A Ruin Model with Compound Poisson Income and Dependence Between Claim Sizes and Claim Intervals

Transcription

1 Acta Mathematicae Applicatae Sinica, English Series Vol. 3, No. 2 (25) DOI:.7/s & Acta Mathema cae Applicatae Sinica, English Series The Editorial Office of AMAS & Springer-Verlag Berlin Heidelberg 25 A Ruin Model with Compound Poisson Income and Dependence Between Claim Sizes and Claim Intervals Yuan-yuan HAO,2, Hu YANG, Department of Statistics and Actuarial Science, Chongqing University, Chongqing 433, China ( yh@cqu.edu.cn) 2 College of Mathematics, Chongqing Normal University, Chongqing 433, China Abstract We consider a ruin model with random income and dependence between claim sizes and claim intervals. In this paper, we extend the determinate premium income into a compound Poisson process and assume that the distribution of the time between two claim occurrences depends on the previous claim size. Given the premium size is exponentially distributed, the (Gerber-Shiu) discounted penalty functions is derived. Finally, we consider a similar model. Keywords discounted penalty function; laplace transform; ruin model; dependence 2 MR Subject Classification 9B3; 6J25 Introduction The classical compound Poisson risk model to describe the surplus process of an insurance portfolio relies on the assumption of independence between claim sizes and claim intervals. As the time went by, this assumption is often too restrictive and there is a need for more general models. Albrecher and Boxma 2 discussed a risk model in which the Poisson arrival rate of next claim is determined by the previous claim size. Essentially, it is a Markov-dependent risk model 3. In addition, Zhou and Cai 5 have considered the dependent risk model with diffusion. And they obtained the explicit expression of the ruin probability. Boudreault et.al. 8 have given a risk model with the reverse dependent structure (i.e. the distribution of the next claim size depends on the last interclaim time). For these risk models, it is explicitly that the premiums are assumed to be received at a constant rate over time. Bao 5 discussed a ruin model, in which the premium is no longer a linear function of time but a Poisson process. Essentially, it is a dependent risk model 3. Bao and Ye 7 studied the discounted penalty function in a class of delayed renewal risk model with random income. Labb and Sendova 2 considered a risk model where both premiums and claims follow compound Poisson processes. For a dependent ruin model, Zhang and Yang 4 derived the discounted penalty penalty function. With this extension, the surplus process has two-side jumps instead of one-side jump only in classic risk model. And for the risk model with two-side jumps, Chi 9 analyzed the discounted penalty function. In this paper a generalization of a ruin model is considered in Section 2, where the income is compound Poisson process and the distribution of the claim intervals depends on the previous Manuscript received January 9, 2. Revised June, 2. Supported by the National Natural Science Foundation of China (No.4265), National Social Science Fund of China (3BTJ8), Scientic and Technological Research Program of Chongqing Municipal Education Commission (No.KJ452, No.KJ3658) and the Fundamental Research Funds for the Central Universities (No. CDJXS8) Corresponding author.

2 446 Y.Y. Hao, H. YANG claim size. In Section 3, the Laplace transforms of the discounted penalty functions are discussed and an example has been shown. In Section 4, we consider a similar ruin model. 2 Model Let us consider the following ruin model U(t) which is an insurance portfolio M(t) N(t) U(t) = u + Y j X i, (2.) j= where u = U() is the initial capital and Y j is the jth premium income with distribution function G( ) and mean σ. M(t) is a Poisson process with intensity. N(t) is a claimnumber process representing the number claims up to time t with interclaim times {W i }. {X i } are assumed to be independent and identically distributed (i.i.d.) with distribution function F(x) = F(x), mean µ. We assume the claim occurrence process to be of the following Markov type: If a claimx i is not less than a threshold B i, then the time until the next claim W i+ is exponentially distributed with rate, otherwise it is exponentially distributed with rate 2. The quantities B i are assumed to be i.i.d. random variables with distribution function B( ) = B( ). Now define T = inf{t : U(t) < } to be the ruin time, Ψ(u) = P(T < U() = u) to be the ruin probability. From 4, We know that µ < Pr(X B) σ + Pr(X < B) σ 2, (2.2) to insure the positive safety loading condition. Let δ be a constant, and ω(x, x 2 ) be a nonnegative measurable function defined on, ) (, ). The discounted penalty function is of the following form m(u) = Ee δt ω(u(t ), U(T) )I(T < ) U() = u, u, (2.3) where I(A) is the indicator function of event A, U(T) is the deficit at ruin and U(T ) is the surplus immediately prior to ruin. If we assume the first claim occurs according to the exponential distribution with rate i, the discounted penalty function is denoted by m i (u). Let Ψ i (u) be the corresponding ruin probability. We will use f( ) to denote the Laplace transform of any function f( ). 3 The Discounted Penalty Function of the Exponential Premium Income In this section, we consider the integral function satisfied by the discounted penalty function. Given that the premium income is exponentially distributed, the Laplace transforms of the discounted penalty function can be obtained. Let V be the time for the first premium. By considering W is exponentially distributed with rate, we have m (u) = Pr(V < W, V dt)e δt m (u + y)dg(y)

