Generalized refracted Lévy process and its application to exit problem

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Model 1

pp. 1–29 (col. fig: NIL)

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Generalized refracted Lévy process and its application to exit problem Kei Noba ∗, Kouji Yano Department of Mathematics, Graduate School of Science, Kyoto University Sakyo-ku, Kyoto 606-8502, Japan Received 30 September 2016; received in revised form 13 March 2018; accepted 18 June 2018 Available online xxxx

Abstract Generalizing Kyprianou–Loeffen’s refracted Lévy processes, we define a new refracted Lévy process which is a Markov process whose positive and negative motions are Lévy processes different from each other. To construct it we utilize the excursion theory. We study its exit problem and the potential measures of the killed processes. We also discuss approximation problem. c 2018 Elsevier B.V. All rights reserved. ⃝

1. Introduction

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Exit problem of a real-valued stochastic process Z = {Z t : t ≥ 0} is the problem to characterize the law of the first time of exiting an interval [b, a] for b < a. In this paper, we are interested in the Laplace transform ) ( + ExZ e−qτa ; τa+ < τb− (1.1) for q ≥ 0 and a starting point x ∈ [b, a], where τa+ = inf{t > 0 : Z t > a} and τb− = inf{t > 0 : Z t < b}.

2 3 4

5

6

(1.2)

∗ Corresponding author.

E-mail addresses: [email protected] (K. Noba), [email protected] (K. Yano). https://doi.org/10.1016/j.spa.2018.06.004 c 2018 Elsevier B.V. All rights reserved. 0304-4149/⃝

Please cite this article in press as: K. Noba, K. Yano, Generalized refracted L´evy process and its application to exit problem, Stochastic Processes and their Applications (2018), https://doi.org/10.1016/j.spa.2018.06.004.

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When Z is a spectrally negative L´evy process Z , it is well known that ( ) W (q) (x − b) + , ExZ e−qτa ; τa+ < τb− = Z(q) W Z (a − b) (q)

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where W Z is the q-scale function of Z . Kyprianou and Loeffen [10] have studied the exit problem when Z was a refracted L´evy process U , which was defined as the strong solution of the stochastic differential equation ∫ t Ut − U0 = X t − X 0 + α 1{Us <0} ds t ≥ 0, (1.3) 0

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where the driving noise X is a spectrally negative L´evy process and α is a positive constant. Define Yt = X t + αt. Then the positive and negative motions of U is given as { X t − X s whenever Ur ≥ 0 for any r ∈ [s, t) Ut − Us = Yt − Ys whenever Ur < 0 for any r ∈ [s, t). They proved that the Laplace transform (1.1) for Z = U takes the form ( ) W (q) (x, b) + EUx e−qτa ; τa+ < τb− = U(q) , WU (a, b)

(1.4)

(q)

12

where the function WU is defined by (q)

13

(q)

WU (x, y) = WY (x − y) + α1(x≥0)

∫

x

(q)

(q)′

W X (x − z)WY (z − y)dz

(1.5)

0 (q)

14 15 16 17 18 19 20 21 22 23

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with WY being the q-scale function of Y . They obtained, in addition, a representation of the potential measures of U using scale functions of X and Y . A spectrally negative L´evy process can be regarded as the capital of an insurance company and applied to evaluate the risk of ruin. Hence it is sometimes called a L´evy insurance risk process. The Kyprianou–Loeffen’s refracted L´evy process U can be regarded as a modified insurance risk process when dividends are being paid out at a rate α during the period it exceeds 0. In this paper, we generalize Kyprianou–Loeffen’s refracted L´evy processes. For two L´evy processes X and Y which may have different L´evy exponents, we construct a new refracted process whose positive and negative motions have the same law as X and Y , respectively. More precisely, { If x > 0,(Ut )t≤τ − under PUx is equal in law to(X t )t≤τ − under PxX 0 0 If x < 0,(Ut )t≤τ + under PUx is equal in law to(Yt )t≤τ + under PYx . 0

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0

One may expect that we can characterize the desired process as a solution to the following stochastic differential equation ∫ ∫ Ut − U0 = 1{Us− ≥0} d X s + 1{Us− <0} dYs , (1.6) (0,t]

(0,t] d

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where the driving noises X and Y are supposed to be independent. Although (1.6) for Yt = X t + αt is apparently different from (1.3) because of independence, their solutions are actually equivalent in law. When X has bounded variation paths, we can construct a solution of (1.6) by a simple method of piecing excursions (see [10]); otherwise we do not know existence of a solution of (1.6). When X and Y are compound Poisson processes with positive drifts, uniqueness of the Please cite this article in press as: K. Noba, K. Yano, Generalized refracted L´evy process and its application to exit problem, Stochastic Processes and their Applications (2018), https://doi.org/10.1016/j.spa.2018.06.004.

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solution is easily proved because of the fact that the point 0 is irregular for itself for any solution U ; otherwise we do not know uniqueness of a solution of (1.6). In this paper we utilize the excursion theory instead of a stochastic differential equation. Let X and Y be two spectrally negative L´evy processes. Suppose X has unbounded variation paths and has no Gaussian component. We then define the excursion measure n U by ⎞ ⎛ ⏐ ( ( )) )) ⏐ ⎟ ⎜ 0( ( (1.7) n U F (Ut )t<τ − , (Ut+τ − )t≥0 = n X ⎝EYy F w, (Yt0 )t≥0 ⏐⏐ ⎠, y=X (τ0− ) 0 0

1 2

w=(X (t)) t<τ0−

where n X stands for an excursion measure of X and Yt0 = Yt∧T0 for the stopped process of Y 0 upon hitting zero. We define the stopped process PUx by (1.7) with n X being replaced by PxX . We U can construct a Feller process from n together with the family of stopped processes { therefore } U0 Px . As one of our main theorems, we show the Laplace transform (1.1) for the process x̸=0

(q) WU

Z = U , our new refracted L´evy process, takes the same form as (1.4) where will be defined (q) in Theorem 6.2 in a more complicated form than (1.5). Note that WU ’s will be represented using only Laplace exponents and scale functions of X and Y . Furthermore, we will study the potential measures of U with and without absorbing barriers. We finally discuss approximation problem. Let X and Y be as in the previous paragraph. Let X (n) and Y (n) be the compound Poisson processes with positive drifts obtained from X and Y , respectively, by removing small jumps of magnitude less than n1 . Assuming that X (n) and Y (n) are independent, we construct U (n) as the unique solution of (1.6). We thus show that U (n) converges to our refracted process U in law on the space of c`adl`ag paths equipped with the Skorokhod topology. The organization of the present paper is as follows. In Section 2 we propose some notation and recall preliminary facts about spectrally negative L´evy processes. In Section 3 we calculate several quantities related to excursion measures and scale functions. In Section 4 we recall Kyprianou–Loeffen’s refracted L´evy processes. In Section 5 we define our new refracted L´evy processes. In Section 6 we study the exit problem of our refracted L´evy processes. In Section 7 we calculate the potential measures of our refracted L´evy processes killed upon exiting [b, a]. In Section 8 we study the approximation problem. In the Appendix we make a careful treatment of Markov property of our new process. 2. Notation and preliminaries Let D denote the set of functions ω : [0, ∞) → R which are c`adl`ag. We equip D with the Skorokhod topology. Let B(D) denote the class of Borel sets of D. When we consider a process Z = {Z t : t ≥ 0} = {Z (t) : t ≥ 0}, we always write PxZ for the underlying probability measure for Z starting from x. In addition to the passage times τa+ and τb− defined in (1.2), we sometimes need the hitting time of a point x ∈ R denoted by Tx = inf{t > 0 : Z t = x}. For q > 0, x ∈ R and a non-negative or bounded measurable function f , we write (∫ ∞ ) (q) Z −qt R Z f (x) := Ex e f (Z t )dt . 0 Please cite this article in press as: K. Noba, K. Yano, Generalized refracted L´evy process and its application to exit problem, Stochastic Processes and their Applications (2018), https://doi.org/10.1016/j.spa.2018.06.004.

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2

We write r Z (x, y) for the resolvent density, if it exists, i.e., ∫ (q) (q) r Z (x, y) f (y)dy. R Z f (x) = R

We sometimes settle a lower barrier b < 0 and an upper barrier a > 0. For q > 0, x ∈ R and a non-negative or bounded measurable function f , we write (∫ + − ) τa ∧τb (q;b,a) (q) −qt Z RZ e f (Z t )dt , f (x) := R Z f (x) := Ex 0 (q;b) RZ

f (x) :=

(q) RZ

f (x) :=

ExZ

τb−

(∫

) e

−qt

f (Z t )dt ,

0 (q;a) RZ

f (x) :=

(q) RZ

f (x) := ExZ

(∫

τa+

) e−qt f (Z t )dt .

0 (q;b,a)

3 4 5 6 7 8 9 10 11

12

(q)

(q;b,a)

(q)

(q;b,a)

(q)

We write r Z (x, y) = r Z (x, y) for (x, y) = r Z (x, y), r Z (x, y) = r Z (x, y) and r Z the corresponding densities, if they exist. Let Z be a spectrally negative L´evy process, which is always assumed not to be monotone. Then it is well known that the Laplace exponent ( ) Ψ Z (q) := log E0Z eq Z 1 is finite for all q ≥ 0. We denote its right inverse by Φ Z (θ ) = inf{q ≥ 0 : Ψ Z (q) = θ }, which is finite for all θ ≥ 0. If Z has bounded variation paths, the Laplace exponent is known to necessarily take the form ∫ ( ) Ψ Z (q) = δ Z q − 1 − eqy Π Z (dy) (−∞,0)

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for ∫ some constant δ Z > 0 and some L´evy measure Π Z satisfying Π Z [0, ∞) = 0 and (−∞,0) (1 ∧ |y|) Π Z (dy) < ∞. If Z has unbounded variation paths, the Laplace exponent is known to necessarily take the form ∫ ( ) σ Z2 2 1 − eqy + qy1(−1,0) (y) Π Z (dy) (2.1) q − Ψ Z (q) = γ Z q + 2 (−∞,0) for some constants ∫ ( )γ Z ∈ R and σ Z ≥ 0 and some L´evy measure Π Z satisfying Π Z [0, ∞) = 0 and (−∞,0) 1 ∧ y 2 Π Z (dy) < ∞. (q)

19 20

21

(q)

Definition 2.1. For each q ≥ 0, we define W Z : R → [0, ∞) such that W Z = 0 on (−∞, 0) (q) and W Z on [0, ∞) is continuous satisfying ∫ ∞ 1 (q) e−βx W Z (x)d x = Ψ (β) −q Z 0 (q)

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for all β > Φ Z (q). This function W Z is called the q-scale function of Z . For the proof of unique existence and its basic facts listed below, see, e.g., [9]. For all b < x < a and q ≥ 0, we have ( ) W (q) (x − b) + ExZ e−qτa ; τa+ < τb− = Z(q) (2.2) W Z (a − b) Please cite this article in press as: K. Noba, K. Yano, Generalized refracted L´evy process and its application to exit problem, Stochastic Processes and their Applications (2018), https://doi.org/10.1016/j.spa.2018.06.004.

