the moments of the time of ruin, the surplus before ruin, and the deficit at ruin

Insurance: Mathematics and Economics 27 (2000) 19–44

The moments of the time of ruin, the surplus before ruin,and the deficit at ruin

X. Sheldon Lina,∗, Gordon E. Willmotba Department of Statistics and Actuarial Science, University of Iowa, Iowa City, IA 52242, USA

b Department of Statistics and Actuarial Science, University of Waterloo, Waterloo, Ont., Canada N2L 3G1

Received 1 August 1999; received in revised form 1 December 1999; accepted 20 January 2000

Abstract

In this paper we extend the results in Lin and Willmot (1999 Insurance: Mathematics and Economics 25, 63–84) to propertiesrelated to the joint and marginal moments of the time of ruin, the surplus before the time of ruin, and the deficit at the timeof ruin. We use an approach developed in Lin and Willmot (1999), under which the solution to a defective renewal equationis expressed in terms of a compound geometric tail, to derive explicitly the joint and marginal moments. This approach alsoallows for the establishment of recursive relations between these moments. Examples are given for the cases when the claimsize distribution is exponential, combinations of exponentials and mixtures of Erlangs. © 2000 Elsevier Science B.V. Allrights reserved.

Keywords:Time of ruin; Surplus; Deficit; Compound geometric; Higher moment; Equilibrium distributions; Renewal equation

1. Introduction and background

There has been considerable research interest in the analysis of the distributions of the time of ruin, the surplusbefore the time of ruin, and the deficit at the time of ruin. The identification of these distributions is difficultsince analytic expressions do not exist in most cases. Further, there are in general no closed form solutions for themoments of these distributions. Various approaches have been utilized in order to study the probabilistic propertiesof these distributions. In Gerber et al. (1987) and Dufresne and Gerber (1988) the investigations focus on the analyticexpression of the probability of ruin and the distribution function of the surplus before the time of ruin for the casethat the individual claim size is a combination of exponential distributions or a combination of gamma distributions.Dickson (1992), Dickson and Waters (1992), Dickson and Dos Reis (1996), Willmot and Lin (1998) and Schmidli(1999), discuss analytic properties of the distribution of the surplus before the time of ruin, the distribution ofthe deficit at the time of ruin and their relationship. Dickson and Waters (1991) and Dickson et al. (1995) considerrecursive calculation for these distributions when the individual claim size is fully discrete. Di Lorenzo and Tessitore(1996) propose a numerical approximation for calculation of the distribution of the surplus before the time of ruin.The moment properties of the time of ruin are considered in Delbaen (1990) and in Picard and Lefevre (1998, 1999)

∗ Corresponding author. Tel.:+1-319-335-0730; fax:+1-319-335-3017.E-mail addresses:[email protected] (X.S. Lin), [email protected] (G.E. Willmot).

0167-6687/00/$ – see front matter © 2000 Elsevier Science B.V. All rights reserved.PII: S0167-6687(00)00038-X

20 X.S. Lin, G.E. Willmot / Insurance: Mathematics and Economics 27 (2000) 19–44

(also see Gerber (1979, Chapter 9, Section 3)). Recently, Gerber and Shiu (1997, 1998) study the joint distributionof the time of ruin, the surplus before the time of ruin, and the deficit at the time of ruin by considering an expecteddiscounted penalty involving these three random variables. They show that this expected discounted penalty as afunction of the initial surplus is the solution of a certain (defective) renewal equation. This approach allows for theuse of existing renewal theoretic techniques as well as new techniques developed for renewal equations.

In Lin and Willmot (1999) and Willmot and Lin (2000), a different approach is proposed for solving defectiverenewal equations. Under this approach the solution of a defective renewal equation is expressed in terms of acompound geometric tail. This has the advantage of allowing for the use of analytic properties of a compoundgeometric distribution and for the implementation of exact and approximate results which have been developed forthe tail. Tails of compound geometric distributions have been studied extensively both analytically and numerically.In particular, recursive formulas (e.g., Panjer and Willmot (1992)) are available, as are upper and lower bounds(e.g., Lin (1996)). Exact solutions are sometimes available (e.g., Dufresne and Gerber (1988) and Tijms (1994)).Also, the well known Cramer–Lundberg asymptotic formula (e.g., Gerber (1979)) is available. Second and moreimportantly, for a large class of claim size distributions, the tail of the associated compound geometric distributioncan be approximated very nicely by a combination of two exponential distributions. For example, if the claimsize distribution is new worse than used in convex ordering (NWUC) or new better than used in convex ordering(NBUC),1 a combination of two exponential distributions with proper parameter values will match the probabilitymass at zero and the mean, and has the same asymptotic formula. This combination is referred to as the Tijmsapproximation (see Tijms (1994) and Willmot (1997)). Thus, we can take advantage of the analytic form of theTijms approximation.

In this paper we extend the approach in Lin and Willmot (1999) to properties related to the joint and marginalmoments of the time of ruin, the surplus before the time of ruin, and the deficit at the time of ruin. Althoughthese problems have been considered by various authors, the present approach allows one to express the joint andmarginal moments in terms of compound geometric tails and to reveal recursive relations between these moments.The advantage of this approach becomes apparent when applied to the recursive calculation of the higher momentsat the time of ruin. In the rest of this section, we present the classical continuous time risk model, Gerber and Shiu’sexpected discounted penalty function and its associated renewal equation, and the solution of the renewal equationin terms of a compound geometric tail which is obtained in Lin and Willmot (1999). Section 2 discusses equilibriumdistributions and reliability classifications. Section 3 introduces auxiliary functions which involve the compoundgeometric tail associated with Gerber and Shiu’s renewal equation and the related equilibrium distributions. Theauxiliary functions are necessary for computation of the moments of the deficit and the moments of the time of ruin.Section 4 considers properties related to the moments of the deficit at ruin and Section 5 considers the moments ofthe surplus immediately before ruin. In Section 6, we derive the moments at the time of ruin.

We now begin with the classical continuous time risk model. LetNt be the number of claims from an insuranceportfolio. It is assumed thatNt follows a Poisson process with meanλ. The individual claim sizesX1, X2, . . . , inde-pendent ofNt , are positive, independent and identically distributed random variables with common distribution func-tion (DF)P(x) = 1− P (x) = Pr(X ≤ x), momentspj = ∫ ∞

0 xj dP(x) for j = 0,1,2, . . . , and Laplace–Stieltjestransformp(s) = ∫ ∞

0 e−sx dP(x). The aggregate claims process is{St ; t ≥ 0}, whereSt = X1 +X2 + · · · +XNt(with St = 0 if Nt = 0). The insurer’s surplus process is{Ut ; t ≥ 0} with Ut = u + ct − St , whereu ≥ 0 is theinitial surplus,c = λp1(1 + θ) the premium rate per unit time, andθ > 0 the relative security loading.

Let T = inf {t;U(t) < 0} be the first time that the surplus becomes negative and is called the time of ruin. Theprobabilityψ(u) = Pr{T < ∞} is called the probability of (ultimate) ruin. Two nonnegative random variables inconnection with the time of ruin are the surplus immediately before the time of ruinU(T−), whereT− is the leftlimit of T and the deficit at the time of ruin|U(T )|. For details, see Bowers et al. (1997, Chapter 13), Gerber (1979,Chapters 8 and 9) or De Vylder (1996, Part I, Chapter 7).

1 For their definitions, see Willmot (1997). It is worthwhile to point out that the DFR and IMRL classes and the IFR and DMRL classesintroduced in Section 2 are subclasses of the NWUC class and the NBUC class, respectively.

X.S. Lin, G.E. Willmot / Insurance: Mathematics and Economics 27 (2000) 19–44 21

Letw(x1, x2), 0 ≤ x1, x2 < ∞, be a nonnegative function. Forδ ≥ 0, define

φ(u) = E{e−δT w(U(T−), |U(T )|)I (T < ∞)}, (1.1)

whereI (T < ∞) = 1, T < ∞, and I (T < ∞) = 0 otherwise. The quantityw(U(T−), |U(T )|) can beinterpreted as the penalty at the time of ruin, where the surplus isU(T−) and the deficit is|U(T )|. Thus,φ(u) isthe expected discounted penalty whenδ is interpreted as a force of interest. Since

ψ(u) = E{I (T < ∞)}, (1.2)

φ(u) reduces toψ(u) if δ = 0 andw(x1, x2) = 1. The functionφ(u) is useful both for unifying and generalizingknown results in connection with joint and marginal distributions ofT , U(T−), and|U(T )| (see, e.g., Gerber andShiu (1997, 1998)). Moreover, (1.1) may be viewed in terms of Laplace transforms withδ the argument. Gerberand Shiu (1998) show that the functionφ(u) satisfies the defective renewal equation

φ(u) = λ

c

∫ u

0φ(u− x)

∫ ∞

x

e−ρ(y−x) dP(y)dx + λ

ceρu

∫ ∞

u

e−ρx∫ ∞

x

w(x, y − x)dP(y)dx. (1.3)

In (1.3),ρ = ρ(δ) is the unique nonnegative solution of the equation

cρ − δ = λ− λp(ρ). (1.4)

We assume the functionw(x1, x2) to be such that the second term of (1.3) is finite. It is easy to see that (1.3) hasa unique solution by identifying the Laplace transforms of the equation or applying the principle of contractionmaps. Further discussion of (1.4) and its discrete analogue can be found in Cheng et al. (2000) and Gerber and Shiu(1999).

