Col.lecció d Economia E14/310. Some optimization and decision problems in proportional reinsurance. Anna Castañer M. Mercè Claramunt Maite Mármol

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1 Col.lecció d Economia E14/310 Some optimization and decision problems in proportional reinsurance Anna Castañer M. Mercè Claramunt Maite Mármol

2 UB Economics Working Papers 2014/310 Some optimization and decision problems in proportional reinsurance Abstract: Reinsurance is one of the tools that an insurer can use to mitigate the underwriting risk and then to control its solvency. In this paper, we focus on the proportional reinsurance arrangements and we examine several optimization and decision problems of the insurer with respect to the reinsurance strategy. To this end, we use as decision tools not only the probability of ruin but also the random variable deficit at ruin if ruin occurs. The discounted penalty function Gerber & Shiu, 1998) is employed to calculate as particular cases the probability of ruin and the moments and the distribution function of the deficit at ruin if ruin occurs. We consider the classical risk theory model assuming a Poisson process and an individual claim amount phase-type distributed, modified with a proportional reinsurance with a retention level that is not constant and depends on the level of the surplus. Depending on whether the initial surplus is below or above a threshold level, the discounted penalty function behaves differently. General expressions for this discounted penalty function are obtained, as well as interesting theoretical results and explicit expressions for phase-type 2 distribution. These results are applied in numerical examples of decision problems based on the probability of ruin and on different risk measures of the deficit at ruin if ruin occurs the expectation, the Value at Risk and the Tail Value at Risk). JEL Codes: G22. Keywords: Deficit at ruin, Gerber-Shiu function, Risk measures. Anna Castañer Facultat d'economia i Empresa Universitat de Barcelona M. Mercè Claramunt Facultat d'economia i Empresa Universitat de Barcelona Maite Mármol Facultat d'economia i Empresa Universitat de Barcelona Acknowledgements: The authors thank the Spanish Ministerio de Educación y Ciencia MICINN) for its support grant ECO C03-03).. ISSN

3 1. Introduction The influence of reinsurance strategies on the solvency of the insurer is an important subject and has been widely analyzed in actuarial literature see e.g. Castañer et al. 2010, 2012), Dickson & Waters 1996) and Centeno 1986, 2002, 2005)). Several optimization problems have been considered using different kinds of reinsurance strategies, being the proportional, the excess of loss and the stop-loss the most well-known see Centeno & Simões 2009) and the references therein). One of the main measures used to control solvency is ruin probability, but in this paper we use alsoother measures related to the deficit at ruin if ruin occurs, as its expectation or the Value at Risk VaR) or the Tail Value at Risk TVaR). We study the random variable deficit at ruin if ruin occurs in the classical risk theory model considering a proportional reinsurance arrangement, where the retention level is not constant and depends on the level of the surplus. This type of reinsurance, called threshold proportional reinsurance, has been first defined and studied in Castañer et al. 2010, 2012), and includes, as a particular case, the classical proportional reinsurance with constant retention level. In the classical risk theory model, the surplus, Rt), at a given time t [0, ) is defined as Rt) = u+ct St), with u = R0) 0 being the insurer s initial surplus, St) the aggregate claims and c the instantaneous premium rate. St) is modeled as a compound Poisson process St) = Nt) The claim number process {N t)} t=0 is assumed to be Poisson with parameter λ. Specifically, the corresponding claim inter-arrival times, denoted by {T i } are independent and identically distributed i.i.d.) exponentially distributed random variables with parameter λ, where T 1 denotes the time until the first claim and T i, for i > 1, denotes the time between the i 1)th and ith claims. The random variables {X i } are the positive claim severities, which are i.i.d. random variables with common probability density function f x) and {N t)} t=0 is independent of {X i,i 1}. We assume that the insurer s premium income is received continuously at rate c per unit time, where c = λe[x]1+ρ), and ρ > 0 is the premium loading factor. The time of ruin is T = min{t 0 Rt) < 0}, with T = if Rt) 0 for all t 0. The deficit at ruin if ruin occurs is Y = R + T) and the surplus immediately prior to ruin is R T). The probability of ultimate ruin from initial surplus u is denoted ψu) and defined by X i. ψu) = P [T < R0) = u] = E{IT < ) R0) = u}, 2

