2.1. Discussion of the jump kernel

We have

\begin{equation*}\mathbb{P}\!\left( Y(x) >y\mid X_{t-}=x\right) =K\!\left( x,y\right) = \int_{ ( y, \infty ) } K (x, dz).\end{equation*}

Clearly $K\!\left( x,y\right) $ is a non-increasing function of y for all $y\geq x$ satisfying $K\!\left( x,y\right) =1$ for all $y<x$ . By continuity, this implies that $K\!\left( x,x\right) =1$ , so that the law of $\Delta(x) $ has no atom at $0$ .

In the sequel we concentrate on the separable case

(8) \begin{equation}K\!\left( x,y\right) {=}\frac{k(y) }{k(x) }\text{,}\end{equation}

where $k\,:\, [0, \infty ) \to [ 0, \infty ] $ is any positive non-increasing function. In what follows we suppose that k is continuous and finite on $ (0, \infty)$ .

Fix $z>0$ and assume $y=x+z$ . Then

\begin{equation*}{\bf P}\!\left( Y(x) >y\right) =\frac{k\!\left( x+z\right) }{k\!\left(x\right) } .\end{equation*}

Depending on $k(x) $ , this probability can be a decreasing or an increasing function of x for each z.

Example 2. The case $k\!\left( 0\right) <\infty $ . Without loss of generality, we may take $k\!\left( 0\right) =1$ . Assume that $k(x) ={\bf P}\!\left( Z>x\right) $ for some proper random variable $Z>0$ and that

\begin{equation*}Y(x) \stackrel{d}{=}Z\mid Z>x,\end{equation*}

so that $Y(x) $ is amenable to the truncation of Z above x. Thus

\begin{equation*}{\bf P}\!\left( Y\!\left( X_{t-}\right) >y\mid X_{t-}=x\right) =\frac{{\bf P}\!\left( Z>y,Z>x\right) }{{\bf P}\!\left( Z>x\right) }=\frac{k(y) }{k(x) }, \quad \text{ for }y>x .\end{equation*}

A particular (exponential) choice is

\begin{equation*}k(x) =e^{-\theta x},\qquad \theta >0,\end{equation*}

with ${\bf P}\!\left( Y\!\left( X_{t-}\right) >y\mid X_{t-}=x\right) =e^{-\theta\!\left( y-x\right) }$ depending only on $y-x$ . Another possible choice (Pareto) is $k(x) =\left( 1+x\right) ^{-c}$ , $c>0$ .

Note that $K\!\left( 0,y\right) =k(y) >0$ for all $y>0$ , and $k\!\left(y\right) $ turns out to be the complementary probability distribution function of a jump above y, starting from 0: state 0 is reflecting.

The case $ k (0 ) = \infty$ . Consider $k(x)=\int_{x}^{\infty }\mu \!\left( Z\in dy\right) $ for some positive Radon measure $\mu $ with infinite total mass. In this case,

\begin{equation*}{\bf P}\!\left( Y\!\left( X_{t-}\right) >y\mid X_{t-}=x\right) =\frac{\mu \!\left(Z>y,Z>x\right) }{\mu \!\left( Z>x\right) }=\frac{k(y) }{k\!\left(x\right) },\quad \text{ for }y>x.\end{equation*}

Now, $K\!\left( 0,y\right) =0$ for all $y>0$ and state 0 becomes attracting. An example is $k(x) =x^{-c}$ , $c>0$ , which is not a complementary probability distribution function.

The ratio $k(y)/k(x) $ is thus the conditional probability that a jump is greater than the level y given that it did occur and that it is greater than the level x; see Equation (1) of Eliazar and Klafter [Reference Eliazar and Klafter14] for a similar choice.

Our motivation for choosing the separable form $K\!\left(x,y\right) =k(y)/k(x) $ is that it accounts for the possibility of having state 0 either absorbing or reflecting for upward jumps launched from 0, and also that it can account for either negative or positive feedback of the current population size on the number of incoming immigrants.

Example 3. One can think of many other important and natural choices of $K\!\left( x,y\right)$ , not in the separable class, among which are those for which

\begin{equation*}K\!\left( x,dy\right) =\delta _{Vx}\!\left( dy\right)\end{equation*}

for some random variable $V>1$ . For this class of kernels, state 0 is always attracting. For example, we have the following:

1. 1. Choosing $V=1+E$ where E is exponentially distributed with probability density function $e^{-\theta x}$ , $Y\!\left(x\right) =Vx$ yields

   \begin{equation*}{\bf P}\!\left( Y(x) >y\mid X_{-}=x\right) =K\!\left( x,y\right) = {\bf P}\!\left( \left( 1+E\right) x>y\right) =e^{-\theta \!\left( \frac{y}{x} -1\right) }.\end{equation*}

2. 2. Choosing $V=1+E$ where E is Pareto-distributed with probability density function $\left( 1+x\right) ^{-c}$ , $c>0$ , $Y(x) =Vx$ yields

   \begin{equation*}{\bf P}\!\left( Y(x) >y\mid X_{-}=x\right) =K\!\left( x,y\right) = {\bf P}\!\left( \left( 1+E\right) x>y\right) =\left( y/x\right) ^{-c},\end{equation*}
   both with $K\!\left( 0,y\right) =0$ .

3. 3. If $V\sim \delta _{v}$ , $v>1$ , then $K\!\left( x,y\right) ={\bf 1}\!\left( y/x\leq v\right)$ . The three kernels depend on $y/x$ .

Note that in all three cases, $\partial _{x}K (x, x+z) >0$ , so that the larger x, the larger ${\bf P}\!\left( Y(x) >x+z\right)$ . If the population stays high, the probability of a large number of immigrants will be enhanced. There is positive feedback of x on $\Delta (x)$ , translating to a herd effect.

Remark 1. A consequence of the separability condition of K is the following. Consider a Markov sequence of after-jump positions defined recursively by $Z_{n}=Y\!\left( Z_{n-1}\right) $ , $Z_{0}=x_{0}$ . With $x_{m}>x_{m-1}$ , we have

\begin{equation*}{\bf P}\!\left( Z_{m}>x_{m}\mid Z_{m-1}=x_{m-1}\right) =K\!\left(x_{m-1},x_{m}\right) \text{, }m=1,...,n,\end{equation*}

so that, with $x_{0}<x_{1}<...<x_{n}$ , and under the separability condition on K, the product

\begin{equation*}\prod_{m=1}^{n}{\bf P}\!\left( Z_{m}>x_{m}\mid Z_{m-1}=x_{m-1}\right)=\prod_{m=1}^{n}K\!\left( x_{m-1},x_{m}\right) =\prod_{m=1}^{n}\frac{k\!\left(x_{m}\right) }{k\!\left( x_{m-1}\right) }=\frac{k\!\left( x_{n}\right) }{k\!\left(x_{0}\right) }\end{equation*}

depends only on the initial and terminal states $\left( x_{0},x_{n}\right) $ and not on the full path $\left( x_{0},...,x_{n}\right)$ .

