Runhuan Feng∗ and Yasutaka Shimizu† DOI: 10.1016/j.insmatheco.2014.08.001
Abstract
The Markov additive process (MAP) has become an increasingly popular modeling tool in the applied probability literature. In many applications, quantities of interest are represented as functionals of MAPs and potential measures, also known as resolvent measures, have played a key role in the representations of explicit solutions to these functionals. In this paper, closed-form solutions to potential measures for spectrally negative MAPs are found using a novel approach based on algebraic operations of matrix operators. This approach also provides a connection between results from fluctuation theoretic techniques and those from classical differential equation techniques. In the end, the paper presents a number of applications to ruin-related quantities as well as verification of known results concerning exit problems.
Keywords: Markov additive processes, potential measure, resolvent density, Markov renewal equation, scale matrix, exit problems.
MSC2010: 60J25, 91B30, 60J28.
1
Introduction
In the recent applied probability literature, there are two notably different lines of development for exit problems of L´evy processes and Markov additive processes, with a considerable amount of applications in ruin theory.
Historically, classical first passage time problems are often solved by differential equations and integro-differential equations. Many authors have developed solution methods for various risk models with an increasing use of differential and integral operators. See, for example, Li and Garrido [37], Gerber and Shiu [24], Landriault and Willmot [36], Cheung and Landriault [15], Cai et al. [11], Zhang and Cheung [39], etc. The recent work by Albrecher et al. [2] and Albrecher et al. [1] sheds light on the connection of these applications with the study of integro-differential algebras. See details in Rosenkranz and Regensburger [38].
There is also a growing popularity in applications of fluctuation theoretical techniques to ruin-related problems, which employ entirely different solution methods such as martingales and changes of measures. See for example, Avram, Palmowski and Pistorius [5], Breuer [9, 10], Kyprianou and Zhou [35], etc. Interested readers may consult Kyprianou [32] and Asmussen and Albrecher [4] for comprehensive accounts of these techniques and applications in ruin theory.
Despite the overlap of their research aims, there are relatively few papers that make connec-tions between the two approaches. Some earlier attempts to connect the two approaches can
∗
Department of Mathematics, University of Illinois at Urbana-Champaign, 1409 W. Green Street, M/C 382, Urbana, IL 61801, USA;[email protected]
†
Department of Applied Mathematics, Waseda University 3-4-1, Okubo, Shinjuku, Tokyo 169-8555, Japan. [email protected]
be seen in Biffis and Kyprianou [7], Biffis and Morales [8], and Feng and Shimizu [22] in vari-ous settings for spectrally negative L´evy processes. In this paper, we intend to establish some equivalency of results from fluctuation techniques and those from integro-differential operator techniques in the context of MAP risk models.
The key to analyzing each stochastic process often involves studying its semigroup. It is well known that every transition density function defines a contraction semigroup (c.f. Theorem II.2.1 of [17]). For any Borel measurable function l,
Ptl(x) =Ex[l(Xt)].
In the applied probability literature, it is often important to find explicit expressions for the resolvent operator (the corresponding measure is called the potential measure), which is known to be the Laplace transform of the semigroup.
Rδl(x) =
Z ∞
0
e−δtPtl(x)dt, δ >0.
In the context of ruin theory, different expressions of the resolvent operator are known for the two above-mentioned methods, which elucidates differences between the two theories. We quote an example from spectrally negative L´evy risk models denoted by X ={Xt, t ≥0} with characteristics (µ, σ2, ν) killed when exiting [0,∞).
In the applications of fluctuation theory, solutions are typically represented in terms of the scale function, which is defined via its Laplace transform
Z ∞
0
e−λxW(δ)(x)dx= 1
ψ(λ)−δ,
where ψ(λ) = logE[exp(λX1)]. It is well-known that (c.f. Kyprianou [32, Corollary 8.8])
Rδl(x) =
Z ∞
0
r(δ)(x, y)l(y)dy, (1.1)
where r(δ) is known as the potential density
r(δ)(x, y) =e−ρyW(δ)(x)−W(δ)(x−y), x, y≥0. (1.2) The constant ρ is the unique non-negative root of the equation ψ(ρ)−δ = 0.
In contrast, the exposition of solutions to ruin-related quantities in classic ruin theory often hinges on compound geometric distributions. For brevity, we reiterate a special case in Feng and Shimizu [22] where σ >0 and R01zν(dz)<∞.(The corresponding result forσ = 0 is also provided in the paper.) The associated compound geometric distribution is given by
Gδ(y) = (1−p) + ∞ X k=1 (1−p)pk Z x 0 qδ∗k(y)dy,
where ∗denotes convolution and
p:= 2 σ2 Z ∞ 0 Eβ{Tρν}(y)dy qδ(y) := 2 pσ2Eβ{Tρν}(y).
and β := 2c/σ2+ρ. The two integral operators involved are defined by
Esf(x) :=e−sx Z x 0 esyf(y)dy; Tsν(x) :=esx Z ∞ x e−syν(dy).
It is shown that the potential density has the series representation r(δ)(x, y) = 1 1−p Z [0,x) Bδ(x−z, y)dGδ(z), (1.3) where Bδ(x, y) = 2 σ2(β+ρ)e −βx(eβy−e−ρy), 0< y < x; 2 σ2(β+ρ)(e ρx−e−βx)e−ρy, x≤y.
Note, however, Feng and Shimizu [22] did not just stop at this class of L´evy process for which (1.3) holds but also used an operator-based method to prove (1.2) for the general class of spectrally negative L´evy processes, which was only known previously via fluctuation methods.
To bridge the gap between the two types of results in (1.2) and (1.3), Feng and Shimizu [22] proved their equivalence through the representation of the scale function in terms of the com-pound geometric distribution
W(δ)(x) = 2
σ2(1−p)(β+ρ)
Z
[0,x)
(eρ(x−y)−e−β(x−y))dGδ(y). (1.4)
The primary goals of this paper can be summarized as follows.
1. We aim to identify potential measures for spectrally negative Markov additive processes (MAPs), of which less is known in the literature compared with L´evy processes. It was brought to our attention recently that the expression of the potential density in Theorem 4.2 was also found in Ivanovs [27] with fluctuation methods. Nevertheless, this work is entirely based on operator calculation and gives more explicit expressions in certain cases. The operator-based approach leads to a natural extension of classical ruin methods. 2. We are also interested in the analogue of (1.4) for the MAP risk model, which appears
to be a key quantity in connecting the seemingly different yet equivalent solutions from classical ruin methods and fluctuation methods.
The rest of this paper is organized as follows. We introduce the notation of the MAP risk model in Section 2.1 and present the main results of the paper in Section 2.2, in which major steps of the matrix-operator approach are outlined in Section 3. As a consequence, in Section 4, we derive the potential densities for the MAP risk model in two cases, which are essentially the analogues of (1.2) and (1.3). In Section 5, we present some new results for ruin-related problems and also verify through a few examples that the results produced from the matrix-operator approach are consistent with those from Kyprianou and Palmowski [34].
2
Markov additive risk model
2.1 Models and notation
Let (Ω,F,P) be a probability space on which the following stochastic processes are defined.
A Markov additive process is defined as a bivariate Markov process {(Xt, Jt), t ≥ 0}, where
J ={Jt, t≥0} is a Markov process with a state spaceE and the increments ofX ={Xt, t≥0} are governed by {Jt}in the sense that for all bounded measurable functions f and g,
E[f(Xt+s−Xt)g(Jt+s)|Ft] =E(0,Jt)[f(Xs)g(Js)],
with probability measures {P(x,i), x ∈ R, i ∈ E} under which P(x,i){X0 = x, J0 = i} = 1.
process with no negative jumps is known as a spectrally positive L´evy (SPL) process. In this paper, we consider a fairly general class of risk models known as the MAP risk model in ruin theory. LetJbe an irreducible continuous-time Markov chain with state spaceE ={1,2,· · · , m}
and infinitesimal generator matrix Λ = (λij)i,j∈E with λii = −Pi6=jλij < 0. For Jt = i, the level process X is given by a L´evy process with finite meanXt(i) =µit−St(i)−Z
(i)
t , where S(i) is a subordinator and Z(i) is a mean zero SPL process. In the context of ruin theory, X is often
interpreted as the surplus of an insurance business withµi denoting the rate at which premiums are collected under the economy state i, {St(i)} denoting the aggregate claim costs and {Zt(i)}
denoting a perturbation caused by unexpected losses. Hence X has the L´evy exponent
ψi(s) =µis+ 1 2σ 2 is2+ Z ∞ 0 (e−sz−1)νiS(dz) + Z ∞ 0 (e−sz−1 +sz)νiZ(dz),
whereνiS andνiZ are the corresponding L´evy measures forS(i)and Z(i) respectively, andσi≥0 is the volatility coefficient for the Gaussian component of Z(i). Note that the finite mean of X
implies that R1∞zνiZ(dz) < ∞, which explains why the usual indicator I(z < 1) in the L´evy exponent is removed. Throughout the paper, we assume
−X(i) (i= 1, . . . , m) is not a subordinator.
This condition is naturally satisfied because the premium rate µi>0 for alli∈E. In addition, a jump of J from ito j6=ihas probability qij of triggering a negative jump of X at the same instant. We assume the law of jumps to be absolutely continuous and denote the density of the absolute value of jump size by fij(y) for y >0 almost everywhere. We use the matrix C with (Cij = λij(1−qij))i,j∈E for transition rates without incurring jumps and the matrix D with (Dij =λijqij)i,j∈E for the ones with jumps, the matrix F = (fij). Note that C+D= Λ and
qii= 0 for alli∈E. We further assume that the asymptotic driftof X is positive, i.e. under Pi fori∈E,
lim t→∞
Xt
t >0 a.s., (2.1)
which corresponds to the net profit conditionin the classical ruin literature.
