Discounted probabilities and ruin theory in the compound binomial model

Insurance: Mathematics and Economics 26 (2000) 239 250 Discounted probabilities and ruin theory in the compound binomial model Shixue Cheng a, Hans U. Gerber b,, Elias S.W. Shiu c,1 a School of Information, People s University of China, Beijing 100872, China b Ecole des hautes études commerciales, Université de Lausanne, CH-1015 Lausanne, Switzerland c Department of Statistics and Actuarial Science, The University of Iowa, Iowa City, IA 52242-1409, USA Received 1 June 1998; received in revised form 1 September 1999; accepted 24 November 1999 Abstract The aggregate claims are modeled as a compound binomial process, and the individual claim amounts are integer-valued. We study f(x, y; u), the discounted probability of ruin for an initial surplus u, such that the surplus just before ruin is x and the deficit at ruin is y. This function can be used to calculate the expected present value of a penalty that is due at ruin, and, if it is interpreted as a probability generating function, to obtain certain information about the time of ruin. An explicit formula for f(x, y; 0) is derived. Then it is shown how f(x, y; u) can be expressed in terms of f(x, y; 0) and an auxiliary function h(u) that is the solution of a certain recursive equation and is independent of x and y. As an application, we use the asymptotic expansion of h(u) to obtain an asymptotic formula for f(x, y; u). In this model, certain results can be obtained more easily than in the compound Poisson model and provide additional insight. For the case u=0, expressions for the expected present value of a payment of 1 at ruin and the expected time of ruin (given that ruin occurs) are obtained. A discrete version of Dickson s formula is provided. 2000 Elsevier Science B.V. All rights reserved. Keywords: Risk theory; Ruin probability; Deficit at ruin; Time of ruin; Compound binomial model 1. Introduction Traditionally, most results of risk theory are derived in a continuous-time model in which the aggregate claims are a compound Poisson process. The model appears to be more realistic, the methods are perceived as elegant and sophisticated, and some results seem to be tied to the compound Poisson assumption. In contrast, this paper considers a discrete-time model, where the aggregate claims are modeled as a compound binomial process, and where the possible claim amounts are integral multiples of the annual premium. With relative simple (but perhaps also aesthetic) methods, attractive results can be derived in this model. These results are of an independent interest, and they provide a better understanding of the analogous results in the continuous-time model. In fact, the latter can be viewed as limiting cases of the former. In this sense, any given result in the discrete-time model is stronger than the corresponding result in the continuous-time model. Corresponding author. Tel.: +41-21-692-3371; fax: +41-21-692-3305. E-mail addresses: hgerber@hec.unil.ch (H.U. Gerber), eshiu@stat.uiowa.edu (E.S.W. Shiu) 1 Tel.: +319-335-2580; fax: +319-335-3017. 0167-6687/00/$ see front matter 2000 Elsevier Science B.V. All rights reserved. PII: S0167-6687(99)00053-0

240 S. Cheng et al. / Insurance: Mathematics and Economics 26 (2000) 239 250 In Section 3 we consider the discounted power martingale. For a given discount factor, the base must satisfy Lundberg s fundamental equation, which has always one solution between zero and one, and a larger, second solution under some regularity conditions for the claim amount distribution. The main goal of the paper is to study the function f(x, y; u), which is the discounted probability of ruin, given an initial surplus of u, such that the surplus just before ruin is x, and the deficit at ruin is y. With this function, the expected present value of a penalty due at ruin can be calculated. In Section 4, an explicit formula for f(x, y; 0) is obtained. In Section 5, we show that f(x, y; u) can be expressed in a transparent fashion through a function h(u), which does not depend on x or y, and is the solution of a certain recursive equation. In Section 6, an asymptotic formula for h(u) is derived and used to obtain an asymptotic formula for f(x, y; u). Applications include an explicit expression for the expected discounted value of a payment of 1 at the time of ruin, a discrete-time version of Dickson s formula, formulas for the expected time of ruin, as well as some asymptotic results. 