Dividend payments with a threshold strategy in the compound Poisson risk model perturbed by diffusion

Insurance: Mathematics and Economics 40 (2007) 509–523 www.elsevier.com/locate/ime

Dividend payments with a threshold strategy in the compound Poisson risk model perturbed by diffusionI Ning Wan Department of Applied Mathematics, Tongji University, Shanghai 200092, China Received December 2005; received in revised form August 2006; accepted 11 August 2006

Abstract In the absence of dividends, the surplus of an insurance company is modelled by a compound Poisson process perturbed by diffusion. Dividends are paid at a constant rate whenever the modified surplus is above the threshold, otherwise no dividends are paid. Two integro-differential equations for the expected discounted dividend payments prior to ruin are derived and closed-form solutions are given. Accordingly, the Gerber–Shiu expected discounted penalty function and some ruin related functionals, the probability of ultimate ruin, the time of ruin and the surplus before ruin and the deficit at ruin, are considered and their analytic expressions are given by general solution formulas. Finally the moment-generating function of the total discounted dividends until ruin is discussed. c 2006 Elsevier B.V. All rights reserved.

Keywords: Compound Poisson model; Diffusion; Threshold strategy; Dividend payments; Gerber–Shiu discounted penalty function; Ruin related functionals

1. Introduction Dividend strategy for insurance risk models were first proposed by De Finetti (1957) to reflect more realistically the surplus cash flows in an insurance portfolio, and he found that the optimal strategy must be a barrier strategy. From then on, barrier strategies have been studied in a number of papers and books, including Lin et al. (2003), Gerber and Shiu (2004), Li (2005). But if such a barrier strategy is applied, ultimate ruin of the company is certain. These considerations lead to the idea of imposing restrictions on the nature of the dividend stream, resulting in optimization problems with additional constraints. Jeanblanc-Picqu´e and Shiryaev (1995) and Asmussen and Taksar (1997) postulated a bounded dividend rate, that is, that the dividends paid per unit time should not exceed an upper bound. They show that the optimal dividend strategy is now a generalized barrier strategy, which is called a threshold strategy. According to such a strategy, dividends are paid at a constant rate whenever the modified surplus is above the threshold, and no dividends are paid whenever the modified surplus is below the threshold. Gerber and Shiu (2006) present some down-to-earth calculations about this

I Project supported by CNCF No: 10471106.

E-mail address: [email protected]. c 2006 Elsevier B.V. All rights reserved. 0167-6687/$ - see front matter doi:10.1016/j.insmatheco.2006.08.002

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by modelling the surplus of the company by a Wiener process. Gerber and Shiu (2005), Lin and Pavlova (2006) study the problem in the classical setting—that the aggregate claims are modelled as a compound Poisson process. In this paper we consider the following classical surplus process perturbed by a diffusion X (t) = x + ct + σ W (t) −

N (t) X

Zi ,

t ≥ 0,

(1.1)

i=1

where {N (t); t ≥ 0} is a Poisson process with parameter λ, denoting the total number of claims from an insurance portfolio. Z i , i = 1, . . ., independent of {N (t); t ≥ 0}, are positive i.i.d. random variables with distribution function P(z) = P(Z ≤ z) and density function p(z). {W (t); t ≥ 0} is a standard Wiener process that is independent of the P N (t) aggregate claims process S(t) := i=1 Z i and σ > 0 is the dispersion parameter. In the above model, x = X (0) ≥ 0 is the initial surplus, c = λµ1 (1 + θ ) is the premium rate per unit time, and θ > 0 is the relative security loading factor. The insurance company will pay dividends to its shareholders. For t ≥ 0, let D(t) denote the aggregate dividends paid by time t. Thus, X˜ (t) = X (t) − D(t)

(1.2)

is the company’s surplus at time t. We call X˜ (t) the modified surplus. Let δ > 0 be the force of interest for valuation, and let Dx,b denote the present value of all dividends until time of ruin Tb , Tb

Z Dx,b =

e−δt dD(t).

(1.3)

0

We shall assume that the company pays dividends according to the following strategy governed by parameters b > 0 and α > 0. Whenever the modified surplus is below the level b, no dividends are paid. However, when the modified surplus is above b, dividends are paid continuously at a constant rate α. Thus, the threshold b plays the role of a breakpoint or a regime-switching boundary. With I (·) denoting the indicator function, an alternative expression for Dx,b is Dx,b = α

Tb

Z

e−δt I ( X˜ (t) > b) dt.

(1.4)

0

For x ≥ 0, we use the symbol V (x; b) to denote the expectation of Dx,b , V (x; b) = E[Dx,b |X (0) = x].

