The Compound Poisson Surplus Model with Interest and Liquid Reserves: Analysis of the Gerber Shiu Discounted Penalty Function

Transcription

1 Methodol Comput Appl Probab (29) 11: DOI 1.17/s The Compound Poisson Surplus Model with Interest Liquid Reserves: Analysis of the Gerber Shiu Discounted Penalty Function Jun Cai Runhuan Feng Gordon E. Willmot Received: 1 January 27 / Revised: 25 June 27 / Accepted: 7 September 27 / Published online: 1 November 27 Springer Science + Business Media, LLC 27 Abstract We modify the compound Poisson surplus model for an insurer by including liquid reserves interest on the surplus. When the surplus of an insurer is below a fixed level, the surplus is kept as liquid reserves, which do not earn interest. When the surplus attains the level, the excess of the surplus over the level will receive interest at a constant rate. If the level goes to infinity, the modified model is reduced to the classical compound Poisson risk model. If the level is set to zero, the modified model becomes the compound Poisson risk model with interest. We study ruin probability other quantities related to ruin in the modified compound Poisson surplus model by the Gerber Shiu function discuss the impact of interest liquid reserves on the ruin probability, the deficit at ruin, other ruin quantities. First, we derive a system of integro-differential equations for the Gerber Shiu function. By solving the system of equations, we obtain the general solution for the Gerber Shiu function. Then, we give the exact solutions for the Gerber Shiu function when the initial surplus is equal to the liquid reserve level or equal to zero. These solutions are the key to the exact solution for the Gerber Shiu function in general cases. As applications, we derive the exact solution for the zero discounted Gerber Shiu function when claim sizes are exponentially distributed the exact solution for the ruin probability when claim sizes have Erlang(2) distributions. Finally, we use numerical examples to illustrate the impact of interest liquid reserves on the ruin probability. J. Cai (B) R. Feng G. E. Willmot Department of Statistics Actuarial Science, University of Waterloo, Waterloo, Ontario N2L 3G1, Canada jcai@math.uwaterloo.ca R. Feng r3feng@math.uwaterloo.ca G. E. Willmot gewillmo@math.uwaterloo.ca

2 42 Methodol Comput Appl Probab (29) 11: Keywords Ruin probability Deficit at ruin Surplus just before ruin Gerber Shiu function Interest force Liquid reserve Defective renewal equation Volterra equation of the second kind AMS 2 Subject Classification Primary 91B3; Secondary 91B7 1 Introduction We consider a surplus model for an insurer. Assume that the initial surplus of an insurer at time is u. LetN(t) be the number of claims to be incurred by the insurer in the time interval (, t {N(t), t } be a homogeneous Poisson process with intensity λ>. Let the size or amount of the nth claim be Y n {Y n, n 1} be a sequence of i.i.d. nonnegative rom variables with common distribution function P(x) = 1 P(x), which satisfies P() = has a mean μ = P(x)dx >. Furthermore, assume that {Y n, n 1} {N(t), t } are independent the insurer receives premiums continuously at a constant rate of c >. Thus, the accumulated amount of claims up to time t is Z (t) = N(t) n=1 Y n the surplus of the insurer at time t is u + ct Z (t). (1.1) Furthermore, if all the positive surplus of the insurer receives interest at an interest force r >, or equivalently, at an interest rate e r 1 >, then the surplus of the insurer at time t becomes ue rt + c s t t e r(t x) dz(x), (1.2) s t = (e rt 1)/r is the accumulated value at time t of one unit of money invested at time. The surplus model (1.1) is the classical risk model without any investment. The surplus model (1.2) is a risk model with interest, in which all the positive surplus of an insurer is invested into a risk-free market. Model (1.2) has been studied extensively in the literature. Sundt Teugels (1995, 1997) gave a detailed study of ruin probability in model (1.2). Yang Zhang (21a,b) studied the distributions of the surplus just before ruin the deficit at ruin, respectively, in model (1.2). Further, Yang Zhang (21c) discussed the joint distribution of the surplus just before ruin the deficit at ruin in model (1.2). To adapt a surplus model closer to reality, we employ the idea from Embrechts Schmidli (1994). Assume that an insurer is allowed to make investment only if the insurer s surplus attains a level of only the excess of the surplus over the level will receive interest at the interest force r. The surplus below the level is kept as liquid reserves, which do not earn interest. Let U(t) denote the surplus of an insurer at time t with the liquid reserve level the interest force r. Then, the surplus process {U(t), t } satisfies the following stochastic differential equations: { du(t) = cdt + ru(t)dt dz(t) if U(t), (1.3) du(t) = cdt dz(t) if U(t) <, with U() = u.

