DOI 10.1007/s00780-007-0045-5
An ODE approach for the expected discounted penalty
at ruin in a jump-diffusion model
Yu-Ting Chen· Cheng-Few Lee · Yuan-Chung Sheu
Received: 14 June 2006 / Accepted: 11 April 2007 / Published online: 31 May 2007
© Springer-Verlag 2007
Abstract Under the assumption that the asset value follows a phase-type
jump-diffusion, we show that the expected discounted penalty satisfies an ODE and obtain a general form for the expected discounted penalty. In particular, if only downward jumps are allowed, we get an explicit formula in terms of the penalty function and jump distribution. On the other hand, if the downward jump distribution is a mixture of exponential distributions (and upward jumps are determined by a general Lévy measure), we obtain closed-form solutions for the expected discounted penalty. As an application, we work out an example in Leland’s structural model with jumps. For earlier and related results, see Gerber and Landry [Insur. Math. Econ. 22:263– 276,1998], Hilberink and Rogers [Finance Stoch. 6:237–263,2002], Asmussen et al. [Stoch. Proc. Appl. 109:79–111,2004], and Kyprianou and Surya [Finance Stoch. 11:131–152,2007].
Keywords Jump-diffusion· Expected discounted penalty · Phase-type distribution ·
Optimal capital structure
JEL Classification G12· C60 · C61 · C65
Mathematics Subject Classification (2000) 60J75· 91B28 · 91B30 · 91B70
Y.-T. Chen· C.-F. Lee
Institute of Finance, National Chiao Tung University, Hsinchu, Taiwan C.-F. Lee
Department of Finance, Rutgers University, New Brunswick, NJ, USA Y.-C. Sheu (
)Department of Applied Mathematics, National Chiao Tung University, Hsinchu, Taiwan e-mail: sheu@math.nctu.edu.tw
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 premiums are received, and Z= (Zt; t ≥ 0) is a compound Poisson process which represents the aggregate claims between time 0 and t . (Note that in the insurance context, X has only downward 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 [17] considered the expected discounted penalty
Φ(x)= Ex
e−rτg(Xτ)
. (1.2)
(Here r≥ 0 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). For a general penalty scheme g, (1.2) represents the amount payable at ruin, and it depends on the deficit at ruin. For more results and related problems, see [1].
Gerber [15] extended the classical model (1.1) by adding an independent diffusion. Then the surplus process takes the form
Xt= x + ct + σ Wt− Zt, t≥ 0. (1.3)
Here σ > 0, and W = (Wt; t ≥ 0) is a standard Brownian motion independent of
Z. In this case, ruin may be caused by oscillation (that is, Xτ = 0) or by a claim (that is, Xτ <0). Dufresne and Gerber [13] studied the probability of ruin caused by oscillation and the probability of ruin caused by a claim. Moreover, as in [16,17] con-sidered the expected discounted penalty (1.2). They heuristically derived the integro-differential equation for Φ and showed that Φ satisfies a renewal integral equation. As an application of the renewal equation, they determined the optimal exercise strategy for a perpetual American put option under the assumptions that the log price of the stock is of the form (1.3) and only downward jumps of X are allowed. On the other hand, by a different approach, Mordecki [25] considered optimal exercise strategies and perpetual options for exponential Lévy models. In terms of the supremum and infimum processes of the Lévy process, he derived closed formulas for the optimal exercise strategies and prices of perpetual American call and put options. In partic-ular, if X is the independent difference of a spectrally positive Levy process and a compound Poisson process whose jump distribution is a mixture of exponential dis-tributions, Mordecki [25] gave explicit formulas for optimal exercise strategies and prices of perpetual American put options. (Similar results also hold for call options.) For related results on American option pricing, see also [2,5,6,8,9], and others.
In addition to option pricing theory, the expected discounted penalty also is of im-portance in the structural form modeling of credit risk. The structural form modeling includes, based on (exogenous) default caused by an insufficiency of assets relative to liabilities, the classic Black–Scholes–Merton model of corporate debt pricing and Leland’s structural model, for which (endogenous) default occurs when the issuer’s
assets reach a level so small that the issuer finds it optimal to declare bankruptcy. All the aforementioned models of credit risk have relied on diffusion processes to model the evolution of the assets. However, while the diffusion approach is mathematically tractable and has inputs and parameters of the models observable and estimable, it cannot capture the basic features of credit risk observed empirically.
There are many extensions of the Black–Scholes–Merton model and Leland’s model. One example is [18] in which the model of Leland [22] or Leland and Toft [23] was extended by adding downward jumps in the dynamics of the firm’s assets. If the log value of the assets follows a jump-diffusion as in (1.3) with only downward jumps, Hilberink and Rogers gave explicit formulas for the values of the debt and the value of the firm up to Fourier transforms. Without closed-form solutions for these two values, by imposing smooth-pasting condition, they surprisingly determined the optimal default boundary in terms of the Wiener–Hopf factors of a Lévy process. Also, by numerical inversion for Fourier transforms, they presented some interesting results and discussed their interpretation. For recent works and related results, see [4,11,21].
In this paper we study the function Φ defined in (1.2). Here X is a jump-diffusion of the form (1.3), and the jump distribution for X is a two-sided phase-type distri-bution. The main results of the paper are outlined as follows. We first show that Φ satisfies an integro-differential equation (Theorem2.4) and then derive an ordinary differential equation for Φ (Theorem3.5). Based on the ODE, we show in Proposition 3.6 that the function Φ can be written as a linear combination of known exponential functions. Moreover, if only downward jumps are allowed, we calculate any higher-order (right-hand) derivative of Φ at zero in terms of the penalty function g and jump distribution. As a consequence, we obtain an explicit formula for Φ when there are only downward jumps for X (Theorem3.7). If the downward jumps have a mixed exponential distribution and the upward jumps have a general distribution, we also obtain an explicit formula for Φ in Theorem4.3. As hints for possible finance and insurance applications of our results, we provide some examples. In particular, in the setup of Leland’s model with jumps, we determine the optimal endogenous default and obtain the equity, debt, and firm values in closed-form formulas in Examples3.9
and4.6.
The plan of the rest of the paper is as follows. In Sect.2we introduce the process
X and derive an integro-differential equation for Φ. Section3 recalls the definition of phase-type distributions and presents our ODE approach for Φ. In Sect.4we con-sider a general Lévy process X which is a difference of a spectrally positive Lévy process and a compound Poisson process with only upward jumps. Moreover, if the jump distribution for the compound Poisson process is a mixture of exponential dis-tributions, based on the results in Sect.3, we conjecture the solution form of Φ and by using the Feynman-Kac formula we verify our conjecture. Section 5 concludes the paper. Notation, proofs of lemmas, propositions, and technical results are relegated to appendices.
2 Integro-differential equation
To start with, we specify a Lévy process that we consider in this paper unless otherwise stated. We are given a probability space (Ω,F, P) on which there are
a standard Brownian motion W = (Wt; t ≥ 0) and a compound Poisson process
Z= (Zt=
Nt
n=1Yn; t ≥ 0). Here the Poisson process N = (Nt; t ≥ 0) has parameter
λ >0 and the random variables (Yn; n ∈ N) are independent and identically distrib-uted. We assume further that the distribution F of Y1has a bounded density f that is continuous onR\{0}. In addition, W, N, and(Yn)are assumed to be independent. For every x∈ R, let Pxbe the law of the process
Xt= X0+ ct + σ Wt− Zt, (2.1)
where c∈ R, σ > 0, and X0= x. Write P0forP and Ex[Z] =
Z(ω) dPx(ω)for a random variable Z. For every ζ∈ iR, we have
E0 eζ X1= eψ (ζ ), (2.2) where ψ (ζ )= Dζ2+ cζ + λ e−ζydF (y)− λ (2.3) and D=σ 2 2 .
