2.3.1 Maximising the expected profit of the reinsurer

We denote the profit of the reinsurer by A, so that

$$A{\equals}\,(1{\plus}\rho )\def\Bbb{\tf="Macopen"}{\Bbb E}[f(X)]{\minus}f(X)$$

Denote the objective function of this model by

$$l(\rho )\,{\equals}\,\def\Bbb{\tf="Macopen"}{\Bbb E}[A]{\equals}\,\rho {\Bbb E}[f(X)]$$

Then the optimal reinsurance problem can be stated as

(9) $$\mathop {{\rm max}}\limits_\rho l(\rho ){\equals}\,\mathop {{\rm max}}\limits_\rho \rho \def\Bbb{\tf="Macopen"}{\Bbb E}[f(X)]$$

We remark that f in (9) implicitly depends on ρ as described by Theorem 1, so that l(ρ) is indeed a complicated non-linear function of ρ.

We now apply the results about the value range of the safety loading ρ derived in section 2.2 and solve the optimisation problem (9) accordingly.

Case 1: For aλ>1, we have $$\rho \in\big( {{1 \over \gamma }{\minus}1,{1 \over \gamma }e^{{a\lambda {\minus}1}} {\minus}1} \big)$$ and the ceded loss function should be f(x)=(x−d*)₊. we can write the objective function in this case as

$$l_{1} (\rho ){\equals}\,\rho \def\Bbb{\tf="Macopen"}{\Bbb E}\left[ {X\,{\minus}\,d^{{\asterisk}} } \right]_{\!{\plus}} {\equals}\,{\rho \over {\lambda (1{\plus}\rho )}}$$

where we can see l ₁(ρ) is increasing and concave over ρ. Hence l ₁(ρ) will go to its supremum value as ρ goes to $${1 \over \gamma }e^{{a\lambda {\minus}1}} {\minus}1$$ . Since ρ cannot reach $${1 \over \gamma }e^{{a\lambda {\minus}1}} {\minus}1$$ , the maximum of l ₁(ρ) cannot be achieved. Thus $$l_{1} (\rho )\in\left( {{{1{\minus}\gamma } \over \lambda },{1 \over \lambda }{\minus}{\gamma \over \lambda }e^{{1{\minus}a\lambda }} } \right)$$ , where $$\rho \in\big( {{1 \over \gamma }{\minus}1,{1 \over \gamma }e^{{a\lambda {\minus}1}} {\minus}1} \big)$$ . To conclude, we have

$$\mathop {{\rm sup}\,}\limits_\rho l_{1} (\rho )\,{\equals}\,{1 \over \lambda }{\minus}{\gamma \over \lambda }e^{{1{\minus}a\lambda }} $$

$${\rm as}\quad \rho \nearrow{1 \over \gamma }e^{{a\lambda {\minus}1}} {\minus}1$$

Case 2: For aλ>1, we have $$\rho \,{\equals}\,{1 \over \gamma }e^{{a\lambda {\minus}1}} {\minus}1$$ and the ceded loss function should be f(x)=c(x−d*)₊, c∈[0.1]. We write the objective function in this case as

$$l_{2} (\rho )\,{\equals}\,cl_{1} (\rho )\,{\equals}\,c\rho \def\Bbb{\tf="Macopen"}{\Bbb E}\left[ {X{\minus}d^{{\asterisk}} } \right]_{{\plus}} {\equals}\,{{c\rho } \over {\lambda (1{\plus}\rho )}}$$

In this case, the safety loading ρ has only one possible value, so we have the maximum l ₂(ρ) as follows:

$$\mathop {{\rm max}\,}\limits_\rho l_{2} (\rho )\,{\equals}\,c\left( {{1 \over \lambda }{\minus}{\gamma \over \lambda }e^{{1{\minus}a\lambda }} } \right)$$

$${\rm at}\;\;\rho \,{\equals}\,{1 \over \gamma }e^{{a\lambda {\minus}1}} {\minus}1$$

Now we compare the supremum or maximum values in Case 1 and Case 2. In both cases we have aλ>1, and c∈[0, 1], so $$c\left( {{1 \over \lambda }{\minus}{\gamma \over \lambda }e^{{1{\minus}a\lambda }} } \right)\leq {1 \over \lambda }{\minus}{\gamma \over \lambda }e^{{1{\minus}a\lambda }} $$ . Then we obtain that

(10) $$\mathop {{\rm max}\,}\limits_\rho l_{2} (\rho )\leq \mathop {{\rm sup}}\limits_\rho \,l_{1} (\rho )$$

Case 3: If aλ>1, we have $$0\,\lt\,\rho \leq {1 \over \gamma }{\minus}1$$ . If γ<aλ≤1, we have $$0\,\lt\,\rho \,\lt\,{{a\lambda } \over \gamma }{\minus}1$$ . The ceded loss function chosen by the insurer in this case is given by f(x)=x, x≥0. We can write the objective function in this case as

$$l_{3} (\rho ){\equals}\,\rho \def\Bbb{\tf="Macopen"}{\Bbb E}[X]{\equals}\,{\gamma \over \lambda }\rho $$

which is linear and increasing over ρ. If aλ>1, l ₃(ρ) achieves its maximum at $$\rho \,{\equals}\,{1 \over \gamma }{\minus}1$$ , and we have $$l_{3} (\rho )\in\left( {\left. {0,{{1{\minus}\gamma } \over \lambda }} \right]} \right.$$ for $$\rho \in\big( {\left. {0,{1 \over \gamma }{\minus}1} \big]} \right.$$ . On the other hand, if γ<aλ≤1, l ₃(ρ) will go to its supremum as ρ approaches $${{a\lambda } \over \gamma }{\minus}1$$ , and we have s $$l_{3} (\rho )\in\left( {0,{{a\lambda {\minus}\gamma } \over \lambda }} \right)$$ when $$\rho \in\big( {0,{{a\lambda } \over \gamma }{\minus}1} \big)$$ . To sum up

$$\mathop {{\rm sup}\,}\limits_\rho l_{3} (\rho )\left\{ {\matrix{ {\equals}\, \hfill &#x0026; {{{1{\minus}\gamma } \over \lambda },\quad a\lambda \,\gt\,1} \hfill \cr {\equals}\, \hfill &#x0026; {{{a\lambda {\minus}\gamma } \over \lambda },\quad \gamma \,\lt\,a\lambda \leq 1} \hfill \cr } } \right.$$

$${\rm when}\quad \rho \left\{ {\matrix{ {\equals}\, \hfill &#x0026; {{1 \over \gamma }{\minus}1,\quad a\lambda \,\gt\,1} \hfill \cr \nearrow \hfill &#x0026; {{{a\lambda } \over \gamma }{\minus}1,\quad \gamma \,\lt\,a\lambda \leq 1} \hfill \cr } } \right.$$

Case 4: For γ<aλ≤1, we have the safety loading $$\rho \,{\equals}\,{{a\lambda } \over \gamma }{\minus}1$$ . The ceded loss function used in this case should be in the form of f(x)=cx, c∈[0, 1]. We denote the objective function in this case by

$$l_{4} (\rho ){\equals}\,cl_{3} (\rho ){\equals}\,c\rho \def\Bbb{\tf="Macopen"}{\Bbb E}[X]{\equals}\,{{c\rho \gamma } \over \lambda }$$

Since the safety loading can only equal $${{a\lambda } \over \gamma }{\minus}1$$ , we have

$$\mathop {{\rm max}}\limits_\rho \,l_{4} (\rho )\,{\equals}\,{{c(a\lambda {\minus}\gamma )} \over \lambda }$$

$${\rm at}\quad \rho \,{\equals}\,{{a\lambda } \over \gamma }{\minus}1$$

Now we compare the results derived in Case 3 and Case 4 when γ<aλ≤1. Since c∈[0, 1], we have $${{c(a\lambda {\minus}\gamma )} \over \lambda }\leq {{a\lambda {\minus}\gamma } \over \lambda }$$ . Then we can state that

(11) $$\mathop {{\rm max}\,}\limits_\rho l_{4} (\rho )\leq \mathop {{\rm sup}}\limits_\rho \,l_{3} (\rho )$$

Case 5: For other cases, the ceded loss function should be f(x)≡0. Consequently, the corresponding objective function l ₅(ρ)≡0, and hence the maximum value of l ₅(ρ) will also be 0:

$$\mathop {{\rm max}\,}\limits_\rho l_{5} (\rho )\,{\equals}\,0$$

$${\rm for}\,{\rm all}\,{\rm other}\,\rho $$

Now we have solved the optimisation problem based on the five different cases and we have also compared the maximum objective functions. Figure 1 shows the relationship between the safety loading ρ and the objective function l(ρ) by combining the results obtained for these five cases.

Figure 1 The expected profit of the reinsurer facing one risk.

The results shows that when aλ>1, the optimal expected profit of the reinsurer may not be attainable: as ρ goes to $${1 \over \gamma }e^{{a\lambda {\minus}1}} {\minus}1$$ , the expected profit will increase accordingly but may not reach a maximum value. The reason is that in Case 2, the insurer is indifferent to any constant c∈[0, 1], and hence the reinsurer cannot precisely predict which c the insurer will choose. This creates a possible jump between l ₁ and l ₂ at $$\rho \,{\equals}\,{1 \over \gamma }e^{{a\lambda {\minus}1}} {\minus}1$$ .

When γ<aλ≤1, the optimal expected profit of the reinsurer again may not be obtainable. As ρ goes to $${{a\lambda } \over \gamma }{\minus}1$$ , the expected profit of the reinsurer will increase accordingly but there could be a jump between l ₃ and l ₄ because of the arbitrariness of the constant c in Case 4.

