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Acta Mathematica Scientia 2006,26B(2):321–330 www.wipm.ac.cn/publish/ ON THE RUIN FUNCTIONS FOR A CORRELATED AGGREGATE CLAIMS MODEL WITH POISSON AND ERLANG RISK PROCESSES∗ Liu Yan ( School of Mathematics and Statistics, Wuhan University, Wuhan 430072, China E-mail: yanliu@whu.edu.cn; ywqyj@yahoo.com.cn; yijunhu@public.wh.hb.cn ¢¥ ) Yang Wenquan( ¦¤£ ) Hu Yijun (  §¡ ) Abstract This article considers a risk model as in Yuen et al. (2002). Under this model the two claim number processes are correlated. Claim occurrence of both classes relate to Poisson and Erlang processes. The formulae is derived for the distribution of the surplus immediately before ruin, for the distribution of the surplus immediately after ruin and the joint distribution of the surplus immediately before and after ruin. The asymptotic property of these ruin functions is also investigated. Key words Correlated aggregate claims, Poisson process, Erlang process, ruin functions 2000 MR Subject Classification 91B30 1 Introduction In recent years, many authors studied various aspects of the so-called correlated aggregate claims model. In this model, the two claim number processes are correlated by the incorporation of a common component into the two claim number processes. Ambagaspitiya (1998) considered a general method of constructing a vector of p (p ≥ 2) dependent claim numbers from a vector of independent random variables, and derived formulae to compute the correlated aggregate claims distribution for the corresponding common component of p dependent classes of business. Cossette and Marceau (2000) used a discrete-time approach to study how such a dependence relation affects the finite-time ruin probabilities and adjustment coefficient. On the other hand, Erlang distribution is one of the most commonly used distributions in queueing theory which is closely related to risk theory, see for example, Asmussen (1987, 1989), and Tak´cs (1962). Dickson (1998) showed how the methods that have been applied to a derive results for the compound Poisson model can be adapted to derive results for a class of risk processes in which claims occur as an Erlang process. Dickson and Hipp (1998) considered the infinite time survival probability as a compound geometric random variable under the Erlang(2) risk model. Sun and Yang (2004) derived the integro-differential equation and Laplace May 3, 2004; revised May 13, 2004. This work was supported in part by the National Natural Science Foundation of China (10071058, 70273029) and the Ministry of Education of China. ∗ Received 322 ACTA MATHEMATICA SCIENTIA Vol.26 Ser.B transform of the joint distributions of the surplus immediately before and after ruin for Erlang(2) risk processes. Yuen et al. (2002) derived explicit expressions for the ultimate survival (ruin) probabilities for a correlated aggregate claims model with Poisson and Erlang risk processes when the claim