Parisian excursion with capital injection for draw-down reflected Levy insurance risk process
PPARISIAN EXCURSION WITH CAPITAL INJECTION FOR DRAW-DOWNREFLECTED L´EVY INSURANCE RISK PROCESS
BUDHI SURYA, WENYUAN WANG, XIANGHUA ZHAO AND XIAOWEN ZHOU
Abstract.
This paper discusses Parisian ruin problem with capital injection for L´evy insurancerisk process. Capital injection takes place at the draw-down time of the surplus process whenit drops below a pre-specified function of its last record maximum. The capital is continuouslypaid to keep the surplus above the draw-down level until either the surplus process goes above therecord high or a Parisian type ruin occurs, which is announced at the first instance the surplusprocess has undergone an excursion below the record for an independent exponential period of timeconsecutively since the time the capital was first injected. Some distributional identities concerningthe excursion are presented. Firstly, we give the Parisian ruin probability and the joint Laplacetransform (possibly killed at the first passage time above a fixed level of the surplus process) ofthe ruin time, surplus position at ruin, and the total capital injection at ruin. Secondly, we obtainthe q -potential measure of the surplus process killed at Parisian ruin. Finally, we give expectedpresent value of the total discounted capital payments up to the Parisian ruin time. The resultsare derived using recent developments in fluctuation and excursion theory of spectrally negativeL´evy process and are presented semi explicitly in terms of the scale function of the L´evy process.Some numerical examples are given to facilitate the analysis of the impact of initial surplus andfrequency of observation period to the ruin probability and to the expected total capital injection. Introduction
Parisian ruin has been actively studied since its introduction by Chesney et al. (1997). Incontrary to the classical ruin model of Cram´er-Lundberg in which case ruin/default occurs atthe first instance the underlying (surplus) process crossing below a threshold, Parisian ruin isannounced at the first time the process has undertaken an excursion below the default level for afixed consecutive period of time. It has been applied in finance, among others, by Francois andMorellec (2004), Broadie et al. (2007), and recently by Antill and Grenadier (2019), in particularfor studying firm’s optimal capital structure in the presence of Chapters 7 and 11 (default andreorganization proceeding). The default level may be determined endogenously in the sense ofLeland and Toft (1996) by maximizing the firm’s equity value. Under Chapter 11 the firm isgranted a dilution period during which the firm is given the opportunity to operate and reorganizeitself until its asset value goes above the default level, or otherwise liquidated. In their recent work,Palmowski et al. (2020) revisit the Leland-Toft model under spectrally negative L´evy process,discussed earlier in Hilberink and Rogers (2002) and Kyprianou and Surya (2007), by consideringinformation of the firm’s asset only available periodically at Poisson time. They showed using theresults of Albrecher et al. (2016) that under Poisson observation of the firm’s assets, nonzero creditspreads holds for firm with lower initial endowment than default level, and in particular, the defaulttime corresponds to the Parisian ruin time with an independent exponential excursion period.
Mathematics Subject Classification.
Primary: 60G51; Secondary: 60E10, 60J35.
Key words and phrases.
Spectrally negative L´evy process, reflected process, draw-down time, potential measure,excursion theory, risk process, Parisian ruin, capital injection. a r X i v : . [ q -f i n . M F ] M a y BUDHI SURYA, WENYUAN WANG, XIANGHUA ZHAO AND XIAOWEN ZHOU
It was introduced to insurance/actuarial science literature started by the work of Dassios andWu (2008) for compound Poisson process, then extended to spectrally negative L´evy process sub-sequently by Czarna and Palmowski (2011), Loeffen et al. (2013), among others. Further distri-butional identities concerning Parisian ruin with independent exponential excursion period werediscussed in Baurdoux et al. (2016). The results are generalized to fixed excursion period for aclass of penalty functions by Loeffen et al. (2018) which identify known results on Parisian ruin.It is worth mentioning that the Parisian excursion discussed in the above literature gets startedat the first passage below a threshold (default level) of the underlying process. In the past decadessome discussions have been developed towards risk protection mechanism against certain financialassets’ outperformance over their last record maximum, also referred to as high-water mark ordraw-down. We refer interested readers to the works by Goetzmann et al. (2003) and Agarwalet al. (2009). Default is triggered when the underlying process has gone below a specified levelfrom its last record maximum. Distributional identities regarding first-passage above a thresholdfor draw-down process were presented in Avram et al. (2004, 2007) and used for pricing Russianoptions under randomized maturity, and solving optimal dividend problem where the cumulativepaid dividend is given by the running supremum of the surplus process. First-passage identities fordraw-down process were later extended to a more general form of threshold boundary for e.g. byZhou (2007), Wang and Zhou (2018), and Li et al. (2019). Parisian excursion below a fixed levelfrom the last record maximum of the surplus process was considered in Surya (2019).The introduction of capital injection to the firm may prevent the firm from going default at thetime its asset value decreases below a certain threshold and the firm needs to meet its commitmentto pay dividends to the shareholders. We refer among others to Kulenko and Schmidli (2008), Taoet al. (2011), Avanzi et al. (2011), Bayraktar et al. (2013), and Wang et al. (2019).In this paper we study Parisian ruin from the last record maximum of surplus process with capitalinjection. We assume there is no dividend payment in the model. As the source of uncertainty inthe surplus process is the downward jumps of the L´evy insurance risk process. Capital is providedto the firm by the stakeholder as soon as the surplus process goes below a given function of thelast record maximum of the surplus process. It is continuously provided to the firm to keep thesurplus above the draw-down level until the surplus goes back to above the last record or the ruinoccurs. If the surplus process has stayed below the record since the first capital injection longerthan an independent exponential period of time, the firm faces the credit event and is liquidated.Distributional identities concerning the Parisian ruin are presented.The rest of the paper is arranged as follows. In Section 2 we introduce the model and formulatethe problems we are interested. The main results and proofs are presented in Section 3. Section 4presents some numerical examples of the main results. In particular, they are presented to analyzethe impact of observation frequency and initial value of surplus to various shapes of ruin probabilityand expected nett present value of the total capital injection. Section 5 concludes this paper.2.
