Directory UMM :Data Elmu:jurnal:I:Insurance Mathematics And Economics:Vol26.Issue2-3.2000:

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Discounted probabilities and ruin theory in the compound

binomial model

Shixue Cheng

, Hans U. Gerber

b,∗

_{, Elias S.W. Shiu}

c,1 a_{School of Information, People’s University of China, Beijing 100872, China} b_{Ecole des hautes études commerciales, Université de Lausanne, CH-1015 Lausanne, Switzerland} c_{Department of Statistics and Actuarial Science, The University of Iowa, Iowa City, IA 52242-1409, USA}

Received 1 June 1998; received in revised form 1 September 1999; accepted 24 November 1999 Abstract

The aggregate claims are modeled as a compound binomial process, and the individual claim amounts are integer-valued. We studyf(x,y;u), the “discounted” probability of ruin for an initial surplusu, such that the surplus just before ruin isxand the deficit at ruin isy. This function can be used to calculate the expected present value of a penalty that is due at ruin, and, if it is interpreted as a probability generating function, to obtain certain information about the time of ruin. An explicit formula forf(x,y; 0) is derived. Then it is shown howf(x,y;u) can be expressed in terms off(x,y; 0) and an auxiliary functionh(u) that is the solution of a certain recursive equation and is independent ofxandy. As an application, we use the asymptotic expansion ofh(u) to obtain an asymptotic formula forf(x,y;u). In this model, certain results can be obtained more easily than in the compound Poisson model and provide additional insight. For the caseu=0, expressions for the expected present value of a payment of 1 at ruin and the expected time of ruin (given that ruin occurs) are obtained. A discrete version of Dickson’s formula is provided. © 2000 Elsevier Science B.V. All rights reserved.

Keywords:Risk theory; Ruin probability; Deficit at ruin; Time of ruin; Compound binomial model

1. Introduction

Traditionally, most results of risk theory are derived in a continuous-time model in which the aggregate claims are a compound Poisson process. The model appears to be more “realistic”, the methods are perceived as elegant and sophisticated, and some results seem to be tied to the compound Poisson assumption. In contrast, this paper considers a discrete-time model, where the aggregate claims are modeled as a compound binomial process, and where the possible claim amounts are integral multiples of the annual premium. With relative simple (but perhaps also aesthetic) methods, attractive results can be derived in this model. These results are of an independent interest, and they provide a better understanding of the analogous results in the continuous-time model. In fact, the latter can be viewed as limiting cases of the former. In this sense, any given result in the discrete-time model is stronger than the corresponding result in the continuous-time model.

∗_{Corresponding author. Tel.:}₊_{41-21-692-3371; fax:}₊_{41-21-692-3305.}

E-mail addresses:hgerber@hec.unil.ch (H.U. Gerber), eshiu@stat.uiowa.edu (E.S.W. Shiu) 1_Tel.:₊_{319-335-2580; fax:}₊_{319-335-3017.}

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In Section 3 we consider the discounted power martingale. For a given discount factor, the base must satisfy Lundberg’s fundamental equation, which has always one solution between zero and one, and a larger, second solution under some regularity conditions for the claim amount distribution. The main goal of the paper is to study the functionf(x,y; u), which is the “discounted” probability of ruin, given an initial surplus ofu, such that the surplus just before ruin isx, and the deficit at ruin isy. With this function, the expected present value of a penalty due at ruin can be calculated. In Section 4, an explicit formula forf(x,y; 0) is obtained. In Section 5, we show that

f(x,y;u) can be expressed in a transparent fashion through a functionh(u), which does not depend onxory, and is the solution of a certain recursive equation. In Section 6, an asymptotic formula forh(u) is derived and used to obtain an asymptotic formula forf(x,y;u). Applications include an explicit expression for the expected discounted value of a payment of 1 at the time of ruin, a discrete-time version of Dickson’s formula, formulas for the expected time of ruin, as well as some asymptotic results.

2. The compound binomial model

We consider a discrete-time model, in which the number of insurance claims is governed by a binomial process

N(t),t=0, 1, 2,. . . In each time period, the probability of a claim isq, 0<q<1, and the probability of no claim is 1−q. The claim occurrences in different time periods are independent events. The individual claim amountsX1,

X2,X3,. . . are mutually independent, identically distributed, positive and integer-valued random variables; they are independent of the binomial process{N(t)}. PutX=X1and let

p(x)=Pr(X=x), x =1,2,3, . . . (2.1)

be the common probability function of the individual claim amounts. (The value ofp(x) is zero ifxis not a positive integer.) This is called acompound binomialmodel, and has been considered by Gerber (1988), Shiu (1989), Willmot (1993), Dickson (1994), DeVylder (1996) (Chapter 10), DeVylder and Marceau (1996) (Section 2), and Cheng and Wu (1998a,b).

The compound binomial model can also be used to model the case where there can be more than one claim in each time period. Then we assume that the total claims in each time period are mutually independent, identically distributed and integer-valued random variables. LetYjdenote the sum of the claims in periodj. We consider

Pr(Yj =0)=1−q, Pr(Yj =y)=qp(y), y =1,2,3, . . . To avoid confusion, we shall not refer to this interpretation in this paper.