3 A ruin model with compound Poisson income and dependence between claim sizes and claim intervals Pr(V > W, W dt)e δt m (u x)pr(x B)dF(x) + m 2 (u x)pr(x < B)dF(x) + ω(u, x u)df(x) u = e (++δ)t dt m (u + y)dg(y) + e (++δ)t u dt m (u x)b(x)df(x) + m 2 (u x) B(x)dF(x) + ω(u) = m (u + y)dg(y) + m (u x)b(x)df(x) + + δ + + δ + m 2 (u x) B(x)dF(x) + ω(u), (3.) where Let we have ω(u) = u ω(u, x u)df(x). γ (x) = B(x)f(x), γ 2 (x) = B(x)f(x), A i (u) = m i (u + y)dg(y), m (u) = + + δ A (u) + m (u x)γ (x)dx + + δ + m 2 (u x)γ 2 (x)dx + ω(u). (3.2) Similarly, we have m 2 (u) = δ A 2 2(u) + m (u x)γ (x)dx δ + m 2 (u x)γ 2 (x)dx + ω(u). (3.3) Taking Laplace transforms of (3.2) and (3.3), we obtain m (s) = + + δã(s) + m (s) γ (s) + m 2 (s) γ 2 (s) + ω(s), + + δ (3.4) 2 m 2 (s) = δã2(s) + m (s) γ (s) + m 2 (s) γ 2 (s) + ω(s) δ (3.5) Suppose that the premium incomes Y j s are exponentially distributed, i.e., G(y) = e y σ, y >. Now we introduce the Dickson-Hipp operator T r studied in. For r C having a nonnegative real part, the operator T r defined on an integral function h, T r h(x) = x e r(y x) h(y)dy, x. (3.6)

4 448 Y.Y. Hao, H. YANG When x =, it is the usual Laplace transform. The operator T r is commutative, i.e. T r T s h(x) = T s T r h(x), and furthermore T r T s h(x) = T rh(x) T s h(x), r s. s r Let Re(s) > σ and by using Dickson-Hipp operator in Ãi(s), we have à i (s) = = σ = σ = σ e su m i (u + y) y σ e σ dydu e su u m i (t)e (t u) σ dtdu e su T σ m i(u)du = σ T st σ m i() T s m i () T m i() σ σ s = m i(s) m i ( σs Applying the above result in (3.4) and (3.5), we have + + δ = ω(s) + + δ δ σs γ (s) m (s) + + δ + + δ m ( σs 2 γ 2 (s) δ. (3.7) γ 2 (s) + + δ m 2(s) σs, (3.8) m 2 (s) 2 γ (s) δ m (s) = 2 ω(s) δ δ m 2( σs. (3.9) Simplifying (3.8) and (3.9), we could obtain m (s) = h (s) ω(s) χ 2(s) m ( ( ++δ)( σs) γ 2(s) m 2( χ (s)χ 2 (s) ( ++δ)( 2++δ)( σs) 2 γ(s) γ2(s) ( ++δ)( 2++δ) m 2 (s) = h 2(s) ω(s) χ (s) m 2( ( 2++δ)( σs) 2 γ (s) m ( χ (s)χ 2 (s) ( ++δ)( 2++δ)( σs) 2 γ(s) γ2(s) ( ++δ)( 2++δ), (3.), (3.) where χ i (s) = ( i + + δ)( σs) i γ i (s), i =, 2, i + + δ h (s) = + + δ ( + + δ)( δ)( σs), 2 h 2 (s) = δ 2 ( + + δ)( δ)( σs). To solve m i (u), we need to find m i ( σ), i =, 2. Here we will consider the roots of the denominator of (3.) and (3.), or equally the roots of the following equation: χ (s)χ 2 (s) 2 γ (s) γ 2 (s) =. (3.2) ( + + δ)( δ)

5 A ruin model with compound Poisson income and dependence between claim sizes and claim intervals 449 Lemma. For δ >, the denominators of Equations (3.) and (3.) have exactly 2 roots, say, ρ (δ), ρ 2 (δ) in the right half complex plane, Re(ρ i (δ)) >, i =, 2. Proof. The left side of the Equation (3.2) can be rewritten as which is equivalent to 2 2 σs χ i (s)( σs) = 2 γ (s) γ 2 (s)( σs) 2 ( + + δ)( δ), i + + δ = γ (s)( σs) σs ( + + δ) + 2 γ 2 (s)( σs) ( δ) σs δ + + δ. (3.3) Let r > be a sufficiently large number, and C r denote a collection which contains a right semicircle with radius and the imaginary axis running from ir to ir. For s on the imaginary axis, we have and for s on the semicircle, we have for ε >, when r is sufficiently large. Specially, we use { + δ ε = min, + δ }, 2 i i σs <, i =, 2, i + δ ( i + + δ)σs i i σs i + δ ( i + + δ)σs = i i + + δ σ s i+δ < + ε, i =, 2, ( s i++δ)σ σ s i+δ ( s < i ( + ε). i + + δ i++δ)σ Now, we prove the rightside of Equation (3.3) is no more than the leftside. γ (s)( σs) σs + 2 γ 2 (s)( σs) δ σs ( + + δ) δ ( δ) = σs γ (s)( σs) i + + δ + δ ( + + δ)σs + 2 γ 2 (s)( σs) 2 + δ ( δ)σs 2 ( ) σs γ (s) σs i + + δ + δ ( + + δ)σs + γ 2 (s) 2 2 σs 2 + δ ( δ)σs 2 < ( σs γ (s) + γ 2 (s) ) i + + δ 2 ( σs γ () + γ 2 () ) i + + δ 2 = σs. i + + δ