For the purpose of our analysis, the renewal equation (1.3) may be rewritten (Lin and Willmot (1999)) as follows:

φ(u) = 1

1 + β

∫ u

0φ(u− x)dG(x)+ 1

1 + βH(u), u ≥ 0, (1.5)

where the DFG(x) = 1 − G(x) is given by

G(x) = P (x)− eρx∫ ∞x

e−ρy dP(y)

ρ∫ ∞

0 e−ρyP (y)dy, (1.6)

β = (1 + θ)p1∫ ∞0 e−ρyP (y)dy

− 1, (1.7)

H(u) = eρu∫ ∞u

e−ρx∫ ∞xw(x, y − x)dP(y)dx∫ ∞

0 e−ρyP (y)dy. (1.8)

The solution of (1.5) exists uniquely and can be proved using Laplace transformation or the principle of contractionmaps.

An alternative expression forG(x) is given by

G(x) =∫ ∞

0 e−ρyP (x + y)dy∫ ∞0 e−ρyP (y)dy

, x ≥ 0. (1.9)

Furthermore,G(x) is differentiable and

G′(x) = eρx∫ ∞x

e−ρy dP(y)∫ ∞0 e−ρyP (y)dy

, x ≥ 0. (1.10)


Define now the associated compound geometric DFK(u) = 1 − K(u) by

K(u) =∞∑n=1

β

1 + β

(1

1 + β

)nG∗n(u), u ≥ 0, (1.11)

whereG∗n(u) is the tail of then-fold convolution ofG(u). It is easy to see thatK(u) is the solution to the followingrenewal equation:

K(u) = 1

1 + β

∫ u

0K(u− x)dG(x)+ 1

1 + βG(u), u ≥ 0. (1.12)

K(u)may also be viewed as the Laplace transform at the time of ruinT . To see this, we letw(x1, x2) = 1. It followsimmediately from (1.8) and (1.9) thatH(u) = G(u). Thus, withw(x1, x2) = 1, φ(u) is the Laplace transform ofT andK(u) = φ(u).

We now look at the special case whenδ = 0. It is clear that

G(x) = 1

p1

∫ x

0P (y)dy. (1.13)

Thus, in this case,

K(u) = ψ(u). (1.14)

We now state a theorem which shows that the solution to (1.5) can be expressed in terms ofK(u).

Theorem 1.1(Lin and Willmot (1999)).The solutionφ(u) to (1.5)may be expressed as

φ(u) = 1

β

∫ u

0H(u− x)dK(x)+ 1

1 + βH(u), (1.15)

φ(u) = − 1

β

∫ u

0K(u− x)dH(x)− H(0)

βK(u)+ 1

βH(u). (1.16)

If H(u) is differentiable, thenφ(u) may be expressed as

φ(u) = − 1

β

∫ u

0K(u− x)H ′(x)dx − H(0)

βK(u)+ 1

βH(u), u ≥ 0. (1.17)

The proof of Theorem 1.1 and a detailed discussion ofG(x), H(u) andK(u) are given in Lin and Willmot (1999).

2. Equilibrium distributions and reliability classifications

Equilibrium distributions associated with a given distribution function and reliability classifications play animportant role in our analysis. Many results in the following sections heavily depend on equilibrium distributions.Reliability classifications enable us to derive upper and lower bounds for functions of the deficit.

In what follows, we briefly discuss equilibrium distributions and some reliability classes. LetP(x), x > 0, be adistribution function. The equilibrium DFP1(x) = 1− P1(x) of P(x) is defined byP1(x) = ∫ x

0 P (y)dy/p1. In thepresent contextP1(x) may be interpreted as the DF of the amount of the drop in the surplus level, given that thereis a drop (e.g., Bowers et al. (1997, Chapter 12)). The DFP(x) is said to be decreasing (increasing) failure rate orDFR (IFR) if P (x + y)/P (x) is nondecreasing (nonincreasing) inx for fixedy ≥ 0. A larger class of distributionsis the new worse (better) than used or NWU (NBU) class withP (x + y) ≥ (≤)P (x)P (y) for all x ≥ 0 andy ≥ 0.


The mean residual lifetime ofP(x) is defined by

rP (x) =∫ ∞xP (y)dy

P (x)= rP (0)

P1(x)

P (x), x ≥ 0. (2.1)

Another class larger than the DFR (IFR) class is the increasing (decreasing) mean residual lifetime or IMRL (DMRL)class for whichrP (x) is nondecreasing (nonincreasing) forx ≥ 0. Similarly,P(x) is new worse (better) than usedin expectation or NWUE (NBUE) ifrP (x) ≥ (≤)rP (0), or equivalentlyP1(x) ≥ (≤)P (x), x ≥ 0. See Fagiuoliand Pellerey (1994) and Lin and Willmot (1999) and references therein for further details on these classifications.

The notion of higher order equilibrium DFs is also needed in what follows. LetP2(x) = 1 − P2(x) be theequilibrium DF ofP1(x). The DFP2(x) is called the second-order equilibrium DF ofP(x). Similarly, we definethenth order equilibrium DF ofP(x) by

Pn(x) = 1 − Pn(x) =∫ x

0 Pn−1(y)dy∫ ∞0 Pn−1(y)dy

, n = 1,2, . . . , (2.2)

whereP0(x) = 1 − P0(x) = P (x). It can be shown that the mean of the DFPn(x), n = 1,2, . . . , is given by∫ ∞

0Pn(x)dx = pn+1

(n+ 1)pn, (2.3)

and that the relation between the DFsPn(x) andP(x) is given by

Pn(x) = 1

pn

∫ ∞

x

(y − x)n dP(y). (2.4)

See Hesselager et al. (1998) and references therein for further details.The connection of the equilibrium DFsPn(x) of P(x) and the equilibrium DFsGn(x) ofG0(x) = G(x), where

P(x) is the individual claim size DF andG(x) is the associated “claim size” DF to the renewal equation (1.5) cannow be established.

It is easy to see from (1.9) that in the special case whenδ = 0, G(x) = P1(x) sinceρ = 0. Thus,Gn(x) =Pn+1(x) in this case. Letµn(ρ) be thenth moment ofG(x), i.e.

µn(ρ) =∫ ∞

0xn dG(x). (2.5)

Thus,

µn(0) =∫ ∞

0xn dP1(x) = 1

p1

∫ ∞

0xnP (x)dx = pn+1

(n+ 1)p1. (2.6)

Forρ > 0, we have

µn(ρ) = µn(0)

∫ ∞0 e−ρx dPn+1(x)∫ ∞

0 e−ρx dP1(x). (2.7)

The equilibrium DFGn(x) of G(x) can be expressed in terms of the equilibrium DFPn(x). In fact we have,analogous to (1.9),

Gn(x) =∫ ∞

0 e−ρyPn(x + y)dy∫ ∞0 e−ρyPn(y)dy

. (2.8)

The derivation of (2.7) and (2.8) can be found in Section 3 of Lin and Willmot (1999).


3. Convolution of equilibrium distributions and compound geometric tails

In this section we introduce two auxiliary functions which involve the convolution of the compound geometrictail K(u) and thenth equilibrium tailGn(x) of the “claim size” distributionG(x) = G0(x). Both functions arenecessary in computing the moments of the (discounted) deficit and the moments at the time of ruin as well as otherimportant quantities in classical ruin theory, as will become clear in Sections 4 and 6.