A matrix is said to bepositive stable if all of its eigenvalues have positive real parts. For a positive stable m×mmatrix S and any bounded vector function f,
L{f}(S) =
Z ∞
0
e−Sxf(x)dx.
Similarly, for any positive stable matrix S, the following quantities are well-defined.
ψS(S) = Z ∞ 0 (e−Sz−I)∆νS(dz); ψL(S) = Z ∞ 0 (e−Sz−I+Sz)∆νZ(dz),
where I is an m×m identity matrix, ∆νS = diag(ν1S,· · ·, νmS) and all other ∆’s are defined
similarly. For brevity, we often write ψS+L := ψS +ψL and ∆ν = ∆νS + ∆νZ. The symbol
∗ stands for the convolution, i.e., for any two conformable matrix functions (Aik)i∈E,k∈E0 and
(Bkj)k∈E0,j∈E00, (A∗B)ij =Pk∈E0
R∞
0 Aik(x−z)Bkj(z)dz. The notation◦denotes the Hadamard
product (entrywise multiplication). All vectors are considered column vectors by default and >
denotes the transpose operation.
According to Proposition XI.2.2 of Asmussen and Albrecher [4], we can find the following matrix moment generating function for Xt with
E(0,i)[esXt;Jt=j] =
eK(sI)t
where the matrix function K(·) is defined for any m×m positive stable matrixS,
K(S) :=C+S2∆σ2/2+S∆µ+ψS(S) +ψL(S) +L{D◦F}(S).
Lemma 2.1. The generalized Lundberg equation
det[K(sI)−δI] = 0, (2.2) has m roots (counting multiplicity) in the region ofCRe>0.
Proof. Since−Xi (i= 1, . . . , m) is not a subordinator, Theorems 1 and 2 of Ivanovs et al. [28] yield the result under the condition (2.1).
Lemma 2.2. Suppose that the equation (2.2) has m distinct solutions, say ρj (j = 1, . . . , m). Then there exists an m×m matrixR whose eigenvalues are ρj (j= 1, . . . , m) such that
K(R)−δI = 0. (2.3)
In particular, such R is positive stable and diagonalizable.
The proof can be found in Appendix B. The uniqueness ofRis not proven, but any such R
would do in the following solution representations.
2.2 A summary of main results
As a generalization of (1.1), we are interested in the potential (resolvent) density r(δ) of the MAP (X, J). For a vector of measurable functions l(x) = (l(x,1),· · · , l(x, m))>, consider the following vector function
H(x) :=Ex Z τ0 0 e−δtl(Xt, Jt)dt ,
where Ex = (E(x,1),· · · ,E(x,m))> and τ0 = inf{t:Xt<0}. Formally, we write
H(x) =Ex Z ∞ 0 e−δtl(Xt, Jt)I(t < τ0)dt = Z ∞ 0 Ex[l(Xt, Jt)I(t < τ0)]dt= Z ∞ 0 e−δtPtl(x)dt,
where{Pt, t≥0}is the semigroup of the MAP and determines the potential densityr(δ)through
H(x) =Rδl(x) =
Z ∞
0
r(δ)(x, y)l(y)dy.
(1) General case:
The representation of the potential density relies on a key quantity, which is known as the
δ-scale matrix, sayW(δ), given via the Laplace transform
Z ∞
0
e−rxW(δ)(x)dx= (K(rI)−δI)−1. (2.4)
The existence of the scale matrix is shown in Ivanovs and Palmowski [29]. Note, however, one can also define the unique scale matrix using a matrix differential equation, which is equivalent to the definition via the Laplace transform (2.4) (See Section 3.3). We will show in Sections 3 and 4 that under the assumptions
(A1) E(x,i)R0τ0e−δt|l(Xt, Jt)|dt
(A2) H(x) = (H(x, i))i∈E has bounded second and third derivatives forx∈(0,∞); (A3) ∆σ2 is non-singular;
(A4) δ >0 or (δ= 0 and C is non-singular); (A5) Eq. (2.2) has m distinct roots,
the resolvent density is given by
r(δ)(x, y) =W(δ)(x)e−Ry−W(δ)(x−y), x, y≥0,
withRbeing a positive stable and diagonalizable matrix solution to K(R)−δI = 0.We learned that this result also appeared in Ivanovs [27, Theorem 1], which was derived from the use of occupation densities.
(2) Particular case:
A subclass of risk models that often appear in the ruin literature is the difference of a Brownian motion with positive drift and a subordinator, where the subordinator represents aggregate claims and the Brownian motion corresponds to fluctuations due to unexpected losses or gains. So we consider in the framework of MAP risk models the particular case where (A6) R01z∆ν(dz)<∞.
Under assumptions (A1)−(A6), we shall prove that the resolvent density has a more explicit form r(δ)(x, y) = Z [0,x) G(du)∆2/σ2B(x−u, y) ! ,
where the matrix function G is defined in (3.12), Q is a solution to the Lyapunov equation
P Q+QR=I withP =R+ ∆2µ/σ2,and
B(x, y) =
e−P x(eP yQ−Qe−Ry), 0< y < x; (QeRu−e−P xQ)e−Ry, y > x.
The existence ofQ can be seen in Remark 4.1. We will also show that the scale matrix admits the series expansion
W(δ)(x) =
Z
[0,x)
G(dy)∆2/σ2(QeR(x−y)−e−P(x−y)Q), x≥0.
Remark 2.1. Although we did not prove (A2) in this paper, it is expected that the differentiability conditions can be proven at least in the case of (A3). Similar arguments were made with jump-diffusion models in the presence of a Brownian motion in Feng [21, Appendix C].
3
Matrix operator solution methods
We focus on the following quantity through which potential measures are obtained.
H(x, i) =E(x,i) Z τ0 0 e−δtl(Xt, Jt)dt , x≥0, (3.1)
for some measurable function or generalized function l : R×E 7→ R∪ {∞}. The quantity is
interpreted as the expected present value of running costs up to ruin in Cai et al. [11]. The time of ruin can be extended to the time of default τd= inf{t:Xt< d}.However, such an extension is trivial for a spatially homogeneous model such as the MAP. Some examples of the bivariate cost function l can be seen in Cheung and Feng [14], Feng [19, 20].
3.1 Matrix integro-differential equation
We first state a rather standard result for the matrix integro-differential equation satisfied byH. However, we are unable to pinpoint a reference to this result in the existing literature. Thus, a proof is given in Appendix B to make the paper self-contained. Interested readers may consult Jacobsen [30] for more information on the infinitesimal generator of a MAP.
Lemma 3.1. Suppose that l(x) = (l(x, i))i∈E is continuous inx∈(0,∞) except for a countable set of discontinuities Dsuch that (A1),(A2) hold true. Then H satisfies the following integro-differential equation of matrix form:
(A −δ)H(x) +l(x) = 0, (3.2) for x∈(0,∞)∩Dc, where A is the infinitesimal generator
AH(x) = ∆σ2/2H00(x) + ∆µH0(x) +CH(x) + Z ∞ 0 ∆νS(dz){H(x−z)−H(x)} + Z ∞ 0 ∆νZ(dz){H(x−z)−H(x) +zH0(x)}+ (D◦F)∗H(x).
Remark 3.1. It is easy to show that sinceH is assumed to possess bounded derivatives, it must be bounded by a polynomial, a property that is used frequently in the subsequent theorems. See, for example, Lemmas A.1 and A.2.
We seek solutions to the vector function H in two cases. The first solution is based on a Markov renewal equation, which is analogous to the scalar defective renewal equation in classic ruin theory. The second solution generalizes the first solution and appears to be consistent with what would be expected from fluctuation theory. To better understand the proofs of Theorems 3.1 and 3.2, interested readers may check the scalar cases in Feng [21] and Feng and Shimizu [22].
3.2 Solution via a Markov renewal equation
It is shown in Feng and Shimizu [22] that a potential measure for a spectrally negative L´evy process with characteristics (µ, σ2, ν) can be obtained through a defective renewal equation if
and only if the Levy measure ν satisfiesR01zν(dz)<∞.Similarly, we can show that a potential measure for the MAP under consideration can be obtained through a Markov renewal equation if and only if (A6) is satisfied. Under this condition, the L´evy processes{Z(i), i∈E}are restricted
to Brownian motions, subordinators or their independent sums. In the case of subordinators, they can be combined with S(i). Hence without loss of generality, we simply consider in this subsection that ∆νZ ≡0,in which case {Z(i), i∈E} reduce to Brownian motions.
Define the matrix operatorES with a matrix functionh for which the integral exists
ESh(x) =e−Sx
Z x
0
eSyh(y)dy,
and the matrix Dickson-Hipp operator
TSh(x) =eSx
Z ∞
x
e−Syh(y)dy.
In the case where ∆ν is a matrix of measures, it is interpreted that
TS∆ν(x) =eSx
Z ∞
x
Theorem 3.1. Suppose that(A1)−(A4)are true. ThenH= (H(x, i))>i∈E satisfies the following Markov renewal equation:
H(x) = (G∗H) (x) +U(x), x≥0, (3.3) where
G= ∆−σ21/2EP{TR∆ν+TR(D◦F)}; U = ∆
−1
σ2/2EPTRl; (3.4)
with R given in (2.3) andP = ∆2µ/σ2+R.