(ψ is called the characteristic exponent of X.) Moreover, the infinitesimal generator
Lof X has a domain containingC02(R) and, for any h ∈ C02(R),
Lh(x)= Dh(x)+ ch(x)+ λ
h(x− y) dF (y) − λh(x). (2.4) (For details, see [3].) On the other hand, 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}, (2.5)
is an (Ft)-stopping time. Let τ= τ(−∞,0].
From now on, we fix a bounded Borel penalty function g on (−∞, 0] and let the function Φ be given by Φ(x)= Ex e−rτg(Xτ) , x∈ R. (2.6)
Note that Φ(x)= g(x) for x ≤ 0 and, in the insurance literature, Φ is called the
expected discounted penalty for the penalty function g.
In [16] and [30], the following regularities were used implicitly. For a rigorous proof and related works, see [7,10].
Theorem 2.1 The function Φ in (2.6) has the following properties: 1. For all r≥ 0, Φ ∈ Cb1(R+)∩ C2(R++).
Lemma 2.2 The process{e−r(τ∧t)Φ(Xτ∧t); t ≥ 0} is a (Px,Ft∧τ)-martingale and, for all t , Ex e−rτg(Xτ)Fτ∧t = e−r(t∧τ)Φ(Xτ∧t).
Hence, for any (Ft∧τ)-stopping time η,
Ex
e−r(t∧η∧τ)Φ(Xt∧τ∧η)= Φ(x).
Proof Please refer to AppendixA.
We define Lh(x) by expression (2.4) for all functions h onR such that h, h,and the integral in (2.4) exist at x.
Lemma 2.3 The function LΦ is inC(R++).
Proof Please refer to AppendixA.
Theorem 2.4 The function Φ satisfies the integro-differential equation
(L− r)Φ(x) = 0, x > 0. (2.7)
If r > 0, or r= 0 and E[X1] > 0, we have Φ ∈ C0,b2 (R++).
Proof Assume that (L− r)Φ(x) = 0 for all x > 0. Then we have
Φ(x)= −c DΦ (x)− λ D Φ(x− y) dF (y) +λ+ r D Φ(x).
The second statement follows from this and from Theorem2.1.
To establish (2.7), we fix x > 0 and let ∈ (0, x). By Theorem2.1, we have that
Φ∈ Cb1(R+)∩ C2(R++). Since the behavior of Φnear 0+ and +∞ is unclear, we stop the process X at the time ˜τ, where ˜τ is the entry time of (−∞, ] ∪ [x + 1, ∞). Then ˜τ is an (Ft)-stopping time. Moreover, since ˜τ ≤ τ , we see that ˜τ ∧ t is an
(Fτ∧t)-stopping time. By Lemma 2.2, we have Ex[e−r(t∧ ˜τ)Φ(X˜τ∧t)] = Φ(x). In AppendixA, we also prove that
Ex e−r(t∧ ˜τ)Φ(X˜τ∧t)= Ex ˜τ∧t 0 e−ru(L− r)Φ(Xu) du + Φ(x). (2.8)
From these we get
Ex t∧ ˜τ 0 e−ru(L− r)Φ(Xu) du = 0. (2.9)
On the other hand, by Lemma2.3,
Ex
sup u<t∧ ˜τ
Therefore, by (2.9), we obtain (L− r)Φ(x) =1 tEx t∧ ˜τ 0 e−ru(L− r)Φ(Xu) du − (L − r)Φ(x) ≤ Ex sup u<t∧ ˜τ e−ru(L− r)Φ(Xu)− (L − r)Φ(X0) +(L− r)Φ(x)Ex[t − ˜τ ∧ t] t .
Note that ˜τ > 0 Px-a.s. and hence t − ˜τ ∧ t = 0 for all sufficiently small t. Since further 0≤ t− ˜τ∧tt ≤ 2, we have Ex[t− ˜τ∧tt ] → 0 as t → 0+. Together with (2.10) and the last inequality, we establish that
(L− r)Φ(x) ≤lim sup t→0+ 1tEx t∧ ˜τ 0 e−ru(L− r)Φ(Xu) du − (L − r)Φ(x) = 0.
The proof is complete.
3 Explicit formula for Φ: the ODE method
In this section, we consider phase-type jump distributions and the boundary value problem
(L− r)Φ = 0 in R++,
Φ= g onR−, (3.1)
where L is defined by (2.4) and r≥ 0. Our main purpose is to show that if Φ satisfies (3.1), then it satisfies an ODE onR++. From this we obtain a general form of Φ and derive an explicit formula for Φ under some technical conditions. Based on these results, we consider an example in credit risk modeling (Example3.9). To begin with, we recall the definition of phase-type distributions.
Definition 3.1 Assume B is an N × N nonsingular subintensity matrix, that is,
bij ≥ 0 for i = j, bii≤ 0, and b= −Be∈ RN+\{0}. Here, e = [1 1 · · · 1] and
0= [0 0 · · · 0]. Let α be an N-dimensional probability function. The probability
dis-tribution function F with the density function
f (x)=
αexBb, x >0,
0, x≤ 0,
is called a phase-type distribution. We denote this distribution by PH(α, B). We say that the representation (α, B) is minimal if there do not exist N0< N, α of di-mension N0, and a nonsingular subintensity matrix B of dimension N0 such that
We consider the process X defined in (2.1) and assume that its jump distribution
F has the probability density function f given by
f (x)=
pf(+)(x), x >0,
qf(−)(−x), x < 0, (3.2)
where p+ q = 1, p, q ∈ R+,and f(±)are of PH(α±, B±). Here B+and B−are not necessarily of the same dimension. We write I for the identity matrices of the same dimensions as those of B+and B−if there is no confusion.
Remark 3.2
1. A summary of analytic facts of phase-type distributions is given in AppendixB. It is worth noting that the phase-type distributions are dense in the set of all prob-ability distributions onR++. Special cases of phase-type distributions include ex-ponential distributions, Gamma distributions with integer parameter, and mixtures of exponential distributions. For details, see [1] or [26].
2. Asmussen et al. [2] considered a Lévy process X as in (2.1) except that Z =
(Zt; t ≥ 0) is given by Zt= Nt+ n=1 Yn+− Nt− n=1 Yn−, (3.3)
where N±are both Poisson processes, and (Yn+)(resp. (Yn−)) are independently and identically distributed with distribution PH(α+, B+) (resp. PH(α−, B−)). They also assumed that W , N+, N−, (Yn+), and (Yn−)are independent. Their processes have the same characteristic exponent as ours, and therefore the finite-dimensional distributions of their processes coincide with those of ours. Since the finite-dimensional distributions determine the law of the process, our definition of
Xis sufficient.
By assumption (3.2) and TheoremB.2, the characteristic exponent ψ in (2.2) is given by
ψ (ζ )= Dζ2+ cζ + λψ1(ζ )− λ, ζ ∈ iR, (3.4) where ψ1(ζ )=e−ζyf (y) dyis of the form
ψ1(ζ )= pα+(ζ I− B+)−1b++ qα−(−ζ I − B−)−1b−. (3.5) Since the right-hand side of (3.5) is a rational function in ζ onC (see TheoremB.2), the right-hand side of (3.4) actually is a rational function onC with a finite number of poles inC\iR. Accordingly, we consider ψ and ψ1onC as analytic functions except at the poles inC\iR.
