MSTA Modern Stochastics: Theory and Applications 2351-6054 2351-6046 2351-6046 VTeXMokslininkų g. 2A, 08412 Vilnius, Lithuania MSTA89 10.15559/17-MSTA89 Research Article The risk model with stochastic premiums, dependence and a threshold dividend strategy RagulinaOlenaragulina.olena@gmail.com Taras Shevchenko National University of Kyiv Department of Probability Theory, Statistics and Actuarial Mathematics 64 Volodymyrska Str., 01601 Kyiv, Ukraine 2017 812201744315351 492017 21102017 8112017 © 2017 The Author(s). Published by VTeX2017 Open access article under the CC BY license.

The paper deals with a generalization of the risk model with stochastic premiums where dependence structures between claim sizes and inter-claim times as well as premium sizes and inter-premium times are modeled by Farlie–Gumbel–Morgenstern copulas. In addition, dividends are paid to its shareholders according to a threshold dividend strategy. We derive integral and integro-differential equations for the Gerber–Shiu function and the expected discounted dividend payments until ruin. Next, we concentrate on the detailed investigation of the model in the case of exponentially distributed claim and premium sizes. In particular, we find explicit formulas for the ruin probability in the model without either dividend payments or dependence as well as for the expected discounted dividend payments in the model without dependence. Finally, numerical illustrations are presented.

 Risk model with stochastic premiums Farlie–Gumbel–Morgenstern copula threshold strategy Gerber–Shiu function expected discounted dividend payments ruin probability integro-differential equation 91B30 60G55 62P05 Introduction

In the actuarial literature, a lot of attention is paid to the investigation of the ruin measures such as the ruin probability, the surplus prior to ruin and the deficit at ruin (see, e.g., [6, 34, 39] and references therein). A unified approach to the study of these risk measures together by combining them into one function was proposed by Gerber and Shiu [22], who introduced the expected discounted penalty function for the classical risk model. The so-called Gerber–Shiu function has been investigated further by many authors (see, e.g., [13, 23, 29, 42, 44]) in more general risk models. In those risk models, claim sizes and inter-claim times are assumed to be mutually independent, which simplifies the investigation of the ruin measures. Nevertheless, this assumption has been proved to be very restrictive in some real applications. For instance, in modelling damages caused by natural catastrophic events, the intensity of the catastrophe and the time elapsed since the last catastrophe are expected to be dependent [10, 38]. That is why more and more authors have concentrated on the investigation of risk models with dependence between claim sizes and inter-claim times recently.

Albrecher and Boxma [1] consider a generalization of the classical risk model, where the distribution of the time between two claims depends on the previous claim size (see also [2] for an extension). Albrecher and Teugels [3] apply the random walk approach and allow the inter-claim time and its subsequent claim size to be dependent according to an arbitrary structure. Boudreault, Cossette, Landriault and Marceau [11] consider a particular dependence structure between the inter-claim time and the subsequent claim size and derive the defective renewal equation satisfied by the expected discounted penalty function. In [33], the authors study the ruin probability in a model where the time between two claim occurrences determines the distribution of the next claim size. The ruin probability in a model with independent but not necessarily identically distributed claim sizes and inter-claim times is investigated in [5] (see also references therein).

Cossette, Marceau and Marri [15, 16] deal with an extension of the classical compound Poisson risk model where a dependence structure between the claim size and the inter-claim time is introduced through a Farlie–Gumbel–Morgenstern copula and its generalization. They derive the integro-differential equation and the Laplace transform of the Gerber–Shiu discounted penalty function and concentrate on exponentially distributed claim sizes. Zhang and Yang [47] extend these results to the compound Poisson risk model perturbed by a Brownian motion. In [12, 24, 46], the authors deal with the Sparre Andersen risk model where the inter-claim times follow the Erlang distribution and extend results obtained in [15, 16].

In all these papers, the dependence structure between the claim sizes and the inter-claim times is described by the Farlie–Gumbel–Morgenstern copula. This copula is often used in applications to introduce dependence structures due to its tractability and simplicity. It allows positive and negative dependence as well as independence. Nevertheless, the Farlie–Gumbel–Morgenstern copula has been shown to be somewhat limited since it does not allow the modeling of high dependencies. Indeed, its dependence parameter is θ∈[−1,1] , so its Spearman’s rho and Kendall’s tau are ρθ=θ/3∈[−1/3,1/3] and τθ=2θ/9∈[−2/9,2/9] , respectively (see, e.g., [7, 8, 37] and references therein). This limited range of dependence restricts the usefulness of this copula for modeling. Note that the dependence parameter θ can be easily estimated from real data due to the simple relations between it and the measures of association ρθ and τθ . For more information on the Farlie–Gumbel–Morgenstern copula, we refer to, e.g., [21, 37] (see also [16] and references therein for applications of this copula). Despite the popularity of the Farlie–Gumbel–Morgenstern copula, other copulas have been used in risk theory, for instance, an Archimedean copula [4], a Gaussian copula [19] and a Spearman copula [25].

Risk models where an insurance company pays dividends to its shareholders are of great interest in risk theory. Dividend strategies for insurance risk models were first proposed by De Finetti [20], who dealt with a binomial model. Barrier strategies for the classical risk model and its different generalizations have been studied in a number of papers (see, e.g., [14, 27, 28, 30, 31, 43, 45]). For optimal dividend problems in insurance risk models, see the monograph by Schmidli [40] and references therein.

Cossette, Marceau and Marri [17, 18] consider the classical risk process with a constant dividend barrier and a dependence structure between claim sizes and inter-claim times introduced through the Farlie–Gumbel–Morgenstern copula. They analyze the Gerber–Shiu function and the expected discounted dividend payments and then concentrate on exponentially distributed claim sizes investigating the impact of the dependence on ruin quantities. The same model is studied in [32], where, in particular, the authors show that the solution to the integro-differential equation for the Gerber–Shiu function is a linear combination of the Gerber–Shiu function with no barrier and the solution to the associated homogeneous integro-differential equation. For some earlier results in this direction, see also [26].

Shi, Liu and Zhang [41] consider the compound Poisson risk model with a threshold dividend strategy and a dependence structure modeled by the Farlie–Gumbel–Morgenstern copula. They derive integro-differential equations for the Gerber–Shiu function and the expected discounted dividend payments paid until ruin as well as renewal equations for these functions, which are used to obtain explicit formulas for them.

The present paper deals with a generalization of the risk model with stochastic premiums introduced and studied in [9] (see also [34]). In contrast to the classical compound Poisson risk model, where premiums arrive with constant intensity and are not random, in this risk model premiums also form a compound Poisson process, i.e. they arrive at random times and their sizes are also random (see also [35, 36] for a generalization of the classical risk model where an insurance company gets additional funds whenever a claim arrives). In [9], claim sizes and inter-claim times are assumed to be mutually independent, and the same assumption is made concerning premium arrivals. In this paper, we suppose that the dependence structures between claim sizes and inter-claim times as well as premium sizes and inter-premium times are modeled by the Farlie–Gumbel–Morgenstern copulas, which allows positive and negative dependence as well as independence. In addition, we suppose that the insurance company pays dividends to its shareholders according to a threshold dividend strategy. To be more precise, this implies that when the surplus is below some fixed threshold, no dividends are paid, and when the surplus exceeds or equals the threshold, dividends are paid continuously at some constant rate. Our subjects of investigation are the Gerber–Shiu function, a special case of which is the ruin probability, and the expected discounted dividend payments until ruin.

The rest of the paper is organized as follows. In Section 2, we describe the risk model we deal with. In Section 3, we derive integral and integro-differential equations for the Gerber–Shiu function. In Section 4, we obtain corresponding equations for the expected discounted dividend payments until ruin. Section 5 deals with exponentially distributed claim and premium sizes in some special cases of the model. Namely, we consider the ruin probability in the model without either dividend payments or dependence, and the expected discounted dividend payments in the model without dependence. In these simpler models, we reduce the integral and integro-differential equations derived in Sections 2 and 3 to linear differential equations and find explicit solutions to these equations. Section 6 provides some numerical illustrations.

 Description of the model

Let (Ω,F,P) be a probability space satisfying the usual conditions, and let all the stochastic objects we use below be defined on it.

In the risk model with stochastic premiums introduced in [9] (see also [34]), claim sizes form a sequence (Yi)i≥1 of non-negative independent and identically distributed (i.i.d.) random variables (r.v.’s) with cumulative distribution function (c.d.f.) FY(y)=P[Yi≤y] . The number of claims on the time interval [0,t] is a Poisson process (Nt)t≥0 with constant intensity λ>0 0$]]>. Next, premium sizes form a sequence (Y¯i)i≥1 of non-negative i.i.d. r.v.’s with c.d.f. F¯Y¯(y)=P[Y¯i≤y] . The number of premiums on the time interval [0,t] is a Poisson process (N¯t)t≥0 with constant intensity λ¯>0 0$]]>. Thus, the total claims and premiums on [0,t] equal ∑i=1NtYi and ∑i=1N¯tY¯i , respectively. We set ∑i=10Yi=0 if Nt=0 , and ∑i=10Y¯i=0 if N¯t=0 . In what follows, we also assume that the r.v.’s (Yi)i≥1 have a probability density function (p.d.f.) fY(y) and a finite expectation μ>0 0$]]>, and the r.v.’s (Y¯i)i≥1 have a probability density function (p.d.f.) fY¯(y) and a finite expectation μ¯>0 0$]]>.

We denote a non-negative initial surplus of the insurance company by x. Let Xt(x) be its surplus at time t provided that the initial surplus is x. Then the surplus process (Xt(x))t≥0 follows the equation Xt(x)=x+ ∑i=1N¯tY¯i− ∑i=1NtYi,t≥0.

In [9], the r.v.’s (Yi)i≥1 and (Y¯i)i≥1 , and the processes (Nt)t≥0 and (N¯t)t≥0 are assumed to be mutually independent. In this paper, we suppose that the claim sizes (Yi)i≥1 and the inter-claim times are not independent but with a dependence structure modeled by a Farlie–Gumbel–Morgenstern copula, and we make the same assumption concerning premium arrivals.

To be more precise, let (Ti)i≥1 be a sequence of inter-arrival times of (Nt)t≥0 . In particular, T1 is the time of the first claim. Thus, (Ti)i≥1 are i.i.d. r.v.’s with p.d.f. fT(t)=λe−λt . We assume that (Yi,Ti)i≥1 are i.i.d. random vectors and for every fixed i≥1 , the dependence structure between Yi and Ti is modeled by a Farlie–Gumbel–Morgenstern copula with parameter θ∈[−1,1] , i.e. CθFGM(u1,u2)=u1u2+θu1u2(1−u1)(1−u2),u1,u2∈[0,1] (see, e.g., [21, 37] for more information on copulas). In other words, a claim size depends on the time elapsed from the previous claim. Therefore, the bivariate c.d.f. of (Yi,Ti)i≥1 is defined by FY,T(y,t)=CθFGM(FY(y),FT(t))=FY(y)FT(t)+θFY(y)FT(t)(1−FY(y))(1−FT(t)),y≥0,t≥0. The corresponding bivariate p.d.f. of (Yi,Ti)i≥1 is given by fY,T(y,t)=fY(y)fT(t)+θfY(y)fT(t)(1−2FY(y))(1−2FT(t))=λe−λtfY(y)+θhY(y)(2λe−2λt−λe−λt),y≥0,t≥0, where hY(y)=fY(y)(1−2FY(y)) , y≥0 . Note that the case θ=0 corresponds to the situation where the claim sizes and the inter-claim times are independent.

