# The Exit Time and the Dividend Value Function for

Authors: Peng Li; Chuancun Yin; Ming Zhou
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

We investigate the exit times from an interval for a general one-dimensional time-homogeneous diffusion process and their applications to the dividend problem in risk theory. Specifically, we first use Dynkin’s formula to derive the ordinary differential equations satisfied by the Laplace transform of the exit times. Then, as some examples, we solve the closed-form expression of the Laplace transform of the exit times for several popular diffusions, which are commonly used in modelling of finance and insurance market. Most interestingly, as the applications of the exit times, we create the connect between the dividend value function and the Laplace transform of the exit times. Both the barrier and threshold dividend value function are clearly expressed in terms of the Laplace transform of the exit times.


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## Body

## 1. Introduction

Diffusion processes have extensive applications in economics, finance, queueing, mathematical biology, and electric engineering. See, for example, [1–4] and the references therein. The main tool for studying various properties of diffusion is the result on exit times from an interval. Motivated by Yin et al. in [5], who considered the exit problems for jump processes with applications to dividend problems. In this paper, we consider the Laplace transforms of some random variables involving the exit time for the general one-dimension diffusion processes with applications to dividend problems.LetU={Ut,t≥0} be a one-dimensional time-homogeneous diffusion process, which is defined by the following stochastic differential equation:
(1)dUt=μ(Ut)dt+σ(Ut)dBt,U0=u∈(a,b),
where Bt is a Brownian motion and a<b are constants. It is well known that under certain conditions on the coefficients μ(u) and σ(u), the SDE (1) has a unique strong solution for each starting point. The solution Ut is a time-homogeneous strong Markov process with infinitesimal generator as follows:
(2)𝒜g(u)=12σ2(u)g′′(u)+μ(u)g′(u),u∈(a,b),
for any twice continuously differentiable function g.Define(3)τa-=inf{t≥0:Ut≤a},τb+=inf{t≥0:Ut≥b},(4)τab=τa-∧τb+.Forδ>0, we consider the following Laplace transforms:
(5)φ1(u)=Eu[e-δτa-,τa-<τb+],(6)φ2(u)=Eu[e-δτb+,τb+<τa-],(7)φ(u)=Eu[e-δτab]=φ1(u)+φ2(u).We study the differential equations satisfied by the Laplace transforms and some applications of the popular dividend strategy in risk theory.The rest of the paper is organized as follows. Section2 studies the Laplace transforms of exit times and considers some popular diffusions. Some applications in the calculation of dividend value functions for the barrier strategy and the threshold strategy are considered in Section 3.


