On Regularities of Expected Discounted Penalty at Ruin in
Two-Sided Jump-Diffusion Model
Yu-Ting Chen, Institute of Finance
National Chiao Tung University, Hsinchu, Taiwan Cheng-Few Lee, Department of Finance Rutgers University, New Brunswick, NJ, USA Yuan-Chung Sheu∗, Department of Applied Mathematics
National Chiao Tung University, Hsinchu, Taiwan.
Abstract
The expected discounted penalty with downside jumps has been extensively studied in Gerber and Shiu(1998), Gerber and Landry(1998), Tsai and Wilmott(2002) and oth-ers. In this paper, we study the expected discounted penalty in a perturbed compound Poisson model with two sided jumps. We show that it is always twice continuously differentiable provided that the jump size distribution has a bounded continuous(a.e.) density. Moreover, under some minor conditions, both its first and second derivatives vanish at infinity. Next, based on Boyarchenko and Levendorskiˇi(2002), we derive an integro-differential equation for the expected discounted penalty. When the jump size distribution is exponential, as an application of the integro-differential equation, we obtain an explicit formula for the expected discounted penalty. We close this paper by giving a generalization of the renewal integral equation in Gerber and Landry(1998).
Key words: jump diffusion, integro-differential equation, expected discounted penalty,
renewal integral equation, perpetual American put option
Mathematics Subject Classification(2000): 60G40 Running Title: Expected Discounted Penalty at Ruin
∗Postal address: Department of Mathematics, University of Washington, Seattle, WA 98195, USA. Email
address: [email protected]. (The author is currently visiting the department of mathematics, Univer-sity of Washington, Seattle, USA.)
1
Introduction
In the classical model of ruin theory, the process
Xt= x + ct − Zt, t ≥ 0, (1.1)
stands for the surplus process of an insurance company. Here, x > 0 is the initial surplus,
c > 0 is the rate at which the premiums are received, and Z = (Zt; t ∈ R+) is a compound
Poisson process which represents the aggregate claims between time 0 and t. (And hence Z has only upward jumps.) Ruin is the event that Xt≤ 0 for some t ≥ 0. Let τ be the time
of ruin and Xτ the negative surplus when ruin occurs. Given a penalty scheme g, Gerber
and Shiu(1998) considered the expected discounted penalty Φ(x) = E£e−rτg(Xτ)
¤
. (1.2)
Here, r is the risk-free rate and we use the convention that e−r·(+∞)= 0. By taking g ≡ 1
and r = 0, the ruin probability is a special case of (1.2). See Asmussen(2000) for related problems and results.
Gerber(1970) extended the classical model (1.1) by adding an independent diffusion so that the surplus process is a jump diffusion:
Xt= x + ct + σWt− Zt, t ≥ 0. (1.3)
Here σ > 0 and W = (Wt; t ≥ 0) is a standard Brownian motion that is independent of Z.
And this diffusion part adds an uncertainty to premium income or an additional uncertainty to aggregate claims. In this case ruin may be caused by oscillation (that is, Xτ = 0) or by a
claim (that is, Xτ < 0). Dufresne and Gerber(1991) studied the probability of ruin caused
by oscillation and the probability of ruin caused by a claim. Moreover, as in Gerber and Shiu(1998), Gerber and Landry(1998) considered the discounted expected penalty (1.2). In Gerber and Landry(1998), they began with an analysis of the surplus process in an infinitesimal amount of time and then heuristically derived an integro-differential equation for Φ under the assumption that Φ is twice continuously differentiable. Next, Gerber and Landry(1998) implicitly assumed that the derivatives of Φ are bounded and used the integro-differential equation to show that Φ satisfies a renewal integral equation. Based on this integral equation, they investigated the asymptotic behavior of Φ by means of classical renewal theory. Also, as an application of the integral equation, they obtained the optimal exercise strategy of a perpetual American put option. For related works, see Wang and Wu(2000) for a proof of continuous differentiability and intego-differential equation for a special penalty function, Tsai and Wilmott(2001) for a generalization of Gerber and Shiu(1998) and Gerber and Landry(1998). See also Cai(2004) for a proof of second order continuous differentiability of expected discounted penalty under compound Poisson model with stochastic interest rate.
In a recent paper, Cai and Yang(2005) discussed the continuous differentiability of ex-pected discounted penalty which is commonly assumed in insurance literature. They also showed that the ruin probability under stochastic interest rate indeed satisfies second order continuous differentiability if the density of jump distribution is continuous. As a byproduct, they gave a rigorous proof of the integro-differential equation for ruin probability. However, the result of Cai and Yang(2005) did not cover the case of general penalty function in the model of Gerber and Landry(1998), and they did not study the asymptotic behavior of Φ
and its derivatives at infinity. This still leaves the theoretical question unanswered whether the regularities, not just continuous differentiability, for expected discounted penalty as-sumed in Gerber and Landry(1998) and others do hold. On the other hand, it is worth noting that many problems in options pricing or credit risk modelling can be formulated in terms of expected discounted penalty. (Then it is nature to consider two-sided jump distributions.) For related works in this direction, see Asmussen et al.(2004), Boyarchenko and Levendorskii(2002), Chan(2005), Chen et al.(2006), Hilberink and Rogers(2002), Kou and Wang(2004), Mordecki(2002) and many others.
In this paper, we study the expected discounted penalty in a perturbed compound Poisson model with two sided jumps. We first use the distribution of Brownian motion stopped at exponential time to give an integral equation of simple structure(Theorem 2.1). If the penalty function is bounded and the jump distribution has a bounded continuous (a.e.) density, we show in Theorem 3.1 that the expected discounted penalty is always twice continuously differentiable. Moreover both its first and second order derivatives vanish at infinity if the expected discounted penalty is.(This is the case if the surplus process has a strictly positive total drift or the risk force is strictly positive(Proposition 3.1).) Next, we recall in Section 4 the potential theory for L´evy process and use the result of Boyarchenko and Levendorskiˇi(2002) to show that the expected discounted penalty for a large class of L´evy processes is a weak solution to an integro-differential equation(Theorem 4.2) and is a strong one if it is twice continuously differentiable(Theorem 4.3). This justifies the derivation of integro-differential equation in both Gerber and Shiu(1998) and Gerber and Landry(1998)(in which only downward jumps for X are allowed.) When the jump size distribution is exponential, based on the integro-differential equation, we get an ODE for the expected discounted penalty. By solving the ODE, we obtain an explicit formula for Φ. (See equation (5.10).) In particular, as an example, we have closed form solution( equation (5.14)) for any L-stopped perpetual American put option. Based on different approaches, similar results were obtained earlier in Asmussen et al.(2004) and Mordecki(2002). For an extension of this ODE approach, we refer to Chen et al.(2006). In the last section, under some assumptions, we show that the expected discounted penalty function satisfies the renewal (Gerber-Landry) integral equation(Theorem 6.1). This result generalizes Theorem 3 in Gerber and Landry(1998). See also Tsai and Wilmott(2002) for more general penalty scheme g(Xτ −, Xτ). As shown in Gerber and Landry(1998) and many others in insurance
literature, the generalized Gerber-Landry integral equation will serve as an interesting result for further study of expected discounted penalty, if the underlying process is a general two-sided jump diffusion.
2
Preliminaries
Let (Ω, F, P) be a probability space. On (Ω, F, P), there are a standard Brownian motion
W = (Wt; t ∈ R+) and a compound Poisson process Z = (Zt; t ∈ R+) with Zt=
PNt
n=1Yn.
Here N = (Nt; t ∈ R+) is a Poisson process with parameter λ > 0 and (Yn; n ∈ N)
are independent and identically distributed with a common distribution F . We assume R
{0}dF (y) = 0 and W, N and (Yn) are independent. For every x ∈ R, let Px be the law of
the process
Xt= X0+ ct + σWt− Zt, t ≥ 0, (2.1)
where c ∈ R, σ > 0 and X0 = x. Let P0 = P and write Ex[Z] =
R
Z(ω)dPx(ω) for a random
variable Z. Also, set Xc
Let (Ft) be the usual augmentation of the natural filtration of X. Then for every Borel
set A, the entry time of A by X,
τA= inf{t ≥ 0; Xt∈ A}, (2.2)
is an (Ft)−stopping time. (See Chapter III Theorem 2.17 in Revuz and Yor(2005).) And
so is the time T²= inf{t ≥ 0; |Xt− Xt−| > ²}. (See Sato(1996) page 276.) Throughout this
paper, we consider the function
Φ(x) = Ex
£
e−rτg(Xτ)
¤
. (2.3)
Here, r ≥ 0(we adopt the convention that e−r·+∞ = 0), τ = τ
(−∞,0] and g : R− → R+ is a
bounded nonnegative Borel function. In the sequel, we write ˆx = xσ for every x ∈ R and J as the first event time of the Poisson process N . Note that J = infn∈NT1/n and hence J is an (Ft)−stopping time.
