# On a Dual Model with Barrier Strategy

Authors: Yuzhen Wen; Chuancun Yin
Journal: Journal of Applied Mathematics
 (2012)

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## Abstract

We consider the dual of the generalized Erlang(n) risk model with a barrier dividend strategy. We derive integro-differential equations with boundary conditions satisfied by the expectation of the sum of discounted dividends until ruin and the moment-generating function of the discounted dividend payments until ruin, respectively. The results are illustrated by several examples.


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## Body

## 1.  Introduction

Many interesting results have been obtained on a model that is dual to the classical insurance risk model in recent years. See, for example, Albrecher et al. [1], Ng [2], Avanzi et al. [3], and Avanzi and Gerber [4]. In the classical dual model (see Grandell [5]), the surplus at time t is
(1.1)U(t)=u-ct+∑k=1N(t)‍Zk≡u-ct+S(t),t≥0,
where u and c are constants, u≥0 is the initial surplus and c>0 is the rate of expenses, S(t)=∑k=1N(t)Zk is the aggregate positive gains process, {Zk}k≥1 is a sequence of independent and identically distributed nonnegative random variables with a common probability distribution function P(x), and {N(t)} is a Poisson process with rate λ. Moreover, it is assumed that {N(t)} and {Zk}k≥1 are independent. In (1.1), the expected increment of the surplus per unit time is
(1.2)μ=E[S(1)]-c=λ∫0+∞‍xp(x)dx-c.
It is assumed that μ>0.In the model (1.1), the premium rate is negative, causing the surplus to decrease. Claims, on the other hand, cause the surplus to jump up. Thus the premium rate should be viewed as an expense rate and claims should be viewed as profits or gains. While not very popular in insurance mathematics, this model has appeared in various literature (see Cramér [6], Seal [7], Tákacs [8], and the references cited therein). In Avanzi et al. [3], the authors studied the expected total discounted dividends until ruin for the dual model under the barrier strategy by means of integro-differential equations.Recently, the research to models with two-sided jumps has been attracting a lot of attention in applied probability. For example, Perry et al. [9] studied the one- and two-sided first exit problems for a compound Poisson process with negative and positive jumps and linear deterministic decrease between jumps and assumed that the jumps have hyperexponential distributions. Kou and Wang [10] used a double exponential jump diffusion process to model the asset return. Asmussen et al. [11] considered the stock price models as an exponential Lévy process with phase-type jumps in two directions. For some related work see, among others, Jacobsen [12], Dong and Wang [13], Dong and Wang [14], Cai et al. [15], Zhang et al. [16], Chi and Lin [17], Cai and Kou [18], and the references therein.Motivated by some related work mentioned above, we consider a more general risk process. We will assume that the number of gains up to timet is an ordinary renewal process:
(1.3)N(t)=max⁡{k≥1:W1+W2+⋯+Wk≤t},
where the random variables {Wi}i≥1 are independent and identically generalized Erlang(n)-distributed, that is, the Wi’s are distributed as the sum of n independent and exponentially distributed random variables:
(1.4)Wi=ξ1+ξ2+⋯+ξn,i=1,2,…,
where ξj(j=1,2,…,n) may have different exponential parameters λj>0. We also assumed that the jumps are two-sided. The upward jumps can be interpreted as the random gains of the company, while the downward jumps are interpreted as the random loss of the company. The common density of the jumps is given by
(1.5)p(x)=pp1(x)I{x≥0}+qp2(-x)I{x<0},
where p1(x) and p2(x) are two arbitrary probability density functions on [0,∞) and p,q≥0 are two constants such that p+q=1. Denoted are the probability distribution functions of p1(x) and p2(x), respectively, by P1(x) and P2(x).We then consider the modification of the surplus process by a barrier strategy with a barrierb. Whenever the surplus exceeds the barrier, the excess is paid out immediately as a dividend. But when U(t) is below b, no dividends are paid. The modified surplus at time t is given by
(1.6)X(t)=U(t)-D(t),t≥0,
where D(t) denote the aggregate dividends paid between time 0 and time t, that is,
(1.7)D(t)=(max⁡0≤s≤tU(s)-b)+.
Let
(1.8)T=inf⁡{t≥0:X(t)<0}
be the time of ruin for the modified surplus X, and let
(1.9)D=∫0T‍e-δtdD(t)
be the sum of the discounted dividend payments, where δ>0 is the force of interest for valuation.In this paper, we consider the expectation and the moment-generating function of the sum of the discounted dividends until ruin. In Section2, we derive an integro-differential equation with boundary conditions for the expectation of the discounted dividends until ruin. In Section 3, we obtain an integro-differential equation with boundary conditions for the moment-generating function of the discounted dividend payments until ruin.


