on reinsurance and investment for large insurance portfolios

Insurance: Mathematics and Economics 42 (2008) 434–444www.elsevier.com/locate/ime

On reinsurance and investment for large insurance portfolios

Shangzhen Luoa,∗, Michael Taksarb, Allanus Tsoib

a Department of Mathematics, University of Northern Iowa, Cedar Falls, IA 50614, USAb Department of Mathematics, University of Missouri, Columbia, MO 60211, USA

Received January 2007; received in revised form March 2007; accepted 27 April 2007

Abstract

We consider a problem of optimal reinsurance and investment for an insurance company whose surplus is governed by a linear diffusion. Thecompany’s risk (and simultaneously its potential profit) is reduced through reinsurance, while in addition the company invests its surplus in afinancial market. Our main goal is to find an optimal reinsurance–investment policy which minimizes the probability of ruin. More specifically,in this paper we consider the case of proportional reinsurance, and investment in a Black–Scholes market with one risk-free asset (bond, or bankaccount) and one risky asset (stock). We apply stochastic control theory to solve this problem. It transpires that the qualitative nature of thesolution depends significantly on the interplay between the exogenous parameters and the constraints that we impose on the investment, such asthe presence or absence of shortselling and/or borrowing. In each case we solve the corresponding Hamilton–Jacobi–Bellman equation and find aclosed-form expression for the minimal ruin probability as well as the optimal reinsurance–investment policy.c© 2007 Elsevier B.V. All rights reserved.

PACS: IM52; IE53; IB91

Keywords: Ruin probability; Stochastic control; Black–Scholes model; Hamilton–Jacobi–Bellman equation; Proportional reinsurance

1. Introduction

One of the fundamental objectives that an insurancecompany pursues is the minimization of its ruin probability,i.e. the probability of the event that the surplus becomes non-positive. In this paper, the insurance company is allowed totake reinsurance and/or invest its capital in a Black–Scholesmarket. Some related problems have been dealt with throughoptimization and stochastic control techniques (see Browne(1997), Emanuel et al. (1975), Hipp and Plum (2000), Schmidli(2001), Schimidli (2002), Taksar and Markussen (2003)). Therehave been two major types of mathematical model for thesurplus of the insurance company. In the first type of model,the surplus process is represented by the classical risk process,i.e. the Cramer–Lundberg model. In the second type, the surplusprocess is represented by a diffusion, i.e. one uses a diffusionapproximation for the classical Cramer–Lundberg risk process.

∗ Corresponding author. Fax: +1 319 273 2546.E-mail addresses: [email protected] (S. Luo), [email protected]

(M. Taksar), [email protected] (A. Tsoi).

0167-6687/$ - see front matter c© 2007 Elsevier B.V. All rights reserved.doi:10.1016/j.insmatheco.2007.04.002

The second type is used usually when one deals with largeinsurance portfolios, where an individual claim is relativelysmall compared to the size of the surplus. In Schimidli(2002), one considers the model under which reinsurance andinvestment are allowed, the surplus process is modeled by theclassical risk process, and a numerical procedure for solving theHamilton–Jacobi–Bellman (HJB) equation is given. In Taksarand Markussen (2003), the surplus process is modeled by adiffusion process, reinsurance is allowed and there is no controlon investment activities, where all the surplus is invested inone financial asset. There, a closed-form solution, includingruin probability and optimal reinsurance policy, is obtained bysolving the HJB equation analytically.

In this paper we consider proportional reinsurance under theBlack–Scholes environment, in which dynamical portfolio re-adjustment is allowed. Through stochastic control techniqueswe provide a complete solution to the correspondingoptimization problem. As a result we find explicit expressionsfor the optimal ruin probability and the structure of the optimalcontrol policies. This solution provides us with an opportunityto observe the sensitivity of the optimal ruin probability

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http://dx.doi.org/10.1016/j.insmatheco.2007.04.002

S. Luo et al. / Insurance: Mathematics and Economics 42 (2008) 434–444 435

and the optimal control to the exogenous parameters andconstraints on investment. As we will see, the change of theexogenous parameters and investment constraints results inrather significant qualitative changes of the optimal controlpolicies and the minimal ruin probability.

In proportional reinsurance, the reinsurer is required topay a certain fraction of each claim, while in return thecedent (insurer) diverts the same or a larger fraction of allthe premiums to the reinsurer. If the safety loading of thereinsurer and the cedent are the same; that is, if the fractionof the premium diverted to the reinsurer is the same as thefraction of each claim covered by the reinsurer, then the contractis called a cheap reinsurance; if the safety loading of thereinsurer is higher than that of the cedent, such a contract iscalled a noncheap reinsurance. In this paper, we will workon the diffusion approximation model (see Emanuel et al.(1975), Garrido (1989), Højgaard and Taksar (1998), Taksarand Markussen (2003)). If both reinsurance and investment areabsent, the dynamics of the surplus is given by

dRt = µdt + σ0dw(0)t ;

R0 = x

where w(0)· is a standard Brownian motion. The constant x with

x > 0 is the initial surplus, while the constants µ > 0 and σ0 >

0 are the exogenous parameters. Suppose now that we takeproportional reinsurance into consideration. The proportionalreinsurance level is associated with the value (1 − a), where0 ≤ a ≤ 1 is called the risk exposure. If the risk exposure ofthe company is fixed, then the cedent pays 100a% of each claimwhile the rest 100(1 − a)% is paid by the reinsurer. To this end,the cedent diverts part of the premiums to the reinsurer at therate of λ(1 − a) with λ ≥ µ. As has already been mentioned,when λ = µ, the reinsurance is called cheap, while if λ > µ, itis called noncheap. The corresponding diffusion approximationdynamics for the surplus process becomes:

dRt = [µ − (1 − a)λ]dt + aσ0dw(0)t ;

R0 = x .

In addition, we use b to denote the fraction of the surplus beinginvested in a risky asset, whose price process is governed by theclassical Black–Scholes dynamics:

dSt = r1St dt + σ1St dw(1)t .

