# Behaviors near explosion of nonlinear CSBPs with regularly varying
mechanisms

Bo Li , Clément Foucart and Xiaowen Zhou School of Mathematics and LPMC,
Nankai University, Tianjin 300071, PR China. E-mail<EMAIL_ADDRESS>cnCMAP,
Ecole Polytechnique, IP Paris, Palaiseau, France. Email:
<EMAIL_ADDRESS>Department of Mathematics and Statistics,
Concordia University, Montreal, Canada H3G 1M8. Email:
<EMAIL_ADDRESS>
We study the explosion phenomenon of nonlinear continuous-state branching
processes (nonlinear CSBPs). First an explicit integral test for explosion is
designed when the rate function does not increase too fast. We then exhibit
three different regimes of explosion when the branching mechanism and the rate
function are regularly varying respectively at $0$ and $\infty$ with indices
$\alpha$ and $\beta$ such that $0\leq\alpha\leq\beta$ and $\beta>0$. If
$\alpha>0$ then the renormalisation of the process before its explosion is
linear. When moreover $\alpha\neq\beta$, the limiting distribution is that of
a ratio of two independent random variables whose laws are identified. When
$\alpha=\beta$, the limiting random variable shrinks to a constant. Last, when
$\alpha=0$, i.e. the branching mechanism is slowly varying at $0$, the process
is studied with the help of a nonlinear renormalisation. The limiting
distribution becomes the inverse uniform distribution. This complements
results known in the case of finite mean and provides new insight for the
classical explosive continuous-state branching processes (for which
$\beta=1$).

Keywords: continuous-state branching process, nonlinear time-change,
explosion, spectrally positive Lévy process, perpetual integral, asymptotic
overshoot, regular variation.

MSC2020: 60J25, 60J80.

##  1 Introduction and main results

Continuous-state branching processes (CSBP for short) are processes
representing the total size of a population in which, intuitively, each
individual reproduces independently with the same law. It is well known that
when the offspring distribution has infinite first moment, the process might
explode in finite time, in the sense that large jumps happen so fast that the
process reaches infinity in finite time; see for instance Savits [24] and Grey
[11].

The purpose of this article is to study the renormalisations in law of the
processes prior to their explosion. Instead of only considering classical
CSBPs, we treat the larger class of the so-called nonlinear CSBPs. The latter
have been introduced and studied in recent works; see e.g. Li and Zhou [16]
and the references therein. They are parametrized by two functions: a
branching mechanism $\psi$, which is the Laplace exponent of a spectrally
positive Lévy process (SPLP for short) $\xi:=(\xi(t),t\geq 0)$, and a rate
function $R$, which is assumed to be a positive measurable function on
$(0,\infty)$, bounded away from both $0$ and $\infty$ on every compact subset
of $(0,\infty)$. Equivalently, $1/R$ is measurable positive and locally
bounded. Heuristically, the branching mechanism $\psi$ governs the
reproduction, as in the classical CSBP, and $R$ controls the reproduction
rate.

More explicitly, let $x\in[0,\infty)$ and denote by $\mathbb{P}_{x}$ the law
of the SPLP $\xi$ with $\xi(0)=x$ a.s. We will also write
$\mathbb{P}:=\mathbb{P}_{0}$. For any $a,b\geq 0$, denote the first passage
times of the process $\xi$ by

$\tau_{a}^{-}:=\inf\\{t>0,\xi(t)<a\\}\quad\text{and}\quad\tau_{b}^{+}:=\inf\\{t>0,\xi(t)>b\\}.$
(1.1)

Define the additive functional $\eta$ by

$\eta(t):=\int_{0}^{t}\frac{\mathrm{d}s}{R(\xi(s))}\quad\text{and}\quad\eta^{-1}(s):=\inf\\{t\geq
0:\eta(t)>s\\},$ (1.2)

for $t\in[0,\tau_{0}^{-})$ and $s\in[0,\eta(\tau_{0}^{-}))$ with convention
$\inf\emptyset=\infty$. The nonlinear CSBP $X$, with branching mechanism
$\psi$ and rate function $R$ is defined by time changing $\xi$ as follows: for
any $t\geq 0$,

$X(t):=\left\\{\begin{aligned}
&\,\,\xi\big{(}\eta^{-1}(t)\big{)},&&\text{when}\,\,t\in[0,\eta(\tau^{-}_{0})),\\\
&\,\,\infty,&&\text{when}\,\,\eta(\infty)\leq t<\tau^{-}_{0}=\infty,\\\
&\,\,0,&&\text{when}\,\,\tau^{-}_{0}<\infty,\,\eta(\tau^{-}_{0})<\infty\,\,\text{and}\,\,t\geq\eta(\tau^{-}_{0}).\end{aligned}\right.$
(1.3)

We stress that by definition the process $X$ takes its values in $[0,\infty]$
and both boundaries $0$ and $\infty$ are absorbing.

In order to avoid cumbersome notation, we also denote the law of the process
$X$ started from $x$ by $\mathbb{P}_{x}$. This will not cause any confusion.
The assumption that $1/R$ is positive on compact subsets of $(0,\infty)$
ensures that the time-changed SPLP is well-defined when started from a point
$x\in(0,\infty)$. We also highlight that for any rate function $R$, the
nonlinear CSBP $X$ inherits from $\xi$ the strong Markov property and its
càdlàg sample paths.

The definition (1.3) above is also referred to as a Lamperti-type transform of
the process $\xi$. We call $\xi$ the parent Lévy process of $X$. It is well
known that, if the function $R$ is the identity function, the process $X$
reduces to the classical CSBP with branching mechanism $\psi$, and if $R$ is
an exponential function, the process $X$ is associated to a positive self-
similar Markov process. The two aforementioned transforms are called the
Lamperti transformations in the literature; see e.g. Kyprianou [15, Chapter 12
and Chapter 13]. We also mention that the polynomial case, for which
$R(x)=x^{\beta}$, has been introduced in Li [18].

A nice feature of this class of Markov processes is that the function $R$
allows for a new degree of freedom to exhibit richer long-term behaviors
(extinction, coming down from infinity, explosion, etc) in comparison to the
classical CSBP framework. For instance, Li et al. [17] investigate the role of
the rate function in the extinguishing phenomenon i.e. the convergence of $X$
towards $0$ without reaching it. Foucart et al. [9] treat the case of a parent
Lévy process that either oscillates or drifts towards $-\infty$ and address
the question of whether the time-changed Lévy process $X$ can be started from
infinity, and at what speed it comes down from infinity. Last but not least,
the study of the explosion of the process $X$ and of its behavior “one second”
prior to explosion has been carried out in [16] in the case of a parent Lévy
process with finite mean. The latter assumption is in particular ruling out
the classical CSBPs.

Our main objective is to investigate the case a parent Lévy process with
infinite mean. We first collect some elementary properties of nonlinear CSBPs
that are consequences of the time-change relationship (1.3). We review then
briefly some results known in the setting of a jump measure with finite mean.
We shall also explain why other arguments have to be designed when it is
infinite.

From now on, consider a parent Lévy process $\xi$ with no negative jumps
drifting towards $+\infty$. For any $y>0$ and $a>0$, set

$T^{+}_{b}:=\inf\\{t>0:X_{t}>b\\}\text{ and }T^{-}_{a}:=\inf\\{t>0:X_{t}\leq
a\\}.$

By the assumption of local boundedness of $R$ on $(0,\infty)$, we have for any
$b>x>a>0$, $\mathbb{P}_{x}$-almost surely

$\\{T_{b}^{+}\wedge
T_{a}^{-}<\infty\\}=\\{\tau_{b}^{+}\wedge\tau_{a}^{-}<\infty,\eta(\tau_{b}^{+}\wedge\tau_{a}^{-})<\infty\\}=\\{\tau_{b}^{+}\wedge\tau_{a}^{-}<\infty\\},$

and the latter event has probability one. Define the explosion time of $X$ by

$T_{\infty}^{+}:=\inf\\{t>0:X_{t}=\infty\\}=\underset{b\rightarrow\infty}{\lim}\\!\\!\uparrow
T^{+}_{b}.$

The following identity relating $T^{+}_{\infty}$ for $X$ and $\eta$ for $\xi$
is obtained immediately from the definition (1.3): almost surely on the non-
extinction event, that is, on
$\\{X_{t}\underset{t\rightarrow\infty}{\nrightarrow}0\\}=\\{\tau_{0}^{-}=\infty\\}$,
we have

$T_{\infty}^{+}=\eta(\infty)=\int^{\infty}_{0}\frac{\mathrm{d}s}{R(\xi_{s})}.$
(1.4)

Hence, the event of explosion $\\{T_{\infty}^{+}<\infty\\}$ is equivalent to
the event of finiteness of the perpetual integral $\eta(\infty)$. This is a
classical topic, of independent interest, which has been studied intensively.
We refer the reader for instance to Döring and Kyprianou [7], Kolb and Savov
[14], Li and Zhou [19] and the references therein.

We now sketch the results obtained in [16] under the assumption that the
parent Lévy process $\xi$ has a finite mean, i.e.

$\mathbb{E}[\xi(1)]=-\psi^{\prime}(0)\in(0,\infty),$ (1.5)

and explain the differences with the setting $\mathbb{E}[\xi(1)]=\infty$. It
is established in [16] that the process explodes on the non-extinction event
with positive probability, i.e.
$\mathbb{P}_{x}(\eta(\infty)<\infty;\tau_{0}^{-}=\infty)>0$, see (1.4), if and
only if

$\int^{\infty}\frac{\mathrm{d}y}{R(y)}<\infty.$ (1.6)

Explosion occurs here not due to the jumps but due to drastic speed-up of the
time, i.e. $R(x)$ goes quickly towards $\infty$ as $x\to\infty$, so that the
long time behavior of $\xi$ becomes the short time behavior of $X$. This can
also be explained by the strong law of large numbers for Lévy processes, which
ensures that for any $x\geq 0$, $\mathbb{P}_{x}$-almost surely

$\xi(s)\underset{s\rightarrow\infty}{\sim}s\mathbb{E}[\xi(1)]=-s\psi^{\prime}(0).$

Replacing $\xi(s)$ in the integrand of (1.4) by the equivalent function
$-s\psi^{\prime}(0)$ would then lead to the condition (1.6).

In the same fashion, and still at a heuristic level, on the event of non-
extinction, the nonlinear CSBP $X$ fluctuates, as $t$ converges to
$T^{+}_{\infty}$, along the explosive deterministic path $(x(t))_{t\geq 0}$
satisfying

$\mathrm{d}x(t)=-\psi^{\prime}(0)\cdot R(x(t))\,\mathrm{d}t,\quad x(0)>0,$

and the speeds of explosion, i.e. the behavior of $X(T^{+}_{\infty}-t)$, for
different types of rate functions, have been investigated in [16]. The
techniques designed in [16] are mostly relying on the fact that the process
$\xi$ has a stationary overshoot distribution under condition (1.5). Since the
latter does not exist when $-\psi^{\prime}(0)=\infty$, most of the arguments
explained above break down, including the question of explosion itself (i.e.
the finiteness of $\eta(\infty)$). Moreover, there is no deterministic path
clearly defined around which the process oscillates before its explosion. The
problems of finding a sharp condition for explosion and how a nonlinear CSBP
behaves near its explosion time in the case of infinite mean (hence with very
large jumps) require therefore a study on their own and are the principal aims
of this paper.

Our first result is a necessary and sufficient condition (in terms of an
integral test) for explosion of nonlinear CSBPs when the rate function $R$ is
subject to a certain assumption on its growth. Note that no assumption is made
on the mean in the following theorem. For any monotone function $H$, we denote
by $\mathrm{d}H$ the associated Stieljes measure111We will also use the
notation $H(\mathrm{d}z)$: $\int_{x}^{y}\mathrm{d}H(z)=H(y)-H(z)$ for $x<y$.

######  Theorem 1.1 (Explosion criterion).

If $R$ is increasing and

$\limsup_{x\to\infty}\frac{R(x+a)}{R(x)}<\infty\quad\text{for every $a>0$},$
(1.7)

then for any $x\in(0,\infty)$,

$\mathbb{P}_{x}(T^{+}_{\infty}<\infty)>0\text{ if and only if
}\int^{\infty}\frac{-1}{\psi(1/y)}\mathrm{d}\Big{(}-\frac{1}{R(y)}\Big{)}<\infty.$
(1.8)

Moreover, when (1.8) holds, one has
$\mathbb{P}_{x}(T_{\infty}^{+}<\infty|T_{a}^{-}=\infty)=1$ for any
$a\in(0,x)$.

######  Remark 1.1.

By condition (1.7), function $R$ cannot be rapidly varying. For instance, the
exponential functions are excluded.

######  Remark 1.2.

The setting of a classical CSBP is covered by the theorem. Indeed the identity
function $R(y)=y$, for all $y\geq 0$, satisfies (1.7) and the integral
condition in (1.8) can be rewritten, with the help of change of variable,
$u=1/y$, as $\int_{0}\frac{\mathrm{d}u}{-\psi(u)}<\infty$. We recover here the
classical condition for explosion of a CSBP; see Grey [11] and the forthcoming
Section 3.4 for more details.

The study of the behavior prior to explosion requires to characterize the
asymptotic overshoot $X(T^{+}_{x})-x$ as $x\to\infty$. In the case of infinite
mean, the latter tends to $\infty$ in probability and we have to look for some
explicit scaling limits. We then need to impose some assumption of regular
variations on both the rate function and the distribution of the large jumps
of the parent process $\xi$.

Let $\mathcal{R}_{\beta}$ be the set of regularly varying functions with
parameter $\beta$; see Section 2.2 for more details on those functions. In the
next statements, we shall mainly focus on nonlinear CSBPs with a branching
mechanism and a rate function satisfying the following regular variation
properties:

$\mathbb{H}$: $\quad-\psi\in\mathcal{R}_{\alpha}$ at $0$ and
$R\in\mathcal{R}_{\beta}$ at $\infty$ for some $\alpha\in[0,1],\beta>0$.

For $\alpha\in[0,1)$, we always have $\psi^{\prime}(0)=-\infty$. The case
$\alpha=1$ is a critical case for which both $-\psi^{\prime}(0)=\infty$ and
$-\psi^{\prime}(0)\in(0,\infty)$ are possible. Notice also that any function
$R$ regularly varying near $\infty$ satisfies (1.7). We will see in the next
corollary that the integral test in Theorem 1.1 can be further simplified in
this setting. We then study the explosion behavior according to differential
values of parameters $\alpha$ and $\beta$ when assumption $\mathbb{H}$ holds.

######  Corollary 1.1.

If $R\in\mathcal{R}_{\beta}$ at $\infty$ for some $\beta>0$, then for any
$x\in(0,\infty)$, we have

$\mathbb{P}_{x}(T^{+}_{\infty}<\infty)>0\text{ if and only if
}\mathcal{E}:=\int^{\infty}\frac{-\mathrm{d}y}{yR(y)\psi(1/y)}<\infty.$ (1.9)

If assumption $\mathbb{H}$ is in force, we further have

  1. 1.

if $\alpha>\beta>0$, then
$\mathbb{P}_{x}\big{(}T_{\infty}^{+}<\infty\big{)}=0$;

  2. 2.

if $\alpha=\beta>0$, then
$\mathbb{P}_{x}\big{(}T_{\infty}^{+}<\infty\big{)}>0$ if and only if
$\mathcal{E}<\infty$;

  3. 3.

if $\beta>\alpha\geq 0$, then
$\mathbb{P}_{x}\big{(}T_{\infty}^{+}<\infty\big{)}>0$.

######  Remark 1.3.

The three cases in Corollary 1.1 cover different situations:

  1. 1.

When $1\geq\alpha>\beta>0$, the rate function $R$ is slowing down the jump
rate so much that the explosion is prevented. The process $X$ is in this case
transient and nonexplosive. We shall rule out this case in the next.

  2. 2.

In the critical case $\alpha=\beta$, the power functions in $R(y)$ and
$\psi(1/y)$ cancel out and whether explosion occurs or not depends on the
slowly varying parts in $R$ and $\psi$. For instance, if $\psi(s)=s\log s$
(Neveu’s branching mechanism) and $R(y)=yL(y)$ for some slowly varying
function $L$ at $\infty$, then the process explodes if and only if
$\mathcal{E}=\int^{\infty}\frac{\mathrm{d}y}{yL(y)\log(y)}<\infty$.

  3. 3.

When $\beta>\alpha\geq 0$, the explosion occurs. More precisely, when
$\beta>1$, the rate function itself can cause the explosion since
$\int^{\infty}\frac{\mathrm{d}y}{R(y)}<\infty$. When $\beta\leq 1$ (the
classical CSBP with $\beta=1$ falls into this case), the rate function itself
“can not bring” the process to $\infty$ in finite time since
$\int^{\infty}\frac{\mathrm{d}y}{R(y)}=\infty$, and the parent process is
necessary to have extremely large jumps for explosion to occur.

