Optimal reinsurance problem under fixed cost and exponential preferences
OOptimal reinsurance problemunder fixed cost and exponentialpreferences
Brachetta M. ∗† [email protected] Ceci, C. ‡ [email protected] Abstract
We investigate an optimal reinsurance problem for an insurance company facing a constantfixed cost when the reinsurance contract is signed. The insurer needs to optimally chooseboth the starting time of the reinsurance contract and the retention level in order to maxi-mize the expected utility of terminal wealth. This leads to a mixed optimal control/optimalstopping time problem, which is solved by a two-step procedure: first considering the pure-reinsurance stochastic control problem and next discussing a time-inhomogeneous optimalstopping problem with discontinuous reward. Using the classical Cram´er-Lundberg approxi-mation risk model, we prove that the optimal strategy is deterministic and depends on themodel parameters. In particular, we show that there exists a maximum fixed cost that theinsurer is willing to pay for the contract activation. Finally, we provide some economicalinterpretations and numerical simulations.
Keywords:
Optimal Reinsurance; Mixed Control Problem; Optimal Stopping; TransactionCost.
MSC Classification codes:
1. Introduction
Insurance business requires the transfer of risks from the policyholders to the insurer, who re-ceives a risk premium as a reward. In some cases, it could be convenient to cede these risksto a third party, which is the reinsurance company. From the operational viewpoint, a risk-sharing agreement helps the insurer reducing unexpected losses, stabilizing operating results,increasing business capacity and so on. By means of a reinsurance treaty, the reinsurance com-pany agrees to idemnify the primary insurer (cedent) against all or part of the losses which mayoccur under policies which the latter issued. The cedent will pay a reinsurance premium inexchange for this service. Roughly speaking, this is an insurance for insurers. When subscrib-ing a reinsurance treaty, a natural question is to determine the (optimal) level of the retainedlosses. Optimal reinsurance problems have been intensively studied by many authors under differ-ent criteria, especially through expected utility maximization and ruin probability minimization,see for example [De Finetti, 1940], [B¨uhlmann, 1970], [Gerber, 1969], [Irgens and Paulsen, 2004],[Brachetta and Ceci, 2019] and references therein.The main novelty of this article is that subscription costs are considered. In practice, whenthe agreement is signed, a fixed cost is usually paid in addition to the reinsurance premium. This ∗ Department of Mathematics, Politecnico of Milan, Piazza Leonardo da Vinci, 32, 20133 Milano, Italy. † Corresponding author. ‡ Department of Economics, University of Chieti-Pescara, Viale Pindaro, 42 - 65127 Pescara, Italy. a r X i v : . [ q -f i n . M F ] J a n spect has not been investigated by nearly all the studies, except for [Egami and Young, 2009]and [Li et al., 2015]. In the former work the authors discussed the reinsurance problem subject toa fixed cost for buying reinsurance and a time delay in completing the reinsurance transaction.They solved the problem considering a performance criterion with linear current reward andshowed that it is optimal to buy reinsurance when the surplus lies in a bounded interval dependingon the delay time. In the latter paper, under the criterion of minimizing the ruin probability,the original problem is reduced to a time-homogeneous optimal stopping problem. In particular,the authors show that the fixed cost forces the insurer to postpone buying reinsurance until thesurplus process hits a certain level.Hence the presence of a fixed cost is closely related to the possibility of postponing thesubscription of the reinsurance agreement. This, in turn, involves an optimal stopping problem,which is attached to the optimal choice of the retention level, which is a well known stochasticcontrol problem. The novelty of our paper consists in considering this mixed stochastic controlproblem under the criterion of maximizing the expected utility of terminal wealth. The strategyof the insurance company consists of the retention level of a proportional reinsurance and thesubscription timing. When the contract is signed, a given fixed cost is paid and the optimalretention level is applied. We use a diffusion approximation to model the insurer’s surplusprocess (see [Schmidli, 2018]). The insurance company has exponential preferences and is allowedto invest in a risk-less bond.As already mentioned, this setup leads to a combined problem of optimal stopping andstochastic control with finite horizon, which we will solve by a two-step procedure. First, weprovide the solution of the pure reinsurance problem (with starting time equal to zero). Next,we discuss an optimal stopping time problem with a suitable reward function depending on thevalue function of the pure reinsurance problem. Differently to [Egami and Young, 2009] and[Li et al., 2015], the associated optimal stopping problem turns out to be time-inhomogeneousand with discontinuous stopping reward with respect to the time. We provide an explicit solution,also showing that the optimal stopping time is deterministic. Moreover, we find that only twocases possible, depending on the model parameters. When the fixed cost is greater than a suitablethreshold (whose analytical expression is available), the optimal choice is not to subscribe thereinsurance; otherwise, the insurer immediately subscribes the contract.The paper is organized as follows. In Section 2, we describe the model and formulate theproblem as a mixed stochastic control problem, that is a problem which involves both optimalcontrol and stopping. In Section 3 we discuss the pure reinsurance problem (without stopping) bysolving the associated Halmilton-Jacobi-Bellman equation. Section 4 is devoted to the reductionof the original (mixed) problem to a suitable optimal stopping problem, which is then investigatedin Section 5. Here we provide a Verification Theorem and we solve the associated variationalinequality. In Section 6 we give the explicit solution to the original problem and we discusssome economic implications of our results. Finally, in Section 7 some numerical simulations areperformed in order to better understand the economic interpretation of our findings.
