# Some State-Specific Exit Probabilities in a Markov-Modulated Risk Model

Authors: Jingchao Li; Shuanming Li
Journal: Mathematical Problems in Engineering
 (2020)

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## Abstract

In this paper, we study some state-specific one-sided exit probabilities in a Markov-modulated risk process including the probability that ruin occurs without or with the surplus visiting certain states; the probability that ruin occurs without or with a claim occurring in certain states; the probability that the surplus attains a target level without or with visiting certain states; and the probability that the surplus attains a target level without or with a claim occurring in certain states. We also investigate the corresponding two-sided first exit probabilities without (or with) the surplus visiting certain states or without (or with) claims occurring in certain states. All these probabilities can be expressed elegantly in terms of some modified matrix scale functions which are easily computable.


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## Body

## 1. Introduction

In this paper, we consider a Markov-modulated risk model which has been studied by a number of authors since the important paper proposed by Asmussen [1]. In such a model, both the claim amounts and the claim arrivals are influenced by an external environment process which we denote by Jt,t≥0. For example, it is well known that the weather or climate conditions have impacts on automobile, property, and casualty insurance claims. This process is assumed to be a homogeneous, irreducible, and recurrent Markov process on a finite state space E=1,2,…,m. The process has intensity matrix A=αi,ji,j=1m, with αi,i:=−αi for i∈E, and has stationary distribution π⟶=π1,π2,…,πm.LetNt denote the number of claims occurring in 0,t. If Js=i for all s in a small interval t,t+h, then Nt+h−Nt represents the number of claims occurring in that interval and has a Poisson distribution with parameter λi>0. The process Nt,t≥0 is a Markov-modulated Poisson process, and it can also be viewed as a Poisson process with its parameter driven by the external environment process Jt,t≥0.Given thatJt=i, we assume that individual claim amounts have distribution function Fi and probability density function fi, with Laplace transform f^is=∫0∞e−sxfixdx and finite mean μii∈E.Then, the surplus processUt,t≥0 in our Markov-modulated risk model is given by(1)Ut=u+∑i=1m∫0tIJs=idUis,t≥0,where u≥0 is the initial surplus and I is the indication function, and for i∈E, let Uitt≥0 be the surplus process in the classical risk model with premium rate ci, Poisson parameter λi, and individual claim amount distribution Fi. The positive loading condition is also assumed to hold, i.e.,(2)∑i=1mπici−λiμi>0.We next defineTu=inft≥0,Ut<0 to be the time of ruin, with Tu=∞ if Ut≥0 for all t≥0, and(3)ψiu=PTu<∞J0=i,i∈E,to be the probability of ruin given that initial environment state i.LetNu≔NTu be the number of claims and Su≔STu=∑n=1NTuXn be the aggregate claims by the time of ruin, respectively, where Xn is the amount of the nth claim. Now, Tu, Nu, and Su can be decomposed as(4)Tu=∑l=1mTu,l,Nu=∑l=1mNu,l,Su=∑l=1mSu,l,where Tu,l is the duration that the surplus process spends in state l∈E until the time of ruin, Nu,l is the number of claims that occur in state l until the time of ruin, and Su,l is the aggregate claims that occur in state l until the time of ruin.For1≤l1<l2<⋯<lk≤m, let Ek=l1,l2,…,lk⊂E be a substate space. We define(5)Tu,Ek=∑j=1kTu,lj,Nu,Ek=∑j=1kNu,lj,Su,Ek=∑j=1kSu,lj,to be duration that the surplus spent in substate space Ek=l1,l2,…,lk, the number of claims occurred in Ek, and the aggregate claims occurred in Ek, respectively, by the time of ruin. In particular, if Ek=E, then Tu,Ek,Nu,Ek, and Su,Ek recover Tu,Nu, and Su; if k=1 and Ek=l, then Tu,Ek,Nu,Ek, and Su,Ek simplify to Tu,l,Nu,l, and Su,l, respectively.The Markov-modulated risk models have been studied extensively in the literature. References include the studies of Asmussen et al. [2]; Bäuerle [3]; Schmidli [4]; Snoussi [5]; Lu and Li [6]; Lu [7]; Ng and Yang [8]; and Li and Lu [9]; and Li and Lu [10]. An excellent review paper of the Markov-modulated Poisson process is the paper by Fischer and Meier-Hellstern [11]. Li et al. [12] study the distributions of Tu and Nu for a risk model with Markovian arrival process (MAP). In the study by Li et al. [13], the one-sided exit time (time of ruin or the first hitting time), the number of claims, and the aggregate claims by the exit time are decomposed into m components, respectively, according to the underlying Markov state. Joint Laplace transforms and probability generating functions of all these decomposed quantities are analyzed via matrix scale functions and matrix analysis methods. The motivation of our work is that the insurer would consider a collection of similar conditions as a whole and risk information based on the subset of states would be helpful for insurers to better understand and control their risks for a particular subset of interest and make strategic and contingency plans accordingly.In this paper, we investigate some state-specific one-sided and two-sided first exit probabilities using several versions of the matrix scale functions for the risk model. To the best knowledge of the authors, this is the first paper to study exit probabilities of the subset of states by using different forms of the matrix scale functions. We start in Section2 by defining these modified or dimension-reduced matrix scale functions for the Markov-modulated risk model as well as for the classical risk model. We study the probability of ruin without the surplus visiting the state-state space Ek and the probability of ruin without claims occurring in Ek in Sections 3 and 4, respectively. The probability of reaching a target level without the surplus visiting Ek and the probability of reaching a target level without claims occurring in Ek are discussed in Sections 5 and 6, respectively. In Section 7, we study the corresponding two-sided first exit probabilities without the surplus visiting Ek or without claims occurring in Ek. Some numerical examples are demonstrated in Section 8.