Spectrally negative L´evy process and its reflected processes
Write X ≡ { X ( t ); t ≥ } , defined on a probability space (Ω , F , ( F t ) t ≥ , P ) with probability laws { P x ; x ∈ R } denoting the family of probability laws of X such that X = x , with P = P , and naturalfiltration {F t ; t ≥ } of X satisfying the usual conditions of right-continuity and completeness. Inparticular, we exclude a spectrally negative L´evy process that is not a purely increasing linear driftor the negative of a subordinator. The L´evy-Itˆo sample paths decomposition of the L´evy processis given by X t = µt + σB t + (cid:90) t (cid:90) { x< − } x N (d x, d s ) + (cid:90) t (cid:90) {− ≤ x< } x (cid:0) N (d x, d s ) − ν (d x )d s (cid:1) , (1) ARISIAN EXCURSION FOR DRAW-DOWN REFLECTED L´EVY INSURANCE RISK PROCESS 3 where µ ∈ R , σ ≥ B t ) t ≥ is standard Brownian motion, whilst N (d x, d t ) denotes the Poissonrandom measure associated with the jumps process ∆ X t := X t − X t − of X . This Poisson randommeasure has compensator given by ν (d x )d t , where ν is the L´evy measure satisfying the condition:(2) (cid:90) −∞ (1 ∧ x ) ν (d x ) < ∞ . In the expression (1), X may define the surplus process of an insurance firm in which µ representsthe premium rate charged on the insurance holder and σB t is of volatile trading uncertainties whichresults from the firm investing in financial market. The other two jump components correspondto the compensated small claims and uncompensated big claims from the insurance holder whosedistribution is given by the L´evy measure ν satisfying the integral constraint (2). In the classicalmodel of Cramer-Lundberg risk process with positive drift c > c = µ − (cid:82) −∞ x { x> − } ν (d x ) with ν (d x ) = βF (d x ) for Poisson claim arrival intensity β > F of the claim size.Denote the running supremum process X ≡ { sup ≤ s ≤ t X ( s ) , t ≥ } with X (0) = x under P x . Givena value a ∈ R , the process X reflected from below at the level a is defined as X ( t ) − ( X ( t ) − a ) ∧ , t ≥ X ( t ) := inf ≤ s ≤ t X ( s ) with X (0) = x under P x , denotes the running infimum process. Let { Y ( t ) , t ≥ } be the process X reflected from below at the level 0 (cf., Pistorius (2004)).The draw-down time associated to a draw-down function ξ on ( −∞ , ∞ ) satisfying ξ ( x ) < x, x ∈ ( −∞ , ∞ ), the ξ -draw-down time in short, is defined as τ ξ ≡ τ ξ ( X ) := inf { t ≥ X ( t ) < ξ ( X ( t )) } with the convention inf ∅ := ∞ . We define the process X reflected at the ξ -draw-down time τ ξ as X ( t ) − [ τ ξ , ∞ ) ( t ) (cid:18) inf τ ξ ≤ s ≤ t X ( s ) − ξ ( X ( τ ξ )) (cid:19) ∧ , t ≥ , where we call ξ ( X ( τ ξ )) the draw-down level at the draw-down time τ ξ .We now define the draw-down reflected process U for X . Intuitively, the process U initially agreeswith X until the first draw-down time of U . Then it starts to evolve according to X reflected atthe draw-down level until the next draw-down time of U when it is reflected at the draw-down levelagain, and so on. Then given that U ( s ) = U ( s ) := sup ≤ t ≤ s U ( t ), the process { U ( t ); t ≥ s } evolveswithout reflection until the next draw-down time τ ξ ; and given that U ( s ) > U ( s ), the process { U ( t ); t ≥ s } is reflected from below at the current draw-down level ξ ( U ( s )) until it comes back tothe level U ( s ). Note that the process U is not a Markov process in general, but the process ( U, U )is Markovian. Write P x,y and E x,y for the law of ( U, U ) such that U (0) = x and U (0) = y . Forsimplicity, denote P x = P x,x and E x = E x,x .To be more precise, define T := 0 and U ( T ) := X (0). Suppose first that for n ≥ U ( t ) hasbeen defined on [0 , T n ] for T n < ∞ , n ≥
1. Let X n +1 be an independent copy of X starting at U ( T n ) and U n +1 be the process X n +1 reflected at its ξ -draw-down time τ ξ ( X n +1 ). If τ ξ ( X n +1 ) = ∞ ,let T n +1 := ∞ , and if τ ξ ( X n +1 ) < ∞ , let T n +1 := T n + inf { t ≥ U n +1 ( t ) > X n +1 ( τ ξ ( X n +1 )) } , where X n +1 ( t ) := sup ≤ s ≤ t X n +1 ( s ). Observe that T n +1 < ∞ if τ ξ ( X n +1 ) < ∞ . Then define U ( T n + t ) := U n +1 ( t ) for t ∈ [0 , T n +1 − T n ) and U ( T n +1 ) := U n +1 ( T n +1 − T n ) if T n +1 < ∞ . Suppose now that U ( t ) has been defined on [0 , T n = ∞ ) for n ≥
0. For convenience, let T n +1 := ∞ .For the well-definedness of the process U , we are referred to Wang and Zhou (2019). BUDHI SURYA, WENYUAN WANG, XIANGHUA ZHAO AND XIAOWEN ZHOU
For the process X , define its first up-crossing time of level b ∈ ( −∞ , ∞ ) and first down-crossingtime of level c ∈ ( −∞ , ∞ ), respectively, by τ + b := inf { t ≥ X ( t ) > b } and τ − c := inf { t ≥ X ( t ) < c } . For the processes Y and U , their first up-crossing times of b ∈ ( −∞ , ∞ ) are defined respectivelyby σ + b := inf { t ≥ Y ( t ) > b } and κ + b := inf { t ≥ U ( t ) > b } . The Parisian ruin time θ λ of the process U is defined as N λ := inf { n ≥ T n − T n − − τ ξ ( X n ) > e ( n ) λ } , θ λ := T N λ − + τ ξ ( X N λ ) + e ( N λ ) λ , where { e ( n ) λ ; n ≥ } is a sequence of i.i.d. exponential randoms with parameter λ >
0, and isindependent of X .Due to the absence of positive jumps, it is therefore sensible to define ψ ( θ ) := ln E x (cid:16) e θ ( X − x ) (cid:17) = µθ + 12 σ θ + (cid:90) ( −∞ , (cid:16) e θx − − θx ( − , ( x ) (cid:17) ν (d x ) , It is known that ψ ( θ ) is finite for θ ∈ [0 , ∞ ) in which case it is strictly convex and infinitelydifferentiable. As in Bertoin (1996), the q -scale functions { W q ; q ≥ } of X are defined as follows.For each q ≥ W q : [0 , ∞ ) → [0 , ∞ ) is the unique strictly increasing and continuous function withLaplace transform (cid:90) ∞ e − θx W q ( x )d x = 1 ψ ( θ ) − q , for θ > Φ( q ) , (3)where Φ( q ) is the largest solution of the equation ψ ( θ ) = q . Further define W q ( x ) = 0 for x < W for the 0-scale function W .It is known that W q (0) = 0 if and only if process X has sample paths of unbounded variation. If X has sample paths of unbounded variation, or if X has sample paths of bounded variation and theL´evy measure has no atoms, then the scale function W q is continuously differentiable over (0 , ∞ ).By Loeffen (2008), if X has a L´evy measure which has a completely monotone density, then W q is twice continuously differentiable over (0 , ∞ ) when X is of unbounded variation. Moreover, ifprocess X has a nontrivial Gaussian component, then W q is twice continuously differentiable over(0 , ∞ ). Below is an example of the scale function W q for downward jump-diffusion process withexponentially distributed jumps, which will be used for numerical examples discussed in Section 4. Example 2.1.
