# The Ornstein-Uhlenbeck-Type Model with a Hybrid Dividend Strategy

Authors: Dan Zhu; Chuancun Yin
Journal: Journal of Applied Mathematics
 (2013)

---

## Abstract

We consider the Ornstein-Uhlenbeck-type model. We first introduce the model and then find the ordinary differential equations and boundary conditions satisfied by the dividend functions; closed-form solutions for the dividend value functions are given. We also study the distribution of the time value of ruin. Furthermore, the moments and moment-generating functions of total discounted dividends until ruin are discussed.


---

## Body

## 1. Introduction

In recent years, the dividend problem has gained a lot of attention in the actuarial literature. Dividend strategies for insurance risk model were first proposed by de Finetti [1], who considered a discrete time random walk with step size ±1 and found that the optimal dividend strategy must be a barrier strategy. From then on, the problem of optimal dividend strategy has been studied in continuous time, for example, Asmussen and Taksar [2], Albrecher et al. [3], Gao and Yin [4], Gerber and Shiu [5, 6], Wan [7] and so on. The optimal dividend problem in a compound Poisson model was studied by Gerber and Shiu [8]. Optimal dividend in an Ornstein-Uhlenbeck-type model with credit and debit interest was considered in Cai et al. [9]. For a class of compound Poisson process perturbed by diffusion with a threshold dividend strategy, the expected discounted penalty function has been studied by Wan [7]. The perturbed Sparre Andersen model with a threshold dividend strategy was settled by Gao and Yin [10].Recently, the multilayer dividend strategy as an extension of the threshold dividend strategy has drawn many authors attention. For example, the perturbed Sparre Andersen and compound Poisson risk models with multilayer dividend strategy have been studied by Yang and Zhang [11, 12]. The integrodifferential equations for the expected discounted penalty function were derived and solved; when the claims are subexponentially distributed, the asymptotic formula for ruin probability is obtained. The Ornstein-Uhlenbeck-type model is a very important model in applied probability which has recently gained a lot of attention. See, for example, Cai et al. [9] and Fang and Wu [13]. More general, Wong and Zhao [14] consider the optimal dividends and bankruptcy in an Ornstein-Uhlenbeck process with the surplus-dependent credit/debit interest rate. Motivated by the above work, in this paper, we consider a hybrid dividend strategy which combined a barrier strategy with a threshold strategy in an Ornstein-Uhlenbeck-type model. For simplicity, we consider only one threshold and one barrier.The remainder of the paper is organized as follows. In Section2, we describe the model and discuss the dividend functions until ruin, in Section 3, we give the limit of dividends level, and in Section 4 we get the expression by Laplace transform of ruin time. The partial differential equation with boundary conditions satisfied by the moments and moment-generating function is proved in Section 5.


