Directory UMM :Data Elmu:jurnal:I:Insurance Mathematics And Economics:Vol27.Issue1.2000:

The moments of the time of ruin, the surplus before ruin,

and the deficit at ruin

X. Sheldon Lin

a,∗

, Gordon E. Willmot

b

a_{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 properties related to the joint and marginal moments of the time of ruin, the surplus before the time of ruin, and the deficit at the time of ruin. We use an approach developed in Lin and Willmot (1999), under which the solution to a defective renewal equation is expressed in terms of a compound geometric tail, to derive explicitly the joint and marginal moments. This approach also allows for the establishment of recursive relations between these moments. Examples are given for the cases when the claim size distribution is exponential, combinations of exponentials and mixtures of Erlangs. © 2000 Elsevier Science B.V. All rights 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 surplus before the time of ruin, and the deficit at the time of ruin. The identification of these distributions is difficult since analytic expressions do not exist in most cases. Further, there are in general no closed form solutions for the moments of these distributions. Various approaches have been utilized in order to study the probabilistic properties of these distributions. In Gerber et al. (1987) and Dufresne and Gerber (1988) the investigations focus on the analytic expression of the probability of ruin and the distribution function of the surplus before the time of ruin for the case that 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 of the deficit at the time of ruin and their relationship. Dickson and Waters (1991) and Dickson et al. (1995) consider recursive 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: shlin@stat.uiowa.edu (X.S. Lin), gewillmo@setosa.uwaterloo.ca (G.E. Willmot). 0167-6687/00/$ – see front matter © 2000 Elsevier Science B.V. All rights reserved.

(also see Gerber (1979, Chapter 9, Section 3)). Recently, Gerber and Shiu (1997, 1998) study the joint distribution of the time of ruin, the surplus before the time of ruin, and the deficit at the time of ruin by considering an expected discounted penalty involving these three random variables. They show that this expected discounted penalty as a function of the initial surplus is the solution of a certain (defective) renewal equation. This approach allows for the use 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 defective renewal equations. Under this approach the solution of a defective renewal equation is expressed in terms of a compound geometric tail. This has the advantage of allowing for the use of analytic properties of a compound geometric distribution and for the implementation of exact and approximate results which have been developed for the 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 more importantly, for a large class of claim size distributions, the tail of the associated compound geometric distribution can be approximated very nicely by a combination of two exponential distributions. For example, if the claim size 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 probability mass at zero and the mean, and has the same asymptotic formula. This combination is referred to as the Tijms approximation (see Tijms (1994) and Willmot (1997)). Thus, we can take advantage of the analytic form of the Tijms approximation.

In this paper we extend the approach in Lin and Willmot (1999) to properties related to the joint and marginal moments of the time of ruin, the surplus before the time of ruin, and the deficit at the time of ruin. Although these problems have been considered by various authors, the present approach allows one to express the joint and marginal 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 moments at the time of ruin. In the rest of this section, we present the classical continuous time risk model, Gerber and Shiu’s expected discounted penalty function and its associated renewal equation, and the solution of the renewal equation in terms of a compound geometric tail which is obtained in Lin and Willmot (1999). Section 2 discusses equilibrium distributions and reliability classifications. Section 3 introduces auxiliary functions which involve the compound geometric tail associated with Gerber and Shiu’s renewal equation and the related equilibrium distributions. The auxiliary 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 of the 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. LetNtbe the number of claims from an insurance

portfolio. It is assumed thatNtfollows 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 =R

∞

0 x

j_{dP (x)forj =0,1,2, . . ., and Laplace–Stieltjes}

transformp(s)˜ =R∞ 0 e

−sx_{dP (x). The aggregate claims process is{S}

t;t ≥0}, whereSt =X1+X2+ · · · +XNt

(withSt =0 ifNt =0). The insurer’s surplus process is{Ut;t ≥0}withUt =u+ct−St, whereu≥0 is the

initial surplus,c=λp1(1+θ )the premium rate per unit time, andθ >0 the relative security loading.

LetT =inf{t;U (t ) <0}be the first time that the surplus becomes negative and is called the time of ruin. The probabilityψ(u)=Pr{T <∞}is called the probability of (ultimate) ruin. Two nonnegative random variables in connection with the time of ruin are the surplus immediately before the time of ruinU (T−), whereT−is the left limit ofT 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 classes}

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

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

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

ψ(u)=E{I (T <∞)}, (1.2)

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

φ (u)= λ c

Z u 0

φ (u−x) Z ∞

x

e−ρ(y−x)dP (y)dx+λ ce

ρuZ ∞ u

e−ρx Z ∞

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) has

a unique solution by identifying the Laplace transforms of the equation or applying the principle of contraction maps. 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 Z u

0

φ (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)−eρxR∞

x e

−ρy_{dP (y)}

ρR₀∞e−ρy_{P (y)}¯ _dy , (1.6)

β =_R∞(1+θ )p1

0 e−ρyP (y)¯ dy

−1, (1.7)

H (u)= e

ρuR∞

u e

−ρxR∞

x w(x, y−x)dP (y)dx

R∞

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 contraction maps.

An alternative expression forG(x)is given by

¯ G(x)=

R∞ 0 e

−ρy_{P (x¯ +y)dy}

R∞

0 e−ρyP (y)¯ dy

, x ≥0. (1.9)

Furthermore,G(x)is differentiable and

G′(x)= e

ρxR∞

x e

−ρy_{dP (y)}

R∞

0 e−ρyP (y)¯ dy

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

¯ K(u)=

∞

X

n=1 β

1+β

₁

1+β n

¯

G∗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 following renewal equation:

¯

K(u)= 1

1+β Z u

0 ¯

K(u−x)dG(x)+ 1

1+β ¯

G(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 follows

immediately from (1.8) and (1.9) thatH (u)= ¯G(u). Thus, withw(x1, x2)=1,φ (u)is the Laplace transform of T andK(u)¯ =φ (u).

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

G(x)= 1 p1

Z x 0

¯

P (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

Z u 0

H (u−x)dK(x)+ 1

1+βH (u), (1.15)

φ (u)= −1 β

Z u

0 ¯

K(u−x)dH (x)−H (0) β

¯ K(u)+ 1

βH (u). (1.16)

IfH (u)is differentiable, thenφ (u)may be expressed as

φ (u)= −1 β

Z u

0 ¯

K(u−x)H′(x)dx−H (0) β

¯ K(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 an important 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 a distribution function. The equilibrium DFP1(x)=1− ¯P1(x)ofP (x)is defined byP1(x)=Rx

0P (y)¯ dy/p1. In the

present contextP1(x)may be interpreted as the DF of the amount of the drop in the surplus level, given that there is a drop (e.g., Bowers et al. (1997, Chapter 12)). The DFP (x)is said to be decreasing (increasing) failure rate or DFR (IFR) ifP (x¯ +y)/P (x)¯ is nondecreasing (nonincreasing) inxfor fixedy ≥0. A larger class of distributions is the new worse (better) than used or NWU (NBU) class withP (x¯ +y)≥(≤)P (x)¯ P (y)¯ for allx ≥0 andy ≥0.