Proof. We denote the entrywise differential operatorD:= (d/dx)I, and
Hν(x) = Z ∞ 0 ∆νS(dz){H(x−z)−H(x)} + Z ∞ 0 ∆νZ(dz){H(x−z)−H(x) +zH0(x)}+ (D◦F)∗H(x). (3.5)
We rewrite the equation (3.2) in terms of the operators:
(B− D)(A+D)∆σ2/2H(x) ={Hν +l}(x), x >0, (3.6)
whereAandB are bothm×mmatrices satisfying thatA−B = ∆2µ/σ2 andBA= (δI−C)∆−1
σ2/2.
We choose B to be the positive stable solution to the following quadratic matrix equation in
m×mmatrix Q,
Q2+Q∆2µ/σ2 −(δI−C)∆−1
σ2/2 = 0.
Since C is a subintensity matrix, it is known that the eigenvalues of C either are 0 or have negative real parts. The assumption (A4) implies that δI−C is nonsingular and hence Lemma C.1 shows the existence and uniqueness of such a matrix B. We setA:=B+ ∆2µ/σ2 and hence
BA=B2+B∆2µ/σ2 = (δI−C)∆−1
σ2/2.
Since ∆σ2 is non-singular, we must have H(0) = 0. Now we can use the properties (A.1)
and (A.4) to obtain
∆σ2/2H=EAY, (3.7)
where
Y =TB{Hν+l}.
With the matrix R which is a solution to (2.3), we can show using the property (A.3) that
Y = TR{Hν+l} − TR(B−R)TB{Hν +l}=TR{Hν +l} − TR(B−R)Y. We now apply property (A.5) on the second term to get
Y = TR{Hν+l} − TR[R(B−R) + (B−R)A]EAY + (B−R)EAY. (3.8) Since R is diagonalizable and positive stable, we can expand the first term using (A.9).
Y = TR[ψS+L(R) +L{D◦F}(R)]H+ [TR∆ν+TR{D◦F}]∗H+TRl
In view of (3.7), we can simplify
TR[R(B−R) + (B−R)A]EAY =TR[(δI−C)−R2∆σ2/2−R∆µ]H. (3.10)
Since R satisfies the matrix equation (2.3) which implies that
(δI−C)−R2∆σ2/2−R∆µ=ψS+L(R) +L{D◦F}(R),
we obtain from (3.9) and (3.10) that
Y = [TR∆ν+TR{D◦F}]∗H+TRl+ (B−R)EAY.
We choose anotherm×m matrixP and use the property (A.2) on (3.7) to obtain ∆σ2/2H=EPY +EP(P−A)EAY, and that EPY =EPnTR∆ν+TR{D◦F} o ∗H+EPTRl+EP(B−R)EAY. Let P =A−B+R= ∆µ∆−σ21/2+R,
Then we shall arrive at the renewal equation. ∆σ2/2H =EP
n
TR∆ν +TR{D◦F}
o
∗H+EPTRl.
Because ∆σ2 is non-singular, we obtain the Markov renewal equation (3.3).
Corollary 3.1. Suppose that ∆σ2 = 0, (A1),(A2),(A4),(A5)are true. Then H = (H(x, i))>i∈E
satisfies the following Markov renewal equation:
H(x) = (g∗H) (x) +U(x), x≥0, (3.11) where R is determined by (2.3) and
g= ∆−µ1{TR∆ν+TR(D◦F)}; U = ∆−µ1TRl. The proof of Corollary 3.1 can be found in Appendix B.
Solving Markov renewal equations (3.3) and (3.11), we obtain the first representation of a solution to the functionH, of a form typically seen in classical ruin theory.
Corollary 3.2. In both Theorem 3.1 and Corollary 3.1, the solution toH is given by
H(x) =
Z
[0,x)
G(dy)U(x−y), x≥0,
where g is defined in (3.4)and
G(x) =I+
∞
X
k=1
g∗k(x), x≥0. (3.12)
Proof. It is proven in Appendix D that Rx
0 g(y)dy is a semi-Markov matrix. Thus the result
3.3 Solution via the scale matrix
Here we discuss the general case where
Z 1
0
z∆ν(dz)≤ ∞. (3.13)
Note that the Markov renewal equation approach is no longer applicable when the equality is true in (3.13).
The discussion of the existence of the scale matrix in (2.4) is provided in Ivanovs [27, Theorem 7.1] and its differentiability is provided in Kyprianou and Palmowski [34], although there are slight discrepancies in their definitions of scale matrix. For the sake of completeness, we provide some supporting arguments based on a matrix differential equation.
Here we assume ∆σ2/2 is non-degenerate. There exist a unique matrix solution W(δ)(x) to
the homogeneous integro-differential equation
AW(δ)= 0, (3.14)
with boundary conditions
W(δ)0(0) = ∆2/σ2, W(δ)(0) = 0. (3.15)
Similar boundary conditions for the scalar case can be seen in Chan et al. [13]. The proof of existence and uniqueness requires some technical but standard arguments using the Banach fixed point theorem.
We can show that W(δ) has precisely the Laplace transform of what is defined as a scale
matrix. Denote the first and second integrals in (3.14) by I1(x) andI2(x) respectively. Thus,
LI1(r) = Z ∞ 0 ∆νZ(dz) Z ∞ 0 e−rx[W(δ)(x−z)−W(δ)(x)]dx = Z ∞ 0 ∆νZ(dz) Z ∞ 0 e−r(y+z)W(δ)(y)dy− Z ∞ 0 e−ryW(δ)(y)dy = Z ∞ 0 (e−rz−1)∆νZ(dz) Z ∞ 0 e−ryW(δ)(y)dy ,
where we use the fact that W(δ)(x) = 0 for x <0 in the penultimate equality. Similarly,
LI2(r) = Z ∞ 0 ∆νZ(dz) Z ∞ 0 e−rx[W(δ)(x−z)−W(δ)(x) +zW(δ)0(x)]dx = Z ∞ 0 ∆νZ(dz) Z ∞ 0 e−r(y+z)W(δ)(y)dy− Z ∞ 0 e−ryW(δ)(y)dy+rz Z ∞ 0 e−rxW(δ)(x)dx = Z ∞ 0 (e−rz−1 +rz)∆νZ(dz) Z ∞ 0 e−ryW(δ)(y)dy ,
where we use the boundary condition W(δ)(0) = 0 while applying integration by parts. Thus, taking the Laplace transform on both sides of (3.14) gives
(K(rI)−δI)L{W(δ)}(s)−∆σ2/2(rW(δ)(0) +W(δ)
0
(0))−∆µW(δ)(0) = 0. Using the boundary conditions (3.15) we obtain the desired expression (2.4).
With slight abuse of notation, we define a measureW(δ) by
W(δ)[0, x) =W(δ)(0) +
Z x
0
Whether it is a measure or a function should be clear from the context. In the case wherem >1,
W(δ)(dx) could be a matrix of signed measures. However the following derivations require only that the scale matrix is of bounded variation entrywise so that the Lebesgue-Stieltjes integrals are well-defined. The bounded variation follows from the almost everywhere differentiability of
W(δ),which is true if every X(i) is of unbounded variation. (See Ivanovs and Palmowski [29], the remark of Theorem 5.) In this paper, since (A3) is assumed for simplicity, the integral with respect to a measure W(δ)(dx) is well-defined.
We present the second representation of a solution to the function H in terms of the scale matrix, of a form typically observed in fluctuation theory. Note, however, the proof is a contin-uation of the matrix-operator approach used in Section 3.2.
Theorem 3.2. Suppose that (A1)−(A4) hold true. Then it follows for all x >0 that
H(x) =
Z x
0
Π(dy)TRl(x−y), (3.17) where the measure Π is defined by
Π(dx) =W(δ)(dx)−W(δ)(x)Rdx. (3.18) Proof. Note that in this case ∆σ2/2is non-singular and the equality (3.8) in the proof of Theorem
3.1 holds true even if R1
0 z∆ν(dz) = ∞. Taking Laplace transforms on both sides of (3.8) and
using (A.8) on the first term gives for eachr >0,
LY(r) = (rI−R)−1 h ψS+L(R) +L{D◦F}(R)−ψS+L(rI)− L{D◦F}(rI) i · LH(r) +L[TR{N(R)H}](r) +TRL{l}(r) −L{TR[R(B−R) + (B−R)A]∆σ2/2H}(r) + (B−R)∆σ2/2LH(r). (3.19)
We shall simplify the components of (3.19) separately. Note that
L[TR{N(R)H}](r)− L{TR[R(B−R) + (B−R)A]∆σ2/2H}(r)
= L[TR{N(R)H}](r)− L{TR[(δI−C)−R∆µ−R2∆σ2/2]H}(r)
= L[TR{(K(R)−δI)H}](r) = 0.
It is easy to verify that
LY(r) = (rI+A)∆σ2/2LH(r).
Recall that A−B = ∆2µ/σ2 and BA= (δI−C)∆−1
σ2/2.Thus, (rI−R)−1{K(rI)−δI}={(rI+A)∆σ2/2−(B−R)∆σ2/2} −(rI−R)−1 h ψS+L(R) +L{D◦F}(R)−ψS+L(rI)− L{D◦F}(rI) i .
Combining all the simplified components of (3.19), we obtain the following equation
(rI−R)−1{K(rI)−δI} · LH(r) =L{TRl}(r), (3.20) which produces (3.17) provided that the Laplace-Stieltjes transform of Π is given by
Z ∞
0
e−rxΠ(dx) = [K(rI)−δI]−1(rI−R).
4
Potential Measure
As consequences of the previous analysis of solutions to H, we can obtain two representations of potential measures respectively under the conditions (A6) and (3.13). For notational brevity, we shall writeEx(Y) as a shorthand for (E(x,1)(Y),E(x,2)(Y),· · ·,E(x,m)(Y))>.
