# Ruin Probabilities in the Mixed Claim Frequency Risk Models

Authors: Zhao Xiaoqin; Chuangxia Huang
Journal: Mathematical Problems in Engineering
 (2014)

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## Abstract

We consider two mixed claim frequency risk models. Some important probabilistic properties are obtained by probability-theory methods. Some important results about ruin probabilities are obtained by martingale approach.


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## Body

## 1. Introduction

Mixture models are a fundamental tool in applied statistics, for most mixture models, including the widely used mixtures of Gaussians and hidden Markov models (HMMs); the current practice relies on the Expectation-Maximization (EM) algorithm, a local search heuristic for maximum likelihood estimation; an efficient method of moments approach to parameter estimation for a broad class of high-dimensional mixture models with many components was developed [1–3]. Finite mixture models have a long history in statistics; a detailed review of mixture models and models-based clustering was provided in [4], for a finite mixture of regressions model, [5, 6] develop an efficient EM algorithm for numerical optimization with provable convergence properties.In this paper we consider two mixture models. In Section2 we set up the binomial-Poisson model, and some important results about ruin probabilities are obtained by martingale approach.In Section3 we also set up another Poisson-dualistic model; in this section we obtain some important probabilistic properties and estimates for ruin probability.In the classical risk model, the surplus of an insurance company over the interval(0,t] is X(t), which is defined by
(1)X(t)=ct-∑k=1N(t)Zk,
where N(t) is to be interpreted as the number of claims on the company during the interval (0,t]; assume that {N(t);t≥0} is an homogeneous Poisson process with intensity λ. In the complex problems of actual operation in insurance business, insurance company classifies the risk by its some characteristics, but the claim frequency λ of the individual policy which has been classified into the same kind of portfolio may be different; that is, this is the so-called nonhomogeneity. For a nonhomogeneous portfolio, we can assume that λ is a random variable; thus the mixed Poisson distribution model can be derived.In general, just as reported in [6] if the number for claim N is a discrete distribution with parameter α and the distribution sequence is
(2)P(N=k∣α)=p(k;α)
and the parameter α is random variable or random vector, its probability distribution function is
(3)α~G(α).
Then the corresponding risk model is a mixed claim frequency risk model.


## 2. Binomial-Poisson Model

### 2.1. The Setting-Up of the Model

Definition 1.
SupposeM={M(t);t≥0} is an homogeneous Poisson process with intensity λ. Consider M(t)=m; then the conditional distribution of N={N(t);t≥0} is a binomial distribution with parameters m, p. A sequence {Zk}1∞ is an independent and identically distributed nonnegative random variable, having the common distribution function F, with F(0)=0, mean value μ, and variance σ2; the above random process and random sequence are mutually independent; c,p(c>0,0<p<1) are both constants.
Let(4)X(t)=ct-∑k=1N(t)Zk.
Then the process defined by (4) is a binomial-Poisson mixed claim frequency risk model.
The conditional moment generating function ofN is
(5)MN(s∣m)=(pes+q)m,(q=1-p).
Therefore
(6)MN(t)(s)=EesN(t)=E[E(esN(t)∣M(t))]=∑m=0∞(pes+q)me-λt(λt)mm!=exp⁡[λpt(es-1)].
Then {N(t);t≥0} is a homogeneous Poisson process with intensity λp. Thus the model defined by (4) is a classical risk model.


### 2.2. The Meaning of the Model in the Insurance Practice

Suppose thatM(t) is the number of accidents during the interval (0,t]; the number of claims per accident L is 0-1 distribution with p; that is, P(L=1)=p, P(L=0)=1-p (for example, in the deductible insurance, the probability of loss exceeding the amount of deductible is p); then the number of claims N(t) during the interval (0,t] is a conditional binomial distribution with parameters m,p. Obviously the process {N(t);t≥0} is a p-sparse process of the process {M(t);t≥0}. {N(t);t≥0} is a homogeneous Poisson process with intensity λp, if only {M(t);t≥0} is a homogeneous Poisson process with intensity λ. c is the premium rate, {Zk}1∞ is the size of claim amount per accident, and ∑k=1N(t)Zk is the total claim process.


### 2.3. Several Conclusions about the Ultimate Ruin Probability

We can get easily(7)E[X(t)]=(c-λpμ)t.
In order to stabilize the operation of company, we should ensure that premiums received in a unit of time meet c>λpμ.The relative safety loadingρ is defined by
(8)ρ=cλpμ-1.We can now define the ruin probabilityψ(u) of a company facing the risk process (4) and having initial capital u. Consider
(9)ψ(u)=P{u+X(t)<0forsomet>0}.Let(10)h(r)=∫0∞erzdF(z)-1.We assume that there existsr∞>0 such that h(r)↑+∞ when r↑r∞, that is, a light-tailed distribution F.Theorem 2.
Ifρ>0, ψ(0)=λpμ/c.Theorem 3.
Let amount claimed sequence{Zk,k=1,2,…} with exponential distribution with mean μ; then
(11)ψ(u)=11+ρe-ρu/μ(1+ρ),
where ρ is given by (8).Theorem 4.
Consider(12)lim⁡u→∞eRuψ(u)=ρμh′(R)-c/λp,
where R is the positive solution of h(r)=cr/λp, which is called adjustment coefficient. And ρ is given by (8).Theorem 5.
Ruin probability:(13)ψ(u)≤e-Ru,
where R is the positive solution of h(r)=cr/λp, which is called adjustment coefficient.The above theorem can be derived directly by the corresponding results in classical risk theory [7].