We also assume that the premium received in each time period is one. We do not necessarily make the assumption that the premiums contain a positivesecurity loading, i.e.

1−qE(X) >0, (2.2)

may not hold.

Let the initial surplus beu, which is a nonnegative integer. Fort=0, 1, 2,. . ., the surplus at timetis

U (t )=u+t−[X1+X2+ · · · +XN (t )]. (2.3)

“Ruin” is the event thatU(t)≤0 for somet≥1. We suppose thatp(1)<1 so that ruin is possible. Let

T =inf{t ≥1 : U (t )≤0} (2.4)

denote thetime of ruin. We are interested in the joint probability distribution of the time of ruin,T, the surplus just before ruin,U(T−1), and the surplus at ruin,U(T). Forx,y=0, 1, 2,. . .,t=1, 2, 3,. . ., define

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Note thatf(0,y,t;u) can be different from 0 only ift=1 andu=0; obviously

f (0, y,1;0)=qp(y+1). (2.6)

Letvbe a discount factor (0<v<1). Our primary goal is to explore the “discounted” probability f (x, y;u)=

∞ X

t=1

vtf (x, y, t;u). (2.7)

Using the functionf(x,y;u), we can calculate the expected discounted value of a “penalty” which is due at ruin and may depend on the surpluses just before and at ruin. We may also viewvas a parameter; then (2.7) is the formula of a (probability) generating function.

3. Lundberg’s fundamental equation

We seek a numberr>0 such that{vtr−U (t )}is a martingale. This is the condition that

E[vtr−U (t )|U (0)=u]=r−u. (3.1)

Because

E[r−U (t )|U (0)=u]=r−u−t{qE[rX]+(1−q)}t, (3.2) the martingale condition is

φ (r)=v−1, (3.3)

where

φ (r)=qE[rX−1]+(1−q)r−1, r >0. (3.4)

We call (3.3)Lundberg’s fundamental equationto honor the Swedish actuary Lundberg, who had pointed out that the compound Poisson version of (3.3) is “fundamental to the whole of collective risk theory” (Lundberg, 1932, p. 144). Becauseφ(r) is a strictly convex function, Eq. (3.3) has at most two roots. Also,φ(1)=1<v−1, andφ(r) tends to∞forr→0+; hence, Eq. (3.3) has a solutionr=ρ∈(0, 1). Furthermore, under some regularity conditions for the tail of the probability functionp(x), Eq. (3.3) has another solutionr=R>1.

For a first application, letxbe an integer,x>U(0)=u, and consider

Tx=inf{t : U (t )≥x}, (3.5)

the first time the surplus rises to the levelx. Note that the inequality in definition (3.5) can be replaced by an equality, because the process{U(t)}is skip-free upwards. We claim that

E[vTx_{I (T}

x<∞)|U (0)=u]=ρx−u, (3.6)

whereIdenotes the indicator function, i.e.I(A)=1 ifAis true andI(A)=0 ifAis false. To show this, we observe that{vtρ−U (t )}is a positive martingale that is bounded above byρ−x fort<Tx. By applying theoptional sampling

theorem, we have

ρ−u=E[vTx_ρ−U (Tx)_{I (T}

x<∞)|U (0)=u]=E[vTxI (Tx<∞)|U (0)=u]ρ−x, (3.7) from which (3.6) follows.

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For a second application, suppose that (3.3) has the other rootR>1. Then{vtR−U (t )}is a positive martingale that is bounded above by 1 fort<T. By the optional sampling theorem, we obtain

R−u=E[vTR−U (T )I (T <∞)|U (0)=u]. (3.8)

In general, it is difficult to simplify the expression on the right-hand side of (3.8), becauseT andU(T) are not independent, and the distribution ofU(T) is unknown. There are two noteworthy exceptions. One is the case where all claims are of size 2,X≡≡2. ThenU(T)=0, providedu>0. Hence it follows from (3.8) that, foru=1, 2, 3,. . .,

E[vTI (T <∞)|U (0)=u]=R−u, (3.9)

where R= 1+

1−4q(1−q)v2

2qv (3.10)

by (3.3). The formula foru=0 will be given in formula (4.8), which is a general result. We note that (3.10) can be found in Feller (1968, p. 350) The other exception is the case of a geometric claim amount distribution,

p(x)=(1−c)cx−1, x=1,2,3, . . . (3.11)

Then, givenT<∞,−U(T) is independent ofT, and (3.8) becomes

R−u=E[vTI (T <∞)|U (0)=u]E[R−U (T )|T <∞]. (3.12) Because, givenT<∞,U(T) has the geometric distribution

(1−c)cx, x =0,1,2, . . . we have

E[R−U (T )|T <∞]= 1−c

1−cR. (3.13)

It follows from (3.12) and (3.13) that E[vTI (T <∞)|U (0)=u]=1−cR

1−c R −u_.

