The compound binomial risk model with delayed claims and random income

Mathematical and Computer Modelling 55 (2012) 1315–1323

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The compound binomial risk model with delayed claims and random income Zhenhua Bao a , He Liu b,∗ a

School of Mathematics, Liaoning Normal University, Dalian 116029, China

b

School of Mathematics, Physics and Biology Engineering, Inner Mongolia University of Science and Technology, Baotou 014010, China

article

info

Article history: Received 8 June 2011 Received in revised form 8 October 2011 Accepted 10 October 2011 Keywords: Main claim By-claim Generating function Recursive equation Numerical results

abstract A compound binomial risk model with delayed claims and random premium income is investigated. We derive recursive equations for both the probability of ultimate ruin and the joint distribution of the surplus one period prior to ruin and the deficit at ruin. Finally, we give an alternative expression for the probability of ultimate ruin. © 2011 Elsevier Ltd. All rights reserved.

1. Introduction The classical compound binomial risk model has been studied extensively in actuarial literature. For examples [1–3], among many others. Note that the deterministic premium rate and the independence between the claims and interclaim times are assumed in this well-known risk model. Yuen and Guo [4] consider a specific dependence structure between the claim sizes and interclaim times, in which each claim causes a by-claim but the occurrence of the by-claim may be delayed. They give recursive formulas for the finite time ruin probabilities. Xiao and Guo [5] further investigate the joint distribution of the surplus immediately prior to ruin and deficit at ruin. The readers are referred to [4] for the interpretation on these kind of dependencies. In the present paper we consider the compound binomial model with time-correlated claims and stochastic income. The motivation for the stochastic income is that the insurance company may have lump sums of income. To the best of our knowledge, Boucherie et al. [6] first added a compound Poisson process with positive jumps to the classical Cramér–Lundberg risk model to describe the stochastic income. Recently, similar topics were discussed by many authors, see [7–12] for more details. It is well-known that the compound binomial risk model is the discrete analogue corresponding to the classical Cramér–Lundberg risk model. Along the similar lines as the Cramér–Lundberg risk model with random income, we extend the risk model studied in [4] by adding another binomial process to capture the uncertainty of the customer’s arrivals and payments. Throughout, denote by N the set of nature numbers and N+ = N/{0}. We assume that the premium income process is a binomial process {Mk , k ∈ N} with parameter q0 (0 < q0 ≤ 1). More precisely, in any time period, we denote by q0 the probability that a premium of size 1 is received and p0 = 1 − q0 the probability that no premium is received.

∗

Corresponding author. E-mail addresses: [email protected] (Z. Bao), [email protected] (H. Liu).

0895-7177/$ – see front matter © 2011 Elsevier Ltd. All rights reserved. doi:10.1016/j.mcm.2011.10.009

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Z. Bao, H. Liu / Mathematical and Computer Modelling 55 (2012) 1315–1323

It is assumed that there are two types of insurance claims, namely the main claims and the by-claims. In any time period, the probability of a main claim is q (0 < q < 1) and the probability of no main claim is p = 1 − q, thus the number of main claims up to time k is also a binomial process {Nk , k ∈ N} with parameter q. Independence is assumed between {Mk , k ∈ N} and {Nk , k ∈ N}. The main claims {Xj , j ∈ N+ }, independent of {Nk , k ∈ N}, are independent and identically distributed (i.i.d.) positive and integer valued random variables with the same distribution as the generic random variable X and probability mass function∑ (pmf) fX (x), x ∈ N+ . ∑ The corresponding probability generating function (pgf) and mean of X are defined as ∞ ∞ x  fX (z ) = x=1 xfX (x), respectively. Suppose that each main claim induces a by-claim, which occurs x=1 z fX (x) and µX = simultaneously with the main claim with probability θ (0 ≤ θ ≤ 1), or the by-claim may be delayed one period later with probability 1 − θ . The by-claims {Yj , j ∈ N+ } are also i.i.d. random variables distributed as a positive and integer valued random variable Y with pmf, pgf and mean denoted by fY (y), fY (z ) and µY , respectively. Then the total claim amount process is given by Sk = SkX + SkY ,

k ∈ N+ ,

where SkX and SkY are the total main claims and by-claims, respectively, in the first k time periods. And the corresponding surplus of the insurer at time k is Uk = u + Mk − Sk ,

k ∈ N,

(1)

where the initial surplus U0 = u is a non-negative integer. It is easy to see that EMk = kq0

and

ESk+1 = kq(µX + µY ) + qµX + qθ µY .