(1.5)

The classical risk model perturbed by a diffusion was first introduced by Gerber (1970) and has been further studied by many authors during the last few years; e.g., Dufresne and Gerber (1991), Gerber and Landry (1998), Wang and Wu (2000), Chiu and Yin (2003), Li (2005) and so on. This paper is to study the threshold strategy in the case of the compound Poisson risk model perturbed by diffusion, presenting some analytic results by a Green function method. In Section 2, two integro-differential equations and their closed-form expressions for the value of a threshold strategy with an arbitrary parameter b are obtained. In Section 3, another interesting function—the Gerber–Shiu expected discounted penalty function is discussed. In Sections 4 and 5 the analytical results of Section 3 are applied to the probability of ultimate ruin, the Laplace transform of the time of ruin. The moment-generating function of the total discounted dividends until ruin is considered in Section 6. Finally in the Appendix, a general method to solve the above kind of integro-differential equations is presented and solution formulas are obtained. 2. The expected discounted dividend payments In this section, we will give the integro-differential equations and the renewal equations satisfied by the expected discounted dividend payments V (x; b), then its closed-form solutions. Firstly, we derive the two integro-differential equations according to (1.1)–(1.5).

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It is easy to see that V (x; b) behaves differently with different initial surplus. Hence, for notational convenience, we set  V1 (x; b), 0 ≤ x ≤ b V (x; b) = V2 (x; b), b ≤ x < ∞. Then we have Theorem 2.1. For 0 < x < b, V (x; b) satisfies the following homogeneous integro-differential equation: Z x σ 2 00 V1 (x − z; b) p(z) dz = 0, V1 (x; b) + cV10 (x; b) − (λ + δ)V1 (x; b) + λ 2 0 and for x > b, V (x; b) satisfies the following nonhomogeneous integro-differential equation: Z x−b σ 2 00 0 V (x; b) + (c − α)V2 (x; b) − (λ + δ)V2 (x; b) + λ V2 (x − z; b) p(z) dz 2 2 0 Z x +λ V1 (x − z; b) p(z) dz + α = 0.

(2.1)

(2.2)

x−b

Proof. We adopt a similar approach to those used in Dufresne and Gerber (1991) and Gerber and Landry (1998). Consider the infinitesimal interval from 0 to dt. Conditioning, one obtains that when 0 ≤ x ≤ b, V1 (x; b) = e−δdt {P(T1 > dt)E[V1 (x + cdt + σ W (dt); b)] + P(T1 ≤ dt)E[V1 (x + cdt + σ W (dt) − Z 1 ; b)]}, when b < x < ∞ V2 (x; b) = e−δdt {αdt + P(T1 > dt)E[V2 (x + (c − α) dt + σ W (dt); b)] + P(T1 ≤ dt) E[E[V2 (x + (c − α) dt + σ W (dt) − Z 1 ; b)|Z 1 ∈ (0, x + (c − α) dt + σ W (dt))] + E[V1 (x + (c − α) dt + σ W (dt) − Z 1 ; b)|Z 1 ∈ (x + (c − α) dt + σ W (dt), ∞)]]}, where T1 is the first time that the claim happens. Since e−δdt = 1 − δdt + o(dt), P(T1 > dt) = 1 − λdt + o(dt),

P(T1 ≤ dt) = λdt + o(dt).

By Taylor’s expansion and some careful calculations, letting dt → 0, we get that the integro-differential equations (2.1) and (2.2) hold.  Furthermore, if X (0) = 0, ruin is immediate and no dividend is paid; if X (0) → ∞, ruin does not happen all the time and dividends are always paid at a constant rate α. So we have the boundary conditions V1 (0; b) = 0,

(2.3)

α lim V2 (x; b) = , x→∞ δ

(2.4)

with the continuity of V (x; b) and V 0 (x; b) V1 (b−; b) = V2 (b+; b),

(2.5)

V10 (b−; b)

(2.6)

=

V20 (b+; b).

Here we interpret (2.6). As we all know, the model (1.1) can be seen as a limitation of a compound Poisson process, the proof of which can be found in Li (2005). Using the same notation as in Li (2005), i.e., set c = σ 2 / P N (t) and λ = σ 2 / 2 , when  → 0 the surplus process X  = x + (c + c )t − i=1 Z i −  N (t) converges weakly to the surplus process (1.1), where N (t) is a Poisson process with parameter λ > 0. According to (5.4) in Gerber and Shiu (2005), in the compound Poisson model, (c + c )V10 (b−; b) = (c + c − α)V20 (b+; b) + α, since c, α, V10 (b−; b) and V20 (b+; b) are bounded, when  → 0 (2.6) holds.

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From (2.1) to (2.6), we get that the solutions V (x; b) satisfy the following renewal equations, details can be seen in the Appendix where we solve this kind of integro-differential equation in general cases. For 0 < x < b, Z b V1 (u; b)G 0 (x, u) du, (2.7) V1 (x; b) = F(x) + λˆ 0

and for b < x < ∞, α V2 (x; b) = + Fα (x) + λˆ δ

∞

Z

V2 (u; b)G 1 (x, u) du + λˆ

b

Z

V1 (u; b)G 2 (x, u) du

(2.8)

