An Exponential Cox-Ingersoll-Ross Process as Discounting Factor
aa r X i v : . [ q -f i n . M F ] A ug An Exponential Cox–Ingersoll–Ross Process as DiscountingFactor
Julia Eisenberg ∗ and Yuliya Mishura † TU Wien/University of Liverpool Taras Shevchenko National University of Kyiv
Abstract
We consider an economic agent (a household or an insurance company) modelling itssurplus process by a deterministic process or by a Brownian motion with drift. Thegoal is to maximise the expected discounted spendings/dividend payments, giventhat the discounting factor is given by an exponential CIR process.In the deterministic case, we are able to find explicit expressions for the optimalstrategy and the value function.For the Brownian motion case, we offer a method allowing to show that for a smallvolatility the optimal strategy is a constant-barrier strategy.
Key words:
Hamilton–Jacobi–Bellman equation, Cox–Ingersoll–Ross process, div-idends, Brownian risk model, consumption.
Primary 93E20Secondary 91B30, 60K10
An insurance company’s credit rating indicates its ability to pay customer’s claims. Abad credit rating can affect company’s business plan, growth potential or even survivalchances if new finance is needed to fulfil the capital requirements prescribed by SolvencyII. The rating process run by a credit rating agency includes quantitative and qualitativeanalysis, where cash flow is one of the most important factors. A particular attention ispaid to dividend payments, which are commonly believed to indicate company’s financialhealth. Searching for the optimal strategy maximising the value of expected discounteddividends under different constraints and in different setups has been a popular problemin actuarial mathematics for a long time. The papers by Shreve et al. [14], Asmussenand Taksar [3], Azcue and Muler [4] are just some examples. For a detailed review we ∗ [email protected] † [email protected] a = 0 . b = 0 . δ = 0 .
09 (left picture), δ = 0 .
045 (middle) and δ = 0 .
02 (right).as a discounting factor in the context of consumption/dividend maximisation problems.Despite the fact that the value function depends on two variables – the surplus and thediscounting process – we are able to find explicit expressions for the optimal strategyand the value function in the deterministic income case and (under some restrictions onthe underlying CIR) in the case of Brownian risk model.It will be of major importance for the understanding of the paper to remind the readeron some properties and results connected to CIR processes. Accordingly, we organisedthe paper as follows: in the next subsection we give an overview over CIR processes. Forthe convenience of reading, we postpone the technical proofs to the appendix.In Section 2, we consider the case of a deterministic, linear in time income process, whichcan be interpreted as the income of an individual or household. There, we will distinguishbetween two different cases concerning the parameters of the considered CIR process andgive explicit expressions for the optimal strategy and the value function. Here, we solvethe problem of dividend maximisation for special parameters of the underlying CIRprocess. Conclusion at the end of Sections 2 gives an overview over the possible futureresearch directions. Some technical proofs are given in the appendix, Section 3.
For the sake of clarity of presentation, we postpone the most proofs of this subsectionto the appendix, Section 3. Here and in the following we use the common notation: P [ · | Y = y ] = P y [ · ] and E [ · | Y = y ] = E y [ · ] for any stochastic process { Y t } . In theremainder of the paper we let r = { r t } be a Cox–Ingersoll–Ross (CIR) processd r t = ( ar t + b ) d t + δ √ r t d W t , (1)where a , b and δ are positive constants and W = { W t } is a standard Brownian motion.Due, for example, to [10], CIR processes have the strong Markov property. We define M ( r, t ) := E r [ e − r t ] . r t with initial value r is given by f ( y ) := c ( t ) e − u ( t,r ) − v ( t,y ) (cid:16) v ( t, y ) u ( t, r ) (cid:17) q/ I q (cid:0) p u ( t, r ) v ( t, y ) (cid:1) , (2)where I q ( x ) = ∞ P m =0 1 m !Γ( m + q +1) (cid:16) x (cid:17) m + q is the modified Bessel function of the first kindand c ( t ) := 2 a ( e at − δ , q := 2 bδ − ,u ( t, r ) := c ( t ) re at , v ( t, y ) := c ( t ) y. Also, one has M ( r, t ) = E r [ e − r t ] = e − abδ t β ( t ) bδ · e − rβ ( t ) , (3)where β ( t ) := δ a + (cid:0) − δ a (cid:1) e − at . Lemma 1.1
In the case δ ≤ a , the function M ( r, t ) is strictly decreasing in t and the process { e − r t } is a supermartingale. Proof:
Using that β ′ ( t ) = a (cid:0) − δ a (cid:1) e − at β ( t ) >
0, we obtain M t ( r, t ) = M ( r, t ) n − bβ ( t ) − rae − at (cid:0) − δ a (cid:1) β ( t ) o < r ∈ R + . The supermartingale property follows immediately due to the Markovproperty and the structure of M . (cid:3) Lemma 1.2
Due to [12, p. 282], the function M solves the partial differential equation ( ar + b ) M r ( r, t ) + δ r M rr ( r, t ) − M t ( r, t ) = 0 . The below lemma ensures the well-definiteness of the problems we are going to consider.