2.2. The infinitesimal generator

In what follows we always work with separable kernels. Moreover, we write $X_t ( x) $ for the process given in (7) to emphasize the dependence on the starting point x; that is, $X_t (x) $ designates the process with the above dynamics (7) and satisfying $ X_0 ( x) = x$ . If the value of the starting point x is not important, we shall also write $ X_t $ instead of $ X_t ( x)$ .

Under the separability condition, the infinitesimal generator of $X_{t}$ acting on bounded smooth test functions u takes the following simple form:

(9) \begin{equation}\left( \mathcal{G}u\right) \left( x\right) =-\alpha (x)u^{\prime }\left( x\right) +\frac{\beta(x) }{k(x) }\int_{x}^{\infty }k(y) u^{\prime }\left( y\right) dy , \qquad x \geq 0.\end{equation}

2.3. Relationship between decay–surge and growth–collapse processes

In this subsection, we exhibit a natural relationship between decay–surge population models, as studied here, and growth–collapse models as developed in Boxma et al. [Reference Boxma, Perry, Stadje and Zacks4], Goncalves et al. [Reference Goncalves, Huillet and Löcherbach16], Gripenberg [Reference Gripenberg17], and Hanson and Tuckwell [Reference Hanson and Tuckwell18]. Growth–collapse models describe deterministic population growth where at random jump times the population undergoes a catastrophe and falls to a random fraction of its previous size. More precisely, the generator of a growth–collapse process, having parameters $ ( \tilde \alpha, \tilde \beta, \tilde h )$ , is given for all smooth test functions by

(10) \begin{equation}\big(\tilde {\mathcal G} u\big) (x) = \tilde \alpha (x) u^{\prime} ( x) - \tilde \beta ( x)/\tilde h(x) \int_0^x u^{\prime}(y)\tilde h(y) dy , \qquad x \geq 0.\end{equation}

In the above formula, $ \tilde \alpha, \tilde \beta $ are continuous and positive functions on $ (0, \infty)$ , and $ \tilde h $ is positive and non-decreasing on $ (0, \infty)$ .

In what follows, consider a decay–surge process $X_{t}$ defined by the triple $\left( \alpha ,\beta ,k\right) $ and let $\widetilde{X}_{t}=1/X_{t}$ .

Proposition 1. The process $\widetilde{X}_{t}$ is a growth–collapse process as studied in [Reference Goncalves, Huillet and Löcherbach16] with triple $\left( \widetilde{\alpha },\widetilde{\beta },\widetilde{h}\right) $ given by

\begin{equation*}\widetilde{\alpha }(x) =x^{2}\alpha (1/x) , \qquad\widetilde{\beta }(x) =\beta (1/x), \qquad\widetilde{h}(x) =k(1/x), \qquad x > 0.\end{equation*}

Proof. Let u be any smooth test function, and study $u ( \tilde X_t) = u \circ g ( X_t) $ with $g ( x) = 1/x $ . By Itô’s formula for processes with jumps,

\begin{equation*} u \big( \tilde X_t \big) = u \circ g (X_t ) = u \big( \tilde X_0 \big) + \int_0^t {\mathcal G} (u \circ g ) (X_{t^{\prime}}) dt^{\prime} + M_t, \end{equation*}

where $M_t$ is a local martingale. We obtain

\begin{align*}{\mathcal G} (u \circ g ) (x) & = -\alpha(x)(u \circ g)^{\prime}(x)+ \frac{\beta(x)}{k(x)}\int_x^{\infty} (u \circ g)^{\prime}(y) k(y) dy \\ & = \frac{1}{x^2}\alpha(x)u^{\prime}\Big(\frac{1}{x}\Big) + \frac{\beta(x)}{k(x)}\int_x^{\infty} u^{\prime}\Big(\frac{1}{y}\Big) k(y) \frac{-dy}{y^2}.\end{align*}

Using the change of variable $y=1/x$ , this last expression can be rewritten as

\begin{equation*}\frac{1}{x^2}\alpha(x)u^{\prime}\Big(\frac{1}{x}\Big) -\frac{\beta(x)}{k(x)}\int_0^{1/x} u^{\prime}(z) k\Big(\frac{1}{z}\Big) dz= \widetilde{\alpha }(y)u^{\prime}(y)-\frac{\widetilde{\beta }(y)}{\widetilde{h}(y)}\int_{0}^{y}u^{\prime}(t)\widetilde{h}(t)dt,\end{equation*}

which is the generator of the process $\widetilde{X}_{t}$ .

In what follows we speak about the above relation between the decay–surge (DS) process X and the growth–collapse (GC) process $ \tilde X$ as the DS–GC duality. Some simple properties of the process X follow directly from the above duality, as we show next. Of course, the above duality does only hold up to the first time one of the two processes leaves the interior $ (0, \infty ) $ of its state space. Therefore, particular attention has to be paid to state 0 for $X_t$ , or equivalently to state $ + \infty $ for $\tilde X_t$ . Most of our results will only hold true under conditions ensuring that, starting from $ x > 0$ , the process $X_t$ will not hit 0 in finite time.

Another important difference between the two processes is that the simple transformation $ x \mapsto 1/x$ maps a priori unbounded sample paths $ X_t $ into bounded ones $ \tilde X_t $ (starting from $ \tilde X_0 = 1/x$ , almost surely, $ \tilde X_t \le \tilde x_t ( 1/x)$ —a relation which does not hold for X).

2.4. First consequences of the DS–GC duality

Given $X_{0}=x > 0$ , the first jump times both of the DS process $X_t$ , starting from x, and of the GC process $ \tilde X_t$ , starting from $ 1/x$ , coincide and are given by

\begin{equation*}T_{x}=\inf \big\{ t > 0 \,:\, X_t \neq X_{t-} | X_0 = x \big\}= \tilde T_{\frac1x}= \inf \left\{ t>0\,:\,\tilde X_{t}\neq \tilde X_{t-}| \tilde X_{0}=\frac1x\right\}.\end{equation*}

Introducing

\begin{equation*}\Gamma(x) \,:\!=\,\int_1^{x}\gamma(y) dy, \quad \mbox{ where } \gamma(x) \,:\!=\,\beta (x)/\alpha(x) , \ x > 0 ,\end{equation*}

and the corresponding quantity associated to the process $ \tilde X_t$ ,

\begin{equation*} \widetilde{\Gamma }(x) = \int_1^x \tilde \gamma(y)dy , \qquad \tilde \gamma (x)= \tilde \beta (x)/ \tilde \alpha(x) , \qquad x > 0,\end{equation*}

clearly $\widetilde{\Gamma }(x) =-\Gamma \!\left( 1/x\right) $ for all $ x > 0$ .