LetP0(ζ )be the “minimal” polynomial with leading coefficient 1 such that the zeros ofP0(ζ )coincide with the poles of ψ1(ζ ), counting their multiplicity. Write
P1(ζ )= P0(ζ )
Then the zeros of ψ(ζ )− r coincide with those of the polynomial P1(ζ ), counting their multiplicity.
On the other hand, the infinitesimal generator of the process X takes the form
Lh(x)= Dh(x)+ ch(x)+ λpα+TB+
+h(x)b++ λqα−TB−−h(x)b−− λh(x) (3.7)
for all h∈ C20(R). Here, for a nonsingular subintensity matrix B, the matrix-valued
operators TB±are defined on the set of bounded measurable functions h by
TB±h(x)=
R±
h(x− y)e±Bydy. (3.8)
(When we perform integration and differentiation with respect to a matrix of contin-uous parameter, these operations are meant to be performed termwise.)
The following is a further refinement of Theorem2.4.
Proposition 3.3 Assume that the jump density f is like in (3.2), and E[X1] > 0 if
r= 0. Then Φ is in C0,b∞(R++)and, for k≥ 0, we have the recursive formula
Φ(k+2)(x)= − c DΦ (k+1)(x)+(λ+ r) D Φ (k)(x)− λ DEk(x), (3.9) where Ek(x)= pα+ Bk+TB+ +Φ(x)+ k−1 j=0 Bj+Φ(k−1−j)(x) b+ + qα− (−B−)kTB− −Φ(x)+ k−1 j=0 (−1)j+1Bj−Φ(k−1−j)(x) b−. (3.10) (Here Φ(0)(x)= Φ(x).)
Proof Please refer to AppendixA.
Next we transform the integro-differential equations into ordinary differential equations when the two sides of the jump distribution are both phase-type distrib-utions. Before giving the general result, we treat a simple case by direct calculation.
Example 3.4 We consider the model (2.1) with the jump distribution dF (y)=
ηe−ηy1y>0dyfor some η > 0. Then
ψ (ζ )= Dζ2+ cζ + λ η η+ ζ − λ and Lh(x)= Dh(x)+ ch(x)+ λ ∞ 0 h(x− y)ηe−ηydy− λh(x).
Note that0∞h(x− y)e−ηydy= e−ηx−∞x h(y)eηydyand d dx + η e−ηx x −∞h(y)e ηydy = h(x).
Hence, by Theorem2.4, Φ satisfies the ODE 0= d dx + η (L− r)Φ(x) = DΦ(x)+ (Dη + c)Φ(x)+ (cη − λ − r)Φ(x)− rηΦ(x).
On the other hand, we haveP0(ζ )= η + ζ and
P1(ζ )= Dζ2(ζ+ η) + cζ(ζ + η) + λη − (λ + r)(ζ + η)
= Dζ3+ (Dη + c)ζ2+ (cη − λ − r)ζ − rη. Therefore Φ satisfies an ODE with the characteristic polynomialP1.
Theorem 3.5 Let D1be the differential operator with the characteristic polynomial
P1given by (3.6). Then D1Φ≡ 0 on R++.
Proof Let L2= L2(R) be the space of square-integrable functions defined on R, and setf1, f2 =
f1(x)f2(x) dx. For an operator A on L2, write A∗ for its adjoint (i.e.,Ak, h = k, A∗h for all k, h in L2). By integration by parts, we havek, h =
−k, h whenever k or h has compact support. Write T h(x) =h(x− y)f (y) dy. By a change of variables and Fubini’s theorem, we have
T h, k = k(x) h(x− y)f (y) dy dx = h(z) k(z− y)f (−y) dy dz =h, T∗k,
where T∗k(x)≡k(x− y)f (−y) dy.
Let D0 be the differential operator with the characteristic polynomial P0 and
φ∈ Cc∞(R++)a test function. Recall that, by Theorem2.4, (L− r)Φ ≡ 0 on R++. Therefore we have 0=D0(L− r)Φ, φ =Φ,L∗− rD0∗φ. (3.11) Here L∗is given by L∗h(x)= Dh(x)− ch(x)+ λT∗h(x)− λh(x). Set LD= L − r − λT . Then L∗− rD∗0φ= λT∗D0∗φ+ L∗DD0∗φ= λT∗D∗0φ+ (D0LD)∗φ. (3.12) Write D2as the differential operator corresponding to the polynomialP2(ζ )=
D2∗φare in L2∩ L1and have the same Fourier transforms. (Here L1= L1(R) is the space of integrable functions onR.)
First, we show that T∗D0∗φand D∗2φ are in L1∩ L2. Since φ∈ Cc∞(R++), we obtain D2∗φ∈ L1∩ L2. Also,
T∗
D0∗φ (x)dx≤ D∗0φ (x− y)dx f (−y) dy ≤D∗0φL
1f L1<∞,
and hence T0∗D0∗φ ∈ L1. We show that T∗D0∗φ∈ L2. Since D∗0φ∈ Cc∞(R++)
and T∗h(x)=h(x − y)f (−y) dy, it suffices to show that T k ∈ L2 for any
k∈ C∞c (R++). Note that, by TheoremB.1, f has tails that decay exponentially and, hence,f2dx <∞. Let k ∈ Cc∞(R++)and write H for the compact support of k. By the Cauchy–Schwarz inequality we get
T k(x)2dx= k(x− y)f (y) dy 2 dx= H k(y)f (x− y) dy 2 dx ≤ k(y)2dy H f (x− y)2dy dx ≤ k2 L2f 2 L2 H dx <∞.
Next, we show that the Fourier transforms F(T∗D0∗φ) and F(D2∗φ) coin-cide, where Fh(θ) =e−2πiθxh(x) dx. Recall that ψ1(ζ )=e−ζyf (y) dy and notice that F(D0∗φ)(θ )= P0(−2πiθ)F(φ)(θ). (See [28], Sect. 5.3.) Since
T∗D∗0φ∈ L1∩ L2, we have, for all θ∈ R,
F(T∗D∗0φ)(θ )= e−2πiθx D0∗φ (x− y)f (−y) dy dx = D0∗φ (x− y)e−2πiθ(x−y)dx e−2πiθyf (−y) dy = ψ1(−2πiθ)P0(−2πiθ)F(φ)(θ) = P2(−2πiθ)F(φ)(θ) = FD2∗φ(θ ).
By the Fourier inversion formula, we deduce that T∗D0∗φ= D2∗φalmost everywhere. By (3.12) and the fact that T∗D0∗φ = D2∗φ a.e., (L∗ − r)D0∗φ = λD2∗φ+ (D0LD)∗φ= D1∗φa.e. Hence, by (3.11), we have
0=Φ, (L∗− r)D0∗φ=Φ, D∗1φ= D1Φ, φ.
Since φ ∈ Cc∞(R++) is arbitrary, we have D1Φ = 0 a.s. on (0, ∞). Since
Φ∈ C0,b∞(R++), D1Φ≡ 0 on R++. This completes the proof. Assume from now on that r > 0. We denote byZ(r−)= (ρir)Si=1the collection of zeros of ψ(ζ )− r (counting their multiplicity) with strictly negative real parts. We
say thatZ(r−)is separable if its members are distinct. We writeZ(−)forZ(r−)and ρi for ρir if these cause no confusion.
Proposition 3.6 There exist polynomials Qi(x)such that Φ(x)=
S
i=1Qi(x)eρix
for x≥ 0.
Proof Please refer to AppendixA.
Let U be the column vector with entries Uj= Φ(j−1)(0) for 1≤ j ≤ S and V the
S× S Vandermonde matrix with Vij = ρji−1. Here Φ(0)(0)= Φ(0) and, for k ≥ 1,
Φ(k)(0) is the kth (right-hand) derivative of Φ at 0.