Consider one-sided jump-diffusion process X with ψ ( θ ) = µθ + σ θ − aθθ + c for all θ ∈ R s.t. θ (cid:54) = − c . It is known that the inverse of the Laplace transform (3) for q > W q ( x ) = e − β x ψ (cid:48) ( − β ) + e − β x ψ (cid:48) ( − β ) + e Φ( q ) x ψ (cid:48) (Φ( q )) , ∀ x ≥ , (4)where − β , − β , and Φ( q ) denotes three roots of ψ ( θ ) = q s.t. − β < − c < − β < < Φ( q ). It isstraightforward to check by taking Laplace transform that W q ( x ) = e Φ( q ) x W Φ( q ) ( x ) where W Φ( q ) ( x ) = e − ( β +Φ( q )) x ψ (cid:48) ( − β ) + e − ( β +Φ( q )) x ψ (cid:48) ( − β ) + 1 ψ (cid:48) (Φ( q )) , ∀ x ≥ , (5)with W Φ( q ) ( x ) = 0 for x <
0. It is known that ψ (cid:48) ( − β ) < ψ (cid:48) ( − β ) < ψ (cid:48) (Φ( u )) > W Φ( q ) ( x ) plays the role of W ( x ) under the Esscher transform of measure P Φ( q ) defined by d P Φ( q ) d P (cid:12)(cid:12)(cid:12) F t = e Φ( q ) X t − qt . It is straight forward to check that W Φ( q ) ( x ) is increasing for x ≥
0, and sois W q ( x ), concave and is bounded from above by 1 /ψ (cid:48) (Φ( q )). ARISIAN EXCURSION FOR DRAW-DOWN REFLECTED L´EVY INSURANCE RISK PROCESS 5 x W q ( x ) (a) W q ( x ) for σ = 0 . x W ( x ) (b) W Φ( q ) ( x ) for σ = 0 . x W q ( x ) (c) W q ( x ) for σ = 0. x W ( x ) (d) W Φ( q ) ( x ) for σ = 0. Figure 1.
Scale function W q ( x ) and W Φ( q ) ( x ) for downward jump-diffusion processwith Laplace exponent ψ ( θ ) = µθ + σ θ − aθθ + c for µ = 0 . , a = 0 . , c = 9 , q = 0 . Z q ( x ) := 1 + q (cid:90) x W q ( z )d z, x ≥ , and Z q ( x, θ ) := e θx (cid:18) − ( ψ ( θ ) − q ) (cid:90) x e − θz W q ( z )d z (cid:19) , θ ≥ , x ≥ , with Z ( x, θ ) := Z ( x, θ ), and W q ( x ) := (cid:90) x W q ( z )d z, q ≥ , x ≥ , and Z q ( x ) := (cid:90) x Z q ( z )d z = x + q (cid:90) x (cid:90) z W q ( w )d w d z, q ≥ , x ≥ . BUDHI SURYA, WENYUAN WANG, XIANGHUA ZHAO AND XIAOWEN ZHOU
In the sequel, without loss of generality we assume X ≡ X . By Li et al. (2017), we have E x (e − qκ + b { κ + b <τ ξ } ) = E x (cid:16) e − qτ + b { τ + b <τ ξ } (cid:17) = exp (cid:32) − (cid:90) bx W (cid:48) q ( ξ ( z )) W q ( ξ ( z )) d z (cid:33) , (6)where ξ ( z ) = z − ξ ( z ). For x ∈ [0 , b ] and q ≥
0, from Proposition 2 in Pistorius (2004) we have E x (e − qσ + b ) = Z q ( x ) Z q ( b ) . (7)By Kyprianou (2006), the resolvent measure corresponding to X is absolutely continuous withrespect to the Lebesgue measure with density given by (cid:90) ∞ e − qt P x ( X ( t ) ∈ d y ; t < τ − c ∧ τ + b )d t = (cid:18) W q ( x − c ) W q ( b − c ) W q ( b − y ) − W q ( x − y ) (cid:19) ( c,b ) ( y )d y, (8)for x ∈ ( c, b ). By Pistorius (2004), the resolvent measure corresponding to Y is also absolutelycontinuous with respect to the Lebesgue measure and has a version of density given by (cid:90) ∞ e − qt P x ( Y ( t ) ∈ d y, t < σ + b )d t = (cid:18) Z q ( x ) Z q ( b ) W q ( b − y ) − W q ( x − y ) (cid:19) [0 ,b ) ( y )d y, (9)where x ∈ [0 , b ).Define the total amount of capital injections made until time t for the draw-down reflected processas R ( t ) := − N − (cid:88) k =1 [ T k − + τ ξ ( X k ) , ∞ ) ( t ) (cid:18) inf τ ξ ( X k ) ≤ s ≤ T k ∧ t − T k − X k ( s ) − ξ ( X k ( τ ξ ( X k ))) (cid:19) ∧ . where N := inf { n : T n = ∞} = inf { n : τ ξ ( X n ) = ∞} .In this paper, we are interested in evaluating:(a) Expectation of the net present value of principal payment of one unit at time κ + b ∧ θ λ : U ξ ( x ; b ) = E x (cid:16) e − q ( κ + b ∧ θ λ ) (cid:17) . (b) The joint Laplace transform of κ + b ∧ θ λ , U ( κ + b ∧ θ λ ) and R ( κ + b ∧ θ λ ), i.e. G ξ ( x ; b ) = E x (cid:16) e − q ( κ + b ∧ θ λ ) + uU ( κ + b ∧ θ λ ) − vR ( κ + b ∧ θ λ ) (cid:17) , q, λ, u, v ∈ R + , b ∈ R , x ∈ ( −∞ , b ] . (c) The potential measure of U involving the Parisian ruin time, i.e. (cid:90) ∞ e − qt P x (cid:0) U ( t ) ∈ d u, t < κ + b ∧ θ λ (cid:1) d t, q, λ ∈ R + , b ∈ R , x, u ∈ ( −∞ , b ] . (d) The expectation of the total discounted capital injections until κ + b ∧ θ λ , i.e. V ξ ( x ; b ) = E x (cid:32)(cid:90) κ + b ∧ θ λ e − qt d R ( t ) (cid:33) , q, λ ∈ R + , b ∈ R , x ∈ ( −∞ , b ] . ARISIAN EXCURSION FOR DRAW-DOWN REFLECTED L´EVY INSURANCE RISK PROCESS 7
