A two-dimensional dividend problem for collaborating companies and an optimal stopping problem

ABSTRACT We consider two insurance companies with wealth processes described by two independent Brownian motions with drift. The goal of the companies is to maximize their expected aggregated discounted dividend payments until ruin. The companies are allowed to help each other by means of transfer payments. But in contrast to Gu et al. [(2018). Optimal dividend strategies of two collaborating businesses in the diffusion approximation model. Mathematics of Operations Research 43(2), 377–398], they are not obliged to do so, if one company faces ruin. We show that the problem is equivalent to a mixture of a one-dimensional singular control problem and an optimal stopping problem. The value function is explicitly constructed and a verification result is proved. Moreover, the optimal strategy is provided as well.


Introduction
Since B. de Finnetti has suggested in his seminal paper (de Finetti 1957) an alternative to the classical ruin probability approach to evaluate an insurance company, namely the maximization of expected discounted dividends paid by the company, there has been a lot of research in this direction. We just mention a few examples.
Gerber proved in Gerber (1969) that, if the underlying wealth process follows the classical Cramer-Lundberg model, it is optimal to use a so called band strategy, which degenerates in the case of exponential individual claims to a barrier strategy, meaning that you have to keep your endowment just below a certain barrier value by paying dividends (see also Schmidli 2008 for a different proof of these facts). Similarly, Asmussen and Taksar (1997) showed that a barrier strategy is optimal in the diffusion approximation case as well. The problem is discussed for more general diffusion processes in Marciniak and Palmowski (2016), and in the case of finite time horizon one can find results in Grandits (2013), Grandits (2014) and Grandits (2015). For a survey on the optimal dividend literature, see e.g. Avanzi (2009) and the monograph (Azcue and Muler 2014).
Although most of the literature concerns the one-dimensional case, there has been some interest in the two-dimensional case recently. For example, Gerber and Shiu (2006) give sufficient conditions on the model parameters for a two-dimensional Brownian motion with drift, under which it is optimal for the two companies to merge. In Albrecher et al. (2017), a two-dimensional Cramer-Lundberg model is considered and, among other things, the value function is characterized as the smallest viscosity supersolution of the corresponding Hamilton Jacobi Bellman (HJB) equation. The paper, which is closest to the present one, is Gu et al. (2018). There the wealth process of the two companies is described by two independent Brownian motions with drift. Collaboration between the companies is allowed, which means that as soon as the wealth of one company hits zero, the other one can help by means of transfer payments (actually transfer payments are allowed at any time). It is also assumed that the solvent company has not only the possibility to help the insolvent one, but it is obliged to do so. Hence, in this model the possible ruin of both companies happens at the same time. It is shown that a barrier strategy, where the barrier value corresponds to the optimal value for the merged companies, is the optimal strategy. This last assumption of the model in Gu et al. (2018) is the one, which we abandon. In our model the solvent company can help the insolvent one, but it is not obliged to do so. (Similarly as in (Gu et al. 2018), transfer payments are possible at any time, so it is allowed in our setting to 'liquidate' one company by transferring its wealth completely to the other one at an arbitrary time.) Our paper can be considered as a singular stochastic control problem, subject to discretionary stopping. Some articles in this direction are Davis and Zervos (1994), Davis and Zariphopoulou (1995), , , Karatzas and Zamfirescu (2006) and Zervos (2003).
The schedule of our paper is as follows. In Section 2, we introduce the model and prove some preliminary properties of the value function. Moreover, we show that our problem can be seen as a one-dimensional mixture of a singular control problem and an optimal stopping problem. The HJB equation has the form of a variational inequality, but with three operators instead of the usual two. In Section 3, we construct the value function explicitly. By this we mean that at most three transition points have to be calculated numerically, and in between these points the value function is the solution of ODE's with constant coefficients. Let us note already here that -in contrast to Gu et al. (2018) -our value function is not C 2 . In fact, it will turn out to be C 1 , but with an a.e. bounded second derivative. In Section 4, we prove a verification theorem and the optimal strategy is also provided. A numerical example concludes the paper.