Lemma 1.3 If a > then the CIR process { r t } fulfils lim t →∞ r t = ∞ a.s. For the proof confer the appendix, Section 3.The usual method for proving a verification theorem is to apply Ito’s formula, and toprove the stochastic integral to be a martingale. Later, we will see that the followingresult provides the necessary martingale argument for the verification theorem.4 emma 1.4
For b < δ let q be given like in (2) then it holds Z ∞ y − bδ e − aδ y d y < ∞ , Z s E r (cid:2) r − q − t (cid:3) d t < ∞ for all s ∈ R + . Proof:
Due to 2 b < δ it holds − < q < Z ∞ y − bδ e − aδ y d y = Z ∞ y (cid:0) − bδ (cid:1) − e − aδ y d y = Γ( − q ) (cid:16) aδ (cid:17) q < ∞ . Further, using (2) and the bounded convergence theorem, we get: E r (cid:2) r − q − t (cid:3) = ∞ X m =0 c ( t ) q +1+2 m e − c ( t ) re at m !Γ( m + q + 1) Z ∞ y m − bδ e − aδ y e − c ( t ) y d y = ∞ X m =0 c ( t ) q +1+2 m e − c ( t ) re at m !Γ( m + q + 1) · Γ( m − q ) (cid:0) aδ + c ( t ) (cid:1) m − q . Due to lim t → c ( t ) q +1+2 m e − c ( t ) re at = 0, the above power series is integrable over (0 , s ) forevery s ∈ R + . (cid:3) Throughout this paper we will use the following notation: for a fixed r ∗ ∈ R + , we define τ := inf { t ≥ r t = r ∗ , r = r ≤ r ∗ } ρ := inf { t ≥ r t = r ∗ , r = r ≥ r ∗ } . Further, we let ψ ( r ) := E r h Z τ e − r s d s i , for r ≤ r ∗ , (4) φ ( r ) := E r h [ ρ< ∞ ] i , for r ≥ r ∗ , (5) φ ( r ) := E r h [ ρ< ∞ ] ρ i for r ≥ r ∗ . (6)Since lim t →∞ r t = ∞ a.s., we know τ < ∞ a.s. Lemma 1.5
The functions ψ ( r ) , φ ( r ) and φ ( r ) solve the differential equations e − r + ( ar + b ) g ′ ( r ) + δ r g ′′ ( r ) = 0 , (7)( ar + b ) g ′ ( r ) + δ r g ′′ ( r ) = 0 , (8)( ar + b ) g ′ ( r ) + δ r g ′′ ( r ) + φ ( r ) = 0 , (9)5 orrespondingly with boundary conditions ψ ( r ∗ ) = 0 and ψ ′ (0) = − b ,φ ( r ∗ ) = 1 and φ ( ∞ ) = 0 ,φ ( r ∗ ) = 0 and φ ( ∞ ) = 0 . For the proof confer the appendix, Section 3.
Before considering an insurance company with surplus process following a Brownianmotion, we look at the problem of consumption maximisation for an individual with adeterministic income. The discounting factor is assumed to be given by an exponentialCIR process, { e − r t } . The filtration {F t } is generated by { r t } So, let the income processof the considered individual or household be given by X t = x + µt , µ > C denote the accumulated consumption process up to time t and the ex-consumptionincome be given by X Ct = x + µt − C t . We call a strategy C admissible if it is adapted to the filtration {F t } , is non-decreasingand fulfils C ≥ X Ct ≥ t ∈ R + , meaning in particular that C t ≤ x + µt . Inthe following, we denote the set of all admissible strategies by A . Our target is to findan optimal consumption strategy, such that E ( r,x ) h Z ∞ e − r s d C s i → max!i.e. the expected discounted consumption is maximised. The following notation will beused throughout this section: V C ( r, x ) := E ( r,x ) h Z ∞ e − r s d C s i ,V ( r, x ) := sup C ∈ A V C ( r, x ) . The Hamilton–Jacobi–Bellman (HJB) equation can be motivated using the standardmethods from stochastic control theory, so that we omit the detailed derivation and justrefer to [13, pp. 98,103] and references therein. The HJB turns out to consist of twopartial differential equations with linear coefficients:max (cid:8) µV x + ( ar + b ) V r + δ r V rr , e − r − V x (cid:9) = 0 . (10)6n particular, Proposition 2.5 below will illustrate that the HJB equation correspondsto the considered problem.To simplify our considerations we introduce the following notation L ( f )( r, x ) = µf x ( r, x ) + ( ar + b ) f r ( r, x ) + δ r f rr ( r, x )for any appropriate function f := R → R .In order to get an idea how the value function and the optimal strategy look like, weconsider first the performance function corresponding to the strategy “maximal spend-ing”. This function is given by H ( r, x ) := xe − r + µ E r h Z ∞ e − r s d s i = xe − r + µ Z ∞ M ( r, s ) d s , (11)with M ( r, t ) = E r [ e − r t ] and using Fubini’s theorem. The function M fulfils M ∈C , ( R ) and solves the differential equation, confer the appendix, Section 3:( ar + b ) M r ( r, t ) + δ r M rr ( r, t ) − M t ( r, t ) = 0 . This in particular means that, using the Leibniz integral rule, we get H r ( r, x ) = − xe − r + µ Z ∞ M r ( r, s ) d s and H rr ( r, x ) = xe − r + µ Z ∞ M rr ( r, s ) d s . Inserting H ( r, x ) into the HJB equation yields on the one hand e − r − H x ( r, x ) = 0 and onthe other hand using R ∞ M s ( r, s ) d s = M ( r, s ) (cid:12)(cid:12)(cid:12) ∞ = − e − r and the differential equationfor M we obtain that: µH x + ( ar + b ) H r + δ r H rr = µe − r + xe − r (cid:0) − ar − b + δ r (cid:1) + µ Z ∞ ( ar + b ) M r ( r, s ) + δ r M rr ( r, s ) d s = µe − r + xe − r (cid:0) − ar − b + δ r (cid:1) + µ Z ∞ M s ( r, s ) d s = µe − r + xe − r (cid:0) − ar − b + δ r (cid:1) − µe − r = xe − r (cid:0) − ar − b + δ r (cid:1) . Note that the sign of the above expression does not depend on x and define for δ > aR := b δ − a . (12)In the following we will consider different combinations of the parameters a , b and δ ,influencing the solution to the HJB equation (10).7 .1 The case δ ≤ a In this case it obviously holds − ar − b + δ r < b and r , which in turn means that H ( r, x ) solves the HJB equation(10).Now, we can formulate the following verification theorem: Theorem 2.1
The function H ( r, x ) is the value function and the strategy C max t := x + µt “ to alwaysspend the maximal possible amount independent of r and x ” is the optimal strategy. We skip the proof, as it goes similar to the proof of the verification theorem in the nextsubsection. a < δ In this case, H ( r, x ) defined in (11) does not solve the HJB equation (10) for r > R .For instance in [13, p. 27] one finds that in order to solve an optimisation problem thereare two ways: to show directly that the value function solves the HJB equation or toguess the optimal strategy and to prove that the corresponding return function solvesthe HJB equation. Here, we will follow the second method.We conjecture that the optimal strategy is of a barrier type, i.e. there is a positiveconstant ¯ r ∈ R + such that it is optimal to wait if r > ¯ r and to immediately spendeverything if r ≤ ¯ r . Since we do not know how the optimal barrier should look like, welet ¯ r ∈ R + be arbitrary but fixed. The corresponding return function consists of twoparts: F ( r, x ) := xe − r + µ E r h Z τ e − r s d s i + ˜ F , r ≤ ¯ rG ( r, x ) := E r h(cid:0) x + µρ (cid:1) [ ρ< ∞ ] i e − ¯ r + ˜ F E r h [ ρ< ∞ ] i , r > ¯ r . where ˜ F is some positive constant whose value should be determined later. It means that F describes the spendings if the initial value r = r ≤ ¯ r , and G describes the waitinguntil r t approaches ¯ r or ∞ .By construction, G (¯ r, x ) = F (¯ r, x ) and G x (¯ r, x ) = F x (¯ r, x ) for all x ∈ R + .The question is whether G and F given above solve the HJB equation (10) on [¯ r, ∞ )and on [0 , ¯ r ] respectively and fulfil G r (¯ r, x ) = F r (¯ r, x ) and G rr (¯ r, x ) = F rr (¯ r, x ) for all x ∈ R + with bounded derivatives F r and G r . F and G In this subsection, we investigate the properties of functions F and G . Using notation(4), we can rewrite F as follows F ( r, x ) = xe − r + µψ ( r ) + ˜ F . F into the HJB equation (10) we obtain on the one hand e − r − F x = 0 and on the other hand, using Lemma 1.5: L ( F )( r, x ) = µe − r + xe − r (cid:0) − ar − b + δ r (cid:1) + µ ( ar + b ) ψ ′ + µ δ r ψ ′′ ( r )= xe − r (cid:0) − ar − b + δ r (cid:1) . Thus, F solves the HJB equation (10) on the set [0 , ¯ r ] × R + , if ¯ r ≤ R , defined in (12).Consider now the function G . Using Definitions (5) and (6), G can be rewritten asfollows G ( r, x ) = xφ ( r ) e − ¯ r + µφ ( r ) e − ¯ r + ˜ F φ ( r ) . We are going to find out under which conditions G solves the HJB equation (10) on theinterval [¯ r, ∞ ). Lemma 1.5 yields L ( G )( r, x ) = µφ ( r ) e − ¯ r + (cid:0) xe − ¯ r + ˜ F (cid:1)n ( ar + b ) φ ′ ( r ) + δ r φ ′′ ( r ) o + µe − ¯ r n ( ar + b ) φ ′ ( r ) + δ r φ ′′ ( r ) o = µφ ( r ) e − ¯ r − µφ ( r ) e − ¯ r = 0 . Therefore, we have to consider e − r − G x ( r, x ) and search for conditions supplying therelation e − r − G x ( r, x ) ≤ r, ∞ ) × R + in order for G to solve the HJB equation (10).First, we prove the following auxiliary result: Lemma 2.2
The function φ ( r ) is decreasing, and there is a unique r ∗ ∈ [0 , R ] such that φ ( r ∗ ) = − φ ′ ( r ∗ ) = 1 and φ ( r ) > − φ ′ ( r ) for r > r ∗ . Proof:
For the proof confer the appendix, Section 3. (cid:3)
The following Lemma considers the expression e − r − G x ( r, x ) if the barrier is given by r ∗ defined above. Lemma 2.3
Let ¯ r = r ∗ , defined in Lemma 2.2. Then, for all r > r ∗ the following inequality holdstrue: G x ( r, x ) = φ ( r ) e − r ∗ > e − r . Proof:
Deriving e r φ ( r ) yields, using Lemma 2.2: (cid:0) e r φ ( r ) (cid:1) ′ = e r φ ( r ) (cid:16) φ ′ ( r ) φ ( r ) (cid:17) > e − r − G x ( r, x ) = e − r e − r ∗ (cid:16) e r ∗ − e r φ ( r ) (cid:17) < e − r e − r ∗ (cid:16) e r ∗ − e r ∗ φ ( r ∗ ) (cid:17) = 0for r > r ∗ . (cid:3) We can conclude that G solves the HJB (10) on [ r ∗ , ∞ ) if ¯ r = r ∗ .9 .2.2 The optimal strategy and verification theorem From now on we assume ¯ r = r ∗ , i.e. φ ′ ( r ∗ ) φ ( r ∗ ) = − F and G , solve the HJB equation (10) on [0 , r ∗ ] andon [ r ∗ , ∞ ) correspondingly. Next, we have to look at the derivatives G r ( r ∗ , x ), F r ( r ∗ , x )and G rr ( r ∗ , x ), F rr ( r ∗ , x ) in order to guarantee a smooth value function. It holds G r ( r ∗ , x ) = xe − r ∗ φ ′ ( r ∗ ) + µφ ′ ( r ∗ ) e − r ∗ + ˜ F φ ′ ( r ∗ )= − xe − r ∗ + µφ ′ ( r ∗ ) e − r ∗ − ˜ F ,F r ( r ∗ , x ) = − xe − r ∗ + µψ ′ ( r ∗ )and G rr ( r ∗ , x ) = xe − r ∗ φ ′′ ( r ∗ ) + µφ ′′ ( r ∗ ) e − r ∗ + ˜ F φ ′′ ( r ∗ ) ,F rr ( r ∗ , x ) = xe − r ∗ + µψ ′′ ( r ∗ ) . Remark 2.4
Choosing F ( r ∗ ,
0) = ˜ F = µφ ′ ( r ∗ ) e − r ∗ − µψ ′ ( r ∗ ) (13) yields G r ( r ∗ , x ) = F r ( r ∗ , x ) for all x ∈ R + . Note that it holds ˜ F ≥ due to the proof ofLemma 1.5, confer appendix, Section 3.Consider the differential equations (7) multiplied by ( − µ ) ; (8) multiplied by ˜ F and (9) multiplied by µe − r ∗ at r ∗ : − µe − r ∗ − ( ar ∗ + b ) µψ ′ ( r ∗ ) − δ r ∗ µψ ′′ ( r ∗ ) = 0 , ( ar ∗ + b ) ˜ F φ ′ ( r ∗ ) + δ r ∗ F φ ′′ ( r ∗ ) = 0 ,e − r ∗ ( ar ∗ + b ) µφ ′ ( r ∗ ) + µe − r ∗ δ r ∗ φ ′′ ( r ∗ ) + µe − r ∗ φ ( r ∗ ) = 0 . Using φ ( r ∗ ) = − φ ′ ( r ∗ ) = 1 and adding the above equations yields ( ar ∗ + b ) (cid:8) µφ ′ ( r ∗ ) e − r ∗ − µψ ′ ( r ∗ ) − ˜ F (cid:9) = − δ r ∗ (cid:8) µφ ′′ ( r ∗ ) e − r ∗ + ˜ F φ ′′ ( r ∗ ) − µψ ′′ ( r ∗ ) (cid:9) . Note that by definition of ˜ F , the lhs of the above equation equals zero, meaning G rr ( r ∗ ,
0) = F rr ( r ∗ , . However, in general it does not hold φ ′′ ( r ∗ ) = 1 . For that reason G rr ( r ∗ , x ) = F rr ( r ∗ , x ) if x = 0 . (cid:4) We formulate the following verification theorem.10 heorem 2.5
The optimal strategy C ∗ is to immediately spend any available amount bigger thanzero if r ≤ r ∗ , i.e. C ∗ t = (cid:0) x + µλ tr ∗ (cid:1) [ λ tr ∗ > , where λ tr ∗ := sup { s ∈ [0 , t ) : r s ≤ r ∗ } with sup {∅} = 0 . The value function V ( r, x ) solves the HJB equation (10) and fulfils V ( r, x ) = v ( r, x ) with v ( r, x ) = ( F ( r, x ) if ( r, x ) ∈ [0 , r ∗ ] × R + G ( r, x ) if ( r, x ) ∈ [ r ∗ , ∞ ) × R + with F ( r ∗ ,
0) = ˜ F given in (13) . Proof:
Note that it holds { λ tr ∗ ≤ u } = (cid:8) sup { s ∈ [0 , t ) : r s ≤ r ∗ } ≤ u (cid:9) = (cid:8) inf u r ∗ (cid:9) Because the running infimum above is F t -measurable, we can conclude that the strategy C ∗ defined above is an admissible strategy.Since F ∈ C , ((0 , r ∗ ) × R + ), G ∈ C , (( r ∗ , ∞ ) × R + ), F ( r ∗ , x ) = G ( r ∗ , x ), F r ( r ∗ , x ) = G r ( r ∗ , x ) and 1I [ r t = r ∗ ] = 1 a.s., we can apply the change-of-variable formula due to [11].Let C be an arbitrary admissible strategy and ˆ X the ex-consumption process under C .Then v ( r t , ˆ X t ) = v ( r, x ) + Z t L ( v )( r s , ˆ X s ) d s + Z t δ √ r s v r ( r s , ˆ X s ) d W s − Z t v x ( r s , ˆ X s ) d C s . Further, we know that v solves the HJB equation (10), meaning L ( v )( r, x ) ≤ e − r − v x ( r, x ) ≤ r, x ) ∈ R . Therefore, v ( r t , ˆ X t ) ≤ v ( r, x ) + Z t δ √ r s v r ( r s , ˆ X s ) d W s − Z t e − r s d C s . The stochastic integral above is a martingale with expectation zero, confer the proof ofLemma 1.5 in Section 3. Taking the expectations on the both sides of the above equality,one gets E ( r,x ) (cid:2) v ( r t , ˆ X t ) (cid:3) ≤ v ( r, x ) − E ( r,x ) h Z t e − r s d C s i . Because lim r →∞ v ( r, x ) = 0 and v ( r, x ) is bounded, by dominated convergence we caninterchange limit and integration and obtain v ( r, x ) ≥ E ( r,x ) h Z ∞ e − r s d C s i Taking the strategy C ∗ yields equality. (cid:3) Thus, the barrier strategy with barrier given by r ∗ , defined in Lemma 2.2, is optimal.The corresponding return function is the value function and solves the HJB equation(10). 11igure 2: Lhs: The value function V ( r, x ), consisting of F ( r, x ) (black) and G ( r, x )(grey). Rhs: Dependence of the barrier r ∗ on δ . Example 2.6