The mean residual lifetime ofP (x)is defined by

rP(x)=

R∞

x P (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 used

in expectation or NWUE (NBUE) ifrP(x)≥(≤)rP(0), or equivalentlyP¯1(x)≥(≤)P (x), x¯ ≥0. See Fagiuoli

and 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. Let P2(x) = 1− ¯P2(x)be the

equilibrium DF ofP1(x). The DFP2(x)is called the second-order equilibrium DF ofP (x). Similarly, we define

thenth order equilibrium DF ofP (x)by

Pn(x)=1− ¯Pn(x)=

Rx

0P¯n−1(y)dy R∞

0 P¯n−1(y)dy

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

whereP¯0(x)=1−P0(x)= ¯P (x). It can be shown that the mean of the DFPn(x), n=1,2, . . ., is given by

Z ∞ 0

¯

Pn(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

Z ∞

x

(y−x)ndP (y). (2.4)

See Hesselager et al. (1998) and references therein for further details.

The connection of the equilibrium DFsPn(x)ofP (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) can now 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)=

Z ∞ 0

xndG(x). (2.5)

Thus,

µn(0)=

Z ∞ 0

xndP1(x)=

1

p1 Z ∞

0

xnP (x)¯ dx= pn+1

(n+1)p1. (2.6)

Forρ >0, we have

µn(ρ)=µn(0)

R∞ 0 e

−ρx_dP n+1(x) R∞

0 e−ρxdP1(x)

. (2.7)

The equilibrium DF Gn(x)of G(x)can be expressed in terms of the equilibrium DF Pn(x). In fact we have,

analogous to (1.9),

¯ Gn(x)=

R∞ 0 e

−ρy_P¯

n(x+y)dy

R∞

0 e−ρyP¯n(y)dy

. (2.8)

3. Convolution of equilibrium distributions and compound geometric tails

In this section we introduce two auxiliary functions which involve the convolution of the compound geometric tailK(u)¯ and thenth equilibrium tailG¯n(x)of the “claim size” distributionG(x)= G0(x). Both functions are

necessary in computing the moments of the (discounted) deficit and the moments at the time of ruin as well as other important 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 follows

from (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)

Z u

0 ¯

K(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)=

Z u 0

¯

K(u−x)G¯n(x)dx+

Z ∞

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)

Z ∞

u

αn(t, ρ)dt−

Z ∞

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)

Z ∞

u

αn(t, ρ)dt =

Z ∞

u

Z t 0

¯

K(t−x)dGn+1(x)dt+

Z ∞

u

¯

Gn+1(t )dt.

By interchanging the order of integration, we have Z ∞

u

Z t 0

¯

K(t−x)dGn+1(x)dt = Z u

0 Z ∞

u

¯

K(t−x)dtdGn+1(x)+ Z ∞

u

Z ∞

x

¯

K(t−x)dtdGn+1(x) =

Z u 0

Z ∞

u−x

¯

K(t )dtdGn+1(x)+ ¯Gn+1(u)

Z ∞ 0

¯ K(t )dt.

Integration by parts yields Z u

0 Z ∞

u−x

¯

K(t )dtdGn+1(x) = − ¯Gn+1(x) Z ∞

u−x

¯ K(t )dt

u

x=0 +

Z u 0

¯

K(u−x)G¯n+1(x)dx = − ¯Gn+1(u)

Z ∞ 0

¯

K(t )dt+ Z ∞

u

¯ K(t )dt+

Z u 0

¯

K(u−x)G¯n+1(x)dx.

Thus 1

gn(ρ)

Z ∞

u

αn(t, ρ)dt =

Z ∞

u

¯ K(t )dt+

Z u 0

¯

K(u−x)G¯n+1(x)dx+ Z ∞

u

¯

Gn+1(t )dt =

Z ∞

u

¯

K(t )dt+αn+1(u, ρ)

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

Theorem 3.1.

α0(u, ρ)=β Z ∞

u

¯

K(x)dx (3.5)

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

αn(u, ρ)=

β µn(ρ)

Z ∞

u

(x−u)nK(x)¯ dx−

n−1 X

j=0

n j

_µ

n−j(ρ)

µn(ρ)

Z ∞

u

(x−u)jK(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

Z ∞

u

¯ K(t )dt−

Z ∞

u

¯

K(t )dt =β Z ∞

u

¯ K(t )dt,

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

u

Z ∞

t

(y−t )jK(y)¯ dydt= 1 j+1

Z ∞

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)=

Z u

0

ψ(u−x)P¯n(x)dx+

Z ∞

u

¯

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

and so

τn(u)=

pn+1 (n+1)pn

Z u

0

ψ(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 Z ∞

u

ψ(x)dx (3.10)

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

τn(u)=

np1θ

pn

Z ∞

u

(x−u)n−1ψ(x)dx−

n−2 X

j=0

n j

_p

n−j

pn

Z ∞

u

(x−u)jψ(x)dx. (3.11)

Proof. Whenδ =0,ρ =0,G(x)=P1(x), and (2.6) holds. ThusP¯n(x)= ¯Gn−1(x). SinceK(u)¯ =ψ(u)in this

case, it follows from (3.3) and (3.8) that

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

It is evident in the proofs of the above theorems that the existence of certain moments of the claim size distribution

P (x)is required. For instance, ifpn+2is finite, thenαn(u, ρ)andτn+1(u) are both finite. In order to focus on

analytical representation of functions of interest, we always assume that required moments are finite throughout the rest of the paper.

In the following examples we demonstrate how to evaluateαn(u, ρ)andτn(u)whenP (x)is a combination of

exponentials or a mixture of Erlangs. The case whenP (x)is a combination of exponentials is considered in Gerber et al. (1987) and Dufresne and Gerber (1988). From a formal standpoint, a combination of two exponentials is very 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 corresponding

approximations for moments at the time of ruin and the deficit of ruin, as is discussed in later sections. In what follows, a combination of two exponentials may be viewed in this context.

Example 3.1 (Combinations of exponentials). The case where P (x) is a combination of exponentials is first considered in Gerber et al. (1987) and Dufresne and Gerber (1988) in which a closed form expression for the probability 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, . . . , P¯n(x)=e−µx. It

follows from (2.8) that

¯

Gn(x)=e−µx.

Thus,µn(ρ)=µn(0)=n!/µnfrom (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 = θ µ

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)=C1e−R1u_+C2e−R2u_.

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

Since Z ∞

u

(x−u)jK(x)¯ dx =C1e−R1u Z ∞

0

xje−R1x_dx+C2e−R2u Z ∞

0

xje−R2x_dx

=C1j!R₁−j−1e−R1u_+C2j!R−j−1

2 e

−R2u_, Theorem 3.1 yields

αn(u, ρ)=C∗1,ne

−R1u_+C∗

2,ne

where

C₁∗_,n=C1 



βn!R−₁n−1 µn(ρ)

−

n−1 X

j=0

n j

_µ

n−j(ρ)j!R

−j−1 1 µn(ρ)



,

C₂∗_,n=C2 



βn!R−₂n−1 µn(ρ)

−

n−1 X

j=0

n j

_µ

n−j(ρ)j!R

−j−1 2 µn(ρ)



.