## 2. Expectation of the Discounted Dividends

Denote byV(u;b) the expectation of the discounted dividends until ruin if the barrier strategy with parameter b is applied:
(2.1)V(u;b)=E[D∣X(0)=u],0≤u≤b.
Note that
(2.2)V(u;b)=u-b+V(b;b),u>b.Let∂/∂u denote the differentiation operator with respect to u. And we define ∏j=21‍·=1.Theorem 2.1.
The functionV(u;b) satisfies the following integro-differential equation:
(2.3)[∏j=1n‍(cλj∂∂u+1+δλj)]V(u;b)=p∫0b-u‍V(u+x;b)p1(x)dx+p∫b-u∞‍[1-P1(x)]dx+pV(b;b)[1-P1(b-u)]+q∫0u‍V(u-x;b)p2(x)dx,0<u<b
with boundary conditions
(2.4)V(0;b)=0,(2.5)[∏j=2k‍(cλj-1∂∂u+1+δλj-1)]∂V(u;b)∂u|u=b=1,k=1,2,…,n.Proof.
We letVj(u;b) denote the expectation of the discounted dividends if the risk process is in state j(j=1,…,n). Eventually, we are interested in V(u;b)=V1(u;b). Conditioning on the occurrence of a (sub-) claim within an infinitesimal time interval, we obtain for 0≤u<b and j=1,…,n-1,
(2.6)Vj(u;b)=e-δdt{P(ξj>dt)Vj(u-cdt;b)+P(ξj≤dt)Vj+1(u-cdt;b)}.
Note that we have
(2.7)e-δdt=1-δdt+o(dt),P(ξj>dt)=1-λjdt+o(dt),P(ξj≤dt)=λjdt+o(dt),Vj(u-cdt;b)=Vj(u;b)-c∂Vj(u;b)∂udt+o(dt).
Substituting these formulas into (2.6), after some careful calculations, we have for j=1,…,n-1(2.8)λjVj+1(u;b)=[(λj+δ)+c∂∂u]Vj(u;b).
For j=n, we have
(2.9)Vn(u;b)=e-δdt[P(ξn>dt)Vn(u-cdt;b)+P(ξn≤dt)∫-∞∞‍V(u+x;b)p(x)dx]+o(dt)=Vn(u;b)-(δ+λn)Vn(u;b)dt-c∂Vn(u;b)∂udt+pλndt×[∫0b-u‍V(u+x;b)p1(x)dx+∫b-u∞‍(u-b+x)p1(x)dx+V(b;b)(1-P1(b-u))]+qλndt∫-u0‍V(u+x;b)p2(-x)dx+o(dt)=Vn(u;b)-(δ+λn)Vn(u;b)dt-c∂Vn(u;b)∂udt+pλndt[∫0b-u‍V(u+x;b)p1(x)dx+∫b-u∞‍[1-P1(x)]dx+V(b;b)(1-P1(b-u))]+qλndt∫0u‍V(u-x;b)p2(x)dx+o(dt),
which leads to
(2.10)[(λn+δ)+c∂∂u]Vn(u;b)=λnp∫0b-u‍V(u+x;b)p1(x)dx+λnp∫b-u∞‍[1-P1(x)]dx+λnpV(b;b)[1-P1(b-u)]+λnq∫0u‍V(u-x;b)p2(x)dx.
It follows from (2.8) that
(2.11)Vj+1(u;b)=[(λj+δ)+c(∂/∂u)λj]Vj(u;b),(j=1,2,…,n-1)
and subsequently
(2.12)Vn(u;b)=[∏j=1n-1‍(λj+δ)+c(∂/∂u)λj]V1(u;b),
which together with (2.10) yields (2.3).
Since the ruin is immediate ifu=0, we have the boundary condition (2.4).
Foru=b, we obtain analogously for j=1,2,…,n-1 that
