambda y}\right) . \end{aligned}$$

Therefore, in the case ${\mathbb {E}}^{(e)}\big [\xi _1\big ]\in (-\infty , 0)$, a simpler condition than (23) is the following

$$\begin{aligned} \int _{0}^{\infty } e^{-\lambda y} \log \left( 1+\frac{1}{\lambda y}\right) {\bar{\mu }}(y)\textrm{d} y<\infty . \end{aligned}$$

We also observe that since the mapping $\lambda \mapsto \Phi _\lambda (u)$ is decreasing, we have that if (22) holds for some $\lambda >0$, then it holds for any ${\tilde{\lambda }}>\lambda$. Moreover, note that $\log (1+1/y\lambda )=\log (1/y\lambda ) + \log (y\lambda +1)$ and since the problem is at zero only $\log (1/y\lambda )$ matters for the finiteness of the above integral.

Our next main result deals with the critical explosive regime. More precisely, we assume that $\xi$ satisfies the so-called Spitzer’s condition, i.e.

Bertoin and Doney in [6] showed that the later condition is equivalent to ${\mathbb {P}}^{(e)} (\xi _s\ge 0) \rightarrow \rho$ as $s\rightarrow \infty$. Recall from (12) that $\kappa (\theta , \lambda )$ is the Laplace exponent of the ascending ladder process $(L^{-1},H)$. Now, according to Theorem IV.12 in [5], Spitzer’s condition (A) is equivalent to $\theta \mapsto \kappa ( \theta , 0)$ being regularly varying at zero with index $\rho \in (0,1)$, that is,

$$\begin{aligned} \lim \limits _{t\downarrow 0} \frac{\kappa (c t, 0)}{\kappa (t, 0)} =c^\rho , \quad \quad \text {for all}\quad \quad c>0. \end{aligned}$$

In addition, the function $\theta \mapsto \kappa ( \theta , 0)$ may always be written in the form

$$\begin{aligned} \kappa (t,0) = t^\rho \, {\widetilde{\ell }}(t), \end{aligned}$$

(24)

where ${\widetilde{\ell }}$ is a slowly varying function at $0^+$, i.e. for all positive constant c, it holds

$$\begin{aligned} \lim \limits _{x\downarrow 0}\frac{{\widetilde{\ell }}(c x)}{{\widetilde{\ell }}(x)}=1. \end{aligned}$$

According to Theorems VI.14 and VI.18 in [5] or Remark 3.5 in [16], Spitzer’s condition determines the asymptotic behaviour of the probability that the Lévy process $\xi$ remains negative. In other words, under Spitzer’s condition (A) and for $x<0$, we have

$$\begin{aligned} \lim \limits _{t\rightarrow \infty } \frac{\sqrt{\pi }}{\kappa (1/t, 0)} {\mathbb {P}}_{x}^{(e)}\big ({\overline{\xi }}_t<0\big ) =\lim \limits _{t\rightarrow \infty } \frac{\sqrt{\pi }}{\kappa (1/t, 0)} \widehat{{\mathbb {P}}}_{-x}^{(e)}\big ({\underline{\xi }}_t> 0\big )= U(-x), \end{aligned}$$

(25)

where we recall that $U(\cdot )$ is the renewal function for the ascending ladder height, defined in (13).

In order to control the effect of the environment on the event of non-explosion, we need other assumptions. The following integrability condition is needed to guarantee the non-explosion of the process in unfavourable environments. Let us assume

In particular, from (14), we observe that the previous condition is satisfied if

$$\begin{aligned} \int _{0^+} z\ln ^2(z) \mu (\textrm{d} z)< \infty . \end{aligned}$$

On the other hand, we shall assume that the pure branching process of the CBLE Z is lower bounded by a stable branching mechanism whose associated CBLE explodes with positive probability. More precisely, we assume that

The above condition is necessary to deal with the functional $v_t(s,\lambda ,\xi )$ and to obtain an upper bound for the speed of non-explosion when the sample paths of the Lévy process have a high running supremum (see Proposition 5.4 for details).

Roughly speaking, our aim is to show, under the above conditions, that the probability of non-explosion varies regularly at $\infty$ with index $\rho$.

Theorem 1.4

(Critical-explosion regime) Suppose that (19) holds and that conditions (A), (B) and (C) are also fulfilled. Then, for any $z>0$, there exists $0< {\mathfrak {C}}(z) <\infty$ such that

$$\begin{aligned} \lim \limits _{t\rightarrow \infty } \frac{1}{\kappa (1/t,0)} {\mathbb {P}}_{z}(Z_t < \infty ) = {\mathfrak {C}}(z). \end{aligned}$$

The previous result provides evidence that the asymptotic behaviour of the non-explosion probability is deeply related to the fluctuations of the Lévy environment $\xi$.

The remainder of this paper is devoted to the proofs of the main results. In Sect. 2, we present the proofs of Theorem 1.1 and Proposition 1.2. In Sect. 3, the proof of Theorem 1.3 is given. In Sect. 4, we introduce continuous-state branching processes in a conditioned Lévy environment. This conditioned version is required to study the long-term behaviour of the non-explosion probability in the critical regime. Section 5 is devoted to the long-term behaviour results for the critical regime.

2 Proofs of Theorem 1.1 and Proposition 1.2

We first deal with the proof of Theorem 1.1 which relies on the extension of the classical Carathéodory’s theorem for ordinary differential equations that we state here for completeness. For its proof, the reader is referred to Theorems 1.1, 2.1 and 2.3 in Person [22].

Theorem 2.1

(Extended Carathéodory’s existence theorem) Let $I=[-b,b]$ with $b>0$. Assume that the function $f: I\times {\mathbb {R}} \rightarrow {\mathbb {R}}$ satisfies the following conditions:

• (i) the mapping $s\mapsto f(s,\theta )$ is measurable for each fixed $\theta \in {\mathbb {R}}$,

• (ii) the mapping $\theta \mapsto f(s,\theta )$ is continuous for each fixed $s \in I$,

• (iii) there exists a Lebesgue-integrable function m on the interval I such that

  $$\begin{aligned} |f(s,\theta )| \le m(s) \big ( 1+|\theta |\big ), \quad (s,\theta )\in I\times {\mathbb {R}}. \end{aligned}$$

Then, there exists an absolutely continuous function u(x) such that

$$\begin{aligned} u(x) =\int _{0}^{x} f(y,u(y))\textrm{d} y, \quad \quad x\in I. \end{aligned}$$

(26)

On the other hand, by integration by parts, we note that the function $|\psi _0|=-\psi _0$ can also be rewritten as follows

$$\begin{aligned} |\psi _0(\lambda )|= \lambda \int _{0}^\infty e^{-\lambda x} {\bar{\mu }}(x)\textrm{d} x, \end{aligned}$$

(27)

where we recall that ${\bar{\mu }}(x)= \mu (x,\infty )$. The previous expression will be useful for what follows.

Proof of Theorem 1.1

The first part of the proof follows from similar arguments as those used in [4] and [19] in the case of finite mean (i.e. when $|\psi _0^\prime (0+)|<\infty$) whenever there is an a.s. solution of the backward differential equation (16). We present its proof for the sake of completeness.

We first deduce an explicit expression for the reweighted process $Z_te^{\xi _t}$, for $t\ge 0$. In order to do so, we consider the function $f(x,y)= xe^{-y}$ and apply Itô’s formula (see, for example, Theorem II.5.1 in [14]). Then observing that $f_x'(x,y) = e^{-y}, f_y'(x,y)= -xe^{-y}, f_{xy}''(x,y)=f_{yx}''(x,y)=-e^{-y}, f_{xx}''(x,y)=0, f_{yy}''(x,y)= xe^{-y}$ and applying directly Theorem II.5.1 in [14], we see

$$\begin{aligned} Z_te^{-\xi _t} =&Z_0 e^{-\xi _0} +\sigma \int _0^t f_x'(Z_s, \xi _s) Z_s\textrm{d} B^{(e)}_s +\sigma \int _0^t f_y'(Z_s, \xi _s) \textrm{d} B^{(e)}_s\\&+(\alpha +\delta )\int _0^t f_x'(Z_s, \xi _s) Z_s\textrm{d} s+ {\overline{\alpha }}\int _0^t f_y'(Z_s, \xi _s)\textrm{d} s+\frac{\sigma ^2}{2}\int _0^t f_{xx}^{''}(Z_s, \xi _s) Z_s^2\textrm{d} s \\&+\frac{\sigma ^2}{2}\int _0^t f_{yy}^{''}(Z_s, \xi _s) \textrm{d} s+\frac{\sigma ^2}{2}\int _0^t f_{xy}^{''}(Z_s, \xi _s) Z_s\textrm{d} s+ \frac{\sigma ^2}{2}\int _0^t f_{yx}^{''}(Z_s, \xi _s) Z_s\textrm{d} s\\&+\int _0^t\int _{[0,\infty )}\int _0^{Z_{s-}}\left( f(Z_{s-}+z, \xi _{s-})-f(Z_{s-}, \xi _{s-})\right) N^{(b)}(\textrm{d} s, \textrm{d} z, \textrm{d} u)\\&+\int _0^t\int _{(-1,1)^c}\left( f(Z_{s-}+Z_{s-}(e^x-1), \xi _{s-}+x)-f(Z_{s-}, \xi _{s-})\right) N^{(e)}(\textrm{d} s, \textrm{d} x)\\&+\int _0^t\int _{(-1,1)}\left( f(Z_{s-}+Z_{s-}(e^x-1), \xi _{s-}+x)-f(Z_{s-}, \xi _{s-})\right) {\widetilde{N}}^{(e)}(\textrm{d} s, \textrm{d} x)\\&+\int _0^t\int _{(-1,1)}\left( f(Z_{s}+Z_{s}(e^x-1), \xi _{s}+x)-f(Z_{s}, \xi _{s})\right. \\&\left. -Z_s(e^x-1)f^{'}_x(Z_s, \xi _{s})-xf^{'}_y(Z_s, \xi _{s})\right) \pi (\textrm{d} x) \textrm{d} s. \end{aligned}$$

