# A Note on First Passage Functionals for Lévy Processes with Jumps of Rational Laplace Transforms

Authors: Djilali Ait-Aoudia
Journal: Abstract and Applied Analysis
 (2016)

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## Abstract

This paper investigates the two-sided first exit problem for a jump process having jumps with rational Laplace transform. The corresponding boundary value problem is solved to obtain an explicit formula for the first passage functional. Also, we derive the distribution of the first passage time to two-sided barriers and the value at the first passage time.


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## Body

## 1. Introduction

One-sided and two-sided exit problems for the compound Poisson processes and jump-diffusion processes with two-sided jumps have been applied widely in a variety of fields. For example, in actuarial mathematics, the problem of first exit from a half-line is of fundamental interest with regard to the classical ruin problem and the expected discount penalty function or the Gerber-Shiu function as well as the expected total; see, for example, Mordecki [1], Xing et al. [2], Zhang et al. [3], Lewis and Mordecki [4], and Avram et al. [5]. In mathematical finance, the first passage time plays a crucial role for the pricing of many path-dependent options and American type options; see, for example, Geman and Yor [6], Bertoin [7], Kyprianou [8], Rogers [9], Avram et al. [10], and Alili and Kyprianou [11]. Recently, Cai [12] investigated the first passage time of hyperexponential jump-diffusion process. Cai and Kou [13] proposed a mixed-exponential jump-diffusion process to model the asset return and found an expression for the joint distribution of the first exit problem for a jump and overshoot for a mixed-exponential jump-diffusion process. Chen et al. [14] and Yin et al. [15] discussed the first passage functional for hyperexponential jump-diffusion process.Motivated by works mentioned above, the main objective of this paper is to study the first exit time of the two-sided first exit problem for jump-diffusion process having jump with rational Laplace transform proposed by Lewis and Mordecki [4]; see also Kuznetsov [16]. This extends recent results obtained in Chen et al. [14, Theorem 2.5] on the hyperexponential jump-diffusion process.The rest of the paper is organized as follows. In Section2, we introduce the jump-diffusion process having jumps with rational Laplace transform. Section 3 concentrates on deriving the joint distribution of first exit time and a nonnegative bounded measurable function of the process value at the first exit time to two flat barriers. Section 4 presents the analytical solution to the pricing problem of standard double-barrier options.


## 2. The Model

A Lévy jump-diffusion processX={Xt,t≥0} is defined as(1)Xt=X0+μt+σWt-∑i=1NtYi,where μ∈R and σ>0 represent the drift and volatility of the diffusion part, respectively, W={Wt,t≥0} is a (standard) Brownian motion, N={Nt,t≥0} is a homogeneous Poisson process with rate λ, and {Yi,i=1,2,…} are independent and identically distributed random variables supported in R∖{0}; moreover, {Wt,t≥0}, {Nt,t≥0} and {Yi,i=1,2,…} are mutually independent; finally, the probability density function (pdf) of Y1 is given by(2)fy=∑j=1m∑i=1mjpijηjiyi-1i-1!e-ηjy1y≥0+∑j=1n∑i=1njqijθji-yi-1i-1!eθjy1y<0,where pij,qij≥0, such that ∑j=1m‍∑i=1mj‍pij+∑j=1n‍∑i=1nj‍qi,j=1, Re(ηj) and Re(θj)>0 and that ηi≠ηj and θi≠θj for all i≠j.Another important tool to establish the key result of the article is the infinitesimal generator ofXt. Note that Xt is a Markovian process and its infinitesimal generator is given by(3)Lhx≔limt↘0EhXt∣X0=x-hxt=μh′x+σ22h′′x+λ∫-∞+∞hx-y-hxfydy,for any bounded and twice continuously differentiable function h.Throughout the rest of the paper, the law ofX such that X0=x is denoted by Px and the corresponding expectation by Ex; we write P and E when x=0. The Lévy exponent of X is given by (4)Gζ=ln⁡Eexp⁡ζXtt=μζ+σ22ζ2+λEe-ζY1-1=μζ+σ22ζ2+λ∑j=1m∑i=1mjpijηjiηj+ζi+∑j=1n∑i=1nj-qijθjiζ-θji-1.Accordingly, G is a rational function on C. The equation G(ζ)-α=0 with >0,σ>0, and μ∈R yields S=M+N+2 zeros with M=∑i=1m‍mi and N=∑j=1n‍mi; see Kuznetsov [16].Let us denote the zeros ofG(ζ)-α in the half-plane Re(ζ)>0Re(ζ)<0 as ρ1,ρ2,…,ρM+1ρM+2,ρM+3,…,ρM+N+2. Also, we assume that all zeros of G(ζ)-α are simple.