Theorem 3.7 Suppose thatZ(−)is separable. Then we have Φ(x)=
S
i=1Qieρix,
where Q= (Q1, Q2, . . . , QS)T is the unique constant column vector satisfying the
system of linear equations V Q= U. Moreover, if there are no upward jumps for X
(i.e., q= 0), then Φ(k)(0) can be obtained explicitly by the recursive formula
Φ(k+2)(0)=1 D −cΦ(k+1)(0)+ (λ + r)Φ(k)(0)− λEk(0) (3.13) with Φ(0)= g(0), Ek(0)= α+ Bk+ ∞ 0 g(−y)eBydy+ k−1 j=0 Bj+Φ(k−1−j)(0) b+, (3.14) and Φ(0)= − c D + ρ ∗ r g(0)+ λ D ∞ 0 dv ∞ v dF (y) e−ρr∗vg(v− y). (3.15)
(Here ρ∗r is a positive real number satisfying ψ (ρr∗)= r.)
Proof By Proposition3.6, we know that Φ(x)=Si=1Qi(x)eρix, where Q
j(x)are polynomials. SinceZ(−)is separable, standard theory of ordinary differential equa-tions gives that Qj(x)must be a constant for all j . For details, see [29], Lessons 20B and 20D.
Simple calculation shows that the constant vector Q satisfies the system of linear equations V Q= U. Since Z(−)is separable, the ρi are distinct and, hence, the Van-dermonde matrix V is invertible (see, e.g., [14], p. 218). Since V is invertible, Q is the unique solution of the system of linear equations. By letting x→ 0+ in (3.9) and (3.10), we obtain (3.13) and (3.14). Formula (3.15) follows from TheoremC.1.
Remark 3.8 It is interesting to compare our results with those of Asmussen et al. [2]. By martingale stopping and Wiener–Hopf factorization, they obtained similar results as ours.
As an application of Theorem3.7, we consider Hilberink and Rogers’ extension of Leland’s model. We first recall the basic setup of Leland’s model and then obtain
the optimal default boundary in explicit form. Furthermore, we provide a procedure to calculate the values of bond, firm, and equity.
Example 3.9 (Optimal capital structure) We shall assume that, under a risk-neutral
measureQ, the value of the firm’s assets is given by Vt= V ect+σ Wt−Zt. The setting of debt issuance and coupon payment follows the convention in [23]. In the time in-terval (t, t+ dt), the firm issues new debt with face value a dt and maturity profile
k(t )= me−mt. With the exponential maturity profile, the face value of the debt matur-ing in (t, t+ dt) is the same as the face value of the newly issued debt. Thus the face value of all pending debt is equal to the constant A= a0∞e−mtdt= a/m. All
bond-holders will receive coupons at rate b until default. All debt is of equal seniority and, once default occurs, the bondholders get the rest of the value, βVτ, after the bank-ruptcy cost (1− β)Vτ. We assume that there is a corporate tax rate δ, and the coupons paid can be offset against tax. We further assume that the default occurs at time
τ = inf{t ≥ 0; Vt≤ L}, where L will be determined optimally later on. As in [18], we assume that the market has a risk-free rate r > 0, and μ > 0 is the proportional rate at which profit is disbursed to investors. Then the total value at time 0 of all debt is
D(V , L)=bA+ mA m+ r EQ 1− e−(m+r)τ+ βEQVτe−(m+r)τ , (3.16)
and the value of the firm is
v(V , L)= V +bAδ r EQ 1− e−rτ− (1 − β)EQVτe−rτ . (3.17)
The value of equity of the firm is given by
S(V , L)= v(V, L) − D(V, L). (3.18)
(See [18] or Remark4.8below.) By imposing the smooth-pasting condition, we cal-culate the optimal default-triggering level L∗that maximizes the value of equity of the firm.
Set Xt = logVt
L and x = log V
L. Note that τ = inf{t ≥ 0; Vt ≤ L} = inf{t ≥ 0; Xt ≤ 0} and, under the risk-neutral measure Q, X is of the form (2.1). Then we have D(V , L)=A(m+ b) m+ r 1− Gm+r(x) + βLHm+r(x) (3.19) and v(V , L)= V +Aδb r 1− Gr(x) − (1 − β)LHr(x), (3.20) where Gs(x)= Ex e−sτ (3.21) and Hs(x)= Ex e−sτeXτ. (3.22)
Hence, we obtain that ∂S ∂V(V , L)= ∂ ∂V v(V , L)− D(V, L) = 1 −δAb r G r(x) 1 V − (1 − β)LH r(x) 1 V +A(m+ b) m+ r G m+r(x) 1 V − βLH m+r(x) 1 V.
As in [18,23], by imposing the smooth pasting condition (i.e., ∂V∂S(V ,L)|V=L= 0), we get the optimal default-triggering level
L∗= δAb r Gr(0)− A(m+b) m+r Gm+r(0) 1− (1 − β)Hr(0)− βHm+r(0). (3.23) Assume further that X has only downward jumps and σ > 0. (Indeed, in this case, Kyprianou and Surya [21] proved that the optimal default triggering level does satisfy the condition of smooth pasting.) By taking g(y)= ey1y
≤0in (3.15), we get Hs(0)= d dxHs(x) x=0 = −c D− ρ ∗ s + λ D ∞ 0 dv ∞ v dF (y) e−ρs∗vev−y = −c D− ρ ∗ s + λ D ∞ 0 dF (y) y 0 e(1−ρs∗)ve−ydv = −c D− ρ ∗ s + λ D(1− ρs∗) ∞ 0 e−ρs∗y− 1 + 1 − e−ydF (y) = −c D− ρ ∗ s + 1 D(1− ρs∗) −Dρs∗2− cρs∗+ s −ψ (1)− D − c = s− ψ(1) D(1− ρs∗) + 1. (3.24)
(Here in the fourth equation we use the fact that ρs∗is a positive real number satisfying
ψ (ρs∗)− s = 0.) Similarly, by taking g(y) = 1 for all y ≤ 0 in (3.15), we get
Gs(0)= d dxGs(x) x=0 = −c D − ρ ∗ s + λ D ∞ 0 dv ∞ v dF (y)e−ρs∗v = −c D − ρ ∗ s + λ D ∞ 0 dF (y) y 0 e−ρs∗vdv = −c D − ρ ∗ s + λ Dρs∗ ∞ 0 1− e−ρs∗ydF (y)
= −c D − ρ ∗ s + 1 Dρs∗ Dρs∗2+ cρs∗− s = −s Dρs∗. (3.25)
Plugging (3.25) and (3.24) into (3.23) and using the fact that ψ(1)= r − μ (since
V = EQ[e−(r−μ)V1] = V e−(r−μ)+ψ(1)), we obtain the optimal default-triggering level L∗= A(m+b) ρm∗+r − δAb ρr∗ βρm∗+μ m+r−1+ (1 − β) μ ρ∗r−1 . (3.26)
Assume that the jump distribution F is of phase-type distribution PH(α, B) on
(0,∞). By using (3.18–3.20), we compute D(V , L), v(V , L) and S(V , L) in terms of Gs(x)and Hs(x). Note that
G(ks +2)(0)= −c DG (k+1) s (0)+ λ+ s D G (k) s (0)− λ DIk(0) (3.27) and Hs(k+2)(0)= −c DH (k+1) s (0)+ λ+ s D H (k) s (0)− λ DJk(0), (3.28) where Ik(0)= α Bk ∞ 0 eBydy+ k−1 j=0 BjG(ks −1−j)(0) bT and Jk(0)= α Bk ∞ 0 e−yeBydy+ k−1 j=0 BjHs(k−1−j)(0) bT.