Therefore, we further assume q0 > q (µX + µY ) . This assumption ensures that E(Mk − Sk ) > 0 for k ∈ N+ and hence the safety loading is positive. Define the time of ruin T = inf{k > 0 : Uk < 0} with T = ∞ if ruin does not occur, then

ψ(u) = Pr {T < ∞ | U0 = u} is the probability of ultimate ruin from initial surplus u. Note that if ruin occurs, UT −1 is the surplus one period prior to ruin and |UT | is the deficit at ruin, then the joint distributions of the surplus one period prior to ruin and the deficit at ruin is defined as

ϕ(u, x, y) = Pr {T < ∞, UT −1 = x, UT = −y | U0 = u} . The rest of the paper is structured as follows. In Section 2, a recursive equation for ϕ(u, x, y) is derived. Then we compute a numerical example to illustrate the application of the recursive equation. In Section 3, both a recursive equation and an alternative expression for ψ(u) are obtained. Numerical examples are also provided. 2. The joint distribution of the surplus one period prior to ruin and the deficit at ruin To handle the surplus process (1), we consider the scenario that a main claim occurs but its associated by-claim is delayed to the next period, then a complementary surplus process is defined as follows: U1,k = u + Mk − Sk −  Y I{k≥1} ,

k ∈ N,

(2)

where  Y has the same distribution as Y and is independent of all other variables. Denote by ϕ1 (u, x, y) the joint distribution of the surplus one period to ruin and the deficit at ruin when the surplus process in (2) is postulated. It is assumed that ∑∞ ∑prior ∞ u=0 ϕ(u, x, y) < ∞, u=0 ϕ1 (u, x, y) < ∞ throughout the rest of the paper. For notational convenience, define hX (m) = p0 fX (m) + q0 fX (m + 1), m ∈ N and hY (n) = p0 fY (n) + q0 fY (n + 1), n ∈ N, then we have the following result for ϕ(0, x, y). Theorem 1. When u = 0, it holds that

ϕ(0, x, y) =

(p + qθ )D(x, y) + q(1 − θ )D1 (x, y) , q0 p(p + qθ )

(3)

with D(x, y) = q(1 − θ )hX (x + y) + qθ

−

hX (m)fY (n),

(4)

m+n=x+y m≥0,n≥1

and D1 (x, y) = phY (x + y) + q(1 − θ )

− m+n=x+y m≥0,n≥1

hX (m)fY (n) + qθ

− m+n+l=x+y m≥0,n≥1,l≥1

hX (m)fY (n)fY (l).

(5)

Z. Bao, H. Liu / Mathematical and Computer Modelling 55 (2012) 1315–1323

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Proof. To derive (3), by considering the occurrence (or not) of the premium income and claims in the next period, we separate the six possible cases as follows: 1. 2. 3. 4. 5. 6.

One premium arrival and no claim occurs; No premium arrival and no claim occurs; One premium arrival, a main claim occurs and its associated by-claim occurs in the same period; No premium arrival, a main claim occurs and its associated by-claim occurs in the same period; One premium arrival, a main claim occurs but its associated by-claim is delayed to the next period; No premium arrival, a main claim occurs but its associated by-claim is delayed to the next period. Using the total probability formula, we get

ϕ(u, x, y) = q0 pϕ(u + 1, x, y) + p0 pϕ(u, x, y)      −  − + q0 qθ ϕ(u + 1 − m − n, x, y)fX (m)fY (n) + I{u=x} fX (m)fY (n)    m+n≤u+1  m+n=x+y+1 m≥1,n≥1 m≥1,n≥1       − − fX (m)fY (n) ϕ(u − m − n, x, y)fX (m)fY (n) + I{u=x} + p0 qθ     m+n≤u m+n=x+y m≥1,n≥1 m≥1,n≥1   − + q0 q(1 − θ) ϕ1 (u + 1 − m, x, y)fX (m) + I{u=x} fX (x + y + 1) 1≤m≤u+1



 −

+ p0 q(1 − θ )

ϕ1 (u − m, x, y)fX (m) + I{u=x} fX (x + y) .

1≤m≤u

Rewrite the above equation as

(1 − p0 p)ϕ(u, x, y) = q0 pϕ(u + 1, x, y) + q(1 − θ)

−

ϕ1 (u − m, x, y)hX (m)

0≤m≤u

+ qθ

−

ϕ(u − m − n, x, y)hX (m)fY (n) + I{u=x} D(x, y).

