When does surplus reach a given target before ruin in the Markov-modulated diffusion model?

Journal of the Korean Statistical Society 39 (2010) 207–219

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When does surplus reach a given target before ruin in the Markov-modulated diffusion model? Hu Yang, Zhimin Zhang ∗ Department of Statistics and Actuarial Science, Chongqing University, People’s Republic of China

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Article history: Received 19 April 2009 Accepted 7 July 2009 Available online 17 July 2009

In this paper, we consider a Markov-modulated diffusion risk model. We study when the surplus reaches a given level b (≥U (0)) before ruin. We show that the Laplace transform of such random time can be expressed via the expected discounted penalty functions. Finally, we modify the Markov-modulated diffusion model by a barrier dividend strategy, and give some expressions for the expected discounted penalty functions and the expected discounted dividend payments. Crown Copyright © 2009 Published by Elsevier B.V. on behalf of The Korean Statistical Society. All rights reserved.

AMS 2000 subject classifications: primary 91B30 secondary 91B70 Keywords: Markov-modulated diffusion risk model Expected discounted penalty function Expected discounted dividend payments Martingale

1. Introduction Asmussen (1989) proposed the Markov-modulated risk model, in which a continuous time homogeneous Markov process

{J (t ), t ≥ 0} is used to regulate the claim arrival intensities and the claim amount distributions. The Markov-modulated risk model, also known as Markovian regime switching model in the finance and actuarial science literature, can be seen as a generalization of the classical compound Poisson model. As pointed out by Asmussen (1989), in health insurance, sojourns of {J (t ), t ≥ 0} could be certain types of epidemics, or in automobile insurance, these could be weather types. Because of its flexibility in modeling the risk process, the study of the Markov-modulated risk model has drawn more and more attention, see for example, Asmussen (1989, 2000), Bäuerle (1996), Bäuerle and Kötter (2007) Lu and Li (2005), Li and Lu (2008), Lu and Tsai (2007) and the references therein. Suppose that the environment process {J (t ), t ≥ 0} is a homogeneous, irreducible and recurrent Pm Markov process with finite state space E = {1, 2, . . . , m}. Denote the intensity matrix of J (t ) by 3 = (αij )m with i,j=1 j=1 αij = 0 and αii := −αi for i ∈ E. Let π = (π1 , . . . , πm ) be the stationary distribution of J (t ). Let {N (t ), t ≥ 0} be the claim number process, {Xn , n = 1, 2, . . .} be a sequence of positive r.v.’s denoting the claim amounts, and {W (t ), t ≥ 0} be a standard Brownian motion starting from zero. We assume that the premium rate, the claim arrival intensity, the claim amount distribution and the diffusion volatility are all regulated by the environment process J (t ). If J (s) = i for all s in a small interval (t , t + h], the premium rate is ci > 0, N (t + h) − N (t ) is assumed to be a Poisson process with intensity λi , the claim amount distribution is Fi with density fi , Laplace transform fˆi (s) and mean µi , and the volatility of the diffusion is σi > 0. The Markov-modulated diffusion risk model {U (t ), t ≥ 0} is given by U (t ) = u +

t

Z

cJ (s) ds − 0

∗

N (t ) X

t

Z

σJ (s) dW (s),

Xi +

i =1

0

Corresponding author. E-mail address: [email protected] (Z. Zhang).

1226-3192/$ – see front matter Crown Copyright © 2009 Published by Elsevier B.V. on behalf of The Korean Statistical Society. All rights reserved. doi:10.1016/j.jkss.2009.07.001

(1.1)

208

H. Yang, Z. Zhang / Journal of the Korean Statistical Society 39 (2010) 207–219

where u ≥ 0 is the initial surplus. If we let Ui (t ), i = 1, 2, . . . , m, be m independent perturbed compound Poisson risk models with premium rate ci , Poisson intensity λi , claim size distribution Fi , diffusion volatility σi and zero initial surplus. Then U (t ) can be expressed as U (t ) = u +

t

Z

I (J (s) = i)dUi (s), 0

where I (A) is an indicator function of event A. By the law of large numbers and standard property of Brownian motion, we have lim

U (t )

t →∞

t

=

m X

πi (ci − λi µi ),

a.s.

i=1

Thus, to guarantee that the surplus process has a positive drift, we assume throughout this paper that the following net-profit condition holds m X

πi (ci − λi µi ) > 0.

(1.2)

i=1

Let Tu = inf{t ≥ 0 : U (t ) ≤ 0, U (0) = u}, or ∞ otherwise, be the time of ruin. Besides the ruin time Tu , we are interested in the hitting time of the Markov-modulated diffusion risk process to a given target before ruin defined as Tˆu,b = inf{0 ≤ t < Tu : U (t ) = b} for 0 ≤ u < b. Since by the net profit condition (1.2), ruin probability is strictly less than 1 and the surplus will become very large over a long time prior to ruin, then insurer may be more interested in the time of the surplus to achieve a certain target before ruin. Moreover, such hitting time has a close relation with the dividend problems in risk theory. For example, when the dividend payments are taken into consideration under a barrier strategy with level b, Tˆu,b becomes the time when to pay the dividends for the first time. Recently, the problems on the compound Poisson surplus process in the presence of an upper barrier has been studied by many authors, see, e.g. Dickson and Gray (1984), Gerber (1990), Picard and Lefevre (1994), Wang and Politis (2002) and Zhou (2004). In this paper, we study when the surplus reaches a certain level prior to ruin in the Markov-modulated diffusion model. We show that the Laplace transform of Tˆu,b can be expressed via the expected discounted penalty functions defined below. The rest of this paper is organized as follows: In Section 2, we revisit the discounted penalty functions and give general expressions for them. In Section 3, we derive the Laplace transform of Tˆu,b , and in Section 4 apply it to derive the expected discounted penalty functions and the expected discounted dividend payments in a Markov-modulated diffusion risk model with a barrier dividend strategy. In Section 5, the solution to a homogeneous integro-differential equation is analyzed in the two-state model, and in Section 6, a numerical example is given. 2. The discounted penalty functions revisited In this section, we derive equations for the expected discounted penalty functions and show that the solutions can be expressed via a solution to a system of homogeneous integro-differential equations. Let w(x1 , x2 ), x1 , x2 ≥ 0, be a nonnegative measurable function, δij be the Kronecker’s symbol. For notational convenience, let Eui (·) = E(·|J (0) = i, U (0) = u). Given that the initial state is i, and ruin occurs in sate j, the expected discounted penalty function is defined for δ ≥ 0

