Dependent risks and excess of loss reinsurance

Insurance: Mathematics and Economics 37 (2005) 229–238

Dependent risks and excess of loss reinsurance Maria de Lourdes Centeno∗ ISEG, Technical University of Lisbon, CEMAPRE, Rua do Quelhas 2, 1200 781 Lisboa, Portugal Received July 2004 ; received in revised form December 2004; accepted 16 December 2004

Abstract In this paper we study, from the insurance point of view, the optimal excess of loss retention limits for two dependent risks. We consider two optimization criteria, which are quite connected. The expected utility of wealth with respect to the exponential utility function and the adjustment coefficient of the retained aggregate claims amount. We consider that the number of claims is generated by a bivariate Poisson distribution. The premium calculation principle used for the excess of loss treaties is the expected value principle. Although the systems of equations, that give the optimal solution for both problems, look quite similar, we will see that the optimal solution is heavily dependent on the criterion chosen. © 2005 Elsevier B.V. All rights reserved. Keywords: Reinsurance; Excess of loss; Expected utility of wealth; Exponential utility function; Adjustment coefficient; Bivariate Poisson; Dependent risks

1. Introduction Although the literature on mathematical risk models incorporating dependent risk structures is increasing rapidly, very few of these contributions deal with the problem in relation to reinsurance. In this paper we consider a risk model involving two risks, which could represent two classes of insurance businesses, dependent through the number of claims. Let {Xij }j=1,2,... be the claim size random variables for risk i, i = 1, 2. We assume that, for i = 1, 2, {Xij }j=1,2,... are i.i.d. random variables with common distribution function Fi and expected value µi . We assume that their moment generating functions exist, that Fi (x) = 0 for x ≤ 0 and that 0 < Fi (x) < 1 for x > 0, with i = 1, 2. Let S1 and S2 be the aggregate claims amounts for the first and second risks, respectively, with Si =

Ni


Xij ,

j=1 ∗

Tel.: +351 213925871; fax: +351 213922782. E-mail address: [email protected].

0167-6687/$ – see front matter © 2005 Elsevier B.V. All rights reserved. doi:10.1016/j.insmatheco.2004.12.001

(1)

M.L. Centeno / Insurance: Mathematics and Economics 37 (2005) 229–238

230



 = 0 where Ni is the number of claims of each risk i in a given period of time. We consider that {X1j }j=1,2,... are independent of {X2j }j=1,2,... and that both are independent of N1 and N2 . We assume that 0 i=1

N1 = K1 + K

and N2 = K2 + K

(2)

where K1 , K2 and K are independent Poisson random variables with parameters λ1 , λ2 and λ, respectively, i.e., (N1 , N2 ) is the bivariate Poisson distribution, constructed by Holgate (1964) (see also Johnson et al., 1997). Hence S1 and S2 are correlated through K. This model was, among others, considered by Wang (1998) on the description of correlated risk portfolios. The dependence between two lines of business arises generally by a common effect, which will affect both the number and the claim size of both lines. However in some situations the severities have little correlation, in which case the model can be applied. The classical case of dependence is natural hazards where usually at least two lines of business are affected - homeowners and motor hull. We address the problem of determining, from the insurer point of view, the optimal excess of loss retention limit Mi for each risk, assuming that the premium calculation principle used for each treaty is the expected value principle with loading coefficient αi > 0. In Section 2, we choose the retention limits in such a way that the expected utility of wealth, with respect to an exponential utility function, is maximized. In Section 3, the optimization criterion is the adjustment coefficient of the retained business. In Section 4, we give an example.