2. The compound binomial model We consider a discrete-time model, in which the number of insurance claims is governed by a binomial process N(t), t=0, 1, 2,... In each time period, the probability of a claim is q, 0<q<1, and the probability of no claim is 1 q. The claim occurrences in different time periods are independent events. The individual claim amounts X 1, X 2, X 3,... are mutually independent, identically distributed, positive and integer-valued random variables; they are independent of the binomial process {N(t)}. Put X=X 1 and let p(x) = Pr(X = x), x = 1, 2, 3,... (2.1) be the common probability function of the individual claim amounts. (The value of p(x) is zero if x is not a positive integer.) This is called a compound binomial model, and has been considered by Gerber (1988), Shiu (1989), Willmot (1993), Dickson (1994), DeVylder (1996) (Chapter 10), DeVylder and Marceau (1996) (Section 2), and Cheng and Wu (1998a,b). The compound binomial model can also be used to model the case where there can be more than one claim in each time period. Then we assume that the total claims in each time period are mutually independent, identically distributed and integer-valued random variables. Let Y j denote the sum of the claims in period j. We consider Pr(Y j = 0) = 1 q, Pr(Y j = y) = qp(y), y = 1, 2, 3,... To avoid confusion, we shall not refer to this interpretation in this paper. We also assume that the premium received in each time period is one. We do not necessarily make the assumption that the premiums contain a positive security loading, i.e. 1 qe(x) > 0, (2.2) may not hold. Let the initial surplus be u, which is a nonnegative integer. For t=0, 1, 2,..., the surplus at time t is U(t) = u + t [X 1 + X 2 + +X N(t) ]. (2.3) Ruin is the event that U(t) 0 for some t 1. We suppose that p(1)<1 so that ruin is possible. Let T = inf{t 1: U(t) 0} (2.4) denote the time of ruin. We are interested in the joint probability distribution of the time of ruin, T, the surplus just before ruin, U(T 1), and the surplus at ruin, U(T). For x, y=0, 1, 2,..., t=1, 2, 3,..., define f(x,y,t; u) = Pr[U(T 1) = x, U(T ) = y, T = t U(0) = u]. (2.5)

S. Cheng et al. / Insurance: Mathematics and Economics 26 (2000) 239 250 241 Note that f(0, y, t; u) can be different from 0 only if t=1 and u=0; obviously f(0,y,1; 0) = qp(y + 1). (2.6) Let v be a discount factor (0<v<1). Our primary goal is to explore the discounted probability f(x,y; u) = v t f(x,y,t; u). (2.7) t=1 Using the function f(x, y; u), we can calculate the expected discounted value of a penalty which is due at ruin and may depend on the surpluses just before and at ruin. We may also view v as a parameter; then (2.7) is the formula of a (probability) generating function. 3. Lundberg s fundamental equation We seek a number r>0 such that {v t r U(t) } is a martingale. This is the condition that E[v t r U(t) U(0) = u] = r u. (3.1) Because E[r U(t) U(0) = u] = r u t {qe[r X ] + (1 q)} t, (3.2) the martingale condition is φ(r) = v 1, (3.3) where φ(r) = qe[r X 1 ] + (1 q)r 1, r > 0. (3.4) We call (3.3) Lundberg s fundamental equation to honor the Swedish actuary Lundberg, who had pointed out that the compound Poisson version of (3.3) is fundamental to the whole of collective risk theory (Lundberg, 1932, p. 144). Because φ(r) is a strictly convex function, Eq. (3.3) has at most two roots. Also, φ(1)=1<v 1, and φ(r) tends to for r 0+; hence, Eq. (3.3) has a solution r=ρ (0, 1). Furthermore, under some regularity conditions for the tail of the probability function p(x), Eq. (3.3) has another solution r=r>1. For a first application, let x be an integer, x>u(0)=u, and consider T x = inf{t : U(t) x}, (3.5) the first time the surplus rises to the level x. Note that the inequality in definition (3.5) can be replaced by an equality, because the process {U(t)} is skip-free upwards. We claim that E[v T x I(T x < ) U(0) = u] = ρ x u, (3.6) where I denotes the indicator function, i.e. I(A)=1 ifa is true and I(A)=0 ifa is false. To show this, we observe that {v t ρ U(t) } is a positive martingale that is bounded above by ρ x for t<t x. By applying the optional sampling theorem, wehave ρ u = E[v T x ρ U(T x) I(T x < ) U(0) = u] = E[v T x I(T x < ) U(0) = u]ρ x, (3.7) from which (3.6) follows.