## 2.1. The Setting-Up of the Model

Definition 1.
SupposeM={M(t);t≥0} is an homogeneous Poisson process with intensity λ. Consider M(t)=m; then the conditional distribution of N={N(t);t≥0} is a binomial distribution with parameters m, p. A sequence {Zk}1∞ is an independent and identically distributed nonnegative random variable, having the common distribution function F, with F(0)=0, mean value μ, and variance σ2; the above random process and random sequence are mutually independent; c,p(c>0,0<p<1) are both constants.
Let(4)X(t)=ct-∑k=1N(t)Zk.
Then the process defined by (4) is a binomial-Poisson mixed claim frequency risk model.
The conditional moment generating function ofN is
(5)MN(s∣m)=(pes+q)m,(q=1-p).
Therefore
(6)MN(t)(s)=EesN(t)=E[E(esN(t)∣M(t))]=∑m=0∞(pes+q)me-λt(λt)mm!=exp⁡[λpt(es-1)].
Then {N(t);t≥0} is a homogeneous Poisson process with intensity λp. Thus the model defined by (4) is a classical risk model.


## 2.2. The Meaning of the Model in the Insurance Practice

Suppose thatM(t) is the number of accidents during the interval (0,t]; the number of claims per accident L is 0-1 distribution with p; that is, P(L=1)=p, P(L=0)=1-p (for example, in the deductible insurance, the probability of loss exceeding the amount of deductible is p); then the number of claims N(t) during the interval (0,t] is a conditional binomial distribution with parameters m,p. Obviously the process {N(t);t≥0} is a p-sparse process of the process {M(t);t≥0}. {N(t);t≥0} is a homogeneous Poisson process with intensity λp, if only {M(t);t≥0} is a homogeneous Poisson process with intensity λ. c is the premium rate, {Zk}1∞ is the size of claim amount per accident, and ∑k=1N(t)Zk is the total claim process.


## 2.3. Several Conclusions about the Ultimate Ruin Probability

We can get easily(7)E[X(t)]=(c-λpμ)t.
In order to stabilize the operation of company, we should ensure that premiums received in a unit of time meet c>λpμ.The relative safety loadingρ is defined by
(8)ρ=cλpμ-1.We can now define the ruin probabilityψ(u) of a company facing the risk process (4) and having initial capital u. Consider
(9)ψ(u)=P{u+X(t)<0forsomet>0}.Let(10)h(r)=∫0∞erzdF(z)-1.We assume that there existsr∞>0 such that h(r)↑+∞ when r↑r∞, that is, a light-tailed distribution F.Theorem 2.
Ifρ>0, ψ(0)=λpμ/c.Theorem 3.
Let amount claimed sequence{Zk,k=1,2,…} with exponential distribution with mean μ; then
(11)ψ(u)=11+ρe-ρu/μ(1+ρ),
where ρ is given by (8).Theorem 4.
Consider(12)lim⁡u→∞eRuψ(u)=ρμh′(R)-c/λp,
where R is the positive solution of h(r)=cr/λp, which is called adjustment coefficient. And ρ is given by (8).Theorem 5.
Ruin probability:(13)ψ(u)≤e-Ru,
where R is the positive solution of h(r)=cr/λp, which is called adjustment coefficient.The above theorem can be derived directly by the corresponding results in classical risk theory [7].


## 3. Poisson-Dualistic Model

### 3.1. The Setting-Up of the Model

Definition 6.
The random variableλ follows two-point distribution; that is, P(λ=λi)=pi, i=1,2, p2=1-p1. The conditional distribution of N={N(t);t≥0} about λ is a homogeneous Poisson process with intensity λ. A sequence {Zk}1∞ is independent and identically distributed nonnegative random variable, having the common distribution function F, with F(0)=0, mean value μ, and variance σ2; {N(t);t≥0} and {Zk} are mutually independent; p1,p2,λ1,λ2,c are both positive real constants.
Let(14)Y(t)=ct-∑k=1N(t)Zk.
Then the random process defined by (14) is a Poisson-dualistic mixed claim frequency risk model.
Obviously,N={N(t);t≥0} is a mixed Poisson process.


### 3.2. The Meaning of the Model in the Insurance Practice

We assume that a portfolio is composed of high risk and low risk insurance policy, where high risk policy is accounting forp1 and low risk policy is accounting for p2=1-p1; the Poisson parameter of these two kinds of policy, respectively, is λ1, λ2 (corresponding high risk cover for λ1, obviously λ1>λ2). c is the premium rate, {Zk}1∞ is the size of claim amount per accident, and ∑k=1N(t)Zk is the total claim process.