Next, let (T¯i)i≥1 be a sequence of inter-arrival times of (N¯t)t≥0 . In particular, T¯1 is the time of the first premium. Therefore, (T¯i)i≥1 are i.i.d. r.v.’s with p.d.f. fT¯(t)=λ¯e−λ¯t . We also suppose that (Y¯i,T¯i)i≥1 are i.i.d. random vectors and for every fixed i≥1 , the dependence structure between Y¯i and T¯i is modeled by a Farlie–Gumbel–Morgenstern copula with parameter θ¯∈[−1,1] . So the bivariate p.d.f. of (Y¯i,T¯i)i≥1 is given by fY¯,T¯(y,t)=λ¯e−λ¯tfY¯(y)+θ¯hY¯(y)(2λ¯e−2λ¯t−λ¯e−λ¯t),y≥0,t≥0, where hY¯(y)=fY¯(y)(1−2FY¯(y)) , y≥0 . The case θ¯=0 corresponds to the situation where the premium sizes and the inter-premium times are independent. The random vectors (Yi,Ti)i≥1 and (Y¯i,T¯i)i≥1 are assumed to be mutually independent.

From (2) and (3) we obtain the conditional p.d.f.’s of the claim and premium sizes: fY|T(y|t)=fY,T(y,t)fT(t)=fY(y)+θhY(y)(2e−λt−1),y≥0,t≥0, and fY¯|T¯(y|t)=fY¯,T¯(y,t)fT¯(t)=fY¯(y)+θ¯hY¯(y)(2e−λ¯t−1),y≥0,t≥0.

Moreover, we suppose that the insurance company pays dividends to its shareholders according to the following threshold dividend strategy. Let b>0 0$]]> be a threshold. When the surplus is below b, no dividends are paid. When the surplus exceeds or equals b, dividends are paid continuously at a rate d>0 0$]]>. Let (Xtb(x))t≥0 denote the modified surplus process under this threshold dividend strategy. Then Xtb(x)=x+ ∑i=1N¯tY¯i− ∑i=1NtYi−d∫0t1(Xsb(x)≥b)ds,t≥0, where 1(·) is the indicator function.

Let (Dt)t≥0 denote the dividend distributing process. For the threshold dividend strategy described above, we have dDt=ddtifXtb(x)≥b,0ifXtb(x)<b.

Next, let τb(x)=inf{t≥0:Xtb(x)<0} be the ruin time for the risk process (Xtb(x))t≥0 defined by (6). In what follows, we omit the dependence on x and write τb instead of τb(x) when no confusion can arise.

For δ0≥0 , the Gerber–Shiu function is defined by m(x,b)=E[e−δ0τbw(Xτb−b(x),|Xτbb(x)|)1(τb<∞)|X0b(x)=x],x≥0, where w(·,·) is a bounded non-negative measurable function, Xτb−b(x) is the surplus immediately before ruin and |Xτbb(x)| is a deficit at ruin. Note that if w(·,·)≡1 and δ0=0 , then m(x,b) becomes the infinite-horizon ruin probability ψ(x)=E[1(τb<∞)|X0b(x)=x].

For δ>0 0$]]>, the expected discounted dividend payments until ruin are defined by v(x,b)=E[∫0τbe−δtdDt|X0b(x)=x],x≥0.

For simplicity of notation, we also write m(x) and v(x) instead of m(x,b) and v(x,b) , respectively, when no confusion can arise. Moreover, we set m(x,b)=m1(x)ifx∈[0,b],m2(x)ifx∈[b,∞), and v(x,b)=v1(x)ifx∈[0,b],v2(x)ifx∈[b,∞). Thus, the functions m1(x) and v1(x) are defined on [0,b] , the functions m2(x) and v2(x) are defined on [b,∞) and we have m1(b)=m2(b) and v1(b)=v2(b) .

 Equations for the Gerber–Shiu function

Let the surplus process (Xtb(x))t≥0 follow (6) under the above assumptions with θ≠0 and θ¯≠0 . Moreover, let the p.d.f.’s fY(y) and fY¯(y) have the derivatives fY′(y) and fY¯′(y) on R+ , which are continuous and bounded on R+ , and let w(u1,u2) have the second derivatives wu1u1″(u1,u2) , wu1u2″(u1,u2) and wu2u2″(u1,u2) on R+2 , which are continuous and bounded on R+2 as functions of two variables. Then the Gerber–Shiu function m(x) satisfies the equations (λ+λ¯+δ0)m1(x)=λ(∫0xm1(x−y)fY(y)dy+∫x∞w(x,y−x)fY(y)dy)+λθ(λ¯+δ0)2λ+λ¯+δ0(∫0xm1(x−y)hY(y)dy+∫x∞w(x,y−x)hY(y)dy)+λ¯(∫0b−xm1(x+y)fY¯(y)dy+∫b−x∞m2(x+y)fY¯(y)dy)+λ¯θ¯(λ+δ0)λ+2λ¯+δ0(∫0b−xm1(x+y)hY¯(y)dy+∫b−x∞m2(x+y)hY¯(y)dy),x∈[0,b], and d3m2‴(x)+(4λ+4λ¯+3δ0)d2m2″(x)+((λ+2λ¯+δ0)(3λ+2λ¯+2δ0)+(2λ+λ¯+δ0)(λ+λ¯+δ0))dm2′(x)+(λ+2λ¯+δ0)(2λ+λ¯+δ0)(λ+λ¯+δ0)m2(x)=(λ+2λ¯+δ0)(2λ+λ¯+δ0)β1(x)+(3λ+3λ¯+2δ0)dβ1′(x)+d2β1″(x)−2(λ+2λ¯+δ0)β2(x)−2dβ2′(x)+2λ¯2θ¯(λ¯−λ)∫0∞m2(x+y)hY¯(y)dy,x∈[b,∞), where β1(x)=λ∫x−bxm1(x−y)(fY(y)+θhY(y))dy+λ∫0x−bm2(x−y)(fY(y)+θhY(y))dy+λ∫x∞w(x,y−x)(fY(y)+θhY(y))dy+λ¯∫0∞m2(x+y)(fY¯(y)+θ¯hY¯(y))dy,x∈[b,∞), and β2(x)=λ2θ(∫x−bxm1(x−y)hY(y)dy+∫0x−bm2(x−y)hY(y)dy+∫x∞w(x,y−x)hY(y)dy)+λ¯2θ¯∫0∞m2(x+y)hY¯(y)dy,x∈[b,∞).

It is easily seen that the time of the first jump of (Xtb(x))t≥0 , T1∧T¯1 , is exponentially distributed with mean 1/(λ+λ¯) . Furthermore, P[T1∧T¯1=T1]=λ/(λ+λ¯) and P[T1∧T¯1=T¯1]=λ¯/(λ+λ¯) .

We first deal with the case x∈[0,b] . Considering the time and the size of the first jump of (Xtb(x))t≥0 and applying the law of total probability we obtain m(x)=∫0∞e−(λ+λ¯)t(λ∫0xe−δ0tm(x−y)fY|T(y|t)dy+λ∫x∞e−δ0tw(x,y−x)fY|T(y|t)dy+λ¯∫0∞e−δ0tm(x+y)fY¯|T¯(y|t)dy)dt,x∈[0,b].

Substituting (4) and (5) into (11) and taking into account (7) give m1(x)=∫0∞e−(λ+λ¯+δ0)t(λ∫0xm1(x−y)(fY(y)+θhY(y)(2e−λt−1))dy+λ∫x∞w(x,y−x)(fY(y)+θhY(y)(2e−λt−1))dy+λ¯∫0b−xm1(x+y)(fY¯(y)+θ¯hY¯(y)(2e−λ¯t−1))dy+λ¯∫b−x∞m2(x+y)(fY¯(y)+θ¯hY¯(y)(2e−λ¯t−1))dy)dt,x∈[0,b].

Separating the integrals on the right-hand side of (12) into integrals w.r.t. either t or y yields m1(x)=λ∫0∞e−(λ+λ¯+δ0)tdt(∫0xm1(x−y)fY(y)dy+∫x∞w(x,y−x)fY(y)dy)+λθ∫0∞e−(λ+λ¯+δ0)t(2e−λt−1)dt(∫0xm1(x−y)hY(y)dy+∫x∞w(x,y−x)hY(y)dy)+λ¯∫0∞e−(λ+λ¯+δ0)tdt(∫0b−xm1(x+y)fY¯(y)dy+∫b−x∞m2(x+y)fY¯(y)dy)+λ¯θ¯∫0∞e−(λ+λ¯+δ0)t(2e−λ¯t−1)dt(∫0b−xm1(x+y)hY¯(y)dy+∫b−x∞m2(x+y)hY¯(y)dy),x∈[0,b].

Taking the integrals w.r.t. t on the right-hand side of (13) we get m1(x)=λλ+λ¯+δ0(∫0xm1(x−y)fY(y)dy+∫x∞w(x,y−x)fY(y)dy)+λθ(λ¯+δ0)(2λ+λ¯+δ0)(λ+λ¯+δ0)(∫0xm1(x−y)hY(y)dy+∫x∞w(x,y−x)hY(y)dy)+λ¯λ+λ¯+δ0(∫0b−xm1(x+y)fY¯(y)dy+∫b−x∞m2(x+y)fY¯(y)dy)+λ¯θ¯(λ+δ0)(λ+2λ¯+δ0)(λ+λ¯+δ0)(∫0b−xm1(x+y)hY¯(y)dy+∫b−x∞m2(x+y)hY¯(y)dy),x∈[0,b], which yields (9).

Let now x∈[b,∞) . Considering the time and the size of the first jump of (Xtb(x))t≥0 and applying the law of total probability we have m(x)=∫0(x−b)/de−(λ+λ¯)t(λ∫0x−dte−δ0tm(x−dt−y)fY|T(y|t)dy+λ∫x−dt∞e−δ0tw(x−dt,y−x−dt)fY|T(y|t)dy+λ¯∫0∞e−δ0tm(x−dt+y)fY¯|T¯(y|t)dy)dt+∫(x−b)/d∞e−(λ+λ¯)t(λ∫0be−δ0tm(b−y)fY|T(y|t)dy+λ∫b∞e−δ0tw(b,y−b)fY|T(y|t)dy+λ¯∫0∞e−δ0tm(b+y)fY¯|T¯(y|t)dy)dt,x∈[b,∞).

Substituting (4) and (5) into (14) and taking into account (7) give m2(x)=I1,2,3(x)+I4,5,6(x),x∈[b,∞), where I1,2,3(x)=∫0(x−b)/de−(λ+λ¯+δ0)t×(λ∫0x−dt−bm2(x−dt−y)(fY(y)+θhY(y)(2e−λt−1))dy+λ∫x−dt−bx−dtm1(x−dt−y)(fY(y)+θhY(y)(2e−λt−1))dy+λ∫x−dt∞w(x−dt,y−x+dt)(fY(y)+θhY(y)(2e−λt−1))dy+λ¯∫0∞m2(x−dt+y)(fY¯(y)+θ¯hY¯(y)(2e−λ¯t−1))dy)dt and I4,5,6(x)=∫(x−b)/d∞e−(λ+λ¯+δ0)t×(λ∫0bm1(b−y)(fY(y)+θhY(y)(2e−λt−1))dy+λ∫b∞w(b,y−b)(fY(y)+θhY(y)(2e−λt−1))dy+λ¯∫0∞m2(b+y)(fY¯(y)+θ¯hY¯(y)(2e−λ¯t−1))dy)dt.

Changing the variable x−dt=s in the outer integral of the expression for I1,2,3(x) yields I1,2,3(x)=1d∫bxe−(λ+λ¯+δ0)(x−s)/d×(λ∫0s−bm2(s−y)(fY(y)+θhY(y)(2e−λ(x−s)/d−1))dy+λ∫s−bsm1(s−y)(fY(y)+θhY(y)(2e−λ(x−s)/d−1))dy+λ∫s∞w(s,y−s)(fY(y)+θhY(y)(2e−λ(x−s)/d−1))dy+λ¯∫0∞m2(s+y)(fY¯(y)+θ¯hY¯(y)(2e−λ¯(x−s)/d−1))dy)ds=1de−(λ+λ¯+δ0)x/dI1(x)+2de−(2λ+λ¯+δ0)x/dI2(x)+2de−(λ+2λ¯+δ0)x/dI3(x),x∈[b,∞), where I1(x)=∫bxe(λ+λ¯+δ0)s/d(λ∫0s−bm2(s−y)(fY(y)−θhY(y))dy+λ∫s−bsm1(s−y)(fY(y)−θhY(y))dy+λ∫s∞w(s,y−s)(fY(y)−θhY(y))dy+λ¯∫0∞m2(s+y)(fY¯(y)−θ¯hY¯(y))dy)ds,I2(x)=λθ∫bxe(2λ+λ¯+δ0)s/d(∫0s−bm2(s−y)hY(y)dy+∫s−bsm1(s−y)hY(y)dy+∫s∞w(s,y−s)hY(y)dy)ds and I3(x)=λ¯θ¯∫bxe(λ+2λ¯+δ0)s/d(∫0∞m2(s+y)hY¯(y)dy)ds.