2.3.2 Maximising expected utility of the reinsurer

For this optimisation model, we suppose that the utility function of the reinsurer is exponential:

$$U(x){\equals}\,{\minus}e^{{{\minus}\theta x}} ,\theta \,\gt\,0$$

which is commonly used in financial economics because of its analytical tractability. The analysis works for other utility functions as well, such as linear utility, quadratic utility. An example using the quadratic utility is provided in Appendix B. The profit of the reinsurer is again denoted by A, so that

$$A{\equals}\,(1{\plus}\rho )\def\Bbb{\tf="Macopen"}\def\Bbb{\tf="Macopen"}{\Bbb E}[f(X)]\,{\minus}\,f(X)$$

We denote the objective function by

$$k(\rho ){\equals}\,\def\Bbb{\tf="Macopen"}{\Bbb E}[U(A)]$$

The optimisation problem we want to solve in this subsection is the following:

(12) $$\mathop {{\rm max}}\limits_\rho \,k(\rho ){\equals}\,\mathop {{\rm max}}\limits_\rho \,\def\Bbb{\tf="Macopen"}{\Bbb E}[U(A)]$$

To solve this problem, we apply the results about the value range of the safety loading ρ derived in section 2.2.

Case 1: For aλ>1, we have $$\rho \in\big( {{1 \over \gamma }{\minus}1,{1 \over \gamma }e^{{a\lambda {\minus}1}} {\minus}1} \big)$$ and the ceded loss function chosen by the insurer is given by f(x)=(x−d*)₊. The profit of the reinsurer hence can be stated as

$$A_{1} {\equals}\,(1{\plus}\rho )\def\Bbb{\tf="Macopen"}{\Bbb E}\left[ {X\,{\minus}\,d^{{\asterisk}} } \right]\!_{{\plus}} \,{\minus}\,\left( {X\,{\minus}\,d^{{\asterisk}} } \right)_{\!{\plus}} {\equals}\,{1 \over \lambda }\,{\minus}\,\left( {X\,{\minus}\,d^{{\asterisk}} } \right)_{\!{\plus}} $$

We derive the objective function as follows:

(13) $$\eqalignno{ k_{1} (\rho ){\equals}\, &#x0026; \def\Bbb{\tf="Macopen"}{\Bbb E}\left[ {U\left( {A_{1} } \right)} \right] \cr {\equals}\, &#x0026; \left[ {1{\minus}\rho ^{{\asterisk}} } \right]\def\Bbb{\tf="Macopen"}{\Bbb E}\big[ {{\minus}e^{{{\minus}{\theta \over \lambda }}} } \big]{\plus}\rho ^{{\asterisk}} \def\Bbb{\tf="Macopen"}{\Bbb E}\left[ {{\minus}e^{{{\minus}\theta \left[ {{1 \over \lambda }{\minus}\left( {X{\minus}d^{{\asterisk}} } \right)_{{\plus}} } \right]}} \left| {X\,\gt\,d^{{\asterisk}} } \right.} \right] \cr {\equals}\, &#x0026; \left( {1{\minus}\rho ^{{\asterisk}} } \right)\big( {{\minus}e^{{{\minus}{\theta \over \lambda }}} } \big){\minus}\rho ^{{\asterisk}} \lambda e^{{{\minus}{\theta \over \lambda }}} {\int}_0^\infty e^{{(\theta {\minus}\lambda )y}} dy $$

From this equation, we have the following conclusion:

• ∙ If θ<λ, $$k_{1} (\rho ){\equals}\,{\minus}e^{{{\minus}{\theta \over \lambda }}} {\plus}{\theta \over {\theta {\minus}\lambda }}e^{{{\minus}{\theta \over \lambda }}} {1 \over {1{\plus}\rho }}$$ .

• ∙ Otherwise, if θ≥λ, k ₁(ρ)=−∞, which is the worst possible result and hence could be ignored.

When θ<λ, k ₁(ρ) is increasing and concave. The value range of k ₁(ρ) is given by

$$k_{1} (\rho )\in\left( {\left( {{{\gamma \theta } \over {\theta {\minus}\lambda }}{\minus}1} \right)e^{{{\minus}{\theta \over \lambda }}} ,{\minus}e^{{{\minus}{\theta \over \lambda }}} {\plus}{{\gamma \theta } \over {\theta \,{\minus}\,\gamma }}e^{{1{\minus}a\lambda {\minus}{\theta \over \lambda }}} } \right)$$

when $$\rho \in\big( {{1 \over \gamma }{\minus}1,{1 \over \gamma }e^{{a\lambda {\minus}1}} {\minus}1} \big)$$ . Since the end points of the interval of ρ cannot be achieved, k ₁(ρ) will go to its supremum as ρ goes to $${1 \over \gamma }e^{{a\lambda {\minus}1}} {\minus}1$$ . To sum up

$$\mathop {{\rm sup}}\limits_\rho \,k_{1} (\rho ){\equals}\,{\minus}e^{{{\minus}{\theta \over \lambda }}} {\plus}{{\gamma \theta } \over {\theta \,{\minus}\,\gamma }}e^{{1{\minus}a\lambda {\minus}{\theta \over \lambda }}} $$

$${\rm when}\quad \rho \nearrow{1 \over \gamma }e^{{a\lambda {\minus}1}} {\minus}1$$

Case 2: For aλ>1, we have $$\rho \,{\equals}\,{1 \over \lambda }e^{{a\lambda {\minus}1}} {\minus}1$$ and the ceded loss function used in this case is f(x)= c(x−d*)₊, c∈[0, 1]. The profit of the reinsurer is denoted by A ₂, then A ₂=cA ₁. The corresponding objective function k ₂ can be derived as follows:

(14) $$\eqalignno{ k_{2} (\rho ){\equals}\, &#x0026; \def\Bbb{\tf="Macopen"}{\Bbb E}\left[ {U\left( {A_{2} } \right)} \right] \cr {\equals}\, &#x0026; \left( {1{\minus}\rho ^{{\asterisk}} } \right)\big( {{\minus}e^{{{\minus}{{c\theta } \over \lambda }}} } \big){\minus}\rho ^{{\asterisk}} \lambda e^{{{\minus}{{c\theta } \over \lambda }}} {\int}_0^\infty e^{{(c\theta {\minus}\lambda )y}} dy $$

From (14), we have the following conclusion:

• ∙ If cθ<λ, $$k_{2} (\rho ){\equals}\,{\minus}e^{{{\minus}{{c\theta } \over \lambda }}} {\plus}{{c\theta } \over {c\theta {\minus}\lambda }}e^{{{\minus}{{c\theta } \over \lambda }}} {1 \over {1{\plus}\rho }}$$ .

• ∙ Otherwise, if cθ≥λ, k ₂(ρ)=−∞, which is the worst possible result and hence could be ignored.

Since in this case the safety loading ρ only have one possible value, we can write down the maximum value k ₂(ρ) directly as follows:

$$\mathop {{\rm max}\,}\limits_\rho k_{2} (\rho ){\equals}\,{\minus}e^{{{\minus}{{c\theta } \over \lambda }}} {\plus}{{c\theta } \over {c\theta \,{\minus}\,\lambda }}e^{{{\minus}{{c\theta } \over \lambda }}} {1 \over {1{\plus}\rho }}$$

$${\rm at}\quad \rho \,{\equals}\,{1 \over \gamma }e^{{a\lambda {\minus}1}} {\minus}1$$

Now we compare the results in Case 1 and Case 2. Define

$$q(\theta ){\equals}\,{\minus}e^{{{\minus}{\theta \over \lambda }}} {\plus}{{\gamma \theta } \over {\theta \,{\minus}\,\gamma }}e^{{1{\minus}a\lambda {\minus}{\theta \over \lambda }}} $$

Since θ<λ, we have q(θ) is increasing in θ. Since cθ≤θ, with c∈[0, 1], we find that q(cθ)≤q(θ). Hence we have the following comparison holds:

(15) $$\mathop {{\rm max}}\limits_\rho \,k_{2} (\rho )\leq \mathop {{\rm sup}}\limits_\rho \,k_{1} (\rho )$$

Case 3: If aλ>1, we have $$0\,\lt\,\rho \leq {1 \over \gamma }{\minus}1$$ . If γ<aλ≤1, we have $$0\,\lt\,\rho \,\lt\,{{a\lambda } \over \gamma }{\minus}1$$ . The ceded loss function used in this case is full insurance f(x)=x. The profit of the reinsurer is denoted by A ₃, and we have

$$A_{3} {\equals}\,(1{\plus}\rho )\def\Bbb{\tf="Macopen"}{\Bbb E}[X]{\minus}X{\equals}\,(1{\plus}\rho ){\gamma \over \lambda }{\minus}X$$

The objective function can be derived as follows:

(16) $$\eqalignno{ k_{3} (\rho ){\equals}\, &#x0026; \def\Bbb{\tf="Macopen"}{\Bbb E}\left[ {U\left( {A_{3} } \right)} \right] \cr &#x0026; \vskip-5pt{\equals}\,{\minus}\gamma \lambda e^{{{\minus}\theta (1{\plus}\rho ){\gamma \over \lambda }}} {\int}_0^\infty e^{{(\theta {\minus}\lambda )x}} dx $$

We then have the following conclusion by analysing (16):

• ∙ If θ<λ, $$k_{3} (\rho ){\equals}\,{{\lambda \gamma } \over {\theta {\minus}\lambda }}e^{{{\minus}\theta (1{\plus}\rho ){\gamma \over \lambda }}} $$ .

• ∙ Otherwise, if θ≥λ, k ₃(ρ)=−∞, which is the worst possible result and hence could be ignored.