0

b

where λˆ , F(x), Fα (x), G 0 (x, u; b), G 1 (x, u; b), G 2 (x, u; b) are defined in the Appendix with 0, V (b; b) instead of α f (x), a(b) and λα δ (1 − P(x − b)), V (b; b) − δ instead of g(x), a1 (b), respectively. Similarly, we can get the analytical expressions of V1 (x; b) and V2 (x; b) from (33) and (34) in the Appendix, we have Z b ∞ X (k) k ˆ V1 (x; b) = F(x) + λ G 0 (x, u; b)F(u) du k=1 ∞ X

V2 (x; b) = Fˆα (x) +

0

λˆ k

∞

Z b

k=1

(2.9) (k) G 1 (x, u; b) Fˆα (u) du

where α Fˆα (x) = + Fα (x) + λˆ δ

b

Z

V1 (u; b)G 2 (x, u) du

0

(1)

G 0 (x, u) = G 0 (x, u) Z b (k) (k−1) G 0 (x, u) = G 0 (x, s)G 0 (s, u) ds,

k = 2, 3, . . .

(2.10)

0

(1)

G 1 (x, u) = G 1 (x, u) Z ∞ (k) (k−1) G 1 (x, u) = G 1 (x, s)G 1 (s, u) ds,

k = 2, 3, . . . .

b

The above expression (2.9) can be rewritten as " # Z b ∞ X (k) V1 (x; b) = V (b; b) fˆ0 (x) + λˆ k G 0 (x, u; b) fˆ0 (u) du , V (b; b)Vˆ1 (x; b) k=1

0

V2 (x; b) = V (b; b)Vg (x; b) + h 1 (x; b) +

∞ X k=1

+ V (b; b)

∞ X k=1

λˆ k

∞

Z b

λˆ k

∞

Z b

(k)

G 1 (x, u; b)h 1 (u; b) du

(k) G 1 (x, u; b)Vg (u; b) du

(2.11)

where f 0 (x) 1 − e−b   Z ∞ α ˆ ˆ h 1 (x) = 1 − g0 (x) + λ (1 − P(y − b)) K (x, y) dy δ b Z b Vg (x) = g0 (x) + λˆ Vˆ1 (u; b)G 2 (x, u) du fˆ0 (x) =

0

and f 0 (x), g0 (x) are defined as (5) and (19) in the Appendix, respectively.

(2.12)

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In view of the continuity condition (2.6), we can get by (2.12), h 01 (b) + V (b; b) =

∞ P

λˆ k

k=1

Vˆ10 (b; b) − Vg0 (b) −

R∞ b ∞ P

k=1

(k)

dG 1 dx

λˆ k

(b, u; b)h 1 (u; b) du

R∞ b

(k)

dG 1 dx

(2.13) (b, u; b)Vg (u; b) du

here (1)

dG i dG i (x, u) = (x, u), i = 0, 1 dx dx Z ∞ (k) dG i dG i (k−1) (x, u) = (x, s)G i (s, u) ds, dx dx b

(2.14) i = 0, 1, k = 2, 3, . . . .

Thus according to (2.11)–(2.14) we get the closed-form solution of V (x; b). Next we will give the numerical algorithm and a numerical result assuming that the individual claim amounts are exponentially distributed, p(z) = µe−µz .

(2.15)

Then by some simple calculation we have  2   Z b σ 2 e−y − c − λ − δ + λ + δ K (x, y) dy f 0 (x) = 1 − e−x + 2 2 σ 0 2 Z 2(c + λ + δ − α − σ2 ) ∞ b−y g0 (x) = eb−x + e Kˆ (x, y) dy σ2 b Z ∞ h 1 (x) = 1 − g0 (x) + λˆ e−µ(y−b) Kˆ (x, y) dy

(2.16)

b

where K (x, y), Kˆ (x, y), λˆ are defined in the Appendix. The algorithm is as follows: for a fixed dividend payment level b, first we compute Vˆ1 (x; b) for 0 < x < b, by this we can get Vg (x); then according to (2.13) V (b; b) is obtained; finally, we obtain the value of V1 (x; b) and V2 (x; b) by (2.11). For x ∈ (0.25, 1), take λ = 5, µ = 5, σ = 0.2, c = 0.3, δ = 0.06, α = 0.1 and the fixed dividend payment level b = 0.5. The value of V (x; b) is as follows, V (0.25) = 0.1564 V (0.45) = 0.7323 V (0.65) = 2.2571 V (0.85) = 3.4408

V (0.3) = 0.2331 V (0.5) = 1.0558 V (0.7) = 2.6024 V (0.9) = 3.6589

V (0.35) = 0.3442 V (0.55) = 1.4742 V (0.75) = 2.9142 V (0.95) = 3.8499

V (0.4) = 0.5041 V (0.6) = 1.8795 V (0.8) = 3.1932 V (1.0) = 4.0160.

3. The expected discounted penalty function In the following we will discuss the famous Gerber–Shiu expected discounted penalty function under the compound Poisson risk model perturbed by diffusion. Denote ϕ(x; b) = E{e−δTb w(X (Tb −), |X (Tb )|)I (Tb < ∞)|X (0) = x},

(3.1)

where δ, Tb are interpreted as the force of interest and the ruin time, respectively, w(x1 , x2 ), x1 ≥ 0, x2 ≥ 0 is a nonnegative function of the surplus immediately before ruin, x1 , and the deficit at ruin, x2 , and I (·) is the indicator function.