Here w(1)· is a standard Brownian motion independent of w

(0)· ,

while r1 > 0 and σ1 > 0 are the exogenous parametersassociated with the stock price stochastic process. The portion(1−b) of the surplus is invested in a risk-free asset with growthrate r0 > 0 and dynamics:

dBt = r0 Bt dt.

We treat the risk exposure a and the fraction b of the surplusbeing invested in the risky asset as control parameters. At anytime t ≥ 0, the fractions a = a(t) and b = b(t) are chosen bythe company. We denote π(·) = (a(·), b(·)). Once the policy

π(·) is chosen, the dynamics of the surplus process becomes

dRπt = [µ − (1 − a(t))λ + r0(1 − b(t))Rπ

t + r1b(t)Rπt ]dt

+ a(t)σ0dw(0)t + b(t)σ1 Rπ

t dw(1)t ; (1.1)

R0 = x .

Ruin is defined as in the event that the surplus becomesnonpositive. Our objective is to minimize the probability of ruinby identifying the optimal reinsurance and investment policyrepresented by π .

There are four types of constraint on the investment: (i) bothshortselling and borrowing are allowed; (ii) neither borrowingnor shortselling is allowed; (iii) borrowing is allowed but notshortselling; (iv) shortselling is allowed but not borrowing. Inthe first case, fraction b(t) of the surplus in the risky assetcan take any real value, i.e. −∞ < b(t) < ∞; For instance,if b(t) > 0, it indicates that the insurer purchases the riskyasset while b(t) < 0 indicates shortselling. In the second case,fraction b(t) is only allowed to be within [0, 1]. However, theconstraint b(t) > 0 indicates no shortselling allowed whileb(t) < 1 indicates no borrowing. We will present the solutionsfor the first two cases in Sections 3 and 4 respectively. In thelast section, we show how the other two cases can be reduced tothe previous ones. Most of the proofs in this paper are omittedand we refer them to Luo et al. (2007), where a full versionincluding the detailed proofs and a special case of zero interestrate is provided.

2. Formulation of the problem, Hamilton–Jacobi–Bellmanequation and verification theorem

We start with a complete probability space (Ω ,F, P),endowed with a filtration G = Gt t≥0, and two independentstandard Brownian motions w(0) and w(1) adapted to G. A(control) policy π(·) = (a(·), b(·)) is said to be admissible ifa(·) and b(·) are predictable with respect to G and for each t ≥ 0the processes a(·) and b(·) satisfy the following constraints

(1) 0 ≤ a(t) ≤ 1,(2) b(t) ∈ A,

where A = (−∞, ∞) in the case when both shortsellingand borrowing are allowed, A = [0, 1] when there is neithershortselling nor borrowing, A = [0, ∞) if borrowing but notshortselling is allowed, and A = (−∞, 1] if shortselling butnot borrowing is allowed. The set of all admissible policies isdenoted by Π . Once a policy π is chosen, the dynamics of thesurplus process is given by (1.1). The ruin time is defined as:

τπ = inft > 0 : Rπt ≤ 0. (2.1)

We define Vπ (x) as the ruin probability under policy π whenthe initial surplus is x . That is,

Vπ (x) = Px (τπ < ∞) = P(τπ < ∞|R0 = x). (2.2)

The objective is to find the optimal value function

V (x) = infπ∈Π

Vπ (x); (2.3)

436 S. Luo et al. / Insurance: Mathematics and Economics 42 (2008) 434–444

and the optimal policy π∗ such that

Vπ∗(x) = V (x). (2.4)

The above problem is equivalent to an infinite time horizoncontrol problem, which we will solve using dynamicprogramming techniques (Fleming and Rishel, 1975; Flemingand Soner, 1993; Whittle, 1983; Yong and Zhou, 1999)).

We start with the associated Hamilton–Jacobi–Bellman(HJB) equation for the value function V :

Theorem 2.1. Assume that V defined by (2.3) is twicecontinuously differentiable on (0, ∞). Then V satisfies thefollowing Hamilton–Jacobi–Bellman equation:

0 = inf0≤a≤1,b∈A

[µ − (1 − a)λ + ((1 − b)r0 + br1)x] V ′(x)

+12(a2σ 2

0 + b2σ 21 x2)V ′′(x)

. (2.5)

The proof of this theorem is standard (see chapter IVin Fleming and Soner (1993) or Taksar and Markussen(2003)). We will seek a decreasing, convex, twice continuouslydifferentiable solution to the Eq. (2.5) subject to the followingboundary conditions

V (0) = 1;

V (∞) = 0.(2.6)

These conditions say that when the surplus vanishes, the ruinprobability is 1, while when the surplus tends to infinity, theruin probability converges to 0. The following verificationtheorem is essential in solving the associated stochastic controlproblem.

Theorem 2.2. Let W ∈ C2 be a decreasing convex solution toHJB equation (2.5) subject to the boundary conditions (2.6).Then the value function V given by (2.3) coincides with W .Furthermore, let (a∗(x), b∗(x)) be such that

0 = [µ − (1 − a∗(x))λ + ((1 − b∗(x))r0 + b∗(x)r1)x]W ′(x)

+12[(a∗(x))2σ 2

0 + (b∗(x))2σ 21 x2

]W ′′(x), (2.7)

for all 0 ≤ x < ∞. Then the policy π∗(·) of the followingfeedback form π∗(s) = (a∗(Rπ∗

s ), b∗(Rπ∗

s )), where Rπ∗

s is thesolution to (1.1), is the optimal policy. That is,

W (x) = V (x) = Vπ∗(x).

The proof of the verification theorem relies on the followinglemma, whose proof is identical to the proof of Lemma 6.1in Taksar and Markussen (2003).

