2. Preliminaries

In this section, we recall some definitions, notations, and helpful lemmas used in this study. For real vectors ${\bf x}=(x_{1},\ldots,x_{n})$ and ${\bf \lambda }=(\lambda _{1},\ldots,\lambda _{n})$, the Hadamard product is denoted as ${\bf \lambda }\circ {\bf x}=(\lambda _{1}x_{1},\ldots,\lambda _{n}x_{n})$ and ${\bf {x}}_{\{i,j\}}$ stands for the subvector of $\bf {x}$ with its $i$th and $j$th entries deleted. Let $\mathbb {R}=(-\infty,+\infty )$ and $\mathbb {R}_+=[0,+\infty )$. Let $x_{1:n}\leq \ldots \leq x_{n:n}$ be the increasing sequence of the components of $\bf {x}$. Denote $\mathcal {I}_+^n=\{\bf {x}:0\leq x_1\leq x_2\leq \ldots \leq x_n\}$ and $\mathcal {D}_+^n=\{\bf {x}: x_1\geq x_2\geq \ldots \geq x_n\geq 0\}.$ All inverse functions are assumed to exist whenever they appear.

It is evident that $\bf {x}\stackrel {{\rm m}}{\succeq }\bf {y}$ implies $\bf {x}\succeq _{\rm w}\bf {y}$, while the reverse statement is not true in general. Majorization order is quite useful in establishing various inequalities arising from actuarial science, applied probability, reliability theory, and so on. The following lemma, implying that the weak submajorization order is preserved under transformation induced by increasing convex functions, is borrowed from Theorem 5.A.2 of Marshall et al. [Reference Marshall, Olkin and Arnold34] and plays a key role in proving the main results.

Lemma 2.2. Consider two real vectors $\bf {x}=(x_1,x_2,\ldots,x_n)\in \mathbb {R}^n$ and $\bf {y}=(y_1,y_2,\ldots,y_n)\in \mathbb {R}^n$. For all increasing convex function $\phi$, it follows that $(x_1,x_2,\ldots,x_n)\succeq _{\rm w}(y_1,y_2,\ldots,y_n)$ implies

$$(\phi(x_1),\phi(x_2),\ldots,\phi(x_n))\succeq_{\rm w}(\phi(y_1),\phi(y_2),\ldots,\phi(y_n)).$$

For more discussions on their properties and applications, one can refer to [Reference Marshall, Olkin and Arnold34].

In the context of applied probability, the stop-loss order is termed as the increasing convex order defined in the sense that $X \leq _{\rm sl} Y$ if and only if $\mathbb {E}[\phi (X)]\leq \mathbb {E}[\phi (Y)]$ for all increasing convex function $\phi$. Henceforth, we shall sometimes adopt the notation “$\leq _{\rm icx}$” as an alternative of the stop-loss order “$\leq _{\rm sl}$.” It is known that the hazard rate order implies the usual stochastic order, which in turn implies the stop-loss order. According to Theorem 4.A.3. of Shaked and Shanthikumar [Reference Shaked and Shanthikumar36], $X\leq _{\rm icx}Y$ if and only if

$$\int_x^\infty \bar{F}(u)\,\mathrm{d} u\leq \int_x^\infty \bar{G}(u)\,\mathrm{d} u,\quad \text{for all } x\in \mathbb{R},$$

or equivalently,

$$\int_{\alpha}^{1}F^{{-}1}_X(t)\,\mathrm{d} t\leq\int_{\alpha}^{1}F^{{-}1}_{Y}(t)\,\mathrm{d} t,\quad \text{for all } 0\leq\alpha\leq 1.$$

Let $\{\tau (1),\ldots,\tau (n)\}$ be any permutation of $\{1,2,\ldots,n\}$ under the permutation operator $\tau$. For any vector $\bf {x}=(x_{1},x_{2},\ldots,x_{n})\in \mathbb {R}^{n}$, we use $\tau (\bf {x})$ to denote the permuted vector $(x_{\tau (1)},x_{\tau (2)},\ldots,x_{\tau (n)})$. Let $\bf {x}_{\downarrow }$ denote the decreasing rearrangement of $\bf {x}$, $\bf {\lambda }_{\uparrow }$ denote the increasing rearrangement of $\bf {\lambda }$, and $\bf {\lambda }_{\downarrow }$ denote the decreasing rearrangement of $\bf {\lambda }$.

The next lemma is indebted to Cheung [Reference Cheung11].

For any $(i,j)$ with $1\leq i< j\leq n$, let $\tau _{ij}(\bf {x})=(x_{1},\ldots,x_{j},\ldots,x_{i},\ldots,x_{n})$ and

\begin{align*} \mathcal{G}_{s}^{i,j}(n)& =\{g(\bf{x}): g(\bf{x})\geq g(\tau_{ij}(\bf{x}))\text{ for any } x_{i}\leq x_{j}\},\\ \mathcal{G}_{l}^{i,j}(n)& =\{g(\bf{x}): g(\bf{x})-g(\tau_{ij}(\bf{x})) \text{ is decreasing in } x_{i}\leq x_{j}\},\\ \mathcal{G}_{r}^{i,j}(n)& =\{g(\bf{x}): g(\bf{x})-g(\tau_{ij}(\bf{x})) \text{ is increasing in } x_{j}\geq x_{i}\}. \end{align*}

The following lemma gives bivariate characterization of RWSAI random vectors, which plays a key role in proving our main results.

For ease of reference, we denote $\bf {0}=(0,\ldots,0)$, $\bf {1}=(1,\ldots,1)$, $\bf {I}=(I_1,\ldots,I_n)$ comprised of $n$ Bernoulli random variables, $p(\bf \lambda )=\mathbb {P}(\bf {I}=\bf \lambda )$,

$$\Lambda_{k}=\{{\bf \lambda}\,|\,\lambda_{i}=0\ {\rm or}\ 1, \ i=1,2,\ldots,n,\ \lambda_{1}+\cdots+\lambda_{n}=k\},\quad k=0,\ldots,n,$$

and let, for $1\leq i\neq j\leq n$ and $k=1,\ldots,n-1$,

\begin{align*} \Lambda_{k}^{i,j}(0,1)& =\{{\bf \lambda}\in\Lambda_{k}\,|\,\lambda_{i}=0,\lambda_{j}=1\},\quad \Lambda_{k}^{i,j}(0,0)=\{{\bf \lambda}\in\Lambda_{k}\,|\,\lambda_{i}=\lambda_{j}=0\},\\ \Lambda_{k}^{i,j}(1,0)& =\{{\bf \lambda}\in\Lambda_{k}\,|\,\lambda_{i}=1,\lambda_{j}=0\},\quad \Lambda_{k}^{i,j}(1,1)=\{{\bf \lambda}\in\Lambda_{k}\,|\,\lambda_{i}=\lambda_{j}=1\}. \end{align*}

It is easy to see that $\Lambda _{1}^{i,j}(1,1)=\Lambda _{n-1}^{i,j}(0,0)=\emptyset$ and

$$\Lambda_{k}=\Lambda_{k}^{i,j}(0,1)\cup\Lambda_{k}^{i,j}(0,0)\cup\Lambda_{k}^{i,j}(1,0)\cup \Lambda_{k}^{i,j}(1,1).$$

Formally, for a random vector $\bf {X}=(X_1,\ldots,X_n)$ with univariate marginal distributions $F_1,\ldots,F_n$ and survival functions $\bar {F}_1,\ldots,\bar {F}_n$, the well-known Sklar's theorem [Reference Sklar37] states that there exist $C:[0,1]^n\mapsto [0,1]$ and $\bar {C}:[0,1]^n\mapsto [0,1]$ such that its distribution function $F$ and survival function $\bar {F}$ can be represented as, for all $\bf {x}=(x_1,\ldots,x_n)$,

$$F(\bf{x})=C(F_{1}(x_{1}),\ldots,F_{n}(x_{n})) \quad\text{and}\quad \bar{F}(\bf{x})=\bar{C}(\bar{F}_{1}(x_{1}),\ldots,\bar{F}_{n}(x_{n})).$$

Then, $C$ and $\bar {C}$ are called the copula and survival copula of $\bf {X}$, respectively. For a decreasing and continuous function $\phi :\mathbb {R}_+\cup \{+\infty \}\mapsto [0,1]$ with $\phi (0)=1$, $\phi (+\infty )=0$ and the pseudo-inverse $\psi :=\phi ^{-1}$, the function

$$C_{\phi}(u_{1},\cdots,u_{n})=\phi(\psi(u_{1})+\cdots+\psi(u_{n})),\quad \text{for all } u_{i}\in[0,1],\ i=1,2,\ldots,n,$$

is called an Archimedean copula with generator $\psi$ if $(-1)^{k}\phi ^{(k)}(x)\geq 0$ for $k=0,\ldots,n-2$ and $(-1)^{n-2}\phi ^{(n-2)}(x)$ is decreasing and convex.