Arguing as in Sections 2.4 and 2.5 of [Reference Goncalves, Huillet and Löcherbach16], a direct consequence of the above duality is the fact that for all $t < t_0 ( x)$ ,

(11) \begin{equation} \mathbb{P}\!\left( T_{x} >t\right) =e^{-\int_{0}^{t}\beta\!\left( x_{s}(x) \right) ds}=e^{-\left[ \Gamma (x)-\Gamma \!\left( x_{t}(x) \right) \right] }.\end{equation}

To ensure that $\mathbb{P}\!\left( T_{x} <\infty \right) =1$ , in accordance with Assumption 1 of [Reference Goncalves, Huillet and Löcherbach16] we will impose the following condition.

Proof. By duality, we have

\begin{equation*} {\mathbb{P}} (X \mbox{ jumps before reaching 0} | X_0 = x ) = {\mathbb{P}} \!\left( \tilde X \mbox{ jumps before reaching $ + \infty $}| \tilde X_0 = \frac1x \right) = 1, \end{equation*}

as has been shown in Section 2.5 of [Reference Goncalves, Huillet and Löcherbach16], and this implies the assertion.

In particular, the only situation where the question of the extinction of the process X (either local or total) makes sense is when $t_0(x)<\infty $ and $\Gamma(0)>-\infty$ .

Example 4. We give an example where finite-time extinction of the process is possible. Suppose $\alpha(x) =\alpha_1 x^{a}$ with $\alpha_1>0$ and $a<1$ . Then $x_{t}\!\left(x\right)$ , started at x, hits 0 in finite time $t_{0}(x)=x^{1-a}/\left[ \alpha_1 \!\left( 1-a\right) \right] $ , with

\begin{equation*}x_{t}(x) =\left( x^{1-a}+\alpha_1 \!\left( a-1\right) t\right)^{1/\left( 1-a\right) };\end{equation*}

see [Reference Goncalves, Huillet and Löcherbach15]. Suppose $\beta(x) =\beta_1 >0$ is constant. Then, with $\gamma_1=\beta_1/\alpha_1>0$ ,

\begin{equation*}\Gamma(x) =\int_{1}^{x}\gamma(y) dy=\frac{\gamma_1 }{1-a}\left( x^{1-a}-1\right)\end{equation*}

with $\Gamma \!\left( 0\right) =-\frac{\gamma_1 }{1-a}>-\infty$ . Assumption 1 is not fulfilled, so X can hit 0 in finite time and there is a positive probability that $T_x =\infty $ . On this last event, the flow $x_{t}(x) $ has all the time necessary to first hit 0 and, if in addition the kernel k is chosen so that $k(0)=\infty$ , to go extinct definitively. The time of extinction $\tau(x) $ of X itself can be deduced from the renewal equation in distribution

\begin{equation*}\tau(x) \stackrel{d}{=}t_{0}(x) {\bf 1}_{\{T_x =\infty \}}+\tau ^{\prime }\left( Y\!\left( x_{T_x }\!\left(x\right) \right) \right) {\bf 1}_{\{T_x <\infty \}},\end{equation*}

where $\tau ^{\prime }$ is a copy of $\tau $ .

We conclude that for this family of models, X itself goes extinct in finite time. This is an interesting regime that we shall not investigate any further.

Let us come back to the discussion of Assumption 1. It follows immediately from Equation (13) in [Reference Goncalves, Huillet and Löcherbach16] that for $x>0$ , under Assumption 1 and supposing that $ t_0 ( x) = \infty $ for all $ x > 0$ ,

\begin{equation*}\mathbb{E}\!\left( T_{x} \right) =e^{-\Gamma(x) }\int_{0}^{x}\frac{dz}{\alpha(z)}e^{\Gamma(z) }.\end{equation*}

Clearly, when $x\rightarrow 0$ , $\mathbb{E}\!\left( T_{x}\right) \sim 1/\alpha(x) $ .

2.5. Classification of state 0

Recall that for all $ x > 0$ ,

\begin{equation*}t_{0}(x{)}=\int_{0}^{x}\frac{dy}{\alpha(y) }\end{equation*}

represents the time required for $x_{t}$ to move from $x>0$ to 0. So the following hold:

\begin{align*}&\text{ If } t_{0}(x{)}<\infty \mbox{ and } \Gamma ( 0 ) > - \infty , &\text{ state }0\text{ is accessible.} \\&\text{ If } t_{0}(x{)}=\infty \mbox{ or } \Gamma ( 0 ) = - \infty , &\quad \text{ state }0\text{ is inaccessible.}\end{align*}

We therefore introduce the following conditions, which apply in the separable case $K(x,y)=k(y)/k(x)$ .

Condition (R): $\beta \!\left( 0\right) >0$ and $K\!\left( 0,y\right) =k(y)/k\!\left( 0\right)>0 $ for some $y>0$ .

Condition (A): $\frac{\beta(0)}{k(0)}k(y)=0 $ for all $y > 0 $ .

State 0 is reflecting if Condition $\left( R\right) $ is satisfied, and it is absorbing if Condition (A) is satisfied.

This leads to four possible combinations for the boundary state 0:

1. 1. Condition (R) is satisfied, and $t_{0}(x)<$ $\infty $ and $ \Gamma (0 ) > - \infty\, :$ regular (reflecting and accessible).

2. 2. Condition (R) is satisfied, and $t_{0}(x)=$ $\infty $ or $ \Gamma ( 0 ) = - \infty\, :$ entrance (reflecting and inaccessible).

3. 3. Condition (A) is satisfied, and $t_{0}(x)<$ $\infty $ and $ \Gamma (0 ) > - \infty\, :$ exit (absorbing and accessible).

4. 4. Condition (A) is satisfied, and $t_{0}(x)=$ $\infty $ or $ \Gamma ( 0 ) = - \infty\, :$ natural (absorbing and inaccessible).

2.6. Speed measure

Suppose now an invariant measure (or speed measure) $\pi \!\left( dy\right) $ exists. Since we supposed $\alpha (x)>0$ for all $x>0$ , we necessarily have $x_{\infty }(x)=0$ for all $x>0$ , and so the support of $\pi $ is $[x_{\infty }(x)=0,\infty)$ . Thanks to our duality relation, by Equation (19) of [Reference Goncalves, Huillet and Löcherbach16] the explicit expression for the speed measure is given by $\pi ( dy) = \pi ( y) dy $ with

(12) \begin{equation}\pi(y) =C\frac{k(y) e^{\Gamma(y) }}{\alpha(y) }, \end{equation}

up to a multiplicative constant $C>0$ . If and only if this function is integrable at 0 and $\infty$ , $\pi(y) $ can be tuned to a probability density function.