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and

1

ExZ

(

e

−qτa+

; τa+

)

<∞ =e

−Φ Z (q)(a−x)

.

(2.3)

It is known that, when Z has bounded variation paths, we have (q)

W Z (0) =

3

1 δZ

(2.4)

for all q ≥ 0. For all y ∈ R and q ≥ 0, we have (q)

(q;b)

(q)

(q)

4

5

r Z (x, y) = Φ Z′ (q)e−Φ Z (q)(y−x) − W Z (x − y), rZ

2

x ∈ R,

(q)

(q)

(q) W Z (a

(q) W Z (x

(x, y) = r Z (x, y) = e−Φ Z (q)(y−b) W Z (x − b) − W Z (x − y), x ∈ [b, ∞),

(2.5)

6

(2.6)

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(q;a) r Z (x,

y) =

(q) r Z (x,

y) = e

−Φ Z (q)(a−x)

− y) −

− y), x ∈ (−∞, a]

10

and

11

(q;b,a)

rZ

(q)

(x, y) = r Z (x, y) =

(q) W Z (x

(q) − b)W Z (a (q) W Z (a − b)

− y)

(q)

− W Z (x − y), x ∈ [b, a].

(2.7)

˜Z for the measure carried on (−∞, 0) × (0, ∞) defined by We write Π ˜Z (du dv) := 1{u<0, v>0} Π Z (du − v)dv. Π

0

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Theorem 2.2. (i) For all 0 < x < ∞, q ≥ 0, and non-negative measurable function f : R2 → [0, ∞), we have ( ) ∫ (q) − Z −qτ0− − − − Ex e f (Z τ , Z τ − ); τ0 < ∞, Z τ < 0 = f (u, v)G Z (x, du dv), 0

12

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0

(q)

where G Z (x, ·) is the measure carried on (−∞, 0) × (0, ∞) defined by (q) G Z (x, du

dv) :=

(q;0) ˜Z (du r Z (x, v)Π

dv).

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(ii) For all 0 < x < a, q ≥ 0, and non-negative measurable function f , we have ) ∫ ( − (q,a) f (u, v)G Z (x, du dv), ExZ e−qτ0 f (Z τ − , Z τ − − ); τ0− < τa+ , Z τ − < 0 = 0

(q,a)

where G Z

(q,a)

GZ

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20

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0

0

(x, ·) is the measure carried on (−∞, 0) × (0, ∞) defined by (q)

(q;0,a)

(x, du dv) = G Z (x, du dv) := r Z

˜Z (du dv). (x, v)Π

22

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We omit the proof of Theorem 2.2 because it can be found in [9, Theorem 10.1 and Exercise (q) (q) 10.6] and also in [10, Theorem 5.5]. These kernels G Z and G Z are called the Gerber–Shiu measures.

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3. Some calculations related to excursion measures and scale functions

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In this section, we make some calculations related to excursion measures and scale functions for a spectrally negative L´evy process X . See [12,1] and [11] for recent studies on a close relation between n X , i.e., the excursion measure of X itself, and the excursion measure of the reflected X . Please cite this article in press as: K. Noba, K. Yano, Generalized refracted L´evy process and its application to exit problem, Stochastic Processes and their Applications (2018), https://doi.org/10.1016/j.spa.2018.06.004.

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We divide the discussion into the two cases of unbounded and of bounded variations.

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(I) We assume that

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X has unbounded variation paths and have no Gaussian component. Since 0 is regular for itself, X has an excursion measure n X away from zero. We impose on n X the following normalization: ( ) 1 1 = Ψ X′ (Φ X (q)) . (3.1) n X 1 − e−qT0 = (q) = ′ Φ X (q) r X (0, 0) Note that n X is carried on the set of c`adl`ag paths stopped upon hitting 0. Note also that n X possesses the Markov property; for example, ( )) 0 ( 0 n X (X s ∈ B1 , X t ∈ B2 ) = n X 1{X s ∈B1 } P XX (s) X t−s ∈ B2 , for all 0 < s < t and B1 , B2 ∈ B(R), where X t0 = X t∧T0 denotes the stopped process of X upon hitting zero. Since X has no Gaussian component, we can see 0 < τ0− < T0 ≤ ∞ or τ0− = T0 = ∞

n X -a.e.

by [12, Theorem 3]. Theorem 3.1. For all a > 0 and q ≥ 0, we have ( ) + 1 n X e−qτa ; τa+ < ∞ = (q) . W X (a) In particular, we have ( ) n X τa+ < ∞ =

(3.2)

1 , W X (a)

(3.3)

where W X := W X(0) . Remark 3.2. The two ways of normalization (3.1) and (3.3) are natural analogies of those for diffusion processes. See [3, (2.5) and Theorem 3.1] and [16, (39) and Theorem 3.1]. Remark 3.3. The left-hand side of (3.2) may admit several other expressions, such as ) ) ( ( ) ( + + + n X e−qτa = n X e−qτa ; τa+ < ∞ = n X e−qτa ; τa+ < τ0− .

(3.4)

+

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The first equality of (3.4) follows from the fact that e−qτa = 0 on {τa+ = ∞}. Since X has no positive jumps, the measure n X is supported on the disjoint union {τa+ < τ0− ≤ ∞} ∪ {τ0− < τa+ = ∞} ∪ {τ0− = τa+ = ∞}. Thus the sets {τa+ < ∞} and {τa+ < τ0− } are equal up to n X -null sets, which yields the second equality of (3.4). The following theorem can be regarded as the Gerber–Shiu measure for the excursion measure (see also [11]). Please cite this article in press as: K. Noba, K. Yano, Generalized refracted L´evy process and its application to exit problem, Stochastic Processes and their Applications (2018), https://doi.org/10.1016/j.spa.2018.06.004.

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Theorem 3.4. For all q ≥ 0 and non-negative measurable function f , we have ( ) ∫ (q) − X −qτ0− n e f (X τ − , X τ − − ); τ0 < ∞ = f (u, v)K X (du dv), 0

0

7 1

(3.5)

2

(q)

where K X is the measure carried on (−∞, 0) × (0, ∞) defined by (q) K X (du

˜X (du dv). dv) = e−Φ X (q)v Π

3

(3.6)

We prove Theorems 3.1 and 3.4 at the same time.

5

Proof of Theorems 3.1 and 3.4. Step.1 We show that the quantity ( ) + (q) c := n X e−qτa ; τa+ < ∞ W X (a)

6

7

does not depend upon a > 0 nor q ≥ 0. First, we prove ( ) ( ) + (q) n X τa+ < ∞ W X (a) = n X e−qτa ; τa+ < ∞ W X (a) for all a > 0 and q ≥ 0. Using the monotone convergence theorem, we have ( ) ( ( ) ) + + + n X e−qτa ; τa+ < ∞ n X e−qτε e−qτa 1{τa+ <∞} ◦ θτε+ ; τε+ < ∞ ( ) ( ( ) ) . = lim + ε↓0 n X τa+ < ∞ n X e−qτε 1{τa+ <∞} ◦ θτε+ ; τε+ < ∞

4

8

9

10

(3.7)

Using the strong Markov property and (2.2), we have ( ) ( ) + + n X e−qτε ; τε+ < ∞ EεX e−qτa 1{τa+ <τ − } W X (a) 0 ( ) ( (3.7) = lim = (q) , ) + + − ε↓0 W X (a) n X e−qτε ; τε+ < ∞ PεX τa < τ0

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12

13

(q)

where we used limε↓0 W X (ε)/W X (ε) = 1 by [5, Lemma 1. (i)]. Second, we prove ( ) ( ) n X τa+1 < ∞ W X (a1 ) = n X τa+2 < ∞ W X (a2 ), for all 0 < a1 < a2 . This identity can be obtained by (( ) ) − X + + X + + n τ < τ ◦ θ ; τ < ∞ n (τa2 < ∞) a2 a1 τa1 0 W X (a1 ) ( + ) = = , + X X W X (a2 ) n (τa1 < ∞) n τa1 < ∞ where we used the strong Markov property and (2.2). (q) KX

Step.2 We show (3.5) with being multiplied by c. Using the monotone convergence theorem and the strong Markov property, we have ( ) ) ( − n X e−qτ0 f X τ − , X τ − − ; τ0− < ∞ 0 ) ) ( (0 ) ( + − − + X −qτε −qτ0 { } ◦ θτε+ ; τε < ∞, τ0 < ∞ (3.8) = lim n e e f X τ − , X τ −− 1 ε↓0

0

0

X − >ε τ0 −

and using (i) of Theorem 2.2 and Step.1, we have ∫ ∞ ∫ c (q) (3.8) = lim (q) W X (ε) dv e−Φ X (q)v f (u, v)Π X (du − v) ε↓0 W (ε) ε (−∞,0) X ∫ (q) = f (u, v)cK X (du, dv).

Please cite this article in press as: K. Noba, K. Yano, Generalized refracted L´evy process and its application to exit problem, Stochastic Processes and their Applications (2018), https://doi.org/10.1016/j.spa.2018.06.004.

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Step.3 We show c = 1. Since X has no Gaussian component, i.e., σ X = 0, differentiating (2.1), we have ∫ ( qy ) ′ Ψ X (q) = q X + ye − y1(−1,0) (y) Π X (dy) (3.9) (−∞,0)

4

5

for all q > 0. Using (3.1), we have on one hand ∫ ( ) n X 1 − e−qT0 = Ψ X′ (Φ X (q)) = γ X +

(

) yeΦ X (q)y − y1(−1,0) (y) Π X (dy).