So we assume that θ<λ, and it is easy to see that k ₃(ρ) is increasing and concave over ρ. Therefore, if aλ>1, the maximum of k ₃(ρ) will be achieved at the largest possible safety loading $$\rho \,{\equals}\,{1 \over \gamma }{\minus}1$$ , and we have $$k_{3} (\rho )\in\big( {\left. {{{\gamma \lambda } \over {\theta {\minus}\lambda }}e^{{{\minus}{{\theta \gamma } \over \lambda }}} ,{{\gamma \lambda } \over {\theta {\minus}\lambda }}e^{{{\minus}{\theta \over \lambda }}} } \big]} \right.$$ when $$\rho \in\big( {\left. {0,{1 \over \gamma }{\minus}1} \big]} \right.$$ . On the other hand, if γ<aλ≤1, k ₃(ρ) will go to its supremum as ρ goes to $${{a\lambda } \over \gamma }{\minus}1$$ , and we have $$k_{3} (\rho )\in\big( {{{\gamma \lambda } \over {\theta {\minus}\lambda }}e^{{{\minus}{{\theta \gamma } \over \lambda }}} ,{{\gamma \lambda } \over {\theta {\minus}\lambda }}e^{{{\minus}a\theta }} } \big)$$ when $$\rho \in\big( {0,{{a\lambda } \over \gamma }{\minus}1} \big)$$ . To sum up, we have

$$\mathop {{\rm sup}\,}\limits_\rho k_{3} (\rho )\left\{ {\matrix{ {\equals}\, \hfill &#x0026; {k_{3} \big( {{1 \over \gamma }{\minus}1} \big){\equals}\,{{\gamma \lambda } \over {\theta \,{\minus}\,\lambda }}e^{{{\minus}{\theta \over \lambda }}} ,\quad a\lambda \,\gt\,1} \hfill \cr {\equals}\, \hfill &#x0026; {k_{3} \big( {{{a\lambda } \over \gamma }{\minus}1} \big){\equals}\,{{\gamma \lambda } \over {\theta \,{\minus}\,\lambda }}e^{{{\minus}a\theta }} ,\quad \gamma \,\lt\,a\lambda \leq 1} \hfill \cr } } \right.$$

$${\rm when}\quad \rho \left\{ {\matrix{ {\equals}\, &#x0026; \hskip-20pt{{1 \over \gamma }{\minus}1,\quad a\lambda \,\gt\,1} \cr \nearrow &#x0026; {{{a\lambda } \over \gamma }{\minus}1,\quad \gamma \,\lt\,a\lambda \leq 1} \cr } } \right$$

Now we compare the results obtained in Case 1 and Case 3 when aλ>1. We have $${{\rm inf}}_\rho k_{1} (\rho ){\equals}\,{{\rm lim}}_{\rho \,\to\,{1 \over \gamma }{\minus}1} k_{1} (\rho ){\equals}\,\left( {{{\gamma \theta } \over {\theta \,{\minus}\,\lambda }}{\minus}1} \right)e^{{{\minus}{\theta \over \lambda }}} $$ and $${{\rm max}}_\rho k_{3} (\rho ){\equals}\,k_{3} \big( {{1 \over \gamma }{\minus}1} \big){\equals}\,{{\gamma \lambda } \over {\theta \,{\minus}\,\lambda }}e^{{{\minus}{\theta \over \lambda }}} $$ . To compare this two values, we have

$$\mathop {{\rm inf}\,}\limits_\rho k_{1} (\rho ){\minus}\mathop {{\rm max}\,}\limits_\rho k_{3} (\rho ){\equals}\,(\gamma {\minus}1)e^{{{\minus}{\theta \over \lambda }}} \,\lt\,0$$

Therefore, it is concluded that

(17) $$\mathop {{\rm inf}\,}\limits_\rho k_{1} (\rho )\,\lt\,\mathop {{\rm max}}\limits_\rho \,k_{3} (\rho )$$

Case 4: For γ<aλ≤1, we have $$\rho \,{\equals}\,{{a\lambda } \over \gamma }{\minus}1$$ . The ceded loss function applied in this case should be f(x)=cx, c∈[0, 1]. The profit of the reinsurer is denoted by A ₄, and we have A ₄=cA ₃. The objective function can be derived as follows:

(18) $$\eqalignno{ k_{4} (\rho ){\equals}\, &#x0026; \def\Bbb{\tf="Macopen"}{\Bbb E}\left[ {U\left( {A_{4} } \right)} \right] \cr \vskip-2pt{\equals}\, &#x0026; \vskip-2pt{\minus}\gamma \lambda e^{{{\minus}c\theta (1{\plus}\rho ){\gamma \over \lambda }}} {\int}_0^\infty e^{{(c\theta {\minus}\lambda )x}} dx $$

Similar to the argument used in Case 3, we have the following conclusion upon analysing (18):

• ∙ If cθ<λ, $$k_{4} (\rho ){\equals}\,{{\gamma \lambda } \over {c\theta {\minus}\lambda }}e^{{{\minus}c\theta (1{\plus}\rho ){\gamma \over \lambda }}} $$ .

• ∙ Otherwise, if cθ≥λ, we have k ₄(ρ)=−∞, which is the worst possible result and hence could be ignored.

Since in this case the safety loading ρ only has one possible value, we can write down the maximum value of k ₄(ρ) directly as follows:

$$\mathop {{\rm max}}\limits_\rho \,k_{4} (\rho ){\equals}\,{{\gamma \lambda } \over {c\theta \,{\minus}\,\lambda }}e^{{{\minus}ac\theta }} $$

$${\rm at}\quad \rho \,{\equals}\,{{a\lambda } \over \gamma }{\minus}1$$

Next we compare the results in Case 3 and Case 4, when γ<aλ≤1. Define

$$r(\theta )\,{\equals}\,{{\gamma \lambda } \over {\theta \,{\minus}\,\lambda }}e^{{{\minus}a\theta }} $$

Then $${{\rm sup}}_\rho k_{3} (\rho ){\equals}\,r(\theta )$$ , and $${\rm max}_{\rho } k_{4} (\rho ){\equals}\,r(c\theta )$$ . As we have assumed that θ<λ, r(θ) is decreasing in θ. Since cθ≤θ, with c∈[0, 1], we find that r(cθ)≥r(θ) . Hence the following comparison holds:

(19) $$\mathop {{\rm max}}\limits_\rho \,k_{4} (\rho )\geq \mathop {{\rm sup}\,}\limits_\rho k_{3} (\rho )$$

Case 5: For all other cases, the ceded loss function should be f(x)≡0. Consequently, the corresponding objective function k ₅(ρ)≡−1, and the maximum value of k ₅(ρ) is also −1:

$$\mathop {{\rm max}}\limits_\rho \,k_{5} (\rho )\,{\equals}\,{\minus}\!1$$

Figure 2 shows the relationship between the safety loading ρ and the objective function k(ρ). The results of all five cases discussed above are combined and compared. Since the comparison between k ₁, k ₃ and k ₅ partly depends on the value of the parameter θ involved in the exponential utility function, we use two black arrows to indicate the uncertain vertical locations of the curves.

Figure 2 The expected utility of the reinsurer facing one risk.

The results show that if aλ>1, the comparison between point B, point D, and k ₅ partly depends on the value of the parameter θ involved in the exponential utility function. So when aλ>1, point D might be lower than point B; point C might be lower than point A; and k ₅(ρ) might be lower than point D. Moreover, discontinuity between point D and point E may or may not exist, due to arbitrariness of the constant c in Case 2.

When γ<aλ≤1, k ₅(ρ) might be lower than C, in this case the optimal expected exponential utility is obtained in Case 4 and value is $${{\gamma \lambda } \over {c\theta \,{\minus}\,\lambda }}e^{{{\minus}ac\theta }} $$ , when ρ is equal to $${{a\lambda } \over \gamma }{\minus}1$$ . It should also be remarked that the apparent discontinuity between point B and point C may or may not exist, depending how the insurer chooses the arbitrary constant c in Case 4.

2.3.3 Minimising VaR of the total loss of the reinsurer

Let Y denote the total loss of the reinsurer, which is given by

$$Y\,{\equals}\,f(X)\,{\minus}\,\delta _{{f,\rho }} (X)$$

The optimal safety loading problem at confidence level 1−β∈(0, 1) can be stated as

$$\mathop {{\rm min}}\limits_\rho \:{\rm VaR}\:_{Y} (\beta )$$

Here we assume β≤α because in general the reinsurer’s risk tolerance level is higher. To simplify our notation, we set $$b\,\colon\!{\equals}\,S_{X}^{{{\minus}1}} (\beta )$$ . Because of the assumption β≤α, we further have b≥a.

We denote the objective function as j(ρ)=VaR _Y (β), and have the following deduction:

$$\eqalignno{ j(\rho ){\equals}\, &#x0026; {\rm VaR}_{Y} (\beta ) \cr {\equals}\, &#x0026; {\rm VaR}_{{f(X)}} (\beta )\,{\minus}\,(1{\plus}\rho )\def\Bbb{\tf="Macopen"}{\Bbb E}[f(X)] \cr {\equals}\, &#x0026; f[{\rm VaR}\:_{X} (\beta )]\,{\minus}\,(1{\plus}\rho )\def\Bbb{\tf="Macopen"}{\Bbb E}[f(X)] \cr {\equals}\, &#x0026; f(b)\,{\minus}\,(1{\plus}\rho )\def\Bbb{\tf="Macopen"}{\Bbb E}[f(X)] $$

Here, the second equality comes from the translational invariance of VaR, and the third equality follows from Theorem 1(a) of Dhaene et al. (Reference Dhaene, Denuit, Goovaerts, Kaas and Vyncke2002).

As in previous sections, we assume that X follows a zero-modified exponential distribution with survival function given by

$$S_{X} (t){\equals}\,\gamma e^{{{\minus}\lambda t}} ,\lambda \,\gt\,0,0\,\lt\,\gamma \,\lt\,1,\:t\,\gt\,0$$

Case 1: For aλ>1, we have $$\rho \in\big( {{1 \over \gamma }{\minus}1,{1 \over \gamma }e^{{a\lambda {\minus}1}} {\minus}1} \big)$$ and the ceded loss function is f(x)=(x−d*)₊. In this case, our objective function is given by

$$\eqalignno{ j_{1} (\rho )\,{\equals}\, &#x0026; f(b){\minus}(1{\plus}\rho )\def\Bbb{\tf="Macopen"}{\Bbb E}[f(X)] \cr {\equals}\, &#x0026; b{\minus}{1 \over \lambda }{\minus}{1 \over \lambda }{\rm ln}[\gamma (1{\plus}\rho )] $$

Since j ₁ is decreasing, j ₁(ρ) will go to its infimum as ρ goes to $${1 \over \gamma }e^{{a\lambda {\minus}1}} {\minus}1$$ . We can then derive the value range of j ₁(ρ):

$$j_{1} (\rho )\in\left( {b{\minus}a,b{\minus}{1 \over \lambda }} \right)$$

when $$\rho \in\big( {{1 \over \gamma }{\minus}1,{1 \over \gamma }e^{{a\lambda {\minus}1}} {\minus}1} \big)$$ . Hence we have

$$\mathop {{\rm inf}}\limits_\rho j_{1} (\rho ){\equals}\,j_{1} \left( {{1 \over \gamma }e^{{a\lambda {\minus}1}} {\minus}1} \right){\equals}\,b{\minus}a$$

$${\rm when}\quad \rho \nearrow{1 \over \gamma }e^{{a\lambda {\minus}1}} {\minus}1$$