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Clearly, the discounted penalty function ϕ(x; b) behaves differently with different initial surplus x. Here for notational convenience, we write  ϕ (x; b), 0 ≤ x ≤ b ϕ(x; b) = 1 ϕ2 (x; b), x > b. By similar derivation to (2.1) and (2.2), we get that ϕ(x; b) satisfies the following integro-differential equations: For 0 < x < b, σ 2 00 ϕ (x; b) + cϕ10 (x; b) − (λ + δ)ϕ1 (x; b) + λ 2 1 Z ∞ +λ w(x, x − z) p(z) dz = 0,

Z

x

ϕ1 (x − z; b) p(z) dz

0

(3.2)

x

and for x > b, Z x−b σ 2 00 0 ϕ2 (x − z; b) p(z) dz ϕ (x; b) + (c − α)ϕ2 (x; b) − (λ + δ)ϕ2 (x; b) + λ 2 2 0 Z x Z ∞ +λ ϕ1 (x − z; b) p(z) dz + λ w(x, x − z) p(z) dz = 0, x−b

(3.3)

x

with the conditions ϕ1 (0; b) = w(0, 0),

(3.4)

lim ϕ2 (x; b) = 0,

(3.5)

x→∞

ϕ1 (b−; b) = ϕ2 (b+; b),

(3.6)

ϕ10 (b−; b)

(3.7)

=

ϕ20 (b+; b).

(3.4)–(3.7) are obvious. If the initial surplus x = 0 then ruin is immediate, i.e., Tb = 0, so X (Tb −) = 0 and |X (Tb )| = 0, thus we have that (3.4) holds; if X (0) → ∞, ruin does not happen all the time, hence Tb = ∞ and (3.5) holds; similar approach to (2.5) and (2.6) can prove that (3.6) and (3.7) hold. From (3.2) and (3.3), we know that the expected discounted penalty function alsoR satisfies the integro-differential ∞ equations (1) and (14) in the Appendix with f (x) = w(0, 0)[δ + λ(1 − P(x))] − λ x w(x, x − z) p(z) dz, a(b) = R∞ ϕ(b) − w(0, 0) and g(x) = −λ x w(x, x − z) p(z) dz, a1 (b) = ϕ(b). Using the solution formulas (2), (15), (33) and (34), we can get the renewal equations: ϕ1 (x; b) = F(x) + λˆ

b

Z

ϕ1 (u; b)G 0 (x, u; b) du

(3.8)

0

ϕ2 (x; b) = Fα (x) + λˆ

∞

Z

ϕ2 (u; b)G 1 (x, u; b) du + λˆ

b

Z

ϕ1 (u; b)G 2 (x, u; b) du

(3.9)

0

b

and analytical expressions satisfied by the expected discounted penalty function: ϕ1 (x; b) = F(x) +

∞ X

λˆ k

0

k=1

ϕ2 (x; b) = Fˆα (x) +

∞ X k=1

b

Z

λˆ k

∞

Z b

(k)

G 0 (x, u; b)F(u) du (k)

G 1 (x, u; b) Fˆα (u) du.

(3.10)

(3.11)

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Using a similar argument as in Section 2, we can get ∞ P

j 0 (b) +

λˆ k

b

k=1

ϕ(b) =

(k)

dG 1 dx

R∞

fˆ0 (b) − gˆ 0 (b) −

∞ P

λˆ k

(b, u; b) j (u) du − hˆ 0 (b)

R∞ b

k=1

(k)

dG 1 dx

.

(3.12)

(b, u; b)g(u) ˆ du

Thus we get the closed-form solution of ϕ(x) as follows, ˆ ϕ1 (x; b) = ϕ(b) fˆ(x) + h(x),

x ∈ (0, b) ! Z ∞ ∞ X (k) G 1 (x, u; b)g(u) ˆ du ϕ2 (x; b) = ϕ(b) g(x) ˆ + λˆ k + j (x) +

λˆ k

∞

Z b

k=1

(3.13)

b

k=1 ∞ X

(k)

G 1 (x, u; b) j (u) du,

x ∈ (b, ∞)

where ϕ(b) is defined in (3.12) and f 0 (x) +

∞ P

λˆ k

R∞ b

k=1

fˆ(x) =

(k)

G 1 (x, u; b) f 0 (u) du

1 − e−b

ˆ h(x) = h(x) +

∞ X

λˆ k

b

Z 0

k=1

(k)

G 0 (x, u; b)h(u) du

Z b w(0, 0) 2 f (x) + f (y)K (x, y) dy 0 1 − e−b σ2 0 Z ∞ g(x) ˆ = g0 (x) + λˆ fˆ(u)G 2 (x, u) du

(3.14)

h(x) = −

b

j (x) =

2 σ2

∞

Z

g(y) Kˆ (x, y)dy + λˆ

ˆ h(u)G 2 (x, u) du.