The following lemma provides sufficient conditions based on stochastic orders and Archimedean copulas to characterize LWSAI and RWSAI random vectors.

The notions of RWSAI and LWSAI can be roughly understood in the sense that a set of random variables are generally positively dependent and ordered in some stochastic sense, which agrees with many practical scenarios in actuarial science. For instance, all business lines in an insurance portfolio are normally positively dependent under external common shock induced by systemic risks such as some severe natural disasters including the earthquake and flood. From the viewpoint of mathematical technicality, it has standard steps on functional characterizations to verify whether or not a random vector is RWSAI or LWSAI according to Theorem 3.2 or 3.3 of You and Li [Reference You and Li38]. Besides, according to Lemma 2.9, we can parametrize RWSAI/LWSAI losses by applying Archimedean copulas and some traditional stochastic orders, which produces a wide class of multivariate distributions containing the commonly used multivariate $t$, multivariate $F$, and multivariate Pareto distributions as special cases; see Hollander et al. [Reference Hollander, Proschan and Sethuraman23]. Therefore, the notions of RWSAI and LWSAI have nice mathematical tractability and can be expediently adapted for modeling various practical insurance scenarios.

A subset $I\subseteq \mathbb {R}^{n}$ is said to be comonotonic if, for any $(x_{1},\ldots,x_{n})\in I$ and $(y_{1},\ldots,y_{n})\in I$, either $x_{i}\leq y_{i}$ for $i=1,\ldots,n$ or $x_{i}\geq y_{i}$ for $i=1,\ldots,n$. A random vector $\bf {X}$ is said to be comonotonic if there is a comonotonic subset $I$ such that $\mathbb {P}(\bf {X}\in I)=1$. The following lemma states that the summation of the comonotonic random variables is maximized via the convex order.

For more study on comonotonicity, interested readers can refer to Dhaene and Goovaerts [Reference Dhaene and Goovaerts20], Kaas et al. [Reference Kaas, Dhaene and Goovaerts26], and Dhaene et al. [Reference Dhaene, Denuit, Goovaerts, Kaas and Vyncke21,Reference Dhaene, Denuit, Goovaerts, Kaas and Vyncke22].

3. Allocations of policy limits without randomization

Let $X_1,X_2,\ldots,X_n$ be $n$ potential losses faced by the insurer. By paying a fix amount of premium to the reinsurance company, we assume that the insurer will be compensated by means of the layer reinsurance contract, that is, the amount of $\min \{(X_i-d_i)_+,l_i\}$ can be obtained from the reinsurer if the random loss $X_i$ is claimed. After buying reinsurance contracts, the insurer is then granted a right to freely allocate a fixed amount of policy limit $l>0$ and/or deductible $d>0$ to each potential risk. If $(l_1,l_2,\ldots,l_n)$ and $(d_1,d_2,\ldots,d_n)$ are the allocated policy limits and deductibles, then we have $l_1+\cdots +l_n=l$, $d_1+\cdots +d_n=d$, $l_i\geq 0$, and $d_i\geq 0$, for $i=1,\ldots,n$. We use $\mathcal {S}_n(l)$ and $\mathcal {S}_n(d)$ to denote the classes of all admissible allocations of policy limits and deductibles, respectively.

This section addresses the problem of allocating policy limits and deductibles by maximizing the expected utility of the terminal wealth for a risk-averse insurer. According to the above statements, we want to consider the following optimization problem

(1)\begin{equation} \max_{\bf{d}\in\mathcal{S}_n(d),\bf{l}\in\mathcal{S}_n(l)} \mathbb{E}\left[u\left(\omega-\left(\sum_{i=1}^nX_i-\sum_{i=1}^n((X_i-d_{i})_+{\wedge} l_i)\right)\right)\right], \end{equation}

where the utility function $u$ is increasing and concave, $\omega >0$ is the initial wealthFootnote ² of the insurer. Let $\tilde {u}(x)=-u(\omega -x)$. Then, the problem boils down to solving

(2)\begin{equation} \min_{\bf{d}\in\mathcal{S}_n(d),\bf{l}\in\mathcal{S}_n(l)} \mathbb{E}\left[\tilde{u}\left(\sum_{i=1}^n\left(X_{i}\wedge d_{i}\right)+\sum_{i=1}^n(X_{i}-d_{i}-l_{i})_+\right)\right], \end{equation}

where $\tilde {u}$ is increasing and convex.

To begin with, let us consider one numerical example illustrating that the optimal allocations of policy limits cannot be presented when the deductibles are fixed and different from each other. The basic idea for the refutation of the ordering result is to replace the integrals of the quantile function of aggregate risks from a bivariate vector by their numerical integrals.

Example 3.1. Consider the Clayton copula described by the generator $\psi (x)={1}/{x^{\theta }}-1$ such that $x\psi '(x)$ is increasing in $x\in (0,1]$, for $\theta >0$. Suppose that $(X_{1},X_{2})$ is a random vector linked by an Archimedean copula with generator function $\psi$ having parameter $\theta =2$, where $X_{1}$ and $X_{2}$ have the exponential distribution with scale parameters $\lambda _{1}=0.9$ and $\lambda _{2}=0.2$, respectively. Then, the joint distribution function of $\bf {X}$ is given by

$$F_{\bf{X}}(x_{1},x_{2})=[(1-e^{{-}0.9x_{1}})^{-\theta}+(1-e^{{-}0.2x_{2}})^{-\theta}].$$

According to Theorem 5.7 of Cai and Wei [Reference Cai and Wei7], we know $(X_{1},X_{2})$ is RWSAI.

Consider the following allocation polices of policy limits $\bf {l}_{1}=(1,2)\in \mathcal {I}_+^2$, $\bf {l}_{2}=(2,1)\in \mathcal {D}_+^2$, $\bf {l}_{3}=(2,8)\in \mathcal {I}_+^2$, $\bf {l}_{4}=(8,2)\in \mathcal {D}_+^2$ for fixed deductibles $\bf {d}=(1,15)\in \mathcal {I}_+^2$. Denote

$$U_{\bf{l}_{r}}=X_{1}\wedge d_{1}+(X_{1}-(d_{1}+l_{r1}))_+{+}X_{2}\wedge d_{2}+(X_{2}-(d_{2}+l_{r2}))_+,$$

where $\bf {l}_{r}=(l_{r1},l_{r2})$, for $r=1,2,3,4$. Let

$$\Delta_{\bf{l}_{r}}(\alpha):=\int_{\alpha}^{1}F_{U_{\bf{l}_{r}}}^{{-}1}(t)\,\mathrm{d} t,\quad \text{for}\ r=1,2,3,4\ \text{and}\ \alpha\in[0,1].$$

Given a sequence of independent and identically distributed (i.i.d.) observations $(X_{i,1},X_{i,2})$'s sampled from the distribution $F_{\bf {X}}$, one can obtain the empirical quantile $\hat {F}^{-1}_{U_{{\bf l}_r}}(t)$ of $F^{-1}_{U_{{\bf l}_r}}(t)$, for $t\in [0,1]$. Hence, we can use

$$\hat{\Delta}_{\bf{l}_{r}}(\alpha)=\int_{\alpha_{j}}^{1}\hat{F}_{U_{\bf{l}_{r}}}^{{-}1}(t)\,\mathrm{d} t$$

to approximate $\Delta _{{\bf l}_r}(\alpha )$. The empirical values of $\Delta _{{\bf l}_r}(\alpha _j)$, for $r=1,2,3,4$, are generated under $(\alpha _{1},\alpha _{2},\ldots,\alpha _{20})=(0,0.05,0.1,\ldots,0.95)$. As seen from Figure 1(a) and (b), $\hat {\Delta }_{\bf {l}_{1}}$ is larger than $\hat {\Delta }_{\bf {l}_{2}}$, while $\hat {\Delta }_{\bf {l}_{3}}$ is smaller than $\hat {\Delta }_{\bf {l}_{4}}$. Due to the strong law of large numbers, it follows that $\Delta _{\bf {l}_{1}}(\alpha )>\Delta _{\bf {l}_{4}}(\alpha )$ while $\Delta _{\bf {l}_{3}}(\alpha )<\Delta _{\bf {l}_{4}}(\alpha )$ for $\alpha \in [0,1]$. Therefore, the orderings among the optimal policy limits can be opposite in problem (2) for different values of deductibles.