(6)

m+n≤u m≥0,n≥1

By (6), summation of u from 0 to ∞ yields

(1 − p0 p)

∞ −

ϕ(u, x, y) = q0 p

u=0

∞ −

ϕ(u, x, y) + q(1 − θ )

u=1

+ qθ ∑

(1 − p0 p)

m≥0

∞ −

hX (m) = 1 and

∞ − −

ϕ(u, x, y) = q0 p

u=0

∑

u =1

ϕ(u − m − n, x, y)hX (m)fY (n) + D(x, y).

m+n≤u m≥0,n≥1

m+n≥1 m≥0,n≥1

∞ −

ϕ1 (u − m, x, y)hX (m)

u=0 0≤m≤u

u =0

By noting that

∞ − −

hX (m)fY (n) = 1, we obtain

ϕ(u, x, y) + q(1 − θ )

∞ −

ϕ1 (u, x, y) + qθ

u =0

∞ −

ϕ(u, x, y) + D(x, y).

(7)

u=0

A similar method can be used to deal with the complementary surplus process (2). For u ∈ N, we have



 −

ϕ1 (u, x, y) = q0 p

ϕ(u + 1 − n, x, y)fY (n) + I{u=x} fY (x + y + 1)

1≤n≤u+1

 + p0 p

 −

ϕ(u − n, x, y)fY (n) + I{u=x} fY (x + y)

1≤n≤u

+ q0 qθ

  

−

  m+n+l≤u+1

m≥1,n≥1,l≥1

ϕ(u + 1 − m − n − l, x, y)fX (m)fY (n)fY (l) + I{u=x}

− m+n+l=x+y+1 m≥1,n≥1,l≥1

fX (m)fY (n)fY (l)

    

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Z. Bao, H. Liu / Mathematical and Computer Modelling 55 (2012) 1315–1323

  

   + p0 qθ ϕ(u − m − n − l, x, y)fX (m)fY (n)fY (l) + I{u=x} fX (m)fY (n)fY (l)    m+n+l≤u  m+n+l=x+y m≥1,n≥1,l≥1 m≥1,n≥1,l≥1      −  − + q0 q(1 − θ ) ϕ1 (u + 1 − m − n, x, y)fX (m)fY (n) + I{u=x} fX (m)fY (n)    m+n≤u+1  m+n=x+y+1 m≥1,n≥1 m≥1,n≥1      −  − ϕ1 (u − m − n, x, y)fX (m)fY (n) + I{u=x} fX (m)fY (n) . + p0 q(1 − θ )     m+n≤u m+n=x+y −

−

m≥1,n≥1

m≥1,n≥1

Rewrite the above equation as

ϕ1 (u, x, y) = p

−

ϕ(u − n, x, y)hY (n) + q(1 − θ )

ϕ1 (u − m − n, x, y)hX (m)fY (n)

m+n≤u m≥0,n≥1

0≤n≤u

+ qθ

−

−

ϕ(u − m − n − l, x, y)hX (m)fY (n)fY (l) + I{u=x} D1 (x, y).

(8)

l+m+n≤u m≥0,n≥1,l≥1

By (8), summing over u from 0 to ∞ yields ∞ −

ϕ1 (u, x, y) = p

∞ −

ϕ(u, x, y) + q(1 − θ )

ϕ1 (u, x, y) + qθ

Finally, combining (7) and (9) yields (3).

∞ −

ϕ(u, x, y) + D1 (x, y).

(9)

u=0

u =0

u =0

u =0

∞ −



Because ϕ(0, x, y) can be calculated by (3), we are ready to give the recursive formula for ϕ(u, x, y). Theorem 2. For u ∈ N+ , the following recursive equation holds for ϕ(u, x, y), q0 pϕ(u, x, y) = (1 − p0 p)ϕ(u − 1, x, y) − q

u − l =2

− q0 qp(1 − θ )ϕ(0, x, y)

−

−

ϕ(u − l, x, y)

hX (m)fY (n)

m+n=l−1 m≥0,n≥1

hX (m)fY (n) + ω(u, x, y),

(10)

m+n=u m≥0,n≥1

with

ω(u, x, y) =

  

−

q(1 − θ )

 

hX (m)fY (n)I{u≥x+2} − I{u=x+1}

  

D(x, y) − q(1 − θ )hX (u − x − 1)I{u≥x+1} D1 (x, y),

 

m+n=u−x−1 m≥0,n≥1

where ϕ(0, x, y), D(x, y) and D1 (x, y) are given by (3)–(5), respectively. Proof. To derive (10), the technique of generating functions will be applied. When 0 ≤ z < 1, denote the generating functions of ϕ and ϕ1 by

 ϕ (z ) =

∞ −

z u ϕ(u, x, y) and  ϕ1 (z ) =

u =0

∞ −

z u ϕ1 (u, x, y).

u=0 u +1

and summing over u from 0 to ∞ yields   (1 − p0 p)z ϕ (z ) = q0 p  ϕ (z ) − ϕ(0, x, y) + qθ z hX (z ) fY (z ) ϕ (z ) + q(1 − θ )z hX (z ) ϕ1 (z ) + z x+1 D(x, y).