Case 2: If aλ>1, we have $$\rho \,{\equals}\,{1 \over \lambda }e^{{a\lambda {\minus}1}} {\minus}1$$ and the ceded loss function is f(x)=c(x−d*)₊, c∈[0.1]. Then objective function equals

$$j_{2} (\rho ){\equals}\,cj_{1} (\rho ){\equals}\,c\left\{ {b{\minus}{1 \over \lambda }{\minus}{1 \over \lambda }{\rm ln}\left[ {\gamma (1{\plus}\rho )} \right]} \right\}$$

In this case, the safety loading ρ only has one possible value, hence we write down the minimum j ₂(ρ) as follows:

$$\mathop {{\rm min}\,}\limits_\rho j_{2} (\rho ){\equals}\,c(b{\minus}a)$$

$${\rm at}\;\;\rho \,{\equals}\,{1 \over \gamma }e^{{a\lambda {\minus}1}} {\minus}1$$

Now we compare the results derived in Case 1 and Case 2. Since c∈[0, 1], we have c(b−a)≤b−a. Then we have the following comparison holds:

(20) $$\mathop {{\rm min}}\limits_\rho \,j_{2} (\rho )\leq \mathop {{\rm inf}}\limits_\rho \,j_{1} (\rho )$$

This could be a strict inequality or simply an equality, depending on how the insurer chooses the constant c∈[0, 1].

Case 3: If aλ>1, we have $$0\,\lt\,\rho \leq {1 \over \gamma }{\minus}1$$ . Otherwise, if γ<aλ≤1, we have $$0\,\lt\,\rho \,\lt\,{{a\lambda } \over \gamma }{\minus}1$$ . In this case, the insurer prefers full insurance such that f(x)=x. The corresponding objective function j ₃(ρ) in this case is given by

$$\eqalignno{ j_{3} (\rho ){\equals}\, &#x0026; f(b)\,{\minus}\,(1{\plus}\rho )\def\Bbb{\tf="Macopen"}{\Bbb E}[f(x)] \cr {\equals}\, &#x0026; b\,{\minus}\,(1{\plus}\rho ){\gamma \over \lambda } $$

Since $$j_{3}^{\,'} (\rho ){\equals}\,{\minus}{\gamma \over \lambda }\,\lt\,0$$ , j ₃ is decreasing over ρ. If aλ>1, j ₃(ρ) will achieve its minimum value at $$\rho \,{\equals}\,{1 \over \gamma }{\minus}1$$ , and we have $$j_{3} (\rho )\in\left( {\left. {b{\minus}{1 \over \lambda },b{\minus}{\gamma \over \lambda }} \right]} \right.$$ when $$\rho \in\big( {\left. {0,{1 \over \gamma }{\minus}1} \big]} \right.$$ . If γ<aλ≤1, j ₃(ρ) will go to its infimum as ρ goes to $${{a\lambda } \over \gamma }{\minus}1$$ , and we have $$j_{3} (\rho )\in\left( {b{\minus}a,b{\minus}{\gamma \over \lambda }} \right)$$ when $$\rho \in\big( {0,{{a\lambda } \over \gamma }{\minus}1} \big)$$ . To sum up,

$$\mathop {{\rm inf}}\limits_\rho \,j_{3} (\rho )\left\{ {\matrix{ {\equals}\, \hfill &#x0026; {j_{3} \big( {{1 \over \gamma }{\minus}1} \big){\equals}\,b{\minus}{1 \over \lambda },\quad a\lambda \,\gt\,1} \hfill \cr {\equals}\, \hfill &#x0026; {j_{3} \big( {{{a\lambda } \over \gamma }{\minus}1} \big){\equals}\,b{\minus}a,\quad \gamma \,\lt\,a\lambda \leq 1} \hfill \cr } } \right.$$

$${\rm when}\quad \rho \left\{ {\matrix{ {\equals}\, \hfill &#x0026; {{1 \over \gamma }{\minus}1,\quad a\lambda \,\gt\,1} \hfill \cr \nearrow \hfill &#x0026; {{{a\lambda } \over \gamma }{\minus}1,\quad \gamma \,\lt\,a\lambda \leq 1} \hfill \cr } } \right.$$

Case 4: When aλ≤1, we have $$\rho \,{\equals}\,{{a\lambda } \over \gamma }{\minus}1$$ and the ceded loss function is f(x)=cx, c∈[0, 1]. The objective function j ₄(ρ) could be derived as follows:

$$\eqalignno{ j_{4} (\rho ){\equals}\, &#x0026; cj_{3} (\rho ) \cr {\equals}\, &#x0026; cb\,{\minus}\,c(1{\plus}\rho ){\gamma \over \lambda } $$

As the safety loading ρ can only take on one possible value, we can write down the minimum j ₄(ρ) as follows:

$$\mathop {{\rm min}\,}\limits_\rho j_{4} (\rho ){\equals}\,c(b{\minus}a)$$

$${\rm at}\quad \rho \,{\equals}\,{{a\lambda } \over \gamma }{\minus}1$$

Now we compare the results derived in Case 3 and Case 4, when γ<aλ≤1. Since c∈[0, 1], we have c(b−a)≤b−a. Then we have the following comparison:

(21) $$\mathop {{\rm min}\,}\limits_\rho j_{4} (\rho )\leq \mathop {{\rm inf}}\limits_\rho \,j_{3} (\rho )$$

Case 5: For all other cases, the ceded loss function should be identically 0. Consequently, the objective function j ₅(ρ)≡0. We can state the result under this circumstance as follows:

$$\mathop {{\rm max}}\limits_\rho \,j_{5} (\rho ){\equals}\,0$$

Figure 3 shows the relationship between the safety loading ρ and the objective function j(ρ). The results of all five cases discussed above are combined and compared.

Figure 3 The value-at-risk of the total loss of the reinsurer facing one risk.

The result shows that in both cases aλ>1 and γ<aλ≤1 if we set ρ to be as large as possible, VaR will go to 0 which is optimal. This analysis makes sense in the real insurance and reinsurance markets, because if the reinsurer set its safety loading ρ as large as possible, then the insurer will buy less and less reinsurance for the price is too high. Once the safety loading ρ is larger than some threshold, the insurer will not seek reinsurance at all, and so the ceded loss function f(x) becomes 0. Although the result itself is not too interesting, we will nevertheless demonstrate how it can be applied in some more realistic optimisation problems in section 5.

3.3.1 Maximising the expected profit of the reinsurer

The profit of the reinsurer is denoted by $$\bar{A},$$ and we have

$$\bar{A}{\equals}\,(1{\plus}\rho )\def\Bbb{\tf="Macopen"}{\Bbb E}\left( {f_{1} \left( {X_{1} } \right){\plus}f_{2} \left( {X_{2} } \right)} \right){\minus}\left[ {f_{1} \left( {X_{1} } \right){\plus}f_{2} \left( {X_{2} } \right)} \right]$$

The objective function is denoted by

$$L(\rho ){\equals}\,\def\Bbb{\tf="Macopen"}{\Bbb E}[\bar{A}]\,{\equals}\,\rho \def\Bbb{\tf="Macopen"}{\Bbb E}\left[ {f_{1} \left( {X_{1} } \right){\plus}f_{2} \left( {X_{2} } \right)} \right]$$

Hence the optimal problem can be stated as

(22) $$\mathop {{\rm max}\,}\limits_\rho L(\rho ){\equals}\,\mathop {{\rm max}\,}\limits_\rho \def\Bbb{\tf="Macopen"}\rho {\Bbb E}\left[ {f_{1} \left( {X_{1} } \right){\plus}f_{2} \left( {X_{2} } \right)} \right]$$

We will apply the results about the value range of the safety loading ρ derived in the previous section to solve this problem.

Case 1: For a ₂ λ>1, we have $$\rho \in\big( {{1 \over \gamma }{\minus}1,{1 \over \gamma }e^{{a_{2} \lambda {\minus}1}} {\minus}1} \big)$$ . The ceded loss functions for insurer 1 and insurer 2 are f ₁(x ₁)=(x ₁−d*)₊ and f ₂(x ₂)=(x ₂−d*)₊, respectively. We can derive that

$$L_{1} (\rho ){\equals}\,2l_{1} (\rho ){\equals}\,{{2\rho } \over {\lambda (1{\plus}\rho )}}$$

which is increasing and concave over ρ. Hence L ₁(ρ) will go to its supremum as ρ goes to $${1 \over \gamma }e^{{a_{2} \lambda {\minus}1}} {\minus}1$$ . Since ρ cannot reach $${1 \over \gamma }e^{{a_{2} \lambda {\minus}1}} {\minus}1$$ , the supremum is not achievable. Moreover, we have $$L_{1} (\rho )\in\left( {2\left( {{{1{\minus}\gamma } \over \lambda }} \right),2\left( {{1 \over \lambda }{\minus}{\gamma \over \lambda }e^{{1{\minus}a_{2} \lambda }} } \right)} \right)$$ when $$\rho \in\big( {{1 \over \gamma }{\minus}1,{1 \over \gamma }e^{{a_{2} \lambda {\minus}1}} {\minus}1} \big)$$ . So we have

$$\mathop {{\rm sup}}\limits_\rho \,L_{1} (\rho ){\equals}\,2\left( {{1 \over \lambda }{\minus}{\gamma \over \lambda }e^{{1{\minus}a_{2} \lambda }} } \right)$$

$${\rm when}\quad \rho \nearrow{1 \over \gamma }e^{{a_{2} \lambda {\minus}1}} {\minus}1$$

Case 2: For a ₂ λ>1, we have $$\rho \,{\equals}\,{1 \over \gamma }e^{{a_{2} \lambda {\minus}1}} {\minus}1$$ . The ceded loss functions for insurer 1 and insurer 2 are f ₁(x ₁)=c(x ₁−d*)₊ and f ₂(x ₂)=c(x ₂−d*)₊, c∈[0, 1], respectively, and we have assumed that the two insurers choose the same constant c for simplicity. We can derive in this case that

$$L_{2} (\rho ){\equals}\,2l_{2} (\rho ){\equals}\,{{2c\rho } \over {\lambda (1{\plus}\rho )}}$$