The model and preliminary results
The wealth of the two companies in question is described by the vector Z t := (X t , Y t ), with initial value Z 0 = (X 0 , Y 0 ) = z = (x, y), positive constants μ 1 , μ 2 , σ 1 , σ 2 and x, y ∈ R + 0 . The process L t := (L (1) t , L (2) t ) is caglad, monotone non-decreasing and F-progressively measurable, where F is the completed natural filtration of the two independent Brownian motions To include payments at time zero, we formally define L 0− = 0. These measurability requirements are standard in singular control theory, and can be found, e.g. in Fleming and Soner (2006, p. 296). The same holds for the process C t : t describe the accumulated dividend payments of company one respectively company two until time t, whereas the process C (1,2) t describes the accumulated transfer payments from company one to two, and similarly for C In Gu et al. (2018), the additional admissibility requirement for the control processes (L t , C t ), (ru) is imposed, where τ (ru) := inf{t ≥ 0|X t < 0, Y t < 0} denotes the moment when both companies are ruined. We replace it by with an arbitrary stopping time τ . This means that the companies help each other only until the stopping time τ ≤ τ (ru) . If τ is chosen as the ruin time of one company, this company gets no help any more at this time. If it is not the ruin time a priori, one transfers all the endowment of one company to the 'better' one at this arbitrary stopping time. Naturally, we assume that the remaining company behaves optimally after time τ . This means that in our optimal control problem the control processes run actually only up to time t = τ , because the optimal solution after this time is well known, namely the optimal 'one-dimensional solution'. This gives the following mixture of singular control and optimal stopping problem, with the objective functional and value function If not otherwise stated, the integral always includes possible jumps at the lower limit, but not at the upper one. β > 0 is a discounting factor and we define where V (1) (s), V (2) (s) are the value functions of the one-dimensional dividend maximization problem for the parameters (μ 1 , σ 1 ), (μ 2 , σ 2 ), respectively. These can be found, e.g. in Jeanblanc-Picqué and Shiryaev (1995) or Asmussen and Taksar (1997), and for latter use we give the one-dimensional result in Theorem 2.1: The value function V (1) (x) for the one-dimensional dividend maximization problem where the endowment process is described by linear Brownian motion with drift μ 1 and volatility σ 1 , is given by the following C 2 (R + 0 ) function.
Finally, b 1 is a positive constant, representing the value of the barrier in the barrier dividend strategy, which is optimal here. One has An analogous result holds clearly for V (2) (y).
Our next Lemma shows that the value function V(x, y) actually depends only on the sum s = x+y. By slight abuse of notation we do not introduce a new symbol for the value function, but write V = V(s).
Lemma 2.1: The value function V of problem (4) depends actually only on s = x + y, i.e. V = V(s).
We introduce now a new admissible strategy by transforming capital from one company to the other at time t = 0, s.t. we have again an endowment (x, y). Then we proceed with the original strategy, i.e. we have, for t > 0 and < 0, For > 0 we simply exchange the indices one and two. Furthermore, we leave (L, τ ) unchanged, i.e. (L,τ ) = (L, τ ). This leads to and since was arbitrary small and the given strategy arbitrary, we conclude Our next result is more or less trivial and shows the monotonicity of the value function.

Lemma 2.2: We have
Proof: Clearly, each strategy (L, C, τ ), which is admissible for a total endowment s, is admissible for s + (if we possibly shift capital from one company to the other at time t = 0, to ensure that both companies start with an endowment at least as good as the original one).
For later use we provide the following.