## 3. Distribution of the First Passage Time to Two Flat Barriers

Defineτ to be the first passage time of Xt to two flat barriers h and H with h<H; that is,(5)τ≔inft≥0:Xt≥horXt≤H.And let(6)ϕx=Exe-ατgXτ,where α>0 and g is nonnegative bounded measurable function.Now, by Feynman-Kac formula (see, e.g., Theorem4.4.2, Karatzas and Shreve [17]) we have that ϕ(x) must satisfy(7)L-αϕx=0inh,H,ϕx=gxon-∞,h∪H,+∞.Our goal in this section is to solve the boundary problem (7) and find explicit formulae for ϕ(x). We first show that ϕ satisfies an integrodifferential equation and then derive an ordinary differential equation for ϕ. Based on the ODE, we show that ϕ can be written as a linear combination of known exponential functions. As a consequence, its explicit form is obtained, for instance, choosing g(x) to be eγx; it is easy to derive the joint distribution of the first passage time of X to two flat barriers and the process value at the first passage time.Now, letP0(ζ)=∏j=1m‍∏i=1mj‍(ζ+ηj)i∏j=1n‍∏i=1nj‍(ζ-θj)i; then P1(ζ)=P0(ζ)(G(ζ)-α) is a polynomial whose zero coincides with those of G(ζ)-α. Also, denote by D the differential operator such that its characteristic polynomial is P1(ζ).The following lemma will be needed for our proof of Proposition2.Lemma 1.
Letd(k) indicate the kth derivative with respect to x of any differentiable function and define(8)Fi,ηj,x=ddx+ηjie-ηjx∫-∞xϕyx-yi-1eηjydywith d/dx+ηj(i) being the generalized Leibniz operator such that(9)ddx+ηji≔∑k=0iikηji-kdk.Then, for all i≥1,(10)Fi,ηj,x=i-1!ϕx.Proof.
We proceed by induction oni. For i=1, we have (11)F1,ηj,x=ddx+ηje-ηjx∫-∞xϕyeηjydy=-ηje-ηjx∫-∞xϕyeηjydy+ϕx+ηje-ηjx∫-∞xϕyeηjydy=ϕx.Moreover, (12)ddx+ηje-ηjx∫-∞xϕyx-yi-1eηjydy=-ηje-ηjx∫-∞xϕyx-yi-1eηjydy+i-1e-ηjx∫-∞xϕyx-yi-2eηjydy+ηje-ηjx∫-∞xϕyx-yi-1eηjydy=i-1e-ηjx∫-∞xϕyx-yi-2eηjydy.It follows inductively that(13)Fi,ηj,x=ddx+ηjie-ηjx∫-∞xϕyx-yi-1eηjydy=ddx+ηji-1ddx+ηje-ηjx∫-∞xϕyx-yi-1eηjydy=i-1ddx+ηji-1e-ηjx∫-∞xϕyx-yi-2eηjydy=i-1Fi-1,ηj,x=i-1!F1,ηj,x=i-1!ϕx,which is the desired result.We may now state the following.Proposition 2.
Suppose a bounded solutionϕ defined on R to the boundary value problem (7) exists. Then on R∖{h,H}, ϕ is infinitely differentiable and satisfies the ODE,(14)Dϕ≡0,onh,H.Hence, on (h,H), ϕ(x)=∑k=1S‍Qkeρkx for some constants Qk.Proof.
Applying the infinitesimal generatorL to the function ϕ, we obtain (15)Lϕx=σ22ϕ′′x+μϕ′x+λ∑i=1m∑j=1mipijηjii-1!e-ηjx∫-∞xϕyx-yi-1eηjydy+λ∑j=1m∑i=1mjqijθjii-1!eθjx∫x+∞ϕyy-xi-1e-θjydy-λϕx.Next, ϕ will be shown to satisfy an ODE. Using Lemma 1, we get, for j=1,2,…,m, i=1,2,…,mj,(16)ddx+ηjie-ηjx∫-∞xx-yi-1ϕyeηjydy=i-1!ϕx.The same computation will yield, for j=1,2,…,n, i=1,2,…,nj, (17)ddx-θjieθjx∫x+∞y-xi-1ϕye-θjydy=-i-1!