(2.13)Vj(b;b)=e-δdt{P(ξj>dt)Vj(b;b)-cdt+P(ξj≤dt)Vj+1(b;b)},
which by Taylor expansion leads to
(2.14)λjVj+1(b;b)=(λj+δ)Vj(b;b)+c.
Comparing these equations with the corresponding ones in (2.8), the continuity of Vj(u;b) at u=b then implies that
(2.15)∂Vj(u;b)∂u|u=b=1,j=1,2,…,n-1.
Similarly, one can verify that (2.15) also holds for j=n. For j=1, (2.15) is equivalent to (2.5) with k=1. It follows from (2.11) that
(2.16)Vk(u;b)=[∏j=2k‍(λj-1+δ)+c(∂/∂u)λj-1]V(u;b),k=2,3,…,n.
Applying the operator ∂/∂u to both sides of the above equation, we get
(2.17)[∏j=2k‍(λj-1+δ)+c(∂/∂u)λj-1]∂V(u;b)∂u|u=b=∂Vk(u;b)∂u|u=b=1,k=2,3,…,n,
which is the boundary condition (2.5).Remark 2.2.
When the gains waiting timeWi have an exponential distribution with k(y)=λe-λy for λ>0, y≥0, q=0, p=1, and p(x)=p1(x), we can get the integro-differential equation of V(u;b):
(2.18)cV′(u;b)+(λ+δ)V(u;b)=λ∫0b-u‍V(u+x;b)p(x)dx+λ∫b-u∞‍[1-P(x)]dx+λV(b;b)[1-P(b-u)].
This is the result (2.3) in Avanzi et al. [3].Example 2.3.
Forn=1, λ1:=λ, let p1(x)=β1e-β1x, p2(x)=β2e-β2x,x>0, where β1 and β2 are two positive constants. Then
(2.19)p(x)=pβ1e-β1xI{x≥0}+qβ2eβ2xI{x<0}.
From (2.3), we have
(2.20)[cλ∂∂u+(1+δλ)]V(u;b)=pβ1eβ1u∫ubV(x;b)e-β1xdx+pβ1e-β1(b-u)+pV(b;b)e-β1(b-u)+qβ2e-β2u∫0uV(x;b)eβ2xdx.
Applying the operator ((∂/∂u)-β1)((∂/∂u)+β2) to both sides of (2.20), we get
(2.21)(∂∂u-β1)(∂∂u+β2)[cλ∂∂u+(1+δλ)]V(u;b)=(qβ2-pβ1)∂V(u;b)∂u-(p+q)β1β2V(u;b).
The characteristic equation of (2.21) is
(2.22)(r-β1)(r+β2)[cλr+(1+δλ)]=(qβ2-pβ1)r-(p+q)β1β2.
That is,
(2.23)cλr+(1+δλ)=pβ1β1-r+qβ2β2+r.
The expression on the left-hand side is a linear function of r, while the expression on the right-hand side is a rational function with poles at r=β1,-β2. By a graphical arguments, it can be verified that the characteristic equation above has exactly three real roots r1,r2,r3 satisfying
(2.24)-∞<r1<-β2<r2<0<r3<β1<∞.
Hence, we set
(2.25)V(u;b)=c1er1u+c2er2u+c3er3u,
where c1=c1(b), c2=c2(b), and c3=c3(b) are constants need to be determined. It follows from (2.4) and (2.5) that we have