Using (3.24), (3.25), and Gs(0)= Hs(0)= 1 in (3.27) and (3.28), we get higher-order derivatives of Gs and Hs at zero. Then, by Theorem3.7, we compute the constant vector Q and, hence, obtain solutions for Gs(x)and Hs(x).
To illustrate our method, we consider the particular case dF (x)= ηe−ηx1x>0dx. Then ψ(ζ ) = Dζ2 + cζ + λη+ζη − λ and, for every s > 0, there exist
ρ1< η < ρ2<0 < ρ∗such that ψ(ρ1)= ψ(ρ2)= ψ(ρ∗)= s (i.e., Z(s−)= {ρ1, ρ2}), and we write ρ∗for ρs∗. Recall that Gs(0)=Dρ−s∗. Hence Gs(x)= Q1eρ1x+ Q2eρ2x, where Q is the unique column vector satisfying the equation
1 1 ρ1ρ2 Q1 Q2 = 1 −s Dρ∗ .
Simple algebra shows that
Q1 Q2 = 1 ρ2− ρ1 ρ2+Dρs∗ −ρ1−Dρs∗ .
Since ρ1ρ2ρ∗=sηD, we get Gs(x)= ρ2+Dρs∗ ρ2− ρ1 eρ1x+−ρ1− s Dρ∗ ρ2− ρ1 eρ2x =ρ2(η+ ρ1) η(ρ2− ρ1) eρ1x+−ρ1(η+ ρ2) η(ρ2− ρ1) eρ1x. (3.29)
Recall that H(0)=D(1s−ψ(1)−ρ∗)+ 1. Hence Hs(x)= P1e ρ1x+ P
2eρ2x, where P is the unique column vector satisfying the equation
1 1 ρ1ρ2 P1 P2 = 1 s−ψ(1) D(1−ρ∗)+ 1 .
Simple algebra shows that
P1 P2 = 1 ρ2− ρ1 ρ2− 1 −D(1s−ψ(1)−ρ∗) 1− ρ1+D(1s−ψ(1)−ρ∗) .
Using the fact that (1+ η)(ψ(1) − s) = D(1 − ρ1)(1− ρ2)(1− ρ∗), we obtain
Hs(x)= ρ2− 1 −D(1s−ψ(1)−ρ∗) ρ2− ρ1 eρ1x+1− ρ1+ s−ψ(1) D(1−ρ∗) ρ2− ρ1 eρ2x =(ρ2− 1)(η + ρ1) (1+ η)(ρ2− ρ1) eρ1x+(1− ρ1)(η+ ρ2) (1+ η)(ρ2− ρ1) eρ1x. (3.30)
Remark 3.10 The optimal level L∗in (3.26) coincides with that in [18]. Our ODE approach also works for the perpetual American put option. Indeed, we determine the optimal stopping times when only downward jumps are allowed. Also we calculate exactly the price of the perpetual American put option.
For general two-sided jump distributions, we get a necessary condition for the co-efficient constant Q. This necessary condition will play an important role in Sect.4.
Proposition 3.11 If (α+, B+)is minimal, Z(−) is separable, and p > 0, then the
constant vector Q for Φ satisfiesSi=1Qi= g(0) and
α+ ∞ 0 g(−y)eB+ydy exB+b+= α+ S i=1 Qi(ρiI− B+)−1 exB+b+, ∀x > 0. (3.31)
4 The constant vector Q: the second method
In this section, we consider a more general process whose upward jumps are de-termined by a general Lévy measure and downward jumps by a compound Poisson process with a mixture of exponential jump distributions. We show that if a vector Q satisfies a system of linear equations, then a conjectured function must be the desired function Φ. To find a solution to the system of equations, instead of directly inverting the system of equations, we exploit the technique in [12] to obtain an explicit solution for Φ. With this explicit formula, we compute especially in Example4.6the optimal default-triggering level. We also provide an example (Example4.7) which shows that we can calculate certain option prices with finite maturity up to a Fourier transform.
Let X= (Xt,{Px}x∈R)be the Lévy processes given by
Xt= X0+ Xt(+)− Zt, t≥ 0. (4.1) Here X(+) = (Xt(+); t ≥ 0) is a Lévy process on R such that it starts at 0 and has a nontrivial diffusion part and no downward jumps, and Z is a compound Poisson process that is independent of X(+) with jump distribution given by
f(+)=PH(α+, B+). As before, underPx, X0= x a.s. Clearly, the process in (4.1) is a generalization of (2.1).
The characteristic exponent ψ of X is given by
ψ (ζ )= Dζ2+ cζ + ∞ 0 eζ z− 1 − ζ z1|z|≤1ν(dz)+ λ ∞ 0 e−ζyf(+)(y) dy− λ,
where ν is an arbitrary Lévy measure on (0,∞) and0∞min(1, z2)ν(dz) <∞.
More-over, the infinitesimal generator of X has a domain containingC02(R) and is given by Lh(x)=Dh(x)+ ch(x)+ ∞ 0 h(x+ z) − h(x) − h(x)z1|z|≤1ν(dz) + λ ∞ 0 h(x− z)f(+)(z) dz− λh(x).
Recall that Φ(x)= Ex[e−rτg(Xτ)], where τ = inf{t ≥ 0; Xt ≤ 0}. The next proposition gives a converse to Theorem2.4.
Proposition 4.1 If φ≡ g on (−∞, 0], φ ∈ C02(R+),and (L− r)φ ≡ 0 on R++, then
φ≡ Φ on R.
Proof Please refer to AppendixA.
We consider below the special case where f(+)(y) is a mixture of exponential distributions. Namely, there exist constants (pj)mj=1 and (ηj)mj=1such that pj >0,
ηj>0, m j=1pj= 1, and f(+)(y)= m j=1 pjηje−ηjy (4.2)
on y > 0 and f(+)(y)= 0 otherwise. Without loss of generality, assume that the ηj are distinct.
It is worth noting that Z(−)= (ρi)Si=1 is separable and S= m + 1 (see [2], Lemma 1(1) or [25], Corollary 2). Now, based on Theorem3.7, we conjecture that
Φ(x)= Ex[e−rτg(Xτ)] =
S
i=1Qieρixfor some Qi. By Proposition3.11, we fur-ther assume that Q satisfies (3.31). More precisely, for f(+)given by (4.2),
m j=1 pjηje−ηjx 0 −∞g(y)e ηjydy= m +1 i=1 Qi m j=1 pj ηje−ηjx ρi+ ηj . (4.3)
(Note that f(+)=PH((p1, . . . , pm),diag(−η1, . . . ,−ηm)).) Recall that
S
i=1Qi =
g(0). Then, by comparing the coefficients of e−ηjx in (4.3), we get the system of
linear equations ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ m +1 i=1 Qi= g(0), m +1 i=1 Qiηj ρi+ ηj = 0 −∞g(y)ηje ηjydy, 1≤ j ≤ m. (4.4)
The following proposition confirms our conjecture.
Proposition 4.2 If Q satisfies (4.4) for some bounded Borel-measurable function g,
then Φ(x)= Ex e−rτg(Xτ) = S i=1 Qieρix, for x≥ 0. (4.5)
Proof Please refer to AppendixA.
In fact, the coefficient matrix of the system of equations (4.4) is invertible. Hence we can write Q in a general matrix form. Instead of the matrix form, we exploit the technique in [12] to get an explicit formula for Q.
