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A note on the compound binomial model with randomized dividend strategy
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Applied Mathematics and Computation 194 (2007) 276–286 www.elsevier.com/locate/amc
A note on the compound binomial model with randomized dividend strategy Zhen-hua Bao School of Mathematics, Liaoning Normal University, Dalian 116029, China
Abstract This paper considers the compound binomial model with randomized decisions on paying dividends. By using two formulas obtained by Tan and Yang [J.Y. Tan, X.Q. Yang, The compound binomial model with randomized decisions on paying dividends, Insurance: Mathematics and Economics 39 (2006) 1–18], two defective renewal equations for the Gerber–Shiu penalty function are derived and solved. The analytic solutions obtained are utilized to derive the probability of ultimate ruin, the deficit distribution at ruin and the distribution of the claim causing ruin. The asymptotic estimate satisfied by the penalty function is discussed in some detail. Ó 2007 Elsevier Inc. All rights reserved. Keywords: Compound binomial model; Defective renewal equation; Dividend strategy; Gerber–Shiu penalty function; Asymptotic estimate
1. Introduction The classical compound binomial risk model is a discrete time risk process with the following features. The premium received in each period is 1. In any period the probability of claim is p ð0 < p < 1Þ, and probability of no claim is q ¼ 1 p. We assume that claims occur at the end of the period and denote by nt ¼ 1 the event where a claim occurs in period ðt 1; t and nt ¼ 0 the event where no claim occurs in the time period ðt 1; t. The claims are independent and identically distributed (i.i.d.) positive integer valued random variables distributed as a random variable X with cumulative distribution function (c.d.f.) F ðxÞ ¼ 1 F ðxÞ and probability function (p.f.) f ðxÞ; x 2 N þ . Where we denote by N the set of nature numbers and N þ ¼ N f0g. Throughðx1Þ out, we assume the equilibrium p.f. of X is f1 ðxÞ ¼ FEðX ; x 2 N þ , and F 1 ðxÞ ¼ 1 F 1 ðxÞ is its c.d.f. For Þ t ¼ 0; 1; . . ., the surplus at time t is U ðtÞ ¼ u þ t
t X
X i ni ;
i¼1
E-mail address: [email protected] 0096-3003/$ - see front matter Ó 2007 Elsevier Inc. All rights reserved. doi:10.1016/j.amc.2007.04.023
ð1Þ
Z.-h. Bao / Applied Mathematics and Computation 194 (2007) 276–286
277
where U ð0Þ ¼ u, which is a non-negative integer. The risk model (1) has been studied extensively since it was introduced by Gerber [6]. Gerber derived a formula for the ruin probability, the distribution of the deficit at ruin and the distribution of the surplus just before ruin in the case that the initial surplus is 0. Shiu [14] derived similar expressions as Gerber’s using different methods. Cheng et al. [2] derived the moment generating function of the time to ruin for zero initial surplus and then they derived recursively the joint distribution of the surplus prior to ruin and the deficit at ruin. Dickson [4] applied elementary methods to derive the joint distribution of the surplus before ruin and the deficit at ruin when the initial reserve is 0. Recently, Gerber and Shiu [7] introduced a discounted penalty function with respect to the time of ruin, the surplus immediately before ruin and the deficit at ruin, which has proven to be a powerful analytical tool. See also Lin and Willmot [11]. For the compound binomial model, Pavlova and Willmot [13] derived a defective renewal equation satisfied by the Gerber–Shiu discounted penalty function. Dividend strategies for insurance risk models were first proposed by De Finetti [3] to reflect more realistically the surplus cash flows in an insurance portfolio. Barrier strategies for the classical compound Poisson risk model have been studied extensively in the literature, including Albrecher et al. [1], Dickson and Waters [5], Gerber and Shiu [7,8], Lin et al. [12], Lin and Pavlova [10], among many others. In Tan and Yang [15], the authors considered the compound binomial model modified by the inclusion of randomized dividend strategies. The insurer will pay a dividend of 1 with a probability q0 ð0 6 q0 < 1Þ in each time period if the surplus is greater than or equal to a non-negative integer b at the beginning of the period. That is, for t ¼ 0; 1; 2; . . ., the surplus process at time t is given by U ðtÞ ¼ u þ t
t X
gk IðU ðk 1Þ P bÞ
k¼1
t X
X k nk ;
ð2Þ
k¼1
where IðEÞ is the indicator functionof E, gk ðk P 1Þ is a series of randomized decision functions Ptan event which are i.i.d. and independent of k¼1 X k nk . In detail, we denote by gk ¼ 1 the event where a dividend of 1 is paid at the time k and gk ¼ 0 the event when no dividend is paid at the time k. We assume P ðgk ¼ 1Þ ¼ q0 ;
P ðgk ¼ 0Þ ¼ p0 ;
where p0 þ q0 ¼ 1. Note that if q0 ¼ 0, the dividend risk model reduces to the classical model without constraints, see Tan and Yang [15] for more motivations about this dividend strategy. In which the authors obtained the recursive formula and asymptotic estimate for the Gerber–Shiu discounted free penalty function. We also assume that the positive security loading condition holds. That is, if we denote by h the relative security loading then h¼
1 pl q0 > 0; pl
ð3Þ
where l ¼ EðX Þ < 1. We now introduce the Gerber–Shiu discounted penalty function. Let T ¼ ft 2 N þ ; U ðtÞ < 0g be the time of ruin in the modified model (2) with T ¼ 1 if ruin does not occur. Note that if ruin occurs, jU ðT Þj is the deficit at ruin and U ðT Þ is the surplus immediately prior to ruin. Denote by mv ðuÞ ¼ E½vT xðU ðT Þ; jU ðT ÞjÞIðT < 1ÞjU ð0Þ ¼ u;
ð4Þ
the Gerber–Shiu expected discounted penalty function. Here, xðu1 ; u2 Þ : N N þ ! N is a non-negative bounded function, 0 < v 6 1 is the discount factor. We remark that there is a slight difference from Tan and Yang’s in the definition (compare Eq. (4) with Eq. (2.11) in Tan and Yang [15]). The aim of this paper is to study the Gerber–Shiu discounted free penalty function for risk model (2). For simplicity, we write mðuÞ ¼ m1 ðuÞ. The rest of the paper is organized as follows: In Section 2, we derive two defective renewal equations satisfied by the penalty function. Analytical solutions of the two renewal equations are presented in Section 3. As applications, we apply the analytical solutions to the probability of ultimate ruin, the deficit at ruin jU ðT Þj and the claim causing ruin U ðT Þ þ jU ðT Þj. In Section 4, we derive the asymptotic estimation for the penalty function, which is simpler and more computable than Tan and Yang’s.