2. Problem formulation2.1. Model formulation
Let
T > , F , P , F ) is a complete probability spaceendowed with a filtration F . = {F t } t ∈ [0 ,T ] satisfying the usual conditions.Let us denote by R = { R t } t ∈ [0 ,T ] the surplus process of an insurance company. There isa wide range of risk models in the actuarial literature, see for instance [Grandell, 1991] and[Schmidli, 2018]. In the Cram´er-Lundberg risk model the claims arrival times are describedby the sequence of claims arrival times { T n } n ≥ , with T n < T n +1 P -a.e. ∀ n ≥
1, while thecorresponding claim sizes are given by { Z n } n ≥ . In particular, the number of occurred claimsup to time t ≥ N t = ∑︂ n =1 { T n ≤ t } , λ >
0, independent of thesequence { Z n } n ≥ . Moreover, { Z n } n ≥ are i.i.d. random variables with common probabilitydistribution function F Z ( z ), z ∈ (0 , + ∞ ), having finite first and second moments denoted by µ > µ >
0, respectively. In this context the surplus process is given by R + ct − N t ∑︂ n =1 Z n , R > , (2.1)where R is the initial capital and c > t ≥ E [︃ N t ∑︂ n =1 Z n ]︃ = λµt and var [︃ N t ∑︂ n =1 Z n ]︃ = λµ t. In this paper we use the diffusion approximation of the Cram´er-Lundberg model (2.1), see forexample [Grandell, 1991]. Precisely, we assume that the surplus process follows this stochasticdifferential equation (SDE): dR t = p dt + σ dW t , R > , where W = { W t } t ∈ [0 ,T ] is a standard Brownian motion, σ = √ λµ and p denotes the insurer’snet profit, that is p = c − µλ . In particular, under the expected value principle we have that c = (1 + θ i ) µλ and hence p = θ i µλ , with θ i > R > dB t = B t Rdt, B = 1 , hence the wealth process X = { X t } t ∈ [0 ,T ] evolves according to dX t = RX t dt + p dt + σ dW t , X = R > . (2.2)The explicit solution of the SDE (2.2) is given by the following equation: X t = R e Rt + ∫︂ t e R ( t − s ) pds + ∫︂ t e R ( t − s ) σ dW s , t ∈ [0 , T ] . (2.3)Now let τ denote an F -stopping time. At time τ the insurer can subscribe a proportionalreinsurance contract with retention level u ∈ [0 , u represents the percentage of retained losses, so that u = 0 means full reinsur-ance, while u = 1 is equivalent to no reinsurance. In order to buy a reinsurance agreement, theprimary insurer pays a reinsurance premium q ( u ) ≥
0. When the reinsurance contract is signedat time t = 0, the Cram´er-Lundberg risk model (2.1) is replaced by the following equation: R + ( c − q ( u )) t − N t ∑︂ n =1 uZ n , R > . Under the expected value principle we have that q ( u ) = (1 + θ )(1 − u ) µλ , u ∈ [0 , θ satistying θ > θ i (preventing the insurer from gaining a risk-freeprofit).Let us denote by R u = { R ut } t ∈ [0 ,T ] the reserve process in the Cram´er-Lundberg approximationassociated with a given reinsurance strategy { u t } t ∈ [0 ,T ] when the reinsurance contract is signedat time t = 0. Following [Eisenberg and Schmidli, 2009], under the expected value principle, R u follows dR ut = ( p − q + qu t ) dt + σ u t dW t , R u = R , (2.4)3here q = θλµ denotes the reinsurer’s net profit. We set q > p (non-cheap reinsurance). Thewealth process under the strategy { u t } t ∈ [0 ,T ] evolves according to this SDE: dX ut = RX ut dt + dR ut , X u = R , (2.5)which admits this explicit representation: X ut = R e Rt + ∫︂ t e R ( t − s ) ( p − q + qu s ) ds + ∫︂ t e R ( t − s ) σ u s dW s . (2.6)We assume that a constant fixed cost K > α = ( τ, { u t } t ∈ [ τ,T ] ), with τ ≤ T . Let H t = I { τ ≤ t } be the indicator process of the contract starting time. For τ < T P -a.s. , the totalwealth X α = { X αt } t ∈ [0 ,T ] associated with a given strategy α is given by dX αt = (1 − H t ) dX t + H t dX ut − KdH t , X α = R > , (2.7)while on the event { τ = T } we have that dX αt = dX t , X α = R > , (2.8)where X satisfies equation (2.2).Equation (2.7) can be written more explicitly as dX αt = {︄ dX t , t < τ, X = R ,dX ut , τ < t ≤ T, X uτ = X τ − K, (2.9)where X and X u satisfy equations (2.2) and (2.5), respectively.In our setting the null reinsurance corresponds to the choice τ = T , P -a.s., to which weassociate the strategy α null = ( T,
1) and X α null t = X t , t ∈ [0 , T ] . The insurers’ objective is to maximize the expected utility of the terminal wealth:sup α ∈A E [︁ U ( X αT ) ]︁ , (2.10)where U : R → [0 , + ∞ ) is the utility function representing the insurer’s preferences and A theclass of admissible strategies (see Definition 2.1 below).We focus on CARA ( Constant Absolute Risk Aversion ) utility functions, whose general ex-pression is given by U ( x ) = 1 − e − ηx , x ∈ R , where η > Definition 2.1 (Admissible strategies) . We denote by A the set of admissible strategies α =( τ, { u t } t ∈ [ τ,T ] ) , where τ is an F -stopping time such that τ ≤ T and { u t } t ∈ [ τ,T ] is an F -predictableprocess with values in [0 , . Let us observe that the null strategy α null = ( T, is included in A .When we want to restrict the controls to the time interval [ t, T ] , we will use the notation A t . roposition 2.1. Let α ∈ A , then E [︁ e − ηX αT ]︁ < + ∞ . Proof.