3 Methodol Comput Appl Probab (29) 11: Mathematically, models (1.1) (1.2) are the special cases of model (1.3) when = =, respectively. In addition, model (1.3) can also be interpreted as follows. Assume that an insurer deposits all positive surplus in a bank account. The account pays interest only when the surplus attains the level interest is paid only on the excess of the surplus over the level at the interest force r. In this paper, we consider ruin probability other quantities related to ruin in model (1.3) discussthe impactofinterestliquid reservesonthesequantities. We study these quantities by defining the Gerber Shiu function for model (1.3). The Gerber Shiu function is the expected value of a discounted penalty function at ruin was introduced by Gerber Shiu (1997, 1998) for model (1.1). The function has become a stard approach for the study of ruin theory in a variety of risk models. The Gerber Shiu function includes ruin probability, the distribution functions of the deficit at ruin the surplus just before ruin, many other interesting quantities related to ruin as special cases. See, for example, Cai Dickson (22) for the Gerber Shiu function in model (1.2) Cai(24) for the Gerber Shiu function in the compound Poisson surplus process with risky investments. To define the Gerber Shiu function for model (1.3), we first introduce the following notations. Let T be the ruin time of the surplus process {U(t), t }, namely, T = min{t : U(t) <} T = if U(t) for all t. Thus, U(T) is the deficit at ruin. We denote the surplus just before ruin by U(T ). Notice that U(T) > U(T ). Then the Gerber Shiu function for model (1.3) is defined as m(u) = E [ e δt w(u(t ), U(T) ) I(T < ) U() = u, (1.4) w(x, y), x, y >, is a nonnegative function denoting the penalty due at ruin; δ is a constant; I(A) is the indicator function of a set A. Throughout this paper, we assume that c >λμ lim u m(u) =. The rest of the paper is organized as follows. In Section 2, wederiveasystem of integral integro-differential equations satisfied by the Gerber Shiu function. By solving the system of equations, we obtain the general solution for the Gerber Shiu function based on the Volterra equations of the second kind defective renewal equations. In Section 3, we give the exact solutions for the Gerber Shiu function when the initial surplus is equal to the liquid reserve level or equal to zero. These solutions are the key to the exact solution for the Gerber Shiu function in general cases. As applications, in Section 4, we give the exact solution for the zero discounted Gerber Shiu function when claim sizes are exponentially distributed, in Section 5, we give the ruin probability with Erlang(2) claim sizes. Finally, in Section 6, we illustrate these results by computing the ruin probability in model (1.3) discuss the effect of interest liquid reserves on the ruin probability. 2 Integral Equations General Solutions Wefirstpoint outthat thesamplepathofthe Gerber Shiufunctionm(u) is different when u u <.Ifu, the surplus process begins with interest income thus curves until the next claim arrives. If u <, the surplus process starts without investment, which means that the surplus increases linearly at the rate c. As it accumulates to the level, the surplus process turns to a curve reflecting interest

4 44 Methodol Comput Appl Probab (29) 11: income on the excess of the surplus over the level. Thus, we distinguish the two cases by writing m(u) = { m1 (u) if u, m 2 (u) if u <. Indeed,wewillseethatm 1 (u) m 2 (u) satisfy different integral integrodifferential equations. A common technique used in ruin theory for deriving integral equations for ruin quantities is to condition on the arrival time the amount of the first claim to obtain equations based on the renewal nature of a surplus process. In risk model (1.3), ruin can occurs only when a claim occurs. The surplus process is regenerated at the time points of claims. Furthermore, the liquid reserve level sets up a benchmark for different scenarios the surplus process might be regenerated with or without investment. Using the technique, we now derive integral equations satisfied by m 1 (u) m 2 (u). In the case u, there are three possibilities when the first claim occurs: (1) if the claim size is smaller than (u )e rt + c s t, then the surplus process is only pulled down to a point with a new initial surplus remaining above the level ; (2) if the claim size is between (u )e rt + c s t (u )e rt + c s t +, ruin does not occur but the surplus process will restart without investment; (3) if the claim size is larger than (u )e rt + c s t +, ruin occurs penalty is exercised. Thus, by denoting h 1 (t) = (u )e rt + c s t + conditioning the time amount of the first claim, we obtain for u, m 1 (u) = [ h1(t) λe λt e δt m 1 (h 1 (t) y)dp(y) h 1(t) h1(t) + e δt m 2 (h 1 (t) y)dp(y) h 1(t) + e δt w(h 1 (t), y h 1 (t))dp(y) dt. (2.1) In the case u <,let t = ( u)/c, which is the time when the surplus attains the level if no claim occurs prior to time t. Thus, there are two possibilities when the first claim occurs before time t : (1) if the claim size is smaller than u + ct, ruin does not occur the surplus process regenerates still without investment; (2) if the claim size is larger than u + ct, ruin occurs. On the other h, there are three possibilities when the first claim occurs after time t : (i) if the claim size is below c s t t, the surplus remains with investment; (ii) if the claim size is between c s t t c s t t +, ruin does not occur but the surplus drops to the case without investment; (iii) if the claim size is larger than c s t t +, the surplus process would break down ruin occurs. Hence, by denoting h 2 (t) = c s t t +

5 Methodol Comput Appl Probab (29) 11: conditioning on the time amount of the first claim, we get for u <, m 2 (u) = t [+ct λe λt e δt m 2 (u + ct y)dp(y) + + u+ct λe λt t h 2(t) Define the function H(x) for x by e δt w(u + ct, y u ct)dp(y) dt [ h2(t) e δt m 1 (h 2 (t) y)dp(y) h2(t) + e δt m 2 (h 2 (t) y)dp(y) h 2(t) + e δt w(h 2 (t), y h 2 (t))dp(y) dt. (2.2) H(x) = x w(x, y x)dp(y), (2.3) which is the expected penalty when ruin occurs the surplus just before ruin is x. Furthermore, denote by γ(x) = x x m 2 (x y)dp(y) + H(x) for x. (2.4) Thus, with the functions H(x) γ(x), Eqs are respectively reduced to, when u, [ h1(t) m 1 (u) = λe (λ+δ)t m 1 (h 1 (t) y)dp(y) + γ(h 1 (t)) dt, (2.5) when u <, m 2 (u) = t λe (λ+δ)t t + λe (λ+δ)t [+ct m 2 (u + ct y)dp(y) + H(u + ct) dt [ h2(t) m 1 (h 2 (t) y)dp(y) + γ(h 2 (t)) dt. (2.6) Then, changing variables of x = h 1 (t) in Eq. 2.5 noticing dx = e rt (r(u ) + c)dt = (r(x ) + c)dt, we obtain when u, m 1 (u) = λ(r(u ) + c) (λ+δ)/r (r(x ) + c) 1 (λ+δ)/r u [ x m 1 (x y)dp(y) + γ(x) dx. (2.7) Furthermore, changing variables of x = u + ct in the first integral with respect to t from to t in Eq. 2.6 with dx = cdt changing variables of x = h 2 (t) in the second