We now present our main results on the “speeds of explosion” for the nonlinear
CSBPs satisfying the assumption $\mathbb{H}$. We will study the asymptotics of
$X(T^{+}_{\infty}-t)$ when $t$ goes to $0+$, given that
$T^{+}_{\infty}<\infty$. Different renormalisations that lead to convergence
in law or convergence in probability will be obtained. The limit law of
$X(T^{+}_{\infty}-t)$ after renormalisation will be expressed with the help of
the following two random variables:

  * •

For $\alpha\in(0,1)$, let $\chi_{\alpha}$ be a random variable taking values
in $[1,\infty)$ with probability law

$\mathbb{P}(\chi_{\alpha}>z)=\frac{\sin\alpha\pi}{\pi}\int_{0}^{1}u^{\alpha-1}(z-u)^{-\alpha}\mathrm{d}u\quad\text{for
$z\geq 1$}.$ (1.10)

We also set $\chi_{\alpha}=1$ a.s. for $\alpha=1$ and $\chi_{\alpha}=\infty$
a.s. for $\alpha=0$.

  * •

For $\alpha\in[0,1]$ and $\beta>\alpha$, let $\varrho_{\alpha,\beta}$ be a
nonnegative random variable with moment generating function given by

$M_{\varrho_{\alpha,\beta}}(\theta):=\mathbb{E}[e^{\theta\varrho_{\alpha,\beta}}]=1+\sum_{n\geq
1}\frac{\theta^{n}}{n!}\Big{(}\prod_{k=1}^{n}\frac{\Gamma(k(\beta-\alpha)+1)}{\Gamma(k(\beta-\alpha)+\alpha)}\Big{)}\quad\text{
for }|\theta|<\frac{1}{\beta}.$ (1.11)

When $\alpha=1$, $\varrho_{\alpha,\beta}\equiv 1$ a.s. and when $\alpha=0$,
$\varrho_{\alpha,\beta}$ is exponentially distributed with parameter
$1/\beta$. It is also notable that when $\beta=1$ and $0\leq\alpha<1$, (1.11)
can be simplified into

$M_{\varrho_{\alpha,1}}(\theta)=\sum_{n\geq
0}\frac{\theta^{n}}{n!}\Gamma(1+n(1-\alpha)),$

and we recover the moment-generating function of the Weibull distribution with
parameter $\frac{1}{1-\alpha}$, namely the random variable $\rho_{\alpha,1}$
has cumulative distribution function

$\mathbb{P}(\rho_{\alpha,1}\leq t)=1-e^{-t^{\frac{1}{1-\alpha}}},\text{ for
all }t\geq 0.$

The laws of $\chi_{\alpha}$ and $\varrho_{\alpha,\beta}$ have already appeared
in the literature and can be expressed with the help of a stable subordinator
through, respectively, its renormalized asymptotic overshoot and a certain
perpetual integral; see the forthcoming Propositions 2.3 and 2.5 for more
details.

From now on, we shall always work on the event of explosion. We write
$\mathbb{P}_{x}(\cdot|T_{\infty}^{+}<\infty)$ for the conditional distribution
given $X(0)=x$ and $T_{\infty}^{+}<\infty$. The convergence in distribution is
denoted by “$\Rightarrow$”. Note that when explosion happens, we always have
$\mathcal{E}<\infty$ and thus

$\varphi(x):=\int_{x}^{\infty}\frac{-\mathrm{d}y}{yR(y)\psi(1/y)}<\infty\quad\text{for
$x>1/p$},$ (1.12)

where $p:=\inf\\{s>0:\psi(s)>0\\}$ is the largest zero of the branching
mechanism (if $\psi(s)<0$ for all $s>0$, $p=\infty$ and $1/p=0$). Background
on the function $\psi$ is provided in Subsection 2.1. Note also that the
inverse function of $\varphi$,

$\varphi^{-1}(t):=\inf\\{x>0,\varphi(x)=t\\},$

is well defined for any $t<\varphi(p^{-1})$.

######  Theorem 1.2 (Speeds of explosion).

Assume that $\mathbb{H}$ holds and $\mathcal{E}<\infty$ in (1.9). We have the
following regimes of weak convergence for different choices of
$\beta\geq\alpha\geq 0$.

  1. 1.

If $\beta>\alpha>0$, then

$\frac{X(T_{\infty}^{+}-t)}{\varphi^{-1}(t)}\bigg{|}_{\mathbb{P}_{1}(\cdot|T_{\infty}^{+}<\infty)}\Rightarrow\frac{1}{\chi_{\alpha}}\times\varrho_{\alpha,\beta}^{\frac{1}{\beta-\alpha}}\quad\text{as
$t\to 0+$},$ (1.13)

where $\chi_{\alpha}$ and $\varrho_{\alpha,\beta}$ are two independent random
variables whose probability laws are specified in (1.10) and (1.11),
respectively.

  2. 2.

If $\beta=\alpha>0$ and $\mathcal{E}<\infty$, then

$\frac{X(T_{\infty}^{+}-t)}{\varphi^{-1}(\Gamma(\alpha)t)}\Big{|}_{\mathbb{P}_{1}(\cdot|T_{\infty}^{+}<\infty)}\rightarrow
1\text{ in probability as $t\to 0+$}.$

  3. 3.

If $\beta>\alpha=0$, then

$\frac{\psi(1/X(T^{+}_{\infty}-t))}{\psi(1/\varphi^{-1}(t))}\bigg{|}_{\mathbb{P}_{1}(\cdot|T_{\infty}^{+}<\infty)}\Rightarrow\frac{1}{\Lambda}\quad\text{as
$t\to 0+$},$ (1.14)

where $\Lambda$ is a uniform random variable on $(0,1)$.

######  Remark 1.4.

The order of the renormalisation, which we call the speed of explosion, is
different in each of the three regimes in Theorem 1.2.

  1. 1.

For $\beta>\alpha>0$, one has $\varphi\in\mathcal{R}_{\alpha-\beta}$ at
$\infty$, so that $\varphi^{-1}\in\mathcal{R}_{\frac{1}{\alpha-\beta}}$ at
$0$. The speed of explosion is thus of the order $t^{\frac{1}{\alpha-\beta}}$.

  2. 2.

For $\beta=\alpha>0$, one can easily check that the function $\varphi$ is
slowly varying at $\infty$. Hence, its inverse $\varphi^{-1}$ is rapidly
varying; see the next remark for some examples.

  3. 3.

For $\beta>\alpha=0$, the convergence in (1.13) actually also holds true. But
since $\chi_{\alpha}=\infty$ a.s., the limit is $0$. The speed of explosion of
$X$ when $\alpha=0$ is thus faster than $\varphi^{-1}$. The nonlinear
renormalisation in (1.14) depends on the form of the slowly varying function
$-\psi$. For instance, if $R(x)\sim x^{\beta}$ and $\psi(1/x)\sim-(1/\log
x)^{r}$ as $x\to\infty$ for some $r>0$, then one can check from (1.14) that

$\frac{\log X(T^{+}_{\infty}-t)}{\log
1/t}\Big{|}_{\mathbb{P}_{1}(\cdot|T_{\infty}^{+}<\infty)}\Rightarrow\frac{1}{\beta\Lambda^{1/r}}\quad\text{as
$t\to 0+$.}$

Equivalently,

$\log\log X(T^{+}_{\infty}-t)-\log\log
1/t\,\Big{|}_{\mathbb{P}_{1}(\cdot|T_{\infty}^{+}<\infty)}\Rightarrow\log
1/\beta+\mathbbm{e}_{r}$ (1.15)

where $\mathbbm{e}_{r}$ is an exponential random variable with parameter $r$
and we have used the fact that
$\mathbbm{e}_{r}\overset{D}{=}\log\frac{1}{\Lambda^{1/r}}$.

######  Remark 1.5.

Classical CSBPs, for which $R(y)=y$ and hence $\beta=1$, can belong to any of
the three different cases covered by Theorem 1.2 according the behaviour of
their branching mechanism $\psi$ near $0$. By the change of variable $u=1/y$
in (1.12), one has $\varphi(x)=\int_{0}^{1/x}\frac{\mathrm{d}u}{-\psi(u)}$ for
all $x>1/p$.

  1. 1.

In Case 1, since $\rho_{\alpha,1}$ has a Weibull distribution with parameter
$\frac{1}{1-\alpha}$, the random variable
$\rho_{\alpha,1}^{\frac{1}{1-\alpha}}$ has a standard exponential law and the
limit law in (1.13) is a mixture of exponential laws with random parameter
$\chi_{\alpha}$. This case is explored further in Section 3.4.

  2. 2.

Case 2 with $\alpha=\beta=1$ contains for instance the branching mechanism
$\psi$ such that $\psi(x)\underset{x\rightarrow 0}{\sim}-x(\log 1/x)^{\gamma}$
with $\gamma>1$. Simple calculations give
$\varphi^{-1}(t)\underset{t\rightarrow
0}{\sim}e^{c_{\gamma}t^{-\frac{1}{\gamma-1}}}$ with
$c_{\gamma}=(\gamma-1)^{-\frac{1}{\gamma-1}}>0$.

  3. 3.

In Case 3, the function $\varphi^{-1}$ can behave differently according to the
slowly varying function $\psi$. However, using the fact that $-\psi$ is
increasing in a neighbourhood of $0$, one can plainly check that
$\varphi(x)\geq-\frac{\psi(1/x)}{x}=:g(x)$. Since $\varphi^{-1}(t)\geq
g^{-1}(t)$, we have

$\frac{X(T_{\infty}^{+}-t)}{g^{-1}(t)}\Big{|}_{\mathbb{P}_{x}(\cdot|T_{\infty}^{+}<\infty)}\rightarrow
0\text{ in probability as $t\to 0+$}.$

Notice that $g^{-1}\in\mathcal{R}_{-1}$ at $0$. So, it is of order $1/t$.
Lastly, note that the first term in the limit (1.15) vanishes in the CSBP
case, since $\beta=1$, and we have an exponential law with parameter $r$.

######  Remark 1.6.

In Case 1 with $\alpha=1$, our result also holds in some cases with finite
mean and coincide with the result obtained in [16, Theorem 3.6-(a)]. Indeed
when $-\psi\in\mathcal{R}_{1}$ and
$\mathbb{E}[\xi_{1}]=-\psi^{\prime}(0)<\infty$, we have
$x\psi(1/x)\to-\psi^{\prime}(0)<\infty$ as $x$ goes to $\infty$. Thus, the
explosive condition in (1.9) reduces to (1.6). Moreover,
$\varphi(x)\underset{x\rightarrow\infty}{\sim}\frac{1}{\mathbb{E}[\xi_{1}]}\int_{x}^{\infty}\frac{\mathrm{d}y}{R(y)}$
and in both Case 1 and Case 3 of Theorem 1.2 for $\alpha=1$ and
$-\psi^{\prime}(0)<\infty$, we have

$\frac{X(T_{\infty}^{+}-t)}{\varphi^{-1}(t)}\rightarrow 1\text{ in probability
as $t\to 0+$}.$

No classical branching process falls into this setting.

The rest of the paper is organized as follows. In Section 2, we gather
fundamental facts on the spectrally positive Lévy processes, their scale
functions and their potential measure. We then recall elements on regularly
varying functions and state some results about the overshoot of a SPLP with a
regularly varying Laplace exponent. A representation of the random variable
$\rho_{\alpha,\beta}$ with the help of a stable subordinator is also given. We
prove these results as well as Theorems 1.1 and 1.2 in Section 3.

Notation: Throughout the paper, we write $\displaystyle f(x)\sim g(x)$ if
$f(x)/g(x)\to 1$, $f(x)=o(g(x))$ if $f(x)/g(x)\to 0$, $f(x)=O(g(x))$ if
$\limsup_{x\to\infty}f(x)/g(x)<\infty$ and finally $f(x)\asymp g(x)$ if
$f(x)/g(x)$ is bounded away from both $0$ and $\infty$. For any Borel measure
$\nu$ on $(0,\infty)$ we write
$\bar{\nu}(z):=\int_{z}^{\infty}\nu(\mathrm{d}y)$ for all $z>0$. We denote by
$\overset{\text{D}}{=}$ the equality in distribution. Throughout the paper we
take the convention $\inf\emptyset=\infty$.

##  2 Preliminaries

###  2.1 Spectrally positive Lévy processes and their scale functions

Let $\xi$ be a spectrally positive Lévy process, i.e. a Lévy process with no
negative jumps. Its Laplace exponent is well defined and takes the Lévy-
Khintchine form, for $s\geq 0$,

$\psi(s)=\log\mathbb{E}\big{[}e^{-s\xi(1)}\big{]}=\gamma
s+\frac{\sigma^{2}}{2}s^{2}+\int_{0}^{\infty}\big{(}e^{-sz}-1+sz\mathbf{1}(z\leq
1)\big{)}\nu(\mathrm{d}z),$ (2.1)

where $\gamma\in\mathbb{R},\sigma\geq 0$ are constants and $\nu$ is the Lévy
measure, namely, it is a $\sigma$-finite measure on $(0,\infty)$ satisfying
$\int_{0}^{\infty}(1\wedge z^{2})\nu(\mathrm{d}z)<\infty$.

  * •

When $\xi$ is not a subordinator, one has $\psi(\infty)=\infty$ and the
process $\xi$ may in general oscillates or drifts towards +$\infty$ or
$-\infty$. As we are interested in the explosion, we shall focus on the case
of a process $\xi$ drifting towards $+\infty$. This is equivalent to the
assumption $\mathbb{E}(\xi(1))=-\psi^{\prime}(0)\in(0,\infty]$. In the non-
subordinator case, the right-continuous inverse of $\psi$ defined by
$\phi(q):=\inf\\{s>0,\psi(s)>q\\}$ is well-defined for all $q>0$. Recall also
that we set $p=\phi(0)$. Another equivalent condition for the process to drift
towards $+\infty$ is $p>0$.

  * •

When $\xi$ is a subordinator, i.e. $\xi$ has monotone nondecreasing sample
paths, the function $\psi$ is non-positive and takes the form

$\psi(s)=-\delta s+\int_{0}^{\infty}\big{(}e^{-sz}-1\big{)}\nu(\mathrm{d}z),$
(2.2)

with $\delta\geq 0$ is the drift and the measure $\nu$ satisfies
$\int_{0}^{\infty}(1\wedge z)\nu(\mathrm{d}z)<\infty$. In this setting,
$\psi(\infty)=-\infty$ and it is more convenient to work with the function
$-\psi$ which satisfies $-\log\mathbb{E}[e^{-s\xi(1)}]=-\psi(s)$. In the
subordinator case, $\phi$ is not defined on $\mathbb{R}^{+}$ and in particular
$p=\phi(0)=\inf\emptyset=\infty$.

The study of the explosion of the process $X$ is more involved when the parent
Lévy process $\xi$ is not a subordinator, i.e. $0<p<\infty$. This is mainly
due to the fluctuations of the process which makes it more difficult to follow
its position and to compute its potential measure.

Non-subordinator case: $0<p<\infty$. In order to handle the non-subordinator
case, we shall need some properties of fluctuation theory. Most of the
literature on fluctuations of spectrally one-sided Lévy processes concerns the
spectrally negative cases, that is to say, processes with no positive jumps.
We rewrite below some classical fluctuation identities in the spectrally
positive setting. They can all be found using the fact that if $\xi$ is a SPLP
then its dual process $-\xi$ is spectrally negative.

Scale function $W$ plays a key role in the fluctuation theory and will be one
of our main tool later on. It is defined as the unique continuous increasing
function on $\mathbb{R}^{+}$ satisfying

$\int_{0}^{\infty}e^{-sz}W(z)\mathrm{d}z=\frac{1}{\psi(s)},\quad\text{for
$s>p$},$ (2.3)

with the convention $W(z)=0$ for $z<0$. We refer the readers to [2] and [15]
for more detailed discussions on the spectrally negative Lévy process and its
scale functions. We shall also use the notation $W(\mathrm{d}z)$ for the
Stieljes measure associated to $W$.

Recall the first passage times of $\xi$, $\tau_{a}^{-}$ and $\tau_{b}^{+}$,
see (1.1). For any $x\in(a,b)$, one has the identities

$\mathbb{P}_{x}(\tau_{a}^{-}\leq\tau_{b}^{+})=\frac{W(b-x)}{W(b-a)}\quad\text{and}\quad\mathbb{P}_{x}(\tau_{a}^{-}<\infty)=\underset{b\rightarrow\infty}{\lim}\mathbb{P}_{x}(\tau_{a}^{-}\leq\tau_{b}^{+})=e^{-p(x-a)}.$
(2.4)

The following expression of the potential measure of $\xi$ killed on leaving
the interval $[a,\infty)$ can be found for instance in [15, Corollary 8.8].
For any $a\in\mathbb{R}$ and $x,y\in(a,\infty)$,

$U_{a}(x,\mathrm{d}y):=\int_{0}^{\infty}\mathbb{P}_{x}(\xi(t)\in\mathrm{d}y,t<\tau_{a}^{-})\mathrm{d}t=\big{(}e^{p(a-x)}W(y-a)-W(y-x)\big{)}\mathrm{d}y,$
(2.5)

for any $x\geq a$ and where we recall that $p=\phi(0)$. Define for any
$y\in\mathbb{R}$

$\kappa(y):=e^{py}\phi^{\prime}(0)-W(y).$ (2.6)

Recall that $W(y)=0$ for $y<0$ so that $\kappa(y)=e^{py}\phi^{\prime}(0)$ for
$y<0$. One has

$U_{a}(x,\mathrm{d}y)=\big{(}\kappa(y-x)-e^{p(a-x)}\kappa(y-a)\big{)}\mathrm{d}y.$
(2.7)

Similarly, under the assumption $p>0$, the process is transient and its
potential measure has a density bounded on $[0,\infty)$, see [15, Corollary
8.9]. When $\xi$ starts from $0$, the latter is given by

$U(\mathrm{d}y):=\int_{0}^{\infty}\mathbb{P}(\xi(t)\in\mathrm{d}y)\mathrm{d}t=\kappa(y)\mathrm{d}y$
(2.8)

and we call $\kappa$ the potential density of $\xi$.