## 2. Notation and Preliminaries

We define the followingm×m matrices:(6)c=diagc1,c2,…,cm,λ=diagλ1,λ2,…,λm,fx=diagf1x,f2x,…,fmx,Fx=diagF1x,F2x,…,Fmx=I−F¯x,I=diag1,1,…,1.LetEk=l1,l2,…,lk⊂E and for any m×m matrix X=xi,ji,j∈E, we define the following four matrices:(7)X11=xi,ji,j∈Ek,X22=xi,ji,j∈E/Ek,X12=xi,ji∈Ek,j∈E/Ek,X21=xi,ji∈E/Ek,j∈Ek.Here,X22 is an m−k×m−k matrix obtained by deleting the li-th row and the li-th column of X, for i∈Ek. We remark that if Ek=E, then X11=XandX22=X12=X21=∅.


### 2.1. Matrix Scale Function and Its Modified and Dimension-Reduced Versions

In this section, we define a matrix-form scale function and its several modified or reduced forms which play important roles in the rest of the paper.Definition 1.
DefineWu to be a matrix function such that(8)W^s=∫0∞e−suWudu=Bs−1,where(9)Bs=sI−c−1λ−A+c−1λf^s.
Here,Wu is called the matrix-form scale function for the Markov-modulated risk model with W0=lims⟶∞sW^s=lims⟶∞sBs−1=I. A similar matrix-form scale function can be found in Li et al.’s study [13].Definition 2.
DefineW11u and W22u to be a k×k matrix function and an m−k×m−k matrix function, respectively, such that(10)W^11s=∫0∞e−suW11udu=B11s−1,W^22s=∫0∞e−suW22udu=B22s−1,where(11)B11s=sI11−c11−1λ11−A11+c11−1λ11f^11s,B22s=sI22−c22−1λ22−A22+c22−1λ22f^22s.Remark 1.
(1)
W11u and W22u are called reduced matrix-form scale functions for the Markov-modulated risk model with(12)W110=lims⟶∞sW^11s=lims⟶∞sB11s−1=I11,and similarly,W220=I22.(2)
IfEk=E, then W11=Wu; if Ek=∅, then W22=Wu.(3)
IfEk=i, then B11s=s−λi+αi/ci+λif^is/ci, and W11u=ηαiu is the αi-scale function defined in the next section for the classical risk model.(4)
ForEk⊂E, it follows from the matrix generalization of Rouche’s Theorem (De Smit [14] or Li and Ren [15]) that DetB11s=0 has k roots with positive real parts, say ρ11,1,ρ11,2,…,ρ11,k. Similarly, DetB22s=0 has m−k roots with positive real parts, say ρ22,1,ρ22,2,…,ρ22,m−k.Definition 3.
ForEk=l1,l2,…,lk⊂E, we define Vu,Ek to be an m×m matrix function such that(13)V^s;Ek=∫0∞e−suVu;Ekdu=Bs;Ek−1,where(14)Bs;Ek=sI−c−1λ−A+c−1λI−IEkf^s,where IEk is an m×m diagonal matrix with the i-th entry being 1, for i∈Ek, and all other entries being zeros.Remark 2.
(1)
Vu;Ek is called a modified matrix scale function for the Markov-modulated risk model with(15)V0;Ek=lims⟶∞sV^s;Ek=lims⟶∞sBs;Ek−1=I.(2)
IfEk=∅, then Vu;∅=Wu.(3)
ForEk⊂E, it follows from the matrix generalization of Rouche’s Theorem (De Smit [14] or Li and Ren [15]) that DetBs;Ek=0 has m roots with positive real parts, say ρ1Ek,ρ2Ek,…,ρmEk.
To evaluate the matrix scale functionsWu,W11u,W22u, and Vu;Ek, we need to make assumptions about the claim size distributions. If we assume that each f^is is a rational function, then we can show that each element of W^s=Bs−1=B∗s/DetBs is a rational function which is readily inverted to obtain Wu by partial fractions. The same are true for W11u,W22u, and Vu;Ek.


### 2.2. Scale Function in the Classical Risk Model

Letηαu with ηα0=1 be a function such that its Laplace transform η^αs=∫0∞e−suηαudu is given by(16)η^αs=1s−λ+α/c+λf^s/c.Remark 3.
(i)
ηαu is called the α-scale function (apart from a multiplicative constant) in the classical risk model with premium rate c, Poisson arrival rates λ, and claim amounts density fx.(ii)
Chan et al. [16] show that ηαu has the form of(17)ηαu=eραu−Ψαu1−Ψα0,u≥0,whereΨαu has three different forms and ρ=ρα with ρ0=0 is the unique positive solution of the following generalized Lundberg’s equation:(18)λ+α−cs=λf^s.In particular, ifα=0,Ψ0u=ψu simplifies to the ruin probability for the classical risk model and η0u=1−ψu/1−ψ0.(iii)
If we replacec by ck, replace λ by λk, replace fx by fkx, and replace α by αk, then we write ηαu as ηαku and write ρ=ρα as ρk=ρkαk, for k=1,2,…,m.