6 46 Methodol Comput Appl Probab (29) 11: integral with respect to t from t to in Eq. 2.6 with dx = ce r(t t) dt = (r(x ) + c)dt, we obtain when u <, m 2 (u) = λ [ x c e(λ+δ)u/c e (λ+δ)x/c m 2 (x y)dp(y) + H(x) dx u ( ) 1 (λ+δ)/r + λc (λ+δ)/r e (λ+δ)( u)/c r(x ) + c [ x m 1 (x y)dp(y) + γ(x) dx. (2.8) Thus, by differentiating Eqs , we give integro-differential equations for m 1 (u) m 2 (u) in the following theorem. Theorem 2.1 When u, m 1 (u)= λ + δ r(u ) + c m λ 1(u) r(u ) + c when u <, m 2 (u) = λ + δ m 2 (u) λ [ c c [ m 1 (u y)dp(y)+γ(u) (2.9) m 2 (u y)dp(y) + H(u). (2.1) It is easy to see from setting u = in Eqs that m 1 (u) m 2 (u) are continuous with m 1 () = m 2 (). (2.11) We point out that the integro-differential equation (2.1)form 2 (u) is independent of m 1 (u). However, the solution of m 2 (u) is subject to the boundary condition of Eq. 2.11, which is involved with m 1 (). Replacing u by t in Eq. 2.9 then rearranging it, we see that for t, [ t (r(t ) + c)m 1 (t) = (λ + δ)m 1(t) λ m 1 (t y)dp(y) + γ(t). Integrating the above equation from to u with respect to t yields, for u, (r(u ) + c)m 1 (u) cm 1 () r [ = (λ + δ) m 1 (t)dt λ = (λ + δ) which means that for u, t [ m 1 (t)dt λ P(u y)m 1 (y)dy + m 1 (u) = l 1 (u) + m 1 (t)dt m 1 (t y)dp(y)dt + γ(t)dt γ(t)dt, k(u, t)m 1 (t)dt, (2.12)

7 Methodol Comput Appl Probab (29) 11: l 1 (u) = cm 1() r(u ) + c λ r(u ) + c γ(t)dt (2.13) r + δ + λ P(u t) k(u, t) =. r(u ) + c The integral equation (2.12) is of the Volterra equation of the second kind. Note that l 1 (u) is continuous in u k(u, t) is continuous in u t when P(x) is a continuous distribution function. Thus, by Theorem 3.3 of Linz (1985), we conclude that when P(x) is a continuous distribution function, the unique solution of m 1 (u) is expressed as, for u, m 1 (u) = l 1 (u) + K(u, s)l 1 (s)ds, (2.14) the resolvent kernel K(u, s) = k m (u, s), u > s k m (u, s) = s m=1 k(u, t)k m 1 (t, s)dt, m = 2, 3,, u > s with k 1 (u, s) = k(u, s). Furthermore, note that the function l 1 (u) is involved with m 1 () γ(u) hence m 2 (u). Thus, once m 1 () m 2 (u) are determined, the exact solution for m 1 (u) is given by Eq To obtain a general solution for m 2 (u), we replace u by x in Eq. 2.1 then integrate both sides of the equation from to u with respect to x. Thus, we obtain for u <, m 2 (u) m 2 () = λ + δ c = λ + δ c which implies that for u <, m 2 (x)dx λ [ x c m 2 (x)dx λ [ P(y)m 2 (u y)dy + c m 2 (u) = l 2 (u) + l 2 (u) = m 2 () λ c m 2 (x y)dp(y)dx + H(x)dx H(x)dx, ν(u, t)m 2 (t)dt, (2.15) ν(u, t) = λ c P(u t) + δ c. H(y)dy (2.16)

8 48 Methodol Comput Appl Probab (29) 11: Again, the integral equation 2.15 is of the Volterra equation of the second kind. Notice that l 2 (u) is continuous in u < ν(u, t) is continuous in t u < when P(x) is a continuous distribution function. Therefore, Theorem 3.3 of Linz (1985) meansthatwhenp(x) is a continuous distribution function, the unique solution for m 2 (u) is expressed as, for u <, the resolvent kernel ν m (u, s) = s m 2 (u) = l 2 (u) + V(u, s) = V(u, s)l 2 (s)ds, (2.17) ν m (u, s), u > s m=1 ν(u, t)ν m 1 (t, s)dt, m = 2, 3,, u > s with ν 1 (u, s) = ν(u, s). Note that l 2 (u) in Eq involves only one unknown constant m 2 (). However, by setting u = in Eq using Eqs , we have m 1 ()=m 2 () λ H(y)dy+m 2 () V(, s)ds λ s V(, s) H(y)dyds, c c which implies that m 2 () = m 1() + (λ/c) H(y)dy + (λ/c) V(, s) s H(y)dyds 1 + V(, s)ds. (2.18) Therefore, the exact solution for m 2 (u) is available by Eq once m 1 () is given. In addition, the solution of Eq for m 1 (u) is subject to m 1 () m 2 (u). However, when we substitute the general solution of Eq for m 2 (u) in Eq. 2.14, the solution of Eq for m 1 (u) is subject to only one unknown constant m 1 (). Hence the solution to m 1 () is the key to obtaining the exact solutions of m 1 (u) m 2 (u). 3 Exact Solutions to m 1 () m 2 () for General Claim Sizes We can derive the exact solutions of m 1 () m 2 () inthecaseofδ =. Thiscase includes many important functions such as ruin probability, the distribution functions of the deficit at ruin, the surplus just before ruin, the claim amount causing ruin. To do so, letting δ =, x = u, (x) = m 1 (x + ) in Eq. 2.12, we obtain for x, (x) = c () rx + c λ rx + c x γ(t + )dt + x k (x, t)(t)dt, (3.1)

