Escape probabilities of compound renewal processes with drift

TL;DR

This paper analyzes the escape probabilities of compound renewal processes with drift, providing integral equation solutions and explicit formulas for specific cases like Erlang and hypo-exponential arrivals, with applications in actuarial science.

Contribution

It introduces a unified approach to compute escape probabilities for compound renewal processes with drift, including new solutions for Erlang, hypo-exponential, and rational Laplace transform cases.

Findings

Explicit solutions for escape probabilities with Erlang and hypo-exponential arrivals

Identification of solvable cases with two-sided jumps

Connection to scale functions of diffusion processes

Abstract

We consider the problem of determining escape probabilities from an interval of a general compound renewal process with drift. This problem is reduced to the solution of a certain integral equation. In an actuarial situation where only negative jumps arise we give a general solution for escape and survival probabilities under Erlang$(n)$ and hypo-exponential arrivals. These ideas are generalized to the class of arrival distributions having rational Laplace transforms. In a general situation with two-sided jumps we also identify important families of solvable cases. A parallelism with the "scale function" of diffusion processes is drawn.

Tables

Table 1. Table 1: The EP N b ​ ( x ) subscript 𝑁 𝑏 𝑥 N_{b}(x) in terms of the decomposition ( 3.21 ) and the characteristics of F 𝐹 F and H 𝐻 H . Recall that the classes ℳ ℳ \mathcal{M} and ℋ ℋ \mathcal{H} are defined in ( 5.1 ) and ( 6.15 ). Symbol ⋆ ⋆ \star denotes a non-null component. We also list the IE appropriate to particular cases and the solution.

$q_1$	$q_2$	$p_1$	$p_2$	Jumps	$F$	$H$	IE	$N_b\left(x\right)$	$N_b\left(x\right)$
$\star$	$\star$	0	$0$	One-sided	$ℰ\left(\lambda \right)$	$-$	(4.1)	(4.4)	$\pi \left(x\right)/\pi \left(b\right)$
$\star$	$\star$	0	$0$	One-sided	Hypoexp.	$-$	(4.17)	(4.29)	$\Delta (x,b)/\Delta (b,b)$
$\star$	$\star$	0	$0$	One-sided	$\in ℳ$	$-$	(5.10)	(5.19)	$\Delta (x,b)/\Delta (b,b)$
$\star$	$\star$	0	$\star$	Two-sided	$\in ℳ$	$-$	(6.1)	(6.3)	$\Delta (x,b)/\Delta (b,b)$
$0$	$\star$	0	$\star$	Two-sided	$-$	$-$	(3.23)	(3.22)	$\pi \left(b-x\right)$
$0$	$\star$	$\star$	$\star$	Two-sided	$-$	$-$	(6.9)	(6.11)	$\pi \left(b-x\right)$
$\star$	$\star$	$\star$	$\star$	Two-sided	$ℰ\left(\lambda \right)$		(3.1)	$-$
$\star$	$\star$	$\star$	$\star$	Two-sided	$ℰ\left(\lambda \right)$	$\in \mathscr{H}$		(6.23)	$\Delta (x,b)/\Delta (b,b)$
$\star$	$\star$	$\star$	$\star$	Two-sided	$-$	$-$	(3.19)	$-$

Equations

$$X_t=x+ct+Y_t,Y_t=\sum _{n=0}^{N_t}{J}_n,$$

$$N:(0,b)\times\mathbb{R}^{+}\to[0,1],\ (x,b)\mapsto N_{b}(x)=\mathbb{P}\Big{(}\tau^{b}<\tau^{0}|X_{0}=x\Big{)}$$

$$\underset{x\to b^{-}}{\lim }{N}_b(x)\equiv N_b(b^{-})=1$$

$$\{{\tau }_1\ge t_b\}\cup \{{\tau }_1<t_b,J_1\ge b-x\}\subset {\mathbf{U}}_b^x\subset \{{\tau }_1<t_b,J_1\le -b{\}}^c\text{ and }{\mathbf{U}}_b^x\subset {\mathbf{U}}_b^y$$

$$\infty =X_{{\tau }^{\infty }}\equiv x+c{\tau }^{\infty }+J_1+\cdots +J_{N_{{\tau }^{\infty }}}\text{ a.s. which implies }{N}_{{\tau }^{\infty }}=\infty$$

$$\mathbb{E}{e}^{-s({\tau }_1+\cdots +{\tau }_{N_{{\tau }^{\infty }}})}=\underset{n\to \infty }{\lim }(\mathbb{E}{e}^{-s{\tau }_1}{)}^n=0,s>0$$

$$\underset{b\to \infty }{\lim }\{{\tau }^b<{\tau }^0\}=\{{\tau }^0=\infty \}\text{ a.s. }$$

$$\underset{b\to\infty}{\lim}N_{b}(x)=\underset{b\to\infty}{\lim}\mathbb{P}(\mathbf{U}_{b})=\mathbb{P}(\underset{b\to\infty}{\lim}\mathbf{U}_{b})=\mathbb{P}\Big{(}X_{t}>0,\forall t\Big{)}\equiv S(x)$$

$$\mathbb{P}\Big{(}\tau^{b+d}<\tau^{a+d}|X_{0}=x+d\Big{)}=\mathbb{P}\Big{(}\tau^{b}<\tau^{a}|X_{0}=x\Big{)}=N_{b-a}(x-a)$$

$$\mathbb{P}\Big{(}\tilde{\tau}^{b}<\tilde{\tau}^{0}|X_{0}=x\Big{)}=1-N_{b}(b-x)$$

$$\mathbb{P}\Big{(}X_{t}\in dz|X_{r}=y\Big{)}=\delta(z-y-c(t-r))dz$$

$$\mathbb{P}\Big{(}X_{\mathbf{t}_{n}}\leq z|X_{\mathbf{t}_{n}^{-}}=y\Big{)}=\mathbb{P}\Big{(}J_{n}\leq z-y\Big{)}=H(z-y)$$

$$N(x)=1-N(b-x)$$

$$N(x):={\mathbb{P}}^x({\mathbf{U}}_b)={\mathbb{P}}^{\theta (x)}(\theta \circ {\mathbf{U}}_b)=1-{\mathbb{P}}^{b-x}({\mathbf{U}}_b)=1-N(b-x)$$

$$\mathbb{P}\Big{(}\tau^{b}<\tau^{0}|X_{r}=x\Big{)}=\mathbb{P}\Big{(}\tau^{b}<\tau^{0}|X_{0}=x\Big{)}$$

$$N_{b}(x,r,z):=\mathbb{P}\Big{(}\tau^{b}<\tau^{0}|X_{r}=x,B_{r}^{-}=z\Big{)}$$

$$\mathbb{P}\Big{(}\tau^{b}<\tau^{0}|\mathcal{F}_{r}\Big{)}=\mathbb{P}\Big{(}\tau^{b}<\tau^{0}|X_{r},B_{r}^{-}\Big{)}=N_{b}(X_{r},r,B_{r}^{-})$$

$$\mathcal{F}_{r}=\sigma\big{(}X_{s},s\leq r\big{)}=\sigma\Big{(}B_{r}^{-},N_{r},\mathbf{t}_{0},\dots\mathbf{t}_{N_{r}},J_{1},\dots,J_{N_{r}}\Big{)}$$

$$\text{ and }\mathcal{N}_{r}=\sigma\big{(}N_{s},s\leq r\big{)}=\sigma\Big{(}N_{r},\mathbf{t}_{0},\dots\mathbf{t}_{N_{r}}\Big{)}$$

$$X_{t+{\mathbf{t}}_n}=x+c(t+{\mathbf{t}}_n)+\sum _{{\mathbf{t}}_j\in (0,{\mathbf{t}}_n+t]}{J}_j=X_{{\mathbf{t}}_n}+ct+\sum _{{\mathbf{t}}_j\in ({\mathbf{t}}_n,{\mathbf{t}}_n+t]}{J}_j$$

$${\mathcal{A}}_r=\{X_{s_1}=x_1,\dots X_{s_n}=x_n,B_r^{-}=z,r-z\le s_1<\dots s_n\le r\}$$

$$\mathcal{\sigma}(\mathcal{A}_{r})=\sigma\big{(}X_{r},B_{r}^{-}\big{)}$$

$$\mathbb{P}\Big{(}\tau^{b}<\tau^{0}|\sigma(\mathcal{A}_{r})\vee\mathcal{F}_{r-B_{r}^{-}}\Big{)}=\mathbb{P}\Big{(}\tau^{b}<\tau^{0}|\sigma(\mathcal{A}_{r})\Big{)}=\mathbb{P}\Big{(}\tau^{b}<\tau^{0}|X_{r},B_{r}^{-}\Big{)}\square$$

$$\Big{(}\lambda-c\partial_{x}\Big{)}N(x)=\lambda\bar{H}(b-x)+\lambda\int_{-x}^{b-x}N(x+y)dH(y),\ -\infty<x<\infty$$

$$\text{Besides }\mathbb{P}\Big{(}\tau^{b}<\tau^{0}|\mathcal{F}_{r}\Big{)}=\mathbb{P}\Big{(}\tau^{b}<\tau^{0}|X_{r}\Big{)}:=N(X_{r})$$

$$\Psi\mapsto\mathbf{G}\Psi(x)=c\partial_{x}\Psi(x)+\lambda\int_{\mathbb{R}}\big{(}\Psi(x+y)-\Psi(x)\big{)}dH(y)$$

$$\mathbb{P}^{x}\Big{(}X_{{\bm{\tau}}}\in[b,\infty)\Big{)}=N(x)\mathbf{1}_{(0,b)}(x)+\mathbf{1}_{[b,\infty)}(x)+0\mathbf{1}_{(-\infty,0)}(x)\equiv\mathbf{N}(x)$$

$$\frac{\partial \Psi }{\partial t}+\mathbf{G}\Psi (t,y)=m(t,y),y\in (0,b)\text{ and }$$

$$\Psi (t,y)=n(t,y),y\notin (0,b)$$

$$M_{t}:=\Psi(t,X_{t})]-\Psi(0,x)-\int_{0}^{t}\Big{(}\frac{\partial}{\partial s}+\mathbf{G}\Big{)}\Psi(s,X_{s})\,\mathrm{d}s.$$

Full text

60G55,

60K05,

60J75,

Escape and ruin probabilities, renewal reward process,

Integral equations,

arXiv:0000.0000 \startlocaldefs

\endlocaldefs

,

and

t1 Spanish Agencia Estatal de Investigación FIS2016-78904 and European Fondo Europeo de Desarrollo Regional (AEI/FEDER, UE) C3-2-P t2 Agència de Gestió d’Ajuts Universitaris i de Recerca (AGAUR), 2017SGR1064.

1 Introduction

The problem of determining escape probabilities from an interval of general diffusions is a classical issue solved in terms of the “scale and speed functions” (see [1, 2, 3] for an overview). Unfortunately, no such well established theory exists for compound renewal processes with drift. Concretely we consider a random process $(X_{t})_{t\geq 0}$ on a probability space $(\Omega,\mathcal{G},\mathbb{P})$ whose dynamics combines uniform motion with speed $c\in\mathbb{R}$ and sudden jumps $J_{n}\in\mathbb{R}$ at time epochs $\mathbf{t}_{n},n\in\mathbb{N}$ triggered by a renewal process $(N_{t})_{t\geq t_{0}}$, where $N_{t}=\#\{\mathbf{t_{n}}\in\mathbb{T}:0<\mathbf{t_{n}}\leq t\}$ counts the number of “events” $\mathbf{t}_{n},n=1,\dots\infty$ “observed” in the time window $(0,t]$ and $\mathbb{T}=\{\mathbf{t}_{0},\mathbf{t}_{1},\dots\mathbf{t}_{n},\dots\}\subset\mathbb{R}^{+}$. We define $\mathbf{t}_{0}=0$ and $x=X_{t_{0}}$. Thus

[TABLE]

When $c=0$ the resulting “renewal reward process” $X_{t}=\sum_{n=0}^{N_{t}}J_{n}$ has a prominent role in reliability and system maintenance. It also describes earthquake shocks [4] or stock markets where sudden price changes are allowed, [5].