Separating the integrals in the expression for I4,5,6(x) into integrals w.r.t. either t or y we get I4,5,6(x)=λ∫(x−b)/d∞e−(λ+λ¯+δ0)tdt(∫0bm1(b−y)fY(y)dy+∫b∞w(b,y−b)fY(y)dy)+λθ∫(x−b)/d∞e−(λ+λ¯+δ0)t(2e−λt−1)dt(∫0bm1(b−y)hY(y)dy+∫b∞w(b,y−b)hY(y)dy)+λ¯∫(x−b)/d∞e−(λ+λ¯+δ0)tdt∫0∞m2(x+y)fY¯(y)dy+λ¯θ¯∫(x−b)/d∞e−(λ+λ¯+δ0)t(2e−λ¯t−1)dt∫0∞m2(b+y)hY¯(y)dy.

Taking the integrals w.r.t. t on the right-hand side of (17) we obtain I4,5,6(x)=I4(x)+I5(x)+I6(x),x∈[b,∞), where I4(x)=e−(λ+λ¯+δ0)(x−b)/dλ+λ¯+δ0×(λ∫0bm1(b−y)fY(y)dy+λ∫b∞w(b,y−b)fY(y)dy−λθ∫0bm1(b−y)hY(y)dy−λθ∫b∞w(b,y−b)hY(y)dy+λ¯∫0∞m2(b+y)fY¯(y)dy−λ¯θ¯∫0∞m2(b+y)hY¯(y)dy),I5(x)=2λθe−(2λ+λ¯+δ0)(x−b)/d2λ+λ¯+δ0×(∫0bm1(b−y)hY(y)dy+∫b∞w(b,y−b)hY(y)dy) and I6(x)=2λ¯θ¯e−(λ+2λ¯+δ0)(x−b)/dλ+2λ¯+δ0∫0∞m2(b+y)hY¯(y)dy.

Thus, substituting (16) and (18) into (15) we have m2(x)=1de−(λ+λ¯+δ0)x/dI1(x)+2de−(2λ+λ¯+δ0)x/dI2(x)+2de−(λ+2λ¯+δ0)x/dI3(x)+I4(x)+I5(x)+I6(x),x∈[b,∞).

It is easily seen from (9) that m1(x) is continuous on [0,b] , and from (19) we conclude that m2(x) is continuous on [b,∞) . Indeed, the right-hand sides of (9) and (19) are continuous on [0,b] and [b,∞) , respectively, and so are the left-hand sides. Therefore, from (19) we deduce that m2(x) is differentiable on [b,∞) . Differentiating (19) gives m2′(x)=−λ+λ¯+δ0d2e−(λ+λ¯+δ0)x/dI1(x)−2(2λ+λ¯+δ0)d2e−(2λ+λ¯+δ0)x/dI2(x)−2(λ+2λ¯+δ0)d2e−(λ+2λ¯+δ0)x/dI3(x)−λ+λ¯+δ0dI4(x)−2λ+λ¯+δ0dI5(x)−λ+2λ¯+δ0dI6(x)+1dβ1(x),x∈[b,∞), where the function β1(x) is defined in the assertion of the theorem.

Multiplying (19) by (λ+λ¯+δ0)/d and adding (20) we get dm2′(x)+(λ+λ¯+δ0)m2(x)=−2λde−(2λ+λ¯+δ0)x/dI2(x)−2λ¯de−(λ+2λ¯+δ0)x/dI3(x)−λI5(x)−λ¯I6(x)+β1(x),x∈[b,∞).

Since m1(x) and m2(x) are continuous and bounded on [0,b] and [b,∞) , respectively, and w(u1,u2) is continuous and bounded on R+2 as a function of two variables, from (21) we conclude that so is m2′(x) on [b,∞) . Taking into account that fY′(y) and fY¯′(y) are continuous and bounded on R+ , and wu1′(u1,u2) and wu2′(u1,u2) are continuous and bounded on R+2 , from (9) we conclude that so is m1′(x) on [0,b] . Hence, β1(x) is differentiable on [b,∞) . From this and (21) it follows that m2(x) is twice differentiable on [b,∞) . Differentiating (21) gives dm2″(x)+(λ+λ¯+δ0)m2′(x)=2λ(2λ+λ¯+δ0)d2e−(2λ+λ¯+δ0)x/dI2(x)+2λ¯(λ+2λ¯+δ0)d2e−(λ+2λ¯+δ0)x/dI3(x)+λ(2λ+λ¯+δ0)dI5(x)+λ¯(λ+2λ¯+δ0)dI6(x)+β1′(x)−2dβ2(x),x∈[b,∞), where β1′(x)=λ∫x−bxm1′(x−y)(fY(y)+θhY(y))dy+λ∫0x−bm2′(x−y)(fY(y)+θhY(y))dy+λ∫x∞(wu1′(x,y−x)−wu2′(x,y−x))(fY(y)+θhY(y))dy+λ¯∫0∞m2′(x+y)(fY¯(y)+θ¯hY¯(y))dy+λ(m1(0)−w(x,0))(fY(x)+θhY(x)),x∈[b,∞), which is continuous and bounded on [b,∞) , and the function β2(x) is defined in the assertion of the theorem. Here wu1′(·,·) and wu2′(·,·) stand for the partial derivatives of w(u1,u2) w.r.t. the first and the second variables, respectively.

Multiplying (21) by (2λ+λ¯+δ0)/d and adding (22) we obtain d2m2″(x)+(3λ+2λ¯+2δ0)dm2′(x)+(2λ+λ¯+δ0)(λ+λ¯+δ0)m2(x)=2λ¯(λ¯−λ)de−(λ+2λ¯+δ0)x/dI3(x)+λ¯(λ¯−λ)I6(x)+(2λ+λ¯+δ0)β1(x)+dβ1′(x)−2β2(x),x∈[b,∞).

It is easily seen from (23) that m2″(x) is continuous and bounded on [b,∞) . Taking into account that fY′(y) and fY¯′(y) are continuous and bounded on R+ , and wu1u1″(u1,u2) , wu1u2″(u1,u2) and wu2u2″(u1,u2) are continuous and bounded on R+2 , from (9) we conclude that so is m1″(x) on [0,b] . Hence, β1(x) is twice differentiable on [b,∞) . Moreover, applying similar arguments shows that β2(x) is differentiable on [b,∞) . From this and (23) it follows that m2(x) has the third derivative on [b,∞) . Differentiating (23) gives d2m2‴(x)+(3λ+2λ¯+2δ0)dm2″(x)+(2λ+λ¯+δ0)(λ+λ¯+δ0)m2′(x)=−2λ¯(λ¯−λ)(λ+2λ¯+δ0)d2e−(λ+2λ¯+δ0)x/dI3(x)−λ¯(λ¯−λ)(λ+2λ¯+δ0)dI6(x)+(2λ+λ¯+δ0)β1′(x)+dβ1″(x)−2β2′(x)+2λ¯2θ¯(λ¯−λ)d∫0∞m2(x+y)hY¯(y)dy,x∈[b,∞).

Multiplying (23) by (λ+2λ¯+δ0)/d and adding (24) yield (10), which completes the proof.  □

To solve equations (9) and (10), we need some boundary conditions. The first one is m1(b)=m2(b) . Next, using standard considerations (see, e.g., [34, 36, 39]) we can show that limx→∞m2(x)=0 provided that the net profit condition holds. Finally, we can substitute x=b into the intermediate equations (e.g., equation (21)) to get additional boundary conditions involving derivatives of m2(x) . Furthermore, equations (9) and (10) are not solvable analytically in the general case, so we can use, for instance, numerical methods. Nevertheless, we can give explicit expressions for m(x) in some particular cases (see Section 5). The uniqueness of the required solutions to these equations should be justified in each case.

The corresponding model without dividend payments is obtained by b→∞ . In this case, the Gerber–Shiu function m(x) satisfies the integral equation (λ+λ¯+δ0)m(x)=λ(∫0xm(u)fY(x−u)du+∫0∞w(x,u)fY(x+u)du)+λθ(λ¯+δ0)2λ+λ¯+δ0(∫0xm(u)hY(x−u)du+∫0∞w(x,u)hY(x+u)du)+λ¯∫x∞m(u)fY¯(u−x)du+λ¯θ¯(λ+δ0)λ+2λ¯+δ0∫x∞m(u)hY¯(u−x)du,x∈[0,∞). Note that equation (25) for the ruin probability coincides with the equation derived in [9] (see also [34]) if θ=0 and θ¯=0 .

In Theorem 1, we assume that θ≠0 and θ¯≠0 . Otherwise, we do not need to differentiate (19) three times and can obtain equations not involving the third derivative of m2(x) instead of (10).

Thus, if θ=0 and θ¯=0 , from (21) we have dm2′(x)+(λ+λ¯+δ0)m2(x)=β1(x),x∈[b,∞).

Next, if θ≠0 and θ¯=0 , from (23) we have d2m2″(x)+(3λ+2λ¯+2δ0)dm2′(x)+(2λ+λ¯+δ0)(λ+λ¯+δ0)m2(x)=(2λ+λ¯+δ0)β1(x)+dβ1′(x)−2β2(x),x∈[b,∞).

Finally, if θ=0 and θ¯≠0 , multiplying (21) by (λ+2λ¯+δ0)/d and adding (22) we get d2m2″(x)+(2λ+3λ¯+2δ0)dm2′(x)+(λ+2λ¯+δ0)(λ+λ¯+δ0)m2(x)=(λ+2λ¯+δ0)β1(x)+dβ1′(x)−2β2(x),x∈[b,∞). Note that to obtain (26)–(28), it is enough to have weaker smoothness assumptions on fY(y) , fY¯(y) and w(u1,u2) .

Equation (9) holds in all possible cases. Furthermore, (9) involves no derivatives and holds under weaker assumptions than (10). To be more precise, we do not need the differentiability of fY(y) , fY¯(y) and w(u1,u2) to get (9).

 Equations for the expected discounted dividend payments until ruin

Let the surplus process (Xtb(x))t≥0 follow (6) under the above assumptions with θ≠0 and θ¯≠0 . Moreover, let the p.d.f.’s fY(y) and fY¯(y) have the derivatives fY′(y) and fY¯′(y) on R+ , which are continuous and bounded on R+ . Then the expected discounted dividend payments until ruin v(x) satisfy the equations (λ+λ¯+δ)v1(x)=λ∫0xv1(x−y)fY(y)dy+λθ(λ¯+δ)2λ+λ¯+δ∫0xv1(x−y)hY(y)dy+λ¯(∫0b−xv1(x+y)fY¯(y)dy+∫b−x∞v2(x+y)fY¯(y)dy)+λ¯θ¯(λ+δ)λ+2λ¯+δ(∫0b−xv1(x+y)hY¯(y)dy+∫b−x∞v2(x+y)hY¯(y)dy),x∈[0,b], and d3v2‴(x)+(4λ+4λ¯+3δ)d2v2″(x)+((λ+2λ¯+δ)(3λ+2λ¯+2δ)+(2λ+λ¯+δ)(λ+λ¯+δ))dv2′(x)+(λ+2λ¯+δ)(2λ+λ¯+δ)(λ+λ¯+δ)v2(x)=(λ+2λ¯+δ)(2λ+λ¯+δ)β3(x)+(3λ+3λ¯+2δ)dβ3′(x)+d2β3″(x)−2(λ+2λ¯+δ)β4(x)−2dβ4′(x)+(λ+2λ¯+δ)(2λ+λ¯+δ)d+2λ¯2θ¯(λ¯−λ)∫0∞v2(x+y)hY¯(y)dy,x∈[b,∞), where β3(x)=λ∫x−bxv1(x−y)(fY(y)+θhY(y))dy+λ∫0x−bv2(x−y)(fY(y)+θhY(y))dy+λ¯∫0∞v2(x+y)(fY¯(y)+θ¯hY¯(y))dy,x∈[b,∞), and β4(x)=λ2θ(∫x−bxv1(x−y)hY(y)dy+∫0x−bv2(x−y)hY(y)dy)+λ¯2θ¯∫x∞v2(u)hY¯(u−x)du,x∈[b,∞).