Defineg−1(ρ) = p1, µ0(ρ) = 1, andgn(ρ) to be the mean of thenth equilibrium DFGn(x) ofG(x). It followsfrom (2.3) that

gn(ρ) = µn+1(ρ)

(n+ 1)µn(ρ), n = 0,1,2, . . . (3.1)

Also define

αn(u, ρ) = gn(ρ)

{∫ u

0K(u− x)dGn+1(x)+ Gn+1(u)

}, n = −1,0,1,2, . . . (3.2)

Evidently,α−1(u, ρ) = p1(1 + β)K(u) from (1.12), and

αn(u, ρ) =∫ u

0K(u− x)Gn(x)dx +

∫ ∞

u

Gn(x)dx, n = 0,1,2, . . . (3.3)

We will derive an alternative representation forαn(u, ρ), beginning with the following recursive relation.

Lemma 3.1.

αn+1(u, ρ) = 1

gn(ρ)

∫ ∞

u

αn(t, ρ)dt −∫ ∞

u

K(t)dt, n = −1,0,1,2, . . . (3.4)

Proof. Forn = −1,0,1,2, . . . , one has from (3.2) that

1

gn(ρ)

∫ ∞

u

αn(t, ρ)dt =∫ ∞

u

∫ t

0K(t − x)dGn+1(x)dt +

∫ ∞

u

Gn+1(t)dt.

By interchanging the order of integration, we have∫ ∞

u

∫ t

0K(t − x)dGn+1(x)dt =

∫ u

0

∫ ∞

u

K(t − x)dt dGn+1(x)+∫ ∞

u

∫ ∞

x

K(t − x)dt dGn+1(x)

=∫ u

0

∫ ∞

u−xK(t)dt dGn+1(x)+ Gn+1(u)

∫ ∞

0K(t)dt.

Integration by parts yields∫ u

0

∫ ∞

u−xK(t)dt dGn+1(x) = −Gn+1(x)

∫ ∞

u−xK(t)dt

∣∣∣∣u

x=0+

∫ u

0K(u− x)Gn+1(x)dx

= −Gn+1(u)

∫ ∞

0K(t)dt +

∫ ∞

u

K(t)dt +∫ u

0K(u− x)Gn+1(x)dx.

Thus

1

gn(ρ)

∫ ∞

u

αn(t, ρ)dt =∫ ∞

u

K(t)dt +∫ u

0K(u− x)Gn+1(x)dx +

∫ ∞

u

Gn+1(t)dt

=∫ ∞

u

K(t)dt + αn+1(u, ρ) �

using (3.3).


We have the following alternative representation forαn(u, ρ) from the lemma.

Theorem 3.1.

α0(u, ρ) = β

∫ ∞

u

K(x)dx (3.5)

and forn = 1,2,3, . . . ,

αn(u, ρ) = β

µn(ρ)

∫ ∞

u

(x − u)nK(x)dx −n−1∑j=0

(n

j

)µn−j (ρ)µn(ρ)

∫ ∞

u

(x − u)j K(x)dx. (3.6)

Proof. Sinceα−1(u, ρ) = p1(1 + β)K(u), (3.4) yields withn = −1,

α0(u, ρ) = p1(1 + β)

p1

∫ ∞

u

K(t)dt −∫ ∞

u

K(t)dt = β

∫ ∞

u

K(t)dt,

proving (3.5). Forj ≥ 0, one has by interchanging the order of integration

∫ ∞

u

∫ ∞

t

(y − t)j K(y)dy dt = 1

j + 1

∫ ∞

u

(y − u)j+1K(y)dy. (3.7)

With the help of (3.7) and Lemma 3.1, it is straightforward to verify that (3.6) holds by induction onn. �

In the special case whenδ = 0, it is convenient to define

τn(u) =∫ u

0ψ(u− x)Pn(x)dx +

∫ ∞

u

Pn(x)dx, n = 1,2,3, . . . , (3.8)

and so

τn(u) = pn+1

(n+ 1)pn

{∫ u

0ψ(u− x)dPn+1(x)+ Pn+1(u)

}, n = 1,2,3, . . . (3.9)

We have the following corollary.

Corollary 3.1.

τ1(u) = θ

∫ ∞

u

ψ(x)dx (3.10)

and forn = 2,3,4, . . . ,

τn(u) = np1θ

pn

∫ ∞

u

(x − u)n−1ψ(x)dx −n−2∑j=0

(n

j

)pn−jpn

∫ ∞

u

(x − u)jψ(x)dx. (3.11)

Proof. Whenδ = 0, ρ = 0,G(x) = P1(x), and (2.6) holds. ThusPn(x) = Gn−1(x). SinceK(u) = ψ(u) in thiscase, it follows from (3.3) and (3.8) that

τn(u) = αn−1(u,0), n = 1,2,3, . . . (3.12)

Sinceβ = θ whenρ = 0, (3.5) becomes (3.10). Then (3.11) follows directly from (2.6) and (3.6). �


It is evident in the proofs of the above theorems that the existence of certain moments of the claim size distributionP(x) is required. For instance, ifpn+2 is finite, thenαn(u, ρ) andτn+1(u) are both finite. In order to focus onanalytical representation of functions of interest, we always assume that required moments are finite throughout therest of the paper.

In the following examples we demonstrate how to evaluateαn(u, ρ) andτn(u) whenP(x) is a combination ofexponentials or a mixture of Erlangs. The case whenP(x) is a combination of exponentials is considered in Gerberet al. (1987) and Dufresne and Gerber (1988). From a formal standpoint, a combination of two exponentials isvery important since it may be viewed as providing a Tijms approximation to the relevant quantity (see Willmot(1997) and references therein). In particular, Tijms approximations forαn(u, ρ) andτn(u) allow for correspondingapproximations for moments at the time of ruin and the deficit of ruin, as is discussed in later sections. In whatfollows, a combination of two exponentials may be viewed in this context.

Example 3.1 (Combinations of exponentials). The case whereP(x) is a combination of exponentials is firstconsidered in Gerber et al. (1987) and Dufresne and Gerber (1988) in which a closed form expression for theprobability of ruin and the distribution of the deficit at the time of ruin are derived.

We begin with an exponential claim size DFP(x) = e−µx . In this case, for alln = 0,1, . . . , Pn(x) = e−µx . Itfollows from (2.8) that

Gn(x) = e−µx.

Thus,µn(ρ) = µn(0) = n!/µn from (2.7). Moreover,

K(u) = 1

1 + βe−Ru, (3.13)

where

1

1 + β= µ

ρ + µ

1

1 + θ,

R = βµ

1 + β= θµ

1 + θ+ ρ

ρ + µ

µ

1 + θ

(see Example 4.1 of Lin and Willmot (1999)). A direct computation from (3.2) yields

αn(u, ρ) = 1

µe−Ru. (3.14)

We now assume that the functionK(u) is of the form

K(u) = C1 e−R1u + C2 e−R2u.

This is the case whenP(x) is a combination of two exponentials or a Tijms approximation is used. A detaileddiscussion of this is also given in Example 4.1 of Lin and Willmot (1999).

Since∫ ∞

u

(x − u)j K(x)dx = C1 e−R1u

∫ ∞

0xj e−R1x dx + C2 e−R2u

∫ ∞

0xj e−R2x dx

= C1j !R−j−11 e−R1u + C2j !R−j−1

2 e−R2u,

Theorem 3.1 yields

αn(u, ρ) = C∗1,n e−R1u + C∗

2,n e−R2u, (3.15)


where

C∗1,n = C1

βn!R−n−1

1

µn(ρ)−n−1∑j=0

(n

j

)µn−j (ρ)j !R−j−1

1

µn(ρ)

,

C∗2,n = C2

βn!R−n−1

2

µn(ρ)−n−1∑j=0

(n

j

)µn−j (ρ)j !R−j−1

2

µn(ρ)

.

With ρ = 0, we obtain from (2.6) that

τn(u) = C∗∗1,n e−R1u + C∗∗

2,n e−R2u, (3.16)

C∗∗1,n = n!C1R

−n1

pn

p1θ −

n−1∑j=1

pj+1Rj

1

(j + 1)!

,

C∗∗2,n = n!C2R

−n2

pn

p1θ −

n−1∑j=1

pj+1Rj

2

(j + 1)!

.

For a general linear combination of exponentials, the derivation is similar and is omitted here.

Mixtures of Erlangs are an important distributional class in modeling insurance losses for the reason that anycontinuous distribution on(0,∞) may be approximated arbitrarily accurately by a distribution of this type (seeTijms (1994, pp. 162–164)).

Example 3.2(Mixtures of Erlangs). The density function of a mixture of Erlangs is given by

P ′(x) =r∑k=1

qkµ(µx)k−1 e−µx

(k − 1)!, x ≥ 0,

where{q1, q2, . . . , qr} is a probability distribution. Example 4.2 of Lin and Willmot (1999) shows that

K(u) = e−µu∞∑j=0

Cj(µu)j

j !, u ≥ 0,

where

Cj = 1

1 + β

j∑k=1

q∗k Cj−k + 1

1 + β

∞∑k=j+1

q∗k , j = 1,2,3, . . . .