The prototype model of risk theory to describe the cash flow $(X_{t})_{t\geq t_{0}}$ at an insurance company results when a drift $c>0$ is incorporated to account for the constant premium’s rate. By contrast claims arrive according to a renewal reward process $Y_{t}=\sum_{n=0}^{N_{t}}J_{n}$ with arrivals $\mathbf{t}_{n},\mathbf{t}_{n}<\mathbf{t}_{n+1}$ and sizes (or ”severities”) $J_{n}<0$. This classical risk reserve process was first introduced by Cramer-Lundberg under Poissonian arrivals $N_{t}\sim\mathcal{P}(\lambda t)$ [6] and generalized to general renewals by Sparre-Andersen, cf. [7]. It is usually complemented with the “net profit condition” $c\mathbb{E}\tau_{1}+\mathbb{E}J_{1}>0$ (NPC) $-$see [8, 9] for general background. Even such simplified situation is far from trivial and during the last two decades substantial research has been devoted to this topic: Ruin probabilities with Poisson arrivals have been studied in [6]. Under Erlang $\Gamma(2,\lambda)$ arrivals they can be represented as a compound geometric random variable, cf. [10, 11]. See also [12, 13]. The distribution of the time to ruin under Erlang times is considered in[14, 15, 16, 17]. Ruin probabilities under more general settings like Lévy and stable processes appear in the interesting papers [18, 19]. See also[20, 21, 22, 23, 24]. However far less is known about two-barrier exit probabilities even under the assumption $cJ_{1}<0$.

In this work such one-sided jump restriction $cJ_{j}<0$ is not required. To the best of our knowledge little is known about such general models even though they occur naturally in other physical contexts: Energy dissipation in defective nonlinear optical fibers is described by (1.1) where $c>0$ and $J_{j}>0$ account for Energy losses due to damping and inhomogeneities respectively(see [25]). Further motivation is given by the description of temporally aggregated rainfall in meteorology and hydrology contexts, see [26]. Here $(X_{t})$ measures rainfall accumulated at a dam with $J_{n}\geq 0$ representing rainfall intensity from the $j-$th shower while a term $ct\in\mathbb{R}$ accounts for the overall constant water inflow rate due to the opposite effects of evaporation, water consumption and melting of ice and inflow of water (hence both cases $c>0,<0$ might appear). In a different context, (1.1) also models the dynamics of snow depth on mountain hillsides, see [27]. Here both positive and negative jumps may occur due to snowfalls and, respectively, avalanches. The drift term $c<0$ accounts for snow melting during no-snow days. Finally, the space dynamics of bacteria and several other living organisms is described by a renewal process with a linear drift term, see [28]. Exit times are naturally related to the question of whether certain levels will be attained.

This paper is structured as follows. Let $a<x_{0}<b$ be two fixed levels and call $\tau^{a}=\inf\{t>0:X_{t}\leq a\}$, $\tau^{b}=\inf\{t>0:X_{t}\geq b\}>0$. We study two-barrier escape probabilities \mathbb{P}^{x}\Big{(}\tau^{b}<\tau^{a}\Big{)}, the probability that starting from $x\in(a,b)$ the process (1.1) exits $(a,b)$ via the upper barrier, when both positive and negative jumps occur. By means of renewal arguments we show that the basic EP solves a certain linear Fredholm integral equation (IE) with non-constant coefficients, cf. eq. (3.19). (We use $\mathbb{P}^{x}(A)\equiv\mathbb{P}(A|X_{0}=x),A\in\mathcal{F}_{\infty}$). We note that for pure jump Markov processes escape probabilities (EP) have been considered by extension of Feller ideas and Dynkin’s formula. Some ideas in this regard appear in [29, 30, 31]. However the difficulty of the resulting Dirichlet problem has prevented much progress for the solution (nevertheless, in a remarkable paper Bertoin ([32]) considers exit probabilities for one sided (i.e. without positive jumps) stable Lévy processes). The formalism of Feller-Markov semigroups is not generally applicable here since (1.1) is not Markov (nevertheless such theory is briefly used). Such lack of Markovianess implies that relevant probabilities depend on the accessible information. We also study how *accumulated information affects more general EP * of the form \mathbb{P}\Big{(}\tau^{b}<\tau^{0}\mid\mathcal{F}_{r}\Big{)} where $\mathcal{F}_{r}\equiv\sigma(X_{t},t\leq r)$ is the information $\sigma-$ field and $r$ is an arbitrary epoch of time.

Once established that all EP are codified in terms of the solution of a Fredholm IE we devote our interest to obtaining solutions for the previous IE. Unfortunately, in a general situation a closed form solution is not possible. Thus we attempt to classify the variety of cases that may arise (see table 1) and clarify the role of different jump contributions. For ample classes of data we derive simplified equations and give the corresponding solution (sections 4-6). Due to its importance, a great deal of interest is devoted to the case where support $(cJ_{1})\subset(-\infty,0)-$ the risk model. Under Poisson arrivals we give (section 4) a general solution for the EP. We find the factorization, see (4.3) \mathbb{P}^{x}\Big{(}\tau^{b}<\tau^{a}\Big{)}=\pi(x-a)/\pi(b-a) for a certain $\pi:\mathbb{R}^{+}\to\mathbb{R}$ that depends on the jump distribution. $\pi$ can be identified with the survival probability $S(x)$ when it exists.

The analysis is then extended to hypo-exponential arrivals (sums of $n$ independent exponential variables with different rates) and hence in particular to Erlang arrivals $\tau_{1}\sim\Gamma(n,\lambda)$. We show that \mathbb{P}^{x}\Big{(}\tau^{b}<\tau^{0}\Big{)}=\frac{\Delta(x,b)}{\Delta(b,b)} with $\Delta(x,b)=\det{\bm{\Theta}}(x,b)$ for a certain matrix $({\bm{\Theta}})_{n\times n}$.

In section 5 we develop a formalism to deal with an ample class of arrival distributions. We prove that if $\tau_{1}\sim F$ has rational Laplace transform and support $J_{1}\subset(-\infty,0)$ one can derive an equivalent integro-differential equation, amenable to Laplace transform. We discuss how to incorporate appropriate boundary conditions at $x=b$ that pin down the EP. The previous representation still holds with a far more complicated matrix ${\bm{\Theta}}$. These ideas generalize to a double barrier situation previous studies regarding the survival probability $-$which could be recovered letting $b\to\infty$.

Section 6 considers the problem of solving the corresponding IE under a general situation where jumps can take both signs. We identify important cases where such task can be accomplished:

When support $J_{1}\subset(-\infty,-b)\cup[0,\infty)$ the EP can be determined in closed form, regardless the distribution of arrival times and jump sizes.

The case support $J_{1}\subset(-\infty,-0)\cup[b-x,\infty)$ is also remarkable: the solution is a natural generalization of the risk model of sections 4,5. Finally under Poissonian arrivals and jumps with rational characteristic function the EP satisfies a simple ordinary differential equation. In Table 1 we summarize our results for the EP. Note how scale functions only appear in the very particular case when $(X)$ is Lévy with negative jumps. Ideas for mean escape times off $(a,b)$ appear in [25, 33].

The appendices are devoted to establish several facts that codify densities in terms of differential equations. We assume familiarity with Schwartz tempered distribution theory and Banach’s fixed point theorems.

Given $X_{0}=x$ then $t_{b}=(b-x)/c$ is the time remaining to reach the boundary $b$ when no jumps happen. Besides $A^{c}$ is the complementary event of $A\in\Omega$. Given the cdf $F$, $\bar{F}=1-F$ denotes its tail; if $g:\mathbb{R}\to\mathbb{R}$ is Borel measurable and and $\mu_{F}$ the Lebesgue-Stieltjes measure associated to $F$ we set $\int_{\mathbb{R}}g(x)dF(x):=\int_{\mathbb{R}}g(x)d\mu_{F}(x)$. The Laplace transform (LT) of $g:\mathbb{R}^{+}\to\mathbb{R}$ is denoted as $\mathcal{L}(g)(s)\equiv\hat{g}(s):=\int_{0}^{\infty}g(x)e^{-sx}dx$.

2 Effect of the accumulated information on escape probabilities

2.1 General properties

Here we study properties of \mathbb{P}\Big{(}\tau^{b}<\tau^{a}|X_{0}=x\Big{)}\equiv N(x;a;b) in terms of parameters $a<x<b$. We make the natural assumptions on $(X_{t})_{t\geq 0}$ (see (1.1))

Assumption 1**.**

(A1) Interarrival times $\tau_{n}\equiv\mathbf{t}_{n}-\mathbf{t}_{n-1}>0,n=1,\dots\infty$ are i.i.d. nonnegative r.v. with common distribution function $F$, namely $\tau_{n}\overset{\text{i.i.d}}{\sim}F$.

Assumption 2**.**

(A2) $(J_{n})\stackrel{{\scriptstyle\text{i.i.d.}}}{{\sim}}H$ define an i.i.d sequence with a common cdf $H$.

Assumption 3**.**

(A3) The sequence $(J_{1},\dots J_{n},\dots)$ is independent of the underlying renewal process $(N_{t})_{t\geq t_{0}}$.

Assumption 4**.**

(A4) Process $(N_{t})_{t\geq t_{0}}$ has filtration \mathcal{N}_{t}=\sigma\big{(}N_{s},s\leq t\big{)} while $(X_{t})_{t\geq t_{0}}$ has filtration \mathcal{F}_{t}=\sigma\big{(}X_{s},s\leq t\big{)}.

Proposition 1**.**

Let $c>0$, $\tau^{0}\equiv\tau^{0}_{-}=\inf\{t>0:X_{t}\leq 0\}$, $\tau^{b}\equiv\tau^{b}_{+}=\inf\{t>0:X_{t}\geq b\}>0$. The function $N_{b}(x)$ defined by

[TABLE]

is monotone in both variables. If $t_{b}=(b-x)/c$ it satisfies

[TABLE] 2. 2.

$N_{b}(x)\leq N_{b}(y)\leq N_{b}(b^{-})=1,\ 0<x\leq y<b$** 3. 3.

N_{b^{\prime}}(x)\geq N_{b}(x)\geq N_{b\to\infty}(x)=\mathbb{P}\Big{(}X_{t}>0,\forall t\Big{)}:=S(x),\ x<b^{\prime}\leq b** 4. 4.

\mathbb{P}\Big{(}\tau^{0}<\tau^{b}|X_{0}=x\Big{)}=1-N_{b}(x)** 5. 5.

$0<1-F(t_{b}^{-})H(b^{-}-x)\leq N_{b}(x)\leq 1-F(t_{b}^{-})H(-b)$** 6. 6.

If assumptions A5,A6 below hold $x\mapsto N_{b}(x)$ is continuous.

Proof. Let $\mathbf{U}_{b}^{x}=\{\tau^{b}<\tau^{0}\}$ be the event that, starting from $x$ at $t=0$, the process escapes $(0,b)$ through the upper end. Note that $\{\tau^{b}<\tau^{0}\}=\{X_{{\bm{\tau}}}\in[b,\infty)\}$ where ${\bm{\tau}}\equiv\tau^{0}\wedge\tau^{b}<\infty$ w.p. $1$. As $x$ grows so it does $(X_{t})$, see (1.1) and hence the sequence $x\mapsto\{\mathbf{U}_{b}^{x}\}$ is increasing while $b\mapsto\{\mathbf{U}_{b}^{x}\}$ decreases as $b$ grows. Clearly for $c>0,x<y$

[TABLE]

which implies (5). Letting $x\to b^{-}$ then $F(t_{b}^{-})H(b^{-}-x)\to 0$ and (2.2) follows.