In the following, we show that Φ satisfies an integral equation. First, we compute some functionals of τ and Xτ.
Lemma 2.1 For x > 0, we have Ex £ e−rτ; τ < J¤= e−bΓx and Ex £ e−rτg(Xτ); τ ≥ J ¤ = Z dF (y) Z ∞ 0 dwΦ(w − y) · λ Γe ˆ c(w−x)e−|w−x|Γ−λ Γe ˆ c(w−x)−(w+x)Γ ¸ . (2.4) Here Γ = ˆc +pˆc2+ 2(λ + r) > 0. (2.5)
Proof. Fix x > 0. Write Tα,β = inf{t ≥ 0; αt + Wt= β} and let h(s; α, β) be the density
of Tα,β under P. (See Borodin and Salminen(2002) page 295 formula 2.0.2.) Clearly we
have Ex[e−rτ; τ < J] = E £ e−rTc,−ˆˆ x; T ˆ c,−ˆx< J ¤
. Hence, by independence of W and J and the fact that J is an exponential random variable with mean 1λ, we get
Ex £ e−rτ; τ < J¤= Z ∞ 0 dtλe−λt Z t 0 dse−rsh(s; ˆc, −ˆx) = Z ∞ 0 ds Z ∞ s
dtλe−λte−rsh(s; ˆc, −ˆx) (Fubini’s Theorem)
= Z ∞ 0 dse−(r+λ)sh(s; ˆc, −ˆx) =E0 h e−(r+λ)Tˆc,−ˆx i = exp n −ˆcˆx − ˆxpˆc2+ 2(λ + r)o,
where the last equality follows from Borodin and Salminen(2002) page 295 formula 2.0.1. This proves the first equation.
To prove the second equation, we observe that {τ ≥ J} ⊇ {mins≤JXsc> 0}. Also, Px · τ ≥ J, min 0≤s≤JX c s ≤ 0 ¸ =Px · τ ≥ J, min 0≤s≤JX c s = 0 ¸ =Px[τ ≥ J, XJc = 0] ≤ Px[XJc = 0] = 0,
by the independence of J and W . These imply {τ ≥ J} = {mins≤JXsc> 0}, Px−a.s. From
this and Strong Markov Property, we have Ex £ e−rτg(Xτ); τ ≥ J ¤ =Ex £ e−rJΦ(XJ); τ ≥ J ¤ =Ex · e−rJΦ(XJc− Y1); min s≤JX c s > 0 ¸ .
By the independence of W , J and Y1, this gives
Ex £ e−rτg(Xτ); τ ≥ J ¤ = λ λ + r Z dF (y) Z dt(λ + r)e−(λ+r)tEx · Φ(Xtc− y); min s≤t X c s > 0 ¸ = λ λ + r Z dF (y)Ex · Φ(XJc0− y); min s≤J0X c s > 0 ¸ , (2.6)
where J0 is an exponential random variable with mean 1
λ+r and is independent of Xc.
To complete the proof, we calculate the density of λ+rλ Px
£
mins≤J0Xsc> 0, Xc
J0 ∈ dw
¤ . First, note that Px
£
mins≤J0Xsc> 0, XJc0 ≤ z
¤
= 0 for all z ≤ 0. Second, observe that Px · min s≤J0X c s ≤ 0, XJc0 ≤ 0 ¸ = Px[XJc0 ≤ 0] .
And hence for all z > 0,
λ λ + rPx · min s≤J0X c s > 0, XJc0 ≤ z ¸ = λ λ + r µ Px[XJc0 ≤ z] − Px · min s≤J0X c s ≤ 0, XJc0 1 ≤ z ¸¶ = λ λ + r µ Px[0 < XJc0 ≤ z] − Px · min s≤J0X c s ≤ 0, 0 < XJc0 ≤ z ¸¶ = Z z 0 · λ Γe ˆ c(w−x)e−|w−x|Γ− λ Γe ˆ c(w−x)−(w+x)Γ ¸ dw, (2.7) where the last equation follows from Borodin and Salminen(2002) page 250 formula 1.0.5 and page 252 formula 1.2.5 and Γ is given by (2.5). These give us the desired density on R+ and hence on R. Plugging this density into (2.6) gives (2.4). ¤
Now, we give an integral equation for Φ.
Theorem 2.1 For x ≥ 0, Φ satisfies the following integral equation Φ(x) = e−bΓxg(0) +λ ΓH(x) − λ Γe −(ˆc+Γ)x Z dF (y) Z ∞ 0 dwΦ(w − y)eˆcw−Γw, (2.8)
where Γ is given by (2.5) and H(x) = Z dF (y) Z ∞ 0 dwΦ(w − y)eˆc(w−x)−|w−x|Γ. (2.9)
Proof. Write Φ(x) = g(0)Ex[e−rτ1(τ < J)] + Ex[e−rτg(Xτ)1(τ ≥ J)]. And then the
3
Regularity of Expected Discounted Penalty At Ruin
Lemma 3.1 Suppose that F has a bounded density f that is continuous a.e. on R. Then
x 7→R dF (y)Φ(x − y) is continuous on R.
Proof. Fix x ∈ R. For any z, write Z Φ(z − y)dF (y) = Z ∞ z g(z − y)dF (y) + Z z −∞ Φ(z − y)dF (y).
Let ² > 0 and find M > x such that R(−∞,x−M +1]∪[x+M −1,∞)dF (y) < ². Then for all |x − z| < 1/2, ¯ ¯ ¯ ¯ Z Φ(z − y)dF (y) − Z Φ(x − y)dF (y) ¯ ¯ ¯ ¯ ≤2kΦk∞² + ¯ ¯ ¯ ¯ Z z+M z g(z − y)dF (y) − Z x+M x g(x − y)dF (y) ¯ ¯ ¯ ¯ + ¯ ¯ ¯ ¯ Z z z−M Φ(z − y)dF (y) − Z x x−M Φ(x − y)dF (y) ¯ ¯ ¯ ¯ ≤2kΦk∞² + ¯ ¯ ¯ ¯ Z 0 −M g(y)[f (z − y) − f (x − y)]dy ¯ ¯ ¯ ¯ + ¯ ¯ ¯ ¯ Z M 0 Φ(y)[f (z − y) − f (x − y)]dy ¯ ¯ ¯ ¯ . Since f , g and Φ are both bounded and f is continuous a.e. on R, by Dominated Convergence Theorem, we have lim sup z↓x ¯ ¯ ¯ ¯ Z Φ(z − y)dF (y) − Z Φ(x − y)dF (y) ¯ ¯ ¯ ¯ ≤ 2kΦk∞².
Since the last inequality holds for all ² > 0, the proof is completed. ¤ Lemma 3.2 Suppose F has a bounded density f that is continuous a.e. on R. The function
H(x) defined in (2.9) is twice continuously differentiable on R+ = [0, ∞). Moreover, its
first order and second order derivatives are given by H0(x) = − (ˆc + Γ)e−(ˆc+Γ)x Z x 0 dw Z dF (y)Φ(w − y)eˆcw+Γw − (ˆc − Γ)e−(ˆc−Γ)x Z ∞ x dw Z dF (y)Φ(w − y)ecw−Γwˆ (3.1) and H00(x) =(ˆc + Γ)2e−(ˆc+Γ)x Z x 0 dw Z dF (y)Φ(w − y)eˆcw+Γw + (ˆc − Γ)2e−(ˆc−Γ)x Z ∞ x dw Z dF (y)Φ(w − y)eˆcw−Γw− 2Γ Z dF (y)Φ(x − y). (3.2) Proof. Using the definition of H and Fubini’s Theorem, we write
H(x) = e−(ˆc+Γ)x Z x 0 dw Z dF (y)Φ(w − y)eˆcw+Γw+ e−(ˆc−Γ)x Z ∞ x dw Z dF (y)Φ(w − y)eˆcw−Γw.