(2.26)c1+c2+c3=0,c1r1er1b+c2r2er2b+c3r3er3b=1.
Substituting (2.25) into (2.20), and since this equation must be satisfied for u=b, we have
(2.27)c1[(cλr1+1+δλ-p)er1b-qβ2er1b-e-β2br1+β2]+c2[(cλr2+1+δλ-p)er2b-qβ2er2b-e-β2br2+β2]+c3[(cλr3+1+δλ-p)er3b-qβ2er3b-e-β2br3+β2]-pβ1=0,
which can be rewritten as
(2.28)Δ1c1+Δ2c2+Δ3c3=pβ1,
where
(2.29)Δ1=pr1β1-r1er1b+qβ2β2+r1e-β2b,Δ2=pr2β1-r2er2b+qβ2β2+r2e-β2b,Δ3=pr3β1-r3er3b+qβ2β2+r3e-β2b.
Solving system (2.26) and (2.28) gives c1=A1/A,c2=A2/A,andc3=A3/A, where
(2.30)A=pr1r2(r2-r1)(β1-r1)(β1-r2)e(r1+r2)b+pr1r3(r1-r3)(β1-r1)(β1-r3)e(r1+r3)b+pr2r3(r3-r2)(β1-r2)(β1-r3)e(r2+r3)b+qβ2(r3er3b-r2er2br1+β2+r1er1b-r3er3br2+β2+r2er2b-r1er1br3+β2)e-β2b,A1=pr22β1(β1-r2)er2b-pr32β1(β1-r3)er3b+qβ2(1r2+β2-1r3+β2)e-β2b,A2=-pr12β1(β1-r1)er1b+pr32β1(β1-r3)er3b+qβ2(1r3+β2-1r1+β2)e-β2b,A3=pr12β1(β1-r1)er1b-pr22β1(β1-r2)er2b+qβ2(1r1+β2-1r2+β2)e-β2b.Example 2.4.
Forn=2,q=0,p=1, and p(x)=p1(x)=βe-βx, β>0, x≥0, we have
(2.31)[∏j=12‍(cλj∂∂u+(1+δλj))]V(u;b)=βeβu∫ub‍V(x;b)e-βxdx+1βe-β(b-u)+V(b;b)e-β(b-u).
Applying the operator (∂/∂u)-β to both sides of (2.31), we get
(2.32)(∂∂u-β)∏j=12‍(cλj∂∂u+(1+δλj))V(u;b)=-βV(u;b),
from which we get the characteristic equation
(2.33)(cr+λ1+δ)(cr+λ2+δ)λ1λ2=ββ-r.
By a graphical arguments, it can be verified that the characteristic equation above has exactly three real roots  r1,r2,r3 satisfying
(2.34)r1<r2<0<r3<β.
Hence we get(2.35)V(u,b)=c1er1u+c2er2u+c3er3u,
where c1=c1(b), c2=c2(b), and c3=c3(b) are constants. It follows from (2.4) and (2.5) that
(2.36)c1+c2+c3=0,(2.37)c1r1er1b+c2r2er2b+c3r3er3b=1.
Substituting (2.35) into (2.31), and because this equation must be satisfied for all 0≤u≤b, the sum of the coefficients of e-β(u-b) must be zero. Therefore,
(2.38)c1r1r1-βer1b+c2r2r2-βer2b+c3r3r3-βer3b+1β=0.
It follows from (2.36)–(2.38) that
(2.39)c1=B1B,c2=B2B,c3=B3B,
where
(2.40)B=β(r3-β)(r1-r2)r1r2e(r1+r2)b+β(r2-β)(r3-r1)r1r3e(r1+r3)b+β(r1-β)(r2-r3)r2r3e(r2+r3)b,B1=(r1-β)(r3-β)r22er2b-(r1-β)(r2-β)r32er3b,B2=(r1-β)(r2-β)r32er3b-(r2-β)(r3-β)r12er1b,B3=(r2-β)(r3-β)r12er1b-(r1-β)(r3-β)r22er2b.