φij (u) = Eui [e−δTu w(U (Tu −), |U (Tu )|); Tu < ∞, J (Tu = j)], where U (Tu −) and |U (Tu )| are the surplus immediately before ruin and the deficit at ruin. Note that ruin can be caused by either oscillations or a claim. We can decompose φij (u) as follows:

φij (u) = φd,ij (u) + φw,ij (u), where

φd,ij (u) = w(0, 0)Eui [e−δTu ; Tu < ∞, U (Tu ) = 0, J (Tu ) = j] with φd,ij (0) = w(0, 0)δij , is the expected discounted penalty function at ruin when ruin is due to oscillations, and

φw,ij (u) = Eui [e−δTu w(U (Tu −), |U (Tu )|); Tu < ∞, U (Tu ) < 0, J (Tu ) = j] with φw,ij (0) = 0, is the expected discounted penalty function at ruin caused by a claim. Without loss of generality, we set w(0, 0) = 1 throughout this paper. The expected discounted penalty function has become an effective tool to study various ruin related quantities. One ruin related quantity of interest in this paper is the discounted density of the deficit at ruin, gij (u; y), satisfying gij (u; y)dy = Eui [e−δ Tu ; |U (Tu )| ∈ dy, Tu < ∞, U (Tu ) < 0, J (Tu ) = j]. which can be obtained by setting w(x1 , x2 ) = I (x2 = y) in φw,ij (u). For the univariate case (m = 1), we denote the expected discounted penalty functions by φd (u), φw (u), and denote the discounted density function of the deficit at ruin by g (u; y).

H. Yang, Z. Zhang / Journal of the Korean Statistical Society 39 (2010) 207–219

209

By using the same arguments as in Lu and Tsai (2007), we can obtain the following integro-differential equations for

φd,ij (u) and φw,ij (u), i, j ∈ E,

(λi + δ)φd,ij (u) = Di φd00,ij (u) + ci φd0 ,ij (u) +

m X

αik φd,kj (u) + λi

m X

φd,ij (u − x)dFi (x),

(2.1)

0

k=1

00 0 (λi + δ)φw,ij (u) = Di φw, ij (u) + ci φw,ij (u) +

u

Z

αik φw,kj (u) + λi

u

Z

k=1

 φw,ij (u − x)dFi (x) + δij ωi (u) ,

(2.2)

0

R∞

where Di = σi2 /2, ωi (u) = u w(u, x − u)dFi (x). Denote the differentiation operator by D , and put

φd (u) = (φd,ij (u))m i,j=1 ,

φw (u) = (φw,ij (u))m i,j=1 ,

G(x) = diag(λ1 f1 (x), . . . , λm fm (x)),

ξ(u) = diag(λ1 ω1 (u), . . . , λm ωm (u)), Aδ (D ) = diag(D1 D + c1 D − λ1 − δ, . . . , Dm D 2 + cm D − λm − δ) + 3. 2

Then we can rewrite (2.1) and (2.2) in the following matrix form: u

Z

Aδ (D )φd (u) +

G(x)φd (u − x)dx = 0,

(2.3)

G(x)φw (u − x)dx + ξ(u) = 0,

(2.4)

0

Aδ (D )φw (u) +

u

Z 0

with boundary conditions φd (0) = I, φw (0) = 0, where I is the identity matrix, and 0 is the zero matrix. We now apply Laplace transforms to solve Eqs. (2.3) and (2.4). Hereafter, we denote the Laplace transform of a function 0 m 0 by adding a hat on the corresponding letter. Let φ0d (0) = (φd0 ,i,j (0))m i,j=1 , φw (0) = (φw,i,j (0))i,j=1 , D = diag(D1 , D2 , . . . , Dm ), C = diag(c1 , c2 , . . . , cm ). After taking Laplace transforms in (2.3) and (2.4), we obtain

[Aδ (s) + Gˆ (s)]φˆ d (s) = Dφ0d (0) − χ(s), [Aδ (s) + Gˆ (s)]φˆ w (s) =

Dφ0w (0)

ˆ s), − ξ(

(2.5) (2.6)

where χ(s) = −Ds − C. Let v(u) = (vij (u))m i,j=1 be a particular solution of the following homogeneous integro-differential equation Aδ (D )v(u) +

u

Z

G(x)v(u − x)dx = 0

(2.7)

0

with boundary conditions v(0) = 0, v0 (0) = I. Then applying Laplace transform to (2.7) gives

ˆ (s)]−1 D, vˆ (s) = [Aδ (s) + G

(2.8)

which together with (2.5) and (2.6) gives

  φˆ d (s) = vˆ (s) φ0d (0) + D−1 C + vˆ (s)s, ˆ s). φˆ w (s) = vˆ (s)φ0w (0) − vˆ (s)D−1 ξ( By inverting the Laplace transforms in the above two formulas, one finds

  φd (u) = v(u) φ0d (0) + D−1 C + v0 (u), Z u φw (u) = v(u)φ0w (0) − v(u − x)ξ(x)dxD−1 .