By Lemma 3.1, w 7→R dF (y)Φ(w −y) is continuous on R. Hence, by Fundamental Theorem
of Calculus, H is differentiable on (0, ∞) and its first order derivative is given by (3.1). Similarly, by (3.1), H0 is continuously differentiable on (0, ∞) and its derivative is given by
(3.2). Note that H0(0+) and H00(0+) can be calculated directly by definition and we omit
the details. ¤
Lemma 3.3 If limx→∞Φ(x) = 0 and F has a bounded density which is continuous a.e.,
then lim x→∞e −(ˆc−Γ)x Z ∞ x dw Z dF (y)Φ(w − y)eˆcw−Γw= 0 and lim x→∞e −(ˆc+Γ)x Z x 0 dw Z dF (y)Φ(w − y)eˆcw+Γw= 0.
Proof. By the integral equation (2.8) and Lemma 3.2, we deduce that Φ is a continu-ously differentiable function on R+. So, by L’Hˆospital’s rule and Fundamental Theorem of
Calculus, lim x→∞e −(ˆc−Γ)x Z ∞ x dw Z
dF (y)Φ(w − y)eˆcw−Γw= lim
x→∞
−e(ˆc−Γ)xRdF (y)Φ(x − y)
(ˆc − Γ)e(ˆc−Γ)x = 0,
where the last equality follows from the assumption that limx→∞Φ(x) = 0 and Dominated
Convergence Theorem. The fact that limx→∞e−(ˆc+Γ)x
Rx
0 dw
R
dF (y)Φ(w − y)eˆcw+Γw = 0
follows similarly. ¤
Theorem 3.1 (1) For any jump distribution F on R such that R{0}dF (y) = 1, we have
Φ ∈ Cb(R+). (2) If the jump distribution has a bounded density that is continuous a.e. on
R, then Φ ∈ C2
b(R+). Moreover, if limx→∞Φ(x) = 0, then Φ ∈ C02(R+).
Proof. By (2.8) and the boundedness of Φ, we obtain that Φ ∈ Cb(R+). This proves (1).
To prove (2), we assume that the jump distribution F has a bounded density f that is continuous a.e. on R. Clearly we have
e−bΓx− λ Γe −(ˆc+Γ)x Z dF (y) Z ∞ 0 dwΦ(w − y)ecw−Γwˆ ∈ C02(R+).
By (2.8), it suffices to show that H defined in (2.9) is in C2
b(R+). Note that Γ − ˆc > 0.
Hence, by Lemma 3.2, for all x ∈ R+,
|H0(x)| ≤(ˆc + Γ)kΦk∞e−(ˆc+Γ)x Z x 0 dwe(ˆc+Γ)w+ (Γ − ˆc)kΦk∞e−(ˆc−Γ)x Z ∞ x dwe(ˆc−Γ)w =kΦk∞ h e−(ˆc+Γ)x ³ e(ˆc+Γ)x− 1 ´ + e−(ˆc−Γ)xe(ˆc−Γ)x i ≤ 3kΦk∞< ∞. (3.3)
This shows that H(x) ∈ C1
b(R+). Since
¯
¯RdF (y)Φ(x − y)¯¯ ≤ kΦk∞ for all x ∈ R+, we deduce from the above estimate and (3.2) that H ∈ Cb2(R+) as well.
Assume further that limx→∞Φ(x) = 0. By Lemma 3.3 and Eq. (3.1), we get that H ∈ C1
0(R+). Similarly, by Lemma 3.3 and Equations (3.1) and (3.2), we get that H ∈ C02(R+).
Proposition 3.1 If r > 0 or r = 0 and EX1> 0, then limx→∞Φ(x) = 0.
Proof. Write Φ(x) = E [e−rτxg(x + X
τx)], where τx= inf{t ≥ 0; ct + σWt− Zt≤ −x}.
First consider the case that r = 0 and EX1 > 0. By the assumption that EX1 > 0
and the Law of Large Numbers for L´evy processes(see Sato[1999] page 246 and page 247), we have limt→∞Xt = ∞ P−a.s. This gives limx→∞1[τx<∞] = 0 P-a.s. By Dominated Convergence Theorem, we get that limx→∞Φ(x) = 0.
Next, consider the case that r > 0. Note that τx is nondecreasing and unbounded. This
implies that τx ↑ ∞ as x → ∞ P0−a.s. By Dominated Convergence Theorem again, the
result follows. ¤
4
The Integro-Differential Equation
We recall here some works of Boyarchenko and Levendorskiˇi[BL02] in the setting of gen-eral L´evy processes and establish its connections with the integro-differential equation of expected discounted penalty.
We consider a family (X, {Px}x∈Rd) of L´evy processes on Rd. Here under Px, X0 = x
a.s.. Assume that X under P0 has characteristic triple (A, ν, γ). Then for every t ∈ R+,
E0
h eihz,Xti
i
= e−tΨ(z), z ∈ Rd,
where Ψ is of the form: Ψ(z) = 1
2hz, Azi − ihγ, zi + Z
Rd
³
1 − eihz,yi+ ihz, yi1{y;|y|≤1}(y) ´
ν(dy). (4.1) Here γ ∈ Rd, A is a symmetric nonnegative definite d × d matrix, and ν is a Borel measure on Rd satisfying
ν({0}) = 0,
Z
Rd(|x|
2∧ 1)ν(dx) < ∞. (4.2)
(As before, we write Ex for the expectation with respect to the probability measure Px and
reserve E for E0.) The function Ψ is often called the characteristic exponent of X.
Write C0(Rd) as the collection of continuous functions f on Rd such that f vanish at
infinity. In addition, write C2
0(Rd) as the collection of all twice continuously differentiable
functions f on Rd such that f and its partial derivatives with order ≤ 2 are all in C
0(Rd).
Then the infinitesimal generator L of X has a domain containing C02(Rd) and for any f ∈
C2 0(Rd), Lf (x) = 1 2 d X j,k=1 Ajk ∂ 2f ∂xj∂xk (x) + d X j=1 γj∂x∂f j(x) + Z Rd f(x + y) − f(x) −Xd j=1 yj ∂f ∂xj(x)1{y;|y|≤1} ν(dy). (4.3) Consider the family of resolvent operators {Ur; r ≥ 0} of X, that is, for every
nonnega-tive measurable function f ,
Urf (x) = Ex ·Z ∞ 0 e−rtf (Xt)dt ¸ .
Corresponding to the family of resolvent operators is the family of measures {Ur(x, ·); r ≥
0, x ∈ Rd} given by:
Ur(x, B) = Ur1B(x), B ∈ B(Rd).
We assume that Ur(x, ·) is absolutely continuous for x = 0 and for some r ≥ 0(hence for all r ≥ 0 and for all x). Then there exists a unique r−co-excessive function ur such that
for all nonnegative Borel measurable f
Urf (x) =
Z
ur(y − x)f (y)dy. (4.4) (For details, see Bertoin(1996).)
Let B be a closed set in Rd. Set T
B = inf{t ≥ 0; Xt∈ B}, the first entrance time of B.
Define for any g ∈ L∞(B) and r ≥ 0, PBrg(x) = Ex £ e−rTBg(X TB) ¤ . (4.5)
(Clearly, if d = 1 and B = (−∞, 0], then Pr
Bg(x) = Φ(x). ) In particular, if g ≡ 1 on B, we
have the following.(See Bertoin(1996) Theorem II.7.)
Theorem 4.1 Let r > 0 and B be a closed set. Then there exists a unique Radon measure
µB supported on B(that is, µB(Bc) = 0) such that
Z Rd e Ur(y, dx)µB(dy) = Ex £ e−rTB¤dx (x ∈ Rd).
Here eUr is the resolvent operator for the dual process eX = −X.
Corollary 4.1 Pr
B1(x) =
R
ur(y − x)µ
B(dy) for almost every x.