## 2. The Model

Consider the following surplus process:(1)dUt=(μ+ρUt)dt+σdWt,t≥0,
where ρ>0 is the force of interest, μ>0 is the drift coefficient, σ>0 is the diffusion coefficient, and {Wt}t≥0 is the standard Brownian motion. We will assume that the company pays dividends according to the following strategy governed by parameters b2>b1>0 and α>0. Whenever the modified surplus is below the level b1, no dividends are paid. However, when the modified surplus is above b1 and below the b2, dividends are paid continuously at a constant rate α. When the modified surplus is above b2, dividends are completely paid. For t≥0, let
(2)D(t)=D1(t)+D2(t)
denote the aggregate dividend paid by time t, where D1(t) and D2(t) are caused by the different parts of dividends, respectively. Thus,
(3)Ut~=Ut-D(t)
is the modified surplus at time t. Let δ>0 be the force of interest for valuation; in this paper, we assume that ρ<δ. Let I(A) be the indicator function of event A and let D denote the present value of all dividends until ruin
(4)D=∫0T‍e-δtdD1(t)+∫0T‍e-δtdD2(t),
where
(5)T=inf{t≥0:Ut~≤0},
is the time of ruin, and
(6)D1(t)=α∫0t‍I(b1≤Us~≤b2)ds,D2(t)=η-b2,η=max0≤s≤t{Us-D1(s)}.Forx≥0, we use the symbol V(x;b1,b2) to denote the expectation of D. That is,
(7)V(x;b1,b2)=E[D∣U0~=x]=Ex[D].Define the random times(8)τ0-=inf(t≥0:Ut≤0),τb1+=inf(t≥0:Ut≥b1),τb1-=inf(t≥0:Ut~≤b1),τb2+=inf(t≥0:Ut~≥b2),τ=inf(t≥0:Ut~=b1orUt~=b2),with the conventioninf∅=∞.Lemma 1.
Assume thatV(x) is twice continuously differentiable on (0,b1)∪(b1,b2)∪(b2,∞). For 0<x≤b1, V(x) satisfies the following ordinary differential equation:
(9)12σ2V′′(x)+(μ+ρx)V′(x)-δV(x)=0;
for b1<x≤b2, V(x) satisfies the following ordinary differential equation:
(10)12σ2V′′(x)+(μ+ρx-α)V′(x)-δV(x)=0;
for x>b2, V(x) satisfies the following equation:
(11)V(x)=x-b2+V(b2),
with boundary conditions
(12)V(0;b1,b2)=0,V′(b2;b1,b2)=1.Proof.
By virtually the same arguments as in Yin and Wen [15], we can prove (9) and (10). The boundary conditions can be derived the same as in Gerber and Shiu [6] or Cai et al. [9].
The ordinary differential equation (9) has two positive independent solutions f1, f2 such that f1 is strictly decreasing and f2 is strictly increasing (see e.g., [16]). Let f3, f4 be such solution for the ordinary differential equation (10), where f3 is strictly decreasing and f4 is strictly increasing. In Cai et al. [9], the authors pointed out that these independent solutions are given by
(13)f1(x)=exp{-1ρσ2(μ+ρx)2}×U(12+δ2ρ,12,1ρσ2(μ+ρx)2),(14)f2(x)=(μ+ρx)exp{-1ρσ2(μ+ρx)2}×M(1+δ2ρ,32,1ρσ2(μ+ρx)2),(15)f3(x)=exp{-1ρσ2(μ-α+ρx)2}×U(12+δ2ρ,12,1ρσ2(μ-α+ρx)2),(16)f4(x)=(μ-α+ρx)exp{-1ρσ2(μ-α+ρx)2}×M(1+δ2ρ,32,1ρσ2(μ-α+ρx)2),