(2.9) (2.10)

0

Note that we still have to determine two unknown matrices φ0d (0) and φ0w (0) appearing in (2.9) and (2.10). To this end, we consider the roots of the following characteristic equation

Aδ (s) + Gˆ (s) = 0,

(2.11)

ˆ (s)| is the determinant of the matrix Aδ (s) + Gˆ (s). where |Aδ (s) + G Theorem 1. (i) For δ > 0, the characteristic equation (2.11) has exactly m roots, say ρ1 (δ), . . . , ρm (δ), on the right half complex plane.(ii) For δ = 0, the characteristic equation (2.11) has exactly m − 1 roots, say ρ1 (0), ρ2 (0), . . . , ρm−1 (0), on the right half complex plane, and a mth root ρm (0) = 0.

210

H. Yang, Z. Zhang / Journal of the Korean Statistical Society 39 (2010) 207–219

Proof. We follow the idea in the proof of Proposition 2.1 of Albrecher and Boxma (2005). Let P(s) = diag(D1 s2 + c1 s − δ − λ1 + α11 , . . . , Dm s2 + cm s − δ − λm + αmm ), and for 0 ≤ t ≤ 1, let

ˆ (s) − P(s)). P(s, t ) = P(s) + t (Aδ (s) + G ˆ (s) = P(s, 1). For i = 1, . . . , m, the equation Di s2 + ci s − δ − λi + αii = 0 has the following Obviously, we have Aδ (s) + G two roots s+ i =

−ci +

q

ci2 + 4Di (δ + λi − αii ) 2Di

> 0,

s− i =

−ci −

q

ci2 + 4Di (δ + λi − αii ) 2Di

< 0.

+ + + Let Cδ denote the circle with its center at (s+ max , 0) and radius smax , where smax = max1≤i≤m si , and let C δ denote the area + of {s : Re(s) ≥ 0, |s − s+ | ≥ s } . max max For 0 ≤ t ≤ 1, s ∈ C δ , the matrix P(s, t ) is diagonally dominant because

|Di s2 + ci s − δ − λi + αii + t λi fˆi (s)| ≥ |Di s2 + ci s − δ − λi + αii | − t λi |fˆi (s)| + ˆ = |Di (s − s− i )(s − si )| − t λi |fi (s)| + + + ≥ Di |s− i |(|s − smax + smax − si |) − t λi + + + ≥ Di |s− i |(|s − smax | − (smax − si )) − t λi + ≥ Di |s− i si | − t λi = δ + λi − αii − t λi m X αik . > −αii ≥ t k=1,k6=i

The diagonal dominance implies that P(s, t ) is nonsingular for s ∈ C δ . Now let f (t ) be the number of zeros of |P(s, t )| = 0 inside Cδ . Then we have f (t ) =

1 2π i

∂ |P(s, t )| ∂s

Z Cδ

|P(s, t )|

ds.

Here f (t ) is a continuous function on [0, 1], integer-valued, and hence constant. Since f (0) = m, this implies that also f (1) = m. The proof of (i) is completed. To prove (ii), we let δ → 0 and conclude from (i) that there exist exactly m roots inside or on C0 . Obviously, 0 is a root of Eq. (2.11) when δ = 0. Thus, in order to prove (ii), it suffices to show that 0 is a simple root, then 1(λ) = diag(λ1 , λ2 , . . . , λm ). Then

ˆ (s) = 3 + Ds2 + Cs − (1(λ) − Gˆ (s)). A0 (s) + G ˆ (s) can be seen as a perturbation of 3. Since the environment process {J (t )} is irreducible, For |s| sufficiently small, A0 (s) + G E(s) be the eigenvalue and right eigenvector of A0 (s) + Gˆ (s) such that ν(0) = 0, 0 is a simple eigenvalue of 3. Let ν(s) and d E(0) = 1, where 1 is a column vector of 1s. Differentiating both sides of the following equation d [A0 (s) + Gˆ (s)]dE(s) = ν(s)dE(s), and then setting s = 0 yields

(C − 1(λ)µ)1 + 3dE0 (0) = ν 0 (0)1, where µ = diag(µ1 , µ2 , . . . , µm ). Pre-multiplying both sides of the above equation by π yields m P

ν (0) = 0

πi (ci − λi µi )

i=1 m P

>0

(2.12)

πi

i=1

thanks to the net profit condition (1.2). Let 1(ν) = (ν1 , ν2 , . . . , νm ) be the eigenvalue matrix of 3 such that ν1 = 0 and νi 6= 0 for i = 2, 3, . . . , m. Accordingly, ˆ (s) with 1(ν(0)) = 1(ν). Differentiating the let 1(ν(s)) = (ν1 (s), ν2 (s), . . . , νm (s)) be the eigenvalue matrix of A0 (s) + G following identity w.r.t. s

|A0 (s) + Gˆ (s)| =

m Y i =1

νi (s),

H. Yang, Z. Zhang / Journal of the Korean Statistical Society 39 (2010) 207–219

211

and then setting s = 0, one finds

∂ ˆ |A0 (s) + G(s)| = ν10 (0)ν2 · · · νm 6= 0 ∂s s =0 due to (2.12). This completes the proof of (ii).