Lemma 2.1. Let

ηNπ = inft > 0 : Rπ

t ≥ N , (2.8)

and

τ Nπ = min(τπ , ηN

π ) ≡ inft > 0 : Rπt 6∈ [0, N ] (2.9)

then for any N > 0 and any policy π

P(τ Nπ < ∞) = 1. (2.10)

Remark 2.1. We note that Lemma 2.1 is indispensable forproving the verification theorem. Otherwise we can have amodel in which the optimal value function can be identicallyzero for every nonzero initial surplus. For example, in the caseof cheap reinsurance, i.e. when λ = µ, we can let a = 0 andb = 0 so that dRπ

t = 0. I.e. Rπt = x for all t ≥ 0. This means

that the insurer can pass all the premiums to the reinsurer andthe reinsurer will cover all the claims, while the insurer keepsthe initial surplus x as cash forever. Hence the ruin probabilityis 0 in this case.

3. The case with both shortselling and borrowing

In this case the control b can take on any real value. The HJBequation is

0 = inf0≤a≤1,−∞<b<∞

[µ − (1 − a)λ

+ ((1 − b)r0 + br1)x]V ′(x)

+12(a2σ 2

0 + b2σ 21 x2)V ′′(x)

. (3.1)

We modify this equation by replacing the control b by β = bx .This control corresponds to the total amount of money investedon the risky asset rather than to the fraction of the surplus. Thedynamics of the controlled process Rt is given by the sameEq. (1.1) with b(t)Rπ

t replaced by β(t). The corresponding HJBequation becomes

0 = inf0≤a≤1,−∞<β<∞

[µ − (1 − a)λ

+ r0x + β(r1 − r0)]V′(x)

+12(a2σ 2

0 + β2σ 21 )V ′′(x)

. (3.2)

To solve the HJB equation we will need several lemmas whichwe formulate here. Put

x∗=

λ − µ

r0; (3.3)

Lemma 3.1. If the initial surplus satisfies x ≥ x∗, then V (x)

= 0.

Remark 3.1. The lemma above says that if the insurancecompany has a sufficiently large initial surplus, then the interestobtained by investing all the surplus on the risk-free bondcan cover the shortfall between the premiums received andthe amount needed to pay to the reinsurance company whichcovers 100% of each claim. Hence the surplus of the insurancecompany will never vanish.

Remark 3.2. In the sequel, when considering the cases withr0 > 0 we will seek a solution to HJB equation (2.5) on(0, x∗) automatically extending this solution to (0, ∞) bysetting V (x) = 0 for x ≥ x∗. Such an extension mightresult in a function V which is twice continuously differentiableeverywhere except the point x∗. At x∗ the function V andits first derivative are continuous, while the second derivative


might have a discontinuity of the first order. This function maynot formally satisfy the condition of the verification theorem(Theorem 2.2); however, the proof of this theorem is based onthe application of Ito’s formula, which is valid for the functionswhose second derivative has a finite number of discontinuities.Therefore, we can apply Theorem 2.2 to V in the same way asif it had been a C2 function. Thus, when we will write that Vis a C2 function everywhere except the point x∗, we will meanthat V is a C1 function whose second derivative is continuouseverywhere except the point x∗ where the second derivative hasa discontinuity of the first order.

Next, we consider HJB equation (3.2) and we try to find aconvex, decreasing, twice continuously differentiable solutionwhich satisfies the boundary conditions V (0) = 1 and V (x∗) =

0 on the set:

O = x : 0 < x < x∗. (3.4)

For any C2 decreasing convex function W , denote

aW (x) = −λW ′(x)

σ 20 W ′′(x)

,

bW (x) = −(r1 − r0)W ′(x)

xσ 21 W ′′(x)

,

βW (x) = xbW (x).

(3.5)

The expressions above will be used to find the minimizers ofthe HJB equations.

Lemma 3.2. Suppose V is a decreasing, convex and twicecontinuously differentiable function on O, and let

OV 1 = 0 < x < x∗: aV (x) < 1. (3.6)

If V is a solution to

−(λ − µ − r0x)V ′(x)

V ′′(x)

−12

(λ2

σ 20

+(r1 − r0)

2

σ 21

)(V ′(x)

V ′′(x)

)2

= 0, (3.7)

then V solves HJB equation (3.2) on OV 1 and the converse ofthis statement also holds.

Similarly we can obtain:

Lemma 3.3. Suppose V is a decreasing, convex, twicecontinuously differentiable function on O, and let

OV 2 = 0 < x < x∗: aV (x) ≥ 1. (3.8)


12σ 2

0 + (µ + r0x)V ′(x)

V ′′(x)−

12

(r1 − r0)2

σ 21

(V ′(x)

V ′′(x)

)2

= 0, (3.9)

then V solves HJB equation (3.2) on OV 2 and vice-versa.

We proceed with solving HJB equation (3.2). There are twocases to be considered separately:

(1) µ < λ < µ +

√µ2 +

σ 20

σ 21(r1 − r0)2;

(2) λ ≥ µ +

√µ2 +

σ 20

σ 21(r1 − r0)2.

3.1. The case µ < λ < µ +

√µ2 +

σ 20

σ 21(r1 − r0)2

Suppose V solves HJB equation (3.2), then on OV 1,Lemma 3.2 implies

V ′(x)

V ′′(x)=

1g1(x)

, (3.10)

where

g1(x) =

λ2

σ 20

+(r1−r0)

2

σ 21

2(µ − λ + r0x). (3.11)

Noticing aV (x) < 1 on OV 1, from (3.5) and (3.10) we havex > x0, where

x0 =

λ2− 2µλ −

σ 20

σ 21(r1 − r0)

2

2λr0; (3.12)

Since x0 < 0 under λ < µ +

√µ2 +

σ 20

σ 21(r1 − r0)2, Thus,

OV 1 = O. Note that OV 2 is empty. Solving (3.10) withboundary conditions V (0) = 1 and V (x∗) = 0, the discussionin this section yields (by the verification theorem):

Theorem 3.1. If λ < µ +

√µ2 +

σ 20

σ 21(r1 − r0)2, then the

minimum ruin function is given by

V (x) =

[λ − µ − r0x

λ − µ

]1+λ2

2r0σ20

+(r1−r0)2

2r0σ21 , 0 < x < x∗

0, x ≥ x∗,

(3.13)

which is a decreasing convex C2 function everywhere except atx∗. The optimal risk exposure feedback function is given by

a∗(x) =

2λ(λ − µ − r0x)

λ2 +σ 2

0 (r1−r0)2

σ 21

, 0 < x < x∗

0, x ≥ x∗,

(3.14)

and the optimal investment feedback function is given by

β∗(x) =

2(r1 − r0)(λ − µ − r0x)

λ2σ 21

σ 20

+ (r1 − r0)2, 0 < x < x∗

0, x ≥ x∗.