Remark 6. (i) When $k(x) =e^{-\kappa _{1}x}$ , $\kappa _{1}>0$ , $\alpha(x) =\alpha _{1}x$ , and $\beta(x) =\beta_{1}>0$ constant, $\Gamma(y) =\gamma _{1}\log y$ , $\gamma_{1}=\beta _{1}/\alpha _{1}$ , and

\begin{equation*}\pi(y) =Cy^{\gamma _{1}-1}e^{-\kappa _{1}y},\end{equation*}

a Gamma $\left( \gamma _{1},\kappa _{1}\right) $ distribution. This result is well known, corresponding to the linear decay–surge model (a jump version of the damped Langevin equation) having an invariant (integrable) probability density; see Malrieu [Reference Malrieu26]. We shall show later that the corresponding process X is positive recurrent.

(ii) A less obvious power-law example is as follows. Assume $\alpha(x) =\alpha _{1}x^{a}$ ( $a>1$ ) and $\beta (x)=\beta _{1}x^{b}$ , $\alpha _{1},\beta _{1}>0$ , so that

\begin{equation*}\Gamma \!\left(y\right) =\frac{\gamma _{1}}{b-a+1}y^{b-a+1}.\end{equation*}

We have $\Gamma \!\left(0\right) =-\infty $ if we assume $b-a+1=-\theta $ with $\theta >0$ ; hence $\Gamma(y) =-\frac{\gamma _{1}}{\theta }y^{-\theta }$ . Taking $k\!\left(y\right) =e^{-\kappa _{1}y^{\eta }}$ , $\kappa _{1},\eta >0$ ,

\begin{equation*}\pi(y) =Cy^{-a}e^{-\left( \kappa _{1}y^{\eta }+\frac{\gamma _{1}}{\theta }y^{-\theta }\right) },\end{equation*}

which is integrable at both $y=0$ and $y=\infty $ . As a special case, if $a=2$ and $b=0$ (constant jump rate $\beta(x) $ ), $\eta =1$ ,

\begin{equation*}\pi(y) =Cy^{-2}e^{-\left( \kappa _{1}y+\gamma _{1}y^{-1}\right)},\end{equation*}

an inverse Gaussian density.

(iii) In Eliazar and Klafter [Reference Eliazar and Klafter13, Reference Eliazar and Klafter14], a special case of our model was introduced for which $k(y) =\beta(y) $ . In such cases,

\begin{equation*}\pi(y) =C\gamma(y) e^{\Gamma(y) },\end{equation*}

so that

\begin{equation*}\int_{0}^{x}\pi(y) dy=C\!\left( e^{\Gamma (x)}-e^{\Gamma \!\left( 0\right) }\right) =Ce^{\Gamma(x) },\end{equation*}

under the assumption $\Gamma \!\left( 0\right) =-\infty $ . If in addition $\Gamma \!\left( \infty \right) <\infty $ , $\pi(y) $ can be tuned to a probability density.

3.1. Hitting times

Fix $a < x < b$ . In what follows we shall be interested in hitting times of the positions a and b, starting from x, under the condition $\Gamma( \infty ) = \infty$ . Because of the asymmetric structure of the process (continuous motion downwards and up-moves by jumps only), these times are given by

\begin{equation*}\tau_{x, b } = \inf \{ t > 0 \,:\, X_t = b \} = \inf \{ t > 0 \,:\, X_t \le b ,X_{t- } > b \}\end{equation*}

and

\begin{equation*}\tau_{x, a } = \inf \{ t > 0 \,:\, X_t = a \} = \inf \{ t > 0 \ \,:\, \ X_t \le a \} .\end{equation*}

Obviously, $\tau_{a,a} = \tau_{b, b } = 0$ .

Let $T=\tau _{x,a}\wedge \tau _{x,b}$ . In contrast to the study of processes with continuous trajectories, it is not clear that $T<\infty $ almost surely. Indeed, starting from x, the process could jump across the barrier of height b before hitting a and then never enter the interval [0, b] again. So we suppose in the sequel that $T<\infty $ almost surely. Then

\begin{equation*}\mathbb{P}\!\left( \tau _{x,a}<\tau _{x,b}\right) +\mathbb{P}\!\left( \tau_{x,b}<\tau _{x,a}\right) =1.\end{equation*}

Proposition 4. Suppose $\Gamma (\infty )=\infty $ and that Assumption 2 does not hold. Let $0<a<x<b<\infty $ and suppose that $T= \tau _{x,a}\wedge \tau _{x,b}<\infty $ almost surely. Then

(18) \begin{equation}\mathbb{P}\!\left( \tau _{x,a}<\tau _{x,b}\right) = \frac{\int_{x}^{b}\frac{\gamma(y) }{k(y) }e^{{-}\Gamma(y) }dy}{\int_{a}^{b}\frac{\gamma(y) }{k(y) }e^{{-}\Gamma(y) }dy}. \end{equation}

Proof. Under the above assumptions, $s_1 (X_{t})$ is a local martingale and the stopped martingale $M_{t}=s_1(X_{T\wedge t})$ is bounded, which follows from the fact that $X_{T\wedge t}\geq a$ together with the observation that $s_1$ is decreasing, implying that $M_{t}\leq s_1(a)$ . Therefore from the stopping theorem we have

\begin{equation*}s_1 (x)=\mathbb{E}(M_0)=\mathbb{E}(s_1(X_0))=\mathbb{E}(M_T).\end{equation*}

Moreover,

\begin{equation*}\mathbb{E}(M_T)=s_1(a)\mathbb{P}(T=\tau _{x,a})+s_1(b)\mathbb{P}(T=\tau_{x,b}).\end{equation*}

But $\{T=\tau _{x,a}\}=\{\tau _{x,a}<\tau _{x,b}\}$ and $\{T=\tau_{x,b}\}=\{\tau _{x,b}<\tau _{x,a}\}$ , so that

\begin{equation*}s_1(x)=s_1(a)\mathbb{P}(\tau _{x,a}<\tau _{x,b})+s_1 (b)\mathbb{P}(\tau_{x,b}<\tau _{x,a}).\end{equation*}

Remark 9. Let $\tau _{x,[b,\infty )}=\inf \{t>0\,:\,X_{t}\geq b\}$ and $\tau_{x,[0,a]}=\inf \{t>0\,:\,X_{t}\le a\}$ be the entrance times to the intervals $[b,\infty )$ and $[0,a]$ . Observe that by the structure of the process, namely the continuity of the downward motion,

\begin{equation*}\{\tau _{x,a}<\tau _{x,b}\}\subset \{\tau _{x,a}<\tau _{x,[b,\infty)}\}\subset \{\tau _{x,a}<\tau _{x,b}\}.\end{equation*}

Indeed, the second inclusion is trivial since $\tau _{x,[b,\infty )}\le \tau_{x,b}$ . The first inclusion follows from the fact that it is not possible to jump across b and then hit a without touching b. Observing moreover that for $ a < x $ , $ \tau_{x, a} = \tau_{ x, [0, a ] }$ , (18) can therefore be rewritten as

(19) \begin{equation}\mathbb{P}\!\left( \tau _{x,[0,a]}<\tau _{x,[b,\infty )}\right) =\frac{s_{1}\left( x\right) -s_{1}\left( b\right) }{s_{1}\left( a\right)-s_{1}\left( b\right) } . \end{equation}

Now suppose that the process does not explode in finite time; that is, during each finite time interval, almost surely, only a finite number of jumps appear. In this case $\tau_{x,[b,\infty )}\overset{}{\to }+\infty $ as $b\to \infty$ . Then, letting $ b \to \infty $ in (19), we obtain for any $a>0$

(20) \begin{equation}\mathbb{P}\!\left( \tau _{x,a}<\infty \right) ={\frac{s_{1}(x)}{s_{1}(a)}}=\frac{\int_{x}^{\infty }\frac{\gamma(y) }{k(y) }e^{{-}\Gamma(y) }dy}{\int_{a}^{\infty }\frac{\gamma(y)}{k(y) }e^{{-}\Gamma(y) }dy}<1, \end{equation}

since $ \int_a^x \frac{\gamma(y)}{k(y) }e^{{-}\Gamma(y) }dy > 0 $ by the assumptions on k and $ \gamma$ .