4.1 Series representation
Theorem 4.1. Under the assumptions(A1)−(A6), for any measurable functionl(x) = (l(x, i))i∈E,
Ex Z τ0 0 e−δtl(Xt, Jt)dt = Z ∞ 0 r(δ)(x, y)l(y)dy, (4.1) where r(δ) is given by r(δ)(x, y) = Z [0,x) G(du)∆−σ21/2B(x−u, y) ! ,
which is the density of potential measure for X killed on exiting [0,∞), and Q is a solution to the Lyapunov equation P Q+QR=I,
B(x, y) =
e−P u(eP yQ−Qe−Ry), 0< y < u; (QeRu−e−P uQ)e−Ry, y > u.
Proof. Since ∆σ2 is nonsingular, the point x = 0 is regular: P(τ0 = 0) = 1. Hence, H(0) = 0
when l is measurable, and it follows from Corollary 3.2 that
H(x) =
Z
[0,x)
G(dy)∆−σ21/2EPTRl(x−y), x≥0.
Note thatP Q+QR=I. Using the identity (A.6), we must have
EPTRl(x) = e−P x Z x 0 eP yQ−Qe−Ryl(y)dy + Z ∞ x QeRx−e−P xQe−Ryl(y)dy= Z ∞ 0 B(x, y)l(y)dy.
Changing the order of integration yields the desired solution.
Remark 4.1. According to Theorem 4.4.6 of Horn and Johnson [26, p270], the equationP Q+
QR =I has a unique solution Q if and only if P and R have no common eigenvalues. This is clearly the case when ∆2µ/σ2 6= 0. In this case, the vector of stacked columns of Q denoted by
vecQ is determined by
vecQ= (I⊗P+R>⊗I)−1vecI,
where ⊗ denotes the Kronecker product.
Interestingly, the Lyapunov equation seems to play an important role in the representations of many exit problems for MAPs. A similar Lyapunov equation appears in Albrecher and Ivanovs [3] for the survival probability of a MAP risk model with Poissonian observations.
4.2 Scale matrix representation
Theorem 4.2. Under the assumptions(A1)−(A5), for any measurable functionl(x) = (l(x, i))i∈E,
Ex Z τ0 0 e−δtl(Xt, Jt)dt = Z ∞ 0 r(δ)(x, y)l(y)dy,
where r(δ)(x, y) =W(δ)(x)e−Ry−W(δ)(x−y),for x, y≥0,which is the density of the potential measure for X killed on exiting [0,∞).
Proof. We first assume that all assumptions of Theorem 3.2 are satisfied. Since ∆σ2 is
non-singular, we must have H(0) = 0.Hence,
H(x) = Z x 0 Π(dy)TRl(x−y) = Z x 0 h W(δ)(dy)−W(δ)(y)RdyieR(x−y) Z ∞ x−y e−Rzl(z)dz = Z x 0 Z x x−z W(δ)(dy)e−Ry−W(δ)(y)Re−Rydy eR(x−z)l(z)dz + Z ∞ x Z x 0 W(δ)(dy)e−Ry−W(δ)(y)Re−Rydy eR(x−z)l(z)dz.
Note thatW(δ)(dy) =W(δ)0(y) dy on (x−z, x).Thus, using integration by parts, we obtain
Z x x−z W(δ)(dy)e−Ry−W(δ)(y)Re−Rydy = Z x x−z W(δ)(dy)e−Ry+W(δ)(x)e−Rx−W(δ)(x−z)e−R(x−z)− Z x x−z dW(δ)(y)e−Ry = W(δ)(x)e−Rx−W(δ)(x−z)e−R(x−z).
Similarly, because of (3.16), we have
Z x 0 W(δ)(dy)e−Ry−W(δ)(y)Re−Rydy = Z x 0 W(δ)(dy)e−Ry+W(δ)(x)e−Rx−W(δ)(0)− Z x 0 dW(δ)(y)e−Ry=W(δ)(x)e−Rx.
Using the fact that W(δ)(z) = 0 when z <0, we have proved the desired expression.
The two representations of the resolvent density for a spectrally negative MAP are gener-alizations of those for a spectrally negative L´evy process shown in the introduction. The next theorem establishes the equivalence of the two representations.
Theorem 4.3. Under the assumptions (A3)−(A6),
W(δ)(x) =
Z
[0,x)
G(dy)∆−σ21/2(Qe
R(x−y)−e−P(x−y)Q), x≥0.
Proof. Denote the expression on the right-hand side by RHS and take Laplace transform.
L{RHS}(r) = Z ∞ [0,∞) e−ryG(dy)∆σ−21/2[Q(rI−R) −1−(rI+P)−1Q] = Z ∞ [0,∞) e−ryG(dy)∆σ−21/2(rI+P) −1(rI−R)−1,
with the last equality from the fact that P Q+QR=I.It is easy to show that Z ∞ 0 e−ryTR∆ν(y)dy = (R−rI)−1[ψS(rI)−ψS(R)]. Thus, we obtain Z [0,∞) e−ryG(dy) = I− Z [0,∞) e−ryG(y)dy !−1 = I−∆−σ21/2(P+rI) −1(R−rI)−1hψ S(rI)−ψS(R) +L{D◦F}(rI)− L{D◦F}(R) i−1 = (δI−K(rI))−1(R−rI)(P+rI)∆σ2/2, since (R−rI)(P +rI)∆σ2/2− h ψS(rI)−ψS(R) +L{D◦F}(rI)− L{D◦F}(R) i = h R2∆σ2/2+R∆µ+ψS(R) +L{D◦F}(R) +C i −hr2∆σ2/2+r∆µ+ψS(rI) +L{D◦F}(rI) +C i =δI−K(rI).
Therefore, simple substitutions yieldL{RHS}=L{W(δ)}= (K(rI)−δI)−1,which confirms the equality as the scale matrix is uniquely determined by its Laplace transform.
5
Applications
Quantities of interest in ruin theory are often functionals of underlying risk processes, many of which are special cases of the function H. By identifying the corresponding l, we can obtain explicit solutions to all such functionals by Theorems 4.1 and 4.2.
5.1 Total claim costs up to ruin
The expected present value (EPV) of penalty at ruin, known as the Gerber-Shiu function, is the classical tool used to analyze the joint distribution of the surplus prior to ruin, the deficit at ruin as well as the time of ruin. It is known in many risk models driven by Markov processes that the Gerber-Shiu function can be retrieved from the EPV of total claim costs up to ruin, which by itself is a special case of function H. We can easily draw a similar connection among these quantities in the context of Markov additive processes.
A natural extension of the EPV of total claim costs up to ruin in the Markov additive risk model is given by M(x) =Ex X 0≤t≤τ0 e−δt$(Jt−,Jt)(Xt−, Xt)I(|∆Xt|>0) ,
where$(i,j)(x, y) takes into account the cost of every claim depending on the state of MAP (x, i) prior to the claim and the state (y, j) immediately following the claim. We write ∆$(x, y) = diag($(i,i)(x, y))i∈E and$= ($(i,j)(x, y))i,j∈E.
Proposition 5.1. M can be obtained from H by letting
l(x) = Z ∞ 0 ∆$(x, x−z)∆ν(dz) + Z ∞ 0 D◦$(x, x−z)◦F(z)dz 1. (5.1)
Proof. Consider the case J0 =i. Let T1 = inf{t :Jt 6= i}. Using the strong Markov property, we obtain Mi(x) = E(x,i) h X 0≤t<T1 e−δt$(Jt−,Jt)(Xt−, Xt)I(∆Xt>0) +e−δT1$ (JT1−,JT1)(XT1−, XT1)I(∆XT1 >0) +e −δT1M JT1(XT1) i . (5.2) Note thatXt=Xt(i) and Jt=ifor 0≤t < T1. Thus,
E(x,i) X 0≤t<T1 e−δt$(Jt−,Jt)(Xt−, Xt)I(∆Xt>0) =E(x,i) Z T1 0 e−δt Z ∞ 0 $(i,i)(Xt, Xt−z)νiS+Z(dz)dt , (5.3)
where we used the compensation formula (cf. Section 4.3.2 of Kyprianou [32] or Bertoin [6], page 7). Similarly, we can show
E(x,i) h e−δT1$ (JT1−,JT1)(XT1−, XT1)I(∆XT1 >0) i = E(x,i) Z T1 0 e−δtX j∈E λijqij Z ∞ 0 $(i,j)(Xt, Xt−z)fij(z)dzdt . (5.4)
In view of (5.2), (5.3) and (5.4), we obtain
Mi(x) =E(x,i) h e−δT1M JT1(XT1) i +E(x,i) Z T1 0 e−δt Z ∞ 0 $(i,i)(Xt, Xt−z)νiS+Z(dz) + X j∈E λijqij Z ∞ 0 $(i,j)(Xt, Xt−z)fij(z)dz dt . (5.5) Applying the strong Markov property to H(x, i), we also obtain
H(x, i) =E(x,i) Z T1 0 e−δtl(Xt)dt +E(x,i)[e−δT1H(XT1, JT1)]. (5.6)
It follows immediately from (5.5) and (5.6) that M is a special case of H with (5.1).