We can calculate the value function explicitly: ψ ( r ) = 2 δ Z r ∗ x y − bδ e − aδ y Z y e − z z bδ − e aδ z d z d y,φ ( r ) = 1 R ∞ r ∗ y − bδ e − aδ y d y Z ∞ r y − bδ e − aδ y d y,φ ( r ) = 2 δ R ∞ r ∗ y − bδ e − aδ y R yr ∗ φ ( z ) z bδ − e aδ z d z d y R ∞ r ∗ y − bδ e − aδ y d y Z rr ∗ y − bδ e − aδ y d y − δ Z rr ∗ y − bδ e − aδ y Z yr ∗ φ ( z ) z bδ − e aδ z d z d y. For well-definiteness of φ and φ confer the proof of Lemma 1.7 in the appendix, Section3. Let a = 0 . , b = 0 . , δ = 0 . and µ = 0 . . The both functions, F and G areillustrated in Figure 2. (cid:4) -barrier strategy In the following, we are going to discuss a very special strategy for the case 2 b < δ :“spending just if the underlying CIR hits zero”. In fact, we know from [7] that if 2 b < δ ,a CIR process can attain zero with a positive probability. With growing volatility, theprobability to “dive” and touch zero increases.In the previous subsection, we have shown that the value function solves the HJBequation (10) and that the optimal strategy is a barrier strategy with a constant barrier r ∗ fulfilling r ∗ ≤ R , with R given in (12). By the structure of R , it holds R → δ → ∞ . This brings up the question whether the optimal barrier could be equal to zerofor some big values of δ . 12et δ < ∞ be fixed and fulfil δ > { a, b } , meaning that the probability for { r t } to hit zero is positive. Due to subsection 2.1 the function H defined in (11) is not thevalue function.Consider the strategy C with C t := (cid:0) x + µλ t (cid:1) [ λ t > , λ t = sup { s ∈ [0 , t ) : r s = 0 } withsup {∅} = 0 “ to spend the maximal possible amount only if r t = 0”.Letting again ρ := inf { t ≥ r t = 0 , r > } , we define ˆ φ ( r ) := E r [ ρ < ∞ ] and ˆ φ ( r ) := E r [ ρ [ ρ < ∞ ] ]. Lemma 2.7
The functions ˆ φ ( r ) and ˆ φ ( r ) solve the differential equations (8) and (9) on (0 , ∞ ) correspondingly. The function φ ( r ) is finite on (0 , ∞ ) and it holds ˆ φ ( r ) = 1 R ∞ y − bδ e − aδ y d y Z ∞ x y − bδ e − aδ y d y, ˆ φ ( r ) = 2 δ R ∞ y − bδ e − aδ y R y φ ( z ) z bδ − e aδ z d z d y R ∞ y − bδ e − aδ y d y Z r y − bδ e − aδ y d y − δ Z r y − bδ e − aδ y Z y φ ( z ) z bδ − e aδ z d z d y Proof:
Due to Lemma 1.4 it holds R ∞ y − bδ e − aδ y d y < ∞ . Due to Lemma 1.5, thefunctions ˆ φ ( r ) = E r [ ρ < ∞ ] and ˆ φ ( r ) = E r [ ρ [ ρ< ∞ ] ] solve the differential equations (8)and (9) on (0 , ∞ ) correspondingly and the function φ is finite. (cid:3) Define further λ := sup { t ≥ r t = 0 } , i.e. λ is the last exit time from zero before the { r t } approaches ∞ . It is clear thatusing C , in case r = 0 one spends everything immediately, saves money until { r t } approaches zero for the next time and spends everything there. The game ends at time λ defined above. Lemma 2.8
Let δ > b , then λ < ∞ a.s. and E [ λ ] = Z ∞ t (cid:0) e at − (cid:1) − bδ R ∞ (cid:0) e az − (cid:1) − bδ d z d t < ∞ . Proof:
For the proof confer the appendix, Section 3. (cid:3)
Now, we can write down the return function corresponding to the strategy C : V ( r, x ) := E r (cid:2)(cid:0) x + µρ + ˜ V (cid:1) [ ρ < ∞ ] (cid:3) = ( x + ˜ V ) ˆ φ ( r ) + µ ˆ φ ( r ) , where ˜ V = V (0 ,
0) = µ E [ λ ]. 13 roposition 2.9 The strategy C cannot be optimal for any δ < ∞ . Proof:
Recall that the function ˆ φ ( r ) is given byˆ φ ( r ) = 1 R ∞ y − bδ e − aδ y d y Z ∞ x y − bδ e − aδ y d y . This means in particular ˆ φ (0) = 1 and lim r → ˆ φ ′ ( r ) = −∞ giving e − r − V x ( r, x ) = e − r − φ ( r ) > r ∈ (0 , ε ) and some ε >
0. This means that V does not solve the HJB (10) on(0 , ε ) × R + . Since in the previous subsection we have shown that the value functionsolves the HJB, we can conclude that C will never be optimal. (cid:3) In this subsection, we add complexity to our model by assuming that the underlyingsurplus (previously called income) process is given by a Brownian motion with drift.Since it is unrealistic to assume strong random fluctuations in the income of an individualor household, we change the economic interpretation from maximising the consumptionof an individual to the maximising of dividends of an insurance company. The differenceto the previous case comes up also in the fact that we stop our considerations when thesurplus becomes negative (ruins). Taking into consideration the ruin time destroys thelinear dependence of the value function on the surplus. In general, the return functioncorresponding to a constant barrier strategy will have a representation as a power serieswith non-linear functions as summands. Therefore - using the chess terminology - inorder to keep the problem in check, we assume for this subsection δ = 2 a .Technically, we consider an insurance company whose surplus is given by a Brownianmotion with drift X t = x + µt + σB t , where { B t } is a standard Brownian motion and µ, σ > t are given by C t , yielding for theex-dividend surplus X C : X Ct = x + µt + σB t − C t . The consideration will be stopped at the ruin time τ C of X C . Let further { W t } , theBrownian motion driving the discounting CIR process (1), be independent of { B t } , andthe underlying filtration {F t } be the filtration generated by the pair { W t , B t } . We calla strategy C admissible if C t is adapted to {F t } , C ≥ X Ct ≥ t ≥