Theorem 4.3 Assume that X satisfies (4.1) and f(+)(y) is given by (4.2). Then,
for any bounded Borel function g: R−→ R, we have Φ(x) = Ex[e−rτg(Xτ)] =
S i=1QieρixonR+, where, for 1≤ h ≤ m + 1, Qh= 1 ρh m+1 =1, =h(−ρh+ ρ) m +1 j=1 Rj m+1 k=1 (ηj+ ρk) m +1 i=1,i =j −ρh− ηi ηj− ηi (4.6) and Rj= g(0) − ∞ 0 g(−y)ηje−ηjydy. (4.7) (Here we set ηm+1= 0.)
Proof Clearly, the system (4.4) is equivalent to the system of linear equations m +1 i=1 Qiρi ρi+ ηj = R j, 1≤ j ≤ m + 1, (4.8)
where the Rj are given by (4.7). Since the system (4.4) admits at most one solution, so does (4.8).
Consider the rational function
H (x)= m +1 j=1 Rj m +1 k=1 ηj+ ρk x+ ρk m +1 i=1,i =j x− ηi ηj− ηi . (4.9)
Note that, for each summand in (4.9), the numerator is a polynomial of degree m, and the denominator is a polynomial of degree m+ 1. By the principle of partial fraction decomposition, there exist constants Dh, 1≤ h ≤ m + 1, such that
m +1 j=1 Rj m +1 k=1 ηj+ ρk x+ ρk m +1 i=1,i =j x− ηi ηj− ηi = H (x) = m +1 i=1 Diρi ρi+ x . (4.10)
By multiplying both sides of (4.10) by x+ ρh and then setting x= −ρh, we get that Dh is given by the right-hand side of (4.6). On the other hand, since
m+1 i=1,i =j ηh−ηi ηj−ηi = 0 for all h = j, m +1 i=1 Diρi ρi+ ηh = H (ηh )= Rh m +1 k=1 ηh+ ρk ηh+ ρk m +1 i=1,i =h ηh− ηi ηh− ηi = Rh , 1≤ h ≤ m + 1.
That is, D is a solution to (4.8). Since the solution to (4.8) is unique and (4.4) and (4.8) are equivalent, Q= D is the unique solution of (4.4). By Proposition 4.2, we have Φ≡Si=1QieρixonR
+.
Remark 4.4 In fact, by approximation in (4.6) and (4.7), one sees that the con-clusion of Theorem 4.3 still holds if g is any Borel function on R− such that
0
−∞|g(y)|eηjydy <∞ for all 1 ≤ j ≤ m.
Example 4.5 We consider some special g.
(1) Assume that g≡ 1(−∞,y) for some y≤ 0. Then Rj = −eηjy for 1≤ j ≤ m and Rm+1= 0. Hence, Ex e−rτ1Xτ<y = m+1 h=1 −1 ρh m+1 =1, =h(−ρh+ ρ) × m j=1 eηjy m+1 k=1 (ηj+ ρk) m +1 i=1,i =j −ρh− ηi ηj− ηi eρhx.
(2) Assume that g≡ 1{0}. Then Rj= 1 for all j. Hence, Ex e−rτ1{0}(Xτ) = m +1 h=1 1 ρh m+1 =1, =h(−ρh+ ρ) × m +1 j=1 m+1 k=1 (ηj+ ρk) m +1 i=1,i =j −ρh− ηi ηj− ηi eρhx. (Cf. [20].)
Example 4.6 (Optimal capital structure, continued) We use the notation and setting
of Example 3.9, except that the process X is of the form (4.1) (also assume that
σ >0) and Z has a mixed exponential jump distribution (4.2). For every s > 0, let
ρis,1≤ i ≤ m + 1, be the negative real solutions to ψ(z) − s = 0. As before, we write
ρi for ρis if this causes no confusion. By taking g≡ 1 in (4.7), we get Rj = 0 for 1≤ j ≤ m and Rm+1= 1. By Theorem4.3, we get
Gs(x)= m +1 k=1 ρk m+1 h=1 1 ρh m+1 =1, =h(−ρh+ ρ) m i=1 ρh+ ηi ηi eρhx (4.11) and, hence, Gs(0)= m +1 k=1 ρk m+1 h=1 1 m+1 =1, =h(−ρh+ ρ) m i=1 ρh+ ηi ηi . (4.12)
Similarly, by taking g(y)= ey for y≤ 0 in (4.7), we get Rj =1+η1
j for 1≤ j ≤
m+ 1. By Theorem4.3again, we get
Hs(x)= m +1 h=1 1 ρh m+1 =1, =h(−ρh+ ρ) × m+1 j=1 1 1+ ηj m+1 k=1 (ηj+ ρk) m +1 i=1,i =j −ρh− ηi ηj− ηi eρhx (4.13) and, hence, Hs(0)= m +1 h=1 1 m+1 =1, =h(−ρh+ ρ) × m+1 j=1 1 1+ ηj m+1 k=1 (ηj+ ρk) m +1 i=1,i =j −ρh− ηi ηj− ηi . (4.14)
Plugging formulas (4.12) and (4.14) into (3.23), we get the explicit solution for the optimal default triggering level L∗(under the smooth pasting assumption). Similarly, plugging (4.11) and (4.13) into (3.19), (3.20), and (3.18) respectively, we get closed formulas for the value of debt, the value of the firm, and hence the value of equity.
Example 4.7 (Option pricing) A first-touch digital is a digital contract paying $1
when and if a prespecified event occurs. Consider the first-touch digital, or
touch-and-out option, which pays $1 if the stock price S falls below the level H from
above. If the stock price turns out to stay above the level H up to the expiration time
T >0, the claim expires worthless. It can be interpreted as an American-put-like option with a digital payoff. Since a payoff is the same for all levels of the stock price below the barrier, it is clearly optimal to exercise the option once the level H is crossed. We consider a general framework. Let S be the price process of a security satisfying S= H eX under a risk-neutral probability measure, where X is given by (4.1), and H > 0 is a given constant threshold. Assume the existence of a constant risk-free rate r > 0. Consider a derivative with maturity T on this security whose payoff structure is given by
1. a bounded function k(SτH)at the time τH≤ T , where τH= inf{t; St≤ H },
2. a constant payment A at the time T if S does not cross the boundary H till matu-rity.
Let x= log(S0/H ), g(y)= k(H ey)for y≤ 0, and τ = inf{t; Xt ≤ 0} as usual. Then by the risk-neutral pricing formula the risk-neutral price of this derivative is given by V(S0, T )=Ex e−rτg(Xτ)1[τ≤T ] + Ex e−rTA1[τ>T ] =Ex e−rτg(Xτ)1[τ≤T ] + e−rTA1− P x[τ ≤ T ] . (4.15)
To find the unknown quantitiesEx[e−rτg(Xτ)1[τ≤T ]] and Px[τ ≤ T ], we take their Laplace transforms with respect to time T . By Fubini’s theorem, for every β > 0,
∞ 0 e−βTEx e−rτg(Xτ)1[τ≤T ] dT=1 βEx e−(r+β)τg(Xτ) (4.16) and ∞ 0 e−βTPx[τ ≤ T ] dT = 1 βEx e−βτ. (4.17)
The right-hand sides of both equations can be written explicitly using Theorem4.3. Hence we have an expression for the valueV(S0, T )up to a Fourier transform.
For the first-touch digital, we have k≡ 1 and A = 0. Hence, using (4.16) and (4.11) with s= r + β, we deduce that the Laplace transform of the option price with respect to T is given by ∞ 0 e−βTV(S0, T ) dT = 1 βEx e−(r+β)τ = 1 β m +1 k=1 ρk m+1 h=1 1 ρh m+1 =1, =h(−ρh+ ρ) m i=1 ρh+ ηi ηi S0 H ρh . (4.18)
Remark 4.8 It is worth noting that the time zero value of debt in Example3.9is of the form A0∞me−mTB(V , L, T ) dT, where B(V , L, T ) is the time zero value of a bond issued at zero with face value 1 and maturity T . The price formula (3.16) follows by similar arguments as in Example4.7.