The function $\kappa$ is not explicit in general. There is however an
equivalence near $0$ of its Laplace transform.

######  Lemma 2.1.

Suppose that $0<p<\infty$. We have as $s\to 0+$

$\hat{\kappa}(s):=\int_{0}^{\infty}e^{-sy}\kappa(y)\mathrm{d}y=\int_{0}^{\infty}e^{-sy}U(\mathrm{d}y)\sim\frac{-1}{\psi(s)}.$
(2.9)

###### Proof.

One can calculate directly that

$\hat{\kappa}(s)=\int_{0}^{\infty}e^{-sy}\kappa(y)\mathrm{d}y=\frac{\phi^{\prime}(0)}{s-p}-\frac{1}{\psi(s)}=\left(1-\frac{\phi^{\prime}(0)\psi(s)}{s-p}\right)\frac{1}{-\psi(s)}\sim\frac{-1}{\psi(s)}\text{\
at $0$},$

where the identity above holds first for $s>p$, and then for all $s>0$ by
analytic extension and the equivalence follows readily since $\psi(0)=0$. ∎

Subordinator case: $p=\infty$. Assume now that $\xi$ is a subordinator. For
any $x\geq 0$, define the potential measure

$U(x,\mathrm{d}y):=\int_{0}^{\infty}\mathbb{P}_{x}(\xi(t)\in\mathrm{d}y)\mathrm{d}t\text{
and }U(\mathrm{d}y):=U(0,\mathrm{d}y).$ (2.10)

One has, see e.g. [2, Chapter III, page 74], for all $s>0$,

$\hat{U}(s):=\int_{0}^{\infty}e^{-sy}U(\mathrm{d}y)=\mathbb{E}\left[\int_{0}^{\infty}e^{-s\xi(t)}\mathrm{d}t\right]=\frac{1}{-\psi(s)}.$
(2.11)

Moreover, it is easily checked that

$\hat{U}(x,s):=\int_{x}^{\infty}e^{-sy}U(x,\mathrm{d}y)=e^{-sx}\int_{0}^{\infty}e^{-sz}U(\mathrm{d}z)\underset{s\rightarrow
0+}{\sim}\hat{U}(s).$

We stress that contrary to the previous setting, the measure $U$ may not have
a density. However, in both cases the measure on $(0,\infty)$,
$\mathbf{1}(y>0)U(\mathrm{d}y)$ has a Laplace transform satisfying (2.9).

###  2.2 Regularly varying functions and Laplace exponents

A function $f$ is regularly varying at $\infty$ with index
$\gamma\in\mathbb{R}$, written $f\in\mathcal{R}_{\gamma}$ at $\infty$, if it
is a positive function defined on a neighbourhood $[A,\infty)$ of $\infty$
satisfying

$f(\lambda x)\big{/}f(x)\to\lambda^{\gamma}\quad\text{as $x\to\infty$ for all
$\lambda>0$}.$

In particular, $f$ is called slowly varying at $\infty$ if $\gamma=0$.
Similarly, a function $f$ is regularly varying at $0$ with index $\gamma$ if
and only if $x\mapsto f(1/x)\in\mathcal{R}_{-\gamma}$ at $\infty$. Note that
$f\in\mathcal{R}_{\gamma}$ if and only if $x\mapsto
x^{-\gamma}f(x)\in\mathcal{R}_{0}$.

Many properties of regularly varying functions will be used in the proofs of
the main results. Specifically, we shall appeal to the uniform convergence
theorem (UCT for short), see Bingham et al. [5, Theorem 1.5.1], the Karamata
and Tauberian theorems, see [5, Theorem 1.5.11 and Theorem 1.7.1] and the
monotone density theorem, [5, Theorem 1.7.2]. We recall the Karamata theorem
and the Tauberian theorem for convenience of the reader.

######  Proposition 2.1 (Karamata’s Theorem).

Let $f\in\mathcal{R}_{\rho}$ at $\infty$ for $\rho\in\mathbb{R}$ be locally
bounded on $[A,\infty)$.

  1. 1.

For any $\sigma\geq-(1+\rho)$, we have

$x^{\sigma+1}f(x)\bigg{/}\int_{A}^{x}t^{\sigma}f(t)\mathrm{d}t\underset{x\rightarrow\infty}{\to}1+\rho+\sigma;$

  2. 2.

For any $\sigma<-(1+\rho)$ (and for $\sigma=-(\rho+1)$ if
$\int^{\infty}t^{-\sigma}f(t)\mathrm{d}t<\infty$), we have

$x^{\sigma+1}f(x)\bigg{/}\int_{x}^{\infty}t^{\sigma}f(t)\mathrm{d}t\underset{x\rightarrow\infty}{\to}-\big{(}1+\rho+\sigma\big{)}.$

######  Proposition 2.2 (Karamata’s Tauberian Theorem).

Let $\mu$ be a positive measure on $\mathbb{R}^{+}$ and $l\in\mathcal{R}_{0}$.
Put $\mu(x):=\mu(0,x]$ for $x>0$. Then for $c,\rho\geq 0$,

$\mu(x)\sim\frac{cx^{\rho}l(x)}{\Gamma(1+\rho)}\quad\text{if and only
if}\quad\hat{\mu}(s):=\int_{0}^{\infty}e^{-sy}\mu(dy)\sim cs^{-\rho}l(1/s)$

as $x\to\infty$ and $s\to 0+$ respectively. For $c=0$ the left hand side is
interpreted as $\mu(x)=o(x^{\rho}l(x))$ and the right hand side is interpreted
similarly.

The following lemmas are consequences of the propositions above. The first one
guarantees that if $f\in\mathcal{R}_{-\gamma}$ with negative index, then it is
equivalent to a decreasing and differentiable function.

######  Lemma 2.2.

If $f\in\mathcal{R}_{-\gamma}$ at $\infty$ for some $\gamma>0$, we have

$f(x)\sim
f^{\downarrow}(x):=\gamma\int_{x}^{\infty}\frac{f(y)}{y}\,\mathrm{d}y\quad\text{as
$x\to\infty$}.$ (2.12)

Notice that the function $f^{\downarrow}$ is decreasing.

The next lemma allows us, among other things, to construct a regularly varying
Laplace exponent $\psi$ from the Lévy measure $\nu$ in (2.1). Recall that
$\bar{\nu}$ denotes the tail of $\nu$.

######  Lemma 2.3.

Suppose that $-\psi^{\prime}(0+)=\infty$. For any $\alpha\in[0,1]$,

$-\psi(s)\sim s^{\alpha}l(1/s)\text{\ at $0+$\quad if and only
if\quad}\int_{1}^{x}\bar{\nu}(z)\mathrm{d}z\sim\frac{x^{1-\alpha}l(x)}{\Gamma(2-\alpha)}\in\mathcal{R}_{1-\alpha}\text{\
at $\infty$},$ (2.13)

where $l\in\mathcal{R}_{0}$ at $\infty$.

For $\alpha\in[0,1)$, (2.13) is equivalent to
$\displaystyle\bar{\nu}(x)\sim\frac{x^{-\alpha}l(x)}{\Gamma(1-\alpha)}\in\mathcal{R}_{-\alpha}$
at $\infty$.

###### Proof.

It follows from (2.1) that when $\mathbb{E}[\xi(1)]=-\psi^{\prime}(0)=\infty$,

$-\psi(s)\sim\int_{1}^{\infty}(1-e^{-sy})\nu(\mathrm{d}y)=s\int_{0}^{\infty}e^{-sz}\bar{\nu}(1\vee
z)\mathrm{d}z,\text{ as }s\to 0+.$

Applying Proposition 2.2 we have (2.13). If $\alpha\in[0,1)$, we further have
from (2.13) that

$\displaystyle\int_{1}^{x}\bar{\nu}(z)\mathrm{d}z\sim\frac{x^{1-\alpha}l(x)}{\Gamma(2-\alpha)}$
at $\infty$ if and only if
$\displaystyle\bar{\nu}(x)\sim\frac{x^{-\alpha}l(x)}{\Gamma(1-\alpha)}\in\mathcal{R}_{-\alpha}$
at $\infty$,

applying the monotone density theorem. ∎

Note that if $-\psi\in\mathcal{R}_{\alpha}$ then $\psi$ is negative in a
neighbourhood of $0$ and necessarily $p\in(0,\infty]$, so that the Lévy
process $\xi$, with exponent $\psi$, drifts towards $+\infty$ a.s.. Another
important application of the Tauberian theorem is the regularly varying
property for the integral $\int_{0}^{x}U(\mathrm{d}y)$ with $U$ the potential
measure of $\xi$; see (2.8) and (2.11).

######  Lemma 2.4 (Regular variation of the potential measure).

Suppose $-\psi\in\mathcal{R}_{\alpha}$ at $0$. Then in both cases $0<p<\infty$
and $p=\infty$, we have

$\int_{0}^{x}U(\mathrm{d}y)\sim\frac{-1}{\psi(1/x)}\frac{1}{\Gamma(1+\alpha)}\in\mathcal{R}_{\alpha}\text{\
at $\infty$}.$ (2.14)

###### Proof.

The result in the subordinator case is a direct consequence of Proposition
2.2. In the non-subordinator case, one has
$\int_{0}^{x}U(\mathrm{d}y)=\int_{0}^{x}\kappa(y)\mathrm{d}y$ and by (2.9),
$\hat{\kappa}(s)\sim\frac{-1}{\psi(s)}\in\mathcal{R}_{-\alpha}\text{\ at
$0$}$. Thus, (2.14) follows from Proposition 2.2. ∎

###  2.3 On the probability laws of $\chi_{\alpha}$ and
$\varrho_{\alpha,\beta}$

To study the explosive behaviour of $X$, it is necessary to understand the
asymptotic distribution of the overshoots for the parent process $\xi$. In the
case of infinite mean, we need $-\psi$ to be regularly varying to take use of
the scaling property. The following result is known for a subordinator, we
refer to Bertoin [2, Exercice III.4]; see also Doney and Maller [6] for a more
detailed discussion on the convergence of $\xi(\tau_{x}^{+})/x\rightarrow 1$
either in probability or almost surely.

######  Proposition 2.3.

Let $\alpha\in[0,1]$, suppose that $p>0$ and $-\psi\in\mathcal{R}_{\alpha}$ at
$0$. Then

$x^{-1}\xi(\tau_{x}^{+})\Rightarrow\chi_{\alpha}\quad\text{under $\mathbb{P}$
as $x\to\infty$},$ (2.15)

where $\chi_{\alpha}$ is a random variable whose distribution is specified in
(1.10). The convergence (2.15) also holds under
$\mathbb{P}_{1}(\cdot|\tau_{a}^{-}=\infty)$ for every $a\in(0,1)$.

The proof of Proposition 2.3 is provided in Section 3.1.1. In the case
$\alpha=0$, one observes by combining [2, Theorem III.6] with Proposition 2.3
that $\xi(\tau_{x}^{+}-)=o(x)$ and $x=o(\xi(\tau_{x}^{+}))$ in probability
when $x\to\infty$. In this extreme case, when the process upcrosses an
arbitrary large level $x$, it overshoots so far above it that $x$ is
negligible in comparison to the position of the process after the overshoot.

We shall establish in this direction the following proposition.

######  Proposition 2.4.

If $-\psi\in\mathcal{R}_{0}$ at $0$, we have

$\psi(1/\xi(\tau_{x}^{+}))\big{/}\psi(1/x)\Rightarrow\Lambda,\quad\text{under
$\mathbb{P}$ as $x\to\infty$},$ (2.16)

where $\Lambda$ is a uniform random variable on $(0,1)$. The convergence
(2.16) also holds under $\mathbb{P}_{1}(\cdot|\tau_{a}^{-}=\infty)$ for every
$a\in(0,1)$.

Such a result is not really new. It appears under another form in Klüppelberg
et al. [13, Theorem 4.4]. Assumptions of the latter theorem are however not
satisfied here and we will give a short proof of the proposition which suits
our framework. The nonlinear renormalisation is also reminiscent of that
obtained in some functional limit theorems for subordinators with slowly
varying exponent; see for instance [20], [12] and [10]. The proof of
Proposition 2.4 is in Section 3.3.4. This will follow from arguments designed
to treat the Case 2 of Theorem 1.2.

As mentioned in Section 1, the random variable $\varrho_{\alpha,\beta}$,
defined in (1.11), can be encountered in different contexts. The next
proposition states that the law of $\varrho_{\alpha,\beta}$ is indeed that of
some perpetual integral for an $\alpha$-stable subordinator.

######  Proposition 2.5.

For $\beta>\alpha>0$, let $S:=(S(t),t\geq 0)$ be a stable subordinator started
from $S(0)=1$ with Laplace exponent $\psi(s)=-c_{0}s^{\alpha}$ for some
$\alpha\in(0,1)$. Then for any $n\geq 1$,

$\mathbb{E}_{1}\bigg{[}\Big{(}\int_{0}^{\infty}\frac{\mathrm{d}t}{S(t)^{\beta}}\Big{)}^{n}\bigg{]}=\frac{1}{c_{0}(\beta-\alpha)}\prod_{k=1}^{n}\frac{\Gamma\big{(}k(\beta-\alpha)+1\big{)}}{\Gamma\big{(}k(\beta-\alpha)+\alpha\big{)}}.$
(2.17)

Moreover, the following large deviation estimate holds:

$\limsup_{t\to\infty}t^{-\frac{1}{1-\alpha}}\log\mathbb{P}_{1}\bigg{(}\int_{0}^{\infty}\frac{\mathrm{d}u}{S(u)^{\beta}}>t\bigg{)}=-(1-\alpha)c_{0}^{\frac{1}{1-\alpha}}(\beta-\alpha)^{\frac{\alpha}{1-\alpha}}.$
(2.18)

For the case $c_{0}(\beta-\alpha)=1$ we have by (1.11)

$\varrho_{\alpha,\beta}\overset{\text{D}}{=}\int_{0}^{\infty}\frac{\mathrm{d}t}{S(t)^{\beta}}.$

######  Remark 2.1.

The perpetual integral $\int_{0}^{\infty}\frac{\mathrm{d}t}{S(t)^{\beta}}$
under $\mathbb{P}_{1}$ can also be seen as the explosion time of the nonlinear
CSBP started from $1$, with rate function $R(x)=x^{\beta}$, and parent Lévy
process the subordinator $S$.

##  3 Proofs of the main results

This section is dedicated to the proofs of our main results. We first
establish the criterion of explosion stated in Theorem 1.1. Corollary 1.1 on
the regularly varying cases then follows easily. We next prove Proposition 2.3
on the asymptotic overshoot distribution of $\xi$ and Proposition 2.5 which
provides a representation of $\varrho_{\alpha,\beta}$ in terms of a perpetual
integral of a stable subordinator. Then we investigate the residual explosion
time $T_{\infty}^{+}-T_{x}^{+}$ for large $x$ by estimating its moments and
establish the convergences of $X(T_{\infty}^{+}-t)$ in Theorem 1.2 case by
case.

###  3.1 Proof of Proposition 2.3 and Proposition 2.5

####  3.1.1 Proof of Proposition 2.3 (study of the overshoot)

Assume first that $\xi$ is not a subordinator. Denote the running supremum of
the process $\xi$ by $\bar{\xi}(t):=\sup\\{\xi(s),s\leq t\\}$, and call
$H:=(H(t))_{t\geq 0}$ the ladder height process, that is,

$H(t)=\bar{\xi}(L^{-1}(t))$ (3.1)

for the local time $L$ of $\bar{\xi}-\xi$ at $0$. By definition of the latter,
$L^{-1}(t)$ is a zero of $\bar{\xi}-\xi$; see [2, Proposition IV.7(iii)] and
one has therefore almost surely for all $t\geq 0$,

$H(t)=\bar{\xi}(L^{-1}(t))=\bar{\xi}(L^{-1}(t))-\xi(L^{-1}(t))+\xi(L^{-1}(t))=\xi(L^{-1}(t)).$

Moreover, when the process $\xi$ goes above level $x$ for the first time, a
new supremum is reached and we thus have

$\mathbb{P}\big{(}\bar{\xi}(\tau_{x}^{+}-)\in\mathrm{d}u,\xi(\tau_{x}^{+})\in\mathrm{d}v\big{)}=\mathbb{P}\big{(}H(\sigma_{x}^{+}-)\in\mathrm{d}u,H(\sigma_{x}^{+})\in\mathrm{d}v\big{)}$
(3.2)

where $\sigma_{x}^{+}$ denotes the first passage time above $x$ of the process
$H$. We are thus left to study the overshoot of $(H(t))_{t\geq 0}$; see for
instance Doney and Maller [6] and [16, Lemma 2.5] where this technique is also
applied.