where M and U are called the confluent hypergeometric functions of the first and second kind, respectively.
Denote(17)h(x)=f1(0)f2(x)-f2(0)f1(x),g(x)=f3(b2)f4(x)-f4(b2)f3(x),k(x)=f3(x)f4(b1)-f3(b1)f4(x).
The expressions of the expected discounted dividend payments are given by Theorem2.Theorem 2.
(i) For0≤x≤b1,
(18)V(x;b1,b2)=(+k′(b1)g′(b1)h(b1)])-1(-αg′(b1)k′(b2)+k′(b1)g(b1)δ+αk′(b1)g′(b1))×(+k′(b1)g′(b1)h(b1)])-1δ[h′(b1)g(b1)k′(b2)-g′(b1)h(b1)k′(b2)+k′(b1)g′(b1)h(b1)]+k′(b1)g′(b1)h(b1)])-1)-1)h(x).
(ii) Forb1<x≤b2,
(19)V(x;b1,b2)=αδ+k(x)k′(b2)+([(g(x)k′(b2)-g′(b2)k(x))][δh(b1)k′(b1)-αh′(b1)k′(b2)]×[(g(x)k′(b2)-g′(b2)k(x))])×(k′(b2)δ[h′(b1)g(b1)k′(b2)-g′(b1)h(b1)k′(b2)+k′(b1)g′(b1)h(b1)])-1.Proof.
When0<x≤b1, by the strong Markov property of the process Ut, we have
(20)Ex[D]=Ex[DI(τb1+<τ0-)]+Ex[DI(τb1+>τ0-)]=Ex[DI(τb1+<τ0-)]=Ex[e-δτb1+I(τb1+<τ0-)]V(b1;b1,b2).
For0<x≤b1, using Itô's formula or Dynkin's formula as in Li et al. [17], we find that f(x):=Ex[e-δτb1+I(τb1+<τ0-)] satisfies the ordinary differential equation (9), with boundary conditions f(b1)=1,  f(0)=0. Assume that the solution of the equation is f(x)=c1f1(x)+c2f2(x); from the boundary conditions, we obtain
(21)c1=-f2(0)f1(0)f2(b1)-f2(0)f1(b1),c2=f1(0)f1(0)f2(b1)-f2(0)f1(b1).
So we have(22)Ex[e-δτb1+I(τb1+<τ0-)]=h(x)h(b1).
Similarly, whenb1<x≤b2,
(23)Ex[D]=Ex[∫0τ‍αe-δtdt]+Ex[e-δt]V(U~τ;b1,b2)=Ex[(-αδ)(e-δτ-1)]+Ex[e-δτI(τb1-<τb2+)]V(b1;b1,b2)+Ex[e-δτI(τb2+<τb1-)]V(b2;b1,b2).
Forb1<x≤b2, using Itô's formula or Dynkin's formula, we have that f(x):=Ex[e-δτb1-I(τb1-<τb2+)] satisfies the ordinary differential equation (10) with boundary conditions f(b1)=1,  f(b2)=0. Assume that the solution of the equation is f(x)=c3f3(x)+c4f4(x); from the boundary conditions, we obtain
(24)c3=-f4(b2)f3(b2)f4(b1)-f4(b2)f3(b1),c4=f3(b2)f3(b2)f4(b1)-f4(b2)f3(b1),
so we have
(25)Ex[e-δτb1-I(τb1-<τb2+)]=g(x)g(b1).
Similarly,f(x):=Ex[e-δτb2+I(τb2+<τb1-)] satisfies (10) with boundary conditions f(b1)=0,  f(b2)=1, so we get
(26)Ex[e-δτb2+I(τb2+<τb1-)]=f3(x)f4(b1)-f3(b1)f4(x)g(b1)=k(x)g(b1),
where
(27)g(b1)=k(b2).
For0<x≤b1, dividends will be payable if the surplus process without ruin and reaches b1, so that
(28)V(x;b1,b2)=Ex[e-δτb1+I(τb1+<τ0-)]V(b1;b1,b2)=h(x)h(b1)V(b1;b1,b2).
Forb1<x≤b2, dividends are paid continuously at a constant rate α, so we obtain
(29)V(x;b1,b2)=αδ+Ex[e-δτI(τb1-<τb2+)](V(b1;b1,b2)-αδ)+Ex[e-δτI(τb2+<τb1-)](V(b2;b1,b2)-αδ)=αδ+g(x)g(b1)(V(b1;b1,b2)-αδ)+k(x)g(b1)(V(b2;b1,b2)-αδ).
UsingV′(b2;b1,b2)=1 and V′(b1+;b1,b2)=V′(b1-;b1,b2), we get
(30)g′(b2)g(b1)(V(b1;b1,b2)-αδ)+k′(b2)g(b1)(V(b2;b1,b2)-αδ)=1,h′(b1)h(b1)V(b1;b1,b2)=g′(b1)g(b1)(V(b1;b1,b2)-αδ)+k′(b1)g(b1)(V(b2;b1,b2)-αδ).
With some careful calculations, we obtainV(b1;b1,b2) and V(b2;b1,b2), so we get the results (18) and (19).