## 2. Laplace Transform

In this section, we consider the Laplace transform of the exit time for the general diffusion process{Ut;t≥0} defined by (1).Theorem 1.
The functionφ1(u) defined by (5) satisfies the following differential equation:
(8)12σ2(u)φ1′′(u)+μ(u)φ1′(u)=δφ1(u),u∈(a,b),
with the boundary conditions φ1(a)=1, φ1(b)=0.Proof.
We assume thatf(u) is twice continuously differentiable and satisfies the following differential equation:
(9)𝒜f(u)=δf(u),u∈(a,b),(10)f(a)=1,f(b)=0.
Applying Dynkin’s formula to h(t,Ut)=e-δtf(Ut), we obtain
(11)Eu[h(t,Ut)]=h(0,u)+Eu(∫0t‍(𝒜-δ)h(s,Us)ds)=f(u)+Eu(∫0t‍e-δs(𝒜-δ)f(Us)ds).
Since τab<∞ is a stopping time, it follows from the optional sampling theorem that
(12)Eu[e-δ(τab∧t)f(U(τab∧t))]=f(u)+Eu(∫0(τab∧t)‍e-δs(𝒜-δ)f(Us)ds∫0(τab∧t)‍),
and letting t→∞, we get
(13)Eu[e-δτabf(Uτab)]=f(u)+Eu(∫0τab‍e-δs(𝒜-δ)f(Us)ds).
By the definitions of τab, we get
(14)Eu[e-δτabf(Uτab)]=f(a)Eu[e-δτa-;τa-<τb+]+f(b)Eu[e-δτb+;τb+<τa-].
Substituting (9) and (10) into (13) and (14), we get
(15)f(u)=Eu[e-δτa-;τa-<τb+]=φ1(u).
This completes the proof.Theorem 2.
The functionφ2(u)=Eu[e-δτb+;τb+<τa-] satisfies the following differential equation:
(16)12σ2(u)φ2′′(u)+μ(u)φ2′(u)=δφ2(u),u∈(a,b),
with the boundary conditions φ2(a)=0, φ2(b)=1.Proof.
The proof of this theorem is similar to that of Theorem1. We first assume that f(u) is twice continuously differentiable and satisfies the following differential equation:
(17)𝒜f(u)=δf(u),u∈(a,b),f(a)=0,f(b)=1.
Applying Dynkin’s formula to e-δtf(Ut), and after the same discussion as of Theorem 1, we also can obtain (13) and (14). Substituting (17) into (13) and (14), we get
(18)f(u)=Eu[e-δτb+;τb+<τa-]=φ2(u).
This completes the proof.According to the definition of (7), we can lead to the following theorem from Theorems 1 and 2.Theorem 3.
The functionφ(u)=Eu[e-δτab] satisfies the following differential equation:
(19)12σ2(u)φ′′(u)+μ(u)φ′(u)=δφ(u),u∈(a,b),
with the boundary conditions φ(a)=1, φ(b)=1.Now, we consider some examples.Example 4.
The Bessel process:dUt=((d-1)/2Ut)dt+dBt, where d>1 is a real number. We assume that b>a>0 in this process.First, we consider the following differential equation:(20)12f′′(u)+d-12uf′(u)=δf(u).
It is well known that the increasing and decreasing solutions are, respectively, as follows:
(21)f+(u)=u-vIv(2δu),f-(u)=u-vKv(2δu),
where v=(d-2)/2, and Iv(·) and Kv(·) are the usual modified Bessel functions.Then, from Theorem1, we can give φ1(u) as follows:
(22)φ1(u)=C1f+(u)+C2f-(u),φ1(a)=1,φ1(b)=0,
where the constants C1 and C2 are to be determined. From the boundary conditions (22), we can obtain the expression of the constants C1 and C2 as follows:
(23)C1=avKv(2δb)Iv(2δa)Kv(2δb)-Iv(2δb)Kv(2δa),C2=avIv(2δb)Iv(2δb)Kv(2δa)-Iv(2δa)Kv(2δb).
So, we get
(24)φ1(u)=(au)vKv(2δb)Iv(2δu)-Iv(2δb)Kv(2δu)Iv(2δa)Kv(2δb)-Iv(2δb)Kv(2δa).According to Theorem2 and (21), we can give φ2(u) as follows:
(25)φ2(u)=C3f+(u)+C4f-(u),φ2(a)=0,φ2(b)=1,
where the constants C3 and C4 are to be determined. From the boundary conditions (25), we can determine the constants and obtain the expression of φ2(u) as follows:
(26)φ2(u)=(bu)vKv(2δa)Iv(2δu)-Iv(2δa)Kv(2δu)Iv(2δb)Kv(2δa)-Iv(2δa)Kv(2δb).According to Theorem3, the expression of φ(u) can be obtained from solving the following differential equation:
(27)12φ′′(u)+d-12uφ′(u)=δφ(u),φ(a)=1,φ(b)=1.