Proof. Let A be any Borel set. Recall by (4.4), eUr(y, A) = R
Aeur(x − y)dx. Hence, by Theorem 4.1, Z A Ex[e−rTB]dx = Z e
Ur(y, A)µB(dy) =
Z Z A e ur(x − y)dxµB(dy) = Z A Z ur(y − x)µB(dy)dx,
where the last equality follows from Fubini’s Theorem and the fact that ur(x) = eur(−x).
Since A is arbitrary, we have completed the proof. ¤ Note that since eX has characteristic triple (A, eν, −γ) with eν(B) = ν(−B), the
infinites-imal generator eL of eX is given by, for f ∈ C2 0(Rd), e Lf (x) = 1 2 d X j,k=1 Ajk ∂2f ∂xj∂xk (x) − d X j=1 γj ∂f ∂xj (x) + Z Rd f(x − y) − f(x) + d X j=1 yj∂x∂f j(x)1{y;|y|≤1}(y) ν(dy). (4.6) We write φ ∈ C∞
c (Rd) if φ is infinitely differentiable and has a compact support.
Lemma 4.1 If φ ∈ C∞
Proof. Assume φ ∈ C∞ c (Rd) and write G(x) = Z φ(x − y) − φ(x) + 1{y;|y|≤1} d X j=1 yj∂x∂φ j(x) ν(dy).
By (4.6), it suffices to show that G ∈ L1(Rd). Assume φ vanishes outside F = B(0, R).
We show separatelyRF|G(x)|dx < ∞ and RFc|G(x)|dx < ∞. By Hille-Yosida’s Theorem,
we have (r − eL)φ ∈ C0(Rd)(see Sato(1996) Theorem 31.3 and Theorem 31.5) and hence
G ∈ C0(Rd). Since the Lebesgue measure m(F ) of F is finite and G ∈ C0(Rd), it follows
that
Z
F
|G(x)|dx ≤ m(F )kGk∞< ∞.
It remains to show thatRFc|G(x)|dx < ∞. For every x ∈ Fc, we have
G(x) = Z φ(x − y)ν(dy). This gives Z Fc|G(x)|dx ≤ Z Fcdx Z |φ(x − y)| ν(dy) = Z Fcdx Z |y|≤1 |φ(x − y)|ν(dy) + Z Fcdx Z |y|>1 |φ(x − y)|ν(dy).
SinceR|y|>1ν(dy) < ∞ by (4.2),
Z Fc dx Z |y|>1 |φ(x − y)|ν(dy) ≤ Z |y|>1 ν(dy) Z |φ(x − y)|dx = Z |y|>1 ν(dy) Z |φ(x)|dx < ∞.
Note that, since φ ∈ Cc∞(Rd), the second order derivatives of φ are uniformly bounded. So, for all x, y ∈ Rd, ¯ ¯ ¯ ¯ ¯ ¯φ(x − y) − φ(x) + d X j=1 yj∂x∂φ j (x) ¯ ¯ ¯ ¯ ¯ ¯≤ 1 2|y| 2 Xd j,k=1 ° ° ° ° ∂ 2φ ∂xj∂xk ° ° ° ° ∞ . (4.7) by Taylor’s Theorem. That is,
¯ ¯ ¯ ¯ ¯ ¯φ(x − y) − φ(x) + d X j=1 yj∂x∂φ j(x) ¯ ¯ ¯ ¯ ¯ ¯≤ C|y|
2, for some constant C > 0.
Also we have φ = 0 on Fc and x − y ∈ F with |y| ≤ 1 imply that x ∈ B(0, R + 1). Hence,
we get Z Fc dx Z |y|≤1 |φ(x − y)|ν(dy) = Z Fc dx Z |y|≤1 1F(x − y) ¯ ¯ ¯ ¯ ¯ ¯φ(x − y) − φ(x) + d X j=1 yj∂x∂φ j(x) ¯ ¯ ¯ ¯ ¯ ¯ν(dy) ≤ C Z Fcdx Z |y|≤1 1B(0,R+1)(x)|y|2ν(dy) = C Z Fc∩B(0,R+1)dx Z |y|≤1 |y|2ν(dy) < ∞,
Hence, we obtainRFc|G(x)|dx < ∞. ¤
To give the following definition, we write hf1, f2i =R f1(x)f2(x)dx. Definition 4.1 Let r ≥ 0, B be a closed set in Rd and g ∈ L
∞(B). We say that a bounded Borel measurable function h : Rd→ R is a weak solution of the boundary value problem:
½
(r − L)h = 0, in Bc,
h = g, on B. (4.8)
if h(x) = g(x) for any x ∈ B and, for any φ ∈ Cc∞(Bc),
hh, (r − eL)φi = 0.
Also, we say that h is a strong solution of the boundary problem (4.8) if h(x) = g(x) for all x ∈ B, h ∈ C2(Bc) and
(r − L)h = 0 on Bc.
The following theorem is a version of Theorem 2.1 in Boyarchenko and Levendorski(2002). For the convenience of readers, we give a sketch of proof.
Theorem 4.2 Assume that the characteristic exponent Ψ of X admits an analytic
contin-uation into a tube domain Rd+ iO, where O is an open set in Rd containing 0. Let B be a closed set in Rd. Then for any r ≥ 0 and g ∈ L
∞(B), PBrg is a weak solution of the boundary value problem (4.8).
Sketch of Proof: We consider firstly g ≡ 1 on B. We show that for r > 0, Pr
B1 is a weak
solution to (4.8). For all r > 0 and φ ∈ C∞
c (Bc), we have, by Corollary 4.1, Z PBr1(x)(r − eL)φ(x)dx = Z µZ ur(y − x)dµB(y) ¶ (r − eL)φ(x)dx. Since ur(z) = eur(−z), we obtain, by (4.4), Z PBr1(x)(r − eL)φ(x)dx = Z dµB(y) µZ ur(y − x)(r − eL)φ(x)dx ¶ = Z dµB(y) µZ e ur(x − y)(r − eL)φ(x)dx ¶ = Z e Ur(r − eL)φ(y)dµB(y).
By Hille-Yosida Theorem, eUr(r − eL) = I. So,
Z PBr1(x)(r − eL)φ(x)dx = Z e Ur(r − eL)φ(x)dµB(x) = Z φ(x)dµB(x).
Since φ vanishes on B and µB is supported on B, this gives hPBr1, (r − eL)φi = 0.
Therefore Pr
B1 is a weak solution to the boundary value problem (4.8) for any r > 0.
The assumption that the characteristic exponent admits an analytic continutation implies that Pr
Bg is a weak solution to the boundary value problem (4.8) for any r > 0 and any g ∈ L∞(B).
Finally, we complete the proof by showing that Pr
Bg is a weak solution to (4.8) for r = 0
and g ∈ L∞(B). Since {Pr
Bg; r ≥ 0} is uniformly bounded and by Lemma 4.1, eLφ ∈ L1,
{(r − eL)φ · Pr
Bg; 0 ≤ r ≤ 1} are dominated by an integrable function. Therefore, by the
Dominated Convergence Theorem, 0 = lim
r↓0+hP r
Bg, (r − eL)φi = hPB0g, −eLφi.
This implies that PB0g is a weak solution to (4.8) for r = 0. ¤
Remark 4.1. In Boyarchenko and Levendorski(2002), Remark 2.4.(d), they mentioned that an approximation scheme is possible to relax the analytic continuation assumption. ¤
Moreover, we have the following:
Theorem 4.3 Given r ≥ 0 and B a closed set in Rd. Suppose Pr
Bg is a weak solution of (4.8) and PBrg ∈ C2(Bc) ∩ C(Bc). Then Pr
Bg is a strong solution to (4.8) in the following two cases:
1. g is bounded continuous on B and ν is a finite measure.
2. g is a bounded Borel measurable function on B and ν is a finite measure with a bounded density that is continuous a.e..