(3.10)

(−∞,0)

On the other hand, using the monotone convergence theorem and the strong Markov property, we have ( ) n X 1 − e−qT0 ( ) ( ) =n X τ0− = ∞ + n X 1 − e−qT0 ; τ0− < ∞ ( ) ( ) ( ) ( ) − =n X τ1+ < ∞ lim E1X τ p+ < τ0− + n X 1 − e−qτ0 E XX (τ − ) e−qT0 ; τ0− < ∞ . (3.11) p↑∞

6

7

8

9

0

Using (2.2), (2.3), Step.1 and Step.2, we have ( ) ( ) − − W X (1) + n X 1 − e−qτ0 +Φ X (q)X (τ0 ) ; τ0− < ∞ (3.11) =n X τ1+ < ∞ lim p↑∞ W X ( p) ∫ ( −Φ (0)v ) 1 ˜X (du dv). =c e X − eΦ X (q)(u−v) Π +c W X (∞) Since it is known that ⎧ ( ) 1 ⎨ P lim X t = ∞ = 1 W X (∞) = Ψ X′ (0+) t↑∞ ⎩ ∞ otherwise (see e.g., [9, pp. 247]), we have ∫ ( ) (3.12) = c Ψ X′ (0+) ∨ 0 + c (−∞,0)

10 11

12

∫ Π X (du)

−u (

) e−Φ X (0)v − eΦ X (q)u dv.

(3.12)

(3.13)

0

We divide the remainder of the proof into two parts. (i) Suppose Ψ X′ (0+) > 0. In this case, we have Φ X (0) = 0 and so ∫ ( Φ (q)u ) (3.13) = cΨ X′ (0+) + c ue X − u Π X (du).

(3.14)

(−∞,0) 13

14

Using (3.9), we have ( ∫ (3.14) = c γ X +

(

) ) ueΦ X (q)u − u1(−1,0) (u) Π X (du) .

(−∞,0) 15 16

17

Using (3.10), we obtain c = 1. (ii) Suppose Ψ X′ (0+) ≤ 0. In this case, we have ( ) ∫ 1 1 (3.13) = c ueΦ X (q)u + − eΦ X (0)u Π X (du). Φ X (0) Φ X (0) (−∞,0)

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Since Ψ (Φ(0)) = 0 and by (2.1) with σ X = 0, we have ( ) ∫ ( Φ (q)u ) (3.15) = c γ X + ue X − u1(−1,0) (u) Π X (du) .

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Using (3.10), we obtain c = 1. Thus the proof is complete. □

18

(3.15)

(−∞,0)

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We need the following two lemmas for later use.

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Lemma 3.5. For all a > 0, q ≥ 0 and non-negative measurable function f we have ( ( ) ) ∫ (q,a) − X −qτ0− + − − n e f X τ , X τ − ; τ0 < τa = f (u, v)K X (du dv), 0

(q,a)

(q,a)

3

0

is the measure carried on (−∞, 0) × (0, ∞) defined by

where K X KX

2

4

(q)

W X (a − v) ˜ Π X (du dv) (q) W X (a)

(q)

(du dv) = K X (du dv) :=

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The proof is parallel to that of (i) of Theorem 2.2, so that we omit it.

6

Lemma 3.6. For all q ≥ 0 and non-negative measurable function f , we have ) ∫ (∫ − ∞ τ0 ∧T0 −qt X e−Φ X (q)y f (y)dy. e f (X t ) dt = n

7

8

0

0

Proof. Using the monotone convergence theorem, we have ) (∫ − ) (∫ − τ0 ∧T0 τ0 ∧T0 −qt X −qt + X e f (X t )dt = lim n e f (X t )dt; τε < ∞ n ε↓0

0

τε+

9

(3.16)

10

and using the strong Markov property, we have ) ) (∫ − ( τ0 ∧T0 −qt + X −qτε+ e f (X t )dt ◦ θτε+ ; τε < ∞ (3.16) = lim n e ε↓0

0

(

= lim n X e

−qτε+

ε↓0

1

= lim ε↓0

(q)

W X (ε)

)

; τε+ < ∞ EεX

EεX

(∫

τ0− ∧T0

(∫

τ0− ∧T0

) e−qt f (X t ) dt

0

) e

−qt

f (X t ) dt ,

(3.17)

0

where in (3.17) we used Theorem 3.1. Using (2.6) with b = 0, we obtain (∫ − ) ∫ ∫ ∞ τ0 ∞ 1 X −qt −Φ X (q)y e f (X t ) dt ≤ f (y)e−Φ X (q)y dy. f (y)e dy ≤ (q) Eε 0 0 ε W X (ε) By the monotone convergence theorem, the proof is complete. □ (II) We assume that X has bounded variation paths. Note that in this case 0 is irregular for itself. We write

11

12

13

14 15

0

n X = δ X P0X . Then we have (q) ( ) ( ) + + W (0) 1 0 n X e−qτa ; τa+ < ∞ = δ X E0X e−qτa ; τa+ < ∞ = δ X X(q) = (q) , W X (a) W X (a) Please cite this article in press as: K. Noba, K. Yano, Generalized refracted L´evy process and its application to exit problem, Stochastic Processes and their Applications (2018), https://doi.org/10.1016/j.spa.2018.06.004.

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where we used (2.2) and (2.4). Thus we see that Theorem 3.1 still holds in this case. Theorem 3.4, Lemmas 3.5 and 3.6 still hold as they are by a similar argument. In particular, we obtain ( ) ) 1 0( = Ψ X′ (Φ X (q)). n X 1 − e−qT0 = δ X E0X 1 − e−qT0 = ′ Φ X (q)

4

which may be regarded as the counterpart of the normalization (3.1) in the unbounded case.

5

4. Kyprianou–Loeffen’s refracted Lévy processes

6 7 8

9 10

11

Let us recall some results of Kyprianou–Loeffen [10]. We fix a constant α > 0 and let X be a general spectrally negative L´evy process, which may possibly have Gaussian component. Set Yt = X t + αt. Note that 0 < δ X < δ X + α = δY if X has bounded variation paths. Theorem 4.1 ([10]). For a fixed starting point U0 = x ∈ R, there exists a unique strong solution to (1.3). Let U be a solution to Kyprianou–Loeffen’s stochastic differential equation (1.3). (q)

12

13

14

15

16

17

18

19

20

21

Theorem 4.2 ([10]). For all x ∈ [b, a] and q ≥ 0, we have (1.4) where WU is defined by (1.5). They also calculated the potential densities with and without barriers. Theorem 4.3 ([10]). For all x ∈ [b, a], q > 0, we have ⎧ (q) ⎪ WU (x, b) (q) (q) ⎪ ⎪ y ∈ (0, a] ⎪ ⎨ W (q) (a, b) W X (a − y) − W X (x − y) (q) U r U (x, y) = ⎪ W (q) (x, b) (q) ⎪ (q) U ⎪ ⎪ WU (a, y) − WU (x, y) y ∈ [b, 0], ⎩ (q) WU (a, b) ⎧ (q) ⎪ W (x, b) −Φ X (q)(y−b) (q) ⎪ ⎪ U − W X (x − y) y ∈ (0, ∞) ⎪ ⎨ α H (q) (b; b) e (q) U r U (x, y) = (q) ⎪ H U (y; b) (q) ⎪ (q) ⎪ ⎪ WU (x, b) − WU (x, y) y ∈ [b, 0] ⎩ (q) H U (b; b) ∫∞ (q) (q)′ with H U (y; b) = 0 e−Φ X (q)(z−b) WY (z − y)dz, ⎧ (q) ⎪ ⎪ ⎪ H U (x; b) W (q) (a − y) − W (q) (x − y) ⎪ y ∈ (0, a] ⎪ X ⎨ H (q) (a; b) X (q) U r U (x, y) = (q) ⎪ ⎪ H (x; b) (q) (q) ⎪ ⎪ U WU (a, y) − WU (x, y) y ∈ [b, 0] ⎪ ⎩ (q) H U (a; b) ∫x (q) (q) with H U (x; b) = eΦY (q)(x−b) + αΦY (q) 0 eΦY (q)(z−b) W X (x − z)dz, and ⎧ (q) (q) ⎨ HU (x, y; b) − W X (x − y) ∫ ∞ (q) rU (x, y) = (q)′ (q) ⎩ HU(q) (x, y; b) e−ΦY (q)(z−b) WY (z − y)dz − WU (x, y)

y ∈ (0, a] y ∈ [b, 0]

0 (q)

22

(q)

(q)

Y (q) −Φ X (q)y with HU (x, y; b) = eΦY (q)b H U (x; b) Φ X (q)−Φ e , where WU has been given in (1.5). ΦY (q)

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11

5. Generalization of refracted Lévy processes

1

We now generalize Kyprianou–Loeffen’s refracted L´evy processes. We assume that X and Y are spectrally negative L´evy processes. We assume, in addition, that X has no Gaussian component whenever X has unbounded variation paths.

2 3

4

U0

In the unbounded variation case, we define the law of the stopped process Px by ⎛ ⎞ ⏐ ( ( )) ⏐ ( ( )) 0 ⎜ 0 ⎟ PUx F (Ut )t<τ − , (Ut+τ − )t≥0 = PxX ⎝EYy F w, (Yt0 )t≥0 ⏐⏐ ⎠ x ̸= 0 y=X (τ − ) 0 0 0 w=(X (t)) t<τ0−

and the excursion measure n U by ⎛ ⏐ ( ( )) ( ( )) ⏐ U 0 X⎜ Y0 n F (Ut )t<τ − , (Ut+τ − )t≥0 = n ⎝E y F w, (Yt )t≥0 ⏐⏐ 0

0

⎞ y=X (τ0− ) w=(X (t)) t<τ0−

⎟ ⎠

for all non-negative measurable functional F, where Yt0 = Yt∧T0 denotes the stopped process of Y upon hitting zero. Thus, we appeal to the excursion theory (see Appendix), to construct the strong Markov process U without stagnancy at 0 (that is, RU(1) 1{0} = 0) from n U together with 0 {PUx }x̸=0 . In the bounded variation case, we define U as a solution of (1.3) constructed connecting X and Y mutually (this argument is similar as [10]). When X and Y are compound Poisson processes, uniqueness of the solution of (1.3) is easily proved. We write X0

n X = δ X P0

U0

n U = δ X P0 .

and

5 6 7 8 9 10 11

12

Then we obtain (1.7) as a formula. Therefore we can do a simultaneous discussion in between the two cases of bounded and of unbounded variation.