Here it is assumed thatC0 = (1 + β)−1, and

q∗k =

∑rj=kqj

(µµ+ρ

)j−k∑rj=1qj

∑j−1i=0

(µµ+ρ

)i ,for k = 1,2, . . . , r, andq∗

k = 0, for k > r.


Now, for anyi = 0,1, . . . , andj = 0,1, . . . ,

∫ ∞

u

(x − u)j(µx)i

i!e−µx dx = µi

i!e−µu

∫ ∞

0xj (x + u)i e−µx dx = µi

i!e−µu

i∑k=0

(i

k

)uk

∫ ∞

0xj+i−k e−µx dx

= µ−j−1 e−µui∑

k=0

(j + i − k)!

(i − k)!

(µu)k

k!.

Thus,

αn(u, ρ) = e−µu∞∑k=0

C∗k,n

(µu)k

k!, (3.17)

where

C∗k,n =

∞∑i=0

Ck+i

βµ−n−1

µn(ρ)

(n+ i)!

i!−n−1∑j=0

(n

j

)µn−j (ρ)µ−j−1

µn(ρ)

(j + i)!

i!

.

In the rest of this paper, we turn to evaluation ofφ(u) for a particular weight functionw(x, y).

4. The deficit at the time of ruin

In this section we consider in more detail the (possibly discounted) deficit at the time of ruin beginning with themoments. We are focusing on their analytic solutions. For numerical algorithms, see Dickson et al. (1995) and thereferences therein.

Consider the special casew(x1, x2) = xk2, wherek is a positive integer. A relatively simple representation existsfor φ(u) in (1.1).

Theorem 4.1. For k = 1,2,3, . . . ,

E{e−δT |U(T )|k∣∣∣ T < ∞} = 1

βψ(u)[kµk−1(ρ)αk−1(u, ρ)− µk(ρ)K(u)], (4.1)

whereµn(ρ) = ∫ ∞0 xn dG(x) is given in (2.7) andαn(u, ρ) is given inTheorem 3.1.

Proof. Using (2.4) and (2.8), Eq. (1.8) becomes, withw(x1, x2) = xk2,

H(u) = eρu∫ ∞u

e−ρx∫ ∞x(y − x)k dP(y)dx∫ ∞

0 e−ρyP (y)dy= pk eρu

∫ ∞u

e−ρxPk(x)dx

p1∫ ∞

0 e−ρy dP1(y)= pk

∫ ∞0 e−ρyPk(y)dy

p1∫ ∞

0 e−ρy dP1(y)Gk(u).

Furthermore, using (2.3),

H(u) = pk+1∫ ∞

0 e−ρy dPk+1(y)

(k + 1)p1∫ ∞

0 e−ρy dP1(y)Gk(u),

and from (2.6) and (2.7), we obtain the simple result

H(u) = µk(ρ)Gk(u). (4.2)


Then, using (3.1), sinceGk(u) = ∫ ∞uGk−1(x)dx/gk−1(ρ), we obtain

H ′(u) = − µk(ρ)

gk−1(ρ)Gk−1(u) = −kµk−1(ρ)Gk−1(u),

and from (1.17),

φ(u) = kµk−1(ρ)

β

∫ u

0K(u− x)Gk−1(x)dx + µk(ρ)

β

∫ ∞

u

Gk−1(x)

gk−1(ρ)dx − µk(ρ)

βK(u)

= kµk−1(ρ)

β

{∫ u

0K(u− x)Gk−1(x)dx +

∫ ∞

u

Gk−1(x)dx

}− µk(ρ)

βK(u)

= kµk−1(ρ)

βαk−1(u, ρ)− µk(ρ)

βK(u),

using (3.3). That is,

E{e−δT |U(T )|k I (T < ∞)} = kµk−1(ρ)

βαk−1(u, ρ)− µk(ρ)

βK(u), (4.3)

and division byψ(u) gives (4.1). �

Whenk = 1, with the help of (3.5), (4.3) reduces to

E{e−δT∣∣∣U(T )|I (T < ∞)} =

∫ ∞

u

K(x)dx − µ1(ρ)

βK(u). (4.4)

Whenδ = 0 we have the following important corollary.

Corollary 4.1. For k = 1,2,3, . . . ,

E{|U(T )|k∣∣∣ T < ∞} = pk

p1θ

τk(u)

ψ(u)− pk+1

(k + 1)p1θ, (4.5)

whereτk(u) is given inCorollary 3.1.

Proof. Whenδ = 0, K(u) = ψ(u), αk−1(u,0) = τk(u), µn(0) is given by (2.6), andβ = θ . Then (4.1) reducesto (4.5). �

Whenk = 1, (4.5) reduces to (using (3.10))

E{|U(T )|∣∣∣ T < ∞} =

∫ ∞

u

ψ(x)

ψ(u)dx − p2

2p1θ. (4.6)

The second moment of the deficit, hence the variance of the deficit, can also easily be evaluated from (4.5).As demonstrated in Examples 3.1 and 3.2,ψ(u) andτk(u) can be computed explicitly whenP(x) is a combination

of exponentials, a mixture of Erlangs, or approximated using a Tijms approximation. Thus, with the help of (4.5)the moments of the deficit|U(T )| are also obtainable as shown in the following examples.

Example 4.1(Combinations of exponentials). We again begin with an exponential. In this case,

E{|U(T )|k∣∣∣ T < ∞} = pk

p1θ

τk(u)

ψ(u)− pk+1

(k + 1)p1θ= k!µ−k

µ−1θ

µ−1 e−Ru

(1 + θ)−1 e−Ru − (k + 1)!µ−k−1

(k + 1)µ−1θ= k!µ−k,

which agrees with the fact that in the exponential case the deficit is also exponentially distributed and it is independentof the time of ruin.


Next, we examine a combination of two exponentials. This is also the case when a Tijms approximation is used.It follows from (3.16) that

E{|U(T )|k∣∣∣ T < ∞} = pk

p1θ

τk(u)

ψ(u)− pk+1

(k + 1)p1θ= C1,k e−R1u + C2,k e−R2u

C1 e−R1u + C2 e−R2u, (4.7)

where

C1,k = k!C1R−k1

p1θ

p1θ −

k∑j=1

pj+1Rj

1

(j + 1)!

,

C2,k = k!C2R−k2

p1θ

p1θ −

k∑j=1

pj+1Rj

2

(j + 1)!

.

Example 4.2(Mixtures of Erlangs). For a mixture of Erlangs claim size distribution, it follows immediately fromExample 3.2 that thekth unconditional discounted moment is

E{e−δT |U(T )|kI (T < ∞)} = e−µu∞∑j=0

Cj,k(µu)j

j !, (4.8)

where

Cj,k = 1

β[kµk−1(ρ)C

∗j,k−1 − µk(ρ)Cj ].

The constantsC∗j,k−1 andCj are given in Example 3.1.

We remark that when the individual claim size is a combination of exponentials or a mixture of Erlangs, thedensity function of the deficit is of the same type but with new weights. These weights are expressed as functions ofthe initial surplusu. As a result, the moments of the deficit can be calculated accordingly (for details, see Willmot(2000)). The method we employed in this section provides a different approach in calculating the moments of thedeficit. Under this method, the moments of the deficit are calculated directly. The coefficients in (4.7) and (4.8)are independent of the initial surplusu, which allows for the calculation of the moments of the deficit for differentvalues ofu without reevaluating the distribution of the deficit. Another advantage of this approach is that we areable to evaluate the expected discounted deficit using (4.1), which the approach in Willmot (2000) cannot. Next,we consider bounds for functions of the deficit.

Theorem 4.2. If P(x) is IMRL (DMRL), then fork = 1,2,3, . . . ,

E{e−δT |U(T )|k I (T < ∞)} ≥ (≤)µk(ρ)K(u). (4.9)

Proof. By Theorem 3.2(b) of Lin and Willmot (1999),G(x) is IMRL (DMRL). This implies (Fagiuoli and Pellerey,1994) that thej th equilibrium DFGj(x) is NWUE (NBUE), i.e.,Gj+1(x) ≥ (≤)Gj (x). Then by (4.2),

H(u) = µk(ρ)Gk(u) ≥ (≤)µk(ρ)Gk−1(u) ≥ (≤) · · · ≥ (≤)µk(ρ)G(u).Thus, Theorem 4.1 of Lin and Willmot (1999) applies, which shows that ifH(u) ≤ (≥)cG(u), thenφ(u) ≤(≥)cK(u), and the result is proved. �

To derive a simple bound for more general functions of the deficit, we recall a result in Lin and Willmot (1999).