Figure 1. $\hat {\Delta }_{\bf {l}_{r}}(\alpha _{j})$ for $j=1,2,\ldots,20$ and $r=1,2,3,4$. (a) $\hat {\Delta }_{\bf {l}_{1}}-\hat {\Delta }_{\bf {l}_{2}}$ and (b) $\hat {\Delta }_{\bf {l}_{3}}-\hat {\Delta }_{\bf {l}_{4}}$.

As per Example 3.1, the orderings among optimal allocations of policy limits can be opposite when the deductibles are different from each other, which will bring difficulties in seeking for the explicit ordering configuration of optimal allocation policies under such setting. On account of technicality in the proof, we shall study the allocation problem of policy limits by fixing $d_i$'s as $\bar {d}$ such that $d_i=\bar {d}=d/n$.

The following result depicts the orderings among the best allocations of policy limits, which extends the result of Proposition 1 in Cheung [Reference Cheung11] to the case of dependent risks under $\bar {d}=0$.

Proof. Let $\bf {l}=(l_{1},\ldots,l_{i},\ldots,l_{j},\ldots l_{n})$ be any admissible allocation vector with $l_{i}\leq l_{j}$, for $1\leq i< j\leq n$. Denote the objective function in (2) as $\pi (\bf {l})$. Then, it suffices to show that $\pi (\bf {l})\leq \pi (\tau _{ij}(\bf {l}))$. The RWSAI property of $\bf {X}$ guarantees that $[(X_{i},X_{j})\,|\,\bf {X}_{\{i,j\}}]$ is also RWSAI. Given $\bf {X}_{\{i,j\}}=\bf {x}_{\{i,j\}}$, let us consider the following two functions:

$$g_{2}(x_{i},x_{j})=\tilde{u}(x_i\wedge \bar{d}+(x_i-(\bar{d}+l_j))_+{+}x_j\wedge \bar{d}+(x_j-(\bar{d}+l_i))_+{+}T_{ij})$$

and

$$g_{1}(x_{i},x_{j})=\tilde{u}(x_i\wedge \bar{d}+(x_i-(\bar{d}+l_i))_+{+}x_j\wedge \bar{d}+(x_j-(\bar{d}+l_j))_+{+}T_{ij}),$$

where $T_{ij}=\sum _{r\neq i,j}^{n}(x_{r}\wedge \bar {d}+(x_{r}-(\bar {d}+l_{r}))_+)$. For $x'_{j}\ge x_{j}\ge x_{i}$, we denote

\begin{align*} L_{1}& =x_{i}\wedge \bar{d}+(x_{i}-(\bar{d}+l_{j}))_+{+}x_{j}\wedge \bar{d}+(x_{j}-(\bar{d}+l_{i}))_+{+}T_{ij},\\ L_{2}& =x_{i}\wedge \bar{d}+(x_{i}-(\bar{d}+l_{i}))_+{+}x_{j}\wedge \bar{d}+(x_{j}-(\bar{d}+l_{j}))_+{+}T_{ij},\\ L'_{1}& =x_{i}\wedge \bar{d}+(x_{i}-(\bar{d}+l_{j}))_+{+}x'_{j}\wedge \bar{d}+(x'_{j}-(\bar{d}+l_{i}))_+{+}T_{ij},\\ L'_{2}& =x_{i}\wedge \bar{d}+(x_{i}-(\bar{d}+l_{i}))_+{+}x'_{j}\wedge \bar{d}+(x'_{j}-(\bar{d}+l_{j}))_+{+}T_{ij}. \end{align*}

It is obvious that $L'_{1}\geq L_{1}$. By Lemma 2.5, we know that $(x_i-(\bar {d}+l_i))_++(x_j-(\bar {d}+l_j))_+$ is an AD function in $((x_i,x_j),(l_i,l_j))$, which implies that $L'_{1}\geq L'_{2}$ and $L'_{1}+L_{2}\geq L'_{2}+L_{1}$. Thus, we have $(L'_{1},L_{2})\succeq _{\rm w}(L'_{2},L_{1})$. Upon applying Lemma 2.2, we have $(\tilde {u}(L'_{1}),\tilde {u}(L_{2}))\succeq _{\rm w}(\tilde {u}(L'_{2}),\tilde {u}(L_{1}))$ according to the increasingness and convexity of $\tilde {u}$. Hence,

$$(g_{2}(x_{i},x'_{j}),g_{1}(x_{i},x_{j}))\succeq_{\rm w}(g_{1}(x_{i},x'_{j}),g_{2}(x_{i},x_{j})),$$

which implies that $g_{2}(x_{i},x'_{j})-g_{1}(x_{i},x'_{j})\geq g_{2}(x_{i},x_{j})-g_{1}(x_{i},x_{j})$, that is, $g_{2}(x_{i},x_{j})-g_{1}(x_{i},x_{j})$ is increasing in $x_{j}\geq x_{i}$.

On the other hand, it is plain that $g_{2}(x_{i},x_{j})+g_{2}(x_{j},x_{i})=g_{1}(x_{i},x_{j})+g_{1}(x_{j},x_{i})$ for any $x_{j}\geq x_{i}$. Thus, from Lemma 2.7, it follows that

(3)\begin{align} & \mathbb{E}[\tilde{u}(X_{i}\wedge \bar{d}+(X_{i}-\bar{d}-l_{i})_+{+}X_{j}\wedge \bar{d}+(X_{j}-\bar{d}-l_{j})_+\nonumber\\ & \qquad +(\bf{X}\wedge \bar{\bf d}+(\bf{X}-\bar{\bf d}-\bf{l})_+)_{\{i,j\}})\,|\,\bf{X}_{\{i,j\}}=\bf{x}_{\{i,j\}}]\nonumber\\ & \quad \leq \mathbb{E}[\tilde{u}(X_{i}\wedge \bar{d}+(X_{i}-\bar{d}-l_{j})_+{+}X_{j}\wedge \bar{d}+(X_{j}-\bar{d}-l_{i})_+\nonumber\\ & \qquad +(\bf{X}\wedge \bar{\bf d}+(\bf{X}-\bar{\bf d}-\bf{l})_+)_{\{i,j\}})\,|\,\bf{X}_{\{i,j\}}=\bf{x}_{\{i,j\}}], \end{align}

where $\bar {\bf d}=(\bar {d},\ldots,\bar {d})$. By applying double expectation formula on inequality (3), we have

\begin{align*} \pi(\bf{l})& = \mathbb{E}[\tilde{u}(X_{i}\wedge \bar{d}+(X_{i}-\bar{d}-l_{i})_+{+}X_{j}\wedge \bar{d}+(X_{j}-\bar{d}-l_{j})_+\\ & \quad+(\bf{X}\wedge \bar{\bf d}+(\bf{X}-\bar{\bf d}-\bf{l})_+)_{\{i,j\}})] \\ & \leq \mathbb{E}[\tilde{u}(X_{i}\wedge \bar{d}+(X_{i}-\bar{d}-l_{j})_+{+}X_{j}\wedge \bar{d}+(X_{j}-\bar{d}-l_{i})_+ \\ & \quad+(\bf{X}\wedge \bar{\bf d}+(\bf{X}-\bar{\bf d}-\bf{l})_+)_{\{i,j\}})] \\ & =\pi(\tau_{ij}(\bf{l})), \end{align*}

which yields the desired inequality.

The next example illustrates the result of Theorem 3.2.

Example 3.3. Under the setup of Example 3.1, we consider the following two polices $\bf {l}_{5}=(2,10)$, $\bf {l}_{6}=(10,2)$ by fixing $\bar {d}=6$. As observed in Figure 2, it holds that $\hat {\Delta }_{\bf {l}_{5}}(\alpha )\leq \hat {\Delta }_{\bf {l}_{6}}(\alpha )$ for $\alpha \in [0,1]$, which indicates that $\Delta _{\bf {l}_{5}}(\alpha )\leq \Delta _{\bf {l}_{6}}(\alpha )$ for $\alpha \in [0,1]$ by applying the strong law of large numbers, which validates the finding in Theorem 3.2.

Figure 2. Plots of $\hat {\Delta }_{\bf {l}_{r}}(\alpha _{j})$ for $j=1,2,\ldots,20$ and $r=5,6$.

According to Theorem 3.2, one may ask that whether or not the explicit configuration of the optimal allocation vector in $\mathcal {I}_+^{n}$ could be obtained in terms of the majorization order. The following example provides a negative answer.