According to [2, Theorem VII.4], $H$ is a subordinator with Laplace exponent
given by

$\Phi(s):=\frac{-1}{t}\log\mathbb{E}\big{[}\exp(-sH(t))\big{]}=\frac{\psi(s)}{s-p}\quad\text{for
all}\quad s>0.$ (3.3)

Even though we do not need it later, for completeness we also identify the
drift and the Lévy measure of $H$. It follows from (2.1) that

$\displaystyle\Phi(s)=\frac{\psi(s)-\psi(p)}{s-p}=$ $\displaystyle\
\Big{(}\frac{\sigma^{2}}{2}(s+p)+\gamma\Big{)}+\int_{0}^{\infty}\Big{(}\frac{e^{(p-s)z}-1}{s-p}e^{-pz}+z\mathbf{1}(z<1)\Big{)}\nu(\mathrm{d}z)$
(3.4) $\displaystyle=$ $\displaystyle\
\Big{(}\frac{\sigma^{2}}{2}s+\frac{\sigma^{2}}{2}p+\gamma\Big{)}+\int_{0}^{\infty}\int_{0}^{z}\Big{(}\mathbf{1}(z<1)-e^{-pz+(p-s)y}\Big{)}\mathrm{d}y\nu(\mathrm{d}z).$

Since $p>0$ and $\psi(p)=0$, we have for $s=0$

$0=\Big{(}\frac{\sigma^{2}}{2}p+\gamma\Big{)}+\int_{0}^{\infty}\int_{0}^{z}\Big{(}\mathbf{1}(z<1)-e^{-pz+py}\Big{)}\mathrm{d}y\nu(\mathrm{d}z).$

Therefore, substituting the above identity into (3.4) gives

$\displaystyle\Phi(s)=\frac{\psi(s)}{s-p}=$ $\displaystyle\
\frac{\sigma^{2}}{2}s+\int_{0}^{\infty}\int_{0}^{z}\Big{(}e^{-pz+py}-e^{-pz+(p-s)y}\Big{)}\nu(\mathrm{d}z)\mathrm{d}y$
(3.5) $\displaystyle=$ $\displaystyle\
\frac{\sigma^{2}}{2}s+\int_{0}^{\infty}\big{(}1-e^{-sy}\big{)}\Big{(}\int_{y}^{\infty}e^{p(y-z)}\nu(\mathrm{d}z)\Big{)}\mathrm{d}y$
$\displaystyle=:$ $\displaystyle\
\frac{\sigma^{2}}{2}s+\int_{0}^{\infty}\big{(}1-e^{-sy}\big{)}\Pi(\mathrm{d}y)=s\Big{(}\frac{\sigma^{2}}{2}+\int_{0}^{\infty}e^{-sz}\bar{\Pi}(z)\mathrm{d}z\Big{)},$

where $\bar{\Pi}(z):=\int_{z}^{\infty}\Pi(\mathrm{d}u)$; see [15, Exercice
6.5].

We next study the limit in law of $\xi(\tau_{x}^{+})/x=H(\sigma_{x}^{+})/x$ as
$x$ goes to $\infty$. Recall our assumption $-\psi\in\mathcal{R}_{\alpha}$ at
$0$. The formula in (3.3) entails that the subordinator $H$ has also a Laplace
exponent regularly varying at $0$ with index $\alpha$. Note that the study of
scaling limit in such a subordinator setting is the purpose of [2, Exercice
III.3-4]. We provide a solution below.

Denote by $\mathcal{U}$ the potential measure of $H$, that is,

$\mathcal{U}(x):=\mathbb{E}\Big{[}\int_{0}^{\infty}\mathbf{1}\big{(}H(t)\leq
x\big{)}\mathrm{d}t\Big{]}.$

Then

$\hat{\mathcal{U}}(s):=\int_{0}^{\infty}e^{-sy}\mathcal{U}(\mathrm{d}y)=\frac{1}{\Phi(s)}\text{
for all }s>0;$

see [2, Page 74]. According to [2, Proposition III.2, Theorem III. 5], one has

$\mathbb{P}\big{(}H(\sigma_{x}^{+}-)\in\mathrm{d}u,H(\sigma_{x}^{+})\in\mathrm{d}v\big{)}=\mathcal{U}(\mathrm{d}u)\Pi(\mathrm{d}v-u)\quad\text{and}\quad\mathbb{P}\big{(}H(\sigma_{x}^{+})=x\big{)}=\frac{\sigma^{2}}{2}u(x)$
(3.6)

where $u$ is the continuous and positive version of the density of
$\mathcal{U}$ when $\sigma>0$. Similarly to [2, Lemma III.7], the following
calculations follow from (3.6).

$\displaystyle\int_{0}^{\infty}e^{-qt}\mathbb{E}\big{[}e^{-\theta
x^{-1}\xi(\tau_{xt}^{+})}\big{]}\mathrm{d}t=$ $\displaystyle\
\iiint_{u+v>xt\geq
u}e^{-qt-\frac{\theta}{x}(u+v)}\mathcal{U}(\mathrm{d}u)\Pi(\mathrm{d}v)\mathrm{d}t+\int_{0}^{\infty}e^{-(q+\theta)t}\frac{\sigma^{2}}{2}u(xt)\mathrm{d}t$
$\displaystyle=$ $\displaystyle\
\frac{1}{q}\iint_{u,v>0}\mathcal{U}(\mathrm{d}u)\Pi(\mathrm{d}v)e^{-\frac{\theta}{x}(u+v)}\big{(}e^{-\frac{q}{x}u}-e^{-\frac{q}{x}(u+v)}\big{)}+\frac{\sigma^{2}}{2x}\hat{\mathcal{U}}\left(\frac{q+\theta}{x}\right)$
$\displaystyle=$ $\displaystyle\
\frac{1}{q}\hat{\mathcal{U}}\Big{(}\frac{q+\theta}{x}\Big{)}\left(\Phi\Big{(}\frac{q+\theta}{x}\Big{)}-\Phi\Big{(}\frac{\theta}{x}\Big{)}-\frac{\sigma^{2}}{2}\frac{q}{x}\right)+\frac{\sigma^{2}}{2x}\hat{\mathcal{U}}\big{(}\frac{q+\theta}{x}\big{)}$
$\displaystyle=$ $\displaystyle\
\frac{1}{q}\Big{(}1-\frac{\Phi(\theta/x)}{\Phi((\theta+q)/x)}\Big{)}.$ (3.7)

For $\alpha\in(0,1)$,

$\lim_{x\to\infty}\int_{0}^{\infty}e^{-qt}\mathbb{E}\Big{[}e^{-\theta
x^{-1}\xi(\tau_{xt}^{+})}\Big{]}\mathrm{d}t=\frac{1}{q}\Big{(}1-\Big{(}\frac{\theta}{\theta+q}\Big{)}^{\alpha}\Big{)}=\int_{0}^{\infty}e^{-qt}\Big{(}\int_{t}^{\infty}\frac{\theta^{\alpha}u^{\alpha-1}}{\Gamma(\alpha)}e^{-\theta
u}\mathrm{d}u\Big{)}\mathrm{d}t.$

Further note that $x^{-1}\xi(\tau_{xt}^{+})$ is increasing in $t>0$. So, the
function $\mathbb{E}\big{[}e^{-\theta x^{-1}\xi(\tau_{xt}^{+})}\big{]}$ is
decreasing in $t$. We then obtain by integration by parts that

$\lim_{x\to\infty}\int_{0}^{\infty}e^{-qt}\mathrm{d}\bigg{(}1-\mathbb{E}\Big{[}e^{-\theta
x^{-1}\xi(\tau_{xt}^{+})}\Big{]}\bigg{)}=\int_{0}^{\infty}e^{-qt}\mathrm{d}\Big{(}\int_{0}^{t}\frac{\theta^{\alpha}u^{\alpha-1}}{\Gamma(\alpha)}e^{-\theta
u}\mathrm{d}u\Big{)}.$

which implies that for a.e. $t>0$, and then for all $t>0$ by continuity,

$\lim_{x\to\infty}\left(1-\mathbb{E}\Big{[}e^{-\theta
x^{-1}\xi(\tau_{xt}^{+})}\Big{]}\right)=\int_{0}^{t}\frac{\theta^{\alpha}u^{\alpha-1}}{\Gamma(\alpha)}e^{-\theta
u}\mathrm{d}u.$

In particular, for $t=1$

$\displaystyle\lim_{x\to\infty}$ $\displaystyle\mathbb{E}\Big{[}1-e^{-\theta
x^{-1}\xi(\tau_{x}^{+})}\Big{]}=\theta\times\theta^{\alpha-1}\int_{0}^{1}\frac{u^{\alpha-1}}{\Gamma(\alpha)}e^{-\theta
u}\mathrm{d}u$ $\displaystyle=$ $\displaystyle\
\theta\int_{0}^{\infty}\frac{v^{-\alpha}e^{-\theta
v}}{\Gamma(1-\alpha)}\mathrm{d}v\int_{0}^{1}\frac{u^{\alpha-1}e^{-\theta
u}}{\Gamma(\alpha)}\mathrm{d}u=\theta\int_{0}^{\infty}e^{-\theta
z}\mathrm{d}z\int_{0}^{1\wedge
z}\frac{u^{\alpha-1}(z-u)^{-\alpha}}{\Gamma(\alpha)\Gamma(1-\alpha)}\mathrm{d}u$
$\displaystyle=$ $\displaystyle\ 1-e^{-\theta}+\int_{1}^{\infty}\theta
e^{-\theta
z}\mathrm{d}z\int_{0}^{1}\frac{u^{\alpha-1}(z-u)^{-\alpha}}{\Gamma(\alpha)\Gamma(1-\alpha)}\mathrm{d}u=\int_{0}^{\infty}\theta
e^{-\theta z}\mathbb{P}(\chi_{\alpha}>z)\mathrm{d}z$

where we recall (1.10), the identity
$\frac{1}{\Gamma(\alpha)\Gamma(1-\alpha)}=\frac{\sin\alpha\pi}{\pi}$ and we
make use of the facts that

$\int_{0}^{1\wedge
z}\frac{u^{\alpha-1}(z-u)^{-\alpha}}{\Gamma(\alpha)\Gamma(1-\alpha)}\mathrm{d}u=1\text{
and }\mathbb{P}(\chi_{\alpha}>z)=1,\,\,\forall z<1.$

The desired result is proved. The results for the degenerate cases of
$\alpha=0$ or $1$ can also be derived from (3.1.1).

If $\xi$ is a subordinator, i.e. $-\psi$ takes the form (2.2), one can
directly obtain the result from the same calculation in (3.1.1) by replacing
in (3.6), $\Phi$, $\mathcal{U}$ and the constant $\sigma^{2}/2$, respectively,
with the Laplace exponent of $\xi$, $-\psi$, its potential measure $U$ in
(2.10) and its drift $\delta$. Heuristically, in the subordinator case, the
ladder height process is simply the process itself.

We now verify that the convergence in law
$x^{-1}\xi(\tau_{x}^{+})\Longrightarrow\chi_{\alpha}$ as $x$ goes to $\infty$
also holds under $\mathbb{P}_{1}(\cdot|\tau_{a}^{-}=\infty)$ for every
$1>a>0$. Let $f$ be bounded continuous function, write

$\mathbb{E}_{1}\big{[}f\big{(}\xi(\tau_{x}^{+})/x\big{)};\tau^{-}_{a}=\infty\big{]}=\mathbb{E}_{1}\big{[}f\big{(}\xi(\tau_{x}^{+})/x\big{)}\big{]}-\mathbb{E}_{1}\big{[}f\big{(}\xi(\tau_{x}^{+})/x\big{)};\tau^{-}_{a}<\infty\big{]}.$
(3.8)

By the Markov property at time $\tau^{-}_{a}$,

$\displaystyle\mathbb{E}_{1}\big{[}f(\big{(}\xi(\tau_{x}^{+})/x\big{)};\tau^{-}_{a}<\infty\big{]}$
$\displaystyle=\
\mathbb{E}_{1}\big{[}f\big{(}\xi(\tau_{x}^{+})/x\big{)};\tau^{-}_{a}<\tau_{x}^{+},\tau^{-}_{a}<\infty\big{]}+\mathbb{E}_{1}\big{[}f\big{(}\xi(\tau_{x}^{+})/x\big{)};\tau_{x}^{+}<\tau^{-}_{a}<\infty\big{]}$
$\displaystyle=\
\mathbb{E}_{a}\big{[}f\big{(}\xi(\tau_{x}^{+})/x)\big{)}\big{]}\cdot\mathbb{P}_{1}\big{(}\tau_{a}^{-}<\tau_{x}^{+}\big{)}+\mathbb{E}_{1}\big{[}f\big{(}\xi(\tau_{x}^{+})/x\big{)};\tau_{x}^{+}<\tau_{a}^{-}<\infty\big{]}$
(3.9)
$\displaystyle\underset{x\rightarrow\infty}{\longrightarrow}\mathbb{E}\big{[}f(\chi_{\alpha})\big{]}\cdot\mathbb{P}_{1}(\tau_{a}^{-}<\infty),$

where we use the convergence
$x^{-1}\xi(\tau^{+}_{x})\Longrightarrow\chi_{\alpha}$ previously established,
the fact that $\tau_{x}^{+}\underset{x\rightarrow\infty}{\rightarrow}\infty$
a.s. and the boundedness of $f$. The latter in particular entails that the
second term in (3.1.1) vanishes as it is bounded as follows.

$\big{\lvert}\mathbb{E}_{1}\big{[}f\big{(}\xi(\tau_{x}^{+})/x\big{)};\tau_{x}^{+}<\tau_{a}^{-}<\infty\big{]}\big{\lvert}\leq||f||_{\infty}\cdot\mathbb{P}_{1}(\tau_{x}^{+}<\tau_{a}^{-}<\infty)\leq||f||_{\infty}\cdot\mathbb{P}_{x}(\tau_{a}^{-}<\infty)=||f||_{\infty}\cdot
e^{p(a-x)}.$

Going back to (3.8), we get indeed

$\mathbb{E}_{1}\big{[}f\big{(}\xi(\tau_{x}^{+})/x\big{)}\big{|}\tau^{-}_{a}=\infty\big{]}\underset{x\rightarrow\infty}{\longrightarrow}\mathbb{E}\big{[}f(\chi_{\alpha})\big{]}.$

####  3.1.2 Proof of Proposition 2.5 (study of $\varrho_{\alpha,\beta}$)

###### Proof.

Let $S$ be an $\alpha$-stable subordinator started from $1$ with
$\psi(s)=-c_{0}s^{\alpha}$ for some $\alpha\in(0,1)$. Applying the second
Lamperti transform, see [15, Theorem 13.1], we have

$S(t)=\exp\big{(}\zeta\big{(}I^{-1}(t)\big{)}\big{)}$

where by a direct calculation one can show that $\zeta$ is an SPLP started
from $0$, with Laplace exponent

$\Psi(s):=-\log\mathbb{E}\big{[}e^{-s\zeta(1)}\big{]}=\frac{c_{0}s}{\Gamma(1-\alpha)}\int_{0}^{\infty}e^{-sz}\big{(}e^{z}-1\big{)}^{-\alpha}\,dz=c_{0}\frac{\Gamma(s+\alpha)}{\Gamma(s)}$
(3.10)

and where $I(t):=\int_{0}^{t}e^{\alpha\zeta(s)}\,\mathrm{d}s$ and $I^{-1}$
denotes its inverse function. Then, for $\beta>\alpha$, we have

$\iota(\infty):=\int_{0}^{\infty}S(t)^{-\beta}\,\mathrm{d}t=\int_{0}^{\infty}e^{-\beta\zeta(I^{-1}(t))}\,\mathrm{d}t=\int_{0}^{\infty}e^{-\beta\zeta(s)}\,\mathrm{d}I(s)=\int_{0}^{\infty}e^{(\alpha-\beta)\zeta(s)}\,\mathrm{d}s.$

It is proved by Bertoin and Yor [3] that, for all $r>0$

$b_{r}:=\mathbb{E}_{1}\big{[}\iota(\infty)^{r}\big{]}=\frac{r}{\Psi(r(\beta-\alpha))}\mathbb{E}_{1}\big{[}\iota(\infty)^{r-1}\big{]}=\frac{r}{\Psi(r(\beta-\alpha))}b_{r-1}.$

In particular, taking $r=n$ and making use of (3.10) gives the expression
(2.17) for the moment generating function of $\iota(\infty)$.