## 3. The Special Dividends Strategy

In this section, we consider the limit of dividends level. Letb2→∞; by the expressions of k and g, we have
(31)g(b1)~-f3(b1)f4(b2),g′(b1)~-f′(b1)f4(b2),k′(b2)~-f3(b1)f4′(b2),
where
(32)f4(b2)⟶∞,f4′(b2)⟶∞.Substituting the above expressions into (18) and (19), and setting
(33)A=-αf3′(b1)f4(b2)f3(b1)f4′(b2),B=δk′(b1)f3(b1)f4(b2),C=αk′(b1)f3′(b1)f4(b2),D=h′(b1)f3(b1)f4′(b2)f3(b1)f4(b2),E=f3′(b1)f4(b2)f3(b1)f4′(b2),F=f3′(b1)f4(b2)k′(b1)h(b1),
we obtain, for 0<x≤b1,
(34)V(x;b1,b2)=A-B-Cδ[D-E-F]h(x).Let(35)A′=δh(b1)k′(b1)+αh′(b1)f3(b1)f4′(b2),B′=-(g(x)f3(b1)f4′(b2)-g′(b2)k(x)),C′=h′(b1)f32(b1)f4(b2)f4′(b2),D′=k′(b1)f4(b2)h(b1)f3(b1)f4′(b2),E′=k′2(b1)f4(b2)h(b1);
we obtain, for b1<x≤∞,
(36)V(x;b1,b2)=αδ+k(x)-f3(b1)f4′(b2)-A′B′f3(b1)f4′(b2)δ[C′-D′-E′].Then dividing numerator and denominator of (34) and (36) by f4(b2)f4′(b2), we get the expected discounted dividend payments for the threshold strategy
(37)V(x;b)={αδf3′(b)h(x)f3′(b)h(b)-f3(b)h′(b),if0<x≤b1,αδ+αδf3(x)h′(b)f3′(b)h(b)-f3(b)h′(b),ifx>b2,
which is (15) in Fang and Wu [13].Remark 3.
Similarly, whenb1=b2=b, we get the expected discounted dividend payments for the barrier strategy
(38)V(x;b)={h(x)h′(b),if0<x≤b,x-b+h(b)h′(b),ifx>b,
which is (9) and (10) in Cai et al. [9].