(3.15)

3.2. The case λ ≥ µ +

√µ2 +

σ 20

σ 21(r1 − r0)2

Suppose V is a solution to (3.2). Then on OV 1 the conditionaV (x) < 1 implies x > x0, where x0 is non-negative under λ ≥


µ +

√µ2 +

σ 20

σ 21(r1 − r0)2. Thus, OV 1 = x : x0 < x < x∗

.

Then a solution to (3.10) with boundary condition V (x∗) = 0is of the following form:

V1(x) = eC1

∫ x∗

xe−

∫ x∗

u g1(v)dvdu, (3.16)

where C1 is a constant to be determined later.Next, we find an expression for the solution to (3.2) on

OV 2. If V solves (3.9), by Lemma 3.3, on OV 2, V satisfiesV ′(x)V ′′(x)

=1

g2(x), where

g2(x) =

(r1−r0)2

σ 21

µ + r0x −

√(µ + r0x)2 +

σ 20

σ 21(r1 − r0)2

if r0 6= r1,

−2(µ + r0x)

σ 20

if r0 = r1.

(3.17)

Thus, aV (x) ≥ 1 is equivalent to x ≤ x0 and hence OV 2 =

0 < x < x0. The solution to (3.9) on OV 2 with boundarycondition V (0) = 1 is of the following form:

V2(x) = 1 − eC2

∫ x

0e∫ u

0 g2(v)dvdu, (3.18)

where C2 is a constant to be determined later. To determine thefree constants, we apply a smooth fit at x0 by setting

V1(x0) = V2(x0); V ′

1(x0) = V ′

2(x0).

Solving the above two equations for C1 and C2 results in

C1 = C2 +

∫ x0

0g2(x)dx +

∫ x∗

x01g1(x)dx;

C2 = − ln

[exp

∫ x0

0g2(x)dx +

∫ x∗

x0

g1(x)dx

×

∫ x∗

x0

e−∫ x∗

u g1(v)dvdu +

∫ x0

0e∫ u

0 g2(v)dvdu

]. (3.19)

Lemma 3.4. V ′′

1 (x0) = V ′′

2 (x0).

The verification theorem and Remark 3.2 lead to thefollowing theorem:

Theorem 3.2. If λ ≥ µ +

√µ2 +

σ 20

σ 21(r1 − r0)2, the minimum

ruin probability function V is given by

V (x) =

1 − eC2

∫ x

0e∫ u

0 g2(v)dvdu, 0 < x ≤ x0,

eC1

∫ x∗

xe−

∫ x∗

u g1(v)dvdu, x0 < x < x∗,

0, x ≥ x∗.

which is a decreasing convex C2 function everywhere except atx∗. Here x0 is given by (3.12), g1(x) by (3.11), g2(x) by (3.17),and C1, C2 by (3.19). The optimal risk exposure feedback

function is

a∗(x) =

1, 0 < x ≤ x0,2λ(λ − µ − r0x)

λ2 +(r1−r0)

2σ 20

σ 21

, x0 < x < x∗,

0, x ≥ x∗.

The optimal investment feedback function is:

β∗(x) =

−

r1 − r0

σ 21 g2(x)

, 0 < x ≤ x0,

2(r1 − r0)(λ − µ − r0x)

λ2σ 21

σ 20

+ (r1 − r0)2, x0 < x < x∗,

0, x ≥ x∗.

4. The case with neither shortselling nor borrowing

In this case, the range A of the control functional b(·)

is [0, 1]. The HJB equation for the optimal ruin probabilityfunction V becomes:

inf0≤a≤1,0≤b≤1

[µ − (1 − a)λ + ((1 − b)r0 + br1)x] V ′(x)

+12(a2σ 2

0 + b2σ 21 x2)V ′′(x)

= 0. (4.1)

Next, we state several lemmas in order to obtain the solution ofthis equation.

4.1. Several lemmas

Lemma 4.1. Suppose V is a twice continuously differentiabledecreasing, convex function on O. Let OV 1 = x ∈ O :

aV (x) < 1, bV (x) < 1. If V is a solution to

V ′(x)

V ′′(x)= −

2(λ − µ − r0x)

λ2

σ 20

+(r1−r0)

2

σ 21

, (4.2)

then V solves HJB equation (4.1) on OV 1 and vice versa.

Lemma 4.2. Suppose V is a twice continuously differentiabledecreasing, convex function on O and let OV 2 = x ∈ O :

aV (x) < 1, bV (x) ≥ 1. If V is a solution to

λ2

σ 20

(V ′(x)

V ′′(x)

)2

+ 2(λ − µ − r1x)V ′(x)

V ′′(x)− σ 2

1 x2= 0, (4.3)

then V solves HJB equation (4.1) on OV 2 and vice versa.

Similarly, we have the following two lemmas:

Lemma 4.3. Suppose V is a twice continuously differentiabledecreasing, convex function on O and let

OV 3 = x ∈ O : aV (x) ≥ 1, bV (x) < 1. (4.4)


(r1 − r0)2

σ 21

(V ′(x)

V ′′(x)

)2

− 2(µ + r0x)V ′(x)

V ′′(x)− σ 2

0 = 0, (4.5)

then V solves Eq. (4.1) on OV 3 and vice versa.