Furthermore, from the definition of φ(u)=φ1(u)+φ2(u), we also can get the expression of φ(u). The two methods can lead to the same results as follows:
(28)φ(u)=(1u)v×([avKv(2δb)-bvKv(2δa)]+[bvIv(2δa)-avIv(2δb)]Kv(2δu)[avKv(2δb)-bvKv(2δa)]Iv(2δu)+[bvIv(2δa)-avIv(2δb)]Kv(2δu))×((2δa)Iv(2δa)Kv(2δb)-Iv(2δb)Kv(2δa))-1.Example 5 (the square root process (see [6])).
(29)
d
U
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We assume thatb>a>0 and consider the following differential equation:
(30)12σ2u2f′′(u)+(vk-vu)f′(u)=δf(u),δ>0.
We assume that (2ν/σ2)k is not an integer, the two linear independent solutions are
(31)f+(u)=M(δv,2vσ2k,2vσ2u),f-(u)=U(δv,2vσ2k,2vσ2u),
where M and U are the confluent hypergeometric functions of the first and second kinds, respectively. Then, as the way at used in Example 4, and from Theorems 1 and 2, we get that the expressions of φ1(u) and φ2(u) are as follows:
(32)φ1(u)=g(b,u)-g(u,b)g(b,a)-g(a,b),φ2(u)=g(a,u)-g(u,a)g(a,b)-g(b,a),
where
(33)g(x,y)=U(δv,2vσ2k,2vσ2x)M(δv,2vσ2k,2vσ2y).
So, we can get
(34)φ(u)=g(a,u)-g(b,u)+g(u,b)-g(u,a)g(a,b)-g(b,a).Example 6 (the Ornstein-Uhlenbeck process (see [7])).
(35)
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The Ornstein-Uhlenbeck process above is the only process that is simultaneously Gaussian, Markov, and stationary, and has been discussed extensively, see, for example [2–4, 8].
We consider the following differential equation:(36)12σ2f′′(u)+v(k-u)f′(u)=δf(u).
In the case ofk=0, σ=1, the two independent solutions to
(37)12f′′(u)-vuf′(u)=δf(u)
are
(38)f+(u)=H-δ/v(-vu)=2-δ/2ve(1/2)vu2D-δ/v(-2vu),f-(u)=H-δ/v(vu)=2-δ/2ve(1/2)vu2D-δ/v(2vu),
where Hv(·) and Dv(·) are, respectively, the Hermite and parabolic functions. We obtain the expressions of φ1(u) and φ2(u) as follows:
(39)φ1(u)=e(1/2)v(u2-a2)h(u,b)-h(b,u)h(a,b)-h(b,a),φ2(u)=e(1/2)v(u2-b2)h(a,u)-h(u,a)h(a,b)-h(b,a),
where
(40)h(x,y)=D-δ/v(-2vx)D-δ/v(2vy).
By the definition of φ(u), we get
(41)φ(u)=(e(1/2)v(u2-a2)(h(u,b)-h(b,u))+e(1/2)v(u2-b2)(h(a,u)-h(u,a)))×(h(a,b)-h(b,a)e(1/2)v(u2-a2))-1.
For the generalk and σ, the two independent solutions of (36) are, respectively, as follows:
(42)f+(u)=H-δ/v(-vσ(u-k))=2-δ/2ve(1/2)(v/σ2)(u-k)2D-δ/v(-2vσ(u-k)),f-(u)=H-δ/v(vσ(u-k))=2-δ/2ve(1/2)(v/σ2)(u-k)2D-δ/v(2vσ(u-k)).
Then, we obtain the expressions of φ1(u) and φ2(u) as follows:
(43)φ1(u)=e(1/2)(v/σ2)[(u-k)2-(a-k)2]h0(u,b)-h0(b,u)h0(a,b)-h0(b,a),φ2(u)=e(1/2)(v/σ2)[(u-k)2-(b-k)2]h0(a,u)-h0(u,a)h0(a,b)-h0(b,a),
where
(44)h0(x,y)=D-δ/v(-2vσ(x-k))D-δ/v(2vσ(y-k)).
Finally, we get
(45)φ(u)=e(1/2)(v/σ2)[(u-k)2-(a-k)2]h0(u,b)-h0(b,u)h0(a,b)-h0(b,a)+e(1/2)(v/σ2)[(u-k)2-(b-k)2]h0(a,u)-h0(u,a)h0(a,b)-h0(b,a).Example 7 (the Gompertz Brownian motion process (see [9])).
(46)
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We assume that b>a>0.Now, consider the differential equation(47)12σ2u2f′′(u)+vu(lnk-lnUt)f′(u)=δf(u),δ>0.
It is well known that the increasing and decreasing solutions are, respectively, as follows:
(48)f+(u)=M(δ2v,12,vσ2(lnuk+σ22v)2),f-(u)=U(δ2v,12,vσ2(lnuk+σ22v)2),
where M and U, as in Example 5, are the first and second Kummer’s function, respectively. From the boundary conditions φ1(a)=1 and φ1(b)=0, we get
(49)φ1(u)=g0(b,u)-g0(u,b)g0(b,a)-g0(a,b),
where
(50)g0(x,y)=U(δ2v,12,vσ2(lnxk+σ22v)2)×M(δ2v,12,vσ2(lnyk+σ22v)2).
From the boundary conditions φ2(a)=0 and φ2(b)=1, we get
(51)φ2(u)=g0(u,a)-g0(a,u)g0(b,a)-g0(a,b).
Then, we obtain
(52)φ(u)=g0(b,u)-g0(a,u)+g0(u,a)-g0(u,b)g0(b,a)-g0(a,b).