As a consequence, we obtain the following.

Example 6. Choose $\alpha(x) =x^{2}$ , $\beta \!\left(x\right) =1+x^{2}$ , so that $\gamma(x) =1+1/x^{2}$ and $\Gamma(x) =x-1/x $ with $\Gamma \!\left(0\right) =-\infty $ and $\Gamma \!\left( \infty \right) =\infty$ .

Choose also $k(x) =e^{-x/2}$ . Then Assumption 2 is violated: the survival function of the big upward jumps decays too slowly in comparison to the decay of $ e^{ - \Gamma}$ . The speed density is

\begin{equation*}\pi(x) =C\frac{k(x) e^{\Gamma(x) }}{\alpha(x) }=Cx^{-2}e^{x/2 -1/x},\end{equation*}

which is integrable at 0 but not at $ \infty$ . The process explodes (has an infinite number of upward jumps in finite time) with positive probability.

3.2. First moments of hitting times

Let $a > 0$ . We are looking for positive solutions of

\begin{equation*}\left( \mathcal{G}\phi _{a}\right) \left( x\right) =-1, x \geq a,\end{equation*}

with boundary condition $\phi _{a}\left( a\right) =0$ . The above is equivalent to

\begin{equation*}\left( \mathcal{G}\phi _{a}\right) \left( x\right) =-\alpha (x)\phi _{a}^{\prime}(x) +\frac{\beta(x) }{k\!\left(x\right) }\int_{x}^{\infty }k(y) \phi _{a}^{\prime }\!\left(y\right) dy=-1.\end{equation*}

This is also

\begin{equation*}-k(x) \phi _{a}^{\prime}(x) +\gamma (x)\int_{x}^{\infty }k(y) \phi _{a}^{\prime}(y) dy=-\frac{k(x) }{\alpha(x) }.\end{equation*}

Putting $U(x) \,:\!=\,\int_{x}^{\infty }k(y) \phi_{a}^{\prime}(y) dy$ , the latter integro-differential equation reads

\begin{equation*}U^{\prime}(x) =-\gamma(x) U(x) -\frac{k(x) }{\alpha(x) }. \end{equation*}

Supposing that $\int^{+\infty} \pi (y ) dy < \infty $ (recall (12)), this leads to

\begin{align*}U(x) & = e^{-\Gamma(x) }\int_{x}^{\infty }e^{\Gamma(y) }\frac{k(y) }{\alpha(y) }dy , \\-U^{\prime}(x) & = \gamma(x) e^{-\Gamma \!\left(x\right) }\int_{x}^{\infty }e^{\Gamma(y) }\frac{k\!\left(y\right) }{\alpha(y) }dy+\frac{k(x) }{\alpha \!\left(x\right) }=k(x) \phi _{a}^{\prime }\left( x\right) ,\end{align*}

so that

(21) \begin{align} \phi _{a}\left( x\right) & = \int_{a}^{x}dy\frac{\gamma(y) }{k(y) }e^{-\Gamma(y) }\int_{y}^{\infty }e^{\Gamma\!\left( z\right) }\frac{k\!\left( z\right) }{\alpha(z) }dz+\int_{a}^{x}\frac{dy}{\alpha(y) } \nonumber \\& = \int_{a}^{\infty }dz\pi(z) \left[ s_1\!\left( a\right) -s_1\!\left( x\wedge z\right) \right] +\int_{a}^{x}\frac{dy}{\alpha \!\left(y\right) }.\end{align}

Notice that $[ a, \infty ) \ni x \mapsto \phi_a ( x) $ is non-decreasing and that $\phi_a ( x) < \infty $ for all $x > a > 0$ under our assumptions. Dynkin’s formula implies that for all $x > a $ and all $t \geq 0$ ,

(22) \begin{equation} \mathbb{E}_x ( t \wedge \tau_{x, a } )= \phi_a ( x)- \mathbb{E}_x \big( \phi_a \big(X_{t \wedge \tau_{x, a } } \big) \big).\end{equation}

In particular, since $\phi_a ( \cdot ) \geq 0$ ,

\begin{equation*}\mathbb{E}_x ( t \wedge \tau_{x, a } ) \le \phi_a ( x) < \infty ,\end{equation*}

so that we may let $t \to \infty $ in the above inequality to obtain by monotone convergence that

\begin{equation*}\mathbb{E}_x ( \tau_{x, a } ) \le \phi_a ( x) < \infty .\end{equation*}

Next we obtain from (22), using Fatou’s lemma, that

\begin{align*}\mathbb{E}_x ( \tau_{x, a } ) = \lim_{ t \to \infty } \mathbb{E}_x ( t\wedge \tau_{x, a } ) = \phi_a ( x) - \lim_{ t \to \infty } \mathbb{E}_x \big(\phi_a \big( X_{t \wedge \tau_{x, a } } \big) \big) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\\geq \phi_a ( x) -\mathbb{E}_x \Big( \liminf_{ t \to \infty } \phi_a \big( X_{t\wedge \tau_{x, a } } \big) \Big) = \phi_ a (x) ,\end{align*}

where we have used that $\liminf_{ t \to \infty } \phi_a ( X_{t \wedge\tau_{x, a } } ) = \phi_a ( a) = 0$ .

As a consequence we have just shown the following.

Remark 10. The last term $\int_{a}^{x}\frac{dy}{\alpha(y) }$ in the right-hand side of the expression for $\phi _{a}(x) $ in (21) is the time needed for the deterministic flow to first hit a starting from $x>a$ , which is a lower bound of $\phi _{a}(x) $ . Considering the tail function of the speed density $\pi(y) $ , namely

\begin{equation*}\overline{\pi }(y) \,:\!=\, \int_{y}^{\infty }e^{\Gamma(z) }\frac{k\!\left( z\right) }{\alpha(z) }dz,\end{equation*}

the first term in the expression for $\phi _{a}(x) $ is

\begin{equation*}\int_{a}^{x}-ds_{1}(y) \overline{\pi }(y) =-\left[s_{1}(y) \overline{\pi }(y) \right]_{a}^{x}-\int_{a}^{x}s_{1}(y) \pi(y) dy,\end{equation*}

emphasizing the importance of the pair $\left( s_{1}\!\left( \cdot \right),\pi \!\left( \cdot \right) \right) $ in the evaluation of $\phi _{a}\!\left(x\right) $ . If a is a small critical value below which the population can be considered in danger, this is the mean value of a ‘quasi-extinction’ event when the initial size of the population was x.