### 3.3. Some Probabilistic Properties of Model

The probabilistic properties of the number of claimsN(t) are as follows.Property 1.
Claim frequency distribution is(15)P(N(t)=k)=p1e-λ1t(λ1t)kk!+p2e-λ2t(λ2t)kk!,hhhhhhhhhhhhhhhhhhhhhhhhlk=0,1,2,….Proof.
The above distribution can be derived by(16)P(N(t)=k)=∫ΩP(N(t)=k∣λ)dG(λ).
The proof is ended.Property 2.
The mean and variance ofN(t) are
(17)E[N(t)]=(λ1p1+λ2p2)tVar⁡[N(t)]=(λ1-λ2)2p1p2t2+(λ1p1+λ2p2)t.Proof.
Consider(18)E[N(t)]=E[E(N(t)∣λ)]=E(λt)=(λ1p1+λ2p2)tVar⁡[N(t)]=Var⁡[E(N(t)∣λ)]+E[Var⁡(N(t)∣λ)]=Var⁡(λt)+E(λt)=[λ12p1+λ22p2-(λ1p1+λ2p2)2]t2+(λ1p1+λ2p2)t=(λ1-λ2)2p1p2t2+(λ1p1+λ2p2)t.
The proof is ended.Since the mean and variance of the Poisson distribution are always equal, if the sample variance of the portfolio’s claim of a random variable is greater than the number of the sample mean, we can conclude that the existence of this policy combination of a degree of nonhomogeneity, and becauseVar⁡[N(t)]-E[N(t)]=Var⁡(λt) can reflect the variance of the degree of nonhomogeneity of portfolio, if the variance is much more greater than the mean, that is, the bigger (λ1-λ2)2p1p2 is, the more serious the nonhomogeneity is.Property 3.
The moment generating function ofN(t) is
(19)MN(t)(r)=p1exp⁡[λ1t(er-1)]+p2exp⁡[λ2t(er-1)].Proof.
Consider(20)MN(t)(r)=EerN(t)=E[E(erN(t)∣λ)]=E[exp⁡(λt(er-1))]=p1exp⁡[λ1t(er-1)]+p2exp⁡[λ2t(er-1)].
The proof is ended.Theorem 7.
Assume that the random variableλ follows two-point distribution; that is, P(λ=λi)=pi, i=1,2, and p2=1-p1. The conditional distribution of N={N(t);t≥0} about λ is a homogeneous Poisson process with intensity λ. Then the process N has stationary increments.Proof.
Consider(21)P(N(t+h)-N(t)=k)=E[P(N(t+h)-N(t)=k∣λ)]=E[e-λh(λh)kk!]=p1e-λ1h(λ1h)kk!+p2e-λ2h(λ2h)kk!,k=0,1,2,….
The proof is ended.Theorem 8.
Assume that the random variableλ follows two-point distribution; that is, P(λ=λi)=pi, i=1,2, p2=1-p1. The conditional distribution of N={N(t);t≥0} about λ is a homogeneous Poisson process with intensity λ if and only if the interval {Tn} of {N(t);t≥0} about λ has conditional independence and is the same as the exponential distribution with parameter λ; thus {Tn} follow the following mixed exponential distribution:
(22)fTn(t)=p1λ1e-λ1t+p2λ2e-λ2t,t>0.Proof.
We only need to prove that when the conditional distribution of{Tn} about λ is exponentially distributed, {Tn} follow the mixed exponential distribution. Then, when t>0, the distribution function of {Tn} is
(23)FTn(t)=P(Tn≤t)=E[P(Tn≤t∣λ)]=E[1-e-λt]=p1(1-e-λ1t)+p2(1-e-λ2t).
Thus the density function of {Tn} is
(24)fTn(t)=p1λ1e-λ1t+p2λ2e-λ2t,t>0.
The proof is ended.Theorem 9.
It occurN(t) claims over (0,t]; then the occurrence probability of intensity λ=λ1 for claim is
(25)P(λ=λ1∣N(t)=n)=p1e-λ1t(λ1t)np1e-λ1t(λ1t)n+p2e-λ2t(λ2t)n.Proof.
It can be derived by(26)P(λ=λ1∣N(t)=n)=P(N(t)=n∣λ=λ1)P(λ=λ1)P(N(t)=n).
The proof is ended.Thus we can get(27)P(λ=λ2∣N(t)=n)=1-p1e-λ1t(λ1t)np1e-λ1t(λ1t)n+p2e-λ2t(λ2t)n=p2e-λ2t(λ2t)np1e-λ1t(λ1t)n+p2e-λ2t(λ2t)n.Theorem 10.
Assume thatU(t) is the time interval from the moment t to the next claim; then the conditional distribution of U(t) is
(28)P(U(t)≤x∣N(t)=n)=p1e-λ1t(λ1t)n(1-e-λ1x)+p2e-λ2t(λ2t)n(1-e-λ2x)p1e-λ1t(λ1t)n+p2e-λ2t(λ2t)n.Proof.
ObviouslyU(t) is the occurrence time after the moment t, when λ is a constant; then U(t) follows the exponential distribution with λ. Then the result can be derived by
(29)P(U(t)≤x∣N(t)=n)=P(U(t)≤x∣N(t)=n,λ=λ1)×P(λ=λ1∣N(t)=n)+P(U(t)≤x∣N(t)=n,λ=λ2)×P(λ=λ2∣N(t)=n)=p1e-λ1t(λ1t)n(1-e-λ1x)+p2e-λ2t(λ2t)n(1-e-λ2x)p1e-λ1t(λ1t)n+p2e-λ2t(λ2t)n.
The proof is ended.The probabilistic properties of total amount of claimsS(t)=∑k=1N(t)Zk are as follows.Property 4.
The mean and variance of total amount of claimsS(t)=∑k=1N(t)Zk are
(30)E[S(t)]=(λ1p1+λ2p2)μtVar⁡[S(t)]=μ2(λ1-λ2)2p1p2t2+(μ2+σ2)(λ1p1+λ2p2)t.Proof.
Consider(31)E[S(t)]=E[N(t)]EZk=(λ1p1+λ2p2)μtVar⁡[S(t)]=Var⁡[E(S(t)∣N(t))]+E[Var⁡(S(t)∣N(t))]=Var⁡[N(t)EZk]+E[N(t)Var⁡Zk]=μ2Var⁡[N(t)]+σ2E[N(t)]=μ2[(λ1-λ2)2p1p2t2+(λ1p1+λ2p2)t]+σ2(λ1p1+λ2p2)t=μ2(λ1-λ2)2p1p2t2+(μ2+σ2)(λ1p1+λ2p2)t.
The proof is ended.Property 5.
The moment generating function of total amount of claimsS(t)=∑k=1N(t)Zk is
(32)MS(t)(r)=p1exp⁡[λ1t(MZ(r)-1)]+p2exp⁡[λ2t(MZ(r)-1)],
where MZ(r) is the moment generation function of the individual claim amount.Proof.
We can get the following from (19):
(33)MS(t)(r)=EerS(t)=E[(MZ(r))N(t)]=MN(t)(ln⁡MZ(r))=p1exp⁡[λ1t(MZ(r)-1)]+p2exp⁡[λ2t(MZ(r)-1)].
The proof is ended.


