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In this paper, we investigate the optimization of mutual proportional reinsurance—a mutual reserve system that is in- tended for the collective reinsurance needs of homogeneous mutual members, such as P&I Clubs in marine mutual in- surance and reserve banks in the US Federal Reserve, where a mutual member is both an insurer and an insured. Compared to general (non-mutual) insurance models, which involve one-sided impulse control (i.e., either downside or upside impulse) of the underlying insurance reserve process that is required to be positive, a mutual insurance differs in allowing two-sided impulse control (i.e., both downside and upside impulse), coupled with the classical proportional control of reinsurance. We prove that a special band-type impulse control (*a*, A, B, b) with *a*=0 and *a*<A<B<b, coupled with a proportional reinsurance policy (classical control), is optimal when the objective is to minimize the total maintenance cost. That is, when the reserve position reaches a lower boundary of a=0, the reserve should immedi- ately be raised to level A; when the reserve reaches an upper boundary of b, it should immediately be reduced to a level B. An interesting finding produced by the study reported in this paper is that there exists a situation such that if the up- side fixed cost is relatively large in comparison to a finite threshold, then the optimal band control is reduced to a downside only (i.e., dividend payment only) control in the form of (0, 0; B, b) with *a*=A=0. In this case, it is opti- mal for the mutual insurance firm to go bankrupt as soon as its reserve level reaches zero, rather than to jump restart by calling for additional contingent funds. This finding partially explains why many mutual insurance companies, that were once quite popular in the financial markets, are either disappeared or converted to non-mutual ones.

This study is motivated by the long lasting while still strong-going success of marine mutual insurance (e.g., a P&I club of ship owners), the longest in the insurance business with a history of over 200 years; and by the still unanswered question why the once popular mutual insurance firms almost disappeared after 1990’s, either went bankrupt or converted to commercial (non-mutual) insurance business. A client of a mutual insurance is both insurer and insured, under a unique contingent regulation scheme of two-way impulse control (termed band control) which we give a brief description as follows. A mutual insurance firm is allowed: 1) to make calls for contingent injection from its clients once the reserve level is considered too low, as opposed to a commercial (non-mutual) insurance firm which would go bankrupt once its reserve becomes depleted; 2) or to make refunds (pay dividends) to its clients if the reserve is considered too high. We shall note that it is also permissible to adjust premium rate and liability policy in the continuous time horizon in conjunction of the band control as just mentioned, which in this case is referred to as a mixed band control of insurance reserves (i.e., a band type impulse control mixed with time-continuous classical control of a reserve process). Although in practice, mutual insurance engages mixed band control, most studies in the current literature are focused on the band control only, without classical control (e.g., with given constant premium rate, see [1,2] for reference). In this paper we consider mixed band control of mutual reinsurance which provides insurance to other mutual insurance firms, such as P&I clubs, of which each insures a number of ship-owners.

Reinsurance has been long investigated as an intrinsic part of commercial insurance, of which the mainstream modeling framework is profit maximization with the one-sided impulse control of an underlying reserve process. There are two types of one-sided impulse control: downside-only impulse control (such as a dividend payment) with a fixed cost (e.g., Cadenillas et al. [

The mutual proportional reinsurance model developed in this paper is a generalization of the proportional reinsurance models (e.g., Cadenillas et al. [

A pure two-sided impulse control problem (i.e., without a classical rate control) was investigated by Constantinides [

The remainder of the paper is organized as follows. In Section 2, we formulate the mathematical model and specify the HJB equation and the QVI of the corresponding stochastic control problem. We solve the QVI for the optimal value function in Section 3. In Section 3.2, we characterize and analyze the threshold. In Section 4, we prove the verification theorem and verify the optimal control. Finally, we make concluding remarks in Section 5.