ϕx.Now, since σ>0 and (L-α)ϕ≡0 then, thanks to Proposition 3.3 in the work of Chen et al. [18], ϕ is infinitely differentiable on (h,H) and(18)0=∏j=1m∏i=1mjddx+ηji∏j=1n∏i=1njddx-θjiL-αϕx=∏j=1m∏i=1mjddx+ηji∏j=1n∏i=1njddx-θjiσ22d2dx2+μddx-λ-αϕx+λ∑j=1m∑i=1mj∏k=1,k≠jm∏i=1mjddx+ηkipijηjii-1!i-1!ϕx-λ∑j=1n∑i=1nj∏k=1,k≠jn∏i=1njddx-θkiqijθjii-1!i-1!ϕx.Hence, (18) transforms the integrodifferential equation (L-α)ϕ≡0 into an ODE:D^ϕ≡0, where D^ is high order differential operator.
To complete the proof,D^ must be shown to coincide with D. To show that the characteristic polynomials of D and D^ will suffice, write P^(ζ) as the characteristic polynomial of D^. Then, by (18), P^ is given by (19)P^ζ=∏j=1m∏i=1mjζ+ηji∏j=1n∏i=1njζ-θjiμζ+σ22ζ2+λ∑j=1m∑i=1mjpijηjiζ+ηji+∑j=1n∑i=1nj-qijθjiζ-θii-1-α=P0ζGζ-α.This equation reveals that the characteristic polynomial P1(ζ) of D equals that, P^(ζ), of D^, which completes the proof.Proposition 3.
Suppose thatϕ is a bounded solution to the boundary value problem (7) and, on (h,H),ϕ(x)=∑k=1S‍Qkeρkx for some constants Qk. Then the constant vector Q satisfies the equation(20)AQ=Vg,where A is S×S nonsingular matrix given by(21)
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Hand Vg=Vg,1(j,i),j=1,2…,m,i=1,2…,mj,g(h),Vg,2(j,i),j=1,2…,n,i=1,2…,nj,g(H),(24)Vg,1j,i=∫-∞0gy+hηji-yi-1eηjyi-1!dy,(25)Vg,2j,i=∫0+∞gy+Hθjiyi-1e-θjyi-1!dy.Proof.
SinceL-αϕ=0 and ϕ(x)=∑k=1S‍Qkeρkx on (h,H), which entails(26)0=L-αϕx=σ22ϕ′′x+μϕ′x+λ∫-∞+∞ϕx-yfydy-λ+αϕx=∑k=1SQkeρkxσ22ρk2+μρk-λ+α+λ∫-∞+∞ϕx-yfydy,furthermore(27)∫-∞+∞ϕx-yfydy=∫-∞h​+∫H+∞​gyfx-ydy+∫x-Hx-hϕx-yfydy=∑j=1m∑i=1mjpije-ηjx-h∫-∞0ηjii-1!gy+hx-h-yi-1eηjydy+∑j=1n∑i=1njqijeθjx-H∫0+∞θjii-1!gy+Hy+H-xi-1e-θjydy+∑j=1m∑i=1mjpij∑k=1SQkeρkxηjii-1!∫0x-hyi-1e-ηj+ρkydy+∑j=1m∑i=1mjqij∑k=1SQkeρkxθjii-1!∫x-H0yi-1eθj-ρkydy.Now, since (a±y)i-1=∑l=0i-1‍i-1l(±y)l(a)i-1-l,(28)∫0ayi-1e-βydy=β-iΓi,aβ=β-ii-1!1-e-aβ∑l=0i-1aβll!with Γi,x being the incomplete gamma function (see Gradshteyn and Ryzhik [19, page 342]).
Consequently, by combining (26) and (27) and taking into account the fact that G(ρk)-α=0 for all k=1,2,…,S, we obtain(29)0=∑j=1m∑i=1mjpije-ηjx-h∫-∞0ηjii-1!gy+h∑l=0i-1i-1lx-hl-yi-1-leηjydy+∑j=1n∑i=1njqijeθjx-H∫0+∞θjii-1!gy+H∑l=0i-1i-1lH-xlyi-1-le-θjydy+∑k=1SQkeρkh∑j=1m∑i=1mj-pije-ηjx-hηjiηj+ρki∑l=0i-1ηj+ρkx-hll!+∑k=1SQkeρkH∑j=1m∑i=1mj-qijeθjx-Hθjiρk-θji∑l=0i-1θj-ρkx-Hll!.Comparing e-ηj(x-h) and eθj(x-H) yields (20), which entails the desired result.Lemma 4.
For a given value ofα>0 the matrix A given by (21) is invertible.Proof.