Using equations (2.6) and (2.9), we have that E [︁ e − ηX αT ]︁ = E [︁ e − ηX T I { τ = T } ]︁ + E [︁ e − η ( X τ − K ) e R ( T − τ ) e − η ∫︁ Tτ e R ( T − s ) ( p − q + qu s ) ds e − η ∫︁ Tτ e R ( T − s ) σ u s dW s I { τ 3. The pure reinsurance problem In order to have a self-contained article, in this section we briefly investigate a pure reinsuranceproblem, which corresponds to the problem (2.10) with fixed starting time t = 0. Precisely, wedeal with inf u ∈U E [︁ e − ηX uT ]︁ , where U denotes the class of admissible strategies u = { u t } t ∈ [0 ,T ] , which are all the F -predictableprocesses with values in [0 , V ¯ ( t, x ) the value function associated to thisproblem, that is V ¯ ( t, x ) = inf u ∈U t E [︁ e − ηX u,t,xT ]︁ , ( t, x ) ∈ [0 , T ] × R , (3.1)with U t denoting the restriction of U to the time interval [ t, T ] and { X u,t,xs } s ∈ [ t,T ] denotes theprocess satisying equation (2.5) with initial data ( t, x ) ∈ [0 , T ] × R . It is well known that the5alue function (3.1) can be characterized as a classical solution to the associated Hamilton-Jacobi-Bellman (HJB) equation: {︄ min u ∈ [0 , L u V ¯ ( t, x ) = 0 , ∀ ( t, x ) ∈ [0 , T ) × R V ¯ ( T, x ) = e − ηx ∀ x ∈ R , (3.2)where, using equations (2.4) and (2.5), the generator of the Markov process X u is given by L u f ( t, x ) = ∂f∂t ( t, x ) + ( Rx + p − q + qu ) ∂f∂x ( t, x ) + 12 σ u ∂ f∂x ( t, x ) , f ∈ C , ((0 , T ) × R ) . Under the ansatz V ¯ ( t, x ) = e − ηxe R ( T − t ) ϕ ( t ), the HJB equation reads as ϕ ′ ( t ) + Ψ( t ) ϕ ( t ) = 0 , ϕ ( T ) = 1 , where Ψ( t ) = min u ∈ [0 , {− ηe R ( T − t ) ( p − q + qu ) + 12 σ u η e R ( T − t ) } . Solving the minimization problem we find the unique minimizer: u ∗ ( t ) = qησ e − R ( T − t ) ∨ , t ∈ [0 , T ] . Under the additional condition q < ησ , (3.3) u ∗ ( t ) simplifies to u ∗ ( t ) = qησ e − R ( T − t ) ∈ (0 , , t ∈ [0 , T ] . (3.4)Using this expression we readily obtain thatΨ( t ) = ηe R ( T − t ) ( q − p ) − q σ . (3.5)By classical verification arguments, we can verify that the value function given in (3.1) takes thisform: V ¯ ( t, x ) = e − ηxe R ( T − t ) e ∫︁ Tt Ψ( s ) ds = e − ηxe R ( T − t ) e η ( q − p ) R ( e R ( T − t ) − e − q σ ( T − t ) , (3.6)and, under the condition (3.3), equation (3.4) provides an optimal reinsurance strategy. Remark 3.1. Comparing the optimal strategy u ∗ ( s ) , s ∈ [ t, T ] , to the null reinsurance u ( s ) = 1 , s ∈ [ t, T ] , by means of (3.1) we get that V ¯ ( t, x ) ≤ E [︁ e − ηX t,xT ]︁ ∀ ( t, x ) ∈ [0 , T ] × R . (3.7) Moreover, by equation (2.2) we have that g ( t, x ) . = E [︁ e − ηX t,xT ]︁ = E [︁ e − ηxe R ( T − t ) e − η ∫︁ Tt e R ( T − s ) pds e − η ∫︁ Tt e R ( T − s ) σ dW s ]︁ = e − ηxe R ( T − t ) e − ηR ( e R ( T − t ) − p e R η σ ( e R ( T − t ) − . (3.8) Defining h ( t ) . = ηe R ( T − t ) (︁ ηe R ( T − t ) σ − p )︁ , (3.9) we can write g ( t, x ) = e − ηxe R ( T − t ) e ∫︁ Tt h ( s ) ds . (3.10) Hence, using (3.5) and (3.6) , inequality (3.7) reads as ∫︂ Tt [Ψ( s ) − h ( s )] ds = ηqR ( e R ( T − t ) − − q σ ( T − t ) − R η σ ( e R ( T − t ) − ≤ ∀ t ∈ [0 , T ] . (3.11)6 . Reduction to an optimal stopping problem We can show that the mixed stochastic control problem (2.11) can be reduced to an optimalstopping problem. Let us denote by T t,T is the set of F -stopping times τ such that t ≤ τ ≤ T . Theorem 4.1. We have that V ( t, x ) = inf τ ∈T t,T E [︁ V ¯ ( τ, X t,xτ − K ) I { τ From the theory of optimal stopping (see, for instance [Øksendal, 2003]), whenthe cost function G ( t, x ) is continuous and the value function W ( t, x ) = inf τ ∈T t,T E [︁ G ( τ, X t,xτ )] , t ∈ [0 , T ] × R is sufficiently regular, it can be characterized as a solution to the following variational inequality: min {L W ( t, x ) , G ( t, x ) − W ( t, x ) } = 0 , ( t, x ) ∈ (0 , T ) × R . (5.3) This is a free-boundary problem, whose solution is the function W ( t, x ) and the so-called contin-uation region, which is defined as C = { ( t, x ) ∈ (0 , T ) × R : W ( t, x ) < G ( t, x ) } . (5.4) Moreover, it is known that the first exit time of the process X t,x from the region C τ ∗ t,x . = inf { s ∈ [ t, T ] : ( s, X t,xs ) / ∈ C} . provides an optimal stopping time.In our optimal stopping problem (4.1) , the cost function is V ¯ ( t, x − K ) I { t Theorem 5.1 (Verification Theorem) . Let φ : [0 , T ] × R → R be a function satisfying the as-sumptions below and C (the continuation region) be defined by C = { ( t, x ) ∈ (0 , T ) × R : φ ( t, x ) < V ¯ ( t, x − K ) } . (5.5) Suppose that the following conditions are satisfied.1. There exists t ∗ ∈ [0 , T ) such that C = ( t ∗ , T ) × R . . φ ∈ C ([0 , T ] × R ) , φ is C w.r.t t in (0 , t ∗ ) and ( t ∗ , T ) , separately, and C w.r.t. x ∈ R ;3. φ ( t, x ) ≤ V ¯ ( t, x − K ) ∀ ( t, x ) ∈ [0 , T ] × R and φ ( T, x ) = e − ηx ∀ x ∈ R ;4. φ is a solution to the following variational inequality {︄ L φ ( t, x ) ≥ ∀ ( t, x ) ∈ (0 , t ∗ ) × R L φ ( t, x ) = 0 ∀ ( t, x ) ∈ C = ( t ∗ , T ) × R . (5.6) 5. the family { φ ( τ, X τ ); τ ∈ T ,T } is uniformly integrable.Moreover, let τ ∗ t,x the first exit time from the region C of the process X t,x , that is τ ∗ t,x . = inf { s ∈ [ t, T ] : ( s, X t,xs ) / ∈ C} . with the convention τ ∗ t,x = T if the set on the right-hand side is empty.Then φ ( t, x ) = V ( t, x ) on [0 , T ] × R and τ ∗ t,x is an optimal stopping time for problem (4.1) .Proof. For any ( t, x ) ∈ [0 , T ) × R let us take the sequence of stopping times { τ n } n ≥ such that τ n . = inf { s ≥ t | | X t,xs | ≥ n } . We first prove that, ∀ τ ∈ T t,T φ ( t, x ) ≤ E [ φ ( τ ∧ τ n , X t,xτ ∧ τ n )] , ∀ ( t, x ) ∈ [0 , T ) × R . (5.7)Due to the specific form of the continuation region we have two cases. If t ≥ t ∗ , since φ ∈ C , (( t ∗ , T ) × R ), applying Dynkin’s formula we get that for any arbitrary stopping time τ ∈ T t,T φ ( t, x ) = E [ φ ( τ ∧ τ n , X t,xτ ∧ τ n )] − E [︃∫︂ τ ∧ τ n t L φ ( s, X t,xs ) ds ]︃ = E [ φ ( τ ∧ τ n , X t,xτ ∧ τ n )] . If t < t ∗ , we have again by Dynkin’s formula, since φ ∈ C , ((0 , t ∗ ) × R ), that φ ( t, x ) = E [ φ ( τ ∧ τ n ∧ t ∗ , X t,xτ ∧ τ n ∧ t ∗ )] − E [︃∫︂ τ ∧ τ n ∧ t ∗ t L φ ( s, X t,xs ) ds ]︃ ≤ E [ φ ( τ ∧ τ n ∧ t ∗ , X t,xτ ∧ τ n ∧ t ∗ )]and, similarly, since φ ∈ C , (( t ∗ , T ) × R ), E [ φ ( τ ∧ τ n ∧ t ∗ , X t,xτ ∧ τ n ∧ t ∗ )] = E [ φ ( τ ∧ τ n , X t,xτ ∧ τ n )] − E [︃∫︂ τ ∧ τ n τ ∧ τ n ∧ t ∗ L φ ( s, X t,xs ) ds ]︃ = E [ φ ( τ ∧ τ n , X t,xτ ∧ τ n )] , hence (5.7) is proved.Now letting n → + ∞ in (5.7), recalling that φ ∈ C ([0 , T ] × R ) and using Fatou Lemma weget that φ ( t, x ) ≤ E [ φ ( τ, X t,xτ )] ≤ E [ V ¯ ( τ, X t,xτ − K ) I { τ 9. If the stopping region is not empty, for t = 0, we have that φ (0 , x ) = V ¯ (0 , x − K ) ∀ x ∈ R , otherwise by continuity of both the functions if φ (0 , x ) > V ¯ (0 , x − K ) (or φ (0 , x ) Let g as defined in equation (3.10) . The families { V ¯ ( τ, X τ − K ); τ ∈ T ,T } and { g ( τ, X τ ); τ ∈ T ,T } are uniformly integrable.Proof. Recalling that V ¯ ( t, x ) ≤ g ( t, x ) by (3.7), we have that V ¯ ( t, x − k ) ≤ e ηKe RT g ( t, x ), hencethe statement follows by the uniformly integrability of the family { g ( τ, X τ ) : τ ∈ T ,T } . It is well known that if for any arbitrary δ > τ ∈ T ,T E [ g ( τ, X τ ) δ ] < + ∞ , then the proof is complete. To this end, we observe that E [ g ( τ, X τ ) δ ] = E [ e (1+ δ ) ∫︁ Tτ h ( s ) ds e − η (1+ δ ) e R ( T − τ ) X τ ] ≤ e R (1+ δ ) η σ e RT e − η (1+ δ ) R e RT E [ e − η (1+ δ ) ∫︁ τ e R ( T − s ) σ dW s ] ≤ e R (1+ δ ) η σ e RT e − η (1+ δ ) R e RT e R (1+ δ ) η σ ( e RT − < + ∞ . The guess for the continuation region C given in the assumption 1 . of the Verification Theoremfollows by the next result. Lemma 5.2. The set A . = { ( t, x ) ∈ (0 , T ) × R : L V ¯ ( t, x − K ) < } (5.8) is included in the continuation region, that is A ⊆ C = { ( t, x ) ∈ (0 , T ) × R : V ( t, x ) < V ¯ ( t, x − K ) } . Moreover, the following equation holds: A = ( t A , T ) × R , where t A . = 0 ∨ [︃ T − R log (︃ q + RK + √︁ ( q + RK ) − q ησ )︃]︃ ∧ T. (5.9) In particular, only three cases are possible, depending on the model parameters: . if y ∗ := q + RK + √︁ ( q + RK ) − q ησ ≥ e RT , then t A = 0 and L V ¯ ( t, x − K ) < ∀ ( t, x ) ∈ (0 , T ) × R , so that A = (0 , T ) × R , implyingthat C = (0 , T ) × R ;2. if < y ∗ = q + RK + √︁ ( q + RK ) − q ησ < e RT , then < t A < T and L V ¯ ( t, x − K ) < ∀ ( t, x ) ∈ ( t A , T ) × R ; in this case A = ( t A , T ) × R ;3. if y ∗ = q + RK + √︁ ( q + RK ) − q ησ ≤ , then t A = T and L V ¯ ( t, x − K ) ≥ ∀ ( t, x ) ∈ (0 , T ) × R , so that A = ∅ .Proof. First let us observe that V ¯ ( t, x − K ) ∈ C , ((0 , T ) × R ) ∩ C ([0 , T ] × R ) and the family { V ¯ ( τ, X τ − K ); τ ∈ T ,T } is uniformly integrable by Lemma 5.1. Now choose ( t ¯ , x ¯) ∈ A , let B ⊂ A be a neighborhood of ( t ¯ , x ¯) with τ B < T , where τ B denotes the first exit time of X t ¯ ,x ¯ from B . Then by Dynkin’s formula V ¯ ( t ¯ , x ¯ − K ) = E [ V ¯ ( τ B , X t ¯ ,x ¯ τ B − K )] − E [︃∫︂ τ B t ¯ L V ¯ ( s, X t ¯ ,x ¯ s − K ) ds ]︃ > E [ V ¯ ( τ B , X t ¯ ,x ¯ τ B − K )] ≥ V ( t ¯ , x ¯) . Hence ( t ¯ , x ¯) ∈ C and A ⊆ C .Next, recalling (5.2), we have that L V ¯ ( t, x − K ) = V ¯ ( t, x − K ) (︁ − Ψ( t ) − ηe R ( T − t ) ( p + RK ) + 12 η e R ( T − t ) σ )︁ , so that L V ¯ ( t, x − K ) < t ) > η e R ( T − t ) σ − ηe R ( T − t ) ( p + RK ) , (5.10)that is, using (3.5), 12 η e R ( T − t ) σ − ηe R ( T − t ) ( q + RK ) + 12 q σ < . Using a change of variable z = e R ( T − t ) , we can rewrite the inequality as12 η σ z − η ( q + RK ) z + 12 q σ < . Since η [( q + K ) − q ] > q + RK − √︁ ( q + RK ) − q ησ < z < q + RK + √︁ ( q + RK ) − q ησ . Recalling (3.3), we can verify that q + RK − √︁ ( q + RK ) − q ησ < qησ < , 11o that the inequality reads as t A = T − R log (︃ q + RK + √︁ ( q + RK ) − q ησ )︃ < t < T. Depending on the model parameters, we can see that only the three cases above are possible.Equivalently, L V ¯ ( t, x − K ) < t A < t < T . Remark 5.2. As consequence of Lemma 5.2, recalling (3.9) , in Cases 1 and 2, that is when ≤ t A < T , we have that Ψ( t ) − h ( t ) + ηRKe R ( T − t ) > , ∀ t > t A , see equation (5.10) , which implies, ∀ t ≥ t A ∫︂ Tt (Ψ( s ) − h ( s )) ds + ∫︂ Tt ηRKe R ( T − s ) ds > , equivalently ∫︂ Tt (Ψ( s ) − h ( s )) ds + ηKe R ( T − t ) > ηK for all t ∈ [ t A , T ) . In Case 3, that is when t A = T , since Ψ( t ) − h ( t ) + ηRKe R ( T − t ) < , ∀ t ∈ [0 , T ) we have that ∫︂ Tt (Ψ( s ) − h ( s )) ds + ηKe R ( T − t ) < ηK, for all t ∈ [0 , T ) . We need the following preliminary result to provide an explicit expression for the valuefunction of the problem (4.1). Lemma 5.3. The function ˜︁ V ( t, x ) = Cg ( t, x ) , ( t, x ) ∈ (0 , T ) × R , with C any positive constantand g as given in equation (3.10) , is a solution to the PDE L ˜︁ V ( t, x ) = 0 , ( t, x ) ∈ (0 , T ) × R .In particular, g is a