## 3. Moment-Generating Function of the Discounted Dividends

We denote moment-generating function ofD by
(3.1)M(u,y,b)=E[eyD∣X(0)=u].
Let ∂/∂y denote the differentiation operator with respect to y and correspondingly ∂/∂u the differentiation operator with respect to u.Theorem 3.1.
The moment-generating functionM(u,y,b)(0<u<b) satisfies the following integro-differential equation:
(3.2)[∏j=1n‍(δyλj∂∂y+cλj∂∂u+1)]M(u,y,b)=p∫0∞M(u+x,y,b)p1(x)dx+q∫0uM(u-x,y,b)p2(x)dx+q[1-P2(u)],
with boundary conditions
(3.3)[∏j=2k(δy∂∂y+c∂∂u+λj-1)‍]∂M(u,y,b)∂u|u=b=y[∏j=2k‍(δy∂∂y+c∂∂u+λj-1)]M(u,y,b)|u=b,k=1,2,…,n.Proof.
As in Albrecher et al. [19], let Mj(u,y,b) denote the moment-generating function of D if the risk process is in state j(j=1,…,n-1). Eventually, we are interested in M(u,y,b):=M1(u,y,b). Conditioning on the occurrence of a (sub-)claim within an infinitesimal time interval, we obtain for 0≤u<b and   j=1,…,n-1,
(3.4)Mj(u,y,b)=P(ξj>dt)Mj(u-cdt,ye-δdt,b)+P(ξj≤dt)Mj+1(u-cdt,ye-δdt,b).
It follows from (3.4) that
(3.5)λjMj+1(u,y,b)=[λj+δy∂∂y+c∂∂u]Mj(u,y,b).
For j=n, we have
(3.6)[δy∂∂y+c∂∂u+λn]Mn(u,y,b)=pλn∫0∞‍M1(u+x,y,b)p1(x)dx+qλn∫-u0‍M1(u+x,y,b)p2(-x)dx+qλn[1-P2(u)].
It follows from (3.5) that we have
(3.7)(∏j=1n-1‍λj)Mn(u,y,b)=[∏j=1n-1‍(δy∂∂y+c∂∂u+λj)]M1(u,y,b),
which together with (3.6) yields (3.2).
Foru=b, we obtain analogously for j=1,2,…,n-1(3.8)Mj(b,y,b)=(1-λjdt)e-ycdtMj(b,ye-δdt,b)+λjdte-ycdtMj+1(b,ye-δdt,b)+o(dt)
which leads to
(3.9)λjMj+1(b,y,b)=(λj+yc)Mj(b,y,b)+yδ∂Mj∂y(b,y,b).
Comparing these equations with the corresponding equations in (3.5), the continuity of Mj(u,y,b) at u=b implies
(3.10)∂Mj(u,y,b)∂u|u=b=yMj(b,y,b),j=1,2,…,n-1.
Similarly, we can show that (3.10) holds true for j=n. For j=1, (3.10) is equivalent to (3.3) for k=1. Now it just remains to express equations (3.10) for j=2,…,n in terms of M1=M, which is done by virtue of (3.9).Form∈ℕ, we denote the mth moment of D by
(3.11)Wm(u;b)=E[Dm∣X(0)=u].Theorem 3.2 :.
Themth moment Wm(u;b)(0<u<b) satisfies the following integro-differential equation
(3.12)[∏j=1n‍(δmλj+cλj∂∂u+1)]Wm(u;b)=p∫0b-u‍Wm(u+x;b)p1(x)dx+p∑i=0m‍(mi)Wi(b;b)∫b-u∞‍(u-b+x)m-ip1(x)dx+q∫0u‍Wm(u-x;b)p2(x)dx
with boundary conditions
(3.13)[∏j=2k‍(δm+c∂∂u+λj-1)]∂Wm(u;b)∂u|u=b=m[∏j=2k‍(δ(m-1)+c∂∂u+λj-1)]Wm-1(u;b)|u=b,k=1,2,…,n.Proof.
Since(3.14)M(u,y,b)=1+∑m=1∞‍ymm!Wm(u,b),
the result follows if we equate the coefficients of ym(m=0,1,2,…) in (3.2) and (3.3).Remark 3.3.
We remark that whenm=1, W0(u;b)=1, and W1(u;b)=V(u;b), we reobtained the result of Theorem 2.1; when n=1, p=0, and q=1, (3.12) reduces to (2.3) of Cheung and Drekic [20]; when p=0, q=1, (3.2) and (3.3) reduce to (2) and (3) of Albrecher et al. [19], and (3.12) and (3.13) reduce to (9) and (10) of Albrecher et al. [19].



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