Lemma 4.4. Suppose V is a twice continuously differentiabledecreasing, convex function on O, and let

OV 4 = x ∈ O : aV (x) ≥ 1, bV (x) ≥ 1. (4.6)


V ′(x)

V ′′(x)= −

σ 20 + σ 2

1 x2

2(µ + r1x), (4.7)

then V solves Eq. (4.1) on OV 4 and vice versa.

The lemmas above will be applied to obtain a decreasing andconvex C2 solution V to Eq. (4.1). In what follows, we supposethat such a function V exists, and then consider seperately thebehavior of V on the sets OV 1, OV 2, OV 3 and OV 4 to deriveexpressions of V on them. We then conclude that V is theoptimal solution by the verification theorem.

In the next four subsections, we consider the following fourcases, respectively:

(1) r1 > r0 and µ < λ ≤ 2µ;

(2) r1 > r0 and 2µ < λ < µ +

√µ2 + (r2

1 − r20 )

σ 20

σ 21

;

(3) r1 > r0 and λ ≥ µ +

√µ2 + (r2

1 − r20 )

σ 20

σ 21

;

(4) r0 ≥ r1.

4.2. The case r1 > r0 and µ < λ ≤ 2µ

Suppose that V solves HJB equation (4.1). According toLemma 4.1, on OV 1 the function V satisfies

V ′(x)

V ′′(x)=

1f1(x)

, (4.8)

where

f1(x) = −

(r1−r0)2

σ 21

+λ2

σ 20

2(λ − µ − r0x); (4.9)

From (3.5) and (4.8), we see that aV (x) < 1 yields x > x3,where

x3 =1

2λr0

[λ2

− 2µλ − (r1 − r0)2 σ 2

0

σ 21

](4.10)

which is negative under λ < µ +

√µ2 + (r1 − r0)2 σ 2

0σ 2

1. From

(3.5) and (4.8), bV (x) < 1 is equivalent to x > x1, where

x1 =2(r1 − r0)(λ − µ)

r21 − r2

0 + λ2 σ 21

σ 20

. (4.11)

Hence OV 1 = x1 < x < x∗. The solution to the Eq. (4.8)

with the boundary condition V (x∗) = 0, has the followingform:

V1,1(x) =

∫ x∗

xe−

∫ x∗

x f1(v)dv+D1 du, (4.12)

where D1 is a constant that will be determined later.

It follows from Lemma 4.2 that, for x ∈ OV 2,

V ′(x)

V ′′(x)=

1f2(x)

, (4.13)

where

f2(x) = −

λ2

σ 20

λ − µ − r1x +

√(µ − λ + r1x)2 + λ2 σ 2

1σ 2

0x2

. (4.14)

From (3.5) and (4.13), aV (x) < 1 is equivalent to x2 < x < x2,where

x2, x2 =

r1 ∓

√r2

1 − λ(λ − 2µ)σ 2

1σ 2

0

λσ 2

1σ 2

0

. (4.15)

Likewise, bV (x) ≥ 1 is equivalent to 0 ≤ x ≤ x1. Noticex2 ≥ x1 and x2 < 0 under µ < λ ≤ 2µ, thus OV 2 = 0 < x ≤

x1. The solution to Eq. (4.13) onOV 2 with boundary conditionV (0) = 1 has the following form:

V1,2(x) = 1 −

∫ x

0e∫ u

0 f2(v)dv+D2 du. (4.16)

Note that OV 1 ∪ OV 2 = O, thus OV 3 = OV 4 = ∅. Next, weglue our solution at x1 by setting

V1,1(x1) = V1,2(x1); V ′

1,1(x1) = V ′

1,2(x1).

As before, V ′′

1,1(x1) = V ′′

1,2(x1) will follow from theseequalities. Solving for D1 and D2, we find

D1 = D2 +

∫ x∗

x1

f1(x)dx +

∫ x1

0f2(x)dx;

D2 = − ln

exp

∫ x∗

x1

f1(x)dx +

∫ x1

0f2(x)dx

×

∫ x∗

x1

e−∫ x∗

u f1(v)dvdu +

∫ x1

0e∫ u

0 f2(v)dvdu

. (4.17)

In summary we have the following theorem.

Theorem 4.1. If r1 > r0 and λ ≤ 2µ, the minimal ruinprobability function V is given by

V (x) =

1 −

∫ x

0e∫ u

0 f2(v)dv+D2 du, 0 < x ≤ x1∫ x∗

xe−

∫ x∗

x f1(v)dv+D1 du, x1 < x ≤ x∗

0, x > x∗,

which is a decreasing, convex, C2 function everywhere exceptat x∗. The constants D1 and D2 are given in (4.17); f1 and f2

are defined in (4.9) and (4.14), respectively, and x1 is definedby (4.11). The optimal risk exposure feedback function is given


by

a∗(x) =

λ − µ − r1x +

√(µ − λ + r1x)2 + λ2 σ 2

1σ 2

0x2

λ,

x ≤ x1

2λ(λ − µ − r0x)

(r1 − r0)2 σ 20

σ 21

+ λ2, x1 < x ≤ x∗

0, x > x∗,

and the optimal investment feedback function is

b∗(x) =

1, 0 < x ≤ x12(r1 − r0)(λ − µ − r0x)

x

((r1 − r0)2 + λ2 σ 2

1σ 2

0

) , x1 < x ≤ x∗

0, x > x∗.

4.3. The case r1 > r0 and 2µ < λ < µ +

√µ2 + (r2

1 − r20 )

σ 20

σ 21

Suppose that V solves the HJB equation. To find OV 1,similarly to Section 4.2, the condition aV (x) < 1 impliesx > x3, and bV (x) < 1 is equivalent to x > x1; Thus,OV 1 = (x1, x∗) by the following lemma:

Lemma 4.5. If r1 > r0 and λ < µ+

√µ2 + (r2

1 − r20 )

σ 20

σ 21

, then

x1 > x3.