3.3. Mean first hitting time of 0

Suppose that $ \Gamma ( 0 ) > - \infty$ . Then for flows $x_{t}(x) $ that go extinct in finite time $t_0(x) $ , under the condition that $\int^{+\infty }\pi (y)dy<\infty$ , one can let $a\rightarrow 0$ in the expression for $\phi _{a}(x) $ to obtain

\begin{equation*}\phi _{0}(x) =\int_{0}^{x}dy\frac{\gamma(y) }{k(y) }e^{-\Gamma(y) }\int_{y}^{\infty }e^{\Gamma\!\left( z\right) }\frac{k\!\left( z\right) }{\alpha(z) }dz+\int_{0}^{x}\frac{dy}{\alpha(y) },\end{equation*}

which is the expected time to eventual extinction of X starting from x; that is, $\phi _{0}(x) =\mathbb{E}\tau _{x,0}$ .

The last term $\int_{0}^{x}\frac{dy}{\alpha(y) }=t_0\!\left(x\right) <\infty $ in the expression for $\phi _{0}(x) $ is the time needed for the deterministic flow to first hit 0 starting from $x>0$ , which is a lower bound of $\phi _{0}(x) $ .

Notice that under the conditions $ t_0 ( x) < \infty $ and $ k(0 ) < \infty$ , $\int_{0 }\pi (y)dy<\infty$ , so that $ \pi$ can be tuned into a probability. It is easy to see that $ \Gamma ( 0 ) > - \infty $ then implies that $ \phi_0 ( x) < \infty$ .

4.1. Irreducibility and Harris recurrence

In this section we impose the assumption that $ \Gamma ( \infty ) = \infty $ , so that non-trivial scale functions do exist. We also suppose that Assumption 2 holds, since otherwise the process either is transient at $ \infty $ or explodes in finite time. Then the function V introduced in Proposition 7 is a Lyapunov function. This is almost the Harris recurrence of the process; all we need to show is an irreducibility property that we are going to check now.

Theorem 1. Suppose that we are in the separable case, that $k \in C^1 $ , and that 0 is inaccessible, that is, $t_0 ( x) = \infty $ for all x. Then every compact set $ C \subset ] 0, \infty [ $ is ‘petite’ in the sense of Meyn and Tweedie. More precisely, there exist $ t > 0$ , $ \alpha \in (0, 1 ) $ , and a probability measure $ \nu $ on $ ( \mathbb{R}_+ , {\mathcal{B} }( \mathbb{R}_+) )$ such that

\begin{equation*} P_t (x, dy ) \geq \alpha \mathbf{1}_C (x) \nu (dy ) .\end{equation*}

Proof. Suppose without loss of generality that $ C = [a, b ] $ with $ 0 < a < b$ . Fix any $ t > 0$ . The idea of our construction is to impose that all processes $ X_s ( x)$ , $a \le x \le b $ , have one single common jump during $ [0, t ]$ . Indeed, notice that for each $ x \in C$ , the jump rate of $ X_s ( x) $ is given by $ \beta ( x_s(x) ) $ taking values in a compact set $ [ \beta_* , \beta^* ] $ , where $ \beta_* = \min \{ \beta ( x_s ( x ) ), 0 \le s \le t, x \in C \} $ and $ \beta^* = \max \{ \beta ( x_s ( x ) ), 0 \le s \le t, x \in C \}$ . Notice that $ 0 < \beta_* < \beta^* < \infty $ since $ \beta $ is supposed to be positive on $ (0, \infty )$ . We then construct all processes $ X_s (x) $ , $s \le t$ , $x \in C $ , using the same underlying Poisson random measure $ M$ . It thus suffices to impose that E holds, where

\begin{equation*} E = \{ M ( [0, t - \varepsilon] \times [ 0 , \beta^* ] ) = 0 , M ( [t- \varepsilon, t] \times [0, \beta_* ] ) = 1, M ( [t- \varepsilon, t] \times ] \beta_*, \beta^* ] ) = 0\} .\end{equation*}

Indeed, the above implies that up to time $ t - \varepsilon$ , none of the processes $ X_s ( x) $ , $x \in C$ , jumps. The second and third assumption imply moreover that the unique jump time, call it S, of M within $ [ t- \varepsilon, t ] \times [ 0 , \beta^* ] $ is a common jump of all processes. For each value of $x \in C$ , the associated process X (x) then chooses a new after-jump position y according to

(23) \begin{equation} \frac{1}{k( x_S (x)) }| k^{\prime}(y)| dy \mathbf{1}_{ \{ y \geq x_S ( x) \} }.\end{equation}

Case 1. Suppose k is strictly decreasing, that is, $|k^{\prime}| (y ) > 0 $ for all y. Fix then any open ball $ B \subset [ b, \infty [ $ and notice that $\mathbf{1}_{ \{ y \geq x_S ( x) \} } \geq \mathbf{1}_B ( y )$ , since $ b \geq x_S (x)$ . Moreover, since k is decreasing, $ 1/ k (x_S (x) ) \geq 1/ k ( x_t (a) )$ . Therefore, the transition density given in (23) can be bounded below, independently of x, by

\begin{equation*} \frac{1}{k( x_t (a)) } \mathbf{1}_B ( y ) | k^{\prime}(y)| dy = p \tilde \nu ( dy ) , \end{equation*}

where $\tilde \nu ( dy ) = c \mathbf{1}_B ( y ) | k^{\prime}(y)| dy $ , normalized to be a probability density, and where $ p = \frac{1}{k( x_t (a)) } / c$ . In other words, on the event E, with probability p, all particles choose a new and common position $ y \sim \tilde \nu ( dy ) $ and couple.

Case 2. Suppose $| k^{\prime}| $ is different from 0 on a ball B (but not necessarily on the whole state space). We suppose without loss of generality that B has compact closure. Then it suffices to take t sufficiently large in the first step so that $ x_{t- \varepsilon } (b) < \inf B$ . Indeed, this implies once more that $ \mathbf{1}_B ( y ) \le \mathbf{1}_{ \{ y \geq x_s ( x) \}} $ for all $ x \in C $ and for s the unique common jump time.

