Insurance: Mathematics and Economics 58 (2014) 1–13
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Insurance: Mathematics and Economics journal homepage: www.elsevier.com/locate/ime
On dividend strategies with non-exponential discounting Qian Zhao a,b , Jiaqin Wei b,∗ , Rongming Wang a,c a
School of Finance and Statistics, East China Normal University, Shanghai, 200241, China
b
Department of Applied Finance and Actuarial Studies, Faculty of Business and Economics, Macquarie University, Sydney, NSW 2109, Australia
c
Research Center of International Finance and Risk Management, East China Normal University, Shanghai, 200241, China
highlights • The dividend maximization problem is studied with a non-constant discount rate. • An equilibrium HJB-equation is considered for this time-inconsistent control problem. • Equilibrium strategies and value functions are obtained in two examples.
article
info
Article history: Received November 2012 Received in revised form May 2014 Accepted 2 June 2014 Keywords: Dividend strategies Non-exponential discounting Time inconsistence Equilibrium strategies Equilibrium HJB-equation
abstract In this paper, we study the dividend maximization problem with a non-constant discount rate in a diffusion risk model. We assume that the dividends can only be paid at a bounded rate and restrict ourselves to Markov strategies. This is a time inconsistent control problem. The equilibrium HJB-equation is given and the verification theorem is proven for a general discount function. Considering a mixture of exponential discount functions and a pseudo-exponential discount function, we get equilibrium dividend strategies and the corresponding equilibrium value functions by solving the equilibrium HJB-equations. © 2014 Elsevier B.V. All rights reserved.
1. Introduction Since it was proposed by De Finetti (1957), the optimization of dividend payments has been investigated by many researchers under various risk models. This problem is usually phrased as the management’s problem of determining optimal timing and size of dividend payments in the presence of bankruptcy risk. For more literature on this problem, we refer the reader to a recent survey paper by Avanzi (2009). In the very rich literature, a common assumption is that the discount rate is constant over time so the discount function is exponential. However, some empirical studies of human behavior suggest that the assumption of constant discount rate is unrealistic, see, e.g., Thaler (1981), Ainslie (1992) and Loewenstein and Prelec (1992). Indeed, there is experimental evidence that people are impatient about choices in the short term but are more patient when choosing between long-term alternatives. More precisely,
∗
Corresponding author. Tel.: +61 2 9850 1056. E-mail addresses:
[email protected] (Q. Zhao),
[email protected] (J. Wei),
[email protected] (R. Wang). http://dx.doi.org/10.1016/j.insmatheco.2014.06.001 0167-6687/© 2014 Elsevier B.V. All rights reserved.
events in the near future tend to be discounted at a higher rate than events that occur in the long run. Considering such an effect, individual behavior is best described by the hyperbolic discounting (see Phelps and Pollak, 1968), which has been extensively studied in the areas of microeconomics, macroeconomics, and behavioral finance, such as Laibson (1997) and Barro (1999) among others. However, difficulties arise when we try to solve an optimal control problem with a non-constant discount rate by the standard dynamic programming approach. In fact, the standard optimal control techniques give rise to time inconsistent strategies, i.e., a strategy that is optimal for the initial time may be not optimal later. This is the so-called time inconsistent control problem and the classical dynamic programming principle does no longer hold. Strotz (1955) studies the time inconsistent problem within a game theoretic framework by using Nash equilibrium points. They seek the equilibrium policy as the solution of a subgame-perfect equilibrium where player t can be viewed as the future incarnation of the decision-maker at time t. Recently, there is an increasing attention in the time inconsistent control problem due to the practical applications in economics and finance. A modified HJB equation is derived in Marín-Solano
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Q. Zhao et al. / Insurance: Mathematics and Economics 58 (2014) 1–13
and Navas (2010) which solves an optimal consumption and investment problem with the non-constant discount rate for both naive and sophisticated agents. A similar problem is also considered by another approach in Ekeland and Lazrak (2006) and Ekeland and Pirvu (2008), who provide the precise definition of the equilibrium concept in continuous time for the first time. They characterize the equilibrium policies through the solutions of a flow of a BSDE, and they show, for a special form of the discount factor, that this BSDE reduces to a system of two ODEs which has a solution. Considering the hyperbolic discounting, Ekeland et al. (2012) study the portfolio management problem for an investor who is allowed to consume and take out life insurance, and they characterize the equilibrium strategy by an integral equation. Following this definition of the equilibrium strategy, Björk and Murgoci (2010) studied the time-inconsistent control problem in a general Markov framework, and derived the equilibrium HJB-equation together with the verification theorem. Björk et al. (2014) studied Markowitz’s optimal portfolio problem with statedependent risk aversion by utilizing the equilibrium HJB-equation obtained in Björk and Murgoci (2010). In this paper, we revisit the dividend maximization problem with a general discount function in a diffusion risk model. We assume that the dividends can only be paid at a bounded rate and restrict ourselves to Markov strategies. We use the equilibrium HJB-equation to solve this problem. In contrast to the papers mentioned above which consider a fixed time horizon or an infinite time horizon, in the dividend problem the ruin risk should be taken into account and the time horizon is a random variable (the time of ruin). Thus, the equilibrium HJB-equation given in this paper looks different to the one obtained in Björk and Murgoci (2010). We first give the equilibrium HJB-equation which is motivated by Yong (2012) and the verification theorem for a general discount function. Then we solve the equilibrium HJB-equation for two special non-exponential discount functions: a mixture of exponential discount functions and a pseudo-exponential discount function. For more details about these discount functions, we refer the reader to Ekeland and Lazrak (2006) and Ekeland and Pirvu (2008). Under the mixture of exponential discount functions, our results show that if the bound of the dividend rate is small enough, then the equilibrium strategy is to always pay the maximal dividend rate; otherwise, the equilibrium strategy is to pay the maximal dividend rate when the surplus is above a barrier and pay nothing when the surplus is below the barrier. Given some conditions, the results are similar under the pseudo-exponential discount function. These features of the equilibrium dividend strategies are similar to the optimal strategies obtained in Asmussen and Taksar (1997) who consider the exponential discounting in the diffusion risk model. The remainder of this paper is organized as follows. The dividend problem and the definition of an equilibrium strategy are given in Section 2. The equilibrium HJB-equation and a verification theorem are presented in Section 3. In Section 4, we study two cases with a mixture of exponential discount functions and a pseudo-exponential discount function. 2. The model In the case of no control, the surplus process is assumed to follow dXt = µdt + σ dWt ,
t ≥ 0,
where µ, σ are positive constants and {Wt }t ≥0 is a onedimensional standard Brownian motion on a filtered probability space Ω , F , {Ft }t ≥0 , P satisfying the usual hypotheses. The filtration {Ft }t ≥0 is completed and generated by {Wt }t ≥0 .
A dividend strategy is described by a stochastic process {lt }t ≥0 . Here, lt ≥ 0 is the rate of dividend payout at time t which is assumed to be bounded by a constant M > 0. We restrict ourselves to the feedback control strategies (Markov strategies), i.e. at time t, the control lt is given by lt = π (t , x), where x is the surplus level at time t and the control law π : [0, ∞) × [0, ∞) → [0, M ] is a Borel measurable function. In Section 4, we need to distinguish the cases with M < µ and M ≥ µ, and to be more careful with the former case when we verify the strategy conjectured is indeed an equilibrium strategy (see Corollaries 4.3 and 4.8). When applying the control law π , we denote by {Xtπ }t ≥0 the controlled risk process. Considering the controlled system starting from the initial time t ∈ [0, ∞), {Xsπ } evolves according to
dXsπ = µds + σ dWs − π (s, Xsπ )ds, Xtπ = x.