Withρ=0, we obtain from (2.6) that

τn(u)=C1∗∗,ne

−R1u_+C∗∗

2,ne

−R2u_{, (3.16)}

C₁∗∗_,n=n!C1R

−n

1 pn



p1θ−

n−1 X

j=1 pj+1R₁j (j+1)!



,

C₂∗∗_,n=n!C2R

−n

2 pn



p1θ−

n−1 X

j=1 pj+1R₂j (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 any continuous distribution on(0,∞)may be approximated arbitrarily accurately by a distribution of this type (see Tijms (1994, pp. 162–164)).

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

P′(x)=

r

X

k=1 qk

µ(µx)k−1e−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

∞

X

j=0 ¯ Cj

(µu)j

j! , u≥0, where

¯ Cj =

1 1+β

j

X

k=1

q_k∗C¯j−k+

1 1+β

∞

X

k=j+1

q_k∗, j =1,2,3, . . . .

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

q_k∗= Pr

j=kqj

µ µ+ρ

j−k

Pr

j=1qjPj

−1

i=0

µ µ+ρ

i,

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

u

(x−u)j(µx)

i

i! e

−µx_{dx =}µi

i! e

−µuZ ∞

0

xj(x+u)ie−µxdx =µ

i

i! e

−µu i

X

k=0

i k

uk

Z ∞ 0

xj+i−ke−µxdx

=µ−j−1e−µu

i

X

k=0

(j+i−k)!

(i−k)!

(µu)k k! . Thus,

αn(u, ρ)=e−µu

∞

X

k=0 ¯ C_k,n∗ (µu)

k

k! , (3.17)

where

¯ C_k,n∗ =

∞

X

i=0 ¯ Ck+i





βµ−n−1 µn(ρ)

(n+i)!

i! −

n−1 X

j=0

n j

_µ

n−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 the moments. We are focusing on their analytic solutions. For numerical algorithms, see Dickson et al. (1995) and the references therein.

Consider the special casew(x1, x2)=x₂k, wherekis a positive integer. A relatively simple representation exists forφ (u)in (1.1).

Theorem 4.1. Fork=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)=

R∞ 0 x

n_{dG(x)is given in (2.7) andα}

n(u, ρ)is given in Theorem 3.1.

Proof. Using (2.4) and (2.8), Eq. (1.8) becomes, withw(x1, x2)=x2k, H (u)= e

ρuR∞

u e

−ρxR∞

x (y−x)kdP (y)dx

R∞

0 e−ρyP (y)¯ dy

=pke

ρuR∞

u e

−ρx_P¯ k(x)dx

p1 R∞

0 e−ρydP1(y)

= pk R∞

0 e

−ρy_P¯ k(y)dy

p1 R∞

0 e−ρydP1(y) ¯ Gk(u).

Furthermore, using (2.3),

H (u)= pk+1 R∞

0 e

−ρy_dP k+1(y) (k+1)p1R

∞

0 e−ρydP1(y) ¯ Gk(u),

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

Then, using (3.1), sinceG¯k(u)=R_u∞G¯k−1(x)dx/gk−1(ρ), we obtain

H′(u)= − µk(ρ) gk−1(ρ)

¯

Gk−1(u)= −kµk−1(ρ)G¯k−1(u),

and from (1.17),

φ (u) = kµk−1(ρ) β

Z u

0 ¯

K(u−x)G¯k−1(x)dx+ µk(ρ)

β Z ∞

u

¯ Gk−1(x)

gk−1(ρ)

dx−µk(ρ) β

¯ K(u) = kµk−1(ρ)

β

Z u

0 ¯

K(u−x)G¯k−1(x)dx+

Z ∞

u

¯

Gk−1(x)dx

−µk(ρ) β K(u)¯ = kµk−1(ρ)

β αk−1(u, ρ)− µk(ρ)

β K(u),¯

using (3.3). That is,

E{e−δT|U (T )|kI (T <∞)} = kµk−1(ρ)

β αk−1(u, ρ)− µk(ρ)

β ¯

K(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 <∞)} = Z ∞

u

¯

K(x)dx−µ1(ρ) β

¯

K(u). (4.4)

Whenδ =0 we have the following important corollary.

Corollary 4.1. Fork=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 in Corollary 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) reduces

to (4.5).

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

E{|U (T )|

T <∞} = Z ∞

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−1_θ

µ−1e−Ru

(1+θ )−1_e−Ru −

(k+1)!µ−k−1 (k+1)µ−1_θ =k!µ

−k_,

which agrees with the fact that in the exponential case the deficit is also exponentially distributed and it is independent of 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

= C1ˆ ,ke

−R1u_{+ ˆC2}

,ke−R2u

C1e−R1u+C2e−R2u

, (4.7)

where

ˆ C1,k=

k!C1R−₁k p1θ



p1θ−

k

X

j=1 pj+1R₁j (j+1)!



,

ˆ C2,k=

k!C2R−₂k p1θ



p1θ−

k

X

j=1 pj+1Rj₂ (j+1)!



.

Example 4.2 (Mixtures of Erlangs). For a mixture of Erlangs claim size distribution, it follows immediately from

Example 3.2 that thekth unconditional discounted moment is

E{e−δT|U (T )|kI (T <∞)} =e−µu

∞

X

j=0 ˆ Cj,k

(µu)j

j! , (4.8)

where

ˆ Cj,k=

1

β[kµk−1(ρ) ¯

C_j,k∗ ₋₁−µk(ρ)C¯j].

The constantsC¯_j,k∗ ₋₁andC¯j are given in Example 3.1.

We remark that when the individual claim size is a combination of exponentials or a mixture of Erlangs, the density function of the deficit is of the same type but with new weights. These weights are expressed as functions of the 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 the deficit. 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 different values ofuwithout reevaluating the distribution of the deficit. Another advantage of this approach is that we are able 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. IfP (x)is IMRL (DMRL), then fork=1,2,3, . . .,

E{e−δT|U (T )|kI (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 thejth equilibrium DFGj(x)is NWUE (NBUE), i.e.,G¯j+1(x)≥(≤)G¯j(x). Then by (4.2),

H (u)=µk(ρ)G¯k(u)≥(≤)µk(ρ)G¯k−1(u)≥(≤)· · · ≥(≤)µk(ρ)G(u).¯

Thus, Theorem 4.1 of Lin and Willmot (1999) applies, which shows that ifH (u) ≤ (≥)cG(u)¯ , then φ (u) ≤

(≥)cK(u)¯ , and the result is proved.