The classical Cramer-Lundberg model of an insurance reserve (surplus) is described via a compound Poisson process:

where is the amount of the surplus available at time, quantity represents the premium rate, is the Poisson process of incoming claims and is the size of the ith claim. This surplus process can be approximated by a diffusion process with drift and diffusion coefficient, where is the intensity of the Poisson process. We assume that the insurer always sets (i.e.). Thus, with no control, the reserve process is described by

where is a standard Brownian motion.

We start with a probability space, that is endowed with information filtration and a standard Brownian motion on adapted to. Two types of controls are used in this model. The first is related to the ability to directly control its reserve by raising cash from or making refunds to members at any particular time. The second is related to the mutual insurance firm’s ability to delegate all or part of its risk to a reinsurance company, simultaneously reducing the incoming premium (all or part of which is in this case channeled to the reinsurance company). In this model, we consider a proportional reinsurance scheme. This type of scheme corresponds to the original insurer paying u fraction of the original claim. The premium rate coming to the original insurer is simultaneously reduced by the same fraction. The reinsurance rate can be chosen dynamically depending on the situation.

Mathematically, control U takes a triple form:

where is a predictable process with respect to, the random variables constitute an increasing sequence of stopping times with respect to, and is a sequence of -measurable random variables,.

The meaning of these controls is as follows. The quantity represents the fraction (reinsurance share) of the premium and risk (loss) incurred by the mutual insurance at time. Suppose that is chosen at time. Then, in the diffusion approximation (2.1), drift and diffusion coefficient are reduced by factor (see Cadenillas et al. [

The fact that the process is adapted to information filtration means that any decision has to be made on the basis of past rather than the future information. The stopping times represent the times when the ith intervention to change the reserve level is made. If, then the decision is to raise cash by calling the members/ clients. If, then the decision is to make a refund. The fact that is a stopping time and is measurable also indicates that the decisions concerning when to make a contingent call and how much cash to raise are made on the basis of only past information. The same applies to the refund decisions.

Once control U is chosen, the dynamics of the reserve process becomes:

Define the ruin time as

Control is called admissible for initial position if, for for any,

and if

We denote the set of all admissible controls by.

The meaning of admissibility is as follows. At any time the decision to make a refund is made, the refund amount cannot exceed the available reserve. As can be seen in the following, if this condition is not satisfied, then one can always achieve a cost equal to, simply by making an infinitely large refund. The second condition of admissibility is a rather natural technical condition of integrability.

The objective in this model is to minimize the operational cost and the lost opportunity to invest the money in the market. Cost function is defined as

Here, and denote the positive and negative components of, that is, and . The costs associated with refunds are of a different nature. A contingent call always increases the total cost, whereas a refund decreases it. However fixed set-up costs and are incurred regardless of of the size of a contingent call or a refund. In addition, when the call is made and the cash is raised, there is a proportional cost associated with the amount raised. The constant represents the amount of cash that needs to be raised in order for one dollar to be added to the reserve. If the reserve is used for a refund, then a part of it may be charged as tax. The constant represents the amount actually received by the shareholders for each dollar taken from the reserve.

Given a discount rate r, the cost functional associated with the control U is defined as

The objective is to find the value function,

and optimal control, such that

For each, define the infinitesimal generator. For any twice continuously differentiable function

Let be the inf-convolution operator, defined as

Definition 2.1. The QVI of the control problem are

and

together with the tightness condition

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In this model, the application of the control that is related to calls and refunds results in a jump in the reserve process. This type of model is considered in the framework of the so-called impulse control. Because we also have a control whose application changes the drift and the diffusion coefficient of the controlled process, the resulting mathematical problem becomes a mixed regular-impulse control problem (e.g., Cadenillas et al. [

We conjecture that, in our case, the optimal intervention (impulse control) component of the problem is also of the band type. Moreover, as the following analysis implicitly shows, we can narrow our search for the optimal policy to a special band-type control, where the level associated with the contingent calls is set to zero. Therefore, only three of the four band-type policy parameters remain unknown. After finding these parameters (and determining the optimal drift/diffusion control in the continuation region), we will see that the cost function associated with this policy satisfies the QVI.