0

b

(k)

b

Z

(k)

G 0 (x, u), G 0 (x, u), G 1 (x, u), G 1 (x, u) are defined as (35) in the Appendix with f (x), a(b), g(x), a1 (b) defined in the above. 4. The probability of ultimate ruin Based on the analytical results of the expected discounted penalty function, the probability of ultimate ruin is discussed in this section. As is well known, many important ruin related functionals can be obtained from the discounted penalty function. If setting δ = 0 and w(x1 , x2 ) = 1 for all x1 , x2 ≥ 0, then ϕ(x; b) becomes the probability of ultimate ruin, which we denote by ψ(x; b) with initial surplus x and the threshold b. From (3.8) to (3.11), we have the following result:  Z b ∞ X  (k) k  ˆ λ G 0 (x, u; b)Fψ (u) du, 0 < x < b,  ψ1 (x; b) = Fψ (x) + ψ(x; b) =

k=1

where Fψ (x) =

ϕ(b) − 1 f 0 (x), 1 − e−b

0

k=1

∞ X    λˆ k ψ2 (x; b) = Fˆψ (x) +

∞

Z b

(k) G 1 (x, u; b) Fˆψ (u) du,

(4.1)

b
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Z b Z b 2 (1 − e−u )G 0 (x, u) du, f 1 (y)K (x, y)dy − λˆ f 0 (x) = 1 − e−x + 2 σ 0 0  2  Z x σ f 1 (x) = e−x ez p(z) dz + λ − λP(x), −c−λ+λ 2 0 Z ∞ (1 − P(y)) Kˆ (x, y) dy, Fˆψ (x) = ϕ(b)g0 (x) − λˆ b Z ∞ Z ∞ 2 g0 (x) = eb−x + 2 eb−u G 1 (x, u) du, g1 (y) Kˆ (x, y) dy − λˆ σ b b   Z x−b σ2 g1 (x) = eb−x c − α + λ − ez p(z) dz . −λ 2 0 5. The time of ruin This section discusses another particular case of ϕ(x; b) by setting δ ≥ 0 and w(x1 , x2 ) = 1 for all x1 , x2 ≥ 0. Denote L(x; b) = E[e−δTb |X (0) = x],

(5.1)

where x = X (0) is the initial surplus. This is the expected present value of a payment 1 at the time of ruin, and at the same time, the Laplace transform of the probability density function of Tb . As a function of the initial surplus x, L(x; b) satisfies the second-order integro-differential equations: For 0 < x < b, Z x σ 2 00 L (x; b) + cL 0 (x; b) − (λ + δ)L(x; b) + λ L(x − z; b) p(z) dz + λ(1 − P(x)) = 0, (5.2) 2 0 and for b < x < ∞ Z x σ 2 00 0 L (x; b) + (c − α)L (x; b) − (λ + δ)L(x; b) + λ L(x − z; b) p(z) dz + λ(1 − P(x)) = 0. (5.3) 2 0 Rx R x−b We know that 0 L(x − z; b) p(z) dz in (5.3) should be divided into two parts, 0 L(x − z; b) p(z) dz and Rx x−b L(x − z; b) p(z) dz like the former, here for simplicity, we just write them together. Setting w(x1 , x2 ) = 1 in (3.4) and (3.5), we get the conditions of L(x; b). lim L(x; b) = 0.

(5.4)

x→∞

L(0; b) = 1.

(5.5)

It is easy to see that (3.1) and (3.2) belong to the type of (1) and (14), we just need to replace f (x), a(b) in (1) with δ, L(b; b) − 1 and replace g(x), a1 (b) in (14) with −λ(1 − P(x)), L(b; b) and we can express L(x; b) as the follows,  Z b   ˆ L(u; b)G 0 (x, u) du, 0 < x < b  F(x) + λ Z b Z0 ∞ L(x; b) = (5.6)    Fα (x) + λˆ L(u; b)G 1 (x, u) du + λˆ L(u; b)G 2 (x, u) du, b < x < ∞ b

0

where the parameters of F(x), Fα (x), G 0 (x, u), G 1 (x, u) and G 2 (x, u) have been changed as the above. We can get the analytical expressions of (5.6) in the same way as (33) and (34) with f (x), a(b), g(x), a1 (b) changed. 6. The moments and the moment-generating function of Dx,b In this section the moments and the moment-generating function of the cumulative dividend payments are discussed. We adopt similar approach and notations to the one utilized in Section 8 of Gerber and Shiu (2006).