The proof is similar to the proof of Theorem 1, so we omit detailed considerations.

Let x∈[0,b] . Considering the time and the size of the first jump of (Xtb(x))t≥0 and applying the law of total probability, we obtain v(x)=∫0∞e−(λ+λ¯)t(λ∫0xe−δtv(x−y)fY|T(y|t)dy+λ¯∫0∞e−δtv(x+y)fY¯|T¯(y|t)dy)dt,x∈[0,b].

Comparing (31) with (11) and applying arguments similar to those in the proof of Theorem 1 yield (29).

Let now x∈[b,∞) . By the law of total probability, we have v(x)=∫0(x−b)/de−(λ+λ¯)t((λ+λ¯)∫0tde−δsds+λ∫0x−dte−δtv(x−dt−y)fY|T(y|t)dy+λ¯∫0∞e−δtv(x−dt+y)fY¯|T¯(y|t)dy)dt+∫(x−b)/d∞e−(λ+λ¯)t((λ+λ¯)∫0(x−b)/dde−δsds+λ∫0be−δtv(b−y)fY|T(y|t)dy+λ¯∫0∞e−δtv(b+y)fY¯|T¯(y|t)dy)dt,x∈[b,∞).

Taking into account that ∫0(x−b)/d(λ+λ¯)e−(λ+λ¯)t∫0tde−δsdsdt=(λ+λ¯)dδ∫0(x−b)/de−(λ+λ¯)t(1−e−δt)dt=dδ(δλ+λ¯+δ−e−(λ+λ¯)(x−b)/d+λ+λ¯λ+λ¯+δe−(λ+λ¯+δ)(x−b)/d) and ∫(x−b)/d∞(λ+λ¯)e−(λ+λ¯)t∫0(x−b)/dde−δsdsdt=(λ+λ¯)dδ(1−e−δ(x−b)/d)∫(x−b)/d∞e−(λ+λ¯)tdt=dδ(e−(λ+λ¯)(x−b)/d−e−(λ+λ¯+δ)(x−b)/d), substituting (4) and (5) into (32) and using (8) give v2(x)=I7,8,9(x)+I10,11,12(x)+dλ+λ¯+δ(1−e−(λ+λ¯+δ)(x−b)/d),x∈[b,∞), where I7,8,9(x)=∫0(x−b)/de−(λ+λ¯+δ)t×(λ∫0x−dt−bv2(x−dt−y)(fY(y)+θhY(y)(2e−λt−1))dy+λ∫x−dt−bx−dtv1(x−dt−y)(fY(y)+θhY(y)(2e−λt−1))dy+λ¯∫0∞v2(x−dt+y)(fY¯(y)+θ¯hY¯(y)(2e−λ¯t−1))dy)dt and I10,11,12(x)=∫(x−b)/d∞e−(λ+λ¯+δ)t×(λ∫0bv1(b−y)(fY(y)+θhY(y)(2e−λt−1))dy+λ¯∫0∞v2(b+y)(fY¯(y)+θ¯hY¯(y)(2e−λ¯t−1))dy)dt.

Changing the variable x−dt=s in the outer integral of the expression for I7,8,9(x) yields I7,8,9(x)=1de−(λ+λ¯+δ)x/dI7(x)+2de−(2λ+λ¯+δ)x/dI8(x)+2de−(λ+2λ¯+δ)x/dI9(x),x∈[b,∞), where I7(x)=∫bxe(λ+λ¯+δ)s/d(λ∫0s−bv2(s−y)(fY(y)−θhY(y))dy+λ∫s−bsv1(s−y)(fY(y)−θhY(y))dy+λ¯∫0∞v2(s+y)(fY¯(y)−θ¯hY¯(y))dy)ds,I8(x)=λθ∫bxe(2λ+λ¯+δ)s/d(∫0s−bv2(s−y)hY(y)dy+∫s−bsv1(s−y)hY(y)dy)ds and I9(x)=λ¯θ¯∫bxe(λ+2λ¯+δ)s/d(∫0∞v2(s+y)hY¯(y)dy)ds.

Separating the integrals in the expression for I10,11,12(x) into integrals w.r.t. either t or y and then taking the integrals w.r.t. t we obtain I10,11,12(x)=I10(x)+I11(x)+I12(x),x∈[b,∞), where I10(x)=e−(λ+λ¯+δ)(x−b)/dλ+λ¯+δ×(λ∫0bv1(b−y)fY(y)dy−λθ∫0bv1(b−y)hY(y)dy+λ¯∫0∞v2(b+y)fY¯(y)dy−λ¯θ¯∫0∞v2(b+y)hY¯(y)dy),I11(x)=2λθe−(2λ+λ¯+δ)(x−b)/d2λ+λ¯+δ∫0bv1(b−y)hY(y)dy and I12(x)=2λ¯θ¯e−(λ+2λ¯+δ)(x−b)/dλ+2λ¯+δ∫0∞v2(b+y)hY¯(y)dy.

Thus, substituting (34) and (35) into (33) we have v2(x)=1de−(λ+λ¯+δ)x/dI7(x)+2de−(2λ+λ¯+δ)x/dI8(x)+2de−(λ+2λ¯+δ)x/dI9(x)+I10(x)+I11(x)+I12(x)+dλ+λ¯+δ(1−e−(λ+λ¯+δ)(x−b)/d),x∈[b,∞).

It is easily seen from (29) that v1(x) is continuous on [0,b] , and from (36) we conclude that v2(x) is continuous on [b,∞) . Hence, from (36) we deduce that v2(x) is differentiable on [b,∞) . Differentiating (36) gives v2′(x)=−λ+λ¯+δd2e−(λ+λ¯+δ)x/dI7(x)−2(2λ+λ¯+δ)d2e−(2λ+λ¯+δ)x/dI8(x)−2(λ+2λ¯+δ)d2e−(λ+2λ¯+δ)x/dI9(x)−λ+λ¯+δdI10(x)−2λ+λ¯+δdI11(x)−λ+2λ¯+δdI12(x)+1dβ3(x)+e−(λ+λ¯+δ)(x−b)/d,x∈[b,∞), where the function β3(x) is defined in the assertion of the theorem.

Multiplying (36) by (λ+λ¯+δ)/d and adding (37) we get dv2′(x)+(λ+λ¯+δ)v2(x)=−2λde−(2λ+λ¯+δ)x/dI8(x)−2λ¯de−(λ+2λ¯+δ)x/dI9(x)−λI11(x)−λ¯I12(x)+β3(x)+d,x∈[b,∞).

Since v1(x) and v2(x) are continuous and bounded on [0,b] and [b,∞) , respectively, from (38) we conclude that so is v2′(x) on [b,∞) . Taking into account that fY′(y) and fY¯′(y) are continuous and bounded on R+ , from (29) we conclude that so is v1′(x) on [0,b] . Hence, β3(x) is differentiable on [b,∞) . From this and (38) it follows that v2(x) is twice differentiable on [b,∞) . Differentiating (38) gives dv2″(x)+(λ+λ¯+δ)v2′(x)=2λ(2λ+λ¯+δ)d2e−(2λ+λ¯+δ)x/dI8(x)+2λ¯(λ+2λ¯+δ)d2e−(λ+2λ¯+δ)x/dI9(x)+λ(2λ+λ¯+δ)dI11(x)+λ¯(λ+2λ¯+δ)dI12(x)+β3′(x)−2dβ4(x),x∈[b,∞), where β3′(x)=λ∫x−bxv1′(x−y)(fY(y)+θhY(y))dy+λ∫0x−bv2′(x−y)(fY(y)+θhY(y))dy+λ¯∫0∞v2′(x+y)(fY¯(y)+θ¯hY¯(y))dy+λv1(0)(fY(x)+θhY(x)),x∈[b,∞), which is continuous and bounded on [b,∞) , and the function β4(x) is defined in the assertion of the theorem.

Multiplying (38) by (2λ+λ¯+δ)/d and adding (39) we obtain d2v2″(x)+(3λ+2λ¯+2δ)dv2′(x)+(2λ+λ¯+δ)(λ+λ¯+δ)v2(x)=2λ¯(λ¯−λ)de−(λ+2λ¯+δ)x/dI9(x)+λ¯(λ¯−λ)I12(x)+(2λ+λ¯+δ)β3(x)+dβ3′(x)−2β4(x)+(2λ+λ¯+δ)d,x∈[b,∞).

It is easily seen from (40) that v2″(x) is continuous and bounded on [b,∞) . Taking into account that fY′(y) and fY¯′(y) are continuous and bounded on R+ , from (29) we conclude that so is v1″(x) on [0,b] . Hence, β3(x) is twice differentiable on [b,∞) . Moreover, applying similar arguments shows that β4(x) is differentiable on [b,∞) . From this and (40) it follows that v2(x) has the third derivative on [b,∞) . Differentiating (40) gives d2v2‴(x)+(3λ+2λ¯+2δ)dv2″(x)+(2λ+λ¯+δ)(λ+λ¯+δ)v2′(x)=−2λ¯(λ¯−λ)(λ+2λ¯+δ)d2e−(λ+2λ¯+δ)x/dI9(x)−λ¯(λ¯−λ)(λ+2λ¯+δ)dI12(x)+(2λ+λ¯+δ)β3′(x)+dβ3″(x)−2β4′(x)+2λ¯2θ¯(λ¯−λ)d∫0∞v2(x+y)hY¯(y)dy,x∈[b,∞).

Multiplying (40) by (λ+2λ¯+δ0)/d and adding (41) yield (30), which completes the proof.  □

To solve equations (29) and (30), we use the following boundary conditions. First of all, we have v1(b)=v2(b) . Next, if the net profit condition holds, applying arguments similar to those in [40, p. 70] we can show that limx→∞v2(x)=d/δ . Moreover, we can substitute x=b into the intermediate equations (e.g., equation (38)) to get additional boundary conditions involving derivatives of v2(x) . The uniqueness of the required solutions should also be justified. If θ=0 and θ¯=0 , we can find explicit solutions to the equations (see Section 5).

If at least one of the parameters θ and θ¯ is equal to 0, we do not need to differentiate (36) three times and can obtain equations not involving the third derivative of v2(x) instead of (30).

Thus, if θ=0 and θ¯=0 , from (38) we have dv2′(x)+(λ+λ¯+δ0)v2(x)=β3(x)+d,x∈[b,∞).

If θ≠0 and θ¯=0 , from (40) we get d2v2″(x)+(3λ+2λ¯+2δ0)dv2′(x)+(2λ+λ¯+δ)(λ+λ¯+δ)v2(x)=(2λ+λ¯+δ)β3(x)+dβ3′(x)−2β4(x)+(2λ+λ¯+δ)d,x∈[b,∞).

If θ=0 and θ¯≠0 , multiplying (38) by (λ+2λ¯+δ0)/d and adding (39) we have d2v2″(x)+(2λ+3λ¯+2δ)dv2′(x)+(λ+2λ¯+δ)(λ+λ¯+δ)v2(x)=(λ+2λ¯+δ)β3(x)+dβ3′(x)−2β4(x)+(λ+2λ¯+δ)d,x∈[b,∞). To obtain (42)–(44), it is enough to have weaker smoothness assumptions on fY(y) and fY¯(y) .