5 Concluding remarks
The expected discounted penalty is a generalized notion of the ruin probability in the insurance literature. It has been widely studied and generalized since Gerber and Shiu [17]. On the other hand, this function has been a major concern in the pricing of perpetual financial securities and the Laplace transforms of securities with finite maturity, and examples include option theory and credit risk modeling.
While empirical studies have been indicating the failure of diffusion models, the seminal paper of Merton [24] has received more attention in recent years. However, unlike the diffusion case in which many functionals are available, this is no longer the case for jump-diffusion processes, especially in the first-passage models. This is where the difficulty in the pricing of securities under a jump-diffusion process arises. In this paper, we consider the jump-diffusion process as in [2]. In the case where the jump distribution is a two-sided phase-type one, by a Fourier transform argument, we transform the integro-differential equation of the expected discounted penalty into a homogeneous ordinary differential equation. The present method could possibly be extended to transform more integro-differential equations into ordinary differential equations and hence give an alternative approach to compute prices in jump-diffusion models.
Next, by ODE theory, we know that the solution for the expected discounted penalty is a linear combination of some known exponential functions. Moreover, by using the limit behavior of the expected discounted penalty and the integro-differential equation, we can determine these coefficients (under some conditions). All these distinguish not only our approach from that of Asmussen et al. [2] but also itself from the classical method to solve differential equations in which the knowledge of boundary values is required. This could be applied to solve other functionals once we have transformed their integro-differential into an ordinary differential equation.
Our results in fact provide explicit solutions to a large amount of existing pric-ing problems in finance. Uspric-ing our closed-form solutions, especially Theorem4.3, the Laplace transforms of securities with finite maturity have explicit solutions as mentioned in Example4.7, and we show an example by considering a touch-and-out option. In addition, the pricing problems of perpetual securities in jump-diffusion models have improved answers. For example, both the value of shareholders and the optimal bankruptcy level (by using the smooth-pasting condition) in the optimal capital structure problem have exact solutions. We worked out these in Example3.9
under the setup of Hilberink and Rogers [18], which is an extension of Leland and Toft [23]. Moreover, in Example4.6, we obtained a closed-form solution even in a two-sided jump case. By using our closed-form solution as criterion, many structural form models in credit risk can be reconsidered and modified to see whether more phenomena in empirical studies can be captured by a jump-diffusion model.
Acknowledgements We thank two anonymous referees for their comments on an early draft of this paper. We are also grateful to the editor for his useful comments.
Appendix A Notation and proofs
Notation We writeR+= {x ∈ R; x ≥ 0}, R++= {x ∈ R; x > 0}, and analogously R−andR−−.
Let I be an interval inR and n ∈ N. We introduce the following function spaces: 1. C(I) is the space of real-valued continuous functions f on I .
2. Cb(I )is the space of bounded continuous functions f on I .
3. C0(I ) is the space of continuous functions f on I with limx→∞f (x)= 0 and limx→−∞f (x)= 0, provided that I is not bounded above (resp. below). Also, C0,b(I )= C0(I )∩ Cb(I ).
4. Cc(I )is the space of continuous functions f on I with compact supports. 5. Cn(I )is the space consisting of f ∈ C(I) with f(n)∈ C(I). Here, on I◦, f(n)is
the usual nth derivative. If x is the left- (resp. right-)hand boundary point of I and is in I , f(n)(x)is the nth right- (resp. left-)hand derivative at x. C0n(I ), Cbn(I ),
Cn
c(I ),andC0,bn (I )are defined similarly.
6. Cc∞(I )=!kCkc(I ),C0,b∞(I )=!kC0,bk (I ), andC∞0 (I )=!kC0k(I ).
Proof of Lemma2.2 Fix t≥ 0. On {t ≥ τ}, using the local property of conditional expectation (see [19], Lemma 6.2) and the fact that Φ= g on R−, we have
Ex e−rτg(Xτ)Fτ∧t = Ex e−rτg(Xτ)Fτ = e−rτΦ(X τ). Similarly, on{t < τ}, Ex e−rτg(Xτ)Fτ∧t = Ex e−rτg(Xτ)Ft .
By the strong Markov property of X, on{t < τ},
Ex e−rτg(Xτ)Fτ∧t = e−rtEXt e−rτg(Xτ) = e−rtΦ(Xt).
The proof is complete.
Proof of Lemma2.3 Since Φ∈ C2(R
++), it suffices to show that, as y→ x,
Φ(y− z) dF (z) →
Φ(x− z) dF (z).
For any y > 0, write
Φ(y− z) dF (z) = ∞ y g(y− z) dF (z) + y −∞Φ(y− z) dF (z).
Let > 0 and find M > x such that(−∞,x−M+1]∪[x+M−1,∞)dF (z) < . Then, for all y > 0 such that|x − y| ≤ 1/2,
Φ(y− z) dF (z) − Φ(x− z) dF (z) ≤ 2g∞ + y+M y g(y− z) dF (z) − x+M x g(x− z) dF (z) + y y−M Φ(y− z) dF (z) − x x−M Φ(x− z) dF (z) ≤ 2g∞ + 0 −Mg(z) f (y− z) − f (x − z)dz + M 0 Φ(z)f (y− z) − f (x − z)dz,
where f is the density function for F . Since g and Φ are bounded and f is continuous onR\{0}, by dominated convergence, we have
lim sup y→x Φ(y− z) dF (z) − Φ(x− z) dF (z) ≤ 2g∞ .
Since > 0 is arbitrary, the proof is completed. To prove (2.8), we need the following lemmas and Dynkin’s formula.
Lemma A.1 For every bounded Borel-measurable function g onR−, there exists a
sequence of uniformly bounded functions (gn)inC02(R−)such that gn→ g
Lebesgue-a.e. onR−and gn(0)= g(0) for all n.
Proof Write dm for Lebesgue measure on R. Let h be a normal density, and set
dG(y)= h(y)dm. Then, by the strict positivity of h, dG and dm are equivalent.
Fix > 0. By dominated convergence, we can find N large enough such that we have−∞0 |g1[−N,0]− g| dG < . Also, since g1[−N,0]is bounded and in L1(R, dm),
a modification of the proof of Theorem2.4in [28], Chap. 2, shows that we can find a sequence of uniformly bounded C20(R−)-functions (gn)such that gn → g1[−N,0]
dm-a.e. Since dG and dm are equivalent, gn→ g1[0,N] dG-a.e. Hence, by domi-nated convergence, we can pick n large such that|gn− g1[−N,0]| dG < . We have obtained that−∞0 |gn− g| dG < 2 . And this implies that there exists a sequence of C02(R−)-functions (gn)such that gn→ g dG-a.e. on R−. Since dG and dm are equivalent, we see that gn→ g dm-a.e. on R−. In addition, it is clear that we can pick (gn)to be uniformly bounded.
Now, for each n, define aC02(R−)-function hn such that hn= gnon (−∞, 1/n),
hn(0)= g(0) and (hn) is uniformly bounded. Then it is obvious that (hn)is the
desired sequence. This completes the proof.
Lemma A.2 The following distributions are absolutely continuous with respect to Lebesgue measure:
(1) The distributionPx[X˜τ<0, X˜τ∈ ·], where X is defined by (2.1), and ˜τ is defined
as in the proof of Theorem2.4.