We remark that Theorem 1 generalizes the corresponding result in Lu and Tsai (2007), where only the case for the twostate model (m = 2) is proved. In the rest of this paper, we assume the m roots are distinct and denote them by ρ1 , . . . , ρm for simplicity. For the univariate case, we denote the unique positive root of (2.11) by ρ . ˆ (ρi )| = 0 for i = 1, . . . , m, then Now we return to determine the unknown matrices φ0d (0) and φ0w (0). Since |Aδ (ρi ) + G

ˆ (ρi ). Let hEi T be the left eigenvector with respect to 0. Setting s = ρi in (2.5) and 0 is an eigenvalue of the matrix Aδ (ρi ) + G Ei T , we obtain (2.6) and then pre-multiplying by h Ei T [Aδ (ρi ) + Gˆ (ρi )]φd (ρi ) = hETi Dφ0d (0) − hETi χ(ρi ) = 0E T , h

(2.13)

ˆ i ) = 0E T . Ei T [Aδ (ρi ) + Gˆ (ρi )]φw (ρi ) = hETi Dφ0w (0) − hETi ξ(ρ h

(2.14)

and After solving Eqs. (2.13) and (2.14), we obtain

E1 , . . . , hEm )T , then Corollary 1. Let H = (h 

ET1 χ(ρ1 ) h



 .. , . EhTm χ(ρm )  T  ˆ 1) E1 ξ(ρ h   .. φ0w (0) = (HD)−1  . . T ˆ E ξ(ρm ) h  φ0d (0) = (HD)−1 

(2.15)

(2.16)

m

We see from (2.9) and (2.10) that the function v(u) plays an important role in expressing the discounted penalty functions. The Laplace transform (2.8) for v(u) can be inverted in most cases, especially when the claim size density fi (i ∈ E ) has a rational Laplace transform, i.e. fˆi (s) can be expressed as a fraction with numerator and denominator being polynomials. 3. The hitting time to a given target before ruin In this section, we study the problem when the surplus first reaches a given target b > 0 before ruin in the Markovmodulated diffusion risk model. Given a level b ≥ U (0) = u, define Tu,b = inf{t ≥ 0 : U (t ) = b} to be the time when the surplus process U (t ) reaches level b for the first time. For δ ≥ 0, define Rij (u; b) = Eui [e−δ Tu,b ; J (Tu,b ) = j],

i, j ∈ E

to be the Laplace transform of Tu,b given that the initial state is i and the surplus first reaches level b at state j. Let R(u; b) = (Rij (u; b))Ni,j=1 . Obviously, we have R(u; b) = R(0; b − u) since U (t ) has conditional stationary independent increments. Now we construct a martingale to derive the expression for R(u; b). Without loss of generality, we set U (0) = 0. Define the matrix Q(t , s) with elements being qij (t , s) = E0i [e−δ t +sU (t ) ; J (t ) = j],

i, j ∈ E ,

where s is such that |fˆi (s)| < ∞ for i = 1, 2, . . . , m. Similar to the method used in Lemma 2.1 in Asmussen and Kella (2000), we obtain by conditioning on the time when the environment state first changes and recalling that Ui (t ) is the perturbed compound Poisson risk model defined in Section 1, qij (t , s) = e−αi t E0i [e−δ t +sUi (t ) ]δij +

t

Z

αi e−αi v

0

= e(ϕi (s)−αi )t δij +

k6=i

" =e

(ϕi (s)−αi )t

t

XZ

δij +

k6=i

αi

E0i [e−δv+sUi (v) ]qkj (t − v, s)dv

αik e(ϕi (s)−αi )v qkj (t − v, s)dv

0

#

t

XZ k6=j

X αik

αik e 0

where ϕi (s) = Di s2 + ci s − δ − λi (1 − fˆi (s)).

(αi −ϕi (s))v

qkj (v, s)dv ,

(3.1)

212

H. Yang, Z. Zhang / Journal of the Korean Statistical Society 39 (2010) 207–219

Dividing both sides of the above equation by e(ϕi (s)−αi )t and differentiating w.r.t. t leads to m X ∂ qij (t , s) = qij (t , s)ϕi (s) + αik qkj (t , s), ∂t k =1

or in matrix form

∂ ˆ (s))Q(t , s). Q(t , s) = (Aδ (s) + G ∂t

(3.2)

ˆ Q(t , s) = e(Aδ (s)+G(s))t .

(3.3)

Solving the above equation with the initial condition Q(0, s) = I yields

ˆ (s) with Ek(s) being the normalized right eigenvector such that πEk(s) = Let κ(s) be the eigenvalue of the matrix Aδ (s) + G 1. Let ki (s) be the ith element of E k(s), then similar to Lemma 2.1 in Asmussen and Kella (2000), we can check that {e−δt +sU (t )−t κ(s) kJ (t ) (s)} is a martingale with respect to the natural filtration Ft generalized by {U (t ), J (t )}. By Theorem 1, ˆ (ρi )| = 0. Therefore, we can choose κ(ρi ) = 0 to be the eigenvalue of the matrix Aδ (ρi ) + Gˆ (ρi ). we know that |Aδ (ρi ) + G Accordingly, the eigenvector E k(ρi ) with respect to 0 can be determined by the following equations (Aδ (ρi ) + Gˆ (ρi ))Ek(ρi ) = 0E,

πEk(ρi ) = 1.

(3.4)

Theorem 2. Let K(ρ) = (E k(ρ1 ), . . . , E k(ρm )), 1(ρ) = diag(ρ1 , . . . , ρm ), M(ρ) = K(ρ)1(ρ)K−1 (ρ). Then R(u; b) = e−M(ρ)(b−u) .