solution to the PDE with boundary condition g ( T, x ) = e − ηx ∀ x ∈ R .Proof. Using the ansatz ˜︁ V ( t, x ) = e − ηxe R ( T − t ) γ ( t ), we can reduce the PDE L ˜︁ V ( t, x ) = 0 to thefollowing equation: e − ηxe R ( T − t ) γ ′ ( t ) − ηe R ( T − t ) V ( t, x ) p + 12 η e R ( T − t ) V ( t, x ) σ = 0 , which is equivalent to this ODE: γ ′ ( t ) + h ( t ) γ ( t ) = 0 , ( t, x ) ∈ (0 , T ) × R , where the function h is given in (3.9).Since the solution of the ODE is γ ( t ) = C e ∫︁ Tt h ( s ) ds , we get the expression of ˜︁ V as above.Finally, setting C = 1, g satisfies the PDE above with the terminal condition g ( T, x ) = e − ηx ∀ x ∈ R .Before proving the main result of this section, which is Theorem 5.2, we compare g ( t, x ),given in (3.10), with V ¯ ( t, x − K ). Lemma 5.4. Let H ( t ) = ∫︂ Tt (Ψ( s ) − h ( s )) ds + ηKe R ( T − t ) , t ∈ [0 , T ] , (5.11) then we distinguish two cases: . if H (0) ≥ , then g ( t, x ) < V ¯ ( t, x − K ) ∀ ( t, x ) ∈ (0 , T ] × R ;2. if H (0) < , then there exists t ∗ ∈ (0 , t A ) such that g ( t ∗ , x ) = V ¯ ( t ∗ , x − K ) ∀ x ∈ R and g ( t, x ) < V ¯ ( t, x − K ) ∀ ( t, x ) ∈ ( t ∗ , T ] × R .Proof. Let us observe that the inequality g ( t, x ) < V ¯ ( t, x − K ) writes as e − ηxe R ( T − t ) e ∫︁ Tt h ( s ) ds < e − η ( x − K ) e R ( T − t ) e ∫︁ Tt Ψ( s ) ds , that is e ∫︁ Tt (Ψ( s ) − h ( s )) ds e ηKe R ( T − t ) > ⇔ ∫︂ Tt (Ψ( s ) − h ( s )) ds + ηKe R ( T − t ) = H ( t ) > . We distinguish three cases:(i) when 0 ≤ t A < T , we have that H ( t ) ≥ ηK > ∀ t > t A by Remark 5.2 and it easy to verifythat H is increasing in [0 , t A ], while it is decreasing in [ t A , T ]. Hence, it takes the maximumvalue at t = t A . As a consequence, if H (0) ≥ H ( t ) > ∀ t ∈ (0 , T ], being H ( T ) = ηK > H (0) < t ∗ ∈ (0 , t A ) such that H ( t ∗ ) = 0, that is g ( t ∗ , x ) = V ¯ ( t ∗ , x − K ) ∀ x ∈ R , and H ( t ) > ∀ ( t, x ) ∈ ( t ∗ , T ], that is g ( t, x ) < V ¯ ( t, x − K ) ∀ ( t, x ) ∈ ( t ∗ , T ] × R ;(ii) when t A = T , by Lemma 5.2 we get that H is increasing in [0 , T ] and we can repeatthe same arguments as in the previous case to distinguish the two casese H (0) ≥ H (0) < 0, obtaining the same results;(iii) when t A = 0, by Remark 5.2 we know that H is decreasing in [0 , T ], so that H ( t ) ≥ ηK > ∀ t ∈ [0 , T ], that is g ( t, x ) < V ¯ ( t, x − K ), ∀ ( t, x ) ∈ (0 , T ] × R . Moreover, in this case H (0) ≥ Proposition 5.1. Let C = { ( t, x ) ∈ (0 , T ) × R : V ( t, x ) < V ¯ ( t, x − K ) } . (5.12) Then we distinguish two cases:1. if H (0) ≥ , then C = (0 , T ) × R ,2. if H (0) < , then ( t ∗ , T ) × R ⊆ C , where t ∗ ∈ (0 , t A ) is the unique solution to equation H ( t ) = ∫︂ Tt (Ψ( s ) − h ( s )) ds + ηKe R ( T − t ) = 0 . Proof. We apply Lemma 5.4. In Case 1, we have that V ( t, x ) ≤ g ( t, x ) < V ¯ ( t, x − K ) ∀ ( t, x ) ∈ (0 , T ) × R , that is C = (0 , T ) × R . In Case 2, we have that V ( t, x ) ≤ g ( t, x ) < V ¯ ( t, x − K ) ∀ ( t, x ) ∈ ( t ∗ , T ) × R , which implies ( t ∗ , T ) × R ⊆ C , and this concludes the proof.Now we are ready for the main result of this section. Theorem 5.2. Let H be given in (5.11) . The solution of the optimal stopping problem (4.1) takes different forms, depending on the model parameters. Precisely, we have two cases:1. if H (0) = ∫︁ T ( ψ ( s ) − h ( s )) ds + ηKe RT ≥ , then the continuation region is C = (0 , T ) × R ,the value function is V ( t, x ) = g ( t, x ) = e − ηxe R ( T − t ) e ∫︁ Tt h ( s ) ds , ( t, x ) ∈ [0 , T ] × R and τ ∗ t,x = T is an optimal stopping time; . if H (0) = ∫︁ T ( ψ ( s ) − h ( s )) ds + ηKe RT < , then C = ( t ∗ , T ) × R , where t ∗ ∈ ( t A , T ) is theunique solution to H ( t ) = 0 , the value function is V ( t, x ) = {︄ V ¯ ( t, x − K ) = e − η ( x − K ) e R ( T − t ) e ∫︁ Tt Ψ( s ) ds ( t, x ) ∈ [0 , t ∗ ] × R g ( t, x ) = E [︁ e − ηX t,xT ] = e − ηxe R ( T − t ) e ∫︁ Tt h ( s ) ds ( t, x ) ∈ ( t ∗ , T ] × R (5.13) and τ ∗ t,x , given by τ ∗ t,x = {︄ t ( t, x ) ∈ [0 , t ∗ ] × R T ( t, x ) ∈ ( t ∗ , T ] × R , (5.14) is an optimal stopping time.Proof. We prove the two cases separately, applying Theorem 5.1 in each one. Case 1 The continuation region is C = (0 , T ) × R by Proposition 5.1, hence assumption 1 of Theorem5.1 is fulfilled. Moreover, τ ∗ t,x = T . Observing that g ( t, x ) = e − ηxe R ( T − t ) e ∫︁ Tt h ( s ) ds ∈ C , ((0 , T ) × R ) ∩ C ([0 , T ] × R ) , the assumption 2 of Theorem 5.1 is clearly matched. The assumption 3 is implied by Lemma5.4. Moreover, the variational inequality (5.6) (assumption 4) is fulfilled by Lemma 5.3. Finally,by Lemma 5.1 the last condition in Theorem 5.1 is fulfilled. Case 2 C = ( t ∗ , T ) × R clearly satisfies the first assumption of Theorem 5.1. Taking φ ( t, x ) = {︄ V ¯ ( t, x − K ) = e − η ( x − K ) e R ( T − t ) e ∫︁ Tt Ψ( s ) ds ( t, x ) ∈ [0 , t ∗ ] × R g ( t, x ) = E [︁ e − ηX t,xT ] = e − ηxe R ( T − t ) e ∫︁ Tt h ( s ) ds ( t, x ) ∈ ( t ∗ , T ] × R , observing that Lemma 5.4 ensures the existence of t ∗ ∈ (0 , t A ) such that g ( t ∗ , x ) = V ¯ ( t ∗ , x − K )when H (0) < 0, the smoothness conditions of the second assumption are matched. Moreover,according to Lemma 5.4, g ( t, x ) < V ¯ ( t, x − K ) ∀ ( t, x ) ∈ ( t ∗ , T ] and the assumption 3 is fulfilled.That the variational inequality (5.6) is satisfied by φ is a consequence of the results of Section3 and of Lemma 5.3. Finally, Lemma 5.1 implies the fifth assumption of Theorem 5.1 and theproof is complete. 6. Solution to the original problem As a direct consequence of the results obtained in the previous section and Theorem 4.1, weprovide an explicit solution to the optimal reinsurance problem under fixed cost given in (2.11). Theorem 6.1. Let us define K ∗ = − qR (1 − e − RT ) + 12 q ησ T e − RT + 14 R ησ ( e RT − e − RT ) > . (6.1) Two cases are possible, depending on the model parameters:1. if K ≥ K ∗ , then the value function given in (2.11) is V ( t, x ) = g ( t, x ) = E [︁ e − ηX t,xT ] and the optimal strategy is α ∗ = ( T, , that is no reinsurance is purchased; . if K < K ∗ , then the value function is V ( t, x ) = {︄ V ¯ ( t, x − K ) = e − η ( x − K ) e R ( T − t ) e ∫︁ Tt Ψ( s ) ds ( t, x ) ∈ [0 , t ∗ ] × R g ( t, x ) = E [︁ e − ηX t,xT ] = e − ηxe R ( T − t ) e ∫︁ Tt h ( s ) ds ( t, x ) ∈ ( t ∗ , T ] × R , where t ∗ ∈ (0 , T ) is the unique solution to the equation η ( qR + K ) e R ( T − t ) − q σ ( T − t ) − R η σ ( e R ( T − t ) − − ηqR = 0 , and the optimal strategy is α ∗ = ( τ ∗ t,x , { qησ e − R ( T − s ) } s ∈ [ τ ∗ t,x ,T ] ) , with τ ∗ t,x given in (5.14) .Proof. Let us observe that, using Remark 3.1, H ( t ) = ∫︂ Tt (Ψ( s ) − h ( s )) ds + ηKe R ( T − t ) = ηqR ( e R ( T − t ) − − q σ ( T − t ) − R η σ ( e R ( T − t ) − 1) + ηKe R ( T − t ) . and the condition H (0) ≥ K ≥ − η e − RT ∫︂ T (Ψ( s ) − h ( s )) ds = K ∗ , while the condition H (0) < K < K ∗ . That K ∗ > K ∗ > K ∗ , then the insurer will not buy protection from her.In Case 2, at any time t ∈ [0 , T ], the insurer immediately subscribes the reinsurance agreementif the time instant t ∗ has not passed, applying the optimal retention level from that moment on;otherwise, if t > t ∗ , no reinsurance will be bought.We notice that it is never optimal to wait for buying reinsurance. That is, it is convenienteither