9 Methodol Comput Appl Probab (29) 11: k r + λ P(x t) (x, t) = k(x +, t + ) =. rx + c Equation 3.1 is the same as Eq. 2.6 of Cai Dickson (22) if we denote γ(t + ) in Eq. 3.1 by A(t) as in Eq. 2.6 of Cai Dickson (22). Also, note that r in Eq. 3.1 is denoted by δ in Eq. 2.6 of Cai Dickson (22). We denote by γ(t) = ˆp 1 (t) = e tx γ(x)dx, (3.2) e tx dp 1 (x) = e tx P(x)dx/μ. (3.3) Thus, by Eqs of Cai Dickson (22), we obtain ( λm A β(rt) exp ct + λμ ) t φ 1(rs)ds dt () = c ( exp ct + λμ ), t φ 1(rs)ds dt m A = β(t) = 1 φ 1 (t) = m A γ(t + )dt, e tx γ(x + )dx = et γ(t) m A, e tx dp 1 (x) = ˆp 1 (t). Therefore, m 1 () = () implies that λ ( γ(rt) exp ct + rt + λμ ) t ˆp 1(rs)ds dt m 1 () = c ( exp ct + λμ ). (3.4) t ˆp 1(rs)ds dt Recall that when δ =, the integral equation (2.15) becomes a defective renewal equation, namely, for u <, m 2 (u) = l 2 (u) + λμ c m 2 (u y)dp 1 (y), (3.5) P 1 (y) = y P(x)dx/μ is the equilibrium distribution function of the distribution P. Thus, by Theorem of Resnick (22), the unique solution to m 2 (u) of Eq. 3.5 is given by, for u <, m 2 (u) = l 2 (u y)dg(y), (3.6)

10 41 Methodol Comput Appl Probab (29) 11: G(y) = ( λμ c n= ) n P n 1 (y) (3.7) is a bounded continuous increasing function P1 n is the n-fold convolution of the distribution P 1 with itself. Note that (1 λμ/c)g(x) is a compound geometric distribution function. We point out that the integral in Eq. 3.6 with respect to G(y) is on the interval [, u. ByEqs.2.11, 2.16,3.6, we have ( m 1 () = m 2 () λ x ) H(y)dy dg(x). c Hence, m 2 () = m 1() + (λ/c) x H(y)dy dg(x). (3.8) G() Therefore, the exact solution for m 2 (u) is available by Eq. 3.6 once m 1 () is given. We point out that the function γ(t) in Eq. 3.4 involves the function m 2 (t), whichin turn depends on m 1 (). Hence, we can substitute γ(t) with an expression involving m 1 () into Eq. 3.4 obtain the exact solution for m 1 () as follows. By Eqs , we have, γ(s) = e sx x x H(s) = m 2 (x y)dp(y)dx + H(s), (3.9) By Eqs , we obtain for u <, e sx H(x)dx. (3.1) m 2 (u) = m 2 ()G(u) B(u), (3.11) B(u) = λ c y Thus, by Eqs. 3.9, 3.11,3.8, we have, γ(s) = m 2 () e sx x e sx x x x H(x)dx dg(y). (3.12) G(x y)dp(y)dx B(x y)dp(y)dx+ H(s) = m 1 ()a(s) + b(s), (3.13)

11 Methodol Comput Appl Probab (29) 11: a(s) = b(s) = λ c e sx x x x G(x y)dp(y)dx, (3.14) G() H(y)dydG(x) e sx x x Substituting Eq in Eq. 3.4 gives λ (m 1()a(rt) + b(rt)) exp m 1 () = c exp G() e sx x x G(x y)dp(y)dx B(x y)dp(y)dx+ H(s). (3.15) ( ct + rt + λμ t ) ( ct + λμ t ˆp 1(rs)ds ) ˆp 1(rs)ds dt. dt Solving the above equation gives the exact solution for m 1 (). Then, the exact solution for m 2 () comes from Eq We present these exact solutions in the following theorem. Theorem 3.1 Let δ =. Then, m 1 () = ( ct + rt + λμ t ) ˆp 1(rs)ds dt ) dt λ b(rt) exp ( ) c λ a(rt) e rt exp ( ct + λμ t ˆp 1(rs)ds (3.16) λ ( b(rt) exp ct + rt + λμ ) t ˆp 1(rs)ds dt m 2 () = G() ( ) c λ a(rt) e rt exp ( ct + λμ ) t ˆp 1(rs)ds dt + (λ/c) x H(y)dy dg(x), (3.17) G() the functions a(s) b(s) are given in Eqs , respectively. The two exact solutions are useful when one determines the exact solutions for the Gerber Shiu function from its general solutions. We will illustrate the applications of Theorem 3.1 in the following sections. 4 Exact Solution to the Zero Discounted Gerber Shiu Function with Exponential Claim Sizes In this section, we assume that P(x) = 1 e βx, x >, β>, namely, the claim size distribution P is an exponential distribution with mean μ = 1/β.