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2. Defective renewal equations for the penalty function In this section, we derive two renewal equations for the discounted free penalty function mðuÞ: one for the initial surplus below the barrier and the other for the initial surplus above the barrier. Throughout the entire paper we will use curly capital letters to denote the corresponding generating functions. From Eqs. (3.6) and (3.7) of Tan and Yang [15] with slight modification, we know that mðuÞ ¼ qmðu þ 1Þ þ p
uþ1 X
mðu þ 1 kÞf ðkÞ þ pxðu þ 1Þ;
0 6 u < b;
ð5Þ
u P b;
ð6Þ
k¼1
mðuÞ ¼ q½q0 mðuÞ þ p0 mðu þ 1Þ þ pp0
uþ1 X
mðu þ 1 kÞf ðkÞ
k¼1
þ pq0
u X
mðu kÞf ðkÞ þ pp0 xðu þ 1Þ þ pq0 xðuÞ;
k¼1
where xðuÞ ¼
1 X
xðu; k uÞf ðkÞ:
k¼uþ1
Now, we deal with Eq. (6) first. Eq. (6) can be rewritten as qp0 mðu þ 1Þ ¼ ð1 qq0 ÞmðuÞ pp0
uþ1 X
mðu þ 1 kÞf ðkÞ
k¼1
pq0
u X
mðu kÞf ðkÞ pp0 xðu þ 1Þ pq0 xðuÞ;
u P b:
ð7Þ
k¼1
After multiplying by zu and summing over u from 0 to 1 Eq. (7) yields qp0 ½MðzÞ mð0Þ ¼ ð1 qq0 ÞzMðzÞ pp0 MðzÞFðzÞ pq0 zMðzÞFðzÞ pp0 ðWðzÞ xð0ÞÞ pq0 zWðzÞ; or equivalently ½qp0 ð1 qq0 Þz þ pp0 FðzÞ þ pq0 zFðzÞMðzÞ ¼ qp0 mð0Þ þ pp0 xð0Þ pp0 WðzÞ pq0 zWðzÞ:
ð8Þ
It is easy to see that z = 1 is the root to the equation qp0 ð1 qq0 Þz þ pp0 FðzÞ þ pq0 zFðzÞ ¼ 0: Therefore, to determine the constant qp0 mð0Þ þ pp0 xð0Þ, we set z = 1 in (8) to obtain qp0 mð0Þ þ pp0 xð0Þ ¼ pp0 Wð1Þ þ pq0 Wð1Þ ¼ pWð1Þ:
ð9Þ
Substituting (9) into (8) results in ½qp0 ð1 qq0 Þz þ pp0 FðzÞ þ pq0 zFðzÞMðzÞ ¼ pp0 ½Wð1Þ WðzÞ þ pq0 ½Wð1Þ zWðzÞ:
ð10Þ
We subtract qp0 ð1 qq0 Þ þ pp0 Fð1Þ þ pq0 Fð1Þ ¼ 0 from the first term on the left hand side of (8) to obtain ½ð1 qq0 Þð1 zÞ pp0 ðFð1Þ FðzÞÞ pq0 ðFð1Þ zFðzÞÞMðzÞ ¼ pp0 ½Wð1Þ WðzÞ þ pq0 ½Wð1Þ zWðzÞ that is MðzÞ ¼
p Fð1Þ FðzÞ Fð1Þ zFðzÞ Wð1Þ WðzÞ Wð1Þ zWðzÞ p0 MðzÞ þ q0 MðzÞ þ p0 þ q0 : 1 qq0 1z 1z 1z 1z ð11Þ
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279
Now for any function aðxÞ; x 2 N , with generating function AðzÞ, one has (see [13]) 1 1 X AðtÞ AðzÞ X ¼ zu tiu1 aðiÞ: tz u¼0 i¼uþ1 Also, one has easily that 1 1 X tAðtÞ zAðzÞ X ¼ zu tiu aðiÞ: tz i¼u u¼0 Thus, equating the coefficients of zu in (11) we obtain ( ) u 1 1 X X X p mðuÞ ¼ mðu iÞ½p0 F ðiÞ þ q0 F ði 1Þ þ p0 xðiÞ þ q0 xðiÞ 1 qq0 i¼0 i¼u i¼uþ1 ( ) u 1 X X p ¼ mðu iÞ½p0 F ðiÞ þ q0 F ði 1Þ þ xðiÞ p0 xðuÞ : 1 qq0 i¼0 i¼u
ð12Þ
We now demonstrate the renewal equation (12) is defective. In fact, by the positive relative security condition (3), we have 1 X p pl þ pq0 ½p F ðiÞ þ q0 F ði 1Þ ¼ < 1: 1 qq0 i¼0 0 1 qq0 For the case of 0 6 u < b, we have by (5) that q½MðzÞ mð0Þ ¼ zMðzÞ pMðzÞFðzÞ p½WðzÞ xð0Þ; or equivalently ½q z þ pFðzÞMðzÞ ¼ qmð0Þ p½WðzÞ xð0Þ:
ð13Þ
Let z = 1 in (13) we have qmð0Þ ¼ p½Wð1Þ xð0Þ; then by imitating the proof above we get Fð1Þ FðzÞ Wð1Þ WðzÞ MðzÞ þ p : 1z 1z Equating the coefficients of zu in (14) we obtain u 1 X X mðu iÞF ðiÞ þ p xðiÞ: mðuÞ ¼ p MðzÞ ¼ p
i¼0
ð14Þ
ð15Þ
i¼uþ1
In this case 1 X F ðiÞ ¼ pl < 1 q0 6 1; p i¼0
which means renewal equation (15) is defective. All above derivations are summarized by the theorem below. Theorem 2.1. For u 2 N , the penalty function mðuÞ satisfies: If u P b ( ) u 1 X X p mðuÞ ¼ mðu iÞ½p0 F ðiÞ þ q0 F ði 1Þ þ xðiÞ p0 xðuÞ 1 qq0 i¼0 i¼u with
" # 1 p X mð0Þ ¼ xðiÞ p0 xð0Þ : qp0 i¼0
ð16Þ
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Z.-h. Bao / Applied Mathematics and Computation 194 (2007) 276–286
If 0 6 u < b mðuÞ ¼ p
u X
mðu iÞF ðiÞ þ p
i¼0
with
1 X
ð17Þ
xðiÞ
i¼uþ1
" # 1 p X mð0Þ ¼ xðiÞ xð0Þ : q i¼0
Remark. If q0 ¼ 0, which is the case of compound binomial risk model without constraints, then Eq. (16) 1 reduces to Eq. (17). If we denote by g ¼ pl 1 > 0, then (17) can be rewritten as mðuÞ ¼
u 1 X 1 X mðu iÞf1 ði þ 1Þ þ p xðiÞ; 1 þ g i¼0 i¼uþ1
which is consistent with Theorem 4.1 in Pavlova and Willmot [13]. 3. Solutions of the renewal equations and applications In this section, we obtain analytical expressions for the penalty function, which allow us to compute explicitly the important quantities of interest in risk theory. For this, we introduce two associated p.f.s as follows: g1 ðiÞ ¼
F ðiÞ ; l
g2 ðiÞ ¼
1 ½p F ðiÞ þ q0 F ði 1Þ; l þ q0 0
i 2 N:
It is easy to see that gk ðk ¼ 1; 2Þ are proper p.f.s. Denote by Gk ðxÞ ¼ 1 Gk ðxÞ ðk ¼ 1; 2Þ the corresponding c.d.f.s. Now we introduce two parameters bk ðk ¼ 1; 2Þ as 1 pðl þ q0 Þ ¼ ; 1 þ b2 1 qq0
1 ¼ pl; 1 þ b1 or equivalently b1 ¼
1 pl ; pl
b2 ¼
p0 pl : pðl þ q0 Þ
Then the renewal equations in Theorem 2.1 may be expressed in terms of gk and bk ðk ¼ 1; 2Þ as 8 u P 1 1 > > mðu iÞg1 ðiÞ þ 1þb s1 ðuÞ if 0 6 u < b; < 1þb1 1 i¼0 mðuÞ ¼ u P > 1 1 > : 1þb mðu iÞg2 ðiÞ þ 1þb s2 ðuÞ if u P b; 2 1
ð18Þ
i¼0
where 1 1 X s1 ðuÞ ¼ xðiÞ; l i¼uþ1
" # 1 X 1 s2 ðuÞ ¼ xðiÞ p0 xðuÞ : l þ q0 i¼u
Define the associated compound geometric c.d.f.s H k ðxÞ ¼ 1 H k ðxÞ ðk ¼ 1; 2Þ by n 1 X bk 1 H k ðuÞ ¼ Gn u 2 N; k ðuÞ; 1 þ b k 1 þ bk n¼1 where Gn k is the tail of n-fold convolution of Gk with itself. Assume that hk ðk ¼ 1; 2Þ are the corresponding bk p.f.s with hk ð0Þ ¼ 1þb . k
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281
The defective renewal functions can be solved by using generating functions, namely Theorem 3.1. For u 2 N , the solutions of the renewal equations in Theorem 2.1 can be expressed as 8 P u 2 1 > > s1 ðu iÞh1 ðiÞ þ pql s1 ðuÞ if 0 6 u < b; < b1 i¼0 mðuÞ ¼ u P > p2 ðlþq0 Þ > : b1 s2 ðu iÞh2 ðiÞ þ ð1qq s ðuÞ if u P b: Þqp 2 2 0
i¼0
ð19Þ
0
Proof. It is not hard to see that Hk ðzÞ ¼
bk =ð1 þ bk Þ b g ð0Þ=ð1 þ bk Þ k k : 1 Gk ðzÞ=ð1 þ bk Þ 1 þ bk gk ð0Þ
ð20Þ
From (18) we know that MðzÞ ¼
1=ð1 þ bk Þ Tk ðzÞ: 1 Gk ðzÞ=ð1 þ bk Þ
ð21Þ
Combine (20) and (21) we obtain 1 b g ð0Þ=ð1 þ bk Þ MðzÞ ¼ Hk ðzÞ þ k k Tk ðzÞ; bk 1 þ bk gk ð0Þ by equating the coefficients of zu in (22) yields (19) after some algebra.