By replacing all the factors in the above identity and recalling the definition of ${\overline{\alpha }}$ below identity (5), we obtain

$$\begin{aligned} Z_te^{-\xi _t} = Z_0e^{-\xi _0} + \int _{0}^{t} \int _{[0, \infty )} \int _{0}^{Z_{s-}} ze^{-\xi _{s-}}N^{(b)}(\textrm{d}s, \textrm{d}z, \textrm{d} u). \end{aligned}$$

(28)

Next, we fix $\lambda \ge 0$ and $t\ge s\ge 0$. We assume for now that $(v_t(s,\lambda , \xi ), s\in [0,t])$ is solution of the backward differential equation (16). For simplicity on exposition, we denote by $H_t(s)= \exp \{-Z_se^{-\xi _s}v_t(s,\lambda , \xi )\}$ and we apply again Itô’s formula to the processes $(Z_se^{-\xi _s}, s\le t)$ and $(v_t(s,\lambda , \xi ), s\le t)$ with the function $f(x,y)=e^{-xy}$, that is,

$$\begin{aligned} H_t(t)&= H_t(s) + \int _{s}^{t}f_y'(Z_{r}e^{\xi _{r}}, v_t(r, \lambda ,\xi )) \textrm{d}(v_t(r, \lambda ,\xi )) \\&\quad + \int _{s}^{t} \int _{[0,\infty )} \int _{0}^{Z_{s-}} \left( f\Big ((Z_{r-}+z)e^{-\xi _{r-}}, v_t(r,\lambda ,\xi )\Big ) \right. \\&\quad - f(Z_{r-}e^{\xi _{r-}}, v_t(r, \lambda ,\xi )) \Big ) N^{(b)}(\textrm{d}r, \textrm{d}z, \textrm{d} u). \end{aligned}$$

Now using (16), we get

$$\begin{aligned} H_t(t)&= H_t(s) - \int _{s}^{t} H_t(r) Z_r e^{-\xi _r} e^{\xi _r} \psi _0\big (v_t(r,\lambda ,\xi )e^{-\xi _r}\big ) \textrm{d} r \\&\quad + \int _{s}^{t} \int _{[0,\infty )} \int _{0}^{Z_{r-}} H_t(r-) \Big (e^{-ze^{-\xi _{r-}}v_t(r,\lambda , \xi )}-1 \Big ) N^{(b)}(\textrm{d}r, \textrm{d}z, \textrm{d} u) \\&= H_t(s) - \int _{s}^{t} \int _{[0,\infty )}H_t(r) Z_r \Big (e^{-ze^{-\xi _{r}}v_t(r,\lambda , \xi )}-1 \Big ) \mu (\textrm{d} z)\textrm{d} r \\&\quad + \int _{s}^{t} \int _{[0,\infty )} \int _{0}^{Z_{r-}} H_t(r-) \Big (e^{-ze^{-\xi _{r-}}v_t(r,\lambda , \xi )}-1 \Big ) N^{(b)}(\textrm{d}r, \textrm{d}z, \textrm{d} u) \\&= H_t(s) - \int _{s}^{t} \int _{[0,\infty )} \int _{0}^{Z_{s-}} H_t(r-) \Big (1-e^{-ze^{-\xi _{r-}}v_t(r,\lambda , \xi )} \Big ) {\widetilde{N}}^{(b)}(\textrm{d}s, \textrm{d} z, \textrm{d} r), \end{aligned}$$

where ${\widetilde{N}}^{(b)}(\textrm{d} s, \textrm{d} z, \textrm{d}r)$ denotes the compensated version of $N^{(b)}(\textrm{d} s, \textrm{d} z, \textrm{d}r)$. By taking conditional expectations in both sides, we get

$$\begin{aligned}{\mathbb {E}}_{(z,x)}\left[ H_t(t) \Big | \xi , {\mathcal {F}}^{(b)}_s\right] = H_t(s),\end{aligned}$$

as expected.

In order to show the existence of Eq. (16), we will appeal to the extended version of Carathéodory’s existence Theorem 2.1. Fix $\omega \in \Omega ^{(e)}$ and $t, \lambda \ge 0$. Denote by $f: [0,t] \times {\mathbb {R}} \rightarrow [-\infty ,0]$ the following function

$$\begin{aligned} f(s,\theta )= e^{\xi _s(\omega )}\psi _0\big (\theta e^{-\xi _s(\omega )}\big ). \end{aligned}$$

In the following, we omit the notation $\omega$ for the sake of brevity. First, we observe that the mapping $s\mapsto f(s,\theta )$ is measurable for each fixed $\theta \in {\mathbb {R}}$. Indeed, the latter follows from the fact that $\xi =(\xi _s,\ s \ge 0)$ possesses càdlàg paths and is $({\mathcal {F}}^{(e)}_t)_{t\ge 0}$-adapted. More precisely, the application $(s,\omega )\mapsto \xi _s$ is ${\mathcal {B}}([0,t])\otimes {\mathcal {F}}^{(e)}_t$-measurable implying that the mapping $s\mapsto f(s, \theta )$ is ${\mathcal {B}}([0,t])$-measurable for each fixed $\theta \in {\mathbb {R}}$ and $\omega \in \Omega ^{(e)}$. Furthermore, we have that the function $\theta \mapsto f(s,\theta )$ is continuous for each fixed $s\in [0,t]$ since $\psi _0$ is continuous. Therefore, according to Theorem 2.1, the proof is completed once we have shown that there exists an integrable function m on [0, t], such that for any $(s,\theta ) \in [0,t]\times {\mathbb {R}}$

$$\begin{aligned} |f(s,\theta )|\le m(s)\big (1+ |\theta |\big ). \end{aligned}$$

Note that for any $(s,\theta ) \in [0,t]\times {\mathbb {R}}$, we have

$$\begin{aligned} |f(s,\theta )|= & {} \big |e^{\xi _s}\psi _0\big (\theta e^{-\xi _s}\big )\big | = e^{\xi _s}|\psi _0\big ( \theta e^{-\xi _s}\big )|{\textbf{1}}_{\{\xi _s\ge 0\}} + e^{\xi _s}|\psi _0\big (\theta e^{-\xi _s}\big )|{\textbf{1}}_{\{\xi _s< 0\}}\\\le & {} e^{\xi _s}|\psi _0(\theta )|{\textbf{1}}_{\{\xi _s\ge 0\}} +e^{\xi _s}|\psi _0\big (\theta e^{-\xi _s}\big )|{\textbf{1}}_{\{\xi _s < 0\}}, \end{aligned}$$

where in the last inequality, we have used that $|\psi _0|$ is an increasing function. Now, since $|\psi _0|$ is a concave function, it is well known that for any $\theta >0$ and $k>1$, we have $|\psi _0(\theta )|\le k|\psi _0(\theta /k)|$ (see, for instance, the proof of [5, Proposition III. 1]). In particular, this inequality implies

$$\begin{aligned} |f(s,\theta )| \le e^{\xi _s}|\psi _0(\theta )|{\textbf{1}}_{\{\xi _s\ge 0\}} +|\psi _0(\theta )|{\textbf{1}}_{\{\xi _s< 0\}} \le \max \{e^{\xi _s}{\textbf{1}}_{\{\xi _s\ge 0\}}, {\textbf{1}}_{\{\xi _s < 0\}}\}|\psi _0(\theta )|. \end{aligned}$$

On the other hand, since $|\psi _0|$ is a Bernstein function, it is well known that there exists $\texttt {c}, d>0$, such that $|\psi _0(\theta )|\le \texttt {c}+d\theta$ for any $\theta \ge 0$ (see, for instance, Corollary 3.8 in [23]). It turns out that

$$\begin{aligned} |f(s,\theta )|\le & {} m(s) \big (1 + |\theta |\big ),\quad \quad \text {for all} \quad \quad (s,\theta ) \in [0,t]\times {\mathbb {R}}, \end{aligned}$$

where

$$\begin{aligned} m(s):=(\texttt {c}\vee d)\max \{e^{\xi _s}{\textbf{1}}_{\{\xi _s\ge 0\}}, {\textbf{1}}_{\{\xi _s < 0\}}\}, \quad \quad s \in [0,t]. \end{aligned}$$

Note that m is an integrable function on [0, t] since the Lévy process $\xi$ has càdlàg paths. Finally, thanks to Theorem 2.1, there exists an a.s. solution of (16). $\square$

The functional $v_t(s,\lambda ,\xi )$ has useful monotonicity properties as it is stated in the following lemma. In the forthcoming sections, we will make use of these properties.

Lemma 2.2

For any $\lambda \ge 0$ and $t\ge 0$, the mapping $s \mapsto v_t(s,\lambda , \xi )$ is decreasing on [0, t]. For any $s\in [0,t]$, the mapping $\lambda \mapsto v_t(s,\lambda ,\xi )$ is increasing on $[0,\infty )$.