Furthermore, on OV 1, the minimal ruin probability functionis also of the form (4.12), and we express it as:

V2,1(x) =

∫ x∗

xe−

∫ x∗

x f1(v)dv+D3 du. (4.18)

Similarly to the previous subsection, we see that on OV 2, thecondition aV (x) < 1 leads to x2 < x < x2. Furthermore, thecondition bV (x) ≥ 1 yields 0 ≤ x ≤ x1. Then, by Lemma 4.6given below, we obtain OV 2 = (x2, x1].

Lemma 4.6. If r1 > r0 and 2µ < λ < µ +√µ2 + (r2

1 − r20 )

σ 20

σ 21

, x2 < x1 < x2.

The solution V to (4.13) has the following form:

V2,2(x) = D5 −

∫ x

0e∫ u

0 f2(v)dv+D4 du, (4.19)

where D4 and D5 are constants to be determined later.Next, we identify OV 3. From Lemma 4.3, V solves

V ′(x)

V ′′(x)=

1f3(x)

(4.20)

on OV 3, where

f3(x) =

(r1−r0)2

σ 21

µ + r0x −

√(µ + r0x)2 + (r1 − r0)2 σ 2

0σ 2

1

. (4.21)

Notice that aV (x) ≥ 1 is equivalent to x ≤ x3 and bV (x) < 1is equivalent to x > x4, where

x4 =

−µ +

√µ2 + (r2

1 − r20 )

σ 20

σ 21

r1 + r0. (4.22)

We conclude OV 3 = ∅ by the following lemma

Lemma 4.7. If λ < µ +

√µ2 + (r2

1 − r20 )

σ 20

σ 21

, then x3 < x4.

Now we proceed with finding OV 4. From Lemma 4.4, wesee that on OV 4

V ′(x)

V ′′(x)=

1f4(x)

, (4.23)

where

f4(x) = −2(µ + r1x)

σ 20 + σ 2

1 x2. (4.24)

Therefore, aV (x) ≥ 1 yields OV 4 ⊆ (0, x2] ∪ [x2, x∗).Furthermore, bV (x) ≥ 1 gives 0 ≤ x ≤ x4 where x4 isgiven by (4.22). Thus, we have OV 4 = (0, x2] by the followingLemma 4.8:

Lemma 4.8. If r1 > r0 and 2µ < λ < µ +√µ2 + (r2

1 − r20 )

σ 20

σ 21

, x2 < x4 < x2.

The solution to (4.23) with boundary condition V (0) = 1 hasthe form:

V2,3(x) = 1 −

∫ x

0e∫ u

0 f4(v)dv+D6 du, (4.25)

where D6 is a constant to be determined. To glue our solutionsat x1 and x2, we set

V2,1(x1) = V2,2(x1); V ′

2,1(x1) = V ′

2,2(x1);

V2,2(x2) = V2,3(x2); V ′

2,2(x2) = V ′

2,3(x2).(4.26)

Solving these equations, we obtain

D4 = − ln∫ x1

x2

e∫ u

x2f2(v)dvdu + e

∫ x∗

x1f1(x)dx+

∫ x1x2

f2(x)dx

×

∫ x∗

x1

e−∫ x1

u f1(v)dvdu

+ e−∫ x2

0 f4(x)dx∫ x2

0e∫ u

0 f4(v)dvdu

;

D3 = D4 +

∫ x∗

x1

f1(x)dx +

∫ x1

x2

f2(x)dx;

D5 = eD4

[∫ x1

x2

e∫ u

x2f2(v)dvdu + e

∫ x∗

x1f1(x)dx+

∫ x1x2

f2(x)dx

×

∫ x∗

x1

e−∫ x∗

u f1(v)dvdu

];

D6 = D4 −

∫ x2

0f4(x)dx . (4.27)


As before, V ′′

2,1(x1) = V ′′

2,2(x1) and V ′′

2,2(x2) = V ′′

2,3(x2)

follow from (4.26). Thus, from the verification theorem andRemark 3.2, we have:

Theorem 4.2. If r1 > r0 and 2µ < λ < µ +√µ2 + (r2

1 − r20 )

σ 20

σ 21

, the minimal ruin probability function V

is given by

V (x) =

1 −

∫ x

0e∫ u

0 f4(v)dv+D6 du, 0 < x ≤ x2

D5 −

∫ x

0e∫ u

0 f2(v)dv+D4 du, x2 < x ≤ x1∫ x∗

xe−

∫ x∗

x f1(v)dv+D3 du, x1 < x ≤ x∗

0, x > x∗,

(4.28)

which is a decreasing convex C2 function everywhere exceptat x∗. The constants D3, D4, D5 and D6 are given in (4.27);f1, f2, f4 are defined in (4.9), (4.14) and (4.24), respectively;x1 and x2 are defined by (4.11) and (4.15), respectively. Theoptimal risk exposure feedback function is given by

a∗(x) =

1, 0 < x ≤ x2

λ − µ − r1x +

√(µ − λ + r1x)2 + λ2 σ 2

1σ 2

0x2

λx2 < x ≤ x1

2λ(λ − µ − r0x)

(r1 − r0)2 σ 20

σ 21

+ λ2, x1 < x ≤ x∗

0, x > x∗,


b∗(x) =

1, 0 < x ≤ x12(r1 − r0)(λ − µ − r0x)

x

[(r1 − r0)2 + λ2 σ 2

1σ 2

0

] , x1 < x ≤ x∗

0, x > x∗.

4.4. The case r1 > r0 and λ ≥ µ +

√µ2 + (r2

1 − r20 )

σ 20

σ 21

Suppose V solves HJB equation (4.1). To findOV 1, similarlyto Section 4.2, we see that aV (x) < 1 yields x > x3, andbV (x) < 1 yields x > x1. As in Lemma 4.5, x3 ≥ x1 holds

under λ ≥ µ +

√µ2 + (r2

1 − r20 )

σ 20

σ 21

. Thus, OV 1 = (x3, x∗).