Lemma 4.1. If

η(x) = 1

P (x)

∫ ∞

x

w(x, y − x)dP(y) ≤ (≥)c, (4.10)

where0< c < ∞, thenφ(u) ≤ (≥)cK(u).

With the help of Lemma 4.1, we have the following simple bound.

Theorem 4.3. Suppose thatA(y) = 1 − A(y) is a DF on(0,∞) satisfying

P (x + y)

P (x)≥ (≤)A(y), x ≥ 0, y ≥ 0. (4.11)

Then ifw(x) is nondecreasing in x,

E{e−δT w(|U(T )|)I (T < ∞)} ≥ (≤){∫ ∞

0w(y)dA(y)

}K(u), u ≥ 0. (4.12)

Proof. Let Zx for x ≥ 0 be a random variable such that Pr(Zx > y) = P (x + y)/P (x). Then withw(x1, x2) =w(x2), (4.10) may be expressed asη(x) = E{w(Zx)}. Then by (4.11) and Ross (1996, p. 405), ifw(x) is nonde-creasing, thenη(x) ≥ (≤)∫ ∞

0 w(y)dA(y). �

It is clear from the proof of Theorem 4.3 that if (4.11) holds andw(x) is nonincreasing inx, then inequality (4.12)is reversed.

The key to the application of Theorem 4.3 is to identify the DFA(y) for a givenP(x). If P(x) is NWU (NBU),then (4.11) holds withA(y) = P(y). For more details on the choice ofP(x), see Willmot and Lin (1997).

It is worth noting that Eq. (4.12) may be written whenδ = 0 as

E{w(|U(T )|)∣∣∣ T < ∞} ≥ (≤)

∫ ∞

0w(y)dA(y). (4.13)

In the special case whenw(x1, x2) = xk2 andP(x) is DFR (IFR), from Ross (1996, p. 405) and Theorem 3.5,µk(ρ) ≥ (≤)pk, and Theorem 4.2 gives a tighter bound than Theorem 4.3.

If P(x) = 1 − exp(−x/p1) andw(x1, x2) = w(x2), thenη(x) = ∫ ∞0 w(y)dP(y) independent ofx and

H(u) = {∫ ∞0 w(y)dP(y)}G(u), implying thatφ(u) = {∫ ∞

0 w(y)dP(y)}K(u). Thus, withw(y) = e−sy ,

E{exp[−δT − s|U(T )|]I (T < ∞)} = (1 + p1s)−1E{e−δT I (T < ∞)}. (4.14)

Eq. (4.14) may be viewed as the joint Laplace transform of the (defective) distribution ofT and|U(T )|, implyingthatT and|U(T )| are independent (conditional onT < ∞), and that|U(T )| is exponential with meanp1. Thisresult is given by Gerber (1979, p. 138). The Laplace transform of the deficit for arbitrary claim size DF can also beevaluated in a similar manner. In that case, we simply letw(x1, x2) = e−sx2. However, the resulting formula willbe complicated, except when the claim size is exponentially distributed as discussed above.

5. Surplus before the time of ruin

We now consider joint moments of the surplusU(T−) before ruin and the deficit|U(T )| at ruin may be obtainedwith δ = 0 andw(x, y) = xjyk. Then (1.8) becomes, using (2.4),

H(u) = 1

p1

∫ ∞

u

xj∫ ∞

x

(y − x)k dP(y)dx = pk

p1

∫ ∞

u

xj Pk(x)dx, u ≥ 0. (5.1)


Thus,

H ′(u) = −pkp1uj Pk(u),

and

H(0) = pk+1

(k + 1)p1

∫ ∞

0xj dPk+1(x) = k!j !pk+j+1

(k + j + 1)!p1

using Hesselager et al. (1998). Therefore, from (1.17)

E{{U(T−)}j |U(T )|k I (T < ∞)} = pk

p1θ

{∫ u

0ψ(u− x)xj Pk(x)dx +

∫ ∞

u

xj Pk(x)dx

}

−{

k!j !pk+j+1

(k + j + 1)!p1θ

}ψ(u). (5.2)

Moments ofU(T−) are obtainable from (5.2) withk = 0. That is, forj = 1,2,3, . . .

E{{U(T−)}j∣∣∣ T < ∞} = 1

θψ(u)

{∫ u

0ψ(u− x)xj dP1(x)+

∫ ∞

u

xj dP1(x)

}− pj+1

(j + 1)p1θ. (5.3)

The integrals in (5.3) are easily evaluated when the claim size distribution is a combination of exponentials or amixture of Erlangs. We omit the details.

The defective density of the surplusU(T−) is also easily obtainable. Letδ = 0, w(x1, x2) = e−sx1 in (1.8).Thus,

H(u) =∫ ∞

u

e−sx dP1(x),

and (1.17) yields

E{e−sU(T−)I (T < ∞)} = 1

θ

∫ u

0ψ(u− x)e−sx dP1(x)+ 1

θ

∫ ∞

u

e−sx dP1(x)− ψ(u)

θ

∫ ∞

0e−sx dP1(x)

= 1

θ

∫ u

0[ψ(u− x)− ψ(u)] e−sx dP1(x)+ 1

θ

∫ ∞

u

[1 − ψ(u)] e−sx dP1(x). (5.4)

Hence, by the uniqueness of the Laplace transform, the defective density of the surplusU(T−) before the time ofruin is

f (x) ={(1/θ)[(1 − P(x))/p1][ψ(u− x)− ψ(u)], x ≤ u,

(1/θ)[(1 − P(x))/p1][1 − ψ(u)], x > u,(5.5)

which is in agreement with Dickson (1992).

6. Moments involving the time of ruin

In this section we employ the interpretation ofφ(u) in (1.1) as a Laplace transform with argumentδ to find momentsinvolvingT . Our approach is to differentiate (1.5) with respect toδ and show that the corresponding moment satisfiesanother defective renewal equation. However, to be able to differentiate (1.5), we need the following lemma.

Lemma 6.1. Suppose thatφ(u, δ) is the solution of the following renewal equation:

c

λφ(u, δ) =

∫ u

0φ(u− x, δ)m(x, ρ)dx +M(u, δ), (6.1)


whereρ = ρ(δ) is the nonnegative solution of(1.4)and

m(x, ρ) =∫ ∞

x

e−ρ(y−x) dP(y), x ≥ 0. (6.2)

If the second momentp2 of P(x) is finite,M(u, δ) is a differentiable function with respect toδ on [0, δ0) for someδ0 > 0, andM1(u, δ) = ∂M(u, δ)/∂δ is continuous in u, thenφ(u, δ) is differentiable with respect toδ on [0, δ0)andφ1(u, δ) = ∂φ(u, δ)/∂δ is the solution of the following renewal equation:

c

λφ1(u, δ) =

∫ u

0φ1(u− x, δ)m(x, ρ)dx +

∫ u

0φ(u− x, δ)m1(x, δ)dx +M1(u, δ), (6.3)

wherem1(x, ρ) = ∂m(x, ρ)/∂δ.

Proof. Obviously, (6.3) is a defective renewal equation and hence has a unique solution which we denote asφ1(u, δ).Define

B(u, δ,1δ) = φ(u, δ +1δ)− φ(u, δ)

1δ− φ1(u, δ).

Then, we have

c

λB(u, δ,1δ) =

∫ u

0B(u− x, δ,1δ)m(x, ρ(δ +1δ))dx + C(u, δ,1δ),

where

C(u, δ,1δ) =∫ u

0φ(u− x, δ)

[m(x, ρ(δ +1δ))−m(x, ρ)

1δ−m1(x, δ)

]dx

+M(u, δ +1δ)−M(u, δ)

1δ−M1(u, δ).

Let

J (u, δ,1δ) = sup0≤x≤u

|B(u, δ,1δ)|.

Then,

c

λJ (u, δ,1δ) ≤ J (u, δ,1δ)

∫ u

0m(x, ρ(δ +1δ))dx + |C(u, δ,1δ)|

≤ J (u, δ,1δ)

∫ u

0P (x)dx + |C(u, δ,1δ)| ≤ J (u, δ,1δ)p1 + |C(u, δ,1δ)|.

Thus,

J (u, δ,1δ) ≤ 1

θp1|C(u, δ,1δ)|.