## 2.1. Matrix Scale Function and Its Modified and Dimension-Reduced Versions

In this section, we define a matrix-form scale function and its several modified or reduced forms which play important roles in the rest of the paper.Definition 1.
DefineWu to be a matrix function such that(8)W^s=∫0∞e−suWudu=Bs−1,where(9)Bs=sI−c−1λ−A+c−1λf^s.
Here,Wu is called the matrix-form scale function for the Markov-modulated risk model with W0=lims⟶∞sW^s=lims⟶∞sBs−1=I. A similar matrix-form scale function can be found in Li et al.’s study [13].Definition 2.
DefineW11u and W22u to be a k×k matrix function and an m−k×m−k matrix function, respectively, such that(10)W^11s=∫0∞e−suW11udu=B11s−1,W^22s=∫0∞e−suW22udu=B22s−1,where(11)B11s=sI11−c11−1λ11−A11+c11−1λ11f^11s,B22s=sI22−c22−1λ22−A22+c22−1λ22f^22s.Remark 1.
(1)
W11u and W22u are called reduced matrix-form scale functions for the Markov-modulated risk model with(12)W110=lims⟶∞sW^11s=lims⟶∞sB11s−1=I11,and similarly,W220=I22.(2)
IfEk=E, then W11=Wu; if Ek=∅, then W22=Wu.(3)
IfEk=i, then B11s=s−λi+αi/ci+λif^is/ci, and W11u=ηαiu is the αi-scale function defined in the next section for the classical risk model.(4)
ForEk⊂E, it follows from the matrix generalization of Rouche’s Theorem (De Smit [14] or Li and Ren [15]) that DetB11s=0 has k roots with positive real parts, say ρ11,1,ρ11,2,…,ρ11,k. Similarly, DetB22s=0 has m−k roots with positive real parts, say ρ22,1,ρ22,2,…,ρ22,m−k.Definition 3.
ForEk=l1,l2,…,lk⊂E, we define Vu,Ek to be an m×m matrix function such that(13)V^s;Ek=∫0∞e−suVu;Ekdu=Bs;Ek−1,where(14)Bs;Ek=sI−c−1λ−A+c−1λI−IEkf^s,where IEk is an m×m diagonal matrix with the i-th entry being 1, for i∈Ek, and all other entries being zeros.Remark 2.
(1)
Vu;Ek is called a modified matrix scale function for the Markov-modulated risk model with(15)V0;Ek=lims⟶∞sV^s;Ek=lims⟶∞sBs;Ek−1=I.(2)
IfEk=∅, then Vu;∅=Wu.(3)
ForEk⊂E, it follows from the matrix generalization of Rouche’s Theorem (De Smit [14] or Li and Ren [15]) that DetBs;Ek=0 has m roots with positive real parts, say ρ1Ek,ρ2Ek,…,ρmEk.
To evaluate the matrix scale functionsWu,W11u,W22u, and Vu;Ek, we need to make assumptions about the claim size distributions. If we assume that each f^is is a rational function, then we can show that each element of W^s=Bs−1=B∗s/DetBs is a rational function which is readily inverted to obtain Wu by partial fractions. The same are true for W11u,W22u, and Vu;Ek.


## 2.2. Scale Function in the Classical Risk Model

Letηαu with ηα0=1 be a function such that its Laplace transform η^αs=∫0∞e−suηαudu is given by(16)η^αs=1s−λ+α/c+λf^s/c.Remark 3.
(i)
ηαu is called the α-scale function (apart from a multiplicative constant) in the classical risk model with premium rate c, Poisson arrival rates λ, and claim amounts density fx.(ii)
Chan et al. [16] show that ηαu has the form of(17)ηαu=eραu−Ψαu1−Ψα0,u≥0,whereΨαu has three different forms and ρ=ρα with ρ0=0 is the unique positive solution of the following generalized Lundberg’s equation:(18)λ+α−cs=λf^s.In particular, ifα=0,Ψ0u=ψu simplifies to the ruin probability for the classical risk model and η0u=1−ψu/1−ψ0.(iii)
If we replacec by ck, replace λ by λk, replace fx by fkx, and replace α by αk, then we write ηαu as ηαku and write ρ=ρα as ρk=ρkαk, for k=1,2,…,m.


## 3. Probability of Ruin without the Surplus Visiting States inEk

First, we define fori∈E,(19)ψiu;Ek=PTu<∞,Jt∉Ek,0≤t≤TuJ0=i=PTu<∞,Tu,Ek=0J0=i,to be the probability of ruin without the external environmental Markov process visiting the substate space Ek prior to ruin.Remark 4.
(1)
It follows from the definition above thatψiu;Ek=0 if i∈Ek.(2)
ψiu;k is the probability of ruin without the surplus visiting state k, for k≠i, given J0=i.(3)
ψiu;E\Ek is the probability of ruin with the surplus ever visiting every state in Ek.(4)
ψiu;E\i is the probability of ruin with the surplus only visiting the initial state i.(5)
IfEk=∅, then ψiu;∅=ψiu is the probability of ruin given that the initial state is i.
Conditioning on the events that may occur in an infinitesimal interval, we can obtain the following IDE forψiu;Ek:(20)ciψi′u;Ek=λiψiu;Ek−λi∫0uψiu−x;Ekfixdx−λiF¯iu−∑j=1,j∉Ekmαi,jψju;Ek,i∈E\Ek.
Letψ⟶u;Ek=ψiu;Ek,i∈E\Ek⊤ be an m−k×1 column vector. Equation (20) can be rewritten in matrix form as(21)ψ⟶′u;Ek=c22−1λ22−A22ψ⟶u;Ek=−c22−1λ22∫0uf22xψ⟶u−x;Ekdx+F¯22u1⟶.
Denoteψ⟶^s;Ek=∫0∞e−suψ⟶u;Ekdu. Taking Laplace transforms of both sides of equation (21) gives(22)B22sψ⟶^s;Ek=ψ⟶0;Ek−c22−1λ22F¯^22s1⟶,where B22s=sI22−c22−1λ22−A22+c22−1λ22f^22s. Use the same arguments as in the study by Li et al. [13] to find ψ⟶0;Ek. Let q⟶22,i⊤ be a 1×m−k row vector such that q⟶22,i⊤B22ρ22,i=0⟶⊤, for i=1,2,…,m−k. Setting s=ρ22,i and left-multiplying both sides of (22) by the vector q⟶22,i⊤, we have(23)0⟶⊤=q⟶22,i⊤B22ρ22,iψ⟶^ρ22,i;Ek=q⟶22,i⊤ψ⟶0;Ek−q⟶22,i⊤c22−1λ22F¯^22ρ22,i1⟶,i=1,2,…,m−k.
Sincec22,λ22,andF¯22u are all diagonal matrixes,(24)q⟶22,i⊤ψ⟶0;Ek=q⟶22,i⊤c22−1λ22F¯^22ρ22,i1⟶=q⟶22,i⊤c22−1λ22∫0∞e−ρ22,iuF¯22xdx1⟶=∫0∞e−ρ22,iuq⟶22,i⊤F¯22xdxc22−1λ221⟶.
In matrix form, we have(25)ψ⟶0;Ek=∫0∞e−xΘ22F¯22xdxc22−1λ221⟶,where Θ22=q22−1ρ22q22, ρ22=diagρ22,1,ρ22,2,…,ρ22,m−k, and(26)q22=q⟶22,1,q⟶22,2,…,q⟶22,m−k⊤,with q⟶22,i⊤ being a 1×m−k row vector such that q⟶22,i⊤B22ρ22,i=0⟶⊤, for i=1,2,…,m−k.