Remark 5.1. Similar arguments show that the following quantities are all special cases ofH. 1. Expected present value (EPV) of total costs due to regime changes (a transition in the
Markov chain J): MC(x) = (MC 1 (x),· · ·, MmC(x))> where MiC(x) =E(x,i) X 0≤t≤τ0 e−δt$(Jt−,Jt)(Xt−, Xt)I(|∆Xt|>0, Jt−6=Jt) ,
corresponds to the cost function
l(x) = Z ∞ 0 D◦$(x, x−z)◦F(z)dz 1;
2. EPV of total costs due to jumps caused by the subordinator within each regime: MD(x) = (M1D(x),· · · , MmD(x))> where MiD(x) =E(x,i) X 0≤t≤τ0 e−δt$(Jt−,Jt)(Xt−, Xt)I(|∆Xt|= ∆SJt(t)>0, Jt−=Jt) ,
corresponds to the cost function
l(x) = Z ∞ 0 ∆$(x, x−z)∆νS(dz) 1;
3. EPV of total costs due to jumps caused by the SPL process within each regime: MD(x) = (M1D(x),· · · , MmD(x))> where MiD(x) =E(x,i) X 0≤t≤τ0 e−δt$(Jt−,Jt)(Xt−, Xt)I(|∆Xt|= ∆ZJt(t)>0, Jt−=Jt) ,
corresponds to the cost function
l(x) = Z ∞ 0 ∆$(x, x−z)∆νZ(dz) 1;
4. The Gerber-Shiu function: m·j(x) = (m1j(x),· · · , mmj(x))> for any j∈E, where
mij(x) =E(x,i) h e−δτ0w (Jτ0−,Jτ0)(Xτ0−,|Xτ0|)I(τ0<∞, Jτ0 =j) i ,
corresponds to the cost function
l(x) = Z ∞ x ∆w(x, z−x)∆ν(dz) ·j + Z ∞ x D◦w(x, z−x)◦F(z)dz ·j + w(0,0)δ0(x) ·j, (5.7) where ∆w(x, y) = diag(w(i,i)(x, y))i∈E and w(x, y) = (w(i,j)(x, y))i,j∈E.
Remark 5.2. In the case wherew(0,0) = 0, we apply Theorem 4.2 with the cost function (5.7) to obtain the solution to the Gerber-Shiu function m(x) = (mij(x))i,j∈E
m(x) = Z ∞ 0 Z ∞ 0 R(δ)(x, y)h∆w(y, u)∆ν(du+y) +D◦w(y, u)◦F(u+y)du i dy.
This formula is a generalization of Theorem 4.1 of Asmussen and Albrecher[4, p.385].
5.2 Total operating costs up to ruin with ruin occurring
In many papers on MAP risk models, the state of the Markov chainJ at the time of ruinτ0 is
specified in the definition of functionals. In doing so, they implicitly assume thatτ0<∞, which
is restricted to the cases where ruin does occur. Here we want to point out that such functionals are also special cases ofH. For example, defineU(x) = (Uij(x))i,j∈E where
Uij(x) =E(x,i) " Z τ0 0 e−δtl∗Jt(Xt)dt ! I(τ0 <∞, Jτ0 =j) # .
Proposition 5.2. U can be obtained fromH by letting l= (l∗◦L·j)j∈E,where L(x) = Z ∞ 0 Z ∞ 0 R(0)(x, y)h∆ν(du+y) +D◦F(u+y)du i dy. (5.8)
Proof. We can use the strong Markov property to representU in the form ofH.
Uij(x) = E(x,i) " Z ∞ 0 e−δtl∗Jt(Xt)I{Xs≥0∀0≤s≤t}I{τ0<∞,Jτ0=j}dt !# = E(x,i) Z ∞ 0 E(x,i) n e−δtl∗Jt(Xt)I{Xs≥0∀0≤s≤t}I{τ0<∞,Jτ0=j} Ft o dt = E(x,i) Z ∞ 0 e−δtl∗Jt(Xt)I{Xs≥0∀0≤s≤t}E(x,i) n I{τ0<∞,Jτ0=j} Ft o dt = E(x,i) Z τ0 0 e−δtlJ∗t(Xt)PJt,j(Xt)dt ,
where Lij(x) =P(x,i){τ <∞, Jτ =j} is the ruin probability. The expression (5.8) follows from Remark 5.2 by letting w(x, y) = 1, which is a matrix of all 1’s.
5.3 Connections to exit problems
Kyprianou and Palmowski [34] established many identities regarding exit problems of MAPs. The paper employs time reversal arguments and the Asmussen-Kella martingale, with an anal-ogy to a similar technical treatment for a spectrally negative Levy process in Kyprianou and Palmowski [33]. As by-products of results from Section 5.2, we are able to reaffirm some of their identities. Although not as comprehensive as the approach of Kyprianou and Palmowski [34], our approach is based merely on elementary matrix operator identities. It should also be noted that no assumption on the stationary distribution ofJ is made in this paper, whereas the time reversal arguments in Kyprianou and Palmowski [34] requiredJ0to have stationary distribution.
This matrix operator approach avoids the stationary distribution assumption so that delayed risk models in which J0 has an arbitrary distribution can be analyzed in the same framework.
First, we provide a result for the Laplace transform of the time of ruin (interpreted as the discounted probability of ruin) defined by V(x) = (Vij)i,j∈E,where
Vij(x) =E(x,i)[e−δτ0I(τ0<∞, Jτ0 =j)].
Theorem 5.1. The Laplace-Stieltjes transform of V is given by
Z ∞
0
e−rxdV(x) = (K(r)−δI)−1(I−rR−1)(δI−Λ). (5.9)
Proof. Clearly, the function Vij can be retrieved from the Gerber-Shiu function mij by letting
wij(x, y) = 1. In view of (3.20), LV(r) = R∞ 0 e −ryV(y)dy must satisfy (rI−R)−1{K(rI)−δI} · LV(r) =L{TRA}(r) + ∆σ2/2V(0), where V(0) =I and A(x) :=A1(x) +A2(x) +A3(x) := Z ∞ x ∆ν(dz) + Z ∞ x D◦F(z)dz+δ0(x)I.
Because of (A.3), we obtain
We claim that LA(R)− LA(rI) = (rR)−1 " RψS+L(rI) +L{D◦F}(rI)−D −rψS+L(R) +L{D◦F}(R)−D # . (5.11) It follows that LA(R)− LA(rI) + (rI−R)1 2∆σ2 = (rR)−1 " R r∆µ+ψS+L(rI) +L{D◦F}(rI)−D −r R∆µ+ψS+L(R) +L{D◦F}(R)−D # + (rR)−1 1 2Rr 2∆ σ2− 1 2rR 2∆ σ2 = (rR)−1 h RK(r)−R(C+D)−rK(R) +r(C+D) i = (rR)−1 h RK(r)−RΛ−rK(R) +rΛ i .
Therefore, it follows from (B.6) that
LV(r) = (K(r)−δI)−1(rR)−1 h RK(r)−RΛ−rK(R) +rΛ i . In other words, Z ∞ 0 e−rxdV(x) = r Z ∞ 0 e−rxV(x)dx−V(0) = (K(r)−δI)−1R−1hRK(r)−RΛ−rK(R) +rΛi−I = (K(r)−δI)−1R−1 h δR−RΛ−rK(R) +rΛ i .
which leads to (5.9) after rearrangement.
Last, we give a proof of (5.11). For any positive stable matrixS, we have that
LA2(S) = Z ∞ 0 Z z x e−SxD◦F(dz) = Z ∞ 0 S−1(I−e−Sz)D◦F(dz) =S−1[D− L{D◦F}(S)]. Hence, LA2(R)− LA2(rI) = (Rr)−1 " R Z ∞ 0 e−rzD◦F(dz)−D −r Z ∞ 0 e−RzD◦F(dz)−D # .
For any positive stable matrixS and >0, we also have that
Z ∞ e−Sx Z ∞ x ∆ν(dz)dx = Z ∞ Z z e−Sxdx∆ν(dz) = S−1e−S Z ∞ ∆ν(dz)−S−1 Z ∞ e−Sz∆ν(dz).
Note thatνiS+L is a finite measure on [,∞) for alli∈E. Define ˜l1(r) :=R∞e−rxl1(x)dx.Then ˜ l1(R)−˜l1(rI) = (Rr)−1 " re−R Z ∞ ∆ν(dz)−Re−r Z ∞ ∆ν(dz) +R Z ∞ e−rz∆ν(dz)−r Z ∞ e−Rz∆ν(dz) +R Z ∞ (e−rz+rz−1)∆ν(dz)−r Z ∞ (e−Rz+Rz−I)∆ν(dz) # = (Rr)−1 " r(e−R+R−I) Z ∞ ∆ν(dz)−R(e−r+r−1) Z ∞ ∆ν(dz) +R Z ∞ (e−rz+rz−1)∆ν(dz)−r Z ∞ (e−Rz+Rz−I)∆ν(dz) # .
Noticing that |2I(z≥)| ≤z2 and thatR01z2νS+L(dz)<∞, we see that
(e−r+r−1) Z ∞ ∆νS(dz)≤ r2 2 Z 1 0 2I(z≥) ∆ν(dz)→0, →0,
by the Lebesgue convergence theorem, which yields (5.11).