Proof. Write H(x) = Pr
Bg(x). For any multi-index α = (α1, · · · , αd), set |α| =
Pd j=1αj and write Dαf (x) = µ ∂ ∂x1 ¶α1 · · · µ ∂ ∂xd ¶αd f (x). By partial integration, if φ ∈ Cc∞(Bc), (−1)|α| Z Dαφ(x)f (x)dx = Z φ(x)Dαf (x)dx, ∀f ∈ C|α|(Bc). From this, one immediately gets
Z 1 2 d X j,k=1 Ajk ∂ 2φ ∂xj∂xk − d X j=1 γj∂x∂φ j (x)H(x)dx = Z 1 2 d X j,k=1 Ajk ∂ 2H ∂xj∂xk + d X j=1 γj∂x∂H j (x)φ(x)dx. (4.9)
On the other hand, we have Z dx Z ν(dy)H(x) φ(x − y) − φ(x) + 1{y;|y|≤1} d X j=1 yj∂x∂φ j (x) = Z ν(dy) Z dxH(x) φ(x − y) − φ(x) + 1{y;|y|≤1} d X j=1 yj∂x∂φ j (x) = Z ν(dy) Z dxφ(x) H(x + y) − H(x) − 1{y;|y|≤1} d X j=1 yj∂x∂H j(x) .
Then applying the Fubini’s Theorem again gives Z dx Z ν(dy)H(x) φ(x − y) − φ(x) + 1{y;|y|≤1} d X j=1 yj ∂φ ∂xj (x) = Z dx Z ν(dy)φ(x) H(x + y) − H(x) − 1{y;|y|≤1} d X j=1 yj∂x∂H j (x) . (4.10) Now, combining (4.9) and (4.10) gives that hH, eLφi = hLH, φi. Hence,
h(r − L)H, φi = hH, (r − eL)φi = 0, ∀φ ∈ Cc∞(Bc). This implies that (r − L)H = 0 a.s. on Bc.
To show that (r − L)H(x) = 0 for every x ∈ Bc, it suffices to show that (r − L)H is continuous on Bc. Since H is in C2(Bc), by (4.3), we need only show that
Z H(x + y) − H(x) − 1{y;|y|≤1} d X j=1 yj∂H ∂xj (x) ν(dy) (4.11) is continuous on Bc. First, assume that g is continuous and ν is a finite measure. Since H ∈ C2(Bc) ∩ C(Bc), H = g on ∂B by the definition of H, and g ∈ C(B), we have H is
continuous on Rd. So, using the assumption ν is a finite measure, we conclude that the
function in (4.11) is continuous on Bc. This proves (1). Second, assume that g is bounded
and ν is a finite measure with an a.e. bounded continuous density. Then a modification of Lemma 3.1 also gives that the function R H(x + y)ν(dy) in (4.11) is continuous. This
proves (2).
¤ In the following two sections, we provide applications of integro-differential equation for Φ.
5
Closed Form Solution for Expected Discounted Penalty
From now on, we consider the process X in the setting of section 2 and Φ as defined in (1.2). Recall that r ≥ 0, τ is the first entry time of B = (−∞, 0] and g is a bounded function on (−∞, 0]. We assume further that F has a bounded density that is continuous a.e. in R. By Theorem 4.2 and Remark 4.1, Φ is a weak solution to the boundary value problem (4.8). Moreover, by Theorem 3.1 and Theorem 4.3, we have the integro-differential equation
σ2 2 Φ 00(x) + cΦ0(x) + λ Z x −∞ Φ(x − y)dF (y) + λ Z ∞ x g(x − y)dF (y) − (λ + r)Φ(x) = 0. (5.1) In particular, if gD(x) = g(x)1(x = 0) and ΦD ≡ Pr
BgD, then ΦD(z) = 0 for z < 0 and
hence (5.1) becomes σ2 2 Φ 00 D(z) + cΦ0D(z) + λ Z z −∞ ΦD(z − y)dF (y) − (λ + r)ΦD(z) = 0, ∀z > 0. (5.2)
(This is Eq.(6) in Gerber and Landry(1998) in which X has only downward jumps and the integro-differential equation was derived heuristically .) On the other hand, if we set
gJ(x) = g(x)1(x < 0) and ΦJ = Pr
BgJ, then ΦJ(z) = g(z) for z < 0 and hence (5.1) becomes σ2 2 Φ 00 J(z) + cΦ0J(z) + λ Z z 0 ΦJ(z − y)dF (y) + λ Z ∞ z g(z − y)dF (y) − (λ + r)ΦJ(z) = 0. (5.3) (C.f. the last equation followed by Eq.(2.4) in Tsai and Wilmott(2002).)
Consider the case that dF (y) = ηe−ηy1(y > 0)dy for some η > 0. Then (5.1) becomes σ2 2 Φ 00(x) + cΦ0(x) + λ Z x 0 Φ(y)ηe−η(x−y)dy + λ Z 0 −∞ g(y)ηe−η(x−y)dy − (λ + r)Φ(x) = 0. (5.4) Differentiating both sides of (5.4) yields
0 =σ2 2 Φ 000(x) + cΦ00(x) + ληΦ(x) − λη2e−ηx Z x 0 Φ(y)eηydy − λη2e−ηx Z 0 −∞ g(y)eηydy − (λ + r)Φ0(x). By (5.4), we can rewrite the last equation as
0 =σ2 2 Φ 000(x) + cΦ00(x) + ληΦ(x) + η µ σ2 2 Φ 00(x) + cΦ0(x) − (λ + r)Φ(x) ¶ − (λ + r)Φ0(x). Define the differential operator
A = σ2 2 d3 dx3 + µ c + ησ2 2 ¶ d2 dx2 + (cη − λ − r) d dx − ηr. (5.5)
Then Φ satisfies the ordinary differential equation
AΦ = 0 on (0, ∞). (5.6)
Next, we consider the boundary conditions for Φ. By the definition of Φ, we have
Φ(0) = g(0) (5.7)
Also by Proposition 3.1, we have
lim
x→∞Φ(x) = 0. (5.8)
Clearly, (5.4) gives another boundary condition of Φ:
σ2
2 Φ
00(0+) + cΦ0(0+) = (λ + r)g(0) − λ
Z 0
−∞
Then as shown in Appendix B, if r > 0, Φ takes the following form Φ(x) = PBrg(x) = h (λ + r)g(0) − λR−∞0 g(y)ηeηydy i − ³ σ2 2 ρ22+ cρ2 ´ g(0) σ2 2(ρ21− ρ22) − c(ρ2− ρ1) eρ1x + h (λ + r)g(0) − λR−∞0 g(y)ηeηydyi−³σ2 2 ρ21+ cρ1 ´ g(0) σ2 2(ρ22− ρ21) − c(ρ1− ρ2) eρ2x, x ∈ R +. (5.10) Here ρ1 and ρ2 are the two strictly negative zeros of the characteristic polynomial of A and ρ2 < ρ1 < 0. Now, if g is an nonnegative function such that
R0
−∞g(y)ηeηydy < ∞, by
approximation of g by bounded functions, we see that (5.10) still holds for g. In the following, we consider an example in option pricing.
Example 5.1 (Perpetual American Put Option) Assume that the price process (St; t ≥ 0) of a stock follows (under a risk-neutral probability measure) an exponential jump diffusion process:
St= s exp {ct + σWt− Zt} , t ∈ R+,
where s > 0 is a constant. Consider a perpetual American put option
V∗(s) = sup τ E £ e−rτ(K − Sτ)+|S0 = s ¤ .
Here the supremum is taken over all stopping times. Mordecki(2002) shows that there exists a constant L∗ independent of s such that
V∗(s) = E£e−rτL∗(K − S τL∗) +|S 0 = s ¤ .
Here, we define, for all constant L > 0, the L-stopped strategy
τL= inf{t ≥ 0; St≤ L}. (5.11)
Therefore, it is tempted as a preliminary step to consider for any constant L > 0 an
L−stopped perpetual American put option whose no arbitrage price is given by V (s, L) = E[e−rτL(K − S
τL)+|S0 = s]. (5.12)
Fix L > 0 and set
Xt= log St− log L = X0+ ct + σWt− Zt,
where X0 = log s − log L. Then (5.11) becomes
τL= inf {t ≥ 0; Xt≤ 0} , (5.13)
and the L-stopped perpetual American put option price becomes
V (s, L) = E h e−rτL¡K − LeXτL¢+|X0= x i , x = log s − log L. Clearly, V (s, L) = ΦL(x) = PBrgL(x) = E[e−rτLg L(XτL)|X0= x],
where gL(y) = (K − Ley)+, x = log s − log L and B = (−∞, 0].