Since the safety loading ρ has only one possible value, so we have

$$\mathop {{\rm max}}\limits_\rho \,L_{2} (\rho ){\equals}\,2c\left( {{1 \over \lambda }{\minus}{\gamma \over \lambda }e^{{1{\minus}a\lambda }} } \right)$$

$${\rm at}\quad \rho \,{\equals}\,{1 \over \gamma }e^{{a_{2} \lambda {\minus}1}} {\minus}1$$

We now make a comparison between the results in Case 1 and Case 2 as follows:

$$\mathop {{\rm max}\,}\limits_\rho L_{2} (\rho )\leq \mathop {{\rm sup}}\limits_\rho \,L_{1} (\rho )$$

Case 3: If a ₂ λ>1, we have $$\rho \in\big( {\left. {0,{1 \over \gamma }{\minus}1} \big]} \right.$$ . If γ<a ₂ λ≤1, we have $$\rho \in\big( {0,{{a_{2} \lambda } \over \gamma }{\minus}1} \big)$$ . The ceded loss functions for insurer 1 and insurer 2 are f ₁(x ₁)=x ₁ and f ₂(x ₂)=x ₂, respectively. From this, we obtain

$$L_{3} (\rho ){\equals}\,2l_{3} (\rho ){\equals}\,2{\gamma \over \lambda }\rho $$

which is linear and increasing. Hence if a ₂ λ>1, the maximum L ₃(ρ) will be achieved at the largest possible safety loading $$\rho \,{\equals}\,{1 \over \gamma }{\minus}1$$ , and we have $$L_{3} (\rho )\in\left( {\left. {0,2\left( {{{1{\minus}\gamma } \over \lambda }} \right)} \right]} \right.$$ when $$\rho \in\big( {\left. {0,{1 \over \gamma }{\minus}1} \big]} \right.$$ . On the other hand, if γ<a ₂ λ≤1, L ₃(ρ) will go to its supremum as ρ goes to $${{a_{2} \lambda } \over \gamma }{\minus}1$$ , and we have $$L_{3} (\rho )\in\left( {0,2\left( {{{a_{2} \lambda {\minus}\gamma } \over \lambda }} \right)} \right)$$ when $$\rho \in\big( {0,{{a_{2} \lambda } \over \gamma }{\minus}1} \big)$$ . To sum up

$$\mathop {{\rm sup}}\limits_\rho \,L_{3} (\rho )\left\{ {\matrix{ {{\equals}\,2\left( {{{1{\minus}\gamma } \over \lambda }} \right),\quad a_{2} \lambda \,\gt\,1} \hfill \cr {{\equals}\,2\left( {{{a_{2} \lambda {\minus}\gamma } \over \lambda }} \right),\quad \gamma \,\lt\,a_{2} \lambda \leq 1} \hfill \cr } } \right.$$

$${\rm when}\quad \rho \left\{ {\matrix{ {\equals}\, &#x0026; \hskip-18pt{{1 \over \gamma }{\minus}1,\quad a_{2} \lambda \,\gt\,1} \cr \nearrow &#x0026; {{{a_{2} \lambda } \over \gamma }{\minus}1,\quad \gamma \,\lt\,a_{2} \lambda \leq 1} \cr } } \right$$

Case 4: For γ<a ₂ λ≤1, we have $$\rho \,{\equals}\,{{a_{2} \lambda } \over \gamma }{\minus}1$$ . The ceded loss functions for insurer 1 and insurer 2 are f ₁(x ₁)=cx ₁ and f ₂(x ₂)=cx ₂, c∈[0, 1], respectively. Again, we assume that the same c is chosen by the two insurers for simplicity. In this case

$$L_{4} (\rho ){\equals}\,2l_{4} (\rho ){\equals}\,{{2c\rho \gamma } \over \lambda }$$

Since the safety loading can only be $${{a_{2} \lambda } \over \gamma }{\minus}1$$ , we have

$$\mathop {{\rm max}\,}\limits_\rho L_{4} (\rho ){\equals}\,{{2c(a_{2} \lambda {\minus}\gamma )} \over \lambda }$$

$${\rm at}\quad \rho \,{\equals}\,{{a_{2} \lambda } \over \gamma }{\minus}1$$

From simple computation, we can compare the optimal values in Case 3 and Case 4 when γ<a ₂ λ≤1:

$$\mathop {{\rm max}}\limits_\rho \,L_{4} (\rho )\leq \mathop {{\rm sup}}\limits_\rho \,L_{3} (\rho )$$

Case 5: For a ₂ λ>1, we have $$\rho \,{\equals}\,{1 \over \gamma }e^{{a_{2} \lambda {\minus}1}} {\minus}1$$ . The ceded loss functions for insurer 1 and insurer 2 are f ₁(x ₁)=(x ₁−d*)₊ and f ₂(x ₂)=c(x ₂−d*)₊, c∈[0, 1], respectively, and the corresponding objective function is given by

$$L_{5} (\rho ){\equals}\,(1{\plus}c)l_{1} (\rho ){\equals}\,{{(1{\plus}c)\rho } \over {\lambda (1{\plus}\rho )}}$$

Since the safety loading ρ can only take on one possible value, so we have

$$\mathop {{\rm max}}\limits_\rho \,L_{5} (\rho ){\equals}\,(1{\plus}c)\left( {{1 \over \lambda }{\minus}{\gamma \over \lambda }e^{{1{\minus}a_{2} \lambda }} } \right)$$

$${\rm at}\quad \rho \,{\equals}\,{1 \over \gamma }e^{{a_{2} \lambda {\minus}1}} {\minus}1$$

Now we compare the optimal values in Case 1, Case 2 and Case 5:

$$\mathop {{\rm max}}\limits_\rho \,L_{2} (\rho )\leq \mathop {{\rm max}}\limits_\rho \,L_{5} (\rho )\leq \mathop {{\rm sup}}\limits_\rho \,L_{1} (\rho )$$

Case 6: For γ<a ₂ λ≤1, we have $$\rho \,{\equals}\,{{a_{2} \lambda } \over \gamma }{\minus}1$$ . The ceded loss functions for insurer 1 and insurer 2 are f ₁(x ₁)=x ₁ and f ₂(x ₂)=cx ₂, c∈[0, 1], respectively. The objective function is given by

$$L_{6} (\rho ){\equals}\,(1{\plus}c)l_{3} (\rho ){\equals}\,(1{\plus}c){\gamma \over \lambda }\rho $$

and we have

$$\mathop {{\rm max}\,}\limits_\rho L_{6} (\rho ){\equals}\,(1{\plus}c){{a_{2} \lambda {\minus}\gamma } \over \lambda }$$

$${\rm at}\quad \rho \,{\equals}\,{{a_{2} \lambda } \over \gamma }{\minus}1$$

We can then make comparisons between the results in Case 3, Case 4 and Case 6 when γ<a ₂ λ≤1:

$$\mathop {{\rm max}}\limits_\rho \,L_{4} (\rho )\leq \mathop {{\rm max}}\limits_\rho \,L_{6} (\rho )\leq \mathop {{\rm sup}}\limits_\rho \,L_{3} (\rho )$$

Case 7: For all other cases, f ₁=f ₂≡0 and thus

$$\mathop {{\rm max}}\limits_\rho \,L_{7} (\rho ){\equals}\,0$$

Figure 4 shows the relationship between the safety loading ρ and the objective function L(ρ) by combining the results obtained for the seven cases discussed above.

Figure 4 The expected profit of the reinsurer facing two risk.

The results shows that when a ₂ λ>1, the optimal expected profit of the reinsurer may not be obtainable, depending on how the constant c is chosen in Case 2 and Case 5. If c is chosen to be 1 in any one of the cases, then point D or point E will coincide with point C, and the maximum expected profit can be achieved at $$\rho \,{\equals}\,{1 \over \gamma }e^{{a_{2} \lambda {\minus}1}} {\minus}1$$ . When γ<a ₂ λ≤1, the situation is similar. If in either Case 4 or Case 6, c is chosen to be 1, then point C or point D will coincide with point B, and the optimal expected profit of the reinsurer is achieved at $$\rho \,{\equals}\,{{a_{2} \lambda } \over \gamma }{\minus}1$$ .

3.3.2 Maximising expected utility of the reinsurer

As in section 3, we assume that the utility function of the reinsurer is given by U(x)=−e ^−θx , θ>0. The profit of the reinsurer is denoted by $$\bar{A},$$ and

$$\bar{A}{\equals}\,(1{\plus}\rho )\def\Bbb{\tf="Macopen"}{\Bbb E}\left( {f_{1} \left( {X_{1} } \right){\plus}f_{2} \left( {X_{2} } \right)} \right){\minus}\left[ {f_{1} \left( {X_{1} } \right){\plus}f_{2} \left( {X_{2} } \right)} \right]$$

The reinsurer wants to select the optimal ρ so that $$K(\rho )\,\,\colon\!{\equals}\,\def\Bbb{\tf="Macopen"}{\Bbb E}[U(\bar{A})]$$ can be maximised:

(23) $$\mathop {{\rm max}\,}\limits_\rho K(\rho ){\equals}\,\mathop {{\rm max}}\limits_\rho \,\def\Bbb{\tf="Macopen"}{\Bbb E}[U(\bar{A})]$$

3.3.2.1 Comonotonic Assumption

Suppose that $$\big( {\hat{X}_{1} ,\hat{X}_{2} } \big)$$ is a comonotonic copy of (X ₁, X ₂). As both ceded loss functions f ₁ and f ₂ are increasing, $$\big( {f_{1} \big( {\hat{X}_{1} } \big),f_{2} \big( {\hat{X}_{2} } \big)} \big)$$ is also comonotonic.

The following lemma in Dhaene et al. (Reference Dhaene, Denuit, Goovaerts, Kaas and Vyncke2002) indicates that sum of comonotonic random variables is the largest in the sense of convex order.