Example 3.4. Under the setup of Example 3.1, we consider $\bf {l}_{7}=(3,5)\in \mathcal {I}_+^2$, $\bf {l}_{8}=(1,7)\in \mathcal {I}_+^2$, $\bf {l}_{9}=(49,51)\in \mathcal {I}_+^2$ and $\bf {l}_{10}=(1,99)\in \mathcal {I}_+^2$. It is plain that $\bf {l}_{7}\stackrel {{\rm m}}{\preceq }\bf {l}_{8}$ and $\bf {l}_{9}\stackrel {{\rm m}}{\preceq }\bf {l}_{10}$. As observed from Figure 3(a), $\hat {\Delta }_{\bf {l}_{7}}$ is larger than $\hat {\Delta }_{\bf {l}_{8}}$, while $\hat {\Delta }_{\bf {l}_{9}}$ is smaller than $\hat {\Delta }_{\bf {l}_{10}}$ in accordance with Figure 3(b). Thus, the best allocation of policy limits in $\bf {l}\in \mathcal {I}_+^{n}$ cannot be obtained in terms of the majorization order. The optimal allocation strategy might highly depend on the total amount of policy limit as well as the joint distribution. Therefore, some numerical algorithms should be developed to obtain the exact solution of problem (2), which is left for future research.

Figure 3. Plots of $\hat {\Delta }_{\bf {l}_{r}}(\alpha _{j})$ for $j=1,2,\ldots,20$ and $r=7,8,9,10$. (a) $\hat {\Delta }_{\bf {l}_{7}}-\hat {\Delta }_{\bf {l}_{8}}$ and (b) $\hat {\Delta }_{\bf {l}_{9}}-\hat {\Delta }_{\bf {l}_{10}}$.

4. Allocation of policy limits with randomized reinsurance treaties

Albrecher and Cani [Reference Albrecher and Cani1] discussed randomized reinsurance treaties for insurers to conduct efficient risk management. They argued that a randomized reinsurance treaty might be interesting for the insurer to transfer the risk to the reinsurer. In this section, we study the allocation problem of policy limits in randomized layer reinsurance contracts.

Through paying a fixed amount of premium to the reinsurer, the insurer will be compensated by means of the randomized layer reinsurance contract, that is, the amount of $I_{i}\min \{(X_i-d_i)_+,l_i\}$ will be transferred to the reinsurer if the $i$th claim is realized, where $I_i$ is a Bernoulli random variable indicating the willingness to provide the compensation for the $i$th business line, for $i=1,\ldots,n$. Suppose that $\bf {X}$ is RWSAI and $\bf {I}$ is a LWSAI Bernoulli random vector. Consider the following optimization problem

(4) \begin{equation} \max_{\bf{d}\in\mathcal{S}_n(d),\bf{l}\in\mathcal{S}_n(l)} \mathbb{E}\left[u\left(\omega-\left(\sum_{i=1}^nX_i-\sum_{i=1}^nI_{i}\min\{(X_i-d_{i})_+,l_i\}\right)\right)\right], \end{equation}

where $u$ is the utility function of the insurer and is increasing and concave. Let $\tilde {u}(x)=-u(\omega -x)$. Then, the problem (4) boils down to solving

(5) \begin{equation} \min_{\bf{d}\in\mathcal{S}_n(d),\bf{l}\in\mathcal{S}_n(l)}\mathbb{E} \left[\tilde{u}\left(\sum_{i=1}^nX_i-\sum_{i=1}^nI_{i}\min\{(X_i-d_{i})_+,l_i\}\right)\right], \end{equation}

where $\tilde {u}$ is increasing and convex.

First, we present an example to show that there is no certain answer on the orderings among the optimal allocations of policy limits when the deductibles are fixed and different from each other.

Example 4.1. Set $p((0,0))=0.15$, $p((0,1))=0.5$, $p((1,0))=0.2$, and $p((1,1))=0.15$. It can be verified that $(I_{1},I_{2})$ is LWSAI. Under the setup of Example 3.1, we consider $\bf {d}=(1,15)\in \mathcal {I}_+^2$, $\bf {l}_{1}=(1,2)\in \mathcal {I}_+^2$, $\bf {l}_{2}=(2,1)\in \mathcal {D}_+^2$, $\bf {l}_{3}=(2,8)\in \mathcal {I}_+^2$, and $\bf {l}_{4}=(8,2)\in \mathcal {D}_+^2$. Let

$$U^{I}_{\bf{l}_{r}}=X_{1}-I_{1}((X_{1}-d_{1})_+{\wedge} l_{r1})+X_{2}-I_{2}((X_{2}-d_{2})_+{\wedge} l_{r2}),$$

where $\bf {l}_{r}=(l_{r1},l_{r2})$, for $r=1,2,3,4$. The values of $\hat {\Delta }^{I}_{\bf {l}_1}(\alpha _{j})$, $\hat {\Delta }^{I}_{\bf {l}_2}(\alpha _j)$, $\hat {\Delta }^{I}_{\bf {l}_3}(\alpha _j)$, and $\hat {\Delta }^{I}_{\bf {l}_4}(\alpha _j)$ are plotted with respect to $\alpha _{j}$, for $j=1,2,\ldots,20$. As seen in Figure 4(a), $\hat {\Delta }^{I}_{\bf {l}_{1}}$ is larger than $\hat {\Delta }^{I}_{\bf {l}_{2}}$, while it is observed from Figure 4(b) that $\hat {\Delta }^{I}_{\bf {l}_{3}}$ is smaller than $\hat {\Delta }^{I}_{\bf {l}_{4}}$. By using the strong law of large numbers, the orderings among the optimal policy limits cannot be determined when the deductibles are different.

Figure 4. Plots of (a) $\hat {\Delta }^{I}_{\bf {l}_{1}}(\alpha _{j})-\hat {\Delta }^{I}_{\bf {l}_{2}}(\alpha _{j})$ and (b) $\hat {\Delta }^{I}_{\bf {l}_{3}}(\alpha _{j})-\hat {\Delta }^{I}_{\bf {l}_{4}}(\alpha _{j})$, for $j=1,2,\ldots,20$.

In the sequel, we shall consider optimal allocation strategies of policy limits when the deductibles are fixed as $d_i=\bar {d}=d/n$.

Proof. First, let us introduce some notations to simplify our subsequent discussion:

\begin{align*} & \bf{1}\circ\bf{X}-{\bf \lambda}\circ(({\bf X}-\bar{\bf d})_+{\wedge}{\bf l})=\sum_{i=1}^{n}x_{i}-\sum_{i=1}^{n}\lambda_{i}((x_{i}-\bar{d})_+{\wedge} l_{i}),\\ & A_{i,j}=(\bf{1}\circ\bf{X}-\bf{\lambda}\circ(({\bf X}-\bar{\bf d})_+{\wedge}\bf{l}))_{\{i,j\}}=\sum_{r\neq i,j}^{n}X_{r}-\sum_{r\neq i,j}^{n}\lambda_r((X_{r}-\bar{d})_+{\wedge} l_{r}), \\ & a_{i,j}=(\bf{1}\circ\bf{x}-\bf{\lambda}\circ(({\bf x}-\bar{\bf d})_+{\wedge}\bf{l}))_{\{i,j\}}=\sum_{r\neq i,j}^{n}x_{r}-\sum_{r\neq i,j}^{n}\lambda_r((x_{r}-\bar{d})_+{\wedge} l_{r}),\\ & C_{i,j}=(\bf{1}\circ\bf{X}-\bf{1}\circ(({\bf X}-\bar{\bf d})_+{\wedge}\bf{l}))_{\{i,j\}}=\sum_{r\neq i,j}^{n}X_{r}-\sum_{r\neq i,j}^{n}((X_{r}-\bar{d})_+{\wedge} l_{r}), \\ & c_{i,j}=(\bf{1}\circ\bf{x}-\bf{1}\circ(({\bf x}-\bar{\bf d})_+{\wedge}\bf{l}))_{\{i,j\}}=\sum_{r\neq i,j}^{n}x_{r}-\sum_{r\neq i,j}^{n}((x_{r}-\bar{d})_+{\wedge} l_{r}),\\ & b_{1,2}=(\bf{\lambda}\circ (\bf{x}\wedge \bf{l}))_{\{1,2\}}=\sum_{r\neq1,2}^n\lambda_r (x_{r}\wedge l_{r}). \end{align*}