ð22Þ h
We now proceed with some important special cases in the rest of the section. First, assume xðu1 ; u2 Þ ¼ 1, then mðuÞ becomes the probability of ultimate ruin, which we denote by wðuÞ. From Theorem 3.1 we have the following: Corollary 3.1. For any u 2 N , the probability of ultimate ruin for risk model (2) is given by 8 u P > p2 l 1 > if 0 6 u < b; > b > 1 i¼0 F 1 ðu þ 1 iÞh1 ðiÞ þ q F 1 ðu þ 1Þ > > < u P l wðuÞ ¼ ½F 1 ðu þ 1 iÞ þ q0 f1 ðu þ 1 iÞh2 ðiÞ > ðlþq Þb 0 2 > i¼0 > > > > : þ p2 l ½F 1 ðu þ 1Þ þ q f1 ðu þ 1Þ if u P b 0 ð1qq Þqp 0
with
(p wð0Þ ¼
ðl 1Þ
q p qp0
ðl p0 Þ
ð23Þ
0
if b > 0; if b ¼ 0:
Proof. For 0 6 u < b, s1 ðuÞ ¼
1 1 1 1 1 X X 1 X 1 X 1 X xði; j iÞf ðjÞ ¼ f ðjÞ ¼ F ðiÞ ¼ F 1 ðu þ 1Þ: l i¼uþ1 j¼iþ1 l i¼uþ1 j¼iþ1 l i¼uþ1
In the case of u P b, by (24) we have " # 1 X 1 l q s2 ðuÞ ¼ xðiÞ þ q0 xðuÞ ¼ s1 ðuÞ þ 0 xðuÞ l þ q0 i¼uþ1 l þ q0 l l ¼ ½F 1 ðu þ 1Þ þ q0 f1 ðu þ 1Þ; l þ q0 then (23) follows from (24) and (25) directly.
h
ð24Þ
ð25Þ
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Remark. If q0 ¼ 0, then wð0Þ ¼ pq ðl 1Þ for any b P 0, which is the probability of ultimate ruin when the initial surplus is 0 for compound binomial model (1) (see [14,4]). We now consider another special case involving the deficit at ruin. For y 2 N þ , let F ðyjuÞ ¼ P fjU ðT Þj 6 y; T < 1jU ð0Þ ¼ ug, we have Corollary 3.2. For any u 2 N and y 2 N þ , the defective cumulative distribution function F ðyjuÞ for risk model (2) satisfies 8 u P > 1 > ½F 1 ðu þ 1 iÞ F 1 ðu þ y þ 1 iÞh1 ðiÞ > b1 > > i¼0 > > > > > p2 l > > if 0 6 u < b; > þ q ½F 1 ðu þ 1Þ F 1 ðu þ y þ 1Þ > > < u P F ðyjuÞ ¼ 1 fl½F 1 ðu þ 1 iÞ F 1 ðu þ y þ 1 iÞ > ðlþq0 Þb2 > > i¼0 > > > > > > þq0 ½F ðu iÞ F ðu þ y iÞgh2 ðiÞ > > > > > > : þ p2 fl½F 1 ðu þ 1Þ F 1 ðu þ y þ 1Þ þ q0 ½F ðuÞ F ðu þ yÞg if u P b ð1qq Þqp 0
0
with (p F ðyj0Þ ¼
q
½lF 1 ðy þ 1Þ 1
p qp0
if b > 0;
pq0 ½lF 1 ðy þ 1Þ 1 þ qp F ðyÞ 0
if b ¼ 0:
Proof. Let xðu1 ; u2 Þ ¼ Iðu2 6 yÞ, then mðuÞ ¼ F ðyjuÞ. If 0 6 u < b, we have s1 ðuÞ ¼
¼
iþy 1 1 1 1 X X 1 X 1 X 1 X xði; j iÞf ðjÞ ¼ f ðjÞ ¼ ½F ðiÞ F ði þ yÞ l i¼uþ1 j¼iþ1 l i¼uþ1 j¼iþ1 l i¼uþ1 1 X
½f1 ði þ 1Þ f1 ði þ y þ 1Þ ¼ F 1 ðu þ 1Þ F 1 ðu þ y þ 1Þ:
i¼uþ1
For u P b, similar to Eq. (25) we have " # 1 X 1 l q0 s2 ðuÞ ¼ xðiÞ þ q0 xðuÞ ¼ s1 ðuÞ þ xðuÞ l þ q0 i¼uþ1 l þ q0 l þ q0 ¼
l q0 s1 ðuÞ þ ½F ðuÞ F ðu þ yÞ; l þ q0 l þ q0
then Corollary 3.2 follows.
h
Remark. If b = 0, it is easy to see that F ðyj0Þ is equal to Eq. (5.9) in Tan and Yang [15] after some simple calculations. If q0 ¼ 0, then F ðyj0Þ ¼ pq ðlF 1 ðy þ 1Þ 1Þ for any b P 0, which is the deficit distribution when the initial surplus is 0 for compound binomial model (1). When ruin occurs to a claim, U ðT Þ þ jU ðT Þj is the amount of the claim causing ruin, which has a different distribution than the amount of an arbitrary claim (since it is large enough to cause ruin). Now we derive explicit expression for the distribution of U ðT Þ þ jU ðT Þj in the following result. Denote by F ðsjuÞ ¼ E½IðU ðT Þ þ jU ðT Þj 6 sÞIðT < 1ÞjU ð0Þ ¼ u:
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283
Corollary 3.3. For any u 2 N and s P u þ 2, the defective distribution F ðsjuÞ of U ðT Þ þ jU ðT Þj is 8 P u 1 > ½F 1 ðu þ 1 iÞ F 1 ðs þ 1Þ þ ðu i sÞf1 ðs þ 1Þh1 ðiÞ > > b 1 > i¼0 > > > 2 > > > þ pql ½F 1 ðu þ 1Þ F 1 ðs þ 1Þ þ ðu sÞf1 ðs þ 1Þ > < u P F ðsjuÞ ¼ 1 fl½F 1 ðu þ 1 iÞ F 1 ðs þ 1Þ þ ðu i sÞf1 ðs þ 1Þ > Þb ðlþq > 0 2 > i¼0 > > > > > > þq0 ½F ðu iÞ F ðsÞgh2 ðiÞ > > 2 : þ ð1qqp Þqp fl½F 1 ðu þ 1Þ F 1 ðs þ 1Þ þ ðu sÞf1 ðs þ 1Þ þ q0 ½F ðuÞ F ðsÞg; 0
0
if 0 6 u < b;
if u P b ð26Þ
with
(p F ðsj0Þ ¼
fl½F 1 ðs þ 1Þ sf1 ðs þ 1Þ 1g
q p qp0
fl½F 1 ðs þ 1Þ sf1 ðs þ 1Þ 1 þ q0 F ðsÞg
if b > 0; if b ¼ 0:
Proof. Let xðu1 ; u2 Þ ¼ Iðu1 þ u2 6 sÞ, then mðuÞ ¼ F ðsjuÞ. For s P u þ 2, imitating the proof of Corollary 4.2 in Pavlova and Willmot [13] we have s1 ðuÞ ¼ F 1 ðu þ 1Þ F 1 ðs þ 1Þ þ ðu sÞf1 ðs þ 1Þ; similar to (25), for any u P b, we have l q0 l q0 s2 ðuÞ ¼ s1 ðuÞ þ xðuÞ ¼ s1 ðuÞ þ ½F ðuÞ F ðsÞ: l þ q0 l þ q0 l þ q0 l þ q0
ð27Þ
ð28Þ
Then (26) follows from (27) and (28). h Remark. If q0 ¼ 0, then F ðsj0Þ ¼ pq fl½F 1 ðs þ 1Þ sf1 ðs þ 1Þ 1g for any b P 0, which is the distribution of the claim causing ruin when the initial surplus is 0 for compound binomial model (1). 4. Asymptotic formulas revisited Since the function mðuÞ is the solution of a renewal equation whenever 0 6 u < b or u P b, its asymptotic behavior can then examined by renewal theory. Let fak ; k ¼ 0; 1; 2; . . .g and fbk ; k ¼ 0; 1; 2; . . .g be two non-negative sequences. Consider the defective renewal equation n X unk ak þ bn ; n ¼ 0; 1; . . . un ¼ k¼0
Seek a real number R such that 1 X Rk ak ¼ 1: k¼0
P P ^ bk ¼ Rk bk for simplicity, then if 1 ak < 1 and 1 Denote by ^ ak ¼ Rk ak and ^ k¼0 k^ k¼0 bk < 1, the renewal theorem [9, Chapter 3] states that P1 ^ bk ð29Þ lim Rn un ¼ P1k¼0 : n!1 k^ k¼0 ak Tan and Yang [15] investigated the asymptotic estimation for mðuÞ in detail. Let R > 1 be the root to the equation ðq0 r þ p0 ÞðpFðrÞ þ qÞ ¼ r:
ð30Þ
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Z.-h. Bao / Applied Mathematics and Computation 194 (2007) 276–286
As discussed in Tan and Yang [15], R necessarily exists. They obtained the following result: mðuÞ CRu
ðu ! 1Þ;
ð31Þ
where C¼
Pb ½Rbþ1 q0 R p0 qmð0Þ q0 ðR 1Þ u¼1 mðu 1ÞRu þ p½p0 J ðbÞ þ q0 LðbÞ ; P1 pðR 1Þ k¼1 kRk ½p0 F ðkÞ þ q0 F ðk 1Þ