The derivation of the value function is similar to [

for. Also

for.

Assume that minimizes the function in foregoing equation. If then

provided that the right-hand side of (3.17) belongs to. (Note that if, then (3.16) cannot be satisfied and we exclude from consideration).

Substituting (3.17) into (3.16), we get

The general solution for (3.18) is

where and are free constants to be determined later, and

It is easy to see that. From (3.17), we obtain the expression for (provided that which will be verified later):

Note that the solution of (3.18) coincides with the solution of (3.16) only in the region where

From this expression, we conjecture that there is a switching point such that when. As, by virtue of the equation (3.21), we obtain the following expression for:

For,; and the corresponding differential equation becomes

The general solution for (3.23) is

where

with. Standard arguments show that

For (see e.g., Cadenillas et al. [

However, if bankruptcy is allowed and no intervention is initiated when the process reaches 0, then the boundary condition at 0 becomes straightforward: (see Cadenillas and Zapatero [

We seek the solution by finding a function such that

To find the free constants in the expressions for an and to paste different pieces of the solution together we apply the principle of smooth fit by making the value and the first derivatives to be continuous at the switching points and b,

where is defined by (3.22).

(It should be noted that the function, which is constructed from (3.32)-(3.34) subject to conditions (3.28)- (3.31) and (3.35)-(3.37), corresponds to the case in which the optimal policy leads to). We begin by constructing such a function. The main technique is not to consider the function itself, but rather first to construct.

The form of is shown in

From and (3.17), we have

. By the continuity on and at, and by (3.23), we have. From this relation and (3.24), we have

Let. Then, , and we can write

We can easily get the inequalities:

From (3.22), we get

From, and from the continuity of at, we obtain the expression for:

Let. (Obviously, since.) Now, we can write in terms of and:

What remains is to determine and. Once these constants are found, we have and, and thus. Let

where.

Note that if and, then it is easy to show that for, ,

and,

. Therefore, is decreasing on and is concave on. In the remainder of this section, we find and and complete the construction of the function. We do this in an implicit manner by adopting an auxiliary problem in which no contingent calls are allowed and by using the optimal value function of that problem to construct the function.

Let’s consider a slightly different problem in which only those controls for which on the right-hand side of (2.2) are negative allowed. This problem is similar to that considered in Cadenillas et al. [

The same arguments as those above show that we can make the conjecture that the function should be sought as a solution to (3.39)-(3.44) below.

where.

First note that a general solution to (3.39), (3.41) is, where is the same as in (3.20) and is a free constant, and a general solution to (3.40) is , where and are the same as in (3.25), (3.26).

To solve our auxiliary problem we apply the same technique as that used in Cadenillas et al. [

this expression, constants and are chosen in such a way that and

. That is, the functions and are continuous at. (Note that is a derivative of on and is a derivative of on.) We next examine the family of functions, where. We seek such that becomes the derivative of the optimal value function.

To this end, we start by finding points and such that and

. Note that is a concave function, which is easily checked by differentiation. Let, the point at which the maximum of is achieved (it is easy to see that

, whereas, which shows that

exists; in view of the fact that, it is unique). It is obvious by virtue of the concavity of that, for any, and exist.

We now consider. Informally, is the area under the graph of and above the horizontal line. It is obvious that is a continuous function of. For, we have; Let

Then,

is the optimal value function of the auxiliary problem (see

We employ the function obtained in the previous subsection to construct the derivative of the optimal val-

uefunction. The main idea is to consider and try to find such that. The optimal value function will then be sought in the form of. To this end, we need the following proposition.

Proposition 3.1. Suppose that satisfies (3.39) (satisfies (3.40)); then, for any S, the function satisfies the same equation on the interval shifted by S to the left.

The proof of this proposition is straightforward.

From (3.45), we can see that has a singularity at 0 with. The concavity of on

implies that is increasing on and decreasing on (recall that is constant on). Therefore, there exists unique such that. Define

Note that decreases to at 0 at the order of (see (3.45)); therefore, is integrable at 0 and, as a result,.