2. Maximizing the expected utility As it was proved in Centeno (1988) for an exponential utility function, u(x) = − exp(−βx),

(3)

an insurance company maximizes (having the retention limits as decision variables) its expected utility of wealth with respect to a portfolio of n independent risks if and only if the expected utility of wealth with respect to each of the risks is maximized, which will be achieved for Mi =

1 ln(1 + αi ), i = 1, . . . , n β

(4)

where β = −u /u is the insurer’s risk aversion coefficient. We will study in Result 2.2, what happens for the case n = 2 when the risks are no longer independent, but obey to the model considered in the previous section. Summarising, our aim is to calculate (M1 , M2 ) in order to maximize the expected utility of wealth net of reinsurance, with respect to an exponential utility function, i.e., Maximize E{− exp[−β(P1 + P2 − (Pˆ 1 (M1 ) + Pˆ 2 (M2 ) + Sˆ 1 (M1 ) + Sˆ 2 (M2 )))]}, M1 ,M2

(5)

where Pi , i = 1, 2, is the insurance premium (net of expenses) of risk i, Pˆ i (Mi ) is the reinsurance premium of risk i, i.e.,  Pˆ i (Mi ) = (1 + αi )(λi + λ)

∞

Mi

(1 − Fi (x)) dx

(6)

M.L. Centeno / Insurance: Mathematics and Economics 37 (2005) 229–238

231

and Sˆ i (Mi ) =

Ni


min(Xij , Mi ).

j=1

The following lemma will be helpful for the proof of Result 2.2. Lemma 2.1. The expected utility of wealth is a well behaved function of the retentions, in the sence that the Hessian matrix is negative definite, whenever the gradient is null. Proof. The moment generating function of the total retained aggregate claims is E{exp[β(Sˆ 1 (M1 ) + Sˆ 2 (M2 ))]} = E{E[exp[β(Sˆ 1 (M1 ) + Sˆ 2 (M2 ))]|N1 , N2 ]} = E{(MXˆ 1 (M1 ) (β))N1 (MXˆ 2 (M2 ) (β))N2 } = exp[(λ1 + λ)(MXˆ 1 (M1 ) (β) − 1) + (λ2 + λ)(MXˆ 2 (M2 ) (β) − 1) + λ(MXˆ 1 (M1 ) (β) − 1)(MXˆ 2 (M2 ) (β) − 1)].

(7)

The last equality follows from the probability generating function of the bivariate Poisson (see Johnson et al., 1997 p. 126). MXˆ i (Mi ) (β) is the moment generating function of the retained individual claims amounts of risk i, ˆ i (Mi ) = max(Xi , Mi ), where Xi is identically distributed to Xij , which is is to say X  Mi  Mi MXˆ i (Mi ) (β) = exp(βx) dFi (x) + exp(βMi )(1 − Fi (Mi )) = 1 + β exp(βx)(1 − Fi (x)) dx, (8) 0

0

because we have assumed that Fi (0) = 0. Hence maximizing (5) is equivalent to minimizing  ∞  h(M1 , M2 ) = (1 + α1 )(λ1 + λ) (1 − F1 (x)) dx + (1 + α2 )(λ2 + λ)  ×

M1

M1

0

 ×

0

M1



exp(βx)(1 − F1 (x)) dx + (λ2 + λ)  exp(βx)(1 − F1 (x)) dx

∞

M2

M2

(1 − F2 (x)) dx + (λ1 + λ)

exp(βx)(1 − F2 (x)) dx + βλ

0 M2

exp(βx)(1 − F2 (x)) dx,

(9)

0

and we will show that the Hessian matrix of h(M1 , M2 ) is positive definite in all the points where the first derivatives are null. Differentiating (9) with respect M1 and M2 we obtain  M2 ∂h = (1 − F1 (M1 )){(λ1 + λ)[exp(βM1 ) − (1 + α1 )] + βλ exp(βM1 ) exp(βx)(1 − F2 (x)) dx} (10) ∂M1 0 and ∂h = (1 − F2 (M2 )){(λ2 + λ)[exp(βM2 ) − (1 + α2 )] + βλ exp(βM2 ) ∂M2



M1

exp(βx)(1 − F1 (x)) dx}.