### 3.1. Explicit Expression in terms of the Dimension-Reduced Matrix Scale Function

Rearranging (22) gives(27)ψ⟶^s;Ek=B22s−1ψ⟶0;Ek−c22−1λ22F¯^22s1⟶,as the Laplace transform of W22u is B22s−1; then, by inverting (27), it gives(28)ψ⟶u;Ek=W22uψ⟶0;Ek−∫0uW22u−xF¯22xdxc22−1λ221⟶=∫0∞R22u,xF¯22xdxc22−1λ221⟶,u>0,where W22u is the modified matrix version of the scale function defined in Section 2 and(29)R22u,x=W22ue−Θ22x−W22u−x,0≤x≤u,W22ue−Θ22x,x>u.Remark 5.
(1)
Following the same reasoning, we have(30)ψ⟶u;E\Ek=∫0∞R11u,xF¯11xdxc11−1λ111⟶,u>0,ψ⟶0;E\Ek=∫0∞e−xΘ11F¯11xdxc11−1λ111⟶,whereW11u is the dimension-reduced matrix scale function defined in Section 2, and(31)R11u,x=W11ue−Θ11x−W11u−x,0≤x≤u,W11ue−Θ11x,x>u,Θ11=q11−1ρ11q11,ρ11=diagρ11,1,ρ11,2,…,ρ11,k,q11=q⟶11,1,q⟶11,2,…,q⟶11,m−k⊤,withq⟶11,i⊤ being a 1×k row vector such that q⟶11,i⊤B11ρ11,i=0⟶⊤, for i=1,2,…,k.(2)
In particular, we have(32)ψi0;E\i=λici∫0∞e−ρiuF¯iudu,where ρi is the unique positive solution of(33)λi+αi−cis=λif^is,i=1,2,…,m.
Equation (30) reduces to the following for u>0:(34)ψiu;E\i=ηαiuψi0;E\i−λici∫0uηαiu−xF¯ixdx,i=1,2,…,m,where ψiu;E\i is the probability of ruin with the surplus only visiting the initial state i.


### 3.2. Exponential Claim Distributions

We assume thatfix=βie−βix with βi>0,x>0,i∈E, i.e., fx=βe−βx, with β=diagβ1,β2,…,βm. Differentiating equation (21), we obtain a 2nd-order matrix differential equation as follows:(35)ψ⟶′′u;Ek+β22−c22−1λ22−A22ψ⟶′u;Ek+c22−1β22A22ψ⟶u;Ek=0⟶,with limu⟶∞ψ⟶u;Ek=0⟶ and limu⟶0ψ⟶u;Ek=ψ⟶0;Ek. The solution to this matrix differential equation is(36)ψ⟶u;Ek=e−Ruψ⟶0;Ek,u>0,where ψ⟶0;Ek is given in (25) and R is an m−k×m−k matrix satisfying the following equation:(37)R2−β22−c22−1λ22−A22R+c22−1β22A22=0.Remark 6.
(i)
The solution to (37) is not unique, and the one whose eigenvalues have positive real parts is our R(ii)
A study of approaches to solve quadratic matrix equations is conducted in Higham and Kim [17]
In particular, ifm=2 and E1=2, then it follows from equation (36) that(38)ψ1u;2=e−R1uψ10;2,u>0,where R1 is the positive solution of the following quadratic equation:(39)c1s2−c1β1−λ1+α1s−β1α1=0,and ψ10;2 is given by (32) as follows:(40)ψ10;2=λ1c1∫0∞e−ρ1xF¯1xdx=λ1c1∫0∞e−ρ1xe−β1xdx=λ1c1ρ1+β1,with ρ1 satisfying equation λ1+α1−c1s=λ1f^1s. We comment that −ρ1 is the negative solution to equation (39).


## 3.1. Explicit Expression in terms of the Dimension-Reduced Matrix Scale Function

Rearranging (22) gives(27)ψ⟶^s;Ek=B22s−1ψ⟶0;Ek−c22−1λ22F¯^22s1⟶,as the Laplace transform of W22u is B22s−1; then, by inverting (27), it gives(28)ψ⟶u;Ek=W22uψ⟶0;Ek−∫0uW22u−xF¯22xdxc22−1λ221⟶=∫0∞R22u,xF¯22xdxc22−1λ221⟶,u>0,where W22u is the modified matrix version of the scale function defined in Section 2 and(29)R22u,x=W22ue−Θ22x−W22u−x,0≤x≤u,W22ue−Θ22x,x>u.Remark 5.
(1)
Following the same reasoning, we have(30)ψ⟶u;E\Ek=∫0∞R11u,xF¯11xdxc11−1λ111⟶,u>0,ψ⟶0;E\Ek=∫0∞e−xΘ11F¯11xdxc11−1λ111⟶,whereW11u is the dimension-reduced matrix scale function defined in Section 2, and(31)R11u,x=W11ue−Θ11x−W11u−x,0≤x≤u,W11ue−Θ11x,x>u,Θ11=q11−1ρ11q11,ρ11=diagρ11,1,ρ11,2,…,ρ11,k,q11=q⟶11,1,q⟶11,2,…,q⟶11,m−k⊤,withq⟶11,i⊤ being a 1×k row vector such that q⟶11,i⊤B11ρ11,i=0⟶⊤, for i=1,2,…,k.(2)
In particular, we have(32)ψi0;E\i=λici∫0∞e−ρiuF¯iudu,where ρi is the unique positive solution of(33)λi+αi−cis=λif^is,i=1,2,…,m.
Equation (30) reduces to the following for u>0:(34)ψiu;E\i=ηαiuψi0;E\i−λici∫0uηαiu−xF¯ixdx,i=1,2,…,m,where ψiu;E\i is the probability of ruin with the surplus only visiting the initial state i.