0, theset of admissible strategies will be denoted by B .As a risk measure we consider the value of expected discounted dividends, where thedividends are discounted by an CIR process (1).14e define the return function corresponding to some admissible strategy C to be V C ( r, x ) = E ( r,x ) h R τ C e − r s d C s i and let V ( r, x ) = sup C ∈ B V C ( r, x ) . The HJB equation corresponding to the problem can be derived similarly to [13, pp.98,103]: max (cid:8) µV x + σ V xx + ( ar + b ) V r + arV rr , e − r − V x (cid:9) = 0 . (14)In this setup, we conjecture that the optimal strategy will be of a barrier type with aconstant barrier for the surplus process. It means, we pay any capital bigger than thebarrier, independent of { r t } . Define now the following auxiliary quantities: θ := − µ + p µ + 2 σ bσ , ζ := − µ − p µ + 2 σ bσ , ̺ := ln (cid:0) b − µζ (cid:1) − ln (cid:0) b − µθ (cid:1) θ − ζ . Lemma 2.10
The return function V ̺ ( r, x ) corresponding to the constant barrier strategy ̺ is givenby V ̺ ( r, x ) = ( F ( r, x ) : x ≥ ̺G ( r, x ) : x ≤ ̺, where F ( r, x ) := (cid:0) x − ̺ + µb (cid:1) e − r , if x ≥ ̺G ( r, x ) := e − r e θx − e ζx θe θ̺ − ζe ζ̺ , if x ≤ ̺. The functions F and G fulfil: • F ( r, ̺ ) = G ( r, ̺ ) , F r ( r, ̺ ) = G r ( r, ̺ ) , F rr ( r, ̺ ) = G rr ( r, ̺ ) ; • F x ( r, ̺ ) = e − r = G x ( r, ̺ ) and F xx ( r, ̺ ) = 0 = G xx ( r, ̺ ) for all r ∈ R + ; • G ( r, x ) solves the partial differential equation µf x + σ f xx + ( ar + b ) f r + arf rr = 0 and fulfils G x ( r, x ) ≥ e − r for all r ∈ R + and x ∈ [0 , ̺ ] ; Proof:
For the proof confer [14], Lemma 2.1 and Corollary 2.2. (cid:3) roposition 2.11 The optimal dividend strategy C ∗ is to pay any capital larger than ̺ , i.e. C ∗ t = max n sup ≤ s ≤ τ ∗ ∧ t (cid:16) x + µs + σW s (cid:17) − ̺ ; 0 o , where τ ∗ is the ruin time. The value function V ( r, x ) is given by F on [ ̺, ∞ ) , by G on [0 , ̺ ] and solves the HJB equation (14) . Proof:
Using Lemma 2.10, the proof follows closely the proof in [3], see also [13, p.104]. (cid:3)
Thus, if δ = 2 a , i.e. e bt e − r t is a martingale (confer Lemma 1.1 and Definition (3)),the optimisation problem can be reduced to the classical dividend optimisation problemwith a constant discounting rate, described in [3]. For the deterministic income, we considered different cases dependent on the relationbetween the parameters δ and a . In both cases, the optimal strategy turns out to beof a barrier type, i.e. it is optimal to spend all available money only if the process { r t } is below a certain level, otherwise it is optimal to wait.If the volatility coefficient δ is relatively small, i.e. δ ≤ a , then the paths are going“nearly deterministically” to infinity, meaning that e − r t is a supermartingale. Therefore,the optimal barrier is lying at infinity, and it is always optimal to spend the maximalpossible amount.If 2 a < δ , the process { r t } , moving δ from 2 a upwards shifts the optimal barrier from ∞ to 0, see Figure 2. It means, the higher the volatility the likely the process can hit alower level. It makes sense to wait until the discounting process attains “small” values,and spend the saved amount there.Finally, we showed that the strategy “spendings only if r t = 0” is never optimal, i.e. theoptimal barrier is always greater than zero.Note that the above results strongly differ from the case of integrated Ornstein-Uhlenbeck(OU) discounting. There, see [8], it was optimal to wait if the interest rate was below acertain level and to start consuming otherwise. The reason for swapping of the payingbehaviour in the case of an CIR discounting roots in the fact that an OU process canattain negative values. For more details confer [8].In the Brownian risk model, the case 2 a = δ has not been considered and is a subjectto future research. We conjecture that the optimal strategy there will be of a barriertype with a non-constant barrier depending on the underlying CIR process.16 Appendix