We next show $\tau^{\infty}:=\underset{b\to\infty}{\lim}\tau^{b}=\infty$ a.s. Indeed if $\mathbb{P}(\tau^{\infty}<\infty)>0$ then on $\{\tau^{\infty}<\infty\}$

[TABLE]

namely, the process $(N_{t})$ explodes in finite time. Besides

[TABLE]

This implies $\tau_{1}+\dots+\tau_{N_{\tau^{\infty}}}=\infty$ w.p. $1$ a contradiction. Hence $\lim\tau^{\infty}=\infty$ and

[TABLE]

Sequential continuity of probabilities gives (we drop below the index $x$)

[TABLE]

Finally since ${\bm{\tau}}<\infty$ w.p. $1$. then, up to a null set, $\{\tau^{0}<\tau^{b}\}=\mathbf{U}_{b}^{c}$. (The proof of Item 6 is deferred to Appendix A).

2.1.1 Symmetry properties of the escape probability

Proposition 2**.**

For $x\in(a,b)$, $d\in\mathbb{R}$

[TABLE] 2. 2.

The “reversed” process $\tilde{X}_{t}=x+\tilde{c}t+\sum_{n=0}^{N_{t}}\tilde{J}_{n}$, where $\tilde{c}:=-c$ and $\tilde{J}_{n}:=-J_{n}$ satisfies

[TABLE]

Proof. Note that if there are no jumps in the time interval $(r,t)$ then

[TABLE]

where $\delta$ is Dirac delta with a mass at $y+c(t-r)$. Besides

[TABLE]

Hence it is clear that $(X_{t})$ is a spatially homogeneous process $-\mathbb{E}^{y}(g(X_{t}))=\mathbb{E}^{0}(g(y+X_{t})),\forall y-$ so, conditional on starting at $x$ the distribution of $\tau^{b}$ can only depend on $b-x$. Choosing $d=-a$ (2.4) is obtained. (2.5) follows noting that $\tilde{X}$ is obtained reflecting sample paths of $X$ over the line $X=x$, and hence $\tilde{\tau}^{b}=\tau^{2x-b},\tilde{\tau}^{a}=\tau^{2x-a}$. We finish using item (4).

Remark 1**.**

The invariance of $(X_{t})_{t\geq 0}$ under the group of all space translations and reflections permits with no loss of generality to suppose that $a=0$ and that $c>0$.

Proposition 3**.**

Suppose $c=0$. Then condition (2.2) needs not to hold as the limit $N(b)\equiv\underset{x\to b^{-}}{\lim}\underset{c\to 0}{\lim}N(x)\leq\underset{c\to 0}{\lim}\underset{x\to b^{-}}{\lim}N(b)$ need not commute. Besides

For all $x$ the bounds hold $\bar{H}(b^{-}-x)\leq N(x)\leq\bar{H}(-x)$ 2. 2.

When severities have a symmetric distribution $H(x)=1-H(-x)$ then

[TABLE]

Proof. Noting that $\{J_{1}\geq b-x\}\subset\mathbf{U}_{b}^{x}\subset\{J_{1}>-x\}$ item 1 follows. Besides if $J_{1}$ has symmetric distribution then the law of $X$ must be invariant under reflection from the axis $z=b/2$: \mathbb{P}^{x}\Big{(}X_{t}\in B\Big{)}=\mathbb{P}^{\theta(x)}\Big{(}X_{t}\in\theta\circ B\Big{)}where we call $\theta$ such reflection. Since $\theta\circ\{\tau_{b}<\tau_{0}\}=\{\tau_{0}<\tau_{b}\}=\{\tau_{b}<\tau_{0}\}^{c}$ we have

[TABLE]

2.2 Effect of the past

Here we study how EP are affected by the information collected. Let $r$ be a given epoch of time (the “present” or ‘starting” time $r$). Clearly

[TABLE]

*whenever * $r\in\mathbb{T}=\{\mathbf{t}_{0},\mathbf{t}_{1},\dots\mathbf{t}_{n},\dots\}$, the random set of all arrival times. However (2.7)*does not extend to arbitrary present * since $(X_{t})_{t\geq t_{0}}$ needs not being time-homogenous nor Markovian. Hence the escape probabilities depend on “starting” time $r$ and on which information is accessible. In this situation there is no real reason to fix our attention in \mathbb{P}\Big{(}\tau^{b}<\tau^{0}|X_{0}=x\Big{)} as accumulated information plays a central role. We are interested in \mathbb{P}\Big{(}\tau^{b}<\tau^{0}|\mathcal{F}_{r}\Big{)} conditional on the information at time $r$, \mathcal{F}_{r}=\sigma\Big{(}X_{s},s\leq r\Big{)}.

Given the present $r$, the backward and forward recurrence life $r\pm B_{r}^{\pm}$ mark the epochs of time at which the next and last jump occurred: $B_{r}^{-}=r-\mathbf{t}_{N_{r}}$ and $B_{r}^{+}=\mathbf{t}_{N_{r}+1}-r$. For $x\in\mathbb{R},z\geq 0$ we introduce

[TABLE]

Proposition 4**.**

For an epoch $r$, \mathbb{P}\Big{(}\tau^{b}<\tau^{0}|\mathcal{F}_{r}\Big{)} depends only on the information contained in $X_{r}$ and $B_{r}^{-}$; ulterior information from the past is irrelevant. Concretely,

[TABLE]

Proof. Note first that

[TABLE]

[TABLE]

(If $N_{r}=0$ we define $J_{0}=\mathbf{t}_{0}=0$). Clearly unless $F\sim\mathcal{E}(\lambda)$ neither $(X_{t})$ nor $(N_{t})$ are Markovian. Nevertheless in view of assumptions A2,A3 and that

[TABLE]

it follows that given the past of the process up to time $\mathbf{t}_{n}\in\mathbb{T}$ the future $(X_{t+\mathbf{t}_{n}})$ is conditionally distributed as (1.1) starting at $X_{\mathbf{t}_{n}}$ and is independent of the past: $\mathbb{E}^{x}(g(X_{t+\mathbf{t}_{n}}))=\mathbb{E}^{X_{\mathbf{t}_{n}}}(g(X_{t}))$; besides $Z_{n}:=X_{\mathbf{t}_{n}},n=1,\dots\infty$ is a Markov chain. This suggests some underlying simplicity. Indeed, the history previous to the last jump is not relevant for the future evolution of process $X$. At the epoch $r$ the essential history consists only of those events of the form $\{X_{s}=x,B_{r}^{-}=z,0\leq r-z\leq s\leq r\}$. More correctly, let us define $\mathcal{A}_{r}$ as the class of events

[TABLE]

for some $n\in\mathbb{N}$, $x_{1},\dots,x_{n}$ and $s_{1}<\dots s_{n}\leq r$. Then, assumptions $A1-A4$ imply that conditional on $\sigma(\mathcal{A}_{r})\equiv\mathcal{F}_{r}^{\prime}$, the future evolution of $(X)$ is independent of $\mathcal{F}_{r}$. In addition, given $X_{r}=x$ and $B_{r}^{-}=z$, say, then $X_{s}=x-c(r-s)$ for all $r-z\leq s\leq r$, i.e. the “relevant” past gets determined. (The relevant past $\mathcal{N}_{r}$ of $(N)$ requires knowledge of both $B_{r}^{-}$ and $N_{r}$ but this does not change the argument. See (3.15) below). Hence we have

[TABLE]

Note that $\mathcal{F}_{r}$ is obtained by joining the sigma algebras containing the information prior and after the last arrival: $\mathcal{F}_{r}=\sigma(\mathcal{A}_{r})\vee\mathcal{F}_{r-B_{r}^{-}}$. Hence conditional independence gives

[TABLE]

3 Integral equations for the escape probability

3.1 Poissonian jumps

We consider first the simpler case of Poisson arrivals.

Theorem 1**.**

Suppose $c\neq 0$ and that $\tau_{1}\sim\mathcal{E}(\lambda)$. Then $N(x)$ solves (2.2) and

[TABLE]

[TABLE]

Proof. Here we take advantage that under Poisson arrivals (1.1) is a Lévy-Markov process whose infinitesimal generator $\mathbf{G}$ acts on any $\Psi\in L^{\infty}\cap C^{1}(\mathbb{R})$ in the domain of $\mathbf{G}$ via

[TABLE]

Suppose we allow $(X)$ to start at arbitrary $x$. If $x\in\mathbb{R}-(0,b)$ then ${\bm{\tau}}\equiv\tau_{0}\wedge\tau_{b}=0$ as escape occurs instantly. Note also that $\{\tau^{b}<\tau^{0}\}=\{X_{{\bm{\tau}}}\in[b,\infty)\}$. This insight yields

[TABLE]

where $\mathbf{N}:\mathbb{R}\to\mathbb{R}$ is an extension of $N$ from $(0,b)$ to $\mathbb{R}$. Given $m(t,y),n(t,y)$ let $\Psi(t,y)\in L^{\infty}\cap C^{1}(\mathbb{R})$ solve the Dirichlet boundary problem

[TABLE]

[TABLE]

Additionally if $\Psi$ is in the domain of $\mathbf{G}$ Dynkin’s formula yields that the process

[TABLE]

is a martingale ([9]). Hence $\mathbb{E}\Psi({\bm{\tau}},X_{{\bm{\tau}}})=0$ since ${\bm{\tau}}\equiv\tau_{0}\wedge\tau_{b}$ is a stopping time. Thus for $\Psi\in L^{\infty}\cap C^{1}(\mathbb{R})$ solving (3.5) we have

[TABLE]

[TABLE]

Take $m(t,y)=0$ and $n(y)\equiv\mathbf{1}_{y\in[b,\infty)}$. (3.8) and (3.4) give

[TABLE]

Thus $\mathbf{N}(x)=\Psi(x)$ where $\mathbf{N}$ solves $\mathbf{G}\mathbf{N}(x)=0$ and $\mathbf{G}$ is (3.3). By insertion (3.1) follows $\square$

We consider now the general case $\tau_{n}\overset{iid}{\sim}F$. Here the above theory does not hold since (1.1) is not Markov. We resort to renewal arguments by sharpening the result (2.3) and ideas of section (2.2). (Note that we drop the $b-$ dependence in $(\mathbf{U}_{b}^{x})_{0\leq x\leq b}$ and simply write $\mathbf{U}^{x}\equiv\{\tau_{b}<\tau_{0}\}$).

Theorem 2**.**

$\mathbf{U}^{x}\equiv\{\tau_{b}<\tau_{0}\}$* satisfies*

[TABLE]

i..e it can be decomposed as a disjoint union where $\tilde{\mathbf{U}}^{z}$ satisfies for given $z$

[TABLE] 2. 2.

Let $\mathbf{U}_{r}^{x}$ be the event that $(X)$ exits through the upper barrier when $X_{r}=x$ and $r>0$ is the present time. Call $J\equiv J_{N_{r}+1}$. Then $\mathbf{U}_{r}^{x}$ can be decomposed as the disjoint union $\mathbf{U}_{r}^{x}=\mathbf{U}_{r}^{x,(1)}\sqcup\mathbf{U}_{r}^{x,(2)}\sqcup\mathbf{U}_{r}^{x,(3)}\text{ where }$

[TABLE]

[TABLE]

is made up of independent events.

If $r+B_{r}^{+}\equiv\mathbf{t}\in\mathbb{T}$ then $\tilde{\mathbf{U}}=\mathbf{U}_{\mathbf{t}}^{X_{\mathbf{t}}}$ is conditionally independent of $\mathcal{F}_{\mathbf{t}}$ given $X_{\mathbf{t}}$ and

[TABLE]

Proof. We prove (3.12) since then (3.10) follows letting $r=0$ and noting that $B_{0}^{+}=\tau_{1}$. After time $r$, given $X_{r}=x$, four excluding possibilities unfold, depending on the evolution up to the first arrival: $s\mapsto X_{s},r\leq s\leq r+B_{r}^{+}$

$B_{r}^{+}\geq t_{b}$. Then $X_{r+t_{b}}=b$.

If, by contrast, $B_{r}^{+}<t_{b}$ then a jump $J$ occurs at $r+B_{r}^{+}$ prior to escape. Then 2. 2.