Denote $\eta (\bf {l}):=\mathbb {E}[\tilde {u}(\bf {1}\circ \bf {X}-{\bf \lambda }\circ (({\bf X}-\bar {\bf d})_+\wedge {\bf l}))]$. It suffices to show that $\eta (\bf {l})\leq \eta (\tau _{ij}(\bf {l}))$. Denote

\begin{align*} B_{1}& =\sum_{i=1}^n X_{i}-\sum_{i=1}^n((X_{i}-\bar{d})_+{\wedge} l_{i}),\\ B_{2}& =X_{i}+X_{j}-((X_{i}-\bar{d})_+{\wedge} l_{j})-((X_{j}-\bar{d})_+{\wedge} l_{i})+C_{i,j},\\ B_{3}& =\sum_{i=1}^n X_{i}-\sum_{i=1}^n\lambda_i((X_{i}-\bar{d})_+{\wedge} l_{i}),\\ B_{4}& =X_{i}+X_{j}-\lambda_{i}((X_{i}-\bar{d})_+{\wedge} l_{j})-\lambda_{j}((X_{j}-\bar{d})_+{\wedge} l_{i})+A_{i,j}. \end{align*}

It can be checked that

(6)\begin{align} \eta(\bf{l})& =\sum_{k=0}^{n}\sum_{{\bf \lambda}\in\Lambda_{k}}\mathbb{E}[\tilde{u}(\bf{1}\circ\bf{X}-\bf{I}\circ((\bf{X}-\bar{\bf d})_+{\wedge} \bf{l}))\mid{\bf I}={\bf \lambda}]p({\bf \lambda})\nonumber\\ & =p({\bf 0})\mathbb{E}\left[\tilde{u}\left(\sum_{i=1}^n X_{i}\right)\right]+p({\bf 1})\mathbb{E}[\tilde{u}(B_{1})]+\sum_{k=1}^{n-1}\sum_{{\bf \lambda}\in\Lambda_{k}}p({\bf \lambda})\mathbb{E}[\tilde{u}(B_{3}))]\qquad \end{align}

and

$$\eta(\tau_{ij}(\bf{l}))=p({\bf 0})\mathbb{E}\left[\tilde{u}\left(\sum_{i=1}^n X_{i}\right)\right]+p({\bf 1})\mathbb{E}[\tilde{u}(B_{2})]+\sum_{k=1}^{n-1}\sum_{{\bf \lambda}\in\Lambda_{k}}p({\bf \lambda})\mathbb{E}[\tilde{u}(B_{4})].$$

Then, we have

(7)\begin{equation} \eta(\bf{l})-\eta(\tau_{ij}(\bf{l}))=\sum_{k=1}^{n-1}\sum_{{\bf \lambda}\in\Lambda_{k}}p({\bf \lambda})\{\mathbb{E}[\tilde{u}(B_{3})]-\mathbb{E}[\tilde{u}(B_{4})]\}+p({\bf 1})\{\mathbb{E}[\tilde{u}(B_{1})]-\mathbb{E}[\tilde{u}(B_{2})]\}. \end{equation}

By Theorem 3.2, we have

(8)\begin{equation} \mathbb{E}[\tilde{u}(B_{1})]\leq \mathbb{E}[\tilde{u}(B_{2})]. \end{equation}

On the other hand, for any ${\bf \lambda }\in \Lambda ^{i,j}_{k}(0,0)$, $k=1,2,\ldots,n-1$, it holds that

(9)\begin{equation} \mathbb{E}[\tilde{u}(B_{3})]= \mathbb{E}[\tilde{u}(B_{4})]. \end{equation}

For ${\bf \lambda }\in \Lambda ^{i,j}_{k}(1,1)$, we have

(10)\begin{equation} \mathbb{E}[\tilde{u}(B_{3})]\leq \mathbb{E}[\tilde{u}(B_{4})]. \end{equation}

Denote

\begin{align*} B_{5}& =X_{i}+X_{j}-((X_{j}-\bar{d})_+{\wedge} l_{j})+A_{i,j}, \quad B_{6}=X_{i}+X_{j}-((X_{j}-\bar{d})_+{\wedge} l_{i})+A_{i,j},\\ B_{7}& =X_{i}+X_{j}-((X_{i}-\bar{d})_+{\wedge} l_{i})+A_{i,j}, \quad B_{8}=X_{i}+X_{j}-((X_{i}-\bar{d})_+{\wedge} l_{j})+A_{i,j}. \end{align*}

By applying (8), (9), and (10) to (7), we have

\begin{align*} & \eta(\bf{l})-\eta(\tau_{ij}(\bf{l}))\nonumber\\ & \quad \leq \sum_{k=1}^{n-1}\left\{\sum_{{\bf \lambda}\in\Lambda^{i,j}_{k}(0,1)}p({\bf \lambda})(\mathbb{E}[\tilde{u}(B_{3})]-\mathbb{E}[\tilde{u}(B_{4})])+\sum_{{\bf \lambda}\in\Lambda^{i,j}_{k}(1,0)}p({\bf \lambda})(\mathbb{E}[\tilde{u}(B_{3})]-\mathbb{E}[\tilde{u}(B_{4})])\right\}\nonumber\\ & \quad = \sum_{k=1}^{n-1}\left\{\sum_{{\bf \lambda}\in\Lambda^{i,j}_{k}(0,1)}p({\bf \lambda})(\mathbb{E}[\tilde{u}(B_{5})]-\mathbb{E}[\tilde{u}(B_{6})])+\sum_{{\bf \lambda}\in\Lambda^{i,j}_{k}(0,1)}p(\tau_{ij}({\bf \lambda}))(\mathbb{E}[\tilde{u}(B_{7})]-\mathbb{E}[\tilde{u}(B_{8})])\right\}. \end{align*}

Since $\bf {X}$ is RWSAI, we know that $[(X_{i},X_{j})\,|\,\bf {X}_{\{i,j\}}]$ is RWSAI. Given $\bf {X}_{\{i,j\}}=\bf {x}_{\{i,j\}}$, for $x'_{j}\ge x_{j}\ge x_{i}$, $l_{i}\leq l_{j}$, we denote

\begin{align*} b_{5}& =x_{i}+x_{j}-((x_{j}-\bar{d})_+{\wedge} l_{j})+a_{i,j}, \quad b_{6}=x_{i}+x_{j}-((x_{j}-\bar{d})_+{\wedge} l_{i})+a_{i,j},\\ b_{7}& =x_{i}+x_{j}-((x_{i}-\bar{d})_+{\wedge} l_{i})+a_{i,j}, \quad b_{8}=x_{i}+x_{j}-((x_{i}-\bar{d})_+{\wedge} l_{j})+a_{i,j},\\ b'_{5}& =x_{i}+x'_{j}-((x'_{j}-\bar{d})_+{\wedge} l_{j})+a_{i,j}, \quad b'_{6}=x_{i}+x'_{j}-((x'_{j}-\bar{d})_+{\wedge} l_{i})+a_{i,j},\\ b'_{7}& =x_{i}+x'_{j}-((x_{i}-\bar{d})_+{\wedge} l_{i})+a_{i,j},\quad b'_{8}=x_{i}+x'_{j}-((x_{i}-\bar{d})_+{\wedge} l_{j})+a_{i,j}, \end{align*}

$h_{2}(x_{i},x_{j})=\tilde {u}(b_{6})$ and $h_{1}(x_{i},x_{j})=\tilde {u}(b_{5})$. Then, it is easy to check that $h_{2}(x_{j},x_{i})=\tilde {u}(b_{7})$ and $h_{1}(x_{j},x_{i})=\tilde {u}(b_{8})$. By using Lemma 2.5, we know $-\sum _{i=1}^{n}((x_i-\bar {d})_+\wedge l_{i})$ is an AD function, which implies that

(11)\begin{equation} -((x'_{j}-\bar{d})_+{\wedge} l_{i})-((x_{j}-\bar{d})_+{\wedge} l_{j}) \geq{-}((x'_{j}-\bar{d})_+{\wedge} l_{j})-((x_{j}-\bar{d})_+{\wedge} l_{i}). \end{equation}

Hence, one has $b'_{6}+b_{5}\geq b'_{5}+b_{6}$. By using $x=(x\wedge d)+(x-d)_+$, we have $x_{i}+x'_{j}-((x'_{j}-\bar {d})_+\wedge l_{i})+a_{i,j}=(x'_{j}-\bar {d}-l_{i})_++(x'_{j}\wedge \bar {d})+x_{i}+a_{i,j}$, which further implies $b'_{6}\geq b'_{5}$, $b'_{6}\geq b_{6}$, and thus $(b'_{6},b_{5})\succeq _{\rm w}(b'_{5},b_{6})$. Now, upon applying Lemma 2.2, we have $(\tilde {u}(b'_{6}),\tilde {u}(b_{5}))\succeq _{\rm w}(\tilde {u}(b'_{5}),\tilde {u}(b_{6}))$, which further implies $\tilde {u}(b'_{6})-\tilde {u}(b'_{5})\geq \tilde {u}(b_{6})-\tilde {u}(b_{5})$. Then, $h_{2}(x_{i},x_{j})-h_{1}(x_{i},x_{j})$ is increasing in $x_{j}\geq x_{i}$ for any $x_i$.