sizes are exponentially distributed, and examined the asymptotic property of the ruin probability for the corresponding risk model with general size distributions. In this article we consider the same risk model as in Yuen et al. (2002). Under the assumed risk model the claim number processes involve Poisson and Erlang processes. We derive the formulae for the distribution of the surplus immediately before ruin, for the distribution of the surplus immediately after ruin and the joint distribution of the surplus immediately before and after ruin. The asymptotic property of these ruin functions is also investigated. The article is organized as follows. Section 2 gives a detailed description of the model, and transforms the correlated aggregate claims model into another risk process with two independent classes of business. Section 3 presents the formulae for the ruin functions. In Section 4, the asymptotic results for the ruin functions are given. Section 5 gives all the proofs. 2 Model Set-up and Model Transformation We start with the description of the risk model involving two dependent classes of insurance business, in which we are interested in this article. Let {Xn ; n ≥ 1} be claim size random variables for the first class with common distribution function FX and finite mean µX , {Yn ; n ≥ 1} be those for the second class with common distribution function FY and finite mean µY . We assume that {Xn ; n ≥ 1} and {Yn ; n ≥ 1} are independent. Then the aggregate claim size process is given by N1 (t) N2 (t) S(t) = i=1 Xi + i=1 Yi , (2.1) where Ni (t) is the claim number process for class i, i = 1, 2. We assume that {Xn ; n ≥ 1} and {Yn ; n ≥ 1} are two sequences of independent claim size random variables, and that they are independent of N1 (t) and N2 (t). {N1 (t); t ≥ 0} and {N2 (t); t ≥ 0} are correlated in the way that N1 (t) = M1 (t) + M (t) and N2 (t) = M2 (t) + M (t) (2.2) where M1 (t), M2 (t) and M (t) are three independent renewal processes. We define the surplus process U (t) = u + ct − S(t), where u > 0 is the initial surplus and c > 0 is the premium rate. Let T denote the time to ruin, so that   inf{t : U (t) < 0}, T =  ∞, if U (t)≥0 for all t > 0. ψ(u) = P (T < ∞|U (0) = u), (2.3) Then the probability of ultimate ruin from initial surplus u is defined as No.2 Liu et al: RUIN FUNCTIONS FOR A CORRELATED AGGREGATE CLAIMS MODEL 323 the probability that the surplus immediately prior to ruin is smaller than x > 0 from the initial reserve u as B(u, x) = P (T < ∞, U (T −)≤x|U (0) = u), the probability that the deficit (negative surplus immediately after ruin) is smaller than y > 0 from the initial reserve u as G(u, y) = P (T < ∞, −U (T )≤y|U (0) = u), and the joint probability that the surplus prior to ruin is smaller than x and the surplus after ruin is larger than −y from the initial reserve u as J(u; x, y) = P (T < ∞, −U (T )≤y, U (T −)≤x|U (0) = u). We define the following auxiliary probability distribution: A(u; x, y) = P (T < ∞, −U (T ) > x, U (T −) > y|U (0) = u). It is easy to see that ψ(u) = A(u; 0, 0), B(u, x) = ψ(u) − A(u; 0, x) = A(u; 0, 0) − A(u; 0, x), G(u, y) = ψ(u) − A(u; y, 0) = A(u; 0, 0) − A(u; y, 0), J(u; x, y) = ψ(u) − A(u; 0, x) − A(u; y, 0) + A(u; y, x) = A(u; 0, 0) − A(u; 0, x) − A(u; y, 0) + A(u; y, x). Hence all the four ruin functions can be obtained from A(u; x, y). Assume that Mi (t) is the Poisson process with parameter λi (i = 1, 2), and M (t) is an Erlang (2) process with parameter λ, that is, the number of claim inter-arrival times for M (t) has the density function K(t) = λ2 te−λt , for t > 0. From (2.1) and (2.2), the surplus process (2.3) can be rewritten as M12 M (t) U ′ (t) = u + ct − i=1 Xi′ − i=1 Yi′ , (2.4) ′ ′ where {Xn ; n ≥ 1} and {Yn ; n ≥ 1} are independent random variables, and M12 (t) = M1 (t) + M2 (t) is still a Poisson process with parameter λ1 +λ2 . Furthermore, Xi′ and Yi′ are independent of M12 (t) and M (t), and their distribution functions are given by FX ′ (x) = λ1 λ2 FX (x) + FY (x) and FY ′ (x) = FX ∗ FY (x), λ1 + λ2 λ1 + λ2 respectively, where FX ∗ FY stands for the convolution of FX and FY (see Yuen et al. (2002, Page 206)). Since the transformed process U ′ (t) and the original process U (t) are identically distributed, the process U (t) can be examined via U ′ (t). 324 In view of the fact that M12 ACTA MATHEMATICA SCIENTIA Vol.26 Ser.B E i=1 Xi′ we see that the positive relative security loading condition implies that c > (λ1 + λ2 )µX ′ + λµY ′ . 2 (2.5) = (λ1 + λ2 )tµX ′ and E   M(t) i=1 Yi′  =  λ 1 t− 1 − e−2λt 2 4 µY ′ , Let T1 , T2 , · · · be the inter-arrival times for Xi′ . Then these times are independent and exponentially distributed with mean (λ1 + λ2 )−1 . For Yi , the inter-arrival times L1 , L2 , · · · are independent and have Erlang (2, λ) distribution. Equivalently, L1 = L11 + L12 , L2 = L21 + L22 , · · · , where L11 , L12 , L21 , L22 , · · · are independent exponential random variables with mean λ−1 . In order to derive the formulae for the above ruin functions, we consider a slight change in the distribution L1 . Instead of being a sum of L11 and L12 , let L1 = L12 . That is, L1 is exponentially distributed with mean λ−1 and Li still follows the same Erlang(2, λ) distribution for i = 2, 3, · · ·. We denote the corresponding ruin functions by A1 (u; x, y), B1 (u, x), G1 (u, y) and J1 (u; x, y). 3 Formulae for the Ruin Functions In this section the formulae for the ruin functions are stated as follows. Theorem 3.1 For any u > y > 0 and u > x > 0, we have A(u; x, y) = λ1 + λ2 c + ψ(u) = λ c +∞ +∞ u (1 − FX ′ (z))dz + u+x 0 A(u − z; x, y)(1 − FX ′ (z))dz (A1 (z; x, y) − A(z; x, y))dz; u +∞ u λ1 + λ2 c + λ c +∞ (1 − FX ′ (z))dz + u 0 ψ(u − z)(1 − FX ′ (z))dz (ψ1 (z) − ψ(z))dz; u u B(u, x) = λ1 + λ2 c λ1 + λ2 G(u, y) = c + λ c +∞ B(u − z, x)(1 − FX ′ (z))dz + 0 u+y u λ c +∞ (B1 (z, x) − B(z, x))dz; u (1 − FX ′ (z))dz + u 0 G(u − z, y)(1 − FX ′ (z))dz (G1 (z, y) − G(z, y))dz; u u J(u; x, y) = λ1 + λ2 c J(u − z; x, y)(1 − FX ′ (z))dz + 0 λ c +∞ (J1 (z; x, y) − J(z; x, y))dz. u Remark 3.1 The formula for ψ(u) is just the equation (4.3) of Yuen et al. (2002). No.2 Liu et al: RUIN FUNCTIONS FOR A CORRELATED AGGREGATE CLAIMS MODEL 325 4 Asymptotic Result for the Ruin Functions Before giving the main result of this section, we first introduce a key assumption (see Yuen et al. (2002, Page 213)). Key Assumption We assume that there exists r1 > 0 and r2 > 0 such that h1 (r) ↑ ∞ as r ↑ r1 and h2 (r) ↑ ∞ as r ↑ r2 , where +∞ h1 (r) = 0 +∞ erz dFX ′ (z) − 1, erz dFY ′ (z) − 1. 0 h2 (r) = The main result in this section is the following theorem. Theorem 4.1 Under the key assumption, u→∞ lim eRu A(u; x, y) + A1 (u; x, y) ρ c ≤ ; 2 1 + ρ (λ1 + λ2 )h(1) (R) + 2−1 λh(1) (R) − c 1 2 ψ(u) + ψ1 (u) ρ c lim eRu ; ≤ (1) u→∞ 2 1 + ρ (λ1 + λ2 )h (R) + 2−1 λh(1) (R) − c 1 2 u→∞ lim e Ru B(u, x) + B1 (u, x) ρ c ; ≤ 2 1 + ρ (λ1 + λ2 )h(1) (R) + 2−1 λh(1) (R) − c 1 2 1 2 G(u, y) + G1 (u, y) ρ c lim eRu ; ≤ (1) u→∞ 2 1 + ρ (λ1 + λ2 )h (R) + 2−1 λh(1) (R) − c u→∞ lim e Ru J(u; x, y) + J1 (u; x, y) ρ 2c , ≤ 2 1 + ρ (λ1 + λ2 )h(1) (R) + 2−1 λh(1) (R) − c 1 2 where ρ= −1 (λ1 + λ2 )µX ′ + 2−1 λµY ′ is the relative security loading, and R is the positive solution of the equation 1 (λ1 + λ2 )h1 (r) + λh2 (r) = cr. 2 Remark 4.1 The second inequality above is just the one of Yuen et al. (2002, Page 213). c 5 Proofs Results in Sections 3 and 4 will be proved in this section. Proof of Theorem 3.1 We condition on the time W = min{T1 , L11 }. If W = L11 = t, then no claim occurs in (0, t]. On the other hand, if W = T1 = t, there is a claim at time t and no claim before t. Hence we have +∞ A(u; x, y) = 0 P (W = t, W = L11 )A1 (u + ct; x, y)dt +∞ u+ct + 0 +∞ P (W = t, W = T1 ) 0 A(u + ct − z; x, y)dFX ′ (z)dt +∞ + 0 P (W = t, W = T1 )I(u+ct>y) u+ct+x dFX ′ (z)dt. 326 Note that ACTA MATHEMATICA SCIENTIA Vol.26 Ser.B P (W = T1 ) = P (T1 < L11 ) = P (W = L11 ) = P (T1 > L11 ) λ1 +λ2 , λ = λ, λ P (W > t|W = T1 ) = P (W > t|W = L11 ) = e−λt , where λ = λ1 + λ2 + λ. Using these probabilities, we obtain A(u; x, y) = λ λ + +∞ λe−λt A1 (u + ct; x, y)dt 0 +∞ u+ct λ1 + λ2 λ λ1 + λ2 + λ λe−λt 0 +∞ 0 A(u + ct − z; x, y)dFX ′ (z)dt +∞ λe−λt I(u+ct>y) 0 u+ct+x dFX ′ (z)dt. Putting s = u + ct yields +∞ cA(u; x, y) = λ u e− λ(s−u) c A1 (s; x, y)ds e− λ(s−u) c +∞ s +(λ1 + λ2 ) u +∞ A(s − z; x, y)dFX ′ (z)ds 0 +∞ +(λ1 + λ2 ) u e− λ(s−u) c I(s>y) s+x dFX ′ (z)ds. Differentiating with respect to u we get cA(1) (u; x, y) = −λA1 (u; x, y) + λ u λ c +∞ e− u λ(s−u) c A1 (s; x, y)ds −(λ1 + λ2 ) 0 A(u − z; x, y)dFX ′ (z) +∞ λ + (λ1 + λ2 ) c e− u λ(s−u) c s A(s − z; x, y)dFX ′ (z)ds 0 −(λ1 + λ2 )I(u>y) (1 − FX ′ (u + x)) λ + (λ1 + λ2 ) c +∞ (5.1) e− u λ(s−u) c I(s>y) (1 − FX ′ (s + x))ds u = −λA1 (u; x, y) − (λ1 + λ2 ) 0 A(u − z; x, y)dFX ′ (z) −(λ1 + λ2 )I(u>y) (1 − FX ′ (u + x)) + λA(u; x, y). Integrating (5.1) both sides from 0 to