We also briefly recall concepts in excursion theory for the reflected process { X ( t ) − X ( t ); t ≥ } ,and we refer to Bertoin (1996) for more details. For x ∈ ( −∞ , ∞ ), the process { L ( t ) := X ( t ) − x, t ≥ } serves as a local time at 0 for the Markov process { X ( t ) − X ( t ); t ≥ } under P x . Let thecorresponding inverse local time be defined as L − ( t ) := inf { s ≥ L ( s ) > t } = sup { s ≥ L ( s ) ≤ t } . Further let L − ( t − ) := lim s ↑ t L − ( s ). Define a Poisson point process { ( t, e t ); t ≥ } as e t ( s ) := X ( L − ( t )) − X ( L − ( t − ) + s ) , s ∈ (0 , L − ( t ) − L − ( t − )] , whenever the lifetime of e t is positive, i.e. L − ( t ) − L − ( t − ) >
0. Whenever L − ( t ) − L − ( t − ) = 0,define e t := Υ with Υ being an additional isolated point. A result of Itˆo states that e is aPoisson point process with characteristic measure n if { X ( t ) − X ( t ); t ≥ } is recurrent; otherwise { e t ; t ≤ L ( ∞ ) } is a Poisson point process stopped at the first excursion of infinite lifetime. Here, n is a measure on the space E of excursions, i.e. the space E of c`adl`ag functions f satisfying f : (0 , ζ ) → (0 , ∞ ) for some ζ = ζ ( f ) ∈ (0 , ∞ ] ,f : { ζ } → (0 , ∞ ) if ζ < ∞ , where ζ = ζ ( f ) is the excursion length or lifetime; see Definition 6.13 of Kyprianou (2006) for thedefinition of E . Denote by ε ( · ), or ε for short, a generic excursion belonging to the space E ofcanonical excursions. The excursion height of a canonical excursion ε is denoted by ε = sup t ∈ [0 ,ζ ] ε ( t ).The first passage time of a canonical excursion ε is defined by ρ + b ≡ ρ + b ( ε ) := inf { t ∈ [0 , ζ ] : ε ( t ) > b } , with the convention inf ∅ := ζ .Denote by ε g the excursion (away from 0) with left-end point g for the reflected process { X ( t ) − X ( t ); t ≥ } , and by ζ g and ε g the excursion’s lifetime and the excursion’s height, respectively; seeSection IV.4 of Bertoin (1996). 3. Main results
In this section we present the main results concerning the general draw-down reflected process U with Parisian stopping. For preparation, we recall the following result of Wang and Zhou (2019). Lemma 3.1.
Given θ, q ∈ (0 , ∞ ) and measurable function φ : ( −∞ , ∞ ) → ( −∞ , ∞ ) , we have E x (cid:16) e − qτ ξ e θX ( τ ξ ) φ (cid:0) X ( τ ξ ) (cid:1) ; τ ξ < τ + b (cid:17) = (cid:90) bx φ ( s ) e θξ ( s ) exp (cid:32) − (cid:90) sx W (cid:48) q ( ξ ( z )) W q ( ξ ( z )) d z (cid:33) × (cid:32) W (cid:48) q ( ξ ( s )) W q ( ξ ( s )) Z q ( ξ ( s ) , θ ) − θZ q ( ξ ( s ) , θ ) − ( q − ψ ( θ )) W q ( ξ ( s )) (cid:33) d s, x ∈ ( −∞ , b ] . (10) In particular, we have E x (cid:0) e − qτ ξ φ (cid:0) X ( τ ξ ) (cid:1) ; τ ξ < τ + b (cid:1) = (cid:90) bx φ ( s ) exp (cid:32) − (cid:90) sx W (cid:48) q ( ξ ( z )) W q ( ξ ( z )) d z (cid:33) × (cid:32) W (cid:48) q ( ξ ( s )) W q ( ξ ( s )) Z q ( ξ ( s )) − qW q ( ξ ( s )) (cid:33) d s, x ∈ ( −∞ , b ] , (11) BUDHI SURYA, WENYUAN WANG, XIANGHUA ZHAO AND XIAOWEN ZHOU and E x (cid:16) e θX ( τ ξ ) φ (cid:0) X ( τ ξ ) (cid:1) ; τ ξ < τ + b (cid:17) = (cid:90) bx φ ( s ) e θξ ( s ) exp (cid:18) − (cid:90) sx W (cid:48) ( ξ ( z )) W ( ξ ( z )) d z (cid:19) × (cid:18) W (cid:48) ( ξ ( s )) W ( ξ ( s )) Z ( ξ ( s ) , θ ) − θZ ( ξ ( s ) , θ ) + ψ ( θ ) W ( ξ ( s )) (cid:19) d s, x ∈ ( −∞ , b ] , (12) and E x (cid:0) e − qτ ξ (cid:0) ξ (cid:0) X ( τ ξ ) (cid:1) − X ( τ ξ ) (cid:1) ; τ ξ < τ + b (cid:1) = (cid:90) bx exp (cid:32) − (cid:90) sx W (cid:48) q ( ξ ( z )) W q ( ξ ( z )) d z (cid:33) × (cid:18) Z q ( ξ ( s )) − ψ (cid:48) (0+) W q ( ξ ( s )) − Z q ( ξ ( s )) − ψ (cid:48) (0+) W q ( ξ ( s )) W q ( ξ ( s )) W (cid:48) q ( ξ ( s )) (cid:19) d s, x ∈ ( −∞ , b ] . (13)We start with the Laplace transform of the upper exiting time for the process U . Proposition 3.1.