We are now going to study the tail behaviour of $\iota(\infty)$. Let
$\gamma_{0}(1-\alpha)=1$ and
$z_{0}=(1-\alpha)c_{0}^{\gamma}(\beta-\alpha)^{\alpha\gamma}$. We will show
that

$\mathbb{E}_{1}\Big{[}\exp\Big{(}z\iota(\infty)^{\gamma}\Big{)}\Big{]}=\left\\{\begin{array}[]{ll}<\infty&\text{if
$\gamma<\gamma_{0}$};\\\ <\infty&\text{if $\gamma=\gamma_{0}$ and
$z<z_{0}$};\\\ =\infty&\text{if $\gamma=\gamma_{0}$ and either $z>z_{0}$ or
$\gamma>\gamma_{0}$}.\end{array}\right.$ (3.11)

To prove (3.11), we make use of the asymptotic equivalence

$\frac{b_{r}}{b_{r-1}}=\frac{r}{\Psi\big{(}r(\beta-\alpha)\big{)}}=\frac{1}{c_{0}(\beta-\alpha)}\frac{\Gamma\big{(}r(\beta-\alpha)+1\big{)}}{\Gamma\big{(}r(\beta-\alpha)+\alpha\big{)}}\sim\frac{r^{1-\alpha}}{c_{0}(\beta-\alpha)^{\alpha}}\quad\text{as
$r\to\infty$};$ (3.12)

see e.g. [1, 6.1.47] and the following version of Hölder’s inequality

$(b_{p})^{\frac{q-r}{p-r}}(b_{r})^{\frac{p-q}{p-r}}\geq
b_{\frac{p(q-r)+r(p-q)}{p-r}}=b_{q}\quad\text{for all $p\geq q\geq r>1$}.$
(3.13)

For each fixed $\gamma>0$, denoting $m=1+[\gamma]\geq 2$ we have from (3.12)
that

$a_{n}:=\mathbb{E}_{1}\big{[}(\iota(\infty))^{n\gamma}\big{]}=b_{n\gamma}=\prod_{k=0}^{m-2}\frac{b_{n\gamma-k}}{b_{n\gamma-k-1}}b_{n\gamma-m+1}=\prod_{k=0}^{m-1}\frac{b_{n\gamma-k}}{b_{n\gamma-k-1}}b_{n\gamma-m}.$

Taking $(p,q,r)=(n\gamma-m+1,(n-1)\gamma,n\gamma-m)$ in (3.13), we have from
he above that

$\displaystyle a_{n}=$ $\displaystyle\
\Big{(}\prod_{k=0}^{m-2}\frac{b_{n\gamma-k}}{b_{n\gamma-k-1}}b_{n\gamma-m+1}\Big{)}^{\frac{q-r}{p-r}}\Big{(}\prod_{k=0}^{m-1}\frac{b_{n\gamma-k}}{b_{n\gamma-k-1}}b_{n\gamma-m}\Big{)}^{\frac{p-q}{p-r}}$
$\displaystyle=$ $\displaystyle\
\prod_{k=0}^{m-2}\frac{b_{n\gamma-k}}{b_{n\gamma-k-1}}\Big{(}\frac{b_{n\gamma-m+1}}{b_{n\gamma-m}}\Big{)}^{1+\gamma-m}\big{(}b_{p}\big{)}^{\frac{q-r}{p-r}}\big{(}b_{r}\big{)}^{\frac{p-q}{p-r}}$
$\displaystyle\geq$ $\displaystyle\
\prod_{k=0}^{m-2}\frac{b_{n\gamma-k}}{b_{n\gamma-k-1}}\Big{(}\frac{b_{n\gamma-m+1}}{b_{n\gamma-m}}\Big{)}^{1+\gamma-m}b_{q}=\prod_{k=0}^{m-2}\frac{b_{n\gamma-k}}{b_{n\gamma-k-1}}\Big{(}\frac{b_{n\gamma-m+1}}{b_{n\gamma-m}}\Big{)}^{1+\gamma-m}a_{n-1}.$

Making use of (3.12) and the fact that $\gamma,m$ are constants, we have as
$n\to\infty$

$\frac{a_{n}}{a_{n-1}}\geq\prod_{k=0}^{m-2}\frac{b_{n\gamma-k}}{b_{n\gamma-k-1}}\Big{(}\frac{b_{n\gamma-m+1}}{b_{n\gamma-m}}\Big{)}^{1+\gamma-m}\sim
n^{(1-\alpha)\gamma}\cdot\Big{(}\frac{\gamma^{1-\alpha}}{c_{0}(\beta-\alpha)^{\alpha}}\Big{)}^{\gamma}.$
(3.14)

On the other hand, taking $(p,q,r)=(n\gamma,n\gamma-1,(n-1)\gamma)$ in (3.13),
we have

$b_{n\gamma-1}=b_{q}\leq(b_{p})^{\frac{\gamma-1}{\gamma}}(b_{r})^{\frac{1}{\gamma}}=b_{n\gamma}\cdot\big{(}a_{n-1}/a_{n}\big{)}^{1/\gamma}$

which together with (3.12) gives

$\frac{a_{n}}{a_{n-1}}\leq\bigg{(}\frac{b_{n\gamma}}{b_{n\gamma-1}}\bigg{)}^{\gamma}\sim
n^{(1-\alpha)\gamma}\cdot\Big{(}\frac{\gamma^{1-\alpha}}{c_{0}(\beta-\alpha)^{\alpha}}\Big{)}^{\gamma}.$
(3.15)

Combining (3.14) and (3.15) one has that the Taylor expansion

$\mathbb{E}\Big{[}e^{z\iota(\infty)^{\gamma}}\big{]}=1+\sum_{n\geq
1}\frac{z^{n}}{n!}\mathbb{E}\Big{[}\iota(\infty)^{n\gamma}\Big{]}=\sum_{n\geq
0}\frac{a_{n}}{n!}z^{n}$

is finite for some $z>0$ only if $\gamma(1-\alpha)\leq 1$. Moreover, for
$\gamma=\gamma_{0}=\frac{1}{1-\alpha}$, the series above has a convergence
radius of $c_{0}^{\gamma}(\beta-\alpha)^{\alpha\gamma}\gamma_{0}^{-1}=z_{0}$.

Finally, combining Rolski et al. [22, Theorem 2.3.1] and (3.11), we obtain the
large deviation estimate

$\underset{t\rightarrow\infty}{\limsup}\,t^{-\gamma_{0}}\log\mathbb{P}(\iota(\infty)>t)=-z_{0}$

which gives (2.18). ∎

###  3.2 Proofs of Theorem 1.1 and Corollary 1.1

We proceed here with the proof of the integral test for explosion given in
(1.8) (in terms of rate function $R$ and Laplace exponent $\psi$).

####  3.2.1 Proof of Theorem 1.1 (explosion criterion)

Recall that the parent Lévy process $\xi$ drifts towards $+\infty$, hence the
event $\\{T_{a}^{-}=\infty\\}$ has positive probability.

Our proof is based on the following equivalences, see [19, Theorem 2.2] and
Kolb and Savov [14, Theorem 2.1]: for any $x>0$,

$\displaystyle\mathbb{P}_{x}\big{(}T_{\infty}^{+}<\infty\big{)}>0$
$\displaystyle\Longleftrightarrow\mathbb{P}_{x}\big{(}T_{\infty}^{+}<\infty|T_{a}^{-}=\infty\big{)}=1\text{
for any }a\in(0,x)$
$\displaystyle\Longleftrightarrow\mathbb{E}_{x}\big{(}\eta(\tau_{a}^{-})\big{)}<\infty\text{
for any }a\in(0,x).$ (3.16)

Notice that

$\mathbb{E}_{x}\big{(}\eta(\tau_{a}^{-})\big{)}=U_{a}(1/R)(x)$

where we recall $U_{a}(x,\mathrm{d}y)$ denotes the potential measure of the
process killed when it is below level $a$. See (2.5) for the non-subordinator
case. We are going to show that when $R$ is increasing and has a growth
satisfying (1.7), the condition $U_{a}(1/R)(x)<\infty$ is equivalent to our
more explicit condition (1.8).

We first treat the simple case where $\xi$ is a subordinator, recalling that
in this case $\tau_{a}^{-}=\infty$, $\mathbb{P}_{x}$-a.s. for all $x>a$.
Notice that its Laplace exponent is $-\psi$. One has by definition for any
$x>a>0$,

$U_{a}\big{(}1/R\big{)}(x)=U\big{(}1/R\big{)}(x)=\int_{0}^{\infty}\frac{U(\mathrm{d}z)}{R(x+z)}$

with $U(\mathrm{d}z)$ the potential measure of the subordinator $\xi$, see
(2.11). Recall the condition (1.7):

$\limsup_{z\to\infty}\frac{R(z+x)}{R(z)}<\infty\text{ for every $x>0$},$

and the assumption that $R$ is increasing. One has
$\frac{1}{R(z+x)}\asymp\frac{1}{R(z)}$. Hence
$U_{a}\big{(}1/R\big{)}(x)<\infty$ if and only if
$\int^{\infty}\frac{U(\mathrm{d}z)}{R(z)}<\infty$. Setting $U(y):=U([0,y])$
for any $y\geq 0$, we see, by applying Fubini-Tonelli’s theorem, that

$\int^{\infty}\frac{U(\mathrm{d}z)}{R(z)}<\infty\Longleftrightarrow\int^{\infty}U(y)\mathrm{d}\big{(}-\frac{1}{R(y)}\big{)}<\infty.$

By [2, Proposition III.1], one has the estimate
$U(y)\asymp\frac{1}{-\psi(1/x)}$ and we get the necessary and sufficient
condition. The fact that explosion is almost sure is a consequence of [14,
Theorem 2.1].

We now assume that $\xi$ is not a subordinator. The study in this case is more
involved as the Lévy process may go below any level $a$. Define
$W_{p}(y):=e^{-py}W(y)$ for any $y\geq 0$. This function is the scale function
of the Lévy process $\xi$ under the new measure $\mathbb{P}^{(p)}$ satisfying
$\displaystyle\frac{d\mathbb{P}^{(p)}}{d\mathbb{P}}\Big{|}_{\mathcal{F}_{t}}=e^{-p\xi(t)}$
for all $t\geq 0$; see e.g. [15, Chapter 3.3]. Note in particular that $W_{p}$
is nondecreasing. It can be checked directly from (2.3) that

$\int_{0}^{\infty}e^{-sy}e^{py}W_{p}(\mathrm{d}y)=\int_{0}^{\infty}e^{-sy}\big{(}W(\mathrm{d}y)-pW(y)\mathrm{d}y\big{)}=\frac{s-p}{\psi(s)}=\frac{1}{\Phi(s)}=\hat{\mathcal{U}}(s)$

for any $s>p$, where $\mathcal{U}$ and $\Phi$ are defined in the proof of
Proposition 2.3, and $\hat{\mathcal{U}}$ denotes the Laplace transform of
$\mathcal{U}$. Then [2, Proposition III.1] ensures that for all large $x$

$\int_{0}^{x}e^{py}W_{p}(\mathrm{d}y)=\mathcal{U}(x)\asymp\frac{1}{\Phi(1/x)}=\frac{1/x-p}{\psi(1/x)}.$
(3.17)

On the other hand, we have from (2.5)

$\displaystyle e^{px}U_{a}(1/R)(x)=$ $\displaystyle\
e^{px}\int_{a}^{\infty}\frac{\mathrm{d}y}{R(y)}\big{(}e^{p(a-x)}W(y-a)-W(y-x)\big{)}$
(3.18) $\displaystyle=$ $\displaystyle\
\int_{a}^{\infty}\frac{e^{py}}{R(y)}\big{(}W_{p}(y-a)-W_{p}(y-x)\big{)}\mathrm{d}y$
$\displaystyle=$ $\displaystyle\
\int_{a}^{\infty}\frac{e^{py}\mathrm{d}y}{R(y)}\int_{y-x}^{y-a}W_{p}(\mathrm{d}z)$
$\displaystyle=$ $\displaystyle\
\int_{[0,\infty)}e^{pz}W_{p}(\mathrm{d}z)\Big{(}\int_{a}^{x}\frac{e^{py}\mathrm{d}y}{R(z+y)}\Big{)}$
$\displaystyle=$
$\displaystyle\int_{[0,\infty)}\mathcal{U}(\mathrm{d}z)\Big{(}\int_{a}^{x}\frac{e^{py}\mathrm{d}y}{R(z+y)}\Big{)}.$

where in the third line we use the fact that $W_{p}$ is nondecreasing and
where in the second to the last equality, we have applied Fubini-Tonelli’s
theorem and performed a change of variable. Using the condition (1.7) and the
assumption that $R$ is increasing, it is plain that

$\int_{a}^{x}\frac{e^{py}\mathrm{d}y}{R(z+y)}\asymp\frac{1}{R(z)}\int_{a}^{x}e^{py}\mathrm{d}y\quad\text{for
all large $z$ and fixed $x>a>0$},$ (3.19)

Plugging this into (3.18), we see that

$U_{a}\big{(}1/R\big{)}(x)<\infty\quad\text{ if and only
if}\quad\int^{\infty}\frac{\mathcal{U}(\mathrm{d}z)}{R(z)}<\infty.$ (3.20)

Moreover, by Fubini-Tonelli’s theorem and the estimate (3.17), we see that

$\int^{\infty}\frac{\mathcal{U}(\mathrm{d}z)}{R(z)}<\infty\Longleftrightarrow\int^{\infty}\mathcal{U}(y)\mathrm{d}\big{(}-\frac{1}{R(y)}\big{)}<\infty\Longleftrightarrow\int^{\infty}\frac{-1}{\psi(1/y)}\mathrm{d}\big{(}-\frac{1}{R(y)}\big{)}<\infty.$
(3.21)

This finishes the proof.∎

####  3.2.2 Proof of Corollary 1.1 (application in the regularly varying case)

For $R\in\mathcal{R}_{\beta}$ at $\infty$, by Lemma 2.2 we have

$\frac{1}{R(x)}\sim\beta\int_{x}^{\infty}\frac{\mathrm{d}z}{zR(z)}\quad\text{as
$x\to\infty$}.$ (3.22)

In addition, (1.7) holds since $R(x+a)\sim R(x)$ as $x$ goes to $\infty$.
Applying the identities in (3.21) with $1/R$ replaced by the tail integral
above proves the first statement of Corollary 1.1. The classification of
explosion according to $\alpha$ and $\beta$ when
$-\psi\in\mathcal{R}_{\alpha}$ near $0$ follows plainly by studying whether
the test integral is finite or not.∎

###  3.3 Proof of Theorem 1.2

Until the end of this subsection we focus on the case $\beta\geq\alpha$ and
(1.9) hold for which explosion can occur.

####  3.3.1 Renormalization in law of the explosion time

Case $\beta>\alpha>0$. The following lemma will be applied to show the weak
convergence of $T_{\infty}^{+}$ for $X$ starting from large initial values.
Its proof is a bit technical and is deferred to the Appendix. Recall from
Section 2 that by definition

$U_{a}f(x):=\mathbb{E}_{x}\bigg{[}\int_{0}^{\infty}f(\xi_{t})\mathbf{1}(t<\tau_{a}^{-})\mathrm{d}t\bigg{]}$

for any bounded measurable function $f$ and any $a>0$.

######  Lemma 3.1 (Key equivalence).

Suppose that $p\in(0,\infty]$, $-\psi\in\mathcal{R}_{\alpha}$ at $0$, and
$f\in\mathcal{R}_{-\gamma}$ at $\infty$ for some $\gamma>\alpha\geq 0$. For
any $a>0$, we have

$U_{a}f(x)\sim\frac{\Gamma(\gamma-\alpha+1)}{\Gamma(\gamma)}\int_{x}^{\infty}\frac{f(y)}{-y\psi(1/y)}\mathrm{d}y\quad\text{as
$x\to\infty$}.$

The next lemma provides a renormalisation in law of the explosion time when
the initial value of the process goes to $\infty$.

######  Lemma 3.2.

Suppose that $\mathbb{H}$ hold and $\beta>\alpha\geq 0$. We have for any
$a>0$,

$\frac{\eta(\infty)}{\varphi(x)}\bigg{|}_{\mathbb{P}_{x}(\cdot|\tau_{a}^{-}=\infty)}\Rightarrow\varrho_{\alpha,\beta}\quad\text{as
$x\to\infty$}$

where $\varrho_{\alpha,\beta}$ is the random variable whose distribution is
characterised by (1.11).

