getdoc8844. 139KB Jun 04 2011 12:04:37 AM

in PROBABILITY

A NOTE ON

R

-BALAYAGES OF MATRIX-EXPONENTIAL LÉVY

PRO-CESSES

YU-TING CHEN

Institute of Mathematics, Academia Sinica, Taipei, Taiwan

email: ythc_kh@yahoo.com.tw

YUAN-CHUNG SHEU1

Department of Applied Mathematics, National Chiao-Tung University, Hsinchu, Taiwan

email: sheu@math.nctu.edu.tw

Submitted27 Oct 2008, accepted in final form25 Mar 2009 AMS 2000 Subject classification: 60J75, 91B28, 91B30, 91B70

Keywords: Lévy process, matrix-exponential distribution, first exit, balayage, ruin theory

Abstract

In this note we give semi-explicit solutions for r-balayages of matrix-exponential Lévy processes. To this end, we turn to an identity for the joint Laplace transform of the first entry time and the undershoot and a semi-explicit solution of the negative Wiener-Hopf factor. Our result is closely related to the works by Mordecki in[11], Asmussen, Avram and Pistorius in[3], Chen, Lee and Sheu in[7], and many others.

1

Introduction

Given a Lévy processX = (Xt,t≥0)onR. Forx∈R, denote byPxthe law ofX under whichX0= x and write simplyP₀=P. Letψbe the Laplace exponent ofX, that is,E[eiξX1_{] =expψ(iξ)}

forξ∈R. For every killing rate r≥0, consider ther-balayage operatorPrdefined by

Prg(x)≡E_xe−rτg(X_τ), g∈ B_b(R₋), (1.1)

with the convention that r· ∞ = ∞. Here, by writing R₋ = (−∞, 0], we take τ = inf{t ≥

0;X_t ∈R₋} and B_b(R₋)as the set of all bounded Borel measurable functions defined on R₋. Solutions of r-balayages have been a major concern not only in probability theory itself but also in applied sciences such as mathematical finance and insurance mathematics (see[8]and[9]). It is, however, an undeniable fact that only few Lévy processes allow solutions for theirr-balayages. Besides the classical case of Lévy processes with no negative jumps, various authors have found that r-balayages have semi-explicit solutions when the triplet (r,g,X)is well chosen. See the works by Mordecki in [11], Avram and Usabel in [4], Asmussen, Avram and Pistorius in[3], Alili and Kyprianou in[1], Boyarchenko and Levendorskiˇi in[6], Chen, Lee and Sheu in[7], and

1_{THIS WORK WAS PARTIALLY SUPPORTED BY NATIONAL CENTER FOR THEORETICAL SCIENCE MATHEMATICS} DIVISION AND NSC 97-2628-M-009-014, TAIWAN.

many others. In particular, the salient feature of their choices of the underlying processX is that

X is taken from the class of matrix-exponential Lévy processes. A natural question then arises. We ask whether the class of matrix-exponential Lévy processes allow semi-explicit solutions for

r-balayages for all bounded measurable functionsgand allr≥0.

We say that a Lévy process is amatrix-exponential Lévy processif its downward Lévy measure is a finite measure and has a rational Laplace transform. In other words, the upward Lévy measure of−X is proportional to a matrix-exponential distributiond Fgiven by

d F(y) =

n

X

j=1

nj

X

k=1

Aj,kyk−1e−λjyd y, y>0. (1.2)

Note that the parametersA_j,k andλ_j are not necessarily real; see[10]. The two dense classes of distributions on(0,∞)– phase-type distributions and distributions which arelinearcombinations of exponential distributions – are both subclasses of matrix-exponential distributions. See [2]

Theorem 4.2 and[5]. In particular, by Proposition 1 in[3], matrix-exponential Lévy processes are dense in the class of Lévy processes.

The aim of this paper is to show that r-balayages admit semi-explicit solutions wheneverX is a matrix-exponential Lévy process, gis a bounded Borel measurable function onR_−andr>0. The case thatr=0 will also be considered under some restrictions on the moment ofX. In this paper, we obtain the desired semi-explicit solutions ofr-balayages with the help of a recent result on the negative Wiener-Hopf factor by Lewis and Mordecki in[10]and an identity for the joint Laplace transform of the first exit time and the undershoot by Alili and Kyprianou in[1]. The results are stated in Theorem 2.2. In this way, we overcome the difficulties of either giving a probabilistic interpretation for the jump distribution or transforming the generator of X into an differential operator. The former approach was treated by Asmussen, Avram and Pistorius in[3]for the case that (1.2) is phase-type, while the latter approach was expatiated by Chen, Lee and Sheu in[7]

for the case that the Lévy measure is a two-sided phase-type distributions. It is worth noting that in[3], the authors showed that a semi-explicit solution of anr-balayage can be determined implicitly by a system of linear equations. At the end of the paper, by Lagrange’s interpolation, we demonstrate the consistency of our result with an existing formula. In addition, an application of the formula of ther-balayage to another passage functional ofX is briefly discussed.

2

r

-Balayages and Negative Wiener-Hopf Factor

LetX(+)be a Lévy process with no negative jumps andZbe a compound Poisson process with jump rateλ >0 whose jump distribution is a matrix-exponential distribution given by (1.2). Assume thatX(+)andZare independent. Consider amatrix-exponential Lévy processX given by

X_t =X(+)_t −Z_t, t≥0. (2.1)

It follows from (1.2) and the Lévy-Khintchine formula that the Laplace exponent ψ takes the following form:

ψ(z) =σ 2

2 z

2_+µz+ Z

R₊

€

ez y−1−z y1y≤1 Š

Π(d y) +λ(R(z)−1). (2.2)

Here,σ≥0,µ∈R,Πis a Radon measure onR₊satisfyingΠ({0}) =0 andR_R

+1∧y

2_{Π(d y)<∞,}

λ > 0, andRis a rational function such thatR(z)≡R∞ 0 e

right-hand side of (2.2) admits an analytic continuation into{z;ℜz <0} except at the poles of

Rand is continuous upiR. As a result, we can extendψby (2.2) and the resulting extension is analytic on{z;ℜz<0}except at the poles ofR.