to immediately sign the contract, or not to subscribe at all.In particular, at the starting time t = 0, given an initial wealth R > 0, we have these cases:1. if K ≥ K ∗ , then α ∗ = ( T, K < K ∗ , then α ∗ = (0 , { qησ e − R ( T − s ) } s ∈ [0 ,T ] , that is the optimal choice for the insurerconsists in stipulating the contract at the initial time, selecting the optimal retention level(as in the pure reinsurance problem).By the expression (6.1) we can show that K ∗ is increasing with respect to η and σ , whileit is decreasing with respect to q . More details will be given in the next section by means ofnumerical simulations.Another relevant result for the reinsurance company is the following. Proposition 6.1. For any fixed cost K > there exists q ∗ ∈ (0 , + ∞ ) (depending on K ) suchthat1. if q > q ∗ , then V ( t, x ) = g ( t, x ) = E [︁ e − ηX t,xT ] and α ∗ = ( T, , that is no reinsurance is purchased; . otherwise V ( t, x ) = {︄ V ¯ ( t, x − K ) = e − η ( x − K ) e R ( T − t ) e ∫︁ Tt Ψ( s ) ds ( t, x ) ∈ [0 , t ∗ ] × R g ( t, x ) = E [︁ e − ηX t,xT ] = e − ηxe R ( T − t ) e ∫︁ Tt h ( s ) ds ( t, x ) ∈ ( t ∗ , T ] × R , where t ∗ ∈ (0 , T ) is the unique solution to the equation η ( qR + K ) e R ( T − t ) − q σ ( T − t ) − R η σ ( e R ( T − t ) − − ηqR = 0 , and α ∗ = ( τ ∗ t,x , { qησ e − R ( T − s ) } s ∈ [ τ ∗ t,x ,T ] ) , with τ ∗ t,x is given in (5.14) .Proof. Following Theorem 6.1 and its proof, we can write the condition H (0) ≤ T σ q + (1 − e RT ) ηR q + η σ R ( e RT − − ηKe RT ≥ . To simplify our computations, let us consider this inequality for any q ∈ R . The discriminant ∆must be positive, otherwise the existence of K ∗ > q , = ησ T (︁ e RT − R ± √ ∆ )︁ , q < q . Since q > ησ ( e RT − RT > ησ , only q is relevant because of the condition (3.3). That q ∈ (0 , + ∞ ) is a consequence of theexistence of K ∗ > q was not positive, then H (0) > q > q ∗ = q concludes the proof.The last result is interesting for the reinsurer. In Section 3 we have already stated that thecondition q < ησ (see equation (3.3)) is required in order that the reinsurance agreement isdesirable. In presence of a fixed initial cost, now we know that there exists a threshold q ∗ , whichis smaller than ησ , such that the insurer will never subscribe the contract if q > q ∗ . Remark 6.1. Recalling that q = θλµ (see Section 2, we can give a deeper interpretation of theprevious result. Indeed, we have proven the existence of a maximum safety loading θ ∗ > , whichcannot be exceeded by the reinsurer, otherwise the reinsurance contract will not be subscribed. 7. Numerical simulations In this section we use some numerical simulations in order to further investigate the resultsobtained in Section 6. Unless otherwise specified, all the simulations are performed according tothe parameters of Table 1 below. Parameter Value T η . σ . q . R . Table 1: Model parameters. a) Reinsurer’s net profit (b) Risk aversion (c) Volatility (d) Time horizon Figure 1: Sensitivity analyses with respect to the model parameters. We have previously illustrated how the threshold K ∗ in equation (6.1) is relevant for theinsurer as well as for the reinsurer. Figure 1 shows how this threshold is influenced by the modelparameters. As expected, if the reinsurer increases her net profit (for example, increasing thesafety loading), then the fixed cost should decrease, see Figure 1a. When the insurer is morerisk averse, she is willing to pay a higher fixed cost, see Figure 1b. The same applies when thepotential losses increase, that is σ is high, as in Figure 1c. 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