Conclusion. In any of the above cases, let $ \bar b \,:\!=\, \sup \{ x \,:\, x \in B \} < \infty $ and restrict the set E to

\begin{equation*} E^{\prime} = E \cap \{ M ( [ t - \varepsilon ] \times ] \beta_* , \bar b ] = 0 \} .\end{equation*}

Putting $ \alpha \,:\!=\, p \, \Pr ( E^{\prime} ) $ and

\begin{equation*} \nu (dy ) = \int_{ t _ \varepsilon }^t {\mathcal{L}} ( S | E^{\prime} ) (ds) \int \tilde \nu (dz) \delta_{ x_{ t- s} (z) } (dy )\end{equation*}

then allows us to conclude.

As a consequence of the above considerations we obtain the following theorem.

Example 9. A meaningful recurrent example consists of choosing $\beta \!\left(x\right) =\beta _{1}/x$ , $\beta _{1}>0$ (the surge rate decreases like $1/x $ ), $\alpha(x) =\alpha _{1}/x $ (finite-time extinction of $x_{t}$ ), $\alpha _{1}>0$ , and $k(y) =e^{-y}$ . In this case, all compact sets are ‘petite’. Moreover, with $\gamma _{1}=\beta _{1}/\alpha _{1}$ , $\Gamma(x) =\gamma_{1}x$ and

\begin{equation*}\pi(y) =\frac{y}{\alpha _{1}}e^{\left(\gamma_{1}-1\right)y},\end{equation*}

which can be tuned into a probability density if $\gamma _{1} < 1 $ . This also implies that Assumption 2 is satisfied, so that s can be used to define a Lyapunov function. The associated process X is positive recurrent if $ \gamma_1 < 1$ , null-recurrent if $ \gamma_1 = 1$ .

7.1. Hawkes processes

In the section we study the particular case $\beta \!\left(x\right) =\beta _{1}x$ , $\beta _{1}>0$ (the surge rate increases linearly with x), $\alpha(x) =\alpha _{1}x$ , $\alpha _{1}>0$ (exponentially declining population), and $k(y) =e^{-y}$ . In this case, with $\gamma _{1}=\beta _{1}/\alpha _{1}$ , $\Gamma (x)=\gamma _{1}x$ .

In this case, we remark that $\Gamma (0)=0>-\infty$ . Therefore there is a strictly positive probability that the process will never jump (in which case it is attracted to 0). However, we have $t_{0}(x)=\infty$ , so the process never hits 0 in finite time. Finally, $\beta (0)=0$ implies that state 0 is natural (absorbing and inaccessible).

Note that for this model,

\begin{equation*}\pi(y) =\frac{1}{\alpha _{1}y}e^{\left( \gamma _{1}-1\right) y}\end{equation*}

and we may take a version of the scale function given by

\begin{equation*}s (x) =\frac{\gamma _{1}}{1-\gamma _{1}}[e^{-\left( \gamma _{1}-1\right) y}]_{0}^{x } = \frac{\gamma _{1}}{1-\gamma _{1}} \left( e^{\left( 1 - \gamma _{1} \right) x}- 1 \right) .\end{equation*}

Clearly, Assumption 2 is satisfied if and only if $\gamma _{1}< 1$ . We call the case $ \gamma_1 < 1 $ subcritical, the case $ \gamma_1 > 1 $ supercritical, and the case $ \gamma_1 = 1 $ critical.

Supercritical case. It can be shown that the process does not explode almost surely, so that it is transient in this case (see Proposition 5). The speed density is integrable neither at 0 nor at $\infty $ .

Critical and subcritical case. If $\gamma_1 < 1$ , then $\pi $ is integrable at $+\infty $ , and we find

\begin{equation*}\phi _{a}(x) =-\left[ s(y) \overline{\pi }\left( y\right) \right] _{a}^{x}-\int_{a}^{x}s(y) \pi(y) dy+\frac{1}{\alpha _{1}}\log \frac{x}{a},\end{equation*}

where (with Ei the exponential integral function)

\begin{equation*}\int_{a}^{x}s(y) \pi(y) dy=\frac{\gamma _{1}}{\alpha _{1}\!\left( 1 - \gamma _{1} \right) }\left[ \ln \!\left( \frac{x}{a} \right) - \text{Ei}\!\left( \left( \gamma_{1}-1\right) x\right) + \text{Ei}\!\left( \left( \gamma _{1} -1\right) a\right) \right] .\end{equation*}

In the critical case $\gamma_1 = 1 $ , the hitting time of a will be finite without having finite expectation.

In both the critical and subcritical cases, that is, when $\gamma _{1}\leq 1$ , the process $X_t $ converges to 0 as $t \to \infty $ , as we shall show now.

Owing to the additive structure of the underlying deterministic flow and the exponential jump kernel, we have the explicit representation

(25) \begin{equation} X_t = e^{ - \alpha_1 t } x + \sum_{ n \geq 1 \,:\, S_n \le t } e^{ - \alpha_1(t- S_n ) } Y_n,\end{equation}

where the $(Y_n)_{n \geq 1 } $ are independent and identically distributed exponential random variables with mean 1, such that for all n, $Y_n $ is independent of $S_k$ , $k \le n $ , and of $Y_k$ , $k < n$ . Finally, in (25), the process $X_t$ jumps at rate $\beta_1 X_{t- }$ .

The above system is a linear Hawkes process without immigration, with kernel function $h( t) = e^{ - \alpha_1 t } $ and with random jump heights $(Y_n)_{n \geq 1 } $ (see [Reference Hawkes22]; see also [Reference Brémaud and Massoulié5]). Such a Hawkes process can be interpreted as an inhomogeneous Poisson process with branching. Indeed, the additive structure in (25) suggests the following construction:

• At time 0, we start with a Poisson process having time-dependent rate $\beta _{1}e^{-\alpha _{1}t}x$ .

• At each jump time S of this process, a new (time-inhomogeneous) Poisson process is born and added to the existing one. This new process has intensity $\beta _{1}e^{-\alpha _{1}(t-S)}Y$ , where Y is exponentially distributed with parameter 1, independent of what has happened before. We call the jumps of this newborn Poisson process jumps of generation $1$ .

• At each jump time of generation 1, another time-inhomogeneous Poisson process is born, of the same type, independently of anything else that has happened before. This gives rise to jumps of generation $2$ .

• The above procedure is iterated until it eventually stops, since the remaining Poisson processes do not jump any more.

The total number of jumps of any of the offspring Poisson processes is given by

\begin{equation*}\beta_1 \mathbb{E } (Y) \int_{S}^\infty e^{ - \alpha_1 (t- S ) } dt =\gamma_1.\end{equation*}

So we see that whenever $\gamma_1 \le 1 $ , we are considering a subcritical or critical Galton–Watson process which goes extinct almost surely, after a finite number of reproduction events. This extinction event is equivalent to the fact that the total number of jumps in the system is finite almost surely, so that after the last jump, $X_t$ just converges to 0 (without, however, ever reaching it). Notice that in the subcritical case $\gamma_1 < 1$ , the speed density is integrable at $ \infty$ , while it is not at 0, corresponding to absorption in $0$ .