4 where IA) = 1 if A occurs and IA) = 0 otherwise. Gerber & Shiu 1998) introduced the Gerber-Shiu discounted penalty function φu), [ φu) = E e δt w R T), R + T) ) ] IT < ) R0) = u, 1.1) being δ 0 the discounted factor, and wl,j), l 0, j > 0, the penalty function, so that φu) is the expected discounted penalty payable at ruin. This function is known to satisfy a defective renewal equation Gerber & Shiu, 1998, Li & Garrido, 2004, Willmot, 2007). This function can be used to study the traditional quantities of interest in classical ruin theory, such as ruin probability, time of ruin or deficit at ruin. Therefore, depending on the penalty function wl, j), we can obtain different interpretations for the Gerber-Shiu function: i) For wl,j) = 1, [ ] φu) = E e δt IT < ) R0) = u, i.e. the Laplace transform of the time of ruin being δ the parameter. In addition, if we consider δ = 0, the ultimate ruin probability is obtained φu) = ψu). ii) For wl,j) = j m and m 1, φu) = E [ ] e δt j m IT < ) R0) = u, and dividing this Gerber-Shiu function by the probability of ruin, the ordinary discounted moments of the deficit at ruin if ruin occurs are obtained, α m Y) = E[ e δt j m IT < ) R0) = u ] ψu). 1.2) If we let δ = 0 in 1.2), the ordinary moments of the deficit at ruin if ruin occurs are obtained. iii) For wl,j) = Ij y), φu) = E [ ] e δt Ij y)it < ) R0) = u, and dividing by the probability of ruin we obtain the distribution function of the discounted deficit at ruin if ruin occurs, F Y y) = E[ e δt Ij y)it < ) R0) = u ] ψu). 1.3) If we let δ = 0, the distribution function of the deficit at ruin if ruin occurs is obtained. 3

5 The threshold proportional reinsurance strategy is a dynamic strategy with a retention level that is not constant and depends on the level of the surplus, Rt). A retention level k 1 is applied whenever the reserves are less than a threshold b 0, and a retention level k 2 is applied in the other case. Then, the premium income retained is c 1 and c 2, respectively. We consider that the retention levels give new positive security loadings for the insurer, i.e. the net profit condition is always fulfilled. Then, we can define ρ 1 = ρ R ρ R ρ k 1 factor of reinsurer. and ρ 2 = ρ R ρ R ρ k 2, being ρ R the loading The main objective of this paper is to study the effect of the threshold proportional reinsurance on the probability of ruin and on the other risk measures related with the deficit at ruin. We also perform a comparative analysis with the proportional reinsurance. Our results can assist the insurer in his reinsurance decision process concerning solvency related optimality problems in reinsurance can be found for instance in Dimitrova & Kaishev 2010), Kaishev & Dimitrova 2006) and Castañer et al. 2013)). After this introduction, the paper is organized as follows. In the next section some assumptions and some preliminaries are included. In Section 3.1, we present some general results for the Gerber-Shiu function for the ruin probability, the ordinary moments and the distribution function of the deficit at ruin if ruin occurs, when the individual claim amount follows a general phase-type distribution. In Section 3.2, we assume a phase-type 2 distribution and analyzed the previous results. Then, the explicit expressions are obtained. An interesting result about the distribution of the deficit at ruin if ruin occurs in a model with a threshold reinsurance is demonstrated in Theorem 3.4. In Section 4, some optimization and decision problems of the reinsurance strategy are presented. In this analysis, the ruin probability and the deficit at ruin if ruin occurs are used as decision tools for the insurer. This section includes some numerical examples. Section 5 closes the paper giving some concluding remarks. 2. Assumptions and preliminaries In this paper we analyze the deficit at ruin in the classical risk theory model assuming a compound Poisson process for the aggregate claims and a phase-type distribution for the individual claim amount, when the insurer considers a threshold proportional reinsurance. In this model with threshold proportional reinsurance strategy, the discounted penalty function 1.1) behaves differently, depending on whether initial surplus u is below or above the level b. Hence, for 4

6 notational convenience, we write φ 1 u), 0 u < b, φu) = φ 2 u), u b. In Castañer et al. 2010) a theorem for the integro-differential equation for the Gerber-Shiu function 1.1) is obtained in a Poisson model for the claim process. We include this theorem in order that the paper is self-contained, taking into account that we will use it in the next sections. Theorem 2.1. The discounted penalty function φu) in a Poisson process model satisfies the integro-differential equations φ φ 1 u), 0 u < b, u) = 2.1) φ 2 u), u b, where φ 1u) = λ+δ φ 1 u) λ u k 1 φ 1 u k 1 x)dfx) λ ξ 1 u), c 1 c 1 0 c [ 1 φ 2u) = λ+δ φ 2 u) λ u b k 2 φ 2 u k 2 x)dfx) 2.2) c 2 c 2 + u k 2 u b k 2 φ 1 u k 2 x)dfx) 0 ] λ c 2 ξ 2 u), and ξ 1 u) = wu,k 1 x u)fx)dx, ξ 2 u) = u k 1 wu,k 2 x u)fx)dx. 2.3) u k 2 Let wr T), R + T) ) be a non-negative function of R T) > 0, the surplus immediately before ruin, and R + T) > 0 the surplus at ruin. As we focus our analysis on the deficit at ruin if ruin occurs, we will consider only a specific subgroup of penalty functions WD = {wl,j) = j m,wl,j) = Ij y),wl,j) = 1} with m > 0. We assume that the individual claim amount follows a phase-type distribution P Hα, S). Key results of modern theory of phase-type distributions including theoretical properties, characterization and applications can be found in Neuts 1981, 1989), O Cinneide 1990), Latouche & Ramaswami 1999) or Asmussen 2003). Most of the original applications of phase-type distributions are in queuing theory, but these kind of distributions are widely used in risk theory 5