242 S. Cheng et al. / Insurance: Mathematics and Economics 26 (2000) 239 250 For a second application, suppose that (3.3) has the other root R>1. Then {v t R U(t) } is a positive martingale that is bounded above by 1 for t<t. By the optional sampling theorem, we obtain R u = E[v T R U(T) I(T < ) U(0) = u]. (3.8) In general, it is difficult to simplify the expression on the right-hand side of (3.8), because T and U(T) are not independent, and the distribution of U(T) is unknown. There are two noteworthy exceptions. One is the case where all claims are of size 2, X 2. Then U(T)=0, provided u>0. Hence it follows from (3.8) that, for u=1, 2, 3,..., where E[v T I(T < ) U(0) = u] = R u, (3.9) R = 1 + 1 4q(1 q)v 2 2qv (3.10) by (3.3). The formula for u=0 will be given in formula (4.8), which is a general result. We note that (3.10) can be found in Feller (1968, p. 350) The other exception is the case of a geometric claim amount distribution, p(x) = (1 c)c x 1, x = 1, 2, 3,... (3.11) Then, given T<, U(T) is independent of T, and (3.8) becomes R u = E[v T I(T < ) U(0) = u]e[r U(T) T< ]. (3.12) Because, given T<, U(T) has the geometric distribution we have (1 c)c x, x = 0, 1, 2,... E[R U(T) T< ] = 1 c 1 cr. (3.13) It follows from (3.12) and (3.13) that E[v T I(T < ) U(0) = u] = 1 cr 1 c R u. (3.14) Here, (3.3) leads also to a quadratic equation, with which we can determine R. Remarks. 1. Assume that the second root R of Lundberg s equation (3.3) exists. Consider both roots as functions of v, ρ=ρ(v) and R=R(v). Note that φ(1)=1 and φ (1) = qe[x] 1, (3.15) which is the negative of the security loading. If the security loading is negative, then φ (1)>0. Hence the convex function φ has its minimum in the interval (0, 1), from which it follows that: lim v 1 ρ(v) < 1 = lim R(v). (3.16) v 1 If the security loading is zero, then φ (1)=0. Thus φ has its minimum at 1 and lim ρ(v) = 1 = lim R(v). (3.17) v 1 v 1

S. Cheng et al. / Insurance: Mathematics and Economics 26 (2000) 239 250 243 If the security loading is positive, then φ (1)<0, and lim ρ(v) = 1 < lim R(v). (3.18) v 1 v 1 The corresponding relationships for the compound Poisson case can be found in Section3 of Gerber and Shiu (1999). 2. In formulas such as (3.6) and (3.8), the indicator random variables I(T x < ) and I(T< ) can be dropped, because v<1. However, leaving them in facilitates the discussion of limiting results for v 1. From (3.6) we have Pr[T x < U(0) = u] = ρ(1) x u. (3.19) It then follows from (3.17) and (3.18) that, if the security loading is nonnegative, the right-hand side of (3.19) is 1, i.e. the surplus will rise to the level x with certainty. If the security loading is negative, then ρ(1)<1 according to (3.16), and hence (3.19) is now not a trivial result. On the other hand, as v 1 (3.8) becomes R(1) u = E[R(1) U(T) I(T < ) U(0) = u]. (3.20) If the security loading is nonpositive, it follows from (3.16) and (3.17) that R(1)=1, and hence T< with probability 1, i.e. ruin is certain. If the security loading is positive, then R(1)>1 according to (3.18), and ψ(u) = Pr[T < U(0) = u] = R(1) u E[R(1) U(T) T<, U(0) = u]. (3.21) 4. The key formula The following result is for U(0)=u=0. It is the discrete counterpart of formula (3.12) in Gerber and Shiu (1997) and formula (3.3) in Gerber and Shiu (1998a). Theorem 1. For x=0, 1, 2,..., y=0, 1, 2,..., f(x,y; 0) = qvρ x p(x + y + 1). (4.1) Proof. By (2.7), the left-hand side of (4.1) is v t f(x,y,t; 0). t=1 (4.2) According to definition (2.5), f(x, y, t; 0) is the following product of three probabilities: Pr[U(1) >0, U(2) >0,...,U(t 2) >0, U(t 1) = x U(0) = 0]qp(x + y + 1). (4.3) It follows from a simple duality argument which can be visualized by rotating the graph of the surplus process from time 0 to time t 1by180 (Feller, 1971, Section XII.2) that the conditional probability in (4.3) is the same as Pr[U(1) <x, U(2)<x,...,U(t 2) <x, U(t 1) = x U(0) = 0]. (4.4) Multiplying (4.3) with v t and summing over t, recalling that T x is the first passage time of the surplus process at level x, and applying (3.6) with u=0, we have

244 S. Cheng et al. / Insurance: Mathematics and Economics 26 (2000) 239 250 v t Pr[U(1) >0, U(2) >0,...,U(t 2) >0, U(t 1) = x U(0) = 0]qp(x + y + 1) t=1 = vqp(x + y + 1) v t 1 Pr[U(1) <x, U(2)<x,...,U(t 2) <x, U(t 1) = x U(0) = 0] t=1 = vqp(x + y + 1)E[v T x U(0) = 0] = vqp(x + y + 1)ρ x, (4.5) which is indeed the right-hand side of (4.1). As a check for (4.1), we consider the special case of x=0, f(0,y; 0) = qvp(y + 1), which can be verified using (2.6). We note that DeVylder and Goovaerts (1998) have given an interesting proof of the compound Poisson version of Theorem 1 using the Lagrange expansion formula. Also, Theorem 1 can be generalized to the case where there are m types of individual claims, which are independent and at most one of which can occur in a time period; for such results in the compound Poisson model, see Gerber and Shiu (1999). Option-pricing applications of the compound Poisson version of Theorem 1 can be found in Gerber and Shiu (1998b, 1999). As an illustration of Theorem 1, we calculate the expected present value of 1, payable at the time of ruin, if u=0: E[v T I(T < ) U(0) = 0] = x=0 k=x+1 k=1 f(x,y; 0) = qv x=0y=0 x=0 ρ x k=x+1 p(k) (4.6) by (4.1). By interchanging the order of summation and then noting that ρ is a solution of (3.3), we obtain ρ x p(k) = 1 (1 ρ k )p(k) = 1 1 ρ 1 ρ (1 E[ρX ]) = 1 ( 1 1 ρ q ρ ). (4.7) qv Substituting (4.7) in (4.6) yields E[v T I(T < ) U(0) = 0] = v ρ 1 ρ. (4.8) Note that (4.8) can be interpreted as the probability generating function of the time-of-ruin random variable T. An explicit expansion of the right-hand side of (4.8) in powers of v is possible in special cases such as the two examples in the last section; see also Section XI.3 of Feller (1968), where our v is s. Note that formula (4.8) can be rewritten as E[v T I(T < ) U(0) = 0] = 1 1 v 1 ρ, (4.9) the compound Poisson version of which is formula (3.9) in Gerber and Shiu (1998a) and formula (4.11) in Gerber and Shiu (1999). Let us first assume a negative loading. Then ruin is certain, i.e. I(T< ) 1. Consider ρ as a function of v, ρ=ρ(v). Then ρ(1)<1 by (3.16). Differentiating (4.8) and setting v=1, we obtain E[T U(0) = 0] = 1 1 ρ(1). (4.10) Note that, as the security loading tends to zero, the quantity ρ(1) tends to 1 according to (3.17), and hence the expectation of T goes to infinity; in the special case of a symmetric random walk (X 2 in our model), this is the well-known result that the time of return to the origin has an infinite expectation.