Forr>0, lete_r be an independent exponential random variable with mean 1/r. Recall that the negative Wiener-Hopf factorφ−

r is defined by

φ−_r(z) =E

exp

z inf

0≤s≤er

Xs

, ℜz≥0.

IfE[(X₁)−]<∞andE[X₁]∈(0,∞], then inf0≤s<∞Xs>−∞almost surely by the strong law of large numbers for Lévy processes. (See[12]Theorem 36.5 and Theorem 36.6.) It follows that limr↓0φ−r(z)exists for eachz∈Cwithℜz≥0, and hence we can define

φ₀−(z)¬ E

•

exp

 z inf

0≤s<∞Xs ‹˜

=lim r↓0φ

−

r(z). (2.3)

For convenience, we will slightly abuse the notation to write inf_0≤s≤e0Xsfor inf0≤s<∞Xs.From now

on, we always assume that r≥0and thatE[(X₁)−_]<∞andE[X

1]∈(0,∞]for the case in which r=0.For matrix-sexponential Lévy processes, the negative Wiener-Hopf factor has a semi-explicit solution. We quote the following results by Lewis and Mordecki in[10]. See also[3].

Theorem 2.1. (1) The equationψ−r=0has solutions in{z;ℜz<0}. Moreover, if(ρ_j; 1≤j≤m) are the distinct zeros ofψ−r=0in{z;ℜz <0}and m_j is the multiplicity ofρ_j, thenPm_j=1m_jis equal to n when X(+)_{is a subordinator and n+1when X}(+)_{is not a subordinator. Here, n=}Pn

j=1nj.

(2)φ−

r is a rational function given by

φ−_r(z) = Qm

j=1(−ρj)mj

Qn j=1λ

nj

j

P(z)

Q(z). (2.4)

Here, P and Q are monic polynomials given respectively by P(z) =

n

Y

j=1

(z+λ_j)nj_{, (2.5)}

Q(z) =

m

Y

j=1

(z−ρ_j)mj_{. (2.6)}

(3) The distribution ofinf0≤s≤erXsis given by

P

inf

0≤s≤er

X_s∈d y

= Qm

j=1(−ρj)mj

Qn j=1λ

nj

j

 U

0δ0(d y) +  

m

X

j=1

mj

X

k=1

U j,mj−k

(−y)k−1 (k−1)!e

−ρjy



1(−∞,0)(y)d y

 .

Here, the constantsU

0andUj,kare defined by the partial fraction decomposition ofφ−r:

φ_r−(z)≡ Qm

j=1(−ρj)mj

Qn j=1λ

nj

j

 U

0+

m

X

j=1

mj

X

k=1

U j,mj−k

(z−ρ_j)k

 

We now sketch our plan to find semi-explicit solutions for r-balayages. Recall that a function g

defined on Ris a Schwartz function, denoted by g ∈ S, if sup_x|xjg(ℓ)(x)|<∞ for all j,ℓ≥0. For properties of Schwartz functions, see[13]. Since the class S is sufficiently large, we may obtain, in principle, solutions of general r-balayages by suitable approximation once we have found solutions ofr-balayages of Schwartz functions. Set

ar(x,z)≡Ex

”

e−rτ+zXτ—_{, ℜz≥0.}

For everyg∈ S, it follows from the Fourier inversion formula that

Prg(x) = Z∞

−∞

a_r(x, 2πiζ)Fg(ζ)dζ, (2.7)

with

Fg(ζ) = Z∞

−∞

e−2πi yζg(y)d y

being the Fourier transform ofg. Define a Fourier multiplier operatorT_xr:S −→L2by FTr

xg(ζ)≡ar(x, 2πiζ)Fg(ζ). (2.8)

By (2.7), it follows that Pr_{g(x) =}R∞

−∞FT

r

xg(ζ)dζ. If the Fourier inversion formula is applicable in a suitable sense, we should have

Prg(x) =T_xrg(0).

Hence, ifarhas a form which allows us to identifyTxrsemi-explicitly by (2.8), the desired solutions ofr-balayages of Schwartz functions will fall right into our lap.

To carry out the plan, we first recall an identity forar which states that for allzwithℜz≥0

ar(x,z) =

ez xE

h

exp€zinf0≤s≤erXs

Š

1[inf0≤s≤erXs<−x]

i

E”exp€zinf0≤s≤erXs

Š— . (2.9)

(See[1]Corollary 2.) We can use Theorem 2.1 to calculatear.

Lemma 2.1. For any x>0and any z∈Csuch thatℜz≥0,

ar(x,z) = m

X

j=1

m_Xj−1

k=0

xmj−1−k_eρjx

(m_j−1−k)! k

X

ℓ=0

U_j,k−ℓQj,ℓ(z)

P(z) . Here, Qj,ℓ(z) = (z−ρj)−(ℓ+1)Q(z)for0≤ℓ≤mj−1.

Proof.Using Theorem 2.1, we get that E_x”_e−rτ+zXτ—₌Q(z)

P(z)

m

X

j=1

mj

X

k=1

U_j,m

j−k

k−1 X

ℓ=0

xk−ℓ−1_eρjx

(k−ℓ−1)!(z−ρ_j)ℓ+1

=Q(z) P(z)

m

X

j=1

mXj−1

k=0

U j,mj−k−1

k

X

ℓ=0

xk−ℓeρjx

(k−ℓ)!(z−ρ_j)ℓ+1

=Q(z) P(z)

m

X

j=1

m_Xj−1

k=0

U_j,m

j−k−1

k

X

ℓ=0

xℓ_eρjx

ℓ!(z−ρj)k−ℓ+1

=Q(z) P(z)

m

X

j=1

m_Xj−1

ℓ=0 xℓ

ℓ! m_Xj−1

k=ℓ U

j,mj−k−1e

ρjx

(z−ρ_j)k−ℓ+1

=Q(z) P(z)

m

X

j=1

mXj−1

ℓ=0

xmj−1−ℓ

(m_j−1−ℓ)! mXj−1

k=mj−1−ℓ

U_j,m

j−k−1e

ρjx

(z−ρ_j)k−(mj−1−ℓ)+1

=Q(z) P(z)

m

X

j=1

m_Xj−1

ℓ=0

xmj−1−ℓ

(m_j−1−ℓ)! ℓ

X

k=0

U

j,ℓ−keρjx

(z−ρ_j)k+1.