7 in the last years. In Asmussen & Albrecher 2010) many applications in this field can be found. Other important references on phase-type distributions in risk theory context include Asmussen & Rolski 1992), Rolski et al. 1999), Asmussen & Bladt 1996), Bladt 2005) or Drekic et al. 2004). We present a brief overview of phase-type distributions and their properties. Phase-type distributions: We consider a continuous time Markov chain with a single absorbing state 0 and N transient states. The row vector α contains the probabilities α j that the process starts in the various transient states j = 1,2,...,N. If the probability of starting the process in the absorbing state is zero, N j=1 α j = 1. Then, αe T = 1 where e T is a column vector of ones with n 1 elements. The infinitesimal generator Q for the continuous time Markov chain is given by Q = 0 0, S 0 S where S is the matrix of transition rates among the transient states and S 0 is the column vector of absorption rates into state 0 from the transient states. Necessarily, S 0 = Se T, and S is an N N matrix whose diagonal entries are negative and whose other entries are nonnegative. Under these assumptions, the distribution of time X until the process reaches the absorbing state is said to be phase-type distributed and is denoted P Hα, S) with distribution F X x) = 1 αexpsx)e T for x 0, density function f X x) = αexpsx)s 0 for x > 0 and ordinary moments α m X) = 1) m m!αs m e T, being exp ) the matrix exponential. N, The Laplace transform of the density function f X t) = 0 e tx f X x)dx is rational of degree f X t) = at) bt), with at) = N a i t i, a 0 = 0, bt) = N b i t i, b 0 = 1 and f X x) satisfies the linear differential equation i=0 i=0 N i=0 b i f i) X x) = ) The finite mixture of phase-type distributions is phase-type distributed. Let X i, i = 1,...,k distributed as PHα i,s i ), and Y = I i X i being k I i = 1, and P I i = 1) = p i then Y is PHα,S) with 6

8 S S α = p 1 α 1,...,p k α k ) and S = ) S k Phase-type distributions with N = 2 are phase-type distributions with λ αλ S =, 2.6) βµ µ where λ,µ > 0 and 0 α, β 1, αβ < 1 with a density satisfying 2.4), i.e., f x)+b 1 f x)+ b 2 f x) = 0. If b 2 < 0, the exponential distribution is obtained and all other phase-type distributions are obtained when b 2 > 0. In Dickson & Hipp 2000) it is demonstrated that any phase-type distribution with N = 2 is either an hyper-exponential distribution or a linear combination of an exponential and an Erlang2) with the same scalar parameter. Then, following Dickson & Hipp 2000), a phasetype distribution with N = 2, can be always expressed in standardized form with α 1,α 2 ), S = a 1 a 2, a 1,a 2,a 4 0, being a 1 = β 1, a 4 = β 2 and a 2 = 0 for the hyperexponentialβ 0 a 4 1,β 2 )distributionanda 1 = a 2 = a 4 = β foralinearcombinationofanexponentialβ) and an Erlang2,β). It is easy to prove that the following relations are fulfilled for the standardized form, b 1 = a 1+a 4 a 1 a 4 and b 2 = 1 a 1 a 4 and that the density function using the standardized form is f x) = α 1 a 1 a 2 )e a 1x +α 2 a 4 e a 4x +α 1 a 4 a 2 xe a 1x. 3. Gerber-Shiu function with X phase-type 2 distributed In order to find the expression of the Gerber-Shiu function if the individual claim amount follows a phase-type distribution we first need to obtain the ordinary differential equation from the integro-differential equation included in Theorem 2.1 and then, solve it. In this section, we present some results that are general and useful for any phase-type, the ordinary differential equation and the expression for ξ s u) for s = 1,2. Next, we obtain the explicit expression for the Gerber-Shiu function for N = General results In Theorem 3.1, we present the ordinary differential equation for the Gerber-Shiu function in a Poisson process model.it is general with respect the three specific expressions included in W D. 7