## 3. Applications to Dividend Value Function

### 3.1. Barrier Strategy

In this subsection, we consider the barrier strategy for dividend payments which are discussed in various model, see, for example, [10–13]. More specifically, we assume that the company pays dividends according to the following strategy governed by parameter b>0. Whenever the surplus is above the level b, the excess will be paid as dividends, and when the surplus is below b nothing is paid out. We denote the aggregate dividends paid in the time interval [0,t] by Dr(t), the modified risk process by Ur(t)=Ut-Dr(t), the ruin time by Tr=inf{t≥0:Ur(t)=0}, and the present value of all dividends until ruin Tr by Dr=∫0Tr‍e-δtdDr(t), here, δ>0 is the discount factor, and the expectation of Dr by
(53)Vr(u,b)=Eu[Dr].Now, we want to derive the dividend value function by the Laplace transform of exit time. We denote(54)τ0-=inf{t≥0:Ut≤0},(55)τ0b=τ0-∧τb+,(56)ψ2(u)=E[e-δτb+;τb+<τ0-],
where τb+ is defined by (3). Let a=0 in the function φ2(u) be defined by (6), we get the definition of ψ2(u). So, we get the following lemma from Theorem 2.Lemma 8.
The functionψ2(u) defined by (56) satisfies the following differential equation:
(57)12σ2(u)ψ2′′(u)+μ(u)ψ2′(u)=δψ2(u),u∈(0,b),
with the boundary conditions ψ2(0)=0, ψ2(b)=1.Then, we have the following theorem.Theorem 9.
For0≤u≤b, one has
(58)Vr(u,b)=ψ2(u)ψ2′(b),
where ψ2(u) is defined by (56).Proof.
The one-dimensional diffusion model defined by (1) is a time-homogeneous strong Markov process. Then, when 0<u<b, we have
(59)Vr(u,b)=Eu(∫0Tr‍e-δtdDr(t))=Eu(∫0τ0b‍e-δtdDr(t)+∫τ0bTr‍e-δtdDr(t))=Eu(∫τ0bTr‍e-δtdDr(t))=Eu[e-δτ0b(∫0Tr‍e-δtdDr(t)∘θτ0b)]=Eu[e-δτ0bEUτ0b(∫0Tr‍e-δtdDr(t))]=Eu[e-δτ0bVr(Uτ0b,b)],
where θτ0b is the shift operator. By the definition of τ0b, we get
(60)Eu[eτ0bVr(Uτ0b,b)]=Vr(0,b)Eu[e-δτ0-;τ0-<τb+]+Vr(b,b)Eu[e-δτb+;τb+<τ0-]=Vr(b,b)ψ2(u).
From (59) and (60), we obtain
(61)Vr(u,b)=Vr(b,b)ψ2(u),
where ψ2(u) can be determined from Lemma 8. From [11], for the barrier strategy, we have the following boundary condition:
(62)∂V(u,b)∂u|u=b=1.
Then, we have
(63)Vr(b,b)=1ψ2′(b).
So, we get the result. This completes the proof.Now, we consider the examples discussed in Section2.Example 10.
The Bessel process discussed in Example4. Reference [14] gives the following helpful formulas:
(64)I0(u)≈1,Iv(u)≈(u/2)2Γ(1+v)(v≠-1,-2,…),K0(u)≈-lnu,Kv(u)≈Γ(v)2(u2)-v(Re(v)>0).
Letting a=0 in Example 4, we get ψ2(u) from φ2(u) as follows:
(65)ψ2(u)=(bu)vIv(2δu)Iv(2δb).