and J ðbÞ; LðbÞ are two functions depending on the penalty function xðu1 ; u2 Þ (see Eq. (4.7) in [15]). We remark that the symbol C itself exhibit the dependence on mðuÞ and the formula is cumbersome. It is not very good for computation. In this section, we will give another asymptotic formula in terms of the renewal equation obtained in Section 2. For k ¼ 0; 1; . . ., let p ak ¼ ½p F ðkÞ þ q0 F ðk 1Þ; 1 qq0 0 " # 1 X p xðiÞ þ q0 xðkÞ : bk ¼ 1 qq0 i¼kþ1 Thus we seek R > 1 such that " # 1 1 X X p p Rk F ðkÞ þ q0 Rk F ðk 1Þ ¼ 1: 1 qq0 0 k¼0 k¼0
ð32Þ
After some simple arrangement we know that Eq. (32) is equal to Eq. (30). We call R > 1 the adjustment coefficient. Note that " # 1 1 1 X X X p k k k kR ak ¼ p kR F ðkÞ þ q0 kR F ðk 1Þ 1 qq0 0 k¼1 k¼1 k¼1 " ! # 1 1 1 1 X X X X p ¼ p kRk f ðiÞ f ðkÞ þ q0 kRk f ðiÞ 1 qq0 0 k¼1 i¼k i¼k k¼1 " # " # 1 1 1 1 i X X X X X p p k k k 0 ¼ kR f ðiÞ p0 kR f ðkÞ ¼ f ðiÞ kR p0 RF ðRÞ 1 qq0 k¼1 1 qq0 i¼1 i¼k k¼1 k¼1 " # 1 1 X p R 1 X 0 i iþ1 ¼ f ðiÞð1 R Þ iR f ðiÞ p0 RF ðRÞ 1 qq0 ð1 RÞ2 i¼1 1 R i¼1 " # p R R2 F0 ðRÞ p0 RF0 ðRÞ ¼ ð1 FðRÞÞ 1 qq0 ð1 RÞ2 1R " # p R R 0 þ p0 RF ðRÞ : ¼ ð1 FðRÞÞ ð33Þ 1 qq0 ð1 RÞ2 1R Similarly, we have 1 X k¼1
" # " # 1 1 1 1 1 1 X X X X X X p p k k k k R bk ¼ R xðiÞ þ q0 R xðkÞ ¼ R xðiÞ p0 R xðkÞ 1 qq0 k¼0 i¼kþ1 1 qq0 k¼0 i¼k k¼0 k¼0 " # " # 1 i 1 X X p p 1 X k iþ1 ¼ xðiÞ R p0 WðRÞ ¼ xðiÞð1 R Þ p0 WðRÞ 1 qq0 i¼0 1 qq0 1 R i¼0 k¼0 " # 1 p 1 X R þ p0 WðRÞ : ¼ xðiÞ 1 qq0 1 R i¼0 1R k
ð34Þ
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285
By (29), (33), (34) and Theorem 2.1, it follows that Theorem 4.1. The asymptotic estimate for the penalty function mðuÞ is P1 1R ðp0 þ q0 RÞWðRÞ i¼0 xðiÞ mðuÞ Ru R ð1 RÞðp0 þ q0 RÞF0 ðRÞ þ FðRÞ 1
ðu ! 1Þ:
ð35Þ
It is obvious that Eq. (35) is more convenient to compute than Eq. (31). We now give some examples of some important ruin quantities in risk theory to illustrate applications of the asymptotic estimate for mðuÞ derived in Theorem 4.1. Example 4.1. Consider the special case where xðu1 ; u2 Þ ¼ 1, then mðuÞ ¼ wðuÞ, which is the ultimate ruin probability for risk model (2). In this case WðRÞ ¼
1 ½1 FðRÞ 1R
ð36Þ
and 1 X
xðiÞ ¼ l;
ð37Þ
i¼0
then by (35)–(37) we have wðuÞ
ðp0 þ q0 RÞð1 FðRÞÞ ð1 RÞl Ru Rfð1 RÞðp0 þ q0 RÞF0 ðRÞ þ FðRÞ 1g
ðu ! 1Þ:
When q0 ¼ 0, which is the case of the compound binomial model (1) without constraint. Then we have w ðuÞ
ð1 FðRÞÞ ð1 RÞl Ru Rfð1 RÞF0 ðRÞ þ FðRÞ 1g
ðu ! 1Þ:
Let us consider the case of a zero-truncated geometrical individual claim distribution. Namely, we assume 1 . After some simple algebra, we know that f ðxÞ ¼ ð1 /Þ/x1 ; x 2 N þ , then F ðxÞ ¼ /x and EðX Þ ¼ 1/ FðRÞ ¼
ð1 /ÞR ; 1 R/
F0 ðRÞ ¼
1/ ð1 R/Þ
2
;
and the root greater than 1 to (30) is qp0 ; R¼ ðp0 qq0 Þ/ þ q0 then we can give the explicit expression for wðuÞ and w ðuÞ. For example, we have u p/ / ðu ! 1Þ: w ðuÞ qð1 /Þ q Example 4.2. Consider the case where xðu1 ; u2 Þ ¼ Iðu2 6 yÞ; y 2 N þ , then mðuÞ ¼ F ðyjuÞ, the deficit distribution at ruin for risk model (2). In this case X 1 Ry ½1 FðRÞ þ Ry Ri F ðiÞ 1R i¼0 y1
WðRÞ ¼
ð38Þ
and 1 X i¼0
xðiÞ ¼
y1 X i¼0
F ðiÞ:
ð39Þ
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Z.-h. Bao / Applied Mathematics and Computation 194 (2007) 276–286
By (35), (38) and (39) we have as u ! 1 F ðyjuÞ
Py1 Py1 ðp0 þ q0 RÞð1 FðRÞÞð1 Ry Þ þ ð1 RÞ½Ry i¼0 Ri F ðiÞ i¼0 F ðiÞ u R : Rfð1 RÞðp0 þ q0 RÞF0 ðRÞ þ FðRÞ 1g
When q0 ¼ 0, we then have
Py1 Py1 ð1 FðRÞÞð1 Ry Þ þ ð1 RÞ½Ry i¼0 Ri F ðiÞ i¼0 F ðiÞ u R : F ðyjuÞ Rfð1 RÞF0 ðRÞ þ FðRÞ 1g
In the case of an zero-truncated geometrical individual claim, we have the following simple formula: u p/ð1 /y Þ / F ðuÞ ðu ! 1Þ: qð1 /Þ q References [1] H. Albrecher, J. Hartinger, R. Tichy, On the distribution of dividend payments and the discounted penalty function in a risk model with linear dividend barrier, Scandinavian Actuarial Journal 2 (2005) 103–126. [2] S. Cheng, H.U. Gerber, E.S.W. Shiu, Discounted probabilities and ruin theory in the compound binomial model, Insurance: Mathematics and Economics 26 (2000) 239–250. [3] B. De Finetti, Su un’impostazione alternativa della teoria collettiva del rischio, in: Proceedings of the Transactions of the XV International Congress of Actuaries, vol. 2, 1957, pp. 433–443. [4] D.C.M. Dickson, Some comments on the compound binomial model, ASTIN Bulletin 24 (1994) 33–45. [5] D.C.M. Dickson, H. Waters, Some optimal dividend problems, ASTIN Bulletin 34 (2004) 49–74. [6] H.U. Gerber, Mathematical fun with the compound binomial process, ASTIN Bulletin 18 (1988) 161–168. [7] H.U. Gerber, E.S.W. Shiu, On the time value of ruin, North American Actuarial Journal 2 (1998) 48–78. [8] H.U. Gerber, E.S.W. Shiu, On optimal dividend strategies in the compound Poisson model 2005. Preprint. [9] S. Karlin, H.M. Taylor, A First Course in Stochastic Processes, Academic Press, New York, 1975. [10] X. Lin, K.P. Pavlova, The compound Poisson risk model with a threshold dividend strategy, Insurance: Mathematics and Economics 38 (2006) 57–80. [11] X. Lin, G.E. Willmot, Analysis of a defective renewal equation arising in ruin theory, Insurance: Mathematics and Economics 25 (1999) 63–84. [12] X. Lin, G.E. Willmot, S. Drekic, The classical risk model with a constant dividend barrier: analysis of the Gerber–Shiu discounted penalty function, Insurance: Mathematics and Economics 33 (2003) 551–566. [13] K.P. Pavlova, G.E. Willmot, The discrete stationary renewal risk model and the Gerber–Shiu discounted penalty function, Insurance: Mathematics and Economics 35 (2004) 267–277. [14] E.S.W. Shiu, Probability of eventual ruin in the compound binomial model, ASTIN Bulletin 19 (1989) 179–190. [15] J.Y. Tan, X.Q. Yang, The compound binomial model with randomized decisions on paying dividends, Insurance: Mathematics and Economics 39 (2006) 1–18.