s ≥ t,
(2.1)
Let
τtπ := inf s ≥ t : Xsπ ≤ 0 be the time of ruin under the control law π . Without loss of generality, we assume that Xsπ ≡ 0 for s ≥ τtπ . Let h : [0, ∞) → [0, ∞) be a discount function which satisfies ∞ h(0) = 1, h(t ) ≥ 0 and 0 h(t )dt < ∞. Furthermore, h is assumed to be continuously differentiable on [0, ∞) and h′ (t ) ≤ 0. Definition 2.1. A control law π is said to be admissible if it satisfies: 0 ≤ π (t , x) ≤ M for all (t , x) ∈ [0, ∞) × [0, ∞), π (t , 0) ≡ 0 for all t ∈ [0, ∞). We denote by Π the set of all admissible control laws. For a given admissible control law π and an initial state (t , x) ∈ [0, ∞) × [0, ∞), we define the return function V π by π τt π π V (t , x) = Et ,x h(z − t )π (z , Xz )dz , t
where Et ,x [·] is the expectation conditioned on the event {Xtπ = x}. Note that for any admissible strategy π ∈ Π , we have τtπ
E t ,x
h(z − t )π (z , X π ) dz z
t
∞
h(t )dt < ∞,
≤M
∀(t , x) ∈ [0, ∞) × [0, ∞),
(2.2)
0
which means the performance functions V π (t , x) are well-defined for all admissible strategies. In classical risk theory, the optimal dividend strategy, denoted by π ∗ , is an admissible strategy such that V π (t , x) = sup V π (t , x). ∗
π∈Π
However, in our settings, this optimization problem is timeinconsistent in the sense that the Bellman optimality principle fails. Similar to Ekeland and Pirvu (2008) and Björk and Murgoci (2010), we view the entire problem as a non-cooperative game and look for Nash equilibria for the game. More specifically, we consider a game with one player for each time t, where player t can be regarded as the future incarnation of the decision maker at time t. Given state (t , x), player t will choose a control action π (t , x), and she/he wants to maximize the functional V π (t , x). In the continuous-time model, Ekeland and Lazrak (2006) and Ekeland and Pirvu (2008) give the precise definition of this equilibrium strategy for the first time. Intuitively, equilibrium strategies are the strategies such that, given that they will be implemented in the future, it is optimal to implement them right now.
Q. Zhao et al. / Insurance: Mathematics and Economics 58 (2014) 1–13
3
Definition 2.2. Choose a control law πˆ ∈ Π , a fixed l ∈ [0, M ] and a fixed real number ϵ > 0. For any fixed initial point (t , x) ∈ [0, ∞) × [0, ∞), we define the control law π ϵ by
Theorem 3.2 (Verification Theorem). Assume that there exists a bounded function c (s, t , x) ∈ C 0,1,2 (D [0, ∞) × [0, ∞)) which solves the equilibrium HJB-equation in Definition 3.1. Let
0, π (s, y) = l, πˆ (s, y),
∂ 2c ∂c πˆ (t , x) := φ t , t , (t , t , x), 2 (t , t , x) , ∂x ∂x
for s ∈ [t , ∞), y = 0; for s ∈ [t , t + ϵ], y ∈ (0, ∞); for s ∈ [t + ϵ, ∞), y ∈ (0, ∞).
ϵ
(3.4)
and
If ϵ
lim inf
V πˆ (t , x) − V π (t , x)
ϵ
ϵ→0
V (t , x) := c (t , t , x).
≥ 0,
for all l ∈ [0, M ], we say that πˆ is an equilibrium control law. And the equilibrium value function V is defined by V (t , x) = V πˆ (t , x).
(2.3)
In the following section, we will first give the equilibrium HJB-equation for the equilibrium value function V , and then prove a verification theorem. 3. The equilibrium Hamilton–Jacobi–Bellman equation In this section, we consider the objective function having the form π
V (t , x) = Et ,x
τtπ
C t , z , π (z , Xz ) dz ,
π
(3.1)
t
where C t , z , π (z , Xzπ ) = h(z − t )π (z , Xzπ ), for z ≥ t.
If for any (s, t , x) ∈ D [0, ∞) × [0, ∞) it holds that lim c (s, τn , Xτπˆn ) = 0,
n→∞
a.s.,
(3.6)
where τn = n ∧ τtπˆ , n ≥ t , n = 1, 2, . . . , and X πˆ is the unique solution to the SDE (2.1) with π replaced by πˆ and initial state (t , x), then πˆ given by (3.4) is an equilibrium control law, and V given by (3.5) is the corresponding equilibrium value function. Proof. We give the proof in two steps: 1. We show that V is the value function corresponding to πˆ , i.e., V (t , x) = V πˆ (t , x); 2. We prove that πˆ is indeed the equilibrium control law which is defined by Definition 2.2. Step 1. With (3.4), we rewrite (3.2) as
Lπˆ c (s, t , x) + C (s, t , πˆ (t , x)) = 0, (s, t , x) ∈ D [0, ∞) × (0, ∞), c (s, t , 0) = 0, ∀(s, t ) ∈ D [0, ∞),
(3.7)
For all π ∈ Π and any real valued function f (t , x) ∈ C 1,2 ([0, ∞) × (0, ∞)), which means that the partial derivatives ∂f ∂f ∂2f , , ∂ t ∂ x ∂ x2
exist and are continuous on [0, ∞)×(0, ∞), we define the infinitesimal generator Lπ by
Lπ f (t , x) =
(3.5)
∂f ∂f 1 ∂ 2f (t , x) + (µ − π (t , x)) (t , x) + σ 2 2 (t , x). ∂t ∂x 2 ∂x
where the operator Lπˆ applies to the function c (s, ·, ·). By (3.7), applying Dynkin’s formula to the function c (s, ·, ·) yields that
c (s, t , x) = Et ,x c s, τn , Xτπˆn
τn − E t ,x Lπˆ c s, z , Xzπˆ dz t = Et ,x c s, τn , Xτπˆn τn + E t ,x C (s, z , πˆ (z , Xzπˆ ))dz .
Let D [0, ∞) := {(s, t ) | 0 ≤ s ≤ t < ∞} and C 0,1,2 (D [0, ∞) × [0, ∞)) be the set of all functions defined on D [0, ∞) × [0, ∞) which are continuous with respect to the first variable, continuously differentiable with respect to the second variable and twice continuously differentiable with respect to the third variable. The following equilibrium HJB-equation is motivated by Eq. (4.77) of Yong (2012) and the proof of Theorem 3.2.
∈ C 0,1,2 D [0, ∞) × [0, ∞) , the equilibrium HJB-equation is given by ∂c ∂c (s, t , x) + H s, t , φ t , t , (t , t , x), ∂t ∂x ∂ 2c ∂c ∂2c (3.2) (t , t , x) , ∂ x (s, t , x), ∂ x2 (s, t , x) = 0, ∂ x2 ∀(s, t , x) ∈ D [0, ∞) × (0, ∞), c (s, t , 0) = 0, ∀(s, t ) ∈ D [0, ∞), Definition 3.1. For a function c (s, t , x)
1 2 σ P + (µ − l)p + C (s, t , l), 2 φ(s, t , p, P ) = arg max H (s, t , ·, p, P ),
t
Recalling Definition 2.1 of admissible strategies (see also (2.2)), for given s ≤ t, we have τtπˆ
E t ,x t
πˆ C s, z , πˆ (z , Xz ) dz < ∞, ∀(t , x) ∈ [0, ∞) × [0, ∞).
Since c (·, ·, ·) is bounded, by (3.6), letting n → ∞ and applying the dominated convergence theorem yields c ( s , t , x ) = E t ,x
τtπˆ
πˆ
h(z − s)πˆ (z , Xz )dz ,
t
(s, t , x) ∈ D [0, ∞) × [0, ∞).
where
H (s, t , l, p, P ) =
(3.8)
Thus, we have (3.3)
for (s, t , l, p, P ) ∈ D [0, ∞) × [0, M ] × R2 . Since the equilibrium HJB-equation given in Definition 3.1 is informal, we are now giving a verification theorem for motivating it.