9 This is an important result that implies that if we are analyzing the probability of ruin or the deficit at ruin if ruin occurs, the structure of the solution will be the same. Then, where is the difference? The difference is included in the expression of ξ s u) for s = 1,2. In Theorem 3.2, we present a general expression of ξ s u). Theorem 3.1. If the individual claim amount is distributed as a PHα,S) and wl,j) = wj), φu) is the solution of the ordinary differential equation, φ N+1) s u) = δ c s k N s b N ) λ+δ φ s u)+ b ) N 1 φ N) s u) c s k s b N + N 1 j=1 λ c s b N 1 k N j s N 1 h=j+1 λ c s f N 1 j) 0)+ b j 1 k s b N λ+δ)b j c s b N 3.1) b h f h j 1) 0) φ j) s u), where φ s u), s = 1,2 being s = 1 for 0 < u < b and s = 2 for u > b. The proof of Theorem 3.1 is included in Appendix A. Theorem 3.2. Let the individual claim amount X PHα,S). Then ξ s u) = H s u) G s, s = 1,2, being H s u) = 1 F Xs u)), S = S k s, X s PHα,S ) and for Z PHα, S ) and α = αexps u) H su). α m Z) if wl,j) = j m, m 1, G s = F Z y) if wl,j) = Ij y), 1 if wl,j) = 1, Proof. Taking into account 2.3), if the penalty function is equal to 1, assuming that X PHα,S) ξ s u) = u ks where S = S k s and X s PHα,S ). f X x)dx = αexps u)e T = 1 F Xs u) = H s u), 3.2) 8

10 If the penalty function is j m, then ξ s u) = = Let α = αexps u) H su), then u ks 0 ξ s u) = H s u) k s x u) m f X x)dx = z m αexps u)exps z) S e T) dz, s = 1, z m α exps z) S e T) dz, s = 1,2, z m 1 ) u+z f X dz 3.3) k s k s and taking into account that α exps z) S e T) = f Z z) being Z PHα,S ), ξ s u) = H s u) 1) m m!α S ) m e T = H s u)α m Z), s = 1,2. 3.4) If the penalty function is Ij y), ξ s u) = u+y ks u ks df X x) = 1 αexps u+y))e T 1+αexpS u)e T = αexps u)e T αexps u)exps y)e T, s = 1,2, 3.5) and considering α and H s u), ξ s u) = H s u) 1 α exps y)e T) = H s u)f Z y), s = 1,2. 3.6) 3.2. Results for N = 2 From Theorem 3.2, we derive the following corollary, that gives the expression of ξ s u), s = 1,2 assuming a PHα,S) with N = 2 expressed in standardized form. Corollary 3.3. From Theorem 3.2, if the individual claim amount is PHα,S) with N = 2 expressed in standardized form, being a i,s the elements of S = S k s, s = 1,2, For the penalty functions ξ s u) = C s) 1 e a 1,su +C s) 2 e a 4,su +C s) 3 ue a 1,su, s = 1,2. i) wl,j) = 1: C s) 1 = α 1, C s) 2 = α 2 and C s) 3 = α 1 a 2,s. m! m 1 = α 1 ii) wl,j) = j m : C s) a m 1,s +m! a 2,s a m+1 1,s a i 1,s a i 4,s ), C s) 2 = α 2 m! a m 4,s and C s) m! 3 = α 1 a 2,s a m. 4,s 9

11 iii) wl,j) = Ij y): C s) 1 = α 1 1 e a 1,sy ), C s) 2 = α 2 1 e a 4,sy ) α 1 a 2,s ye a 1,sy and C s) 3 = α 1 a 2,s 1 e a 1,sy ). Proof. Let X PHα,S), being S a matrix expressed in standardized form, and being a i,s the elements of S = S k s,s = 1,2. For wl,j) = 1, from 3.2), if we let a i,s = a i k s, ξ s u) = α 1,α 2 ) e a 1,su a 2,s ue a 1,su e T = α 1 e a1,su +α 2 e a4,su +α 1 a 2,s ue a1,su. 0 e a 4,su For wl,j) = j m, from 3.4), and substituting α = αexps u) H su), ξ s u) = m! 1) m αexps u)s ) m e T = m! 1) m α 1,α 2 ) e a 1,su a 2,s ue a 1,su 0 e a 4,su a 1,s a 2,s 0 a 4,s m e T, being then m! ξ s u) = α 1 a 1 a 2 0 a 4 a m 1,s +m! a 2,s a m+1 1,s m m = 1) m a i 1,sa i 4,s And for wl,j) = Ij y), from 3.6) and using α, ) 1 a m 1,s a 2,s a m+1 1,s m 1 0 a m 4,s a i 1,s a i 4,s, e a1,su m! +α 2 a m e a4,su m! +α 1 4,s a m a 2,s ue a1,su. 4,s ξ s u) = αexps u)e T αexps y)exps u)e T = α 1 1 e a 1,s y ) e a 1,su + α 2 1 e a 4,s y ) α 1 a 2,s ye a 1,sy ) e a 4,su + α 1 1 e a 1,s y ) a 2,s ue a 1,su. Then, the corollary is proved. Obviously, from Corollary 3.3, it is possible to obtain the particular cases included in phasetype 2 distributions. If we consider the hyper-exponentialβ 1,β 2 ) distribution, a 1 = β 1, a 2 = 0 and a 4 = β 2. Then, for the penalty function equal to 1, C s) 1 = α 1, C s) 2 = α 2 and C s) 3 = 0; for the penalty function equal to j m, C s) ks 1 = α m m! 1 β m 1, C s) 2 = α 2 k m s m! β m 2 and C s) 3 = 0; and for penalty ) function equal to Ij y), C s) 1 = α 1 1 e β 1 ks ), y C s) 2 = α 2 1 e β 2 ks y and C s) 3 = 0. 10