## 3.2. Exponential Claim Distributions

We assume thatfix=βie−βix with βi>0,x>0,i∈E, i.e., fx=βe−βx, with β=diagβ1,β2,…,βm. Differentiating equation (21), we obtain a 2nd-order matrix differential equation as follows:(35)ψ⟶′′u;Ek+β22−c22−1λ22−A22ψ⟶′u;Ek+c22−1β22A22ψ⟶u;Ek=0⟶,with limu⟶∞ψ⟶u;Ek=0⟶ and limu⟶0ψ⟶u;Ek=ψ⟶0;Ek. The solution to this matrix differential equation is(36)ψ⟶u;Ek=e−Ruψ⟶0;Ek,u>0,where ψ⟶0;Ek is given in (25) and R is an m−k×m−k matrix satisfying the following equation:(37)R2−β22−c22−1λ22−A22R+c22−1β22A22=0.Remark 6.
(i)
The solution to (37) is not unique, and the one whose eigenvalues have positive real parts is our R(ii)
A study of approaches to solve quadratic matrix equations is conducted in Higham and Kim [17]
In particular, ifm=2 and E1=2, then it follows from equation (36) that(38)ψ1u;2=e−R1uψ10;2,u>0,where R1 is the positive solution of the following quadratic equation:(39)c1s2−c1β1−λ1+α1s−β1α1=0,and ψ10;2 is given by (32) as follows:(40)ψ10;2=λ1c1∫0∞e−ρ1xF¯1xdx=λ1c1∫0∞e−ρ1xe−β1xdx=λ1c1ρ1+β1,with ρ1 satisfying equation λ1+α1−c1s=λ1f^1s. We comment that −ρ1 is the negative solution to equation (39).


## 4. Probability of Ruin without a Claim Occurring in States inEk

ForEk=l1,l2,…,lk and for i∈E, let(41)χiu;Ek=PTu<∞,Nu,Ek=0J0=ibe the probability of ruin without a claim occurring in Ek by the time of ruin given that J0=i.Remark 7.
(1)
χiu;k is the probability of ruin without a claim occurring in state k, for k∈E, given J0=i(2)
χiu;E\Ek is the probability of ruin with claims occurring in states in Ek(3)
χiu;E\i is the probability of ruin with claims only occurring when the process is in the state i(4)
IfEk=∅, then χiu;∅ recovers the ruin probability ψiu(5)
Note thatχiu;Ek−ψiu;Ek is the probability of ruin with the surplus visiting Ek, but without claims occurring in Ek
Conditioning on the events that may occur in an infinitesimal interval, we can obtain the following IDE forχiu;Ek:(42)ciχi′u;Ek=λiχiu;Ek−∑j=1mαi,jχju;Ek−λi∫0uχiu−x;Ekfixdx−λiF¯iu,i∉Ek,ciχi′u;Ek=λiχiu;Ek−∑j=1mαi,jχju;Ek,i∈Ek.
Letχ⟶u;Ek=χiu;Ek;i∈E⊤ be an m×1 column vector. In the matrix form, we have(43)χ⟶′u;Ek=c−1λ−Aχ⟶u;Ek−c−1λI−IEk∫0ufxχ⟶u−x;Ekdx−c−1λI−IEkF¯u1⟶.
Taking Laplace transforms of both sides of (43), we have(44)Bs;Ekχ⟶^s;Ek=χ⟶0;Ek−c−1λI−IEkF¯^s1⟶,where Bs;Ek=sI−c−1λ−A+c−1λI−IEkf^s. Then,(45)χ⟶0;Ek=∫0∞e−xΘEkF¯xdxc−1λI−IEk1⟶,χ⟶u;Ek=Vu;Ekχ⟶0;Ek−∫0uVu−x;EkF¯xdxc−1λI−IEk1⟶=∫0∞Ru,x;EkF¯xdxc−1λI−IEk1⟶,where Vu;Ek is a modified matrix scale function defined in Section 2. And,(46)Ru,x;Ek=Vu;Eke−xΘEk−Vu−x;Ek,0≤x≤u,Vu;Eke−xΘEk,x>u,ΘEk=QEk−1diagρ1Ek,ρ2Ek,…,ρmEkQEk,QEk=q⟶Ek,1,q⟶Ek,2,…,q⟶Ek,m⊤,with q⟶Ek,i⊤ being a 1×mrow vector such that q⟶Ek,i⊤BρiEk;Ek=0⟶⊤, for i=1,2,…,m.
If we assume claim amounts are exponentially distributed, i.e.,fx=βe−βx, with β=diagβ1,β2,…,βm, then by differentiating equation (43), we can obtain a 2nd-order matrix differential equation for χ⟶u;Ek whose solution is given by(47)χ⟶u;Ek=e−Δuχ⟶0;Ek,u>0,where Δ is the solution to the following quadratic matrix equation:(48)Δ2−β−c−1λ−AΔ+c−1βλIEk−A=0.
We remark that the solutions to the equation above is not unique and the one with positive eigenvalues is ourΔ.