(3.14) Here, (3.3) leads also to a quadratic equation, with which we can determineR.

Remarks.

1. Assume that the second root R of Lundberg’s equation(3.3)exists. Consider both roots as functions ofv,ρ=ρ(v)

and R=R(v).Note thatφ(1)=1and

φ′(1)=qE[X]−1, (3.15)

which is the negative of the security loading. If the security loading is negative, thenφ′₍₁₎_>₀_{. Hence the convex}

functionφhas its minimum in the interval(0, 1), from which it follows that:

lim

v→1ρ(v) <1=limv→1R(v). (3.16)

If the security loading is zero, thenφ′(1)=0. Thusφhas its minimum at1and

lim

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If the security loading is positive, thenφ′(1)<0, and

lim

v→1ρ(v)=1<vlim→1R(v). (3.18)

The corresponding relationships for the compound Poisson case can be found in Section3ofGerber and Shiu

(1999).

2. In formulas such as(3.6)and(3.8),the indicator random variables I(Tx<∞)and I(T<∞)can be dropped,

becausev<1.However, leaving them in facilitates the discussion of limiting results forv→1.From(3.6)we have

Pr[Tx<∞|U (0)=u]=ρ(1)x−u. (3.19)

It then follows from(3.17)and(3.18)that, if the security loading is nonnegative, the right-hand side of(3.19)is

1, i.e. the surplus will rise to the level x with certainty. If the security loading is negative, thenρ(1)<1according to(3.16),and hence(3.19)is now not a trivial result. On the other hand, asv→1 (3.8)becomes

R(1)−u=E[R(1)−U (T )I (T <∞)|U (0)=u]. (3.20)

If the security loading is nonpositive, it follows from(3.16) and(3.17)that R(1)=1, and hence T<∞ with probability1, i.e. ruin is certain. If the security loading is positive, then R(1)>1according to(3.18), and

ψ(u)=Pr[T <∞|U (0)=u]= R(1)

−u

E[R(1)−U (T )_|_{T <}_∞_{, U (}₀₎₌_u_]. (3.21)

4. The key formula

The following result is forU(0)=u=0. It is the discrete counterpart of formula (3.12) in Gerber and Shiu (1997) and formula (3.3) in Gerber and Shiu (1998a).

Theorem 1. For x=0, 1, 2,. . .,y=0, 1, 2,. . .,

f (x, y;0)=qvρxp(x+y+1). (4.1)

Proof. By (2.7), the left-hand side of (4.1) is ∞

t=1

vtf (x, y, t;0). (4.2)

According to definition (2.5),f(x,y,t; 0) is the following product of three probabilities:

Pr[U (1) >0, U (2) >0, . . . , U (t−2) >0, U (t−1)=x|U (0)=0]qp(x+y+1). (4.3) It follows from a simple duality argument which can be visualized by rotating the graph of the surplus process from time 0 to timet−1 by 180◦_{(Feller, 1971, Section XII.2) that the conditional probability in (4.3) is the same as}

Pr[U (1) < x, U (2) < x, . . . , U (t−2) < x, U (t−1)=x|U (0)=0]. (4.4) Multiplying (4.3) withvt and summing overt, recalling thatTxis the first passage time of the surplus process at levelx, and applying (3.6) withu=0, we have

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∞ X

t=1

vtPr[U (1) >0, U (2) >0, . . . , U (t−2) >0, U (t−1)=x|U (0)=0]qp(x+y+1)

=vqp(x+y+1) ∞ X

t=1

vt−1Pr[U (1) < x, U (2) < x, . . . , U (t−2) < x, U (t−1)=x|U (0)=0]

=vqp(x+y+1)E[vTx_|_{U (}₀₎₌_0]₌_v_qp_(x₊_y₊₁_)ρx_, _(4.5)

which is indeed the right-hand side of (4.1).

As a check for (4.1), we consider the special case ofx=0, f (0, y;0)=qvp(y+1),

which can be verified using (2.6). We note that DeVylder and Goovaerts (1998) have given an interesting proof of the compound Poisson version of Theorem 1 using the Lagrange expansion formula. Also, Theorem 1 can be generalized to the case where there aremtypes of individual claims, which are independent and at most one of which can occur in a time period; for such results in the compound Poisson model, see Gerber and Shiu (1999). Option-pricing applications of the compound Poisson version of Theorem 1 can be found in Gerber and Shiu (1998b, 1999).