Since $x_{j}\geq x_{i}$ and $l_{j}\geq l_{i}$, we have $b_{6}\leq b_{7}$, $b_{5}\leq b_{8}$, $b_{7}\geq b_{8}$, and $b_{6}+b_{7}\geq b_{5}+b_{8}$, which indicates that $(b_{6},b_{7})\succeq _{\rm w}(b_{5},b_{8})$. Upon applying Lemma 2.2, we have $(\tilde {u}(b_{6}),\tilde {u}(b_{7}))\succeq _{\rm w}(\tilde {u}(b_{5}),\tilde {u}(b_{8}))$, which further implies $\tilde {u}(b_{6})+\tilde {u}(b_{7})\geq \tilde {u}(b_{5})+\tilde {u}(b_{8})$. Thus $h_{2}(x_{i},x_{j})+h_{2}(x_{j},x_{i})\geq h_{1}(x_{i},x_{j})+h_{1}(x_{j},x_{i})$ for any $x_{j}\geq x_{i}$.

By Lemma 2.7, we have $\mathbb {E}[\tilde {u}(h_{1}(x_{i},x_{j}))]\leq \mathbb {E}[\tilde {u}(h_{2}(x_{i},x_{j}]$, that is, $\mathbb {E}[\tilde {u}(B_{5})]\leq \mathbb {E}[\tilde {u}(B_{6})]$. Then, from Lemma 2.8, we have

(12)\begin{equation} \eta(\bf{l})-\eta(\tau_{ij}(\bf{l}))\leq\sum_{k=1}^{n-1}\sum_{{\bf \lambda}\in\Lambda^{i,j}_{k}(0,1)}p(\tau_{ij}({\bf \lambda}))\{\mathbb{E}[\tilde{u}(B_{5})]-\mathbb{E}[\tilde{u}(B_{6})] +\mathbb{E}[\tilde{u}(B_{7})]-\mathbb{E}[\tilde{u}(B_{8})]\}. \end{equation}

In what follows, we show the non-positivity of

(13)\begin{equation} \Delta_{1}:=\mathbb{E}[\tilde{u}(B_{5})]-\mathbb{E}[\tilde{u}(B_{6})] +\mathbb{E}[\tilde{u}(B_{7})]-\mathbb{E}[\tilde{u}(B_{8})]. \end{equation}

Denote $f_{2}(x_{i},x_{j})=\tilde {u}(b_{6})+\tilde {u}(b_{8})$ and $f_{1}(x_{i},x_{j})=\tilde {u}(b_{5})+\tilde {u}(b_{7})$. For ease of presentation, we denote

$$M_{1}=b'_{6},\ M_{2}=b'_{8},\ M_{3}=b_{5},\ M_{4}=b_{7},\ M'_{1}=b_{6},\ M'_{2}=b_{8},\ M'_{3}=b'_{5},\ M'_{4}=b'_{7}.$$

In light of (11), we firstly have

(14)\begin{equation} M_{1}+M_{2}+M_{3}+M_{4} \geq M'_{1}+M'_{2}+M'_{3}+M'_{4}. \end{equation}

The following three cases are considered:

Case 1: $l_{i} \leq (x_{i}-\bar {d})_+\leq (x_{j}-\bar {d})_+$. For this case, we have $M_{1}\geq M_{4}\geq M_{3}$, $M_{1}\geq M_{2}\geq M_{3}$, $M'_{4}\geq M'_{1}\geq M'_{2}$, and $M'_{4}\geq M'_{3}$. Based on these relations, we can obtain the following possible inequalities:

(15a)\begin{align} M_{1}\geq M_{4}\geq M_{2}\geq M_{3}, \quad M'_{4}\geq M'_{1}\geq M'_{3}\geq M'_{2}; \end{align}

(15b)\begin{align} M_{1}\geq M_{2}\geq M_{4}\geq M_{3}, \quad M'_{4}\geq M'_{3}\geq M'_{1}\geq M'_{2}; \end{align}

(15c)\begin{align} M_{1}\geq M_{4}\geq M_{2}\geq M_{3}, \quad M'_{4}\geq M'_{1}\geq M'_{2}\geq M'_{3}; \end{align}

(15d)\begin{align} M_{1}\geq M_{2}\geq M_{4}\geq M_{3}, \quad M'_{4}\geq M'_{1}\geq M'_{3}\geq M'_{2}; \end{align}

(15e)\begin{align} M_{1}\geq M_{2}\geq M_{4}\geq M_{3}, \quad M'_{4}\geq M'_{1}\geq M'_{2}\geq M'_{3}; \end{align}

(15f)\begin{align} M_{1}\geq M_{4}=M_{2}\geq M_{3}, \quad M'_{4}\geq M'_{3}= M'_{1}\geq M'_{2}. \end{align}

Note that $M_{1}=M'_{4}$, $M_{4}=M'_{1}$, $M_{2}\geq M'_{3}$, and $M_{2}\geq M'_{2}$. Then, $M_{1}+M_{2}\geq M'_{4}+M'_{3}$, $M_{1}+M_{4}+M_{2}\geq M'_{4}+M'_{1}+M'_{3}$, and $M_{1}+M_{4}+M_{2}\geq M'_{4}+M'_{1}+M'_{2}$. Then by (14), we know that (15a), (15b), (15c), and (15f) contribute to $(M_{1},M_{2},M_{3},M_{4})\succeq _{\rm w}(M'_{1},M'_{2},M'_{3},M'_{4})$. For (15d) and (15e), it can be seen that $M_{2}\geq M_{4}=M'_{1}$, and thus $M_{1}+M_{2}\geq M'_{4}+M'_{1}$, $M_{1}+M_{2}+M_{4}\geq M'_{4}+M'_{1}+M'_{3}$, and $M_{1}+M_{2}+M_{4}\geq M'_{4}+M'_{1}+M'_{2}$. By (14), we can also get that (15d) and (15e) lead to $(M_{1},M_{2},M_{3},M_{4})\succeq _{\rm w}(M'_{1},M'_{2},M'_{3},M'_{4})$.

Case 2: $(x_{i}-\bar {d})_+\leq l_{i}\leq (x_{j}-\bar {d})_+$. Under this case, we have $M_{2}\geq M_{4}\geq M_{3}$, $M_{2}\geq M_{1}\geq M_{3}$, $M'_{4}\geq M'_{2}\geq M'_{1}$, and $M'_{4}\geq M'_{3}$. The possible inequalities are summarized as follows:

\begin{align*} M_{2}\geq M_{4}\geq M_{1}\geq M_{3}, \qquad M'_{4}\geq M'_{2}\geq M'_{1}\geq M'_{3};\\ M_{2}\geq M_{1}\geq M_{4}\geq M_{3}, \qquad M'_{4}\geq M'_{2}\geq M'_{1}\geq M'_{3};\\ M_{2}\geq M_{1}\geq M_{4}\geq M_{3}, \qquad M'_{4}\geq M'_{3}\geq M'_{2}\geq M'_{1};\\ M_{2}\geq M_{4}\geq M_{1}\geq M_{3}, \qquad M'_{4}\geq M'_{2}\geq M'_{3}\geq M'_{1};\\ M_{2}\geq M_{1}\geq M_{4}\geq M_{3}, \qquad M'_{4}\geq M'_{2}\geq M'_{3}\geq M'_{1};\\ M_{2}\geq M_{4} = M_{1}\geq M_{3}, \qquad M'_{4}\geq M'_{3} = M'_{2}\geq M'_{1}. \end{align*}

Note that $M_{1}\geq M'_{1}$, $M_{1}\geq M'_{3}$, $M_{2}=M'_{4}$, $M_{3}\leq M'_{1}$, and $M_{4}=M'_{2}$. Similar with the discussions in Case 1, one can get $(M_{1},M_{2},M_{3},M_{4})\succeq _{\rm w}(M'_{1},M'_{2},M'_{3},M'_{4})$.