Proof of Lemma 1.3
Assume there exists a set A ∈ F with P [ A ] > t →∞ r t = B < ∞ on A . Then, there is asequence t n → ∞ as n → ∞ such that lim n →∞ r t n = B on A . By Lebesgue’s dominated convergencetheorem and using lim t →∞ E [ e − r t ] = lim t →∞ M ( r, t ) = 0, confer (3) for definition of M , we obtain0 = lim n →∞ E r [ e − r tn ] ≥ lim n →∞ E [ e − r tn A ] = e − B P [ A ] > . The last inequality is a contradiction proving our claim. (cid:3)
Proof of Lemma 1.5
Part I:
Due to [15, p. 127], the differential equation e − r + ( ar + b ) g ′ ( r ) + δ r g ′′ ( r ) = 0has twice continuously differentiable solutions on [0 , r ∗ ]. A general solution to the above differ-ential equation is given by g ′ ( r ) = (cid:16) − δ Z y − δ bδ e ( aδ − ) y d y + C (cid:17) e − aδ r r − bδ . Therefore, in order to have g ′ (0) > −∞ we must define g ′ ( r ) = (cid:16) − δ Z r y (cid:0) bδ − (cid:1) e (cid:0) aδ − (cid:1) y d y (cid:17) r − bδ e − aδ r . Now, letting r → r → g ′ ( r ) = − b . Let ˜ ψ ( r ) denote the unique solution with boundary conditions ˜ ψ ( r ∗ ) = 0 and ˜ ψ ′ (0) = − b . Inthis case it holds lim r →∞ r ˜ ψ ′′ ( r ) = 0 , which means ˜ ψ ′′ ( r ) ∈ o ( r ) for r →
0. Thus, we can apply Ito’s formula on ˜ ψ ( r τ ∧ t ):˜ ψ ( r τ ∧ t ) = ˜ ψ ( r ) + Z τ ∧ t ( ar s + b ) ˜ ψ ′ ( r s ) + δ r s ψ ′′ ( r s ) d s + Z τ ∧ t δ √ r s ˜ ψ ′ ( r s ) d W s . Since ˜ ψ ′ is bounded, the stochastic integral is a martingale with expectation zero. Therefore,taking the expectations on the both sides and letting t → ∞ (interchanging of expectations andlimit is possible due to the bounded convergence theorem) we obtain˜ ψ ( r ) = E r h Z τ e − r s d s i = ψ ( r ) . art II: It is clear that if 2 b ≥ δ and r ∗ = 0, we have φ ( r ) = 1I { } . Therefore, we just need to considerthe remaining cases. Differential equation (8) has a unique solution on ( r ∗ , ∞ ), say ˜ φ ( r ), withboundary conditions ˜ φ ( r ∗ ) = 1 and ˜ φ ( ∞ ) = 0:˜ φ ( r ) = 1 R ∞ r ∗ y − bδ e − aδ y d y Z ∞ r y − bδ e − aδ y d y . Applying Ito’s formula on ˜ φ yields˜ φ ( r ρ ∧ t ) = ˜ φ ( r ) + Z ρ ∧ t ( ar + b ) ˜ φ ′ ( r s ) + δ r φ ′′ ( r s ) d s + Z ρ ∧ t δ √ r s ˜ φ ′ ( r s ) d W s . (15)If r ∗ >
0, then √ r ρ ∧ s ˜ φ ′ ( r ρ ∧ s ) is bounded and the stochastic integral is a martingale withexpectation zero.If r ∗ = 0 and 2 b < δ then: (cid:16) √ r s ˜ φ ′ ( r s ) (cid:17) = r − bδ s e − aδ r s (cid:16) R ∞ y − bδ e − aδ y d y (cid:17) . Note that R ∞ y − bδ e − aδ y d y < ∞ and Z t E h(cid:16) √ r s ˜ φ ′ ( r s ) (cid:17) i d s < ∞ for all t ∈ R + due to Lemma 1.4. Then due to [12, p. 130 Corollary 1.25], the stochastic integralin (15) is a martingale with expectation zero. Applying the expectations and letting t go toinfinity in (15), one obtains ˜ φ ( r ) = E h [ ρ< ∞ ] i = φ ( r ) . Part III: If r ∗ = 0 and 2 b ≥ δ then obviously φ ( r ) ≡
0. Consider now the remaining cases.Differential equation (9) has a unique solution˜ φ ( r ) = 2 δ R ∞ r ∗ y − bδ e − aδ y R yr ∗ φ ( z ) z bδ − e aδ z d z d y R ∞ r ∗ y − bδ e − aδ y d y Z rr ∗ y − bδ e − aδ y d y − δ Z rr ∗ y − bδ e − aδ y Z yr ∗ φ ( z ) z bδ − e aδ z d z d y with boundary conditions ˜ φ ( r ∗ ) = 0 = ˜ φ ( ∞ ). Note that for r ∗ > φ given above: Z ∞ r ∗ y − bδ e − aδ y Z yr ∗ φ ( z ) z bδ − e aδ z d z d y ≤ r ∗ φ ( r ∗ ) < ∞ . Let now r ∗ = 0 and 2 b < δ . Then, applying partial integration for the inner integral and usingthat the negative part is smaller than zero, we get Z ∞ y − bδ e − aδ y Z y φ ( z ) z bδ − e aδ z d z d y ≤ δ b Z ∞ y − bδ e − aδ h y bδ e aδ φ ( y ) + y i < ∞ . he finiteness of the integral above follows from the properties of the Gamma distribution.Further, it is easy to see that ˜ φ ′ ( r ∗ ) >
0. Since ˜ φ solves the differential equation (9), it followsimmediately ˜ φ ′′ ( r ) < φ ′ ( r ) = 0. This means in particular that after becoming negative, thederivative ˜ φ ′ ( r ) remains negative. Therefore ˜ φ ( ∞ ) = 0 implies ˜ φ ( r ) ≥ φ ( r ) it holds˜ φ ( r ρ ∧ t ) = ˜ φ ( r ) + Z ρ ∧ t ( ar s + b ) ˜ φ ′ ( r s ) + δ r s φ ′′ ( r s ) d s + Z ρ ∧ t δ √ r s ˜ φ ′ ( r s ) d W s . Similar to Part II, using Lemma 1.4 and [12, p. 130 Corollary 1.25] one can show that thestochastic integral above is a martingale with expectation zero. Applying the expectations yields E h ˜ φ ( r ρ ∧ t ) i = ˜ φ ( r ) − E h Z ρ ∧ t φ ( r s ) d s i . Note that applying Fubini’s theorem on the expectation on the rhs, one obtains E h Z ρ ∧ t φ ( r s ) d s i = Z ∞ E h [0 ≤ s ≤ ρ ∧ t ] φ ( r s ) i d s = Z ∞ E h [0 ≤ s ≤ ρ ∧ t ] P [ ρ < ∞| r s ] i d s = Z ∞ E h [0 ≤ s ≤ ρ ∧ t ] E [1I [ ρ< ∞ ] |F s ] i d s = Z ∞ E h E (cid:2) [0 ≤ s ≤ ρ ∧ t ] [ ρ< ∞ ] |F s (cid:3)i d s = E h Z ρ ∧ t [ ρ< ∞ ] d s i = E h [ ρ< ∞ ] ρ ∧ t i . Letting t → ∞ , yields ˜ φ ( r ) = E h [ ρ< ∞ ] ρ i = φ ( r ) . Note that ψ ′ ( r ) < φ ′ ( r ∗ ) ≥ φ ( r ∗ ) = 0 and φ ( r ) ≥ φ ′ ( r ∗ ) ≥