Proof

Recall that $\psi _0(\theta )\le 0$. Then, from the backward differential equation (16), we see that the function $s \mapsto v_t(s,\lambda ,\xi )$ is decreasing on [0, t]. Further, from (17) we observe that the mapping $\lambda \mapsto v_t(s,\lambda ,\xi )$ is increasing on $[0,\infty )$ as required. $\square$

We conclude this section with the proof of Proposition 1.2. Before we do so, let us make an important remark about the non-explosion and extinction probabilities. It is easy to deduce, by letting $\lambda \downarrow 0$ in (17) and with the help of the Monotone Convergence Theorem, that the non-explosion probability is given by

$$\begin{aligned} {\mathbb {P}}_{(z,x)}\big (Z_t<\infty \ | \ \xi \big ) = \exp \left\{ -z\lim \limits _{\lambda \downarrow 0} v_t(0,\lambda e^{-x},\xi -x)\right\} , \quad z, t >0, \quad x\in {\mathbb {R}}. \nonumber \\ \end{aligned}$$

(29)

With this in hand, we may now observe that the process Z is conservative if and only if

$$\begin{aligned} \lim \limits _{\lambda \downarrow 0} v_t(0,\lambda e^{-x},\xi -x)=0, \quad \quad \text {for all} \quad \quad t>0. \end{aligned}$$

(30)

On the other hand, we also observe that, by letting $\lambda \uparrow \infty$ in (17) and using again the Monotone Convergence Theorem, the probability of extinction is such that

$$\begin{aligned} {\mathbb {P}}_{(z,x)}\big (Z_t=0 \ | \ \xi \big ) = \exp \left\{ -z\lim \limits _{\lambda \uparrow \infty } v_t(0,\lambda e^{-x},\xi -x)\right\} , \quad z, t >0, \quad x\in {\mathbb {R}}. \end{aligned}$$

Moreover, from Lemma 2.2 since $v_t(t,\lambda e^{-x},\xi -x)=\lambda e^{-x}$, we have that

$$\begin{aligned} v_t(0,\lambda e^{-x},\xi -x)\ge \lambda e^{-x}, \qquad \text {for }\quad \lambda , t> 0, \end{aligned}$$

which combined with the previous identity clearly implies that

$$\begin{aligned} {\mathbb {P}}_{(z,x)}\big (Z_t>0 \ | \ \xi \big ) =1,\quad \quad \text {for all} \quad \quad z, t>0. \end{aligned}$$

(31)

Proof of Proposition 1.2

Fix $t>0$ and recall from (10) that $({\underline{\xi }}_t, t\ge 0)$ and $({\overline{\xi }}_t, t\ge 0)$ denote the running infimum and supremum of the process $\xi$, respectively. First, we assume that the pure branching mechanism $\psi _0$ satisfies (18), that is to say,

$$\begin{aligned} \int _{0+}\frac{1}{|\psi _0(z)|}\textrm{d} z = \int _{0+}\frac{1}{-\psi _0(z)}\textrm{d} z = \infty . \end{aligned}$$

From Theorem 1.1, we see that the backward differential equation (16) can be rewritten as follows

$$\begin{aligned} -t=\int _{0}^{t} \frac{\textrm{d} v_t(s,\lambda ,\xi )}{-e^{\xi _s}\psi _0\big (e^{-\xi _s}v_t(s,\lambda ,\xi )\big )} = \int _{0}^{t} \frac{\textrm{d} v_t(s,\lambda ,\xi )}{e^{\xi _s}|\psi _0\big (e^{-\xi _s}v_t(s,\lambda ,\xi )\big )|}. \end{aligned}$$

Now, we recall that $|\psi _0|$ is an increasing and nonnegative function. Then appealing to the definition of the running infimum and supremum of $\xi$, we observe that the following inequality holds

$$\begin{aligned}\begin{aligned} -t = \int _{0}^{t} \frac{\textrm{d}v_t(s,\lambda ,\xi )}{e^{\xi _s}|\psi _0\big (e^{-\xi _s}v_t(s,\lambda ,\xi )\big )|}&\le \int _{0}^{t} \frac{\textrm{d}v_t(s,\lambda ,\xi )}{e^{{\underline{\xi }}_t} |\psi _0\big (e^{-{\overline{\xi }}_t}v_t(s,\lambda ,\xi )\big )|}\\ {}&= \frac{e^{{\overline{\xi }}_t}}{e^{{\underline{\xi }}_t}} \int _{e^{-{\overline{\xi }}_t} v_t(0,\lambda , \xi )}^{e^{-{\overline{\xi }}_t} \lambda } \frac{1}{|\psi _0(z)|}\textrm{d} z, \end{aligned} \end{aligned}$$

where in the last equality we have used the change of variables $z= e^{-{\overline{\xi }}_t}v_t(s,\lambda ,\xi )$. Next, letting $\lambda \downarrow 0$ in the previous inequality, we get

$$\begin{aligned} \frac{e^{{\overline{\xi }}_t}}{e^{{\underline{\xi }}_t}} \int _{0}^{e^{-{\overline{\xi }}_t} \lim \limits _{\lambda \downarrow 0}v_t(0,\lambda ,\xi )} \frac{1}{|\psi _0(z)|}\textrm{d} z \le t. \end{aligned}$$

(32)

Thus, taking into account our assumption, we are forced to conclude that

$$\begin{aligned} \lim _{\lambda \downarrow 0} v_t(0,\lambda ,\xi )=0. \end{aligned}$$

(33)

In other words, the process is conservative.

On the other hand, we assume that the process is conservative or equivalently that (33) holds. We will proceed by contradiction, and thus, we suppose that

$$\begin{aligned} \int _{0+}\frac{1}{|\psi _0(z)|}\textrm{d} z < \infty . \end{aligned}$$

Similar to the above arguments, we deduce that

$$\begin{aligned}\begin{aligned} -t = \int _{0}^{t} \frac{\textrm{d} v_t(s,\lambda ,\xi )}{-e^{\xi _s}\psi _0\big (e^{-\xi _s}v_t(s,\lambda ,\xi )\big )}&\ge \int _{0}^{t} \frac{\textrm{d} v_t(s,\lambda ,\xi )}{e^{{\overline{\xi }}_t} |\psi _0\big (e^{-{\underline{\xi }}_t}v_t(s,\lambda ,\xi )\big )|}\\ {}&= \frac{e^{{\underline{\xi }}_t}}{e^{{\overline{\xi }}_t}} \int _{e^{-{\underline{\xi }}_t} v_t(0,\lambda , \xi )}^{e^{-{\underline{\xi }}_t}\lambda } \frac{1}{|\psi _0(z)|}\textrm{d} z. \end{aligned}\end{aligned}$$

Taking $\epsilon >0$ sufficiently small, we see

$$\begin{aligned} t \le \frac{e^{{\underline{\xi }}_t}}{e^{{\overline{\xi }}_t}} \int _{e^{-{\underline{\xi }}_t} \lambda }^{\epsilon } \frac{1}{|\psi _0(z)|}\textrm{d} z - \frac{e^{{\underline{\xi }}_t}}{e^{{\overline{\xi }}_t}} \int _{e^{-{\underline{\xi }}_t} v_t(0,\lambda , \xi )}^{\epsilon } \frac{1}{|\psi _0(z)|}\textrm{d} z. \end{aligned}$$

Hence, by taking $\lambda \downarrow 0$ in the above inequality and using (33), we have $t\le 0$, which is a contradiction. Therefore, we deduce that the pure branching mechanism $\psi _0$ satisfies (18). $\square$

3 Subcritical Explosive Regime: Proof of Theorem 1.3

The proof of Theorem 1.3 follows similar ideas as those used in the proof of Proposition 3 in Palau and Pardo [19].

Proof of Theorem 1.3

Let $z>0$ and $x\in {\mathbb {R}}$. We begin by observing, from (4) or (29) that ${\mathbb {P}}_{(z,x)}(Z_t<\infty )$ does not depend of the initial value x of the Lévy process $\xi$. More precisely, from (4) we can observe that $Z_t$ depends only on the initial condition $Z_0=z$. Thus, again from (29), we see

$$\begin{aligned} \lim \limits _{t\rightarrow \infty } {\mathbb {P}}_{z}(Z_t<\infty ) = \lim \limits _{t\rightarrow \infty } {\mathbb {E}}^{(e)}_{x}\Big [\exp \left\{ -zv_t(0,0,\xi -x)\right\} \Big ]. \end{aligned}$$

Hence, as soon as we can establish that

$$\begin{aligned} \lim _{t\rightarrow \infty } v_t(0,0,\xi -x) < \infty , \quad {\mathbb {P}}_{x}^{(e)}-\text {a.s.}, \end{aligned}$$

our proof is completed.

From Lemma 2.2, we see that the mapping $s\mapsto v_t(s,\lambda e^{-x}, \xi -x)$ is decreasing and since $v_t(t,\lambda e^{-x},\xi - x)=\lambda e^{-x}$, we have $v_t(s,\lambda e^{-x},\xi -x)\ge \lambda e^{-x}$ for all $s\in [0,t]$. It follows that, for $s\in [0,t]$ and $\lambda > 0$,

$$\begin{aligned}\begin{aligned} \frac{\partial }{\partial s} v_t(s,\lambda e^{-x}, \xi -x)&=e^{\xi _s-x} \psi _0\big (v_t(s,\lambda e^{-x},\xi -x)e^{-\xi _s+x}\big )\\ {}&= -v_t(s,\lambda e^{-x},\xi -x)\int _{0}^\infty \exp \{-v_t(s,\lambda e^{-x}, \xi - x)e^{-\xi _s+x} z\} {\bar{\mu }}(z)\textrm{d}z \\ {}&\ge -v_t(s,\lambda e^{-x},\xi -x)\int _{0}^\infty \exp \{- \lambda e^{-x} e^{-\xi _s+x} z\} {\bar{\mu }}(z)\textrm{d} z. \end{aligned}\end{aligned}$$

Therefore by integrating,

$$\begin{aligned} \log v_t(0,\lambda e^{-x},\xi -x)\le \log (\lambda e^{-x}) +\int _{0}^{t} \int _{0}^\infty \exp \{- \lambda e^{-\xi _s} z\} {\bar{\mu }}(z)\textrm{d} z \textrm{d} s. \end{aligned}$$

Moreover, from Lemma 2.2, we have that the mapping $\lambda e^{-x} \mapsto v_t(s,\lambda e^{-x},\xi -x)$ is increasing. It turns out that

$$\begin{aligned} v_t(0,0,\xi -x)\le v_t(0,\lambda e^{-x},\xi -x)\le \lambda e^{-x} \exp \left( \int _{0}^{t} \int _{0}^\infty \exp \{- \lambda e^{-\xi _s} z\} {\bar{\mu }}(z)\textrm{d} z \textrm{d} s\right) . \nonumber \\ \end{aligned}$$

(34)

Using the definition of $\Phi _\lambda$ in (20), we get

$$\begin{aligned} \lim \limits _{t\rightarrow \infty }v_t(0,0,\xi -x) \le \lambda e^{-x}\exp \left( \int _{0}^{\infty }\Phi _\lambda (-\xi _s) \textrm{d} s\right) . \end{aligned}$$

(35)

