2000 Mathematics Subject Classification: 60K10, 62P05.The compound Poisson risk models are widely used in practice. In this paper the counting process in the insurance risk model is a compound Poisson process. The model is called Compound Compound Poisson Risk Model. Some basic properties and ruin probability are given. We analyze the model under the proportional reinsurance. The optimal retention level and the corresponding adjustment coefficient are obtained. The particular case of the Pólya-Aeppli risk model is discussed.This paper is partially supported by Sofia University grant 221/2008

Minkova, Leda D.

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Serdica Math. J. 35 (2009), 301–310COMPOUND COMPOUND POISSON RISK MODEL*Leda D. MinkovaCommunicated by S. T. RachevAbstract. The compound Poisson risk models are widely used in practice.In this paper the counting process in the insurance risk model is a compoundPoisson process. The model is called Compound Compound Poisson RiskModel. Some basic properties and ruin probability are given. We analyzethe model under the proportional reinsurance. The optimal retention leveland the corresponding adjustment coefficient are obtained. The particularcase of the Po´lya-Aeppli risk model is discussed.1. Introduction. Assume that the standard model of an insurancecompany, called risk process {X(t), t ≥ 0} is given by*This paper is partially supported by Sofia University grant 221/2008.2000 Mathematics Subject Classification: 60K10, 62P05.Key words: Compound Poisson process, Po´lya-Aeppli risk model, ruin probability, Crame´r-Lundberg approximation.302 Leda D. Minkova(1) X(t) = ct− S(t).Here c is a premium income per unit time and S(t) represents the aggre-gate amount of claims up to time t. The process S(t) is a compound process,given by(2) S(t) =N(t)∑i=1Zi,(0∑1= 0),where N(t) is the counting process, {Zi}∞i=1 is a sequence of independent identi-cally distributed, positive random variables, independent of N(t) with Zi repre-senting the ith claim amount. We assume that the individual claim amounts havea continuous distribution with distribution function F , F (0) = 0, and mean valueµ = EZ1 < ∞. In the classical risk model the process N(t) is a homogeneousPoisson process, see for instance [1] and [7]. In this paper we suppose that thecounting process N(t) is a compound Poisson process with discrete compoundingdistribution, i.e. N(t) =N1(t)∑i=1Yi, where Y1, Y2, . . . are independent identicallydistributed random variables, independent of N1(t) and N1(t) ∼ Po(λt). Let Ydenotes the compounding random variable with probability generating function(PGF) ψ(s) = EsY and EY = m1 and EY2 = m2. Then the PGF of the countingprocess is given byPN(t)(s) = e−λt[1−ψ(s)]and the process (1) is called a Compound Compound Poisson risk model (CCPoisson risk model). It is easy to find that EN(t) = λm1t, ES(t) = λµm1t andthe safety loading coefficientθ =(c− λµm1)tλµm1t=cλµm1− 1 > 0.Denote byτ(u) = inf{t > 0, u+X(t) ≤ 0}the time to ruin of a company having initial capital u. We let τ = ∞, if for allt > 0 u+X(t) > 0.Compound Compound Poisson Risk Model 303The probability of ruin in the infinite horizon time isΨ(u) = P (τ(u) <∞)and in the finite horizon caseΨ(u, t) = P (τ(u) ≤ t).In Section 2 and Section 3 the CC Poisson risk model and martingale approach arediscussed. In Section 4, we analyze the model under the proportional reinsurancein the case of small claims. The retention level maximizing the adjustment coef-ficient is obtained. The effect of reinsurance on the ruin probability is discussed.The particular case of Po´lya-Aeppli risk model is given.2. Martingales for the CC Poisson risk model. Let us denoteby (FXt ) the natural filtration generated by any stochastic process X(t). (FXt )is the smallest complete filtration to which X(t) is adapted.Let us denote by LSZ(r) =∫ ∞0e−rxdF (x) the Laplace-Stieltjes trans-form of any random variable Z with distribution function F (x) and MZ(r) =LSZ(−r) the moment generating function (MGF) of Z.Lemma 1. For the CC Poisson risk modelEe−rX(t) = eg(r)t,where(3) g(r) = λ[ψ(MZ(r))− 1]− cr.P r o o f. Let us