Applied Mathematics and Computation 194 (2007) 276–286 www.elsevier.com/locate/amc
A note on the compound binomial model with randomized dividend strategy Zhen-hua Bao School of Mathematics, Liaoning Normal University, Dalian 116029, China
Abstract This paper considers the compound binomial model with randomized decisions on paying dividends. By using two formulas obtained by Tan and Yang [J.Y. Tan, X.Q. Yang, The compound binomial model with randomized decisions on paying dividends, Insurance: Mathematics and Economics 39 (2006) 1–18], two defective renewal equations for the Gerber–Shiu penalty function are derived and solved. The analytic solutions obtained are utilized to derive the probability of ultimate ruin, the deficit distribution at ruin and the distribution of the claim causing ruin. The asymptotic estimate satisfied by the penalty function is discussed in some detail. Ó 2007 Elsevier Inc. All rights reserved. Keywords: Compound binomial model; Defective renewal equation; Dividend strategy; Gerber–Shiu penalty function; Asymptotic estimate
1. Introduction The classical compound binomial risk model is a discrete time risk process with the following features. The premium received in each period is 1. In any period the probability of claim is p ð0 < p < 1Þ, and probability of no claim is q ¼ 1 p. We assume that claims occur at the end of the period and denote by nt ¼ 1 the event where a claim occurs in period ðt 1; t and nt ¼ 0 the event where no claim occurs in the time period ðt 1; t. The claims are independent and identically distributed (i.i.d.) positive integer valued random variables distributed as a random variable X with cumulative distribution function (c.d.f.) F ðxÞ ¼ 1 F ðxÞ and probability function (p.f.) f ðxÞ; x 2 N þ . Where we denote by N the set of nature numbers and N þ ¼ N f0g. Throughðx1Þ out, we assume the equilibrium p.f. of X is f1 ðxÞ ¼ FEðX ; x 2 N þ , and F 1 ðxÞ ¼ 1 F 1 ðxÞ is its c.d.f. For Þ t ¼ 0; 1; . . ., the surplus at time t is U ðtÞ ¼ u þ t
t X
X i ni ;
i¼1
E-mail address: [email protected] 0096-3003/$ - see front matter Ó 2007 Elsevier Inc. All rights reserved. doi:10.1016/j.amc.2007.04.023
ð1Þ
Z.-h. Bao / Applied Mathematics and Computation 194 (2007) 276–286
277
where U ð0Þ ¼ u, which is a non-negative integer. The risk model (1) has been studied extensively since it was introduced by Gerber [6]. Gerber derived a formula for the ruin probability, the distribution of the deficit at ruin and the distribution of the surplus just before ruin in the case that the initial surplus is 0. Shiu [14] derived similar expressions as Gerber’s using different methods. Cheng et al. [2] derived the moment generating function of the time to ruin for zero initial surplus and then they derived recursively the joint distribution of the surplus prior to ruin and the deficit at ruin. Dickson [4] applied elementary methods to derive the joint distribution of the surplus before ruin and the deficit at ruin when the initial reserve is 0. Recently, Gerber and Shiu [7] introduced a discounted penalty function with respect to the time of ruin, the surplus immediately before ruin and the deficit at ruin, which has proven to be a powerful analytical tool. See also Lin and Willmot [11]. For the compound binomial model, Pavlova and Willmot [13] derived a defective renewal equation satisfied by the Gerber–Shiu discounted penalty function. Dividend strategies for insurance risk models were first proposed by De Finetti [3] to reflect more realistically the surplus cash flows in an insurance portfolio. Barrier strategies for the classical compound Poisson risk model have been studied extensively in the literature, including Albrecher et al. [1], Dickson and Waters [5], Gerber and Shiu [7,8], Lin et al. [12], Lin and Pavlova [10], among many others. In Tan and Yang [15], the authors considered the compound binomial model modified by the inclusion of randomized dividend strategies. The insurer will pay a dividend of 1 with a probability q0 ð0 6 q0 < 1Þ in each time period if the surplus is greater than or equal to a non-negative integer b at the beginning of the period. That is, for t ¼ 0; 1; 2; . . ., the surplus process at time t is given by U ðtÞ ¼ u þ t
t X
gk IðU ðk 1Þ P bÞ
k¼1
t X
X k nk ;
ð2Þ
k¼1
where IðEÞ is the indicator functionof E, gk ðk P 1Þ is a series of randomized decision functions Ptan event which are i.i.d. and independent of k¼1 X k nk . In detail, we denote by gk ¼ 1 the event where a dividend of 1 is paid at the time k and gk ¼ 0 the event when no dividend is paid at the time k. We assume P ðgk ¼ 1Þ ¼ q0 ;
P ðgk ¼ 0Þ ¼ p0 ;
where p0 þ q0 ¼ 1. Note that if q0 ¼ 0, the dividend risk model reduces to the classical model without constraints, see Tan and Yang [15] for more motivations about this dividend strategy. In which the authors obtained the recursive formula and asymptotic estimate for the Gerber–Shiu discounted free penalty function. We also assume that the positive security loading condition holds. That is, if we denote by h the relative security loading then h¼
1 pl q0 > 0; pl
ð3Þ
where l ¼ EðX Þ < 1. We now introduce the Gerber–Shiu discounted penalty function. Let T ¼ ft 2 N þ ; U ðtÞ < 0g be the time of ruin in the modified model (2) with T ¼ 1 if ruin does not occur. Note that if ruin occurs, jU ðT Þj is the deficit at ruin and U ðT Þ is the surplus immediately prior to ruin. Denote by mv ðuÞ ¼ E½vT xðU ðT Þ; jU ðT ÞjÞIðT < 1ÞjU ð0Þ ¼ u;
ð4Þ
the Gerber–Shiu expected discounted penalty function. Here, xðu1 ; u2 Þ : N N þ ! N is a non-negative bounded function, 0 < v 6 1 is the discount factor. We remark that there is a slight difference from Tan and Yang’s in the definition (compare Eq. (4) with Eq. (2.11) in Tan and Yang [15]). The aim of this paper is to study the Gerber–Shiu discounted free penalty function for risk model (2). For simplicity, we write mðuÞ ¼ m1 ðuÞ. The rest of the paper is organized as follows: In Section 2, we derive two defective renewal equations satisfied by the penalty function. Analytical solutions of the two renewal equations are presented in Section 3. As applications, we apply the analytical solutions to the probability of ultimate ruin, the deficit at ruin jU ðT Þj and the claim causing ruin U ðT Þ þ jU ðT Þj. In Section 4, we derive the asymptotic estimation for the penalty function, which is simpler and more computable than Tan and Yang’s.