### 3.4. Estimation of Lundberg Exponential Upper Bounds for the Ultimate Ruin Probability

Obviously,E[Y(t)]=[c-(λ1p1+λ2p2)μ]t.It seems very natural to assume the premium rate per unit of timec>(λ1p1+λ2p2)μ. Further, let c>λ1μ, in order to stabilize the operation of the company.The relative safety loading:(34)ρ=c(λ1p1+λ2p2)μ-1.
Obviously,
(35)ρ>cλ1μ-1=^ρ1.ρ
1 can be seen as the corresponding relative safety loading of portfolio consisted by high risk policy. Obviously in terms of premium rate and the average individual claim amount are equal, the relative safety loading of the corresponding portfolio of model (14) should be greater.Letu be the initial capital; then the ruin moment is defined as
(36)Tu=inf⁡{t≥0∣u+Y(t)<0}.
Obviously Tu is a FY-stopping time.The finite time ruin probability is(37)ψ(u,t)=P(Tu<t).
The ultimate ruin probability is
(38)ψ(u)=P(Tu<∞).
Let
(39)h(r)=MZ(r)-1.
We assume that there exists r∞>0, such that h(r)↑+∞ when r↑r∞. It is easy to be seen that h(0)=0, h′(r)>0, and h′′(r)>0 and that h(r) is continuous on [0,r∞) (where r∞≤+∞).Easy to verify, the risk process defined by (14) is a right continuous process and has the following properties:(i)
Y(0)=0 P-a.s.;(ii)
Y has stationary and independent increments about λ;(iii)
E[Y(t)]=[c-(λ1p1+λ2p2)μ]t>0.Let(40)F∞λ=σ(λ,t≥0),Ft=F∞λ∨FtY.
Note that F0=F∞λ.We can easily get(41)E[e-rY(t)∣F∞λ]=E{exp⁡[-r(ct-S(t))]∣F∞λ}=exp⁡(-rct)·E[exp⁡(rS(t))∣F∞λ]=exp⁡[t(λh(r)-rc)].
From the hypothesis of function h(r), we know ∃r>0, s.t. E[e-rY(t)∣F∞λ]<∞.From property (ii) ofY, we have
(42)exp⁡[t(λh(r)-rc)]=E[e-rY(t)∣F∞λ]=E[e-r(Y(t)-Y(s))·e-rY(s)∣F∞λ]=E[e-r(Y(t)-Y(s))∣F∞λ]E[e-rY(s)∣F∞λ].
Thus
(43)E[e-r(Y(t)-Y(s))∣F∞λ]=exp⁡[(λh(r)-rc)(t-s)].Theorem 11.
Let(44)Mu(t)=exp⁡[-r(u+Y(t))]exp⁡[t(λh(r)-rc)].
Thus Mu={Mu(t);t≥0} is an F-martingale.Proof.
We get from (43) that
(45)E[Mu(t)∣Fs]=E[exp⁡(-r(u+Y(t)))exp⁡(t(λh(r)-rc))∣Fs]=E[exp⁡(-r(u+Y(s)))exp⁡(s(λh(r)-rc))·exp⁡(-r(Y(t)-Y(s)))exp⁡[(λh(r)-rc)(t-s)]∣Fs]=Mu(s)E[exp⁡(-r(Y(t)-Y(s)))exp⁡[(λh(r)-rc)(t-s)]∣Fs]=Mu(s).
The proof is ended.Chooset0<∞ and consider t0∧Tu which is a bounded F-stopping time. We get from Doob’s stopping theorem that
(46)e-ru=Mu(0)=EF0[Mu(t0∧Tu)]=EF0[Mu(t0∧Tu)∣Tu≤t0]PF0(Tu≤t0)+EF0[Mu(t0∧Tu)∣Tu>t0]PF0(Tu>t0)≥EF0[Mu(t0∧Tu)∣Tu≤t0]PF0(Tu≤t0)=EF0[Mu(Tu)∣Tu≤t0]PF0(Tu≤t0).
Using u+Y(Tu)≤0 on {Tu<∞}, the lower bound was shown to be given by
(47)EF0[Mu(Tu)∣Tu≤t0]=EF0[exp⁡(-r(u+Y(Tu)))exp⁡(Tu(λh(r)-rc))∣Tu≤t0]≥EF0[1exp⁡(Tu(λh(r)-rc))∣Tu≤t0]≥inf⁡0≤t≤t0[1exp⁡(t(λh(r)-rc))].
Thus we have
(48)PF0(Tu≤t0)≤e-ruEF0[Mu(Tu)∣Tu≤t0]≤e-rusup⁡0≤t≤t0{exp⁡[t(λh(r)-rc)]}.
and, by taking expectation,
(49)ψ(u,t0)=P(Tu≤t0)≤e-ruE{sup⁡0≤t≤t0exp⁡[t(λh(r)-rc)]}.
When t0→∞ in the above equation, we get
(50)ψ(u)≤e-ruE{sup⁡t≥0exp⁡[t(λh(r)-rc)]}.We now want to chooser as large as possible under the restriction E{sup⁡t≥0exp⁡[t(λh(r)-rc)]}<∞. Let R denote that value, named adjustment coefficient of surplus process (14).Since(51)E{sup⁡t≥0exp⁡[t(λh(r)-rc)]}=p1sup⁡t≥0exp⁡[t(λ1h(r)-rc)]+p2sup⁡t≥0exp⁡[t(λ2h(r)-rc)]=^p1sup⁡t≥0etg1(r)+p2sup⁡t≥0etg2(r),
where gi(r)=λih(r)-rc, i=1,2, the two terms of the right side in the above equality are both positive, which corresponds to restrict
(52)g1(r)≤0,g2(r)≤0.
Since
(53)g1(r)≥g2(r),
thus we just need to restrict
(54)g1(r)≤0.
Then we have
(55)R=sup⁡{r∣g1(r)≤0}.Theorem 12.
The ultimate ruin probability meets the inequality(56)ψ(u)≤e-Ru,
where R is the only positive solution of g1(r)=0, named adjustment coefficient.Proof.
Forg1(r)=λ1h(r)-rc, we have
(57)g1(0)=0,g1′(r)=λ1h′(r)-c.
Therefore
(58)g1′(0)=λ1h′(0)-c=λ1μ-c<0.
Then there exists small enough Δr∈(0,r∞) such that g1(Δr)<0. Further g1(r)→+∞ when r∞<+∞ and r↑r∞.
Whenr∞=+∞, since MZ(r)=1+rμ1+r2μ2/2+⋯>r2μ2/2, where μj is the jth moment of the claim amount {Zk}, then
(59)g1(r)=λ1MZ(r)-λ1-rc>λ1μ22r2-λ1-rc⟶+∞,(r⟶+∞).
Thus there must exist r*∈(0,r∞) such that g1(r*)=0. And from g1′′(r)=λ1h′′(r)>0, we know g1(r) is a lower convex function.
Theng1(r)<0 for 0<r<r* and g1(r)>0 for r*<r<r∞.
Therefore there exists the only positive solution ofg1(r)=0; let r* denote that value. From (55) we know R=r*, so we get ψ(u)≤e-Ru.
The proof is ended.