12 If we consider a linear combination of an exponentialβ) and an Erlang2,β), a 1 = a 2 = a 4 = β. Then, for the penalty function equal to 1, C s) 1 = α 1, C s) 2 = α 2 and C s) 3 = α 1 β k s ; for the penalty function equal to j m, C s) ks 1 = α m m+1)! 1 β, C s) k m s 2 = α m m! 2 β and C s) k m s 3 = α m 1 m! 1 ; and β m 1 for the penalty function equal to Ij y), C s) 1 = α 1 1 e β y) ks, C s) 2 = α 2 1 e β y) ks α 1 β k s ye β ks y and C s) 3 = α 1 β k s 1 e β ks y). If, in addition, we consider α 1 = 1 and α 2 = 0, then we get the Erlang2, β) distribution. The exponential distribution is not a phase-type 2, but it can be obtained considering that α 1 = 0 and α 2 = 1. Once we have obtained the different expressions of ξ s u), we can solve the integro-differential equation for the Gerber-Shiu function. From 3.1), if N = 2, φ s u) = + λ+δ b ) 1 φ s u) 3.7) c s k s b 2 λ+δ)b1 b 0 c s k s b 2 ksb 2 λ ) f 0) φ 2 c s k su)+ δ s c s ksb 2 φ s u), 2 where φ s u), s = 1,2 being s = 1 for 0 < u < b and s = 2 for u > b. In order to solve 3.7) we obtain the characteristic equation for 0 u < b, and for u b, λ+δ r 3 b ) 1 r 2 λ+δ)b1 b 0 c 1 k 1 b 2 c 1 k 1 b 2 k1 2b λ ) f 0) r δ 2 c 1 k 1 c 1 k1 2b 2 λ+δ r 3 b ) 1 r 2 λ+δ)b1 b 0 c 2 k 2 b 2 c 2 k 2 b 2 k2 2b λ ) f 0) r δ 2 c 2 k 2 c 2 k2 2b 2 = 0, = 0, with r i, i = 1,...,6, real and distinct roots of the characteristic equations, so φ 1 u) = 3 F i e riu, 0 u < b, φu) = φ 2 u) = 6 F i e riu, u b. i=4 3.8) To obtain the ruin probability, and the moments and the deficit at ruin not their present values) we have to consider δ = 0, so r 3 = r 6 = 0. Then, to determine F i, i = 1,...,6, we need six equations. One equation is obtained from lim φu) = 0, that gives F 6 = 0. Another equation comes from the continuity condition u φ 1 u) u=b = φ 2 b). 3.9) The other four conditions are obtained substituting 3.8) in 2.2), integrating and rearranging terms, considering Corollary 3.3, and taking into account the values of a 1, a 2 and a 4. For 11

13 the hyper-exponentialβ 1,β 2 ) distribution, let us define h 1 = C1) 1 α 1, h 2 = C1) 2 α 2, h 3 = C2) 1 h 4 = C2) 2 α 2, then the four equations are α 1 and β 1 5 i=4 β 2 5 i=4 F i e β 1 3 β 2 3 r i + β ) 1 b k 2 r i k 2 +β 1 +β 1 r F i e i + β 2 )b k 2 +β 2 r i k 2 +β 2 F i r i k 1 +β 1 = h 1, F i r i k 1 +β 2 = h 2, 3.10) 3 3 F i 1 e F i 1 e r i + β ) ) 1 b k 2 r i k 2 +β 1 = h 3, r i + β ) ) 2 b k 2 r i k 2 +β 2 = h 4. For the linear combination of an exponentialβ) and an Erlang2,β), let us define h 1 = C 1) 1 +C 1) 2, h 2 = C1) 3 k 1 α 1 β, h 3 = C 2) α 2 β α 1β 2 b k 2 ) 5 i=4 +α 1 β 2 3 β 5 i= C 2) 3 k 2 α 1 β 2 and h 4 = C2) α 1 F i β 2 r i k 1 +β) r i + β ) b k2 F i e r i k 2 +β +α 1 β 2 r i + F i 1 e β ) ) b k2 3, then the four equations are α 2 F i β r i k 1 +β = h 1, F i β r i k 1 +β = h 2, 3.11) 5 r i k 2 +β) 2 +α 2 β ) r F i e i + β b k2 3 r i k 2 +β +β 3 r i + β ) b r k2 F i e + α 1β 2 i + β ) b k2 b F i e r i k 2 +β) 2 k 2 r i k 2 +β i=4 r i + F i 1 e β ) ) b k2 3 r i + F i 1 e β ) ) b k2 r i k 2 +β = h 3, r i k 2 +β = h 4. Let us rewrite the linear equation system formed by 3.9) and 3.10) or 3.11) in matrix form, A F = H, being F the vector of unknowns, F = F j ) j=1,...,5, considering H the vector of independent terms H = 0,h 1,h 2,h 3,h 4 ), and A the matrix of the coefficients of the system. Solving the system we obtain F = A 1 H, so F j = 4 h i f ji, j = 1,...,5, being f ji the elements of the matrix A 1. From 3.8), φ 1 u) = 3 F i e riu = 4 h z φu) = z=1 φ 2 u) = 5 F i e riu = 4 h z i=4 z=1 3 f iz e riu = 4 h z c z u), 0 u < b, z=1 5 f iz e riu = 4 h z d z u), u b. i=4 z=1 3.12) 12