Lemma 4.1. If

η(x)= 1 ¯ P (x)

Z ∞

x

w(x, y−x)dP (y)≤(≥)c, (4.10)

where 0< 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−δTw(|U (T )|)I (T <∞)} ≥(≤)

Z ∞

0

w(y)dA(y)

¯

K(u), u≥0. (4.12)

Proof. LetZxforx ≥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)≥(≤)R∞

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). IfP (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 <∞} ≥(≤) Z ∞

0

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

In the special case whenw(x1, x2) = x₂k 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)and w(x1, x2) = w(x2), then η(x) = R

∞

0 w(y)dP (y) independent ofx and H (u)= {R∞

0 w(y)dP (y)} ¯G(u), implying thatφ (u)= { R∞

0 w(y)dP (y)} ¯K(u). Thus, withw(y)=e

−sy_,

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

Eq. (4.14) may be viewed as the joint Laplace transform of the (defective) distribution ofT and|U (T )|, implying thatT and|U (T )|are independent (conditional onT < ∞), and that|U (T )|is exponential with meanp1. This

result is given by Gerber (1979, p. 138). The Laplace transform of the deficit for arbitrary claim size DF can also be evaluated in a similar manner. In that case, we simply letw(x1, x2)=e−sx2. However, the resulting formula will

be 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 obtained withδ=0 andw(x, y)=xjyk. Then (1.8) becomes, using (2.4),

H (u)= 1 p1

Z ∞

u

xj Z ∞

x

(y−x)kdP (y)dx= pk p1

Z ∞

u

Thus,

H′(u)= −pk p1

ujP¯k(u),

and

H (0)= pk+1 (k+1)p1

Z ∞ 0

xjdPk+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 )|kI (T <∞)} = pk p1θ

Z u

0

ψ(u−x)xjP¯k(x)dx+

Z ∞

u

xjP¯k(x)dx

−

_k!j!p

k+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

θ ψ(u) Z u

0

ψ(u−x)xjdP1(x)+ Z ∞

u

xjdP1(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 a mixture 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)= Z ∞

u

e−sxdP1(x),

and (1.17) yields

E{e−sU (T−)I (T <∞)} = 1 θ

Z u 0

ψ(u−x)e−sxdP1(x)+

1

θ Z ∞

u

e−sxdP1(x)− ψ(u)

θ Z ∞

0

e−sxdP1(x) =1

θ Z u

0

[ψ(u−x)−ψ(u)] e−sxdP1(x)+1 θ

Z ∞

u

[1−ψ(u)] e−sxdP1(x). (5.4) Hence, by the uniqueness of the Laplace transform, the defective density of the surplusU (T−)before the time of ruin 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 argumentδto find moments involvingT. Our approach is to differentiate (1.5) with respect toδand show that the corresponding moment satisfies another 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)= Z u

0

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

m(x, ρ)= Z ∞

x

e−ρ(y−x)dP (y), x ≥0. (6.2)

If the second momentp2ofP (x)is finite,M(u, δ)is a differentiable function with respect toδon [0, δ0)for some

δ0>0, andM1(u, δ)=∂M(u, δ)/∂δis continuous in u, thenφ (u, δ)is differentiable with respect toδon [0, δ0)

andφ1(u, δ)=∂φ (u, δ)/∂δis the solution of the following renewal equation: c

λφ1(u, δ)= Z u

0

φ1(u−x, δ)m(x, ρ)dx+ Z u

0

φ (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)= Z u

0

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

where

C(u, δ, 1δ)= Z u

0

φ (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)= sup

0≤x≤u

|B(u, δ, 1δ)|.

Then,

c

λJ (u, δ, 1δ) ≤J (u, δ, 1δ) Z u

0

m(x, ρ(δ+1δ))dx+ |C(u, δ, 1δ)| ≤J (u, δ, 1δ)

Z u

0 ¯

P (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

lim

1δ→0

φ (u, δ+1δ)−φ (u, δ)

1δ =φ1(u, δ),

It is obvious that lim

1δ→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)

=

Z ∞

x

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

1δ dP (y)

=

1δ Z ∞

x

{[ρ′(δ1)]2exp[−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)=

Z u 0

φ (u−x)m(x, ρ)dx+ Z ∞

u

exp[−ρ(δ)(x−u)] Z ∞

x

w(x, y−x)dP (y)dx. (6.4) Thus, Lemma 6.1 yields

c λ

∂

∂δφ (u) = Z u

0 _∂

∂δφ (u−x)

m(x, ρ)dx+ρ′(δ) Z u

0

φ (u−x) ∂

∂ρm(x, ρ)dx −ρ′(δ)

Z ∞

u

(x−u)exp[−ρ(δ)(x−u)] Z ∞

x

w(x, y−x)dP (y)dx. (6.5)

Next, define

H (u,0)= 1 p1

Z ∞

u

Z ∞

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 Z u

0

φ1(u−x)dP1(x)+ 1

1+θH1(u), (6.8)

where

H1(u)=

1

λp1θ Z u

0

ψ(u−x)H (x,0)dx+ 1 λp2₁θ

Z ∞

u

(x−u) Z ∞

x

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= − Z ∞

x

(y−x)dP (y)= −p1P¯1(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) = Z u

0

φ1(u−x)P (x)¯ dx+ 1 λθ

Z u 0

φ (u−x,0)P1(x)¯ dx + 1

λp1θ Z ∞

u

(x−u) Z ∞

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 Z u

0

φ (u−x,0)dP1(x)+

1

1+θH (u,0) (6.13)

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

˜ ψ(s)=

Z ∞ 0

e−sxψ(x)dx = 1

1+θ

_{1− ˜p} 1(s)

s 1−

1 1+θp˜1(s)

−1

, (6.14)

wherep˜1(s)= R∞

0 e

−sx_dP

1(x). Then from (6.13), ˜

φ(s,0)= Z ∞

0

e−suφ (u,0)du= 1

1+θH (s,˜ 0)

1− 1

1+θp1(s)˜ −1

withH (s,˜ 0)=R∞ 0 e

−su_{H (u,0)du. Therefore,s}−1_{{1− ˜p}

1(s)} ˜φ(s,0)= ˜H (s,0)ψ(s)˜ , implying that Z u

0

φ (u−x,0)P¯1(x)dx = Z u

0

ψ(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. Fork=0,1,2, . . ., the functionφ1,k(u)satisfies the defective renewal equation

φ1,k(u)=

1 1+θ

Z u

0

φ1,k(u−x)dP1(x)+

pk+1 (k+1)cp1θ

τk+1(u), (6.17)

and is given explicitly by

φ1,k(u) =

pk

λp₁2θ2 Z u

0

ψ(u−x)τk(x)dx+

pk+1 (k+1)λp₁2θ2

τk+1(u)−

Z u 0

ψ(u−x)ψ(x)dx

− pk+2

(k+1)(k+2)λp₁2θ2ψ(u), (6.18)

Proof. Withδ =0 andw(x, y)=yk, Eqs. (2.4) and (6.6) yieldH (u,0)=[pk+1/(k+1)p1]P¯k+1(u), and

Z ∞

u

(x−u) Z ∞

x

(y−x)kdP (y)dx =pk

Z ∞

u

(x−u)P¯k(x)dx

= pk+1 k+1

Z ∞

u

(x−u)dPk+1(x)=

pk+1 k+1

Z ∞

u

¯

Pk+1(x)dx.