### 3.5. Boundary of the Adjustment CoefficientR

The exact value of adjustment coefficientR generally cannot be determined by g1(r)=0, but because of its great significance to estimate the upper bound of the ruin probability, we estimate the boundary of the adjustment coefficient R.Theorem 13.
The adjustment coefficientR meets the inequality
(60)R<2(c-λ1μ)λ1(μ2+σ2).Proof.
From [7] we know R<2(c-λ1μ)/λ1μ2; then we can get the conclusion.
The proof is ended.Form the above theorem and the expression of the relative safety loadingsρ, ρ1, we get
(61)R<2(c-λ1μ)λ1(μ2+σ2)=2ρ1λ1μλ1(μ2+σ2)=2ρ1μμ2+σ2<2ρμμ2+σ2.The result shows that the upper bound ofR can be only defined by the one- or two-order moment of individual claim amount and the relative safety loading; while having nothing to do with the Poisson parameter, this case have its convenience in use.Theorem 14 (see [7]).
If the individual claim amount{Zk} has the upper bound W, that is, Zk≤W, we have
(62)R>1Wln⁡cλ1μ=1Wln⁡(1+ρ1).The result indicates that, under the conditions of the theorem, the lower bound ofR can be only defined by the relative safety loading ρ1 and the upper bound of individual claim amount.When the exact value ofR cannot be determined by the equation g1(r)=0, we can solve its approximate solution by the numerical method, and using these two boundaries as the initial values for iteration, we can quickly find the approximate solution satisfying requirements of certain accuracy.


### 3.6. The Probability of Survival

Except for a few special cases, generally, complicated calculation is needed to get the adjustment coefficient values. Through the following discussion on integrodifferential equations satisfied by survival probability, we can avoid the computation of the adjustment coefficient.LetΦ(u)=1-ψ(u); then Φ(u) indicates survival probability.Since the number of claim process{N(t);t≥0} about λ is a conditional renewal process and will not be ruined during the period of (0,S1), then we get
(63)Φ(u)=E[P(Y(t)≥0,∀t≥0∣S1,Z1,λ)]=E{E[Φ(u+cS1-Z1)∣λ]}=E{∫0∞λe-λt∫0u+ctΦ(u+ct-z)dF(z)dt}=∫0∞(p1λ1e-λ1t+p2λ2e-λ2t)∫0u+ctΦ(u+ct-z)dF(z)dt.Let(64)x=u+ct.
Then we have
(65)Φ(u)=p1λ1ceλ1u/c∫u∞e-λ1x/c∫0xΦ(x-z)dF(z)dx+p2λ2ceλ2u/c∫u∞e-λ2x/c∫0xΦ(x-z)dF(z)dx.
Let two terms of the hand side in the above equality be, respectively, Φ1(u) and Φ2(u); then Φ(u)=Φ1(u)+Φ2(u).Theorem 15.
The integrodifferential equations assured byΦ1(u) and Φ2(u) are
(66)Φ1′(u)=λ1cΦ1(u)-p1λ1c∫0u[Φ1(u-z)+Φ2(u-z)]dF(z)Φ2′(u)=λ2cΦ2(u)-p2λ2c∫0u[Φ1(u-z)+Φ2(u-z)]dF(z).Proof.
From (65) and the hypothesis of Φ1(u) and Φ2(u), by taking derivation, we can get the above result.
The proof is ended.Φ
1
(
u
) and Φ2(u) are calculated using the above equations; thus the expression of Φ(u) can be obtained. According to the above method to obtain the survival probability, we can immediately get the expression of ruin probability, and then the calculation of the adjustment coefficient can be avoided.


## 3.1. The Setting-Up of the Model

Definition 6.
The random variableλ follows two-point distribution; that is, P(λ=λi)=pi, i=1,2, p2=1-p1. The conditional distribution of N={N(t);t≥0} about λ is a homogeneous Poisson process with intensity λ. A sequence {Zk}1∞ is independent and identically distributed nonnegative random variable, having the common distribution function F, with F(0)=0, mean value μ, and variance σ2; {N(t);t≥0} and {Zk} are mutually independent; p1,p2,λ1,λ2,c are both positive real constants.
Let(14)Y(t)=ct-∑k=1N(t)Zk.
Then the random process defined by (14) is a Poisson-dualistic mixed claim frequency risk model.
Obviously,N={N(t);t≥0} is a mixed Poisson process.


## 3.2. The Meaning of the Model in the Insurance Practice

We assume that a portfolio is composed of high risk and low risk insurance policy, where high risk policy is accounting forp1 and low risk policy is accounting for p2=1-p1; the Poisson parameter of these two kinds of policy, respectively, is λ1, λ2 (corresponding high risk cover for λ1, obviously λ1>λ2). c is the premium rate, {Zk}1∞ is the size of claim amount per accident, and ∑k=1N(t)Zk is the total claim process.