Lemma 1 Suppose that the random vector $$\big( {\hat{X}_{1} ,...,\hat{X}_{n} } \big)\in{\cal F}\big( {F_{1} ,...,F_{n} } \big)$$ is comonotonic, then

$$X_{1} {\plus} \cdots {\plus}X_{n} \leq _{{cx}} \hat{X}_{1} {\plus} \cdots {\plus}\hat{X}_{2} $$

for any random vector $$\left( {X_{1} ,...,X_{n} } \right)\in{\cal F}\left( {F_{1} ,...,F_{n} } \right)$$ .

Then we have

$$f_{1} \left( {X_{1} } \right){\plus}f_{2} \left( {X_{2} } \right)\leq _{{cx}} f_{1} \big( {\hat{X}_{1} } \big){\plus}f_{2} \big( {\hat{X}_{2} } \big)$$

which implies that

$$\delta \left( {f_{1} \left( {X_{1} } \right){\plus}f_{2} \left( {X_{2} } \right)} \right){\minus}\left[ {f_{1} \left( {X_{1} } \right){\plus}f_{2} \left( {X_{2} } \right)} \right]\leq _{{cx}} \delta \big( {f_{1} \big( {\hat{X}_{1} } \big){\plus}f\big( {\hat{X}_{2} } \big)} \right){\minus}\left[ {f_{1} \big( {\hat{X}_{1} } \big){\plus}f\big( {\hat{X}_{2} } \big)} \right]$$

Since −U is convex, we have

$$\def\Bbb{\tf="Macopen"}{\Bbb E}\left[ {U\left( {\delta \left( {f_{1} \left( {X_{1} } \right){\plus}f_{2} \left( {X_{2} } \right)} \right){\minus}\left[ {f_{1} \left( {X_{1} } \right){\plus}f_{2} \left( {X_{2} } \right)} \right]} \right)} \right]\geq \def\Bbb{\tf="Macopen"}{\Bbb E}\left[ {U\left( {\delta \big( {f_{1} \big( {\hat{X}_{1} } \big){\plus}f\big( {\hat{X}_{2} } \big)} \big){\minus}\left[ {f_{1} \big( {\hat{X}_{1} } \big){\plus}f\big( {\hat{X}_{2} } \big)} \right]} \right)} \right]$$

Theorem 2 When $$(\hat{X}_{1} ,\hat{X}_{2} )$$ is comonotonic, the expected utility of the profit of the reinsurer would achieve its minimum value. That is

$${\rm max}\,\,{\rm min}\,\def\Bbb{\tf="Macopen"}{\Bbb E}\left[ {U(\bar{A})} \right]{\equals}\,{\rm max}\,\def\Bbb{\tf="Macopen"}{\Bbb E}\left[ {U\left( {\delta \big( {f_{1} \big( {\hat{X}_{1} } \big){\plus}f_{2} \big( {\hat{X}_{2} } \big)} \big){\minus}\left[ {f_{1} \big( {\hat{X}_{1} } \big){\plus}f_{2} \big( {\hat{X}_{2} } \big)} \right]} \right)} \right]$$

where the minimum on the left is taken over all possible dependence structures among X ₁ and X ₂.

Similar argument for risks under unknown dependency structures can be found in Cheung (Reference Cheung2006). From this theorem, it is reasonable, in view of the unknown dependence structure, to assume that (X ₁, X ₂) is comonotonic. Now we solve the optimisation problem (23) based on the value ranges of the safety loading ρ derived in section 3.2.

Case 1: For a ₂ λ>1, we have $$\rho \in\big( {{1 \over \gamma }{\minus}1,{1 \over \gamma }e^{{a_{2} \lambda {\minus}1}} {\minus}1} \big)$$ . The ceded loss functions chosen by insurer 1 and insurer 2 are f ₁(x ₁)=(x ₁−d*)₊ and f ₂(x ₂)=(x ₂−d*)₊, respectively. The profit of the reinsurer is denoted by $$\bar{A}_{1} $$ , and we have

$$\bar{A}_{1} {\equals}\,2(1{\plus}\rho )\def\Bbb{\tf="Macopen"}{\Bbb E}\left[ {f_{1} \left( {X_{1} } \right)} \right]{\minus}2f_{1} \left( {X_{1} } \right){\equals}\,2A_{1} $$

Therefore

$$K_{1} (\rho ){\equals}\,\def\Bbb{\tf="Macopen"}{\Bbb E}\left[ {U\left( {\bar{A}_{1} } \right)} \right]{\equals}\,\def\Bbb{\tf="Macopen"}{\Bbb E}\left[ {U\left( {2A_{1} } \right)} \right]{\equals}\,\def\Bbb{\tf="Macopen"}{\Bbb E}\left[ {{\minus}e^{{{\minus}2\theta A_{1} }} } \right]{\equals}\,{\minus}e^{{{\minus}{{2\theta } \over \lambda }}} {\plus}{{2\theta } \over {2\theta \,{\minus}\,\lambda }}e^{{{\minus}{{2\theta } \over \lambda }}} {1 \over {1{\plus}\rho }}$$

where 2θ<λ, and K ₁ is increasing and concave. Moreover

$$K_{1} (\rho )\in\left( {\left( {{{2\gamma \theta } \over {2\theta \,{\minus}\,\lambda }}{\minus}1} \right)e^{{{\minus}{{2\theta } \over \lambda }}} ,{\minus}e^{{{\minus}{{2\theta } \over \lambda }}} {\plus}{{2\gamma \theta } \over {2\theta {\minus}\gamma }}e^{{1{\minus}a_{2} \lambda {\minus}{{2\theta } \over \lambda }}} } \right)$$

when $$\rho \in\big( {{1 \over \gamma }{\minus}1,{1 \over \gamma }e^{{a_{2} \lambda {\minus}1}} {\minus}1} \big)$$ . Since the end points of the interval of ρ are not included, the maximum K ₁(ρ) will go to its supremum as ρ goes to $${1 \over \gamma }e^{{a_{2} \lambda {\minus}1}} {\minus}1$$ . Therefore, we have

$$\mathop {{\rm sup}}\limits_\rho \,K_{1} (\rho ){\equals}\,{\minus}e^{{{\minus}{{2\theta } \over \lambda }}} {\plus}{{2\theta \gamma } \over {2\theta {\minus}\gamma }}e^{{1{\minus}a_{2} \lambda {\minus}{{2\theta } \over \lambda }}} $$

$${\rm when}\quad \rho \nearrow{1 \over \gamma }e^{{a_{2} \lambda {\minus}1}} {\minus}1$$

Case 2: For a ₂ λ>1, we have $$\rho \,{\equals}\,{1 \over \gamma }e^{{a_{2} \lambda {\minus}1}} {\minus}1$$ . The ceded loss functions for insurer 1 and insurer 2 are f ₁(x ₁)=c(x ₁−d*)₊ and f ₂(x ₂)=c(x ₂−d*)₊, c∈[0, 1], respectively. The profit of the reinsurer is denoted by $$\bar{A}_{2} $$ , and we have $$\bar{A}_{2} {\equals}\,2A_{2} {\equals}\,2cA_{1} $$ . The objective function is given by

$$K_{2} (\rho ){\equals}\,\def\Bbb{\tf="Macopen"}{\Bbb E}\left[ {U\left( {2A_{2} } \right)} \right]{\equals}\,\def\Bbb{\tf="Macopen"}{\Bbb E}\left[ {U\left( {2cA_{1} } \right)} \right]{\equals}\,{\minus}e^{{{{{\minus}2c\theta } \over \lambda }}} {\plus}{{2c\theta } \over {2c\theta \,{\minus}\,\lambda }}e^{{{{{\minus}2c\theta } \over \lambda }}} {1 \over {1{\plus}\rho }}$$

where 2cθ<λ. Hence

$$\mathop {{\rm max}\,}\limits_\rho K_{2} (\rho ){\equals}\,{\minus}e^{{{{{\minus}2c\theta } \over \lambda }}} {\plus}{{2c\theta \gamma } \over {2c\theta \,{\minus}\,\lambda }}e^{{1{\minus}a\lambda {\minus}{{2c\theta } \over \lambda }}} $$

$${\rm at}\quad \rho \,{\equals}\,{1 \over \gamma }e^{{a_{2} \lambda {\minus}1}} {\minus}1$$

We can make a comparison between the results in Case 1 and Case 2:

$$\mathop {{\rm max}\,}\limits_\rho K_{2} (\rho )\leq \mathop {{\rm sup}\,}\limits_\rho K_{1} (\rho )$$

Case 3: If a ₂ λ>1, we have $$\rho \in\big( {\left. {0,{1 \over \gamma }{\minus}1} \big]} \right.$$ . If γ<a ₂ λ≤1, we have $$\rho \in\big( {0,{{a_{2} \lambda } \over \gamma }{\minus}1} \big)$$ . The ceded loss functions for insurer 1 and insurer 2 are f ₁(x ₁)=x ₁ and f ₂(x ₂)=x ₂, respectively. The profit of the reinsurer is denoted by $$\bar{A}_{3} $$ , and we have $$\bar{A}_{3} {\equals}\,2A_{3} $$ . By applying the deductions in section 2.3.2, we assume that 2θ<λ and the objective function is given by

$$K_{3} (\rho ){\equals}\,\def\Bbb{\tf="Macopen"}{\Bbb E}\left[ {U\left( {2A_{3} } \right)} \right]{\equals}\,{{\gamma \lambda } \over {2\theta \,{\minus}\,\lambda }}e^{{{\minus}2\theta (1{\plus}\rho ){\gamma \over \lambda }}} $$

It is easy to check that K ₃ is increasing and concave in ρ. Therefore if a ₂ λ>1, the maximum K ₃(ρ) will be achieved at the largest possible safety loading $$\rho \,{\equals}\,{1 \over \gamma }{\minus}1$$ , and we have $$K_{3} (\rho )\in\big( {\left. {{{\gamma \lambda } \over {2\theta \,{\minus}\,\lambda }}e^{{{\minus}{{2\theta \gamma } \over \lambda }}} ,{{\gamma \lambda } \over {2\theta \,{\minus}\,\lambda }}e^{{{\minus}{{2\theta } \over \lambda }}} } \bgi]} \right.$$ when $$\rho \in\big( {\left. {0,{1 \over \gamma }{\minus}1} \big]} \right.$$ .