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2. Defective renewal equations for the penalty function In this section, we derive two renewal equations for the discounted free penalty function mðuÞ: one for the initial surplus below the barrier and the other for the initial surplus above the barrier. Throughout the entire paper we will use curly capital letters to denote the corresponding generating functions. From Eqs. (3.6) and (3.7) of Tan and Yang [15] with slight modification, we know that mðuÞ ¼ qmðu þ 1Þ þ p
uþ1 X
mðu þ 1 kÞf ðkÞ þ pxðu þ 1Þ;
0 6 u < b;
ð5Þ
u P b;
ð6Þ
k¼1
mðuÞ ¼ q½q0 mðuÞ þ p0 mðu þ 1Þ þ pp0
uþ1 X
mðu þ 1 kÞf ðkÞ
k¼1
þ pq0
u X
mðu kÞf ðkÞ þ pp0 xðu þ 1Þ þ pq0 xðuÞ;
k¼1
where xðuÞ ¼
1 X
xðu; k uÞf ðkÞ:
k¼uþ1
Now, we deal with Eq. (6) first. Eq. (6) can be rewritten as qp0 mðu þ 1Þ ¼ ð1 qq0 ÞmðuÞ pp0
uþ1 X
mðu þ 1 kÞf ðkÞ
k¼1
pq0
u X
mðu kÞf ðkÞ pp0 xðu þ 1Þ pq0 xðuÞ;
u P b:
ð7Þ
k¼1
After multiplying by zu and summing over u from 0 to 1 Eq. (7) yields qp0 ½MðzÞ mð0Þ ¼ ð1 qq0 ÞzMðzÞ pp0 MðzÞFðzÞ pq0 zMðzÞFðzÞ pp0 ðWðzÞ xð0ÞÞ pq0 zWðzÞ; or equivalently ½qp0 ð1 qq0 Þz þ pp0 FðzÞ þ pq0 zFðzÞMðzÞ ¼ qp0 mð0Þ þ pp0 xð0Þ pp0 WðzÞ pq0 zWðzÞ:
ð8Þ
It is easy to see that z = 1 is the root to the equation qp0 ð1 qq0 Þz þ pp0 FðzÞ þ pq0 zFðzÞ ¼ 0: Therefore, to determine the constant qp0 mð0Þ þ pp0 xð0Þ, we set z = 1 in (8) to obtain qp0 mð0Þ þ pp0 xð0Þ ¼ pp0 Wð1Þ þ pq0 Wð1Þ ¼ pWð1Þ:
ð9Þ
Substituting (9) into (8) results in ½qp0 ð1 qq0 Þz þ pp0 FðzÞ þ pq0 zFðzÞMðzÞ ¼ pp0 ½Wð1Þ WðzÞ þ pq0 ½Wð1Þ zWðzÞ:
ð10Þ
We subtract qp0 ð1 qq0 Þ þ pp0 Fð1Þ þ pq0 Fð1Þ ¼ 0 from the first term on the left hand side of (8) to obtain ½ð1 qq0 Þð1 zÞ pp0 ðFð1Þ FðzÞÞ pq0 ðFð1Þ zFðzÞÞMðzÞ ¼ pp0 ½Wð1Þ WðzÞ þ pq0 ½Wð1Þ zWðzÞ that is MðzÞ ¼
p Fð1Þ FðzÞ Fð1Þ zFðzÞ Wð1Þ WðzÞ Wð1Þ zWðzÞ p0 MðzÞ þ q0 MðzÞ þ p0 þ q0 : 1 qq0 1z 1z 1z 1z ð11Þ
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279
Now for any function aðxÞ; x 2 N , with generating function AðzÞ, one has (see [13]) 1 1 X AðtÞ AðzÞ X ¼ zu tiu1 aðiÞ: tz u¼0 i¼uþ1 Also, one has easily that 1 1 X tAðtÞ zAðzÞ X ¼ zu tiu aðiÞ: tz i¼u u¼0 Thus, equating the coefficients of zu in (11) we obtain ( ) u 1 1 X X X p mðuÞ ¼ mðu iÞ½p0 F ðiÞ þ q0 F ði 1Þ þ p0 xðiÞ þ q0 xðiÞ 1 qq0 i¼0 i¼u i¼uþ1 ( ) u 1 X X p ¼ mðu iÞ½p0 F ðiÞ þ q0 F ði 1Þ þ xðiÞ p0 xðuÞ : 1 qq0 i¼0 i¼u
ð12Þ
We now demonstrate the renewal equation (12) is defective. In fact, by the positive relative security condition (3), we have 1 X p pl þ pq0 ½p F ðiÞ þ q0 F ði 1Þ ¼ < 1: 1 qq0 i¼0 0 1 qq0 For the case of 0 6 u < b, we have by (5) that q½MðzÞ mð0Þ ¼ zMðzÞ pMðzÞFðzÞ p½WðzÞ xð0Þ; or equivalently ½q z þ pFðzÞMðzÞ ¼ qmð0Þ p½WðzÞ xð0Þ:
ð13Þ
Let z = 1 in (13) we have qmð0Þ ¼ p½Wð1Þ xð0Þ; then by imitating the proof above we get Fð1Þ FðzÞ Wð1Þ WðzÞ MðzÞ þ p : 1z 1z Equating the coefficients of zu in (14) we obtain u 1 X X mðu iÞF ðiÞ þ p xðiÞ: mðuÞ ¼ p MðzÞ ¼ p
i¼0
ð14Þ
ð15Þ
i¼uþ1
In this case 1 X F ðiÞ ¼ pl < 1 q0 6 1; p i¼0
which means renewal equation (15) is defective. All above derivations are summarized by the theorem below. Theorem 2.1. For u 2 N , the penalty function mðuÞ satisfies: If u P b ( ) u 1 X X p mðuÞ ¼ mðu iÞ½p0 F ðiÞ þ q0 F ði 1Þ þ xðiÞ p0 xðuÞ 1 qq0 i¼0 i¼u with
" # 1 p X mð0Þ ¼ xðiÞ p0 xð0Þ : qp0 i¼0
ð16Þ
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Z.-h. Bao / Applied Mathematics and Computation 194 (2007) 276–286
If 0 6 u < b mðuÞ ¼ p
u X
mðu iÞF ðiÞ þ p
i¼0
with
1 X
ð17Þ
xðiÞ
i¼uþ1
" # 1 p X mð0Þ ¼ xðiÞ xð0Þ : q i¼0
Remark. If q0 ¼ 0, which is the case of compound binomial risk model without constraints, then Eq. (16) 1 reduces to Eq. (17). If we denote by g ¼ pl 1 > 0, then (17) can be rewritten as mðuÞ ¼
u 1 X 1 X mðu iÞf1 ði þ 1Þ þ p xðiÞ; 1 þ g i¼0 i¼uþ1
which is consistent with Theorem 4.1 in Pavlova and Willmot [13]. 3. Solutions of the renewal equations and applications In this section, we obtain analytical expressions for the penalty function, which allow us to compute explicitly the important quantities of interest in risk theory. For this, we introduce two associated p.f.s as follows: g1 ðiÞ ¼
F ðiÞ ; l
g2 ðiÞ ¼
1 ½p F ðiÞ þ q0 F ði 1Þ; l þ q0 0
i 2 N:
It is easy to see that gk ðk ¼ 1; 2Þ are proper p.f.s. Denote by Gk ðxÞ ¼ 1 Gk ðxÞ ðk ¼ 1; 2Þ the corresponding c.d.f.s. Now we introduce two parameters bk ðk ¼ 1; 2Þ as 1 pðl þ q0 Þ ¼ ; 1 þ b2 1 qq0
1 ¼ pl; 1 þ b1 or equivalently b1 ¼
1 pl ; pl
b2 ¼
p0 pl : pðl þ q0 Þ
Then the renewal equations in Theorem 2.1 may be expressed in terms of gk and bk ðk ¼ 1; 2Þ as 8 u P 1 1 > > mðu iÞg1 ðiÞ þ 1þb s1 ðuÞ if 0 6 u < b; < 1þb1 1 i¼0 mðuÞ ¼ u P > 1 1 > : 1þb mðu iÞg2 ðiÞ þ 1þb s2 ðuÞ if u P b; 2 1
ð18Þ
i¼0
where 1 1 X s1 ðuÞ ¼ xðiÞ; l i¼uþ1
" # 1 X 1 s2 ðuÞ ¼ xðiÞ p0 xðuÞ : l þ q0 i¼u
Define the associated compound geometric c.d.f.s H k ðxÞ ¼ 1 H k ðxÞ ðk ¼ 1; 2Þ by n 1 X bk 1 H k ðuÞ ¼ Gn u 2 N; k ðuÞ; 1 þ b k 1 þ bk n¼1 where Gn k is the tail of n-fold convolution of Gk with itself. Assume that hk ðk ¼ 1; 2Þ are the corresponding bk p.f.s with hk ð0Þ ¼ 1þb . k
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281
The defective renewal functions can be solved by using generating functions, namely Theorem 3.1. For u 2 N , the solutions of the renewal equations in Theorem 2.1 can be expressed as 8 P u 2 1 > > s1 ðu iÞh1 ðiÞ þ pql s1 ðuÞ if 0 6 u < b; < b1 i¼0 mðuÞ ¼ u P > p2 ðlþq0 Þ > : b1 s2 ðu iÞh2 ðiÞ þ ð1qq s ðuÞ if u P b: Þqp 2 2 0
i¼0
ð19Þ
0
Proof. It is not hard to see that Hk ðzÞ ¼
bk =ð1 þ bk Þ b g ð0Þ=ð1 þ bk Þ k k : 1 Gk ðzÞ=ð1 þ bk Þ 1 þ bk gk ð0Þ
ð20Þ
From (18) we know that MðzÞ ¼
1=ð1 þ bk Þ Tk ðzÞ: 1 Gk ðzÞ=ð1 þ bk Þ
ð21Þ
Combine (20) and (21) we obtain 1 b g ð0Þ=ð1 þ bk Þ MðzÞ ¼ Hk ðzÞ þ k k Tk ðzÞ; bk 1 þ bk gk ð0Þ by equating the coefficients of zu in (22) yields (19) after some algebra.