0. On the other hand, if ˜ r := inf { r ≥ ψ ′ ( r ) ≥ } ≤ r ∗ it must hold ψ ′′ (˜ r ) < ψ solves the differentialequation (7). Since, it is a contradiction we can conclude ψ ′ ( r ∗ ) < r ∈ [0 , r ∗ ]. (cid:3) Proof of Lemma 2.2
First note that it obviously holds φ ′ <
0. We can solve the differential equation (8) explicitlyand obtain φ ′ ( r ) = − r − bδ e − aδ r R ∞ r ∗ y − bδ e − aδ y d y . (16)Consider now φ ′ ( r ) φ ( r ) = − r − bδ e − aδ r R ∞ r y − bδ e − aδ y d y . Deriving φ ′ ( r ) φ ( r ) with respect to r yields (cid:16) φ ′ ( r ) φ ( r ) (cid:17) ′ = − φ ′ ( r ) φ ( r ) n − φ ′′ ( r ) φ ′ ( r ) + φ ′ ( r ) φ ( r ) o . Using (16) we obtain φ ′′ ( r ) φ ′ ( r ) = − aδ − bδ r , et for simplicity h ( r ) := φ ′ ( r ) φ ( r ) . Then h ′ ( r ) = − h ( r ) n aδ + 2 bδ r + h ( r ) o = − h ( r ) n h ( r ) o − h ( r ) n aδ + 2 bδ r − o . Note that − h ( r ) > aδ + bδ r − < r > R . That is, if for some ˆ r > R it holds h (ˆ r ) = − h ′ (ˆ r ) <
0, meaning that h (ˆ r ) < − r > ˆ r . But this is a contradiction tolim r →∞ h ( r ) = lim r →∞ φ ′′ ( r ) φ ′ ( r ) = − aδ > − . Therefore, it must hold h ( r ) > − r > R .Because lim r → h ( r ) = −∞ , by the intermediate value theorem there must be an r ∗ ∈ (0 , R ] suchthat h ( r ∗ ) = − (cid:3) Proof of Lemma 2.8
Note that λ y is not a stopping time, because { λ y ≤ t } / ∈ F t .Further, we know from lemma 1.3 and Borodin & Salminen [5, p. 27] that λ y < ∞ a.s. with P r [0 < λ y ≤ t ] = Z t p ( u ; r, y ) G ( y, y ) d u where p ( t ; r, y ) is the transition density of { r t } with respect to the speed measure m with density m ′ of { r t } (for the exact formula for m and m ′ confer [10, p. 366] formula (1.4); the differentialequation for m ′ can be found in [5, p. 18]) and G α ( r, y ) is the Green function with G ( y, y ) = Z ∞ p ( t ; y, y ) d t . Let g ( t ; r, y ) be the density of { r t } with respect to the Lebesgue measure. Then g ( t ; y, y ) = p ( t ; y, y ) m ′ ( y ) . Therefore, using the above formula for G ( y, y ): P y [0 < λ y ≤ t ] = Z t g ( u ; y, y ) R ∞ g ( z ; y, y ) d z d u . According to (2) the density g ( t ; y, y ) is given by g ( t ; y, y ) = c ( t ) e − u ( t,y ) − v ( t,y ) (cid:16) v ( t, y ) u ( t, y ) (cid:17) q/ I q (2 p u ( t, y ) v ( t, y )) , and I q is modified Bessel function of the first kind of order q . Using this explicit representationand I q (2 c ( t ) ye at/ ) = ( c ( t ) e at/ ) q y q ∞ P m =0 ( c ( t ) ye at/ ) m m !Γ( m + q +1) , we obtain g ( t ; y, y ) R ∞ g ( z ; y, y ) d z = c ( t ) e − c ( t ) y ( e at +1) e − aqt/ I q (cid:0) c ( t ) ye at/ (cid:1)R ∞ c ( z ) e − c ( z ) y ( e az +1) e − aqz/ I q (cid:0) c ( z ) ye az/ (cid:1) d z = c ( t ) e − c ( t ) y ( e at +1) e − aqt/ ( c ( t ) e at/ ) q ∞ P m =0 ( c ( t ) ye at/ ) m m !Γ( m + q +1) R ∞ c ( z ) e − c ( z ) y ( e az +1) e − aqz/ ( c ( z ) e az/ ) q ∞ P m =0 ( c ( z ) ye az/ ) m m !Γ( m + q +1) d z y bounded convergence theorem, we can let y go to zero and obtain g ( t ; 0 , R ∞ g ( z ; 0 ,
0) d z = c ( t ) e − aqt/ ( c ( t ) e at/ ) q R ∞ c ( z ) e − aqz/ ( c ( z ) e az/ ) q d z = c ( t ) q +1 R ∞ c ( z ) q +1 d z . Note that indeed it holds by partial integration and using − < q < (cid:16) aδ (cid:17) − q − Z ∞ c ( z ) q +1 d z = Z ∞ (cid:0) e az − (cid:1) − q − d z = 1 − aq Z ∞ (cid:0) e az − (cid:1) − q e − az d z = 1 − aq Z ∞ e aqz (cid:0) e az − (cid:1) − q e − (1+ q ) az d z ≤ − aq Z ∞ e − (1+ q ) az d z = 1 − qa (1 + q ) < ∞ . The last inequality follows because e aqz (cid:0) e az − (cid:1) − q ≤ − − q <
0. With similar argumentsone obtains E [ λ ] = Z ∞ t g ( t ; 0 , R ∞ g ( z ; 0 ,
0) d z d t = Z ∞ t (cid:0) e at − (cid:1) − q − R ∞ (cid:0) e az − (cid:1) − q − d z d t < ∞ . (cid:3) Acknowledgments
The research of the first author was funded by the Austrian Science Fund (FWF), Project numberV 603-N35. Also, the first author would like to thank the University of Liverpool for supportand cooperation.
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