$B_{r}^{+}<t_{b},J>0$ and $X_{r+B_{r}^{+}}\equiv x+cB_{r}^{+}+J\geq b$. Then $X_{r+B_{r}^{+}}\geq b$. 3. 3.

$B_{r}^{+}<t_{b}$ and $0\leq x+cB_{r}^{+}+J<b$. After the “first” renewal $r+B_{r}^{+}$ the process starts at $X_{r+B_{r}^{+}}\equiv x^{\prime}\in(0,b)$ and will exit through the upper barrier if $\mathbf{U}_{\mathbf{t}}^{X_{\mathbf{t}}}$ occurs.

In all these cases escape will occur through the upper barrier. 4. 4.

$B_{r}^{+}<t_{b},J<0$ and $X_{r+B_{r}^{+}}<0$. Then $(X)$ escapes through the lower barrier.

This implies (3.12) where $\tilde{\mathbf{U}}\equiv\mathbf{U}_{\mathbf{t}}^{X_{\mathbf{t}}}$ where $\mathbf{t}\equiv r+B_{r}^{+}\in\mathbb{T},z\equiv X_{\mathbf{t}}$. We now see (3.12) and (3.13). Let $z=X_{\mathbf{t}},t^{\prime}_{b}=(b-z)/c$. Then

[TABLE]

By contrast

[TABLE]

The result follows since assumption A3 implies \sigma\Big{(}\tau_{N_{r}+1},J_{N_{r}+1}\Big{)}\perp\kern-6.0pt\perp\sigma\Big{(}\tau_{n},J_{n},n\geq N_{r}+2\Big{)} where $\mathcal{G}\perp\kern-6.0pt\perp\mathcal{F}$ denotes independence of $\sigma-$ fields. Note also that $\tau_{N_{r}+2}\overset{F}{=}\tau_{1},$ $J_{N_{r}+2}\overset{H}{=}J_{1},\dots$ have the same law (see A1, A2); hence

[TABLE]

Actually \mathbb{P}\Big{(}\tau^{b}<\tau^{0}|\mathcal{F}_{r}\Big{)}, given by (2.9), is retrieved once $N_{b}(x)$ is known.

In the sequel we make the mild and convenient assumptions

Assumption 5**.**

(A5) $t_{b}$ is not a mass of $F$.

Assumption 6**.**

(A6) L:=\mathbb{P}\Big{(}\tau_{1}\leq b/c,J_{1}\in(-b,b)\Big{)}<1

Assumption 7**.**

(A7)$F$ has a density $f$ and $H$ has a density $h$.

Thanks to A5 we avoid the messy distinction between $F(t_{b}^{-})$ and $F(t_{b})$ while A6 guarantees that the IE (3.19) below satisfies a fixed point condition. Assumption A7 is unnecessary at this stage, but will be convenient when we take up the task of solving (3.19) (sections 4-6).

We start considering the conditional distributions of the Markov process $t\mapsto B_{t}^{+}$.

Lemma 1**.**

For any epoch $r$, $B_{r}^{+}$ is conditionally independent of $X_{r}$ and the history $\mathcal{F}_{r}$ given $B_{r}^{-}$. Besides

[TABLE]

[TABLE]

Proof. Given $B_{r}^{-}$ assumption A1 with random $n(\omega)\equiv N_{r}(\omega)$ yield

[TABLE]

Thus $B_{r}^{+}$ is conditionally independent of the history $\mathcal{N}_{r}$ given $B_{r}^{-}$. $(X)$ contains information on $B_{r}^{-}$; hence $B_{r}^{+},X_{r}$ are not independent, but they are given $B_{r}^{-}$. It follows that

[TABLE]

[TABLE]

[TABLE]

Theorem 3**.**

$N_{b}(x,r,z)$* and \mathbb{P}\Big{(}\tau^{b}<\tau^{0}|\mathcal{F}_{r}\Big{)} follow from $N_{b}(x)$ via*

[TABLE]

[TABLE]

In particular $N_{b}(X_{r},r,B_{r}^{-})\equiv N_{b}(X_{r},B_{r}^{-})$ is independent of the present epoch $r$.

In the sequel to ease notation we set $J\equiv J_{N_{r}+1}$ while $\vec{X}=\vec{x}$ stands for $X_{r}=x,B_{r}^{-}=z$. It follows from (2.9) and Proposition (4) that

[TABLE]

where $N_{b}^{j}(x,r,z)$ denote the probabilities of the different terms appearing in the RHS of (3.12). Hence (3.15) gives N_{b}^{1}(x,r,z)=\mathbb{P}\Big{(}B_{r}^{+}>t_{b}|B_{r}^{-}=z\Big{)}=\bar{F}(z+t_{b})/\bar{F}(z).

We evaluate next the probability of $\mathbf{U}_{r}^{x,(3)}$ using the tower property as

[TABLE]

[TABLE]

[TABLE]

[TABLE]

Besides if $B_{r}^{+}=l\leq t_{b},J=y$ then $X_{r+B_{r}^{+}}=x+cl+y$. Hence (3.13) gives

[TABLE]

[TABLE]

y conditional independence (Lemma 1) and assumptions A1-A3 we have

[TABLE]

[TABLE]

Hence eqs. (3.15), (3.17) and (3.18) give

[TABLE]

3.2 Integral equations, general case

We have seen how to codify EP conditional in the history in terms of the basic object $N_{b}(x)$. Here (3.1) is not valid. We now derive integral equations for this object.

Theorem 4**.**

Suppose assumptions A1-A4 hold. The function $x\mapsto N_{b}(x)\equiv N(x)$ of (2.1) satisfies for $0\leq x\leq b$ the IE

[TABLE]

Proof. If $r=0$ is $z=B_{0}^{-}=0,B_{0}^{+}=\tau_{1},N_{b}(x,0,0)=N_{b}(x)\text{ and }$ \mathbb{P}\Big{(}B_{0}^{+}>l\big{|}B_{0}^{-}=0\Big{)}=\mathbb{P}(\tau_{1}>l)=\bar{F}(l). Inserting these values in (3.16) we find (3.19).

Corollary 1**.**

(3.1) and (3.2) hold if either $\tau_{1}$ is exponential or $c=0$. If $r\in\mathbb{T}$ is one of the jump times then (3.2) holds.

Proof. If $c=0$ then $t_{b}=\infty$ and both (3.19) and (3.16) simplify to

[TABLE]

Finally, if $r\in\mathbb{T}$ $B_{r}^{-}=0$ and \mathbb{P}\Big{(}B_{r}^{+}>l\big{|}B_{r}^{-}=0\Big{)}=\bar{F}(l)\quad\square

To clarify the role of different contributions appearing in (3.19) we decompose $H$ a convex combination of *proper * distribution functions $H_{j,\pm}$:

[TABLE]

[TABLE]

[TABLE]

Proposition 5**.**

If support $H=(-\infty,-b]\cup[b-x,\infty)$, i.e. $p_{1}=q_{1}=0$, then

[TABLE] 2. 2.

Let $N_{b}(x)$, $\tilde{N}_{b}(x)$ be the EP corresponding to jump cdf’s $H$ and, respectively, $\tilde{H}$. If $\tilde{H}=H$ on $(-b,b-x)$ and $p_{2}=\tilde{p}_{2}$ then $N_{b}(x)=\tilde{N}_{b}(x)$.

Proof. Using that $H(y)=q_{1}+q_{2}+p_{1}H_{1+}(y)\text{ if }0\leq y<b-x,$

$\ H(y)=q_{1}H_{1-}(y)+q_{2}\text{ if }-b<y\leq 0\text{ and that }$

[TABLE]

(3.19) reads

[TABLE]

[TABLE]

Setting $p_{1}=q_{1}=0$ (3.22) follows. Alternatively, note that when $p_{1}=q_{1}=0$ (3.10) reads $\{\tau^{b}<\tau_{0}\}=\{\tau_{1}\geq t_{b}\}\cup\{\tau_{1}<t_{b},J_{1}>0\}$. (3.22) follows trivially.

The structure of (3.23) shows that that if $\tilde{H}_{1\pm}=H_{1\pm}$ and $p_{2}=\tilde{p}_{2}$ then $N$ and $\tilde{N}$ solve the same equation and hence $N=\tilde{N}$.

Remark 2**.**

In Appendix A we prove that under the mild assumption A6 (3.19) involves a Lipschitz continuous operator with Lipschitz constant $L<1$ and hence has a unique solution. Hence the problem of obtaining the EP is codified in a one-to-one way into solving such linear Fredholm IE. Since generically a closed form solution is not possible to help a reader place properly the situation we indicate the decomposition (3.21) of $H$ as $\vec{p}=(q_{1},q_{2},p_{1},p_{2})$. We study assumptions that render (3.19) solvable in terms of the factors present in decomposition (3.21), denoted $\vec{p}\equiv(q_{1},q_{2},p_{1},p_{2})$. Case $p_{1}=q_{1}=0$ is trivial, cf. (3.22).

4 Actuarial case: $\vec{p}=(\star,\star,0,0)$

Here we discuss the situation of the risk model when only negative jumps are allowed. This corresponds to $p_{1}=p_{2}=0$, or $\vec{p}=(\star,\star,0,0)$ where $\star$ marks non-null components. We shall determine a large class of arrivals distributions under which this equation is solvable. Concretely A7 is modified to

Assumption 8**.**

(A8) Supp. $J_{1}\subset(-\infty,0)$. Besides $\tau_{1}$ and $-J_{1}$ have densities $f_{\tau_{1}}(t)\equiv f(t)$ and $f_{-J_{1}}\equiv h$.

We can write (3.19) as

[TABLE]

4.1 Poisson arrivals

Here we consider the risk model with Poisson arrivals $f(t)=\lambda e^{-\lambda t}$. Eq. (3.1) yiedls that $N$ satisfies also the integro-differential equation

[TABLE]

The key observation is that—unlike (4.1)— (4.2) has constant coefficients and is well suited to Laplace transformation by appropriate extension to the entire line $0\leq x<\infty$. If $\Upsilon(x)$ denotes a *general * solution to (4.2) its Laplace transform (LT) must satisfy

[TABLE]

where $\rho=\lambda/c$ and $\Upsilon_{0}\equiv\Upsilon(0)$. By the Laplace inversion formula $\Upsilon(x)$ can be recovered as $\Upsilon(x)\equiv\Upsilon_{0}\pi(x)-$where

[TABLE]

and the line $s=i\beta,\beta>0$ lies to the right of all singularities, namely we take the standard Bromwich contour. We prove below (see (4.7)) that $\pi(0)=1$ and hence the consistency condition: $\Upsilon_{0}\equiv\Upsilon(x=0)$ is identically satisfied. Hence $\Upsilon(x)$ defines a one-parameter family of solutions labeled by the free constant $\Upsilon_{0}$. To retrieve $N_{b}(x)$ we must require the extra boundary condition (2.2). This gives:

Theorem 5**.**

Suppose A8 holds, $\tau_{1}\sim\mathcal{E}(\lambda)$ and $\pi(x)$ is given by (4.3). Then the EP is

[TABLE]

We now elaborate further on the meaning of $\pi(x)$ and recover the ruin probability.

Proposition 6**.**

$\pi(0)=1$** 2. 2.

Suppose $h(x):=\Phi(x-b)\theta(x-b)$ vanishes for $0\leq x<b$. Then $\pi(x)=e^{\rho x}$ and (see (3.22)) $N(x)=e^{\rho(x-b)}$. Here and elsewhere $\theta(x)=\mathbf{1}_{x\geq 0}$ is the Heaviside function. 3. 3.

Let $m:=-\mathbb{E}J_{1}$. If $\rho m\geq 1$ the survival probability vanishes: $S(x):=N_{b\to\infty}(x)=0$.