consider the random sum from the right hand side of (1)St =N(t)∑k=1Zk,304 Leda D. Minkovawhere N(t) is a compound Poisson process, independent of Zk, k = 1, 2, . . .. St isa CC Poisson process and the MGF is given byMSt(r) = PN(t)(MZ(r)) = e−λt[1−ψ(MZ (r))].For the Laplace-Stieltjes transform of X(t) we have the followingLSX(t)(r) = Ee−rX(t) = Ee−r[ct−S(t)] = e−rctEerS(t) == e−rctPN(t)(MZ(r)) = e−rcte−λt[1−ψ(MZ (r))] = eg(r)t,where g(r) is given by (3). From the martingale theory we get the followingLemma 2. For all r ∈ R the processM(t) = e−rX(t)−g(r)t, t ≥ 0is an FXt -martingale, provided that MZ(r) <∞.3. Martingale approach to the CC Poisson risk model. Usingthe martingale properties of M(t), we will give some useful inequalities for theruin probability.Proposition 1. Let r > 0. For the ruin probabilities of the CC Poissonrisk model we have the following resultsi) Ψ(u, t) ≤ e−ru sup0≤s≤teg(r)s, 0 ≤ t <∞ii) Ψ(u) ≤ e−ru sups≥0eg(r)s.iii) If the Lundberg exponent R exists, then R is the unique strictly positivesolution ofCompound Compound Poisson Risk Model 305(4) λ[ψ(MZ(r))− 1]− cr = 0and(5) Ψ(u) ≤ e−Ru.P r o o f. i) Since at the time of ruin X(τ) ≤ −u, for any t < ∞, themartingale stopping time theorem yields the following1 =M(0) = EM(t ∧ τ)= E[M(t ∧ τ)|τ ≤ t]P (τ ≤ t) + E[M(t ∧ τ)|τ > t]P (τ > t)≥ E[M(t ∧ τ)|τ ≤ t]P (τ ≤ t) = E[e−rX(τ)−g(r)τ |τ ≤ t]P (τ ≤ t)≥ eruE[e−g(r)τ |τ ≤ t]P (τ ≤ t),from which(6) P (τ ≤ t) ≤ e−ruE[e−g(r)τ |τ ≤ t] .The statement i) follows from the above relation.ii) follows immediately from i) when t→∞.iii) Under the condition (4) (g(r) = 0), (5) follows from (6). Let R be apositive solution of g(r) = 0. Because g(0) = 0, g′(0) = λψ′(1)µ − c = λm1µ −c < 0, g′′(r) = λψ′′(MZ(r))(M′Z(r))2 + λψ′(MZ(r))M′′Z(r) and g′′(0) = λ[m2µ2 +m1V ar(Z)] > 0, there is at most one strictly positive solution of the equationg(r) = 0. Remark 1. The condition (4) is known as Crame´r condition and (5) asLundberg inequality.4. Reinsurance. Suppose the insurer has the possibility to chooseproportional reinsurance with retention level b ∈ [0, 1]. This means that the306 Leda D. Minkovainsurer pays bZ of a claim. The premium rate for the reinsurance is(1 + η)(1 − b)λµm1,where η > 0 is the relative safety loading, defined by the reinsurance company.We consider the case η > θ, i.e. the reinsurance is more expensive than firstinsurance. The premium rate for the insurer is[(1 + θ)− (1 + η)(1 − b)]λµm1 = [b(1 + η)− (η − θ)]λµm1,and the surplus process becomes(7) U(t, b) = u+ [b(1 + η)− (η − θ)]λµm1t−N(t)∑k=1bZk,In order that the net profit condition is fulfilled we need[b(1 + η)− (η − θ)]λµm1bλµm1> 1,i.e.b > 1− θη.The adjustment coefficient R(b) under proportional reinsurance solves the equa-tion:(8) λ[ψ(MZ(br))− 1]− [b(1 + η)− (η − θ)]λµm1r = 0.Let Ψ(u, b) denote the probability of ultimate ruin when the proportionalreinsurance is chosen. ThenΨ(u, b) = P (U(t, b) < 0 for some t > 0).Our objective is to find the retention level that minimizes Ψ(u, b). Theproblem related to classical risk model is given in [8]. According the Lundberginequality (7), the retention level will be optimal, if the corresponding Lundbergexponent R is maximal. We know that there is a unique b ∈ [0, 1] where themaximum is attained. If the maximiser b > 1 we know from the uni-modalitythat the optimal b is 1, i.e. no reinsurance is chosen.Compound Compound Poisson Risk Model 307The next result gives the optimal retention level b and maximal adjust-ment coefficient R(b). Similar result is obtained by Hald and Schmidli [2] for theclassical risk model.Lemma 3. The solution of (8) is given by(9) R(b(r)) =1− ψ(MZ(r)) + (1 + η)µm1r(η − θ)µm1 ,where r(b) is invertible function.P r o o f. Assume that r(b) = bR((b)), where R(b) will be the maximalvalue of the adjustment coefficient and r(b) is invertible. If we consider thefunction r → b(r), it follows that(10) b(r) =(η − θ)µm1r1− ψ(MZ(r)) + (1 + η)µm1r .Now R(b(r)) =rb(r)in details is given by (9). The next result gives a way to calculate b and R(b).Theorem 1. Assume that MZ(r) < ∞. Suppose there is a (unique)solution r to(11) ψ′(MZ(r))M′Z(r)− µm1(1 + η) = 0.Then r > 0, the maximal value of R(b(r)) and the retention level b(r) are givenby (9) and (10).P r o o f. The necessary condition for maximizing the value of the adjust-ment coefficient is given by equation (11).Because R′(b(0)) =ηη − θ > 0, the function R(b(r)) is strictly increasingin 0. The second derivative in zero R′′(b(0)) = −(m2 −m1)µ2 +m1EZ2(η − θ)µm1 < 0shows that R(b(r)) is strictly concave. Consequently, the function R(b(r)) has anunique maximum in r, which is the solution of (11). The retention level is givenby (10). 