On the other hand, if γ<a ₂ λ≤1, K ₃(ρ) will go to its supremum when ρ goes to $${{a_{2} \lambda } \over \gamma }{\minus}1$$ , and we have $$K_{3} (\rho )\in\big( {{{\gamma \lambda } \over {2\theta \,{\minus}\,\lambda }}e^{{{\minus}{{2\theta \gamma } \over \lambda }}} ,{{\gamma \lambda } \over {2\theta \,{\minus}\,\lambda }}e^{{{\minus}2a_{2} \theta }} } \big)$$ when $$\rho \in\big( {0,{{a_{2} \lambda } \over \gamma }{\minus}1} \big)$$ . To sum up, we obtain:

$$\mathop {{\rm sup}}\limits_\rho \,K_{3} (\rho )\left\{ {\matrix{ {\equals}\, \hfill &#x0026; {K_{3} \big( {{1 \over \gamma }{\minus}1} \big){\equals}\,{{\gamma \lambda } \over {2\theta \,{\minus}\,\lambda }}e^{{{\minus}{{2\theta } \over \lambda }}} ,\quad a_{2} \lambda \,\gt\,1} \hfill \cr {\equals}\, \hfill &#x0026; {K_{3} \big( {{{a_{2} \lambda } \over \gamma }{\minus}1} \big){\equals}\,{{\gamma \lambda } \over {2\theta \,{\minus}\,\lambda }}e^{{{\minus}2a_{2} \theta }} ,\quad \gamma \,\lt\,a_{2} \lambda \leq 1} \hfill \cr } } \right.$$

$${\rm when}\quad \rho \left\{ {\matrix{ {\equals}\, &#x0026; \hskip-18pt{{1 \over \gamma }{\minus}1,\quad a_{2} \lambda \,\gt\,1} \cr \nearrow &#x0026; {{{a_{2} \lambda } \over \gamma }{\minus}1,\quad \gamma \,\lt\,a_{2} \lambda \leq 1} \cr } } \right.$$

To compare the results of Case 1 and Case 3, when a ₂ λ>1, we have

$$\mathop {{\rm inf}}\limits_\rho K_{1} (\rho )\,\lt\,\mathop {{\rm max}}\limits_\rho K_{3} (\rho )$$

Case 4: For γ<a ₂ λ≤1, we have $$\rho \,{\equals}\,{{a_{2} \lambda } \over \gamma }{\minus}1$$ . The ceded loss functions for insurer 1 and insurer 2 are f ₁(x ₁)=cx ₁ and f ₂(x ₂)=cx ₂, c∈[0, 1]. The profit of the reinsurer is denoted by $$\bar{A}_{4} $$ , and we have $$\bar{A}_{4} {\equals}\,2A_{4} {\equals}\,2cA_{3} $$ . By applying the deductions in section 2.3.2, we assume that 2cθ<λ, and the objective function is given by

$$K_{4} (\rho ){\equals}\,\def\Bbb{\tf="Macopen"}{\Bbb E}\left[ {U\left( {2cA_{3} } \right)} \right]{\equals}\,{{\gamma \lambda } \over {2c\theta \,{\minus}\,\lambda }}e^{{{\minus}2\theta (1{\plus}\rho ){\gamma \over \lambda }}} $$

Since in this case the safety loading ρ has only one possible value, we have

$$\mathop {{\rm max}}\limits_\rho \,K_{4} (\rho ){\equals}\,{{\gamma \lambda } \over {2c\theta \,{\minus}\,\lambda }}e^{{{\minus}2c\theta a_{2} }} $$

$${\rm at}\quad \rho \,{\equals}\,{{a_{2} \lambda } \over \gamma }{\minus}1$$

We can make a comparison between the results in Case 3 and Case 4 directly by applying (19) when γ<a ₂ λ≤1:

$$\mathop {{\rm max}\,}\limits_\rho K_{4} (\rho )\geq \mathop {{\rm sup}}\limits_\rho \,K_{3} (\rho )$$

Case 5: For a ₂ λ>1, we have $$\rho \,{\equals}\,{1 \over \gamma }e^{{a_{2} \lambda {\minus}1}} {\minus}1$$ . The ceded loss functions for insurer 1 and insurer 2 are given by f ₁(x ₁)=(x ₁−d*)₊ and f ₂(x ₂)=c(x ₂−d*)₊, c∈[0, 1], respectively. The profit of the reinsurer is denoted by $$\bar{A}_{5} $$ , and we have

$$\bar{A}_{5} {\equals}\,(1{\plus}c)A_{1} $$

Using a similar argument, we assume that (1+c)θ<λ and the corresponding objective function is given by

$$K_{5} (\rho ){\equals}\,{\minus}e^{{{\minus}{{(1{\plus}c)\theta } \over \lambda }}} {\plus}{{(1{\plus}c)\gamma \theta } \over {(1{\plus}c)\theta \,{\minus}\,\lambda }}e^{{{\minus}{{(1{\plus}c)\theta } \over \lambda }}} {1 \over {1{\plus}\rho }}$$

Therefore,

$$\mathop {{\rm max}}\limits_\rho \,K_{5} (\rho ){\equals}\,{\minus}e^{{{\minus}{{(1{\plus}c)\theta } \over \lambda }}} {\plus}{{(1{\plus}c)\theta \gamma } \over {(1{\plus}c)\theta {\minus}\gamma }}e^{{1{\minus}a_{2} \lambda {\minus}{{(1{\plus}c)\theta } \over \lambda }}} $$

$${\rm at}\;\;\rho \,{\equals}\,{1 \over \gamma }e^{{a_{2} \lambda {\minus}1}} {\minus}1$$

We can compare the optimal values in Case 1, Case 2 and Case 5 by applying (15):

$$\mathop {{\rm max}\,}\limits_\rho K_{2} (\rho )\leq \mathop {{\rm max}\,}\limits_\rho K_{5} (\rho )\leq \mathop {{\rm sup}}\limits_\rho \,K_{1} (\rho )$$

Case 6: For a ₂ λ≤1, we have $$\rho \,{\equals}\,{{a_{2} \lambda } \over \gamma }{\minus}1$$ . The ceded loss functions adopted by insurer 1 and insurer 2 are f ₁(x ₁)=x ₁ and f ₂(x ₂)=cx ₂, c∈[0, 1], respectively. The profit of the reinsurer is denoted by $$\bar{A}_{6} $$ , and we have $$\bar{A}_{6} {\equals}\,(1{\plus}c)A_{3} $$ . With a similar argument used above, we assume that (1+c)θ<λ and the objective function equals

$$K_{6} (\rho ){\equals}\,{{\gamma \lambda } \over {(1{\plus}c)\theta \,{\minus}\,\lambda }}e^{{{\minus}(1{\plus}c)\theta (1{\plus}\rho ){\gamma \over \lambda }}} $$

Moreover, we have

$$\mathop {{\rm max}\,}\limits_\rho K_{6} (\rho ){\equals}\,{{\gamma \lambda } \over {(1{\plus}c)\theta \,{\minus}\,\lambda }}e^{{{\minus}(1{\plus}c)\theta a_{2} }} $$

$${\rm at}\quad \rho \,{\equals}\,{{a_{2} \lambda } \over \gamma }{\minus}1$$

We can make comparisons among the results in Case 3, Case 4 and Case 6 by applying (19) when γ<a ₂ λ≤1:

$$\mathop {{\rm max}}\limits_\rho \,K_{4} (\rho )\geq \mathop {{\rm max}}\limits_\rho \,K_{6} (\rho )\geq \mathop {{\rm sup}}\limits_\rho \,K_{3} (\rho )$$

Case 7: For all other cases, the ceded loss functions of the two insurers will be absolutely 0: f ₁=f ₂≡0. By simple calculation, we obtain

$$\mathop {{\rm max}\,}\limits_\rho K_{7} (\rho ){\equals}\,{\minus}1$$

Figure 5 shows the relationship between the safety loading ρ and the objective function K(ρ). The results of all seven cases discussed above are combined and compared. Since the comparison between K ₁, K ₃ and K ₇ partly depends on the value of the parameter θ involved in the utility function, we use black arrows to indicate the uncertain vertical locations of the curves.

Figure 5 The expected utility of the reinsurer facing two risk.

The results shows that if a ₂ λ>1, the comparison between K ₁, K ₃ and K ₇ partly depends on the value of the parameter θ involved in the exponential utility function. So when aλ>1, point D might be lower than point B; point C might be lower than point A; and K ₇(ρ) might be lower than point D. When γ<aλ≤1, K ₇(ρ) might be lower than point C. Moreover, point E or point F may coincide with point D if the constant c in Case 5 or Case 2 is set at 1. If γ<a ₂ λ≤1, the optimal expected exponential utility is obtained either in Case 4 or Case 7, and the reinsurer can always choose $$\rho \,{\equals}\,{{a_{2} \lambda } \over \gamma }{\minus}1$$ .

3.3.3 Minimising VaR of the total loss of the reinsurer

Following the last subsection, we continue to assume that the losses X ₁ and X ₂ are comonotonic, despite the fact that VaR is not necessarily subadditive. The comonotonic assumption can simplify many computations involving VaR.

The total loss of the reinsurer is given by

$$Y\,{\equals}\,f_{1} \left( {X_{1} } \right){\plus}f_{2} \left( {X_{2} } \right){\minus}(1{\plus}\rho )\def\Bbb{\tf="Macopen"}{\Bbb E}\left[ {f_{1} \left( {X_{1} } \right){\plus}f_{2} \left( {X_{2} } \right)} \right]$$

By applying Lemma 6, we get

$$J(\rho )\,{\equals}\,{\rm VaR}_{Y} (\beta ){\equals}\,f_{1} (b){\plus}f_{2} (b){\minus}(1{\plus}\rho )\def\Bbb{\tf="Macopen"}{\Bbb E}\left[ {f_{1} \left( {X_{1} } \right){\plus}f_{2} \left( {X_{2} } \right)} \right]$$

where β is the confidence level of the reinsurer and we suppose that β≤α ₁ and β≤α ₂, which implies that b≥a ₁ and b≥a ₂. The objective of the reinsurer is to minimise J over ρ.