(3.5)

Proof. Obviously, Tu,b is a finite stopping time a.s. because the process U (t ) with skip-free upward jumps has a positive drift due to the net profit condition. For 0 ≤ t ≤ Tu,b , we have U (t ) ≤ b. Thus, the stopped martingale {e−δ(t ∧Tu,b )+ρn U (t ∧Tu,b ) kJ (t ∧Tu,b ) (ρn )} is bounded. Consequently, we can apply the optional sampling theorem to obtain the following result ki (ρn )eρn u = Eui [e−δ(t ∧Tu,b )+ρn U (t ∧Tu,b ) kJ (t ∧Tu,b ) (ρn )]. After letting t → ∞ and applying dominated convergence theorem, the right-hand side of the above equation becomes

Eui [e−δ Tu,b +ρn U (Tu,b ) kJ (Tu,b ) (ρn )] =

m X

Eui [e−δ Tu,b +ρn b kj (ρn ); J (Tu,b ) = j].

j =1

Then we have ki (ρn )eρn (u−b) =

m X

Rij (u; b)kj (ρn ),

n = 1, 2, . . . , m.

j =1

Solving the above linear equations yields (3.5).



For the special case m = 1, (3.5) becomes

E[e−δ Tu,b |U (0) = u] = e−ρ(b−u) . Now we study Tˆu,b , the hitting time to the given level b before ruin, defined in Section 1. For δ ≥ 0, define B(u; b) = (Bij (u; b))m i,j=1 with elements given by ˆ

Bij (u; b) = Eui [e−δ Tu,b ; J (Tˆu,b ) = j],

0 ≤ u ≤ b,

which is the Laplace transform of Tˆu,b evaluated at δ . The function Bij (u; b) can also be interpreted as the expected discounted payment of one dollar that is due when the surplus first reaches the level b at state j before ruin provided the surplus starts from initial state i and initial surplus u. Theorem 3. The matrix B(u; b) can be expressed as



B(u; b) = e

M(ρ)u

− φd (u) −

∞

Z

g(u; y)e

−M(ρ)y

 dy

0

e

M(ρ)b

− φd (b) −

∞

Z

g(b; y)e

−M(ρ)(y)

−1 dy

.

(3.6)

0

Proof. Without loss of generality, we consider the surplus process U (t ) defined on a canonical space Ω . Let θ be a shift operator, i.e. θt (ω) = ωt +· , t > 0, ω ∈ Ω . Note that Tu = Tu I (Tu < Tu,b ) + (Tu,b + Tb ◦ θTu,b )I (Tu > Tu,b )

= Tu I (Tu < Tu,b ) + [Tˆu,b + Tb ◦ θTˆu,b ].

H. Yang, Z. Zhang / Journal of the Korean Statistical Society 39 (2010) 207–219

213

On the one hand, by the strong Markov property, we have for y > 0 gij (u; y)dy = Eui [e−δ Tu ; −U (Tu ) ∈ dy, J (Tu ) = j]

= Eui [e−δTu ; Tu < Tu,b , −U (Tu ) ∈ dy, J (Tu ) = j] + Eui [e−δTu ; Tu > Tu,b , −U (Tu ) ∈ dy, J (Tu ) = j] = Eui [e−δTu ; Tu < Tu,b , −U (Tu ) ∈ dy, J (Tu ) = j] + Eui [e − U (Tˆu,b + Tb ◦ θ ˆ ) ∈ dy, J (Tˆu,b + Tb ◦ θ ˆ ) = j] =

[e

−δ Tu

u,b

)

;

Tu,b

Tu,b

Eui

−δ(Tˆu,b +Tb ◦θTˆ

; Tu < Tu,b , −U (Tu ) ∈ dy, J (Tu ) = j]

m

+

X

ˆ

Eui [e−δ Tu,b ; J (Tˆu,b ) = k]Ebk [e−δ Tb ; −U (Tb ) ∈ dy, J (Tb ) = j].

(3.7)

k=1

Put W(u; b, y) = (Wij (u; b, y))m i,j=1 with Wij (u; b, y)dy = Eui [e−δ Tu ; Tu < Tu,b , −U (Tu ) ∈ dy, J (Tu ) = j]. Then (3.7) can be rewritten as g(u; y) = W(u; b, y) + B(u; b)g(b; y).

(3.8)

Similarly, we can obtain

φd (u) = Z(u; b, 0) + B(u; b)φd (b),

(3.9)

where Z(u; b, 0) = (Zij (u; b, 0))m i,j=1 with Zij (u; b, 0) = Eui [e−δ Tu ; Tu < Tu,b , U (Tu ) = 0, J (Tu ) = j]. On the other hand, we have Rij (u; b) = Eui [e−δ Tu,b ; J (Tu,b ) = j]

= Eui [e−δTu,b ; Tu > Tu,b , J (Tu,b ) = j] + Eui [e−δTu,b ; Tu < Tu,b , J (Tu,b ) = j] = Bij (u; b) + Eui [e−δ(Tu +T0,b ◦θTu ) ; Tu < Tu,b , U (Tu ) = 0, J (Tu + T0,b ◦ θTu ) = j] Z ∞ + Eui [e−δ(Tu +T−y,b ◦θTu ) ; Tu < Tu,b , −U (Tu ) ∈ dy, J (Tu + T−y,b ◦ θTu ) = j] 0

= Bij (u; b) +

m X

Eui [e−δ Tu ; Tu < Tu,b , U (Tu ) = 0, J (Tu ) = k]E0k [e−δ T0,b ; J (T0,b ) = j]

k=1

+

m Z X k=1

∞ −y

Eui [e−δ Tu ; Tu < Tu,b , −U (Tu ) ∈ dy, J (Tu ) = k]Ek [e−δ T−y,b ; J (T−y,b ) = j] 0

= Bij (u; b) +

m X

Zik (u; b, 0)Rkj (0; b) +

k=1

m Z X k=1

∞

Wik (u; b, y)Rkj (0; y + b)dy, 0

which together with (3.8) and (3.9) yields R(u; b) = B(u; b) + Z(u; b, 0)R(0; b) +

∞

Z

W(u; b, y)R(0; b + y)dy 0

= B(u; b) + [φd (u) − B(u; b)φd (b)]R(0; b) +

∞

Z

[g(u; y) − B(u; b)g(b; y)]R(0; b + y)dy.

(3.10)

0

After solving Eq. (3.10), we obtain (3.6).



When m = 1, we set the penalty function w(x1 , x2 ) = e−ρ x2 and define

ψ(u) := E[e−δTu +ρ U (Tu ) ; Tu < ∞|U (0) = u] = φd (u) +

∞

Z

g (u; y)e−ρ y dy. 0

Accordingly, (3.6) becomes ˆ

Eu [e−δ Tu,b ] =

eρ u − ψ(u) eρ b − ψ(b)

,

which coincides with formula (6.25) in Gerber and Shiu (1998) for the classical compound Poisson model.

(3.11)

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H. Yang, Z. Zhang / Journal of the Korean Statistical Society 39 (2010) 207–219

4. Analysis of the Markov-modulated diffusion model with barrier dividend strategy In this section, we apply the constant dividend barrier strategy to the Markov-modulated diffusion model (1.1), and denote the modified surplus process by Ub (t ) for b > 0. Define the running maximum of U (t ) by M (t ) = max U (s). 0≤s≤t

Then the total dividend payments up to time t is given by D(t ) = (M (t ) − b)+ = max{M (t ) − b, 0}, and the surplus at time t is Ub (t ) = U (t ) − D(t ). Let = inf{t ≥ 0 : Ub (t ) ≤ 0, Ub (0) = u}, or ∞ otherwise, be the ruin time for Ub (t ). Denote the expected discounted penalty functions for Ub (t ) by Tub

b

φd,ij (u; b) = Eui [e−δTu ; Tub < ∞, Ub (Tub ) = 0, J (Tub ) = j], b

φw,ij (u; b) = Eui [e−δTu w(Ub (Tub −), |Ub (Tub )|); Tub < ∞, Ub (Tub ) < 0, J (Tub ) = j]. For δ ≥ 0, denote the discounted dividend payments prior to ruin by Tub

Z Du,b =

e−δ s dD(s),

0

and define Vij (u; b) = Eui [Du,b ; J (Tub ) = j] to be the expected discounted dividend payments before ruin. m m Let φd (u; b) = (φd,ij (u; b))m i,j=1 , φw (u; b) = (φw,ij (u; b))i,j=1 , V(u; b) = (Vij (u; b))i,j=1 . When m = 1, we denote these functions by φd (u; b), φw (u; b), V (u; b). Now we derive expressions for φd (u; b), φw (u; b) and V(u; b). Theorem 4. For 0 < u < b, the expected discounted penalty functions φd (u; b) and φw (u; b) can be expressed as

−1 ∂ φ0d (b), B(u; b) ∂u u=b −1  ∂ φ0w (b). B(u; b) φw (u; b) = φw (u) − B(u; b) ∂u u=b φd (u; b) = φd (u) − B(u; b)



Proof. For 0 < u < b, we have b

φd,ij (u; b) = Eui [e−δTu ; Tub < ∞, Ub (Tub ) = 0, J (Tub ) = j] b

= Eui [e−δTu ; Tub < ∞, Tub < Tˆu,b , Ub (Tub ) = 0, J (Tub ) = j] b + Eui [e−δTu ; Tub < ∞, Tub > Tˆu,b , Ub (Tub ) = 0, J (Tub ) = j] = I1 + I2 .

By the strong Markov property, we have I2 =

m X

Bik (u; b)φd,kj (b; b).

k=1

Now we consider I1 . When Ub (t ) stays below level b, it has the same sample path as that of U (t ). Thus, I1 = Eui [e−δ Tu ; Tu < ∞, Tu < Tˆu,b , U (Tu ) = 0, J (Tu ) = j]

= Eui [e−δTu ; Tu < ∞, U (Tu ) = 0, J (Tu ) = j] − Eui [e−δTu ; Tu < ∞, Tˆu,b < Tu , U (Tu ) = 0, J (Tu ) = j] m X = φd,ij (u) − Bik (u; b)φd,kj (b) k=1

by the strong Markov property.

(4.1)

(4.2)

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215

Combining the above results, we obtain

φd,ij (u; b) = φd,ij (u) −

m X

Bik (u; b)[φd,kj (b) − φd,kj (b; b)].

k=1

Rewriting the above formula in matrix form yields

φd (u; b) = φd (u) − B(u; b)[φd (b) − φd (b; b)].

(4.3) ∂ φ (u; b)|u=b− ∂ u d,ij

Now we remove the unknown quantity φd (b; b) in (4.3). It is easy to show that = 0 by an argument similar to that to derive (8) in Gerber, Lin, and Yang (2006). Thus, after taking (left) derivatives on both sides of (4.3) and setting u = b, we obtain

φd (b) − φd (b; b) =



−1 ∂ B(u; b) φ0d (b), ∂u u =b

which together with (4.3) yields (4.1). By using arguments similar to above, we can prove (4.2).



When m = 1, we obtain by (3.11)

φd (u; b) = φd (u) −

eρ u − ψ(u)

ρ eρ b − ψ 0 (b)

φw (u; b) = φw (u) −

φd0 (b),

eρ u − ψ(u)

ρ eρ b − ψ 0 ( b )

(4.4)

φw0 (b).