If $\rho m<1$ then $\underset{x\to\infty}{\lim}\pi(x)=(1-\rho m)^{-1}$ and $S(x)$ satisfies

[TABLE]

The EP satisfies also

[TABLE]

Proof. We can evaluate $\pi(0)$ as

[TABLE]

Indeed, since $\hat{h}(s)$ is analytic on $\mathbb{C}^{+}:=\{s=s_{R}+is_{I}\in\mathbb{C}:s_{R}\geq 0\}$, the second integrand is $O(1/s^{2})$ and is analytic on $s_{R}\geq 0$. We close the contour with a very large semi-circle in the right half-plane whereupon it vanishes by Cauchy’s theorem.

For item (2) note that $\hat{h}(s)=\hat{\Phi}(s)e^{-bs}$ and hence

[TABLE]

We close Bromwich contour on the on the right half-plane if $j\geq 1-$ whereupon the integral vanishes by analicity. If $j=0$ we close on the left. The residue theorem gives $\int_{\beta-i\infty}^{\beta+i\infty}ds(2\pi is)^{-1}s^{k-n}=\delta_{nk}$ and hence

[TABLE]

For (3) note that the following conditions are equivalent C1: $\pi(x)$ is bounded. C2: $S(x)>0$ (see (4.4)). C3: The net profit condition (NPC) holds (see [8]):

[TABLE]

C4: All non-null zeroes of Lundberg equation (LE) $s=(1-\hat{h}(s))\rho$ have negative real parts (a complete discussion of this aspect is performed in [15, 16]). Hence if $\rho m<1$ we can appeal to dominated convergence and the Tauberian final value theorem for Laplace transforms ([37]) to find $\underset{s\to 0}{\lim}s(\hat{h}(s)-1)=m\equiv-\mathbb{E}J_{1}>0$ and

[TABLE]

The survival probability follows recalling $S(x):=N_{b\to\infty}(x)$.

Remark 3**.**

While result (4.5) for survival probabilities is well known the connection with escape probabilities (4.4) is a different matter. In a seminal paper Bertoin [32] introduces a similar factorization for stable one-sided Lévy process in terms of the Lévy measure. This line of approach is continued in [34, 35, 36]. We note that representation (4.6) does not hold if either assumption $\tau_{1}\sim\mathcal{E}(\lambda)$ or $cJ_{1}<0$ is dropped as we show below. Note also that *the scale function of a difussion * (cf. [1, 37]) is any strictly increasing $s:(a,b)\to\mathbb{R}$ such that

[TABLE]

Remark 4**.**

An analogue of the Pollaczek-Khinchine formula for escape probabilities follows by series expansion in the Laplace integral (4.3)

[TABLE]

4.1.1 Several examples

Example 1**.**

Exponential jumps $-J_{1}\sim\mathcal{E}(\gamma)$. Then $N_{b}(x)=\frac{1+\gamma x}{1+\gamma b}$ when $\rho=\gamma$ and

[TABLE]

Example 2**.**

*Rational jumps. *Suppose that $h(s)$ has rational LT $\hat{h}(s)$. Then $\pi(x)$ is given by sums of terms of the form $\pi(x)\approx\sum_{i}\gamma_{i}e^{s_{i}x}$ where $s_{i}\in\mathcal{R}$ are the zeroes of the denominator.

Example 3**.**

*Gamma jumps. *When $-J_{1}\sim\Gamma(\alpha,\gamma),\alpha\notin\mathbb{N}$ then \hat{h}(s)=\Big{(}\frac{\gamma}{\gamma+s}\Big{)}^{\alpha}- and hence $\hat{\pi}(s)-$ are non-rational and have a branch cut singularity on $(-\gamma,0)$. Nevertheless in particular cases it is possible to perform the inversion of (4.3). Here we suppose $\alpha=1/2$. We rationalize $\hat{\pi}(s)$ to the convenient form

[TABLE]

Let $2s_{\pm}=2\rho-\gamma\pm\sqrt{\gamma(\gamma+4\rho)}$ and $\xi_{\pm}\equiv s_{\pm}+\gamma$. Using a partial fraction expansion lengthy calculations yield $\pi(x)=S(x)/(1-\rho m)$ where $S$ is given in terms of erf function as

[TABLE]

[TABLE]

Example 4**.**

*Constant jumps $-J_{1}=y_{1}\in\mathbb{R}^{+}$.*This corresponds to an actuarial situation where policy holders have the right to a *fixed predetermined * compensation $y_{1}>0$ per claim. Hence $\hat{h}(s)=e^{-y_{1}s}$ and

[TABLE]

[TABLE]

[TABLE]

The integral is evaluated by series expansion, where term-wise we close Bromwich contour with a large half circle on the left or right half $s-$plane. Only when $x-ky_{1}\geq 0$ we can close on the right half-plane and pick a pole at $s=0$. Call $k_{1}\equiv\lfloor x/y_{1}\rfloor$. The residue follows from

[TABLE]

[TABLE]

Hence $\pi(x)=C(x,-\rho,1)=e^{\rho x}\tilde{\pi}(x)\text{ where }$

[TABLE]

As $x\to\infty$ is $k_{1}\to\infty$ and convergence of the series is unclear. Note first that $\tilde{\pi}(x)$ involves an alternating series whose general term $a_{k}\downarrow 0$ monotonically as $k\to\infty$. Thus $\tilde{\pi}(x)$ converges as $x\to\infty$. The convergence of $\pi(x\to\infty)$ is delicate: the ratio test shows that it requires $y_{1}\rho e^{1-y_{1}\rho}<1\Leftrightarrow 1-\rho y_{1}>0$. In this case NPC and (4.9) hold. One has $\underset{x\to\infty}{\lim}\pi(x)=(1-y_{1}\rho)^{-1}$

4.2 Erlang $\Gamma(n,\lambda)$ and hypo-exponential arrivals

Here we generalize the previous results *to the actuarial model under hypo-exponential arrivals. * That is, A8 holds and there exist parameters $0<\lambda_{1}\leq\dots\lambda_{n}$ such that $f_{\tau_{1}}\equiv f$ satisfies

[TABLE]

Remark 5**.**

This distribution corresponds to a sum $\tau_{1}=X_{1}+\dots X_{n}$ of $n$ independent variables $X_{i}\sim\mathcal{E}(\lambda_{i})$. Interesting particular cases are

$\lambda_{1}=\dots=\lambda_{n}$: this yields Erlang distribution $f_{\tau_{1}}(t)=\lambda_{1}(\lambda_{1}t)^{n-1}e^{-\lambda_{1}t}/(n-1)!$. 2. 2.

$\lambda_{j}=j\lambda_{1},\forall j$. Here $\mathbb{P}(\tau_{1}\leq t)=(1-e^{-\lambda_{1}t})^{n}$, the order statistics sampled from an exponential distribution. 3. 3.

Under strict generic inequalities the density is the Lagrange combination

[TABLE]

This situation modeled by (4.1) is not solvable as stands; nevertheless one can transform it to an equivalent, but simpler integro-differential equation with appropriate BCs that generalize (2.2). Using results of section (5) along with remarks (8),(9) we obtain

Theorem 6**.**

Suppose A8 holds where $f_{\tau_{1}}$ is given by (4.15). Let $Q(s)\equiv\prod_{j}(\lambda_{j}+s)$ and $Q_{0}=\prod_{j}\lambda_{j}$. Then, for $\ 0\leq x<b$, $N_{b}(x)$ solves the integro-differential equation

[TABLE]

with BCs at the end-point $x=b$

[TABLE]

In particular, if $f_{\tau_{1}}(t)=\lambda(\lambda t)^{n-1}e^{-\lambda t}/(n-1)!\sim\Gamma(n,\lambda)$ for some $n\in\mathbb{N},\lambda>0$

[TABLE]

Remark 6**.**

Note that -unlike (4.1)- (4.17) has by itself not a unique solution so appropriate BCs are required to pin down the EP. Eqs. (4.17) and (4.18) define a final value problem which needs not be well posed. Section 5 elaborates on their derivation under a fairly general framework (see (5.10) and remark (8)).

We next construct in explicit form the solution to (4.17). To this end we consider the extension from $(0,b)$ to $\mathbb{R}^{+}$ and deprive it of boundary conditions. Using the known properties of Laplace transformation $\mathcal{L}$:

[TABLE]

we obtain that any continuous solution $\Upsilon(x)$ of (4.19) on $\mathbb{R}^{+}$ must satisfy

[TABLE]

[TABLE]

[TABLE]

Hence

[TABLE]

By inversion we have that $\Upsilon(x)$ can be written in terms of $n$ arbitrary constants $\alpha_{j}$ and a fundamental solution $\pi(x)$ as:

[TABLE]

Note that it can be written in the suggestive way (compare with (6.11))

[TABLE]

We have obtained a bundle of solutions $\Upsilon(x)$ parametrized by initial values $\partial_{x}^{k}\Upsilon(0)$. The EP *should * follow by imposing the BCs (4.18) for the values $N^{(k)}(b)$ at $x=b$. It is unclear that the procedure works as this problem needs not be well-posed. We now prove that this is indeed the case. Evaluation of (4.22) and its derivatives at $x=b$ implies that the constants $\alpha_{j},j=0,\dots,n-1$ must satisfy the linear system $\mathbf{A}(b,b)\vec{\alpha}=\vec{1}$ where $\vec{1}=(1,0,\dots,0)^{\dagger}$:

[TABLE]

Here all matrix elements $\mathbf{a}_{ij}=\pi^{(i+j)}\equiv\partial_{x}^{i+j}\pi(x=b),i,j=0,\dots,n-1$ are evaluated at $x=b$. Besides the Wronskian of the functions $\pi,\pi^{(1)},\dots\pi^{(n-1)}$ at $x=b$ is

[TABLE]

Let $\delta_{j}(b)$ be the determinant of the matrix obtained substituting the $j-$th column of the matrix $\mathbf{A}(b,b)$ by the column vector $\vec{1}$. Cramer’s rule gives

[TABLE]

We introduce the $n\times n$ matrix $\mathbf{A}(x,b)$ and $\Delta(x,b)\equiv\det\mathbf{A}(x,b)$ via

[TABLE]

Note that here all matrix elements are evaluated at $x=b$ except for those at the first row. Let $\mathbf{A}_{0j}$ be the $(0,j)$ minor of $\mathbf{A}(x,b)$, the determinant of the matrix that arises deleting the $0-$ row and $j-$ th column. Then$\mathbf{A}_{0j}=(-1)^{j}\delta_{j}$. By row expansion

[TABLE]

Self-consistency of this procedure requires that $\partial^{j}\Delta(0,b)=\delta_{jn}\Delta(b,b)$. We skip the proof which follows using $\pi^{(j)}(0)=\delta_{j,n-1}$ . The following result summarizes the above.

Theorem 7**.**

Suppose A8 holds and $f_{\tau_{1}}$ is given by (4.15)(in particular, $\tau_{1}\sim\Gamma(n,\lambda)$ for some $n\in\mathbb{N},\lambda>0$). Then

The integro-differential equation (4.17) has general solution $\Upsilon(x)=\sum_{j=0}^{n-1}\alpha_{j}\partial_{x}^{j}\pi(x)$ where $\alpha_{j},j=1,\dots n$ are arbitrary constants and $\pi(x)$ is given by (4.22) or (4.23). 2. 2.

The escape probability is given in terms of Wronskian determinants (4.25) as

[TABLE]

Remark 7**.**

The above result could be used to obtain survival probabilities by letting $S(x)=\underset{b\to\infty}{\lim}\det\mathbf{A}(x,b)/\det\mathbf{A}(b,b)$. This will be the subject of a future work.

5 Risk model under rational arrival times

Denote by $\mathcal{M}$ the class of densities $f$ having rational Laplace transform (LT) $\hat{f}$:

[TABLE]

where $Q,R$ are co-prime polynomials of orders $m\equiv\deg(R)<\deg(Q)=n$:

[TABLE]

The characterization of such class is not straightforward: a criteria in in terms of complete monotonicity and unimodality was given by Feller [37] and Bernstein; this approach is pursued in [38]). Obviously $Q(0)=R(0)\equiv a_{0}\neq 0$ and roots of $Q$ must be located in the negative real axis. Besides, with no loss of generality, $a_{n}=1$.