V (t , x) := c (t , t , x) = V πˆ (t , x). Step 2. For a given l ∈ [0, M ], and a fixed real number ϵ > 0, we define π ϵ by Definition 2.2. For simplicity, we denote by X ϵ the path under the control law π ϵ . Without loss of generality, we consider the case where ϵ is sufficiently small such that t + ϵ
4
Q. Zhao et al. / Insurance: Mathematics and Economics 58 (2014) 1–13
πϵ t
ϵ
ϵ ϵ 3. If τtπˆ ≥ τtπ , noting that πˆ s, Xsϵ ≡ 0 for s ≥ τtπ , we have
∧ τtπˆ a.s. By the definition of V πˆ and V π , we obtain πˆ τt πˆ πϵ V (t , x) − V (t , x) = Et ,x C t , s, πˆ s, Xsπˆ ds <τ
τtπ
−
C t , s, π
ϵ
s, X s
(h (s − t ) − h (s − t − ϵ)) πˆ s, Xs ds
ϵ
t +ϵ
ϵ πˆ ϵ h(s − t ) πˆ s, Xs − π s, Xs = Et ,x ds t + Et ,x V πˆ t + ϵ, Xtπˆ+ϵ − V πˆ t + ϵ, Xtϵ+ϵ πˆ τt πˆ + Et ,x (h (s − t ) − h (s − t − ϵ)) πˆ s, Xs ds τtπ
lim
(h (s − t ) − h (s − t − ϵ)) πˆ s, Xs ds
. (3.9)
in three parts separately:
∞
h(s − t ) πˆ s, Xsπˆ − π ϵ s, Xsϵ
ds
πˆ
dV πˆ u, Xuϵ
ϵ
(h (s − t ) − h (s − t − ϵ)) πˆ s, Xs ds
(h (s − t ) − h (s − t − ϵ))
τtπˆ
Et ,x
πˆ
(h (s − t ) − h (s − t − ϵ)) πˆ s, Xs τtπ
− E t ,x
ds
ϵ
(h (s − t ) − h (s − t − ϵ)) πˆ s, Xs ds
t +ϵ
2
πˆ
τtπˆ
≥ E t ,x
(h (s − t ) − h (s − t − ϵ))
t +ϵ
ϵ πˆ × πˆ s, Xs − πˆ s, Xs ds . ∞
ϵ ∂ V πˆ (t , x) ∂ V πˆ (t , x) 1 2 ∂ 2 V πˆ (t , x) = + (µ − l) + σ ∂x 2 ∂ x2 ∂t ϵ
τtπˆ
and
= Lπ V πˆ (t , x) ϵ = Lπ V (t , x) .
Therefore, we always have
πˆ
= Lπˆ V πˆ (t , x) = Lπˆ V (t , x) ,
lim
ϵ
t +ϵ
πˆ
ϵ→0
ϵ πˆ × πˆ s, Xs − πˆ s, Xs ds .
u, Xu
ϵ
t +ϵ
∂ V (t , x) 1 2 ∂ V (t , x) ∂ V (t , x) + µ − πˆ (t , x) + σ ∂ t ∂x 2 ∂ x2
t +ϵ t
(h (s − t ) − h (s − t − ϵ))
τtπ
≥ E t ,x
t
ϵ
ϵ
Applying the Itô’s formula, we get
E t ,x
(h (s − t ) − h (s − t − ϵ)) πˆ s, Xs ds
− E t ,x
t
=
ϵ
t +ϵ
πˆ
ϵ
t +ϵ
= Et ,x V πˆ t + ϵ, Xtπˆ+ϵ − V πˆ (t , x) − Et ,x V πˆ t + ϵ, Xtϵ+ϵ − V πˆ (t , x) t +ϵ t +ϵ = Et ,x dV πˆ u, Xuπˆ − E t ,x dV πˆ u, Xuϵ .
lim
ds
πˆ π If τtϵ ≤ τt , it follows from h (s − t ) − h (s − t − ϵ) ≤ 0 and πˆ s, Xs ≥ 0 that πˆ τt πˆ Et ,x (h (s − t ) − h (s − t − ϵ)) πˆ s, Xs ds
t + ϵ, Xtπˆ+ϵ − V πˆ t + ϵ, Xtϵ+ϵ
ϵ→0
ϵ πˆ × πˆ s, Xs − πˆ s, Xs ds .
2. We rewrite the second part on the right-side of Eq. (3.9) by
dV
τtπˆ
= E t ,x
ϵ
= πˆ (t , x) − π ϵ (t , x).
t +ϵ t
t +ϵ
ϵ→0
τtπ
− E t ,x
1. Noting that 0 h(t )dt < ∞, l and πˆ are bounded and applying the dominated convergence theorem, we get t +ϵ t
πˆ
(h (s − t ) − h (s − t − ϵ)) πˆ s, Xs
ϵ
ϵ V πˆ (t ,x)−V π (t ,x)
t +ϵ
Here π( ˆ s, Xsϵ ) and πˆ (s, Xsπˆ ) are the equilibrium control processes associated with the paths of X ϵ and X πˆ , respectively. According to Eq. (3.9), we now consider the limitation ϵ
τtπˆ
Et ,x
t +ϵ
Et ,x
(h (s − t ) − h (s − t − ϵ)) πˆ s, Xs ds .
t +ϵ
ϵ
− E t ,x
ϵ
Thus,
t +ϵ
Et ,x V πˆ
t +ϵ
E t ,x
τtπˆ
= E t ,x
ds
t
limϵ→0
ϵ
t +ϵ
ϵ
ϵ
E t ,x
t
τtπ
Furthermore, noting that πˆ is bounded and 0 h(s)ds < ∞, by the dominated convergence theorem, we get πˆ τt [h (s − t ) − h (s − t − ϵ)] πˆ s, Xsπˆ − πˆ s, Xsϵ ds Et ,x t +ϵ = 0. lim ϵ→0 ϵ From (3.9) and the above three steps, we obtain that ϵ
V πˆ (t , x) − V π (t , x)
≥ Lπˆ V (t , x) + C t , t , πˆ (t , x) ϵ→0 ϵ ϵ − Lπ V (t , x) + C (t , t , π ϵ (t , x)) . (3.10) lim
Q. Zhao et al. / Insurance: Mathematics and Economics 58 (2014) 1–13
It follows from (3.3) and (3.4) that
We consider the following ansatz:
Lπˆ V (t , x) + C t , t , πˆ (t , x)
c (s, t , x) =
π
= sup {(L V ) (t , x) + C (t , t , π (t , x))} . Therefore, (3.10) and (3.11) imply that lim
ϵ
≥ 0.
This completes the proof.
4. Solutions to two special cases In this section, we try to find a solution of the equilibrium HJB-equation in Definition 3.1 for specific discount functions. First of all, we make a conjecture of an equilibrium strategy for a general discount function. Since 1 H (s, t , l, p, P ) = σ 2 P + (µ − l)p + C (s, t , l) 2 1 = σ 2 P + µp + [h(t − s) − p]l, 2 we have
φ(s, t , p, P ) =
0, M,
ωi e−δi (t −s) Vi (x),
(s, t , x) ∈ D [0, ∞) × [0, ∞),
(4.4)
where the functions Vi (x), i = 1, 2, . . . , N, are given by the system of ODEs
ϵ
ϵ→0
N i =1
(3.11)
π ∈Π
V πˆ (t , x) − V π (t , x)
5
if p ≥ h(t − s), if p < h(t − s).
We assume that there exists a constant b ≥ 0 such that ∂∂ cx (t , t , x) ≥ 1, if 0 ≤ x < b, and ∂∂ cx (t , t , x) < 1, if x ≥ b. Thus, from Theorem 3.2, the equilibrium strategy would be given by
∂ 2c ∂c π( ˆ t , x) = φ t , t , (t , t , x), 2 (t , t , x) ∂x ∂x 0, if 0 ≤ x < b, = M , if x ≥ b.
(4.1)
Then the equilibrium HJB-equation (3.2) becomes
1 2 ∂ 2c ∂c (s, t , x) ( s, t , x) + σ ∂t 2 ∂ x2 ∂c + µ (s, t , x) = 0, (s, t , x) ∈ D [0, ∞) × (0, b), ∂x ∂c 1 ∂ 2c ∂c (s, t , x) + σ 2 2 (s, t , x) + (µ − M ) (s, t , x) ∂ t 2 ∂ x ∂x + h(t − s)M = 0, (s, t , x) ∈ D [0, ∞) × [b, ∞), c (s, t , 0) = 0, ∀(s, t ) ∈ D [0, ∞).