13 14

Theorem 5.1. For all q > 0 and non-negative measurable function f with f (0) = 0, we have (∫ T0 ) (q) U −qt NU f :=n e f (X t )dt ∫ ∞ 0 ∫ (q) (q) = e−Φ X (q)y f (y)dy + RY 0 f (u)K X (du dv). (5.1) 0

Consequently we have (q)

(q)

RU f (0) = (q)

NU f (q)

q NU 1

,

(5.2)

(q)

(q)

RU f (x) = RY 0 f (x) + eΦY (q)x RU f (0), x < 0,

(5.3)

and

15

(q)

(q;0)

RU f (x) = R X

∫ f (x) +

(q)

(q)

RU f (u)G X (x, du dv), x > 0,

(5.4)

where

16

17

(q;0)

RX

f (x) = ExX

(∫

τ0−

) e−qt f (X t )dt .

0

Please cite this article in press as: K. Noba, K. Yano, Generalized refracted L´evy process and its application to exit problem, Stochastic Processes and their Applications (2018), https://doi.org/10.1016/j.spa.2018.06.004.

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∫T ∫ τ− ∫ T (q) (q) Proof. Let us calculate NU f . Since 0 0 = 0 0 + τ −0 , we have that NU f is equal to 0 (∫ − ) (∫ T0 ) ) ( τ0 − U −qt e−qt f (Ut )dt ◦ θτ − ; τ0− < ∞ . n e f (Ut )dt + n U e−qτ0

3 4 5 6

7

0

0

0

Using Lemma 3.6 and Theorem 3.4, we obtain (5.1). (q) Let us prove (5.2). Note that the finiteness of NU 1 will be proved in Lemma A.4. When X has unbounded variation paths, the formula (5.2) can be found, e.g., in [14, pp. 423]. Suppose X has bounded variation paths. We denote T0(0) = 0 and define { } T0(n) = inf t > T0(n−1) : X t = 0 recursively for all n ∈ N. Then we have (∫ (n+1) ) ∞ T0 ∑ (q) (n) U −qt E0 e f (Ut )dt; T0 < ∞ RU f (0) = (n)

=

n=0 ∞ ∑

T0

EU0

(

e

) −qT0 n

T0

(∫

e

−qt

) f (Ut )dt

0

n=0

=

EU0

) e−qt f (Ut )dt ) . ( 0 ∫ T0 −qt qEU0 e dt 0

EU0

0

(∫ T0 0

0

8 9

10

Since we write n U = δ X PU0 , we obtain (5.2). The remainder of the proof is straightforward. □ The following theorem shows the choice of n U leads to a normalization similar to (3.1). Theorem 5.2. For all q > 0, we have ( ) 1 n U 1 − e−qT0 = (q) rU (0, 0+) ∫ ) ) ( Φ (0)u ( ′ ˜X (du dv). e Y − eΦY (q)u−Φ X (q)v Π = Ψ X (0) ∨ 0 +

11

12

(5.5) (5.6)

Proof. By (5.2) of Theorem 5.1, we have ) ( ∫ 1 (q) (q) (q) −Φ X (q)y (du dv) . e 1 + r (u, y)K rU (0, y) = (y>0) X (q) Y0 q NU 1 (q)

13

Since rY 0 (u, y) = 0 for u < 0 and y > 0, we have (q)

14

15

16

rU (0, 0+) =

1 (q)

q NU 1

.

On the other hand, we have ( ) (q) q NU 1 = n U 1 − e−qT0 (q)

17 18

by the definition of NU . Thus we obtain (5.5). The other expression (5.6) can be proved easily by a similar argument to (3.13). □ Please cite this article in press as: K. Noba, K. Yano, Generalized refracted L´evy process and its application to exit problem, Stochastic Processes and their Applications (2018), https://doi.org/10.1016/j.spa.2018.06.004.

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13

6. Exit problem of generalized refracted Lévy processes We prepare a general formula.

1

2

Lemma 6.1. Let Z be a standard process with no positive jumps without stagnancy at 0 (i.e., R (1) Z 1{0} = 0). If 0 is regular for itself, then ( ) + ( ) n Z e−qτa ; τa+ < ∞ + ( ) (6.1) E0Z e−qτa ; τa+ < τb− = n Z 1 − e−qT0 1{τa+ =∞,τ − =∞}

3 4

5

b

for all a > 0 > b and q ≥ 0, where n denotes an excursion measure away from 0. If 0 is 0 irregular for itself, the identity (6.1) still holds where n Z denotes a constant multiple of P0Z . Z

6 7

Proof. It is sufficient to prove (6.1) only when q > 0. We assume first that 0 is regular for itself. Let p denote a Poisson point process defined on the probability space (Ω , F, P) with ∑ characteristic measure n Z . Set η(s) = u≤s T0 ( p(u)). Note that η will be the inverse local time at 0 for the process constructed from the excursions, which equals in law to Z under P0Z . For E ∈ B(D), we write κ E = inf{s ≥ 0 : p(s) ∈ E}. We let A = {τa+ < ∞} ∪ {τb− < ∞} and we denote by ε ∗ = p(κ A ) the first excursion belonging to A. Then we have ( ) ( ) + + ∗ E0Z e−qτa ; τa+ < τb− =E e−qη(κ A −) e−qτa (ε ) ( ) Z −qτ + ( −qη(κ −) ) n e a ; A A (6.2) =E e n Z (A) ( ) Z −qτ + + ( −qη(κ −) ) n e a ; τa < ∞ A =E e (6.3) n Z (A) where E denotes the expectation with respect to P. Note that in (6.2) we used the renewal property + of the Poisson point process and in (6.3) we used the fact that e−qτa = 0 on {τa+ ∑ = ∞, τb− < ∞}. We write p Ac for p restricted to excursions belonging to Ac and write η Ac (s) = u≤s T0 ( p Ac (u)). Since η(κ A −) = η Ac (κ A ) where η Ac and κ A are independent, we have ∫ ∞ ( ) ( ) Z E e−qη(κ A −) =n Z (A) e−n (A)t E e−qη Ac (t) dt ∫0 ∞ ( ) Z =n Z (A) e−n (A)t exp(−tn Z (1 − e−qT0 ; Ac )) dt 0

n Z (A) ). = Z( n 1 − e−qT0 1 Ac Thus we obtain (6.1). We second assume that 0 is irregular for itself. Using the notation of the proof of Theorem 5.1, we have ∞ ( ) ∑ ( ( )) + − Z −qτa+ + E0 e ; τa < τb = E0Z e−qτa ; T0(n) < τa+ < T0(n+1) ∧ τb− n=0

=

∞ ∑

) )n 0 ( + 0( E0Z e−qT0 ; τa+ = ∞, τb− = ∞ E0Z e−qτa ; τa+ < ∞

n=0 Please cite this article in press as: K. Noba, K. Yano, Generalized refracted L´evy process and its application to exit problem, Stochastic Processes and their Applications (2018), https://doi.org/10.1016/j.spa.2018.06.004.

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( ) + 0 E0Z e−qτa ; τa+ < ∞ = ). 0( 1 − E0Z e−qT0 ; τa+ = ∞, τb− = ∞ 1

Thus we obtain (6.1). □

2

Theorem 6.2. For all x ∈ [b, a] and q ≥ 0, we have

3

( ) W (q) (x, b) + EUx e−qτa ; τa+ < τb− = U(q) , WU (a, b)

(6.4)

(q)

where the function WU (x, y) is defined as follows: for x ∈ (0, ∞), ( ) (q) (q) (q) WU (x, y) =W X (x)WY (−y) Ψ X′ (0) ∨ 0 ∫ ( (q) ) (q) (q) (q) ˜X (du dv) + W X (x)WY (−y)eΦY (0)u − WY (u − y)W X (x − v) Π 4

and for x ∈ (−∞, 0], (q)

5

6 7

8

(q)

WU (x, y) = WY (x − y). Proof. We discuss the cases of unbounded and of bounded variation at the same time. ( two ) + We calculate EU0 e−qτa ; τa+ < τb− . Using Lemma 6.1, we have ( ) + ( ) n U e−qτa ; τa+ < ∞ + ( ). EU0 e−qτa ; τa+ < τb− = n U 1 − e−qT0 1{τa+ =∞,τ − =∞} b

9

10

11

Using Theorem 3.1, we can rewrite the numerator as ( ) ( ) + + n U e−qτa ; τa+ < ∞ = n X e−qτa ; τa+ < ∞ =

1 (q) W X (a)

.

) ( We divide the denominator n U 1 − e−qT0 1{τa+ =∞,τ − =∞} into the following sum: b

12

n

U

(

1−e

−qT0

)

+n

U

(

e

−qT0

; {τa+

< ∞} ∩

{τb−

) ( ) = ∞} + n U e−qT0 ; τb− < ∞ .

Let us compute these expectations. For the second term, we have ( ) n U e−qT0 ; {τa+ < ∞} ∩ {τb− = ∞} ) ) ( ( + =n U e−qτa e−qT0 1{τ − =∞} ◦ θτa+ ; τa+ < ∞ b ( ) ( ( ) ) − X −qτa+ + =n e ; τa < ∞ EaX e−qτ0 EYX (τ − ) e−qT0 1{T0 <τ − } ; τ0− < ∞ b 0 ∫ (q) 1 WY (u − b) (q) = (q) G X (a, du dv), (q) W X (a) WY (−b) 13

14

(6.5)

where in (6.5) we used Theorem 3.1, (i) of Theorem 2.2 and (2.2). For the third term, we have ( ) ( ) ( −qT − ) − n U e−qT0 ; τb− < ∞ = n U e−qτ0 EU e 0 ; τb < T0 ; τ0− < ∞ . (6.6) U (τ − ) 0

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Using Theorem 3.4, we have ∫ ( ) ˜X (du dv) (6.6) = e−Φ X (q)v EuY e−qT0 ; τb− < T0 Π ) ( ∫ (q) WY (u − b) ˜ −Φ X (q)v ΦY (q)u = e e − Π X (dudv), (q) WY (−b)

15

(6.7)

where in (6.7) we used (2.2) and (2.3). Therefore, using (5.6), we obtain (q) ( ) WU (a, b) 1 n U 1 − e−qT0 1{τa+ =∞,τ − =∞} = (q) (q) b W X (a) WU (0, b)

1

(6.8)

and we obtain (6.4) for x = 0. For all x < 0, we have ( ) ( ) ( ) + + + EUx e−qτa ; τa+ < τb− = EYx e−qτ0 ; τ0+ < τb− EU0 e−qτa ; τa+ < τb− .