S. Cheng et al. / Insurance: Mathematics and Economics 26 (2000) 239 250 245 For the remainder of this section, we consider the more important case where the security loading is positive. Setting v=1 in (4.6) yields an expression for the probability of ruin for u=0: ψ(0) = q p(k) = qe[x]. (4.11) x=0k=x+1 This is formula (7) in Gerber (1988). In terms of the relative security loading θ, formula (4.11) is ψ(0) = 1 1 + θ, (4.12) which is a classical result in the continuous-time compound Poisson model and can be viewed as an extension of the ballot theorem (Feller, 1968, p. 69). As a check, let us verify (4.11) starting with (4.8). Using the rule of Bernoulli Hôpital, we obtain ψ(0) = 1 1 ρ (1). (4.13) To determine ρ (1), we replace r by ρ(v) in (3.3): φ(ρ(v)) = 1 v. (4.14) Using the chain rule, we differentiate with respect to v and set v=1 to see that φ (ρ(1))ρ (1) = 1. It follows from (3.18) and (3.15) that ρ (1) = 1 φ (1) = 1 1 qe[x]. (4.15) Substituting (4.15) in (4.13) we obtain (4.11) again. Let us now calculate the expected time of ruin, given that ruin occurs. First we differentiate (4.6) with respect to v and set v=1. This yields E[TI(T < ) U(0) = 0] = q p(k) + ρ (1) x p(k) = ψ(0) + qρ (1)E[X(X 1)]. 2 x=0k=x+1 x=0 k=x+1 After dividing by ψ(0) and applying (4.11) and (4.15), we obtain the desired result: E[X(X 1)] E[T T <, U(0) = 0] = 1 + 2E[X](1 qe[x]). (4.16) Hence, somewhat surprisingly, the conditional expected time of ruin, given that ruin occurs, is the shorter, the larger the security loading. This result, which is in contradistinction to (4.10), can be interpreted as follows. The larger the security loading, the faster the surplus is expected to grow as time passes. A larger surplus, in turn, means a smaller chance for ruin. Thus, given that ruin is to occur, it must occur earlier (in some sense) for the case of a larger security loading. We can use (4.9) to obtain higher moments of T.Forn=1, 2, 3,..., the factorial moment E[T(T 1)...(T n + 1)I (T < ) U(0) = 0] (4.17)

246 S. Cheng et al. / Insurance: Mathematics and Economics 26 (2000) 239 250 is the nth derivative of (1 v)/[1 ρ(v)] with respect to v at v=1. Because ρ(v) 1 v 1 = 1 k! ρ(k) (1)(v 1) k 1, (4.18) k=1 the factorial moment is expressed in terms of ρ (k) (1) up to k=n+1. We remark that in the continuous time model, the moments of the time to ruin have been examined by Picard and Lefèvre (1998). 5. Recursive formulas For an arbitrary initial surplus u>0, it is possible to determine f(x, y; u) by a recursive procedure. Given U(0)=u, we consider the first time t, t 1, when U(t) u. The discounted probability of this event with the surplus being of the amount u y at that time is g(y) = f(x,y; 0) = qv ρ x p(x + y + 1) (5.1) x=0 x=0 by (4.1). For u>0, we evaluate f(x, y; u) by conditioning on the first time when the surplus drops to or below its initial value u. We need to distinguish two cases. If u>x, the event that ruin occurs at this time makes no contribution to the quantity f(x, y; u). Hence we have u 1 f(x,y; u) = f(x,y; u z)g(z), u = x + 1,x+ 2,... (5.2) If u x, the event that ruin occurs at this time contributes the quantity f(x u, y+u;0)tof(x, y; u). Thus u 1 f(x,y; u) = f(x,y; u z)g(z) + f(x u, y + u; 0), u = 1, 2,...,x. (5.3) Formulas (5.2) and (5.3) are discrete analogs of (2.11) and (2.10) in Gerber and Shiu (1997). By (4.1) f(x u, y + u; 0) = qvρ x u p(x + y + 1) = ρ u f(x,y; 0), u x. Hence we can combine (5.2) and (5.3) as u 1 f(x,y; u) = f(x,y; u z)g(z) + f(x,y; 0)ρ u I(u x), (5.4) which corresponds to (4.1) of Gerber and Shiu (1997). Here, u=1, 2, 3,..., x=1, 2, 3,..., and y=0, 1, 2,... For given values of x and y, we can calculate f(x, y; u) recursively by (5.4). The following result gives a more elegant way to obtain f(x, y; u): Theorem 2. Let h(u) be defined as the solution of u 1 h(u) = h(u z)g(z) + ρ u, u = 1, 2, 3,... (5.5) Then, for x=1, 2, 3,..., and y=0, 1, 2,..., f(x,y; u) = f(x,y; 0)[h(u) ρ x h(u x)i(u>x)], u = 1, 2, 3,... (5.6)