## 3.3. Some Probabilistic Properties of Model

The probabilistic properties of the number of claimsN(t) are as follows.Property 1.
Claim frequency distribution is(15)P(N(t)=k)=p1e-λ1t(λ1t)kk!+p2e-λ2t(λ2t)kk!,hhhhhhhhhhhhhhhhhhhhhhhhlk=0,1,2,….Proof.
The above distribution can be derived by(16)P(N(t)=k)=∫ΩP(N(t)=k∣λ)dG(λ).
The proof is ended.Property 2.
The mean and variance ofN(t) are
(17)E[N(t)]=(λ1p1+λ2p2)tVar⁡[N(t)]=(λ1-λ2)2p1p2t2+(λ1p1+λ2p2)t.Proof.
Consider(18)E[N(t)]=E[E(N(t)∣λ)]=E(λt)=(λ1p1+λ2p2)tVar⁡[N(t)]=Var⁡[E(N(t)∣λ)]+E[Var⁡(N(t)∣λ)]=Var⁡(λt)+E(λt)=[λ12p1+λ22p2-(λ1p1+λ2p2)2]t2+(λ1p1+λ2p2)t=(λ1-λ2)2p1p2t2+(λ1p1+λ2p2)t.
The proof is ended.Since the mean and variance of the Poisson distribution are always equal, if the sample variance of the portfolio’s claim of a random variable is greater than the number of the sample mean, we can conclude that the existence of this policy combination of a degree of nonhomogeneity, and becauseVar⁡[N(t)]-E[N(t)]=Var⁡(λt) can reflect the variance of the degree of nonhomogeneity of portfolio, if the variance is much more greater than the mean, that is, the bigger (λ1-λ2)2p1p2 is, the more serious the nonhomogeneity is.Property 3.
The moment generating function ofN(t) is
(19)MN(t)(r)=p1exp⁡[λ1t(er-1)]+p2exp⁡[λ2t(er-1)].Proof.
Consider(20)MN(t)(r)=EerN(t)=E[E(erN(t)∣λ)]=E[exp⁡(λt(er-1))]=p1exp⁡[λ1t(er-1)]+p2exp⁡[λ2t(er-1)].
The proof is ended.Theorem 7.
Assume that the random variableλ follows two-point distribution; that is, P(λ=λi)=pi, i=1,2, and p2=1-p1. The conditional distribution of N={N(t);t≥0} about λ is a homogeneous Poisson process with intensity λ. Then the process N has stationary increments.Proof.
Consider(21)P(N(t+h)-N(t)=k)=E[P(N(t+h)-N(t)=k∣λ)]=E[e-λh(λh)kk!]=p1e-λ1h(λ1h)kk!+p2e-λ2h(λ2h)kk!,k=0,1,2,….
The proof is ended.Theorem 8.
Assume that the random variableλ follows two-point distribution; that is, P(λ=λi)=pi, i=1,2, p2=1-p1. The conditional distribution of N={N(t);t≥0} about λ is a homogeneous Poisson process with intensity λ if and only if the interval {Tn} of {N(t);t≥0} about λ has conditional independence and is the same as the exponential distribution with parameter λ; thus {Tn} follow the following mixed exponential distribution:
(22)fTn(t)=p1λ1e-λ1t+p2λ2e-λ2t,t>0.Proof.
We only need to prove that when the conditional distribution of{Tn} about λ is exponentially distributed, {Tn} follow the mixed exponential distribution. Then, when t>0, the distribution function of {Tn} is
(23)FTn(t)=P(Tn≤t)=E[P(Tn≤t∣λ)]=E[1-e-λt]=p1(1-e-λ1t)+p2(1-e-λ2t).
Thus the density function of {Tn} is
(24)fTn(t)=p1λ1e-λ1t+p2λ2e-λ2t,t>0.
The proof is ended.Theorem 9.
It occurN(t) claims over (0,t]; then the occurrence probability of intensity λ=λ1 for claim is
(25)P(λ=λ1∣N(t)=n)=p1e-λ1t(λ1t)np1e-λ1t(λ1t)n+p2e-λ2t(λ2t)n.Proof.
It can be derived by(26)P(λ=λ1∣N(t)=n)=P(N(t)=n∣λ=λ1)P(λ=λ1)P(N(t)=n).
The proof is ended.Thus we can get(27)P(λ=λ2∣N(t)=n)=1-p1e-λ1t(λ1t)np1e-λ1t(λ1t)n+p2e-λ2t(λ2t)n=p2e-λ2t(λ2t)np1e-λ1t(λ1t)n+p2e-λ2t(λ2t)n.Theorem 10.
Assume thatU(t) is the time interval from the moment t to the next claim; then the conditional distribution of U(t) is
(28)P(U(t)≤x∣N(t)=n)=p1e-λ1t(λ1t)n(1-e-λ1x)+p2e-λ2t(λ2t)n(1-e-λ2x)p1e-λ1t(λ1t)n+p2e-λ2t(λ2t)n.Proof.
ObviouslyU(t) is the occurrence time after the moment t, when λ is a constant; then U(t) follows the exponential distribution with λ. Then the result can be derived by
(29)P(U(t)≤x∣N(t)=n)=P(U(t)≤x∣N(t)=n,λ=λ1)×P(λ=λ1∣N(t)=n)+P(U(t)≤x∣N(t)=n,λ=λ2)×P(λ=λ2∣N(t)=n)=p1e-λ1t(λ1t)n(1-e-λ1x)+p2e-λ2t(λ2t)n(1-e-λ2x)p1e-λ1t(λ1t)n+p2e-λ2t(λ2t)n.
The proof is ended.The probabilistic properties of total amount of claimsS(t)=∑k=1N(t)Zk are as follows.Property 4.
The mean and variance of total amount of claimsS(t)=∑k=1N(t)Zk are
(30)E[S(t)]=(λ1p1+λ2p2)μtVar⁡[S(t)]=μ2(λ1-λ2)2p1p2t2+(μ2+σ2)(λ1p1+λ2p2)t.Proof.
Consider(31)E[S(t)]=E[N(t)]EZk=(λ1p1+λ2p2)μtVar⁡[S(t)]=Var⁡[E(S(t)∣N(t))]+E[Var⁡(S(t)∣N(t))]=Var⁡[N(t)EZk]+E[N(t)Var⁡Zk]=μ2Var⁡[N(t)]+σ2E[N(t)]=μ2[(λ1-λ2)2p1p2t2+(λ1p1+λ2p2)t]+σ2(λ1p1+λ2p2)t=μ2(λ1-λ2)2p1p2t2+(μ2+σ2)(λ1p1+λ2p2)t.
The proof is ended.Property 5.
The moment generating function of total amount of claimsS(t)=∑k=1N(t)Zk is
(32)MS(t)(r)=p1exp⁡[λ1t(MZ(r)-1)]+p2exp⁡[λ2t(MZ(r)-1)],
where MZ(r) is the moment generation function of the individual claim amount.Proof.
We can get the following from (19):
(33)MS(t)(r)=EerS(t)=E[(MZ(r))N(t)]=MN(t)(ln⁡MZ(r))=p1exp⁡[λ1t(MZ(r)-1)]+p2exp⁡[λ2t(MZ(r)-1)].
The proof is ended.