## 5. Probability of Hitting a Target Level without the Surplus Ever Visiting States inEk

Foru≤b, let Tub=mint≥0,Ut=b be the first time when the surplus reaches level b. We define(49)ψi,ju,b;Ek=PJt∉Ek,JTub=j,0≤t≤TubJ0=i,i,j∉Ek,to be the probability of hitting a target level b at state j without the surplus ever visiting substate space Ek, given that the initial state is i∉Ek.Remark 8.
(1)
ψi,ju,b;∅ is the probability of hitting a target level b at state j given that the initial state is i(2)
ψi,ju,b;E\i,j is the probability of hitting a target level b at state j from the initial state i without the surplus visiting any other states
By conditioning on the events that may occur in an infinitesimal interval, we obtain the following IDE:(50)ψ′u,b;Ek=c22−1λ22−A22ψu,b;Ek−c22−1λ22∫0∞f22yψu−y;b;Ekdx,where ψu,b;Ek=ψi,ju,b;Ek;i,j∈E\Ek is an m−k×m−k matrix.
The following three properties are satisfied byψi,j:Property 1:ψi,jb,b;Ek=Ii=jProperty 2:ψi,ju,b;Ek is a function of b−uProperty 3:ψi,ju,b;Ek=∑l∉Ekψi,lu,b1;Ekψl,jb1,b;Ek, for u≤b1≤bThis shows that ψu,b;Ek must have the following form:(51)ψu,b;Ek=e−Γ22b−u,u≤b,where Γ22 is an m−k×m−k matrix and all of its eigenvalues must have positive real parts. It can be showed by substituting (51) into (50) that Γ22 satisfies a matrix equation as follows:(52)Γ22+c22−1λ22∫0∞f22xe−Γ22xdx−c22−1λ22−A22=0,with the following solution:(53)Γ22=H22ρ22H22−1,where H22=h⟶22,1,h⟶22,2,…,h⟶22,m−k, ρ22=diagρ22,1,ρ22,2,…,ρ22,m−k, and ρ22,1,ρ22,2,…,ρ22,m−k are the m−k distinct roots of DetB22s=0. The column vector h⟶22,i is that B22ρ22,ih⟶22,i=0⟶.
In particular, forEk=E\i, the result in (51) simplifies to(54)ψi,iu,b;E\i=e−ρib−u,u≤b,which is the probability of hitting a target level b at state i from the initial state i without the surplus visiting any other states.


## 6. Probability of Hitting a Target Level without a Claim Occurring in States inEk

Foru≤b, define χi,ju,b;Ek to be the probability of hitting a target level b at state j∈E without a claim occurring in states in Ek, given that the initial state is i∈E.Remark 9.
(1)
χi,ju,b;∅=ψi,ju,b;∅ is the probability of hitting a target level b at state j given that the initial state is i(2)
χi,ju,b;E is the probability of hitting a target level b at state j from the initial state i without any claim occurring(3)
χi,ju,b;E\k is the probability of hitting a target level b at state j from the initial state i with claims occurring only in state k
Using the same arguments as in the previous section, we have foru≤b,(55)ciχi,j′u,b;Ek=λiχi,ju,b;Ek−∑k=1mαi,kχk,ju,b;Ek−λi∫0∞χi,ju−x,b;Ekfixdx,i∉Ekciχi,j′u,b;Ek=λiχi,ju,b;Ek−∑k=1mαi,kχk,ju,b;Ek,i∈Ek.
This is written in matrix form as(56)χ′u,b;Ek=c−1λ−Aχu,b;Ek=−c−1λI−IEk∫0∞fxχu−x;b;Ekdx,where χu,b;Ek=χi,ju,b;Ek;i,j∈E is an m×m matrix. Clearly, χi,ju,b;Ek has the same three properties as for ψi,ju,b;Ek stated in Section 5, and we have(57)χu,b;Ek=e−ΓEkb−u,u≤b,where ΓEk is an m×m matrix and all of its eigenvalues must have positive real parts and ΓEk satisfies an matrix equation with the following solution:(58)ΓEk=HEkρEkHEk−1,where HEk=h⟶Ek,1,h⟶Ek,2,…,h⟶Ek,m, ρEk=diagρ1Ek,ρ2Ek,…,ρmEk, and ρ1Ek,ρ2Ek,…,ρmEk are the m distinct roots of DetBs;Ek=0. The column vector h⟶Ek,i is that BρiEk;Ekh⟶Ek,i=0⟶.
In particular, whenEk=E, then equations (55) and (56) simplify to(59)χ′u,b;E=c−1λ−Aχu,b;E,with the following solution:(60)χu,b;E=ec−1A−λb−u,u≤b.


## 7. Two-Sided First Exit Probabilities

In Sections3–6, we have studied the one-sided exit probability without the surplus visiting states in Ek or without a claim occurring in states in Ek. In this section, we aim to investigate the corresponding two-sided first exit probabilities.


### 7.1. Probability of Hitting a Target Level prior to Ruin without the Process Visiting States inEk

For0≤u≤b and i,j∈E, we define(61)ξi,ju,b;Ek=PTub<Tu,JTub=j,Jt∉Ek,0≤t<TubJ0=i,to be the probability of reaching b in state j with neither ruin occurring nor the process visiting states in Ek, given that the initial state is i. Clearly, we have the following cases:(1)
ξi,ju,b;Ek=0, if i∈Ek or j∈Ek(2)
ξi,jb,b;Ek=Ii=j, if i,j∉Ek(3)
IfEk=∅, then ξi,ju,b;∅=ξi,ju,b is the probability of reaching b in state j prior to ruin given that J0=iLetξu,b;Ek=ξi,ju,b;Ek;i,j∈E\Ek be an m−k×m−k matrix. Using the same arguments as in Section 3, we can show that ξu,b;Ek satisfies the following matrix-form IDE:(62)ξ′u,b;Ek=c22−1λ22−A22ξu;Ek=−c22−1λ22∫0uf22xξu−x;Ekdx,0≤u<b,with the boundary condition ξb,b;Ek=I. It can be shown by using the Laplace transform method that W22u satisfies the IDE in (62) for 0≤u<∞, we conclude by noting that ξb,b;Ek=I that(63)ξu,b;Ek=W22uW22b−1,0≤u≤b.