2

3

4

Using (2.2) and (6.4) for x = 0, we have (6.4) for x < 0. For all x > 0, we have ( ) + EUx e−qτa ; τa+ < τb− ( ) ( ( ) ) + − a =EUx e−qτa ; τa+ < τ0− + EUx e−qτ0 e−qτ+ 1{τa+ <τ − } ◦ θτ − ; τ0− < τa+ b

(q)

=

W X (x) (q)

W X (a)

(

+ ExX e

−qτ0−

0

) ( ) a EYX (τ − ) e−qτ+ ; τa+ < τb− ; τ0− < τa+ ,

(6.9)

0

where in (6.9) we used (2.2). Using (6.4) for x < 0 and (ii) of Theorem 2.2, the second term of (6.9) is equal to ∫ (q) WU (u, b) (q;0,a) ˜X (du dv). rX (x, v)Π (q) WU (a, b) Thus we obtain (6.4) for x > 0. The proof is complete. □ Corollary 6.3. For all x ∈ (−∞, a] and q ≥ 0, we have ( ) W (q) (x) U −qτa+ = U(q) Ex e W U (a)

5 6

7

8

9

(6.10)

10

(q)

where the function W U (x) is defined as follows: for x ∈ (0, ∞), ) ( (q) (q) W U (x) =W X (x) Ψ X′ (0) ∨ 0 ∫ ( (q) ) (q) ˜X (du dv) + W X (x)eΦY (0)u − W X (x − v)eΦY (q)u Π and for x ∈ (−∞, 0],

11

(q)

W U (x) = eΦY (q)x .

12

(q)

In particular, W U (x) is a continuous and increasing function of x. Proof. Using the monotone convergence theorem and Theorem 6.2, we have (q) (q) ( ) ( ) + + W (x, b)/WY (−b) EUx e−qτa = lim EUx e−qτa ; τa+ < τb− = lim U(q) . b↓−∞ b↓−∞ W (a, b)/W (q) (−b) U Y Please cite this article in press as: K. Noba, K. Yano, Generalized refracted L´evy process and its application to exit problem, Stochastic Processes and their Applications (2018), https://doi.org/10.1016/j.spa.2018.06.004.

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Using the last equality of [7, pp. 124], we have ) ( (q) (q) (q) lim WU (x, b)/WY (−b) = W U (x), b↓−∞

3 4 5 6

7

and we have (6.10). (q) (q) Next, we prove that W U is increasing and continuous. It is obvious that W U is increasing and (q) continuous on (−∞, 0], since W U (x) = eΦY (q)x . Using the dominated convergence theorem, we have 1 1 (q) ( ( )= ) = 1, lim W U (ε) = lim + + ε↓0 ε↓0 U E0 e−qτε EU0 limε↓0 e−qτε (q)

8

so that we see W U is continuous at 0. Since 1

(q)

9

10 11

12

W U (x) =

(

+

EU0 e−qτx

),

( ) + it is thus sufficient to prove that EU0 e−qτx is decreasing and continuous on (0, ∞). For 0 < x < y, we have ( ) ( ) ( )( ( )) + + + + EU0 e−qτx − EU0 e−qτ y = EU0 e−qτx 1 − EUx e−qτ y ≥ 0. Using (2.2), for x > 0, we have ⏐ ( ) ( )⏐ ( )( ( )) + + + + ⏐ ⏐ lim sup ⏐EU0 e−qτx−ε − EU0 e−qτx+ε ⏐ = lim sup EU0 e−qτx−ε 1 − EUx−ε e−qτx+ε ε↓0 ε↓0 ( )) ( + X + ≤ lim sup 1 − Ex−ε < τ0− e−qτx+ε ; τx+ε ε↓0 ) ( (q) W X (x − ε) = 0. = 1 − lim (q) ε↓0 W (x + ε) X

13 14 15

16

The proof is complete. □ Let C0 denote the set of continuous functions f : R → R which vanish at +∞ and −∞. Note that C0 is a Banach space with respect to the supremum norm ∥ f ∥ = supx∈R | f (x)| for f ∈ C0 . Theorem 6.4. Our generalized refracted L´evy process is a Feller process. (q)

17 18 19 20

21

Proof. Since RU comes from transition operators, it is sufficient to verify the following conditions: (q) (i) For all q > 0, RU is a map  from C0 to C0 .  (q)  (ii) For all f ∈ C0 , limq↑∞ q RU f − f  = 0. (1) The proof of (i) (q) First, we prove that RU f is continuous. Let x ∈ R and ε > 0. Noting that U has no positive jump, we have ⏐ ⏐ ⏐ (q) ⏐ (q) ⏐RU f (x + ε) − RU f (x)⏐ ⏐ (∫ + )⏐ ⏐ ⏐ ⏐ ⏐ ( ) τx+ε + ⏐ (q) ⏐ ⏐ U ⏐ (q) U −qτx+ε −qt ≤ ⏐RU f (x + ε) − Ex e RU f (x + ε)⏐ + ⏐Ex e f (Ut )dt ⏐ ⏐ ⏐ 0 Please cite this article in press as: K. Noba, K. Yano, Generalized refracted L´evy process and its application to exit problem, Stochastic Processes and their Applications (2018), https://doi.org/10.1016/j.spa.2018.06.004.

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⏐( ⏐ ( )) + ⏐ ⏐ (q) ≤ ⏐RU f (x + ε)⏐ 1 − EUx e−qτx+ε + ∥ f ∥ EUx

(∫

+ τx+ε

17

) e−qt dt

0

( ( )) + 2 ≤ ∥ f ∥ 1 − EUx e−qτx+ε . q By Corollary 6.3, we have ( ) (q) W U (x) 2 (6.11) = ∥ f ∥ 1 − (q) → 0 as ε ↓ 0. q W (x + ε)

(6.11) 1

2

U

(q)

Thus we obtain right-continuity of RU f . For the left-continuity we have ⏐ ⏐ ⏐ (q) ⏐ (q) ⏐RU f (x − ε) − RU f (x)⏐ ⏐ (∫ + )⏐ ⏐ ⏐ ⏐ ⏐ ( ) τx + ⏐ U ⏐ ⏐ ⏐ (q) (q) ≤ ⏐Ex−ε e−qτx RU f (x) − RU f (x)⏐ + ⏐EUx−ε e−qt f (Ut )dt ⏐ ⏐ ⏐ 0 and the remainder of its proof is similar to that of the right-continuity. (q) Second, we prove that RU f vanishes at −∞. For x < 0, we may rewrite (5.3) as (q) RU

f (x) =

(q) RY

(q) eΦY (q)x RY

f (x) −

f (0) +

(q) eΦY (q)x RU

3 4

f (0).

5

(q) limx↓−∞ RU

By the Feller property of Y , we see that f (x) = 0. (q) Third, we prove that RU f vanishes at +∞. We may assume without loss of generality that f ≥ 0. For all x > 0, we have ) ((∫ − ∫ ) τ0

(q)

RU f (x) = EUx

∞

+ 0

τ0−

e−qt f (Ut )dt

( ) 1 − (q) (q) ≤ R X f (x) + ExX e−qτ0 RU f (X τ − ) · ∥ f ∥ . 0 q ( ) ( ) − − By the Feller property of X and by the fact that ExX e−qτ0 = E0X e−qτ−x → 0 as x → ∞, we obtain

(q) limx↑∞ RU

f (x) = 0.

sup

7 8

(2) The proof of (ii) Define ωε ( f ; x) =

6

9 10

| f (y) − f (x)| .

11

y:|y−x|≤ε

Let us prove the pointwise convergence: lim

q↑∞

(q) q RU

f (x) = f (x),

For x ∈ R and ε > 0, we have ⏐ ⏐ ⏐ ⏐ (q) ⏐q RU f (x) − f (x)⏐ ((∫ + − ∫ τx+ε ∧τx−ε U ≤ qEx + 0

12

(6.12)

x ∈ R.

∞ + − τx+ε ∧τx−ε

)

) e−qt | f (Ut ) − f (x)| dt

( ) ( ) + − + − ≤ EUx 1 − e−q(τx+ε ∧τx−ε ) ωε ( f ; x) + 2 ∥ f ∥ EUx e−q(τx+ε ∧τx−ε ) . Please cite this article in press as: K. Noba, K. Yano, Generalized refracted L´evy process and its application to exit problem, Stochastic Processes and their Applications (2018), https://doi.org/10.1016/j.spa.2018.06.004.

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⏐ ⏐ ⏐ ⏐ (q) We thus obtain lim supq↑∞ ⏐q RU f (x) − f (x)⏐ ≤ ωε ( f ; x) for all ε > 0, which proves (6.12). By a standard argument with the help of the fact that the dual space of C0 can be identified with ( p) the space of signed measures, we can see that RU (C0 ) is dense in C0 for all p > 0. (1) Let f = RU g for some g ∈ C0 . Using the resolvent equation, we have  1     (q)  (q)  (q)   f − q RU f  = RU g − RU f  ≤ ∥g − f ∥ → 0, as q ↑ ∞. q

7

Since RU(1) (C0 ) is dense in C0 , we obtain claim (ii). The proof is now complete. □

8

7. Potential measure of killed refracted Lévy processes

6

9 10

11

12

13

In this section, we calculate the potential measure of refracted L´evy processes killed on exiting [b, a]. Theorem 7.1. For all x ∈ [b, a] and q ≥ 0, we have ⎧ (q) ⎪ WU (x, b) (q) (q) ⎪ ⎪ ⎪ ⎨ W (q) (a, b) W X (a − y) − W X (x − y), (q) U r U (x, y) = (q) ⎪ W (x, b) (q) ⎪ (q) U ⎪ ⎪ WU (a, y) − WU (x, y), ⎩ (q) WU (a, b)

y ∈ (0, a] (7.1) y ∈ [b, 0).