We recall that in this regime the process $-\xi$ drifts to $\infty$, ${\mathbb {P}}^{(e)}_x$ -a.s. Thus, in order to prove that the integral in (35) is finite, let us introduce $\varsigma =\sup \{t\ge 0: -\xi _t\le 0\}$ and observe that

$$\begin{aligned} \int _{0}^{\infty }\Phi _\lambda (-\xi _s) \textrm{d} s = \int _{0}^{\varsigma }\Phi _\lambda (-\xi _s) \textrm{d} s\ + \ \int _{\varsigma }^{\infty }\Phi _\lambda (-\xi _s) \textrm{d} s. \end{aligned}$$

Since $\varsigma <\infty$, ${\mathbb {P}}^{(e)}_x$ -a.s., it follows that the first integral in the right-hand side above is finite ${\mathbb {P}}^{(e)}_x$-a.s. For the second integral, we may appeal to Theorem 1 in Erickson and Maller [11] or the main result in Kolb and Savov [15]. It is important to note that the main result in [15] extends the result in [11] but both results coincide in this particular case. Despite that the aforementioned theorem is written in terms of Lévy processes drifting to $+\infty$, it is not so difficult to see that it can be rewritten, using duality, in terms of Lévy processes drifting to $-\infty$ which is how we are using it here. More precisely, since $\Phi _\lambda$ is a finite positive, non-constant and non-increasing function on $[0,\infty )$ and condition (22) holds, Theorem 1 in [11] guarantees that

$$\begin{aligned}\int _{\varsigma }^{\infty }\Phi _\lambda (-\xi _s) \textrm{d} s<\infty , \qquad {\mathbb {P}}^{(e)}_x-\text {a.s.}\end{aligned}$$

Furthermore, if ${\mathbb {E}}^{(e)}\big [\xi _1\big ]\in (-\infty ,0)$, then $\lim \limits _{x\rightarrow \infty }A_{\xi }(x)$ is finite. In particular, it follows by integration by parts that the integral condition (22) is equivalent to

$$\begin{aligned}\int _{0}^{\infty } \Phi _\lambda (u)\textrm{d} u <\infty .\end{aligned}$$

Moreover, we have

$$\begin{aligned} \int _{0}^{\infty } \Phi _\lambda (u)\textrm{d} u&= \int _{0}^{\infty } \int _{0}^\infty \exp \{- \lambda e^{u} y\} {\bar{\mu }}(y)\textrm{d} y\textrm{d} u \\&= \int _{0}^{\infty } \int _{1}^{\infty } \frac{\exp \{- \lambda y w \}}{w} \textrm{d} w{\bar{\mu }}(y)\textrm{d} y. \end{aligned}$$

Now by the definition of the exponential integral given in (21), we deduce that condition (22) is equivalent to

$$\begin{aligned}\int _{0}^{\infty } {\texttt {E}_1}(\lambda y) {\bar{\mu }}(y)\textrm{d} y<\infty ,\end{aligned}$$

which concludes the proof. $\square$

4 CSBPs in a Conditioned Lévy Environment

Throughout this section, we shall suppose that the Lévy process $\xi$ satisfies Spitzer’s condition (A) which in particular implies that the process oscillates.

As mentioned earlier, the asymptotic behaviour of the non-explosion probability is related to fluctuations of the Lévy process $\xi$, specially to its running supremum. For our purpose, we need to introduce the definition of a CSBP conditioned to stay negative, which roughly speaking means that the running supremum of the auxiliary Lévy process $\xi$ is negative. Let us therefore spend some time in this section gathering together some of the facts of this conditioned version.

Similarly to the definition of Lévy processes conditioned to stay positive and following a similar strategy as in the discrete framework in Afanasyev et al. [1], we would like to introduce a continuous-state branching process in a Lévy environment conditioned to stay negative as a Doob-h transform. The aforementioned process was first investigated by Bansaye et al. [3] with the aim to study the survival event in a critical Lévy environment.

Lemma 4.1

(Bansaye et. al. [3]) Let $z,x >0$. The process $\{{\widehat{U}}(\xi _t){\textbf{1}}_{\{{\underline{\xi }}_t> 0\}}, t\ge 0\}$ is a martingale with respect to $({\mathcal {F}}_t)_{t\ge 0}$ and under ${\mathbb {P}}_{(z,x)}$.

With this in hand, they introduce the law of a continuous-state branching process in a Lévy environment $\xi$ conditioned to stay positive as follows, for $\Lambda \in {\mathcal {F}}_t$, $z,x>0$,

$$\begin{aligned} {\mathbb {P}}^{\uparrow }_{(z,x)}(\Lambda ):=\frac{1}{{\widehat{U}}(x)}{\mathbb {E}}_{(z,x)}\big [{\widehat{U}}(\xi _t){\textbf{1}}_{\{{\underline{\xi }}_t> 0\}}{\textbf{1}}_\Lambda \big ], \end{aligned}$$

where ${\widehat{U}}$ is the renewal function defined in (13). It is natural therefore to cast an eye on similar issues for the study of non-explosion events in a Lévy environment. In contrast, we introduce here the process Z in a Lévy environment $\xi$ conditioned to stay negative. Recall that ${\widehat{\xi }}$ is the dual process of $\xi$.

Appealing to duality and Lemma 4.1, we can see that the process $\{U(-\xi _t){\textbf{1}}_{\{{\overline{\xi }}_t< 0\}}, t\ge 0\}$ is a martingale with respect to $({\mathcal {F}}_t)_{t\ge 0}$ and under ${\mathbb {P}}_{(z,x)}$ with $z>0$ and $x<0$. Then, we introduce the law of the continuous-state branching process in a Lévy environment $\xi$ conditioned to stay negative, as follows: for $\Lambda \in {\mathcal {F}}_t$ for $z>0$ and $x<0$,

$$\begin{aligned} {\mathbb {P}}^{\downarrow }_{(z,x)}(\Lambda ): =\frac{1}{U(-x)}{\mathbb {E}}_{(z,x)}\big [U(-\xi _t){\textbf{1}}_{\{{\overline{\xi }}_t< 0\}}{\textbf{1}}_\Lambda \big ]. \end{aligned}$$

(36)

Intuitively speaking, ${\mathbb {P}}^{\uparrow }_{(z,x)}$ and ${\mathbb {P}}^{\downarrow }_{(z,x)}$ correspond to the law of $(Z,\xi )$ conditioning the random environment $\xi$ to not enter $(-\infty ,0)$ and $(0,\infty )$, respectively.

The following convergence result is crucial for Theorem 1.4. The proof can be obtained directly using duality and Lemma 3.2 in [3].

Lemma 4.2

Fix $z>0,\ x<0$ and assume that Spitzer’s condition (A) holds. Let $R_s$ be a bounded real-valued ${\mathcal {F}}_s$-measurable random variable. Then,

$$\begin{aligned} \lim \limits _{t\rightarrow \infty } {\mathbb {E}}_{(z,x)} \big [R_s \ | \ {\overline{\xi }}_t < 0 \big ]= {\mathbb {E}}^{ \downarrow }_{(z,x)}\big [R_s\big ]. \end{aligned}$$

More generally, let $(R_t, t\ge 0)$ be a uniformly bounded real-valued process adapted to the filtration $({\mathcal {F}}_t, t\ge 0)$, which converges ${\mathbb {P}}^{\downarrow }_{(z,x)}$-a.s., to some random variable $R_\infty$. Then,

$$\begin{aligned} \lim \limits _{t\rightarrow \infty }{\mathbb {E}}_{(z,x)}\big [R_t \ | \ {\overline{\xi }}_t < 0\big ] = {\mathbb {E}}^{ \downarrow }_{(z,x)}\big [R_\infty \big ]. \end{aligned}$$

Recall from Theorem 1.1 that the quenched law of the process $(Z_t e^{-\xi _t}, t\ge 0)$ is completely characterised by the functional $v_t(s,\lambda ,\xi )$. In the case of conditioned environment, we have a similar result. We formalise this in the following lemma whose proof essentially mimics the steps of Proposition 3.3 in [3] and identity (31).

Lemma 4.3

For each $z>0$, $x<0$ and $\lambda \ge 0$, we have

$$\begin{aligned} {\mathbb {E}}_{(z,x)}^{\downarrow }\Big [\exp \big \{-\lambda Z_t e^{-\xi _t}\big \}\Big ] = {\mathbb {E}}_x^{(e),\downarrow }\Big [\exp \{-z v_t(0,\lambda e^{-x},\xi -x)\}\Big ]. \end{aligned}$$

(37)

In particular,

$$\begin{aligned} {\mathbb {P}}^{\downarrow }_{(z,x)}(Z_t<\infty ) = {\mathbb {E}}^{(e),\downarrow }_{x}\Big [\exp \left\{ -zv_t(0,0,\xi -x)\right\} \Big ], \end{aligned}$$

and

$$\begin{aligned} {\mathbb {P}}^{\downarrow }_{(z,x)}(Z_t>0) = 1. \end{aligned}$$

The following lemma states that, with respect to ${\mathbb {P}}^{\downarrow }_{(z,x)}$, the population has positive probability to be finite forever. In other words, Z has a positive probability to be finite when the running supremum of the Lévy environment is negative. The statement holds under the moment condition (B) of the Lévy measure $\mu$.

Lemma 4.4

Assume that the Lévy process $\xi$ satisfies Spitzer’s condition (A) and condition (B). Then, for $z >0$ and $x<0$, we have

$$\begin{aligned} \lim \limits _{t\rightarrow \infty } {\mathbb {P}}^{\downarrow }_{(z,x)}\left( Z_t<\infty \right) > 0. \end{aligned}$$

Note that such behaviour is similar to the behaviour in the subcritical explosive regime (i.e. when the environment drifts to $-\infty$) given in Theorem 1.3.