Multiplying both sides of (6) by z

(11)

Similarly, from (8) we get

 2  ϕ1 (z ) = p hY (z ) ϕ (z ) + qθ hX (z )  fY (z )  ϕ (z ) + q(1 − θ ) hX (z ) fY (z ) ϕ1 (z ) + z x D1 (x, y).

(12)

Z. Bao, H. Liu / Mathematical and Computer Modelling 55 (2012) 1315–1323

1319

Combining (11) and (12) yields







q0 p + qz hX (z ) fY (z ) − (1 − p0 p)z  ϕ (z ) = q(1 − θ ) z x+1 D(x, y) − q0 pϕ(0, x, y)  hX (z ) fY (z ) − z x+1 D(x, y)



hX (z )D1 (x, y) + q0 pϕ(0, x, y). − q(1 − θ )z x+1

(13)

After some modifications, one could see (13) is equivalent to ∞ −

zu

  

q0 pϕ(u, x, y) − (1 − p0 p)ϕ(u − 1, x, y) + q

 

u =1

u−1 −

ϕ(l, x, y)

l =0

  

−

hX (m)fY (n)

m+n=u−l−1 m≥0,n≥1

    

   u = q(1 − θ ) z D(x, y) hX (m)fY (n) − D1 (x, y)hX (u − x − 1)     m+n=u−x−1 u =x +1 ∞ −

−

m≥0,n≥1

−z x+1 D(x, y) − q0 qp(1 − θ )ϕ(0, x, y)

∞ −

zu

u =0

which yields (10) by comparing the coefficients of z u .

−

hX (m)fY (n),

m+n=u m≥0,n≥1



We remark that Eqs. (14) and (21) in [5] are recovered by (3) and (10) in the present paper with q0 = 1, respectively. The recursive formula (10) is powerful in not only theoretics but also computer programming. The following example is borrowed from [5] to give a comparison with deterministic premium and stochastic premium. Example 1. Suppose X and Y are geometrically distributed. More specifically, fX (m) = (1 − a)am−1 , a ∈ (0, 1) and fY (m) = (1 − b)bm−1 , b ∈ (0, 1) for a ̸= b and m ∈ N+ . By noting that

−

am−1 bn−1 =

ak−1 − bk−1 a−b

m+n=k m≥1,n≥1

−

,

k ∈ N+ ,

ak−1 − (k − 1)abk−2 + (k − 2)bk−1

am−1 bn−1 bl−1 =

(a − b)2

m+n+l=k m≥1,n≥1,l≥1

,

k ∈ N+ ,

which can be easily shown by induction, we can get from (4) and (5) that



D(x, y) = q(1 − θ )(1 − a) q0 +

p 0  x +y (1 − a)(1 − b)  p0  x+y  p 0  x +y  a + qθ q0 + a − q0 + b , a a−b a b

and



D1 (x, y) = p(1 − b) q0 +

p 0  x +y (1 − a)(1 − b)  p0  x+y  p 0  x +y  b + q(1 − θ ) q0 + a − q0 + b b a−b a b

  (1 − a)(1 − b)2  p 0  x +y  p0 q0 + a − q0 (x + y) + (x + y − 1) abx+y−1 2 (a − b) a b    p0 + q0 (x + y − 1) + (x + y − 2) bx+y . + qθ

b

Moreover, we have

ϕ(0, x, y) =

1 q0 p

D(x, y) +

q(1 − θ ) q0 p(p + qθ )

D1 (x, y),

and u  p 0  l −1  (1 − a)(1 − b) − p0  l−1  q0 + a − q0 + b ϕ(u − l, x, y) a−b a b l =2 (1 − a)(1 − b)  p0  u  p0  u  − q0 qp(1 − θ ) q0 + a − q0 + b ϕ(0, x, y) + ω(u, x, y), a−b a b

q0 pϕ(u, x, y) = (1 − p0 p)ϕ(u − 1, x, y) − q

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Z. Bao, H. Liu / Mathematical and Computer Modelling 55 (2012) 1315–1323

Table 1 q0 = 0.9, q = 0.2, a = 1/2, b = 1/3.