(11)

0

Calculating the second derivatives of h(M1 , M2 ) at the points where the first derivatives are zero we get   

 M2 ∂2 h  = β exp(βM1 )(1 − F1 (M1 )) λ1 + λ 1 + β exp(βx)(1 − F2 (x)) dx ∂M12 ∂h/∂M1 =0 0

(12)

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 M1 ∂2 h  = β exp(βM2 )(1 − F2 (M2 )) λ2 + λ 1 + β exp(βx)(1 − F1 (x)) dx ∂M22 ∂h/∂M2=0 0

(13)

∂2 h = βλ exp(βM1 + βM2 )(1 − F1 (M1 ))(1 − F2 (M2 )), ∂M1 M2

(14)

and ∂2 h ∂2 h − ∂M12 ∂M22  =



∂2 h ∂M1 M2

2    

∂h/∂M1 =0,∂h/∂M2 =0

   β exp(βM1 )(1 − F1 (M1 )) λ1 + λ 1 + β 

×



M2

exp(βx)(1 − F2 (x)) dx

0

   β exp(βM2 )(1 − F2 (M2 )) λ2 + λ 1 + β



M1

exp(βx)(1 − F1 (x)) dx

0

− β2 λ2 exp(2β(M1 + M2 ))(1 − F1 (M1 ))2 (1 − F2 (M2 ))2 .

(15)

(15) is a polynomial of second degree in λ. The independent term and the coefficient in λ are positive and we will show that the coefficient of the term in λ2 is also positive, since   β2 exp(β(M1 + M2 ))(1 − F1 (M1 ))(1 − F2 (M2 )) × 1 + β  ×



M2

1+β

exp(βx)(1 − F2 (x)) dx − β2 exp(2β(M1 + M2 ))(1 − F1 (M1 ))2 (1 − F2 (M2 ))2 

= β2 exp(β(M1 + M2 ))(1 − F1 (M1 ))(1 − F2 (M2 )) × ×

 1+β

M1

 1+β

×

exp(βx)(1 − F2 (x)) dx

exp(βx)(1 − F1 (x)) dx − exp(β(M1 + M2 ))(1 − F1 (M1 ))(1 − F2 (M2 )) 

= β2 exp(β(M1 + M2 ))(1 − F1 (M1 ))(1 − F2 (M2 )) × M1

M2

0



0



exp(βx)(1 − F2 (x)) dx

0



0



M2

M2

exp(βx) dF2 (x) + exp(βM2 )(1 − F2 (M2 ))

0





exp(βx) dF1 (x) + exp(βM1 )(1 − F1 (M1 )) − exp(β(M1 + M2 ))(1 − F1 (M1 ))(1 − F2 (M2 ))

0



= β2 exp(β(M1 + M2 ))(1 − F1 (M1 ))(1 − F2 (M2 )) ×  + exp(βM2 )(1 − F2 (M2 )) 0



M2

0 M1

M1

exp(βx) dF2 (x)

exp(βx) dF1 (x) 0



exp(βx) dF1 (x) + exp(βM1 )(1 − F1 (M1 ))

M2

exp(βx) dF2 (x) ,

0

(16)

M.L. Centeno / Insurance: Mathematics and Economics 37 (2005) 229–238

233

which is positive. Hence (12), (13) and (15) are positive, which implies that there will be at most one point (M1 , M2 ) such that  ∂h   =0  ∂M1 (17) ∂h    = 0. ∂M2
Result 2.2. The maximum of the expected utility of wealth of the retained aggregate claims amount, with respect to an exponential utility function, 1. when α1 ≥ β

λ λ1 + λ

and α2 ≥ β

λ λ2 + λ



(1/β) ln(1+α2 )

exp(βx)(1 − F2 (x)) dx,

(18)

exp(βx)(1 − F1 (x)) dx,

(19)

0



(1/β) ln(1+α1 )

0

is attained at the only point (M1 , M2 ) satisfying  1 + α1  exp(βM1 ) = ,   M2  λ  1 + β λ1 +λ 0 exp(βx)(1 − F2 (x)) dx 1 + α2    .  M1  exp(βM2 ) = λ 1 + β λ2 +λ 0 exp(βx)(1 − F1 (x)) dx 2. when α1 < β

λ λ1 + λ



(1/β) ln(1+α2 )

exp(βx)(1 − F2 (x)) dx,

(20)