14 being c z u) = 3 f iz e riu and d z u) = 5 f iz e riu. i=4 From the definition of h z, z = 1,...,4, it is straightforward that in the ruin probability case, h z = 1. Then, from 3.12), φ 1 u) = ψ 1 u) = 4 c z u) and φ 2 u) = ψ 2 u) = 4 d z u). z=1 Theorem 3.4. The deficit at ruin if ruin occurs, Y, is distributed as a phase-type PHτ u),m) where τ u) = P1 z u)) z=1,...,4 being P1 z u) = czu) ψ 1 u) if 0 u < b, and τ u) = P2 zu)) z=1,...,4 being P2 z u) = dzu) ψ 2 u) if u b, and M = T 1 0, 0 T 2 being T s = a 1 α 1 a 2 k s k s 0 a 4 k s, s = 1,2. Proof. The distribution of the deficit at ruin if ruin occurs from 1.3) and δ = 0 is F Y y) = φu) ψu). For 0 u < b, from 3.12), F Y y) = φ 1u) ψ 1 u) = 1 ψ 1 u) 4 h z c z u) = z=1 z=1 4 h z P1 z u). 3.13) Knowing the values of h z, z = 1,...,4, defined in 3.10) and 3.11), and considering the values of C s) i obtain in Corollary 3.3 for wl,j) = Ij y), substituting in 3.13), and grouping terms we ) ) F Y y) = 1 W 1 u) P11 u) W 1 u), P1 2u) W 1 u) expt 1 y)e T W 2 u) P13 u) W 2 u), P1 4u) W 2 u) expt 2 y)e T, 3.14) being Let γ = z=1 W 1 u) = P1 1 u)+p1 2 u), W 2 u) = P1 3 u)+p1 4 u), T s = a 1 k s α 1a 2 k s. 0 a 4 k s ) ) P11 u) W 1 u), P1 2u) W 1 u) and δ = P13 u) W 2 u), P1 4u) W 2 u), then 3.14) can be written as F Y y) = 1 W 1 u)γexpt 1 y)e T W 2 u)δexpt 2 y)e T. Taking into account that T s, s = 1,2 has the structure defined in 2.6), then the distribution of the deficit at ruin if ruin occurs is a mixture of two phase-type distributions, Y 1 PHγ,T 1 ) 13

15 and Y 2 PHδ,T 2 ), being the weights W 1 u) and W 2 u). Then, considering that the finite mixture of a phase-type distribution is a phase-type distribution, from 2.5), the theorem for 0 u < b is proved. For u b, applying a similar process it can be demonstrated that Y PHτ u),m), with τ u) = P2 z u)) z=1,...,4. So, Y is a mixture of two phase-type distributions Y 1 PHµ,T 1 ) ) ) and Y 2 PHυ,T 2 ) with µ = P21 u) V 1 u), P2 2u) V 1 u) and υ = P23 u) V 2 u), P2 4u) V 2 u), being the weights V 1 u) = P2 1 u)+p2 2 u)) and V 2 u) = P2 3 u)+p2 4 u)). Example 3.5. As an example, we calculate the probabilities of ruin and the distribution of the deficit at ruin if ruin occurs assuming a threshold reinsurance strategy with X i Erlang2,β) and the following values for the parameters β = 2, λ = 1, b = 2, k 1 = 0.8, k 2 = 0.45, ρ = 0.15, ρ R = 0.25 and δ = 0. Let us first obtain the ruin probability. We know that, in this case, the independent terms of system 3.11), h z, z = 1,...,4 are equal to one and that the matrix A 1 is A 1 = Then, we have ψ 1 u) = e u e u, ψ 2 u) = e u e u. Let us know consider the deficit at ruin if ruin occurs. From Theorem 3.4, Y is distributed as a phase-type, PHτ u),m), with M = , τu) = 14