(x, y) (0, 1) (0, 5) (3, 5) (5, 3) (5, 5)

u=0

u=1

u=2

u=3

u=4

u=5

θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ

=0 = 0.5 =1 =0 = 0.5 =1 =0 = 0.5 =1 =0 = 0.5 =1 =0 = 0.5 =1

0.1712963 0.1174126 0.0833333

0.0980453 0.0514353 0.0231481

0.0850023 0.0457137 0.0226337

0.0744748 0.0403154 0.0204332

0.0654345 0.0354829 0.0180930

0.0575342 0.0312130 0.0159410

0.0097772 0.0140652 0.0163537

0.0053311 0.0055683 0.0045427

0.0046290 0.0049948 0.0044418

0.0040574 0.0044155 0.0040099

0.0035652 0.0038886 0.0035507

0.0031349 0.0034212 0.0031283

0.0012020 0.0020036 0.0022855

0.0014638 0.0025000 0.0029204

0.0017022 0.0029708 0.0035412

0.0019136 0.0033927 0.0041016

0.0012854 0.0020966 0.0023123

0.0011258 0.0018654 0.0021146

0.0012020 0.0020036 0.0022855

0.0014638 0.0025000 0.0029204

0.0017022 0.0029708 0.0035412

0.0019136 0.0033927 0.0041016

0.0021001 0.0037655 0.0045978

0.0022641 0.0040939 0.0050350

0.0002993 0.0005194 0.0005855

0.0003645 0.0006481 0.0007481

0.0004239 0.0007702 0.0009072

0.0004765 0.0008795 0.0010507

0.0005230 0.0009762 0.0011778

0.0005638 0.0010613 0.0012898

u=6

u=7

u=8

u=9

u = 10

u = 11

θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ

=0 = 0.5 =1 =0 = 0.5 =1 =0 = 0.5 =1 =0 = 0.5 =1 =0 = 0.5 =1

0.0505976 0.0274531 0.0140266

0.0444997 0.0241453 0.0123379

0.0391372 0.0212358 0.0108515

0.0344210 0.0186768 0.0095440

0.0302731 0.0164262 0.0083939

0.0266252 0.0144468 0.0073824

0.0027569 0.0030092 0.0027527

0.0024247 0.0026467 0.0024213

0.0021325 0.0023278 0.0021296

0.0018755 0.0020473 0.0018730

0.0016495 0.0018006 0.0016473

0.0014507 0.0015836 0.0014488

0.0009891 0.0016456 0.0018785

0.0008696 0.0014484 0.0016565

0.0007648 0.0012742 0.0014579

0.0006726 0.0011207 0.0012825

0.0005916 0.0009857 0.0011280

0.0005203 0.0008669 0.0009921

0.0015938 0.0027139 0.0031342

0.0013971 0.0024084 0.0028377

0.0012277 0.0021232 0.0025146

0.0010795 0.0018685 0.0022159

0.0009493 0.0016436 0.0019499

0.0008349 0.0014456 0.0017152

0.0003965 0.0007073 0.0008029

0.0003476 0.0006274 0.0007269

0.0003054 0.0005530 0.0006442

0.0002686 0.0004867 0.0005677

0.0002362 0.0004281 0.0004995

0.0002077 0.0003765 0.0004394

(x, y) (0, 1) (0, 5) (3, 5) (5, 3) (5, 5)

with

(1 − a)(1 − b)  p 0  u −x −1  p 0  u −x −1  q0 + a − q0 + b I{u≥x+2} D(x, y) a−b a b   p0 − I{u=x+1} D(x, y) − q(1 − θ )(1 − a)au−x−1 q0 I{u≥x+1} + I{u≥x+2} D1 (x, y).