###### Proof.

Recall the following recursive relationship between the moments of
$\eta(\infty)$; see e.g. [16, Proposition 4.7]: for any $n\geq 1$ and $x>a>0$

$m_{n}(x):=\
\mathbb{E}_{x}\big{[}\eta^{n}(\infty);\tau_{a}^{-}=\infty\big{]}=n\int_{a}^{\infty}\frac{m_{n-1}(y)}{R(y)}U_{a}(x,\mathrm{d}y),$
(3.23)

with $m_{0}(x):=\mathbb{P}_{x}(\tau_{a}^{-}=\infty)=1-e^{p(a-x)}$. In
particular, for every $x>a$ we have

$\displaystyle m_{1}(x):=$ $\displaystyle\
\mathbb{E}_{x}\big{[}\eta(\infty);\tau_{a}^{-}=\infty\big{]}=\int_{0}^{\infty}f_{1}(y)U_{a}(x,\mathrm{d}y),$
(3.24) $\displaystyle m_{2}(x):=$ $\displaystyle\
\mathbb{E}_{x}\big{[}\eta^{2}(\infty);\tau_{a}^{-}=\infty\big{]}=2\int_{0}^{\infty}f_{2}(y)U_{a}(x,\mathrm{d}y),$
(3.25)

where for $y>a$, we set

$f_{1}(y)=\frac{1-e^{p(a-y)}}{R(y)}\sim\frac{1}{R(y)}\in\mathcal{R}_{-\beta}\quad\text{as
$y\to\infty$ and}\quad f_{2}(y)=\frac{m_{1}(y)}{R(y)}.$

By the assumption $\beta>\alpha$, Lemma 3.1 can be applied directly to (3.24)
to show that

$m_{1}(x)\sim\frac{\Gamma(\beta-\alpha+1)}{\Gamma(\beta)}\int_{x}^{\infty}\frac{-\mathrm{d}y}{yR(y)\psi(1/y)}=:c_{1}\cdot\varphi(x)\in\mathcal{R}_{\alpha-\beta}\text{
at }\infty.$

Therefore, $f_{2}\in\mathcal{R}_{\alpha-2\beta}$ at $\infty$ and
$2\beta-\alpha>\alpha$. Applying Lemma 3.1 to (3.25), we have

$\displaystyle m_{2}(x)\sim$ $\displaystyle\
\frac{\Gamma(2\beta-2\alpha+1)}{\Gamma(2\beta-\alpha)}\int_{x}^{\infty}\frac{2\cdot
m_{1}(y)\mathrm{d}y}{-yR(y)\psi(1/y)}\sim\frac{\Gamma(2\beta-2\alpha+1)}{\Gamma(2\beta-\alpha)}\int_{x}^{\infty}\frac{2c_{1}\cdot\varphi(y)\mathrm{d}y}{-yR(y)\psi(1/y)}$
$\displaystyle=$ $\displaystyle\
\frac{\Gamma(2\beta-2\alpha+1)}{\Gamma(2\beta-\alpha)}c_{1}\cdot\varphi^{2}(x)=:c_{2}\cdot\varphi^{2}(x)\in\mathcal{R}_{2(\alpha-\beta)}\text{
at }\infty.$

Similarly, by induction on $n$ one can also show by (3.23) and Lemma 3.1 that
for $n\geq 3$,

$\displaystyle m_{n}(x)\sim$ $\displaystyle\
\frac{\Gamma(n(\beta-\alpha)+1)}{\Gamma(n(\beta-\alpha)+\alpha)}\int_{x}^{\infty}\frac{n\cdot
m_{n-1}(y)\mathrm{d}y}{-yR(y)\psi(1/y)}\sim\frac{\Gamma(n(\beta-\alpha)+1)}{\Gamma(n(\beta-\alpha)+\alpha)}\int_{x}^{\infty}\frac{nc_{n-1}\cdot\varphi^{n-1}(y)\mathrm{d}y}{-yR(y)\psi(1/y)}$
$\displaystyle=$ $\displaystyle\
\frac{\Gamma(n(\beta-\alpha)+1)}{\Gamma(n(\beta-\alpha)+\alpha)}\cdot
c_{n-1}\cdot\varphi^{n}(x)=:c_{n}\cdot\varphi^{n}(x)\in\mathcal{R}_{n(\alpha-\beta)}\text{
at }\infty.$

Since $\mathbb{P}_{x}(\tau_{a}^{-}=\infty)\to 1$ as $x$ goes to $\infty$, it
follows that

$\mathbb{E}_{x}\Big{[}\Big{(}\frac{\eta(\infty)}{\varphi(x)}\Big{)}^{n}\Big{|}\tau_{a}^{-}=\infty\Big{]}\underset{x\rightarrow\infty}{\rightarrow}\prod_{k=1}^{n}\frac{\Gamma(k(\beta-\alpha)+1)}{\Gamma(k(\beta-\alpha)+\alpha)}=:c_{n}\quad\text{for
every $n\in\mathbb{N}$.}$

If $\alpha=1$, then $c_{n}\equiv 1$ and

$\frac{\eta(\infty)}{\varphi(x)}\bigg{|}_{\mathbb{P}_{x}(\cdot|\tau_{a}^{-}=\infty)}\Rightarrow
1\quad\text{in $L^{2}(\mathbb{P})$ as $x\to\infty$}.$

If $\alpha=0$, then $c_{n}=(\beta-\alpha)^{n}n!=\beta^{n}n!$ and

$1+\sum_{n\geq 1}\frac{c_{n}}{n!}s^{n}=\sum_{n\geq 0}(\beta
s)^{n}=\frac{1}{1-\beta s}\quad\text{for $|s\beta|<1$}.$

If $\alpha\in(0,1)$, then

$\frac{c_{n}}{c_{n-1}}=\frac{\Gamma(n(\beta-\alpha)+1)}{\Gamma(n(\beta-\alpha)+\alpha)}\sim(n(\beta-\alpha))^{1-\alpha}\quad\text{as
$n\to\infty$},$ (3.26)

see e.g. [1, 6.1.47]. One can check directly that the sequence $(c_{n})_{n\geq
1}$ satisfies the Carleman condition on the moments. The latter determines
thus uniquely a positive distribution. Then [4, Theorem 30.2] can be applied
to show the limit. ∎

Case $1\geq\beta=\alpha>0$: For this critical case, Lemma 3.1 cannot be
applied as some integrals may fail to converge; see for example (A.3). In the
following, an analogue version is derived. Recalling that for
$f\in\mathcal{R}_{-\alpha}$ at $\infty$ and $U_{a}f(x)<\infty$, we have from
Corollary 1.1

$\int^{\infty}\frac{-f(y)}{y\psi(1/y)}\mathrm{d}y<\infty.$

######  Lemma 3.3.

Suppose that $p\in(0,\infty]$, $-\psi\in\mathcal{R}_{\alpha}$ at $0$,
$f\in\mathcal{R}_{-\alpha}$ at $\infty$ and (1.9) hold. Then we have

$U_{a}f(x)\sim\frac{1}{\Gamma(\alpha)}\int_{x}^{\infty}\frac{-f(z)}{z\psi(1/z)}\mathrm{d}z\in\mathcal{R}_{0}\quad\text{as
$x\to\infty$.}$

The proof of Lemma 3.3 is deferred to Appendix. We use it to obtain the
following weak convergence of $T_{\infty}^{+}$ for $X$ starting from large
values similarly to Lemma 3.2.

######  Lemma 3.4.

Suppose that $\mathbb{H}$ holds for $\beta=\alpha>0$ and (1.9) holds. Then,
for any $a>0$, we have

$\frac{\eta(\infty)}{\varphi(x)}\bigg{|}_{\mathbb{P}_{x}(\cdot|\tau_{a}^{-}=\infty)}\Rightarrow\frac{1}{\Gamma(\alpha)}\quad\text{in
$L^{2}(\mathbb{P})$ as $x\to\infty$.}$

###### Proof.

By applying Lemma 3.3 to (3.24), we find that

$m_{1}(x)=U_{a}f_{1}(x)\sim\frac{1}{\Gamma(\alpha)}\int_{x}^{\infty}\frac{-\mathrm{d}y}{zR(z)\psi(1/z)}=\frac{\varphi(x)}{\Gamma(\alpha)}.$

The fact $m_{1}(\infty)=0$ implies $U_{a}f_{2}(x)<\infty$ in (3.25). Further
applying Lemma 3.3 gives

$m_{2}(x)=2U_{a}f_{2}(x)\sim\frac{2}{\Gamma(\alpha)}\int_{x}^{\infty}\frac{-m_{1}(z)\mathrm{d}z}{zR(z)\psi(1/z)}\sim\frac{-2}{\Gamma^{2}(\alpha)}\int_{x}^{\infty}\varphi^{\prime}(z)\varphi(z)\mathrm{d}z=\frac{\varphi^{2}(x)}{\Gamma^{2}(\alpha)}$

which proves the assertion. ∎

We are now ready to proceed with the study of the speeds and prove Theorem 1.2
for which the next lemma is crucial.

######  Lemma 3.5 (Lemma 2.1 in [16]).

Assume that $p>0$ and $\mathbb{P}_{1}(T_{\infty}^{+}<\infty)>0$. Set
$\bar{X}(T_{\infty}^{+}-t):=\sup\\{X(s),s\leq T^{+}_{\infty}-t\\}$ for any
$t\in[0,T^{+}_{\infty}]$. One has on the event of explosion

$\bar{X}(T_{\infty}^{+}-t)\underset{t\rightarrow
0}{\sim}X(T_{\infty}^{+}-t)\quad\text{under
$\mathbb{P}_{1}(\cdot|T_{a}^{-}<\infty)$ a.s..}$ (3.27)

###### Proof.

Recall that $\bar{\xi}$ denotes the running supremum process of $\xi$. It has
been established222the finite mean condition is not used for this result in
[16, Lemma 2.1] that when $p>0$, we have

$\xi(t)\big{/}\bar{\xi}(t)\to 1\text{ as $t\to\infty$ $\mathbb{P}$-a.s.}.$

The equivalence (3.27) follows plainly by the time-changing relationship
between $X$ and $\xi$, see (1.3). ∎

The statements in Theorem 1.2 are thus equivalent if one replaces $X$ by the
running supremum. We shall therefore work with $\bar{X}$. We treat the cases
1, 3 and 2 separately in the subsections below.

####  3.3.2 Case 1: $\beta>\alpha>0$

###### Proof of Case 1 in Theorem 1.2.

######  Lemma 3.6.

$\frac{T_{\infty}^{+}-T_{x}^{+}}{\varphi(x)}\bigg{|}_{\mathbb{P}_{1}(\cdot|T_{\infty}^{+}<\infty)}\underset{x\rightarrow\infty}{\Longrightarrow}\begin{cases}&\chi_{\alpha}^{\alpha-\beta}\times\varrho_{\alpha,\beta}\text{
if }\alpha<1\\\ &1\text{ if }\alpha=1.\end{cases}$ (3.28)

###### Proof.

We first carry out the computation under
$\mathbb{P}_{1}(\cdot|T_{a}^{-}=\infty)$, i.e. conditioning on level $a$ not
being reached. The result under $\mathbb{P}_{1}(\cdot|T_{\infty}^{+}<\infty)$
will be obtained by letting $a\to 0+$ in a second time. Recalling the
equivalences (3.2.1). Under the assumptions of Theorem 1.2, for any
$a\in(0,1)$ the following four events coincide a.s. under $\mathbb{P}_{1}$,

$\\{T_{a}^{-}=\infty,T_{\infty}^{+}<\infty\\}=\\{\tau_{a}^{-}=\infty,\eta(\infty)<\infty\\}=\\{\tau_{a}^{-}=\infty\\}=\\{T_{a}^{-}=\infty\\}.$

Set $g(z):=\mathbb{P}_{z}(\eta(\infty)/\varphi(z)>t|\tau_{a}^{-}=\infty)$. We
know by Lemma 3.2 that $g(z)\rightarrow\mathbb{P}(\varrho_{\alpha,\beta}>t)$
as $z$ goes to $\infty$ for any $t>0$. Recall Proposition 2.3 and the fact
that it holds under $\mathbb{P}_{1}(\cdot|\tau_{a}^{-}=\infty)$ for every
$x>0$. Applying the Markov property at time $\tau_{x}^{+}$ under
$\mathbb{P}_{1}(\cdot|\tau_{a}^{-}=\infty)$ and recalling that $\varphi$ is
regularly varying at $\infty$ with index $\alpha-\beta$, we have

$\displaystyle\mathbb{E}_{1}\bigg{[}f\Big{(}\frac{\varphi(\xi(\tau_{x}^{+}))}{\varphi(x)}\Big{)};\frac{\eta(\infty)\circ\theta_{\tau_{x}^{+}}}{\varphi(\xi(\tau_{x}^{+}))}>t\bigg{|}\tau_{a}^{-}=\infty\bigg{]}=$
$\displaystyle\
\mathbb{E}_{1}\bigg{[}f\big{(}\frac{\varphi(\xi(\tau_{x}^{+}))}{\varphi(x)}\big{)}\cdot
g(\xi(\tau_{x}^{+}))\bigg{|}\tau_{a}^{-}=\infty\bigg{]}$
$\displaystyle\underset{x\rightarrow\infty}{\longrightarrow}\mathbb{E}[f(\chi^{\alpha-\beta}_{\alpha})]\cdot\mathbb{P}(\varrho_{\alpha,\beta}>t),$

for every continuous and bounded function $f$, where $\theta_{.}$ denotes the
shift operator. We thus get

$\frac{T_{\infty}^{+}-T_{x}^{+}}{\varphi(x)}\bigg{|}_{\mathbb{P}_{1}(\cdot|T_{a}^{-}=\infty)}=\frac{\varphi\big{(}\xi(\tau_{x}^{+})\big{)}}{\varphi(x)}\cdot\frac{\eta(\infty)\circ\theta_{\tau_{x}^{+}}}{\varphi(\xi(\tau_{x}^{+}))}\bigg{|}_{\mathbb{P}_{1}(\cdot|\tau_{a}^{-}=\infty)}\underset{x\rightarrow\infty}{\Longrightarrow}\chi_{\alpha}^{\alpha-\beta}\times\varrho_{\alpha,\beta}$

where $\chi_{\alpha}$ and $\varrho_{\alpha,\beta}$ are independent.

We now show that the convergence holds true for the process conditioned to
explode. Recall that $\\{T_{\infty}^{+}<\infty\\}=\\{\tau_{0}^{-}=\infty\\}$,
$\mathbb{P}_{1}$-a.s.. one has

$\displaystyle\
\mathbb{E}_{1}\Big{[}f\Big{(}\frac{T_{\infty}^{+}-T_{x}^{+}}{\varphi(x)}\Big{)}\Big{|}T_{\infty}^{+}<\infty\Big{]}$
$\displaystyle=$ $\displaystyle\
\mathbb{E}_{1}\Big{[}f\Big{(}\frac{T_{\infty}^{+}-T_{x}^{+}}{\varphi(x)}\Big{)}\Big{|}T_{a}^{-}=\infty\Big{]}\cdot\mathbb{P}_{1}\big{(}\tau_{a}^{-}=\infty\big{|}\tau_{0}^{-}=\infty\big{)}\
+c_{1}\cdot\mathbb{P}_{1}\big{(}\tau_{a}^{-}<\infty\big{|}\tau_{0}^{-}=\infty\big{)}$
$\displaystyle=$ $\displaystyle\
\mathbb{E}_{1}\Big{[}f\Big{(}\frac{T_{\infty}^{+}-T_{x}^{+}}{\varphi(x)}\Big{)}\Big{|}T_{a}^{-}=\infty\Big{]}+c_{2}\cdot\mathbb{P}_{1}\big{(}\tau_{a}^{-}<\infty\big{|}\tau_{0}^{-}=\infty\big{)}$

where $|c_{1}|\leq||f||_{\infty},|c_{2}|\leq 2||f||_{\infty}$ and we make use
of the fact that $\\{\tau_{a}^{-}=\infty\\}\subset\\{\tau_{0}^{-}=\infty\\}$.
Noticing that the limit of the first term above is independent of $a>0$ and
the second term above is uniformly bounded by

$\mathbb{P}_{1}\big{(}\tau_{a}^{-}<\infty\big{|}\tau_{0}^{-}=\infty\big{)}=\frac{e^{p(a-1)}(1-e^{-pa})}{1-e^{-p}}=\frac{e^{p(a-1)}-e^{-p}}{1-e^{-p}}$

which tends to $0$ as $a$ goes to $0$, one thus have

$\frac{T_{\infty}^{+}-T_{x}^{+}}{\varphi(x)}\bigg{|}_{\mathbb{P}_{1}(\cdot|T_{\infty}^{+}<\infty)}\underset{x\rightarrow\infty}{\Longrightarrow}\chi_{\alpha}^{\alpha-\beta}\times\varrho_{\alpha,\beta}.$
(3.29)

For the case $\alpha=1$, the limit in (3.29) degenerates with
$\chi_{\alpha}=\varrho_{\alpha,\beta}=1$. ∎

We now invert the residual time to explosion in order to study the supremum of
the process.

######  Lemma 3.7.

One has as $t\to 0+$,

$\frac{\bar{X}(T_{\infty}^{+}-t)}{\varphi^{-1}(t)}\bigg{|}_{\mathbb{P}_{1}(\cdot|T_{\infty}^{+}<\infty)}\Rightarrow\begin{cases}&\frac{1}{\chi_{\alpha}}\times\varrho_{\alpha,\beta}^{\frac{1}{\beta-\alpha}}\text{
if }\alpha<1,\\\ &1\text{ if }\alpha=1.\end{cases}$

###### Proof.

For all small $t>0$ and large $x>0$,

$\\{\bar{X}(T^{+}_{\infty}-t)>x\\}\subset\\{T_{\infty}^{+}-T_{x}^{+}\geq
t\\}\subset\\{\bar{X}(T^{+}_{\infty}-t)\geq x\\}.$ (3.30)

For $\lambda>0$, taking $x=\lambda\varphi^{-1}(t)$, and applying (3.29)
together with the fact that $\varphi\in\mathcal{R}_{\alpha-\beta}$, we have

$\displaystyle\mathbb{P}_{1}\big{(}T_{\infty}^{+}-T_{x}^{+}\geq
t\big{|}T_{\infty}^{+}<\infty\big{)}=$ $\displaystyle\
\mathbb{P}_{1}\Big{(}\frac{T_{\infty}^{+}-T_{x}^{+}}{\varphi(x)}\geq\frac{t}{\varphi(x)}=\frac{\varphi(x/\lambda)}{\varphi(x)}\Big{|}T_{\infty}^{+}<\infty\Big{)}$
$\displaystyle\to$ $\displaystyle\
\mathbb{P}\big{(}\chi_{\alpha}^{\alpha-\beta}\times\varrho_{\alpha,\beta}\geq\lambda^{\beta-\alpha}\big{)}$

as $t\to 0+$ (consequently $y\to\infty$). The result follows by substituting
the limit above into (3.30), and the continuity of the limit distribution.