The desired equation then follows from the definition ofQj,ℓ. ƒ

The following proposition is the key step to identifyT_xr.

Proposition 2.1. For anyγ∈C_withℜγ >0, the exponential integral operatorE_γdefined by E_γg(x) =

Z∞ 0

e−γtg(x−t)d t= Zx

−∞

e−γ(x−t)g(t)d t (2.10)

is a bijection on the spaceS. Moreover, for any q∈N, Fg(ζ)

(γ+2πiζ)q =F(Eγ)qg(ζ), ζ∈R. (2.11)

Proof. The caseq=1 for (2.11) follows from a straightforward application of Fourier inversion formula. Recall thatF is a bijection fromS ontoS. Using (2.11) forq=1 and the fact that

Fg∈ S, we see thatF Eγg∈ S and henceEγg∈ S. This shows thatEγis a bijection fromS to

S. Finally, by the fact thatEγis bijective and iteration, we deduce that (2.11) holds for allq≥1.ƒ

Remark 1. (1) It is plain that the operatorEγis well-defined on the set of all bounded Borel measurable functions.

(2) For allq≥1, we have

€

E_γŠqg(x) = Zx

−∞

(x−t)q−1

e−γ(x−t)g(t)d t. (2.12) In particular, forx≤0, the integral depends only on the restriction ofhtoR₋, for anyh∈ Bb+(R).

ƒ

Theorem 2.2. (Main Result)Suppose that g is a bounded measurable function defined onR₋. Then the r-balayage Prg is given by

Prg(x) =

m

X

j=1

mXj−1

k=0  

Pk

ℓ=0Uj,k−ℓVj,ℓ(g)

(m_j−1−k)!



xmj−1−k_eρjx_{. (2.13)}

Here, we firstly define the constantsW(j,ℓ)

p,q by the partial fraction decomposition of Qj,ℓ/P:

Qj,ℓ(2πiζ)

P(2πiζ) =   

δ_0,ℓ+Pn_p=1Pnp

q=1

W(j,ℓ) p,q

(2πiζ+λp)q, if X

(+)_{is not a subordinator,}

Pn p=1

Pnp

q=1

W(j,ℓ)

p,q

(2πiζ+λp)q, if X

(+)_{is a subordinator,} (2.14)

whereδ0,ℓis Kronecker’s delta, and then set the linear functionalsVj,ℓby

(2.15)

V_j,ℓ(h) =   

h(0)δ_0,ℓ+Pn_p=1Pnp

q=1W(pj,,qℓ)

E_λ

p

q

h(0), if X(+)is not a subordinator, Pn

p=1 Pnp

q=1W(pj,,qℓ)

E_λpqh(0), if X(+)is a subordinator.

The formula of each(E_λ

p)

q_{is given by (2.12).}

Proof. We prove the desired conclusion holds whenever g ∈ S. By (2.7) and Lemma 2.1, we obtain

E_xe−rτg(X_τ)=

m

X

j=1

mXj−1

k=0

xmj−1−k_eρjx

(mj−1−k)! k

X

ℓ=0

U j,k−ℓ

Z∞ −∞

Qj,ℓ(2πiζ)

P(2πiζ) Fg(ζ)dζ. (2.16)

By Proposition 2.1 and the Fourier inversion formula, we deduce that the formula (2.13) holds for everyg∈ S.

Next, we show that (2.13) holds for all bounded Borel measurable functions g. Given a bounded Borel measurable function g, choose a uniformly bounded sequence(gq)of Schwartz functions such that gq converges to galmost everywhere on(−∞, 0)and gq(0) =g(0)for allq. (See[7] Lemma A.1 and[14]Chapter 2 Theorem 2.4.) Recall the law ofXτrestricted on(−∞, 0)is abso-lutely continuous underP_x forx>0 (see[7]Lemma A.2). Hence, by passing limit, we see that

Prgq(x)−→Prg(x). On the other hand, note that for eachq≥1, the formula (2.13) holds withg replaced byg_q. If we denote byePr_{gthe function on the right hand side of (2.13), then by Remark} 1 and dominated convergencePr_g

q(x)−→ePrg(x). This givesPrg(x) =ePrg(x), and the proof is

complete. ƒ

Remark 2. (1) To avoid using (2.9), one can first consider the Laplace transform ofr-balayages for g∈ S. Recall that the Pecherskii-Rogzin identity expresses the Laplace transform ofa_r(x,z)

in x in terms of the negative Wiener-Hopf factor. This implies that one may identify the Fourier symbol of the operatorLPrin terms of the negative Wiener-Hopf factor, where the operatorL is the Laplace transform operator. Using a version of Laplace inversion formula and residue calculus, one obtains the same result as in Theorem 2.2.

(2) Take g ≡1, g =1{0} and g = 1(−∞,0). If X stands for the surplus process of an insurance

company, then the resulting balayages represent respectively the discounted ruin probability, the discounted measure of ruin caused by diffusion, and the discounted measure of ruin caused by

jump. ƒ

3

Consistency with an Existing Formula

As in[7], we consider the case that the downward jump distributionF is a convex combination of exponential distributions:

d F(y) =

n

X

j=1

pjλje−λjy1y>0d y,

wherepj>0 andλj>0 for all 1≤ j≤nandλjare distinct. We assume further thatX(+)is not a subordinator.