## 4. The Time Value of Ruin under a Hybrid Dividend Strategy

In this section, we focus on the Laplace transform of the time value of ruin. We assume that dividends are paid according to threshold strategy with parametersb1, α and barrier strategy with parameter b2. Let L(x;b1,b2) denote the Laplace transform of the time value of ruin; for x≥0,
(39)L(x;b1,b2)=E[e-δT∣U0=x].Let(40)ρ(x)=f1(x)f2(b1)-f2(x)f1(b1).Theorem 4.
For0<x≤b1, one has
(41)L(x;b1,b2)=h(x)h(b1)A+ρ(x)h(b1),
and, for b1<x≤b2, one has
(42)L(x;b1,b2)=g(x)g(b1)A+k(x)g(b1)B,
where
(43)A=(k′(b1)h(b1)g(b1)-ρ′(b1)k′(b2)g(b1))×(g′(b2)k′(b1)h(b1)+g(b1)h′(b1)k′(b2)-g′(b1)h(b1)k′(b2))-1,B=(g2(b1)h′(b1)-g(b1)g′(b1)h(b1)+g(b1)g′(b2)ρ′(b1))×(g′(b2)k′(b1)h(b1)+g(b1)h′(b1)k′(b2)-g′(b1)h(b1)k′(b2))-1.Proof.
For0<x≤b1, applying the strong Markov property, we obtain
(44)L(x;b1,b2)=Ex[e-δT]=Ex[e-δTI(τb1+<τ0-)]+Ex[e-δTI(τb1+>τ0-)]=Ex[e-δτb1+I(τb1+<τ0-)]L(b1;b1,b2)+Ex[e-δτ0-I(τb1+>τ0-)]=h(x)h(b1)L(b1;b1,b2)+ρ(x)h(b1).
Similarly, forb1<x≤b2,
(45)L(x;b1,b2)=Ex[e-δT]=Ex[e-δτ]L(x(τ);b1,b2)=Ex[e-δτI(τb1-<τb2+)]L(b1;b1,b2)+Ex[e-δτI(τb1->τb2+)]L(b2;b1,b2)=g(x)g(b1)L(b1;b1,b2)+k(x)g(b1)L(b2;b1,b2).
It can be verified thatL′(b2;b1,b2)=1 and L′(b1+;b1,b2)=L′(b1-;b1,b2), so we get
(46)g′(b2)g(b1)L(b1;b1,b2)+k′(b2)g(b1)L(b2;b1,b2)=1,h′(b1)h(b1)L(b1;b1,b2)+ρ′(x)h(b1)=g′(b1)g(b1)L(b1;b1,b2)+k′(b1)g(b1)L(b2;b1,b2).
From (46) we obtain L(b1;b1,b2) and L(b2;b1,b2), so we get the results (41) and (42).Remark 5.
(1) Letϕ(x;b1,b2) denote the expected discounted penalty at ruin; in this model, the penalty at ruin is a constant M>0, so we get
(47)ϕ(x;b1,b2)=ML(x;b1,b2).
Substitution of (41) and (42) into (47) yields an expression for ϕ(x;b1,b2).
(2) Letψ(x;b1,b2)=P(T<∞∣U0=x) be the probability of ruin. Note that
(48)E[e-δT∣U0=x]=Ex[e-δTI(T<∞)]+Ex[e-δTI(T=∞)]=Ex[e-δTI(T<∞)];
then
(49)ψ(x;b1,b2)=limδ→0L(x;b1,b2)=P(T<∞).
From (14)–(18), we have
(50)limδ→0f1(x)=exp{-1ρσ2(μ+ρx)2}×U(12,12,1ρσ2(μ+ρx)2),limδ→0f2(x)=(μ+ρx)exp{-1ρσ2(μ+ρx)2}×M(1,32,1ρσ2(μ+ρx)2),limδ→0f3(x)=exp{-1ρσ2(μ-α+ρx)2}×U(12,12,1ρσ2(μ-α+ρx)2),limδ→0f4(x)=(μ-α+ρx)exp{-1ρσ2(μ-α+ρx)2}×M(1,32,1ρσ2(μ-α+ρx)2).
Taking limit in (41) and (42) yields ψ(x;b1,b2).