S. Cheng et al. / Insurance: Mathematics and Economics 26 (2000) 239 250 247 Proof. For each positive integer x, we introduce a function of u, γ (x; u), which is defined as the solution of the renewal equation u 1 γ(x; u) = γ(x; u z)g(z) + ρ u I(u x), u = 1, 2, 3,... (5.7) Comparing (5.7) with (5.4), we gather that f(x,y; u) = f(x,y; 0)γ (x; u). (5.8) Thus to prove (5.6), we have to show that γ(x; u) = h(u) ρ x h(u x)i(u>x). (5.9) Because the solution of (5.7) is unique, it suffices to verify that the function on the right-hand side of (5.9) satisfies (5.7). Hence we consider u 1 [h(u z) ρ x h(u z x)i(u z>x)]g(z) + ρ u I(u x), u = 1, 2, 3,... If u x, this is u 1 h(u z)g(z) + ρ u = h(u) = h(u) ρ x h(u x)i(u>x). If u>x, this is u 1 u x 1 h(u z)g(z) ρ x h(u z x)g(z) = [h(u) ρ u ] ρ x [h(u x) ρ (u x) ] = h(u) ρ x h(u x)i(u>x). Remarks. 1. With (5.5) and (5.1), the function h(u) can be determined recursively for u=1, 2, 3,... Alternatively, Eq. (5.5) can be solved in terms of generating functions. 2. For u=1, 2, 3,..., the quantity 1/h(u) has a probabilistic interpretation. For given positive integers u and x, u x, it follows from (5.6) that [ ] 1 f(x,y; 0) = f(x,y; u). (5.10) h(u) Suppose that U(0)=0. Before ruin can occur with U(T 1)=x, x u>0, the surplus process {U(t)} must necessarily have attained the level u (because the premium is 1 per unit time). From this observation and formula (5.10) we see that 1/h(u) can be interpreted as the expected present value of a contingent payment of 1 that is made when the surplus process, with initial value 0, attains the level u for the first time, provided that ruin has not occurred by then. The compound Poisson version of (5.10) is (6.26) of Gerber and Shiu (1998a). 3. It follows from the interpretation above that 1 = (1 q)v. h(1) (5.11)

248 S. Cheng et al. / Insurance: Mathematics and Economics 26 (2000) 239 250 On the other hand, we obtain from (5.5) and (5.1) that h(1) = h(1)g(0) + ρ 1 = h(1)qv ρ x p(x + 1) + ρ 1. (5.12) x=0 These two formulas can be reconciled because ρ satisfies Lundberg s equation (3.3). 4. Assuming a positive security loading, we consider the limiting case v=1 and hence ρ=1. Then it follows from (5.5) that h(u)can be interpreted as the expected number of weak record lows of the surplus process {U(t)} before ruin, with U(0)=u. There is a close connection between h(u) and the probability of ruin ψ(u). Observe that the number of future record lows of the process{u(t)} has at any time a geometric distribution with mean 1/[1 ψ(0)]. The expected number of record lows before ruin is the expected number of record lows starting from time zero minus the expected number of future record lows from the time of ruin. Thus we have h(u) = 1 ψ(u) 1 ψ(0). (5.13) Substituting (5.13) in (5.6), we obtain a discrete version of a formula by Dickson (1992), which is for the compound Poisson model. In that context, Dickson s formula has been generalized by Gerber and Shiu (1997, 1998a) and Theorem 2 suggests that a generalization is also possible in the compound binomial model. 