ð22Þ h
We now proceed with some important special cases in the rest of the section. First, assume xðu1 ; u2 Þ ¼ 1, then mðuÞ becomes the probability of ultimate ruin, which we denote by wðuÞ. From Theorem 3.1 we have the following: Corollary 3.1. For any u 2 N , the probability of ultimate ruin for risk model (2) is given by 8 u P > p2 l 1 > if 0 6 u < b; > b > 1 i¼0 F 1 ðu þ 1 iÞh1 ðiÞ þ q F 1 ðu þ 1Þ > > < u P l wðuÞ ¼ ½F 1 ðu þ 1 iÞ þ q0 f1 ðu þ 1 iÞh2 ðiÞ > ðlþq Þb 0 2 > i¼0 > > > > : þ p2 l ½F 1 ðu þ 1Þ þ q f1 ðu þ 1Þ if u P b 0 ð1qq Þqp 0
with
(p wð0Þ ¼
ðl 1Þ
q p qp0
ðl p0 Þ
ð23Þ
0
if b > 0; if b ¼ 0:
Proof. For 0 6 u < b, s1 ðuÞ ¼
1 1 1 1 1 X X 1 X 1 X 1 X xði; j iÞf ðjÞ ¼ f ðjÞ ¼ F ðiÞ ¼ F 1 ðu þ 1Þ: l i¼uþ1 j¼iþ1 l i¼uþ1 j¼iþ1 l i¼uþ1
In the case of u P b, by (24) we have " # 1 X 1 l q s2 ðuÞ ¼ xðiÞ þ q0 xðuÞ ¼ s1 ðuÞ þ 0 xðuÞ l þ q0 i¼uþ1 l þ q0 l l ¼ ½F 1 ðu þ 1Þ þ q0 f1 ðu þ 1Þ; l þ q0 then (23) follows from (24) and (25) directly.
h
ð24Þ
ð25Þ
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Remark. If q0 ¼ 0, then wð0Þ ¼ pq ðl 1Þ for any b P 0, which is the probability of ultimate ruin when the initial surplus is 0 for compound binomial model (1) (see [14,4]). We now consider another special case involving the deficit at ruin. For y 2 N þ , let F ðyjuÞ ¼ P fjU ðT Þj 6 y; T < 1jU ð0Þ ¼ ug, we have Corollary 3.2. For any u 2 N and y 2 N þ , the defective cumulative distribution function F ðyjuÞ for risk model (2) satisfies 8 u P > 1 > ½F 1 ðu þ 1 iÞ F 1 ðu þ y þ 1 iÞh1 ðiÞ > b1 > > i¼0 > > > > > p2 l > > if 0 6 u < b; > þ q ½F 1 ðu þ 1Þ F 1 ðu þ y þ 1Þ > > < u P F ðyjuÞ ¼ 1 fl½F 1 ðu þ 1 iÞ F 1 ðu þ y þ 1 iÞ > ðlþq0 Þb2 > > i¼0 > > > > > > þq0 ½F ðu iÞ F ðu þ y iÞgh2 ðiÞ > > > > > > : þ p2 fl½F 1 ðu þ 1Þ F 1 ðu þ y þ 1Þ þ q0 ½F ðuÞ F ðu þ yÞg if u P b ð1qq Þqp 0
0
with (p F ðyj0Þ ¼
q
½lF 1 ðy þ 1Þ 1
p qp0
if b > 0;
pq0 ½lF 1 ðy þ 1Þ 1 þ qp F ðyÞ 0
if b ¼ 0:
Proof. Let xðu1 ; u2 Þ ¼ Iðu2 6 yÞ, then mðuÞ ¼ F ðyjuÞ. If 0 6 u < b, we have s1 ðuÞ ¼
¼
iþy 1 1 1 1 X X 1 X 1 X 1 X xði; j iÞf ðjÞ ¼ f ðjÞ ¼ ½F ðiÞ F ði þ yÞ l i¼uþ1 j¼iþ1 l i¼uþ1 j¼iþ1 l i¼uþ1 1 X
½f1 ði þ 1Þ f1 ði þ y þ 1Þ ¼ F 1 ðu þ 1Þ F 1 ðu þ y þ 1Þ:
i¼uþ1
For u P b, similar to Eq. (25) we have " # 1 X 1 l q0 s2 ðuÞ ¼ xðiÞ þ q0 xðuÞ ¼ s1 ðuÞ þ xðuÞ l þ q0 i¼uþ1 l þ q0 l þ q0 ¼
l q0 s1 ðuÞ þ ½F ðuÞ F ðu þ yÞ; l þ q0 l þ q0
then Corollary 3.2 follows.
h
Remark. If b = 0, it is easy to see that F ðyj0Þ is equal to Eq. (5.9) in Tan and Yang [15] after some simple calculations. If q0 ¼ 0, then F ðyj0Þ ¼ pq ðlF 1 ðy þ 1Þ 1Þ for any b P 0, which is the deficit distribution when the initial surplus is 0 for compound binomial model (1). When ruin occurs to a claim, U ðT Þ þ jU ðT Þj is the amount of the claim causing ruin, which has a different distribution than the amount of an arbitrary claim (since it is large enough to cause ruin). Now we derive explicit expression for the distribution of U ðT Þ þ jU ðT Þj in the following result. Denote by F ðsjuÞ ¼ E½IðU ðT Þ þ jU ðT Þj 6 sÞIðT < 1ÞjU ð0Þ ¼ u:
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283
Corollary 3.3. For any u 2 N and s P u þ 2, the defective distribution F ðsjuÞ of U ðT Þ þ jU ðT Þj is 8 P u 1 > ½F 1 ðu þ 1 iÞ F 1 ðs þ 1Þ þ ðu i sÞf1 ðs þ 1Þh1 ðiÞ > > b 1 > i¼0 > > > 2 > > > þ pql ½F 1 ðu þ 1Þ F 1 ðs þ 1Þ þ ðu sÞf1 ðs þ 1Þ > < u P F ðsjuÞ ¼ 1 fl½F 1 ðu þ 1 iÞ F 1 ðs þ 1Þ þ ðu i sÞf1 ðs þ 1Þ > Þb ðlþq > 0 2 > i¼0 > > > > > > þq0 ½F ðu iÞ F ðsÞgh2 ðiÞ > > 2 : þ ð1qqp Þqp fl½F 1 ðu þ 1Þ F 1 ðs þ 1Þ þ ðu sÞf1 ðs þ 1Þ þ q0 ½F ðuÞ F ðsÞg; 0
0
if 0 6 u < b;
if u P b ð26Þ
with
(p F ðsj0Þ ¼
fl½F 1 ðs þ 1Þ sf1 ðs þ 1Þ 1g
q p qp0
fl½F 1 ðs þ 1Þ sf1 ðs þ 1Þ 1 þ q0 F ðsÞg
if b > 0; if b ¼ 0:
Proof. Let xðu1 ; u2 Þ ¼ Iðu1 þ u2 6 sÞ, then mðuÞ ¼ F ðsjuÞ. For s P u þ 2, imitating the proof of Corollary 4.2 in Pavlova and Willmot [13] we have s1 ðuÞ ¼ F 1 ðu þ 1Þ F 1 ðs þ 1Þ þ ðu sÞf1 ðs þ 1Þ; similar to (25), for any u P b, we have l q0 l q0 s2 ðuÞ ¼ s1 ðuÞ þ xðuÞ ¼ s1 ðuÞ þ ½F ðuÞ F ðsÞ: l þ q0 l þ q0 l þ q0 l þ q0
ð27Þ
ð28Þ
Then (26) follows from (27) and (28). h Remark. If q0 ¼ 0, then F ðsj0Þ ¼ pq fl½F 1 ðs þ 1Þ sf1 ðs þ 1Þ 1g for any b P 0, which is the distribution of the claim causing ruin when the initial surplus is 0 for compound binomial model (1). 4. Asymptotic formulas revisited Since the function mðuÞ is the solution of a renewal equation whenever 0 6 u < b or u P b, its asymptotic behavior can then examined by renewal theory. Let fak ; k ¼ 0; 1; 2; . . .g and fbk ; k ¼ 0; 1; 2; . . .g be two non-negative sequences. Consider the defective renewal equation n X unk ak þ bn ; n ¼ 0; 1; . . . un ¼ k¼0
Seek a real number R such that 1 X Rk ak ¼ 1: k¼0
P P ^ bk ¼ Rk bk for simplicity, then if 1 ak < 1 and 1 Denote by ^ ak ¼ Rk ak and ^ k¼0 k^ k¼0 bk < 1, the renewal theorem [9, Chapter 3] states that P1 ^ bk ð29Þ lim Rn un ¼ P1k¼0 : n!1 k^ k¼0 ak Tan and Yang [15] investigated the asymptotic estimation for mðuÞ in detail. Let R > 1 be the root to the equation ðq0 r þ p0 ÞðpFðrÞ þ qÞ ¼ r:
ð30Þ
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As discussed in Tan and Yang [15], R necessarily exists. They obtained the following result: mðuÞ CRu
ðu ! 1Þ;
ð31Þ
where C¼
Pb ½Rbþ1 q0 R p0 qmð0Þ q0 ðR 1Þ u¼1 mðu 1ÞRu þ p½p0 J ðbÞ þ q0 LðbÞ ; P1 pðR 1Þ k¼1 kRk ½p0 F ðkÞ þ q0 F ðk 1Þ