Using the following formula:
(66)Iv′(u)=vuIv(u)+Iv+1(u),
we have
(67)ψ2′(b)=(bu)v2δIv+1(2δb)Iv(2δb).
Then, from Theorem 9, and substituting ψ2(u) and ψ2′(b) into (58), we obtain
(68)Vr(u,b)=Iv(2δu)2δIv+1(2δb).Example 11.
We consider the square root process discussed in Example5. Let a=0 in Example 5, according to M(x,y,0)=1, we obtain
(69)ψ2(u)=1C[U(δv,2vσ2k,0)M(δv,2vσ2k,2vσ2u)-U(δv,2vσ2k,2vσ2u)],
where
(70)C=U(δv,2vσ2k,0)M(δv,2vσ2k,2vσ2b)-U(δv,2vσ2k,2vσ2b).
Using the following helpful formulas (see [14]):
(71)∂M(c1,c2,u)∂u=c1c2M(c1+1,c2+1,u),∂U(c1,c2,u)∂u=-c1U(c1+1,c2+1,u),
we get
(72)ψ2′(b)=1C[δvkU(δv,2vσ2k,0)M(δv+1,2vσ2k+1,2vσ2b)+2δσ2U(δv+1,2vσ2k+1,2vσ2b)].
Finally, we have
(73)Vr(u,b)=[U(δv,2vσ2k,0)M(δv,2vσ2k,2vσ2u)-U(δv,2vσ2k,2vσ2u)]×[δvkU(δv,2vσ2k,0)M(δv+1,2vσ2k+1,2vσ2b)+2δσ2U(δv+1,2vσ2k+1,2vσ2b)]-1.Example 12.
We consider the Ornstein-Uhlenbeck process considered in Example6, and let a=0 in Example 6. From [14], we have the following formulas:
(74)Dv(0)=2v/2πΓ((1/2)-(v/2)),Dv′(u)=-u2Dv(u)+vDv-1(u),
where Γ(x) is the gamma function.In the case ofk=0, σ=1, we have
(75)ψ2(u)=e(v/2)(u2-b2)D-δ/v(2vu)-D-δ/v(-2vu)D-δ/v(2vb)-D-δ/v(-2vb),ψ2′(b)=δ2vD(-δ/v)-1(2vb)+D(-δ/v)-1(-2vb)D-δ/v(-2vb)-D-δ/v(2vb).
Then, we get
(76)Vr(u,b)=1δv2e(v/2)(u2-b2)×D-δ/v(-2vu)-D-δ/v(2vu)D(-δ/v)-1(2vb)+D(-δ/v)-1(-2vb).For the generalk, σ, we have
(77)ψ2(u)=1Ge(v/2σ2)((u-k)2-(b-k)2)×[D-δ/v((2vσ)k)D-δ/v((2vσ)(u-k))-D-δ/v(-(2vσ)k)×D-δ/v((-2vσ)(u-k))],
where
(78)G=D-δ/v((2vσ)k)D-δ/v((2vσ)(b-k))-D-δ/v(-(2vσ)k)D-δ/v(-(2vσ)(u-k)).
We can get
(79)ψ2′(b)=-1Gδσ2v×[D-δ/v((2vσ)k)D(-δ/v)-1((2vσ)(b-k))+D-δ/v(-(2vσ)k)×D(-δ/v)-1(-(2vσ)(b-k))].
Finally, we obtain
(80)Vr(u,b)=1Ge(v/2σ2)((u-k)2-(b-k)2)σδv2×[(-(2vσ)(u-k))D-δ/v(-(2vσ)k)×D-δ/v(-(2vσ)(u-k))-D-δ/v((2vσ)k)×D-δ/v((2vσ)(u-k))]×[((2vσ)(b-k))D-δ/v((2vσ)k)×D(-δ/v)-1((2vσ)(b-k))+D-δ/v(-(2vσ)k)×D(-δ/v)-1(-(2vσ)(b-k))]-1.Example 13.
We consider the Gompertz Brownian motion process discussed in Example7. In [15], the authors point out as u→+∞,
(81)U(c1,c2,u)=u-c1[1+o(|u|-1)],M(c1,c2,u)=Γ(c2)Γ(c1)euuc1-c2[1+o(|u|-1)].
Letting a=0 in Example 7, and using (81), we lead to the expression of ψ2(u) as follows:
(82)ψ2(u)=U(δ2v,12,vσ2(lnuk+σ22v)2)×[U(δ2v,12,vσ2(lnbk+σ22v)2)]-1.
Using (71), we get
(83)ψ2′(b)=-δσ2bkln(bk+σ22v)×U(δ2v+1,12+1,vσ2(lnbk+σ22v)2)×[U(δ2v,12,vσ2(lnbk+σ22v)2)]-1.
Then, from Theorem 9, we have
(84)Vr(u,b)=-σ2bkδU(δ2v,12,vσ2(lnuk+σ22v)2)×[U((lnbk+σ22v)2δ2v+1,12ln(bk+σ22v)×U(δ2v+1,12+1,vσ2(lnbk+σ22v)2)].