Secondly, we can also produce a result regarding the running infimum of the process denoted by Xt:= inf0≤s≤tXs.Define the matrix functionO by
Oij(s) =E(0,i)[esXeδI(Jeδ =j)]. Corollary 5.1. The solution to O is given by
O(s) =δ(K(s)−δI)−1(sR−1−I). (5.12) Proof. Leteδ be an independent exponential random variable with mean 1/δ. Note that
P(x,i)(τ0 < eδ, Jeδ =j) = E (x,i)[I(τ 0 < eδ)E(Xτ0,Jτ0){I(Je δ−τ0 =j)}] = X k∈E P(x,i)(τ0 < eδ, Jτ0 =k)P{J0 =k, Jeδ =j}
with last equality from the memoryless property of eδ and thatJ is independent of X. SinceE(x,i)[e−δτ0I(τ 0<∞, Jτ0 =j)] =P (x,i)(τ 0 < eδ, Jτ0 =j),we see that P(0,i)(−Xeδ > x, Jeδ =j) =P (x,i)(τ 0 < eδ, Jeδ =j) = X k∈E E(x,i)[e−δτ0I(τ0 <∞, Jτ0 =k)]P{J0 =k, Jeδ =j}=δ X k∈E vik(x)(δI−Λ)−kj1. Note that Oij(x) = R∞ 0 e −sxd
P(0,i)(−Xeδ ≤ x, Jeδ =j). Hence, the desired result follows from O(x) =−R∞
0 e
−sxdV(x){δ(δI−Λ)−1}.
Remark 5.3. The formulas (5.9) and (5.12) are in exact agreement with equation (7), (9) of Kyprianou and Palmowski [34]. Note that (9) was derived from (7) in Kyprianou and Pal-mowski [34] using time-reversal and Asmussen-Kella martingale arguments but our proof is much simpler. A quick comparison shows that the matrix Dˆ(q)> (they used q in place of δ) in their paper is effectively the matrix R−1 in this paper. It is worthwhile to note that Kyprianou and Palmowski [34] stated (p.125) “establishing an expression of Λ(q) for spectrally negative MAPs is an open question”. This problem has been addressed in Ivanovs and Palmowski [29] but here we provide an alternative derivation using integral and differential operators.
Acknowledgement
We would like to thank Dr. Jevgenijs Ivanovs for bringing to our attention his recent work [27] on the subject. The invaluable suggestions from Dr. Ivanovs and the anonymous referee for improving the paper are greatly appreciated. This research has been partially supported by the Ministry of Education, Science, Sports and Culture, Grant-in-Aid for Young Scientists (B), no.24740061, 2013–2014, and Japan Science and Technology Agency, CREST.
A
Matrix identities
For brevity, we often use e−S· to indicate the function x7→e−Sx. We suppress the argumentx
whenever it is clear from context.
Lemma A.1. For any matricesS1, S2, and vector function h,
ES1(S1+D)h=H−e
−S1·h(0); (A.1)
ES1(S1−S2)ES2h=ES2H− ES1h; (A.2) TS1(S2−S1)TS2h=TS1H− TS2h, (A.3)
provided that h is of polynomial growth,S1 and S2 are positive stable and compatible in (A.3).
Proof. (A.1): Using the integration by parts entrywise and putting them in matrix form
ES1{S1h+h0}=S1e−S1x Z x 0 eS1yh(y)dy+e−S1x Z x 0 eS1ydh(y) = S1e−S1x Z x 0 eS1yh(y)dy+h−e−S1xh(0)−e−S1x Z x 0 deS1xh(y) =h−e−S1xh(0).
(A.2): Changing the order of integration, we have
ES1(S1−S2)ES2h = Z x 0 e−S1(x−y)(S 1−S2) Z y 0 e−S2(y−z)h(z)dzdy = Z x 0 e−S1x Z x z eS1y(S 1−S2)e−S2ydy eS2zh(z)dz. Note that d dye S1ye−S2y =eS1y(S 1−S2)e−S2y,
which implies that
Z x z eS1y(S 1−S2)e−S2ydy =eS1xe−S2x−eS1ze−S2z. Therefore we get ES1(S1−S2)ES2h= Z x 0 e−S2(x−z)h(z)dz− Z x 0 e−S1(x−z)h(z)dz,
which completes the proof.
(A.3): Substituting w=u+v, we have
TS1(S2−S1)TS2h = Z ∞ 0 Z ∞ 0 e−uS1(S 2−S1)e−vS2h(x+u+v)dudv = Z ∞ 0 Z w 0 e−uS1(S 2−S1)euS2du e−wS2h(x+w)dw.
Note that d due −uS1euS2 =e−uS1(S 2−S1)euS2, which implies Z w 0 e−uS1(S 2−S1)euS2du=e−wS1ewS2−I. Therefore, we get TS1(S2−S1)TS2h= Z ∞ 0 e−wS1h(x+w)dw− Z ∞ 0 e−wS2h(x+w)dw
completing the proof.
Lemma A.2. For any positive stable matricesS1, S2 compatible with a matrixQ, and any vector
function h of polynomial growth, the following identities hold true.
TS1(S1− D)h=h; (A.4)
TS1(S1Q+QS2)ES2h=TS1Qh+QES2h, (A.5) ES1(S1Q+QS2)TS2h=ES1Qh+QTS2h−e
−S2·QLh(S
2). (A.6)
Proof. We first note that, forj= 1,2, lim y→0e
−Sjyh(y) = 0. (A.7)
(A.4): By the integration-by-parts, it follows from (A.7) that
TS1(S1− D)h = e S1x Z ∞ x e−S1yS 1h(y)dy−eS1x Z ∞ x e−S1ydh(y) = eS1x Z ∞ x e−S1yS 1h(y)dy− −h(x) +eS1x Z ∞ x e−S1yS 1h(y)dy =h.
(A.5): We prove similarly by changing the order of integrations
TS1(S1Q+QS2)ES2h = Z ∞ 0 e−S1y(S 1Q+QS2) Z x+y 0 e−S2(x+y−z)h(z)dzdy = Z x 0 Z ∞ 0 e−S1y(S 1Q+QS2)e−S2ydy e−S2(x−z)h(z)dz + Z ∞ x Z ∞ z−x e−S1y(S 1Q+QS2)e−S2ydy e−S2(x−z)h(z)dz.
Because of (A.7), we must have
Z ∞ x e−S1y(S 1Q+QS2)e−S2ydy=e−S1xQe−S2x. Therefore, TS1(S1+S2)ES2h=Q Z x 0 e−S2(x−z)h(z)dz+ Z ∞ x e−S1(z−x)Qh(z)dz.
(A.6): We prove in a similar manner that ES1(S1Q+QS2)TS2h = e−S1x Z x 0 Z z 0 eS1y(S 1Q+QS2)eS2y e−S2zh(z)dz +e−S1x Z ∞ x Z x 0 eS1y(S 1Q+QS2)eS2y e−S2zh(z)dz = e−S1x Z x 0 eS1zQ−Qe−S2zh(z)dz+ Z ∞ x QeS2x−e−S1xQe−S2zh(z)dz,
which yields the identity after rearrangement.
Lemma A.3. Suppose thatS is diagonalizable and positive stable. Then it holds that
L{TSHν}(r) = (rI−S)−1 h ψS+L(S) +L{D◦F}(S)−ψS+L(rI)− L{D◦F}(rI) i · LH(r) +L[TS{N(S)H}](r), (A.8) where the constant matrix N(S) =ψS+L(S) +L{D◦F}(S).Furthermore, if∆νZ ≡0,then
TS{Hν}=TS{N(S)H}+ [TS∆ν+TS{D◦F}]∗H. (A.9) Proof. SinceSis diagonalizable, there must exist a matrix Φ such thatS = Φ−1diag(s1,· · · , sm)Φ wheres1,· · · , sm>0. We denote the three terms on the right-hand side of (3.5) byH1, H2 and
H3 respectively. Firstly, TSH1(u) = eSu Z ∞ u e−Sy Z ∞ 0 ∆νS(dz){H(y−z)−H(y)} dy = eSu Z ∞ u e−Sy Z ∞ 0 n ∆νS(dz)H(y−z)−(I−e−Sz+e−Sz)∆νS(dz)H(y) o dy = eSu Z ∞ u e−Sy Z ∞ 0 n ∆νS(dz)H(y−z)−e−Sz∆νS(dz)H(y) o dy +TS{ψS(S)H}(u).
Hence we need to evaluate the first term which we shall denote byJ1(u).
J1(u) = eSu Z ∞ 0 Z ∞ u dy e−Sy∆νS(dz)H(y−z)− Z ∞ u dy e−S(y+z)∆νS(dz)H(y) = eSu Z ∞ 0 Z ∞ u dy e−Sy∆νS(dz)H(y−z)− Z ∞ u+z dy e−Sy∆νS(dz)H(y−z) = eSu Z ∞ 0 Z u+z u dy e−Sy∆νS(dz)H(y−z) = eSu Z ∞ 0 Z u u−z dx e−S(x+z)∆νS(dz)H(x) = Z u 0 dx eS(u−x) Z ∞ u−x e−Sz∆νS(dz)H(x) = Z u 0 dy eSy Z ∞ y e−Sz∆νS(dz)H(u−y) =T∆ν∗H(u),
where we have used many times the fact that H(x) = 0 whenx <0. To summarize,
or equivalently, L{TSH1}(r) = (rI−S)−1[ψS(S)−ψS(rI)]· LH(r) +L{TS{ψS(S)H}}(r). (A.10) Secondly, TSH2(u) =eSu Z ∞ u e−Sy Z ∞ 0 ∆νZ(dz){H(y−z)−H(y) +zH0(y)} dy = eSu Z ∞ u e−Syh Z ∞ 0 ∆νZ(dz)H(y−z)−(I−e−Sz−Sz+e−Sz +Sz)∆νZ(dz)H(y) +z∆νZ(dz)H0(y) i dy = eSu Z ∞ u e−Syh Z ∞ 0 ∆νZ(dz)H(y−z)−(e−Sz+Sz)∆νZ(dz)H(y) +z∆νZ(dz)H0(y) i dy+TS{ψL(S)H}(u). We simplify the first term denoted by J2(u).