As an illustration of Theorem 1, we calculate the expected present value of 1, payable at the time of ruin, ifu=0: E[vTI (T <∞)|U (0)=0]=

∞ X

x=0 ∞ X

y=0

f (x, y;0)=qv

∞ X

x=0 ρx

∞ X

k=x+1

p(k) (4.6)

by (4.1). By interchanging the order of summation and then noting thatρis a solution of (3.3), we obtain ∞

x=0 ρx

∞ X

k=x+1

p(k)= 1

1−ρ ∞ X

k=1

(1−ρk)p(k)= 1

1−ρ(1−E[ρ

X_]₎₌ 1 1−ρ

₁ q − ρ qv . (4.7)

Substituting (4.7) in (4.6) yields

E[vTI (T <∞)|U (0)=0]=v−ρ

1−ρ. (4.8)

Note that (4.8) can be interpreted as the probability generating function of the time-of-ruin random variableT. An explicit expansion of the right-hand side of (4.8) in powers ofvis possible in special cases such as the two examples in the last section; see also Section XI.3 of Feller (1968), where ourviss. Note that formula (4.8) can be rewritten as

E[vTI (T <∞)|U (0)=0]=1− 1−v

1−ρ, (4.9)

the compound Poisson version of which is formula (3.9) in Gerber and Shiu (1998a) and formula (4.11) in Gerber and Shiu (1999).

Let us first assume a negative loading. Then ruin is certain, i.e.I(T<∞)≡≡1. Consider ρ as a function of v, ρ=ρ(v). Thenρ(1)<1 by (3.16). Differentiating (4.8) and settingv=1, we obtain

E[T|U (0)=0]= 1

1−ρ(1). (4.10)

Note that, as the security loading tends to zero, the quantityρ(1) tends to 1 according to (3.17), and hence the expectation ofTgoes to infinity; in the special case of a symmetric random walk (X≡≡2 in our model), this is the well-known result that the time of return to the origin has an infinite expectation.

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For the remainder of this section, we consider the more important case where the security loading is positive. Settingv=1 in (4.6) yields an expression for the probability of ruin foru=0:

ψ (0)=q ∞ X

x=0 ∞ X

k=x+1

p(k)=qE[X]. (4.11)

This is formula (7) in Gerber (1988). In terms of therelative security loadingθ, formula (4.11) is ψ (0)= 1

1+θ, (4.12)

which is a classical result in the continuous-time compound Poisson model and can be viewed as an extension of theballot theorem(Feller, 1968, p. 69).

As a check, let us verify (4.11) starting with (4.8). Using the rule of Bernoulli–Hôpital, we obtain ψ (0)=1− 1

ρ′₍₁₎. (4.13)

To determineρ′(1), we replacerbyρ(v) in (3.3): φ (ρ(v))=1

v. (4.14)

Using the chain rule, we differentiate with respect tovand setv=1 to see that φ′(ρ(1))ρ′(1)= −1.

It follows from (3.18) and (3.15) that ρ′(1)= −1

φ′₍₁₎ = 1

1−qE[X]. (4.15)

Substituting (4.15) in (4.13) we obtain (4.11) again.

Let us now calculate the expected time of ruin, given that ruin occurs. First we differentiate (4.6) with respect to vand setv=1. This yields

E[TI(T <∞)|U (0)=0]=q 

 ∞ X

x=0 ∞ X

k=x+1

p(k)+ρ′(1) ∞ X

x=0 x

∞ X

k=x+1 p(k)



=ψ (0)+

qρ′(1)E[X(X−1)]

2 .

After dividing byψ(0) and applying (4.11) and (4.15), we obtain the desired result: E[T|T <∞, U (0)=0]=1+ E[X(X−1)]

2E[X](1−qE[X]). (4.16)

Hence, somewhat surprisingly, the conditional expected time of ruin, given that ruin occurs, is the shorter, the larger the security loading. This result, which is in contradistinction to (4.10), can be interpreted as follows. The larger the security loading, the faster the surplus is expected to grow as time passes. A larger surplus, in turn, means a smaller chance for ruin. Thus, given that ruin is to occur, it must occur earlier (in some sense) for the case of a larger security loading.

We can use (4.9) to obtain higher moments ofT. Forn=1, 2, 3,. . ., the factorial moment

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is thenth derivative of−(1−v)/[1−ρ(v)] with respect tovatv=1. Because ρ(v)−1

v−1 = ∞ X

k=1 1 k!ρ

(k)₍₁_)(v₋₁₎k−1_, _(4.18)

the factorial moment is expressed in terms ofρ(k)(1) up tok=n+1. We remark that in the continuous time model, the moments of the time to ruin have been examined by Picard and Lefèvre (1998).

5. Recursive formulas

For an arbitrary initial surplusu>0, it is possible to determinef(x,y;u) by a recursive procedure. GivenU(0)=u, we consider the first timet,t≥1, whenU(t)≤u. The “discounted” probability of this event with the surplus being of the amountu−yat that time is

g(y)=

∞ X

x=0

f (x, y;0)=qv

∞ X

x=0

ρxp(x+y+1) (5.1)

by (4.1). Foru>0, we evaluatef(x,y;u) by conditioning on the first time when the surplus drops to or below its initial valueu. We need to distinguish two cases. Ifu>x, the event that ruin occurs at this time makes no contribution to the quantityf(x,y;u). Hence we have

f (x, y;u)=

u−1 X

z=0

f (x, y;u−z)g(z), u=x+1, x+2, . . . (5.2)

Ifu≤x, the event that ruin occurs at this time contributes the quantityf(x−u,y+u; 0) tof(x,y;u). Thus f (x, y;u)=

u−1 X

z=0

f (x, y;u−z)g(z)+f (x−u, y+u;0), u=1,2, . . . , x. (5.3)

Formulas (5.2) and (5.3) are discrete analogs of (2.11) and (2.10) in Gerber and Shiu (1997). By (4.1) f (x−u, y+u;0)=qvρx−up(x+y+1)=ρ−uf (x, y;0), u≤x.