In this case, we have the followings.

(1) n_j=1 for all 1≤ j≤nandm_j=1 for all 1≤j≤m=n+1. See (1.2) and Theorem 2.1. (2) φ−

r(z) =

Qm j=1(−ρj) Qn

j=1λj

P(z)

Q(z), whereP(z) =

Qn

j=1(z+λj)andQ(z) =

Qn+1

j=1(z−ρj). See Theorem 2.1.

(3) φ−_r(z) =

Qm j=1(−ρj) Qn

j=1λj

Pn+1

j=1

U_j,0

z−ρj, where

U

j,0are given by

U j,0=

Qn

k=1(ρj+λk)

Q

k6=j,1≤k≤n+1(ρj−ρk)

, ∀1≤j≤n+1. (3.1)

See Theorem 2.1.

(4) Q_j,0(z) =Qn_ℓ=+_1,1_ℓ6=j(z−ρℓ)and the constants

W(j,0) k,1

are defined by the equation:

Qj,0(z) P(z) =1+

n

X

k=1

W(j,0) k,1 z+λ_k.

Hence,

W(j,0) k,1 =

Qn+1

ℓ=1,ℓ6=j(−λk−ρℓ)

Qn

ℓ=1,ℓ6=k(−λk+λℓ)

=− Qn+1

ℓ=1,ℓ6=j(λk+ρℓ)

Qn

ℓ=1,ℓ6=k(λk−λℓ)

. (3.2)

See Theorem 2.2.

(5) The linear functionals(V_{j,0; 1≤j≤n+1)are given by} V_{j,0(g) =g(0) +}

n

X

k=1

W(j,0)

k,1 Eλkg(0). (3.3)

(6) Ifgis a bounded measurable function defined onR₋, then we have

Prg(x) =

m

X

j=1 eρjx_U

j,0Vj,0(g), x>0. (3.4)

whereU_j,0andV_j,0(g)are given in (3.1) and (3.3) respectively. See Theorem 2.2.

In the following, we will show that our formula (3.4) coincides with the one in[7]Theorem 4.3. We write Pr_{g(x) =}Pm

j=1e

ρjxT

j,0(g), where Tj,0(g)is given in (4.6) of Chen et al.(2007). We

verify that T_j,0(g) =U_j,0V_j,0(g). By linearity of the operatorPr, it suffices to consider the two cases: (I) g≡1{0}, and (II)g∈ Bb(R)such thatg(0) =0.

In case (I), the formula in[7]Example 4.5 states that, by settingλ_n+1=0,

T j,0(g) =

1

ρ_jQn_k+₌1_1,k6=j(−ρ_j+ρ_k)

n+1 X

k=1  

n+1 Y

ℓ=1

(λ_k+ρℓ) n+1 Y

i=1,i6=k

−ρ_j−λ_i λ_k−λ_i

 

= 1

ρ_jQn_k+₌1_1,k6=j(ρ_j−ρ_k)

n+1 X

k=1  

n+1 Y

ℓ=1

(λk+ρℓ) n+1 Y

i=1,i6=k

ρ_j+λ_i λ_k−λ_i

 

= 1

ρ_jQn_k+₌1_1,k6=j(ρ_j−ρ_k)

n+1 X

k=1  

n+1 Y

ℓ=1,ℓ6=j

(λ_k+ρ_ℓ)

n+1 Y

i=1,i6=k 1

λ_k−λ_i

n+1 Y

i=1,i6=k

(ρ_j+λ_i)(ρ_j+λ_k)  

= Qn+1

i=1(ρj+λi)

Qn+1

k=1,k6=j(ρj−ρk) 1

ρj n+1 X

k=1  

n+1 Y

ℓ=1,ℓ6=j

(λ_k+ρ_ℓ)

n+1 Y

i=1,i6=k 1

λk−λi

 

= Qn

i=1(ρj+λi)

Qn+1

k=1,k6=j(ρj−ρk) n+1 X

k=1  

n+1 Y

ℓ=1,ℓ6=j

(λ_k+ρℓ) n+1 Y

i=1,i6=k 1

λ_k−λ_i

 .

=U_j,0 n+1 X

k=1  

n+1 Y

ℓ=1,ℓ6=j

(λk+ρℓ) n+1 Y

i=1,i6=k 1

λ_k−λ_i



, (3.5)

where the last equality follows from (3.1). SinceV

j,0(g) =1, it suffices to show that

n+1 X

k=1  

n+1 Y

ℓ=1,ℓ6=j

(λ_k+ρ_ℓ)

n+1 Y

i=1,i6=k 1

λ_k−λ_i



=1. (3.6)

Note that, sinceλ_n+1=0, we have

n+1 X

k=1  

n+1 Y

ℓ=1,ℓ6=j

(λ_k+ρ_ℓ)

n+1 Y

i=1,i6=k 1

λ_k−λ_i

 

=

n

X

k=1  

n+1 Y

ℓ=1,ℓ6=j

(λ_k+ρℓ) n

Y

i=1,i6=k 1

λ_k−λ_i

  1

λ_k+

n+1 Y

ℓ=1,ℓ6=j

ρℓ n

Y

i=1

1

−λ_i. (3.7)

ordern+1, we obtain by Lagrange’s interpolation that

Q(−x) =

n+1 X

k=1

Q(−λ_k)

n+2 Y

i=1,i6=k

x−λ_i λ_k−λ_i

=(x+ρ_j)

n+1 X

k=1

Q(−λ_k)

λ_k+ρ_j

n+1 Y

i=1,i6=k

x−λ_i λ_k−λ_i

=(x+ρj)P(−x)x n+1 X

k=1

Qn+1

ℓ=1(λk+ρℓ)