Assume thatAC=0 for some vector C=C1,C2,…,CST. Consider the function V(x)=∑k=1S‍Ckeρkx for x∈(h,H) and V(x)=0, otherwise, with ρ1,…,ρS to be the distinct zeros of the equation G(x)-α=0. Since AC=0 and V(x) is a solution to the boundary value problem,(30)L-αϕx=0,inh,H,ϕx=0,on-∞,h∪H,+∞.From the uniqueness of solutions to the boundary value problem (30), V(x)≡0 in (h,H). Now, since {eρkx,1≤k≤S} are linearly independent then C=0 and A is invertible.In the following,y·z is written for the usual inner product of the vectors y and z in CS and for every real value x, eαρ(x)=[eρ1x,eρ2x,…,eρSx], where ρ1,ρ2,…,ρS are the S=N+M+2 roots of the equation G(ζ)=α. Our main result is the following.Theorem 5.
For anyα≥0 and a nonnegative bounded measurable function g on (h,H)c, the following assertions are equivalent: (a)
ϕ(x)=Ex[e-ατg(Xτ)], where τ≔inft≥0:Xt≥horXt≤H.(b)
The functionϕ(x) solves the boundary problem (7).(c)
The functionϕ(x) is given by the formula (31)ϕx=Qg·eαρx,ifx∈h,H,gx,ifx∉h,H,with Q(g)=A-1Vg and A and Vg are given by formulas (21) and (24), respectively.Proof.
The fact that(b) implies (c) is straightforward consequence of Proposition 3. Conversely, consider the function V(x)=∑k=1S‍Qkeρkx for x∈(h,H) and V(x)=g(x) otherwise, where g is a bounded function on (h,H)c and Qk’s are given constants. Then the same reasoning as in Proposition 3 shows that(32)L-αVx=∑j=1m∑i=1mjpije-ηjx-h∫-∞0ηjii-1!gy+hx-h-yi-1eηjydy+∑j=1n∑i=1njqijeθjx-H∫0+∞θjii-1!gy+Hy+H-xi-1e-θjydy+∑j=1m∑i=1mjpij∑k=1SQkeρkxηjii-1!∫0x-hyi-1e-ηj+ρkydy+∑j=1m∑i=1mjqij∑k=1SQkeρkxθjii-1!∫x-H0yi-1eθj-ρkydy.Thanks to (20), we conclude that (c) implies (b).
Let us finally assume that(a) holds. Then by Feynman-Kac formula, the function ϕ(x) solves the boundary problem (7); hence (b) holds. Conversely, thanks to Proposition 4.1 in the work of Chen et al. [18], the boundary problem (7) has a unique solution; consequently (b) implies (a). The proof is complete.As an illustration of the main result of Theorem5, we can obtain closed-form expressions for the expectations of a variety of functions with respect to τ and Xτ. For instance, choosing g(x)=eγx in the above theorem, we can derive the joint Laplace transform of τ,Xτ, which is presented in the following corollary.Corollary 6.
For anyα>0, γ>0,(33)Exe-ατ+γXτ=Q·eαρx,ifx∈h,H,eγx,ifx∉h,H,where Q=Q1,Q2,…,QST=A-1V(γ), A is given by formula (21), and V(γ) is given by(34)Vγ=eγhγ+ηji,j=1,2…,m,i=1,2…,mj,eγh,eγHγ-θji,j=1,2…,n,i=1,2…,nj,eγH.As another consequence of Theorem5 and Lebesgue’s dominated convergence theorem, we get the following for the asymptotic case when h→-∞ and H→+∞, respectively.Corollary 7.
For two flat barriersh and H(h<H), define the first downwards passage time under h and the first upwards passage time over H by (35)τh+≔inf⁡t≥0:Xt≥h,τH-≔inft≥0:Xt≤H.Then for α>0, one has the following:(36)Exe-ατh+gXτh+=Q1·e1αρx,ifx≥h,gx,ifx<h,Exe-ατH-gXτH-=Q2·e2αρx,ifx≤H,gx,ifx>H,with (37)
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## 4. Pricing Double-Barrier Options