### 3.2. Threshold Strategy

We consider the company pays dividends according to the threshold dividend strategy; that is, dividends are paid at a constant rateα whenever the modified surplus is above the threshold b, and no dividends are paid whenever the modified surplus is below b. For recent publications on threshold strategy, see, for example, [3, 16, 17]. We define the modified risk process by
(85)Ud(t)=Ut-Dd(t),
where Dd(t)=α∫0t‍I(Ud(s)>b)ds. Let Dd denote the present value of all dividends until ruin as follows:
(86)Dd=α∫0Td‍e-δsI(Ud(s)>b)ds,
where Td=inf{t≥0:Ud(t)=0}. We denote by Vd(u,b) the expected discounted value of dividend payments; that is,
(87)Vd(u,b)=Eu[Dd].
We denote
(88)τb-=inf{t:Ud(t)≤b},φ3(u)=E[e-δτb-].We can mimic the discussion of Theorem1 to give the differential equation and the boundary conditions satisfied by φ3(u).Lemma 14.
The functionφ3(u) defined by (88) satisfies the following differential equation:
(89)12σ2(u)φ3′′(u)+(μ(u)-α)φ3′(u)=δφ3(u),u≥b,
with the boundary conditions φ3(b)=1, φ3(∞)=0.We have the following theorem.Theorem 15.
Foru∈[0,b], one has
(90)Vd(u;b)=αδφ3′(b)ψ2(u)φ3′(b)-ψ2′(b),
and, for u>b, one has
(91)Vd(u;b)=αδ+αδψ2′(b)φ3(u)φ3′(b)-ψ2′(b).Proof.
Whenu∈[0,b], in view of the strong Markov property, we obtain
(92)Vd(u;b)=Vd(b;b)ψ2(u).
When u>b, since τb- is a stopping time, it follows from the strong Markov property of Ud(u) that
(93)Vd(u,b)=Eu(∫0τb-‍αe-δtdt)+Eu(∫τb-Td‍αe-δtI(Ud(t)>b)dt)=αδ(1-Eue-δτb-)+Eu[e-δτb-EUτb-(∫0Td‍e-δtdDd(t))]=αδ+(Vd(b,b)-αδ)φ3(u).
Using the continuity of the function Vd′(u;b) at u=b, we get
(94)Vd(b;b)ψ2′(b)=(Vd(b,b)-αδ)φ3′(b).
So, we get
(95)Vd(b;b)=αδφ3′(b)φ3′(b)-ψ2′(b).
Substituting the above expression into (92) and (93), we can get the results (90) and (91). This completes the proof.We just consider the square root process.Example 16.
We consider the square root process discussed in Example11. We first solve the differential equation satisfied by φ3(u) as follows
(96)12σ2u2φ3′′(u)+(vk-vu-α)φ3′(u)=δφ3(u),δ>0.
We assume that 2((vk-α)/σ2) is not an integer, the two linear independent solutions are
(97)φ+(u)=M(δv,2(vk-α)σ2,2vσ2u),φ-(u)=U(δv,2(vk-α)σ2,2vσ2u).
Using (81), and from the boundary conditions φ3(b)=1, φ3(∞)=0, we get
(98)φ3(u)=[U(δv,2(vk-α)σ2,2vσ2u)]×[U(δv,2(vk-α)σ2,2vσ2b)]-1.
From (71), we have
(99)φ3′(b)=-2δσ2[U(δv+1,2(vk-α)σ2+1,2vσ2b)]×[U(δv,2(vk-α)σ2,2vσ2b)]-1.
Furthermore, ψ2(u) and ψ2′(b) have been given in Example 11. From Theorem 15, we obtain
(100)Vd(u,b)=αδU(δv+1,2(vk-α)σ2+1,2vσ2b)×[U(δv,2vσ2k,0)M(δv,2vσ2k,2vσ2u)-U(δv,2vσ2k,2vσ2u)]×{CU(δv+1,2(vk-α)σ2+1,2vσ2b)+U(δv,2(vk-α)σ2,2vσ2b)×[σ22vkU(δv,2vσ2k,0)×M(δv+1,2vσ2k+1,2vσ2b)+U(δv+1,2vσ2k+1,2vσ2b)]}-1,u∈[0,b],
and for u>b, we get
(101)Vd(u,b)=αδ-αδU(δv,2(vk-α)σ2,2vσ2u)×[σ22vkU(δv,2vσ2k,0)×M(δv+1,2vσ2k+1,2vσ2b)+U(δv+1,2vσ2k+1,2vσ2b)]×{CU(δv+1,2(vk-α)σ2+1,2vσ2b)+U(δv,2(vk-α)σ2,2vσ2b)×[σ22vkU(δv,2vσ2k,0)×M(δv+1,2vσ2k+1,2vσ2b)+U(δv+1,2vσ2k+1,2vσ2b)]}-1,
where C is defined in Example 11.