Hence we can combine (5.2) and (5.3) as f (x, y;u)=

u−1 X

z=0

f (x, y;u−z)g(z)+f (x, y;0)ρ−uI (u≤x), (5.4)

which corresponds to (4.1) of Gerber and Shiu (1997). Here,u=1, 2, 3,. . .,x=1, 2, 3,. . ., andy=0, 1, 2,. . . For given values ofxandy, we can calculatef(x,y;u) recursively by (5.4). The following result gives a more elegant way to obtainf(x,y;u):

Theorem 2. Let h(u)be defined as the solution of

h(u)=

u−1 X

z=0

h(u−z)g(z)+ρ−u, u=1,2,3, . . . (5.5)

Then, for x=1, 2, 3,. . .,and y=0, 1, 2,. . .,

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Proof. For each positive integerx, we introduce a function ofu,γ(x;u), which is defined as the solution of the renewal equation

γ (x;u)=

u−1 X

z=0

γ (x;u−z)g(z)+ρ−uI (u≤x), u=1,2,3, . . . (5.7)

Comparing (5.7) with (5.4), we gather that

f (x, y;u)=f (x, y;0)γ (x;u). (5.8)

Thus to prove (5.6), we have to show that

γ (x;u)=h(u)−ρ−xh(u−x)I (u > x). (5.9)

Because the solution of (5.7) is unique, it suffices to verify that the function on the right-hand side of (5.9) satisfies (5.7). Hence we consider

u−1 X

z=0

[h(u−z)−ρ−xh(u−z−x)I (u−z > x)]g(z)+ρ−uI (u≤x), u=1,2,3, . . .

Ifu≤x, this is u−1 X

z=0

h(u−z)g(z)+ρ−u =h(u)=h(u)−ρ−xh(u−x)I (u > x).

Ifu>x, this is u−1 X

z=0

h(u−z)g(z)−ρ−x u−x−1

z=0

h(u−z−x)g(z)=[h(u)−ρ−u]−ρ−x[h(u−x)−ρ−(u−x)]

=h(u)−ρ−xh(u−x)I (u > x).

Remarks.

1. With(5.5)and(5.1), the function h(u)can be determined recursively for u=1, 2, 3,. . . Alternatively,Eq. (5.5)

can be solved in terms of generating functions.

2. For u=1, 2, 3,. . .,the quantity1/h(u)has a probabilistic interpretation. For given positive integers u and x, u≤x, it follows from(5.6)that

f (x, y;0)=

1 h(u)

f (x, y;u). (5.10)

Suppose that U(0)=0.Before ruin can occur with U(T−1)=x, x≥u>0,the surplus process{U(t)}must neces-sarily have attained the level u(because the premium is1per unit time).From this observation and formula

(5.10)we see that1/h(u)can be interpreted as the expected present value of a contingent payment of1that is made when the surplus process, with initial value0, attains the level u for the first time, provided that ruin has not occurred by then. The compound Poisson version of(5.10)is(6.26)ofGerber and Shiu (1998a).

3. It follows from the interpretation above that

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On the other hand, we obtain from(5.5)and(5.1)that

h(1)=h(1)g(0)+ρ−1=h(1)qv

∞ X

x=0

ρxp(x+1)+ρ−1. (5.12)

These two formulas can be reconciled becauseρsatisfies Lundberg’s equation(3.3).

4. Assuming a positive security loading, we consider the limiting casev=1and henceρ=1.Then it follows from

(5.5)that h(u)can be interpreted as the expected number of “weak” record lows of the surplus process{U(t)}

before ruin, with U(0)=u. There is a close connection between h(u)and the probability of ruinψ(u). Observe that the number of future record lows of the process{U(t)}has at any time a geometric distribution with mean

1/[1−ψ(0)]. The expected number of record lows before ruin is the expected number of record lows starting from time zero minus the expected number of future record lows from the time of ruin. Thus we have

h(u)= 1−ψ(u)

1−ψ (0). (5.13)

Substituting (5.13) in(5.6), we obtain a discrete version of a formula by Dickson (1992), which is for the

compound Poisson model. In that context, Dickson’s formula has been generalized byGerber and Shiu (1997,

1998a)and Theorem2suggests that a generalization is also possible in the compound binomial model.

6. Asymptotic formulas

We shall derive a two-term asymptotic formula (u→∞) for the functionh(u), which is defined as the solution of the recursive equation (5.5). Based on this asymptotic formula and Theorem 2, we then obtain an asymptotic formula forf(x,y;u). Asymptotically, the functionh(u) has an exponential growth,

h(u)∼Aρ−u foru→ ∞, (6.1)

where the constantAis such that Aρ−u=

∞ X

z=0

Aρ−(u−z)g(z)+ρ−u, (6.2)

A= 1

1−P∞_z₌₀ρz_g(z). (6.3)

However, substituting (6.1) in (5.6) yields only the trivial result thatf(x,y; u)→0 foru→∞. Hence we need a refinement of (6.1).