Definition 2.1:
hence S t fulfills the SDE dS t = μdt + σ dW t − dL t , with a Brownian motion W t .
In the next remark, we provide a useful property of the value function.
Remark 2.1: We give a lower bound for our value function. Indeed, the nonnegativity of the function g(s), the definition of the time horizon and the class of admissible controls implies that Hence, In an analogous fashion one can establish that demonstrating that the value function dominates also the value of an associated standard optimal stopping problem with payoff g(s).
In the last part of the present section we shall provide a property of the value function V(s), which will be useful in the next section, where we shall construct V and the optimal strategy rather explicitly.
We start with a rewriting of the HJB equation and introduce some new notation, i.e. the HJB equation can be written equivalently as with MV := min{−LV + βV, V − 1}. Furthermore, we introduce the following sets where S is called the stopping region, C the continuation region, and the D's will also be used below to construct the value function.
As final preparatory result we give necessary conditions on the form of the sets D.
Proof: We restrict our proof to the set D (1) . By the definition of the set D (1) , see (8), one knows that on is the solution of the (μ 1 , σ 1 )-problem, the first inequality holds on R + . Therefore, we get the following characterization of D (1) We first consider the interval s ∈ [b 1 , ∞). There one finds by the smooth fit condition at the point b 1 , see e.g. Gerber and Shiu (2006), (2.10), the following explicit formula,

a linear increasing function with zero at s (1)
1 . Let us now introduce the operator L := μ 1 (∂/∂s) We distinguish now several cases.
We assert that in this case one has for all s in (0, b 1 ], giving D (1) = [s (1) 1 , ∞). Assume first that H(0) < 0 holds. We then argue by contradiction. If (10) was false, then H(s) would have a strictly positive maximum, say s max ∈ (0, b 1 ), contradicting (9); (note that H(s max ) = H (s max ) = 0 is impossible, since this would imply H ≡ 0.) If H(0) = 0 holds, then we first note that H (0) = 0 is impossible by the same reasons as above for s max . Moreover, H (0) > 0 would imply a strictly positive maximum, hence again a contradiction. If or the existence of a strictly positive maximum, which leads again to a contradiction. Altogether, we have in Case 1: As above we cannot have any extrema with value zero. Moreover, strictly positive maxima or strictly negative minima for H(s) are again prohibited by Equation (9). The remaining case provides a monotone decreasing function H(s) on [0, b 1 ), with some zero in (0, b 1 ), which we call s 1 , ∞) holds in this case.

Explicit construction of the value function
In this section, we shall give a rather explicit construction of the value function and the optimal strategy. By 'rather explicit' we mean the following. In a first step at most four points on the s-axis have to be determined numerically. Once these points are known, the value function can be found by integrating ODE's with constant coefficients, the solution of which are of course well known. The optimal strategy uses some of these points as barrier values for barrier strategies, see below for details.
We start with the definition of the constant γ .
Note that by Proposition A.1, proven in the Appendix, there exists at most one such 'sign change point' (SCP) γ . Moreover, if γ = 0, we have V (1) (s) ≥ V (2) (s), for all s ≥ 0. Finally, the order of the V (i) 's is chosen w.l.o.g.
Indeed, by definition of g we have certainly g (γ +) − g (γ −) ≥ 0. We now argue by contradiction. So assume g (γ +) = g (γ −); we then have necessarily g (γ +) = g (γ −), because otherwise D(s) := V (1) (s) − V (2) (s) would not change the sign in γ . Hence, there would be a zero of third order of D(s) at γ . But it can be shown with the same methods (basically Descartes rule), as are used in Proposition A.1, that this is impossible.
For the following procedure compare also Figure 1. We continue with the definition of the point s as the solution of g(s) = μ/β, and we assert that holds. Since g is monotone increasing, the assertion is equivalent to μ/β > g(γ ). Now we have showing (11). Next we choose r ∈ R, s.t. V(s + r) = μ/β holds, and moreover, let i.e. we shift the solution V horizontally, until it touches g(s) from above. Note that we have g (s) < V (s + r), for s ∈ [s − , s), with small, since g (s) = 1 and V (s + r) > 1 holds there, implying that is well defined. Finally, define s 2 as follows By construction we have now Observe that we have necessarily s 2 = γ , since we have an upwards kink of g at γ , as noticed in Remark 3.1.
We start now with the construction of the value function and distinguish several cases, depending where the point s 2 lies.
Here we define the value function as We assert now Claim 1: The point s 2 defined in (13) fulfills s 2 ≤ b 2 , and s 2 ∈ D (2) ; hence, we are in the second case of Lemma 2.3, implying even s 2 ≤ s (2) 0 < b 2 , in the notation used there.
Here we are actually back in Case 1, by defining and Claim 1 holds as well for the pointŝ 2 . Case 2.2: V(s + ) > g(s), for all s ∈ [0, γ ]. Let s 3 ∈ (γ , s 2 ), solve the system and call the solution function v (s 3 ) . Furthermore, define and finally the value function as One can find a picture of this procedure in Figure 1.
Analogously to Claim 1 one proves for this case Claim 2: One has s 4 ≤ s 2 < b 1 , s 4 , s 2 ∈ D (1) , and s 4 , s 2 ≤ s (1) This concludes the construction of the function V(s), and the next theorem asserts that this function solves the HJB equation in a certain sense.