## 3.4. Estimation of Lundberg Exponential Upper Bounds for the Ultimate Ruin Probability

Obviously,E[Y(t)]=[c-(λ1p1+λ2p2)μ]t.It seems very natural to assume the premium rate per unit of timec>(λ1p1+λ2p2)μ. Further, let c>λ1μ, in order to stabilize the operation of the company.The relative safety loading:(34)ρ=c(λ1p1+λ2p2)μ-1.
Obviously,
(35)ρ>cλ1μ-1=^ρ1.ρ
1 can be seen as the corresponding relative safety loading of portfolio consisted by high risk policy. Obviously in terms of premium rate and the average individual claim amount are equal, the relative safety loading of the corresponding portfolio of model (14) should be greater.Letu be the initial capital; then the ruin moment is defined as
(36)Tu=inf⁡{t≥0∣u+Y(t)<0}.
Obviously Tu is a FY-stopping time.The finite time ruin probability is(37)ψ(u,t)=P(Tu<t).
The ultimate ruin probability is
(38)ψ(u)=P(Tu<∞).
Let
(39)h(r)=MZ(r)-1.
We assume that there exists r∞>0, such that h(r)↑+∞ when r↑r∞. It is easy to be seen that h(0)=0, h′(r)>0, and h′′(r)>0 and that h(r) is continuous on [0,r∞) (where r∞≤+∞).Easy to verify, the risk process defined by (14) is a right continuous process and has the following properties:(i)
Y(0)=0 P-a.s.;(ii)
Y has stationary and independent increments about λ;(iii)
E[Y(t)]=[c-(λ1p1+λ2p2)μ]t>0.Let(40)F∞λ=σ(λ,t≥0),Ft=F∞λ∨FtY.
Note that F0=F∞λ.We can easily get(41)E[e-rY(t)∣F∞λ]=E{exp⁡[-r(ct-S(t))]∣F∞λ}=exp⁡(-rct)·E[exp⁡(rS(t))∣F∞λ]=exp⁡[t(λh(r)-rc)].
From the hypothesis of function h(r), we know ∃r>0, s.t. E[e-rY(t)∣F∞λ]<∞.From property (ii) ofY, we have
(42)exp⁡[t(λh(r)-rc)]=E[e-rY(t)∣F∞λ]=E[e-r(Y(t)-Y(s))·e-rY(s)∣F∞λ]=E[e-r(Y(t)-Y(s))∣F∞λ]E[e-rY(s)∣F∞λ].
Thus
(43)E[e-r(Y(t)-Y(s))∣F∞λ]=exp⁡[(λh(r)-rc)(t-s)].Theorem 11.
Let(44)Mu(t)=exp⁡[-r(u+Y(t))]exp⁡[t(λh(r)-rc)].
Thus Mu={Mu(t);t≥0} is an F-martingale.Proof.
We get from (43) that
(45)E[Mu(t)∣Fs]=E[exp⁡(-r(u+Y(t)))exp⁡(t(λh(r)-rc))∣Fs]=E[exp⁡(-r(u+Y(s)))exp⁡(s(λh(r)-rc))·exp⁡(-r(Y(t)-Y(s)))exp⁡[(λh(r)-rc)(t-s)]∣Fs]=Mu(s)E[exp⁡(-r(Y(t)-Y(s)))exp⁡[(λh(r)-rc)(t-s)]∣Fs]=Mu(s).
The proof is ended.Chooset0<∞ and consider t0∧Tu which is a bounded F-stopping time. We get from Doob’s stopping theorem that
(46)e-ru=Mu(0)=EF0[Mu(t0∧Tu)]=EF0[Mu(t0∧Tu)∣Tu≤t0]PF0(Tu≤t0)+EF0[Mu(t0∧Tu)∣Tu>t0]PF0(Tu>t0)≥EF0[Mu(t0∧Tu)∣Tu≤t0]PF0(Tu≤t0)=EF0[Mu(Tu)∣Tu≤t0]PF0(Tu≤t0).
Using u+Y(Tu)≤0 on {Tu<∞}, the lower bound was shown to be given by
(47)EF0[Mu(Tu)∣Tu≤t0]=EF0[exp⁡(-r(u+Y(Tu)))exp⁡(Tu(λh(r)-rc))∣Tu≤t0]≥EF0[1exp⁡(Tu(λh(r)-rc))∣Tu≤t0]≥inf⁡0≤t≤t0[1exp⁡(t(λh(r)-rc))].
Thus we have
(48)PF0(Tu≤t0)≤e-ruEF0[Mu(Tu)∣Tu≤t0]≤e-rusup⁡0≤t≤t0{exp⁡[t(λh(r)-rc)]}.
and, by taking expectation,
(49)ψ(u,t0)=P(Tu≤t0)≤e-ruE{sup⁡0≤t≤t0exp⁡[t(λh(r)-rc)]}.
When t0→∞ in the above equation, we get
(50)ψ(u)≤e-ruE{sup⁡t≥0exp⁡[t(λh(r)-rc)]}.We now want to chooser as large as possible under the restriction E{sup⁡t≥0exp⁡[t(λh(r)-rc)]}<∞. Let R denote that value, named adjustment coefficient of surplus process (14).Since(51)E{sup⁡t≥0exp⁡[t(λh(r)-rc)]}=p1sup⁡t≥0exp⁡[t(λ1h(r)-rc)]+p2sup⁡t≥0exp⁡[t(λ2h(r)-rc)]=^p1sup⁡t≥0etg1(r)+p2sup⁡t≥0etg2(r),
where gi(r)=λih(r)-rc, i=1,2, the two terms of the right side in the above equality are both positive, which corresponds to restrict
(52)g1(r)≤0,g2(r)≤0.
Since
(53)g1(r)≥g2(r),
thus we just need to restrict
(54)g1(r)≤0.
Then we have
(55)R=sup⁡{r∣g1(r)≤0}.Theorem 12.
The ultimate ruin probability meets the inequality(56)ψ(u)≤e-Ru,
where R is the only positive solution of g1(r)=0, named adjustment coefficient.Proof.
Forg1(r)=λ1h(r)-rc, we have
(57)g1(0)=0,g1′(r)=λ1h′(r)-c.
Therefore
(58)g1′(0)=λ1h′(0)-c=λ1μ-c<0.
Then there exists small enough Δr∈(0,r∞) such that g1(Δr)<0. Further g1(r)→+∞ when r∞<+∞ and r↑r∞.
Whenr∞=+∞, since MZ(r)=1+rμ1+r2μ2/2+⋯>r2μ2/2, where μj is the jth moment of the claim amount {Zk}, then
(59)g1(r)=λ1MZ(r)-λ1-rc>λ1μ22r2-λ1-rc⟶+∞,(r⟶+∞).
Thus there must exist r*∈(0,r∞) such that g1(r*)=0. And from g1′′(r)=λ1h′′(r)>0, we know g1(r) is a lower convex function.
Theng1(r)<0 for 0<r<r* and g1(r)>0 for r*<r<r∞.
Therefore there exists the only positive solution ofg1(r)=0; let r* denote that value. From (55) we know R=r*, so we get ψ(u)≤e-Ru.
The proof is ended.