12 412 Methodol Comput Appl Probab (29) 11: In view of Eq. 2.4 the exponential density function, Eq. 2.9 is reduced to, for u, ( ) βλe βu m 1 (x)e βx dx + m 2 (x)e βx dx = (λ + δ)m 1 (u) (r(u ) + c)m 1 (u) λh(u). Differentiating the above equation with respect to u, we obtain for u, (r(u ) + c)m 1 (u) + (βr(u ) + r λ δ + cβ)m 1 (u) βδm 1(u) = λ(h (u) + β H(u)). (4.1) Similarly, Eq. 2.1 is reduced to, for u <, λ c e βu βe βy m 2 (y)dy = λ + δ c m 2 (u) m 2 (u) λ c H(u). Differentiating the above equation with respect to u, we obtain for u <, m cβ λ δ 2 (u) + m 2 c (u) δβ c m 2(u) = λ ( H (u) + β H(u) ) /c. (4.2) The general solutions for the second order differential equations (4.1) (4.2) are both available. See, for example, Polyanin Zaitsev (1995). However, the exact solutions for the equations can be determined only when the boundary values of m 1 () m 2 () are given. Hence, in the case of δ = or the zero discounted Gerber Shiu function, we can derive the exact solutions for m 1 (u) m 2 (u) based on the general solutions of m 1 (u) m 2 (u) Theorem 3.1. To do so, we let δ = in the rest of this section. Then, Eqs are reduced to, for u, for u <, m 1 (u) + f (u)m 1 (u) = h(u), (4.3) m cβ λ 2 (u) + m 2 (u) = l(u), (4.4) c f (u) = β + r λ r(u ) + c, h(u) = λ ( H (u) + β H(u) ), (4.5) r(u ) + c l(u) = λ ( H (u) + β H(u) ). (4.6) c By formula on page 212 of Polyanin Zaitsev (1995), we know that the general solutions to Eqs are given by, for u, x ) m 1 (u) = C 1 + e (C F(x) 2 + e F(t) h(t)dt dx, (4.7)

13 Methodol Comput Appl Probab (29) 11: for u <, m 2 (u) = C 3 + e S(x) (C 4 + C 1, C 2, C 3,C 4 are arbitrary constants, F(x) = x x ) e S(t) l(t)dt dx, (4.8) f (y)dy = β(x ) + (1 λ/r) ln (1 + r(x )/c), (4.9) x cβ λ S(x) = dy = (β λ/c)x. (4.1) c Then, we can obtain the exact solutions for m 1 (u) m 2 (u) by using Theorem 3.1 the general solutions of Eqs In doing so, letting u be in Eq. 4.7 immediately gives C 1 = m 1 (). (4.11) Furthermore, letting u in Eq. 4.7 by m 1 ( ) =, we find C 2 = m 1() e F(x) x ef(t) h(t)dtdx. (4.12) e F(x) dx To determine C 3 C 4,wesetu = in Eq. 4.8 obtain Then, by Eqs. 2.11, 4.8,4.13, we have which yields m 1 () = m 2 () + C 4 e S(x) dx + C 4 = m 1() m 2 () C 3 = m 2 (). (4.13) e S(x) x e S(x) dx x e S(x) e S(t) l(t)dtdx, es(t) l(t)dtdx. (4.14) Hence, the exact solutions for m 1 (u) m 2 (u) come from the general solutions of Eqs formulas ( ). We state these exact solutions in the following theorem. Theorem 4.1 Let δ = P(x) be an exponential distribution with mean 1/β. Then for u, m 1 (u) = m 1 () m 1() + e F(x) x ef(t) h(t)dtdx u e F(x) dx + e F(x) dx x e F(x) e F(t) h(t)dtdx, (4.15)

14 414 Methodol Comput Appl Probab (29) 11: for u <, m 2 (u) = m 2 () + m 1() m 2 () e (β λ/c)x x e(β λ/c)t l(t)dtdx ( 1 e (β λ/c)u ) 1 e (β λ/c) x + e (β λ/c)x e (β λ/c)t l(t)dtdx, (4.16) m 1 () m 2 () are given in Theorem Ruin Probability We denote the ruin probability with the exponential claim size distribution by ψ(u) = Pr{T < U() = u}. Using Theorem 4.1, we obtain the explicit solution to the ruin probability. In doing so, we notice that the function H(x) defined in Eq. 2.3 is reduced to H(x) = P(x) = e βx, x >, which, together with Eqs , implies that h(u) = l(u) =. Furthermore, by Eqs. 3.7, 3.3, 3.1, 3.12, 3.14, 3.15, we find that ( 1 G(y) = 1 λ ) 1 λ/(cβ) cβ e (β λ/c)y, y, (4.17) ˆp 1 (s) = β β + s, (4.18) H(s) = 1 β + s e (β+s), (4.19) B(y) = λ cβ λ λ cβ λ e (β λ/c)y, y (4.2) a(s) = cβe (β+s)( e β e λ/c) (β + s)[cβ λe (β λ/c), (4.21) (λ βc)e (β+s) b(s) = (β + s)(λe β βce λ/c ). (4.22) Thus, by Eqs , we can obtain ψ() ψ() correspondingly. Hence, by Theorem 4.1, we give the exact solution for the ruin probability with exponential claim sizes in the following theorem. Theorem 4.2 Let P(x) be an exponential distribution function with mean 1/β. Then, the ruin probability in risk model (1.3) is given by, for u, ψ(u) = ψ() u (1 + r(x )/c) 1+λ/r e β(x ) dx (1 + r(x, (4.23) )/c) 1+λ/r e β(x ) dx for u <, ψ(u) = ψ() ψ() e (β λ/c) ψ() ψ() + 1 e (β λ/c) 1 e (β λ/c) e (β λ/c)u, (4.24)