Theorem 3.1: The function V(s), defined by (15)-(18), solves the HJB equation
pointwise, except at points s ∈ ∂S, which are at most three points. Near these points V (s) is locally bounded.
Moreover, the boundary condition V(0) = 0 is fulfilled as well.

The verification result and the optimal strategy
We show in this section that the function, defined in (15)-(18) in the previous section, provides our value function. Moreover, the optimal strategy is given as well. We have the following theorem, the proof of which is an adaption of the proof of Theorem 10.4.1 of Øksendal (2003). (Alternatively, one could use Peskir (2005), where an Ito formula under weak assumptions is proved.) Theorem 4.1: Let G := R + .
(a) Assume we have a function V : 0, a.s,where τ G is the exit time of S t for the set G.Moreover, using the notation defined in (8), we assume Moreover, the optimal strategy can be described in the following way. Description of the optimal strategy. Case A: V(s) ≥ g(s), for all s ≥ 0.
Here it is optimal never to stop the sum process S t prematurely, i.e. in terms of the original companies: the solvent company has to support the insolvent one until both companies are ruined. Moreover, for the sum process a dividend strategy with barrier b has to be used.
Let us remark that, since the transfer processes between the two companies does not influence the sum process, this optimal strategy can be realized in many different ways. E.g. one possibility would be that the dividends are always paid out by company one (which is supported by company two, if in trouble.) This is the strategy described in Gu et al. (2018, p. 10) . This remark applies, mutatis mutandis, in the other cases below.
Case B: Here one should use the barrier strategy with barrier value b, until we reach the stopping time τ := inf{t ≥ 0 | S t ≤ s 2 } (s 2 defined in (13)); at time t = τ one should liquidate company 2 and run company 1 in the optimal way, i.e. with barrier strategy b 1 .
In the following we mean by liquidating one company that one should run the remaining one in the optimal way. Case C.1: {V(s) < g(s)} = ∅, γ > 0, s 2 < γ .
Here one should run the barrier strategy with barrier value b, until we reach the stopping time τ := inf{t ≥ 0 | S t ≤ s 2 }; at time t = τ one should liquidate company 1. The optimal strategy is: s > s 2 : use a barrier strategy with level b until τ := inf{t ≥ 0 | S t ≤ s 2 }; at time t = τ liquidate company 2.
Proof of (a). We start the proof with the remark that, as in Øksendal (2003), Theorem 10.4.1, we have the following approximation result for our candidate function V .
We can find a sequence of functions Øksendal 2003, (D.17) We get now for an arbitrary admissible strategy by Dynkin's formula Going to the limit j → ∞ gives, employing (α)-(δ), where we have used (ii) and (v) in the last inequality. Finally, letting R → ∞ on the right-hand side, we get by Fatou's Lemma concluding the proof of (a). Proof of (b). This is easy to check Proof of (c). We confine here to the Case 2.2 (see formula (18)) in the construction of the value function and Case C.2.b for the optimal strategy, since this is the most complicated case, and the other cases work analogously.
We first note that by Theorem 3.1 our function V(s) fulfills the HJB-equation pointwise except at three points.
Case 1: s > s 2 . Let V j be the approximating sequence for our candidate function V , assured above. Furthermore, let L τ * := inf{t > 0|S t = s 2 } (C * t is, as remarked above, one possible transfer process, leading to (L * , τ * )).
Obviously τ * is a.s. finite, and let τ k,R := inf{t > 0|S t = s 2 + 1/k} ∧ R, for k large enough, s.t. s 2 + 1/k < s holds. Clearly, for k, R → ∞, τ k,R tends to τ * a.s. Again by Dynkin's formula we have Using our approximation result and letting j → ∞ provides where in the last equality we have used the construction of the value function and the optimal strategy. Hence, we have and after k, R → ∞ we end up using dominated, respectively monotone convergence, with Here we use τ * = 0, L * t ≡ 0, giving Case 3: s ∈ (s 5 , s 4 ). We set L * t ≡ 0 and τ * := inf{t > 0|S t = s 5 or S t = s 4 }. Again τ * is a.s. finite, and we use the approximating stopping times τ k,R := inf{t > 0|S t = s 5 + 1/k or S t = s 4 − 1/k} ∧ R, for k large enough.
In the same way as in Case 1 one gets Case 4: s ∈ [0, s 5 ].
As in Case 2 we use τ * = 0, L * t ≡ 0, and get concluding our proof.