## 3.1. Barrier Strategy

In this subsection, we consider the barrier strategy for dividend payments which are discussed in various model, see, for example, [10–13]. More specifically, we assume that the company pays dividends according to the following strategy governed by parameter b>0. Whenever the surplus is above the level b, the excess will be paid as dividends, and when the surplus is below b nothing is paid out. We denote the aggregate dividends paid in the time interval [0,t] by Dr(t), the modified risk process by Ur(t)=Ut-Dr(t), the ruin time by Tr=inf{t≥0:Ur(t)=0}, and the present value of all dividends until ruin Tr by Dr=∫0Tr‍e-δtdDr(t), here, δ>0 is the discount factor, and the expectation of Dr by
(53)Vr(u,b)=Eu[Dr].Now, we want to derive the dividend value function by the Laplace transform of exit time. We denote(54)τ0-=inf{t≥0:Ut≤0},(55)τ0b=τ0-∧τb+,(56)ψ2(u)=E[e-δτb+;τb+<τ0-],
where τb+ is defined by (3). Let a=0 in the function φ2(u) be defined by (6), we get the definition of ψ2(u). So, we get the following lemma from Theorem 2.Lemma 8.
The functionψ2(u) defined by (56) satisfies the following differential equation:
(57)12σ2(u)ψ2′′(u)+μ(u)ψ2′(u)=δψ2(u),u∈(0,b),
with the boundary conditions ψ2(0)=0, ψ2(b)=1.Then, we have the following theorem.Theorem 9.
For0≤u≤b, one has
(58)Vr(u,b)=ψ2(u)ψ2′(b),
where ψ2(u) is defined by (56).Proof.
The one-dimensional diffusion model defined by (1) is a time-homogeneous strong Markov process. Then, when 0<u<b, we have
(59)Vr(u,b)=Eu(∫0Tr‍e-δtdDr(t))=Eu(∫0τ0b‍e-δtdDr(t)+∫τ0bTr‍e-δtdDr(t))=Eu(∫τ0bTr‍e-δtdDr(t))=Eu[e-δτ0b(∫0Tr‍e-δtdDr(t)∘θτ0b)]=Eu[e-δτ0bEUτ0b(∫0Tr‍e-δtdDr(t))]=Eu[e-δτ0bVr(Uτ0b,b)],
where θτ0b is the shift operator. By the definition of τ0b, we get
(60)Eu[eτ0bVr(Uτ0b,b)]=Vr(0,b)Eu[e-δτ0-;τ0-<τb+]+Vr(b,b)Eu[e-δτb+;τb+<τ0-]=Vr(b,b)ψ2(u).
From (59) and (60), we obtain
(61)Vr(u,b)=Vr(b,b)ψ2(u),
where ψ2(u) can be determined from Lemma 8. From [11], for the barrier strategy, we have the following boundary condition:
(62)∂V(u,b)∂u|u=b=1.
Then, we have
(63)Vr(b,b)=1ψ2′(b).
So, we get the result. This completes the proof.Now, we consider the examples discussed in Section2.Example 10.
The Bessel process discussed in Example4. Reference [14] gives the following helpful formulas:
(64)I0(u)≈1,Iv(u)≈(u/2)2Γ(1+v)(v≠-1,-2,…),K0(u)≈-lnu,Kv(u)≈Γ(v)2(u2)-v(Re(v)>0).
Letting a=0 in Example 4, we get ψ2(u) from φ2(u) as follows:
(65)ψ2(u)=(bu)vIv(2δu)Iv(2δb).
Using the following formula:
(66)Iv′(u)=vuIv(u)+Iv+1(u),
we have
(67)ψ2′(b)=(bu)v2δIv+1(2δb)Iv(2δb).
Then, from Theorem 9, and substituting ψ2(u) and ψ2′(b) into (58), we obtain
(68)Vr(u,b)=Iv(2δu)2δIv+1(2δb).Example 11.
We consider the square root process discussed in Example5. Let a=0 in Example 5, according to M(x,y,0)=1, we obtain
(69)ψ2(u)=1C[U(δv,2vσ2k,0)M(δv,2vσ2k,2vσ2u)-U(δv,2vσ2k,2vσ2u)],
where
(70)C=U(δv,2vσ2k,0)M(δv,2vσ2k,2vσ2b)-U(δv,2vσ2k,2vσ2b).
Using the following helpful formulas (see [14]):
(71)∂M(c1,c2,u)∂u=c1c2M(c1+1,c2+1,u),∂U(c1,c2,u)∂u=-c1U(c1+1,c2+1,u),
we get
(72)ψ2′(b)=1C[δvkU(δv,2vσ2k,0)M(δv+1,2vσ2k+1,2vσ2b)+2δσ2U(δv+1,2vσ2k+1,2vσ2b)].
Finally, we have
(73)Vr(u,b)=[U(δv,2vσ2k,0)M(δv,2vσ2k,2vσ2u)-U(δv,2vσ2k,2vσ2u)]×[δvkU(δv,2vσ2k,0)M(δv+1,2vσ2k+1,2vσ2b)+2δσ2U(δv+1,2vσ2k+1,2vσ2b)]-1.Example 12.
We consider the Ornstein-Uhlenbeck process considered in Example6, and let a=0 in Example 6. From [14], we have the following formulas:
(74)Dv(0)=2v/2πΓ((1/2)-(v/2)),Dv′(u)=-u2Dv(u)+vDv-1(u),
where Γ(x) is the gamma function.In the case ofk=0, σ=1, we have
(75)ψ2(u)=e(v/2)(u2-b2)D-δ/v(2vu)-D-δ/v(-2vu)D-δ/v(2vb)-D-δ/v(-2vb),ψ2′(b)=δ2vD(-δ/v)-1(2vb)+D(-δ/v)-1(-2vb)D-δ/v(-2vb)-D-δ/v(2vb).
Then, we get
(76)Vr(u,b)=1δv2e(v/2)(u2-b2)×D-δ/v(-2vu)-D-δ/v(2vu)D(-δ/v)-1(2vb)+D(-δ/v)-1(-2vb).For the generalk, σ, we have
(77)ψ2(u)=1Ge(v/2σ2)((u-k)2-(b-k)2)×[D-δ/v((2vσ)k)D-δ/v((2vσ)(u-k))-D-δ/v(-(2vσ)k)×D-δ/v((-2vσ)(u-k))],
where
(78)G=D-δ/v((2vσ)k)D-δ/v((2vσ)(b-k))-D-δ/v(-(2vσ)k)D-δ/v(-(2vσ)(u-k)).
We can get
(79)ψ2′(b)=-1Gδσ2v×[D-δ/v((2vσ)k)D(-δ/v)-1((2vσ)(b-k))+D-δ/v(-(2vσ)k)×D(-δ/v)-1(-(2vσ)(b-k))].
Finally, we obtain
(80)Vr(u,b)=1Ge(v/2σ2)((u-k)2-(b-k)2)σδv2×[(-(2vσ)(u-k))D-δ/v(-(2vσ)k)×D-δ/v(-(2vσ)(u-k))-D-δ/v((2vσ)k)×D-δ/v((2vσ)(u-k))]×[((2vσ)(b-k))D-δ/v((2vσ)k)×D(-δ/v)-1((2vσ)(b-k))+D-δ/v(-(2vσ)k)×D(-δ/v)-1(-(2vσ)(b-k))]-1.Example 13.
We consider the Gompertz Brownian motion process discussed in Example7. In [15], the authors point out as u→+∞,
(81)U(c1,c2,u)=u-c1[1+o(|u|-1)],M(c1,c2,u)=Γ(c2)Γ(c1)euuc1-c2[1+o(|u|-1)].
Letting a=0 in Example 7, and using (81), we lead to the expression of ψ2(u) as follows:
(82)ψ2(u)=U(δ2v,12,vσ2(lnuk+σ22v)2)×[U(δ2v,12,vσ2(lnbk+σ22v)2)]-1.
Using (71), we get
(83)ψ2′(b)=-δσ2bkln(bk+σ22v)×U(δ2v+1,12+1,vσ2(lnbk+σ22v)2)×[U(δ2v,12,vσ2(lnbk+σ22v)2)]-1.
Then, from Theorem 9, we have
(84)Vr(u,b)=-σ2bkδU(δ2v,12,vσ2(lnuk+σ22v)2)×[U((lnbk+σ22v)2δ2v+1,12ln(bk+σ22v)×U(δ2v+1,12+1,vσ2(lnbk+σ22v)2)].