We now show how our theoretical results can be easily applied to derive pricing formulae for standard double-barrier options. We assume the asset price processSt:t≥0 under the risk-neutral probability measure P is defined as St≔eXt. The log-return process Xt:t≥0 is given by (1) where X0=log(S0) and μ≔r-σ2/2-λE[eY1-1] (i.e., E[e-r-TST]=S0), where r>0 is the risk-free rate. More recently, Cai et al. [20] presented the following.The payoff of a standard double-barrier option is activated (knocked in) or extinguished (knocked out) when the price of the underlying asset crosses barriers. For example, a knock-out call option will not give the holder the payoff of a European call option unless the underlying price remains within a prespecified range before the option matures. More precisely, consider an interval(L,U) and the initial asset price S0 is in it. The holder will receive ST-K+1{τ>T} at maturity T, where (38)τ=inft≥0:St≤UorSt≥L.Under the risk-neutral probability measure, the price of such option with maturity T and strike K is given by (39)e-rTEST-K+1τ>T∣S0.Make a change variable κ≔-log⁡K. Then, the expectation can be represented as(40)Cκ,T≔e-rTExS0eXT-e-κ+1τ>T.Define C^(α,γ) and Δ^(α,γ) as the double Laplace transforms of the price C(κ,T) in (40) and the delta Δ(α,γ)=∂C(κ,T)/∂S0 with respect to T and κ, respectively; that is,(41)C^α,γ=∫0∞∫-∞∞e-γκ-αTCκ,TdκdT,Δ^α,γ=∫0∞∫-∞∞e-γκ-αTΔα,γdκdT.Theorem 8.
For anyγ>0 and α>maxG(γ+1)-r,0(42)C^α,γ=S0γ+1γγ+1Gγ+1-α+rQ·eα+rρx-eγ+1x1logL/S0,logU/S0x,(43)Δ^α,γ=S0γγGγ+1-α+rQ·eα+rρx-eγ+1x1logL/S0,logU/S0x,where Q=A-1V(γ+1) and A associated with r+a is defined as in Theorem 5 and V is given by formula (34).Proof.
Equation (43) is an easy consequence of (42). To prove (43), using an idea of Kou et al. [21] (see also Cai et al. [20]) along with a change of the order of integration and the integral with respect to κ, we obtain(44)C^α,γ=∫0∞dT∫-∞∞e-γκ-αTCκ,Tdκ=Ex∫0τe-α+rTdT∫-logS0+XT∞e-γκS0eXT-e-κdκ=S0γ+1γγ+1Ex∫0τe-α+rT+γ+1XTdT.Now, we suppose that G(γ+1)<α+r and applying Itô’s formula to the process e-(α+r)t+(γ+1)Xt,t≥0, we obtain (45)Mt≔e-α+rt∧τ+γ+1Xt∧τ-eγ+1X0-∫0t∧τe-α+rT-α+reγ+1Xs+Leγ+1XTdT=e-α+rt∧τ+γ+1Xt∧τ-eγ+1X0-Gγ+1-α+r∫0t∧τe-αT+γ+1XTdTis a local martingale starting from M0=0. Since G(γ+1)<α+r, it follows from Fubini’s theorem that (46)E∫0te-α+rT+γ+1XTdT=∫0te-α+rTEeγ+1XTdT=∫0te-α+r+Gγ+1TdT=1-α+r+Gγ+1e-α+r+Gγ+1t-1∀t≥0.It follows from the dominated convergence theorem that {Mt;t≥0} is actually a martingale. In particular(47)Exe-α+rτ+γ+1Xτ-eγ+1x=Gγ+1-α+rEx∫0τe-α+rT+γ+1XTdT.Combining (44) and (47) and applying Corollary 6 we can therefore conclude that(48)C^α,γ=S0γ+1γγ+1Gγ+1-α+rExe-α+rτ+γ+1Xτ-eγ+1x=S0γ+1γγ+1Gγ+1-α+rQ·eα+rρx-eγ+1x1logL/S0,logU/S0x,where Q=A-1V(γ+1) and A associated with r+α is defined as in Theorem 5 and V is given by formula (34), which ends the proof.



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*Source: 5914657-2016-10-24.xml*