Case 1: For a ₂ λ>1, we have $$\rho \in\big( {{1 \over \gamma }{\minus}1,{1 \over \gamma }e^{{a_{2} \lambda {\minus}1}} {\minus}1} \big)$$ . The ceded loss functions for insurer 1 and insurer 2 are f ₁(x ₁)=(x ₁−d*)₊ and f ₂(x ₂)=(x ₂−d*)₊, respectively. By applying the results in (2.3.3), we obtain

$$J_{1} (\rho )\,{\equals}\,2j_{1} (\rho )\,{\equals}\,2\left\{ {b{\minus}{1 \over \lambda }{\minus}{1 \over \lambda }\:{\rm ln}\left[ {\gamma (1{\plus}\rho )} \right]} \right\}$$

Since $$J_{1}^{\,'} (\rho )\,\lt\,0$$ , J ₁ is decreasing over ρ, so J ₁(ρ) will go to its infimum when ρ goes to $${1 \over \gamma }e^{{a_{2} \lambda {\minus}1}} {\minus}1$$ . Moreover, we have $$J_{1} (\rho )\in\left( {2\left( {b{\minus}a_{2} } \right),2\left( {b{\minus}{1 \over \lambda }} \right)} \right)$$ when $$\rho \in\big( {{1 \over \gamma }{\minus}1,{1 \over \gamma }e^{{a_{2} \lambda {\minus}1}} {\minus}1} \big)$$ . Therefore,

$$\mathop {{\rm inf}}\limits_\rho \,J_{1} (\rho ){\equals}\,J_{1} \left( {{1 \over \gamma }e^{{a_{2} \lambda {\minus}1}} {\minus}1} \right){\equals}\,2\left( {b{\minus}a_{2} } \right)$$

$${\rm when}\quad \rho \nearrow{1 \over \gamma }e^{{a_{2} \lambda {\minus}1}} {\minus}1$$

Case 2: For a ₂ λ>1, we have $$\rho \,{\equals}\,{1 \over \gamma }e^{{a_{2} \lambda {\minus}1}} {\minus}1$$ . The ceded loss functions for two insurers are f ₁(x ₁)= c(x ₁−d*)₊, f ₂(x ₂)=c(x ₂−d*)₊, c∈[0, 1], where the same c is assumed. By applying results in (2.3.3), we obtain

$$J_{2} (\rho ){\equals}\,2j_{2} (\rho ){\equals}\,2c\left\{ {b{\minus}{1 \over \lambda }{\minus}{1 \over \lambda }{\rm ln}\left[ {\gamma (1{\plus}\rho )} \right]} \right\}$$

and hence

$$\mathop {{\rm min}\,}\limits_\rho J_{2} (\rho ){\equals}\,2c\left( {b{\minus}a_{2} } \right)$$

$${\rm at}\quad \rho \,{\equals}\,{1 \over \gamma }e^{{a_{2} \lambda {\minus}1}} {\minus}1$$

We can make a comparison between the results in Case 1 and Case 2 by applying (20):

$$\mathop {{\rm min}}\limits_\rho \,J_{2} (\rho )\leq \mathop {{\rm inf}}\limits_\rho \,J_{1} (\rho )$$

Case 3: If a ₂ λ>1, we have $$\rho \in\big( {\left. {0,{1 \over \gamma }{\minus}1} \big]} \right.$$ . If γ<a ₂ λ≤1, $$\rho \in\big( {0,{{a_{2} \lambda } \over \gamma }{\minus}1} \big)$$ . The ceded loss functions for insurer 1 and insurer 2 are f ₁(x ₁)=x ₁, f ₂(x ₂)=x ₂, respectively, and we can easily obtain

$$J_{3} (\rho ){\equals}\,2j_{3} (\rho ){\equals}\,2b{\minus}2(1{\plus}\rho ){\gamma \over \lambda }$$

Since $$J_{3}^{'} (\rho )\,\lt\,0$$ , J ₃ is decreasing over ρ. If a ₂ λ>1, J ₃(ρ) will achieve its minimum value at $$\rho \,{\equals}\,{1 \over \gamma }{\minus}1$$ , and we have $$J_{3} (\rho )\in\left( {\left. {2b{\minus}{2 \over \lambda },2b{\minus}{{2\gamma } \over \lambda }} \right]} \right.$$ when $$\rho \in\big( {\left. {0,{1 \over \gamma }{\minus}1} \big]} \right.$$ . If γ<a ₂ λ≤1, j ₃(ρ) will go to its infimum when ρ goes to $${{a_{2} \lambda } \over \gamma }{\minus}1$$ , and we have $$J_{3} (\rho )\in\left( {2b{\minus}2a_{2} ,2b{\minus}{{2\gamma } \over \lambda }} \right)$$ when $$\rho \in\big( {0,{{a_{2} \lambda } \over \gamma }{\minus}1} \big)$$ . To sum up

$$\mathop {{\rm inf}}\limits_\rho \,J_{3} (\rho )\left\{ {\matrix{ {\equals}\, \hfill &#x0026; {J_{3} \big( {{1 \over \gamma }{\minus}1} \big){\equals}\,2b{\minus}{2 \over \lambda },\quad a_{2} \lambda \,\gt\,1} \hfill \cr {\equals}\, \hfill &#x0026; {J_{3} \big( {{{a_{2} \lambda } \over \gamma }{\minus}1} \big){\equals}\,2b{\minus}2a_{2} ,\quad \gamma \,\lt\,a_{2} \lambda \leq 1} \hfill \cr } } \right.$$

$${\rm when}\quad \rho \left\{ {\matrix{ {\equals}\, \hfill &#x0026; {{1 \over \gamma }{\minus}1,\quad a_{2} \lambda \,\gt\,1} \hfill \cr \nearrow \hfill &#x0026; {{{a_{2} \lambda } \over \gamma }{\minus}1,\quad \gamma \,\lt\,a_{2} \lambda \leq 1} \hfill \cr } } \right.$$

Case 4: For γ<a ₂ λ≤1, we have $$\rho \,{\equals}\,{{a_{2} \lambda } \over \gamma }{\minus}1$$ . The ceded loss functions for insurer 1 and insurer 2 are given by f ₁(x ₁)=cx ₁, f ₂(x ₂)=cx ₂, c∈[0, 1]. We can derive the objective function as

$$J_{4} (\rho ){\equals}\,2j_{4} (\rho ){\equals}\,2cb{\minus}2c(1{\plus}\rho ){\gamma \over \lambda }$$

and hence

$$\mathop {{\rm min}\,}\limits_\rho J_{4} (\rho ){\equals}\,2c(b{\minus}a_{2} )$$

$${\rm at}\quad \rho \,{\equals}\,{{a_{2} \lambda } \over \gamma }{\minus}1$$

We can compare the results in Case 3 and Case 4 by applying ((21)):

$$\mathop {{\rm min}}\limits_\rho \,J_{4} (\rho )\leq \mathop {{\rm inf}}\limits_\rho \,J_{3} (\rho )$$

Case 5: For a ₂ λ>1, we have $$\rho \,{\equals}\,{1 \over \gamma }e^{{a_{2} \lambda {\minus}1}} {\minus}1$$ . The ceded loss functions for insurer 1 and insurer 2 are f ₁(x ₁)=(x ₁−d*)₊ and f ₂(x ₂)=c(x ₂−d*)₊, c∈[0, 1], respectively. Using the same argument, we obtain

$$J_{5} (\rho ){\equals}\,(1{\plus}c)j_{1} (\rho ){\equals}\,(1{\plus}c)\left\{ {b{\minus}{1 \over \lambda }{\minus}{1 \over \lambda }ln\left[ {\gamma (1{\plus}\rho )} \right]} \right\}$$

and thus

$$\mathop {{\rm min}\,}\limits_\rho J_{5} (\rho ){\equals}\,(1{\plus}c)(b{\minus}a_{2} )$$

$${\rm at}\;\;\rho \,{\equals}\,{1 \over \gamma }e^{{a_{2} \lambda {\minus}1}} {\minus}1$$

We can also make comparisons among the results in Case 1, Case 2 and Case 5 by applying (20):

$$\mathop {{\rm min}}\limits_\rho \,J_{2} (\rho )\leq \mathop {{\rm min}}\limits_\rho \,J_{5} (\rho )\leq \mathop {{\rm inf}}\limits_\rho \,J_{1} (\rho )$$

Case 6: For a ₂ λ≤1, we have $$\rho \,{\equals}\,{{a_{2} \lambda } \over \gamma }{\minus}1$$ . The ceded loss functions for insurer 1 and insurer 2 are f ₁(x ₁)=x ₁ and f ₂(x ₂)=cx ₂, c∈[0, 1], respectively. In this case

$$J_{6} (\rho ){\equals}\,(1{\plus}c)j_{3} (\rho ){\equals}\,(1{\plus}c)\left[ {b{\minus}(1{\plus}\rho ){\gamma \over \lambda }} \right]$$

which implies that

$$\mathop {{\rm min}}\limits_\rho \,J_{6} (\rho ){\equals}\,(1{\plus}c)\left( {b{\minus}a_{2} } \right)$$

$${\rm at}\quad \rho \,{\equals}\,{{a_{2} \lambda } \over \gamma }{\minus}1$$

We can make comparisons among the results in Case 3, Case 4 and Case 6 by using (21):

$$\mathop {{\rm min}}\limits_\rho \,J_{4} (\rho )\leq \mathop {{\rm min}\,}\limits_\rho J_{6} (\rho )\leq \mathop {{\rm inf}}\limits_\rho \,J_{3} (\rho )$$

Case 7: For all other cases, the ceded loss functions for the two risks will both be exactly 0: f ₁=f ₂≡0, and hence we have

$$\mathop {{\rm min}\,}\limits_\rho J_{7} (\rho ){\equals}\,0$$

Figure 6 shows the relationship between the safety loading ρ and the objective function J(ρ). The results of all seven cases discussed above are combined and compared.

Figure 6 The value-at-risk of the total loss of the reinsurer facing two risk.

Similar with the results for one risk, our results show that in both cases a ₂ λ>1 and γ<a ₂ λ≤1, if the reinsurer set ρ to be very large, VaR will go to 0 which is optimal. This analysis makes sense in the real reinsurance markets, because if the reinsurer set its safety loading ρ as large as possible, the insurer will purchase less and less reinsurance. Once the safety loading ρ is larger than some threshold, no insurer would buy any reinsurance then the ceded loss function become 0.