(21)

0

is attained at the point (M1 , M2 ) = (0, (1/β) ln(1 + α2 )); 3. and when  (1/β) ln(1+α1 ) λ α2 < β exp(βx)(1 − F1 (x)) dx, λ2 + λ 0

(22)

is attained at the point (M1 , M2 ) = (1/β ln(1 + α1 ), 0). Proof. The previous Lemma asserts that h(M1 , M2 ) is a well behaved function of the retention limits M1 and M2 . Hence the first order conditions for a minimum will be also sufficient. As the retention limits have to be non-negative, the first order conditions for the point (M1 , M2 ) to be a minimum are that  ∂h ∂h   ≥ 0 and M1 =0  ∂M1 ∂M1 (23) ∂h ∂h    ≥ 0 and M2 = 0. ∂M2 ∂M2 The set of points candidates to satisfy these conditions are: the point (M1 , M2 ) satisfying (17) if such a point exists in which case it will be the solution to our problem; the point M1 = 0 and M2 such that ∂h/∂M2 |M1 =0 = 0 if such a point exists (and at that point ∂h/∂M1 would have to be positive); M2 = 0 and M1 such that ∂h/∂M1 |M2 =0 = 0

M.L. Centeno / Insurance: Mathematics and Economics 37 (2005) 229–238

234

if such a point exists (and at that point ∂h/∂M2 would have to be positive); M1 = M2 = 0 if there are no points satisfying the former conditions. We will check that the three first conditions will correspond, respectively to the conditions 1, 2, and 3 stated in the result. 1. Dividing both sides of the equations in (17) by (1 − F1 (M1 )) and (1 − F2 (M2 )), respectively, we obtain (20). The second equation in system (20) is equivalent to 1 M2 = ln β





1 + α2

1 + β(λ/λ2 + λ)

 M1 0

exp(βx)(1 − F1 (x)) dx

.

(24)

Let us consider the function g(M1 , M2 ) = ∂h(M1 , M2 )/∂M1 , given by (10) and let gˆ (M1 ) be the function g(M1 , M2 ) evaluated at the points M2 satisfying (24). Note that  gˆ (0) = −α1 (λ1 + λ) + βλ

(1/β) ln(1+α2 )

exp(βx)(1 − F2 (x)) dx ≤ 0,

(25)

0

by (18). On the other hand  gˆ



1 ln(1 + α1 ) β

> 0.

Hence, there is a value M1 , with 0 < M1 < 1/β ln(1 + α1 ), such that gˆ (M1 ) = 0 and for such a value, the value of M2 , given by (24), is positive, by (19). 2. When (21) holds gˆ (0), given by (25), is nonnegative. For M1 = 0, ∂h/∂M2 = 0 is equivalent to M2 = 1/β ln(1 + α2 ), and part 2. of the result is proved. 3. The proof is analogous to the proof of 2.
It is worth noticing that (21) and (22) can not happen simultaneously, because (21) implies that α1 < α2 and (22) implies that α2 < α1 . This result for two independent risks is obtained considering that λ = 0. In this case (21) and (22) never hold and (20) is equivalent to (4).

3. Maximizing the adjustment coefficient The adjustment coefficient of the retained risk R = R(M1 , M2 ) is defined, as it is well known, by the unique positive root (when such a root exists) of E{exp[R(Sˆ 1 (M1 ) + Sˆ 2 (M2 ))]} = exp[R(P1 + P2 − (Pˆ 1 (M1 ) + Pˆ 2 (M2 )))].

(26)

which is equivalent to G(R; M1 , M2 ) = 0 with

(27)

M.L. Centeno / Insurance: Mathematics and Economics 37 (2005) 229–238

 G(R; M1 , M2 ) = (1 + α1 )(λ1 + λ) 

M1

× 0



+ Rλ

∞

M1

 (1 − F1 (x)) dx + (1 + α2 )(λ2 + λ) 

exp(Rx)(1 − F1 (x)) dx + (λ2 + λ) M1

 exp(Rx)(1 − F1 (x)) dx

0

M2

∞

M2

235

(1 − F2 (x)) dx + (λ1 + λ)

exp(Rx)(1 − F2 (x)) dx

0 M2

exp(Rx)(1 − F2 (x)) dx − (P1 + P2 ).

