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1、The Erlang(2) risk model with a two-step premium rateSun Jing-yunDa Gao-feng(School of Mathematics and statistics, Lanzhou University, Lanzhou 730000, China)Abstract: In this paper, we consider a compound renewal (Sparre Andersen) risk process with a two-step premium rate in which the claim waiting
2、times are Erlang(2) distributed. Anintegro-differential equation with certain boundary condition for Gerber-Shiu function is derived and solved, and use this result we obtain the explicit result about the Laplace transform of the time of ruin and ruin probability when the claim sizes are exponential
3、ly distributed.Keywords: Compound renewal process, Erlang(2) distribution, Integro-differential equation, Gerber-Shiu dicounted penalty function, Time of ruin, Two-step premiumCLC Number: O211.62AMS(2000)Subject Classification: 60K05,65G30 91B30Document code: A1IntroductionIn classical risk process
4、the claims occur as a poisson process. However, in practice, many claims occur according to a more general renewal process.In recent years, the compound re- newal (Sparre Andersen) process with claim inter-arrival times distributed as Erlang(2) (that is a Gamma(2,) distribution) has been studied ext
5、ensively. There are some articles discussing ruin problems based on the Erlang(2) process, for example, Dickson1, Dickson and Hipp2, Cheng and Tang3, Li and Garrido4(for the Erlang(n) process), and references therein.We consider a Sparre Andersen surplus processU(t) = u + ct N(t)Xi=1Xi,t 0,(1.1)wher
6、e U(t) is the surplus of the insurer at time t, u = U(0) is the initial surplus, c is the premium rate per unit time, the random variable Xirepresents the ith claim, they are i.i.d.and with common probability distribution P and density p. Denote by e p(s) =R0esxp(x)dx its Laplace transform. N(t) is
7、the number of claims up to time t, let Tii=1be a sequence of i.i.d.randomE-mail: E-mail: Biography: SUN Jing-yun, male, Han, Gansu, postgraduate, major in: stochastic processes and risk theory.1http:/ variables representing the times between claim with common probability density function k(t) = 2tet
8、, t 0(that is Erlang(2) distribution).Further assume that Tiand Xiare independent and cETi EXi for all i.Li and Garrido5 studied the penalty function for the Erlang(n) risk process with a constant dividend barrier. Both Lin and Pavlova6 and Zhang et al7 have studied the penalty function for classica
9、l risk model with a two-step premium rate. Inspired by them, we consider the discounted penalty function for Erlang(2) risk process with two-step premium rate. That is, for an insurers surplus process, denote b 0 the constant barrier level, and c1 0 the annual premium rate, when the surplus is above
10、 the barrier b, dividends are paid and in this case, the net premium rate after dividend payment is c2( c1) 0, so the surplus process can be denoted by Ub(t),t 0 and it has the following expression:dUb(t) =( c1dt dS(t),Ub(t) b; c2dt dS(t),Ub(t) b;Ub(0) = u,where S(t) =N(t)Pi=1Xi, and ciET1 EX1,i = 1
11、,2. Define ruin time to be Tb= inft;Ub(t) 0,definemb(u) = EeTb(Ub(Tb),|Ub(Tb)|)I(Tb 0L(D)m2,(u) =Zu0m2,(u x)p(x)dx + (u),(2.8)where L(D) = (1 + /)I (c2/)D2=2Pk=0LkDkis an 2th-order linear differentiation op-erator.I and D denote the identity operator and differentiation operator respectively, and (t
12、) =Rt(t,y t)dP(y). IfR0R0(x,y)p(x + y)dxdy b.Theorem 3.1 The discounted penalty function mb(u) satisfies the integro-differential equationm00 b(u) =m00 1(u) =2(+) c1m0 1(u) (+ c1)2m1(u) +2 c21Ru0m1(u y)dP(y)+2c21(u),0 u b,m00 2(u) =2(+) c2m0 2(u) (+ c2)2m2(u) +2 c22Rub 0m2(u y)dP(y)+Ruubm1(u y)dP(y)
13、 +2 c22(u),u b.(3.1)Proof. We employ the basic ideal of Lin and Pavlova6 and Zhang et al7, for 0 u b, conditionon the time and the amount of the first claim, we havemb(u)=m1(u)=Z0k(t)Z0EeTb(U(Tb),|U(Tb)|)I(Tb b, by the same way as deriving (3.2), we havem2(u)=Z0k(t)etZu+c2t0mb(u + c2t y)dP(y) +Zu+c2
14、t(u + c2t,y u c2t)dP(y)dt=Z0k(t)etb(u + c2t)dt =1 c2Zuk(t uc2)e(tu c2) b(t)dt,(3.5)4http:/ then differentiating both side with respect to u yieldsm0 2(u) = 1 c22Zu2e(+)(tu c2) b(t)dt + + c2m2(u)(3.6)and differentiating Eq.(3.6) second time we can yield the integro-differential equation for m2(u).m00
15、 2(u)=2( + ) c2m0 2(u) ( + c2)2m2(u)+2c22Zub0m2(u y)dP(y) +Zuubm1(u y)dP(y) +2 c22(u),u b.(3.7)by which the proof is completed.Note that the Eq.(2.1) of Dickson and Hipp2 is the special case of our above integro-differential equation (3.1). When c1= c,c2= 0, Eq.(3.1) is also in agreement with Eq.(17)(where let n = 2) of Li and Garrido5. By (3.2) and (3.5), we getm2(b+) = lim ub+m2(u) =1 c2Zbk(t bc2)e(tb c2) b(t)dt = m1(b),(3.8)it shows that mb(u) is continuous for u = b. However, the same is not true for m0 b(