(4.2)
∂ Vi 1 2 ∂ 2 Vi σ (x) + µ (x) − δi Vi (x) = 0, 2 ∂x 2 ∂2x 1 2 ∂ Vi ∂ Vi σ (x) + (µ − M ) (x) 2 2 ∂ x ∂x − δi Vi (x) + M = 0, Vi (0) = 0.
x ∈ [0, b), (4.5) x ∈ [b, ∞),
Denote by θ1 (η, c ) and −θ2 (η, c ) the positive and negative roots of the equation 12 σ 2 y2 + ηy − c = 0, respectively. Then
−η + η2 + 2σ 2 c θ1 (η, c ) = , 2 σ 2 2 θ2 (η, c ) = η + η + 2σ c . 2 σ Thus a general solution of Eq. (4.5) has the form
θ (µ,δ )x i Ci1 e 1 + Ci2 e−θ2 (µ,δi )x , M Vi ( x ) = + Ci3 eθ1 (µ−M ,δi )x δ i + Ci4 e−θ2 (µ−M ,δi )x ,
x ∈ [0, b), (4.6) x ∈ [b, ∞),
for i = 1, 2, . . . , N. From Theorem 3.2, we need to find a function c (s, t , x) ∈ C 0,1,2 (D [0, ∞) × [0, ∞)). Thus, in the following we shall find Vi (x), i = 1, 2, . . . , N, which are C 2 functions. Since Vi (0) = 0, and Vi (x) > 0, for all x > 0, we have Ci1 = −Ci2 := Ci > 0, i = 1, 2, . . . , N. Since we are looking for a bounded function c (·, ·, ·) (see Theorem 3.2), we have Ci3 = 0, i = 1, 2, . . . , N. To simplify the notation, let Ci4 := −di , i = 1, 2, . . . , N. Until now, we are not sure whether di is positive or not. Now to find the value of Ci , di , i = 1, 2, . . . , N and b, we use ‘‘the principle of smooth fit’’ to get
V (b+) = V (b−), i = 1, 2, . . . , N , Vi′ (b+) = Vi ′ (b−), i = 1, 2, . . . , N , i i (4.7) ∂c ∂c (t , t , b+) = 1 or equivalently, (t , t , b−) = 1 . ∂x ∂x Therefore by denoting
4.1. A mixture of exponential discount functions Let us consider a case where the dividends are proportionally paid to N inhomogeneous shareholders. The term inhomogeneous refers to the assumption that the shareholders have different discount rates. Then given a control law π , the return function is V π (t , x) =
N
τπ
t
E t ,x
i=1
ωi e−δi t ,
Ci eθi1 b − e−θi2 b =
t
M
δi b
− di e−θi3 b ,
(4.8)
= di θi3 e−θi3 b ,
(4.9)
and
N
N
we can rewrite (4.7) as for i = 1, 2, . . . , N,
Ci θi1 eθi1 b + θi2 e−θi2
ωi e−δi (z −t ) π (z , Xzπ )dz ,
where ωi > 0 satisfying i=1 ωi = 1 is the proportion at which the dividends are paid to the shareholders, δi > 0, i = 1, 2, . . . , N, are the constant discount rates of the shareholders, respectively. In fact, a mixture of exponential discount functions is used in the above example. We consider a discount function defined by h(t ) =
θi1 = θ1 (µ, δi ), θi2 = θ2 (µ, δi ), θi3 = θ2 (µ − M , δi ), i = 1, 2, . . . , N ,
t ≥ 0,
(4.3)
N
ωi Ci θi1 eθi1 b + θi2 e−θi2 b = 1.
From (4.8)–(4.9) we can get Ci and di in the expression of b: Ci =
i =1
where δi > 0, δi ̸= δj , for i ̸= j, and ωi > 0 satisfies
N
i=1
ωi = 1.
(4.10)
i =1
di =
M θi3
δi
M θi3 b e
δi
(θi1 + θi3 ) eθi1 b + (θi2 − θi3 ) e−θi2 b
−1
θi1 eθi1 b + θi2 e−θi2 b , (θi1 + θi3 ) eθi1 b + (θi2 − θi3 ) e−θi2 b
,
(4.11)
(4.12)
6
Q. Zhao et al. / Insurance: Mathematics and Economics 58 (2014) 1–13
for i = 1, 2, . . . , N. Since Ci > 0, we can directly see that di > 0, for i = 1, 2, . . . , N. Substituting Ci into (4.10), we obtain N
δi
i =1
Let F (b) :=
N
ωi
M θi3
δi
i=1
θi1 eθi1 b + θi2 e−θi2 b − 1. (θi1 + θi3 ) eθi1 b + (θi2 − θi3 ) e−θi2 b
N
Lemma 4.1. If positive solution.
i=1
ωi Mδθii3 > 1, then F (b) = 0 has a unique Mθ
i3 > 1 implies that F (0) > 0. Proof. The condition i=1 ωi δi From Lemma 2.1 of Asmussen and Taksar (1997), we know that
N
M
δi
−
1
θi3
−
1
< 0,
θi1
i = 1, 2, . . . , N .
M θi3
θi1 −1 δ θ i i1 + θi3 i=1 N M θi3 θi1 = ωi −1 δi θi1 + θi3 i=1 N θi1 θi3 M 1 1 = ωi − − θi1 + θi3 δi θi3 θi1 i=1 N
ωi
ωi
M θi3
i=1
δi
system (4.8)–(4.10). Proof. (i) It is easy to check the function c (s, t , x) given by (4.13) and b = 0 satisfies the system of ODEs (4.2). Obviously, we have N M ∂c (s, t , 0) = ωi e−δi (t −s) θi3 ≤ 1, ∂x δi i=1
∂c (t , t , x) ≥ 1, ∂x ∂c (t , t , x) < 1, ∂x
(s, t ) ∈ D [0, ∞),
x ∈ [0, b),
(4.15)
x ∈ [b, ∞).
The first and second derivatives of c (s, t , x) given by (4.14) with respect to x are
Furthermore, we have F ′ (b) =
ωi Mδθii3 > 1, then
Thus, ∂∂ cx (t , t , x) < 1, for x ≥ 0, which implies c (·, ·, ·) satisfies the equilibrium HJB-equation (3.2). (ii) Similarly, it is easy to check that b and c (·, ·, ·, ) given by (4.8)–(4.10) and (4.14) satisfy the system of ODEs (4.2). It is sufficient to show
< 0. N
i=1
N M ∂ 2c ( s , t , x ) = − ωi e−δi (t −s) θi32 e−θi3 x < 0, 2 ∂x δi i =1 (s, t , x) ∈ D [0, ∞) × [0, ∞).
Thus, F (+∞) =
N
N ωi e−δi (t −s) Ci i = 1 × eθi1 x − e−θi2 x , x ∈ [0, b), N (4.14) c (s, t , x) = ωi e−δi (t −s) i=1 M − di e−θi3 x , x ∈ [b, ∞), × δi where Ci , di , i = 1, 2, . . . , N, and b is the unique solution to the
θi1 eθi1 b + θi2 e−θi2 b = 1. (θi1 + θi3 ) eθi1 b + (θi2 − θi3 ) e−θi2 b
M θi3
ωi
(ii) If
∆i 2 , θ b i1 (θi1 + θi3 ) e + (θi2 − θi3 ) e−θi2 b
∂c ( t , t , x) ∂x N ωi Ci i=1 = × θi1 eθi1 x + θi2 e−θi2 x , N ωi di θi3 e−θi3 x ,
where
∆i = θi12 eθi1 b − θi22 e−θi2 b × (θi1 + θi3 ) eθi1 b + (θi2 − θi3 ) e−θi2 b − θi1 eθi1 b + θi2 e−θi2 b × θi1 (θi1 + θi3 ) eθi1 b − θi2 (θi2 − θi3 ) e−θi2 b = θi12 (θi2 − θi3 ) − θi22 (θi1 + θi3 )
(t , x) ∈ [0, ∞) × [0, b), (t , x) ∈ [0, ∞) × [b, ∞),
i=1
and
+ θi1 θi2 (θi2 − θi3 ) − θi2 θi1 (θi1 + θi3 )] e(θi1 −θi2 )b = [θi1 (θi2 − θi3 ) − θi2 (θi1 + θi3 )] (θi1 + θi2 ) e(θi1 −θi2 )b = −θi3 (θi1 + θi2 )2 e(θi1 −θi2 )b < 0. Therefore, the equation F (b) = 0 admits a unique solution on (0, ∞).
∂ 2c (t , t , x) ∂ x2 N ωi Ci i=1 = × θi12 eθi1 x − θi22 e−θi2 x , N ωi di θi32 e−θi3 x , −
(t , x) ∈ [0, ∞) × [0, b), (t , x) ∈ [0, ∞) × [b, ∞),
i=1
Theorem 4.2. Given the discount function (4.3), there exists a bounded function c (·, ·, ·) ∈ C 0,1,2 (D [0, ∞) × [0, ∞)) satisfying the equilibrium HJB-equation (3.2).
N
(i) If by
i=1
ωi Mδθii3 ≤ 1, then b = 0 and the function c (·, ·, ·) is given
c (s, t , x) =
N
ωi e
−δi (t −s) M
i=1
x ∈ [0, ∞).