(28)

0

Note the similarity of (28) and (9). R takes the place of β and they differ by a constant. Let E[W(M1 , M2 )] be the insurer’s expected net profit, after reinsurance, i.e.,  ∞ E[W(M1 , M2 )] = P1 + P2 − (λ1 + λ)µ1 − (λ2 + λ)µ2 − α1 (λ1 + λ) (1 − F1 (x)) dx − α2 (λ2 + λ)  ×

M1

∞

M2

(1 − F2 (x)) dx.

(29)

Let ϒ = {(M1 , M2 ) : Mi ≥ 0

and

E[W(M1 , M2 )] > 0}.

(30)

We assume that the moment generating function of Xi exists in the interval (−∞, Qi ), with 0 < Qi ≤ +∞ and that lim MXi (ti ) = +∞.

ti −→Qi

(31)

Let Q = mini=1,2 ζi , with  ζi =

+∞ Qi

if Mi < +∞ if Mi = +∞.

We assume that 2


[Pi − (λi + λ)µi ] > 0

(32)

[Pi − (1 + αi )(λi + λ)µi ] < 0.

(33)

i=1 2
i=1

Lemma 3.1. 1. R(M1 , M2 ) exists if and only if (M1 , M2 ) ∈ ϒ; 2. For any (M1 , M2 ) ∈ ϒ, (∂/∂R)G(R; M1 , M2 ) is positive at R = R(M1 , M2 ). Proof. As it can be easily checked (∂/∂R)G(R; M1 , M2 ) and (∂2 /∂R2 )G(R; M1 , M2 ) are both non-negative for Mi ≥ 0, which implies that G(R; M1 , M2 ) is an increasing and convex function of R. As for fixed (M1 , M2 ), lim G(R; M1 , M2 ) = +∞,

R−→Q

M.L. Centeno / Insurance: Mathematics and Economics 37 (2005) 229–238

236

then G(R; M1 , M2 ) will have a positive root if and only if G(R; M1 , M2 ) is negative at R = 0, which is equivalent to E[W(M1 , M2 )] > 0.
Result 3.2. R(M1 , M2 ) is an unimodal function of the retentions. The maximum of R(M1 , M2 ), 1. when α1 <

P1 + P2 − (λ1 + λ)µ1 − (λ2 + λ)µ2 (λ1 + λ)µ1

and α1 < R

λ λ1 + λ



1/R ln(1+α2 )

(34)

exp(Rx)(1 − F2 (x)) dx,

(35)

0

with R given by the only solution to (27) for (M1 , M2 ) = (0, 1/R ln(1 + α2 )) is attained at the point (M1 , M2 ) = (0, (1/R) ln(1 + α2 )). 2. when α2 <

P1 + P2 − (λ1 + λ)µ1 − (λ2 + λ)µ2 (λ2 + λ)µ2

and α2 < R

λ λ2 + λ



1/R ln(1+α1 )

(36)

exp(Rx)(1 − F1 (x)) dx,

(37)

0

with R given by the only solution to (27) for (M1 , M2 ) = (1/R ln(1 + α1 ), 0) is attained at the point (M1 , M2 ) = (1/R ln(1 + α1 ), 0). 3. if none of conditions imposed in 1. and 2. are fulfilled it is attained at the only point satisfying   M2 1+α1  exp(RM1 ) = ,   1+R λ λ+λ exp(Rx)(1−F2 (x)) dx  0  1 (38)  M1 1+α2 exp(RM2 ) = ,  1+R λ λ+λ exp(Rx)(1−F1 (x)) dx  0  2   G(R; M1 , M2 ) = 0, Proof. Using the implicit function theorem and the previous lemma we only have to notice that ∂G/∂Mi ∂R =− , ∂Mi ∂G/∂R

i = 1, 2

which will be zero if and only if ∂G/∂Mi = 0, and that   ∂2 R  ∂2 G/∂Mi ∂Mj  =− , ∂Mi ∂Mj ∂R/∂Mi =0,∂R/∂Mj =0 ∂G/∂R ∂R/∂Mi =0,∂R/∂Mj =0