Consider the special case that dF (y) = ηe−ηy1(y > 0)dy, η > 0. To apply (5.10), we
need to computeR−∞0 (K − Lez)+ηeηzdz. We have
Z 0 −∞ (K − Lez)+ηeηzdz = Z 0∧log(K/L) −∞ (Kηeηz− Lηe(η+1)z)dz = Z ∞ log+(L/K)(Kηe −ηz− Lηe−(η+1)z)dz = γ(L), where γ(L) = Ke−η log+(L/K)− Lη η + 1e −(η+1) log+(L/K) .
By (5.10), the L−stopped American put option price is given by for all s ∈ R+,
V (s, L) =(λ + r)(K − L) +− λγ(L) −³σ2 2 ρ22+ cρ2 ´ (K − L)+ σ2 2 (ρ21− ρ22) − c(ρ2− ρ1) ³ s L ´ρ1 +(λ + r)(K − L) +− λγ(L) −³σ2 2 ρ21+ cρ1 ´ (K − L)+ σ2 2 (ρ22− ρ21) − c(ρ1− ρ2) ³ s L ´ρ2 , (5.14) where ρ1 and ρ2 are the two strictly negative zeros of the characteristic polynomial of A in (5.5) and ρ2 < ρ1.
¤
6
Gerber-Landry Integral Equation
In this section, under some assumptions, we derive the renewal(Gerber-Landry) integral equation(see (6.20) below) for the expected discounted penalty Φ. Recall the definition of Ψ in (4.1) and write ψ(iz) = −Ψ(z). Then we have
ψ(ζ) = Dζ2+ cζ + λ Z
e−ζydF (y) − λ, (6.1)
where D = σ2/2. We call ψ the exponent of the Laplace transform of X. Set
∆ = sup ½ ζ ∈ R+; Z e−ξydF (y) < ∞, ∀ξ ∈ [0, ζ] ¾ . (6.2)
Throughout this section, we assume that ∆ > 0 and ψ(∆−) > 0. Let r > 0. Since
ψ(0)−r < 0 and ψ is strictly convex on [0, ∆), there exists a unique number ρ∗ ∈ (0, ∆) such
that ψ(ρ∗) − r = 0 and ρ∗ is called the Lundberg’s constant in literatures. An implication of
ρ∗ is that ¡e−rt+ρ∗X
t; t ≥ 0¢is a martingale provided that eρ∗X1 satisfies some integrability
condition. (See Gerber and Landry(1998)). Therefore, in risk neutral pricing(that is, {Vt= eXt; t ≥ 0} is the price process of a security and {e−rtV
t; t ≥ 0} is an (Ft)-martingale under
the risk-neutral probability measure), we have ρ∗ = 1. We write β = c
D+ ρ
∗, α = β + ρ∗, (6.3)
and hρ∗(x) = e−ρ∗xh(x) for any function h. The following proposition gives an expression
Proposition 6.1 The derivative of Φ at 0 is given by Φ0(0+) = ϑ0−Dλ Z 0 −∞ dF (y) Z −y 0 dvΦρ∗(v)e−ρ ∗y , (6.4) where ϑ0 = −βg(0) +Dλ Z ∞ 0 dv Z ∞ v dF (y)e−ρ∗ygρ∗(v − y). (6.5)
Proof. We begin with the equation (5.1) and write v instead of the variable x. Multiplying both sides of this equation by e−ρ∗v
gives e−ρ∗v · D d2 dv2 + c d dv − (λ + r) ¸ Φ(v) + λ Z ∞ −∞ Φ(v − y)e−ρ∗vdF (y) = 0. (6.6) Note that Φ(v) = eρ∗v Φρ∗(v) for v ∈ R. Then Φ0(v) = ρ∗eρ∗vΦρ∗(v) + eρ ∗v Φ0ρ∗(v) and Φ00ρ∗(v) = (ρ∗)2eρ ∗v Φρ∗(v) + 2ρ∗eρ ∗v Φ0ρ∗(v) + eρ ∗v Φ00ρ∗(v). Hence, (6.6) becomes 0 =DΦ00ρ∗(v) + ¡ c + ρ∗σ2¢Φ0ρ∗(v) + λ Z ∞ −∞ Φρ∗(v − y)e−ρ ∗y dF (y) +£D(ρ∗)2+ cρ∗− (λ + r)¤Φρ∗(v).
Recall that ψ(ρ∗) = r. Then
0 =DΦ00ρ∗(v) + ¡ c + ρ∗σ2¢Φ0ρ∗(v) + λ Z ∞ −∞ Φρ∗(v − y)e−ρ ∗y dF (y) − λΦρ∗(v) Z e−ρ∗ydF (y). (6.7) Integrate (6.7) from v = 0 to v = z gives
D£Φ0ρ∗(z) − Φ0ρ∗(0) ¤ +¡c + ρ∗σ2¢[Φρ∗(z) − Φρ∗(0)] + λ Z z 0 dv Z v −∞ dF (y)Φρ∗(v − y)e−ρ ∗y − λ Z z 0 dv Z dF (y)Φρ∗(v)e−ρ ∗y + λ Z z 0 dv Z ∞ v dF (y)gρ∗(v − y)e−ρ ∗y = 0. (6.8) Note that Z z 0 dv Z v −∞ dF (y)Φρ∗(v − y)e−ρ ∗y = Z 0 −∞ dF (y) Z z 0 dvΦρ∗(v − y)e−ρ ∗y + Z z 0 dF (y) Z z y dvΦρ∗(v − y)e−ρ ∗y = Z 0 −∞ dF (y) Z z 0 dvΦρ∗(v − y)e−ρ ∗y + Z z 0 dv Z z−v 0 dF (y)Φρ∗(v)e−ρ ∗y .
So, (6.8) is equivalent to 0 =D£Φ0ρ∗(z) − Φ0ρ∗(0) ¤ +¡c + ρ∗σ2¢[Φρ∗(z) − Φρ∗(0)] + λ " Z 0 −∞ dF (y) Z z 0 dvΦρ∗(v − y)e−ρ ∗y − Z z 0 dv µZ ∞ z−v + Z 0 −∞ ¶ dF (y)Φρ∗(v)e−ρ ∗y + Z z 0 dv Z ∞ v dF (y)e−ρ∗ygρ∗(v − y) # . (6.9)
Recall by Theorem 3.1 that Φ0(z) and Φ(z) tend to zero as z → ∞. Hence, let z → ∞ in
the last equation and we get Φ0ρ∗(0) = 1 D " − (c + ρ∗σ2)g(0) − λ Z 0 −∞ dF (y) Z −y 0 dvΦρ∗(v)e−ρ ∗y + λ Z ∞ 0 dv Z ∞ v dF (y)e−ρ∗ygρ∗(v − y) # . (6.10) Since Φ0(0) = ρ∗g(0) + Φ0 ρ∗(0), we get (6.4). ¤ Proposition 6.2 The function Φ satisfies the equation:
Φ(x) =g(0)e−βx+ lim n,m→∞ µZ n −m Φ(x − v)G(v)dv − e−βx Z n −m Φ(−v)G(v)dv ¶ , x ∈ R+, (6.11) where G = G(+) on [0, ∞), G = G(−) on (−∞, 0) and G(+)(v) =λ D Z v 0 dze−βveαz Z ∞ z dF (y)e−ρ∗y, v > 0, (6.12) and G(−)(v) =λ D Z 0 v dze−βveαz Z z −∞ dF (y)e−ρ∗y, v < 0. (6.13)
Proof. We begin with (6.9). Based on the formula of Φ0
ρ∗(0) in (6.10), (6.9) is simplified to DΦ0ρ∗(z) + ¡ c + ρ∗σ2¢Φρ∗(z) + λU (z) = 0, (6.14) where U = U1+ U2+ U3 and U1(z) = Z 0 −∞ dF (y) Z −y 0 dvΦρ∗(v)e−ρ ∗y + Z 0 −∞ dF (y) Z z 0 dvΦρ∗(v − y)e−ρ ∗y − Z z 0 dv Z 0 −∞ dF (y)Φρ∗(v)e−ρ ∗y = Z 0 −∞ dF (y) Z z−y z dvΦρ∗(v)e−ρ ∗y U2(z) = − Z ∞ 0 dv Z ∞ v dF (y)e−ρ∗ygρ∗(v − y) + Z z 0 dv Z ∞ v dF (y)e−ρ∗ygρ∗(v − y) = − Z ∞ z dv Z ∞ v dF (y)e−ρ∗ygρ∗(v − y), U3(z) = − Z z 0 dv Z ∞ z−v dF (y)Φρ∗(v)e−ρ ∗y .