Equation (29) is true in all possible cases. Since (29) involves no derivatives, it holds under weaker assumptions than (30). To obtain (29), we do not need the differentiability of fY(y) and fY¯(y) .

 Exponentially distributed claim and premium sizes

In this section, we deal with exponentially distributed claim and premium sizes, i.e. fY(y)=1μe−y/μ,hY(y)=2μe−2y/μ−1μe−y/μ,y≥0, and fY¯(y)=1μ¯e−y/μ¯,hY¯(y)=2μ¯e−2y/μ¯−1μ¯e−y/μ¯,y≥0.

The ruin probability in the model without dividend payments

If no dividends are paid, equation (25) for the ruin probability ψ(x) takes the form (λ+λ¯)ψ(x)=λ(∫0xψ(u)fY(x−u)du+∫0∞fY(x+u)du)+λλ¯θ2λ+λ¯(∫0xψ(u)hY(x−u)du+∫0∞hY(x+u)du)+λ¯∫x∞ψ(u)fY¯(u−x)du+λλ¯θ¯λ+2λ¯∫x∞ψ(u)hY¯(u−x)du,x∈[0,∞).

Substituting (45) and (46) into (47) gives (λ+λ¯)ψ(x)=(λ−λλ¯θ2λ+λ¯)I13(x)+λλ¯θ2λ+λ¯I14(x)+(λ¯−λλ¯θ¯λ+2λ¯)I15(x)+λλ¯θ¯λ+2λ¯I16(x)+(λ−λλ¯θ2λ+λ¯)e−x/μ+λλ¯θ2λ+λ¯e−2x/μ,x∈[0,∞), where I13(x)=1μe−x/μ∫0xψ(u)eu/μdu,I14(x)=2μe−2x/μ∫0xψ(u)e2u/μdu,I15(x)=1μ¯ex/μ¯∫x∞ψ(u)e−u/μ¯du,I16(x)=2μ¯e2x/μ¯∫x∞ψ(u)e−2u/μ¯du.

We now show that if either θ=0 or θ¯=0 , integro-differential equation (48) can be reduced to a third-order linear differential equation with constant coefficients.

Let the surplus process (Xt(x))t≥0 follow (1) under the above assumptions, and let claim and premium sizes be exponentially distributed with means μ and μ¯ , respectively.

If θ≠0 and θ¯=0 , then ψ(x) is a solution to the differential equation μ2μ¯(λ+λ¯)(2λ+λ¯)ψ‴(x)+(μμ¯(2λ+3λ¯)(2λ+λ¯)−λμ2(2λ+λ¯)+λλ¯μμ¯θ)ψ″(x)+(2(λ¯μ¯−λμ)(2λ+λ¯)+λλ¯μθ)ψ′(x)=0,x∈[0,∞).

If θ=0 and θ¯≠0 , then ψ(x) is a solution to the differential equation μμ¯2(λ+λ¯)(λ+2λ¯)ψ‴(x)+(−μμ¯(3λ+2λ¯)(λ+2λ¯)+λ¯μ¯2(λ+2λ¯)+λλ¯μμ¯θ¯)ψ″(x)+(2(λμ−λ¯μ¯)(λ+2λ¯)+λλ¯μ¯θ¯)ψ′(x)=0,x∈[0,∞).

First of all, note that I13′(x)=−1μI13(x)+1μψ(x),I14′(x)=−2μI14(x)+2μψ(x),I15′(x)=1μ¯I15(x)−1μ¯ψ(x),I16′(x)=2μ¯I16(x)−2μ¯ψ(x).

From (48), it is easily seen that ψ(x) is differentiable on [0,∞) . Therefore, differentiating (48) yields (λ+λ¯)ψ′(x)=−1μ(λ−λλ¯θ2λ+λ¯)I13(x)−2μλλ¯θ2λ+λ¯I14(x)+1μ¯(λ¯−λλ¯θ¯λ+2λ¯)I15(x)+2μ¯λλ¯θ¯λ+2λ¯I16(x)+(1μ(λ−λλ¯θ2λ+λ¯)+2μλλ¯θ2λ+λ¯−1μ¯(λ¯−λλ¯θ¯λ+2λ¯)−2μ¯λλ¯θ¯λ+2λ¯)ψ(x)−1μ(λ−λλ¯θ2λ+λ¯)e−x/μ−2μλλ¯θ2λ+λ¯e−2x/μ,x∈[0,∞).

Multiplying (51) by μ and adding (48) we obtain μ(λ+λ¯)ψ′(x)+λ¯ψ(x)=−λλ¯θ2λ+λ¯I14(x)+(1+μμ¯)(λ¯−λλ¯θ¯λ+2λ¯)I15(x)+(1+2μμ¯)λλ¯θ¯λ+2λ¯I16(x)+(λλ¯θ2λ+λ¯−μμ¯(λ¯+λλ¯θ¯λ+2λ¯))ψ(x)−λλ¯θ2λ+λ¯e−2x/μ,x∈[0,∞).

From (52), it follows that ψ(x) is twice differentiable on [0,∞) . Differentiating (52) gives μ(λ+λ¯)ψ″(x)+λ¯ψ′(x)=2μλλ¯θ2λ+λ¯I14(x)+1μ¯(1+μμ¯)(λ¯−λλ¯θ¯λ+2λ¯)I15(x)+2μ¯(1+2μμ¯)λλ¯θ¯λ+2λ¯I16(x)+(λλ¯θ2λ+λ¯−μμ¯(λ¯+λλ¯θ¯λ+2λ¯))ψ′(x)−(2μλλ¯θ2λ+λ¯+1μ¯(1+μμ¯)(λ¯−λλ¯θ¯λ+2λ¯)+2μ¯(1+2μμ¯)λλ¯θ¯λ+2λ¯)ψ(x)+2μλλ¯θ2λ+λ¯e−2x/μ,x∈[0,∞).

Multiplying (53) by (−μ¯) and adding (52) we get −μμ¯(λ+λ¯)ψ″(x)+(λμ−λ¯μ¯)ψ′(x)=−(1+2μ¯μ)λλ¯θ2λ+λ¯I14(x)−(1+2μμ¯)λλ¯θ¯λ+2λ¯I16(x)+(−λλ¯μ¯θ2λ+λ¯+λλ¯μθ¯λ+2λ¯)ψ′(x)+((1+2μ¯μ)λλ¯θ2λ+λ¯+(1+2μμ¯)λλ¯θ¯λ+2λ¯)ψ(x)−(1+2μ¯μ)λλ¯θ2λ+λ¯e−2x/μ,x∈[0,∞).

Let now θ≠0 and θ¯=0 . Then (54) takes the form −μμ¯(λ+λ¯)ψ″(x)+(λμ−λ¯μ¯)ψ′(x)=−(1+2μ¯μ)λλ¯θ2λ+λ¯I14(x)−λλ¯μ¯θ2λ+λ¯ψ′(x)+(1+2μ¯μ)λλ¯θ2λ+λ¯ψ(x)−(1+2μ¯μ)λλ¯θ2λ+λ¯e−2x/μ,x∈[0,∞).

From (55), it follows that ψ(x) has the third derivative on [0,∞) . Differentiating (55) yields −μμ¯(λ+λ¯)ψ‴(x)+(λμ−λ¯μ¯)ψ″(x)=2μ(1+2μ¯μ)λλ¯θ2λ+λ¯I14(x)−λλ¯μ¯θ2λ+λ¯ψ″(x)+(1+2μ¯μ)λλ¯θ2λ+λ¯ψ′(x)−2μ(1+2μ¯μ)λλ¯θ2λ+λ¯ψ(x)+2μ(1+2μ¯μ)λλ¯θ2λ+λ¯e−2x/μ,x∈[0,∞).

Multiplying (56) by μ and adding (55) multiplied by 2 we obtain −μ2μ¯(λ+λ¯)ψ‴(x)+(μ(λμ−λ¯μ¯)−2μμ¯(λ+λ¯))ψ″(x)+2(λμ−λ¯μ¯)ψ′(x)=λλ¯μμ¯θ2λ+λ¯ψ″(x)+λλ¯μθ2λ+λ¯ψ′(x),x∈[0,∞), from which (49) follows.

If θ=0 and θ¯≠0 , then (54) takes the form −μμ¯(λ+λ¯)ψ″(x)+(λμ−λ¯μ¯)ψ′(x)=−(1+2μμ¯)λλ¯θ¯λ+2λ¯I16(x)+λλ¯μθ¯λ+2λ¯ψ′(x)+(1+2μμ¯)λλ¯θ¯λ+2λ¯ψ(x),x∈[0,∞).

From (57), it follows that ψ(x) has the third derivative on [0,∞) . Differentiating (57) gives −μμ¯(λ+λ¯)ψ‴(x)+(λμ−λ¯μ¯)ψ″(x)=−2μ¯(1+2μμ¯)λλ¯θ¯λ+2λ¯I16(x)+λλ¯μθ¯λ+2λ¯ψ″(x)+(1+2μμ¯)λλ¯θ¯λ+2λ¯ψ′(x)+2μ¯(1+2μμ¯)λλ¯θ¯λ+2λ¯ψ(x),x∈[0,∞).

Multiplying (58) by (−μ¯) and adding (57) multiplied by 2 we get μμ¯2(λ+λ¯)ψ‴(x)−(μ¯(λμ−λ¯μ¯)+2μμ¯(λ+λ¯))ψ″(x)+2(λμ−λ¯μ¯)ψ′(x)=−λλ¯μμ¯θ¯λ+2λ¯ψ″(x)−λλ¯μ¯θ¯λ+2λ¯ψ′(x),x∈[0,∞), from which (50) follows.  □

To formulate the next theorem, we define the following constants: D1=(μμ¯(2λ+3λ¯)(2λ+λ¯)−λμ2(2λ+λ¯)+λλ¯μμ¯θ)2−4μ2μ¯(λ+λ¯)(2λ+λ¯)(2(λ¯μ¯−λμ)(2λ+λ¯)+λλ¯μθ),z2=−(μμ¯(2λ+3λ¯)(2λ+λ¯)−λμ2(2λ+λ¯)+λλ¯μμ¯θ)+D12μ2μ¯(λ+λ¯)(2λ+λ¯),z3=−(μμ¯(2λ+3λ¯)(2λ+λ¯)−λμ2(2λ+λ¯)+λλ¯μμ¯θ)−D12μ2μ¯(λ+λ¯)(2λ+λ¯) and Δ1=(λ+λ¯−λ¯1−μ¯z2)(μ(λ+λ¯)z3−(λ¯+λ¯μμ¯)μ¯z31−μ¯z3−λλ¯θ2λ+λ¯)−(λ+λ¯−λ¯1−μ¯z3)(μ(λ+λ¯)z2−(λ¯+λ¯μμ¯)μ¯z21−μ¯z2−λλ¯θ2λ+λ¯).

Let the surplus process (Xt(x))t≥0 follow (1) under the above assumptions with θ≠0 and θ¯=0 . Moreover, let claim and premium sizes be exponentially distributed with means μ and μ¯ , respectively, and let λ¯μ¯>λμ \lambda \mu $]]>.

If 2(λ¯μ¯−λμ)(2λ+λ¯)+λλ¯μθ≤0 , then ψ(x)=λ(1−μ¯z3)λ(1−μ¯z3)−λμz3ez3x,x∈[0,∞).

If 2(λ¯μ¯−λμ)(2λ+λ¯)+λλ¯μθ>0 0$]]>, then ψ(x)=C2ez2x+C3ez3x,x∈[0,∞), where the constants C2 and C3 are determined from the system of linear equations (λ+λ¯−λ¯1−μ¯z2)C2+(λ+λ¯−λ¯1−μ¯z3)C3=λ and (μ(λ+λ¯)z2−(λ¯+λ¯μμ¯)μ¯z21−μ¯z2−λλ¯θ2λ+λ¯)C2+(μ(λ+λ¯)z3−(λ¯+λ¯μμ¯)μ¯z31−μ¯z3−λλ¯θ2λ+λ¯)C3=−λλ¯θ2λ+λ¯ provided that Δ1≠0 .