and J ðbÞ; LðbÞ are two functions depending on the penalty function xðu1 ; u2 Þ (see Eq. (4.7) in [15]). We remark that the symbol C itself exhibit the dependence on mðuÞ and the formula is cumbersome. It is not very good for computation. In this section, we will give another asymptotic formula in terms of the renewal equation obtained in Section 2. For k ¼ 0; 1; . . ., let p ak ¼ ½p F ðkÞ þ q0 F ðk 1Þ; 1 qq0 0 " # 1 X p xðiÞ þ q0 xðkÞ : bk ¼ 1 qq0 i¼kþ1 Thus we seek R > 1 such that " # 1 1 X X p p Rk F ðkÞ þ q0 Rk F ðk 1Þ ¼ 1: 1 qq0 0 k¼0 k¼0
ð32Þ
After some simple arrangement we know that Eq. (32) is equal to Eq. (30). We call R > 1 the adjustment coefficient. Note that " # 1 1 1 X X X p k k k kR ak ¼ p kR F ðkÞ þ q0 kR F ðk 1Þ 1 qq0 0 k¼1 k¼1 k¼1 " ! # 1 1 1 1 X X X X p ¼ p kRk f ðiÞ f ðkÞ þ q0 kRk f ðiÞ 1 qq0 0 k¼1 i¼k i¼k k¼1 " # " # 1 1 1 1 i X X X X X p p k k k 0 ¼ kR f ðiÞ p0 kR f ðkÞ ¼ f ðiÞ kR p0 RF ðRÞ 1 qq0 k¼1 1 qq0 i¼1 i¼k k¼1 k¼1 " # 1 1 X p R 1 X 0 i iþ1 ¼ f ðiÞð1 R Þ iR f ðiÞ p0 RF ðRÞ 1 qq0 ð1 RÞ2 i¼1 1 R i¼1 " # p R R2 F0 ðRÞ p0 RF0 ðRÞ ¼ ð1 FðRÞÞ 1 qq0 ð1 RÞ2 1R " # p R R 0 þ p0 RF ðRÞ : ¼ ð1 FðRÞÞ ð33Þ 1 qq0 ð1 RÞ2 1R Similarly, we have 1 X k¼1
" # " # 1 1 1 1 1 1 X X X X X X p p k k k k R bk ¼ R xðiÞ þ q0 R xðkÞ ¼ R xðiÞ p0 R xðkÞ 1 qq0 k¼0 i¼kþ1 1 qq0 k¼0 i¼k k¼0 k¼0 " # " # 1 i 1 X X p p 1 X k iþ1 ¼ xðiÞ R p0 WðRÞ ¼ xðiÞð1 R Þ p0 WðRÞ 1 qq0 i¼0 1 qq0 1 R i¼0 k¼0 " # 1 p 1 X R þ p0 WðRÞ : ¼ xðiÞ 1 qq0 1 R i¼0 1R k
ð34Þ
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285
By (29), (33), (34) and Theorem 2.1, it follows that Theorem 4.1. The asymptotic estimate for the penalty function mðuÞ is P1 1R ðp0 þ q0 RÞWðRÞ i¼0 xðiÞ mðuÞ Ru R ð1 RÞðp0 þ q0 RÞF0 ðRÞ þ FðRÞ 1
ðu ! 1Þ:
ð35Þ
It is obvious that Eq. (35) is more convenient to compute than Eq. (31). We now give some examples of some important ruin quantities in risk theory to illustrate applications of the asymptotic estimate for mðuÞ derived in Theorem 4.1. Example 4.1. Consider the special case where xðu1 ; u2 Þ ¼ 1, then mðuÞ ¼ wðuÞ, which is the ultimate ruin probability for risk model (2). In this case WðRÞ ¼
1 ½1 FðRÞ 1R
ð36Þ
and 1 X
xðiÞ ¼ l;
ð37Þ
i¼0
then by (35)–(37) we have wðuÞ
ðp0 þ q0 RÞð1 FðRÞÞ ð1 RÞl Ru Rfð1 RÞðp0 þ q0 RÞF0 ðRÞ þ FðRÞ 1g
ðu ! 1Þ:
When q0 ¼ 0, which is the case of the compound binomial model (1) without constraint. Then we have w ðuÞ
ð1 FðRÞÞ ð1 RÞl Ru Rfð1 RÞF0 ðRÞ þ FðRÞ 1g
ðu ! 1Þ:
Let us consider the case of a zero-truncated geometrical individual claim distribution. Namely, we assume 1 . After some simple algebra, we know that f ðxÞ ¼ ð1 /Þ/x1 ; x 2 N þ , then F ðxÞ ¼ /x and EðX Þ ¼ 1/ FðRÞ ¼
ð1 /ÞR ; 1 R/
F0 ðRÞ ¼
1/ ð1 R/Þ
2
;
and the root greater than 1 to (30) is qp0 ; R¼ ðp0 qq0 Þ/ þ q0 then we can give the explicit expression for wðuÞ and w ðuÞ. For example, we have u p/ / ðu ! 1Þ: w ðuÞ qð1 /Þ q Example 4.2. Consider the case where xðu1 ; u2 Þ ¼ Iðu2 6 yÞ; y 2 N þ , then mðuÞ ¼ F ðyjuÞ, the deficit distribution at ruin for risk model (2). In this case X 1 Ry ½1 FðRÞ þ Ry Ri F ðiÞ 1R i¼0 y1
WðRÞ ¼
ð38Þ
and 1 X i¼0
xðiÞ ¼
y1 X i¼0
F ðiÞ:
ð39Þ
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Z.-h. Bao / Applied Mathematics and Computation 194 (2007) 276–286
By (35), (38) and (39) we have as u ! 1 F ðyjuÞ
Py1 Py1 ðp0 þ q0 RÞð1 FðRÞÞð1 Ry Þ þ ð1 RÞ½Ry i¼0 Ri F ðiÞ i¼0 F ðiÞ u R : Rfð1 RÞðp0 þ q0 RÞF0 ðRÞ þ FðRÞ 1g
When q0 ¼ 0, we then have
Py1 Py1 ð1 FðRÞÞð1 Ry Þ þ ð1 RÞ½Ry i¼0 Ri F ðiÞ i¼0 F ðiÞ u R : F ðyjuÞ Rfð1 RÞF0 ðRÞ þ FðRÞ 1g
In the case of an zero-truncated geometrical individual claim, we have the following simple formula: u p/ð1 /y Þ / F ðuÞ ðu ! 1Þ: qð1 /Þ q References [1] H. Albrecher, J. Hartinger, R. Tichy, On the distribution of dividend payments and the discounted penalty function in a risk model with linear dividend barrier, Scandinavian Actuarial Journal 2 (2005) 103–126. [2] S. Cheng, H.U. Gerber, E.S.W. Shiu, Discounted probabilities and ruin theory in the compound binomial model, Insurance: Mathematics and Economics 26 (2000) 239–250. [3] B. De Finetti, Su un’impostazione alternativa della teoria collettiva del rischio, in: Proceedings of the Transactions of the XV International Congress of Actuaries, vol. 2, 1957, pp. 433–443. [4] D.C.M. Dickson, Some comments on the compound binomial model, ASTIN Bulletin 24 (1994) 33–45. [5] D.C.M. Dickson, H. Waters, Some optimal dividend problems, ASTIN Bulletin 34 (2004) 49–74. [6] H.U. Gerber, Mathematical fun with the compound binomial process, ASTIN Bulletin 18 (1988) 161–168. [7] H.U. Gerber, E.S.W. Shiu, On the time value of ruin, North American Actuarial Journal 2 (1998) 48–78. [8] H.U. Gerber, E.S.W. Shiu, On optimal dividend strategies in the compound Poisson model 2005. Preprint. [9] S. Karlin, H.M. Taylor, A First Course in Stochastic Processes, Academic Press, New York, 1975. [10] X. Lin, K.P. Pavlova, The compound Poisson risk model with a threshold dividend strategy, Insurance: Mathematics and Economics 38 (2006) 57–80. [11] X. Lin, G.E. Willmot, Analysis of a defective renewal equation arising in ruin theory, Insurance: Mathematics and Economics 25 (1999) 63–84. [12] X. Lin, G.E. Willmot, S. Drekic, The classical risk model with a constant dividend barrier: analysis of the Gerber–Shiu discounted penalty function, Insurance: Mathematics and Economics 33 (2003) 551–566. [13] K.P. Pavlova, G.E. Willmot, The discrete stationary renewal risk model and the Gerber–Shiu discounted penalty function, Insurance: Mathematics and Economics 35 (2004) 267–277. [14] E.S.W. Shiu, Probability of eventual ruin in the compound binomial model, ASTIN Bulletin 19 (1989) 179–190. [15] J.Y. Tan, X.Q. Yang, The compound binomial model with randomized decisions on paying dividends, Insurance: Mathematics and Economics 39 (2006) 1–18.