## 3.2. Threshold Strategy

We consider the company pays dividends according to the threshold dividend strategy; that is, dividends are paid at a constant rateα whenever the modified surplus is above the threshold b, and no dividends are paid whenever the modified surplus is below b. For recent publications on threshold strategy, see, for example, [3, 16, 17]. We define the modified risk process by
(85)Ud(t)=Ut-Dd(t),
where Dd(t)=α∫0t‍I(Ud(s)>b)ds. Let Dd denote the present value of all dividends until ruin as follows:
(86)Dd=α∫0Td‍e-δsI(Ud(s)>b)ds,
where Td=inf{t≥0:Ud(t)=0}. We denote by Vd(u,b) the expected discounted value of dividend payments; that is,
(87)Vd(u,b)=Eu[Dd].
We denote
(88)τb-=inf{t:Ud(t)≤b},φ3(u)=E[e-δτb-].We can mimic the discussion of Theorem1 to give the differential equation and the boundary conditions satisfied by φ3(u).Lemma 14.
The functionφ3(u) defined by (88) satisfies the following differential equation:
(89)12σ2(u)φ3′′(u)+(μ(u)-α)φ3′(u)=δφ3(u),u≥b,
with the boundary conditions φ3(b)=1, φ3(∞)=0.We have the following theorem.Theorem 15.
Foru∈[0,b], one has
(90)Vd(u;b)=αδφ3′(b)ψ2(u)φ3′(b)-ψ2′(b),
and, for u>b, one has
(91)Vd(u;b)=αδ+αδψ2′(b)φ3(u)φ3′(b)-ψ2′(b).Proof.
Whenu∈[0,b], in view of the strong Markov property, we obtain
(92)Vd(u;b)=Vd(b;b)ψ2(u).
When u>b, since τb- is a stopping time, it follows from the strong Markov property of Ud(u) that
(93)Vd(u,b)=Eu(∫0τb-‍αe-δtdt)+Eu(∫τb-Td‍αe-δtI(Ud(t)>b)dt)=αδ(1-Eue-δτb-)+Eu[e-δτb-EUτb-(∫0Td‍e-δtdDd(t))]=αδ+(Vd(b,b)-αδ)φ3(u).
Using the continuity of the function Vd′(u;b) at u=b, we get
(94)Vd(b;b)ψ2′(b)=(Vd(b,b)-αδ)φ3′(b).
So, we get
(95)Vd(b;b)=αδφ3′(b)φ3′(b)-ψ2′(b).
Substituting the above expression into (92) and (93), we can get the results (90) and (91). This completes the proof.We just consider the square root process.Example 16.
We consider the square root process discussed in Example11. We first solve the differential equation satisfied by φ3(u) as follows
(96)12σ2u2φ3′′(u)+(vk-vu-α)φ3′(u)=δφ3(u),δ>0.
We assume that 2((vk-α)/σ2) is not an integer, the two linear independent solutions are
(97)φ+(u)=M(δv,2(vk-α)σ2,2vσ2u),φ-(u)=U(δv,2(vk-α)σ2,2vσ2u).
Using (81), and from the boundary conditions φ3(b)=1, φ3(∞)=0, we get
(98)φ3(u)=[U(δv,2(vk-α)σ2,2vσ2u)]×[U(δv,2(vk-α)σ2,2vσ2b)]-1.
From (71), we have
(99)φ3′(b)=-2δσ2[U(δv+1,2(vk-α)σ2+1,2vσ2b)]×[U(δv,2(vk-α)σ2,2vσ2b)]-1.
Furthermore, ψ2(u) and ψ2′(b) have been given in Example 11. From Theorem 15, we obtain
(100)Vd(u,b)=αδU(δv+1,2(vk-α)σ2+1,2vσ2b)×[U(δv,2vσ2k,0)M(δv,2vσ2k,2vσ2u)-U(δv,2vσ2k,2vσ2u)]×{CU(δv+1,2(vk-α)σ2+1,2vσ2b)+U(δv,2(vk-α)σ2,2vσ2b)×[σ22vkU(δv,2vσ2k,0)×M(δv+1,2vσ2k+1,2vσ2b)+U(δv+1,2vσ2k+1,2vσ2b)]}-1,u∈[0,b],
and for u>b, we get
(101)Vd(u,b)=αδ-αδU(δv,2(vk-α)σ2,2vσ2u)×[σ22vkU(δv,2vσ2k,0)×M(δv+1,2vσ2k+1,2vσ2b)+U(δv+1,2vσ2k+1,2vσ2b)]×{CU(δv+1,2(vk-α)σ2+1,2vσ2b)+U(δv,2(vk-α)σ2,2vσ2b)×[σ22vkU(δv,2vσ2k,0)×M(δv+1,2vσ2k+1,2vσ2b)+U(δv+1,2vσ2k+1,2vσ2b)]}-1,
where C is defined in Example 11.



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