15 Methodol Comput Appl Probab (29) 11: λ(βc λ)e (β λ/c) [ λ/r 1e 1 + r(x )/c β(x ) dx ψ()= λ(βc λ)e (β λ/c) [ λ/r 1e 1 + r(x )/c β(x ) dx + c[βc λe (β λ/c) ψ() = βc λ βc λe ψ() + λ(1 e (β λ/c) ) (β λ/c) βc λe. (β λ/c) We point out that for =, letting in formula (4.24), ψ(u) is reduced to the ruin probability in the classical risk model (1.1), namely for u, ψ(u) = ψ() e (β λ/c)u = λ cβ e (β λ/c)u. (4.25) For =, setting = in formula (4.23), ψ(u) is reduced to the ruin probability in model (1.2), namely for u, ψ(u) = λ u (1 + rx/c)λ/r 1 e βx dx cβ (1 + rx/c) λ/r e βx dx. (4.26) Formula (4.26) is consistent with that on page 136 of Gerber (1979), namely for u, ψ(u) = Ɣ (λ/r, uβ + cβ/r), (4.27) Ɣ (λ/r, cβ/r) + (r/λ)(cβ/r) λ/r e cβ/r Ɣ(α, z) = z y α 1 e y dy, α>, z is the incomplete gamma function. In addition, for >, the formulas for ψ(u) are given in Eqs Distribution of the Deficit at Ruin Let w(x, y) = I(y z), z. The Gerber Shiu function m(u) = m(u, z) defined in Eq. 1.4 is reduced to the (defective) distribution of the deficit at ruin given the initial surplus being u. We denote the (defective) distribution of the deficit at ruin with the exponential claim size distribution by G(u, z) = Pr{ U(T) z, T < U() = u}. (4.28)

16 416 Methodol Comput Appl Probab (29) 11: In this case, we can easily find that H(u) = (1 e βz )e βu,whichleadstothefact that h(u) = l(u) =. In order to implement formulas in Theorem 4.1, we obtain in this case ( 1 G(y) = 1 λ ) 1 λ/(cβ) cβ e (β λ/c)y, y, (4.29) ˆp 1 (s) = β β + s, (4.3) H(s) = 1 e βz e (β+s), (4.31) β + s B(y) = (1 e βz ) ( λ cβ λ λ ) cβ λ e (β λ/c)y, y, (4.32) a(s) = cβe (β+s)( e β e λ/c) (β + s)[cβ λe (β λ/c), (4.33) b(s) = (1 e βz (λ βc)e (β+s) ) (β + s)(λe β βce λ/c ). (4.34) Note that the above functions of G(y), ˆp 1 (s), a(s) are the same as their counterparts of Eqs. 4.17, 4.18, 4.21 for the ruin probability while the above functions of H(u), H(s), B(y), b(s) are equal to their counterparting multiplied by a constant 1 e βz. Thus, by Eqs , we can obtain m 1 () = m 1 (, z) m 2 () = m 2 (, z) correspondingly. Hence, by Theorem 4.1, we give the exact solution for the distribution function of the deficit at ruin in the following theorem. Theorem 4.3 Let P(x) be an exponential distribution function with mean 1/β. Then G(u, z),the(defective) distribution function of the deficit at ruin in risk model Eq. 1.3, is given by, for u, G(u, z) = m 1(, z) u (1 + r(x )/c) 1+λ/r e β(x ) dx (1 + r(x, (4.35) )/c) 1+λ/r e β(x ) dx for u <, G(u, z) = m 1(, z) m 2 (, z) e (β λ/c) + m 2(, z) m 1 (, z) e (β λ/c)u, (4.36) 1 e (β λ/c) 1 e (β λ/c) m 1 (, z) m 2 (, z) can be obtained from Eqs , respectively, with the given functions of Eqs It is easy to see from Theorem 3.1 that m 1 (, z) = (1 e βz )ψ(), m 2 (, z) = (1 e βz )ψ().

17 Methodol Comput Appl Probab (29) 11: Therefore, by Theorems , we conclude that G(u, z) = (1 e βz )ψ(u), u. (4.37) Relationship (4.37) is consistent with that in the classic compound risk model. 5 Exact Solution to the Ruin Probability with Erlang(2) Claim Sizes In this section, we assume that the claim size distribution P is an Erlang(2) distribution with P(y) = 1 e κy κye κy, y >,κ >. We denote the ruin probability with the Erlang(2) claim sizes by ϕ(u) = Pr{T < U() = u}. A key element in determining the ruin probability is the boundary value ϕ() given in Theorem 3.1. The auxiliary functions needed in the theorem are simplified as follows, 1 G(y) = 1 2λ/(κc) ˆp 1 (s) = 1 κ 2 κ + s [ 1 (κ + s 1) 2 s 2 ( κ κ + s κ 2 (s 2 s 1 ) es1u (κ + s 2) 2 s 1 κ 2 (s 1 s 2 ) es2u, (5.1) ) 2, (5.2) H(s) = κs + s + 2κ + κ2 (κ + s) 2 e (κ+s), (5.3) 2λ B(u) = κc 2λ + 2λκ + λs 1 e s1u + 2λκ + λs 2 e s2u. (5.4) c(s 1 s 2 )s 1 c(s 2 s 1 )s 2 Thus, the functions a(s) b(s) defined in Eqs are obtained by Eqs hence ϕ() is obtained by Eq in Theorem 3.1. Using the arguments similar to those for (4.1), we obtain for u, [r(u ) + cϕ (u) +[2r λ + 2κr(u ) + 2κcϕ (u) +[2κ(r λ) + κ 2 r(u ) + κ 2 cϕ (u) =. Let z = u + c/r ϕ (u) = e κz g(z).theng(z) satisfies rz[g (z) 2κg (z) + κ 2 g(z)+[2κrz + 2r λ[g (z) g(z) +[2κ(r λ) + κ 2 rzg(z) =, which implies that rzg (z) + (2r λ)g (z) λκg(z) =. Making another change of variables z = rt 2 /(4λκ) g(z) = t λ/r 1 h(t) yields g (z) = 2λκ ( ) λ r r 1 t λ/r 3 h(t) + 2λκ r tλ/r 2 h (t), g (z) = 4λ2 κ 2 r 2 t 2 [( )( ) ( ) λ λ 2λ r 1 r 3 t λ/r 3 h(t) + r 3 t λ/r 2 h (t) + t λ/r 1 h (t),