Case 3: $(x_{i}-\bar {d})_+\leq (x_{j}-\bar {d})_+ \leq l_{i}$. In this case, we have $M_{2}\geq M_{4}\geq M_{3}$, $M_{2}\geq M_{1}$, $M'_{4}\geq M'_{2}\geq M'_{1}$, and $M'_{4}\geq M'_{3}$. Then, we have the following several possible inequalities:

\begin{align*} M_{2}& \geq M_{4}\geq M_{1}\geq M_{3}, \quad M'_{4}\geq M'_{2}\geq M'_{3}\geq M'_{1};\\ M_{2}& \geq _{1}\geq M_{4}\geq M_{3}, \quad M'_{4}\geq M'_{3}\geq M'_{2}\geq M'_{1};\\ M_{2}& \geq M_{1}\geq M_{4}\geq M_{3}, \quad M'_{4}\geq M'_{2}\geq M'_{3}\geq M'_{1};\\ M_{2}& \geq M_{4}\geq M_{1}\geq M_{3}, \quad M'_{4}\geq M'_{2}\geq M'_{1}\geq M'_{3};\\ M_{2}& \geq M_{1}\geq M_{4}\geq M_{3}, \quad M'_{4}\geq M'_{2}\geq M'_{1}\geq M'_{3};\\ M_{2}& \geq M_{4}\geq M_{3}\geq M_{1}, \quad M'_{4}\geq M'_{2}\geq M'_{1}\geq M'_{3};\\ M_{2}& \geq M_{4} = M_{1}\geq M_{3}, \quad M'_{4}\geq M'_{3} = M'_{2}\geq M'_{1};\\ M_{2}& \geq M_{4}\geq M_{3} = M_{1}, \quad M'_{4}\geq M'_{2}\geq M'_{3} = M'_{1};\\ M_{2}& \geq M_{4} = M_{3} = M_{1}, \quad M'_{4}\geq M'_{3} = M'_{2} = M'_{1}. \end{align*}

Observe that $M_{1}\geq M'_{3}$, $M_{2}=M'_{4}$, $M_{3}=M'_{1}$, and $M_{4}=M'_{2}$. Similarly, it can be checked that all of the above inequalities result in $(M_{1},M_{2},M_{3},M_{4})\succeq _{\rm w}(M'_{1},M'_{2},M'_{3},M'_{4})$.

To sum up, we always have $(M_{1},M_{2},M_{3},M_{4})\succeq _{\rm w}(M'_{1},M'_{2},M'_{3},M'_{4})$, that is, $(b'_{6},b'_{8},b_{5},b_{7})\succeq _{\rm w}(b_{6},b_{8},b'_{5},b'_{7})$. Upon applying Lemma 2.2, we have

$$(\tilde{u}(b'_{6}),\tilde{u}(b'_{8}),\tilde{u}(b_{5}),\tilde{u}(b_{7}))\succeq_{\rm w}(\tilde{u}(b_{6}),\tilde{u}(b_{8}),\tilde{u}(b'_{5}),\tilde{u}(b'_{7})),$$

which in turn implies that

$$\tilde{u}(b'_{6})+\tilde{u}(b'_{8})-\tilde{u}(b'_{5})-\tilde{u}(b'_{7})\geq \tilde{u}(b_{6})+\tilde{u}(b_{8})-\tilde{u}(b_{5})-\tilde{u}(b_{7}).$$

Thus, $f_{2}(x_{i},x_{j})-f_{1}(x_{i},x_{j})$ is increasing in $x_{j}\geq x_{i}$ for any $x_i$. Note that $f_{2}(x_{i},x_{j})+f_{2}(x_{j},x_{i})=f_{1}(x_{i},x_{j})+f_{1}(x_{j},x_{i})$ for any $x_{j}\geq x_{i}$. Upon using Lemma 2.7, we can conclude that $\Delta _{1}\leq 0$. This invokes $\eta (\bf {a})\leq \eta (\tau _{ij}(\bf {a}))$, yielding the desired result.

The next example illustrates the result of Theorem 4.2.

Example 4.3. Under the setup of Example 4.1, we consider $\bf {l}_{3}=(2,8)$, $\bf {l}_{4}=(8,2)$, and $\bar {d}=6$. As displayed in Figure 5, it holds that $\hat {\Delta }^{I}_{\bf {l}_3}(\alpha )\leq \hat {\Delta }^{I}_{\bf {l}_4}(\alpha )$ for $\alpha \in [0,1]$. Therefore, the result of Theorem 4.2 is validated by applying the strong law of large numbers.

Figure 5. Plots of $\hat {\Delta }^{I}_{\bf {l}_{r}}(\alpha _{j})$ for $j=1,2,\ldots,20$ and $r=3,4$.

The next numerical example states that the best allocation vector in $\mathcal {I}_+^n$ cannot be given in terms of the majorization order under the setting of Theorem 4.2.

Example 4.4. Under the setup of Example 4.3, we consider $\bf {l}_{7}=(3,5)\in \mathcal {I}_+^2$, $\bf {l}_{8}=(1,7)\in \mathcal {I}_+^2$, $\bf {l}_{9}=(49,51)\in \mathcal {I}_+^2$, and $\bf {l}_{10}=(1,99)\in \mathcal {I}_+^2$. It is plain that $\bf {l}_{7}\stackrel {{\rm m}}{\preceq }\bf {l}_{8}$ and $\bf {l}_{9}\stackrel {{\rm m}}{\preceq }\bf {l}_{10}$. As seen from Figure 6(a), $\hat {\Delta }^{I}_{\bf {l}_{7}}$ is larger than $\hat {\Delta }^{I}_{\bf {l}_{8}}$, while it is observed in Figure 6(b) that $\hat {\Delta }^{I}_{\bf {l}_{9}}$ is smaller than $\hat {\Delta }^{I}_{\bf {l}_{10}}$. Therefore, by applying the strong law of large numbers, the best allocation vector in $\bf {l}\in \mathcal {I}_+^{n}$ cannot be given in the sense of the majorization order. Numerical algorithms might be resorted to dealing with the optimal allocations.

Figure 6. Plots of $\hat {\Delta }^{I}_{\bf {l}_{7}}(\alpha _{j})-\hat {\Delta }^{I}_{\bf {l}_{8}}(\alpha _{j})$ and $\hat {\Delta }^{I}_{\bf {l}_{9}}(\alpha _{j})-\hat {\Delta }^{I}_{\bf {l}_{10}}(\alpha _{j})$ for $j=1,2,\ldots,20$. (a) $\hat {\Delta }^{I}_{\bf {l}_{7}}-\hat {\Delta }^{I}_{\bf {l}_{8}}$ and (b) $\hat {\Delta }^{I}_{\bf {l}_{9}}-\hat {\Delta }^{I}_{\bf {l}_{10}}$.

In some practical situations, the exact dependence structure among the losses is usually unknown for the insurer. Therefore, one may consider the following robust optimization problem:

(16)\begin{equation} \min_{\bf{l}\in\mathcal{S}(l)}\max_{\bf{X}\in\mathcal{R}(F_1,\ldots,F_n)}\mathbb{E}\left[\tilde{u}\left(\sum_{i=1}^nX_i-\sum_{i=1}^nI_{i}((X_i-\bar{d})_+{\wedge} l_i)\right)\right]. \end{equation}

Being the strongest positive dependence, comonotonicity is commonly used to model the worst dependency structure from the viewpoint of the insurer. The next result shows that the retained loss $\sum _{i=1}^nX_i-\sum _{i=1}^nI_{i}((X_i-\bar {d})_+\wedge l_i)$ is maximized according to the convex order when the claim severities are comonotonic.

Lemma 4.5. Suppose $\bf {X}^{c}=(X_{1}^{c},\ldots,X_{n}^{c})$ is the comonotonic version of $\bf {X}$, and $\bf {I}$ is a Bernoulli random vector. Then, we have

$$\sum_{i=1}^nX_i-\sum_{i=1}^nI_{i}((X_i-\bar{d})_+{\wedge} l_i)\leq_{\rm cx}\sum_{i=1}^nX^c_i-\sum_{i=1}^nI_{i}((X^c_i-\bar{d})_+{\wedge} l_i),$$

for any $\bf {l}=(l_1,\ldots,l_n)\in \mathcal {S}_n(l)$.

Proof. Note that

\begin{align*} & \sum_{i=1}^nx_i-\sum_{i=1}^nI_{i}((x_i-\bar{d})_+{\wedge} l_i)\\ & \quad =\begin{cases} \displaystyle\sum_{j\neq i}^nx_j-\sum_{j\neq i}^nI_{j}((x_j-\bar{d})_+{\wedge} l_j)+(x_i\wedge \bar{d})+(x_i-\bar{d}-l_i), & I_i=1\\ \displaystyle\sum_{j\neq i}^nx_j+x_i, & I_i=0. \end{cases} \end{align*}

It is evident that $\sum _{i=1}^nx_i-\sum _{i=1}^nI_{i}((x_i-\bar {d})_+\wedge l_i)$ is increasing in $x_i$, for $i=1,\ldots,n$. Then by Lemma 2.10, we can get the desired result.