The solution to Eq. (4.8) on OV 1 has the following form:

V3,1(x) =

∫ x∗

xe−

∫ x∗

x f1(v)dv+D7 du, (4.29)

where D7 is a constant to be determined. We can check thatOV 2 = ∅. To find OV 3, we follow the same route as inSection 4.3. The condition aV (x) ≥ 1 is equivalent to x ≤ x3,and bV (x) < 1 results in x > x4. From Lemma 4.7, we

see that x4 ≤ x3 under λ ≥ µ +

√µ2 + (r2

1 − r20 )

σ 20

σ 2 . Thus,
1
OV 3 = (x4, x3]. Solve Eq. (4.20) on OV 3, and V has thefollowing form:

V3,2(x) = D9 −

∫ x

0e∫ u

0 f3(v)dv+D8 du, (4.30)

where D8 and D9 are constants to be determined later.Similarly to the previous subsection, noticing inequality

x4 ≤ x2 holds under λ ≥ µ+

√µ2 + (r2

1 − r20 )

σ 20

σ 21

(Lemma 4.8),

we find OV 4 = (0, x4]. The solution to Eq. (4.23) with theboundary condition V (0) = 1 has the form:

V3,3(x) = 1 −

∫ x

0e∫ u

0 f4(v)dv+D10du, (4.31)

where D10 is a constant to be determined. Now we glue oursolutions at x3 and x4 by setting:

V3,1(x3) = V3,2(x3); V ′

3,1(x3) = V ′

3,2(x3);

V3,2(x4) = V3,3(x4); V ′

3,2(x4) = V ′

3,3(x4).

The result is

D8 = − ln∫ x1

x2

e∫ u

x2f3(v)dvdu + e

∫ x∗

x1f1(x)dx+

∫ x1x2

f3(x)dx

×

∫ x∗

x1

e−∫ x1

u f1(v)dvdu

+ e−∫ x2

0 f4(x)dx∫ x2

0e∫ u

0 f4(v)dvdu

;

D7 = D8 +

∫ x∗

x1

f1(x)dx +

∫ x1

x2

f3(x)dx;

D9 = eD8

[∫ x1

x2

e∫ u

x2f3(v)dvdu + e

∫ x∗

x1f1(x)dx+

∫ x1x2

f3(x)dx

×

∫ x∗

x1

e−∫ x∗

u f1(v)dvdu

];

D10 = D8 −

∫ x2

0f4(x)dx . (4.32)

Thus, we have V ′′

3,1(x3) = V ′′

3,2(x3) and V ′′

3,2(x4) = V ′′

3,3(x4)

and conclude:

Theorem 4.3. If r1 > r0 > 0 and λ ≥ µ +√µ2 + (r2

1 − r20 )

σ 20

σ 21

, the minimal ruin probability function V

is given by

V (x) =

1 −

∫ x

0e∫ u

0 f4(v)dv+D10du, 0 < x ≤ x4

D9 −

∫ x

0e∫ u

0 f3(v)dv+D8 du, x4 < x ≤ x3∫ x∗

xe−

∫ x∗

x f1(v)dv+D7 du, x3 < x ≤ x∗

0, x > x∗,

which is a decreasing, convex C2 function everywhere exceptat x∗. The constants D7, D8, D9 and D10 are given by (4.32);f1, f3, f4 are defined by (4.9), (4.21) and (4.24), respectively;


x3 and x4 are defined by (4.10) and (4.22), respectively. Theoptimal risk exposure feedback function is given by

a∗(x) =

1, 0 < x ≤ x32λ(λ − µ − r0x)

(r1 − r0)2 σ 20

σ 21

+ λ2, x3 < x ≤ x∗

0, x > x∗,


b∗(x) =

1, 0 < x ≤ x4√(µ + r0x)2 + (r1 − r0)2 σ 2

0σ 2

1− (µ + r0x)

(r1 − r0)x,

x4 < x ≤ x3

2(r1 − r0)(λ − µ − r0x)

x

[(r1 − r0)2 + λ2 σ 2

1σ 2

0

] , x3 < x ≤ x∗

0, x > x∗.

4.5. The case r0 ≥ r1

First, from Eq. (3.5), we see that under the assumptionr0 ≥ r1 the inequality bV (x) ≤ 0 is true for any twicecontinuously differentiable decreasing convex function V .Hence the minimizer of HJB equation (4.1) over b ∈ [0, 1] is 0,i.e. b∗(x) = 0. The HJB equation reduces to

0 = inf0≤a≤1

[µ − (1 − a)λ + r0x]V ′(x) +

12

a2σ 20 V ′′(x)

.

(4.33)

In this subsection, we solve HJB equation (4.33) aboveconsidering the following two cases:

(1) µ < λ ≤ 2µ;(2) λ > 2µ.

In the first case when µ < λ ≤ 2µ, we define

OV 5 = x ∈ O : 0 < aV (x) < 1.

Suppose V solves HJB equation (4.33). Then on OV 5, theminimizer is a∗(x) = aV (x). Plugging this into Eq. (4.33) andsimplifying, we obtain

V ′(x)

V ′′(x)= −

2(λ − µ − r0x)

λ2

σ 20

. (4.34)

The condition aV (x) < 1 is equivalent to x > x5, where

x5 =λ − 2µ

2r0. (4.35)

Thus, OV 5 = O since x5 ≤ 0 in this case. Solving Eq. (4.34)with the boundary conditions V (0) = 1 and V (x∗) = 0, weobtain

V (x) =

(λ − µ − r0x

λ − µ

) λ2

2r0σ20 . (4.36)

This can be summarized in the following theorem.