16 e 3.701u e 0.188u, e 3.701u e 0.188u, e 3.701u e 0.188u e 3.701u e 0.188u ) 0 u < b, e 3.701u 0.001e 0.188u, e 3.701u 0.012e 0.188u e 3.701u e 0.188u e 3.701u e 0.188u e 6.646u e 0.080u, e 6.646u e 0.080u, e 6.646u e 0.080u e 6.646u e 0.080u ) u b e 6.646u e 0.080u, e 6.646u e 0.080u e 6.646u e 0.080u e 6.646u e 0.080u For example, for u = 0, ψ 1 0) = , τ 0) = , , , ), F Y y) = y)e 2.5y y)e 4. 4y, α m Y) = m m )m! and for u = 3, ψ 2 3) = , τ 3) = , , , ), F Y y) = y)e 2.5y y)e 4. 4y, α m Y) = m m )m! 4. Influence of threshold) proportional reinsurance on the deficit at ruin if ruin occurs In this section, we quantify the effect on the deficit at ruin if ruin occurs of a proportional reinsurance included the threshold). It is known Drekic et al., 2004) that when the individual claim amount follows a phase-type distribution PHα,S), the deficit at ruin if ruin occurs, Y, is also phase-type distributed with representation PHΠ G,S), where Π G = α +expub), ψu) with B = S + D, D = S 0 α +, S 0 = Se T being α + = λ c αs 1, in the Poisson case. We also have ψu) = α + expub)e T. Then, if the insurer uses a proportional reinsurance contract to reduce the risk, which is the effect on the probability of ruin and on the deficit at ruin if ruin occurs? Let us consider a proportional reinsurance with parameter k, 0 < k 1, such that the retained claim amount for 15

17 the insurer is XR = kx and the retained premium is λe[x]1+ρ) λe[x]1 k)1+ρ R ). We consider that the retention level k gives new positive security loading for the insurer, ρ N = ρ ρ R 1 k) k > 0, i.e., the net profit condition is always fulfilled see Castañer et al. 2007) for more details). Then, if X follows a phase-type distribution PHα,S), XR is also phase-type distributed, PH α, S k). The ruin probability with proportional reinsurance is beingbr = S k S k et PH ΠR G, S k), where ψu) = αs 1 expubr)e T αs 1 e T 1+ρ N ), αs 1. Thedeficitatruinifruinoccurs, Y, isphase-typedistributed, αs 1 e T 1+ρ N ) ΠR G = αs 1 expubr) αs 1 expubr)e T. Hence, the expectation and the variance of the deficit at ruin if ruin occurs can be easily calculated: E[Y] = ΠR G ks 1 e T and V[Y] = 2ΠR G k 2 S 2 e T ΠR G ks 1 e T) 2. The Value at Risk of Y at level p, VaR p [Y] is such that F Y VaR p [Y]) = p, that is VaR p [Y] = F 1 Y p). However, there is no explicit expression for this VaR p [Y], it has to be calculated numerically the package actuar in Dutang et al. 2008) provides functions for phase-type distributions). The Tail Value at Risk of Y at level p, TVaR p [Y], can be calculated from the Value at Risk at the same level considering the following formula Cai & Li, 2005), TVaR p [Y] = VaR p [Y] ΠR GS 1 expvar p [Y] S k )et ΠR G expvar p [Y] S k )et. As an application, we develop the example that has first been used by Gerber et al. 1987) and after by Drekic et al. 2004). They consider an individual claim amount distribution that is an equal mixture of two exponentials at rates 3 and 7 respectively, with Poisson claims at rate λ = 1 and a relative security loading ρ = 0.4. In this case, X is PHα,S), where α = 0.5,0.5), S = 3 0 and B = The ruin probability is ψu) = e u +e 6u The deficit at ruin if ruin occurs, Y, is phase-type distributed, PHΠ G,S), where Π G = 42 7e 5u, ), 6+9e 5u being 48+2e 5u 48+2e 5u and F Y y) = 1 6e5u 7y +42e 5u 3y +9e 7y 7e 3y 2+48e 5u E[Y] = e 5u 21e 5u +504, V[Y] = e 2 5u) 744e 5u 441e 2 5u) e 5u