Proof. We follow the notation of Lemma 6.1 for L, η, κ, etc. Step.1 We calculate in the case x = 0. When X has unbounded variation paths, we have (∫ + − ) τa ∧τb U −qt E0 e f (Ut )dt 0

=EU0

(∫ e

−qs

) 1{s<η(κ A∪B −)} d L(s) n

U

15

16

18

e

f (Ut )dt .

(7.2)

where we used the compensation theorem of the excursion point process. We may rewrite (7.2) using η′ , as ) (∫ ∞ ) (∫ T0 ∧τa+ ∧τ − b U −qη′ (t) U −qt E0 e f (Ut )dt , e 1{t<κ A∪B } dt n 0

0 17

) −qt

0

(0,∞) 14

T0 ∧τa+ ∧τb−

(∫

the first factor of which equals to ∫ ∞ ( ) −tn U 1−e−qT0 ;(A∪B)c −tn U (A∪B) e dt = e 0

1 nU

(

1−

e−qT0 1{(A∪B)c }

).

When X has bounded variation paths, we have (∫ + − ) τa ∧τb −qt U E0 e f (Ut )dt 0

=

∞ ∑ n=0

EU0

(∫

(n+1)

T0

(n)

T0

∧τa+ ∧τb−

) e

−qt

f (Ut )dt; T0(n)

<

τa+

∧

τb−

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=

∞ ∑

)n ( EU0 e−qT0 ; τa+ = ∞, τb− = ∞ EU0

(∫

e−qt f (Ut )dt

0

n=0 U0

(

E0

=

T0 ∧τa+ ∧τb−

)

∫ T0 ∧τa+ ∧τb− 0

e−qt f (Ut )dt

)

), 0( 1 − EU0 e−qT0 ; τa+ = ∞, τb− = ∞

where we used the notation of the proof of Theorem 5.1. Since n U = δ X EU0 , we obtain ( ) ∫ T0 ∧τa+ ∧τb− −qt U (∫ + − ) n e f (Ut )dt τa ∧τb 0 ), EU0 e−qt f (Ut )dt = U ( n 1 − e−qT0 ; τa+ < ∞, τb− = ∞ 0

1

2

which has the same form as in the case of unbounded variation. The denominator has already computed in (6.8). Let us compute the numerator. In the case f = 1(a ′ ,a] for 0 < a ′ < a, we have (∫ ) ( ) T0 ∧τa+ ∧τb− + U −qt n e 1{Ut ∈(a ′ ,a]} dt =n X e−qτa′ ; τa+′ < ∞ EaX′ 0

τa+ ∧τ0−

(∫ ×

) e−qt 1{X t ∈(a ′ ,a]} dt

0

∫

1

=

(q)

W X (a)

(q)

(a ′ ,a]

W X (a − y)dy

(7.3)

where in (7.3) we used Theorem 3.1 and (2.7). Thus we obtain (7.1) for x = 0 and y ∈ (0, a]. In the case f = 1[b,b′ ) for b < b′ < 0, we have ) (∫ T0 ∧τa+ ∧τb− e−qt 1{Ut ∈[b,b′ )} dt nU

3 4

5

0

( =n

X

e

−qτ0−

EYX (τ − ) 0

(∫

τb− ∧T0

) e

−qt

) ; τ0−

1{Yt ∈[b,b′ )} dt

0

<

τa+

.

(7.4)

Using Lemma 3.5, we have ) ∫ b′ (∫ (q) W X (a − v) ˜ (q;b,0) (7.4) = rY (u, y) Π X (du dv) dy. (q) b W X (a)

6

7

(7.5)

Using (6.8), (7.3) and (7.5), we obtain (7.1) for x = 0 and y ∈ [b, 0).

8

9

Step.2 We calculate in the case x < 0. We have (∫ + − ) τa ∧τb EUx e−qt f (Ut )dt 0

=EYx

τ0+ ∧τb−

(∫

) e

−qt

f (Yt )dt

( +

EUx

e

−qτ0+

0

Using (2.5), we have that the first term equals to ∫ 0 (q;b,0) f (y)r Y (x, y)dy.

(∫

τa+ ∧τb−

) e

0

−qt

f (Ut )dt

) ◦

θτ + ; τ0+ 0

<

τb−

. 10

(7.6)

b Please cite this article in press as: K. Noba, K. Yano, Generalized refracted L´evy process and its application to exit problem, Stochastic Processes and their Applications (2018), https://doi.org/10.1016/j.spa.2018.06.004.

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Using (2.2) and Step.1, we have that the second term equals to (∫ + − ) ( ) τa ∧τb + − Y −qτ0+ U −qt Ex e ; τ0 < τb E0 e f (Ut )dt 0

(∫ a (q) WU (x, b) = (q) WU (a, b) 0 (∫ ∫ 0 f (y)

+

(q)

f (y)W X (a − y)dy (q) W X (a

− v)

(q;b,0) × rY (u,

) ) ˜ y)Π X (du dv) dy.

(7.7)

b 1

Using (7.6) and (7.7), we obtain (7.1) for x < 0. Step.3 We calculate in the case x > 0. We have ) (∫ + − τa ∧τb e−qt f (Ut )dt EUx 0

=ExX

τ0− ∧τa+

(∫

) e

−qt

f (X t )dt

( +

EUx

e

−qτ0−

0 2

3

τa+ ∧τb−

(∫

) e

−qt

f (Ut )dt

0

Using (2.7), we have the first term equals to ∫ a (q;0,a) f (y)r X (x, y)dy.

) ◦

θτ − ; τ0− 0

<

τa+

.

(7.8)

0

The second term equals to ( (∫ X −qτ0− U E X (τ − ) Ex e 0

∫ =

EUu

(∫

τa+ ∧τb−

) e

−qt

f (Ut )dt

0

τa+ ∧τb−

) ; τ0−

<

τa+

) e

−qt

(q;0,a)

f (Ut )dt r X

˜X (du dv), (x, v)Π

(7.9)

0 4

5

6

7

where in (7.9) we used (ii) of Theorem 2.2. If f is 0 on (−∞, 0], we have )∫ ( (q) (q) ∞ WU (x, b) W X (x) (q) − (7.9) = f (y)W X (a − y)dy. (q) (q) 0 WU (a, b) W X (a)

(7.10)

From (7.8) and (7.10) we obtain (7.1) for x > 0 and y ∈ (0, a]. If f is 0 on (0, ∞), we have ( (q) ) ∫ ∞ WU (x, b) (q) (q) (7.9) = f (y) WU (a, y) − WU (x, y) dy. (q) 0 WU (a, b)

8

Thus we obtain (7.1) for x > 0 and y ∈ [b, 0). □

9

8. Approximation problem Let Z be a spectrally negative L´evy process. Let Ψ Z denote the Laplace exponent represented by (2.1). For n ∈ N, we define ( ( )) 1 1 Ψ Z (n) (q) = γ Z q − σ Z2 n 2 1 − eq(− n ) + q − n ( ) ∫ qy ( ) − ( + qy1 −1,− 1 (y) Π Z (dy) ) 1−e −∞,− n1

n

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= δ Z (n) q −

∫

(

21

) 1 − eqy Π Z (n) (dy)

(−∞,0)

where δ Z (n) = γ Z + σ Z2 n + Π Z (n) =

∫ (−y)Π Z (dy)

(−1,− n1 ) 1(−∞,− 1 ) Π Z + σ Z2 n 2 δ(− 1 ) . n n

If we denote by Z (n) a L´evy process with Laplace exponent Ψ Z (n) , it is actually a compound Poisson process with positive drift. We note that Ψ Z (n) (q) → Ψ Z (q) for all q ≥ 0, so that we have Z (n) → Z in law on D. More precisely, by Bertoin [2, pp. 210], we see that there exists a coupling of Z (n) ’s such that Z (n) → Z uniformly on compact intervals almost surely, which we will call the uniformly convergent coupling. Let X and Y be spectrally negative L´evy processes and suppose that X has unbounded variation paths and no Gaussian component. For each n ∈ N, let X (n) and Y (n) be independent L´evy processes with Laplace exponents Ψ X (n) and ΨY (n) , respectively. Let U (n) be defined as a unique strong solution of the stochastic differential equation ∫ ∫ Ut(n) = U0(n) + 1{U (n) ≥0} d X s(n) + 1{U (n) <0} dYs(n) . s−

(0,t]

1 2 3 4 5 6 7 8 9

10

s−

(0,t]

(n)

Theorem 8.1. (U (n) , PUx ) converges in distribution to (U, PUx ) for all x ∈ R.

11

We postpone the proof of Theorem 8.1 until the proof of Theorem 8.5.

12

Remark 8.2. We may expect δ X (n) P

X (n)0

→ nX

13

U (n)0

and δ X (n) P

→ nU .

14

The precise statements are as follows: For all bounded continuous function f , we have (∫ T0 ) (∫ T0 ) (n)0 δ X (n) E0X e−qt f (X t(n) )dt → n X e−qt f (X t )dt as n ↑ ∞ 0

15

16

0

and

17

δ X (n) EU0

(n)0

(∫

T0

e−qt f (Ut(n) )dt

)

→ nU

0

T0

(∫

e−qt f (Ut )dt

) as n ↑ ∞.

18

0

The proofs of these formulas are straightforward, so we omit it.

19

Lemma 8.3. For all non-positive x (n) and x satisfying x (n) → x as n ↑ ∞ and for all q > 0 and bounded continuous function f , we have (q)

(q)

RY (n)0 f (x (n) ) → RY 0 f (x) as n ↑ ∞.

(8.1)

Proof. Using the strong Markov property, we have (q) RY (n)0

(n)

f (x ) =

(q) RY (n)

(n)

f (x ) −

+ (n) (q) EYx (n) (e−qτ0 )RY (n)0

20 21

22

23

f (0)

24

(q)

and a similar identity for RY 0 f (x). Using the uniformly convergent coupling and the dominated (q) (q) (q) (q) convergence theorem, we have RY (n) f (x (n) ) = Rx (n) +Y (n) f (0) → Rx+Y f (0) = RY f (x). Since Please cite this article in press as: K. Noba, K. Yano, Generalized refracted L´evy process and its application to exit problem, Stochastic Processes and their Applications (2018), https://doi.org/10.1016/j.spa.2018.06.004.