(4.5)

Now we derive the expected discounted dividend payments V(u; b). Theorem 5. For 0 < u < b, the expected discounted dividend payments V(u; b) can be expressed as

−1 ∂ V(u; b) = B(u; b) . B(u; b) ∂u u=b 

(4.6)

Proof. First, note that no dividend payments are paid when ruin occurs before the surplus reaches level b. Then for 0 < u < b, Vij (u; b) =

Eui

"Z

#

Tub

e

−δ t

dD(t ); Tˆu,b <

e

−δ t

dD(t ); Tˆu,b < Tub , J (Tub ) = j

Tub

, J( ) = j Tub

0

Eui

=

"Z

m X

=

#

Tub Tˆu,b

Bik (u; b)Vkj (b; b)

(4.7)

k=1

by the strong Markov property. That is V(u; b) = B(u; b)V(b; b).

(4.8)

By similar arguments as in Gerber and Shiu (2004), we can obtain the (left) derivative we have V(b; b) =



−1 ∂ B(u; b) , ∂u u =b

which together with (4.8) implies that (4.6) holds.

∂ V (u; b)|u=b− ∂ u ij

= δij . Then from (4.8)

(4.9) 

For the special case m = 1, we obtain by (3.11) V (u; b) =

eρ u − ψ(u)

ρ eρ b − ψ 0 ( b )

,

(4.10)

which is in agreement with formula (7.5) in Gerber and Shiu (1998) for the classical compound Poisson model. Finally, Theorems 4 and 5 give the following relations between the expected discounted penalty functions and the expected discounted dividend payments:

φd (u; b) = φd (u) − V(u; b)φ0d (b). φw (u; b) = φw (u) −

V(u; b)φ0w (b),

which are called dividend-penalty identities (see Gerber et al. (2006)).

(4.11) (4.12)

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5. The solution v(u) in the two-state model In this section, we consider the two-state model (i.e. m = 2) and show how to find the solution v(u) to the homogeneous integro-differential equation (2.7). Assume that the densities f1 , f2 have the following rational Laplace transforms fˆ1 (s) =

qk−1 (s) pk (s)

,

fˆ2 (s) =

ql−1 (s) pl (s)

,

k, l ∈ N+ ,

(5.1)

where pk (s), pl (s) are polynomials of degrees k and l, respectively, while qk−1 (s) and ql−1 (s) satisfying qk−1 (0) = pk (0), ql−1 (0) = pl (0), are polynomials of degrees k − 1 and l − 1, or less, respectively. Furthermore, equations pk (s) = 0 and pl (s) = 0 have roots with only negative real parts. Without loss of generality, assume in what follows that the leading coefficients of pk (s) and pl (s) are 1. Let li (s) = Di s2 + ci s − λi − δ − αi + λi fˆi (s), i = 1, 2. By (2.8), we have

 vˆ (s) := (ˆvij (s))2i,j=1 =

 =

D1 l2 (s) −α2 D1

−α1 D2 D2 l1 (s)



l1 (s)l2 (s) − α1 α2

l11 (s) l21 (s)

l12 (s) l22 (s)

 (5.2)

l1 (s)l2 (s)pk (s)pl (s) − α1 α2 pk (s)pl (s)

where l11 (s) = D1 l2 (s)pk (s)pl (s),

l12 (s) = −α1 D2 pk (s)pl (s),

l21 (s) = −α2 D1 pk (s)pl (s),

l22 (s) = D2 l1 (s)pk (s)pl (s).

It is readily seen that the denominator in (5.2) is a polynomial with leading coefficient D1 D2 and degree k+l+4. By Theorem 1, the denominator can be rewritten as k+l+2

D1 D2 (s − ρ1 )(s − ρ2 )

Y

(s − Ri ),

i =1

where Ri ’s are (complex) numbers with negative real parts. If these k + l + 4 roots are distinct, we have been performing partial fraction,

vˆ ij (s) =

aij1 s − ρ1

+

k+l+2

aij2 s − ρ2

+

X

bijn

n =1

s − Rn

,

i, j = 1, 2,

(5.3)

where aij1 =

lij (ρ1 ) D1 D2 (ρ1 − ρ2 )

,

k+l+2

Q

(ρ1 − Rn )

n=1

aij2 =

lij (ρ2 ) D1 D2 (ρ2 − ρ1 )

,

k+l+2

Q

(ρ2 − Rn )

n=1

bijn =

lij (Rn ) D1 D2 (Rn − ρ1 )(Rn − ρ2 )

k+l+2

Q

. (Rn − Rr )

r =1,r 6=n

Upon inversion of (5.3), we obtain for i, j = 1, 2

vij (u) = aij1 eρ1 u + aij2 eρ2 u +

k+l+2

X

bijn eRn u ,

u ≥ 0.

(5.4)

n=1

We have obtained the general formula for v(u) with elements given by (5.4) in the two-state model when the claim densities are specified by (5.1). For the case m > 2, we can minor the above solution procedure to get v(u), but the calculation may be more complicated.

H. Yang, Z. Zhang / Journal of the Korean Statistical Society 39 (2010) 207–219

217

6. An example In this section, we give an example to illustrate the calculation procedure of some functions. We consider a two-state model with exponential claim size distributions given by F1 (x) = 1 − e−β1 x ,

F2 (x) = 1 − e−β2 x ,

x ≥ 0.

In this case, the characteristic (2.11) reduces to

  α1 α2 (s + β1 )(s + β2 ) = (D1 s2 + c1 s − λ1 − δ − α1 )(s + β1 ) + λ1 β1   × (D2 s2 + c2 s − λ2 − δ − α2 )(s + β2 ) + λ2 β2 .