18 418 Methodol Comput Appl Probab (29) 11: the differentiations are taken with respect to the variables in corresponding brackets. Hence we arrive at the following modified Bessel differential equation h (t) + 1 (λ/r 1)2 t h (t) [1 + h(t) =, t 2 which has two independent solutions, I λ/r 1 (t) K λ/r 1 (t), modified Bessel functions of the first second kinds respectively. See, for example, Andrews (1984). For any real number ν, the modified Bessel functions can be computed by I ν (t) = k= (t/2) 2k+ν k! Ɣ(k + ν + 1), ( π ) I ν (t) I ν (t) K ν (t) =. 2 sin νπ Therefore by transforming back to original variables, omitting constant coefficients letting lim u ϕ(u) =, we obtain the general solution to ϕ(u) is D 1 D 2 are arbitrary constants, Z 1 (u) = Z 2 (u) = u u ϕ(u) = D 1 Z 1 (u) + D 2 Z 2 (u), u, (5.5) e κy (y + c/r) λ/(2r) 1/2 I λ/r 1 [ 2 r e κy (y + c/r) λ/(2r) 1/2 K λ/r 1 [ 2 r λκ(r(y ) + c) dy λκ(r(y ) + c) dy. Similarly, we can show that for u <, ϕ(u) satisfies the following third-order differential equation cϕ (u) + (2κc λ)ϕ (u) + (κ 2 c 2λκ)ϕ (u) =, u <, which admits the general solution ϕ(u) = D 3 + D 4 e s1u + D 5 e s2u, u <, (5.6) D 3, D 4,D 5 are arbitrary constants, s 1 = (λ 2κc) + λ 2 + 4λκc 2c s 2 = (λ 2κc) λ 2 + 4λκc. 2c To obtain the exact solution for ϕ(u), we need to determine the five constants D 1, D 2, D 3, D 4, D 5 in Eqs Notice that the value of ϕ() is determined by substituting Eqs in Eq of Theorem 3.1. Thus, the continuous junction condition ϕ(+) = ϕ( ) = ϕ()

19 Methodol Comput Appl Probab (29) 11: the smooth junction condition ϕ (+) = ϕ ( ) yield the following three equations for the constants D 1 Z 1 () + D 2 Z 2 () = ϕ(), (5.7) D 3 + D 4 e s1 + D 5 e s2 = ϕ(), (5.8) D 1 Z 1 () + D 2 Z 2 () = D 4 s 1 e s1 + D 5 s 2 e s2. (5.9) Furthermore, substituting Eq. 5.6 for m 2 (u) in Eq. 2.1 equating all terms involving ue κu e κu to zero yields κ κ D 3 + D 4 + D 5 = 1 (5.1) κ + s 1 κ + s 2 ( ) κ 2 ( ) κ 2 D 3 + D 4 + D 5 = 1. (5.11) κ + s 1 κ + s 2 Solving the five linear equations ( ), we can determine the five constants D 1, D 2, D 3, D 4, D 5 obtain the exact solution for ϕ(u). We summarize the exact solution for ϕ(u) in the following theorem. Theorem 5.1 Let P(x) be an Erlang(2) distribution with mean 2/κ. Then the ruin probability is given by ϕ(u) = D 1 Z 1 (u) + D 2 Z 2 (u), ϕ(u) = D 3 + D 4 e s1u + D 5 e s2u, u, u <, D 1 = s 1e s1 Z 2 ()D 4 + s 2 e s2 Z 2 ()D 5 ϕ()z 2 () Z 1 ()Z 2() Z 1 ()Z 2 (), D 2 = s 1e s1 Z 1 ()D 4 + s 2 e s2 Z 1 ()D 5 ϕ()z 1 () Z 1 ()Z 2 () Z 1 ()Z, 2() D 3 = 1 D 4 = κ D 4 κ D 5, κ + s 1 κ + s 2 [1 ϕ()κs 2 (κ + s 1 ) 2 κ 3 s 2 κs 2 e s1 (κ + s 1 ) 2 κ 3 s 1 + κs 1 e s2 (κ + s 2 ) 2, [1 ϕ()κs 1 (κ + s 2 ) 2 D 5 = κ 3 s 1 κs 1 e s2 (κ + s 2 ) 2 κ 3 s 2 + κs 2 e s1 (κ + s 1 ). 2 Here, ϕ() is determined by substituting Eqs in Eq of Theorem Impact of Liquid Reserves Interest on the Ruin Probability To illustrate those results derived in the previous sections demonstrate the effect of interest liquid reserves on the Gerber Shiu function, we use Theorems to compute ruin probabilities with exponential Erlang(2) claims sizes for