δi
−θi3 x
1−e
respectively. It is easy to check that ∂∂ cx (t , t , x) > 0, for all (t , x) ∈ [0, ∞) × [0, ∞), which implies that c (t , t , ·) is strictly increasing. Next we show that c (t , t , ·) is a concave function on [0, ∞), i.e. ∂2c (t , t , x) ∂ x2 ∂2c (t , t , x) ∂ x2
< 0, for all (t , x) ∈ [0, ∞) × [0, ∞). First we show that
is continuous at x = b. Apparently, ∂∂ x2c (t , t , x) < 0, for all (t , x) ∈ [0, ∞)×[b, ∞). Recalling (4.4), (4.5) and (4.7), we have
, (4.13)
1 2
σ2
2
N ∂ 2c ∂c ( t , t , b −) = −µ ( t , t , b ) + ωi δi Vi (b), ∂ x2 ∂x i =1
Q. Zhao et al. / Insurance: Mathematics and Economics 58 (2014) 1–13
1 2
σ2
7
∂ c ∂c (t , t , b+) = − (µ − M ) (t , t , b) 2 ∂x ∂x N + ωi δi Vi (b) − M . 2
i =1
Since ∂2c ∂ x2
∂c (t , t , b) ∂x
= 1, we get
∂2c (t , t , b−) ∂ x2
=
(t , t , b). Obviously, for all 0 ≤ x ≤ b, we have
∂2c (t , t , b+) ∂ x2
=
N ∂ 3c ( t , t , x ) = ωi Ci θi13 eθi1 x + θi23 e−θi2 x > 0, 3 ∂x i=1
which means that ∂∂ x2c (t , t , x) ≤ ∂∂ x2c (t , t , b) < 0, for all 0 ≤ x ≤ b. Thus, we proved (4.15). 2
2
Corollary 4.3. Consider the discount function (4.3).
ωi Mδθii3 ≤ 1, then for t ∈ [0, ∞) ∂c ∂ 2c πˆ (t , x) = φ t , t , (t , t , x), 2 (t , t , x) = M , ∂x ∂x x ∈ [0, ∞),
(i) If
N
i =1
is an equilibrium dividend strategy, and V ( t , x) = c ( t , t , x) =
N i=1
ωi
M
δi
1 − e−θi3 x ,
x ∈ [0, ∞),
is the corresponding equilibrium value function. N M θi3 (ii) If > 1, then for t ∈ [0, ∞) i =1 ω i δ i
Fig. 4.1. Equilibrium value functions with a mixture of exponential discount functions.
Example 4.4. Let N = 2, µ = 1, σ = 1, M = 0.8, δ1 = 0.2, δ2 = 0.4. Fig. 4.1 illustrates the equilibrium value functions for the mixture of exponential discount functions with ω1 = 0, 0.4, 0.7 and 1 and ω2 = 1 − ω1 . The barriers are 0.6525, 0.8781, 1.0207 and 1.1452, respectively. The cases with ω1 = 0 and 1 are time consistent and the equilibrium strategies are optimal. 4.2. A pseudo-exponential discount function We now consider a pseudo-exponential discount function defined as h(t ) = (1 + λt )e−δ t ,
t ≥ 0,
(4.16)
∂c ∂ c πˆ (t , x) = φ t , t , (t , t , x), 2 (t , t , x) ∂x ∂x 0, x ∈ [0, b), = M , x ∈ [b, ∞),
where λ > 0, δ > 0 are parameters. We refer the reader to Ekeland and Pirvu (2008) for explanations of this discount function. To ensure h is decreasing, we assume that λ < δ . To simplify the calculations, we shall impose more conditions on λ in the following. We consider the following ansatz:
is an equilibrium dividend strategy, and
c (s, t , x) = e−δ(t −s) {λ(t − s)V3 (x) + V4 (x)} ,
2
(s, t , x) ∈ D [0, ∞) × [0, ∞),
V (t , x) = c (t , t , x)
=
N ωi Ci eθi1 x − e−θi2 x ,
x ∈ [0, b),
N M −θi3 x − di e , ωi δi i =1
x ∈ [b, ∞),
i =1
is the corresponding equilibrium value function. Here Ci , di , i = 1, 2, . . . , N, and b is the unique solution to the system (4.8)– (4.10). Proof. By Theorems 3.2 and 4.2, it is sufficient to verify (3.6). If M ≥ µ , in both cases (i) and (ii), it is well known that P τtπˆ < ∞ = 1 (see, e.g. Gerber and Shiu (2006)). Since c (s, t , 0) = 0 for all (s, t ) ∈ D [0, ∞), we get (3.6). If M < µ, in both cases (i) and (ii), we have P τtπˆ = ∞ > 0 and X πˆπˆ = +∞ on {τtπˆ = ∞}. τt
However, for any s ∈ [0, ∞) we have limt →∞,x→∞ c (s, t , x) = 0. Thus, we still have (3.6). Under the mixture of exponential discount functions, Corollary 4.3 shows that if the bound of the dividend rate is small enough, then the equilibrium strategy is to always pay the maximal dividend rate; otherwise, the equilibrium strategy is to pay the maximal dividend rate when the surplus is above a barrier while paying nothing when the surplus is below this barrier. These features of the equilibrium dividend strategies are similar to the optimal strategies obtained in Asmussen and Taksar (1997) which considered an exponential discount function in a diffusion risk model.
(4.17)
where V3 (·) and V4 (·) are given by
∂ V3 1 2 ∂ 2 V3 σ (x) + µ (x) − δ V3 (x) = 0, 2 2 ∂ x ∂x 2 1 2 ∂ V3 σ (x) + (µ − M ) 2 ∂ x2 ∂ V3 × (x) − δ V3 (x) + M = 0, ∂x V3 (0) = 0,
x ∈ [0, b), (4.18) x ∈ [b, ∞),
and
1 2 ∂ 2 V4 ∂ V4 σ (x) + µ (x) 2 2 ∂ x ∂x − δ V4 (x) + λV3 (x) = 0, ∂ V4 1 2 ∂ 2 V4 σ (x) + (µ − M ) (x) 2 2 ∂ x ∂x − δ V4 (x) + λV3 (x) + M = 0, V4 (0) = 0,
x ∈ [0, b), (4.19) x ∈ [b, ∞),
respectively. It is easy to check that the function c (·, ·, ·) given by (4.17)–(4.19) satisfies the system (4.2). Recalling the situation we discussed in Section 4.1, Eq. (4.18) has a general solution V3 ( x ) =
θ (µ)x − e−θ2 (µ)x , C e 1 M
δ
− de
−θ2 (µ−M )x
,
x ∈ [0, b), x ∈ [b, ∞),
(4.20)
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Q. Zhao et al. / Insurance: Mathematics and Economics 58 (2014) 1–13
where C > 0, d > 0 are two unknown constants to be determined, θ1 (η) and −θ2 (η) are the positive and negative roots of the equation 12 σ 2 y2 + ηy − δ = 0, respectively. According to ‘‘the principle of smooth fit’’, we have
V3 (b+) = V3 (b−), V3′ (b+) = V3′ (b−),
(4.21)
M θ3
d=
M θ3 b e
(θ1 + θ3 ) eθ1 b + (θ2 − θ3 ) e−θ2 b
δ
δ
(−θ3 D3 − θ3 B3 b + B3 ) e−θ3 b − 1 = 0, i.e., D3 =
1 B3 − eθ3 b − B3 b.
−1
,
(4.22)
(θ1 + θ3 ) eθ1 b + (θ2 − θ3 ) e−θ2 b
(4.23)
+ θ2 B3 e−(θ2 +θ3 )b + 2 (θ1 + θ2 ) θ3 B1 be(θ1 −θ2 )b λ M + θ1 θ3 1 + − (θ1 + θ3 ) eθ1 b δ δ λ M + θ2 θ3 1 + − (θ2 − θ3 ) e−θ2 b , δ δ
θ3 = θ2 (µ − M ).
After obtaining V3 , solving ODE (4.19) yields that V4 (x)
eθ1 x + (D2 + B2 x) e−θ2 x , (D1− B1 x) M λ = 1+ + (D3 + B3 x) e−θ3 x , δ δ
G(b),
G(b) := −θ3 B1 e2θ1 b + θ3 B1 e−2θ2 b + θ1 B3 e(θ1 −θ3 )b
where
θ2 = θ2 (µ),
−1
where
θ1 eθ1 b + θ2 e−θ2 b , (θ1 + θ3 ) eθ1 b + (θ2 − θ3 ) e−θ2 b
θ1 = θ1 (µ),
(4.31)
θ3
Putting (4.28) and (4.29) into the left-hand-side of (4.30), it can be rewritten as
which yields that C =
and
0 ≤ x < b, x ≥ b,
(4.32)
and (4.24)
G(0) = (θ1 + θ2 )
λ λ M + θ 1 + − 1 . 3 µ − M − σ 2 θ3 δ δ
where
λC λC > 0, B2 = < 0, 2 µ + σ θ1 µ − σ 2 θ2 λd B3 = < 0. µ − M − σ 2 θ3
Lemma 4.5. If
B1 =
(4.25)
Applying the principle of smooth fit for determining a candidate b, we obtain
4
4
∂c (t , t , b+) = 1 ∂x
(4.27) ∂c or equivalently, (t , t , b−) = 1 . ∂x
[(θ1 + θ3 ) b + 1] B1 eθ1 b − [(θ2 − θ3 ) b − 1] B1 e−θ2 b + B3 e−θ3 b + θ3 1 + λδ Mδ , (θ1 + θ3 ) eθ1 b + (θ2 − θ3 ) e−θ2 b θ3 b
D3 = e
λ Cˆ − B1 b eθ1 b − Cˆ + B1 b e−θ2 b − 1 + δ
M
δ
− B3 b.