Proof

Let $z>0$ and $x<0$. From Lemma 4.3, we already know the formula,

$$\begin{aligned} \lim \limits _{t\rightarrow \infty } {\mathbb {P}}^{\downarrow }_{(z,x)}(Z_t<\infty ) = \lim \limits _{t\rightarrow \infty } {\mathbb {E}}^{(e),\downarrow }_{x}\Big [\exp \left\{ -zv_t(0,0,\xi -x)\right\} \Big ]. \end{aligned}$$

Then, similarly as in Theorem 1.3, in order to deduce our result we will show that

$$\begin{aligned} \lim _{t\rightarrow \infty } v_t(0,0,\xi -x) < \infty , \qquad {\mathbb {P}}_{x}^{(e),\downarrow }-\text {a.s.} \end{aligned}$$

(38)

We recall from the proof of Theorem 1.3 that for all $\lambda >0$

$$\begin{aligned} \lim \limits _{t\rightarrow \infty }v_t(0,0,\xi -x) \le \lambda e^{-x}\exp \left( \int _{0}^{\infty } \Phi _\lambda (-\xi _s) \textrm{d} s\right) . \end{aligned}$$

Now, if the right-hand side of the above inequality is finite ${\mathbb {P}}_{x}^{(e),\downarrow }-\text {a.s.}$, then (38) holds. The result is thus proved once we show that

$$\begin{aligned} {\mathbb {E}}_{x}^{(e),\downarrow }\left[ \int _{0}^{\infty }\Phi _\lambda (-\xi _s) \textrm{d} s\right] < \infty . \end{aligned}$$

First, with the help of Fubini’s theorem and the definition of the measure ${\mathbb {P}}_{x}^{(e),\downarrow }$ in terms of $\widehat{{\mathbb {P}}}_{x}^{(e)}$ (the law of the dual of $\xi$) we obtain

$$\begin{aligned} \begin{aligned} {\mathbb {E}}_{x}^{(e),\downarrow }\left[ \int _{0}^{\infty }\Phi _\lambda (-\xi _s) \textrm{d} s\right]&= \frac{1}{U(-x)}\int _{0}^{\infty } \widehat{{\mathbb {E}}}^{(e)}_{-x}\big [U(\xi _s)\Phi _\lambda (\xi _s){\textbf{1}}_{\{{\underline{\xi }}_s> 0\}}\big ] \textrm{d} s \\ {}&= \frac{1}{U(-x)} \widehat{{\mathbb {E}}}^{(e)}_{-x}\left[ \int _{0}^{\tau _0^-}U(\xi _s)\Phi _\lambda (\xi _s)\textrm{d} s\right] . \end{aligned} \end{aligned}$$

Now, applying Theorem VI.20 in Bertoin [5] to the dual process ${\widehat{\xi }}=-\xi$ and the function $f(y)=U(y)\Phi _\lambda (y), \ y\ge 0$, we deduce that there exists a constant $k>0$ such that

$$\begin{aligned}{} & {} \widehat{{\mathbb {E}}}^{(e)}_{-x}\left[ \int _{0}^{\tau _0^-}U(\xi _s)\Phi _\lambda (\xi _s)\textrm{d} s\right] \\{} & {} \quad = k \int _{[0,\infty )} \textrm{d} {\widehat{U}}(y)\int _{[0,-x]} \textrm{d}U(z)U(y-x-z)\Phi _\lambda (y-x-z). \end{aligned}$$

For the sake of simplicity, we take $k=1$. (We may choose a normalisation of the local time in order to have $k=1$.) Observe that, for any $z\in [0,-x]$ and $y\ge 0$, we have $y-x-z\le y-x$. Further, since $U(\cdot )$ and $\Phi _\lambda (\cdot )$ are increasing functions, we deduce that $U(y-x-z)\le U(y-x)$ and $\Phi _\lambda (y-x-z)\le \Phi _\lambda (y)$, which implies

$$\begin{aligned} \begin{aligned} \widehat{{\mathbb {E}}}^{(e)}_{-x}\left[ \int _{0}^{\tau _0^-}U(\xi _s)\Phi _\lambda (\xi _s)\textrm{d} s\right]&\le \int _{[0,\infty )} \textrm{d} {\widehat{U}}(y)\int _{[0,-x]} \textrm{d}U(z)U(y-x)\Phi _\lambda (y) \\&= (U(-x)-U(0))\int _{[0,\infty )} \textrm{d} {\widehat{U}}(y)U(y-x)\Phi _\lambda (y). \end{aligned} \end{aligned}$$

Next recall that we may rewrite the function $\Phi _\lambda$ as follows,

$$\begin{aligned} \begin{aligned} \Phi _\lambda (u)&= \int _{0}^\infty \exp \{- \lambda e^{u} z\} {\bar{\mu }}(z)\textrm{d} z \\&= \frac{e^{-u}}{\lambda }\int _{(0,\infty )} \Big (1-\exp \{- \lambda e^{u} z\}\Big ) \mu (\textrm{d}z), \end{aligned} \end{aligned}$$

implying that

$$\begin{aligned} \begin{aligned} \Phi _\lambda (u)&\le \int _{(0,\frac{e^{-u}}{\lambda }]}z \mu (\textrm{d}z)+ \frac{e^{-u}}{\lambda }\int _{(\frac{e^{-u}}{\lambda }, \infty )}\mu (\textrm{d}z). \end{aligned} \end{aligned}$$

Hence,

$$\begin{aligned} \begin{aligned} \int _{[0,\infty )}\textrm{d} {\widehat{U}}(y)U(y-x)\frac{e^{-y}}{\lambda }\int _{(\frac{e^{-y}}{\lambda }, \infty )}\mu (\textrm{d}z)&\le \frac{ {\bar{\mu }}(\lambda ^{-1})}{\lambda }\int _{[0,\infty )}\textrm{d} {\widehat{U}}(y)U(y-x)e^{-y} \\&\le C_1\frac{ {\bar{\mu }}(\lambda ^{-1})}{\lambda } \int _{[0,\infty )}\textrm{d} {\widehat{U}}(y)ye^{-y} \\&=C_1 \frac{{\bar{\mu }}(\lambda ^{-1})}{\lambda } \frac{{\widehat{\kappa }}^\prime (0,1)}{{\widehat{\kappa }}(0,1)}<\infty , \end{aligned} \end{aligned}$$

where in the second inequality and in the last identity we have used, respectively, inequality (14) and identity (15), but in terms of ${\widehat{U}}$ and after taking the first derivative. Now, we consider

$$\begin{aligned} \begin{aligned}&\int _{[0,\infty )}\textrm{d} {\widehat{U}}(y)U(y-x)\int _{(0,\frac{e^{-y}}{\lambda }]}z \mu (\textrm{d}z)\\&\quad \le \int _{(0,\infty )}\mu (\textrm{d}z)z \int _{[0,\infty )}\textrm{d} {\widehat{U}}(y)U(y-x){\textbf{1}}_{\{z<e^{-y}\lambda ^{-1}\}} \\&\quad \le \int _{(0,\lambda ^{-1})}\mu (\textrm{d}z)z \int _{[0,-\ln (z\lambda ))}\textrm{d} {\widehat{U}}(y)U(y-x) \\&\quad \le \int _{(0,\lambda ^{-1})}\mu (\textrm{d}z)z {\widehat{U}}(-\ln (z\lambda ))U(-\ln (z\lambda )-x), \end{aligned} \end{aligned}$$

which is clearly finite from condition (B). Hence putting all pieces together, we obtain

$$\begin{aligned} {\mathbb {E}}_{x}^{(e),\downarrow }\left[ \int _{0}^{\infty }\Phi _\lambda (\xi _s) \textrm{d} s\right] <\infty . \end{aligned}$$

This concludes the proof. $\square$

It is important to note that in Baguley et al. [2] there is a necessary and sufficient condition (integral test) for the finiteness of path integrals for standard Markov processes, see Theorem 2.3 and comments below, in terms of their potential measures. The latter is in line with our necessary condition for the finiteness of

$$\begin{aligned} \int _{0}^{\infty }\Phi _\lambda (-\xi _s) \textrm{d} s, \end{aligned}$$

under ${\mathbb {P}}_{x}^{(e),\downarrow }$.

5 Critical Explosive Regime: Proof of Theorem 1.4

Throughout this section, we shall suppose that the Lévy process $\xi$ satisfies Spitzer’s condition (A).

The strategy of our proof follows similar arguments as in [3], where the extinction event has been considered; that is, we split the event $\{Z_t<\infty \}$ into two events by considering the behaviour of the running supremum of the environment. More precisely, we split the non-explosion event into either unfavourable environments, i.e. when the running supremum is negative, or favourable environments, i.e. when the running supremum is positive.

Before we prove Theorem 1.4, we introduce several useful results. Lemmas 4.2 and 4.4 allow us to establish the following result which describes the limit of the non-explosion probability when the associated environment is conditioned to be negative.

Proposition 5.1

Suppose that conditions (A) and (B) are satisfied. Then for every $z >0$ and $x<0$, there exists $0<c(z,x)<\infty$ such that

$$\begin{aligned} \lim \limits _{t \rightarrow \infty }\frac{1}{\kappa (1/t,0)}{\mathbb {P}}_{(z,x)}\Big (Z_t<\infty , \ {\overline{\xi }}_t < 0\Big ) = c(z,x)U(-x). \end{aligned}$$

(39)

Proof

We begin by defining the decreasing sequence of events $A_t=\{Z_t<\infty \}$ for $t\ge 0$, and also the event $A_\infty = \{\forall t \ge 0,\ Z_t<\infty \}$. Now, observe that

$$\begin{aligned} \lim \limits _{t\rightarrow \infty } A_t = A_\infty . \end{aligned}$$

Let $(R_t:={\textbf{1}}_{A_t},\ t\ge 0)$ be a uniformly bounded process adapted to the filtration $({\mathcal {F}}_t, t\ge 0)$. Note that the process $(R_t, t\ge 0)$ converges ${\mathbb {P}}^\downarrow _{(z,x)}$-a.s. to a random variable $R_\infty ={\textbf{1}}_{A_\infty }$. Then, by appealing to Lemma 4.2, we have

$$\begin{aligned} \lim \limits _{t\rightarrow \infty }{\mathbb {E}}_{(z,x)}\big [R_t\ | \ {\overline{\xi }}_t < 0\big ] = {\mathbb {E}}^{ \downarrow }_{(z,x)}\big [R_\infty \big ]. \end{aligned}$$

(40)