## 3.5. Boundary of the Adjustment CoefficientR

The exact value of adjustment coefficientR generally cannot be determined by g1(r)=0, but because of its great significance to estimate the upper bound of the ruin probability, we estimate the boundary of the adjustment coefficient R.Theorem 13.
The adjustment coefficientR meets the inequality
(60)R<2(c-λ1μ)λ1(μ2+σ2).Proof.
From [7] we know R<2(c-λ1μ)/λ1μ2; then we can get the conclusion.
The proof is ended.Form the above theorem and the expression of the relative safety loadingsρ, ρ1, we get
(61)R<2(c-λ1μ)λ1(μ2+σ2)=2ρ1λ1μλ1(μ2+σ2)=2ρ1μμ2+σ2<2ρμμ2+σ2.The result shows that the upper bound ofR can be only defined by the one- or two-order moment of individual claim amount and the relative safety loading; while having nothing to do with the Poisson parameter, this case have its convenience in use.Theorem 14 (see [7]).
If the individual claim amount{Zk} has the upper bound W, that is, Zk≤W, we have
(62)R>1Wln⁡cλ1μ=1Wln⁡(1+ρ1).The result indicates that, under the conditions of the theorem, the lower bound ofR can be only defined by the relative safety loading ρ1 and the upper bound of individual claim amount.When the exact value ofR cannot be determined by the equation g1(r)=0, we can solve its approximate solution by the numerical method, and using these two boundaries as the initial values for iteration, we can quickly find the approximate solution satisfying requirements of certain accuracy.


## 3.6. The Probability of Survival

Except for a few special cases, generally, complicated calculation is needed to get the adjustment coefficient values. Through the following discussion on integrodifferential equations satisfied by survival probability, we can avoid the computation of the adjustment coefficient.LetΦ(u)=1-ψ(u); then Φ(u) indicates survival probability.Since the number of claim process{N(t);t≥0} about λ is a conditional renewal process and will not be ruined during the period of (0,S1), then we get
(63)Φ(u)=E[P(Y(t)≥0,∀t≥0∣S1,Z1,λ)]=E{E[Φ(u+cS1-Z1)∣λ]}=E{∫0∞λe-λt∫0u+ctΦ(u+ct-z)dF(z)dt}=∫0∞(p1λ1e-λ1t+p2λ2e-λ2t)∫0u+ctΦ(u+ct-z)dF(z)dt.Let(64)x=u+ct.
Then we have
(65)Φ(u)=p1λ1ceλ1u/c∫u∞e-λ1x/c∫0xΦ(x-z)dF(z)dx+p2λ2ceλ2u/c∫u∞e-λ2x/c∫0xΦ(x-z)dF(z)dx.
Let two terms of the hand side in the above equality be, respectively, Φ1(u) and Φ2(u); then Φ(u)=Φ1(u)+Φ2(u).Theorem 15.
The integrodifferential equations assured byΦ1(u) and Φ2(u) are
(66)Φ1′(u)=λ1cΦ1(u)-p1λ1c∫0u[Φ1(u-z)+Φ2(u-z)]dF(z)Φ2′(u)=λ2cΦ2(u)-p2λ2c∫0u[Φ1(u-z)+Φ2(u-z)]dF(z).Proof.
From (65) and the hypothesis of Φ1(u) and Φ2(u), by taking derivation, we can get the above result.
The proof is ended.Φ
1
(
u
) and Φ2(u) are calculated using the above equations; thus the expression of Φ(u) can be obtained. According to the above method to obtain the survival probability, we can immediately get the expression of ruin probability, and then the calculation of the adjustment coefficient can be avoided.



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*Source: 314307-2014-05-29.xml*