Theorem 3. If the second root R>1of Lundberg’s equation(3.3)exists, then there exists a constant C such that,

for u→∞,

Aρ−u−h(u)∼CR−u, (6.4)

f (x, y;u)∼f (x, y;0)C _R

ρ x

−1

R−u. (6.5)

Proof. Note that (6.5) follows directly from (5.6) and (6.4). Thus we only need to prove (6.4). Subtracting (5.5) from (6.2), we have

Aρ−u−h(u)=

u−1 X

z=0

[Aρ−(u−z)−h(u−z)]g(z)+A ∞ X

z=u

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BecauseP∞_u₌₁P∞_z₌_uρ−(u−z)g(z) <∞, Eq. (6.6) is a (defective) renewal equation for the functionAρ−u−h(u) (Feller, 1968, Section XIII.10). It follows from theRenewal Theorem(Feller, 1968, p. 331) and the relation

∞ X

z=0

Rzg(z), (6.7)

that there is a constantCsuch that (6.4) holds. Relation (6.7) is best verified by settingu=0 in (3.8). For a compound Poisson version of (6.5), see (6.68) or (4.21) of Gerber and Shiu (1998a). As an application, we now derive an asymptotic formula for the expected time of ruin, given that ruin is to occur, in the case of a positive security loading. From (6.5), we see that, foru→∞,

E[vTI (T <∞)|U (0)=u]∼C ∞ X

x=1 ∞ X

y=0

f (x, y;0) _R

ρ x

−1

R−u. (6.8)

ConsiderρandRas functions ofv. Then the right-hand side of (6.8) is of the formη(v)R(v)−u, where the function η(v) does not depend onu. Because

E[T|T <∞, U (0)=u]= d

dvln{E[v

T_{I (T <}_∞₎_|_{U (}₀₎₌_u_]_}|

v=1, (6.9)

we have

E[T|T <∞, U (0)=u]∼−R

′₍₁₎

R(1) u foru→ ∞. (6.10)

To determine−R′_(1)/_R_{(1), we multiply (3.3) by}_r_{and replace}_r_by_R₍_v_):

qE[R(v)X]+(1−q)= R(v)

v . (6.11)

Differentiating (6.11) with respect tov, settingv=1 and rearranging yields

−R′(1) R(1) =

qE[XR(1)X−1_]₋₁. (6.12)

Acknowledgements

Elias Shiu gratefully acknowledges the support from the Principal Financial Group Foundation.

References

Cheng, S., Wu, B., 1998a. Classical risk model in fully discrete setting. Technical Report, School of Information, People’s University of China, 14 pp.

Cheng, S., Wu, B., 1998b. The survival probability in finite time in fully discrete risk models. Technical Report, School of Information, People’s University of China, 8 pp.

DeVylder, F.E., 1996. Advanced Risk Theory: A Self-Contained Introduction. Editions de l’Universite de Bruxelles, Brussels. DeVylder, F.E., Goovaerts, M.J., 1998. Discussion of “On the time value of ruin”. North American Actuarial Journal 2 (1), 72–74. DeVylder, F.E., Marceau, E., 1996. Classical numerical ruin probabilities. Scandinavian Actuarial Journal, 109–123.

Dickson, D.C.M., 1992. On the distribution of surplus prior to ruin. Insurance: Mathematics and Economics 11, 191–207. Dickson, D.C.M., 1994. Some comments on the compound binomial model. ASTIN Bulletin 24, 33–45.

Feller, W., 1968. An Introduction to Probability Theory and Its Applications, Vol. 1, 3rd Edition. Wiley, New York. Feller, W., 1971. An Introduction to Probability Theory and Its Applications, Vol. 2, 2nd Edition. Wiley, New York.

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Gerber, H.U., 1988. Mathematical fun with the compound binomial process. ASTIN Bulletin 18, 161–168.

Gerber, H.U., Shiu, E.S.W., 1997. The joint distribution of the time of ruin, the surplus immediately before ruin, and the deficit at ruin. Insurance: Mathematics and Economics 21, 129–137.

Gerber, H.U., Shiu, E.S.W., 1998a. On the time value of ruin. North American Actuarial Journal 2 (1), 48–72; Discussions 72–78.

Gerber, H.U., Shiu, E.S.W., 1998b. Pricing perpetual options for jump processes. North American Actuarial Journal 2 (3), 101–107; Discussions 108–112.

Gerber, H.U., Shiu, E.S.W., 1999. From ruin theory to pricing reset guarantees and perpetual put options. Insurance: Mathematics and Economics 24, 3–14.

Lundberg, F., 1932. Some supplementary researches on the collective risk theory. Skandinavisk Aktuarietidskrift 15, 137–158.