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Let M(x, y; b) = E[e y Dx,b |X (0) = x]

(6.1)

denote the moment-generating function of Dx,b . We assume δ > 0. Then 0 ≤ Dx,b ≤ αδ , and M(x, y; b) exists for all y. In view of the strong Markov property of X (t), we have M(x, y; b) = E[M(X (dt), e−δdt y; b)]. By the former arguments, we can get that the moment-generating function M satisfies the following integrodifferential equation, for 0 < x < b, Z x σ 2 ∂2 M ∂M ∂M +c M(x − z, y; b) p(z) dz + λ(1 − P(x)) = 0, (6.2) − δy − λM(x, y; b) + λ 2 ∂x2 ∂x ∂y 0 and for x > b, we have σ 2 ∂2 M ∂M ∂M + (c − α) − δy + αy M(x, y; b) − λM(x, y; b) 2 ∂x2 ∂x ∂y Z x−b Z x +λ M(x − z, y; b) p(z) dz + λ M(x − z, y; b) p(z) dz + λ(1 − P(x)) = 0. 0

(6.3)

x−b

The boundary conditions are M(0, y; b) = 1

(6.4)

lim M(x, y; b) = e

yα/δ

x→∞

(6.5)

and M(x, y; b) , ∂∂M x are continuous at the junction x = b as a function of x. We set M(x, y; b) = 1 +

∞ X yk Vk (x; b), k! k=1

(6.6)

where k Vk (x; b) = E[Dx,b |X (0) = x]

(6.7)

is the kth moment of Dx,b . Substitution of (6.6) into (6.2) and (6.3), with subsequent comparison of the coefficients of y k , yields the integro-differential equations Z x σ 2 00 Vk (x; b) + cVk0 (x; b) − (λ + kδ)Vk (x; b) + λ Vk (x − z; b) p(z) dz = 0, (6.8) 2 0 for 0 < x < b and σ 2 00 V (x; b) + (c − α)Vk0 (x; b) − (λ + kδ)Vk (x; b) + kαVk−1 (x; b) 2 k Z x +λ Vk (x − z; b) p(z) dz = 0,

(6.9)

0

for x > b. They generalize (2.1) and (2.2), which are for k = 1. The boundary conditions are Vk (0; b) = 0

(6.10)

and lim Vk (x; b) =

x→∞

 α k δ

.

(6.11)

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N. Wan / Insurance: Mathematics and Economics 40 (2007) 509–523

0 From the continuity of M(x, y; b) and ∂∂M x as function of x, we can conclude that Vk (x; b) and Vk (x; b) are continuous at the junction x = b. The integro-differential equations (6.8) and (6.9) also are of the same type as (1) and (14) in the Appendix, respectively, and after some functional changes, we directly get the solution from (2) and (15) in the Appendix. For 0 < x < b, Z b ˆ Vk (u)G 0 (x, u) du. (6.12) Vk (x; b) = Fk (x) + λ 0

For b < x < ∞, Vk (x) =

 α k δ

+ Fˆkα (x) + λˆ

∞

Z

Vk (u)G 1 (x, u) du + λˆ b

b

Z

Vk (u)G 2 (x, u) du

(6.13)

0

where λˆ , G 0 (x, u), G 1 (x, u), G 2 (x, u) are defined as that in the Appendix, and the parameters λ + δ of K (x, y) and Kˆ (x, y) are replaced by λ + kδ.   Z b Z b Vk (b; b) 2 −x −u ˆ Fk (x) = 1−e + 2 f k1 (y)K (x, y) dy − λ (1 − e )G 0 (x, u) du , (6.14) 1 − e−b σ 0 0  2  Z x z −x σ − c − λ − kδ + λ e p(z) dz + λ + kδ − λP(x) (6.15) f k1 (x) = e 2 0  Z ∞  α k  2 g0 (x) + 2 Fkα (x) = Vk (b; b) − g(y) Kˆ (x, y) dy, (6.16) δ σ b Z ∞ Z ∞ 2 g0 (x) = eb−x + 2 g1 (y) Kˆ (x, y) dy − λˆ eb−u G 1 (x, u) du (6.17) σ b b   Z x−b σ2 −λ g1 (x) = eb−x c − α + λ + kδ − ez p(z) dz (6.18) 2 0  α k g(x) = −kαVk−1 (x) + (kδ + λ(1 − P(x − b))). (6.19) δ Acknowledgements The author would like to thank Professor Lishang Jiang for his valuable suggestions and the anonymous referees and the editors for their valuable comments. Appendix In this section, we will show that two kinds of integro-differential equations such as (2.1) and (2.2) are identical to the following two renewal equations in details and show that the solutions of integro-differential equations are unique. Based on these, their closed-form solutions are obtained. Theorem A1. The integro-differential equation  2 Z x  σ 00 v1 (x) + cv10 (x) − (λ + δ)v1 (x) + λ v1 (x − z) p(z) dz = f (x) 0  2 v1 (0) = 0, v1 (b) = a(b) is identical to the renewal equation Z b v1 (x) = F(x) + λˆ v1 (u)G 0 (x, u) du 0

(1)

(2)

N. Wan / Insurance: Mathematics and Economics 40 (2007) 509–523

where λˆ =

519

2λ , σ2

G 0 (x, u) = −

b

Z

K (x, y) p(y − u) dy,

(3)

u

Z b a(b) 2 F(x) = f 0 (x) + 2 f (y)K (x, y) dy, 1 − e−b σ 0 Z b Z b 2 (1 − e−u )G 0 (x, u) du f 0 (x) = 1 − e−x + 2 f 1 (y)K (x, y) dy − λˆ σ 0 0   2 Z x σ f 1 (x) = e−x ez p(z) dz + λ + δ − λP(x) −c−λ−δ+λ 2 0