Hence, lim1δ→0C(u, δ,1δ) = 0 implies that lim1δ→0J (u, δ,1δ) = 0. The latter implies that

lim1δ→0

φ(u, δ +1δ)− φ(u, δ)

1δ= φ1(u, δ),

and we will prove the lemma.


It is obvious that

lim1δ→0

M(u, δ +1δ)−M(u, δ)

1δ−M1(u, δ) = 0.

To show that the first term ofC(u, δ,1δ) goes to 0, we first show that

m(x, ρ(δ +1δ))−m(x, ρ)

1δ−m1(x, δ)

is bounded. It is easy to check that∣∣∣∣m(x, ρ(δ +1δ))−m(x, ρ)

1δ−m1(x, δ)

∣∣∣∣=

∣∣∣∣∫ ∞

x

exp[−ρ(δ +1δ)(y − x)] − exp[−ρ(δ)(y − x)] + ρ′(δ)1δ exp[−ρ(δ)(y − x)]

1δdP(y)

∣∣∣∣=

∣∣∣∣1δ∫ ∞

x

{[ρ′(δ1)]2 exp[−ρ(δ1)(y − x)](y − x)2 − ρ′′(δ1)exp[−ρ(δ1)(y − x)](y − x)} dP(y)

∣∣∣∣≤ 1δ

{p2 max

0≤δ1<δ0[ρ′(δ1)]2 + p1 max

0≤δ1<δ0|ρ′′(δ1)|

}< ∞.

The second equality comes from the Taylor expension of e−ρ(δ+1δ)(y−x) up to the second derivative. Sinceφ(u−x, δ)is continuous and hence is bounded, applying the bounded convergence theorem yields the result. �

We now consider moments involving the time of ruin. It is easy to see that with (6.2), (1.5) can be rewritten as

c

λφ(u) =

∫ u

0φ(u− x)m(x, ρ)dx +

∫ ∞

u

exp[−ρ(δ)(x − u)]∫ ∞

x

w(x, y − x)dP(y)dx. (6.4)

Thus, Lemma 6.1 yields

c

λ

∂

∂δφ(u) =

∫ u

0

{∂

∂δφ(u− x)

}m(x, ρ)dx + ρ′(δ)

∫ u

0φ(u− x)

∂

∂ρm(x, ρ)dx

−ρ′(δ)∫ ∞

u

(x − u)exp[−ρ(δ)(x − u)]∫ ∞

x

w(x, y − x)dP(y)dx. (6.5)

Next, define

H(u,0) = 1

p1

∫ ∞

u

∫ ∞

x

w(x, y − x)dP(y)dx, (6.6)

φ1(u) = E{Tw(U(T−), |U(T )|)I (T < ∞)}. (6.7)

We then have the following theorem.

Theorem 6.1. The functionφ1(u) satisfies the defective renewal equation

φ1(u) = 1

1 + θ

∫ u

0φ1(u− x)dP1(x)+ 1

1 + θH1(u), (6.8)

where

H1(u) = 1

λp1θ

∫ u

0ψ(u− x)H(x,0)dx + 1

λp21θ

∫ ∞

u

(x − u)

∫ ∞

x

w(x, y − x)dP(y)dx. (6.9)


Proof. It follows from (1.1) thatφ1(u) = −(∂/∂δ)φ(u)|δ=0. Also, from (6.2),m(x,0) = P (x), and

∂

∂ρm(x, ρ)|ρ=0 = −

∫ ∞

x

(y − x)dP(y) = −p1P1(x). (6.10)

Then, differentiating (1.4) with respect toδ, one has

ρ′(δ){c + λp′(ρ)} = 1, (6.11)

which implies thatρ′(0) = 1/(λp1θ) sinceρ(0) = 0. Then putδ = 0 into (6.5) to obtain

c

λφ1(u) =

∫ u

0φ1(u− x)P (x)dx + 1

λθ

∫ u

0φ(u− x,0)P1(x)dx

+ 1

λp1θ

∫ ∞

u

(x − u)

∫ ∞

x

w(x, y − x)dP(y)dx. (6.12)

In (6.12),φ(u,0) = φ(u)|δ=0, and thus from (1.5) withδ = 0,

φ(u,0) = 1

1 + θ

∫ u

0φ(u− x,0)dP1(x)+ 1

1 + θH(u,0) (6.13)

with H(u,0) given by (6.6). Now, one has (e.g., Panjer and Willmot (1992))

ψ(s) =∫ ∞

0e−sxψ(x)dx = 1

1 + θ

{1 − p1(s)

s

} {1 − 1

1 + θp1(s)

}−1

, (6.14)

wherep1(s) = ∫ ∞0 e−sx dP1(x). Then from (6.13),

φ(s,0) =∫ ∞

0e−suφ(u,0)du = 1

1 + θH (s,0)

{1 − 1

1 + θp1(s)

}−1

with H (s,0) = ∫ ∞0 e−suH(u,0)du. Therefore,s−1{1 − p1(s)}φ(s,0) = H (s,0)ψ(s), implying that∫ u

0φ(u− x,0)P1(x)dx =

∫ u

0ψ(u− x)H(x,0)dx, (6.15)

and substitution into (6.12) yields (6.8) after multiplication byλ/c. �

Next, consider an important special case. Define

φ1,k(u) = E{T |U(T )|kI (T < ∞)}, k = 0,1,2, . . . . (6.16)

We have the following result.

Theorem 6.2. For k = 0,1,2, . . . , the functionφ1,k(u) satisfies the defective renewal equation

φ1,k(u) = 1

1 + θ

∫ u

0φ1,k(u− x)dP1(x)+ pk+1

(k + 1)cp1θτk+1(u), (6.17)

and is given explicitly by

φ1,k(u) = pk

λp21θ

2

∫ u

0ψ(u− x)τk(x)dx + pk+1

(k + 1)λp21θ

2

{τk+1(u)−

∫ u

0ψ(u− x)ψ(x)dx

}

− pk+2

(k + 1)(k + 2)λp21θ

2ψ(u), (6.18)

with τ0(u) = p1(1 + θ)ψ(u), τ1(u) given by(3.10),andτk(u) given by(3.11)for k = 2,3,4, . . . .


Proof. With δ = 0 andw(x, y) = yk, Eqs. (2.4) and (6.6) yieldH(u,0) = [pk+1/(k + 1)p1]Pk+1(u), and∫ ∞

u

(x − u)

∫ ∞

x

(y − x)k dP(y)dx = pk

∫ ∞

u

(x − u)Pk(x)dx

= pk+1

k + 1

∫ ∞

u

(x − u)dPk+1(x) = pk+1

k + 1

∫ ∞

u

Pk+1(x)dx.

Then substitution into (6.9) yields

H1(u) = pk+1

(k + 1)λp21θ

{∫ u

0ψ(u− x)Pk+1(x)dx +

∫ ∞

u

Pk+1(x)dx

},

and from (3.8) we obtain

H1(u) = pk+1

(k + 1)λp21θτk+1(u). (6.19)

Then (6.17) follows from Theorem 6.1. Also,

H1(0) = pk+1

(k + 1)λp21θ

∫ ∞

0Pk+1(x)dx = pk+2

(k + 1)(k + 2)λp21θ

using (2.3). Now, from (3.4) withρ = 0 and (3.12),

τk+1(u) = (k + 1)pkpk+1

∫ ∞

u

τk(t)dt −∫ ∞

u

ψ(t)dt, (6.20)

which holds fork = 0,1,2, . . . , with τ0(u) = p1(1 + θ)ψ(u). Hence

τ ′k+1(u) = − (k + 1)pk

pk+1τk(u)+ ψ(u),

and from (6.19),H ′1(u) = −{(pk/λp2

1θ)τk(u)− [pk+1/((k + 1)λp21θ)]ψ(u)}. Then from (1.17),

φ1,k(u) = 1

θ

∫ u

0ψ(u− x)

{pk

λp21θτk(x)− pk+1

(k + 1)λp21θψ(x)

}dx

+ pk+1

(k + 1)λp21θ

2τk+1(u)− pk+2

(k + 1)(k + 2)λp21θ

2ψ(u),

which is (6.18). �

The choicek = 0 allows one to obtain the mean time to ruin.