For q, λ ∈ (0 , ∞ ) we have E x (cid:16) e − qκ + b { κ + b <θ λ } (cid:17) = exp (cid:18) − (cid:90) bx (cid:96) ( w )d w (cid:19) , x ∈ ( −∞ , b ] , (14) where (cid:96) ( w ) = W (cid:48) q ( ξ ( w )) W q ( ξ ( w )) (cid:32) − Z q ( ξ ( w )) Z q + λ ( ξ ( w )) (cid:33) + qW q ( ξ ( w )) Z q + λ ( ξ ( w )) . Proof:
Denote by f ( x ) the left hand side of (14). We have f ( x ) = E x (cid:16) e − qκ + b { κ + b <τ ξ } (cid:17) + E x (cid:16) e − qκ + b { τ ξ <κ + b <θ λ } (cid:17) , x ∈ ( −∞ , b ] . (15)Note that by definition, τ ξ < κ + b implies X ( τ ξ ) < b which further implies T < κ + b . Hence, takinguse of (7) and (11), we get for x ∈ ( −∞ , b ] E x (cid:16) e − qκ + b { τ ξ <κ + b <θ λ } (cid:17) = E x (cid:16) e − qκ + b { τ ξ Taking derivative on both sides of (17) with respect to x , we have f (cid:48) ( x ) = W (cid:48) q ( ξ ( x )) W q ( ξ ( x )) f ( x ) − f ( x ) Z q + λ ( ξ ( x )) (cid:32) W (cid:48) q ( ξ ( x )) W q ( ξ ( x )) Z q ( ξ ( x )) − qW q ( ξ ( x )) (cid:33) = (cid:96) ( x ) f ( x ) , x ∈ ( −∞ , b ] . (18)Solving (18) we obtain for x ∈ ( −∞ , b ] f ( x ) = C exp (cid:18) − (cid:90) bx (cid:96) ( w )d w (cid:19) , (19)for some constant C . The boundary condition f ( b ) = 1 together with (19) yields (14). (cid:3) Proposition 3.2. For q, λ ∈ (0 , ∞ ) , we have for x ∈ ( −∞ , b ] , E x (cid:16) e − qθ λ { θ λ <κ + b } (cid:17) = (cid:90) bx exp (cid:16) − (cid:90) yx (cid:96) ( w ) dw (cid:17) (cid:96) ( y )d y, (20) where (cid:96) ( x ) is given in (14), while the function (cid:96) ( x ) is defined by (cid:96) ( x ) = λq + λ (cid:32) − Z q + λ ( ξ ( x )) (cid:33) (cid:32) W (cid:48) q ( ξ ( x )) W q ( ξ ( x )) Z q ( ξ ( x )) − qW q ( ξ ( x )) (cid:33) . Proof. Denote by f ξ ( x ; b ) the left hand side of (20). We have f ξ ( x ; b ) = E x (cid:16) e − qθ λ { θ λ <κ + b ,τ ξ <τ + b } (cid:17) = E x (cid:16) e − qθ λ { θ λ <κ + b ,τ ξ <τ + b ,T − τ ξ >e λ } (cid:17) + E x (cid:16) e − qθ λ { θ λ <κ + b ,τ ξ <τ + b ,T − τ ξ ≤ e λ } (cid:17) . (21)By the strong Markov property of the bi-variate process ( U, U ) one can obtain E (cid:16) e − qθ λ { θ λ <κ + b ,τ ξ <τ + b ,T − τ ξ >e λ } (cid:12)(cid:12)(cid:12) F τ ξ (cid:17) = e − qτ ξ { τ ξ <τ + b } (cid:34) E (cid:16) e − qe λ { e λ <σ + z } (cid:17)(cid:12)(cid:12)(cid:12) z = ξ ( X τξ ) (cid:35) = e − qτ ξ { τ ξ <τ + b } λq + λ (cid:16) − (cid:104) E (cid:16) e − ( q + λ ) σ + z (cid:17)(cid:12)(cid:12)(cid:12) z = ξ ( X τξ ) (cid:105)(cid:17) = e − qτ ξ { τ ξ <τ + b } λq + λ (cid:16) − Z q + λ ( ξ ( X τ ξ )) (cid:17) , (22)and E (cid:16) E (cid:16) e − qθ λ { θ λ <κ + b ,τ ξ <τ + b ,T − τ ξ ≤ e λ } (cid:12)(cid:12)(cid:12) F T (cid:17)(cid:12)(cid:12)(cid:12) F τ ξ (cid:17) = e − qτ ξ { τ ξ <τ + b } (cid:34) E (cid:16) e − qσ + z { σ + z ≤ e λ } (cid:17)(cid:12)(cid:12)(cid:12) z = ξ ( X τξ ) (cid:35) E X τξ (cid:16) e − qθ λ { θ λ <κ + b } (cid:17) = e − qτ ξ { τ ξ <τ + b } f ξ ( X τ ξ ; b ) Z q + λ ( ξ ( X τ ξ )) . (23)Following the two identities (22) and (23), we obtain from (21) and (11) the equation f ξ ( x ; b ) = λq + λ E x (cid:32) e − qτ ξ { τ ξ <τ + b } (cid:16) − Z q + λ ( ξ ( X τ ξ )) (cid:17)(cid:33) + E (cid:32) e − qτ ξ { τ ξ <τ + b } f ξ ( X τ ξ ; b ) Z q + λ ( ξ ( X τ ξ )) (cid:33) = λq + λ (cid:90) bx (cid:104) − Z q + λ ( ξ ( s )) (cid:105) exp (cid:32) − (cid:90) sx W (cid:48) q ( ξ ( z )) W q ( ξ ( z )) dz (cid:33) (cid:32) W (cid:48) q ( ξ ( s )) W q ( ξ ( s )) Z q ( ξ ( s )) − qW q ( ξ ( s )) (cid:33) d s + (cid:90) bx f ξ ( s ; b ) Z q + λ ( ξ ( s )) exp (cid:32) − (cid:90) sx W (cid:48) q ( ξ ( z )) W q ( ξ ( z )) dz (cid:33) (cid:32) W (cid:48) q ( ξ ( s )) W q ( ξ ( s )) Z q ( ξ ( s )) − qW q ( ξ ( s )) (cid:33) d s. After taking partial derivative w.r.t x on both sides, we have by the Leibniz integral rule, f (cid:48) ξ ( x ; b ) = − λq + λ (cid:104) − Z q + λ ( ξ ( x )) (cid:105) (cid:32) W (cid:48) q ( ξ ( x )) W q ( ξ ( x )) Z q ( ξ ( x )) − qW q ( ξ ( x )) (cid:33) − f ξ ( x ; b ) Z q + λ ( ξ ( x )) (cid:32) W (cid:48) q ( ξ ( x )) W q ( ξ ( x )) Z q ( ξ ( x )) − qW q ( ξ ( x )) (cid:33) + W (cid:48) q ( ξ ( x )) W q ( ξ ( x )) f ξ ( x ; b )= (cid:96) ( x ) f ξ ( x ; b ) − (cid:96) ( x ) . (24)Identity (20) follows by solving the differential equation (24) subject to the boundary condition f ξ ( b ; b ) = 0. (cid:3) Corollary 3.1 (Parisian Ruin probability) . For λ ∈ (0 , ∞ ) , the Parisian ruin probability is givenby P x (cid:0) θ λ < ∞ ) = 1 − exp (cid:16) − (cid:90) ∞ x (cid:96) ( w ; λ ) dw (cid:17) , x ∈ R , (25) with (cid:96) ( x ; λ ) = W (cid:48) ( ξ ( x )) W ( ξ ( x )) (cid:18) − Z λ ( ξ ( x )) (cid:19) . Theorem 3.1. For q, λ ∈ (0 , ∞ ) , we have for x ∈ ( −∞ , b ] that U ξ ( x ; b ) = (cid:90) bx exp (cid:16) − (cid:90) yx (cid:96) ( w ) dw (cid:17) (cid:96) ( y )d y + exp (cid:16) − (cid:90) bx (cid:96) ( w ) dw (cid:17) . (26) Proof. The proof follows from combining the two results (14) and (20). (cid:3) The next result gives an expression of the joint Laplace transform concerning κ + b ∧ θ λ . Theorem 3.2. For any q, u, v, λ ∈ (0 , ∞ ) with u ∈ (0 , Φ q + λ ) , we have G ξ ( x ; b ) = (cid:18) e ub − (cid:90) bx (cid:96) ( z ) exp (cid:18)(cid:90) bz (cid:96) ( w )d w (cid:19) d z (cid:19) exp (cid:18) − (cid:90) bx (cid:96) ( w )d w (cid:19) , x ∈ ( −∞ , b ] , (27) where (cid:96) ( w ) = W (cid:48) q ( ξ ( w )) W q ( ξ ( w )) (cid:32) − Z q ( ξ ( w ) , v ) Z q + λ ( ξ ( w ) , v ) (cid:33) + vZ q ( ξ ( w ) , v ) + ( q − ψ ( v )) W q ( ξ ( w )) Z q + λ ( ξ ( w ) , v ) ,(cid:96) ( w ) = − e uξ ( w ) (cid:126) ( ξ ( w )) (cid:32) W (cid:48) q ( ξ ( w )) W q ( ξ ( w )) Z q ( ξ ( w ) , v ) − vZ q ( ξ ( w ) , v ) − ( q − ψ ( v )) W q ( ξ ( w )) (cid:33) , (cid:126) ( w ) = λ Φ q + λ − u (cid:18) W q + λ (0+) + (cid:90) y< e uy +( v − u ) y (cid:0) W (cid:48) q + λ ( − y ) − Φ q + λ W q + λ ( − y ) (cid:1) d y − Z q + λ ( w, v ) (cid:18) W q + λ (0+) e uw ARISIAN EXCURSION FOR DRAW-DOWN REFLECTED L´EVY INSURANCE RISK PROCESS 11 + (cid:90) y Given q, u, v ∈ (0 , ∞ ) and x ∈ ( −∞ , b ], applying the Markov property of the process ( U, U )we have G ξ ( x ; b ) := E x (cid:16) e − q ( κ + b ∧ θ λ ) + uU ( κ + b ∧ θ λ ) − vR ( κ + b ∧ θ λ ) (cid:17) = E x (cid:16) E x (cid:16) e − q ( τ ξ + e (1) λ )+ uU ( τ ξ + e (1) λ ) − vR ( τ ξ + e (1) λ ) { τ ξ <κ + b , T − τ ξ >e (1) λ } (cid:12)(cid:12)(cid:12) F τ ξ (cid:17)(cid:17) + E x (cid:16) E x (cid:16) e − q ( κ + b ∧ θ λ ) + uU ( κ + b ∧ θ λ ) − vR ( κ + b ∧ θ λ ) { τ ξ <κ + b , T − τ ξ ≤ e (1) λ } (cid:12)(cid:12)(cid:12) F T (cid:17)(cid:17) + E x (cid:16) e − qτ + b + uX ( τ + b ) − vR ( τ + b ) { τ + b <τ ξ } (cid:17) = E x (cid:18) e − qτ ξ +( u − v ) ξ ( X ( τ ξ ))+ vX ( τ ξ ) { τ ξ <τ + b } (cid:18) E (cid:16) e − qe λ + uY ( e λ )+ v ( X ( e λ )) { σ + z >e λ } (cid:17)(cid:12)(cid:12)(cid:12) z = ξ ( X ( τ ξ )) (cid:19)(cid:19) + E x (cid:16) e − qτ ξ { τ ξ <κ + b } e − v ( ξ ( X ( τ ξ )) − X ( τ ξ )) G ξ ( X ( τ ξ ); b ) E x (cid:16) e − v ( R ( T ) − R ( τ ξ )) − ( q + λ )( T − τ ξ ) (cid:12)(cid:12)(cid:12) F τ ξ (cid:17)(cid:17) + E x (cid:16) e − qτ + b + uX ( τ + b ) − vR ( τ + b ) { τ + b <τ ξ } (cid:17) = E x (cid:18) e − qτ ξ +( u − v ) ξ ( X ( τ ξ ))+ vX ( τ ξ ) { τ ξ <τ + b } (cid:18) E (cid:16) e − qe λ + uY ( e λ )+ v ( X ( e λ )) { σ + z >e λ } (cid:17)(cid:12)(cid:12)(cid:12) z = ξ ( X ( τ ξ )) (cid:19)(cid:19) + E x (cid:18) e − qτ ξ { τ ξ <κ + b } e vX ( τ ξ ) e − vξ ( X ( τ ξ )) G ξ ( X ( τ ξ ); b ) (cid:18) E (cid:16) e v ( X ( σ + z )) − ( q + λ ) σ + z (cid:17)(cid:12)(cid:12)(cid:12) z = ξ ( X ( τ ξ )) (cid:19)(cid:19) +e ub E x (cid:16) e − qτ + b { τ + b <τ ξ } (cid:17) . (28)In addition, (cid:126) ( z ) := E (cid:16) e − qe λ + uY ( e λ )+ v ( X ( e λ ) ∧ { σ + z >e λ } (cid:17) = E (cid:16) e − qe λ + uY ( e λ )+ v ( X ( e λ ) ∧ (cid:17) − E (cid:16) e − qe λ + uY ( e λ )+ v ( X ( e λ ) ∧ { σ + z For q ∈ (0 , ∞ ) , the resolvent measure of U is absolutely continuous with respect tothe Lebesgue measure with density given by (cid:90) ∞ e − qt P x (cid:0) U ( t ) ∈ d u, t < κ + b ∧ θ λ (cid:1) d t = W q (0) exp (cid:18) − (cid:90) yx (cid:96) ( w )d w (cid:19) ( x,b ) ( u )d u + (cid:90) bx exp (cid:18) − (cid:90) yx (cid:96) ( w )d w (cid:19) (cid:96) ( y, u ) ( ξ ( y ) ,y ) ( u )d y d u, x, u ∈ ( −∞ , b ] , (34) where (cid:96) is defined as in Theorem 3.1, and (cid:96) ( y, u ) = (cid:32) W (cid:48) q ( ξ ( y )) W q ( ξ ( y )) Z q ( ξ ( y )) − qW q ( ξ ( y )) (cid:33) Z λ ( u − ξ ( y )) W q ( y − u ) Z λ ( ξ ( y )) Z q + λ ( ξ ( y ))+ W (cid:48) q ( y − u ) − W (cid:48) q ( ξ ( y )) W q ( ξ ( y )) W q ( y − u ) . Proof: Recall U ( t ) = sup s ∈ [0 ,t ] U ( s ) and let e q be an exponential random variable independent of X . For q > x ≤ b and any continuous, non-negative and bounded function h , let qg ( x ) := (cid:90) ∞ q e − qt E x (cid:0) h ( U ( t )); t < κ + b ∧ θ λ (cid:1) d t = E x (cid:16) h ( X ( e q )) { X ( e q ) For q ∈ (0 , ∞ ) , we have V ξ ( x ; b ) = (cid:90) bx exp (cid:18) − (cid:90) yx (cid:96) ( w )d w (cid:19) (cid:96) ( y )d y, x ∈ ( −∞ , b ] , (46) where (cid:96) is defined as in Theorem 3.1, and (cid:96) ( y ) = Z q ( ξ ( y )) − ψ (cid:48) (0+) W q ( ξ ( y )) − Z q ( ξ ( y )) − ψ (cid:48) (0+) W q ( ξ ( y )) W q ( ξ ( y )) W (cid:48) q ( ξ ( y ))+ Z q + λ ( ξ ( y )) − ψ (cid:48) (0+) W q + λ ( ξ ( y )) Z q + λ ( ξ ( y )) (cid:32) W (cid:48) q ( ξ ( y )) W q ( ξ ( y )) Z q ( ξ ( y )) − qW q ( ξ ( y )) (cid:33) . Proof: For q ∈ (0 , ∞ ) and x ∈ ( −∞ , b ], we have V ξ ( x ; b ) = E x (cid:32)(cid:90) κ + b ∧ θ λ e − qt d R ( t ) (cid:33) = E x (cid:16) e − qτ ξ { τ ξ <τ + b } (cid:0) ξ ( X ( τ ξ )) − X ( τ ξ ) (cid:1)(cid:17) + E x (cid:32) { τ ξ <τ + b } (cid:90) T ∧ θ λ τ ξ + e − qt d R ( t ) (cid:33) + E x (cid:32) { τ ξ <κ + b } (cid:90) κ + b ∧ θ λ T ∧ θ λ e − qt d R ( t ) (cid:33) =: V ( x ; b ) + V ( x ; b ) + V ( x ; b ) . (47)By (13), V ( x ; b ) can be expressed as V ( x ; b ) = (cid:90) bx exp (cid:32) − (cid:90) sx W (cid:48) q ( ξ ( z )) W q ( ξ ( z )) d z (cid:33) (cid:32) Z q ( ξ ( s )) − ψ (cid:48) (0+) W q ( ξ ( s )) − Z q ( ξ ( s )) − ψ (cid:48) (0+) W q ( ξ ( s )) W q ( ξ ( s )) W (cid:48) q ( ξ ( s )) (cid:33) d s. (48) By the Markov property for the reflected process ( U, U ), V ( x ; b ) = E x (cid:32) { τ ξ <τ + b , e (1) λ ≥ T − τ ξ } (cid:90) T τ ξ + e − qt d R ( t ) (cid:33) + E x (cid:32) { τ ξ <τ + b , e (1) λ Denote by V (cid:48) ξ ( x ; b ) the derivative of V ξ ( x ; b ) with respect to its first argument. Combining (47),(48), (50) and (51) we have V (cid:48) ξ ( x ; b ) = W (cid:48) q ( ξ ( x )) W q ( ξ ( x )) V ξ ( x ; b ) − V ξ ( x ; b ) Z q + λ ( ξ ( x )) (cid:32) W (cid:48) q ( ξ ( x )) W q ( ξ ( x )) Z q ( ξ ( x )) − qW q ( ξ ( x )) (cid:33) − (cid:18) Z q ( ξ ( x )) − ψ (cid:48) (0+) W q ( ξ ( x )) − Z q ( ξ ( x )) − ψ (cid:48) (0+) W q ( ξ ( x )) W q ( ξ ( x )) W (cid:48) q ( ξ ( x )) (cid:19) − − ψ (cid:48) (0+) q + λ + Z q + λ ( ξ ( x )) + ψ (cid:48) (0+) q + λ Z q + λ ( ξ ( x )) (cid:32) W (cid:48) q ( ξ ( x )) W q ( ξ ( x )) Z q ( ξ ( x )) − qW q ( ξ ( x )) (cid:33) = (cid:96) ( x ) V ξ ( x ; b ) − (cid:96) ( x ) . (52)Solving (52) with boundary condition V ξ ( b, b ) = 0, we obtain (46). (cid:3) Numerical examples To exemplify the main results, we discuss some numerical examples for one-sided jump-diffusionprocess X with Laplace exponent ψ ( θ ) = µθ + σ θ − aθθ + c for all θ ∈ R s.t. θ (cid:54) = − c . See Example2.1 for the corresponding scale function. We set µ = 0 . a = 0 . c = 9 (on average onceevery two years the firm suffers an instantaneous loss of 10% of its value), and q = 5%.Figures 2 and 3 display various shapes of ruin probability P x ( θ λ < ∞ ) (25) and the expectednett present value V ξ ( x ) := V ξ ( x ; b = ∞ ) of the total amount of required capital injection to thefirm as a function of initial surplus x and the default monitoring frequency λ . The computationwas performed for different forms of drawdown function: ξ ( x ) = Kx , with 0 < K < 1, and ξ ( x ) = min { , Kx } . The former dictates that default is announced and followed by capital injectionas soon as the surplus process has crossed below K % of its last record high, whereas the latterdeals with the case of injecting capital to the firm as soon as the surplus is less than one dollar orbelow K % of the last record high, whichever is smaller. The results are presented for two cases: σ = 0 . σ = 0 (paths with bounded variation).Over all, we observe that the ruin probability P x ( θ λ < ∞ ) decreases as the initial surplus in-creases. This implies that firms with higher initial endowment/surplus has lower probability of ruinthan those with lower value of surplus at the beginning. Furthermore, when σ (cid:54) = 0 in which case thefirm has additional (immediate) exposure to risk, say from investing in financial market, the ruinprobability is higher than those firms which do not have any risk exposure ( σ = 0) other than theclaim from the insurance holder, which arrives at exponential random time. From the sample pathspoint of view, the presence of Brownian motion allows the paths to reach a new maximum leveland then makes an excursion below that level before a jump arrives. As a result, such movementtriggers the observation clock start to run and put the firm into a risky position of getting default.Moreover, we also notice from the figure that the higher the observation frequency λ , the higher theruin probability. This is to say that the larger λ gives the firm lesser time to come out of dilutionperiod during which the firm is in financial distress, resulting in a higher chance of going ruin.The choice of drawdown level ξ ( x ) also determines the shape of the ruin probability. Under thedrawdown level Kx , the ruin probability is higher for larger value of K . This is due to the fact thatthe higher value of K sets the default level higher causing the firm to go default/ruin sooner thanlower value of K , in particular when σ (cid:54) = 0. As min { , Kx } ≤ Kx , default is expected to occurearlier under drawdown level ξ ( x ) = Kx than ξ ( x ) = min { , Kx } for the same reason explained.For the latter, the ruin probability decreases linearly when the surplus is about less than one unit,then decreases at exponential rate when the surplus is larger than that value. Similar observation is observed for σ = 0 with only exception that the curve has lower degree of smoothness than thecase σ (cid:54) = 0 in which case the scale function is twice continuously differentiable over (0 , ∞ ).Similar feature is exhibited by the expected total amount of discounted cash for capital injectionfunction V ξ ( x ). Healthier firm with larger initial surplus x requires less capital injection in totalthan that of unhealthier firm with lower surplus. When the firm has more uncertainties as a resultof investing in financial market, the case σ (cid:54) = 0, the firm requires more capital injection thanthose firm which do not have riskier exposure ( σ = 0) other than the claim from the insuranceholder