By Lemma 1, ψ(x) is a solution to (49). We now find the general solution to (49). Its characteristic equation is μ2μ¯(λ+λ¯)(2λ+λ¯)z3+(μμ¯(2λ+3λ¯)(2λ+λ¯)−λμ2(2λ+λ¯)+λλ¯μμ¯θ)z2+(2(λ¯μ¯−λμ)(2λ+λ¯)+λλ¯μθ)z=0.

It is evident that z1=0 is a solution to (63). Next, we prove that the equation μ2μ¯(λ+λ¯)(2λ+λ¯)z2+(μμ¯(2λ+3λ¯)(2λ+λ¯)−λμ2(2λ+λ¯)+λλ¯μμ¯θ)z+(2(λ¯μ¯−λμ)(2λ+λ¯)+λλ¯μθ)=0 has two real roots. To this end, we show that its discriminant D1 defined before the assertion of the theorem is positive. We have D1/μ2=((2λμ¯+3λ¯μ¯−λμ)(2λ+λ¯)+λλ¯μ¯θ)2−8μ¯(λ+λ¯)(λ¯μ¯−λμ)(2λ+λ¯)2−4λλ¯μμ¯θ(λ+λ¯)(2λ+λ¯)=(2λ+λ¯)2((2λμ¯+3λ¯μ¯−λμ)2−8μ¯(λ+λ¯)(λ¯μ¯−λμ))+2λλ¯μ¯θ(2λ+λ¯)(2λμ¯+3λ¯μ¯−λμ−2μ(λ+λ¯))+(λλ¯μ¯θ)2=(2λ+λ¯)2(2λ(μ+μ¯)+(λ¯μ¯−λμ))2+2λλ¯μ¯θ(2λ+λ¯)(2(μ¯−μ)(λ+λ¯)+(λ¯μ¯+λμ))+(λλ¯μ¯θ)2=(2λ+λ¯)2(λ¯μ¯−λμ)2+2λλ¯μ¯θ(2λ+λ¯)(λ¯μ¯−λμ)+(λλ¯μ¯θ)2+(2λ+λ¯)2(4λ2(μ+μ¯)2+4λ(μ+μ¯)(λ¯μ¯−λμ))+4λλ¯μ¯θ(μ¯−μ)(λ+λ¯)(2λ+λ¯)=(2λ+λ¯)2(λ¯μ¯−λμ)2+2λλ¯μ¯θ(2λ+λ¯)(λ¯μ¯−λμ)+(λλ¯μ¯θ)2+4λ(2λ+λ¯)(λ2(μ+μ¯)2+λ(μ+μ¯)(λ¯μ¯−λμ)+(λ+λ¯)(λ(μ+μ¯)2+(μ+μ¯)(λ¯μ¯−λμ)+λ¯μ¯θ(μ¯−μ))).

Since λ¯μ¯>λμ \lambda \mu $]]>, it suffices to show that λ(μ+μ¯)2+(μ+μ¯)(λ¯μ¯−λμ)+λ¯μ¯θ(μ¯−μ))>0. 0.\]]]>

It is obvious that the minimal value of the expression on the left-hand side of the above inequality is attained when either θ=1 or θ=−1 and equals either λμ¯(μ+μ¯)+2λμ¯2 or λμ¯(μ+μ¯)+2λ¯μμ¯ , respectively. Both these expressions are positive. Thus, D1>0 0$]]> and (64) has two real roots z2 and z3 defined before the assertion of the theorem.

Next, it is easily seen that μμ¯(2λ+3λ¯)(2λ+λ¯)−λμ2(2λ+λ¯)+λλ¯μμ¯θ>0. 0.\]]]>

Indeed, since λ¯μ¯>λμ \lambda \mu $]]>, we have μμ¯(2λ+3λ¯)(2λ+λ¯)−λμ2(2λ+λ¯)+λλ¯μμ¯θ=μ(2λ+λ¯)(2λμ¯+3λ¯μ¯−λμ)+λλ¯μμ¯θ>2μμ¯(2λ+λ¯)(λ+λ¯)−λλ¯μμ¯>3λλ¯μμ¯>0. 2\mu \bar{\mu }(2\lambda +\bar{\lambda })(\lambda +\bar{\lambda })-\lambda \bar{\lambda }\mu \bar{\mu }>3\lambda \bar{\lambda }\mu \bar{\mu }>0.\end{array}\]]]>

Therefore, by Vieta’s theorem applied to (64), z3<0 . Moreover, z2≥0 if 2(λ¯μ¯−λμ)(2λ+λ¯)+λλ¯μθ≤0 , and z2<0 if 2(λ¯μ¯−λμ)(2λ+λ¯)+λλ¯μθ>0 0$]]>.

Thus, the general solution to (49) has the form ψ(x)=C1+C2ez2x+C3ez3x,x∈[0,∞), where C1 , C2 and C3 are some constants. To determine them, we use the following boundary conditions. Firstly, since λ¯μ¯>λμ \lambda \mu $]]>, using standard considerations (see, e.g., [34, 36, 39]) we can easily show that limx→∞ψ(x)=0 . Consequently, C1=0 and C2=0 if 2(λ¯μ¯−λμ)(2λ+λ¯)+λλ¯μθ≤0 , and C1=0 if 2(λ¯μ¯−λμ)(2λ+λ¯)+λλ¯μθ>0 0$]]>. Secondly, we use intermediate equations to find other constants.

Let now 2(λ¯μ¯−λμ)(2λ+λ¯)+λλ¯μθ≤0 . The constant C3 is determined by letting x=0 in (48), i.e. from the equation (λ+λ¯)ψ(0)=λ¯μ¯∫0∞ψ(u)e−u/μ¯du+λ.

Substituting ψ(x)=C3ez3x into (66) gives (λ+λ¯)C3=λ¯1−μ¯z3C3+λ, from which (59) follows immediately.

If 2(λ¯μ¯−λμ)(2λ+λ¯)+λλ¯μθ>0 0$]]>, then the constants C2 and C3 are determined by letting x=0 in (48) and (52), i.e. from (66) and the equation μ(λ+λ¯)ψ′(0)+λ¯ψ(0)=λ¯μ¯(1+μμ¯)∫0∞ψ(u)e−u/μ¯du+(λλ¯θ2λ+λ¯−λ¯μμ¯)ψ(0)−λλ¯θ2λ+λ¯.

Substituting (60) into (66) and (67) yields equations (61) and (62), respectively. The system of equations (61) and (62) has a unique solution provided that Δ1≠0 .

Note that letting x=0 in (54) (and in (52) when C2=0 ) gives no additional information about unknown constants. Nevertheless, the equalities must hold for the values of the constants found from (48) (and (52) when C2≠0 ). Consequently, differential equation (49) has the unique solution given by (59) or (60). Since we have derived (49) from (48) without any additional assumptions, we conclude that the function ψ(x) given by (59) or (60) is a unique solution to (48) satisfying the certain conditions. This guaranties that the solution we have found is the ruin probability and completes the proof.  □

The case θ=0 and θ¯≠0 can be considered in a similar way by finding the required solution to equation (50).

 The ruin probability in the model without dependence

Let now θ=0 and θ¯=0 . We set ψ(x)=ψ1(x)ifx∈[0,b],ψ2(x)ifx∈[b,∞). Then equations (9) and (26) for the ruin probability ψ(x) in the case of exponentially distributed claim and premium sizes take the form (λ+λ¯)ψ1(x)=λμe−x/μ∫0xψ1(u)eu/μdu+λe−x/μ+λ¯μ¯ex/μ¯∫xbψ1(u)e−u/μ¯du+λ¯μ¯ex/μ¯∫b∞ψ2(u)e−u/μ¯du,x∈[0,b], and dψ2′(x)+(λ+λ¯)ψ2(x)=λμe−x/μ∫0bψ1(u)eu/μdu+λμe−x/μ∫bxψ2(u)eu/μdu+λe−x/μ+λ¯μ¯ex/μ¯∫x∞ψ2(u)e−u/μ¯du,x∈[b,∞), respectively.

We now show that integro-differential equations (68) and (69) can be reduced to linear differential equations with constant coefficients.

Let the surplus process (Xtb(x))t≥0 follow (6) under the above assumptions with θ=0 and θ¯=0 , and let claim and premium sizes be exponentially distributed with means μ and μ¯ , respectively. Then ψ1(x) and ψ2(x) are solutions to the differential equations μμ¯(λ+λ¯)ψ1″(x)+(λ¯μ¯−λμ)ψ1′(x)=0,x∈[0,b], and dμμ¯ψ2‴(x)+(dμ¯−dμ+μμ¯(λ+λ¯))ψ2″(x)+(λ¯μ¯−λμ−d)ψ2′(x)=0,x∈[b,∞).

From (68), it is easily seen that ψ(x) is differentiable on [0,b] . Differentiating (68) yields (λ+λ¯)ψ1′(x)=−1μ(λμe−x/μ∫0xψ1(u)eu/μdu+λe−x/μ)+1μ¯(λ¯μ¯ex/μ¯∫xbψ1(u)e−u/μ¯du+λ¯μ¯ex/μ¯∫b∞ψ2(u)e−u/μ¯du)+(λμ−λ¯μ¯)ψ1(x),x∈[0,b].

Multiplying (72) by (−μ¯) and adding (68) we obtain −μ¯(λ+λ¯)ψ1′(x)+λ(1+μ¯μ)ψ1(x)=(1+μ¯μ)(λμe−x/μ∫0xψ1(u)eu/μdu+λe−x/μ),x∈[0,b].

From (73), it follows that ψ(x) is twice differentiable on [0,b] . Differentiating (73) gives −μ¯(λ+λ¯)ψ1″(x)+λ(1+μ¯μ)ψ1′(x)=−1μ(1+μ¯μ)(λμe−x/μ∫0xψ1(u)eu/μdu+λe−x/μ)+λμ(1+μ¯μ)ψ1(x),x∈[0,b].

Multiplying (74) by μ and adding (73) we get (70).

From (69), it is easily seen that ψ(x) is twice differentiable on [b,∞) . Differentiating (69) yields dψ2″(x)+(λ+λ¯)ψ2′(x)=−1μ(λμe−x/μ∫0bψ1(u)eu/μdu+λμe−x/μ∫bxψ2(u)eu/μdu+λe−x/μ)+λ¯μ¯2ex/μ¯∫x∞ψ2(u)e−u/μ¯du+(λμ−λ¯μ¯)ψ2(x),x∈[b,∞).

Multiplying (75) by μ and adding (69) we obtain dμψ2″(x)+(d+μ(λ+λ¯))ψ2′(x)+λ¯(1+μμ¯)ψ2(x)=λ¯μ¯(1+μμ¯)ex/μ¯∫x∞ψ2(u)e−u/μ¯du,x∈[b,∞).

From (76), it follows that ψ(x) has the third derivative on [b,∞) . Differentiating (76) gives dμψ2‴(x)+(d+μ(λ+λ¯))ψ2″(x)+λ¯(1+μμ¯)ψ2′(x)=λ¯μ¯2(1+μμ¯)ex/μ¯∫x∞ψ2(u)e−u/μ¯du−λ¯μ¯(1+μμ¯)ψ2(x),x∈[b,∞).

Multiplying (77) by (−μ¯) and adding (76) we get (71).  □

To formulate the next theorem, we define the following constants: D2=(dμ¯−dμ+μμ¯(λ+λ¯))2−4dμμ¯(λ¯μ¯−λμ−d),z5=λμ−λ¯μ¯μμ¯(λ+λ¯),z7=−(dμ¯−dμ+μμ¯(λ+λ¯))+D22dμμ¯ and z8=−(dμ¯−dμ+μμ¯(λ+λ¯))−D22dμμ¯.