Picard, P., Lefèvre, C., 1998. The moments of ruin time in the classical risk model with discrete claim size distribution. Insurance: Mathematics and Economics 23, 157–172, Corr. 25, 105–107.

Shiu, E.S.W., 1989. The probability of eventual ruin in the compound binomial model. ASTIN Bulletin 19, 179–190.

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For the remainder of this section, we consider the more important case where the security loading is positive. Settingv=1 in (4.6) yields an expression for the probability of ruin foru=0:

ψ (0)=q

∞

x=0 ∞

k=x+1

p(k)=qE[X]. (4.11)

This is formula (7) in Gerber (1988). In terms of therelative security loadingθ, formula (4.11) is ψ (0)= 1

1+θ, (4.12)

which is a classical result in the continuous-time compound Poisson model and can be viewed as an extension of theballot theorem(Feller, 1968, p. 69).

As a check, let us verify (4.11) starting with (4.8). Using the rule of Bernoulli–Hôpital, we obtain ψ (0)=1− 1

ρ′₍₁₎. (4.13)

To determineρ′(1), we replacerbyρ(v) in (3.3): φ (ρ(v))=1

v. (4.14)

Using the chain rule, we differentiate with respect tovand setv=1 to see that φ′(ρ(1))ρ′(1)= −1.

It follows from (3.18) and (3.15) that ρ′(1)= −1

φ′₍₁₎ =

1−qE[X]. (4.15)

Substituting (4.15) in (4.13) we obtain (4.11) again.

Let us now calculate the expected time of ruin, given that ruin occurs. First we differentiate (4.6) with respect to vand setv=1. This yields

E[TI(T <∞)|U (0)=0]=q 



∞

x=0 ∞

k=x+1

p(k)+ρ′(1)

∞

x=0

∞

k=x+1

p(k) 

=ψ (0)+

qρ′(1)E[X(X−1)]

2 .

After dividing byψ(0) and applying (4.11) and (4.15), we obtain the desired result: E[T|T <∞, U (0)=0]=1+ E[X(X−1)]

2E[X](1−qE[X]). (4.16)

We can use (4.9) to obtain higher moments ofT. Forn=1, 2, 3,. . ., the factorial moment

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is thenth derivative of−(1−v)/[1−ρ(v)] with respect tovatv=1. Because ρ(v)−1

v−1 =

∞

k=1

1 k!ρ

(k)_(1)(v₋₁₎k−1_, _(4.18)

the factorial moment is expressed in terms ofρ(k)(1) up tok=n+1. We remark that in the continuous time model, the moments of the time to ruin have been examined by Picard and Lefèvre (1998).

5. Recursive formulas

For an arbitrary initial surplusu>0, it is possible to determinef(x,y;u) by a recursive procedure. GivenU(0)=u, we consider the first timet,t≥1, whenU(t)≤u. The “discounted” probability of this event with the surplus being of the amountu−yat that time is

g(y)= ∞

x=0

f (x, y;0)=qv

∞

x=0

ρxp(x+y+1) (5.1)

f (x, y;u)= u−1

z=0

f (x, y;u−z)g(z), u=x+1, x+2, . . . (5.2)

Ifu≤x, the event that ruin occurs at this time contributes the quantityf(x−u,y+u; 0) tof(x,y;u). Thus f (x, y;u)=

u−1

z=0

f (x, y;u−z)g(z)+f (x−u, y+u;0), u=1,2, . . . , x. (5.3) Formulas (5.2) and (5.3) are discrete analogs of (2.11) and (2.10) in Gerber and Shiu (1997). By (4.1)

f (x−u, y+u;0)=qvρx−up(x+y+1)=ρ−uf (x, y;0), u≤x. Hence we can combine (5.2) and (5.3) as

f (x, y;u)= u−1

z=0

f (x, y;u−z)g(z)+f (x, y;0)ρ−uI (u≤x), (5.4)

Theorem 2. Let h(u)be defined as the solution of

h(u)= u−1

z=0

h(u−z)g(z)+ρ−u, u=1,2,3, . . . (5.5)

Then, for x=1, 2, 3,. . .,and y=0, 1, 2,. . .,

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Proof. For each positive integerx, we introduce a function ofu,γ(x;u), which is defined as the solution of the renewal equation

γ (x;u)= u−1

z=0

γ (x;u−z)g(z)+ρ−uI (u≤x), u=1,2,3, . . . (5.7)

Comparing (5.7) with (5.4), we gather that

f (x, y;u)=f (x, y;0)γ (x;u). (5.8)

Thus to prove (5.6), we have to show that

γ (x;u)=h(u)−ρ−xh(u−x)I (u > x). (5.9)

Because the solution of (5.7) is unique, it suffices to verify that the function on the right-hand side of (5.9) satisfies (5.7). Hence we consider

u−1

z=0

[h(u−z)−ρ−xh(u−z−x)I (u−z > x)]g(z)+ρ−uI (u≤x), u=1,2,3, . . .

Ifu≤x, this is

u−1

z=0

h(u−z)g(z)+ρ−u =h(u)=h(u)−ρ−xh(u−x)I (u > x).