6. Asymptotic formulas We shall derive a two-term asymptotic formula (u ) for the function h(u), which is defined as the solution of the recursive equation (5.5). Based on this asymptotic formula and Theorem 2, we then obtain an asymptotic formula for f(x, y; u). Asymptotically, the function h(u) has an exponential growth, h(u) Aρ u for u, (6.1) where the constant A is such that Aρ u = Aρ (u z) g(z) + ρ u, (6.2) or 1 A = 1 ρ z g(z). (6.3) However, substituting (6.1) in (5.6) yields only the trivial result that f(x, y; u) 0 for u. Hence we need a refinement of (6.1). Theorem 3. If the second root R>1 of Lundberg s equation (3.3) exists, then there exists a constant C such that, for u, Aρ u h(u) CR u, (6.4) [( ) R x f(x,y; u) f(x,y; 0)C 1] R u. (6.5) ρ Proof. Note that (6.5) follows directly from (5.6) and (6.4). Thus we only need to prove (6.4). Subtracting (5.5) from (6.2), we have u 1 Aρ u h(u) = [Aρ (u z) h(u z)]g(z) + A ρ (u z) g(z). (6.6) z=u

S. Cheng et al. / Insurance: Mathematics and Economics 26 (2000) 239 250 249 Because u=1 z=u ρ (u z) g(z) <, Eq. (6.6) is a (defective) renewal equation for the function Aρ u h(u) (Feller, 1968, Section XIII.10). It follows from the Renewal Theorem (Feller, 1968, p. 331) and the relation 1 = R z g(z), that there is a constant C such that (6.4) holds. Relation (6.7) is best verified by setting u=0 in (3.8). For a compound Poisson version of (6.5), see (6.68) or (4.21) of Gerber and Shiu (1998a). As an application, we now derive an asymptotic formula for the expected time of ruin, given that ruin is to occur, in the case of a positive security loading. From (6.5), we see that, for u, [( ) R x E[v T I(T < ) U(0) = u] C f(x,y; 0) 1] R u. (6.8) ρ x=1y=0 Consider ρ and R as functions of v. Then the right-hand side of (6.8) is of the form η(v)r(v) u, where the function η(v) does not depend on u. Because E[T T <, U(0) = u] = d dv ln{e[vt I(T < ) U(0) = u]} v=1, (6.9) we have E[T T <, U(0) = u] R (1) u R(1) for u. (6.10) To determine R (1)/R(1), we multiply (3.3) by r and replace r by R(v): qe[r(v) X ] + (1 q) = R(v) v. (6.11) Differentiating (6.11) with respect to v, setting v=1 and rearranging yields R (1) R(1) = 1 qe[xr(1) X 1 ] 1. (6.12) (6.7) Acknowledgements Elias Shiu gratefully acknowledges the support from the Principal Financial Group Foundation. References Cheng, S., Wu, B., 1998a. Classical risk model in fully discrete setting. Technical Report, School of Information, People s University of China, 14 pp. Cheng, S., Wu, B., 1998b. The survival probability in finite time in fully discrete risk models. Technical Report, School of Information, People s University of China, 8 pp. DeVylder, F.E., 1996. Advanced Risk Theory: A Self-Contained Introduction. Editions de l Universite de Bruxelles, Brussels. DeVylder, F.E., Goovaerts, M.J., 1998. Discussion of On the time value of ruin. North American Actuarial Journal 2 (1), 72 74. DeVylder, F.E., Marceau, E., 1996. Classical numerical ruin probabilities. Scandinavian Actuarial Journal, 109 123. Dickson, D.C.M., 1992. On the distribution of surplus prior to ruin. Insurance: Mathematics and Economics 11, 191 207. Dickson, D.C.M., 1994. Some comments on the compound binomial model. ASTIN Bulletin 24, 33 45. Feller, W., 1968. An Introduction to Probability Theory and Its Applications, Vol. 1, 3rd Edition. Wiley, New York. Feller, W., 1971. An Introduction to Probability Theory and Its Applications, Vol. 2, 2nd Edition. Wiley, New York.

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