(4.28) (4.29)
Furthermore, for a b such that ∂∂ cx (t , t , b+) = ∂∂ cx (t , t , b−) = 1, we have
Cˆ − B1 b θ1 − B1 eθ1 b
+
<λ
θ12
1
µ−M −σ 2 θ3
δ
(θ1 + θ3 ) − θ1 θ3 , + θδ3 + θ32 θδ1 − µ+σ1 2 θ
(θ1 + θ3 )
M
(4.33)
1
then G(b) = 0 has a positive solution. Lemma 4.6. If
λ≤
θ1 + θ2 δ 2 (θ1 + θ3 ) (θ1 + θ2 ) δ 2 ∧ , θ1 + 3θ2 M θ3 2θ1 (θ1 + 2θ2 ) M θ3
(4.34)
then
Cˆ + B1 b θ2 − B1 e−θ2 b − 1 = 0,
1 + (θ1 b + 1) B1 eθ1 b − (θ2 b − 1) B1 e−θ2 b
θ1 eθ1 b + θ2 e−θ2 b
where b > 0 such that G(b) = 0. The proofs of Lemmas 4.5 and 4.6 are shown in Appendices A and B, respectively. Now we show the main result of this subsection in the following theorem. Theorem 4.7. Assume that 0 < λ < δ . Given the discount function (4.16), there exists a function c (·, ·, ·) ∈ C 0,1,2 (D [0, ∞)× [0, ∞)) satisfying the equilibrium HJB-equation (3.2).
−1 θ3 1 (i) If Mδ > θ3 and λ ≤ Mδ − θ3 µ−M −σ , then b = 0 2θ + δ 3 and c (·, ·, ·) is given by (4.17) with V3 (x) =
(4.30) V4 (x) =
i.e., Cˆ =
<
θ3 δ
θ1 Cˆ − 3B1 − θ1 B1 b > 0,
From the first two equations in (4.27), we obtain Cˆ =
−θ
+
Since V4 (0) = 0, we have D1 = −D2 := Cˆ . Also noting that B1 + B2 = 0, we rewrite (4.24) as −θ x θ x Cˆ − B1 x e 1 − Cˆ + B1 x e 2 , 0 ≤ x < b, V4 ( x ) = M (4.26) λ 1+ + (D3 + B3 x) e−θ3 x , x ≥ b. δ δ
V (b+) = V (b−), V4′ (b+) = V4′ (b−),
δ
3 M 1 µ−M −σ 2 θ3
,
M
δ
1 − e−θ3 x ,
λ δ
1+
M
x ∈ [0, ∞), M
+ δ δ λ λ × x− 1+ e−θ3 x , µ − M − σ 2 θ3 δ x ∈ [0, ∞).
(4.35)
Q. Zhao et al. / Insurance: Mathematics and Economics 58 (2014) 1–13
(ii) If (4.33) and (4.34) hold, and b is a positive solution to G(b) = 0, then c (·, ·, ·) is given by (4.17) with V3 (x) =
θx C e 1 − e−θ2 x , M
x ∈ [0, b),
− de , x ∈ [b, ∞), δ θ1 x 1x e Cˆ − B − Cˆ + B1 x e−θ2 x , x ∈ [0, b), (4.36) V4 (x) = λ M 1 + δ δ + (D3 + B3 x) e−θ3 x , x ∈ [b, ∞), where b, C , d, Cˆ , B1 , B3 , D3 is a solution to (4.21) and (4.27). Proof. It is easy to check that the function c (·, ·, ·) given by (4.17)– (4.19) satisfies the system (4.2). To prove c (·, ·, ·) satisfies the equilibrium HJB-equation (3.2), it is sufficient to show
∂c (t , t , x) ≥ 1, ∂x ∂ c (t , t , x) < 1, ∂x
x ∈ [0, b), (4.37) x ∈ [b, ∞).
λ λ + θ3 1 + 2 δ µ − M − σ θ3 δ λ x e−θ3 x − θ3 µ − M − σ 2 θ3 λ λ M + θ 1 + e−θ3 x ≥ 3 δ µ − M − σ 2 θ3 δ M 1 θ3 = + λ + θ e−θ3 x > 0. 3 δ µ − M − σ 2 θ3 δ λ Also note that V3 (0) = V4 (0) = 0 and V4′ (0) = µ−M −σ 2θ 3 λ M +θ3 1 + δ δ ∈ (0, 1]. Recalling the second equation of (4.19), M
V4′′ (x) = θ3 (θ3 D3 − 2B3 + θ3 B3 x) e−θ3 x
≤ θ3 [θ3 (D3 + B3 b) − 2B3 ] e−θ3 x = θ3 −B3 − eθ3 b e−θ3 x λ M < θ3 − − 1 eθ3 (b−x) µ − M − σ 2 θ3 δ Mλ ≤ θ3 θ3 − 1 eθ3 (b−x) . δ δ The last inequality follows from Lemma A.1. Furthermore, by (4.34), we have θ3 Mδ λδ − 1 ≤ 0. Therefore, V4′′ (x) < 0, for x ≥ b. Now we deal with the case when 0 ≤ x < b. It follows from (4.19) and (4.27) that 1 2 1 2
(i) Firstly, we show that the function V4 defined by (4.35) is a concave function. Recalling Lemma A.1 and λ > 0, we obtain V4′ (x) =
and
−θ3 x
9
σ 2 V4′′ (b−) = −µV4′ (b) + δ V4 (b) − λV3 (b), σ 2 V4′′ (b+) = − (µ − M ) V4′ (b) + δ V4 (b) − λV3 (b) − M = −µV4′ (b) + δ V4 (b) − λV3 (b),
which yield that V4′′ (b+) = V4′′ (b−) = V4′′ (b). Furthermore, for 0 ≤ x < b,
V4′′′ (x) = θ12 θ1 Cˆ − 3B1 − θ1 B1 x eθ1 x
+ θ22 θ2 Cˆ − 3B1 + θ2 B1 x e−θ2 x > θ12 θ1 Cˆ − 3B1 − θ1 B1 b eθ1 x + θ22 θ2 Cˆ − 3B1 e−θ2 x . It follows from Lemma 4.6 that if (4.34) holds, then V4′′′ (x) > 0 for 0 ≤ x < b. Since V4′′ (x) is continuous at x = b and V4′′ (b) < 0, we get that V4′′ (x) < 0, for 0 ≤ x < b. Therefore, c (t , t , x) = V4 (x) is a concave function on (0, ∞), which together with (4.27) implies (4.37). Similar to the proof of Corollary 4.3, it is easy to verify (3.6). We have the following corollary immediately by Theorems 3.2 and 4.7.
we have 1 2
σ 2 V4′′ (0) = − (µ − M ) V4′ (0) + δ V4 (0) − λV3 (0) − M = − (µ − M ) V4′ (0) − M = −µV4′ (0) + M V4′ (0) − 1 < 0.
−1 θ3 1 , then for (i) If Mδ > θ3 and λ ≤ Mδ − θ3 µ−M −σ 2θ + δ 3 (t , x) ∈ [0, ∞) × [0, ∞),
Thus, M
2λ
λ 1+ δ
+ θ3 µ − M − σ 2 θ3 λ − θ3 x e−θ3 x µ − M − σ 2 θ3 M λ = V4′′ (0) + θ32 x e−θ3 x < 0. δ µ − M − σ 2 θ3
V4 (x) = −θ3 ′′
δ
Therefore ∂∂ cx (t , t , x) = V4′ (x) ≤ 1 for all x > 0. (ii) For x ≥ b, recalling (4.31), V4′ (x) = (−θ3 D3 + B3 − θ3 B3 x) e−θ3 x
≥ (−θ3 D3 + B3 − θ3 B3 b) e
−θ3 x
= − [θ3 (D3 + B3 b) − B3 ] e−θ3 x = eθ3 (b−x) > 0,
Corollary 4.8. Assume that 0 < λ < δ . Consider the discount function (4.16).