We now establish several results that relate $\mathcal{M}$ with solutions of certain ordinary differential equations (ODEs). The proof is deferred to appendix B.

Lemma 2**.**

A density $f\in\mathcal{M}$ iff it is of class $C^{n}$ on $(0,\infty)$, $n\in\mathbb{N}$ and solves the ODE

[TABLE]

where the initial data $f_{0}^{(k)}\equiv\partial_{x}^{k}f(0),k=0,\dots n-1$ solve the linear system

[TABLE]

Corollary 2**.**

It follows from (5.3) that $a_{0}\bar{F}(t)=\sum_{j=1}^{n}a_{j}\partial_{t}^{j-1}f(t)$

“Vectors” $\vec{f}_{0}\equiv(f^{0}_{0},\dots,f^{(n-1)}_{0})$ and $\vec{b}\equiv(b_{0},\dots,b_{n-1})$ and integer $m:=DegR$ have a direct bearing on the degree of complexity of eq. (5.10) below which governs EPs. Here we analyze their structure.

Lemma 3**.**

Let $j_{*}\equiv\min\{j:f^{(j)}_{0}\neq 0\}$. If $f^{(j)}_{0}=0$ for some $j,0\leq j\leq n-2$ then

[TABLE]

[TABLE]

Thus $\vec{f}_{0}$ and $\vec{b}$ have at least one non-vanishing component $f^{(n-1)}_{0}\neq 0$ and must have the structure (here $\star$ denote a non-null component)

[TABLE]

Remark 8**.**

The above allows a partial classification of densities $f\in\mathcal{M}$ in terms of the integers $n,m$ and $\nu_{n},\nu_{m}$: the number of different roots of $Q$ and $R$ where

[TABLE]

Thus, for given $n$ a total of n\Big{(}n^{2}-n+2\Big{)}/2 sub-cases appear. Some light is shed looking at the extreme cases:

$m=0$. This is the *hypoexponential * distribution previously studied. Besides it is Erlang when $\nu_{n}=1$. Here $\vec{b}=(a_{0},0,\overset{n-1}{\dots}0)$ and $\vec{f}_{0}=(0,\overset{n-1}{\dots}0,a_{0})$. 2. 2.

$m=\nu_{m}=n-1$ and $\nu_{n}=n$. Feller ([37], pp. 439) proves that this corresponds to convex mixture of exponentials under the additional condition

[TABLE]

[TABLE]

5.1 Escape probabilities under arrivals with rational LT

We now study EPs for the risk model when A8 holds and $f\in\mathcal{M}$. Such general case is far more involved but can still be solved analytically by appropriately transforming (4.1) into something amenable to Laplace transformation.

Theorem 8**.**

Suppose that $c>0$, and assumptions 8 and (5.1) hold. Let $\mathbb{Q}$ and $\mathbb{R}$ be the differential operators $\mathbb{Q}\equiv Q(-c\partial_{x}),\mathbb{R}\equiv R(-c\partial_{x})$ and

[TABLE]

The solution $N(x)$ of the integral eq. (4.1) is of class $C^{n}(0,b)$ and must also solve on $0\leq x<b$ the integro-differential equation $\mathbb{Q}N-\mathbb{R}(q)=0$, or

[TABLE]

*and the *$n-$BCs of terminal type (we denote $\xi_{k}\equiv(-1/c)^{k+1}f^{(k)}_{0},f_{0}^{(k)}\equiv\partial_{x}^{k}f(0)$)

[TABLE]

[TABLE] 2. 2.

Let $\alpha_{j},\ j=0\dots n-1$ be free constants and $L$ be the function

[TABLE]

Then a general solution to (5.10) is

[TABLE]

Proof. Operating with $\partial^{j}_{x}$ on (4.1) we find, for $j=1,\dots n$

[TABLE]

[TABLE]

[TABLE]

Letting $x\to b^{-}$ the boundary conditions follow.

With appropriate arrangement of the resulting terms we find after some lengthy calculations

[TABLE]

[TABLE]

[TABLE]

Eq. (5.3) implies that the first two terms vanish: Concretely, $Q(\partial_{t})f=0$ yields $E_{1}=E_{2}=0$. Upon simplification and using (5.4) the third term is,

[TABLE]

[TABLE]

Hence $\mathbb{Q}N(x)=\sum_{m=0}^{n-1}(-c)^{m}b_{m}\partial_{x}^{m}q(x)$ and (5.10) follows.

We solve an auxiliary version of (5.10) extended to $0\leq x<\infty$ deprived of boundary conditions. Laplace transformation yields that any solution $\Upsilon(x)$ must satisfy

[TABLE]

where we introduce

[TABLE]

The initial values $\Upsilon_{0}^{(j-k-1)}\equiv\partial_{x}^{(j-k-1)}\Upsilon(0)$ and $q_{0}^{(j-k-1)}\equiv\partial_{x}^{(j-k-1)}q(0)$ are undefined so far. It follows that

[TABLE]

By inversion we find the general solution (5.13) $\quad\square$

In the general case $m\geq 0$ obtention of the EP is far more involved than that of section 4. We now work the details.

Theorem 9**.**

Suppose that assumption 8 and (5.1) hold. Define

[TABLE]

Recall $\xi_{k}\equiv(-1/c)^{k+1}f^{(k)}_{0},k\geq 0$ and $\xi_{-1}:=-1$. Let $\mathbf{A}=(\mathbf{a}_{ji}),i,\ j=0\dots n-1$ and $(\mathbf{\Theta}(x,b)$ be the $n\times n$ (respectively, $(n+1)\times(n+1)$) matrices with entries

[TABLE]

[TABLE]

Then the EP is

[TABLE]

Remark 9**.**

When $m=0$ the equation for the EP (5.10) simplifies to (4.17).

Additionally Lemma (3) and (5.16) give $\xi_{j}=0,0\leq j<n-1$; hence $\mathbf{a}_{ji}=\pi^{(i+j)}(b)$ and all entries but one of the first column of (5.18) vanish. Besides

[TABLE]

giving $L(s)=Q(-cs)-Q(0)\hat{h}(s)$ and ${\bm{\Theta}}(x,b)$ coincides with (4.27).

Proof. Require (5.10) to satisfy (5.11). Note

[TABLE]

[TABLE]

More generally, it follows from (5.11) that (we denote $m_{i}^{(j-k-1)}\equiv\partial_{x}^{(j-k-1)}m_{i}$)

[TABLE]

[TABLE]

where at this stage we introduce the the $n\times n$ matrice $\mathbf{a}=(\mathbf{a}_{ji}),\mathbf{c}=(\mathbf{c}_{ji})$ with entries

[TABLE]

Defining $\xi_{-1}\equiv f_{0}^{(-1)}\equiv-1$ we have $N(x)\equiv\sum_{i=0}^{n-1}\alpha_{i}\pi^{(i)}$ where $\vec{\alpha}=(\alpha_{0},\dots,\alpha_{n-1})$ solves

[TABLE]

Actually, a good deal more can be said about the solution: By linearity one has

[TABLE]

where $\vec{\alpha}^{(k)}\equiv(\alpha_{0}^{(k)},\dots,\alpha_{n-1}^{(k)}),k=0,\dots n-1$ solves the system

[TABLE]

Cramer’s rule yields that

[TABLE]

where $\mathbf{A}=(\mathbf{a})_{ij}$ and ${\bm{\theta}}_{j}^{k}$ is the $n\times n$ matrix obtained substituting the $j-$th column of $\mathbf{A}$ by the column vector $(\vec{e}_{k})_{m}=\delta_{km}$:

[TABLE]

[TABLE]

Call $\mathbf{A}_{k}$ the matrix that results when the $k-^{th}$ row of $\mathbf{A}$ is substituted by the vector $(\pi(x),\pi^{(1)}(x),\dots\pi^{(n-1)}(x))$, namely

[TABLE]

The Laplace co-factor expansion of this determinant yields that

[TABLE]

A similar co-factor expansion of matrix (5.18) gives

[TABLE]

Hence (5.22)-(5.27) yield

[TABLE]

[TABLE]

We next prove that $\det{\bm{\Theta}}(b,b)=\det\mathbf{A}(b)$. Indeed, $\det{\bm{\Theta}}(b,b)=$

[TABLE]

5.2 Escape probabilities when $n\equiv DegQ=2$

We use the former results to give explicit expressions of EP when Deg $Q\equiv n=2$. Let $0<p<1,q\equiv 1-p$. We have the cases (see remark (8)):

$m=0$, $\nu_{n}=1$ (Erlang distribution): $\hat{f}(s)=\lambda^{2}/(\lambda+s)^{2};\ f(t)=\lambda^{2}te^{-\lambda t}$. 2. 2.

$m=0$, $\nu_{n}=2$ (hypoexponential distribution):

[TABLE] 3. 3.

$m=\nu_{n}=1$ (Mixture of exponential and Erlang):

[TABLE] 4. 4.

$m=1,\nu_{n}=2$ (Convex Mixture or hyperexponential):

[TABLE]

Example 5**.**

*Escape probability under hypoexponential and $\Gamma(2,\lambda)$ distributions *:

Suppose $f_{\tau_{1}}$ is given by (5.28) with $\lambda_{1}\leq\lambda_{2}$. Since $m=0$ theorem (7) , (4.29), (4.22) give

[TABLE]

[TABLE]

Example 6**.**

Hyper-exponential: $f(t)=p\lambda_{1}e^{-\lambda_{1}t}+q\lambda_{2}e^{-\lambda_{2}t}$.

Hence we suppose that $\hat{f}(s)$ is given by (5.30) where and $0<p<1,q=1-p$. Since $m=1$ the situation is considerably more complex, and the full formalism of theorem (9) is required; hence we assume $-J_{1}\sim\mathcal{E}(\gamma)$. Since $R(s)=(p\lambda_{1}+q\lambda_{2})s+\lambda_{1}\lambda_{2},Q(s)=(s+\lambda_{1})(s+\lambda_{2})$ the IE for $N(x)$ reads (see (5.9), (5.10))

[TABLE]

Define $\tilde{\lambda}_{1}\equiv\lambda_{1}/c,\tilde{\lambda}_{2}\equiv\lambda_{2}/c$. The function $\pi(x)$ is retrieved from (5.13) and (5.12) where

[TABLE]

Thus $L(s)$ has poles on $\mathcal{R}=\{0,s_{-},s_{+}\}$ where we define s_{\pm}=\Big{(}-A\pm\sqrt{A^{2}-4B}\Big{)}/2. It follows that (see (5.16))

[TABLE]

[TABLE]

[TABLE]

Besides $f_{0}\equiv\lambda_{1}p+q\lambda_{2}$ . Recalling that $\xi_{0}=-f_{0}/c$ and $\mathbf{a}_{10}=\pi^{\prime)}(b)-\xi_{0}m_{0}(b),$ and setting $\mu\equiv c^{5}(s_{+}-s_{-})s_{+}s_{-}$ we find the EP via (5.19) and (5.18) where

[TABLE]

[TABLE]

5.3 Ideas on the case $n=3$

Suppose Deg $Q\equiv n=3$. Bearing in mind the restrictions (5.7) there are up to 12 possible cases labeled by $(\nu_{n},m,\nu_{m})$ which we do not attempt to classify. Consider however the following interesting cases $f_{1,2}\in\mathcal{M}$

Let $0<p<1$ and $\lambda>0$ and \hat{f}_{1}(s)=\lambda\Big{(}\lambda^{2}+2sp\lambda+s^{2}p\Big{)}/(\lambda+s)^{3} 2. 2.

Take now \lambda>1,\alpha=\Big{(}1\pm\sqrt{\lambda^{3}-1}\Big{)}/\lambda and

[TABLE]

The corresponding densities $f_{1,2}$ have equal integers $m=\nu_{m}=2,\nu_{n}=1$ ($R$ having a pair of complex conjugate roots); nevertheless they are markedly different. Actually,

[TABLE]

6 Two-sided problems

In this section we address the situation where jumps may take both signs. *It is remarkable that (3.23) is still solvable when only one of the conditions $q_{1}=0$ or $p_{1}=0$ is required and the remaining parameters $p_{1},p_{2},q_{1},q_{2}$ * are arbitrary. Recall that $\star$ signifies a non-null component. Table 1 summarizes these results.