Corollary 6.1. The mean time to ruin, given that ruin has occurred, is given by

E(T |T < ∞) = ψ1(u)

ψ(u), (6.21)

whereψ1(u) satisfies the defective renewal equation

ψ1(u) = 1

1 + θ

∫ u

0ψ1(u− x)dP1(x)+ 1

c

∫ ∞

u

ψ(x)dx (6.22)


ψ1(u) = 1

λp1θ

{∫ u

0ψ(u− x)ψ(x)dx +

∫ ∞

u

ψ(x)dx − p2

2p1θψ(u)

}. (6.23)


Proof. With the help of (3.10), (6.17) becomes (6.22) withk = 0 andψ1(u) = φ1,0(u). Similarly, (6.18) reducesto (6.23) using (3.10) andτ0(u) = p1(1 + θ)ψ(u). Then dividing byψ(u) yields (6.21). �

We remark that (6.23) was obtained earlier by Schmidli (1994) with an alternative approach.The conditional covariance betweenT and|U(T )| is obtained withk = 1.

Corollary 6.2.

E{T |U(T )|∣∣∣ T < ∞} = φ1,1(u)

ψ(u), (6.24)

whereφ1,1(u) satisfies the defective renewal equation

φ1,1(u) = 1

1 + θ

∫ u

0φ1,1(u− x)dP1(x)+ 1

c

∫ ∞

u

(x − u)ψ(x)dx − p2

2cp1θ

∫ ∞

u

ψ(x)dx (6.25)


φ1,1(u) = 1

λp1θ

∫ u

0ψ(u− x)

∫ ∞

x

ψ(t)dt dx − p2

2λp21θ

2

∫ u

0ψ(u− x)ψ(x)dx

+ 1

λp1θ

∫ ∞

u

(x − u)ψ(x)dx − p2

2λp21θ

2

∫ ∞

u

ψ(x)dx − p3

6λp21θ

2ψ(u). (6.26)

Proof. Whenk = 1, (6.19) reduces to

H1(u) = 1

λp1

∫ ∞

u

(x − u)ψ(x)dx − p2

2λp21θ

∫ ∞

u

ψ(x)dx

with the help of (3.11). Thus (6.17) becomes (6.25). Similarly, whenk = 1, (6.18) reduces to (6.26) using (3.10)and (3.11). �

The covariance ofT and|U(T )| follows from Corollaries 6.1 and 6.2, and (4.6). In general,φ(u,0) satisfyingthe defective renewal equation (6.13) satisfies

φ(u,0) = E{w(U(T−), |U(T )|)I (T < ∞)},which yields the mean ofw (e.g., from (1.17)). Thus (at least in principle) the covariance ofT andw may beobtained using Theorem 6.1 for arbitraryw(x1, x2). We now turn to evaluation of higher moments at the time ofruin T .

Theorem 6.3. The kth moment at the time of ruin, given that ruin has occurred, is given by

E(T k|T < ∞) = ψk(u)

ψ(u), (6.27)

whereψk(u) satisfies a sequence of defective renewal equations

ψk(u) = 1

1 + θ

∫ u

0ψk(u− x)dP1(x)+ k

c

∫ ∞

u

ψk−1(x)dx (6.28)

for k = 1,2, . . . , withψ0(u) = ψ(u), the probability of ruin. Moreover, eachψk(u) is given recursively by

ψk(u) = k

λp1θ

[∫ u

0ψ(u− x)ψk−1(x)dx +

∫ ∞

u

ψk−1(x)dx − ψ(u)

∫ ∞

0ψk−1(x)dx

](6.29)

in terms ofψk−1(u).


Proof. Recalling thatK(u) = E{e−δT I (T < ∞)}, the Laplace transform ofT , we have

ψk(u) = E{T kI (T < ∞)} = (−1)kdk

dδkK(u)|δ=0.

To prove the theorem we begin with the defective renewal equation

c

λK(u) =

∫ u

0K(u− x)m(x, ρ)dx +

∫ ∞

u

m(x, ρ)dx, (6.30)

wherem(x, ρ) is defined in (6.2). Noting that

m(x, ρ) = G′(x)∫ ∞

0e−ρyP (y)dy,

this equation is obtained from (1.8), (1.9) and (6.4) withw(x1, x2) = 1. We note that since∫ ∞um(x, ρ)dx has the

nth derivative with respect toδ, assuming thatpn is finite. By applying Lemma 6.1, we are able to differentiate(6.30)n times.

Define the Laplace transforms

K(s, ρ) =∫ ∞

0e−sxK(x)dx, m(s, ρ) =

∫ ∞

0e−sxm(x, ρ)dx, g(s) =

∫ ∞

0e−sxdG(x). (6.31)

Then we have from (6.30),

c

λK(s, ρ) = K(s, ρ)m(s, ρ)+ m(0, ρ)− m(s, ρ)

s, (6.32)

and from direct calculation

g(s) = ρ

ρ − s

{p(s)− p(ρ)

1 − p(ρ)

}, (6.33)

wherep(s) = ∫ ∞0 e−sx dP(x).

It follows from (1.4) and (6.33) that

m(s, ρ) = p(s)− p(ρ)

ρ − s= λp(s)+ cρ − δ − λ

λ(ρ − s). (6.34)

With the help of (6.34), (6.32) becomes[cs

(1 − 1

1 + θp1(s)

)− δ

]ρK(s, ρ) =

[1 − 1

1 + θp1(s)

]cρ − δ, (6.35)

where

p1(s) =∫ ∞

0e−sx dP1(x) = 1 − p(s)

p1s.

Also, define the Laplace transform

ψk(s) =∫ ∞

0e−sxψk(x)dx.

An obvious relation is

ψk(s) = (−1)kdk

dδkK(s,0). (6.36)


Eq. (6.28) is equivalent to

s

[1 − 1

1 + θp1(s)

]ψk(s) = k

c

[ψk−1(0)− ψk−1(s)

]. (6.37)

We now prove this identity by induction.For k = 1, this is a restatement of Corollary 6.1. Assume that forn = 1, . . . , k − 1, identity (6.37) holds.

Differentiate Eq. (6.35)k + 1 (k ≥ 1) times atδ = 0. The right-hand side is simply

c

[1 − 1

1 + θp1(s)

]ρ(k+1)(0), (6.38)

whereρ(n) denotes thenth derivative ofρ, while the left-hand side equals

cs

[1 − 1

1 + θp1(s)

]dk+1

dδk+1

[ρ(δ)K(s, ρ)

] ∣∣∣∣∣δ=0

− (k + 1)dk

dδk[ρ(δ)K(s, ρ)]

∣∣∣∣∣δ=0

. (6.39)

Sinceρ(0) = 0,

dn

dδn[ρ(δ)K(s, ρ)]

∣∣∣∣δ=0

=n−1∑j=0

(−1)j(n

j

)ρ(n−j)(0)ψj (s)

for n = k, k + 1. Equating (6.38) and (6.39) yields

s

[1 − 1

1 + θp1(s)

] k∑j=0

(−1)j(k + 1j

)ρ(k−j+1)(0)ψj (s)

= k + 1

c

k−1∑j=0

(−1)j(k

j

)ρ(k−j)(0)ψj (s)+

[1 − 1

1 + θp1(s)

]ρ(k+1)(0). (6.40)

Let s = 0 in (6.40). We have

k + 1

c

k−1∑j=0

(−1)j(k

j

)ρ(k−j)(0)ψj (0)+

[1 − 1

1 + θ

]ρ(k+1)(0) = 0.

Thus, (6.40) can be rewritten as

s

[1 − 1

1 + θp1(s)

] k∑j=0

(−1)j(k + 1j

)ρ(k−j+1)(0)ψj (s)

= −k + 1

c

k−1∑j=0

(−1)j(k

j

)ρ(k−j)(0)[ψj (0)− ψj (s)] + 1

1 + θ[1 − p1(s)]ρ

(k+1)(0). (6.41)

With the fact that

ψ0(s) = ψ(s) = (1 − p1(s))/s(1 + θ)

1 − (1/(1 + θ))p1(s),

the first term in the left-hand side of (6.41) is equal to

1

1 + θ[1 − p1(s)]ρ

(k+1)(0).


Hence,

s

[1 − 1

1 + θp1(s)

] k∑j=1

(−1)j(k + 1j

)ρ(k−j+1)(0)ψj (s)

= −k + 1

c

k−1∑j=0

(−1)j(k

j

)ρ(k−j)(0)[ψj (0)− ψj (s)]. (6.42)

By the induction assumption,

s

[1 − 1

1 + θp1(s)

](−1)j+1

(k + 1j + 1

)ρ(k−j)(0)ψj+1(s) = −k + 1

c(−1)j

(k

j

)ρ(k−j)(0)[ψj (0)− ψj (s)]

for j = 0,1, . . . , k − 2. All but the last terms in both sides of (6.42) are canceled out. Hence (6.42) becomes

s

[1 − 1

1 + θp1(s)

](−1)k(k + 1)ρ′(0)ψk(s) = −k + 1

c(−1)k−1kρ′(0)[ψk−1(0)− ψk−1(s)], (6.43)

which is the same as (6.37).Eq. (6.29) is a direct application of Theorem 1.1 to (6.28), whereG(x)=P1(x)andH(u)=(k/λp1)

∫ ∞uψk−1(x)dx.