(4) (5) (6)

and K (x, y) is defined by (8) and (9). Proof. Let K (x, y) be the solution of the following equation,   σ 2 00   K yy (x, y) − cK y0 (x, y) − (λ + δ)K (x, y) = 0 2 K (x, b) = 0, K (x, 0) = 0    K 0 (x, x+) − K 0 (x, x−) = 1, K (x, x+) = K (x, x−). y y Then we can solve (7) explicitly,  C0 (x)(eβ1 y − eβ2 y ), K (x, y) = C1 (x)eβ1 y + C2 (x)eβ2 y ,

0
(7)

(8)

where e−β1 x − e(β2 −β1 )b−β2 x <0 (β1 − β2 )(e(β2 −β1 )b − 1) e−β2 x − e−β1 x >0 C1 (x) = − (β1 − β2 )(1 − e(β1 −β2 )b ) e−β2 x − e−β1 x C2 (x) = <0 (β1 − β2 )(e(β2 −β1 )b − 1) C0 (x) =

(9)

and β1 > 0 > β2 are two roots of the equation σ2 2 β − cβ − (λ + δ) = 0. 2 Set W (x) = v1 (x) −

a(b) (1 − e−x ) 1 − e−b

then W (x) satisfies the equation  2 Z x a(b) 00 σ W (x) + cW 0 (x; b) − (λ + δ)W (x) = −λ W (x − z) p(z) dz + f (x) + f 1 (x) 1 − e−b 0  2 W (0) = 0, W (b) = 0

(10)

(11)

where f 1 (x) is defined as (6). Multiplying both sides of the Eq. (11) by K (x, y) and integrating from 0 to b, integration by part and in view of (7), we get  Z b Z y Z b 2 a(b) ˆ f 1 (y) K (x, y) dy (12) K (x, y) W (y − z; b) p(z) dzdy + 2 f (y) + W (x) = −λ 1 − e−b σ 0 0 0

520

N. Wan / Insurance: Mathematics and Economics 40 (2007) 509–523

where λˆ =

2λ . σ2

Denote

G 0 (x, u) = −

b

Z

K (x, y) p(y − u) dy,

u

we have W (x) = λˆ

b

Z

W (u)G 0 (x, u) du +

0

2 σ2

b

Z



f (y) +

0

and substituting (10) into (13), we get Z b v1 (u)G 0 (x, u) du v1 (x) = F(x) + λˆ

 a(b) f (y) K (x, y) dy 1 1 − e−b

(13)

0
0

where F(x) is defined in (4). Thus we show that the theorem holds.



Theorem A2. The integro-differential equation  2 Z x−b σ 00  0   v (x) + (c − α)v (x) − (λ + δ)v (x) + λ v2 (x − z) p(z) dz 2  2  2 2 Z 0 x +λ v1 (x − z) p(z) dz = g(x)    x−b   v2 (b) = a1 (b), v2 (∞) = 0 is identical to the renewal equation Z ∞ Z v2 (x) = Fα (x) + λˆ v2 (u)G 1 (x, u) du + λˆ b

where λˆ =

b

v1 (u)G 2 (x, u) du

(14)

(15)

0

2λ , σ2

G 1 (x, u) = −

∞

Z

Kˆ (x, y) p(y − u) dy,

(16)

Kˆ (x, y) p(y − u) dy,

(17)

u

G 2 (x, u) = −

∞

Z b

Z ∞ 2 Fα (x) = a1 (b)g0 (x) + 2 g(y) Kˆ (x, y) dy, σ b Z ∞ Z ∞ 2 g0 (x) = eb−x + 2 eb−u G 1 (x, u) du g1 (y) Kˆ (x, y) dy − λˆ σ b b   Z x−b σ2 b−x z g1 (x) = e c−α+λ+δ− −λ e p(z) dz . 2 0 Proof. Let Kˆ (x, y) be the solution of the following equation,  2 σ 00    Kˆ yy (x, y) − (c − α) Kˆ y0 (x, y) − (λ + δ) Kˆ (x, y) = 0  2 ˆ  K (x, b) = 0, Kˆ (x, ∞) = 0    Kˆ 0 (x, x+) − Kˆ 0 (x, x−) = 1, Kˆ (x, x+) = Kˆ (x, x−). y y Then we can solve (21) explicitly,  C3 (x)eγ1 y + C4 (x)eγ2 y , b < y < x ˆ K (x, y) = C5 (x)eγ2 y , x
(18) (19) (20)

(21)

(22)

N. Wan / Insurance: Mathematics and Economics 40 (2007) 509–523

521

where C3 (x) = −

e−γ1 x <0 γ1 − γ2

e(γ1 −γ2 )b−γ1 x >0 γ1 − γ2 e(γ1 −γ2 )b−γ1 x − e−γ2 x <0 C5 (x) = γ1 − γ2

C4 (x) =

(23)

and γ1 > 0 > γ2 are two roots of the equation σ2 2 γ − (c − α)γ − (λ + δ) = 0. 2 Set W (x) = v2 (x) − a1 (b)eb−x