6.1 Support $H\subset(-\infty,0)\cup[b-x,\infty)$ or $\vec{p}=(\star,\star,0,\star)$.

Here we show that the ideas of sections 3 and 4 carry over to the case support $H\subset(-\infty,0)\cup[b-x,\infty)$. We consider the case when jumps are hypoexponential, which helps to keep the algebra tidy. Generalization to arrivals $f\in\mathcal{M}$ is messy but straightforward.

Theorem 10**.**

Suppose assumptions A1-A7 hold with $\tau_{n}$ given by (4.15) and let $Q(s)\equiv\prod_{j}(\lambda_{j}+s)$. Suppose support $H\subset(-\infty,0)\cup[b-x,\infty)$ where $p=1-q=\mathbb{P}(J_{1}\geq b-x)$ and $dH_{-}(x)=h(-x)dx$. Then

$N_{b}(x)$* satisfies the BCs (4.18) and solves for $\ 0\leq x<b$*

[TABLE] 2. 2.

Let $L(s):=Q(-cs)-qQ_{0}\hat{h}(s)$. Define the fundamental solutions

[TABLE]

Let $\mathbf{A}(x,b)$ be the $n\times n$ matrix (4.27) and $\mathbf{B}(x,b)$ be the $(n+1)\times(n+1)$ matrix

[TABLE]

where $\pi_{b}\equiv\pi(b),\dots$.Then the escape probability is

[TABLE]

Proof. We start noting that $\pi_{-1}(x)=\int_{0}^{x}\pi(y)dy\equiv\pi^{(-1)}(x)$. Besides eq. (3.23) reads

[TABLE]

Eq. (6.1) follows with similar reasonings to that used as for (4.17) and (5.10). Note that using the results in remark (8), (5.15) is modified to

[TABLE]

Laplace transformation of (6.1) yields now

[TABLE]

for certain free constants $\alpha_{j},j=0,\dots n-1$. A general solution is

[TABLE]

To obtain $N(x)$ we require for $k=0,\dots n-1$

[TABLE]

Introducing $\alpha_{-1}:=pQ_{0}$ the above can be written as $\Upsilon(x)=\sum_{j=-1}^{n-1}\alpha_{j}\partial_{x}^{j}\pi(x)$ where the $\alpha$’s satisfy the system of $(n+1)$ equations

[TABLE]

This has the form (5.21) where $\xi_{-1}:=pQ_{0},\xi_{0}:=1,\xi_{j}=0,j\geq 1$. Hence, repeating mutatis-mutandis the arguments in theorem (9) we may obtain the solution given by eqs. (5.18), (5.19), where ${\bm{\Theta}}(x,b)$ is now the $(n+2)\times(n+2)$ matrix with determinant

[TABLE]

[TABLE]

[TABLE]

Clearly $\det\mathbf{B}(b,b)=0$ and $\det{\bm{\Theta}}(b,b)=\det\mathbf{A}(b,b)$ and the result follows.

Corollary 3**.**

Suppose that $\tau_{1}\sim\mathcal{E}(\lambda)$ and support $H\subset(-\infty,0)\cup[b-x,\infty)$ where $dH_{-}(x)=h(-x)dx$ and $p=1-q=\mathbb{P}(J_{1}\geq b-x)$. Then (4.4) is generalized to

[TABLE]

[TABLE]

6.2 support $H\subset(-\infty,-b)\cup[0,\infty)$: $\vec{p}=(\star,0,\star,\star)$

We next show that when $q_{1}=0$* the solution to (3.23) can be given in closed way * for general distribution of jumps and severities.

Theorem 11**.**

Suppose assumptions A1-A7 hold with $\tau_{n}\overset{i.i.d}{\sim}F$ and $J_{n}\overset{i.i.d}{\sim}H$ with densities $F^{\prime}=f$ and $H_{+}^{\prime}\equiv h_{+}$. Suppose that $q_{1}=0$, i.e. supp. $H\subset(-\infty,-b]\cup[0,\infty)$. Let $\pi(x)$ be the solution of the integral equation defined on $0\leq x<\infty$

[TABLE]

$\pi(x)$* has Laplace transform*

[TABLE] 2. 2.

Suppose that $F(b/c)p_{1}<1$. Then $N_{b}(x)$ is given by

[TABLE]

Proof. If $q_{1}=0$ (3.23) reads for $\ 0\leq x\leq b$:

[TABLE]

where $1-q=p\equiv p_{1}+p_{2}=\mathbb{P}(J_{1}>0)$. Setting $f_{c}(x):=f(x/c)$, (6.9) implies

[TABLE]

Recalling that $\mathcal{L}(\bar{F})(s)=(1-\hat{f})/s$ and $\mathcal{L}(f_{c})(s)=c\hat{f}(cs)$ and noting that the above is a repeated convolution we find

[TABLE]

(6.10) follows. For (ii) note that $\Pr(J_{1}>x)=p\bar{H}_{+}(x)$. The key idea is to introduce a new function via $\tilde{N}(x)\equiv N(b-x)$. Eq. (6.12) is transformed to

[TABLE]

[TABLE]

where we used $y\to z=x-cl$ and $N(b-z+y)=\tilde{N}(y-z)$. Thus $\tilde{N}(x)$ satisfies for $0\leq x\leq b$ the same equation as $\pi(x)$ does on $[0,\infty)$, namely (6.9). Since the bound (A.1) guarantees the existence of a unique solution both functions must be the same: $\tilde{N}_{b}(x)=\pi(x)$ for $0\leq x\leq b\quad\square$

We next consider several examples of interest with $q_{1}=0$ and $p=\mathbb{P}(J_{1}>0)$.

Example 7**.**

*Exponential jumps. *Suppose $\tau_{1}\sim\mathcal{E}(\lambda)$, supp. $H_{-}\subset(-\infty,-b]$ with $q=\mathbb{P}(J_{1}<0)$ and $H_{+}\sim\mathcal{E}(\gamma)$. (6.11) gives

[TABLE]

where

[TABLE]

Example 8**.**

*fixed magnitude jumps.*We consider the case $\tau_{1}\sim\mathcal{E}(\lambda)$, $q_{1}=0$ while positive jumps have a *fixed magnitude * $y_{1}$, i.e. $\hat{h}_{+}(s)=e^{-sy_{1}}$ and $p=\mathbb{P}(J_{1}>0)$. We see that

[TABLE]

where $C(y,\rho,p)$ was evaluated in example 4. Note that

[TABLE]

Hence, if we define $\xi_{k}\equiv b-x-ky_{1},\rho\equiv\lambda/c\text{ and }k_{1}\equiv\lfloor(b-x)/y_{1}\rfloor$ we finally have

[TABLE]

Note that $\underset{k\to\infty}{\lim}\sum_{j=0}^{k}(\rho\xi_{k})^{j}e^{-\rho\xi_{k}}/j!=1$. Besides when $y_{1}$ is so large as $b-x<y_{1}$ then $0=k_{1}=k=j$ and (6.14) reduces to (3.22). Finally as $p\to 1$ then $N_{b}(x)\to 1$ as expected,

It is interesting to compare the EP correspnding to $\tau_{1}\sim\mathcal{E}(\lambda)$ and the cases where (i) $H\sim$ Laplace$(0,\gamma)$, (ii) $H_{+}\sim\mathcal{E}(\gamma)$, $\mathbb{P}(J_{1}<0)=\mathbb{P}(J_{1}\leq-b)=1-p$, and (iii) $H_{+}\sim\delta(y-y_{0})$ and $\mathbb{P}(J_{1}<0)=\mathbb{P}(J_{1}\leq-b)=1-p$.

6.2.1 Severities with rational characteristic function

In the spirit of section 5.1 we denote by $\mathcal{H}$ the class of densities $h$ having rational characteristic function (CF), namely

[TABLE]

[TABLE]

are co-prime polynomials with $\deg(R)<\deg(Q)=n$. Besides $(a_{j}),(b_{j})\in\mathbb{R}$, $a_{0}=b_{0}$ and $a_{n}\neq 0,b_{n}=0$. It turns out that for severities $h\in\mathcal{H}$ and also $N_{t}\sim\mathcal{P}(\lambda t)$ (3.1) can be reduced further to an ordinary differential equation (ODE) with boundary conditions at $x=0$. Interesting examples of such class include

Suppose positive (negative) jumps are exponentially distributed with means $\gamma_{\pm}^{-1}$ and let $p:=\mathbb{P}(J_{1}>0),q=1-p$. Then

[TABLE]

[TABLE]

Such double-exponential jump models find application in mathematical finance. It corresponds to the polynomials

[TABLE] 2. 2.

$J_{1}\sim$ Laplace$(0,\gamma)$ is recovered when $p=1/2$, $\gamma_{+}=\gamma_{-}:=\gamma>0$. 3. 3.

The *variance gamma * distribution (VGD) is a widely used model in stochastic finance. If $\sigma,\vartheta\in\mathbb{R},n\in\mathbb{N}$ and a^{2}=\Big{(}2+\frac{\vartheta^{2}}{\sigma^{2}}\Big{)}\sigma^{-2} it is given by

[TABLE]

where $P$ is a certain polynomial with degree $n-1$ and $C$ a normalizing constant.

We first establish the following Lemma, which is proved in Appendix C.

Lemma 4**.**

Assume $h\in\mathcal{H}$ with $\tilde{h}$ given by (6.15). Let $I_{j}\equiv h^{(j)}(0^{+})-h^{(j)}(0^{-})$ be the jump at the origin of $\partial_{x}^{j}h,j\leq n-1$. Then $h(x)$ solves the ODE with boundary conditions at $x=0$

[TABLE]

[TABLE]

Reciprocally if $h$ is a density and solves (6.21), (6.22) then it has a CF given by (6.15) $\square$

We now show that EP can be found solving a simple ODE.

Proposition 7**.**

Suppose assumptions A1-A7 hold with $\tau_{1}\sim\mathcal{E}(\lambda)$ and $h\in\mathcal{H}$ satisfiying (6.15). Let $\mathbb{L}$ be the differential operator \mathbb{L}\equiv\Big{(}Q-R-\rho^{-1}\partial_{x}\circ Q\Big{)}(\partial_{x}). Then $N_{b}(x)$ solves the ODE

[TABLE]

Further, $n_{j}\equiv N^{(j)}(b^{-}),j=0,\dots n-1$ satisfies the linear system of BC$:n_{0}=1$ and

[TABLE]

[TABLE]

Proof. We write the jump distribution as $h(y)=h_{+}(y)\theta(y)+h_{-}(-y)\theta(-y)$ where $\theta(x)=\mathbf{1}_{x\in(0,\infty)}$, the Heaviside function. Since $\tau_{1}\sim\mathcal{E}(\lambda)$ then Eq. (3.1) applies. To keep the algebra tidy we introduce $M(x):=N(x)-\rho^{-1}\partial_{x}N(x)$ and (3.1) reads

[TABLE]

By repeated differentiation we find for $j\geq 1$

[TABLE]

This yields the BC (6.24) sending $x\to b$.

Next, operating with $Q(\partial_{x})$ on the LHS of (3.1) yields that $M$ satisfies

[TABLE]

[TABLE]

Recalling (6.21) we see that several terms cancel as $\sum_{j=0}^{n}a_{j}(-1)^{j}\partial_{z}^{j}h(z-x)\equiv Q(-\partial_{z})h(z-x)=0$. The above simplifies to

[TABLE]

[TABLE]

where we used (6.22). Eq. (6.23) follows since

[TABLE]

We next evaluate the EP for several cases of interest when $\tau_{1}\sim\mathcal{E}(\lambda)$.

Example 9**.**

*Risk model recovered. *To warm up suppose again $-J_{1}\sim\mathcal{E}(\gamma)$ and $J_{1}<0$. This entails (see(6.19)) $R(s)=\gamma,Q(s)=\gamma+s$. From (6.23) the EP is found solving

[TABLE]

One checks easily that (4.11) is the only solution to this ODE and BCs.