## 5. The Moment-Generating Function ofD

In this section, the moment-generating function of the hybrid dividend payments is discussed. We adopt a similar approach to that of Gao and Yin in Section3 [10].Let(51)M(x,y;b1,b2)=E[eyD∣U0=x]
denote the moment-generating function of D, and let
(52)Vm(x;b1,b2)=E[DmU0=x],m=1,2,3,…,
denote the mth moment function. The following theorem provides differential equations for the function M(x,y;b1,b2).Theorem 6.
The moment-generating functionM(x,y;b1,b2) satisfies the partial differential equations
(53)σ22∂2M∂x2+∂M∂x(ρx+μ)-∂M∂yyδ=0,0<x≤b1,σ22∂2M∂x2+∂M∂x(ρx+μ-α)-∂M∂yyδ=0,b1<x≤b2,
with boundary conditions
(54)M(0,y;b1,b2)=1,limx→∞M(x,y;b1,b2)=eyα/δ.Proof.
(i) We first provide the solution ofUt. Consider the SDE
(55)dUt=(μ+ρUt)dt+σdWt;
note that, in the case σ=0, the solution to the ODE is (1/ρ)(ρU0+μ)eρt-μ/ρ. To solve the SDE, consider the process
(56)Y(t)=1ρ(ρU0+μ)e-ρt-μρ.
LetP(t)=Ute-ρt and Q(t)=(μ/ρ)(e-ρt-1), so we have dY(t)=dP(t)+dQ(t),
(57)dP(t)=dUte-ρt=Ut(-ρ)e-ρtdt+e-ρtdUt+d[Ut,e-ρt]=Ut(-ρ)e-ρtdt+e-ρtdUt=Ut(-ρ)e-ρtdt+e-ρt(μ+ρUt)dt+σdWt=e-ρtμdt+e-ρtσdWt,dQ(t)=-e-ρtμdt;
we get dY(t)=e-ρtσdWt, so we obtain
(58)Y(t)=Y(0)+∫0t‍e-ρsσdWs.
Now the solution forUt is
(59)Ut=eρt{U0-μρ(e-ρt-1)+∫0t‍e-ρsσdWs}.
(ii) Now we derive the integrodifferential equations forM(x,y;b1,b2).
For0<x≤b1, we consider a small time interval [0,t], the time t>0 such that the surplus will not reach b1 before t; in view of the strong Markov property of the surplus process {Ut,t≥0}, we have
(60)M(x,y;b1,b2)=E[M(Ut,ye-δt;b1,b2)]+o(t).
By Taylor expansion, we have(61)E[M(Ut,ye-δt;b1,b2)]=E[M(x,y;b1,b2)]+∂M∂x(Ut-x)+∂M∂y(ye-δt-y)+12(∂2M∂x2(Ut-x)2∂2M∂y2(ye-δt-y)2)+∂M∂x∂y(Ut-x)(ye-δt-y)+o(t).
SubtractingM(x,y;b1,b2) from each side of the above equation, dividing by t, and then letting t→0, we achieve
(62)σ22∂2M∂x2+∂M∂xρ(x+μρ)-∂M∂yyδ=0.
Similarly, forb1<x≤b2, we consider a small time interval [0,t], the time t>0 such that the surplus will not reach b2; we have
(63)M(x,y;b1,b2)=E[M(Ut-αt,ye-δt;b1,b2)]+o(t).
By Taylor expansion, we get(64)E[M(Ut-αt,ye-δt;b1,b2)]=E[M(x,y;b1,b2)]+∂M∂x(Ut-x-αt)+∂M∂y(ye-δt-y)+12(∂2M∂x2(Ut-x-αt)2∂2M∂y2(ye-δt-y)2)+∂M∂x∂y(Ut-x-αt)(ye-δt-y)+o(t).
SubtractingM(x,y;b1,b2) from each side of (63), dividing by t, and then letting t→0, we achieve
(65)σ22∂2M∂x2+∂M∂x(ρx+μ-α)-∂M∂yyδ=0.
The proof of boundary conditions is routine. This ends the proof of Theorem6.Theorem 7.
For0<x≤b1, Vm(x;b1,b2) satisfies
(66)σ22Vm′′(x;b1,b2)+(ρx+ρ)Vm′(x;b1,b2)-δmVm(x;b1,b2)=0,
and, for b1<x≤b2, Vm(x;b1,b2) satisfies
(67)σ22Vm′′(x;b1,b2)+(ρx+ρ-α)Vm′(x;b1,b2)-δmVm(x;b1,b2)=0,
with the boundary conditions
(68)Vm(0;b1,b2)=0,Vm′(x;b1,b2)|x=b2=mVm-1(b2;b1,b2).Proof.
Recall thatM(x,y;b1,b2)=E[eyD], and Vm(x;b1,b2)=E[Dm]; using the representation
(69)M(x,y;b1,b2)=1+∑m=1∞‍ymm!Vm(x;b1,b2)
and equating the coefficients of ym in (53) yield the ordinary differential equations (66) and (67).Remark 8.
Whenm=1, we get V1(x;b1,b2)=V(x;b1,b2).



---

*Source: 467948-2013-12-26.xml*