Then substitution into (6.9) yields

H1(u)=

pk+1 (k+1)λp2₁θ

Z u

0

ψ(u−x)P¯k+1(x)dx+ Z ∞

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

Z ∞ 0

¯

Pk+1(x)dx=

pk+2 (k+1)(k+2)λp₁2θ

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

τk+1(u)=

(k+1)pk

pk+1 Z ∞

u

τk(t )dt−

Z ∞

u

ψ(t )dt, (6.20)

which holds fork=0,1,2, . . ., withτ0(u)=p1(1+θ )ψ(u). Hence τ_k′₊₁(u)= −(k+1)pk

pk+1

τk(u)+ψ(u),

and from (6.19),H₁′(u)= −{(pk/λp2₁θ )τk(u)−[pk+1/((k+1)λp₁2θ )]ψ(u)}. Then from (1.17),

φ1,k(u) =

1

θ Z u

0

ψ(u−x) (

pk

λp2₁θτk(x)−

pk+1

(k+1)λp₁2θψ(x) )

dx + pk+1

(k+1)λp2₁θ2τk+1(u)−

pk+2

(k+1)(k+2)λp2₁θ2ψ(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

Z u 0

ψ1(u−x)dP1(x)+

1

c Z ∞

u

ψ(x)dx (6.22)

and is given explicitly by

ψ1(u)=

1

λp1θ Z u

0

ψ(u−x)ψ(x)dx+ Z ∞

u

ψ(x)dx− p2

2p1θ ψ(u)

Proof. With the help of (3.10), (6.17) becomes (6.22) withk=0 andψ1(u)=φ1,0(u). Similarly, (6.18) reduces

to (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

Z u 0

φ1,1(u−x)dP1(x)+

1

c Z ∞

u

(x−u)ψ(x)dx− p2

2cp1θ Z ∞

u

ψ(x)dx (6.25)

and is given explicitly by

φ1,1(u) =

1

λp1θ Z u

0

ψ(u−x) Z ∞

x

ψ(t )dtdx− p2

2λp₁2θ2 Z u

0

ψ(u−x)ψ(x)dx + 1

λp1θ Z ∞

u

(x−u)ψ(x)dx− p2

2λp₁2θ2 Z ∞

u

ψ(x)dx− p3

6λp₁2θ2ψ(u). (6.26)

Proof. Whenk=1, (6.19) reduces to

H1(u)= 1 λp1

Z ∞

u

(x−u)ψ(x)dx− p2

2λp₁2θ Z ∞

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)satisfying the 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 andwmay be obtained using Theorem 6.1 for arbitraryw(x1, x2). We now turn to evaluation of higher moments at the time of

ruinT.

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

E(Tk|T <∞)= ψk(u)

ψ(u), (6.27)

whereψk(u)satisfies a sequence of defective renewal equations

ψk(u)=

1 1+θ

Z u

0

ψk(u−x)dP1(x)+ k c

Z ∞

u

ψk−1(x)dx (6.28)

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

ψk(u)=

k λp1θ

Z u

0

ψ(u−x)ψk−1(x)dx+

Z ∞

u

ψk−1(x)dx−ψ(u)

Z ∞ 0

ψk−1(x)dx

(6.29)

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

ψk(u)=E{TkI (T <∞)} =(−1)k

dk

dδkK(u)|¯ δ=0.

To prove the theorem we begin with the defective renewal equation

c λ

¯ K(u)=

Z u

0 ¯

K(u−x)m(x, ρ)dx+ Z ∞

u

m(x, ρ)dx, (6.30)

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

m(x, ρ)=G′(x) Z ∞

0

e−ρyP (y)¯ dy,

this equation is obtained from (1.8), (1.9) and (6.4) withw(x1, x2)=1. We note that sinceR

∞

u m(x, ρ)dxhas the

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

(6.30)ntimes.

Define the Laplace transforms

˜

K(s, ρ)= Z ∞

0

e−sxK(x)¯ dx, m(s, ρ)˜ = Z ∞

0

e−sxm(x, ρ)dx, g(s)˜ = Z ∞

0

e−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)˜ =R∞ 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+θp˜1(s)

−δ

ρK(s, ρ)˜ =

1− 1

1+θp˜1(s)

cρ−δ, (6.35)

where

˜ p1(s)=

Z ∞ 0

e−sxdP1(x)=

1− ˜p(s) p1s

.

Also, define the Laplace transform

˜ ψk(s)=

Z ∞ 0

e−sxψk(x)dx.

An obvious relation is

˜

ψk(s)=(−1)k

dk

Eq. (6.28) is equivalent to

s

1− 1

1+θp˜1(s)

˜ ψk(s)=

k c h

˜

ψk−1(0)− ˜ψk−1(s) i

. (6.37)

We now prove this identity by induction.

Fork = 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+θp˜1(s)

ρ(k+1)(0), (6.38)

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

cs

1− 1

1+θp˜1(s) _dk+1

dδk+1 h

ρ(δ)K(s, ρ)˜ i _δ=0

−(k+1)d

k

dδk[ρ(δ)K(s, ρ)˜ ]

δ=0

. (6.39)

Sinceρ(0)=0, dn

dδn[ρ(δ)K(s, ρ)˜ ]

_δ=0

=

n−1 X

j=0 (−1)j

n j

ρ(n−j )(0)ψ˜j(s)

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

s

1− 1

1+θp˜1(s) k

X

j=0 (−1)j

k+1

j

ρ(k−j+1)(0)ψ˜j(s)

=k+1 c

k−1 X

j=0 (−1)j

k j

ρ(k−j )(0)ψ˜j(s)+

1− 1

1+θp1(s)˜

ρ(k+1)(0). (6.40)

Lets=0 in (6.40). We have

k+1

c

k−1 X

j=0 (−1)j

k j

ρ(k−j )(0)ψ˜j(0)+

1− 1

1+θ

ρ(k+1)(0)=0.

Thus, (6.40) can be rewritten as

s

1− 1

1+θp˜1(s) k

X

j=0 (−1)j

k+1

j

ρ(k−j+1)(0)ψ˜j(s)

= −k+1 c

k−1 X

j=0 (−1)j

k j

ρ(k−j )(0)[ψ˜j(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 ))p˜1(s) ,

the first term in the left-hand side of (6.41) is equal to 1

1+θ[1− ˜p1(s)]ρ

Hence,

s

1− 1

1+θp˜1(s) k

X

j=1 (−1)j

k+1

j

ρ(k−j+1)(0)ψ˜j(s)

= −k+1 c

k−1 X

j=0 (−1)j

k j

ρ(k−j )(0)[ψ˜j(0)− ˜ψj(s)]. (6.42)

By the induction assumption,

s

1− 1

1+θp˜1(s)

(−1)j+1

k+1

j+1

ρ(k−j )(0)ψ˜j+1(s)= − k+1

c (−1)

j

k j

ρ(k−j )(0)[ψ˜j(0)− ˜ψj(s)]

forj =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)ψ˜k(s)= −

k+1

c (−1)

k−1_kρ′

(0)[ψ˜k−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) R∞

u ψk−1(x)dx. Corollary 6.3. The second moment of the time to ruin, given that ruin has occurred, is given by

E(T2

T <∞)= ψ2(u)