(−x+λ_k)(λ_k+ρ_j)

n+1 Y

i=1,i6=k 1

λ_k−λ_i

RecallQj,0(x) =Q(x)/(x−ρj). Then

Q_j,0(−x) =P(−x)(−x)

n+1 X

k=1 Qn+1

ℓ=1,ℓ6=j(λk+ρℓ)

−x+λ_k

n+1 Y

i=1,i6=k 1

λ_k−λ_i

This implies that, using the fact thatλ_n+1=0 again,

Qj,0(−x) P(−x) =−x

n

X

k=1 Qn+1

ℓ=1,ℓ6=j(λk+ρℓ)

(−x+λk)

n+1 Y

i=1,i6=k 1

λk−λi

−x Qn+1

ℓ=1,ℓ6=jρℓ

−x

n

Y

i=1

1

−λi

=−x

n

X

k=1 Qn+1

ℓ=1,ℓ6=j(λk+ρℓ)

(−x+λ_k)

n

Y

i=1,i6=k 1

λ_k−λ_i

1

λ_k +

n+1 Y

ℓ=1,ℓ6=j

ρ_ℓ

n

Y

i=1

1

−λ_i.

SinceQj,0andPare both monic polynomials of ordern, we see that, by lettingx−→ ∞,

1=

n

X

k=1

n+1 Y

ℓ=1,ℓ6=j

(λ_k+ρ_ℓ)

n

Y

i=1,i6=k 1

λ_k−λ_i

1

λ_k +

n+1 Y

ℓ=1,ℓ6=j

ρ_ℓ

n

Y

i=1

1

(−λ_i)

Using this equation, we deduce from (3.7) that (3.6) holds. We have proved that our result coincides with the one in[7]Example 4.5 forg≡1{0}.

Next, consider case (II). Then the formula (4.6) in[7]Theorem 4.3 states that T_{j,0(g) =} −1

ρ_jQn_k+₌1_1,k6=j(−ρ_j+ρ_k)

n

X

k=1  

n+1 Y

ℓ=1

(λk+ρℓ) n+1 Y

i=1,i6=k

−ρ_j−λ_i λ_k−λ_i

Z0 −∞

g(y)λkeλkyd y

 

=− Qn

i=1(ρj+λi)

Qn+1

k=1,k6=j(ρj−ρk) n

X

k=1  

n+1 Y

ℓ=1,ℓ6=j

(λ_k+ρ_ℓ)

n+1 Y

i=1,i6=k 1

λ_k−λ_i

Z0 −∞

g(y)λ_keλky_{d y}

 

=−U j,0

n

X

k=1  

n+1 Y

ℓ=1,ℓ6=j

(λ_k+ρ_ℓ)

n

Y

i=1,i6=k 1

λk−λi

E_λkg(0)   =U_j,0

n

X

k=1

W(j,0)

k,1 Eλkg(0),

where the last equality follows from (3.2). By (3.3) and the fact that g(0) =0, the last equality givesT_{j,0(g) =}U_j,0V_{j,0(g). The conclusion is that our result coincides with the one in[7]Theorem} 4.3 in case (II) as well.

4

Another First-Passage Functional

Let f ∈ B_b(R)andr>0. Consider the first passage functional

Irf(x)¬ E_x

–Zτ

0

e−rsf(X_s)ds ™

. By strong Markov property, we have

Irf(x) =E_x

–Z∞ 0

e−rsf(X_s)ds ™

−E_x

– e−rτE_X

τ

–Z∞ 0

e−rsf(X_s)ds ™™

=Urf −PrUrf(x), (4.1) whereUr_{is ther-resolvent kernel ofX. Hence, to derive a semi-explicit solution ofI}r_{f, it suffices} to find a semi-explicit solution ofUrf. Suppose that the upper Lévy measure ofX is also matrix-exponential. In this case,X is merely a perturbed compound Poisson process both sides of whose jump distributions are matrix-exponential. We can use Theorem 2.1 to calculate Urf. Recall thater is an exponential random variable independent ofX and has mean 1/r. By Wiener-Hopf factorization of Lévy processes,

Urf(x) =E”_f(x+Xer)

— =

Z

R₊ P

–

sup

0≤s≤er

Xs∈d y

™ Z

R₋ P

inf

0≤s≤er

Xs∈dz

f(x+y+z). (4.2) By Theorem 2.1, one can write down semi-explicit formulas for the distributions of sup0≤s≤erXs

and inf0≤s≤erXs. A semi-explicit solution ofU

r_{f will then follow from (4.2). By (4.1), this further} implies a semi-explicit solution ofIrf. We leave the details to the readers.

References

[1] Alili, L., Kyprianou, A.E. Some remarks on first passage of Lévy processes, the American put and pasting principles, Annals of Applied Probability Vol.15No.3 (2005), 2062-2080. MR2152253

[2] Asmussen, S.Applied Probability and Queues, Vol.51 of Applications of Mathematics (New York). 2nd edition, Springer-Verlag, Berlin, (2003). MR1978607

[3] Asmussen, S., Avram, F., Pistorius, M.R.Russian and American put options under exponential phase-type Lévy modesl, Stochastic Processes and Their Applications Vol.109(2004), 79-111. MR2024845

[4] Avram, F., Usabel, M.Finite time ruin probabilities with one Laplace transform, Insurance, Mathematics and Economics Vol.32(2003), 371-377. MR1994498

[5] Botta, R.F., Harris, C.M.Approximation with generalized hyperexponential distributions: weak convergence results, Queueing Systems Vol.2(1986), 169-190. MR0896234

[6] Boyarchenko, S., Levendorskiˇi, S.American options: the EPV pricing model, Annals of Finance Vol.1(2005), 267-292.

[7] Chen, Y.T., Lee, C.F., Sheu, Y.C.An ODE approach for the expected discounted penalty at ruin in a jump-diffusion model, Finance and Stochastics Vol.11(2007), 323-355. MR2322916

[8] Dixit, A.K., Pyndick, R.S.Investment under Uncertainty. Princeton University Press, New Jer-sey, (1994).