We multiply both sides of (6.14) by the integration factor eαz, where α is defined in (6.3). Then eαz¡DΦ0ρ∗(z) + ¡ c + ρ∗σ2¢Φρ∗(z) ¢ = −λeαzU (z). (6.15) On the other hand,
D(eαzΦρ∗(z))0 = eαz ¡ DΦ0ρ∗(z) + ¡ c + ρ∗σ2¢Φρ∗(z) ¢ So, by (6.15), D(eαzΦρ∗(z))0 = −λeαzU (z).
Integrating both sides of the last equation from z = 0 to z = x yields
DeαxΦρ∗(x) − Dg(0) = −λ
Z x
0
eαzU (z)dz.
Divide both sides of the equation by Deβx and this gives
Φ(x) = g(0)e−βx− λ D
Z x
0
e−βx+αzU (z)dz.
Note by the definition of U ,
Φ(x) = g(0)e−βx+ 3 X j=1 −λ D Z x 0 e−βx+αzUj(z)dz. (6.16)
To complete the proof, we write the integral terms on the right hand side into the desired forms. Using the results of Gerber and Landry(1998) Eq.(10) and Eq.(12), we have
−λ D Z x 0 e−βx+αzU3(z)dz = Z x 0 Φ(v)G(+)(x − v)dv = Z x 0 Φ(x − v)G(+)(v)dv. (6.17)
In Appendix A, we show that
−λ D Z x 0 e−βx+αzU1(z)dz = limm→∞ µZ 0 −m Φ(x − v)G(−)(v)dv − e−βx Z 0 −m Φ(−v)G(−)(v)dv ¶ , (6.18) −λ D Z x 0 e−βx+αzU2(z)dz = lim n→∞ µZ n x g(x − v)G(+)(v)dv − e−βx Z n 0 g(−v)G(+)(v)dv ¶ . (6.19) Our result follows by (6.17)-(6.19) and (6.16).
¤ Remark. (1) If R(−∞,0)dF (y) = 0, then G(−) vanishes and G(y) = G(+)(y)1
y>0. In
addition, if β > 0, we show in Lemma 6.1 that R0∞G(y)dy < ∞. So, (6.11) is reduced to
Eq(26) in Gerber and Landry(1998). It is interesting to recall that in this case, Gerber and Landry (Theorem 1 and Theorem 2) showed that
G(y)dy =
Z
R+
and
e−βx= Ex
£
e−rτ1[Xτ=0,τ <τl]¤,
where τl is the first record low defined by τl= inf{t ≥ 0; ∆Xt< 0, Xt< x}.
(2) G is independent of g. ¤
Without loss of generality, we assume that g ≥ 0 in the rest of this section. We consider the conditions under which the improper integrals in (6.11) converge. Then (6.11) becomes
Φ(x) = g(0)e−βx+ Z ∞ −∞ Φ(x − v)G(v)dv − e−βx Z ∞ −∞ Φ(−v)G(v)dv. (6.20) In this case, we say that Φ satisfies the Gerber-Landry integral equation. To discuss the convergence of the two improper integrals, we consider separately
Z ∞ 0 g(−v)G(+)(v)dv < ∞ (6.21) and Z 0 −∞ Φ(−v)G(−)(v)dv < ∞. (6.22)
The following lemma gives the criterion that (6.21) holds.
Lemma 6.1 Suppose thatR(0,∞)dF (y) > 0. Then β > 0 if and only ifR0∞h(−v)G(+)(v)dv <
∞ for all nonnegative bounded Borel h on (−∞, 0].
Proof. The proof of necessity follows exactly the same lines as Eq.(13)-(16) in Gerber and Landry(1998). We show sufficiency.
Take h ≡ 1. Then the convergence of the integral R0∞G(+)(y)dy implies that
0 = lim v→∞e −βv Z v 0 dzeαz Z ∞ z dF (y)e−ρ∗y.
If β ≤ 0 andR(0,∞)dF (y) > 0, we get a contradiction. Hence we have β > 0. ¤ By the definition of β, β > 0 if and only if D(ρ∗)2+ cρ∗ > 0. The following example
shows that we need not always have β > 0.
Example 1. Fix σ > 0 and c < 0. Pick δ > 0 such that Dx2 < −cx for all x ∈ (0, δ). As
in Kou and Wang(2003), consider the double exponential distribution
F (dy) = pη(+)e−η(+)y1y>0dy + qη(−)eη
(−)y
1y<0dy, (6.23)
where (p, q) ∈ (0, 1), p + q = 1, and η± > 0. Then ∆ = η(−)> 0 and ψ(∆−) = +∞. If we
take η(−)< δ, then ρ∗∈ (0, η−) ⊂ (0, δ) and β = c
D + ρ∗ < 0 by the choice of δ. ¤
Next, we consider the validity of (6.22) and assume that β > 0. Suppose (6.22) holds. The convergence of the integralR−∞0 Φ(−v)G(−)(v)dv implies that
0 = lim v→∞Φ(v)G (−)(−v) = lim v→∞Φ(v)e βv Z 0 −v dzeαz Z z −∞ dF (y)e−ρ∗y. (6.24)
Since β > 0, α = β + ρ∗ > 0 and 0 <R0 −∞dzeαz Rz −∞dF (y)e−ρ ∗y < ∞. By (6.24), lim v→∞Φ(v)e βv = 0.
Conversely, suppose for some small δ > 0, lim v→∞Φ(v)e (β+δ)v= 0. (6.25) Then Z ∞ 0 Φ(v)G(−)(−v)dv ≤ Z ∞ 0 e−δv µ Φ(v)e(β+δ)v Z 0 −∞ dzeαz Z z −∞ dF (y)e−ρ∗y ¶ dv < ∞.
And (6.22) holds. Thus a natural question arises whether (6.25) will hold. Let bψ(x) be the
Laplace exponent of the dual process bX = −X. That is, bψ(x) = ψ(−x). Suppose that, with
obvious notations, b∆ > 0 and bψ( b∆−) > 0. Then as before, there exists a unique bρ∗∈ (0, b∆)
such that bψ( bρ∗) = r.
Lemma 6.2 Suppose that g ≥ 0. For all x ≥ 0, Φ(x) ≤ kgk∞e−cρ∗x.
Proof. The proof follows almost the same as Lemma 13.2.2 and Theorem 13.2.1 in Rol-ski(2002). As noted before,
n
e−rt−cρ∗Xt
; t ≥ 0 o
is a (Ft, P0)−martingale. Fix x > 0. Define
a probability measure P∗ on W t≥0Ftby P∗(A) = Ex h e−rt−cρ∗(Xt−x)1 A i , A ∈ Ft. Note that ecρ∗Xτ
≤ 0. Since τ is an (Ft)−stopping time, we have
Φ(x) = E∗ h g(Xτ)ecρ ∗Xτ ; τ < ∞ i e−cρ∗x≤ kgk∞e−cρ ∗x .
(See Asmussen(2000), Theorem XIII 3.2.) This completes the proof.
¤ The following theorem generalizes Theorems 3 in Gerber and Landry(1998). See also Tsai and Wilmott(2002).
Theorem 6.1 The expected discounted penalty Φ satisfies the Gerber-Landry integral
equa-tion (6.20) in the following cases: (1) R(0,∞)dF (y) = 1 and β > 0;
(2) R(0,∞)dF (y) < 1 and 0 < β < bρ∗, where bρ∗ is the Lundberg’s constant for the dual process bX.
Proof. Assume R(0,∞)dF (y) = 1 and β > 0. Then G(−) = 0. Since β > 0 and Φ is
bounded, by Lemma 6.1, both R0∞Φ(−v)G(+)(v)dv and R∞
Hence Φ satisfies the equation (6.20). Next we consider the case thatR(0,∞)dF (y) < 1 and
0 < β < bρ∗. Take δ = cρ∗−β
2 . Then δ > 0 and, by Lemma 6.2, we have
lim v→∞Φ(v)e (β+δ)v≤ kgk ∞e(−cρ ∗+β+δ)v → 0, v → ∞.