J2(u) =eSu Z ∞ 0 hZ ∞ u e−Sy∆νZ(dz)H(y−z)dy− Z ∞ u e−Sy(e−Sz+Sz)∆νZ(dz)H(y)dy + Z ∞ u e−Syz∆νZ(dz)H0(y)dy i = eSu Z ∞ 0 hZ ∞ u e−Sy∆νZ(dz)H(y−z)dy− Z ∞ u e−S(y+z)∆νZ(dz)H(y)dy−e−Suz∆νZ(dz)H(u) i = eSu Z ∞ 0 hZ u+z u e−Sy∆νZ(dz)H(y−z)dy−e−Suz∆νZ(dz)H(u) i .
In the penultimate equality we use the fact that limy→∞e−Sy∆νZ(dz)H(y) = 0, which follows
from (A.7) and that H is bounded and νiZ is finite on a compact set apart from the origin for all i∈E. Applying Laplace transforms yields
L{J2}(r) = Z ∞ 0 Z ∞ 0 e−(rI+S)u Z u u−z e−S(w+z)∆νZ(dz)H(w)dwdu− Z ∞ 0 e−ruz∆νZ(dz)H(u) = Z ∞ 0 h (rI−S)−1[e−(rI−S)w−e−(rI−S)(w+z)]e−S(w+z)∆νZ(dz)H(w)dw−z∆νZ(dz)LH(r) i = Z ∞ 0 h (rI−S)−1[e−(rw−S)z−e−r(w+z)]∆νZ(dz)H(w)dw−z∆νZ(dz)LH(r) i = Z ∞ 0 [(rI−S)−1[e−Sz∆νZ(dz)LH(r)(r)−e−rz∆νZ(dz)LH(r)(r)]−z∆νZ(dz)LH(r)] = (rI−S)−1 Z ∞ 0 [e−Sz +Sz−I+e−rzI+rzI−I]∆νZ(dz)LH(r) = (rI−S)−1[ψL(S)−ψL(rI)]· LH(r). Thus, L{TSh2}(r) = (rI−S)−1[ψL(S)−ψL(rI)]· LH(r) +L{TS{ψL(S)H}}(r). (A.11) Finally, we see that
TSH3(u) =eSu Z ∞ u e−Sy Z y 0 D◦F(y−w)H(w)dwdy = eSu Z u 0 Z ∞ u−w e−S(y+w)D◦F(y)dyH(w)dw+eSu Z ∞ u Z ∞ w e−SyD◦F(y−w)dyH(w)dw = TS{D◦F} ∗H(u) +TS{L{D◦F}(S)H}(u).
Thus,
L{TSH3}(r) = (rI−S)−1[L{D◦F(S)} − L{D◦F(rI)}]· LH(r) +L[TS{L{D◦F}(S)H}](r). (A.12) Combining (A.10), (A.11) and (A.12) gives the identity (A.8). When ∆νZ ≡0,thenJ2 ≡0 and
hence the identity (A.9) follows immediately.
B
Detailed proofs
Proof of Lemma 2.2: LetFδ(s) :=K(sI)−δI. The assumption means that the multiplicity of ρj is 1, which implies that a Jordan pair (vj, ρj) in the sense of Definition 2 in D’Auria et al. [16] satisfies Fδ(ρ
j)vj = 0, that is, each vj is the first Jordan chain corresponding to ρj. Then, by Theorem 1 in D’Auria et al. [16], we find that
v= [v1, . . . , vm], Γ = diag[ρ1, . . . , ρm] are m×m invertible matrices. Using these, we shall construct
R:= (v>)−1Γv> (B.1) Then v>[K(R)−δI] =v>C+ Γ2v>∆σ2/2+ Γv>∆µ+ Z ∞ 0 (e−zΓ−I)v>∆νS(dz) + Z ∞ 0 (e−zΓ−I+zΓ)v>∆νZ(dz) + Z ∞ 0 e−zΓv>D◦F(z) dz−δv>.
Observing the rows of the above matrix, we see that, for each i= 1, . . . , m,
vi>C+ρ2iv>i ∆σ2/2+ρivi>∆µ+ Z ∞ 0 (e−zρi−1)v> i ∆νS(dz) + Z ∞ 0 (e−zρi−1 +zρ i)vi>∆νZ(dz) + Z ∞ 0 e−zρiv> i D◦F(z) dz−δv > =v>i [K(ρi)−δI] = (0, . . . ,0),
which implies that v>[K(R)−δI] = 0.Sincev> is invertible, we conclude that K(R)−δI = 0. Note thatR is diagonalizable by definition, and is positive stable since Re(ρj)>0.
Proof of Lemma 3.1: Letσ:= inf{t:Jt6=i, J0=i}andSn= inf{t:Xt6∈(x−1/n, x+ 1/n)} for sufficiently large nsuch that x−1/n >0. Using the usual arguments involving the strong Markov property (similar to the proof of Theorem 2.1 of Feng [21]), we can show that
E(x,i)[e−δ(σ∧Sn)H(Xσ∧Sn, Jσ∧Sn)]−H(x, i) =−E (x,i) Z σ∧Sn 0 e−δsl(Xs, Js)ds . (B.2)
Denote the left-hand side of (B.2) by h1 and the RHS by −h2. Moreover, since Xt =Xt(i) for
t∈[0, σ), where {Xt(i), t≥0} is a L´evy process with the characteristic triple (σi, µi, νiS+Z), one can show in a manner similar to Lemma A.1 of Feng [21] that
E(x,i)[e−δ(σ∧Sn)H(X((σi)∧Sn)−, i)] =H(x, i) +E(x,i) Z σ∧Sn 0 e−δs(A −δ)H(Xs(i), i)ds ,
where Ais the infinitesimal generator of X(i) given by Ah(x, i) = σ 2 i 2 h 00(x, i) +µ ih0(x, i) + Z ∞ 0 {h(x−z, i)−h(x, i)}νS(dz) + Z ∞ 0 {h(x−z, i)−h(x, i) +zh0(x, i)}νiZ(dz).
Thus, we obtain that
h1(x, i) =E(x,i) h e−δ(σ∧Sn)H(X σ∧Sn, Jσ∧Sn)−H(X (i) (σ∧Sn)−, i) i +E(x,i) h e−δ(σ∧Sn)H(X(i) (σ∧Sn)−, i) i −H(x, i) =E(x,i) h e−δ(σ∧Sn)H(X σ∧Sn, Jσ∧Sn)−H(X (i) (σ∧Sn)−, i) i +E(x,i) Z σ∧Sn 0 e−δs(A −δ)H(Xs(i), i)ds (B.3)
Note that the first transition in J from i to j can be viewed as the first event of a Poisson random measure Nij independent of X with the intensity measure νij for which νij(dy) =
λijqijfij(y)dy+λij(1−qij)δ0(y)dyfori, j∈E, whereδ0 is the Dirac function on{0}. Therefore,
we have E(x,i)[e−δ(σ∧Sn) H(Xσ∧Sn, Jσ∧Sn)−H(X(σ∧Sn)−, i) ] =X j∈E E(x,i) Z σ∧Sn 0 e−δs Z ∞ 0 H(Xs(i−)−y, j)−H(X (i) s−, i) Nij(ds,dy) +X j∈E E(x,i) Z σ∧Sn 0 e−δsH(Xs(i), j)−H(Xs(i), i)λij(1−qij) ds =X j∈E E(x,i) Z σ∧Sn 0 e−δs Z ∞ 0 H(Xs(i)−y, j)λijqijfij(y)dy+H(Xs(i), j)λij(1−qij) ds =X j∈E DijE(x,i) Z σ∧Sn 0 e−δsfij(Xs(i)−z)H(z, j) dzds +X j∈E CijE(x,i) Z σ∧Sn 0 e−δsH(Xs(i), j) ds . (B.4)
For notational convenience, we let
Bh(x, i) =Ah(x, i) +X j∈E Cijh(x, j) +Dij Z ∞ 0 fij(x−z)h(z, j)dz .
Combining (B.3) and (B.4), we obtain
h1(x, i) =
Z σ∧Sn
0
Hence, on one hand, lim n→∞ h1(x, i) (1−Ex[e−δ(σ∧Sn)])/δ −(B −δI)H(x, i) = lim n→∞ E(x,i) h Rσ∧Sn 0 e −δsn(B −δI)H(X(i) s , Js)− BH(x, i) +δH(x, i) o ds i (1−Ex[e−δ(σ∧Sn)])/δ ≤ lim n→∞ E(x,i) h Rσ∧Sn 0 e −δs (B −δI)H(X (i) s , i)− BH(x, i) +δH(x, i) ds i (1−Ex[e−δ(σ∧Sn)])/δ ≤ lim n→∞y∈(x−1sup/n,x+1/n) {BH(y, i)−δH(y, i)} − BH(x, i) +δH(x, i) = 0,
with the second to last inequality arising from Jensen’s inequality and the last inequality from the fact that (B −δ)H(x, i) is continuous inxand Xs∈(x−1/n, x+ 1/n) which is bounded by a compact set.
On the other hand, lim n→∞ h2(x, i) (1−E(x,i)[e−δ(Sn∧σ)])/δ −l(x, i) ≤ lim n→∞y∈(x−1sup/n,x+1/n)|l(y, i)−l(x, i)|= 0,
due to the continuity ofl(x, i) for x6∈D. Since h1 ≡ −h2 by (B.2), we can conclude that
(B −δI)H(x, i) =−l(x, i), x6∈D, i∈E
which is the desired result (3.5).