Therefore, by using the asymptotic behaviour of the probability that the Lévy process $\xi$ remains negative given in (25), we get

$$\begin{aligned} {\mathbb {P}}_{(z,x)}\Big (Z_t<\infty ,\ {\overline{\xi }}_t< 0\Big )= & {} {\mathbb {E}}_{(z,x)}\big [R_t \ |\ {\overline{\xi }}_t< 0\big ] {\mathbb {P}}_{x}^{(e)}\big ({\overline{\xi }}_t < 0\big )\\\sim & {} c(z,x)U(-x)\kappa (1/t,0), \quad \text {as}\quad t \rightarrow \infty , \end{aligned}$$

where $c(z,x):= {\mathbb {E}}_{(z,x)}^{\downarrow }\big [R_\infty \big ]/\sqrt{\pi }$. Furthermore, from Lemma 4.4, we have

$$\begin{aligned} {\mathbb {E}}_{(z,x)}^{\downarrow }\big [R_\infty \big ] = {\mathbb {P}}_{(z,x)}^\downarrow \big (\forall t \ge 0,\ Z_t<\infty \big ) = \lim \limits _{t\rightarrow \infty } {\mathbb {P}}^{\downarrow }_{(z,x)}(Z_t<\infty ) >0, \end{aligned}$$

which completes the first claim. $\square$

Denote by $\tau ^{-}_x$, the first time that $\xi$ is below $(-\infty , x)$, i.e. $\tau ^{-}_x=\inf \{t\ge 0: \xi _t\le x\}$, for $x<0$. The following result follows directly from the inequality (4.7) in [3] (see the proof of Lemma 4.2), duality, the identity $\widehat{{\mathbb {P}}}^{(e)}( \tau ^{-}_{w}>t-\epsilon )=\widehat{{\mathbb {P}}}^{(e)}_{-w}( {\underline{\xi }}_{t-\epsilon }>0 )$ and the estimate in (25). We present its proof for the sake of completeness.

Lemma 5.2

Let $x<0$ and assume that condition (A) holds. Thus for any $s \le t$, as t and s go to $\infty$, we have

$$\begin{aligned} \begin{aligned} \widehat{{\mathbb {P}}}^{(e)}\big (s<\tau _{x}^-\le t\big )&\le \left( C_2\left( \frac{t}{s}\right) ^{\eta + \rho }-1\right) \widehat{{\mathbb {P}}}^{(e)}\big (\tau _{x}^-> t\big )\\&\le C_3\left( C_2\left( \frac{t}{s}\right) ^{\eta + \rho }-1\right) \kappa (1/t,0) \frac{U(-x)}{\sqrt{\pi }}, \end{aligned} \end{aligned}$$

where $C_2>0$, $C_3>1$ and $\eta >0$.

Proof

Let $x<0$ and $s\le t$. We begin by noting

$$\begin{aligned} \widehat{{\mathbb {P}}}^{(e)}\big (s< \tau ^-_{x}\le t\big )= & {} \widehat{{\mathbb {P}}}^{(e)}\big ( \tau ^-_{x}>s\big )-\widehat{{\mathbb {P}}}^{(e)}\big ( \tau ^-_{x}>t\big ) \nonumber \\= & {} \left( \frac{\widehat{{\mathbb {P}}}^{(e)}\big ( \tau ^-_{x}>s\big )}{\widehat{{\mathbb {P}}}^{(e)}\big ( \tau ^-_{x}>t\big ) }-1 \right) \widehat{{\mathbb {P}}}^{(e)}\big ( \tau ^-_{x}>t\big ). \end{aligned}$$

(41)

Now, recall that under Spitzer’s condition (A), the function $\kappa (\cdot , 0)$ is regularly varying at 0 or more precisely, from (25), we have

$$\begin{aligned} \widehat{{\mathbb {P}}}^{(e)}\big (\tau ^-_{x}> t\big ) \sim \frac{U(-x)}{\sqrt{\pi }} t^{-\rho }\ell (t), \quad \text {as} \quad t \rightarrow \infty , \end{aligned}$$

where $\ell$ is the slowly varying function at $\infty$ defined in (24) as $\ell (t)={\widetilde{\ell }}(1/t)$. Hence, for t and s large enough, we have

$$\begin{aligned} \frac{\widehat{{\mathbb {P}}}^{(e)}\big ( \tau ^-_{x}>s\big )}{\widehat{{\mathbb {P}}}^{(e)}\big ( \tau ^-_{x}>t\big ) } \le C_1 \left( \frac{s}{t}\right) ^{-\rho } \frac{\ell (s)}{\ell (t)}, \end{aligned}$$

where $C_1$ is a positive constant. On the other hand, according to Potter’s theorem in [7] we deduce that for any $A>1$ and $\eta >0$ there exists $t_1=t_1(A,\eta )$ such that

$$\begin{aligned} \frac{\ell (s)}{\ell (t)} \le A \max \left\{ \left( \frac{s}{t}\right) ^{\eta }, \left( \frac{s}{t}\right) ^{-\eta }\right\} , \quad \quad t\ge s \ge t_1. \end{aligned}$$

Therefore, for $t\ge s \ge t_1$

$$\begin{aligned} \frac{\widehat{{\mathbb {P}}}^{(e)}\big ( \tau ^-_{x}>s\big )}{\widehat{{\mathbb {P}}}^{(e)}\big ( \tau ^-_{x}>t\big ) } \le C_2 \left( \frac{s}{t}\right) ^{-\rho }\left( \frac{s}{t}\right) ^{-\eta } = C_2\left( \frac{t}{s}\right) ^{\eta + \rho }, \end{aligned}$$

where $C_2$ is a positive constant. Now plugging the later inequality back into (41), we get, as it was claimed,

$$\begin{aligned} \begin{aligned} \widehat{{\mathbb {P}}}^{(e)}\big (s< \tau ^-_{x}\le t\big )&\le \left( \frac{\widehat{{\mathbb {P}}}^{(e)}\big ( \tau ^-_{x}>s\big )}{\widehat{{\mathbb {P}}}^{(e)}\big ( \tau ^-_{x}>t\big ) }-1 \right) \widehat{{\mathbb {P}}}^{(e)}\big ( \tau ^-_{x}>t\big ) \\&\le C_3\left( C_2\left( \frac{t}{s}\right) ^{\eta + \rho }-1\right) \kappa (1/t,0) \frac{U(-x)}{\sqrt{\pi }}, \qquad \text {as}\quad s, t\rightarrow \infty , \end{aligned} \end{aligned}$$

where $C_3$ is a constant bigger than 1. $\square$

Recall that $\texttt {I}_{0,t}(\beta \xi )$ denotes the exponential functional of the Lévy process $\beta \xi$ defined in (9). Our next result will be useful to control the probability of non-explosion under the event that $\{{\overline{\xi }}_{t}>0\}$.

Lemma 5.3

Let $\beta \in (-1,0)$, $C<0$ and $y>0$ and assume that condition (A) holds. Then, there exists a continuous function $y\mapsto C_{\beta }(y)$ on $(0,\infty )$ such that for t large enough, we have

$$\begin{aligned} \widehat{{\mathbb {E}}}^{(e)} \left[ \exp \Big \{-y (C\beta )^{-1/\beta } { \texttt {I}_{0,t}}(-\beta \xi )^{-1/\beta }\Big \}\right] \le 2C_{\beta }(y) \kappa (1/t, 0). \end{aligned}$$

Further,

$$\begin{aligned} \lim \limits _{y\rightarrow \infty } C_\beta (e^y) =0 \quad \quad \text {and}\quad \quad \lim \limits _{y\rightarrow \infty } y C_\beta (e^y) =0. \end{aligned}$$

Proof

First, we recall from Patie and Savov [21, Theorem 2.20] that, under Spitzer’s condition (A), for any continuous and bounded function f on ${\mathbb {R}}^+$ and any constant $a\in (0,1)$, we have

$$\begin{aligned} \lim \limits _{t\rightarrow \infty } \frac{ \widehat{{\mathbb {E}}}^{(e)}\Big [ \texttt {I}_{0,t}(-\beta \xi )^{-a}f\big (\texttt {I}_{0,t}(-\beta \xi )\big )\Big ]}{\kappa (1/t,0)} = \int _{0}^{\infty } f(x)\vartheta _a(\textrm{d}x), \end{aligned}$$

where $\vartheta _a$ is a positive measure on $(0,\infty )$.

For our purposes, we use $f(x)=x^a \exp (-y (C\beta )^{-1/\beta }x^{-1/\beta })$ which is bounded and continuous. Thus by the latter identity, we deduce that, for any $a\in (0,1)$,

$$\begin{aligned} \lim \limits _{t\rightarrow \infty } \frac{ \widehat{{\mathbb {E}}}^{(e)}\left[ \exp \Big \{-y (C\beta )^{-1/\beta } { \texttt {I}_{0,t}}(-\beta \xi )^{-1/\beta }\Big \}\right] }{\kappa (1/t,0)}= C_\beta (y), \end{aligned}$$

where

$$\begin{aligned} C_\beta (y):= \int _{0}^{\infty } x^a \exp \big \{-y (C\beta )^{-1/\beta }x^{-1/\beta }\big \}\vartheta _a(\textrm{d}x), \end{aligned}$$

(42)

which is clearly continuous. The latter implies that there exists $t_0>0$ such that if $t\ge t_0$

$$\begin{aligned} \widehat{{\mathbb {E}}}^{(e)}\left[ \exp \Big \{-y (C\beta )^{-1/\beta } { \texttt {I}_{0,t}}(-\beta \xi )^{-1/\beta }\Big \}\right] \le 2 C_{\beta }(y) \kappa (1/t, 0), \end{aligned}$$

as expected. Furthermore, with the help of the dominated convergence theorem, we obtain

$$\begin{aligned}\begin{aligned} \lim \limits _{y\rightarrow \infty } y C_\beta (e^y) = 2\int _{0}^{\infty } x^a\lim \limits _{y\rightarrow \infty } y \exp \big \{-e^y (C\beta )^{-1/\beta }x^{-1/\beta }\big \}\vartheta _a(\textrm{d}x) =0 \end{aligned}\end{aligned}$$

and that $C_{\beta }(e^y)\rightarrow 0$, as $y\rightarrow \infty$. This concludes the proof. $\square$

The following result makes precise the statement that only paths of the Lévy process with a very low running supremum give a substantial contribution to the speed of non-explosion.