6 45 Y.Y. Hao, H. YANG It is easy to say that both sides of (3.3) are analytic. So by Rouché s theorem, we know that (3.3) has the same number of roots as the following equation in C r, 2 σs =. i + + δ The above equation has two roots. Then Equation (3.2) also has two roots. The Lemma is proved. Remark. Denote the root with the smaller real part by ρ (δ), then it is easy to see that lim ρ (δ) =. We denote the two roots byρ, ρ 2, for simplicity. δ + If we put ρ, ρ 2, be the roots of Equation (3.2), it must also be zeros of the numerators of (3.) and (3.). Then we could find: h (ρ i ) ω(ρ i ) = χ 2 (ρ i ) m ( ( + + δ)( σρ i ) + γ 2 (ρ i ) m 2 (, i =, 2. (3.4) ( + + δ)( δ)( σρ i ) By solving (3.4), we can get m i (. Then m i(s) can also be obtained. Example. Let =, =.4, 2 =.5, σ =, claim sizes and the threshold distribution are exponentially distributed B(x) = e.5x, F(x) = e x. It is easy to check that the positive safety loading Condition (2.2) is obviously fulfilled. Let δ = and ω(u(t ), U(T) ) =, the ruin probability can be obtained. Equation (3.2) becomes = ( s).4 (.4 s + s +.5 ( s + ) s +.5 s +.5. ).5( s).5.5(s +.5) The roots of the above equation are, , , Then solving (3.4), we could obtain Ψ ( σ) =.43265, Ψ2 ( σ) = Finally, Taking inverse Laplace transform in equations (3.) and (3.) yields Ψ (u) = e u e u, Ψ 2 (u) = e u e u, which are the ruin probabilities in this special case. 4 Another Models In every exact t >, the risk process is in one of the two states i =, 2, corresponding to the rate i of the exponential distribution for the time until the first claim occurs. At the time of a claim occurrence the state of the system may change depending on the corresponding claim size. If a claim X i is smaller than a threshold B i, then the state of the risk process changes, otherwise it doesn t. Also, the thresholds {B i, i } are assumed to be i.i.d. random variables with distribution function B( ). Other conditions are the same as the above model. We assume ( 2µ < σ + ), (4.) 2

7 A ruin model with compound Poisson income and dependence between claim sizes and claim intervals 45 then m i (u) (which is the discounted penalty function with initial capital u, given that the system starts out in state i ) is analogous to the previous section. Then we obtain m (u) = + + δ A (u) δ m (u x)γ (x)dx+ m 2 (u) = δ A 2 2(u) δ m 2 (u x)γ (x)dx+ m 2 (u x)γ 2 (x)dx + ω(u), (4.2) m (u x)γ 2 (x)dx + ω(u). (4.3) We assume premium incomes are exponentially distributed with parameter σ. And then taking Laplace transform, we have where m (s) = h (s) ω(s) ξ 2(s) m ( ( ++δ)( σs) γ 2(s) m 2( ξ (s)ξ 2 (s) ( ++δ)( 2++δ)( σs) 2 γ2(s) γ2(s) ( ++δ)( 2++δ) m 2 (s) = h 2(s) ω(s) ξ (s) m 2( ( 2++δ)( σs) 2 γ 2(s) m ( ξ (s)ξ 2 (s) ( ++δ)( 2++δ)( σs) 2 γ2(s) γ2(s) ( ++δ)( 2++δ) ξ i (s) = i + + δ σs i γ (s), i =, 2, i + + δ h (s) = + + δ + 2 ( σs)( γ (s) γ 2 (s)) ( + + δ)( δ)( σs), h 2 (s) = δ 2 2 ( σs)( γ (s) γ 2 (s)) ( + + δ)( δ)( σs)., (4.4), (4.5) m i (u) is the Laplace transform of the discounted penalty function again. Note that the denominators on the right-hand side of (4.4) and (4.5) coincide too. As in Model, we need to prove the denominators of (4.4) and (4.5) have exactly 2 roots in the right half complex plane. Then, m i (u) also can be obtained. 5 Conclusion In this paper, we study the expected discounted penalty function of two general ruin models. These two models have the same random incomes, the same independent relationship between premiums and the claims process, and so on. Although there are minor differences concerning the dependence. The results derived in this paper can be generalized to similar dependence ruin models. References Adan, I., Kulkarni, V. Single-sever queue with Markov depent interarrival and service times. Queueing Systems, 45: (23) 2 Albrecher, H., Boxma, O.J. A ruin model with dependence between claim sizes and claim intervals. Insurances: Mathematics and Economics, 35: (24)

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