Proof of Corollary 3.1:
When ∆σ2 = 0, (3.2) reduces to
{∆µD−(δI−C)}H(x) +{Hν +l}(x) = 0, x >0, or equivalently,
{(δI−C)∆−µ1−D}∆µH=Hν +l.
Note that the matrix (δI−C)∆−µ1 is positive stable under (A2) . Applying (A.4) gives
∆µH =TS(Hν+l), S := (δI−C)∆−µ1. (B.5) Applying (A.3) and using the identity (B.5) leads to
TS(Hν+l) =TR(Hν +l)− TR(S−R)TS(Hν+l) =TR(Hν +l)− TR(S−R)∆µH. (B.6) We expand the first term on the RHS of (B.6) according to (A.9),
TR(Hν +l) =TR{ψS+L(R) +L{D◦F}(R)}H+ [TR∆ν +TR{D◦F}]∗H+TRl. Note that (2.3) implies
(S−R)∆µ=ψS+L(R) +L{D◦F}(R). Thus, the second term on the RHS of (B.6) is given by
TR(S−R)∆µH =TR{ψS+L(R) +L{D◦F}(R)}H. Therefore,
∆µH=TS(Hν+l) = [TR∆ν +TR{D◦F}]∗H+TRl, which yields (3.11) after pre-multiplying ∆−µ1.
C
Solution to a quadratic matrix equation
Lemma C.1. Suppose that (δI −C)∆σ−21/2 is non-singular. Then, for an m×m matrix X, a
quadratic matrix equation
X2−X∆2µ/σ2 −(δI−C)∆−1
σ2/2 = 0, (C.1)
has a positive stable solution.
Proof. LetY :=X>, and consider the transposed equation of (C.1):
Y2−∆2µ/σ2Y −F>= 0, (C.2)
where F := (δI −C)∆−σ21/2, which is non-singular. Thanks to Theorem 3.2 by Guo [25], if F>
is an M-matrix, then (C.2) has a solution Y that is a non-singular M-matrix. Since it is true for any non-singular M-matrix that its transpose is also a non-singular M-matrix; see, e.g., Theorem 1.1 in Guo [25], if F is anM-matrix then a solution X = Y> is also a non-singular
M-matrix, which implies that the eigenvalues of X have positive real parts by Theorem 1.1 in Guo [25] again. Hence the proof is complete if we show that F is an M-matrix, which is equivalent to stating that there exists some positive diagonal matrix Dsuch thatF D= (γij) is strictly row diagonally dominant:
|γii|> m
X
j=1;j6=i
|γij|, i∈E; (C.3)
see Horn and Johnson [26], p.114, Theorem 2.5.3.13. To show (C.3) for F, we take D= ∆σ2/2
so that F D=δI−C= (γij). Note that, for each i∈E,
|γii|=δ−λii; |γij|=λij(1−qij) (j6=i). Since Pm
j=1λij = 0 by the property of the intensity matrix Λ, it is clear that
|γii| ≥ −λii= m X j=1;j6=i λij > m X j=1;j6=i |γij|,
that is, F is anM-matrix. This completes the proof.
D
Semi-Markov matrix
It is known from Markov renewal theory (c.f. C¸ inlar [12, Theorem 3.13]) that the Markov renewal equation (3.11) has a unique solution if Rx
0 g(y)dy is a semi-Markov matrix, i.e.
Z ∞
0
g(z)dz
1≤1. (D.1)
Although the proof appears technical, the idea is rather straightforward. It is easy to show that the functiongis semi-Markov for a MAP without a Gaussian component. In the general case, we extend the result using the weak convergence arguments developed in Gerber and Landry [23].
In the setting of Theorem 3.1, without loss of generality it is assumedZi(t) =σiBi(t) with
σ2i >0 for i∈E. We first replace the Gaussian components by processes of the form
Zi(k)(t) =d(ik)t− Z t 0 Z ∞ 0 zNi(k)(ds,dz), k >0, (D.2)
where Ni(k) is a Poisson random measure with E[Ni(k)(A, t)] = λ
(k)
i ∆k(A) for any Borel set A on (0,∞) and ∆a is the Dirac delta measure that assign measure one to the point a. In other words, Zi(k) is a Poisson process with intensityλk and fixed jump size ofk. We set
d(ik)= σ 2 i k , λ (k) i = σi2 k2, i∈E.
Then we see that, ask→0, the law of the process{Σ(k):= (Zi(k))i∈E} converges weakly to the law of a Brownian motion {B := (σiBi)i∈E} in D(0,∞) endowed with the Skorokhod topology;
see, e.g., Jacod and Shiryayev [31], Corollary VII.3.6.
Consider the process Xi(k)(t) = u+µit−S(i)(t)−Zi(k)(t) with Zi(k) defined in (D.2), and denote the characteristic exponent matrix of X(k) by
K(k)(s) =C+
n
s∆d(k)+ (e−ks−1)∆λ(k) o
+s∆µ+ψS(s) +L{D◦F}(s).
Our first aim is to find a positive stable matrixR(k)satisfying K(k)(R(k)) =δI, which converges toR given in the proof of Lemma 2.2.
Lemma D.1. For any ρj (j∈E) given in Lemma 2.2, there exists some ρ(jk) that is a solution to det[K(k)(sI)−δI] = 0such that ρ(jk) →ρj,as k→0.
Proof. Due to the fact that X(k) →d X in
D(0,∞) and the choice of d(ik) and λ
(k)
i , it follows that for any bounded subset S ⊂ {z∈C|Re(z)>0}and each i, j∈E that
sup s∈S
|Kij(k)(s)−Kij(s)| →0, k→0,
where K(s) is given in Lemma 2.2 withσi2>0. Now considering analytic functions onC
ζ(k)(s) := det[K(k)(sI)−δI], ζ(s) := det[K(sI)−δI].
We see from the above fact that ζ(k)(s) → ζ(s) as k → 0 uniformly in s ∈ S, and that the equation ζ(k)(s) = 0 hasm roots, say{ρ(k)
i , i= 1, . . . , m}by the same argument as in the proof of Lemma 2.2. Hence, by Hurwitaz’ theorem, we can choose a sequence ρ(ik) in a neighborhood of ρj such that ρ(ik)→ρj ask→0.
Lemma D.2. For each k >0, there exists a positive stable matrix R(k) satisfying K(k)(R(k)) =
δI. LetR be the matrix given in Lemma 2.2. Then lim
k→0R
(k)=R. (D.3)
Proof. Usingρj and ρ(jk) given in Lemmas 2.2 and D.1, respectively, we define
Qkj := [K(ρ(jk)I)−δI]>; Qj := [K(ρjI)−δI]>.
Then it follows that Qkj → Qj (k → 0) entrywise. Since rankQkj < m for all k > 0 from the definition of ρkj, we can takem non-zero vectors
Note that this v(jk) can be taken so that as k → 0, vj(k) 6→ 0, by taking normalized sequence
|v(jk)|= 1. Thus, the entrywise limit ofvj(k) must belong to KerQj because entrywise
Qkjv(jk)≡0, Qkj →Qj. Since ρj’s are distinct, it follows that
dim(KerQj) = 1, j= 1, . . . , m.
Then, we can take a sequence such thatvj(k)→cjvj (k→0) entrywise for somecj 6= 0, wherevj is given in Lemma 2.2: [K(ρjI)−δI]>vj = 0. Using the sequence vj(k), we can define matrices
v(k) = (c−1 1 v
(k)
1 , . . . , c−m1v
(k)
m ), which satisfies that v(k) → v entrywise, where v= (v1, . . . , vm) is given in Lemma 2.2. It follows that
|detv(k)| → |detv|>0, k→0.
Since detv(k) is continuous with respect to all components, there exists somek0 >0 such that
|detv(k)|>0 for any k < k0, which means that v(k) is non-singular for k < k0. We can define
R(k) := (v(k))>−1∆ρ(k)(v(k))>,0 < k < k0, where ρ(k) = (ρ(1k), . . . , ρ (k)
m )>. Thus, (D.3) is proved, and by the same argument as in the last half of the proof of Lemma 2.2, we also see that K(k)(R(k)) =δI.
Let T := inf{t > 0|Xt(k) < 0}, and let FS(k) denote the vector of “discounted” survival functions of the deficit caused by jumps of the subordinator.
FS(k)(x, z) :=Ex[e−δT$JT,JT−(X (k) T−, X (k) T )I(∆X (k) T =SJT(T)>0, JT−=JT)], where $(i,j)(x, y) =I(y <−z), z >0, fori, j∈E.
Then it follows from Remark 5.1(2) and (3.11) that
FS(k)(x, z) =
Z x
0
g∗(y)FS(k)(x−y, z)dy+ ∆−µ+1d(k)TR(k)l, (D.4)
where g∗ is some matrix that is not needed in the following derivation,
l(x) = Z ∞ 0 ∆$(x, x−y)∆νS(dy) = Z ∞ x+z ∆νS(dy) = ∆νS(x+z,∞).
Let x= 0 in (D.4). Thus we note that
FS(k)(0, z) = ∆−µ+1d(k) Z ∞
0
e−R(k)y∆
νS+Z(k)(y+z,∞)dy.
FS(k)(0, z) can be interpreted as the “discounted” survival distribution of first record low caused by the subordinator. Therefore, its corresponding “discounted” density is given by
FS(k)(z) :=− d dzF (k) S (0, z) = ∆ −1 µ+d(k)TR(k)∆ν0(z),
Similarly, we can define the vector of “discounted” survival functions of the deficit caused by jumps of fixed size k.
FZ(k)(x, z) :=Ex[e−δT$JT,JT−(X (k) T−, X (k) T )I(∆X (k) T =Z (k) JT(T) =k, JT−=JT)].