For the case $\alpha=1$,

$\frac{T_{\infty}^{+}-T_{x}^{+}}{\varphi(x)}\bigg{|}_{\mathbb{P}_{1}(\cdot|T_{\infty}^{+}<\infty)}\Rightarrow
1\quad\text{as $x\to\infty$}.$ (3.31)

For arbitrary $\varepsilon>0$, taking $x=(1+\varepsilon)\varphi^{-1}(t)$ in
(3.30), we have

$\displaystyle\ \limsup_{t\to
0+}\mathbb{P}_{1}\big{(}\bar{X}\big{(}T_{\infty}^{+}-t\big{)}>(1+\varepsilon)\varphi^{-1}(t)\big{|}T_{\infty}^{+}<\infty\big{)}$
$\displaystyle\leq$ $\displaystyle\ \limsup_{t\to
0+}\mathbb{P}_{1}\bigg{(}\frac{T_{\infty}^{+}-T_{x}^{+}}{\varphi(x)}\geq\frac{t}{\varphi(x)}=\frac{\varphi(\varphi^{-1}(t))}{\varphi((1+\varepsilon)\varphi^{-1}(t))}\bigg{|}T_{\infty}^{+}<\infty\bigg{)}$
$\displaystyle\leq$ $\displaystyle\
\limsup_{x\to\infty}\mathbb{P}_{1}\bigg{(}\frac{T_{\infty}^{+}-T_{x}^{+}}{\varphi(x)}\geq(1+\varepsilon)^{\frac{\beta-\alpha}{2}}\bigg{|}T_{\infty}^{+}<\infty\bigg{)}=0$

by (3.31) and the fact $\varphi\in\mathcal{R}_{\alpha-\beta}$. Similarly,
taking $x=(1-\varepsilon)\varphi^{-1}(t)$ in (3.30), we have by (3.31),

$\displaystyle\ \liminf_{t\to
0+}\mathbb{P}_{1}\big{(}\bar{X}\big{(}T_{\infty}^{+}-t\big{)}\geq(1-\varepsilon)\varphi^{-1}(t)\big{|}T_{\infty}^{+}<\infty\big{)}$
$\displaystyle\geq$ $\displaystyle\ \liminf_{t\to
0+}\mathbb{P}_{1}\bigg{(}\frac{T_{\infty}^{+}-T_{x}^{+}}{\varphi(x)}\geq\frac{t}{\varphi(x)}=\frac{\varphi(\varphi^{-1}(t))}{\varphi((1-\varepsilon)\varphi^{-1}(t))}\bigg{|}T_{\infty}^{+}<\infty\bigg{)}$
$\displaystyle\geq$ $\displaystyle\
\liminf_{x\to\infty}\mathbb{P}_{1}\bigg{(}\frac{T_{\infty}^{+}-T_{x}^{+}}{\varphi(x)}\geq(1-\varepsilon)^{2(\beta-\alpha)}\bigg{|}T_{\infty}^{+}<\infty\bigg{)}=1,$

which proves the critical case $\alpha=1$. This finishes the proof. ∎

The proof of Theorem 1.2, Case 1: $\beta>\alpha>0$ follows by combining the
above Lemma with Lemma 3.5. ∎

####  3.3.3 Case 2: $1\geq\beta=\alpha>0$

###### Proof of Case 2 in Theorem 1.2.

Applying the Markov property at $\tau_{x}^{+}$, we have

$\frac{T_{\infty}^{+}-T_{x}^{+}}{\varphi(x)}\bigg{|}_{\mathbb{P}_{1}(\cdot|T_{a}^{-}=\infty)}=\frac{\varphi\big{(}\xi(\tau_{x}^{+})\big{)}}{\varphi(x)}\cdot\frac{\eta(\infty)\circ\theta_{\tau_{x}^{+}}}{\varphi(\xi(\tau_{x}^{+}))}\bigg{|}_{\mathbb{P}_{1}(\cdot|\tau_{a}^{-}=\infty)}\Rightarrow
1,$

where we use the fact that in this case $\varphi\in\mathcal{R}_{0}$ at
$\infty$ and $y^{-1}\xi(\tau_{x}^{+})$ has a finite weak limit by Proposition
2.3. The limit above then follows from Lemma 3.4.

The scaling limits under $\mathbb{P}_{1}(\cdot|T_{\infty}^{+}<\infty)$ and
that for $X(T_{\infty}^{+}-t)$ are obtained similarly as in the proof of case
$\alpha=1<\beta$ in (3.31), and details are therefore omitted. ∎

####  3.3.4 Case 3: $\beta>\alpha=0$ and proof of Proposition 2.4

The arguments for $\beta>\alpha=0$ are different. In this case,
$\chi_{\alpha}=\infty$ by Proposition 2.3 and $\varrho_{\alpha,\beta}$ has a
standard exponential law; see Section 1. Therefore, the limiting law in (3.29)
is degenerate at $0$ and the proof of the case $\alpha>0$ is not applicable.

We start by establishing the following lemma. Recall that $\Pi$ is the Lévy
measure of the ladder height process $H$ defined in (3.1) and $\nu$ is that of
$\xi$; see (2.1).

######  Lemma 3.8.

Assume that $-\psi\in\mathcal{R}_{0}$ at $0$. For every $x,s\to\infty$ such
that either $\bar{\Pi}(s)/\bar{\Pi}(x)\to\lambda\in(0,1)$ for $0<p<\infty$ or
$\bar{\nu}(s)/\bar{\nu}(x)\to\lambda\in(0,1)$ for $p=\infty$, we have

$\mathbb{P}\big{(}\xi(\tau_{x}^{+})>s\big{)}\underset{x,s\to\infty}{\longrightarrow}\lambda.$
(3.32)

###### Proof.

If $\xi$ with $0<p<\infty$ is not a non-subordinator, by applying [2, Theorem
III.6] to the ladder height process and (3.6), when $-\psi\in\mathcal{R}_{0}$
at $0$, we have

$\displaystyle\mathbb{P}\big{(}\bar{\xi}(\tau_{x}^{+}-)>\delta
y\big{)}\underset{y\rightarrow\infty}{\to}0,\quad\forall\delta\in(0,1)$ (3.33)
$\displaystyle\text{ and
}\mathbb{P}\big{(}\xi(\tau_{x}^{+})>s,\bar{\xi}\big{(}\tau_{x}^{+}-\big{)}\leq\delta
x\big{)}=\int_{0}^{\delta}\mathcal{U}(x\mathrm{d}u)\bar{\Pi}\big{(}x(1-u)\big{)}\cdot\frac{\bar{\Pi}(s-xu)}{\bar{\Pi}(x(1-u))},\quad\forall
s>x.$ (3.34)

Applying Proposition 2.2 to (3.5), we get that

$\hat{\bar{\Pi}}(\theta)\sim\frac{\psi(\theta)}{-p\theta}\in\mathcal{R}_{-1}\Rightarrow\int_{0}^{x}\bar{\Pi}(y)\mathrm{d}y\sim\frac{x\psi(1/x)}{-p}\in\mathcal{R}_{1}\Rightarrow\bar{\Pi}(x)\sim\frac{\psi(1/x)}{-p}\in\mathcal{R}_{0}$
(3.35)

as $\theta\to 0+$ and $x\to\infty$, respectively, where the monotone density
theorem is applied in the last identity (recall that $\bar{\Pi}$ defined in
(3.5) is continuous and strictly decreasing on $(0,\infty)$).

For every $\lambda\in(0,1)$, let $s,x$ satisfy (3.32). Since
$\bar{\Pi}\in\mathcal{R}_{0}$, we have $s/x\to\infty$ and

$\frac{\bar{\Pi}(s-xu)}{\bar{\Pi}(x(1-u))}=\frac{\bar{\Pi}(s-xu)}{\bar{\Pi}(s)}\frac{\bar{\Pi}(x)}{\bar{\Pi}(x(1-u))}\frac{\bar{\Pi}(s)}{\bar{\Pi}(x)}\to\lambda\quad\text{as
$x\to\infty$},$

where the limit above holds uniformly for $u\in[0,\delta]$ by UCT. Using
(3.34) and the above convergence gives

$\displaystyle\left\lvert\mathbb{P}\big{(}\xi(\tau_{x}^{+})>s,\bar{\xi}(\tau_{x}^{+}-)\leq\delta
x\big{)}-\lambda\cdot\mathbb{P}\big{(}\bar{\xi}(\tau_{x}^{+}-)\leq\delta
x\big{)}\right\lvert$ $\displaystyle\qquad\leq\
\int_{0}^{\delta}\mathcal{U}(x\mathrm{d}u)\bar{\Pi}\big{(}x(1-u)\big{)}\cdot\Big{|}\frac{\bar{\Pi}(s-xu)}{\bar{\Pi}(x(1-u))}-\lambda\Big{|}\to
0,$

which further gives (3.32) by applying (3.33).

The subordinator case ($p=\infty$) follows from the same arguments by working
directly with $\xi$ and its Lévy measure $\nu$ instead of $H$ and $\Pi$. We
omit the details. ∎

We now proceed with the study of the speed of explosion.

###### Proof of Case 3 in Theorem 1.2.

We are going to apply Lemma 3.8 with $x$ and $s$ chosen as certain functions
of $t$, going to $\infty$, as $t\to 0+$. Applying the Markov property at
$\tau_{x}^{+}$ to the process $\xi$ gives

$\displaystyle\
\mathbb{P}_{1}\big{(}T_{\infty}^{+}-T_{x}^{+}<t\big{|}T_{a}^{-}=\infty\big{)}=\mathbb{P}_{1}\big{(}\eta(\infty)\circ\theta_{\tau_{x}^{+}}<t\big{|}\tau_{a}^{-}=\infty\big{)}$
(3.36) $\displaystyle=$ $\displaystyle\
\int_{x}^{\infty}\mathbb{P}_{1}\big{(}\xi(\tau_{x}^{+})\in\,\mathrm{d}z\big{|}\tau_{a}^{-}=\infty\big{)}\mathbb{P}_{z}\big{(}\eta(\infty)<t\big{|}\tau_{a}^{-}=\infty\big{)}$
$\displaystyle=$ $\displaystyle\
\int_{x}^{\infty}\mathbb{P}_{1}\big{(}\xi(\tau_{x}^{+})\in\,\mathrm{d}z\big{|}\tau_{a}^{-}=\infty\big{)}\mathbb{P}_{z}\Big{(}z>\varphi^{-1}\Big{(}t\Big{/}\frac{\eta(\infty)}{\varphi(z)}\Big{)}\Big{|}\tau_{a}^{-}=\infty\Big{)}.$

Since $\varrho_{\alpha,\beta}$ for $\alpha=0$ is exponentially distributed,
for arbitrary $\varepsilon>0$, there are $0<m<M<\infty$ so that for every $x$
large enough

$\mathbb{P}_{z}\Big{(}\frac{\eta(\infty)}{\varphi(z)}\in[m,M]\Big{|}\tau_{a}^{-}=\infty\Big{)}>1-\varepsilon,\quad\forall
z>y.$

Plugging the above into (3.36) and using the monotonicity of $\varphi^{-1}$,
one can check that

$\displaystyle\mathbb{P}_{1}\big{(}T_{\infty}^{+}-T_{x}^{+}<t\big{|}T_{a}^{-}=\infty\big{)}\leq$
$\displaystyle\
\varepsilon+\mathbb{P}_{1}\big{(}\xi(\tau_{x}^{+})>\varphi^{-1}(t/M)\big{|}\tau_{a}^{-}=\infty\big{)},$
(3.37)
$\displaystyle\mathbb{P}_{1}\big{(}T_{\infty}^{+}-T_{x}^{+}<t\big{|}T_{a}^{-}=\infty\big{)}\geq$
$\displaystyle\
(1-\varepsilon)\cdot\mathbb{P}_{1}\big{(}\xi(\tau_{x}^{+})>\varphi^{-1}(t/m)\big{|}\tau_{a}^{-}=\infty\big{)}.$

Let $x=\bar{\Pi}^{-1}(\bar{\Pi}(\varphi^{-1}(t))/\lambda)$ for fixed
$\lambda\in(0,1)$ and $s_{1}=\varphi^{-1}(t/m),s_{2}=\varphi^{-1}(t/M)$ in
(3.32). Since $\bar{\Pi}\in\mathcal{R}_{0}$ and
$\varphi\in\mathcal{R}_{-\beta}$, one can check that as $t\to 0+$

$\varphi^{-1}\in\mathcal{R}_{-1/\beta},\quad\frac{\bar{\Pi}(s_{1})}{\bar{\Pi}(x)}=\lambda\frac{\bar{\Pi}(\varphi^{-1}(t/m))}{\bar{\Pi}(\varphi^{-1}(t))}\to\lambda\quad\text{and}\quad\frac{\bar{\Pi}(s_{2})}{\bar{\Pi}(x)}=\lambda\frac{\bar{\Pi}(\varphi^{-1}(t/M))}{\bar{\Pi}(\varphi^{-1}(t))}\to\lambda.$

Thus, applying Lemma 3.8, (3.32) holds true and one sees from (3.37) that

$\mathbb{P}_{1}\big{(}T_{\infty}^{+}-T_{x}^{+}<t\big{|}T_{a}^{-}=\infty\big{)}\to\lambda\quad\text{as
$t\to 0+$},$

which is continuous in $\lambda\in(0,1)$. Similar to (3.29), one can also have

$\mathbb{P}_{1}\big{(}T_{\infty}^{+}-T_{x}^{+}<t\big{|}T_{\infty}^{+}<\infty\big{)}\to\lambda\quad\text{as
$t\to 0+$},$

Going back to the events in (3.30), we get by applying the latter result that
for every $\lambda\in(0,1)$

$\displaystyle\
\mathbb{P}_{1}\Big{(}\lambda\bar{\Pi}(\bar{X}(T^{+}_{\infty}-t))\geq\bar{\Pi}(\varphi^{-1}(t))\Big{|}T_{\infty}^{+}<\infty\Big{)}$
$\displaystyle=$ $\displaystyle\
\mathbb{P}_{1}\Big{(}\bar{X}(T_{\infty}^{+}-t)\leq\bar{\Pi}^{-1}\big{(}\bar{\Pi}(\varphi^{-1}(t))/\lambda\big{)}\Big{|}T_{\infty}^{+}<\infty\Big{)}\underset{t\rightarrow
0+}{\to}\lambda,$

which together with (3.35) concludes the proof. ∎

Along establishing the nonlinear renormalisation for
$\bar{X}(T^{+}_{\infty}-t)$ in the case $\alpha=0$ above, we have implicitly
addressed a nonlinear renormalisation of the overshoot of the parent Lévy
process $\xi$. We further illustrate this by showing Proposition 2.4.

###### Proof of Proposition 2.4.

For every $\lambda\in(0,1)$, choose $s$ and $x$ such that
$\bar{\Pi}(s)=\lambda\bar{\Pi}(x)$. Lemma 3.8 gives

$\displaystyle\mathbb{P}\big{(}\xi(\tau_{x}^{+})>s\big{)}=$ $\displaystyle\
\mathbb{P}\big{(}\bar{\Pi}(\xi(\tau_{x}^{+}))\leq\bar{\Pi}(s)\big{)}$
$\displaystyle=$ $\displaystyle\
\mathbb{P}\big{(}\bar{\Pi}(\xi(\tau_{x}^{+}))\leq\lambda\bar{\Pi}(x)\big{)}=\mathbb{P}\big{(}\bar{\Pi}(\xi(\tau_{x}^{+}))\big{/}\bar{\Pi}(x)\leq\lambda\big{)}\to\lambda$

as $x\to\infty$. We apply further (3.35) and the proof of the proposition is
completed. The fact that the result also holds conditioned on
$\\{\tau_{0}^{-}=\infty\\}$ follows from similar arguments as in the case
$\alpha>0$ whose details are omitted. ∎

Proof of Theorem 1.2: To conclude, Theorem 1.2 is obtained by combining the
three cases treated above with Lemma 3.5. ∎

In the last section, we present some observations on the classical CSBPs. The
branching property allows one to study directly the explosion time and its
renormalisation without going through a study of the perpetual integral.

###  3.4 The case of CSBP

If the rate function $R$ is the identity function, the Markov process $X$
defined by (1.3) is a classical CSBP with branching mechanism $\psi$. Namely,
it fulfills the branching property: for any $x,y\in(0,\infty)$ and $t\geq 0$
$X_{t}(x+y)\overset{D}{=}X^{1}_{t}(x)+X^{2}_{t}(y)$, where $X^{1}_{t}(x)$,
$X^{2}_{t}(y)$ are independent copies of the process $X$ at time $t$ starting,
respectively, from $x$ and $y$. Notice that when $R$ is not linear, the
branching property does not hold and the techniques below do not apply.

A fundamental consequence of the branching property lies on the fact that for
any $x\in(0,\infty)$ and $t\geq 0$, the Laplace transform of the process at
time $t$ satisfies identity

$\mathbb{E}_{x}(e^{-\lambda X_{t}})=e^{-xu_{t}(\lambda)},\,\,\,\lambda>0$
(3.38)

with $t\mapsto u_{t}(\lambda)$ the solution to the following integral equation

$u_{0}(\lambda)=\lambda\quad\text{ and
}\quad\int_{u_{t}(\lambda)}^{\lambda}\frac{\mathrm{d}u}{\psi(u)}=t.$ (3.39)

A lot of information is available from the identity (3.38) and it is not in
general necessary in this setting to go back to the representation of the
process via a time-changed Lévy process.