A numerical example
We chose here μ 1 = σ 1 = β = 1, as well as μ 2 = 0.5, σ 2 = 0.295. Performing the procedure described in Section 3, yields that we have here Case 1 in the construction of the value function. One gets γ ≈ 0.46 and s 2 ≈ 0.04, and a picture can be found in Figure 2. Concerning the optimal strategy, we have Case C.1 of Theorem 4.1c .
We shall now employ Descartes' sign rule and note that it is (by a shift in the exponent and a continuity argument) also valid for negative and/or irrational exponents. Consequently, D (0) (ρ) has at most two positive zeroes, and since ρ = 1 is an obvious zero, we have at most one zero larger than one.
We note that the number of SCP is certainly less or equal than the number of zeroes, hence D (0) (ρ) has at most one SCP larger than one, and therefore D (0) (s) has at most one positive SCP, concluding Case 1.
For the case b 2 > b 1 , we just note that we have here D (0) (b 1 ) < 0, implying b 1 is less than the first local minimum of D (0) (ρ) (otherwise we would have three zeroes of D (0) (ρ), again a contradiction to Descartes' rule).
Finally, we have our last proposition. Proposition A.1: D(s) has at most one positive SCP.
Proof: We first describe, how one can construct D(s), starting with D (0) (s).
For s ∈ [0, b 1 ∧ b 2 ], they are identical. In the interval [b 1 ∧ b 2 , b 1 ∨ b 2 ], D(s) is either strictly increasing or strictly decreasing. Finally, for s ≥ b 1 ∨ b 2 , D(s) is equal to a constant. Hence we can generate the function D(s) by appending smoothly, i.e. C 2 , a graph at the point (b 1 ∧ b 2 , D (0) (b 1 ∧ b 2 )), which is first strictly monotone and then constant.
A simple geometric consideration proves now the assertion of the Proposition. One has just to perform the 'appending procedure', described in the paragraph above, in all subcases (thereby also using Remark A.1!), we have considered in the proof of Lemma A.1. This reveals that one can generate not more than one SCP in all for D(s).