(2) The distributionPx[Xτ<0, Xτ∈ ·] for x > 0, where X is defined by (4.1).
Proof We only show (2), and the proof of (1) follows similarly. LetN be a Borel
set in (−∞, 0) with Lebesgue measure zero. Write Jnfor the nth jump time of the compound Poisson process Z. Observe that, for x > 0,
Px[Xτ <0, Xτ∈ N ] = ∞
n=1
Px[Xτ<0, Xτ∈ N , τ = Jn].
We first show thatPx[Xτ∈ N , Xτ<0, τ= J1] = 0. Note that since X(+), J1and Y1 are independent, so are XJ(+)
1 and Y1. Hence, we have
Px[Xτ∈ N , Xτ<0, τ= J1] = Px XJ(+) 1 − Y1∈ N , X (+) J1 − Y1<0, τ= J1 ≤ Px XJ(+) 1 − Y1∈ N = Px[z − Y1∈ N ] dG(z),
where G is the distribution of X(J+)
1 . Since Y1has an absolutely continuous
distribu-tion, from the last inequality we deduce thatPx[Xτ∈ N , Xτ<0, τ= J1] = 0. In general, for each n≥ 2, we have by the strong Markov property that
Px[Xτ<0, Xτ∈ N , τ = Jn] = Ex
1τ >Jn−1PXJn−1[Xτ <0, Xτ∈ N , τ = J1]
= 0,
since Jn−1is an (Ft)-stopping time and XJn−1>0 on{τ > Jn−1}. This implies that
Px[Xτ <0, Xτ ∈ ·] is absolutely continuous with respect to Lebesgue measure for
every x > 0.
Theorem A.3 (Dynkin’s formula) Let X= (X(t)) be an Rn-valued jump-diffusion, and let k∈ C20(Rn). If η is a stopping time such thatEx(η) <∞, then
Ex k(Xη) = k(x) + Ex η 0 AkX(s)ds
where A is the generator of X defined by
Ak(x)= lim t→0+ 1 t " Ex
kX(t )− k(x)#(if the limit exists). (See, for example, [27], Exercise 44.22.)
Proof of (2.8) As in the proof of Theorem2.4, let > 0 and ˜τ the first exit time for X from ( , x+1). Pick a sequence of uniformly bounded functions (gn)⊂ C02((−∞, 0]) that converges to g, except on a setN of Lebesgue measure 0 in R−−,and gn(0)=
g(0) for all n. (See LemmaA.1.) Therefore, we may consider a sequence of uniformly bounded functions (Φn; n ≥ 1) that satisfies the following conditions:
C1 Φn= gnon (−∞, 0];
C2 Φn= Φ on (1/n, n);
C3 Φn∈ C02(R).
By (C1), (C2), and the fact that gn→ g except on N , we have that Φn→ Φ pointwise onR except on N .
Now, pick n large such that ( , x+ ) ⊂ (1/n, n). Then Φn(x)= Φ(x) by (C2). By Dynkin’s formula, for all t > 0,
Ex e−r( ˜τ∧t)Φn(X˜τ∧t) = Ex ˜τ∧t 0 e−ru(L− r)Φn(Xu) du + Φ(x). (A.1) Since (Φn)is uniformly bounded, Φn→ Φ pointwise on R+,and Φn→ Φ a.e. on
(−∞, 0), by LemmaA.2(1) and dominated convergence, lim n→∞Ex e−r(t∧ ˜τ)Φn(Xt∧ ˜τ) = Ex e−r(t∧ ˜τ)Φ(X˜τ∧t) . (A.2)
On the other hand, for every u <˜τ ∧ t, Xu∈ ( , x + ), and hence Φn(Xu)= Φ(Xu) for all large n. These give
(L− r)Φ(Xu)− (L − r)Φn(Xu)= Φ(Xu− y) − Φn(Xu− y) dF (y) (A.3) and hence (L− r)Φ(Xu)− Φn(Xu) ≤sup n Φn∞+ g∞ . (A.4)
By dominated convergence, (A.3) and the absolute continuity of F imply, for all
u < t∧ ˜τ,
(L− r)Φn(Xu)→ (L − r)Φ(Xu), n→ ∞. By (A.4) and dominated convergence, we have
lim n→∞Ex ˜τ∧t 0 e−ru(L− r)Φn(Xu) du = Ex t∧ ˜τ 0 e−ru(L− r)Φ(Xu)du . (A.5) By (A.2) and (A.5), letting n→ ∞ for both sides of (A.1), we get (2.8).
Proof of Proposition3.3 Fix x > 0. Since (L− r)Φ(x) = 0, by (3.7), we get that (3.9) holds for k= 0 and, hence, Φ ∈ C0,b2 (R++). Since Φ is continuous at x (Theo-rem2.1), we obtain, by TheoremB.3, that TB++Φ is differentiable at x and
d dxT + B+Φ(x)= B+TB++Φ(x)+ Φ(x)I. Similarly, we have dxdTB− −Φ(x)= −B−T −
B−Φ(x)− Φ(x)I. So, by the
k= 1. Since Φ ∈ C0,b2 (R++)and TB±±Φ∈ C0,b(R++), Φ∈ C0,b(R++)and hence
Φ∈ C0,b3 (R++). A similar argument holds for general k.
Proof of Proposition 3.6 Denote by Z(+) the collection of all solutions to
ψ (ζ )− r = 0 with nonnegative real parts, counting their multiplicity. Since D1Φ≡ 0 onR++and ψ(ζ )− r and P1(ζ )have the same zero set, Φ is of the form
Φ(x)= ρi∈Z(−) Qi(x)eρix+ ρ∈Z(+) Rρ(x)eρ x
for some polynomials Qiand Rρ. (See [29], Theorem 19.3 and Lesson 20.) We show that Rρ= 0 for all ρ ∈ Z(+). Assume that Rρ(x) = 0 for some ρ ∈ Z(+). Let a≥ 0 be the maximum of all real parts of members inZ(+),and let{ρ1, . . . , ρk} be the set of all elements inZ(+)such that(ρj)= a. Now, let m ≥ 0 be the smallest integer such that the order of Rρj is≤ m for all j, and select the ρj such that the order of
Rρ
j is equal to m. Still call the selected members{ρ
1, ρ2, . . . , ρk}. Let aj = 0 be the coefficient of xmin Rρj for all j . Then
Φ(x)x−me−ax= j ajei(ρ j)x+ h(x), where h(x) → 0 as x → ∞. However,jajei(ρ
j)x is the discrete Fourier transform of a nonzero function and
hence is not identically zero, and it has period 2π . So, Φ(x)x−me−ax 0 as x → ∞.
Since Φ∈ C0(R+), this is impossible. Hence, Rρ≡ 0 for all ρ∈ Z(+).
Proof of Proposition3.11 The first statement follows from the fact thatSi=1Qi =
Φ(0)= g(0).
Consider the second statement. First, note that the minimality of the representa-tion (α+, B+)guarantees that the collection of all zeros η∈ C of the rational func-tion r−ψ(η)1 is equal to the collection of all eigenvalues of B+. (For details, see [2], Lemma 1 and the paragraph above it.) Hence, ρi is not an eigenvalue of B+for each
i, and ρiI− B+is invertible for all i. By TheoremB.1(4) andB.2(2), we have
x −∞Φ(x− y)f (y) dy = x 0 Qeρ(x− y)f (y) dy + 0 −∞Q eρ(x− y)f (y) dy = S i=1 Qieρixpα +(ρiI− B+)−1 I− ex(B+−ρiI )b + + S i=1 Qieρixqα −(−ρiI− B−)−1b−. Therefore, by (3.5),