[9] Gerber, H.U., Shiu, E.S.W.On the time value of ruin, North American Actuarial Journal Vol.2

(1998), 48-78. MR1988433

[10] Lewis, A.L., Mordecki, E.Wiener-Hopf factorization for Lévy processes having positive jumps with rational transforms, Journal of Applied Probability Vol.45(2008), 118-134. MR2409315

[11] Mordecki, E.Optimal stopping and perpetual options for Lévy processes, Finance and Stochas-tics Vol.6(2002), 473-493. MR1932381

[12] Sato, K.Lévy processes and infinitely divisible distributions, Cambridge University Press, Cam-bridge, (1999).

[13] Stein, E.M., Shakarchi, R.Fourier Analysis, Princeton Lectures in Analysis I, Princeton Univer-sity Press, New Jersey, (2003). MR1970295

[14] Stein, E.M., Shakarchi, R.Real Analysis, Princeton Lectures in Analysis III, Princeton University Press, New Jersey,(2005). MR2129625

Theorem 2.2. (Main Result)Suppose that g is a bounded measurable function defined onR₋. Then the r-balayage Prg is given by

Prg(x) =

m

X

j=1 mXj−1

k=0

 

Pk

ℓ=0Uj,k−ℓVj,ℓ(g)

(mj−1−k)!



xmj−1−k_eρjx_{. (2.13)}

Here, we firstly define the constantsW(j,ℓ)

p,q by the partial fraction decomposition of Qj,ℓ/P:

Qj,ℓ(2πiζ)

P(2πiζ) =   

δ_0,ℓ+Pn_p=1Pnp q=1

W(j,ℓ)

p,q

(2πiζ+λp)q, if X

(+)_{is not a subordinator,} Pn

p=1

Pnp q=1

W(j,ℓ)

p,q

(2πiζ+λp)q, if X

(+)_{is a subordinator,} (2.14)

whereδ0,ℓis Kronecker’s delta, and then set the linear functionalsVj,ℓby

(2.15)

V_j,ℓ(h) =   

h(0)δ_0,ℓ+Pn_p=1Pnp_q=1W(j,ℓ)

p,q

E_λ

p

q

h(0), if X(+)is not a subordinator,

Pn p=1

Pnp q=1W(p,qj,ℓ)

E_λpqh(0), if X(+)is a subordinator.

The formula of each(E_λ

p)

q_{is given by (2.12).}

Proof. We prove the desired conclusion holds whenever g ∈ S. By (2.7) and Lemma 2.1, we

obtain

E_xe−rτg(X_τ)=

m

X

j=1 mj_X−1

k=0

xmj−1−k_eρjx

(mj−1−k)! k

X

ℓ=0

U

j,k−ℓ

Z∞ −∞

Qj,ℓ(2πiζ)

P(2πiζ) Fg(ζ)dζ. (2.16) By Proposition 2.1 and the Fourier inversion formula, we deduce that the formula (2.13) holds for everyg∈ S.

Next, we show that (2.13) holds for all bounded Borel measurable functions g. Given a bounded Borel measurable function g, choose a uniformly bounded sequence(gq)of Schwartz functions

such that gq converges to galmost everywhere on(−∞, 0)and gq(0) =g(0)for allq. (See[7]

Lemma A.1 and[14]Chapter 2 Theorem 2.4.) Recall the law ofXτrestricted on(−∞, 0)is

abso-lutely continuous underP_x forx>0 (see[7]Lemma A.2). Hence, by passing limit, we see that

Prgq(x)−→Prg(x). On the other hand, note that for eachq≥1, the formula (2.13) holds withg

replaced byg_q. If we denote byePr_{gthe function on the right hand side of (2.13), then by Remark}

1 and dominated convergencePr_g

q(x)−→ePrg(x). This givesPrg(x) =ePrg(x), and the proof is

complete. ƒ

Remark 2. (1) To avoid using (2.9), one can first consider the Laplace transform ofr-balayages for g∈ S. Recall that the Pecherskii-Rogzin identity expresses the Laplace transform ofa_r(x,z) in x in terms of the negative Wiener-Hopf factor. This implies that one may identify the Fourier symbol of the operatorLPrin terms of the negative Wiener-Hopf factor, where the operatorL is the Laplace transform operator. Using a version of Laplace inversion formula and residue calculus, one obtains the same result as in Theorem 2.2.

(2) Take g ≡1, g =1{0} and g = 1(−∞,0). If X stands for the surplus process of an insurance company, then the resulting balayages represent respectively the discounted ruin probability, the discounted measure of ruin caused by diffusion, and the discounted measure of ruin caused by

jump. ƒ

3

Consistency with an Existing Formula

As in[7], we consider the case that the downward jump distributionF is a convex combination of exponential distributions:

d F(y) =

n

X

j=1

pjλje−λjy1y>0d y,

wherepj>0 andλj>0 for all 1≤ j≤nandλjare distinct. We assume further thatX(+)is not

a subordinator.

In this case, we have the followings.

(1) n_j=1 for all 1≤ j≤nandm_j=1 for all 1≤j≤m=n+1. See (1.2) and Theorem 2.1. (2) φ−

r(z) =

Qm j=1(−ρj)

Qn j=1λj

P(z)

Q(z), whereP(z) = Qn

j=1(z+λj)andQ(z) =

Qn+1

j=1(z−ρj). See Theorem

2.1. (3) φ−_r(z) =

Qm j=1(−ρj)

Qn j=1λj

Pn+1 j=1

U_j,0

z−ρj, where

U

j,0are given by

U

j,0=

Qn

k=1(ρj+λk)

Q

k6=j,1≤k≤n+1(ρj−ρk)

, ∀1≤j≤n+1. (3.1)

See Theorem 2.1.