∂c ∂ 2c πˆ (t , x) = φ t , t , (t , t , x), 2 (t , t , x) = M , ∂x ∂x is an equilibrium dividend strategy, and V (t , x) = c (t , t , x)
λ M M = 1+ + δ δ δ λ λ × x − 1 + e−θ3 x , µ − M − σ 2 θ3 δ
is the corresponding equilibrium value function. (ii) If (4.33) and (4.34) hold, then for t ∈ [0, ∞),
∂c ∂ 2c πˆ (t , x) = φ t , t , (t , t , x), 2 (t , t , x) ∂x ∂x 0, x ∈ [0, b), = M , x ∈ [b, ∞),
10
Q. Zhao et al. / Insurance: Mathematics and Economics 58 (2014) 1–13
Proof. Recall that θ1 and θ3 are given by
µ2 + 2 σ 2 δ θ1 = , 2 σ µ − M + (µ − M )2 + 2σ 2 δ θ3 = . σ2 −µ +
Then it follows that
Fig. 4.2. Equilibrium value functions with a pseudo-exponential discount function.
is an equilibrium dividend strategy, and V (t , x) = c (t , t , x)
ˆ C − B x eθ1 x 1 − Cˆ + B1 x e−θ2 x , = λ M 1+ δ δ + (D3 + B3 x) e−θ3 x ,
x ∈ [0, b),
x ∈ [b, ∞),
1 θ1 1 −µ + µ2 + 2σ 2 δ = − − + δ µ + σ 2 θ1 σ 2δ µ2 + 2σ 2 δ −µ + µ2 + 2σ 2 δ µ2 + 2 σ 2 δ − σ 2 δ = σ 2 δ µ2 + 2σ 2 δ −µ µ2 + 2σ 2 δ + µ2 + σ 2 δ = σ 2 δ µ2 + 2σ 2 δ √ 1 2 2 1 2 µ − 2 µ µ + σ 2 δ + 12 µ2 + σ 2 δ 2 2 2 = σ 2 δ µ2 + 2σ 2 δ 2 √ 2 1 2 µ − µ + σ 2δ 2 2 = > 0, σ 2 δ µ2 + 2σ 2 δ and
is the corresponding equilibrium value function. Here b, Cˆ , B1 ,
B3 , D3 is the solution to (4.27).
1
µ − M − σ 2 θ3
Remark 4.9. (i) Since essentially we are looking for a Nash equilibrium which is not unique in general, we did not discuss the uniqueness of the solution of the equation G(b) = 0 in Lemma 4.5. If there is no unique positive solution, one may choose the ‘‘best’’ one by some other criteria, such as minimizing the ruin probability. Furthermore, given some proper conditions we can get the uniqueness of the solution, see Appendix A. (ii) Note that (4.33) and (4.34) are only sufficient conditions (not necessary). So we only showed the results for two special cases in Theorem 4.7, and we are not clear if the results still hold in other cases. If λ is violating the restriction in (i) and (ii), then one may still get the results under some other sufficient conditions. See an example at the end of Appendix B.
Acknowledgments The authors thank the anonymous referee for his/her valuable comments and suggestions to improve on the earlier versions of this paper. This work was supported by Program of Shanghai Subject Chief Scientist (14XD1401600), the 111 Project (B14019), National Natural Science Foundation of China (11231005) and Doctoral Program Foundation of the Ministry of Education of China (20110076110004).
µ−M +
√
2 2
θ
θ
1 3 Lemma A.1. δ1 − µ+σ1 2 θ > 0 and µ−M −σ 2 θ + δ > 0. 1 3
2 (µ − M ) + 12 (µ − M )2 + σ 2 δ > 0. σ 2 δ (µ − M )2 + 2σ 2 δ
Proof of Lemma 4.5. It is easy to check that
G(0) = (θ1 + θ2 ) λ
θ3 + 2 µ − M − σ θ3 δ 1
M
δ
+ θ3
M
δ
−1 .
By Lemma A.1 and (4.33) we have G(0) > 0. Now by (4.22), (4.23) and (4.25), we rewrite G(b) as G(b) = θ3
×
M θ3 λ µ + σ 2 θ1 δ 1
(θ1 + θ3 ) eθ1 b + (θ2 − θ3 ) e−θ2 b
× −e2θ1 b + e−2θ2 b + ×
Appendix A
θ3 1 = − δ (µ − M )2 + 2σ 2 δ
(µ − M )2 + 2σ 2 δ σ 2δ µ − M + (µ − M )2 + 2σ 2 δ (µ − M )2 + 2σ 2 δ − σ 2 δ = σ 2 δ (µ − M )2 + 2σ 2 δ (µ − M ) (µ − M )2 + 2σ 2 δ + (µ − M )2 + σ 2 δ = σ 2 δ (µ − M )2 + 2σ 2 δ √ 1 (µ − M )2 + 2 22 (µ − M ) 12 (µ − M )2 + σ 2 δ + 21 (µ − M )2 + σ 2 δ 2 = σ 2 δ (µ − M )2 + 2σ 2 δ +
=
Example 4.10. Let µ = 1, σ = 1, M = 1, δ = 0.8. Fig. 4.2 shows the equilibrium value functions for pseudo-exponential discount functions with λ = 0, 0.1 and 0.2. The barriers b are 0.3470, 0.4141 and 0.4796, respectively. The case with λ = 0 is time consistent and the equilibrium strategy is optimal in the classical situation.
+
λ M 2 µ − M − σ θ3 δ
1
(θ1 + θ3 ) eθ1 b + (θ2 − θ3 ) e−θ2 b
2 × θ1 eθ1 b + θ2 e−θ2 b + 2 (θ1 + θ2 ) θ3
λ µ + σ 2 θ1
Q. Zhao et al. / Insurance: Mathematics and Economics 58 (2014) 1–13
×
M θ3
1
be(θ1 −θ2 )b
δ (θ1 + θ3 ) eθ1 b + (θ2 − θ3 ) e−θ2 b λ M + θ1 θ3 1 + − (θ1 + θ3 ) eθ1 b δ δ λ M + θ2 θ3 1 + − (θ2 − θ3 ) e−θ2 b δ δ
:=
1
(θ1 + θ3 ) eθ1 b + (θ2 − θ3 ) e−θ2 b
From Lemma 2.1 of Asmussen and Taksar (1997), we have M
δ
g (b),
λ
M θ3 (θ1 −θ2 )b λ be µ + σ 2 θ1 δ
Mθ λ Mθ λ + 2 2 µ + σ θ1 δ µ − M − σ θ3 δ λ M − (θ1 + θ3 ) (θ1 + θ3 ) + θ1 θ3 1 + δ δ M λ + e(θ1 −θ2 )b 2θ1 θ2 µ − M − σ 2 θ3 δ λ M − (θ1 + θ3 ) (θ2 − θ3 ) + θ1 θ3 1 + δ δ λ M + θ2 θ3 1 + − (θ2 − θ3 ) (θ1 + θ3 ) δ δ λ M θ32 λ M θ22 + e−2θ2 b + µ + σ 2 θ1 δ µ − M − σ 2 θ3 δ λ M − (θ2 − θ3 ) (θ2 − θ3 ) + θ2 θ3 1 + δ δ
+ e2θ1 b −
2 3
λ M θ3 (θ1 −θ2 )b be 2 µ + σ θ1 δ M θ1 1 2θ1 b 2 +e λ θ3 − + δ µ + σ 2 θ1 δ 1 θ3 + θ12 + + θ1 θ3 µ − M − σ 2 θ3 δ M 1 1 × (θ1 + θ3 ) − − δ θ3 θ1 M 1 + e(θ1 −θ2 )b λ 2θ1 θ2 δ µ − M − σ 2 θ3 1 + θ3 (2θ1 θ2 − θ1 θ3 + θ2 θ3 ) δ
= 2 (θ1 + θ2 ) θ3
+ θ3
M
δ
(2θ1 θ2 − θ1 θ3 + θ2 θ3 )
− 2 (θ2 − θ3 ) (θ1 + θ3 ) M 1 θ2 + e−2θ2 b λ θ32 − δ µ + σ 2 θ1 δ 1 θ3 + θ22 + 2 µ − M − σ θ3 δ
−
1
θ3
−
1
θ1
< 0.