18 It is straightforward to include in the model a proportional reinsurance. Let consider a retention level k and a security loading of the reinsurer ρ R = 0.5, with 0.2 < k 1. Then, the net security loading for the insurer is ρ N = 0.5k 0.1 k. The ruin probability is ) e 5 54k+N)u k 1+15k) k 4+165k +5N +e 2Nu k 15k k +5N) ψu) = 1+15k)N 4.15) being N = 4 120k +1341k 2. Figure 1 shows the behaviour of the probability of ruin as a function of u and k left-hand side plot) including the level curves in the right-hand side plot. Figure 1: Ruin probability for different values of u and k The deficit at ruin if ruin occurs, Y, is phase-type distributed, PH ΠR G, S k), with ) k +N +e 2Nu k 15k k +N) ΠR G = ), k +5N +e 2Nu k 15k k +5N) ) 3 2 9k +N +e 2Nu k 15k 2 2+9k +N) ) k +5N +e 2Nu k 15k k +5N) and ) k k +e 2Nu k 15k k 29N) 29N E[Y] = ), k +5N +e 2Nu k 15k k +5N) V[Y] = + 2k 2 84e 2Nu k 15k k +375k 2) +M 1160N +k N) ) k +5N +e 2Nu 2 k 15k k +5N) ) 2k 2 e 4Nu k 15k 2 M +1160N 3k N)) k +5N +e 2Nu k 15k k +5N) ) 2, ) 17

19 being M = k 2. We consider now a threshold proportional reinsurance, that is defined by its three parameters b,k 1,k 2 ). In this case, the probability of ruin is given by 3.12), and the deficit at ruin if ruin occurs is phase-type 4 distributed see Theorem 3.4). The explicit expressions of the probability of ruin and the different measures of the deficit at ruin if ruin occurs can then be easily obtained see Example 3.5 for the Erlang2, β)). Optimization problems regarding ruin probability. Which is the best strategy in order to minimize the ruin probability of the insurer? In order to answer this question we solve two optimization problems. Firstly, the insurer only considers the proportional reinsurance option. Let ψk) be the ruin probability when all the variables that influence the probability are fixed except the retention level k, min k, ψk) 4.16) 0.2 < k 1 being 4.15) the expression for the probability of ruin in this case. It can be proved that this optimum exists, but the expressions for the optimal point and the minimum value have not been included for the sake of brevity. In Table 1, we include the results of this minimization for different values of u, being k the minimum point. The expectation, the variance and the Value at Risk and the Tail Value at Risk for different levels p 0.95, 0.99 and 0.995) of the deficit at ruin if ruin occurs can also be found in Table 1 for the optimal k. Table 1: Minimum probabilities of ruin with proportional reinsurance and E[Y], V[Y], VaR p[y] and TVaR p[y] u k ψk ) E[Y] V[Y] VaR 0.95 [Y] TVaR 0.95 [Y] VaR 0.99 [Y] TVaR 0.99 [Y] VaR [Y] TVaR [Y] Table 1 shows that the minimum ruin probability and the optimal retention level decrease as the initial reserves are increased. Considering that the insurer retains precisely that optimal level that minimizes the ruin probability, the measures analyzed in the table show a decrease in all cases when the initial reserves are higher. As a particular case, the results show that if the 18

20 insurer has zero initial reserves, the best option is not to reinsure k = 1) and to retain all the business this result is consistent with the one obtained in Castañer et al. 2012)). The second optimization problem appears when the insurer considers the threshold proportional reinsurance option then, 4.16) is a particular case). Let ψb,k 1,k 2 ) be the ruin probability as a function of b,k 1,k 2 ) when u is considered to be a parameter. Thus, the problem is, min b,k 1,k 2, ψb,k 1,k 2 ) 4.17) 0.2 < k 1 1, 0.2 < k 2 1 where the ruin probability is calculated with 3.12). This problem is solved numerically with the function NMinimize of the software Mathematica. Table 2 includes the optimum b,k1,k 2 ) with the corresponding minimum probability of ruin for different values of u. We have also included E[Y], V[Y], VaR p [Y] and TVaR p [Y] as in Table 1 for proportional reinsurance. Table 2: Minimum probabilities of ruin with threshold proportional reinsurance and E[Y], V[Y], VaR p[y] and TVaR p[y] u b,k1,k 2 ) ψb,k1,k 2 ) E[Y] V[Y] VaR0.95[Y] TVaR0.95[Y] VaR0.99[Y] TVaR0.99[Y] VaR0.995[Y] TVaR0.995[Y] , 1, ) , 1, ) , 1, ) , 1, ) , 1, ) , 1, ) , 1, ) In this second optimization, the results in Table 2 show that the optimal point slightly varies in spite of the increase in the initial level of reserves. However, as expected, the minimal ruin probability decreases when the initial reserves increase. At the optimal point, the behaviour of the expectation, the variance, the VaR and the TVaR is not monotone with respect to the initial reserves. All these risk measures slightly increase, from u = 0, and then slowly decrease as the initial reserves are increased. With the threshold proportional reinsurance, the insurer can always obtain a lower ruin probability than with the proportional one with a constant retention level). In Table 3, the differences of these two minimum probabilities of ruin the first one attained with proportional reinsurance and the second one attained with threshold proportional reinsurance) are shown, 19

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