(4) Q_j,0(z) =Qn_ℓ=+_1,1_ℓ6=j(z−ρℓ)and the constants

W(j,0)

k,1

are defined by the equation:

Qj,0(z) P(z) =1+

n

X

k=1

W(j,0)

k,1 z+λ_k. Hence,

W(j,0)

k,1 =

Qn+1

ℓ=1,ℓ6=j(−λk−ρℓ)

Qn

ℓ=1,ℓ6=k(−λk+λℓ)

=− Qn+1

ℓ=1,ℓ6=j(λk+ρℓ)

Qn

ℓ=1,ℓ6=k(λk−λℓ)

. (3.2)

See Theorem 2.2.

(5) The linear functionals(V_{j,0; 1≤j≤n+1)are given by}

V_{j,0(g) =g(0) +}

n

X

k=1

W(j,0)

k,1 Eλkg(0). (3.3)

(6) Ifgis a bounded measurable function defined onR₋, then we have

Prg(x) =

m

X

j=1 eρjx_U

j,0Vj,0(g), x>0. (3.4)

whereU_j,0andV_j,0(g)are given in (3.1) and (3.3) respectively. See Theorem 2.2.

In the following, we will show that our formula (3.4) coincides with the one in[7]Theorem 4.3. We write Pr_{g(x) =}Pm

j=1e

ρjxT

j,0(g), where Tj,0(g)is given in (4.6) of Chen et al.(2007). We

verify that T_j,0(g) =U_j,0V_j,0(g). By linearity of the operatorPr, it suffices to consider the two cases: (I) g≡1{0}, and (II)g∈ Bb(R)such thatg(0) =0.

In case (I), the formula in[7]Example 4.5 states that, by settingλ_n+1=0,

T

j,0(g) =

1

ρ_jQn_k+₌1_1,k6=j(−ρ_j+ρ_k)

n+1

X

k=1

 

n+1

Y

ℓ=1

(λ_k+ρℓ)

n+1

Y

i=1,i6=k

−ρ_j−λ_i

λ_k−λ_i  

= 1

ρ_jQn_k=+_1,k61 _=j(ρ_j−ρ_k)

n+1

X

k=1

 

n+1

Y

ℓ=1

(λk+ρℓ)

n+1

Y

i=1,i6=k

ρ_j+λ_i λ_k−λ_i

 

= 1

ρ_jQn_k=+_1,k61 _=j(ρ_j−ρ_k)

n+1

X

k=1

 

n+1

Y

ℓ=1,ℓ6=j

(λ_k+ρ_ℓ)

n+1

Y

i=1,i6=k

1 λ_k−λ_i

n+1

Y

i=1,i6=k

(ρ_j+λ_i)(ρ_j+λ_k)  

= Qn+1

i=1(ρj+λi)

Qn+1

k=1,k6=j(ρj−ρk)

1 ρj

n+1

X

k=1

 

n+1

Y

ℓ=1,ℓ6=j

(λ_k+ρ_ℓ)

n+1

Y

i=1,i6=k

1 λk−λi

 

= Qn

i=1(ρj+λi)

Qn+1

k=1,k6=j(ρj−ρk) n+1

X

k=1

 

n+1

Y

ℓ=1,ℓ6=j

(λ_k+ρℓ)

n+1

Y

i=1,i6=k

1 λ_k−λ_i

 .

=U_j,0

n+1

X

k=1

 

n+1

Y

ℓ=1,ℓ6=j

(λk+ρℓ)

n+1

Y

i=1,i6=k

1 λ_k−λ_i



, (3.5)

where the last equality follows from (3.1). SinceV_{j,0(g) =1, it suffices to show that}

n+1

X

k=1

 

n+1

Y

ℓ=1,ℓ6=j

(λ_k+ρ_ℓ)

n+1

Y

i=1,i6=k

1 λ_k−λ_i



=1. (3.6)

Note that, sinceλ_n+1=0, we have

n+1

X

k=1

 

n+1

Y

ℓ=1,ℓ6=j

(λ_k+ρ_ℓ)

n+1

Y

i=1,i6=k

1 λ_k−λ_i

 

=

n

X

k=1

 

n+1

Y

ℓ=1,ℓ6=j

(λ_k+ρℓ)

n

Y

i=1,i6=k

1 λ_k−λ_i

  1

λ_k+

n+1

Y

ℓ=1,ℓ6=j

ρℓ

n

Y

i=1

1

−λ_i. (3.7)

ordern+1, we obtain by Lagrange’s interpolation that

Q(−x) =

n+1

X

k=1

Q(−λ_k)

n+2

Y

i=1,i6=k x−λ_i λ_k−λ_i =(x+ρ_j)

n+1

X

k=1

Q(−λ_k) λ_k+ρ_j

n+1

Y

i=1,i6=k x−λ_i λ_k−λ_i =(x+ρj)P(−x)x

n+1

X

k=1

Qn+1

ℓ=1(λk+ρℓ)

(−x+λ_k)(λ_k+ρ_j)

n+1

Y

i=1,i6=k

1 λ_k−λ_i RecallQj,0(x) =Q(x)/(x−ρj). Then

Qj,0(−x) =P(−x)(−x) n+1

X

k=1

Qn+1

ℓ=1,ℓ6=j(λk+ρℓ)

−x+λ_k

n+1

Y

i=1,i6=k

1 λ_k−λ_i This implies that, using the fact thatλ_n+1=0 again,

Qj,0(−x) P(−x) =−x

n

X

k=1

Qn+1

ℓ=1,ℓ6=j(λk+ρℓ)

(−x+λk)

n+1

Y

i=1,i6=k

1 λk−λi

−x

Qn+1

ℓ=1,ℓ6=jρℓ

−x

n

Y

i=1

1

−λi

=−x n

X

k=1

Qn+1

ℓ=1,ℓ6=j(λk+ρℓ)

(−x+λ_k)

n

Y

i=1,i6=k

1 λ_k−λ_i

1 λ_k +

n+1

Y

ℓ=1,ℓ6=j

ρ_ℓ

n

Y

i=1

1

−λ_i. SinceQj,0andPare both monic polynomials of ordern, we see that, by lettingx−→ ∞,

1=

n

X

k=1 n+1

Y

ℓ=1,ℓ6=j

(λ_k+ρ_ℓ)

n

Y

i=1,i6=k

1 λ_k−λ_i

1 λ_k +

n+1

Y

ℓ=1,ℓ6=j

ρ_ℓ

n

Y

i=1

1 (−λ_i)

Using this equation, we deduce from (3.7) that (3.6) holds. We have proved that our result coincides with the one in[7]Example 4.5 forg≡1{0}.