From this, as noted before, we conclude thatR−∞0 Φ(−v)G(−)(v)dv < ∞. The convergence of
the integralR−∞0 Φ(x−v)G(−)(v)dv can be proved in a similar fashion. Also ifR
(0,∞)dF (y) =
0, then G(+)= 0 and , henceR∞
0 Φ(x − v)G(+)(v)dF (v) =
R∞
0 g(−v)G(+)(v)dF (v) = 0. On
the other hand, ifR(0,∞)dF (y) > 0, then the assumption β > 0 implies, by Lemma 6.2, that
bothR0∞(x−v)G(+)(v)dF (v) andR∞
0 g(−v)G(+)(v)dF (v) converge. The proof is complete.¤
We close this section by giving an example in whichR(0,∞)dF (y) < 1 and 0 < β < bρ∗.
Example 2. Fix σ > 0 and let c = 0. We consider the jump distribution F in (6.23). In addition, we assume that F satisfies the following conditions: (i) η(+) > η(−), (ii) pη(+) >
qη(−), (iii) qη(+) > pη(−). Since c = 0, we have β = ρ∗. We show that ρ∗ < bρ∗. Note
that ρ∗ ∈ (0, ∆) = (0, η(−)) ⊂ (0, η(+)) = (0, e∆) and bψ is strictly convex on (0, e∆). Hence
ρ∗ < bρ∗ if and only if bψ(ρ∗) − r < 0 = ψ(ρ∗) − r(i.e., bψ(ρ∗) < ψ(ρ∗)). Now, we have
1 λ h ψ(ρ∗) − bψ(ρ∗) i = Z e−ρ∗ydF (y) − Z eρ∗ydF (y) = à pη(+) η(+)+ ρ∗ + qη(−) η(−)− ρ∗ ! − à pη(+) η(+)− ρ∗ + qη(−) η(−)+ ρ∗ ! =2ρ ∗©η(+)η(−)£¡qη(+)− pη(−)¢¤+ (ρ∗)2(pη(+)− qη(−))ª £ (η(+))2− (ρ∗)2¤ £(η(−))2− (ρ∗)2¤ ,
Appendix A. Demonstration of (6.18) and (6.19)
A.1. Demonstration of (6.18). We have
−λ D Z x 0 dze−βx+αzU1(z) = −Dλ Z x 0 dze−βx+αz Z 0 −∞ dF (y) Z z−y z dvΦ(v)e−ρ∗(v+y) = − λ D Z x 0 dze−βx+αz µZ x z + Z ∞ x ¶ dvΦ(v)e−ρ∗v Z z−v −∞ dF (y)e−ρ∗y = − λ D Z x 0 dvΦ(v) Z v 0 dze−βx+αz−ρ∗v Z z−v −∞ dF (y)e−ρ∗y − λ D Z ∞ x dvΦ(v) Z x 0 dze−βx+αz−ρ∗v Z z−v −∞ dF (y)e−ρ∗y = − λ D Z x 0 dvΦ(v)e−β(x−v) Z 0 −v dzeαz Z z −∞ dF (y)e−ρ∗y − λ D Z ∞ x dvΦ(v)e−β(x−v) µZ 0 −v − Z 0 x−v ¶ dzeαz Z z −∞ dF (y)e−ρ∗y = lim m→∞ Ã − λ D Z m 0 dvΦ(v)e−β(x−v) Z 0 −v dzeαz Z z −∞ dF (y)e−ρ∗y + λ D Z m x dvΦ(v)e−β(x−v) Z 0 x−v dzeαz Z z −∞ dF (y)e−ρ∗y ! = lim m→∞ µ −e−βx Z 0 −m dvΦ(−v)G(−)(v) + Z 0 −m dvΦ(x − v)G(−)(v) ¶ ,
where the last equation follows from the definition of G(−) in (6.13). So we see that (6.18)
holds.
A.2. Demonstration of (6.19). We have
−λ D Z x 0 dze−βx+αzU2(z) =λ D Z x 0 e−βx+αz Z ∞ z dF (y) Z y z dve−ρ∗vg(v − y) =λ D Z x 0 dze−βx+αz Z ∞ z dF (y) Z y−z 0 dveρ∗v−ρ∗yg(−v) =λ D Z x 0 dze−βx+αz Z ∞ 0 dvg(−v)eρ∗v Z ∞ v+z dF (y)e−ρ∗y =λ D Z ∞ 0 dvg(−v) Z x 0 dze−βx+αz+ρ∗v Z ∞ v+z dF (y)e−ρ∗y =λ D Z ∞ 0 dvg(−v) Z x+v v dze−βx+αz−βv Z ∞ z dF (y)e−ρ∗y =λ D Z ∞ x dvg(x − v) µZ v 0 − Z v−x 0 ¶ dzeαz−βv Z ∞ z dF (y)e−ρ∗y = lim n→∞ µZ n x dvg(x − v)G(+)(v) − e−βx Z n 0 dvg(−v)G(+)(v) ¶ ,
Appendix B. Solutions of Ordinary Differential Equations
We consider the ordinary differential equation(ODE):
Af = 0, on (0, ∞), (B.1)
subject to the boundary conditions
f (0) = K1, (B.2) lim x→∞f (x) = 0, (B.3) σ2 2 f 00(0) + cf0(0) = K 2, (B.4)
for some constants K1, K2 ∈ R. Here, A is given by (5.5).
Lemma B.1 F or r > 0, the characteristic polynomial p(x) of (5.5) has three distinct real zeros ρ2, ρ1, ρ with
ρ2< −η < ρ1 < 0 < ρ. (B.5) Moreover, on (−η, ∞), the zeros of
p1(x) = σ 2 2 x 2+ cx + λ Z ∞ 0
e−xyηe−ηydy − (λ + r)
and p(x) coincide.
Proof. First, note that p(x) = p1(x)(x + η). Hence, on (−η, ∞), p(x) and p1(x) have the
same zero set. Now, we claim that there admits three distinct zeros −ρ2, −ρ1 and ρ of p1(x)
in R such that (B.5) holds. Using differential calculus, one sees that p1 is strictly convex on
(−η, ∞) with p1(0) = −r < 0. Since limx→(−η)+p1(x) = ∞ and limx→∞p1(x) = ∞, there
are two distinct solutions in (−η, 0) and (0, ∞) respectively. Similarly, there is one solution in (−∞, −η). So, the first statement of the lemma follows. This completes the proof. ¤
The general solution for (B.1) is given by
f (x) = Aeρ1x+ Beρ2x+ Ceρx,
for some constants A, B and C. Since limx→∞f (x) = 0 and ρ2< ρ1 < 0 < ρ, we get
C = 0. This gives f (x) = Aeρ1x+ Beρ2x. Clearly, we have ½ f0(x) = ρ 1Aeρ1x+ Bρ2eρ2x, f00(x) = ρ2 1Aeρ1x+ Bρ22eρ2x. and, hence, ½ f0(0) = ρ 1A + Bρ2, f00(0) = ρ2 1A + Bρ22.
By the boundary conditions (B.2) and (B.4) respectively, we get ( A + B = K1 A ³ σ2 2 ρ21+ cρ1 ´ + B ³ σ2 2 ρ22+ cρ2 ´ = K2.
Simple algebra leads to
A = K2− ³ σ2 2 ρ22+ cρ2 ´ K1 σ2 2 (ρ21− ρ22) − c(ρ2− ρ1) and B = K2− ³ σ2 2 ρ21+ cρ1 ´ K1 σ2 2 (ρ22− ρ21) − c(ρ1− ρ2) .
Hence, the solution of (B.1) with the boundary condition (B.2), (B.3) and (B.4) is given by f (x) = K2− ³ σ2 2 ρ22+ cρ2 ´ K1 σ2 2 (ρ21− ρ22) − c(ρ2− ρ1) eρ1x+ K2− ³ σ2 2 ρ21+ cρ1 ´ K1 σ2 2(ρ22− ρ21) − c(ρ1− ρ2) eρ2x.
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