(24)

then W (x) satisfies the equation  2 Z x−b σ 00    W (x) + (c − α)W 0 (x; b) − (λ + δ)W (x) + λ W (x − z) p(z) dz 2 0 Rx = −λ x−b v1 (x − z) p(z) dz + g(x) + a1 (b)g1 (x)    W (b) = 0, W (∞) = 0

(25)

where g1 (x) is defined as (20). Multiplying both sides of the Eq. (25) by Kˆ (x, y) and integrating from b to ∞, integrating by part and in view of (21), we get Z y−b Z ∞ ˆ ˆ W (x) = −λ K (x, y) W (y − z) p(z) dzdy 0

b

−λˆ

∞

Z

Kˆ (x, y) b

where λˆ =

Z

y

v1 (y − z) p(z) dzdy +

y−b

2 σ2

∞

Z

[g(y) + a1 (b)g1 (y)] Kˆ (x, y) dy

(26)

b

2λ . σ2

Denote Z ∞ G 1 (x, u) = − Kˆ (x, y) p(y − u) dy, u Z ∞ G 2 (x, u) = − Kˆ (x, y) p(y − u) dy, b

and we have W (x) = λˆ

∞

Z

W (u)G 1 (x, u) du + λˆ b

∞

Z b

2 v1 (u)G 2 (x, u) du + 2 σ

Substituting (24) into (27), we get Z ∞ Z v2 (x) = Fα (x) + λˆ W (u)G 1 (x, u) du + λˆ b

b

∞

Z

[g(y) + a1 (b)g1 (y)] Kˆ (x, y) dy. (27) b

v1 (u)G 2 (x, u) du. b < x < ∞

0

where Fα (x) is defined in (18). Thus we show that the theorem holds.



Next we will show that the solution of (1) and (14) is unique using the similar method as in Luo (2005). The same results can be obtained for other integro-differential equations. Here we just take one for example. From the above, we know that (1) is identical to the renewal equation (2), therefore, we just need to show that the solution of (2) is unique.

522

N. Wan / Insurance: Mathematics and Economics 40 (2007) 509–523

First we show that λˆ

b

Z

|G 0 (x, u)|du <

0

λ < 1. λ+δ

(28)

From (3), (8) and (9), we know that G(x, u) > 0 for 0 < u < b and λˆ

b

Z

|G 0 (x, u)|du = −λˆ

b

Z 0

0

b

Z

K (x, y) p(y − u) dy ≤ −λˆ

b

Z

K (x, y) dy.

(29)

0

u

Since b

Z

K (x, y) dy =



0



1 1 − β2 β1

[C0 (x)(1 − eβ2 x ) + C2 (x)(eβ2 b − eβ2 x )],

and in view of β2 < 0 and b > 0, we get   Z b  1 1  β2 b K (x, y) dy > − e − eβ2 x (C0 (x) + C2 (x)). β2 β1 0 By (9), C0 (x) + C2 (x) =

−e−β2 x . β1 − β2

So substituting the above results into (29), we have λˆ

b

Z

|G 0 (x, u)|du ≤ −λˆ

b

Z

0

K (x, y) dy <

0

λ < 1. λ+δ

This is true for all δ ≥ 0. Define a map m : C(0, b) → C(0, b): m(v)(x) = λˆ

b

Z

G 0 (x, u)v(u) du + F(x),

u ∈ (0, b)

(30)

0

then (2) satisfies the operator equation v = m(v).

(31)

∀ψ, ϕ ∈ C(0, b), by (28) we have km(ϕ) − m(ψ)k ≤ λˆ sup x

b

Z

|G 0 (x, u)|du · kϕ − ψk <

0

λ kϕ − ψk , λ+δ

(32)

then it follows that (30) only has a fixed point from the compression mapping principal. From the above, we know that the solution of the renewal equations (1) and (14) is unique. So by iteration, we can get the closed-form solutions of equations (1) and (14) from (2) and (15), v1 (x; b) = F(x) +

∞ X

λˆ k

0

k=1

ˆ v2 (x; b) = F(x) +

∞ X k=1

b

Z

λˆ k

∞

Z b

(k)

G 0 (x, u; b)F(u) du, (k)

ˆ G 1 (x, u; b) F(u) du

(33)

(34)

N. Wan / Insurance: Mathematics and Economics 40 (2007) 509–523

523

where λˆ =

2λ , σ2

ˆ F(x) = Fα (x) + λˆ

b

Z

v1 (u; b)G 2 (x, u) du

0

(1)

G 0 (x, u) = G 0 (x, u) Z b (k−1) (k) G 0 (x, s)G 0 (s, u) ds, G 0 (x, u) =

(35) k = 2, 3, . . .

0

(1)

G 1 (x, u) = G 1 (x, u) Z ∞ (k−1) (k) G 1 (x, s)G 1 (s, u) ds, G 1 (x, u) =

k = 2, 3, . . .

b

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