Proposition 5.4

Fix $z>0$, $x<0$ and $\epsilon \in (0,1)$, and suppose that assumptions (A) and (C) are satisfied. Then, we have for $y>x$

$$\begin{aligned} \lim \limits _{y\rightarrow \infty } \limsup _{t\rightarrow \infty } \frac{1}{\kappa (1/t,0)} {\mathbb {P}}_{(z,x)}\Big (Z_t<\infty , \ {\overline{\xi }}_{t-\epsilon } \ge y\Big ) = 0. \end{aligned}$$

(43)

Proof

Fix $z>0, \ x<0$ and $\epsilon \in (0,1)$. We begin by noting that condition (C) allows us to find a lower bound for $v_t(0,0,\xi -x)$ in terms of the exponential functional of $\xi$. Indeed, we observe from the backward differential equation given in (16) that

$$\begin{aligned} \frac{\partial }{\partial s} v_t(s,\lambda e^{-x},\xi -x)\ge & {} C v_t^{1+\beta }(s,\lambda e^{-x},\xi -x)e^{-\beta (\xi _s-x)},\\ \quad v_t(t,\lambda e^{-x}, \xi -x)= & {} \lambda e^{-x}. \end{aligned}$$

Integrating between 0 and t, we get

$$\begin{aligned} \frac{1}{v_t^\beta (0,\lambda e^{-x},\xi -x)} -\frac{1}{(\lambda e^{-x})^\beta } \le C\beta \int _{0}^{t} e^{-\beta (\xi _s-x)}\textrm{d} s, \quad C\beta >0. \end{aligned}$$

Now, letting $\lambda \downarrow 0$ and taking into account that $\beta \in (-1,0)$ and $C<0$, we deduce the following inequality for all $t\ge 0$,

$$\begin{aligned} v_t(0,0,\xi -x)\ge \big (C\beta \texttt {I}_{0,t}(\beta (\xi -x)) \big )^{-1/\beta }, \end{aligned}$$

(44)

where $\texttt {I}_{0,t}(\beta (\xi -x))$ is the exponential functional of the Lévy process $\beta (\xi -x)$. Hence, the quenched non-explosion probability given in (29) may be bounded in terms of this functional. That is to say, for all $t >0$,

$$\begin{aligned} {\mathbb {P}}_{(z,x)}\big (Z_t<\infty \ \big |\big . \ \xi \big ){} & {} = \exp \big \{-z v_t(0,0, \xi - x)\big \}\\{} & {} \le \exp \Big \{-z \big (C\beta \texttt {I}_{0,t}(\beta (\xi -x)) \big )^{-1/\beta }\Big \}. \end{aligned}$$

Therefore conditioning on the environment, we obtain that for any $y>x$,

$$\begin{aligned}{} & {} {\mathbb {P}}_{(z,x)}\Big (Z_t<\infty ,\ {\overline{\xi }}_{t-\epsilon } \ge y\Big ) \nonumber \\{} & {} \quad = {\mathbb {E}}_{x}^{(e)}\Big [ {\mathbb {P}}_{(z,x)}\big (Z_t<\infty \ \big |\big . \ \xi \big ) {\textbf{1}}_{\{{\overline{\xi }}_{t-\epsilon } \ge y\}}\Big ] \nonumber \\{} & {} \quad \le \widehat{{\mathbb {E}}}^{(e)}\Big [\exp \Big \{-z (C\beta )^{-1/\beta } \texttt {I}_{0,t}(-\beta \xi )^{-1/\beta }\Big \} {\textbf{1}}_{\{{\underline{\xi }}_{t-\epsilon } \le x-y\}}\Big ]. \end{aligned}$$

(45)

Let $w=x-y$ and $t_0>0$. Now, we split the event $\{\tau ^{-}_{w}\le t-\epsilon \}$ for $3t_0 < t$ and $0< \epsilon < 1$, as follows

$$\begin{aligned} \{\tau ^{-}_{w}\le t-\epsilon \}= \{0< \tau ^{-}_{w} \le (t-t_0)/2 \} \cup \{(t-t_0)/2 < \tau ^{-}_{w} \le t-\epsilon \}. \end{aligned}$$

By the monotonicity of the mapping $t\mapsto \texttt {I}_{0,t}(-\beta \xi )$, we have that, under the event $\{0 <\tau ^{-}_{w} \le (t-t_0)/2 \}$, the following inequalities hold

$$\begin{aligned} 0<\tau ^{-}_{w} < \tau ^{-}_{w} + \frac{t+t_0}{2} \le t \qquad {\text {and}} \qquad \int _{0}^{t} e^{\beta \xi _s} \textrm{d} s \ge \int _{\tau ^{-}_{w}}^{\tau ^{-}_{w}+ \frac{t+t_0}{2}} e^{\beta \xi _s} \textrm{d} s. \end{aligned}$$

Similarly, under the event $\{(t-t_0)/2 < \tau ^{-}_{w} \le t-\epsilon \}$, we obtain

$$\begin{aligned} \frac{t-t_0}{2}< \tau ^{-}_{w} < \tau ^{-}_{w} + \epsilon \le t \qquad {\text {and}} \qquad \int _{0}^{t} e^{\beta \xi _s} \ge \int _{\tau ^{-}_{w}}^{\tau ^{-}_{w}+\epsilon } e^{\beta \xi _s}. \end{aligned}$$

Next, appealing to the strong Markov property of $\xi$, we deduce

$$\begin{aligned} \begin{aligned} \widehat{{\mathbb {E}}}^{(e)}&\left[ \exp \Big \{-z(C\beta )^{-1/\beta } \texttt {I}_{\tau ^{-}_{w}, \tau ^{-}_{w}+ \frac{t+t_0}{2}}(-\beta \xi )^{-1/\beta }\Big \};\ 0< \tau ^{-}_{w}\le (t-t_0)/2 \right] \\&\le \widehat{{\mathbb {E}}}^{(e)} \left[ \exp \left\{ - ze^{-w}(C\beta )^{-1/\beta }\left( \int _{0}^{\frac{t+t_0}{2}} e^{\beta \big (\xi _{s+\tau ^{-}_{w}} - \xi _{\tau ^{-}_{w}}\big )} \textrm{d} s\right) ^{-1/\beta }\right\} {\textbf{1}}_{\{0 < \tau ^{-}_{w}\le (t-t_0)/2\} } \right] \\&\le \widehat{{\mathbb {E}}}^{(e)} \left[ \exp \Big \{-ze^{-w}(C\beta )^{-1/\beta } \texttt {I}_{0,\frac{t+t_0}{2}}(-\beta \xi )^{-1/\beta }\Big \}\right] . \end{aligned} \end{aligned}$$

Thus from Lemma 5.3, for t sufficiently large, we have

$$\begin{aligned} \widehat{{\mathbb {E}}}^{(e)} \left[ \exp \Big \{-ze^{-w}(C\beta )^{-1/\beta } \texttt {I}_{0,\frac{t+t_0}{2}}(-\beta \xi )^{-1/\beta }\Big \}\right] \le 2C_{\beta }(ze^{-w}) \kappa \left( \frac{2}{t+t_0}, 0\right) , \end{aligned}$$

where $C_\beta (ze^{-w})$ are defined as in (42).

Using the same arguments as above and Lemmas 5.2 and 5.3, we obtain the following sequence of inequalities for t sufficiently large,

$$\begin{aligned} \begin{aligned}&\widehat{{\mathbb {E}}}^{(e)}\Big [\exp \Big \{-z(C\beta )^{-1/\beta } \texttt {I}_{\tau ^{-}_{w}, \tau ^{-}_{w}+ \epsilon }(-\beta \xi )^{-1/\beta }\Big \}; \ (t-t_0)/2< \tau ^{-}_{w} \le t-\epsilon \Big ] \\&\quad \le \widehat{{\mathbb {E}}}^{(e)} \Big [\exp \Big \{-ze^{-w}(C\beta )^{-1/\beta } \texttt {I}_{0, \epsilon }(-\beta \xi )^{-1/\beta }\Big \}\Big ]\widehat{{\mathbb {P}}}^{(e)}\Big ( \frac{t-t_0}{2}< \tau ^{-}_{w} \le t-\epsilon \Big )\\&\quad \le 2C_{\beta }(ze^{-w}) \kappa \left( \frac{1}{\epsilon }, 0\right) C_3\left( C_2 2^{\eta +\rho } \left( 1+ \frac{t_0 - \epsilon }{2t_0}\right) ^{\eta + \rho }-1\right) \kappa \left( \frac{1}{t-\epsilon }, 0 \right) \frac{U(-w)}{\sqrt{\pi }}. \end{aligned} \end{aligned}$$

Hence plugging this back into (45) (similarly as in the proof of Lemma 4.4 in [18]), we get

$$\begin{aligned} \begin{aligned}&\limsup _{t\rightarrow \infty } \frac{1}{\kappa (1/t,0)} {\mathbb {P}}_{(z,x)}\Big (Z_t<\infty , \ {\overline{\xi }}_{t-\epsilon } \ge y\Big ) \\ {}&\quad \le \limsup _{t\rightarrow \infty } \frac{1}{\kappa (1/t,0)} \widehat{{\mathbb {E}}}^{(e)}\Big [\exp \Big \{-z (C\beta )^{-1/\beta } \texttt {I}_{0,t}(-\beta \xi )^{-1/\beta }\Big \}; \ \tau ^{-}_{w}\le t-\epsilon \Big ] \\ {}&\quad \le C_5(ze^{-w}) \limsup _{t\rightarrow \infty } \frac{1}{\kappa (1/t,0)} \left( \kappa \left( \frac{2}{t+t_0}, 0\right) +\kappa \left( \frac{1}{t-\epsilon }, 0\right) \right) , \end{aligned} \end{aligned}$$

where

$$\begin{aligned} C_5(ze^{-w}):= 2 C_{\beta }(ze^{-w})\left( 1 \vee \kappa \left( \frac{1}{\epsilon }, 0\right) C_3 \left( C_2 2^{\eta +\rho } \left( 1+ \frac{t_0 - \epsilon }{2t_0}\right) ^{\eta + \rho }-1\right) \frac{U(-w)}{\sqrt{\pi }}\right) . \end{aligned}$$

Note that the limsup is finite since $\theta \mapsto \kappa ( \theta , 0)$ is a regular varying function at zero. Finally, taking into account that the renewal function U grows at most linearly, i.e. $U(y-x) = {\mathcal {O}}(y-x)$, recalling the asymptotic behaviour of $C_{\beta }(ze^{-w})$ in Lemma 5.3 and letting $y\rightarrow \infty$ we obtain the desired result. $\square$

For every $z >0$ and $x<0$, denote by c(z, x) the constant defined in Proposition 5.1. Our next result provides some useful properties of the mapping $x\mapsto c(z,x)U(-x)$.