####  3.4.1 Explosion criterion

A striking feature in this setting is that the explosion of the process can be
studied directly from (3.38) as follows:

$\mathbb{P}_{x}(T_{\infty}^{+}>t)=\underset{\lambda\rightarrow
0}{\lim}\mathbb{E}_{x}[e^{-\lambda X_{t}}]=e^{-xu_{t}(0+)}$ (3.40)

with $u_{t}(0+):=\underset{\lambda\rightarrow 0}{\lim}u_{t}(\lambda)$. Hence,
the process $X$ explodes with positive probability if and only if
$u_{t}(0+)>0$. From (3.39), one can verify that this is equivalent to the
following integral condition

$\int_{0}\frac{\mathrm{d}u}{|\psi(u)|}<\infty.$ (3.41)

More precisely, when (3.41) holds, $t\mapsto u_{t}(0+)$ satisfies the equation

$\int^{u_{t}(0+)}_{0}\frac{\mathrm{d}u}{-\psi(u)}=t,\text{ for all }t\geq 0.$
(3.42)

The identity (3.40) provides in fact the cumulative distribution function of
the explosion time. Recall $p$ the largest zero of $\psi$ and notice that
$u_{t}(0+)\underset{t\rightarrow\infty}{\longrightarrow}p$. Then

$\mathbb{P}_{x}(T_{\infty}^{+}>t,T_{\infty}^{+}<\infty)=e^{-xu_{t}(0+)}-e^{-xp}.$
(3.43)

We now restudy briefly the speed of explosion and the renormalisation of the
explosion time in the case $-\psi\in\mathcal{R}_{\alpha}$ at $0$ with
$\alpha\in(0,1)$.

####  3.4.2 The speed of explosion

For simplicity, we assume from now on that $X$ has nondecreasing sample paths,
i.e. the parent Lévy process $\xi$ is a subordinator. In this case $p=\infty$
and the explosion event is almost sure when (3.41) holds. Using the identity
$T^{+}_{\infty}=T^{+}_{a/u_{t}(0+)}+T^{+}_{\infty}\circ\theta_{T^{+}_{a/u_{t}(0+)}}$,
the strong Markov property and (3.43), we get

$\displaystyle\mathbb{P}_{1}\big{(}u_{t}(0+)X_{T_{\infty}^{+}-t}>a\big{)}$
$\displaystyle=\mathbb{P}_{1}\big{(}T_{\infty}^{+}-T^{+}_{a/u_{t}(0+)}\geq
t\big{)}$
$\displaystyle=\mathbb{E}_{1}\bigg{[}\mathbb{P}_{X_{T^{+}_{a/u_{t}(0+)}}}\big{(}T^{+}_{\infty}>t\big{)}\bigg{]}$
$\displaystyle=\mathbb{E}_{1}\bigg{[}e^{-X_{T_{a/u_{t}(0+)}^{+}}u_{t}(0+)}\bigg{]}.$

Assume now that $-\psi\in\mathcal{R}_{\alpha}$ at $0$ with $\alpha\in(0,1)$.
By Proposition 2.3,

$a\frac{u_{t}(0+)}{a}X_{T^{+}_{a/u_{t}(0+)}}=a\frac{u_{t}(0+)}{a}\xi_{\tau^{+}_{a/u_{t}(0+)}}\Longrightarrow
a\chi_{\alpha}.$

Therefore, for any $a>0$
$\mathbb{P}(u_{t}(0+)X_{T_{\infty}^{+}-t}>a)\underset{t\rightarrow
0+}{\longrightarrow}\mathbb{E}[e^{-a\chi_{\alpha}}]$. We get finally the
following limit in law

$u_{t}(0+)X_{T_{\infty}^{+}-t}\Longrightarrow\mathbbm{e}_{\chi_{\alpha}}:=\frac{1}{\chi_{\alpha}}\mathbbm{e}_{1}\text{
as }t\to 0+$ (3.44)

where $\mathbbm{e}_{1}$ is a standard exponential random variable independent
of $\chi_{\alpha}$. In other words, $\mathbbm{e}_{\chi_{\alpha}}$ is a mixture
of exponential random variables with random parameter given by
$\chi_{\alpha}$.

####  3.4.3 Renormalisation of the explosion time

We now explain the link between (3.44) and Case 1 of Theorem 1.2. First recall
$\varphi$ in (1.12). By change of variable $y=1/u$, we get
$\varphi(x)=\int_{0}^{1/x}\frac{\mathrm{d}u}{-\psi(u)}$ for all $x$, and we
easily verify with the help of (3.42) that $\varphi^{-1}(t)=1/u_{t}(0+)$ for
all $t\geq 0$. Since by assumption $-\psi\in\mathcal{R}_{\alpha}$ at $0$,
$\varphi\in\mathcal{R}_{-(1-\alpha)}$ at $\infty$ and
$\varphi^{-1}\in\mathcal{R}_{-1/(1-\alpha)}$ at $\infty$, then

$u_{t\varphi(x)}(0+)=\frac{\varphi^{-1}(\varphi(x))}{\varphi^{-1}(t\varphi(x))}\frac{1}{x}\underset{x\rightarrow\infty}{\sim}t^{\frac{1}{1-\alpha}}\frac{1}{x}.$

Therefore, by applying (3.40), we get

$\displaystyle\mathbb{P}_{x}\left(\frac{T^{+}_{\infty}}{\varphi(x)}\leq
t\right)$
$\displaystyle=e^{-xu_{t\varphi(x)}(0+)}\underset{x\rightarrow\infty}{\longrightarrow}e^{-t^{\frac{1}{1-\alpha}}}.$
(3.45)

Hence, we recover the result
$\frac{T^{+}_{\infty}}{\varphi(x)}\Longrightarrow\rho_{\alpha,1}\text{ as
}x\text{ goes to }\infty$ where $\rho_{\alpha,1}$ has a Weibull distribution
with parameter $\frac{1}{1-\alpha}$. In particular,
$\mathbbm{e}_{1}:=\rho_{\alpha,1}^{\frac{1}{1-\alpha}}$ is a standard
exponential random variable and by (3.44)

$u_{t}(0+)X_{T_{\infty}^{+}-t}\Longrightarrow\mathbbm{e}_{\chi_{\alpha}}\overset{D}{=}\frac{1}{\chi_{\alpha}}\rho_{\alpha,1}^{\frac{1}{1-\alpha}}$

which coincides with Case 1 of Theorem 1.2 with $\beta=1$.

######  Remark 3.1.

  1. 1.

We mention that the convergence in (3.45) was noticed by Sagitov [23]; see
also Pakes [21] where some discrete nonlinear branching processes are studied.

  2. 2.

The case $p<\infty$ for which the sample paths of the CSBP are not monotone
could also be handled directly. But it requires to work with the process
conditioned on explosion; see e.g. Fang and Li [8] for a study of the latter.

Acknowledgements Bo Li and Xiaowen Zhou are supported by NSERC
(RGPIN-2016-06704). Clément Foucart is supported by the European Union (ERC,
SINGER, 101054787). Views and opinions expressed are however those of the
authors only and do not necessarily reflect those of the European Union or the
European Research Council. Neither the European Union nor the granting
authority can be held responsible for them.

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##  Appendix A Appendix

This section is dedicated to the proofs of Lemmas 3.1 and 3.3. Recall that for
the test function $f\in\mathcal{R}_{-\gamma}$ with $\gamma>0$ and
$U_{a}f(x)<\infty$ for some $a>0$, we must have from Corollary 1.1,

$\int^{\infty}\frac{-f(y)}{y\psi(1/y)}\mathrm{d}y<\infty.$

In the non-subordinator case, i.e. $0<p<\infty$, applying (2.14), (2.8) and
(2.12), the finiteness of integral above implies

$\int^{\infty}\frac{f(y)\mathrm{d}y}{y}\int_{0}^{y}\kappa(z)\mathrm{d}z<\infty\Rightarrow\int^{\infty}\kappa(z)dz\int_{z}^{\infty}\frac{f(y)}{y}\mathrm{d}y<\infty\Rightarrow\int^{\infty}f(z)\kappa(z)dz<\infty$

where Fubini-Tonelli’s theorem is applied. Thus, by (2.7), for every $x>a>0$

$\displaystyle U_{a}f(x)=$ $\displaystyle\
\int_{a}^{\infty}\mathbb{E}_{x}\big{[}f(\xi(t));t<\tau_{a}^{-}\big{]}\mathrm{d}t=\int_{a}^{\infty}f(y)\big{(}\kappa(y-x)-e^{p(a-x)}\kappa(y-a)\big{)}\mathrm{d}y$
(A.1) $\displaystyle=$ $\displaystyle\
\int_{0}^{\infty}f(x+y)\kappa(y)\mathrm{d}y+\phi^{\prime}(0)\int_{a}^{x}e^{p(y-x)}f(y)\mathrm{d}y-e^{p(a-x)}\int_{0}^{\infty}f(a+y)\kappa(y)\mathrm{d}y,$

where $\kappa(y)=e^{py}\phi^{\prime}(0)$ for $y<0$ in (2.7).

In the subordinator case, i.e. $p=\infty$, one has $\tau_{a}^{-}=\infty$
$\mathbb{P}_{x}$-a.s for $x>a>0$, and
$U_{a}(x,\mathrm{d}y)=U(x,\mathrm{d}y)=\int_{0}^{\infty}\mathbb{P}_{x}(\xi(t)\in\mathrm{d}y)\mathrm{d}t$,
see (2.10). One has

$U_{0}f(x)=\int_{0}^{\infty}f(y)U(x,\mathrm{d}y)=\int_{0}^{\infty}f(x+y)U(\mathrm{d}y).$

Notice that the latter coincides with (A.1) if we understand the last two
integrals to be zero (recall $p=\infty$) and replace $\kappa(y)\mathrm{d}y$ by
$U(\mathrm{d}y)$.

###### Proof of Lemma 3.1.

We start with the non-subordinator case. It is straightforward to check from
(A.1) that $x\to f(\log x)\in\mathcal{R}_{0}$. Therefore, applying Proposition
2.1 we have

$\int_{a}^{x}e^{p(y-x)}f(y)\mathrm{d}y=u^{-p}\int_{e^{a}}^{u}z^{p-1}f(\log
z)\mathrm{d}z\Big{|}_{u=e^{x}}\sim\frac{1}{p}f(x)\in\mathcal{R}_{-\gamma}\text{\
at\ }\infty,$

which together with the fact $e^{-px}=o(f(x))$ gives

$U_{a}f(x)-\int_{0}^{\infty}f(x+y)\kappa(y)\mathrm{d}y=O\big{(}f(x)\big{)}.$
(A.2)

On the other hand, for $s>\alpha$, we have from Fubini-Tonelli’s theorem that

$\displaystyle\ \int_{0}^{\infty}(x+y)^{-s}\kappa(y)\mathrm{d}y=$
$\displaystyle\iint_{y,u>0}\frac{u^{s-1}}{\Gamma(s)}e^{-(x+y)u}\kappa(y)\mathrm{d}y\mathrm{d}u=\int_{0}^{\infty}\frac{u^{s-1}}{\Gamma(s)}\hat{\kappa}(u)e^{-xu}\mathrm{d}u$
(A.3) $\displaystyle\sim$ $\displaystyle\
\frac{\Gamma(s-\alpha)}{\Gamma(s)}x^{-s}\hat{\kappa}\big{(}1/x\big{)}\sim\frac{\Gamma(s-\alpha)}{\Gamma(s)}\frac{-x^{-s}}{\psi(1/x)}\quad\text{as\
}x\to\infty,$

where (2.9) and Proposition 2.2 are used in the second line above. For any
$\varepsilon>0$, applying the UCT to $f(x)x^{s}\in\mathcal{R}_{s-\gamma}$ with
$s\in(\alpha,\gamma)$, we have for $x$ large enough

$\Big{|}\frac{f(x+y)}{f(x)}-\big{(}\frac{x+y}{x}\big{)}^{-\gamma}\Big{|}\leq\varepsilon\big{(}\frac{x+y}{x}\big{)}^{-s}\quad\forall
y>0.$

Thus, we have for such $x>0$

$\int_{0}^{\infty}\frac{f(x+y)}{f(x)}\kappa(y)\mathrm{d}y=\int_{0}^{\infty}\frac{x^{\gamma}\kappa(y)\mathrm{d}y}{(x+y)^{\gamma}}+\varepsilon_{1}(x)\cdot\int_{0}^{\infty}\frac{x^{s}\kappa(y)\mathrm{d}y}{(x+y)^{s}},$

for some $|\varepsilon_{1}(x)|<\varepsilon$. Applying (A.3) to the two
integrals on the right hand side above, one can find that they are of the same
order, which implies

$\int_{0}^{\infty}f(x+y)\kappa(y)\mathrm{d}y\sim\frac{-f(x)}{\psi(1/x)}\frac{\Gamma(\gamma-\alpha)}{\Gamma(\gamma)}\sim\frac{\Gamma(\gamma-\alpha+1)}{\Gamma(\gamma)}\int_{x}^{\infty}\frac{-f(y)}{y\psi(1/y)}\mathrm{d}y\in\mathcal{R}_{\alpha-\gamma}.$

Plugging into (A.2) with the fact $\psi(0)=0$ completes the proof.

In the subordinator case, we omit the details since the proof follows from the
same arguments. In particular (A.3) holds true if one replaces
$\kappa(y)\mathrm{d}y$ by $U(\mathrm{d}y)$ and $\hat{\kappa}(u)$ by
$\hat{U}(u)$. ∎

###### Proof of Lemma 3.3.

In the non-subordinator case, by following the same argument as before, we see
from (A.1) that (A.2) still holds. The lemma will be proved once we have shown
that

$\int_{0}^{\infty}f(x+y)\kappa(y)\mathrm{d}y\sim\int_{x}^{\infty}f(y)\kappa(y)\mathrm{d}y\sim\frac{1}{\Gamma(\alpha)}\int_{x}^{\infty}\frac{-f(y)}{y\psi(1/y)}\mathrm{d}y\in\mathcal{R}_{0}.$
(A.4)

Together with the finiteness of the integrals, we have from (2.14) and (2.12)
that

$\displaystyle\int_{x}^{\infty}\frac{-f(y)}{y\psi(1/y)}\mathrm{d}y\sim\Gamma(1+\alpha)\int_{x}^{\infty}\frac{f(y)}{y}\int_{0}^{y}\kappa(z)\mathrm{d}z\mathrm{d}y\in\mathcal{R}_{0},$
(A.5)
$\displaystyle\int_{x}^{\infty}\frac{f(z)}{z}dz\cdot\int_{0}^{x}\kappa(z)\mathrm{d}z\sim\frac{f(x)}{\alpha}\int_{0}^{x}\kappa(z)\mathrm{d}z=o\Big{(}\int_{x}^{\infty}\frac{f(y)}{y}\int_{0}^{y}\kappa(z)\mathrm{d}z\mathrm{d}y\Big{)},$
(A.6)

where Proposition 2.1 is applied in the last identity above, and entails
further

$\displaystyle\int_{x}^{\infty}\frac{f(y)}{y}\int_{0}^{y}\kappa(z)\mathrm{d}z\mathrm{d}y=$
$\displaystyle\
\int_{x}^{\infty}\frac{f(y)}{y}\mathrm{d}y\cdot\int_{0}^{x}\kappa(z)\mathrm{d}z+\int_{x}^{\infty}\kappa(z)\int_{z}^{\infty}\frac{f(y)}{y}\mathrm{d}y\mathrm{d}z$
$\displaystyle\sim$ $\displaystyle\
\int_{x}^{\infty}\kappa(z)\int_{z}^{\infty}\frac{f(y)}{y}\mathrm{d}y\mathrm{d}z\sim\frac{1}{\alpha}\int_{x}^{\infty}f(z)\kappa(z)\mathrm{d}z.$

Plugging this into (A.5) gives the second asymptotic equivalence in (A.4).

For arbitrary $b>1$, we consider the difference

$\displaystyle\
\int_{0}^{\infty}f(x+y)\kappa(y)\mathrm{d}y-\int_{bx}^{\infty}f(y)\kappa(y)\mathrm{d}y$
$\displaystyle=$ $\displaystyle\
\int_{bx}^{\infty}\Big{(}\frac{f(x+y)}{f(y)}-1\Big{)}f(y)\kappa(y)\mathrm{d}y+\int_{0}^{bx}f(x+y)\kappa(y)\mathrm{d}y=I_{1}+I_{2}.$

By the fact $\int_{x}^{\infty}f(y)\kappa(y)\mathrm{d}y\in\mathcal{R}_{0}$, we
have

$I_{1}\leq\sup_{y>bx}\Big{|}\frac{f(x+y)}{f(y)}-1\Big{|}\int_{bx}^{\infty}f(y)\kappa(y)\mathrm{d}y\sim\Big{(}1-\Big{(}\frac{b}{1+b}\Big{)}^{\alpha}\Big{)}\int_{x}^{\infty}f(y)\kappa(y)\mathrm{d}y.$

Moreover, for fixed $b>1$, $\frac{x+y}{x}\leq b+1$ for $y<bx$ in $I_{2}$,
applying UCT to $f\in\mathcal{R}_{-\alpha}$ one has

$I_{2}\sim
f(x)\int_{0}^{bx}\Big{(}\frac{x}{x+y}\Big{)}^{\alpha}\kappa(y)\mathrm{d}y\leq
f(x)\int_{0}^{bx}\kappa(y)\mathrm{d}y=o\Big{(}\int_{x}^{\infty}f(y)\kappa(y)\mathrm{d}y\Big{)},$

where (A.6) and the fact
$\int_{0}^{x}\kappa(y)\mathrm{d}y\in\mathcal{R}_{\alpha}$ are used. Therefore,
first letting $x\to\infty$ and then letting $b\to\infty$ gives (A.4), and this
completes the proof.

Details are omitted in the subordinator case, this follows from the same
arguments. ∎