Next, consider case (II). Then the formula (4.6) in[7]Theorem 4.3 states that T_{j,0(g) =} −1

ρ_jQn_k+₌1_1,k6=j(−ρ_j+ρ_k)

n

X

k=1

 

n+1

Y

ℓ=1

(λk+ρℓ)

n+1

Y

i=1,i6=k

−ρ_j−λ_i

λ_k−λ_i Z0

−∞

g(y)λkeλkyd y

 

=−

Qn

i=1(ρj+λi)

Qn+1

k=1,k6=j(ρj−ρk) n

X

k=1

 

n+1

Y

ℓ=1,ℓ6=j

(λ_k+ρ_ℓ)

n+1

Y

i=1,i6=k

1 λ_k−λ_i

Z0 −∞

g(y)λ_keλkyd y

 

=−U

j,0 n

X

k=1

 

n+1

Y

ℓ=1,ℓ6=j

(λ_k+ρ_ℓ)

n

Y

i=1,i6=k

1 λk−λi

E_λkg(0)   =U_j,0

n

X

k=1

W(j,0)

k,1 Eλkg(0),

where the last equality follows from (3.2). By (3.3) and the fact that g(0) =0, the last equality givesT_{j,0(g) =}U_j,0V_{j,0(g). The conclusion is that our result coincides with the one in[7]Theorem} 4.3 in case (II) as well.

4

Another First-Passage Functional

Let f ∈ B_b(R)andr>0. Consider the first passage functional

Irf(x)¬ E_x –Zτ

0

e−rsf(Xs)ds

™ . By strong Markov property, we have

Irf(x) =E_x –Z∞

0

e−rsf(X_s)ds

™

−E_x

–

e−rτE_X

τ

–Z∞ 0

e−rsf(X_s)ds

™™

=Urf −PrUrf(x), (4.1) whereUr_{is ther-resolvent kernel ofX. Hence, to derive a semi-explicit solution ofI}r_{f, it suffices}

to find a semi-explicit solution ofUrf. Suppose that the upper Lévy measure ofX is also matrix-exponential. In this case,X is merely a perturbed compound Poisson process both sides of whose jump distributions are matrix-exponential. We can use Theorem 2.1 to calculate Urf. Recall thater is an exponential random variable independent ofX and has mean 1/r. By Wiener-Hopf

factorization of Lévy processes,

Urf(x) =E”_f(x+Xer)

— =

Z

R₊

P –

sup

0≤s≤er

Xs∈d y

™ Z

R₋

P

inf

0≤s≤erXs∈dz

f(x+y+z). (4.2) By Theorem 2.1, one can write down semi-explicit formulas for the distributions of sup0≤s≤erXs

and inf0≤s≤erXs. A semi-explicit solution ofU

r_{f will then follow from (4.2). By (4.1), this further}

implies a semi-explicit solution ofIrf. We leave the details to the readers.

References

[1] Alili, L., Kyprianou, A.E. Some remarks on first passage of Lévy processes, the American put and pasting principles, Annals of Applied Probability Vol.15No.3 (2005), 2062-2080. MR2152253

[2] Asmussen, S.Applied Probability and Queues, Vol.51 of Applications of Mathematics (New York). 2nd edition, Springer-Verlag, Berlin, (2003). MR1978607

[3] Asmussen, S., Avram, F., Pistorius, M.R.Russian and American put options under exponential phase-type Lévy modesl, Stochastic Processes and Their Applications Vol.109(2004), 79-111. MR2024845

[4] Avram, F., Usabel, M.Finite time ruin probabilities with one Laplace transform, Insurance, Mathematics and Economics Vol.32(2003), 371-377. MR1994498

[5] Botta, R.F., Harris, C.M.Approximation with generalized hyperexponential distributions: weak convergence results, Queueing Systems Vol.2(1986), 169-190. MR0896234

[6] Boyarchenko, S., Levendorskiˇi, S.American options: the EPV pricing model, Annals of Finance Vol.1(2005), 267-292.

[7] Chen, Y.T., Lee, C.F., Sheu, Y.C.An ODE approach for the expected discounted penalty at ruin in a jump-diffusion model, Finance and Stochastics Vol.11(2007), 323-355. MR2322916

[8] Dixit, A.K., Pyndick, R.S.Investment under Uncertainty. Princeton University Press, New Jer-sey, (1994).

[9] Gerber, H.U., Shiu, E.S.W.On the time value of ruin, North American Actuarial Journal Vol.2 (1998), 48-78. MR1988433

[10] Lewis, A.L., Mordecki, E.Wiener-Hopf factorization for Lévy processes having positive jumps with rational transforms, Journal of Applied Probability Vol.45(2008), 118-134. MR2409315

[11] Mordecki, E.Optimal stopping and perpetual options for Lévy processes, Finance and Stochas-tics Vol.6(2002), 473-493. MR1932381

[12] Sato, K.Lévy processes and infinitely divisible distributions, Cambridge University Press, Cam-bridge, (1999).

[13] Stein, E.M., Shakarchi, R.Fourier Analysis, Princeton Lectures in Analysis I, Princeton Univer-sity Press, New Jersey, (2003). MR1970295

[14] Stein, E.M., Shakarchi, R.Real Analysis, Princeton Lectures in Analysis III, Princeton University Press, New Jersey,(2005). MR2129625