Let the surplus process (Xtb(x))t≥0 follow (6) under the above assumptions with θ=0 and θ¯=0 , and let claim and premium sizes be exponentially distributed with means μ and μ¯ , respectively, and let λ¯μ¯>λμ+d \lambda \mu +d$]]>. Then we have ψ1(x)=C4+C5ez5x,x∈[0,b], and ψ2(x)=C7ez7x+C8ez8x,x∈[b,∞), where the constants C4 , C5 , C7 and C8 are determined from the system of linear equations (80)–(83): (λeb/μ¯+λ¯)C4+λ+λ¯μ+μ¯(μ¯eb/μ¯+μez5b)C5+λ¯ez7bμ¯z7−1C7+λ¯ez8bμ¯z8−1C8=λeb/μ¯, λ(1+μ¯μ)C4+μ¯(λ+λ¯)μC5=λ(1+μ¯μ), C4+ez5bC5−ez7bC7−ez8bC8=0 and λ(e−b/μ−1)C4+μ¯(λ+λ¯)μ+μ¯(e−b/μ−ez5b)C5+(λ+λ¯+dz7+λ¯μ¯z7−1)ez7bC7+(λ+λ¯+dz8+λ¯μ¯z8−1)ez8bC8=λe−b/μ.

By Lemma 2, ψ1(x) and ψ2(x) are solutions to (70) and (71). We now find the general solutions to these equations.

It is easily seen that the characteristic equation corresponding to (70) has two roots: z4=0 and z5 given before the assertion of the theorem. Hence, (78) is true with some constants C4 and C5 .

The characteristic equation corresponding to (71) has the form dμμ¯z3+(dμ¯−dμ+μμ¯(λ+λ¯))z2+(λ¯μ¯−λμ−d)z=0.

It is obvious that z6=0 is a solution to (84). We now show that the equation dμμ¯z2+(dμ¯−dμ+μμ¯(λ+λ¯))z+(λ¯μ¯−λμ−d)=0. has two negative roots. We first notice that its discriminant D2 defined above is positive. Indeed, we have D2=d2(μ¯−μ)2+μ2μ¯2(λ+λ¯)2+2dμμ¯(λ+λ¯)(μ¯−μ)+4dμμ¯(d+λμ−λ¯μ¯)=d2(μ+μ¯)2+μ2μ¯2(λ+λ¯)2+2dμμ¯(λ−λ¯)(μ+μ¯)=(d(μ+μ¯)+μμ¯(λ−λ¯))2+4λλ¯μ2μ¯2>0. 0.\end{array}\]]]>

Therefore, (85) has two real roots. Next, by the conditions of the theorem, we have λ¯μ¯−λμ−d>0 0\]]]> and dμ¯−dμ+μμ¯(λ+λ¯)=μ(λ¯μ¯−λμ−d)+λμ2+λμμ¯+dμ¯>0, 0,\]]]> which shows that both roots are negative. Consequently, we get ψ2(x)=C6+C7ez7x+C8ez8x,x∈[b,∞), with some constants C6 , C7 and C8 . Moreover, since λ¯μ¯>λμ+d \lambda \mu +d$]]>, using standard considerations (see, e.g., [34, 36, 39]) we can easily show that limx→∞ψ(x)=0 , which yields C6=0 . Thus, we obtain (79).

The constants C4 , C5 , C7 and C8 are determined by letting x=0 in (68) and (73), taking into account that ψ1(b)=ψ2(b) and letting x=b in (69).

Substituting (78) and (79) into (68) and (73) as x=0 and into (69) as x=b and doing some simplifications yield equations (80), (81) and (83), respectively. Substituting (78) and (79) into the equality ψ1(b)=ψ2(b) gives (82).

We denote the determinant of the system of equations (80)–(83) by Δ2 . A standard computation shows that Δ2=d(z7−z8)e(z7+z8)b(λez5b(λ¯μ¯μ−λ)+μ¯2z7z8μ¯2z7z8−μ¯(z7+z8)+1((λ+λ¯)λ¯μ¯μ−λλ¯ez5b(1+μ¯μ))), which is positive. Indeed, z7−z8>0 0$]]> by definition, λ¯μ¯/μ−λ>0 0$]]> by the conditions of the theorem and μ¯2z7z8−μ¯(z7+z8)+1>0 0$]]> since z7<0 and z8<0 . Moreover, since z5<0 , we have (λ+λ¯)λ¯μ¯μ−λλ¯ez5b(1+μ¯μ)>(λ+λ¯)λ¯μ¯μ−λλ¯(1+μ¯μ)=λ¯(λ¯μ¯μ−λ)>0. (\lambda +\bar{\lambda })\frac{\bar{\lambda }\bar{\mu }}{\mu }-\lambda \bar{\lambda }\bigg(1+\frac{\bar{\mu }}{\mu }\bigg)=\bar{\lambda }\bigg(\frac{\bar{\lambda }\bar{\mu }}{\mu }-\lambda \bigg)>0.\]]]>

Thus, since Δ2≠0 , the system of equations (80)–(83) has a unique solution. Furthermore, note that letting x=b in (76) gives no additional information about unknown constants, but the equality in (76) holds for the values of the constants found from the system of equations (80)–(83). Therefore, each of differential equations (70) and (71) has the unique solution given by (78) or (79), respectively. Since we have derived these equations from (68) and (69) without any additional assumptions, we conclude that the functions ψ1(x) and ψ2(x) given by (78) and (79) are unique solutions to (68) and (69) satisfying the certain conditions. This guaranties that the functions ψ1(x) and ψ2(x) we have found coincide with the ruin probability on the intervals [0,b] and [b,∞) , respectively, which completes the proof.  □

 The expected discounted dividend payments until ruin in the model without dependence

We now also assume that θ=0 and θ¯=0 . Then equations (29) and (42) for the expected discounted dividend payments v(x) in the case of exponentially distributed claim and premium sizes take the form (λ+λ¯+δ)v1(x)=λμe−x/μ∫0xv1(u)eu/μdu+λ¯μ¯ex/μ¯∫xbv1(u)e−u/μ¯du+λ¯μ¯ex/μ¯∫b∞v2(u)e−u/μ¯du,x∈[0,b], and dv2′(x)+(λ+λ¯+δ)v2(x)=λμe−x/μ∫0bv1(u)eu/μdu+λμe−x/μ∫bxv2(u)eu/μdu+λ¯μ¯ex/μ¯∫x∞v2(u)e−u/μ¯du+d,x∈[b,∞), respectively.

Lemma 3 below shows that integro-differential equations (86) and (87) can be reduced to linear differential equations with constant coefficients.

Let the surplus process (Xtb(x))t≥0 follow (6) under the above assumptions with θ=0 and θ¯=0 , and let claim and premium sizes be exponentially distributed with means μ and μ¯ , respectively. Then v1(x) and v2(x) are solutions to the differential equations μμ¯(λ+λ¯+δ)v1″(x)+(μ¯(λ¯+δ)−μ(λ+δ))v1′(x)−δv1(x)=0,x∈[0,b], and dμμ¯v2‴(x)+(d(μ¯−μ)+μμ¯(λ+λ¯+δ))v2″(x)+(μ¯(λ¯+δ)−μ(λ+δ)−d)v2′(x)−δv2(x)=−d,x∈[b,∞).

The proof of the lemma is similar to the proof of Lemma 2.

To formulate the next theorem, we define the following constants: D3=(μ¯(λ¯+δ)−μ(λ+δ))2+4δμμ¯(λ+λ¯+δ),D4=−18δdμμ¯(d(μ¯−μ)+μμ¯(λ+λ¯+δ))(μ¯(λ¯+δ)−μ(λ+δ)−d)+4δ(d(μ¯−μ)+μμ¯(λ+λ¯+δ))3+(d(μ¯−μ)+μμ¯(λ+λ¯+δ))2(μ¯(λ¯+δ)−μ(λ+δ)−d)2−4dμμ¯(μ¯(λ¯+δ)−μ(λ+δ)−d)3−27(δdμμ¯)2,z9=−(μ¯(λ¯+δ)−μ(λ+δ))+D32μμ¯(λ+λ¯+δ) and z10=−(μ¯(λ¯+δ)−μ(λ+δ))−D32μμ¯(λ+λ¯+δ).

Let the surplus process (Xtb(x))t≥0 follow (6) under the above assumptions with θ=0 and θ¯=0 , and let claim and premium sizes be exponentially distributed with means μ and μ¯ , respectively, and let λ¯μ¯>λμ+d \lambda \mu +d$]]> and D4>0 0$]]>. Then we have v1(x)=C9ez9x+C10ez10x,x∈[0,b], and v2(x)=C11ez11x+C12ez12x+d/δ,x∈[b,∞), where z11 and z12 are negative roots of the cubic equation dμμ¯z3+(d(μ¯−μ)+μμ¯(λ+λ¯+δ))z2+(μ¯(λ¯+δ)−μ(λ+δ)−d)z−δ=0 and the constants C9 , C10 , C11 and C12 are determined from the system of linear equations (93)–(96): ((λ+λ¯+δ)eb/μ¯−λ¯μ¯z9−1(ez9b−eb/μ¯))C9+((λ+λ¯+δ)eb/μ¯−λ¯μ¯z10−1(ez10b−eb/μ¯))C10+λ¯ez11bμ¯z11−1C11+λ¯ez12bμ¯z12−1C12=dλ¯δ, (λ+δ+λμ¯μ−μ¯z9(λ+λ¯+δ))C9+(λ+δ+λμ¯μ−μ¯z10(λ+λ¯+δ))C10=0, ez9bC9+ez10bC10−ez11bC11−ez12bC12=d/δ and λμz9+1(e−b/μ−ez9b)C9+λμz10+1(e−b/μ−ez10b)C10+(λ+λ¯+δ+dz11+λ¯μ¯z11−1)ez11bC11+(λ+λ¯+δ+dz12+λ¯μ¯z12−1)ez12bC12=−d(1+λδ) provided that its determinant is not equal to 0.

The proof is similar to the proof of Theorem 4, so we omit detailed considerations. By Lemma 3, v1(x) and v2(x) are solutions to (88) and (89).

It is easily seen that D3>0 0$]]>. Hence the characteristic equation corresponding to (88) has two real roots z9 and z10 given before the assertion of the theorem. This yields (90) with some constants C9 and C10 .

The assumption D4>0 0$]]> guarantees that cubic equation (92) has three distinct real roots. Consequently, the general solution to (89) is given by v2(x)=C11ez11x+C12ez12x+C13ez13x,x∈[b,∞) with some constants C11 , C12 and C13 .

By Vieta’s theorem, we conclude that (92) has either two or no negative roots. Since λ¯μ¯>λμ+d \lambda \mu +d$]]>, applying arguments similar to those in [40, p. 70] shows that limx→∞v2(x)=d/δ . Therefore, if (92) had no negative roots, the function v2(x) would be constant, which is impossible. From this we deduce that (92) has two negative roots. We denote them by z11 and z12 . Since z13>0 0$]]>, we get C13=0 , which yields (91).

To determine the constants C9 , C10 , C11 and C12 , we apply considerations similar to those in the proof of Theorem 4 and obtain the system of linear equations (93)–(96), which has a unique solution provided that its determinant is not equal to 0. Finally, applying arguments similar to those in the proof of Theorem 4 guaranties that the functions v1(x) and v2(x) we have found coincide with the expected discounted dividend payments on the intervals [0,b] and [b,∞) .  □

 Numerical illustrations

We now present numerical examples for the results obtained in Section 5. The claim and premium sizes are also assumed to be exponentially distributed. Set λ=0.1 , λ¯=2.3 , μ=3 and μ¯=0.2 .