The qualitative nature of the solution to the original problem depends on the relationship between and; hence, we divide our analysis into two cases.

Consider the following integral

Geometrically this integral represents the area of a curvilinear triangle bounded by the lines, and the graph of the function. Obviously, is a continuous function of. Because

and, there exists an (see

In what follows, we show that is the derivative of the solution to the QVI, inequalities (2.12)-(2.14).

Let

Also let

By virtue of Proposition 3.1, the function satisfies (3.18) on, as well as (3.23) on. In addition, from (3.48), we can see that

From the construction of the function, we can also see that

We also have and similarly, so satisfies all the conditions (3.28)-(3.31).

Theorem 3.1. The function given by (3.56) is a solution to the QVI (2.12)-(2.14).

The proof of this theorem is divided into several propositions.

Proposition 3.2. The function satisfies (3.16) on.

Proof. 1) From the construction of the function v, we have on. Consequently,

on. As satisfies (3.18), it also satisfies (3.16) because these two equations are equivalent whenever (3.16) holds.

2) To prove that (3.16) holds for, it is sufficient to show that for each,

because, in view of Proposition 3.1, the function satisfies (3.23) (that is,). By subtracting (3.58) from the left hand side of (3.23), we can see that (3.58) is equivalent to

for. In view of the continuity of the first and second derivatives, and in view of the fact that (3.16) holds on, we know that (3.59) is true for. As both and on the left-hand side of (3.59) are decreasing, therefore is nonpositive for all.

Note that, where

Proposition 3.3. For each, we have

If, or if, then

Proof. 1) We first prove, that. Supposethat. The function is continuously differentiable. By construction, is increasing on and decreasing on with for. Therefore, the point is the only point such that. Because, we can see that. Therefore, for,

Also,

if

However,

This proves that for all and also shows that

Because for, we know that for any the function is an increasing function of. Therefore, the minimum in the expression for is attained for; as a result,.

2) Consider for. Because for and for, we can see that has a unique minimum at. Therefore,. Thus, if

The foregoing inequality always holds true because, by construction,

For, we note, that, and hence

To complete the proof of Theorem 3.1, we need only show the following.

Proposition 3.4. If, then (2.12) holds.

Proof. It is sufficient to show that for any and any,

From Proposition 3.2 we know that (3.64) is true for. As, we obtain

.

For any, we have

because V is a decreasing function.

This completes the proof of Theorem 3.1.

When we cannot find any S^{*} such that (3.48) is satisfied. In this case, we set (that is, we have, which corresponds to).

Theorem 3.2. If, then is a solution to the QVI (2.12)-(2.14).

To prove this theorem, it is sufficient to prove Propositions 3.2-3.4. The proofs of Propositions 3.2 and 3.4 are identical to the case of, whereas that of Proposition 3.3 requires a slight modification.

Proposition 3.5. For each,

If, then

Proof. The proof that for all and that for is the same as that in Proposition 3.3.

If, then is equivalent to

The foregoing inequality is always true because

by assumption.

If, then still holds, due to the same argument as that in Proposition 3.3.

Remark 3.1. In the case of we have only for, whereas

in contrast to the case when. Also, when, we have, whereas if. Equivalently, in the case that the fixed cost to call for additional funds is relatively large (i.e.,) , the optimal band control is reduced to with. That is, as soon as the reserve reaches zero, it becomes optimal for the mutual insurance firm to go bankrupt, rather than to be restarted by calling for additional funds.

Theorem 4.1. If V is a solution to QVI (2.12)-(2.14), then for any control,

Proof. We prove this inequality when. In this case,. Let be any admissible control defined by (2.2) and process be the corresponding surplus process (2.3), with. Let be its ruin time given by (2.4). Let and. Then,

By convention,. In view of (2.13), we have

Therefore,

and

In view of (2.6), this implies that for any the second sum in (4.68) is bounded by the same integrable random variable independent of.