ω(u, x, y) = q(1 − θ )

a

Let q0 = 0.9, q = 0.2, a = 1/2, b = 1/3, then Table 1 shows some numerical results of ϕ(u, x, y) for θ = 0, θ = 0.5 and θ = 1. Denote by A(u, x, y) the joint distribution of the surplus one period prior to ruin and the deficit at ruin of the risk model described in [5]. By comparing the numerical results both in [5] and the current manuscript, one can observe that ϕ(u, x, y) ≥ A(u, x, y). It accords with the fact that at each period the stochastic income is equal or less than the deterministic income. 3. Ultimate probability of ruin In the following theorem, we show that ψ(u) satisfies a recursive equation for which the starting point corresponds to an explicit expression of ψ(0). Theorem 3. For u = 0, we have

ψ(0) =

q q0 p(p + qθ )

(µX + µY − q0 − q0 p(1 − θ )) ,

(14)

and for u ≥ 1, ψ(u) can be calculated by q0 pψ(u) = (1 − p0 p)ψ(u − 1) − q

u − l =2

− q0 qp(1 − θ )ψ(0)

− m+n=u m≥0,n≥1

ψ(u − l)

−

hX (m)fY (n)

m+n=l−1 m≥0,n≥1

hX (m)fY (n) + qζ (u),

(15)

Z. Bao, H. Liu / Mathematical and Computer Modelling 55 (2012) 1315–1323

1321

with

ζ (u) = (q + pθ )

−

−

hX (m)fY (n) + p(1 − θ )

m+n≤u−1 m≥0,n≥1

hX (m)hY (n) − 1.

m+n≤u−1 m≥0,n≥0

Proof. From (4) and (5), we have ∞ − ∞ −

D(x, y) = qθ (µX + µY − q0 ) + q(1 − θ )(µX − q0 ) = q(µX − q0 ) + qθ µY ,

x=0 y=1

and ∞ − ∞ −

D1 (x, y) = p(µY − q0 ) + qθ (µX + 2µY − q0 ) + q(1 − θ )(µX + µY − q0 )

x=0 y=1

= qµX + (1 + qθ )µY − q0 . Thus by (3) we get

ψ(0) =

∞ ∞ − −

ϕ(0, x, y) =

x=0 y=1

=

q

∞ ∞ 1 −−

q0 p x=0 y=1

D(x, y) +

q(1 − θ )

∞ − ∞ −

q0 p(p + qθ ) x=0 y=1

D1 (x, y)

(µX + µY − q0 − q0 p(1 − θ )) .

q0 p(p + qθ )

Furthermore, summing (10) over (x, y) ∈ N × N+ yields q0 pψ(u) = (1 − p0 p)ψ(u − 1) − q

u −

ψ(u − l)

l =2

+ q(1 − θ )

∞ u −2 − −

− q(1 − θ )

u −1 − ∞ −

hX (m)fY (n)

m+n=l−1 m≥0,n≥1

−

D(x, y)

x =0 y =1

−

hX (m)fY (n) −

∞ −

D(u − 1, y)

y =1

m+n=u−1−x m≥0,n≥1

D1 (x, y)hX (u − x − 1) − q0 qp(1 − θ )ψ(0)

−

hX (m)fY (n).

(16)

m+n=u m≥0,n≥1

x =0 y =1

Moreover, it is not difficult to verify that u −2 − ∞ −

D(x, y)

x=0 y=1

−

hX (m)fY (n) = q

m+n=u−1−x m≥0,n≥1

− m+n≤u−1 m≥0,n≥1

hX (m)hX (n)fY (l)

m+n+l≤u−1 m≥0,n≥0,l≥1

−

− qθ

−

hX (m)fY (n) − q(1 − θ )

hX (m)hX (n)fY (l)fY (k),

m+n+l+k≤u−1 m≥0,n≥0,l≥1,k≥1

∞ −

−

D(u − 1, y) = q − q(1 − θ )

y=1 u −1 − ∞ −

0≤m≤u−1

−

D1 (x, y)hX (u − x − 1) =

x=0 y=1

hX (m)fY (n),

m+n≤u−1 m≥0,n≥1

hX (m) − p

0≤m≤u−1

− qθ

−

hX (m) − qθ

−

hX (m)hY (n) − q(1 − θ )

m+n≤u−1 m≥0,n≥0

−

m+n+l≤u−1 m≥0,n≥0,l≥1

hX (m)hX (n)fY (l)fY (k).

m+n+l+k≤u−1 m≥0,n≥0,l≥1,k≥1

Finally, substituting the above equations into (16), we get (15) immediately.



Example 2. Suppose X and Y are geometric distributed as in Example 1. Then we have

ψ(0) =

q q0 p(p + qθ )



1 1−a

+

1 1−b

 − q0 − q0 p(1 − θ ) ,

−

hX (m)hX (n)fY (l)

1322

Z. Bao, H. Liu / Mathematical and Computer Modelling 55 (2012) 1315–1323

Table 2 q0 = 0.8, q = 0.2, a = 1/2, b = 1/3.