Lemma 5.5

For each $z > 0$ the map $x \mapsto c(z,x)U(-x)$ on $(-\infty ,0)$ is decreasing, strictly positive and bounded. In particular, for each $z>0$, it holds

$$\begin{aligned}\lim \limits _{x\rightarrow -\infty } c(z,x)U(-x)=: B(z)\in (0,\infty ).\end{aligned}$$

Proof

Note that the strictly positivity follows from the facts that the renewal function $U(-x)$ is a strictly positive function on $(-\infty ,0)$ and by Lemma 5.1. Since ${\mathbb {P}}_x({\overline{\xi }}_t<0) \le {\mathbb {P}}_y({\overline{\xi }}<0)$ for $x\ge y$, we observe that the left-hand side on (39) is decreasing in $x<0$, so does the map $x \mapsto c(z,x)U(-x)$. Now, in order to see that the function is bounded from above, we first observe that similarly as in (45), we have

$$\begin{aligned} {\mathbb {P}}_{(z,x)}\Big (Z_t<\infty ,\ {\overline{\xi }}_{t}< 0\Big ){} & {} = {\mathbb {E}}_{x}^{(e)}\Big [ {\mathbb {P}}_{(z,x)}\big (Z_t<\infty \ \big |\big . \ \xi \big ) {\textbf{1}}_{\{{\overline{\xi }}_{t} < 0\}}\Big ] \\{} & {} \le \widehat{{\mathbb {E}}}^{(e)}\Big [\exp \Big \{-z (C\beta )^{-1/\beta } \texttt {I}_{0,t}(-\beta \xi )^{-1/\beta }\Big \} {\textbf{1}}_{\{{\underline{\xi }}_{t} \ge x\}}\Big ] \\ {}{} & {} \le \widehat{{\mathbb {E}}}^{(e)}\Big [\exp \Big \{-z (C\beta )^{-1/\beta } \texttt {I}_{0,t}(-\beta \xi )^{-1/\beta }\Big \}\Big ]. \end{aligned}$$

Hence, appealing to Lemma 5.3, we obtain

$$\begin{aligned} c(z,x)U(-x)= & {} \lim \limits _{t \rightarrow \infty }\frac{1}{\kappa (1/t,0)}{\mathbb {P}}_{(z,x)}\Big (Z_t<\infty , \ {\overline{\xi }}_t < 0\Big )\\\le & {} \lim \limits _{t \rightarrow \infty }\frac{1}{\kappa (1/t,0)}\widehat{{\mathbb {E}}}^{(e)}\Big [\exp \Big \{-z (C\beta )^{-1/\beta } \texttt {I}_{0,t}(-\beta \xi )^{-1/\beta }\Big \}\Big ]\le C_\beta (z), \end{aligned}$$

where $z\mapsto C_{\beta }(z)$ is the continuous function defined in Lemma 5.3. With this in hand, and considering that the mapping $x \mapsto c(z,x)U(-x)$, on $(-\infty ,0)$, is decreasing and strictly positive, we conclude that for every $z>0$ the limit B(z) exists and it is finite and strictly positive. $\square$

With Propositions 5.1 and 5.4 in hand, we may now proceed to the proof of Theorem 1.4. The proof follows the same arguments as those used in Theorem 1.2 in [3], and we provide its proof for the sake of completeness.

Proof of Theorem 1.4

Fix $\varsigma , z >0$, $x<0$ and $\epsilon \in (0,1)$. We begin by observing from (4) that ${\mathbb {P}}_{(z,x)}(Z_t > 0)$ does not depend of the initial value x of the Lévy process $\xi$. From Proposition 5.4, we also observe that we may choose $y>0$ such that for t large enough,

$$\begin{aligned} {\mathbb {P}}_{(z,x)}\Big (Z_t<\infty ,\ {\overline{\xi }}_{t-\epsilon } \ge y\Big ) \le \varsigma {\mathbb {P}}_{(z,x)}\Big (Z_t<\infty , \ {\overline{\xi }}_{t-\epsilon } < y\Big ). \end{aligned}$$

(46)

Now, note that for t large enough, we get $\{Z_{t}<\infty \}\subset \{Z_{t-\epsilon } <\infty \}$ and using the previous inequality, it follows

$$\begin{aligned} {\mathbb {P}}_{z}(Z_t<\infty )= & {} {\mathbb {P}}_{(z,x)}\Big (Z_t<\infty , \ {\overline{\xi }}_{t-\epsilon } \ge y\Big ) + {\mathbb {P}}_{(z,x)}\Big (Z_t<\infty ,\ {\overline{\xi }}_{t-\epsilon }< y\Big ) \\ {}\le & {} (1+\varsigma ) {\mathbb {P}}_{(z,x-y)}\Big (Z_{t-\epsilon }<\infty , \ {\overline{\xi }}_{t-\epsilon } < 0\Big ). \end{aligned}$$

In other words, for every $\varsigma >0$ there exists $y'<0$ such that for t large enough

$$\begin{aligned}&(1-\varsigma )\frac{{\mathbb {P}}_{(z,y')}\Big (Z_t<\infty , \ {\overline{\xi }}_{t}< 0\Big ) }{\kappa (1/t,0)}\\&\quad \le \frac{ {\mathbb {P}}_{z}(Z_t<\infty ) }{\kappa (1/t, 0)} \le (1+\varsigma ) \frac{{\mathbb {P}}_{(z,y')}\Big (Z_{t-\epsilon }<\infty ,\ {\overline{\xi }}_{t-\epsilon } < 0\Big )}{\kappa (1/(t-\epsilon ),0)} \frac{\kappa (1/(t-\epsilon ),0)}{\kappa (1/t,0)}. \end{aligned}$$

Next, using that the function $\kappa (\cdot , 0)$ is regularly varying at 0, and then appealing to Potter’s theorem in Bingham et al. [7], we see that, for any $A>1$ and $\eta >0$,

$$\begin{aligned} \lim \limits _{t \rightarrow \infty } \frac{\kappa (1/(t-\epsilon ),0)}{\kappa (1/t,0)} = \lim \limits _{t \rightarrow \infty } \frac{\ell (t-\epsilon )}{\ell (t)}\left( \frac{t-\epsilon }{t}\right) ^{-\rho } \le \lim \limits _{t \rightarrow \infty } A\left( \frac{t}{t-\epsilon }\right) ^{\rho + \eta } = A. \end{aligned}$$

On the other hand, according to Proposition 5.1, there exists $0<c(z,y')<\infty$ such that

$$\begin{aligned} \lim \limits _{t \rightarrow \infty }\frac{1}{\kappa (1/t,0)}{\mathbb {P}}_{(z,y')}\Big (Z_t<\infty ,\ {\overline{\xi }}_t < 0\Big ) = c(z,y')U(-y'). \end{aligned}$$

Hence, as a consequence of the above facts, we get

$$\begin{aligned} (1-\varsigma ) c(z,y')U(-y') \le \liminf _{t \rightarrow \infty } \frac{{\mathbb {P}}_{z}(Z_t<\infty )}{\kappa (1/t, 0)} \le (1+\varsigma ) c(z,y')U(-y') A. \end{aligned}$$

We observe that $y'$ is a sequence which may depend on $\varsigma$ and z. Further, this sequence $y'$ goes to $-\infty$ as $\varsigma$ goes to 0. Thus, for any sequence $y_{\varsigma }(z)$, we have

$$\begin{aligned}\begin{aligned} 0< (1-\varsigma ) c(z,y_{\varsigma }(z))U(-y_{\varsigma }(z))&\le \liminf _{t \rightarrow \infty } \frac{ {\mathbb {P}}_{z}(Z_t<\infty ) }{\kappa (1/t, 0)} \\ {}&\le (1+\varsigma ) c(z, y_{\varsigma }(z))U(-y_{\varsigma }(z)) A < \infty , \end{aligned}\end{aligned}$$

where the strictly positivity and finiteness in the previous inequality follows from Lemma 5.5. Therefore, using again Lemma 5.5, we get

$$\begin{aligned}\begin{aligned} 0< \limsup _{\varsigma \rightarrow 0} (1-\varsigma ) c(z,y_{\varsigma }(z))&U(-y_{\varsigma }(z)) \le \liminf _{t \rightarrow \infty } \frac{{\mathbb {P}}_{z}(Z_t<\infty )}{\kappa (1/t, 0)}\\&\le \liminf _{\varsigma \rightarrow 0}(1+\varsigma ) c(z,y_{\varsigma }(z))U(-y_{\varsigma }(z)) A < \infty . \end{aligned}\end{aligned}$$

Since A can be taken arbitrary close to 1, we deduce that the inferior and superior limits are equal, finite and positive. That is to say,

$$\begin{aligned} 0< \lim \limits _{t \rightarrow \infty } \frac{{\mathbb {P}}_{z}(Z_t<\infty )}{\kappa (1/t, 0)}={\mathfrak {C}}(z):=\lim _{\varsigma \rightarrow 0} c(z,y_{\varsigma }(z))U(-y_{\varsigma }(z)) < \infty , \end{aligned}$$

which completes the proof. $\square$