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ΨY (n) → ΨY pointwise as n ↑ ∞, we have ΦY (n) → ΦY pointwise as n ↑ ∞ and thus ( ) ( ) + + (n) (n) EYx (n) e−qτ0 = e−ΦY (n) (q)x → e−ΦY (q)x = EYx e−qτ0 .

3

Thus we obtain (8.1). □

4

Theorem 8.4. For all x ∈ R, q > 0 and bounded continuous function f , we have (q)

(q)

RU (n) f (x) → RU f (x) as n ↑ ∞.

5

6 7 8 9

10

Proof. We may assume without loss of generality that 0 ≤ f ≤ 1. We write ρ X := infn∈N Φ X (n) (q) and ρY := infn∈N ΦY (n) (q). Since Φ Z (q) is strictly positive for all spectrally negative L´evy process Z , we have ρ X and ρY are strictly positive. We prove (8.2) for x = 0. By (5.2) and (5.1) of Theorem 5.1, it is sufficient to prove ∫ ∞ ∫ ∞ e−Φ X (q)y f (y)dy (8.3) e−Φ X (n) (q)y f (y)dy → 0

0 11

and ∫

12

(8.2)

(q)

∫

(q)

RY (n)0 f (u)K X (n) (du dv) →

(q)

(q)

RY 0 f (u)K X (du dv).

(8.4)

Using Φ X (n) → Φ X and the dominated convergence theorem, we have (8.3). Let us prove (8.4). Using (3.6) with c = 1 and changing variables, we have ∫ (q) (q) RY (n)0 f (u)K X (n) (du dv) ∫ −u ∫ (q) e−Φ X (n) (q)v RY (n)0 f (u + v)dv (8.5) = Π X (du)1(u<− 1 ) n

(−∞,0)

0

and a similar identity for (Y , X ). We have ⏐ ⏐ ∫ −u ⏐ ⏐ −Φ X (n) (q)v (q) ⏐1 ⏐ e R f (u + v)dv 1 ⏐ (u<− n ) ⏐ Y (n)0 0 (∫ T0 ) ∫ −u Y (n) ≤ e−qt dt dv e−ρ X v Eu+v 0 0 ∫ ) 1 −u −ρ X v ( ≤ e 1 − eΦY (n) (q)(u+v) dv q 0 ∫ ) 1 −u −ρ X v ( ≤ e 1 − eρY (u+v) dv q 0 )( ) 1 ( 1 − eρ X u 1 − eρY u ∈ L 1 (Π X ) . ≤ qρ X Thus we may apply the dominated convergence theorem to obtain ∫ ∫ −u ( ) (q) lim (8.5) = Π X (du) e−Φ X (q)v RY 0 f (u + v) dv, 0

13

14

n↑∞

15 16 17

(q;0)

18

(−∞,0)

0

which shows (8.4). Thus we obtain (8.2) for x = 0. For x < 0, (8.2) is obvious by (5.3) of Theorem 5.1 and Lemma 8.3. We prove (8.2) for x > 0. By (5.4) of Theorem 5.1, it suffices to prove (q;0)

R X (n) f (x) → R X

f (x) as n ↑ ∞

(8.6)

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23

and

1

X (n)

Ex

( ( )) ( )) ( − (q) (n) −qτ0− (q) as n ↑ ∞. e RU (n) f X τ − → ExX e−qτ0 RU f X τ − −

2

0

0

−

Note that e−qτ0 = e−qτ0 1(τ − <∞) a.s. Since X has no Gaussian component, we have

3

0

inf

t∈[0,τ0− (X ))

X t > 0 and X τ − (X ) < 0 a.s. on {τ0− (X ) < ∞}.

4

0

For almost every sample path with τ0− (X ) < ∞ based on the uniformly convergent coupling of [2, pp. 210], we have inf

t∈[0,τ0− (X ))

X t(n)

→

inf

n↑∞ t∈[0,τ − (X )) 0

X t and

X τ(n)− (X ) → n↑∞ 0

X τ − (X ) ,

=

6

7

0

so that we have τ0− (X )

5

8

τ0− (X (n) )

for large n.

9

Therefore we have (8.6) by the dominated convergence theorem. By the strong Markov property, we have ( ( )) − (q) (n) ExX e−qτ0 RU (n) f X τ(n)− 0 ( ) ( ) − − (n) − (n) (n) (q) (q) = ExX e−qτ0 RY 0(n) f (X τ(n)− ) + ExX eΦY (n) (q)X (τ0 )−qτ0 RU (n) f (0). 0

For the first term we have ( ) ( ) − (q) (n) X (n) −qτ0− (q) lim Ex e RY 0(n) f (X τ − ) = ExX e−qτ0 RY 0 f (X τ − ) n↑∞

0

10

11

0

where we used the dominated convergence theorem and Lemma 8.3. For the second term we have ( ) ( ) − − − (n) − (n) (q) (q) lim ExX eΦY (n) (q)X (τ0 )−qτ0 RU (n) f (0) = ExX eΦY (q)X (τ0 )−qτ0 RU f (0)

12 13

14

n↑∞

where we used ΦY (n) → ΦY and (8.2) for x = 0. The proof is now complete. □

15

For a stochastic process Z , t > 0, x ∈ R and positive or bounded measurable function f , we define Z

Pt f (x) :=

ExZ (

f (Z t )) .

16 17

18

Theorem 8.5. For all q > 0, t > 0 and f ∈ C0 , we have (q)

(q)

RU (n) f → RU f uniformly as n ↑ ∞, U (n)

Pt

f → PtU f uniformly as n ↑ ∞.

(8.7) (8.8)

Proof of Theorems 8.1 and 8.5. From (8.7) we can derive (8.8) by using Theorem 6.4 and (n) [13, Theorem 3.4.2]. Using [6, Theorem 19.25], we can conclude that (U (n) , PUx ) converges in distribution to (U, PUx ) for all x ∈ R. Let us prove (8.7). We divide the proof of (8.7) into three steps. (q) Step.1 Let k > 0 be a constant. We prove {W U (n) (x)}n∈N (q) For this, we prove pointwise convergence limn↑∞ W U (n) (x)

is equicontinuous in x ∈ [−k, k]. (q) (q) = W U (x). Since {W U (n) }n∈N is

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increasing and continuous by Corollary 6.3, the pointwise convergence implies convergence in (q) x ∈ [−k, k], thus {W U (n) (x)}n∈N is equicontinuous in x ∈ [−k, k]. The desired convergence is (q) obvious for x ≤ 0 by the definition of W U (x). For x ≤ 0, it suffices to show ( ) ( ) + + (n) lim EU0 e−qτx = EU0 e−qτx (8.9) n↑∞

6

7

8 9 10

11

by Corollary 6.3. By the strong Markov property, we have ) + + 1( (n) (n) (q) (q) 1 − EU0 (e−qτx ) + EU0 (e−qτx )RU (n) 1(−∞,x) (x). RU (n) 1(−∞,x) (0) = q As f − := 1(−∞,x) is not continuous, we take bounded continuous functions f m− and f m+ such (q) (q) that f m− ↑ f − and f m+ ↓ f + := 1(−∞,x] . Using Theorem 8.4, we have RU (n) f m± → RU f m± . It is (q) (q) obvious that RU (n) f ± → RU f ± . Thus we obtain (8.9). Step.2 We may assume without loss of generality that ∥ f ∥ = 1. Let us prove (q)

12

(q)

RU (n) f (x) → RU f (x) uniformly in x ∈ [−k, k]. (q)

13 14 15

16

Since we have the pointwise convergence by Theorem 8.4, it is sufficient to prove {RU (n) f }n∈N is equicontinuous. For all x, y ∈ R with x < y, making a computation similar to (1) of the proof of Theorem 6.4, we have ( ) (q) ⏐ ⏐ 2 W U (n) (x) ⏐ (q) ⏐ (q) . (8.10) ⏐RU (n) f (y) − RU (n) f (x)⏐ ≤ ∥ f ∥ 1 − (q) q W U (n) (y) (q)

17 18

19

Let ε > 0 be a constant. By Step.1 and since infn∈N W U (n) (−k) = infn∈N e−ΦY (n) (q)k > 0, we see that there exists ξ > 0 such that for all x, y ∈ [−k, k] with 0 < y − x < ξ ⏐ ⏐ (q) (q) ⏐ ⏐ (q) sup ⏐W U (n) (y) − W U (n) (x)⏐ ≤ ε inf W U (n) (−k). n∈N

n∈N

20

Then we have (q)

21

ε infn∈N W U (n) (−k) 2 2 (8.10) ≤ ∥ f ∥ ≤ ∥ f ∥ ε, (q) q q W (n) (y) U

(q)

22 23

24

25

(q)

where we used the fact that W U (n) is increasing. Therefore we conclude that {RU (n) f }n∈N is equicontinuous. Step.3 We prove that for any ε > 0 there is k > 0 such that ⏐ ⏐ ⏐ (q) ⏐ sup sup ⏐RU (n) f (x)⏐ < ε.