Let now the conditions of Theorem 3 hold. Then applying this theorem we can calculate the ruin probability for x∈[0,∞) in the model without dividend payments for different values of θ:

if θ=−0.9 , then ψ(x)≈0.929934e−0.022277x−0.006234e−0.744001x ;

if θ=−0.5 , then ψ(x)≈0.817753e−0.059151x−0.009736e−0.712238x ;

if θ=−0.1 , then ψ(x)≈0.698198e−0.100061x−0.003545e−0.676439x ;

if θ=0.1 , then ψ(x)≈0.634275e−0.122565x+0.004545e−0.656490x ;

if θ=0.5 , then ψ(x)≈0.492433e−0.173655x+0.036374e−0.610511x ;

if θ=0.9 , then ψ(x)≈0.309485e−0.239185x+0.111461e−0.550092x .

Table 1 presents the results of calculations for some values of x.

The ruin probabilities in the model without dividend payments for different values of θ

x∖θ -0.9 -0.5 -0.1 0.1 0.5 0.9
0 0.923700 0.808017 0.694653 0.638820 0.528807 0.420945
1 0.906484 0.766009 0.629915 0.563467 0.433686 0.307949
2 0.888003 0.724172 0.570650 0.497607 0.358674 0.228911
5 0.831762 0.608107 0.423229 0.343831 0.208380 0.100718
7 0.795628 0.540438 0.346536 0.268996 0.146531 0.060380
10 0.744222 0.452611 0.256692 0.186208 0.086812 0.028761
15 0.665780 0.336734 0.155646 0.100887 0.036402 0.008589
20 0.595603 0.250520 0.094376 0.054662 0.015276 0.002591
50 0.305291 0.042479 0.004690 0.001383 0.000083 0.000002
70 0.195532 0.013013 0.000634 0.000119 0.000003 0.000000

Next, we denote by ψ0(x) the ruin probability in this model where θ=0 . It is given by ψ0(x)=λ(μ+μ¯)μ¯(λ+λ¯)exp(−(λ¯μ¯−λμ)xμμ¯(λλ¯)),x∈[0,∞) (see [9, 34]). In our example, ψ0(x)≈0.666667e−0.111111x . The values of ψ0(x) for some x are given in Table 2.

Let now the conditions of Theorems 4 and 5 hold. Set additionally b=5 , d=0.1 and δ=0.01 . Applying Theorems 4 and 5 we can calculate the ruin probability ψ(x) and the expected discounted dividend payments until ruin v(x) : ψ1(x)≈0.389315+0.407125e−0.111111x,x∈[0,5],ψ2(x)≈0.809486e−0.051863x−1.24332·1039e−19.281470x,x∈[5,∞);v1(x)≈4.555889e0.049220x−2.296416e−0.140506x,x∈[0,5],v2(x)≈10−9.149114e−0.107684x+4.07834·1040e−19.405407x,x∈[5,∞).

The results of calculations for some values of x are given in Table 2.

The ruin probabilities with and without dividend payments and the expected discounted dividend payments in the model without dependence

x ψ0(x) ψ(x) v(x)
0 0.666667 0.796440 2.259472
1 0.596560 0.753626 2.790339
2 0.533825 0.715315 3.293343
5 0.382502 0.622904 4.689607
7 0.306284 0.563044 5.694612
10 0.219462 0.481915 6.883176
15 0.125917 0.371835 8.180807
20 0.072245 0.286900 8.938193
50 0.002577 0.060536 9.958020
70 0.000279 0.021455 9.995128

The results presented in Tables 1 and 2 show that the positive dependence between the claim sizes and the inter-claim times decreases the ruin probability and the negative dependence increases it. This conclusion seems to be natural. Indeed, in the case of negative dependence, the situation where large claims arrive in short time intervals is more probable, which obviously leads to ruin in the near future. Moreover, it is easily seen from Table 2 that dividend payments substantially increase the ruin probability, which is also an expected conclusion.

 Acknowledgments

The author is deeply grateful to the anonymous referees for careful reading and valuable comments and suggestions, which helped to improve the earlier version of the paper.

 References Albrecher, H., Boxma, O.J.: A ruin model with dependence between claim sizes and claim intervals. Insurance: Mathematics and Economics 35, 245–254 (2004) MR2095888 Albrecher, H., Boxma, O.J.: On the discounted penalty function in a Markov-dependent risk model. Insurance: Mathematics and Economics 37, 650–672 (2005) Albrecher, H., Teugels, J.L.: Exponential behavior in the presence of dependence in risk theory. Journal of Applied Probability 43, 257–273 (2006) MR2225065 Albrecher, H., Constantinescu, C., Loisel, S.: Explicit ruin formulas for models with dependence among risks. Insurance: Mathematics and Economics 48, 265–270 (2011) Andrulytė, I.M., Bernackaitė, E., Kievinaitė, D., Šiaulys, J.: A Lundberg-type inequality for an inhomogeneous renewal risk model. Modern Stochastics: Theory and Applications 2, 173–184 (2015) MR3389589 Asmussen, S., Albrecher, H.: Ruin Probabilities. World Scientific, Singapore (2010) Bekrizadeh, H., Jamshidi, B.: A new class of bivariate copulas: dependence measures and properties. METRON 75, 31–50 (2017) Bekrizadeh, H., Parham, G.A., Zadkarmi, M.R.: The new generalization of Farlie–Gumbel–Morgenstern copulas. Applied Mathematical Sciences 6, 3527–3533 (2012) MR2929551 Boikov, A.V.: The Cramér–Lundberg model with stochastic premium process. Theory of Probability and Its Applications 47, 489–493 (2003) MR1975908 Boudreault, M.: Modeling and pricing earthquake risk. Scor Canada Actuarial Prize (2003) Boudreault, M., Cossette, H., Landriault, D., Marceau, E.: On a risk model with dependence between interclaim arrivals and claim sizes. Scandinavian Actuarial Journal 2006, 265–285 (2006) Chadjiconstantinidis, S., Vrontos, S.: On a renewal risk process with dependence under a Farlie–Gumbel–Morgenstern copula. Scandinavian Actuarial Journal 2014, 125–158 (2014) Cheng, Y., Tang, Q.: Moments of the surplus before ruin and the deficit of ruin in the Erlang(2) risk process. North American Actuarial Journal 7, 1–12 (2003) Chi, Y., Lin, X.S.: On the threshold dividend strategy for a generalized jump-diffusion risk model. Insurance: Mathematics and Economics 48, 326–337 (2011) Cossette, H., Marceau, E., Marri, F.: On the compound Poisson risk model with dependence based on a generalized Farlie–Gumbel–Morgenstern copula. Insurance: Mathematics and Economics 43, 444–455 (2008) Cossette, H., Marceau, E., Marri, F.: Analysis of ruin measures for the classical compound Poisson risk model with dependence. Scandinavian Actuarial Journal 2010, 221–245 (2010) Cossette, H., Marceau, E., Marri, F.: Constant dividend barrier in a risk model with ageneralized Farlie–Gumbel–Morgenstern copula. Methodology and Computing in Applied Probability 13, 487–510 (2011) MR2822392 Cossette, H., Marceau, E., Marri, F.: On a compound Poisson risk model with dependence and in the presence of a constant dividend barrier. Applied Stochastic Models in Business and Industry 30, 82–98 (2014) Czado, C., Kastenmeier, R., Brechmann, E.C., Min, A.: A mixed copula model for insurance claims and claim sizes. Scandinavian Actuarial Journal 2012, 278–305 (2012) De Finetti, B.: Su un’impostazione alternativa dell teoria colletiva del rischio. Transactions of the XV International Congress of Actuaries 2, 433–443 (1957) Denuit, M., Dhaene, J., Goovaerts, M., Kaas, R.: Actuarial Theory for Dependent Risks: Measures, Orders and Models. John Wiley & Sons, Chichester (2005) Gerber, H.U., Shiu, E.S.W.: On the time value of ruin. North American Actuarial Journal 2, 48–72 (1998) Gerber, H.U., Shiu, E.S.W.: The time value of ruin in a Sparre Andersen model. North American Actuarial Journal 9, 49–69 (2005) Gu, C.: The ruin problem of dependent risk model based on copula function. Journal of Chemical and Pharmaceutical Research 5, 234–240 (2013) Heilpern, S.: Ruin measures for a compound Poisson risk model with dependence based on the Spearman copula and the exponential claim sizes. Insurance: Mathematics and Economics 59, 251–257 (2014) Landriault, D.: Constant dividend barrier in a risk model with interclaim-dependent claim sizes. Insurance: Mathematics and Economics 42, 31–38 (2008) Li, B., Wu, R., Song, M.: A renewal jump-diffusion process with threshold dividend strategy. Journal of Computational and Applied Mathematics 228, 41–55 (2009) Li, S., Garrido, J.: On a class of renewal risk models with a constant dividend barrier. Insurance: Mathematics and Economics 35, 691–701 (2004) Li, S., Garrido, J.: On ruin for the Erlang(n) risk process. Insurance: Mathematics and Economics 34, 391–408 (2004) Lin, X.S., Pavlova, K.P.: The compound Poisson risk model with a threshold dividend strategy. Insurance: Mathematics and Economics 38, 57–80 (2006) Lin, X.S., Willmot, G.E., Drekic, S.: The classical risk model with a constant dividend barrier: analysis of the Gerber–Shiu discounted penalty function. Insurance: Mathematics and Economics 33, 551–566 (2003) Liu, D., Liu, Z., Peng, D.: The Gerber-Shiu expected penalty function for the risk model with dependence and a constand dividend barrier. Abstract and Applied Analysis 2014, 730174–7 (2014) MR3246356 Meng, Q., Zhang, X., Guo, J.: On a risk model with dependence between claim sizes and claim intervals. Statistics and Probability Letters 78, 1727–1734 (2008) MR2528548 Mishura, Y., Ragulina, O.: Ruin Probabilities: Smoothness, Bounds, Supermartingale Approach. ISTE Press – Elsevier, London (2016) Mishura, Y., Ragulina, O., Stroev, O.: Practical approaches to the estimation of the ruin probability in a risk model with additional funds. Modern Stochastics: Theory and Applications 1, 167–180 (2014) Mishura, Y.S., Ragulina, O.Y., Stroev, O.M.: Analytic property of infinite-horizon survival probability in a risk model with additional funds. Theory of Probability and Mathematical Statistics 91, 131–143 (2015) Nelsen, R.B.: An Introduction to Copulas. Springer, New York (2006) Nikoloulopoulos, A.K., Karlis, D.: Fitting copulas to bivariate earthquake data: the seismic gap hypothesis revisited. Environmetrics 19, 251–269 (2008) Rolski, T., Schmidli, H., Schmidt, V., Teugels, J.: Stochastic Processes for Insurance and Finance. John Wiley & Sons, Chichester (1999) Schmidli, H.: Stochastic Control in Insurance. Springer, London (2008) Shi, Y., Liu, P., Zhang, C.: On the compound poisson risk model with dependence and a threshold dividend strategy. Statistics and Probability Letters 83, 1998–2006 (2013) MR3079035 Sun, L.-J.: The expected discounted penalty at ruin in the Erlang(2) risk process. Statistics and Probability Letters 72, 205–217 (2005) Wang, W.: The perturbed Sparre Andersen model with interest and a threshold dividend strategy. Methodology and Computing in Applied Probability 17, 251–283 (2015) MR3343407 Xing, Y., Wu, R.: Moments of the time of ruin, surplus before ruin and the deficit at ruin in the Erlang(N) risk process. Acta Mathematicae Applicatae Sinica, English Series 22, 599–606 (2006) Yang, H., Zhang, Z.: Gerber-Shiu discounted penalty function in a Sparre Andersen model with multi-layer dividend strategy. Insurance: Mathematics and Economics 42, 984–991 (2008) Yong, W., Xiang, H.: Differential equations for ruin probability in a special risk model with FGM copula for the claim size and the inter-claim time. Journal of Inequalities and Applications 2012, 156–13 (2012) Zhang, Z., Yang, H.: Gerber–Shiu analysis in a perturbed risk model with dependence between claim sizes and interclaim times. Journal of Computational and Applied Mathematics 235, 1189–1204 (2011)