Theorem 4.4. If r0 ≥ r1 and µ < λ ≤ 2µ, then the minimalruin probability function V is given by

V (x) =

(

λ − µ − r0x

λ − µ

) λ2

2r0σ20 , 0 ≤ x ≤ x∗

0, x > x∗,

(4.37)

which is a decreasing, convex C2 function everywhere except atx∗. The optimal risk exposure feedback function is given by

a∗(x) =

2(λ − µ − r0x)

λ, 0 ≤ x ≤ x∗

0, x > x∗,

and the optimal investment feedback function is b∗(x) = 0.

Next, we consider the second case when λ > 2µ. SupposeV solves the HJB equation. Then, on OV 5 = x ∈ O :

0 < aV (x) < 1, the minimizer over a is a∗(x) = aV (x);The condition aV (x) < 1 is equivalent to x > x5. HenceOV 5 = (x5, x∗). Solving Eq. (4.34) on OV 5 with the boundarycondition V (x∗) = 0, we see that the solution must be of thefollowing form

V4,1(x) = D11

(λ − µ − r0x

λ − µ

) λ2

2r0σ20 , (4.38)

where D11 is a constant to be determined. Define

OV 6 = x ∈ O : aV (x) ≥ 1.

Suppose V solves the HJB equation. On OV 6, we see that theminimizer over a is a∗(x) = 1. Plugging it into HJB equation(4.1) and simplifying, we obtain

V ′(x)

V ′′(x)= −

σ 20

2(µ + r0x). (4.39)

Hence aV (x) ≥ 1 is equivalent to x ≤ x5. As a result OV 6 =

(0, x5]. The solution to (4.39) with the boundary conditionV (0) = 1 yields

V4,2(x) = 1 − D12

[Φ

(√2r0

σ0x +

√2µ

√r0σ0

)

− Φ

( √2µ

√r0σ0

)], (4.40)

where D12 is a positive constant to be determined, and Φ isthe standard normal cumulative distribution function. Next, weglue our solution at x5 by setting

V4,1(x5) = V4,2(x5); V ′

4,1(x5) = V ′

4,2(x5). (4.41)

We omit the explicit expression for these constants. One canlook up the details in Taksar and Markussen (2003). Thediscussion above results in


Theorem 4.5. If r0 ≥ r1 and λ > 2µ, then the minimal ruinprobability function V (x) is given by

V (x) =

1 − D12

[Φ

(√2r0

σ0x +

√2µ

√r0σ0

)

−Φ

( √2µ

√r0σ0

)], 0 < x ≤ x5

D11

(λ − µ − r0x

λ − µ

) λ2

2r0σ20 , x5 < x ≤ x∗

0, x > x∗,

(4.42)

which is a decreasing convex C2 function everywhere exceptat x∗. The constant x5 is given by (4.35), and D11 and D12are the constants which solve (4.41). The optimal risk exposurefeedback function is given by

a∗(x) =

1, 0 < x ≤ x52(λ − µ − r0x)

λ, x5 < x ≤ x∗

0, x > x∗,

and the optimal investment feedback function is b∗(x) = 0.

5. The remaining cases

We have found closed-form solutions on minimizing theruin probability in the cases when either both shortsellingand borrowing are allowed or when neither shortselling norborrowing are permitted. In what follows, we will show howthe remaining two cases

(iii) with borrowing but not shortselling,(iv) without borrowing but with shortselling,

can be reduced to the previous ones.

5.1. The case with borrowing but no shortselling

When borrowing is allowed the control b takes on values in[0, ∞) and HJB equation (2.5) becomes

0 = inf0≤a≤1,b≥0

[µ − (1 − a)λ + ((1 − b)r0 + br1)x]V ′(x)

+12(a2σ 2

0 + b2σ 21 x2)V ′′(x)

. (5.1)

The solution to Eq. (5.1) can be found by considering thefollowing two subcases:

(1) When r1 − r0 > 0, we see that bV (x) ≥ 0 always holds,hence the minimizer is b∗(x) = bV (x). In this case, theinsurer will never shortsell even though it is allowed. Hencethe solution to Eq. (5.1) is the same as the solution to HJBequation (3.1) with optimization being considered over theset b ∈ (−∞, ∞), where both shortselling and borrowingare allowed.

(2) When r1− r0 ≤ 0, we see that bV (x) ≤ 0 always holds. Theinsurer will never purchase the risky asset, therefore willnever borrow although borrowing is allowed here. In thiscase, we have b∗(x) = 0 and the problem can be reduced tothe case under which neither shortselling nor borrowing isallowed, i.e. the solution to HJB equation (5.1) is the sameas the one to HJB equation (4.33).

5.2. The case with no borrowing but with shortselling

When borrowing is not allowed and shortselling is possiblethe control b takes on values in (−∞, 1] and HJB equation (2.5)becomes

0 = inf0≤a≤1,b≤1

[µ − (1 − a)λ + ((1 − b)r0 + br1)x]V ′(x)

+12(a2σ 2

0 + b2σ 21 x2)V ′′(x)

. (5.2)

The solution to HJB equation (5.2) can be found by consideringthe following two subcases:

(1) When r1 − r0 > 0, in this case we see that bV (x) ≥ 0always holds. Therefore the insurer will never shortsellalthough shortselling is allowed. Note that borrowing isnot allowed here. Hence it can be reduced to the case ofneither borrowing nor shortselling allowed. Consequently,the solution to Eq. (5.2) is the same as the solution to HJBequation (4.1) with optimization being considered over theset b ∈ [0, 1].

(2) When r1 − r0 ≤ 0, in this case we always have bV (x) ≤ 0.Hence minimizer over b is b∗(x) = bV (x). This shows thatunder optimal control the insurer will never purchase therisky asset and hence will never borrow even if we assumeborrowing is possible. Hence the solution to HJB Eq. (5.2)is the same as the one to HJB equation (3.1) with bothshortselling and borrowing allowed.

Acknowledgments

The first author’s research is supported by the UNI summerresearch fellowship. The second author’s research has beensupported by the National Science Foundation, Grant NSFDMS-0505435.

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on reinsurance and investment for large insurance portfolios

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