308 Leda D. MinkovaRemark 2. Note that the value R does not depend on c but on therelative safety loadings only.5. Po´lya-Aeppli risk model. The Po´lya-Aeppli risk model is definedin [4]. In this case the counting process N(t) has a Po´lya-Aeppli distribution(see [3] and [5]). The random variables Yi are geometrically distributed withparameter 1 − ρ and m1 = 11− ρ . The Po´lya-Aeppli counting process is theonly compound Poisson process that is also a renewal process. The problem ofreinsurance for the Po´lya-Aeppli risk model is discussed in [6].The Lundberg exponent R of the Po´lya-Aeppli risk model is given by thepositive solution of the equationρcrMZ(r) + λ(MZ(r)− 1)− cr = 0.From the equation one can see that the solution will not be explicit as in the caseof exponentially distributed claims.5.1. Exponentially distributed claims. Here we suppose that theclaim sizes are exponentially distributed with parameter µ > 0, i.e.F ′(z) =1µe− zµ , z ≥ 0.The moment generating function is given byMZ(r) =11− µr , r <1µandM ′Z(r) =µ(1− µr)2 .The Lundberg exponent r in this case is obtained as the positive solutionof the equation(1− ρ)2 µ(1− µr)2 = (1 + η)µ[1− ρ 11− µr]2= 0,Compound Compound Poisson Risk Model 309i.e.r =1µ(1− (1 + η)− 12).From (10) we findb =(η − θ)√1 + η(1 +√1 + η)yieldingR(b) =(1− ρ)[1 −√1 + η]2(η − θ)µ .In this example there are closed form expressions for b and R(b), if b ≤ 1and the retention level does not depend from ρ.Analyzing the optimal retention levels for the Po´lya-Aeppli risk modeland the classical model gives the possibility to compare the ruin probabilities. Inthe case of proportional reinsurance and ρ = 0, the retention level b(r), given by(10) coincides with the retention level for the classical risk model, say b0, obtainedby Hald and Schmidli [2]. It is easy to verify that b(r) < b0. For the maximalvalues R(b0) and R(b(r)) is fulfilledR(b0) ≥ R(b(r)),which means that the Po´lya-Aeppli risk model with proportional reinsurance ismore dangerous than the classical risk model.Acknowledgment. I thank both the associate editor and the refereefor the valuable suggestions and comments.REFERENCES[1] J. Grandell. Aspects of Risk Theory. Springer Verlag, New York, 1991.[2] M. Hald, H. Schmidli. On the maximisation of the adjustment coefficientunder proportional reinsurance. Astin Bull. 34, 1 (2004), 75–83.310 Leda D. Minkova[3] N. L. Johnson, S. Kotz, A. W. Kemp. Univariate Discrete Distributions,2nd ed., Wiley Series in Probability and Mathematical Statistics, 1992.[4] L. D. Minkova. The Po´lya-Aeppli process and ruin problems. J. Appl.Math. Stochastic Anal. No 3 (2004), 221–234.[5] L. D. Minkova. A generalization of the classical discrete distributions,Comm. Statist. Theory Methods 31, 6 (2002), 871–888.[6] L. D. Minkova. Reinsurance by the Po´lya-Aeppli risk model. In: Proceed-ings of the 2008 International Workshop on Applied Probability, July 7–10,2008, Universite de Technologie de Compiegne, France.[7] T. Rolski, H. Schmidli, V. Schmidt, J. Teugels. Stochastic Processesfor Insurance and Finance. John Wiley & Sons, Chichester, 1999.[8] H. Schmidli.Martingales and Insurance Risk. In: 8th International SummerSchool on Probability Theory and Mathematical Statistics, 1996, 155–188.Faculty of Mathematics and InformaticsSofia University “St. Kl. Ohridski”5, J. Bourchier Blvd1164 Sofia, Bulgariae-mail: leda@fmi.uni-sofia.bg Received August 6, 20009

Compound Compound Poisson Risk Model

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Compound Compound Poisson Risk Model

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