u gives A(u; x, y) = A(0; x, y) − λ1 + λ2 c λ1 + λ2 − c + λ c u 0 u u A1 (s; x, y)ds + 0 s 0 λ c u A(s; x, y)ds 0 A(s − z; x, y)d(1 − FX ′ (z))ds I(s>y) (1 − FX ′ (s + x))ds. 0 No.2 Liu et al: RUIN FUNCTIONS FOR A CORRELATED AGGREGATE CLAIMS MODEL 327 By interchanging integral signs and performing integration by parts, we get u 0 u 0 u s A(s − z; x, y)d(1 − FX ′ (z))ds = 0 A(u − z; x, y)(1 − FX ′ (z))dz − 0 A(s; x, y)ds. Consequently, A(u; x, y) = A(0; x, y) + + λ1 + λ2 c u λ c u (A(s; x, y) − A1 (s; x, y))ds 0 A(u − z; x, y)(1 − FX ′ (z))dz − 0 λ1 + λ2 c u+x (1 − FX ′ (s))ds, y+x where u > y > 0. Let u → ∞, then we have 0 = A(0; x, y) + Then A(0; x, y) = Hence A(u; x, y) = + λ1 + λ2 c λ c +∞ λ c +∞ +∞ (A(s; x, y) − A1 (s; x, y))ds − 0 λ1 + λ2 c +∞ +∞ (1 − FX ′ (s))ds. y+x λ c (A1 (s; x, y) − A(s; x, y))ds + 0 λ1 + λ2 c (1 − FX ′ (s))ds. y+x (5.2) (A1 (s; x, y) − A(s; x, y))ds u +∞ u (1 − FX ′ (s))ds + u+x 0 A(u − z; x, y)(1 − FX ′ (z))dz . Given this, we can immediately obtain the formulae for the other ruin functions. The proof of Theorem 3.1 is completed. Proof of Theorem 4.1 We need only to prove the first inequality. Note that +∞ u+ct A1 (u; x, y) = λ 0 e−λt 0 +∞ A(u + ct − z; x, y)dFY ′ (z)dt +∞ +λ 0 e−λt I(u+ct>y) +∞ u+ct+x u+ct dFY ′ (z)dt A1 (u + ct − z; x, y)dFX ′ (z)dt 0 +∞ +(λ1 + λ2 ) 0 +∞ e−λt +(λ1 + λ2 ) 0 e−λt I(u+ct>y) u+ct+x dFX ′ (z)dt. Putting s = u + ct and differentiating with respect to u gives cA1 (u; x, y) = −λ +λ λ c (1) u A(u − z; x, y)dFY ′ (z) 0 +∞ e− u λ(s−u) c s A(s − z; x, y)dFY ′ (z)ds 0 +∞ −λI(u>y) x+u dFY ′ (z) + λ λ c +∞ e− u λ(s−u) c +∞ I(s>y) s+x dFY ′ (z)ds 328 −(λ1 + λ2 ) ACTA MATHEMATICA SCIENTIA u Vol.26 Ser.B A1 (u − z; x, y)dFX ′ (z) 0 +(λ1 + λ2 ) λ c +∞ e− u λ(s−u) c s A1 (s − z; x, y)dFX ′ (z)ds 0 +∞ −(λ1 + λ2 )I(u>y) +(λ1 + λ2 ) u λ c dFX ′ (z) u+x +∞ λ(s−u) e− c I(s>y) u +∞ A1 (s − z; x, y)dFX ′ (z)ds s+x +∞ (5.3) = −λ 0 A(u − z; x, y)dFY ′ (z) − λI(u>y) x+u u dFY ′ (z) −(λ1 + λ2 ) 0 A1 (u − z; x, y)dFX ′ (z) +∞ −(λ1 + λ2 )I(u>y) u+x dFX ′ (z) + λA1 (u; x, y). Integrating (5.3) both sides from 0 to u gives A1 (u; x, y) = A1 (0; x, y) + + − − λ1 + λ2 c λ1 + λ2 c λ1 + λ2 c λ c u u A(u − z; x, y)(1 − FY ′ (z))dz 0 A1 (u − z; x, y)(1 − FX ′ (z))dz − 0 u λ c u A(s; x, y)ds 0 A1 (s; x, y)ds − 0 u+x λ c u+x (1 − FY ′ (z))dz y+x (1 − FX ′ (z))dz + y+x λ c u A1 (s; x, y)ds 0 = A1 (0; x, y) + + − λ1 + λ2 c λ1 + λ2 c λ c u u A(u − z; x, y)(1 − FY ′ (z))dz 0 A1 (u − z; x, y)(1 − FX ′ (z))dz − 0 u+x λ c u+x (1 − FY ′ (z))dz y+x (1 − FX ′ (z))dz + y+x λ c u (A1 (s; x, y) − A(s; x, y))ds, 0 where u > y > 0. Let u → ∞, then we have A1 (0; x, y) = λ1 + λ2 c λ + c Therefore A1 (u; x, y) = λ1 + λ2 c + − u +∞ u +∞ (1 − FX ′ (z))dz y+x +∞ +∞ (5.4) (A1 (s; x, y) − A(s; x, y)ds . (1 − FY ′ (s))ds − y+x 0 (1 − FX ′ (z))dz + u+x +∞ u 0 A1 (u − z; x, y)(1 − FX ′ (z))dz (5.5) λ c (1 − FY ′ (z))dz + u+x +∞ 0 A(u − z; x, y)(1 − FY ′ (z))dz (A1 (s; x, y) − A(s; x, y))ds . No.2 Liu et al: RUIN FUNCTIONS