ψ(u), (6.44)

whereψ2(u)satisfies the defective renewal equation

ψ2(u)=

1 1+θ

Z u 0

ψ2(u−x)dP1(x)+

1

1+θH2(u) (6.45)

with

H2(u)=

2

λ2_p2 1θ

Z u

0

ψ(u−x) Z ∞

x

ψ(t )dtdx+ Z ∞

u

(x−u)ψ(x)dx

(6.46)

and is given explicitly by

ψ2(u) =

2

λ2_p2 1θ2

Z u 0

ψ(u−x) Z x

0

ψ(x−t )ψ (t )dtdx+ 4 λ2_p2

1θ2 Z u

0

ψ(u−x) Z ∞

x

ψ(t )dtdx + 2

λ2_p2 1θ2

Z ∞

u

(x−u)ψ(x)dx− p2 λ2_p3

1θ3 Z u

0

ψ(u−x)ψ(x)dx−2p1p3θ+3p 2 2

6λ2_p4 1θ4

ψ(u). (6.47)

Proof. Since

H2(u)=

2

λp1 Z ∞

u

ψ1(x)dx,

replacingψ1(x)by (6.23) and employing integration by parts on each term yields (6.46). The fact thatR₀∞ψ(x)dx = p2/2p1θis used in the computation. Eq. (6.47) is a direct application of Theorem 1.1. Noting that

H₂′(u)= 2 λ2_p2

1θ

− Z u

0

ψ(u−x)ψ(x)dx− Z ∞

u

ψ(x)dx+ p2

2p1θ ψ(u)

,

We now evaluate the moments at the time of ruin in the case when the claim size is either a combination of exponentials 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 to point out that if the claim size distribution is neither a combination of exponentials nor a mixture of Erlangs, the calculation of higher(n >2)moments could be complicated. However, we may use the Tijms approximation and in 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) = Ce−Ru, whereR=θ µ/(1+θ ). It is not hard to see from (6.29) by induction that

ψk(u)=e−Ru k

X

j=0 ¯ Cj,k

(Ru)j

j! . (6.48)

We now derive a recursive formula for the coefficientsC¯j,k.

From (6.29), Z u

0

ψ(u−x)ψk−1(x)dx=

1 1+θe

−Ru k−1 X

j=0 ¯ Cj,k−1

Z u 0

(Rx)j j! dx =

1

(1+θ )Re

−Ru k

X

j=1 ¯ Cj−1,k−1

(Ru)j j! ,

Z ∞

u

ψk−1(x)dx =R−1

k−1 X

j=0 ¯ Cj,k−1

Z ∞

u

Rj+1xje−Rx

j! dx =R

−1_e−Ru k−1 X

j=0 



k−1 X

i=j

¯ Ci,k−1



 (Ru)j

j!

Z ∞ 0

ψk−1(x)dx =R−1

k−1 X

i=0 ¯ Ci,k−1.

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

¯ C0,k=

k(1+θ ) cµθ

k−1 X

i=0 ¯

Ci,k−1 (6.49)

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

¯ Cj,k=

k(1+θ )2 cµθ2



 1

1+θC¯j−1,k−1+

k−1 X

i=j

¯ Ci,k−1



, (6.50)

withPk−1

i=k =0 andC¯0,0=C. The first two moments can be easily derived using (6.49) and (6.50). ¯

C0,1=

1+θ cµθ C0¯ ,0=

1

cµθ, ¯

C1,1=

(1+θ )2 cµθ2

1 1+θ

¯ C0,0=

1

cµθ2.

Thus,

E{TI(T <∞)} =

₁

cµθ +

1

cµθ2(Ru)

e−Ru,

Also,

¯ C0,2=

2(1+θ ) cµθ [

¯

C0,1+ ¯C1,1]=

2(1+θ )2 c2_µ2_θ3 , ¯

C1,2=

2(1+θ )2 cµθ2

₁

1+θ ¯

C0,1+ ¯C1,1

=2(1+θ )(1+2θ ) c2_µ2_θ4 , ¯

C2,2=

2(1+θ )2 cµθ2

1

1+θC¯1,1=

2(1+θ ) c2_µ2_θ4 . E{T2I (T <∞)} = 2

c2_µ2_θ3

(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)=C1e−R1u_+C2e−R2u_,

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

ψk(u)= k

X

j=0

[Aj,ke−R1u+Bj,ke−R2u]

uj

j!. (6.51)

The coefficientsAj,kandBj,kare 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 Z u

0

ψ(u−x)ψ(x)dx =C₁2ue−R1u_+C2

2ue

−R2u₊ 2C1C2

R2−R1[e

−R1u_−e−R2u_], Z ∞

u

ψ(x)dx= C1 R1

e−R1u₊C2

R2

e−R2u_. Thus,

E{TI(T <∞)} = 1+θ cθ

C₁2ue−R1u_+C2

2ue

−R2u_{+ C}

1 R1

+ 2C1C2 R2−R1

− _C

1 R1

+C2 R2

C1

e−R1u

+ _C

2 R2

− 2C1C2 R2−R1

− _C

1 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

∞

X

j=0 Cj

(µu)j j! .

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

ψk(u)=e−µu

∞

X

j=0 Cj,k

(µu)j

j! , (6.52)

A recursive algorithm forCj,kis now derived. Since

Z u 0

ψ(u−x)ψk−1(x)dx =e−µu

∞

X

j=0

∞

X

i=0

CjCi,k−1 µj+i

j!i! Z u

0

(u−x)jxidx =e−µu

∞

X

j=0

∞

X

i=0

CjCi,k−1

µj+iuj+i+1 j!i!

Z 1 0

(1−t )jtidt

=e−µu

∞

X

j=0

∞

X

i=0

CjCi,k−1

µj+iuj+i+1 (j+i+1)!=µ

−1_e−µu

∞

X

j=1 



j

X

i=1

Cj−iCi−1,k−1 

 (µu)j

j! ,

Z ∞

u

ψk−1(x)dx =µ−1

∞

X

j=0 Cj,k−1

Z ∞

u

µj+1xje−µx

j! dx

=µ−1e−µu

∞

X

j=0 Cj,k−1

j

X

i=0 (µu)i

i! =µ

−1_e−µu

∞

X

j=0 



∞

X

i=j

Ci,k−1 

 (µu)j

j! , Z ∞

0

ψk−1(x)dx =µ−1

∞

X

i=0 Ci,k−1.

Thus, from (6.29) we obtain

C0,k=

k(1+θ )

cµθ (1−C0)

∞

X

i=0

Ci,k−1= k

λθ(1−C0)

∞

X

i=0

Ci,k−1, (6.53)

and forj =1,2, . . .,

Cj,k=

k λθ





j

X

i=1

Cj−iCi−1,k−1+

∞

X

i=j

Ci,k−1−Cj

∞

X

i=0 Ci,k−1



. (6.54)

Acknowledgements

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

References

Bowers, N., Gerber, H., Hickman, J., Jones, D., Nesbitt, C., 1997. Actuarial Mathematics, 2nd Edition. Society of Actuaries, Schaumburg, IL. Cheng, S., Gerber, H.U., Shiu, E.S.W., 2000. Discounted probabilities and ruin theory in the compound binomial model. Insurance: Mathematics

and Economics, 26, 239–250.