An interesting feature of this model is that it can exhibit a phase transition when $\gamma _{1}$ crosses the value 1.

Finally, in the case of a linear Hawkes process with immigration we have $\beta ( x) = \mu + \beta_1 x $ , with $\mu > 0$ . In this case,

\begin{equation*}\pi( y) =\frac{1}{\alpha _{1}}e^{\left( \gamma _{1}-1\right) y } y^{ \mu/\alpha_1 - 1 },\end{equation*}

which is always integrable in 0 and which can be tuned into a probability in the subcritical case $\gamma_1 < 1$ corresponding to positive recurrence.

7.2. Shot-noise processes

Let $h\!\left( t\right) $ , $t\geq 0$ , with $h\!\left( 0\right) =1$ be a causal non-negative non-increasing response function translating the way shocks will attenuate as time passes in a shot-noise process. We assume $h\!\left(t\right) \rightarrow 0$ as $t\rightarrow \infty $ and

(26) \begin{equation}\int_{0}^{\infty }h\!\left( s\right) ds<\infty . \end{equation}

With $X_{0}=x\geq 0$ , consider then the linear shot-noise process

(27) \begin{equation}X_{t}=x+\int_{0}^{t}\int_{{\Bbb R}_{+}}yh\!\left( t-s\right) \mu \!\left(ds,dy\right) , \end{equation}

where, with $\left( S_{n};\,n\geq 1\right) $ being the points of a homogeneous Poisson point process with intensity $\beta$ , $\mu \!\left( ds,dy\right)=\sum_{n\geq 1}\delta _{S_{n}}\left( ds\right) \delta _{\Delta _{n}}\left(dy\right) $ (translating to independence of the shot heights $\Delta _{n}$ and occurrence times $S_{n}$ ). Note that, with $dN_{s}=\sum_{n\geq1}\Delta _{n}\delta _{S_{n}}\left( ds\right)$ , so with $N_{t}=\sum_{n\geq1}\Delta _{n}{\bf 1}_{\{ S_{n}\leq t \}} $ representing a time-homogeneous compound Poisson process with jump amplitudes $\Delta$ ,

(28) \begin{equation}X_{t}=x+\int_{0}^{t}h\!\left( t-s\right) dN_{s} \end{equation}

is a linearly filtered compound Poisson process. In this form, it is clear that $X_{t}$ cannot be Markov unless $h\!\left( t\right) =e^{-\alpha t}$ , $\alpha >0$ . We define

\begin{align*}\nu \!\left( dt,dy\right) & = {\mathbb{ P}}\!\left( S_{n}\in dt,\Delta _{n}\in dy\text{for some }n\geq 1\right) \\& = \beta dt\cdot {\mathbb{ P}}\!\left( \Delta \in dy\right) .\end{align*}

In the sequel, we shall assume without much loss of generality that $x=0$ .

The linear shot-noise process $X_{t}$ has two alternative equivalent representations, emphasizing its superposition characteristics:

\begin{align*}\left( 1\right) \text{ }X_{t} & = \sum_{n\geq 1}\Delta _{n}h\!\left(t-S_{n}\right) {\bf 1}_{\{ S_{n}\leq t\}}, \\\left( 2\right) \text{ }X_{t} & = \sum_{p=1}^{P_{t}}\Delta _{p}h( t-{\mathcal{S}}_{p}(t)),\end{align*}

where $P_t = \sum_{n\geq1} {\bf 1}_{\{ S_{n}\leq t\}}$ .

Both show that $X_{t}$ is the size at t of the whole decay–surge population, summing up all the declining contributions of the sub-families which appeared in the past at jump times (a shot-noise or filtered Poisson process model appearing also in physics and queuing theory; see [Reference Snyder and Miller32] and [Reference Parzen28]). The contributions $\Delta _{p}h\!\left( t-{\mathcal{S}}_{p}\left(t\right) \right) $ , $p=1, ...P_{t}$ , of the $P_{t}$ families to $X_{t}$ are stochastically ordered in decreasing sizes.

In the Markov case, $h\!\left( t\right) =e^{-\alpha t}$ , $t\geq 0$ , $\alpha >0$ , we have

(29) \begin{equation}X_{t}=e^{-\alpha t}\int_{0}^{t}e^{\alpha s}dN_{s}, \end{equation}

so that

\begin{equation*}dX_{t}=-\alpha X_{t}dt+dN_{t},\end{equation*}

showing that $X_{t}$ is a time-homogeneous Markov process driven by $N_{t}$ , known as the classical linear shot-noise. This is clearly the only choice of the response function that makes $X_{t}$ Markov. In that case, by Campbell’s formula (see [Reference Parzen28]),

\begin{align*}\Phi _{t}^{X}\!\left( q\right) &\,:\!=\, {\mathbb{ E}}e^{-qX_{t}}=e^{-\beta\int_{0}^{t}\left( 1-\phi _{\Delta }\left( qe^{-\alpha \!\left( t-s\right)}\right) \right) ds} \\& = e^{-\frac{\beta }{\alpha }\int_{e^{-\alpha t}}^{1}\frac{1-\phi _{\Delta}\left( qu\right) }{u}du} \quad \text{ where }e^{-\alpha s}=u,\end{align*}

with

\begin{equation*}\Phi _{t}^{X}\left( q\right) \rightarrow \Phi _{\infty }^{X}\left( q\right)=e^{-\frac{\beta }{\alpha }\int_{0}^{1}\frac{1-\phi _{\Delta }\left(qu\right) }{u}du} \quad \text{ as }t\rightarrow \infty .\end{equation*}

The simplest explicit case is when $\phi _{\Delta }\left( q\right) =1/\left(1+q/\theta \right) $ (otherwise $\Delta \sim $ Exp $\left( \theta \right) $ ), so that, with $\gamma =\beta /\alpha$ ,

\begin{equation*}\Phi _{\infty }^{X}\left( q\right) =\left( 1+q/\theta \right) ^{-\gamma },\end{equation*}

the Laplace–Stieltjes transform of a Gamma $\left( \gamma ,\theta \right) $ -distributed random variable $X_{\infty }$ , with density

\begin{equation*}\frac{\theta ^{\gamma }}{\Gamma \!\left( \gamma \right) }x^{\gamma-1}e^{-\theta x}\text{, } \qquad x>0.\end{equation*}

This time-homogeneous linear shot-noise process with exponential attenuation function and exponentially distributed jumps is a decay–surge Markov process with triple

\begin{equation*}\left( \alpha(x) =-\alpha x;\text{ }\beta (x)=\beta ;\text{ }k(x) =e^{-\theta x}\right) .\end{equation*}

Shot-noise processes being generically non-Markov, there is no systematic relationship between decay–surge Markov processes and shot-noise processes. In [Reference Eliazar and Klafter14], it is pointed out that decay–surge Markov processes could be related to the maximal process of nonlinear shot noise; see Eliazar and Klafter [Reference Eliazar and Klafter13, Reference Eliazar and Klafter14].