arriving in exponential random time. As explained above, the presence of Brownian motionputs the firm in riskier situation with higher probability of ruin when the surplus process makesan excursion from its last record high. This induces the firm to ask for more financial coverage(injection) to deal with during the dilution period. Moreover, the longer the observation window(the lower the value of λ ), the firm receives more capital injection from stakeholder than otherwise.The choice of drawdown level ξ ( x ) also shifts the curve ξ ( x ). For ξ ( x ) = Kx , the curve is higher forlarger value of K . The upward shift of the curve is attributed by the fact that the larger value of λ increases the default threshold to higher level giving the firm higher chance of default resulting thefirm in asking for more capital to prevent ruin to occur at earlier stage, in particular when σ (cid:54) = 0.As min { , Kx } ≤ Kx , default is expected to occur later for ξ ( x ) = min { , Kx } than that of underthe drawdown level ξ ( x ) = Kx which in turn requiring lesser amount of capital injection. Under ξ ( x ) = min { , Kx } , the function V ξ ( x ) decreases linearly with quite high slope (rate) when thesurplus is about less than one unit, then decreases exponentially at lower rate when the surplus islarger than that value. Overall, unlike the ruin probability P x ( θ λ < ∞ ), the function V ξ ( x ) retainsits convexity in all cases having zero value at infinity implying the function to be positive for all x .The above analyses conclude our numerical study on the ruin probability and nett present valueof the capital injection which summarizes their various shape w.r.t changing the value of observationfrequency λ and initial surplus x . 5. Conclusion This paper presents some distributional identities concerning excursion below the last record highof surplus process, driven by downward jumps L´evy process, with capital injection. Capital injectionis provided to the firm as soon as the process goes below a drawdown level and is continuously paiduntil the process goes above the record or ruin occurs, which is announced at the first time theprocess has undertaken an excursion below the record high longer than an independent exponentialperiod of time. Identities are given explicitly in terms of the scale function of the L´evy process.The latter makes possible to have a fast numerical computation of the identities and do analysison the impact of observation frequency and initial surplus to the ruin probability and the expectednett present value of the required total capital injection. The choice of some drawdown functionswas made to study various shapes of the ruin probability and the nett present value of the capitalinjection. Numerical study shows that the results implied by the model is found to be consistentwith an observation one would have in financial market. We leave this for further research. Acknowledgements Wenyuan Wang and Xianghua Zhao thank Concordia University where the first draft of this paperwas finished during their visits. Wenyuan Wang acknowledges the support of the National NaturalScience Foundation of China (No. 11601197). Wenyuan Wang and Xiaowen Zhou are supported bya National Sciences and Engineering Research Council of Canada grant (No. RGPIN-2016-06704).Xiaowen Zhou is supported by National Natural Science Foundation of China (No. 11771018). ARISIAN EXCURSION FOR DRAW-DOWN REFLECTED L´EVY INSURANCE RISK PROCESS 21 . . . . . . . . x f ( x , l ) l = l = l = l = (a) ξ ( x ) = 0 . x and σ = 0 . . . . . . . x f ( x , l ) l = l = l = l = (b) ξ ( x ) = 0 . x and σ = 0. . . . . . . x f ( x , l ) K = = = = (c) ξ ( x ) = Kx , λ = 0 . 2, and σ = 0 . . . . . . . x f ( x , l ) K = = = = (d) ξ ( x ) = Kx , λ = 0 . 2, and σ = 0. . . . . . x f ( x , l ) l = l = l = l = (e) ξ ( x ) = min(1 , . x ) and σ = 0 . . . . . . x f ( x , l ) l = l = l = l = (f) ξ ( x ) = min(1 , . x ) and σ = 0. Figure 2. Ruin probability P x ( θ λ < ∞ ) for downward jump-diffusion process withLaplace exponent ψ ( θ ) = µθ + σ θ − aθθ + c for µ = 0 . , a = 0 . , c = 9 , q = 0 . . . . . x V ( x , l ) l = l = l = (a) ξ ( x ) = 0 . x and σ = 0 . . . . . . . x V ( x , l ) l = l = l = (b) ξ ( x ) = 0 . x and σ = 0. . . . . x V ( x , l = . ) K = = = = (c) ξ ( x ) = Kx , λ = 0 . 2, and σ = 0 . . . . . . . x V ( x , l = . ) K = = = = (d) ξ ( x ) = Kx , λ = 0 . 2, and σ = 0. . . . . . . x V ( x , l ) l = l = l = (e) ξ ( x ) = min (1 , . x ) and σ = 0 . . . . . . x V ( x , l ) l = l = l = (f) ξ ( x ) = min (1 , . x ) and σ = 0. Figure 3. Expected capital injection V ( x ) := V ξ ( x ; ∞ ) for jump-diffusion processwith Laplace exponent ψ ( θ ) = µθ + σ θ − aθθ + c for µ = 0 . , a = 0 . , c = 9 , q = 0 . ARISIAN EXCURSION FOR DRAW-DOWN REFLECTED L´EVY INSURANCE RISK PROCESS 23 References [1] Agarwal, V., Daniel, N. and Naik, N. 2009. Role of managerial incentives and discretion in hedge fund perfor-mance. 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