(8.11)

x∈(−∞,−k)∪(k,∞) n∈N

For all x < y < 0 we have ⏐ ⏐ ( ) ⏐ ⏐ ⏐ ⏐ (n) ∫ τ y+ ( ) ⏐ ⏐ (q) ⏐ ⏐ U (q) (n) −qt U (n) −qτ y+ e f (Ut )dt + Ex e RU (n) f (y)⏐ ⏐RU (n) f (x)⏐ = ⏐Ex ⏐ ⏐ 0 ( ) + 1 1 (m) e−qτ y ∥ f ∥ . ≤ sup | f (z)| + sup EYx q z
(8.12)

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By the same argument, for all x > y > 0, we have ⏐ 1 ⏐ ( ) − (m) ⏐ ⏐ (q) e−qτ y ∥ f ∥ . ⏐RU (n) f (x)⏐ ≤ sup | f (z)| + sup ExX q z>y m∈N Since f ∈ C0 , there exists k1 > 0 such that 1 sup | f (z)| < qε. 3 |z|>k1

1

(8.13)

2

3

(8.14)

Using the uniformly convergence coupling, we have for x > y > 0 ( ) ( ) ( ) ( ) + + − − (n) Y (n) Y lim E−x e−qτ−y = E−x e−qτ−y and lim ExX e−qτ y = ExX e−qτ y n↑∞

25

n↑∞

4

5

(8.15)

and

6

7

( ) + Y lim E−x e−qτ−y = 0

x↑∞

and

( ) − lim ExX e−qτ y = 0.

x↑∞

By (8.16), there exists k2 > k1 such that ( ) ( + ) ε ε −qτk− −qτ−k X Y 1 1 and E e < E−k e < k2 2 3∥ f∥ 3∥ f∥ By (8.15), there exists N ∈ N such that for all n > N ⏐ (n) ( ) ( + )⏐ ε −qτ + −qτ−k ⏐ ⏐ Y Y 1 ⏐ < e ⏐E−k2 e −k1 − E−k 2 3∥ f∥ and ⏐ (n) ( ) ( )⏐ ε −qτ + −qτ + ⏐ ⏐ X . ⏐Ek2 e k1 − EkX2 e k1 ⏐ < 3∥ f∥ By (8.16) again, there exists k3 > k2 such that for all n ≤ N ( ( ) + ) ε ε −qτ−k −qτk− Y (n) X (n) 1 1 E−k and E e < e < k3 3 3∥ f∥ 3∥ f∥ Thus we obtain ( ) ( ) 2ε 2ε (n) −qτ + −qτ − Y (n) e −k1 < sup E−k and sup EkX3 e k1 < . 3 ∥ ∥ ∥ 3 f 3 f∥ n∈N n∈N

(8.16)

8

9

10

11

12

13

14

15

16

17

(8.17)

By (8.12), (8.13), (8.14) and (8.17), we obtain (8.11). The proof is complete. □ Uncited references [8] Acknowledgments The authors were supported by JSPS-MAEDI Sakura program. The second author was supported by MEXT KAKENHI grant nos. 26800058 and 15H03624. Appendix. Constructing generalized a refracted process by excursions In this section, we show that we can construct from n U a right-continuous strong Markov processes by means of the excursion theory. We need the following theorem which we state without proof. For t ≥ 0, we denote Dt = σ (ω ↦→ ω(s) : s ≤ t). Please cite this article in press as: K. Noba, K. Yano, Generalized refracted L´evy process and its application to exit problem, Stochastic Processes and their Applications (2018), https://doi.org/10.1016/j.spa.2018.06.004.

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1 2

3 4 5 6

7

Theorem A.1 ([15, Theorem 2]). Let (Z 0 , PxZ ) be a R-valued right-continuous strong Markov process stopped at 0. Suppose that a σ -finite measure n on D satisfies the following conditions: (i) (ii) (iii) (iv)

n (is concentrated on D0 := {ω ∈ D : ω(0) = 0, T0 (ω) > 0, ω(t) = 0 for t ≥ T0 }. ) 0 n (D = ∞. ) n 1 − e−T0 < ∞. For all t > 0, A1 ∈ Dt with A1 ⊂ {T0 > t} and A2 ∈ B(D), ∫ ( ) ( 0 ) Z0 n A1 ∩ θt−1 (A2 ) = Z ∈ A2 n(dω) , Pω(t) A1

8 9 10

11 12

13 14

where θt denotes the shift operator. (v) If a measure n ′ on D satisfies n ≥ n ′ ≥ 0 and the counterpart of Condition (iv) for n ′ , then either n ′ (D0 ) = 0 or n ′ (D0 ) = ∞. Then there is a right-continuous strong Markov process Z for which n is an excursion measure 0 away from 0 and (Z 0 , PxZ ) is the stopped process. To construct the strong Markov process U in Section 5, we need to check that U 0 is a rightcontinuous strong Markov process and that n U satisfies conditions of Theorem A.1. 0

15

Lemma A.2. The stopped process (U 0 , PUx ) has the Markov property. 0

Proof. It is obvious that (U 0 , PUx ) = (Y 0 , P0x ) for x < 0 satisfies the Markov property. We 0 thus need to prove that (U 0 , PUx ) satisfies the Markov property for x > 0. Let A1 ∈ Dt with 0 A1 ⊂ {T0 > t} and A2 ∈ B(D). We write A = A1 ∩ θt−1 (A2 ). By the definition of PUx , we have ⎛ ⎞ ⏐ )⏐ ) 0( ⎜ 0( ⎟ PUx U 0 ∈ A =ExX ⎝PYy w ◦ Y 0 ∈ A ⏐⏐ ; τ0− ≤ t ⎠ y=X (τ0− ) w=(X (s)) s<τ0−

⎛

⎞

⏐ )⏐ ⎜ 0( + ExX ⎝PYy w ◦ Y 0 ∈ A ⏐⏐ 16 17

18

y=X (τ0− ) w=(X (s)) s<τ0−

⎟ ; t > τ0− ⎠ ,

(A.1)

where w ◦ w ′ denotes the concatenation of a path w = (ws )s
⎞

⎛ ⎜ =ExX ⎝PYy

( 0

w ◦ Y 0 ∈ A1 , Ys (

) 0 s≥t−u

) ⏐⏐ ∈ A2 ⏐⏐ y=X u w=(X (s))

s
⎟ ; τ0− ≤ t ⎠

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⎜ 0 =ExX ⎝EYy

( ) ⏐⏐ 0( 1{w◦Y 0 ∈A1 } PYy ′ Y 0 ∈ A2 ⏐

(

( 0 ) ⏐⏐ U ∈ A2 ⏐

=EUx

27

⎞

⎛

0 1{U ∈A1 } PUy

y=Ut

0 y ′ =Yt−u

; τ0−

)⏐ ⏐ ⏐ y=X u ⏐ w=(X (s))

s
⎟ ; τ0− ≤ t ⎠

)

≤t .

We can do a similar argument for (A.1). So we obtain ∫ ( 0 ) ) 0( ) 0 ( U0 Px U ∈ A = PUω(t) U 0 ∈ A2 PUx U 0 ∈ dω .

1

2

A1

□

The proof is complete.

3

Lemma A.3. The stopped process U 0 has the strong Markov property.

4

Proof. Fix t > 0. By ) the proof of [4, Theorem 1 of Section 2.3], it is sufficient to prove 0( that x ↦→ EUx f (Ut0 ) is continuous for all bounded continuous function f with f (0) = 0. Continuity at x < 0 is obvious, by the Feller property of Y 0 . Left-continuity at x = 0 is also 0 obvious. Right-continuity at x = 0 follows from the fact that PUy (T0 ∈ ·) → δ0 . Let us consider y→0

continuity at x > 0. ) ) ( ) 0( 0( EUy f (Ut0 ) =EUy f (Ut0 ); τ0− ∧ t < Tx + E yX f (X t ); Tx ≤ t < τ0− ⎛ ⎞ ⏐ ⏐ ( ) 0 0 + E yX ⎝EYy ′ f (Yt−u ) ⏐⏐ ′ ; Tx ≤ τ0− ≤ t ⎠ . y =X (u) u=τ0−

( ) Note that we have P0X lim y→0 Ty = 0 = 1 by the assumption that X is spectrally negative and of bounded variation. Since X and Y 0 have c`adl`ag paths, we have the following identities: ) ( ) 0( EUy f (Ut0 ); τ0− ∧ t < Tx ≤ ∥ f ∥ P0X τ−−x < Tx−y → 0, y→x 2 ( ) ( ) ⏐⏐ ( ) ; Tx ≤ t ∧ τ0− E yX f (X t ); Tx ≤ t < τ0− = E yX ExX f (X t−u ); t < τ0− ⏐ u=Tx ( ) − X → Ex f (X t ); t < τ0 , y→x

⎛ E yX ⎝E

Y0 y′

(

⎞ ⏐ ⏐ ) 0 f (Yt−u ) ⏐⏐ ′ ; Tx ≤ τ0− ≤ t ⎠ y =X (u) u=τ0−

⎛

⎛

= E yX ⎝ExX ⎝E

Y0 y′

(

⎞ ⏐ ⏐ ⏐ ⏐ ) − 0 ⏐ ⎠ f (Yt−u−v ) ⏐ ′ ; τ0 ≤ t ⏐⏐ y =X (u) u=τ0−

⎛ → ExX ⎝E

y→x

Y0 y′

(

⎞ v=Tx

; Tx ≤ τ0− ∧ t ⎠

⎞ ⏐ ⏐ ) 0 f (Yt−u ) ⏐⏐ ′ ; τ0− ≤ t ⎠ . y =X (u) u=τ0−

The proof is now complete. □

5

Lemma A.4. The measure n = n U satisfies Conditions (i), (ii), (iii), (iv) and (v) in Theorem A.1.

6

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Proof. It is obvious by definition that n U satisfies (i) and (ii). Let us prove (iii). By the definition of n U and by (2.3), we have ) ( ( +) ( ) − n U 1 − e−T0 = n X 1 − e−τ0 EYX − e−τ0 1{τ − <∞} 0 τ0 ( ) − ΦY (1)X τ − 0 1 − = n X 1 − e−τ0 e {τ <∞} .

(A.2)

0

( ) ′ We let q ′ = 1 ∨ inf{q > 0 : Φ X (q) > ΦY (1)}. Since n X 1 − e−q T0 is finite, we obtain ( ) Φ X (q ′ )X − τ0 X −q ′ τ0− (A.2) ≤ n 1 − e e 1{τ − <∞} 0 ) ( ( ) ′ + ′ − = n X 1 − e−q τ0 E XX − e−q τ0 1{τ − <∞} < ∞. τ0

0

0

2 3

The proof of (iv) is the same as that of the Markov property of (U 0 , PUx ) for x > 0 in Lemma A.2. Let us prove (v). We define the σ -finite measure n ′′ by ⎛ ⎞ ⏐ ( ( )) )) ⏐ ⎜ 0( ( ⎟ n ′′ F (Ut )t<τ − , (Ut+τ − )t≥0 = n ′ ⎝E yX F w, (X t0 )t≥0 ⏐⏐ ⎠ − 0

0

y=U (τ0 ) w=(U (t)) t<τ0−

8

for all non-negative measurable functional F. Then n ′′ satisfies the Markov property for 0 {PxX }x∈R\{0} . By the definition of n U , we have n X ≥ n ′′ ≥ 0. By [15, Proposition 1], n X satisfies Condition (v) and we obtain either n ′′ (D0 ) = 0 or n ′′ (D0 ) = ∞, which yields we have either n ′ (D0 ) = 0 or n ′ (D0 ) = ∞. The proof is complete. □

9

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