FOR A CORRELATED AGGREGATE CLAIMS MODEL 329 From (5.2) and (5.4), we obtain A(0; x, y) + A1 (0; x, y) = λ c +∞ (1 − FY ′ (z))dz + y+x 2(λ1 + λ2 ) c +∞ (1 − FX ′ (z))dz. y+x From Theorem 3.1 and (5.5), we have A(u; x, y) + A1 (u; x, y) +∞ = u+x u 2(λ1 + λ2 ) λ (1 − FX ′ (z)) + (1 − FY ′ (z)) dz c c λ1 + λ2 λ (1 − FX ′ (z)) + (1 − FY ′ (z)) A(u − z; x, y)dz c c λ1 + λ2 (1 − FX ′ (z))A1 (u − z; x, y)dz c 2(λ1 + λ2 ) λ (1 − FX ′ (z)) + (1 − FY ′ (z)) dz c c λ1 + λ2 λ (1 − FX ′ (z)) + (1 − FY ′ (z)) c 2c + 0 u + 0 +∞ ≤ u+x u + 0 (A(u − z; x, y) + A1 (u − z; x, y))dz. The positive security loading condition implies that +∞ 0 λ1 + λ2 λ (1 − FX ′ (z)) + (1 − FY ′ (z)) dz < 1. c 2c In view of the key assumption, then there exists a R > 0 such that +∞ eRz 0 λ λ1 + λ2 (1 − FX ′ (z)) + (1 − FY ′ (z)) dz = 1. c 2c Therefore, we have the following renewal type of inequality 1 Ru e (A(u; x, y) + A1 (u; x, y)) 2 +∞ λ1 + λ2 λ ≤ eRu (1 − FX ′ (z)) + (1 − FY ′ (z)) dz c 2c u + e Denote λ1 + λ2 λ (1 − FX ′ (z)) + (1 − FY ′ (z)). c 2c Applying the renewal theorem to the right hand side of the above inequality, it follows H(z) = that u→∞ 1 2 u eRz 0 λ1 + λ2 λ (1 − FX ′ (z)) + (1 − FY ′ (z)) c 2c R(u−z) (A(u − z; x, y) + A1 (u − z; x, y))dz. lim e Ru A(u; x, y) + A1 (u; x, y) ≤ 2 +∞ Ru +∞ e H(z)dzdu 0 u . +∞ Rz H(z)dz ze 0 330 From the facts that +∞ ACTA MATHEMATICA SCIENTIA Vol.26 Ser.B +∞ eRu 0 +∞ u (1 − FX ′ (z))dzdu z = 0 (1 − FX ′ (z)) 0 eRu dudz = +∞ 0 1 R +∞ 0 +∞ (1 − FX ′ (z))(eRz − 1)dz 0 1 = R 1 = R +∞ 1 (1 − FX ′ (z))eRz R h1 (R) − µX ′ , R 1 + R eRs dFX ′ (z) − µX ′ zeRz (1 − FX ′ (z))dz 0 = 1 R 1 = R z(1 − FX ′ (z))eRz − +∞ 0 +∞ +∞ − 0 eRz (1 − FX ′ (z))dz + 0 zeRz dFX ′ (z) h1 (R) (1) + h1 (R) , R 1 (λ1 + λ2 )h1 (R) + λh2 (R) = cR, 2 and we get +∞ Ru +∞ e H(z)dzdu 0 u +∞ Rz H(z)dz ze 0 = c ρ , 1 + ρ (λ1 + λ2 )h(1) (R) + 2−1 λh(1) (R) − c 1 2 where ρ= c (λ1 + λ2 )µX ′ + 2−1 λµY ′ −1 is the relative security loading. The proof of Theorem 4.1 is completed. References 1 Ambagaspitiya R S. On the distribution of a sum of correlated aggregate claims. Insurance: Mathematics and Economics, 1998, 23: 15–19 2 Asmussen S. Applied Probability and Queues. New York: Wiley, 1987 3 Asmussen S. Risk theory in a Markovian environment. Scandinavian Actuarial Journal, 1989, 69–100 4 Cossette H, Marceau E. The discrete-time model with correlated class of business. Insurance: Mathematics and Economics, 2000, 26: 133–149 5 Dickson D C M. On a class of renewal risk processes. North American Actuarial Journal, 1998, 2(3): 60–73 6 Dickson D C M, Hipp C. Ruin probabilities for Erlang (2) risk processes. Insurance: Mathematics and Economics, 1998, 22: 251–262 7 Sun L J, Yang H L. On the joint distributions of surplus immediately before ruin and the deficit at ruin for Erlang(2) risk processes. Insurance: Mathematics and Economics, 2004, 34: 121–123 8 Tak´cs L. Introduction to the Theory of Queues. 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