θ θ θ θ

=0 = 0.5 = 0.7 =1

θ θ θ θ

=0 = 0.5 = 0.7 =1

u=0

u=1

u=2

u=4

u=7

u = 11

u = 16

0.8046875 0.8263889 0.8337766 0.8437500

0.7540690 0.7767650 0.7844914 0.7949219

0.7047763 0.7264871 0.7338781 0.7438558

0.6148305 0.6338914 0.6403802 0.6491401

0.5008143 0.5163458 0.5216331 0.5287710

0.3809768 0.3927919 0.3968141 0.4022440

0.2706611 0.2790550 0.2819125 0.2857701

u = 22

u = 29

u = 37

u = 46

u = 56

u = 67

u = 80

0.1795802 0.1851495 0.1870454 0.1896049

0.1112747 0.1147257 0.1159004 0.1174864

0.0643932 0.0663902 0.0670701 0.0679878

0.0348008 0.0358801 0.0362475 0.0367435

0.0175648 0.0181095 0.0182950 0.0185453

0.0082795 0.0085363 0.0086237 0.0087417

0.0034039 0.0035094 0.0035454 0.0035939

Table 3 q0 = 0.9, q = 0.2, a = 1/2, b = 1/3.

θ θ θ θ

=0 = 0.5 = 0.7 =1

θ θ θ θ

=0 = 0.5 = 0.7 =1

u=0

u=1

u=2

u=3

u=5

u=8

u = 11

0.6527778 0.6913580 0.7044917 0.7222222

0.5771605 0.6148834 0.6277252 0.6450617

0.5083162 0.5423716 0.5539649 0.5696159

0.4472260 0.4773811 0.4876466 0.5015051

0.3459783 0.3693561 0.3773144 0.3880582

0.2353728 0.2512794 0.2566944 0.2640047

0.1601253 0.1709467 0.1746305 0.1796038

u = 15

u = 19

u = 24

u = 28

u = 33

u = 38

u = 45

0.0958072 0.1022819 0.1044860 0.1074617

0.0573240 0.0611979 0.0625167 0.0642971

0.0301654 0.0322040 0.0328980 0.0338348

0.0180487 0.0192685 0.0196837 0.0202443

0.0094977 0.0101396 0.0103581 0.0106531

0.0049980 0.0053357 0.0054507 0.0056059

0.0020344 0.0021719 0.0022187 0.0022819

and for u ≥ 1, u  (1 − a)(1 − b) − p0  l−1  p 0  l −1  q0 + a − q0 + b ψ(u − l) a−b a b l=2 (1 − a)(1 − b)  p0  u  p0  u  − q0 qp(1 − θ ) q0 + a − q0 + b ψ(0) + qζ (u), a−b a b

q0 pψ(u) = (1 − p0 p)ψ(u − 1) − q

with

   (1 − a)(1 − b) 1 − bu−1 1 − au − 1 ζ (u) = − (q0 b + p0 ) (q + pθ ) (q0 a + p0 ) a−b 1−a 1−b   1 − p0 bu−1 − q0 bu 1 − p0 au−1 − q0 au − (q0 b + p0 ) − 1. + p(1 − θ ) (q0 a + p0 ) 1−a 1−b The corresponding numerical results for ψ(u) are shown in Tables 2 and 3 with q0 = 0.8 and q0 = 0.9, respectively. Other parameters are q = 0.2, a = 1/2, b = 1/3. The numerical results show that the higher the initial reserve of the insurer, the lower the probability of ultimate ruin. Also, both tables show that ψ(u) is nondecreasing as θ is increasing. Moreover, by comparing Table 2 with Table 3 we find that the lower q0 is, the larger the initial reserve should be set to prevent the insurer from ruin. The remainder of the paper is devoted to derive another expression for ψ(u). To distinguish between cases θ < 1 and θ = 1, denote the surplus processes by Ukθ<1 and Ukθ=1 with corresponding time of ruin T = inf{k > 0 : Ukθ <1 < 0} and W = inf{k > 0 : Ukθ=1 < 0}, respectively. To emphasize the dependence of ruin on the initial reserve u, we will use symbols like Pr(T < ∞ | u), then Pr(W < ∞ | u) = Pr(W < ∞, T < ∞ | u) + Pr(W < ∞, T = ∞ | u)

= Pr(T < ∞ | u) + Pr(W < ∞, T = ∞ | u). That is

ψ(u) = Pr(T < ∞ | u) = Pr(W < ∞ | u) − Pr(W < ∞, T = ∞ | u),

(17)

from which we could see that Pr(W < ∞, T = ∞ | u) is the main concern in order to get ψ(u). Because each of the two events W < ∞, T = ∞





and



θ=1 W < ∞, UW < 0, Ukθ <1 ≥ 0, k ∈ N

implies the other, we have the following conclusions.