(39)

i, j = 1, 2

(40)

Hence, given the analogy between the functions G and h and the previous two Lemmas we can conclude that the adjustment coefficient is a well behaved function of the retentions, i.e., that its Hessian matrix is negative definite whenever the gradient is null. Condition (34) guarantees that R exists for points of the form (0, M2 ) with M2 big enough, because (34) is equivalent to E[W(0, ∞)] > 0, namely at the point (0, M2 ) satisfying M2 = 1/R ln(1 + α2 ). The same kind of reasoning applies to condition (36). The result follows from Result 2.2.

M.L. Centeno / Insurance: Mathematics and Economics 37 (2005) 229–238

237

Note that the result for independent risks proved by Waters (1979) arises as a special case of this when n = 2 by considering λ = 0.


4. Examples Let us consider that  F1 (x) = 1 −

1 1+x

2 , x>0

and F2 (x) = 1 − e−2x − 2xe−2x , x > 0, i.e., that the claim sizes are distributed according to a Pareto (1,2) and a Gamma (2,2), respectively for lines 1 and 2 (both have mean 1). Let P1 + P2 = 1.1E[S1 + S2 ], λ1 = λ2 , α1 = 0.5, α2 = 0.3 and β = 0.05. Columns 2, 3 and 4 of Table 1 give the retentions that maximise the expected utility of wealth while columns 2, 3 and 4 of Table 2 give the retentions that maximize the adjustment coefficient, when λ/(λ1 + λ) is equal to 0, 0.5 and 1, respectively (which correspond to the independent case, correlation coefficient equal to 0.5 and full dependence, respectively). Note that although M1 is always greater than M2 , the probability that the Pareto distribution exceeds M1 is always much higher than the probability that the Gamma distribution exceeds M2 . For instances 1 − F1 (7.5965) = 0.0135318, while 1 − F2 (4.8133) = 0.0007008. Analysing Table 1 we can say that when we maximize the expected utility of wealth, the higher the dependency the smaller the retention limits. This is no longer true when we maximize the adjustment coefficient. The mathematical reason is that R decreases with λ/(λ1 + λ) (on the assumption that λ1 = λ2 ), while the coefficient of risk aversion is fixed. Although the solutions of both problems looked quite similar, they give completely different answers, with quite unexpected results for the adjustment coefficient.

Table 1 Retention limits that maximize (5) i

1 2

λ/(λ1 + λ) 0

0.5

1

8.1093 5.2473

7.5965 4.8133

7.0966 4.4043

0

0.5

1

6.2301 4.0313 0.0651

7.0961 4.4561 0.0533

7.9108 4.8413 0.0455

Table 2 Retention limits that maximize the adjustment coefficient i

1 2 R

λ/(λ1 + λ)

238

M.L. Centeno / Insurance: Mathematics and Economics 37 (2005) 229–238

Acknowledgement This research has been supported by Fundac¸a˜ o para a Ciˆencia e a Tecnologia — FCT/POCTI.

References Centeno, M.L., 1988. The expected utility applied to reinsurance. In: Munier, B.R. (Ed.), Risk, Decision and Rationality. D. Reidel Publishing Company, Holland. Holgate, P., 1964. Estimation for biavariate Poisson distributions. Biometrika 51, 241–245. Johnson, L., Kotz, S., Balakrishnan, N., 1997. Discrete Multivariate Distributions. Wiley. Wang, S., 1998. Aggregation of correlated risk portfolios: Mothels and algorithms, Proceedings of the Casualty Actuarial Society, vol. LXXXV, pp. 848–939. Waters, H., 1979. Excess of loss reinsurance limits. Scand. Acta J. (1) 37–43.