�

Corollary 6.3. The second moment of the time to ruin, given that ruin has occurred, is given by

E(T 2∣∣∣ T < ∞) = ψ2(u)

ψ(u), (6.44)

whereψ2(u) satisfies the defective renewal equation

ψ2(u) = 1

1 + θ

∫ u

0ψ2(u− x)dP1(x)+ 1

1 + θH2(u) (6.45)

with

H2(u) = 2

λ2p21θ

{∫ u

0ψ(u− x)

∫ ∞

x

ψ(t)dt dx +∫ ∞

u

(x − u)ψ(x)dx

}(6.46)


ψ2(u) = 2

λ2p21θ

2

∫ u

0ψ(u− x)

∫ x

0ψ(x − t)ψ(t)dt dx + 4

λ2p21θ

2

∫ u

0ψ(u− x)

∫ ∞

x

ψ(t)dt dx

+ 2

λ2p21θ

2

∫ ∞

u

(x − u)ψ(x)dx − p2

λ2p31θ

3

∫ u

0ψ(u− x)ψ(x)dx − 2p1p3θ + 3p2

2

6λ2p41θ

4ψ(u). (6.47)

Proof. Since

H2(u) = 2

λp1

∫ ∞

u

ψ1(x)dx,

replacingψ1(x) by (6.23) and employing integration by parts on each term yields (6.46). The fact that∫ ∞

0 ψ(x)dx =p2/2p1θ is used in the computation. Eq. (6.47) is a direct application of Theorem 1.1. Noting that

H ′2(u) = 2

λ2p21θ

{−

∫ u

0ψ(u− x)ψ(x)dx −

∫ ∞

u

ψ(x)dx + p2

2p1θψ(u)

},

we have (6.47). �


We now evaluate the moments at the time of ruin in the case when the claim size is either a combination ofexponentials or a mixture of Erlangs. Gerber (1979, p. 138) considers the mean time to ruin in the exponential case.In the next example, formulas are given for higher moments at the time to ruin in the exponential case. We wish topoint out that if the claim size distribution is neither a combination of exponentials nor a mixture of Erlangs, thecalculation of higher(n > 2) moments could be complicated. However, we may use the Tijms approximation andin this case, the approach in Example 6.1 applies.

Example 6.1(Combinations of exponentials). First, letP (x) = e−µx . Then, from (3.13) withδ = 0, ψ(u) =C e−Ru, whereR = θµ/(1 + θ). It is not hard to see from (6.29) by induction that

ψk(u) = e−Ruk∑j=0

Cj,k(Ru)j

j !. (6.48)

We now derive a recursive formula for the coefficientsCj,k.From (6.29),

∫ u

0ψ(u− x)ψk−1(x)dx = 1

1 + θe−Ru

k−1∑j=0

Cj,k−1

∫ u

0

(Rx)j

j !dx = 1

(1 + θ)Re−Ru

k∑j=1

Cj−1,k−1(Ru)j

j !,

∫ ∞

u

ψk−1(x)dx = R−1k−1∑j=0

Cj,k−1

∫ ∞

u

Rj+1xj e−Rx

j !dx = R−1 e−Ru

k−1∑j=0

k−1∑i=jCi,k−1

(Ru)j

j !

∫ ∞

0ψk−1(x)dx = R−1

k−1∑i=0

Ci,k−1.

Thus, by equating the coefficients of(Ru)j /j !, we have

C0,k = k(1 + θ)

cµθ

k−1∑i=0

Ci,k−1 (6.49)

and forj = 1,2, . . . , k,

Cj,k = k(1 + θ)2

cµθ2

1

1 + θCj−1,k−1 +

k−1∑i=jCi,k−1

, (6.50)

with∑k−1i=k = 0 andC0,0 = C. The first two moments can be easily derived using (6.49) and (6.50).

C0,1 = 1 + θ

cµθC0,0 = 1

cµθ,

C1,1 = (1 + θ)2

cµθ2

1

1 + θC0,0 = 1

cµθ2.

Thus,

E{TI(T < ∞)} =[

1

cµθ+ 1

cµθ2(Ru)

]e−Ru,

which is in agreement with Gerber (1979, p.138).


Also,

C0,2 = 2(1 + θ)

cµθ[C0,1 + C1,1] = 2(1 + θ)2

c2µ2θ3,

C1,2 = 2(1 + θ)2

cµθ2

[1

1 + θC0,1 + C1,1

]= 2(1 + θ)(1 + 2θ)

c2µ2θ4,

C2,2 = 2(1 + θ)2

cµθ2

1

1 + θC1,1 = 2(1 + θ)

c2µ2θ4.

E{T 2I (T < ∞)} = 2

c2µ2θ3

[(1 + θ)2 + (1 + θ)(1 + 2θ)

θ(Ru)+ 1 + θ

θ

(Ru)2

2!

]e−Ru.

For a combination of two exponentials,

ψ(u) = C1 e−R1u + C2 e−R2u,

as shown in Example 3.1. The general formula forψk(u) is

ψk(u) =k∑j=0

[Aj,k e−R1u + Bj,k e−R2u]uj

j !. (6.51)

The coefficientsAj,k andBj,k are obtainable recursively, similar to the method used in the exponential case above.We omit the details but give the mean of the time of ruin in this case. It is easy to see that∫ u

0ψ(u− x)ψ(x)dx = C2

1ue−R1u + C22ue−R2u + 2C1C2

R2 − R1[e−R1u − e−R2u],

∫ ∞

u

ψ(x)dx = C1

R1e−R1u + C2

R2e−R2u.

Thus,

E{TI(T < ∞)} = 1 + θ

cθ

{C2

1ue−R1u + C22ue−R2u +

[C1

R1+ 2C1C2

R2 − R1−

(C1

R1+ C2

R2

)C1

]e−R1u

+[C2

R2− 2C1C2

R2 − R1−

(C1

R1+ C2

R2

)C2

]e−R2u

}.

Example 6.2(Mixtures of Erlangs). In the case when a claim size is a mixture of Erlangs, noting thatψ(u) =K(u)|δ=0, we have (Example 4.2 of Lin and Willmot (1999) or Example 3.2 this paper)

ψ(u) = e−µu∞∑j=0

Cj(µu)j

j !.

Thekth moment of the time of ruin is again of this form, i.e.,

ψk(u) = e−µu∞∑j=0

Cj,k(µu)j

j !, (6.52)

whereCj,0 = Cj .


A recursive algorithm forCj,k is now derived. Since∫ u

0ψ(u− x)ψk−1(x)dx

= e−µu∞∑j=0

∞∑i=0

CjCi,k−1µj+i

j !i!

∫ u

0(u− x)j xi dx = e−µu

∞∑j=0

∞∑i=0

CjCi,k−1µj+iuj+i+1

j !i!

∫ 1

0(1 − t)j t i dt

= e−µu∞∑j=0

∞∑i=0

CjCi,k−1µj+iuj+i+1

(j + i + 1)!= µ−1 e−µu

∞∑j=1

j∑i=1

Cj−iCi−1,k−1

(µu)j

j !,

∫ ∞

u

ψk−1(x)dx = µ−1∞∑j=0

Cj,k−1

∫ ∞

u

µj+1xj e−µx

j !dx

= µ−1 e−µu∞∑j=0

Cj,k−1

j∑i=0

(µu)i

i!= µ−1 e−µu

∞∑j=0

∞∑i=jCi,k−1

(µu)j

j !,

∫ ∞

0ψk−1(x)dx = µ−1

∞∑i=0

Ci,k−1.

Thus, from (6.29) we obtain

C0,k = k(1 + θ)

cµθ(1 − C0)

∞∑i=0

Ci,k−1 = k

λθ(1 − C0)

∞∑i=0

Ci,k−1, (6.53)

and forj = 1,2, . . . ,

Cj,k = k

λθ

j∑i=1

Cj−iCi−1,k−1 +∞∑i=jCi,k−1 − Cj

∞∑i=0

Ci,k−1

. (6.54)

Acknowledgements

This work was supported by the Natural Sciences and Engineering Research Council of Canada. The first authoralso acknowledges the financial support by the University of Iowa Old Gold Fellowship. Both authors would liketo thank the anonymous referees for their suggestions.

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