### 7.2. Probability of Ruin without the Surplus Hitting a Certain Levelb and without Visiting States in Ek

For0≤u≤b and i∈E, we define(64)τiu,b;Ek=PTu<Tub,Jt∉Ek,0≤t<TuJ0=i,to be the probability of ruin without the surplus reaching b and without the process visiting states in Ek, given that the initial state is i. Clearly, we have the following cases:(1)
τiu,b;Ek=0, if i∈Ek(2)
τib,b;Ek=0, if i∈E(3)
τiu,b;E\i is the probability of ruin without the surplus hitting a certain level b and without state change from the initial state i prior to ruin(4)
IfEk=∅, then τiu,b;∅=τi,ju,b is the probability of ruin without the surplus reaching b prior to ruin given that J0=iBy considering whether or not the surplus has reachedb prior to ruin, we have the following formula:(65)ψiu;Ek=τiu,b;Ek+∑j∉Ekξi,ju,b;Ekψjb;Ek,i∉Ek.In the matrix form, we have(66)ψ⟶u;Ek=τ⟶u,b;Ek+ξu,b;Ekψ⟶b;Ek,where τ⟶u,b;Ek=τi,ju,b;Ek,i∈E\Ek⊤. This together with (63) gives(67)τ⟶u,b;Ek=ψ⟶u;Ek−W22uW22b−1ψ⟶b;Ek,0≤u≤b.


### 7.3. Probability of Hitting a Certain Level prior to Ruin without Claims Occurring in States inEk

For0≤u≤b and i,j∈E, we define ηi,ju,b;Ek to be the probability of the surplus reaching b in state j without ruin occurring and without claims occurring in states in Ek, given that J0=i. Clearly, we have the following cases:(1)
ηi,jb,b;Ek=Ii=j, if i,j∈E(2)
IfEk=∅, then ηi,ju,b;∅=ξi,ju,b is the probability of reaching b in state j prior to ruin given that J0=iLetηu,b;Ek=ηi,ju,b;Ek;i,j∈E be an m×m matrix. Using the same arguments as in Section 5, we can show that ηu,b;Ek satisfies the following matrix-form IDE:(68)η′u,b;Ek=c−1λ−Aηu,b;Ek=−c−1λI−IEk∫0ufxηu−x,b;Ekdx,0≤u<b,with the boundary condition η′b,b;Ek=I.Using the Laplace transform method, we can show thatVu;Ek satisfies IDE in (68) for 0≤u<∞. This together with ηb,b;Ek=I shows(69)ηu,b;Ek=Vu;EkVb;Ek−1,0≤u≤b.


### 7.4. Probability of Ruin without Hitting a Certain Levelb and without Claims Occurring in Ek

For0≤u≤b and i∈E, we define(70)ζiu,b;Ek=PTu<Tub,Nu,Ek=0J0=ito be the probability of ruin without the surplus reaching b and without claims occurring in any state in Ek given J0=i. Clearly, ζib,b;Ek=0.By considering whether or not the surplus has reachedb prior to ruin, we have the following formula:(71)χiu;Ek=ζiu,b;Ek+∑j∈Eηi,ju,b;Ekχjb;Ek,i∈E.In the matrix form, we have(72)χ⟶u;Ek=ζ⟶u,b;Ek+ηu,b;Ekχ⟶b;Ek,where ζ⟶u,b;Ek=ζi,ju,b;Ek,i∈E⊤. This together with (69) gives(73)ζ⟶u,b;Ek=χ⟶u;Ek−Vu;EkVb;Ek−1χ⟶b;Ek,0≤u≤b.


## 7.1. Probability of Hitting a Target Level prior to Ruin without the Process Visiting States inEk

For0≤u≤b and i,j∈E, we define(61)ξi,ju,b;Ek=PTub<Tu,JTub=j,Jt∉Ek,0≤t<TubJ0=i,to be the probability of reaching b in state j with neither ruin occurring nor the process visiting states in Ek, given that the initial state is i. Clearly, we have the following cases:(1)
ξi,ju,b;Ek=0, if i∈Ek or j∈Ek(2)
ξi,jb,b;Ek=Ii=j, if i,j∉Ek(3)
IfEk=∅, then ξi,ju,b;∅=ξi,ju,b is the probability of reaching b in state j prior to ruin given that J0=iLetξu,b;Ek=ξi,ju,b;Ek;i,j∈E\Ek be an m−k×m−k matrix. Using the same arguments as in Section 3, we can show that ξu,b;Ek satisfies the following matrix-form IDE:(62)ξ′u,b;Ek=c22−1λ22−A22ξu;Ek=−c22−1λ22∫0uf22xξu−x;Ekdx,0≤u<b,with the boundary condition ξb,b;Ek=I. It can be shown by using the Laplace transform method that W22u satisfies the IDE in (62) for 0≤u<∞, we conclude by noting that ξb,b;Ek=I that(63)ξu,b;Ek=W22uW22b−1,0≤u≤b.


## 7.2. Probability of Ruin without the Surplus Hitting a Certain Levelb and without Visiting States in Ek

For0≤u≤b and i∈E, we define(64)τiu,b;Ek=PTu<Tub,Jt∉Ek,0≤t<TuJ0=i,to be the probability of ruin without the surplus reaching b and without the process visiting states in Ek, given that the initial state is i. Clearly, we have the following cases:(1)
τiu,b;Ek=0, if i∈Ek(2)
τib,b;Ek=0, if i∈E(3)
τiu,b;E\i is the probability of ruin without the surplus hitting a certain level b and without state change from the initial state i prior to ruin(4)
IfEk=∅, then τiu,b;∅=τi,ju,b is the probability of ruin without the surplus reaching b prior to ruin given that J0=iBy considering whether or not the surplus has reachedb prior to ruin, we have the following formula:(65)ψiu;Ek=τiu,b;Ek+∑j∉Ekξi,ju,b;Ekψjb;Ek,i∉Ek.In the matrix form, we have(66)ψ⟶u;Ek=τ⟶u,b;Ek+ξu,b;Ekψ⟶b;Ek,where τ⟶u,b;Ek=τi,ju,b;Ek,i∈E\Ek⊤. This together with (63) gives(67)τ⟶u,b;Ek=ψ⟶u;Ek−W22uW22b−1ψ⟶b;Ek,0≤u≤b.