Z. Bao, H. Liu / Mathematical and Computer Modelling 55 (2012) 1315–1323

1323

(i) Denote by XW the main claim at time W for process Ukθ <1 . If YW is the by-claim associated with XW , then YW must be positive and delayed. Otherwise, we could get θ<1 θ=1 UW = UW < 0, θ<1 which contradicts UW ≥ 0. (ii) The premium income at W + 1 is 1. Otherwise, if we denote all the possible claims at time W + 1 for process Ukθ=1 by Z , then it follows that θ<1 θ=1 UW +1 = UW − Z < 0, θ<1 θ=1 which contradicts UW < 0 and +1 ≥ 0. Moreover, by noticing that UW θ<1 θ=1 0 ≤ UW +1 = U W + 1 − Z ,

we conclude that Z = 0. (iii) Because there is no main claim at W + 1, we get θ<1 θ=1 θ =1 0 ≤ UW +1 = UW +1 = UW + 1,

which yields θ=1 UW = −1,

θ<1 UW +1 = 0.

So that Pr W < ∞, T = ∞ | u = Pr







θ=1 W < ∞, UW = −1; YW is delayed; the premium income at time θ <1 W + 1 is 1; no main claim occurs at W + 1; UW +1 = 0, T = ∞



    θ =1 = Pr W < ∞, UW = −1 | u (1 − θ )q0 p 1 − Pr(T < ∞ | u = 0) . Thus, from (17) we have

      θ =1 ψ(u) = Pr W < ∞ | u − q0 p(1 − θ ) 1 − ψ(0) Pr W < ∞, UW = −1 | u . We now derive the following result for ψ(u) by substituting (14) into the above equation. Theorem 4. The probability of ruin for model (1) has the following expression,

    1−θ θ=1 ψ(u) = Pr W < ∞ | u − = −1 | u . (q0 − qµX − qµY ) Pr W < ∞, UW p + qθ

(18)

We remark that the classical compound binomial riskmodel with random income has been discussedin [13]. By using   θ=1 = −1 | u recursively and the recursive formulas given in [13], we can compute Pr W < ∞ | u and Pr W < ∞, UW hence obtain the expression of ψ(u) by (18). Acknowledgments The authors are grateful to an anonymous referee for his/her thoughtful and detailed comments on this paper. This research was supported by National Natural Science Foundation of China (11001114) and Science Research Foundation of Educational Department of Liaoning Province of China (2009A423). References [1] H.U. Gerber, Mathematical fun with the compound binomial process, ASTIN Bulletin 18 (1988) 161–168. [2] G.E. Willmot, Ruin probabilities in the compound binomial model, Insurance: Mathematics and Economics 12 (1993) 133–142. [3] 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. [4] K.C. Yuen, J. Guo, Ruin probabilities for time-correlated claims in the compound binomial model, Insurance: Mathematics and Economics 29 (2001) 47–57. [5] Y. Xiao, J. Guo, The compound binomial risk model with time-correlated claims, Insurance: Mathematics and Economics 41 (2007) 124–133. [6] R.J. Boucherie, O.J. Boxma, K. Sigman, A note on negative customers, GI/G/I workload, and risk processes, Probability in Engineering and Informational System 11 (1997) 305–311. [7] A.V. Boikov, The Cramér–Lundberg model with stochastic premium process, Theory of Probability and Its Applications 47 (2003) 489–493. [8] G. Temnov, Risk process with random income, Journal of Mathematical Sciences 123 (2004) 3780–3794. [9] Z. Bao, The expected discounted penalty at ruin in the risk process with random income, Applied Mathematics and Computation 179 (2006) 559–566. [10] H. Yang, Z. Zhang, On a class of renewal risk model with random income, Applied Stochastic Models in Business and Industry 25 (2009) 678–695. [11] C. Labbé, K.P. Sendova, The expected discounted penalty function under a risk model with stochastic income, Applied Mathematics and Computation 215 (2009) 1852–1867. [12] I. Karnaukh, Risk process with stochastic income and two-step premium rate, Applied Mathematics and Computation 217 (2010) 778–781. [13] S. Fang, P. Zhao, C. Zhang, The expected discounted penalty function at ruin of the discrete risk model with random income, Mathematica Applicata 21 (2008) 771–777.