By Lemma A.1 and (4.33), it is easy to see that the coefficient of e2θ1 b satisfies
where g (b) := 2 (θ1 + θ2 ) θ3
11
M + (θ2 − θ3 ) θ2 θ3 − (θ2 − θ3 ) . δ
2 1
M
δ
θ32 −
1
+
θ1 δ
+ θ12
1
µ + σ 2 θ1 µ − M − σ 2 θ3 1 1 M − < 0. − + θ1 θ3 (θ1 + θ3 ) δ θ3 θ1
+
θ3 δ
Thus, G(∞) < 0. Therefore, the equation G(b) = 0 admits a positive solution. In the following, we make some comments on the uniqueness of b. Let us consider the equation
−1 G(b) = 0, (θ1 + θ3 ) eθ1 b + (θ2 − θ3 ) e−θ2 b which is the original equation satisfied by b, i.e., (4.30). Denote ˆ (b) the left-hand side of the above equation. After tedious by G calculations, we have
3 (θ1 + θ3 ) eθ1 b + (θ2 − θ3 ) e−θ2 b Gˆ ′ (b) = 2 − (θ1 + θ3 ) eθ1 b + (θ2 − θ3 ) e−θ2 b (θ1 + θ2 )2 M θ3 (θ1 −θ2 )b λ be µ + σ 2 θ1 δ θ1 1 M − + e(2θ1 −θ2 )b λ θ3 (θ1 + θ2 ) 2 δ δ µ + σ 2 θ1 1 θ3 × θ3 (θ2 − θ1 − 2θ3 ) − 2 + µ − M − σ 2 θ3 δ θ3 × θ1 (θ1 + θ2 ) − (θ1 + θ3 ) (θ2 − 3θ1 ) δ M 2 2 − (θ1 + θ3 ) θ3 (θ1 + θ2 ) + e(θ1 −2θ2 )b δ M θ1 1 × λ θ3 (θ1 + θ2 ) −2 (θ2 − θ3 ) θ3 − δ δ µ + σ 2 θ1 θ3 1 θ2 (θ1 + θ2 ) −2 + µ − M − σ 2 θ3 δ θ2 1 + 2θ3 − (θ1 + θ3 ) δ µ + σ 2 θ1 θ3 M + (θ2 − θ3 ) (θ1 + θ2 ) − θ32 (θ1 + θ2 )2 (θ2 − θ3 ) . δ δ
× θ3
Obviously, if λ = 0, which is the case with exponential discounting, ˆ ′ (b) < 0. If λ > 0 is sufficient small, then the coefficients of then G e(2θ1 −θ2 )b and e(θ1 −2θ2 )b are negative. If, in additional, assume that θ2 − θ1 − 2θ3 < 0, then the decreasing function
− (θ1 + θ3 ) eθ1 b + (θ2 − θ3 ) e−θ2 b < 0, ˆ ′ (b) which means the coefficients of be(θ1 −θ2 )b < 0. Thus, we get G < 0 which guarantees the uniqueness of b.
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Q. Zhao et al. / Insurance: Mathematics and Economics 58 (2014) 1–13
θ1 − 2θ1 θ2 bB1 e−θ2 b − 2B1 θ1 eθ1 b + B1 (θ1 − 3θ2 ) e−θ2 b θ1 Cˆ − 3B1 − θ1 B1 b = θ1 eθ1 b + θ2 e−θ2 b θ1 b θ1 + B1 −2θ1 e + (−2θ1 θ2 b + θ1 − 3θ2 ) e−θ2 b = θ1 eθ1 b + θ2 e−θ2 b θ1 θ1 + θ3 − 2 µ+σλ 2 θ Mδθ3 e(θ1 +θ2 )b + θ1 (θ2 − θ3 ) + µ+σλ 2 θ Mδθ3 (−2θ1 θ2 b + θ1 − 3θ2 ) 1 1 . = eθ2 b θ1 eθ1 b + θ2 e−θ2 b (θ1 + θ3 ) eθ1 b + (θ2 − θ3 ) e−θ2 b
(B.1)
Box I.
Thus, it follows from q′ (0) ≥ 0 that
Appendix B Proof of Lemma 4.6. Assume that b is a positive solution to G(x) = 0. It follows that Eq. (B.1) is given in Box I. Let
q(b) := θ1 θ1 + θ3 − 2
M θ3 λ e(θ1 +θ2 )b µ + σ 2 θ1 δ
≥2
M θ3 λ + θ1 (θ2 − θ3 ) + 2 µ + σ θ1 δ
M θ3 λ θ1 θ2 > 0. µ + σ 2 θ1 δ
In fact, the result of Lemma 4.6 may also hold under some other conditions rather than (4.34). For example, let us assume
Then
λ M θ3 e(θ1 +θ2 )b µ + σ 2 θ1 δ
θ1 (θ1 + θ3 ) M θ3 − > 0. θ1 + 3θ2 µ + σ 2 θ1 Recalling 0 < λ < δ and 0 < θ1 , θ3 < θ2 , then we have
θ1 + θ3 − 2 λ M θ3 e(θ1 +θ2 )b , µ + σ 2 θ1 δ
and
λ M θ3 + θ1 (θ2 − θ3 ) µ + σ 2 θ1 δ λ M θ3 + (θ1 − 3θ2 ) µ + σ 2 θ1 δ λ M θ3 = θ1 (θ1 + θ2 ) − (θ1 + 3θ2 ) , 2 µ + σ θ1 δ λ M θ3 q′ (0) = θ1 (θ1 + θ2 ) θ1 + θ3 − 2 µ + σ 2 θ1 δ M θ3 λ θ1 θ2 −2 µ + σ 2 θ1 δ λ M θ3 = θ1 (θ1 + θ2 ) (θ1 + θ3 ) − 2θ1 (θ1 + 2θ2 ) . 2 µ + σ θ1 δ
q(0) = θ1 θ1 + θ3 − 2
θ +θ
If λ ≤ θ 1+3θ2 Mδθ ∧ 1 2 3 Lemma A.1 that 2
(θ1 +θ3 )(θ1 +θ2 ) δ 2 2θ1 (θ1 +2θ2 ) M θ3
holds, then it follows from
θ1 + 3θ2 λ M θ3 2 θ1 + θ2 µ + σ θ1 δ θ1 1 ≥ δ (θ1 + θ2 ) − >0 δ µ + σ 2 θ1
q(0) = (θ1 + θ2 ) θ1 −
and similarly, q′ (0) = (θ1 + θ2 ) (θ1 + θ3 )
λ M θ3 (θ1 + θ2 ) (θ1 + θ3 ) µ + σ 2 θ1 δ θ1 1 ≥ δ (θ1 + θ2 ) (θ1 + θ3 ) − > 0. δ µ + σ 2 θ1 × θ1 −
2θ1 (θ1 + 2θ2 )
θ1 Cˆ − 3B1 − θ1 B1 b > 0.
× (−2θ1 θ2 b + θ1 − 3θ2 ) .
λ M θ3 − 2θ1 θ2 , µ + σ 2 θ1 δ q′′ (b) = θ1 (θ1 + θ2 )2 θ1 + θ3 − 2
λ M θ3 2 µ + σ θ1 δ
Therefore, q′′ (b) > 0, q′ (b) > 0, and then q(b) > 0. Finally, it follows from (B.1) that
q′ (b) = θ1 (θ1 + θ2 ) θ1 + θ3 − 2
θ1 (θ1 + θ2 ) θ1 + θ3 − 2
λ M θ3 M θ3 > θ1 + θ3 − 2 µ + σ 2 θ1 δ µ + σ 2 θ1 M θ3 θ1 + θ3 − >2 2 µ + σ 2 θ1 > 0,
which means q′′ (b) > 0. Similarly,
λ M θ3 2 µ + σ θ1 δ M θ3 >0 > θ1 (θ1 + θ3 ) − (θ1 + 3θ2 ) µ + σ 2 θ1
θ1 (θ1 + θ2 ) − (θ1 + 3θ2 )
and
λ M θ3 µ + σ 2 θ1 δ M θ3 > θ1 (θ1 + θ3 ) − (θ1 + θ2 ) µ + σ 2 θ1 M θ3 + θ2 (θ1 + θ3 ) − (θ1 + 3θ2 ) µ + σ 2 θ1 > 0,
(θ1 + θ2 ) (θ1 + θ3 ) − 2 (θ1 + 2θ2 )
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