## 7.3. Probability of Hitting a Certain Level prior to Ruin without Claims Occurring in States inEk

For0≤u≤b and i,j∈E, we define ηi,ju,b;Ek to be the probability of the surplus reaching b in state j without ruin occurring and without claims occurring in states in Ek, given that J0=i. Clearly, we have the following cases:(1)
ηi,jb,b;Ek=Ii=j, if i,j∈E(2)
IfEk=∅, then ηi,ju,b;∅=ξi,ju,b is the probability of reaching b in state j prior to ruin given that J0=iLetηu,b;Ek=ηi,ju,b;Ek;i,j∈E be an m×m matrix. Using the same arguments as in Section 5, we can show that ηu,b;Ek satisfies the following matrix-form IDE:(68)η′u,b;Ek=c−1λ−Aηu,b;Ek=−c−1λI−IEk∫0ufxηu−x,b;Ekdx,0≤u<b,with the boundary condition η′b,b;Ek=I.Using the Laplace transform method, we can show thatVu;Ek satisfies IDE in (68) for 0≤u<∞. This together with ηb,b;Ek=I shows(69)ηu,b;Ek=Vu;EkVb;Ek−1,0≤u≤b.


## 7.4. Probability of Ruin without Hitting a Certain Levelb and without Claims Occurring in Ek

For0≤u≤b and i∈E, we define(70)ζiu,b;Ek=PTu<Tub,Nu,Ek=0J0=ito be the probability of ruin without the surplus reaching b and without claims occurring in any state in Ek given J0=i. Clearly, ζib,b;Ek=0.By considering whether or not the surplus has reachedb prior to ruin, we have the following formula:(71)χiu;Ek=ζiu,b;Ek+∑j∈Eηi,ju,b;Ekχjb;Ek,i∈E.In the matrix form, we have(72)χ⟶u;Ek=ζ⟶u,b;Ek+ηu,b;Ekχ⟶b;Ek,where ζ⟶u,b;Ek=ζi,ju,b;Ek,i∈E⊤. This together with (69) gives(73)ζ⟶u,b;Ek=χ⟶u;Ek−Vu;EkVb;Ek−1χ⟶b;Ek,0≤u≤b.


## 8. Numerical Illustrations

In this section, we give some numerical results of the probabilities that we have defined in previous sections. We consider two-state Markov-Modulated risk model with intensity matrix as follows:(74)A=−141434−34,and π=π1,π2=3/4,1/4. Further assume that claims follow exponential distributions with the following densities:(75)f1x=e−x,f2x=0.5e−0.5x,x>0.Other parameters are set asλ1=1,λ2=2/3,c1=4/3,andc2=5/3, so that positive loading condition holds true. Then, we have expressions for the probability of ruin without the surplus process visiting certain states and the probability of ruin without a claim occurring in certain states:(76)ψ1u;2=0.5346e−0.4654u,ψ2u;1=0.3388e−0.3306u,χ1u,1=0.0542e−0.3184u,χ1u,2=0.3632e−0.3184u,χ2u,1=0.5817e−0.4183u,χ2u,2=0.2064e−0.4183u.Table1 shows the probability of hitting a target level prior to ruin without the process visiting certain states. Table 2 shows the probability of ruin without the surplus hitting a certain level b and without visiting certain states. Table 3 gives the probability of ruin without hitting a certain level b and without claims occurring in certain states with initial states 1.Table 1
The probability of hitting the target levelb prior to ruin without the process visiting certain states.
b=15b=20ξ1u,b;2u=00.0014690.000196u=10.0029830.000398u=50.0177060.002362u=100.1333860.017793ξ2u,b;1u=00.0011321.04995∗10−6u=10.0000692.29747∗10−6u=50.0011060.000037u=100.0332740.001107Table 2
The probability of ruin without the surplus hitting a certain levelb and without visiting certain states.
b=15b=20b=∞τ1u,b;2u=00.5346100.5346110.534611u=10.3356760.3356770.335677u=50.0521660.0521750.052175u=100.0050260.0050910.005092τ2u,b;1u=00.3388130.3388130.338813u=10.2434360.2434360.243436u=50.0648730.0648760.064876u=100.0123430.0124220.012423Table 3
The probability of ruin without hitting a certain levelb and without claims occurring in certain states with initial state 1.
b=15b=20b=∞ζ1u,b;1u=00.0542300.0542310.054231u=10.0394440.0394440.039444u=50.0110290.0110380.011038u=100.0021520.0022450.002247ζ1u,b;2u=00.3632400.3632420.363242u=10.2641920.2641950.264195u=50.0739100.0739340.073934u=100.0148030.0150440.015049We can see from Table1 that the probability of hitting a target level prior to ruin without the process visiting certain states increases as initial surplus u increases and but decreases as target level b increases. As the target level increases, it is less likely for surplus process to hit that target level prior to ruin. Table 2 illustrates that the probability of ruin without the surplus hitting a certain level b and without visiting certain states does not change much as the target level changes, but only when initial surplus level is higher, the difference is slightly more obvious. This is because when initial surplus is small, ruin is more likely to occur at an earlier stage. In addition, Table 3 shows that the probability of ruin without hitting a claim occurring in certain states with different initial states performs a similar pattern as shown in Table 2 for similar reasons. At last, when b goes to infinity in Table 2, probabilities become probabilities of ruin without the surplus visiting certain states, and when b goes to infinity in Table 3, probabilities are probabilities of ruin without claims occurring in certain states with initial states 1.


## 9. Concluding Remarks

In this paper, we study, for a Markov-modulated risk model, some states-specific one-sided exit probabilities and two-sided first exit probabilities. These probabilities can be expressed in terms of a modified or a dimension-reduced matrix scale function.Further research includes the expression of joint and/or marginal Laplace transforms of state-specific occupation time (the duration that the surplus is between an interval and in substate spaceEk by the time of ruin or other exit times) and the number and the aggregate claims occurred during the state-specific occupation time in terms of the matrix-scale functions defined in this paper.



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*Source: 5830245-2020-10-20.xml*