Delbaen, F., 1990. A remark on the moments of ruin time in classical risk theory. Insurance: Mathematics and Economics 9, 121–126. De Vylder, F.E., 1996. Advanced Risk Theory: A Self-contained Introduction. Editions de L’Universite de Bruxelles, Brussels. Dickson, D.C.M., 1992. On the distribution of surplus prior to ruin. Insurance: Mathematics and Economics 11, 191–207.

Eq. (6.28) is equivalent to

s

1

−

1

1

+

θ

p

˜

1

(s)

˜

ψ

k

(s)

=

k

c

h

˜

ψ

k−1

(

0

)

− ˜

ψ

k−1

(s)

i

.

(6.37)

We now prove this identity by induction.

For

k

=

1, this is a restatement of Corollary 6.1. Assume that for

n

=

1

, . . . , k

−

1, identity (6.37) holds.

Differentiate Eq. (6.35)

k

+

1

(k

≥

1

)

times at

δ

=

0. The right-hand side is simply

c

1

−

1

1

+

θ

p

˜

1

(s)

ρ

(k+1)

(

0

),

(6.38)

where

ρ

(n)

denotes the

n

th derivative of

ρ

, while the left-hand side equals

cs

1

−

1

1

+

θ

p

˜

1

(s)

_d

k+1

d

δ

k+1

h

ρ(δ)

K(s, ρ)

˜

i

_δ=0

−

(k

+

1

)

d

k

d

δ

k

[

ρ(δ)

K(s, ρ)

˜

]

δ=0

.

(6.39)

Since

ρ(

0

)

=

0,

d

n

d

δ

n

[

ρ(δ)

K(s, ρ)

˜

]

_δ=0

=

n−1

X

j=0

(

−1

)

j

n

j

ρ

(n−j )

(

0

)

ψ

˜

j

(s)

for

n

=

k, k

+

1. Equating (6.38) and (6.39) yields

s

1

−

1

1

+

θ

p

˜

1

(s)

k

X

j=0

(

−1

)

j

k

+

1

j

ρ

(k−j+1)

(

0

)

ψ

˜

j

(s)

=

k

+

1

c

k−1

X

j=0

(

−1

)

j

k

j

ρ

(k−j )

(

0

)

ψ

˜

j

(s)

+

1

−

1

1

+

θ

p

˜

1

(s)

ρ

(k+1)

(

0

).

(6.40)

Let

s

=

0 in (6.40). We have

k

+

1

c

k−1

X

j=0

(

−1

)

j

k

j

ρ

(k−j )

(

0

)

ψ

˜

j

(

0

)

+

1

−

1

1

+

θ

ρ

(k+1)

(

0

)

=

0

.

Thus, (6.40) can be rewritten as

s

1

−

1

1

+

θ

p

˜

1

(s)

k

X

j=0

(

−1

)

j

k

+

1

j

ρ

(k−j+1)

(

0

)

ψ

˜

j

(s)

= −

k

+

1

c

k−1

X

j=0

(

−1

)

j

k

j

ρ

(k−j )

(

0

)

[

ψ

˜

j

(

0

)

− ˜

ψ

j

(s)

]

+

1

1

+

θ

[1

− ˜

p

1

(s)

]

ρ

(k+1)

₍

₀

_).

_(6.41)

With the fact that

˜

ψ

0

(s)

= ˜

ψ(s)

=

(

1

− ˜

p

1

(s))/s(

1

+

θ )

1

−

(

1

/(

1

+

θ ))

p

˜

1

(s)

,

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

1

1

+

θ

[1

− ˜

p

1

(s)

]

ρ

Hence,

s

1

−

1

1

+

θ

p

˜

1

(s)

k

X

j=1

(

−1

)

j

k

+

1

j

ρ

(k−j+1)

(

0

)

ψ

˜

j

(s)

= −

k

+

1

c

k−1

X

j=0

(

−1

)

j

k

j

ρ

(k−j )

(

0

)

[

ψ

˜

j

(

0

)

− ˜

ψ

j

(s)

]

.

(6.42)

By the induction assumption,

s

1

−

1

1

+

θ

p

˜

1

(s)

(

−1

)

j+1

k

+

1

j

+

1

ρ

(k−j )

(

0

)

ψ

˜

j+1

(s)

= −

k

+

1

c

(

−1

)

j

k

j

ρ

(k−j )

(

0

)

[

ψ

˜

j

(

0

)

− ˜

ψ

j

(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

+

θ

p

˜

1

(s)

(

−1

)

k

(k

+

1

)ρ

′

(

0

)

ψ

˜

k

(s)

= −

k

+

1

c

(

−1

)

k−1

_kρ

′

(

0

)

[

ψ

˜

k−1

(

0

)

− ˜

ψ

k−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), where

G(x)

=

P

1

(x)

and

H (u)

=

(k/λp

1

)

R

∞

u

ψ

k−1

(x)

d

x

.

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

E(T

2

T <

∞

)

=

ψ

2

(u)

ψ(u)

,

(6.44)

where

ψ

2

(u)

satisfies the defective renewal equation

ψ

2

(u)

=

1

1

+

θ

Z

u

0

ψ

2

(u

−

x)

d

P

1

(x)

+

1

1

+

θ

H

2

(u)

(6.45)

with

H

2

(u)

=

2

λ

2

_p

2 1

θ

Z

u

0

ψ(u

−

x)

Z

∞

x

ψ(t )

d

t

d

x

+

Z

∞

u

(x

−

u)ψ(x)

d

x

(6.46)

and is given explicitly by

ψ

2

(u)

=

2

λ

2

_p

2 1

θ

2

Z

u

0

ψ(u

−

x)

Z

x

0

ψ(x

−

t )ψ (t )

d

t

d

x

+

4

λ

2

_p

2 1

θ

2

Z

u

0

ψ(u

−

x)

Z

∞

x

ψ(t )

d

t

d

x

+

2

λ

2

_p

2 1

θ

2

Z

∞

u

(x

−

u)ψ(x)

d

x

−

p

2

λ

2

_p

3 1

θ

3

Z

u

0

ψ(u

−

x)ψ(x)

d

x

−

2

p

1

p

3

θ

+

3

p

2 2

6

λ

2

_p

4

1

θ

4

ψ(u).

(6.47)

Proof. Since

H

2

(u)

=

2

λp

1

Z

∞

u

ψ

1

(x)

d

x,

replacing

ψ

1

(x)

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

R

₀∞

ψ(x)

d

x

=

p

2

/

2

p

1

θ

is used in the computation. Eq. (6.47) is a direct application of Theorem 1.1. Noting that

H

₂′

(u)

=

2

λ

2

_p

2 1

θ

−

Z

u

0

ψ(u

−

x)ψ(x)

d

x

−

Z

∞

u

ψ(x)

d

x

+

p

2

2

p

1

θ

ψ(u)

,

We now evaluate the moments at the time of ruin in the case when the claim size is either a combination of

exponentials 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 to

point out that if the claim size distribution is neither a combination of exponentials nor a mixture of Erlangs, the

calculation of higher

(n >

2

)

moments could be complicated. However, we may use the Tijms approximation and

in this case, the approach in Example 6.1 applies.