20 42 Methodol Comput Appl Probab (29) 11: different values of δ. We are also interested in comparing the ruin probabilities for Erlang(2) claim sizes with those for exponential claim sizes. To do so, we set model parameters as c = 2.2, λ = 1, μ = 2. Thus the parameter β for the exponential claim size distribution is β =.5 the parameter κ for the Erlang(2) claim size distribution is κ = 1. With the distribution parameters, the exponential claim size distribution has the same mean as the Erlang(2) claim size distribution. We use Theorem 4.2 to compute the ruin probabilities ψ(), ψ(15), ψ(3) with different values of the interest force r the liquid reserve level for the exponential claim sizes. The results are given in Tables 1 3. Furthermore, we use Theorem 5.1 to compute the ruin probabilities ϕ(), ϕ(15), ϕ(3) for the Erlang(2) claim sizes with the same values of r as for ψ(), ψ(15), ψ(3). The corresponding ruin probabilities are enclosed in brackets in Tables 1 3. As expected, the tables show that ruin probability decreases as the force of interest increases while ruin probability increases as the liquid reserve level increases. In particular, in Table 1,when, the ruin probabilities with exponential Erlang(2) claim sizes both tend to the ruin probability with the zero initial surplus in the classical compound Poisson risk model, which is λμ/c =.999. Furthermore, from the three tables, we see that for u =, the ruin probabilities with the exponential claim sizes are smaller than those with the Erlang(2) claim sizes. While, for u = 15 u = 3, the ruin probabilities with the exponential claim sizes are larger than those with the Erlang(2) claim sizes. In fact, we can prove that if κ>β, then the ruin probability ψ(u) with exponential claims the ruin probability ϕ(u) with Erlang(2) claim sizes satisfy ϕ(u) lim u ψ(u) =. Thus, for large values of u, ψ(u) >ϕ(u) if κ>β. Table 1 The impact of r on ψ() ϕ() r = = (.87167) (.84999) (.8193) (.79627) (.77734) (.7616) = (.87471) (.85551) (.82917) (.813) (.79467) (.78172) (.8921) (.88438) (.87553) (.872) (.8667) (.8631) (.957) (.89718) (.89353) (.8914) (.88993) (.88882) (.9466) (.93) (.9127) (.929) (.89962) (.89912) (.9783) (.9738) (.9693) (.9667) (.965) (.9637) (.993) (.991) (.9899) (.9898) (.9897) (.9897) (.999) (.999) (.999) (.999) (.999) (.999) =

21 Methodol Comput Appl Probab (29) 11: Table 2 The impact of r on ψ(15) ϕ(15) r = = (.185) (.1179) (.5121) (.2739) (.165) (.15) = (.1876) (.11824) (.5741) (.3196) (.1936) (.1247) (.2594) (.2121) (.14649) (.11489) (.9375) (.7852) (.3739) (.2838) (.25838) (.24353) (.23328) (.22558) (.33591) (.32434) (.31229) (.3544) (.379) (.29733) (.358) (.35487) (.35168) (.3499) (.34871) (.34783) (.36636) (.36622) (.3667) (.36599) (.36594) (.3659) (.36676) (.36676) (.36676) (.36676) (.36676) (.36676) = (.36676) (.36676) (.36676) (.36676) (.36676) (.36676) To prove this result, we denote lim x a(x)/b(x) = 1 by a(x) b(x). Itiswell known that the modified Bessel function I ν (x) satisfies I ν (x) ex 2π x, (6.1) Table 3 The impact of r on ψ(3) ϕ(3) r = = (.2281) (.611) (.8) (.16) (.4) (.2) = (.2439) (.688) (.97) (.21) (.6) (.2) (.4349) (.1923) (.545) (.21) (.88) (.43) (.7367) (.4881) (.2663) (.1662) (.1118) (.791) (.1475) (.8915) (.7291) (.6367) (.574) (.5274) (.13453) (.133) (.1261) (.12361) (.1221) (.1282) (.1458) (.14561) (.14541) (.14531) (.14523) (.14518) (.14634) (.14634) (.14634) (.14634) (.14634) (.14634) = (.14634) (.14634) (.14634) (.14634) (.14634) (.14634)

22 422 Methodol Comput Appl Probab (29) 11: which is independent of ν. See, for example, Eq. 33 (page 74) of Srivastava Kashyap (1982). Thus, K ν (x) lim =. (6.2) x I ν (x) Hence, it is easy to see from Eqs. 5.5, 6.1, 6.2, l Hopital s rule that ϕ(u) D 1 Z 1 (u) D 1 e κy (y + c/r) which implies that ϕ(u) D 1 u u (y +c/r) λ/(2r) 3/4 exp D 1 = D 1. λκr λ/(2r) 1/2 exp Therefore, by Eqs. 6.3, 4.23, l Hopital s rule, we obtain that ϕ(u) lim u ψ(u) = D lim u exp 4π r { 4π r 2 r λκ(r(y ) + c) } λκ(r(y ) + c) dy, { κy+ 2 } λκ(r(y ) + c) dy, (6.3) r (r(u ) + c) λ/(2r)+1/4 { (κ β)u + 2 } λκ(r(u ) + c), (6.4) r for some constant D. Obviously, when κ>β, the limit in Eq. 6.4 is zero. Furthermore, when κ β, the limit in Eq. 6.4 is. Thus, for large values of u, ψ(u) <ϕ(u) if κ β. Acknowledgements We thank the anonymous referee for his/her careful reading of the paper the helpful suggestions that improved the presentation of the paper. This research was supported by the Natural Sciences Engineering Research Council of Canada (NSERC). References L. C. Andrews, Special Functions for Engineers Applied Mathematicians. Macmillan: New York, J. Cai, Ruin probabilities penalty functions with stochastic rates of interest, Stochastic Processes Their Applications vol. 112 pp , 24. J. Cai, D. C. M. Dickson, On the expected discounted penalty function at ruin of a surplus process with interest, Insurance: Mathematics Economics vol. 3 pp , 22. P. Embrechts, H. Schmidli, Ruin estimation for a general insurance risk model, Advances in Applied Probability vol. 26 pp , H. U. Gerber, An Introduction to Mathematical Risk Theory. S. S. Heubner Foundation Monograph Series 8: Philadelphia, H. U. Gerber, E. S. W. Shiu, The joint distribution of the time of ruin, the surplus immediately before ruin, the deficit at ruin, Insurance: Mathematics Economics vol. 21 pp , H. U. Gerber, E. S. W. Shiu, On the time value of ruin, North American Actuarial Journal vol. 2 no. 1 pp , 1998.

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