(6.1)

Furthermore, assume in what follows c1 = c2 = 2.5, λ1 = 1, λ2 = 2, β1 = 0.5, β2 = 1, D1 = 1, D2 = 1.5, δ = 0.2, α1 = 31 , α2 = 23 . Eq. (6.1) gives the following roots: ρ1 = 0.1742805396, ρ2 = 0.5710955938, R1 = −0.212316019, R2 = −0.4627459763, R3 = −2.651169815, R4 = −3.08581099. By (5.4), one obtains

v11 (u) = 0.4000678314e0.1742805396u + 0.09577027198e0.5710955938u − 0.224742542e−0.212316019u − 0.000286143e−0.4627459763u − 0.018253104e−2.651169815u − 0.252556315e−3.08581099u , v12 (u) = 0.2932041595e0.1742805396u − 0.1481161031e0.5710955938u − 0.1421071609e−0.212316019u + 0.00704682257e−0.4627459763u − 0.05605935834e−2.651169815u + 0.04603164038e−3.08581099u , v21 (u) = 0.3909388793e0.1742805396u − 0.1974881376e0.5710955938u − 0.1894762145e−0.212316019u + 0.0093957634e−0.4627459763u − 0.07474581107e−2.651169815u + 0.06137552044e−3.08581099u , v22 (u) = 0.2865136772e0.1742805396u + 0.3054306176e0.5710955938u − 0.1198078775e−0.212316019u − 0.2313888583e−0.4627459763u − 0.2295610803e−2.651169815u − 0.011186479e−3.08581099u . We can choose H to be

λ2 β2  ρ 1 + β2 H= λ2 β2 2 D2 ρ2 + c2 ρ2 − λ2 − δ − α2 + ρ2 + β2 

D2 ρ12 + c2 ρ1 − λ2 − δ − α2 +

−α1 −α1

  .

Then formula (2.15) gives

−2.801863320 φd (0) = 0.174082679 0



0.1973165967 . −2.110179480



(6.2)

Set the penalty function w(x1 , x2 ) = I (x2 = y), then φw (u) becomes the discounted density of the deficit at ruin g(u; y). In this case, we have

λ1 β1 e−β1 y  ˆ s) =  s + β1 ξ(  

0

 0

 , λ2 β2 e−β2 y  s + β2

and by (2.16)

 ∂ 0.6532041706e−0.5y g(u; y) = 0.1205193304e−0.5y ∂u u =0

0.1426027426e−y . 0.9408700685e−y



By formula (2.9) and (2.10), we obtain the discounted penalty functions

φd,11 (u) = 0.09081957642e−0.212316019u + 0.001445517271e−0.4627459763u + 0.0441430575e−2.651169815u + 0.8635918504e−3.08581099u , φd,12 (u) = 0.04885253983e−0.212316019u − 0.006442705633e−0.4627459763u + 0.1698842821e−2.651169815u − 0.2122941166e−3.08581099u , φd,21 (u) = 0.07656827845e−0.212316019u − 0.0474648804e−0.4627459763u + 0.1807642489e−2.651169815u − 0.2098676461e−3.08581099u , φd,22 (u) = 0.04118665857e−0.212316019u + 0.2115521266e−0.4627459763u + 0.6956700982e−2.651169815u + 0.05159111549e−3.08581099u

(6.3)

218

H. Yang, Z. Zhang / Journal of the Korean Statistical Society 39 (2010) 207–219

Fig. 1. The time value of hitting the barrier b = 10.

and

 g11 (u; y) = e−0.5y 0.2266772068e−0.212316019u + 0.004502798035e−0.4627459763u  − 0.02292183946e−2.651169815u − 0.2082581656e−3.08581099u ,  g12 (u; y) = e−y 0.0747952411e−0.212316019u − 0.01089915369e−0.4627459763u  − 0.1006159122e−2.651169815u + 0.03671982461e−3.08581099u ,  g21 (u; y) = e−0.5y 0.191107294e−0.212316019u − 0.1478534866e−0.4627459763u  − 0.09386411716e−2.651169815u + 0.05061030968e−3.08581099u ,  g22 (u; y) = e−y 0.0630584625e−0.212316019u + 0.3578836705e−0.4627459763u  − 0.4120185851e−2.651169815u − 0.008923547989e−3.08581099u . After determining the matrix K(ρ) according to (3.4), we can get eM(ρ)u . Then the explicit expression for B(u; b) can be obtained by (3.6). Now set the barrier b = 10, we have for 0 ≤ u ≤ 10 B11 (u; 10) = 0.1200612544e0.1742805396u + 0.00106489081e0.5710955938u − 0.06458013867e−0.212316019u

+ 0.0008340878018e−0.4627459763u − 0.01088887331e−2.651169815u − 0.04649122113e−3.08581099u , B12 (u; 10) = 0.05822030388e0.1742805396u − 0.00108853072e0.5710955938u − 0.03115013347e−0.212316019u

+ 0.0004578153633e−0.4627459763u − 0.005594036778e−2.651169815u − 0.02084541834e−3.08581099u , B21 (u; 10) = 0.1173216354e0.1742805396u − 0.00219591423e0.5710955938u − 0.05444630152e−0.212316019u

− 0.02738803535e−0.4627459763u − 0.04458954905e−2.651169815u + 0.01129816488e−3.08581099u , B22 (u; 10) = 0.05689180325e0.1742805396u + 0.00224466214e0.5710955938u − 0.02626209225e−0.212316019u

− 0.01503278590e−0.4627459763u − 0.02290738172e−2.651169815u + 0.005065794522e−3.08581099u . Fig. 1 shows the behavior of B(u; 10). Both B11 (u; 10) and B22 (u; 10) increase with respect to u, while B12 (u; 10) and B21 (u; 10) first increase and then decrease with respect to u. Finally, we can calculate the expected discounted dividend payments by formula (4.6). Fig. 2 shows the behavior of V(u; 10). Furthermore, one can calculate the discounted penalty functions φd (u; b) and φw (u; b) by the dividend-penalty identities (4.11) and (4.12).

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219

Fig. 2. The expected discounted dividend payments.

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