Example 6.1 (Combinations of exponentials). First, let

P (x)

¯

=

e

−µx

. Then, from (3.13) with

δ

=

0

, ψ(u)

=

C

e

−Ru

, where

R

=

θ µ/(

1

+

θ )

. It is not hard to see from (6.29) by induction that

ψ

k

(u)

=

e

−Ru k

X

j=0

¯

C

j,k

(

Ru

)

j

j

!

.

(6.48)

We now derive a recursive formula for the coefficients

C

¯

j,k

.

From (6.29),

Z

u

0

ψ(u

−

x)ψ

k−1

(x)

d

x

=

1

1

+

θ

e

−Ru k−1

X

j=0

¯

C

j,k−1

Z

u

0

(

Rx

)

j

j

!

d

x

=

1

(

1

+

θ )R

e

−Ru

k

X

j=1

¯

C

j−1,k−1

(

Ru

)

j

j

!

,

Z

∞

u

ψ

k−1

(x)

d

x

=

R

−1 k−1

X

j=0

¯

C

j,k−1

Z

∞

u

R

j+1

x

j

e

−Rx

j

!

d

x

=

R

−1

_e

−Ru k−1

X

j=0





k−1

X

i=j

¯

C

i,k−1





(

Ru

)

j

j

!

Z

∞

0

ψ

k−1

(x)

d

x

=

R

−1 k−1

X

i=0

¯

C

i,k−1

.

Thus, by equating the coefficients of

(

Ru

)

j

/j

!, we have

¯

C

0,k

=

k(

1

+

θ )

cµθ

k−1

X

i=0

¯

C

i,k−1

(6.49)

and for

j

=

1

,

2

, . . . , k

,

¯

C

j,k

=

k(

1

+

θ )

2

cµθ

2





1

1

+

θ

C

¯

j−1,k−1

+

k−1

X

i=j

¯

C

i,k−1





,

(6.50)

with

P

k−1

i=k

=

0 and

C

¯

0,0

=

C

. The first two moments can be easily derived using (6.49) and (6.50).

¯

C

0,1

=

1

+

θ

cµθ

C

¯

0,0

=

1

cµθ

,

¯

C

1,1

=

(

1

+

θ )

2

cµθ

2

1

1

+

θ

¯

C

0,0

=

1

cµθ

2

.

Thus,

E

{TI

(T <

∞

)

} =

₁

cµθ

+

1

cµθ

2

(

Ru

)

e

−Ru

,

Also,

¯

C

0,2

=

2

(

1

+

θ )

cµθ

[

¯

C

0,1

+ ¯

C

1,1

]

=

2

(

1

+

θ )

2

c

2

_µ

2

_θ

3

,

¯

C

1,2

=

2

(

1

+

θ )

2

cµθ

2

₁

1

+

θ

¯

C

0,1

+ ¯

C

1,1

=

2

(

1

+

θ )(

1

+

2

θ )

c

2

_µ

2

_θ

4

,

¯

C

2,2

=

2

(

1

+

θ )

2

cµθ

2

1

1

+

θ

C

¯

1,1

=

2

(

1

+

θ )

c

2

_µ

2

_θ

4

.

E

{

T

2

I (T <

∞

)

} =

2

c

2

_µ

2

_θ

3

(

1

+

θ )

2

+

(

1

+

θ )(

1

+

2

θ )

θ

(

Ru

)

+

1

+

θ

θ

(

Ru

)

2

2!

e

−Ru

.

For a combination of two exponentials,

ψ(u)

=

C

1

e

−R1u

+

C

2

e

−R2u

,

as shown in Example 3.1. The general formula for

ψ

k

(u)

is

ψ

k

(u)

=

k

X

j=0

[

A

j,k

e

−R1u

+

B

j,k

e

−R2u

]

u

j

j

!

.

(6.51)

The coefficients

A

j,k

and

B

j,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

Z

u

0

ψ(u

−

x)ψ(x)

d

x

=

C

₁2

u

e

−R1u

₊

_C

2 2

u

e

−R2u

₊

2

C

1

C

2

R

2

−

R

1

[e

−R1u

₋

_e

−R2u

_]

_,

Z

∞

u

ψ(x)

d

x

=

C

1

R

1

e

−R1u

₊

C

2

R

2

e

−R2u

_.

Thus,

E

{TI

(T <

∞

)

} =

1

+

θ

cθ

C

₁2

u

e

−R1u

₊

_C

2 2

u

e

−R2u

₊

_C

1

R

1

+

2

C

1

C

2

R

2

−

R

1

−

_C

1

R

1

+

C

2

R

2

C

1

e

−R1u

+

_C

2

R

2

−

2

C

1

C

2

R

2

−

R

1

−

_C

1

R

1

+

C

2

R

2

C

2

e

−R2u

.

Example 6.2 (Mixtures of Erlangs). In the case when a claim size is a mixture of Erlangs, noting that

ψ(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

∞

X

j=0

C

j

(µu)

j

j

!

.

The

k

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

ψ

k

(u)

=

e

−µu

∞

X

j=0

C

j,k

(µu)

j

j

!

,

(6.52)

A recursive algorithm for

C

j,k

is now derived. Since

Z

u

0

ψ(u

−

x)ψ

k−1

(x)

d

x

=

e

−µu

∞

X

j=0

∞

X

i=0

C

j

C

i,k−1

µ

j+i

j

!

i

!

Z

u

0

(u

−

x)

j

x

i

d

x

=

e

−µu

∞

X

j=0

∞

X

i=0

C

j

C

i,k−1

µ

j+i

u

j+i+1

j

!

i

!

Z

1

0

(

1

−

t )

j

t

i

d

t

=

e

−µu

∞

X

j=0

∞

X

i=0

C

j

C

i,k−1

µ

j+i

u

j+i+1

(j

+

i

+

1

)

!

=

µ

−1

_e

−µu

∞

X

j=1





j

X

i=1

C

j−i

C

i−1,k−1





(µu)

j

j

!

,

Z

∞

u

ψ

k−1

(x)

d

x

=

µ

−1

∞

X

j=0

C

j,k−1

Z

∞

u

µ

j+1

x

j

e

−µx

j

!

d

x

=

µ

−1

e

−µu

∞

X

j=0

C

j,k−1 j

X

i=0

(µu)

i

i

!

=

µ

−1

_e

−µu

∞

X

j=0





∞

X

i=j

C

i,k−1





(µu)

j

j

!

,

Z

∞

0

ψ

k−1

(x)

d

x

=

µ

−1

∞

X

i=0

C

i,k−1

.

Thus, from (6.29) we obtain

C

0,k

=

k(

1

+

θ )

cµθ

(

1

−

C

0

)

∞

X

i=0

C

i,k−1

=

k

λθ

(

1

−

C

0

)

∞

X

i=0

C

i,k−1

,

(6.53)

and for

j

=

1

,

2

, . . .

,

C

j,k

=

k

λθ





j

X

i=1

C

j−i

C

i−1,k−1

+

∞

X

i=j

C

i,k−1

−

C

j

∞

X

i=0

C

i,k−1





.

(6.54)

Acknowledgements

This work was supported by the Natural Sciences and Engineering Research Council of Canada. The first author

also acknowledges the financial support by the University of Iowa Old Gold Fellowship. Both authors would like

to thank the anonymous referees for their suggestions.

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