On, process is continuous, and we can apply Ito’s formula to get

From this equation, using (2.6) and standard but rather tedious arguments, we can deduce that the first sum in (4.69) for all is also bounded by the same integrable random variable. Similar arguments show that

and

(see Cadenillas et al. [

and, taking into account the dominated convergence theorem, we can see that the expectation of the first sum on the right-hand side of (4.68) is nonnegative.

From (2.13), we can see that

Substituting this inequality into (4.2) and taking expectations of both sides, we obtain

Letting, and employing (4.72) and the monotone convergence theorem on, we get

As, on, the inequality (4.75) implies (4.67).

Remark 4.1. For the expectation of the stochastic integral on the right hand side of (4.70) to vanish, it is sufficient for its integrand to be bounded. In particular, it is sufficient for to be bounded. This is the case when. When, the function has a singularity at 0. We can, however, apply the same technique, first replacing by and then passing to a limit as. This will yield inequality (4.74), which is all we need for the proof of Theorem 4.1 Let, that is

From (3.46), (3.45), (3.49), and (3.53), we can see that

Consider the process defined as

, with, , and and are defined below sequentially.

If (that is, no exists, such that (3.48) holds), then

where

If (that is, there exists an, such that (3.48) holds), then

where

Remark 4.2. Informally, if, then process is a continuous diffusion process with a drift and diffusion coefficient of, and , respectively, until the times of intervenetion. The times of intervention in this case are the times at which this diffusion process hits the level, which are associated with the refunds of the constant amount of. The time when 0 is hit is the ruin time.

When, the process is a continuous diffusion process with the same drift and diffusion coefficients as above, between the times of intervention. The intervention times are the times at which this process reaches either 0 or. At point 0, the control is set to displace the process to point, which corresponds to raising cash (making a call to shareholders) in the amount of. Reaching the level results in the displacement of the process to the point which, corresponds to making a refund in the amount of.

Theorem 4.2. (The verification theorem) Let be the control described by (4.76) and (4.79)-(4.80). Then,

Proof. In view of (4.67), it is sufficient to show that

Equality (4.76) shows that

From Propositions 3.3 and 3.5, we know that

and if then. Thus, we can repeat the arguments in the proof of Theorem 3.2 and see that, for, all of the inequalities are tight. As a result, we obtain

Because we know that, when, we have and, when, the function V satisfies, the first term on the left-hand side of (4.84) vanishes, and we get (4.82).

The optimal policy in this model has several interesting nontrivial features. The fact that calls should be made only when there is no possibility of waiting any longer (that is when the reserve reaches zero) is supported by intuition. However, the qualitative structure of the optimal policy and its dependence on the model parameters are not as obvious.

It turns out that it is always optimal to pay dividends, no matter what the costs associated with such payments are. However, raising cash may not be optimal when the initial set-up cost is too high. Quantity, which determines the threshold for set-up cost, such that if the cost is higher than this threshold then it is optimal to allow ruin, is in itself determined via an auxiliary problem with a one-sided impulse control. Although there is no closed-form expression for the quantity, it can be determined in an algorithmic manner prior to solving the optimal control problem for the mutual insurance company.

There is one rather curious feature of the optimal solution when. As our analysis shows, in this case, and, the same as is the case when. However, from the construction of the optimal policy, we can see that the two band-type policy is optimal in this case as well. In this borderline case, we thus have two optimal policies, one for which with the lower band equal to and one for which reaching 0 corresponds to ruin and for which. This is a rather unique feature of this particular problem that has not been observed previously.

A natural question arises: what if ruin is explicitly disallowed, and we must find an optimal policy from among those for which. As can be seen from our analysiswe find a solution to this problem for the case of

. However, when this inequality does not hold, then of the stochastic control technique and the HJB equation used in this paper do not work. Another approach should be developed, as can be seen indirectly in the work of Eisenberg [