Remark 10**.**

Notice that the EP for Brownian motion with drift $v\equiv(\gamma-\rho)/2$ satisfies also $\mathbb{L}N=0$ and that its infinitesimal generator is, up to a constant, $\mathbb{L}\equiv\partial_{xx}+(\gamma-\rho)\partial_{x}$.

Example 10**.**

*Laplace distribution. *Suppose that $J_{1}\sim$ Laplace$(0,\gamma)$. It follows that (see eq. (6.19)) $R(s)=\gamma^{2},Q(s)=\gamma^{2}-s^{2}$ and $N_{b}(x)$ must satisfy

[TABLE]

[TABLE]

Inserting $N(x)=\alpha+\beta_{-}e^{s_{-}x}+\beta_{+}e^{s_{+}x}$ where $2s_{\pm}=\rho\pm\sqrt{\rho^{2}+4\gamma^{2}}$ results in a linear system for $\alpha,\beta_{\pm}$. After a considerable amount of algebra the EP simplifies to

[TABLE]

[TABLE]

Note how, despite being a Levy process, the EP does not admit scale functions.

The following result gives the EP. We skip the proof.

Theorem 12**.**

Suppose assumptions A1-A7 hold with $\tau_{1}\sim\mathcal{E}(\lambda)$ and $h\in\mathcal{H}$ satisfies (6.15). Let $s_{k},k=0,\dots n-1$ be the roots of Lundberg equation $0=(1-s/\rho)Q(s)-R(s)$ and suppose they are all simple (Note that $s=0$ is always a root). Define the $n\times n$ matrices $\mathbf{A}=(\mathbf{a}_{jk})$ and $\mathbf{M}=(m_{jk}),\ j,k=0\dots n$ with entries

[TABLE]

[TABLE]

Then the EP is given by (5.19) where $x^{\prime}\equiv x-b$ and ${\bm{\Theta}}(x,b)$ is the $(n+1)\times(n+1)$ matrix

[TABLE]

6.3 The case of zero drift: $c=0$

The drift-less case deserves particular interest for its relevance to reliability theory. Besides several interesting simplifications occur. We reformulate Corollary 1 as

Corollary 4**.**

Suppose assumptions A1-A4 hold and that $c=0$. Then the escape probability is independent of the history and of the arrival distribution $F$. $N$ solves the integral equation (3.20)

We now turn our attention to solving this under appropriate restrictions. If either $q_{1}=0$ or $p_{1}=0$ vanish the solution becomes quite simple.

Theorem 13**.**

Suppose that $c=0$ and let $p=\mathbb{P}(J_{1}>0),q=1-p$.

$\vec{p}=(\star,\star,0,\star)$: Suppose support $H\subset(-\infty,0)\cup[b-x,\infty)$ and that $-J_{1}\mathbf{1}_{J_{1}<0}$ has density $h\equiv h_{-}$ (see (3.21)). The EP is

[TABLE] 2. 2.

$\vec{p}=(\star,0,\star,\star)$: Suppose support $H\subset(-\infty,-b)\cup(0,\infty)$ and that $J_{1}\mathbf{1}_{J_{1}>0}$ has density $h$. The EP is

[TABLE]

Note how this agrees with (2.5).

Proof. The result follows taking a LT on (3.20), which reads, respectively

[TABLE]

[TABLE]

We now consider the case when severities have rational CF. The result follows from those of last section letting $1/\rho=0$.

Proposition 8**.**

Suppose $c=0$ and that severities have rational CF given by (6.15).

Let $L\equiv(Q-R)(s)$ and $\mathbb{L}\equiv L(\partial_{x})$. Then $N(x)$ solves the ODE (6.23) $\mathbb{L}N=0$ and the BC’s (6.24) setting $1/\rho\equiv 0$

6.3.1 Different examples

Example 11**.**

We consider the case support $J_{1}=(-\infty,-b]\cup\{y_{1}\}$, namely positive jumps have a *fixed magnitude * $y_{1}>0$ and negative jumps exceed $b$. Let $p=\mathbb{P}(J_{1}=y_{1})\in(0,1)$. Recalling that $k_{1}\equiv\lfloor(b-x)/y_{1}\rfloor$ and letting $c=0$ (6.14) simplifies to

[TABLE]

Alternatively, note that $(X)$ escapes through $b$ iff the first $k_{1}+1$ jumps are positive.

Example 12**.**

We study EP for the family of CFs

[TABLE]

depending on five parameters $0\leq p\leq 1,q=1-p,\gamma_{\pm}>0$ and $\epsilon_{\pm}=0,\pm 1$ . We denote

[TABLE]

When $-\epsilon_{-}=\epsilon_{+}=1$ the problem is trivial: $N(x)=p$. The case $\epsilon_{-}=0,\epsilon_{+}=1$ is covered by theorem (13): the EP follows from (6.27)

[TABLE]

If $\epsilon_{-}=-1,\epsilon_{+}=0$ the EP follows from (6.28). We find (which agrees with (2.5))

[TABLE]

When $\epsilon_{\pm}=0$ theorem (13) does not apply; nevertheless, since it corresponds to $h\in\mathcal{H}$, viz. (6.17) the ideas of this section do. It follows from (6.18) and (6.19) that

[TABLE]

Here $Q-R(s)=-s^{2}+vs$. Hence the EP can be found solving Eq. (6.23):

[TABLE]

with appropriate BC’s (6.24) with $n=2$ and $1/\rho\equiv 0$. After tedious algebra one finds that the EP under jump density (6.17) $h(x)=p\gamma_{+}e^{-\gamma_{+}x}\theta(x)+q\gamma_{-}e^{\gamma_{-}x}\theta(-x)$ is

[TABLE]

Example 13**.**

We consider different special cases

$p=\frac{\gamma_{-}}{\gamma_{-}+\gamma_{+}}$ or $\tilde{h}(\omega)=\frac{\gamma_{-}\gamma_{+}}{(\gamma_{+}-i\omega)(\gamma_{-}+i\omega)}$. This gives *variance gamma distribution (VGD) * (6.20) with parameters $n=1,\ \sigma^{2}=2\gamma_{-}\gamma_{+}\text{ and }\vartheta=\frac{\gamma_{-}-\gamma_{-}}{\gamma_{-}+\gamma_{+}}$. 2. 2.

$p=\frac{\gamma_{+}}{\gamma_{-}+\gamma_{+}}$. This is a limit case for NPC: here $\mathbb{E}J_{1}=v=0$. Hence the EP (6.35) is ill-defined and must be obtained from scratch; we find

[TABLE] 3. 3.

Letting $\gamma_{+}=\gamma_{-}$ in eq. (6.36) one recovers N(x)=\Big{(}1+\gamma x\Big{)}/\Big{(}2+b\gamma\Big{)}, the EP under $J_{1}\sim$ Laplace$(0,\gamma)$. Note how (2.6) is satisfied.

Remark 11**.**

Notice that the EP has the neat factorization $N_{b}(x)\equiv A(x)/B(b)$. This is interesting, since $(X)$ is a two-sided not necessarily Lévy-Markov process.

Letting $\gamma_{\pm}\to\infty$ with $p=1/2$ and $2v\equiv\gamma_{+}-\gamma_{-}$ constant (6.35) goes into the scale function for Brownian motion with drift $v$. We have

[TABLE]

Actually the infinitesimal generator of BM is $\mathbb{L}\equiv(1/2)\partial_{xx}-v\partial_{x}$ and the EP satisfies also (6.34) $\mathbb{L}N=0$ with *different BCs * $N(b)=1-N(0)=1$. Such remarkable coincidence can be traced to the fact that if $\tilde{h}(\omega\Delta)-1=O(|\omega\Delta|^{\alpha}),\Delta\to 0$ then marginal probabilities for (1.1) converge in a weak sense into a fractional diffusion, in particular to BM.

Appendix A

Theorem 14**.**

Suppose that assumptions 5-7 hold and let $c>0,b<\infty$. Then (3.19) (or (3.23)) has a unique continuous solution $x\mapsto N_{b}(x)$ which satisfies the bound

[TABLE]

where we recall that L:=\mathbb{P}\Big{(}\tau_{1}\leq b/c,J_{1}\in(-b,b)\Big{)}, see assumption A6. Further, (3.19) without forcing term has only the trivial solution.

Proof. We use Banach fixed point theorem with the metric on $C_{0}(0,b)$ induced by the sup-norm on $(0,b)$ denoted as $\|..\|_{\infty}$. Given $N\in C_{0}$ let us introduce the integral operator

[TABLE]

[TABLE]

[TABLE]

Let $N\in B\subset C_{0}(0,b)$, the unit ball in $C_{0}(0,b)$. Note first that

[TABLE]

[TABLE]

Setting for convenience $q_{1}=0,y:=b-x-cl$ and recalling assumption A5 we also have

[TABLE]

[TABLE]

Alternatively, use Scheffe’s theorem. Therefore $\tilde{\mathcal{O}}B\subset B$:$\tilde{\mathcal{O}}$ is a bounded endomorphism of $B$.

More generally for $N_{1},N_{2}\in C_{0}$ linearity implies

[TABLE]

Hence, if Assumptions 5, 6 hold $\mathcal{O}$ is a contraction operator on $C_{0}(0,b)$: Lipschitz continuous with Lipschitz constant $L<1$. Thus it has just one fixed point which satisfies $N(x)=\mathcal{O}N(x):=N^{H}(t_{b})+\tilde{\mathcal{O}}N(x)$, namely (3.23) and

[TABLE]

By Gronwall’s Lemma, (3.23) without forcing term can only have the trivial solution $N=0$.

Remark 12**.**

When either $b\to\infty$ or $c=q_{2}=p_{2}=0$ the Lipschitz constant $L\to 1$, the operator $\mathcal{O}$ is not contractive, only non-expansive and (A.1) blows up. Actually the corresponding IE for survival probabilities has always a multi-parameter family of solutions.

If, by contrast, Assumption A6 holds but A5 is dropped the previous reasoning shows mutatis-mutandis that there exist unique solution but needs not being continuous. Finally, note that Kolmogorov-Riesz theorem proves that $\tilde{\mathcal{O}}$ is a *compact operator * on $C_{0}(0,b)$.

Appendix B

Proof of lemma (2). Suppose $f$ satisfies (5.1) and (5.2) where $Q,R\neq 0$. Choose the initial values $(f^{(k)}_{0})$ to satisfy the system (5.4)- which is possible since the associated system is triangular with determinant $=a_{n}^{n}\neq 0$. Note that in this case \mathcal{L}\Big{(}Q(\partial_{t})f\Big{)}\equiv

[TABLE]

[TABLE]

By uniqueness of Laplace transform is $Q(\partial_{t})f=0$, i.e. $f$ solves (5.3) and (5.4).

Reciprocally, suppose that $f$ solves (5.3) for a certain minimal $Q$ with IC $(f^{(k)}_{0})$; a simple calculation shows that $\hat{f}(s)$ is given by (5.1) where $P(s)=\underset{j=0}{\overset{n-1}{\sum}}b_{j}s^{j}$ and $b_{j}\equiv\sum_{k=0}^{n-j-1}a_{j+k+1}f^{(k)}_{0},j=0,1,\dots n-1$. In particular $b_{0}=a_{0}$ and $\hat{f}(0)=1$. $\ \square$

Proof of Lemma 2. Since $f$ is of class $C^{n}$ then $f^{(j)}$ is bounded and $\underset{t\to 0}{\lim}f^{(j)}(t)=f^{(j)}_{0}=0$ exists, $j\leq n$. If $f^{(j)}_{0}=0$ for some $1\leq j\leq n-1$ the initial value theorem for LT yields

[TABLE]

Besides $b_{k}\equiv\sum_{l=0}^{n-k-1}a_{l+k+1}f^{l}_{0}$ vanishes provided $n-k-1\leq j\Leftrightarrow k\geq n-j-1$.