Ifu>x, this is

u−1

z=0

h(u−z)g(z)−ρ−x

u−x−1

z=0

h(u−z−x)g(z)=[h(u)−ρ−u]−ρ−x[h(u−x)−ρ−(u−x)]

=h(u)−ρ−xh(u−x)I (u > x).

Remarks.

1. With(5.5)and(5.1), the function h(u)can be determined recursively for u=1, 2, 3,. . . Alternatively,Eq. (5.5) can be solved in terms of generating functions.

2. For u=1, 2, 3,. . .,the quantity1/h(u)has a probabilistic interpretation. For given positive integers u and x, u≤x, it follows from(5.6)that

f (x, y;0)=

1 h(u)

f (x, y;u). (5.10)

Suppose that U(0)=0.Before ruin can occur with U(T−1)=x, x≥u>0,the surplus process{U(t)}must neces-sarily have attained the level u(because the premium is1per unit time).From this observation and formula (5.10)we see that1/h(u)can be interpreted as the expected present value of a contingent payment of1that is made when the surplus process, with initial value0, attains the level u for the first time, provided that ruin has not occurred by then. The compound Poisson version of(5.10)is(6.26)ofGerber and Shiu (1998a).

3. It follows from the interpretation above that 1

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On the other hand, we obtain from(5.5)and(5.1)that h(1)=h(1)g(0)+ρ−1=h(1)qv

∞

x=0

ρxp(x+1)+ρ−1. (5.12)

These two formulas can be reconciled becauseρsatisfies Lundberg’s equation(3.3).

4. Assuming a positive security loading, we consider the limiting casev=1and henceρ=1.Then it follows from (5.5)that h(u)can be interpreted as the expected number of “weak” record lows of the surplus process{U(t)} before ruin, with U(0)=u. There is a close connection between h(u)and the probability of ruinψ(u). Observe that the number of future record lows of the process{U(t)}has at any time a geometric distribution with mean 1/[1−ψ(0)]. The expected number of record lows before ruin is the expected number of record lows starting from time zero minus the expected number of future record lows from the time of ruin. Thus we have

h(u)= 1−ψ(u)

1−ψ (0). (5.13)

Substituting (5.13) in(5.6), we obtain a discrete version of a formula by Dickson (1992), which is for the compound Poisson model. In that context, Dickson’s formula has been generalized byGerber and Shiu (1997, 1998a)and Theorem2suggests that a generalization is also possible in the compound binomial model.

6. Asymptotic formulas

We shall derive a two-term asymptotic formula (u→∞) for the functionh(u), which is defined as the solution of the recursive equation (5.5). Based on this asymptotic formula and Theorem 2, we then obtain an asymptotic formula forf(x,y;u). Asymptotically, the functionh(u) has an exponential growth,

h(u)∼Aρ−u foru→ ∞, (6.1)

where the constantAis such that Aρ−u=

∞

z=0

Aρ−(u−z)g(z)+ρ−u, (6.2)

A= 1

1−P∞

z=0ρzg(z)

. (6.3)

However, substituting (6.1) in (5.6) yields only the trivial result thatf(x,y; u)→0 foru→∞. Hence we need a refinement of (6.1).

Theorem 3. If the second root R>1of Lundberg’s equation(3.3)exists, then there exists a constant C such that, for u→∞,

Aρ−u−h(u)∼CR−u, (6.4)

f (x, y;u)∼f (x, y;0)C

ρ x

−1

R−u. (6.5)

Proof. Note that (6.5) follows directly from (5.6) and (6.4). Thus we only need to prove (6.4). Subtracting (5.5) from (6.2), we have

Aρ−u−h(u)= u−1

z=0

[Aρ−(u−z)−h(u−z)]g(z)+A

∞

z=u

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BecauseP∞

u=1

P∞

z=uρ−(u−z)g(z) <∞, Eq. (6.6) is a (defective) renewal equation for the functionAρ−u−h(u)

(Feller, 1968, Section XIII.10). It follows from theRenewal Theorem(Feller, 1968, p. 331) and the relation 1=

∞

z=0

Rzg(z), (6.7)

E[vTI (T <∞)|U (0)=u]∼C

∞

x=1 ∞

y=0

f (x, y;0)

ρ x

−1

R−u. (6.8)

ConsiderρandRas functions ofv. Then the right-hand side of (6.8) is of the formη(v)R(v)−u, where the function η(v) does not depend onu. Because

E[T|T <∞, U (0)=u]= d

dvln{E[v

T_{I (T <}_∞₎_|_{U (0)}₌_u]_}|

v=1, (6.9)

we have

E[T|T <∞, U (0)=u]∼−R ′₍₁₎

R(1) u foru→ ∞. (6.10)

To determine−R′_{(1)/R(1), we multiply (3.3) by}_r_{and replace}_r_by_R(v):

qE[R(v)X]+(1−q)= R(v)

v . (6.11)

Differentiating (6.11) with respect tov, settingv=1 and rearranging yields

−R′(1) R(1) =

qE[XR(1)X−1_]₋₁. (6.12)