In light of Lemma 4.5, problem (16) reduces to

(17)\begin{equation} \min_{\bf{l}\in\mathcal{S}(l)}\mathbb{E}\left[\tilde{u}\left(\sum_{i=1}^nX_i-\sum_{i=1}^nI_{i}((X_i-\bar{d})_+{\wedge} l_i)\right)\right], \end{equation}

where $\bf {X}$ is comonotonic. As per Example 4.4, the best allocation in $\mathcal {I}_+^n$ cannot be given. In the sequel, we consider the worst allocation of policy limits in $\mathcal {D}_+^n$.

Theorem 4.6. Suppose that $\bf {X}$ is comonotonic with $X_{1}\leq _{\rm st}\cdots \leq _{\rm st}X_{n}$ and $\bf {I}$ is LWSAI. For any two allocation vectors $\bf {l}_1, \bf {l}_2\in \mathcal {D}_+^{n}$, $\bf {l}_1\succeq _{\rm m}\bf {l}_2$ implies that

$$\sum_{r=1}^n(X_r-I_r((X_r-\bar{d})_+{\wedge} l_{1r}))\geq_{\rm sl}\sum_{r=1}^n(X_r-I_r((X_r-\bar{d})_+{\wedge} l_{2r})).$$

Proof. Without loss of generality, we set $\bar {d}=0$ for ease of the presentation of the proof. By exploiting a similar proof method as in Theorem 4.2, it is enough to prove

(18)\begin{align} \eta(\bf{l}_1)-\eta(\bf{l_2}) & =\sum_{k=1}^{n-1}\sum_{{\bf \lambda}\in\Lambda_{k}}p({\bf \lambda})\left\{\mathbb{E}\left[\tilde{u}\left(\sum_{i=1}^{n}X_i-\bf{\lambda} \circ(\bf{X}\wedge\bf{l}_1)\right)\right]\right.\nonumber\\ & \quad \left.-\mathbb{E}\left[\tilde{u}\left(\sum_{i=1}^{n}X_i-\bf{\lambda}\circ (\bf{X}\wedge\bf{l}_2)\right)\right]\right\}\nonumber\\ & \quad +p({\bf 1})\left\{\mathbb{E}\left[\tilde{u}\left(\sum_{i=1}^n(X_i-l_{1i})_+\right)\right] -\mathbb{E}\left[\tilde{u}\left(\sum_{i=1}^n(X_i-l_{2i})_+\right)\right]\right\}. \end{align}

By the nature of majorization order, it suffices to prove the non-negativity of (18) under the conditions $l_{11}\geq l_{12}$, $l_{21} \geq l_{22}$, $(l_{11},l_{12})\stackrel {\rm m}{\succeq }(l_{21},l_{22})$, and $l_{1i}=l_{2i}$ for $i=3,\ldots,n$.

Owing to comonotonicity of $(X_{1},X_{2})$ with $X_{1}\leq _{\rm st}X_{2}$, we have $X_{1}$ is smaller than $X_{2}$ almost surely given that $\bf {X}_{\{1,2\}}=\bf {x}_{\{1,2\}}$. Taking the realizations of $X_1$ and $X_2$ as $x_1$ and $x_2$, we know $x_1\leq x_2$ with probability 1. Then, the assumption $(l_{11},l_{12})\stackrel {\rm m}{\succeq }(l_{21},l_{22})$ implies $(x_1-l_{11})_++(x_2-l_{12})_+\geq (x_1-l_{21})_++(x_2-l_{22})_+$ and $(x_1\wedge l_{11})+(x_2\wedge l_{12})\leq (x_1\wedge l_{21})+(x_2\wedge l_{22})$ with probability 1. Therefore, upon using double expectation formula, it follows that

(19)\begin{equation} \mathbb{E}\left[\tilde{u}\left(\sum_{i=1}^n(X_i-l_{1i})_+\right)\right] \geq\mathbb{E}\left[\tilde{u}\left(\sum_{i=1}^n(X_i-l_{2i})_+\right)\right]. \end{equation}

Besides, for any ${\bf \lambda }\in \Lambda ^{1,2}_{k}(0,0)$, $k=1,2,\ldots,n-1$, it holds that

(20)\begin{equation} \mathbb{E}\left[\tilde{u}\left(\sum_{i=1}^{n}X_i-\bf{\lambda} \circ(\bf{X}\wedge\bf{l}_1)\right)\right] =\mathbb{E}\left[\tilde{u}\left(\sum_{i=1}^{n}X_i-\bf{\lambda}\circ (\bf{X}\wedge\bf{l}_2)\right)\right]. \end{equation}

For ${\bf \lambda }\in \Lambda ^{1,2}_{k}(1,1)$, we can obtain

(21)\begin{align} & \mathbb{E}\left[\tilde{u}\left(\sum_{i=1}^{n}X_i-(X_1\wedge l_{11})-(X_2\wedge l_{12})-(\bf{\lambda}\circ(\bf{X}\wedge\bf{l}_1))_{\{1,2\}}\right)\right]\nonumber\\ & \quad \geq\mathbb{E}\left[\tilde{u}\left(\sum_{i=1}^{n}X_i-(X_1\wedge l_{21})-(X_2\wedge l_{22})-(\bf{\lambda}\circ(\bf{X}\wedge\bf{l}_2))_{\{1,2\}}\right)\right]. \end{align}

By using (19), (20), and (21), it can be reached that

(22)\begin{align} & \eta(\bf{l}_1)-\eta(\bf{l}_2)\nonumber\\ & \quad\geq\sum_{k=1}^{n-1}\left\{\sum_{{\bf \lambda}\in\Lambda^{1,2}_{k}(0,1)}p({\bf \lambda})\left(\mathbb{E}\left[\tilde{u}\left(\sum_{i=1}^{n}X_i- \bf{\lambda}\circ(\bf{X}\wedge\bf{l}_1)\right)\right] -\mathbb{E}\left[\tilde{u}\left(\sum_{i=1}^{n}X_i-\bf{\lambda}\circ (\bf{X}\wedge\bf{l}_2)\right)\right]\right)\right. \nonumber\\ & \qquad\left.+\sum_{{\bf \lambda}\in\Lambda^{1,2}_{k}(1,0)}p({\bf \lambda})\left(\mathbb{E}\left[\tilde{u}\left(\sum_{i=1}^{n}X_i-\bf{\lambda} \circ(\bf{X}\wedge\bf{l}_1)\right)\right] -\mathbb{E}\left[\tilde{u}\left(\sum_{i=1}^{n}X_i-\bf{\lambda} \circ(\bf{X}\wedge\bf{l}_2)\right)\right]\right)\right\}\nonumber\\ & \quad= \sum_{k=1}^{n-1}\left\{\sum_{{\bf \lambda}\in\Lambda^{1,2}_{k}(0,1)}p({\bf \lambda})\left(\mathbb{E}\left[\tilde{u}\left(\sum_{i=1}^{n}X_i-(X_2\wedge l_{12})-(\bf{\lambda}\circ(\bf{X}\wedge\bf{l}_1))_{\{1,2\}}\right)\right]\right.\right.\nonumber\\ & \qquad \left.-\mathbb{E}\left[\tilde{u}\left(\sum_{i=1}^{n}X_i-(X_2\wedge l_{22})-(\bf{\lambda}\circ(\bf{X}\wedge\bf{l}_1))_{\{1,2\}}\right)\right]\right)\nonumber\\ & \qquad +\sum_{{\bf \lambda}\in\Lambda^{1,2}_{k}(0,1)}p(\tau_{12}({\bf \lambda}))\left(\mathbb{E}\left[\tilde{u}\left(\sum_{i=1}^{n}X_i-(X_1\wedge l_{11})-(\bf{\lambda}\circ(\bf{X}\wedge\bf{l}_1))_{\{1,2\}}\right)\right]\right.\nonumber\\ & \qquad \left.\left.-\mathbb{E}\left[\tilde{u}\left(\sum_{i=1}^{n}X_i-(X_1\wedge l_{21})-(\bf{\lambda}\circ(\bf{X}\wedge\bf{l}_1))_{\{1,2\}}\right)\right]\right)\right\}\nonumber\\ & \quad\geq \sum_{k=1}^{n-1}\sum_{{\bf \lambda}\in\Lambda^{1,2}_{k}(0,1)}p(\tau_{12}({\bf \lambda}))\Delta_{2}, \end{align}

where (22) is due to Lemma 2.8 with $l_{12}\leq l_{22}$, and

\begin{align*} \Delta_{2}& :=\mathbb{E}\left[\tilde{u}\left(\sum_{i=1}^{n}X_i-(X_2\wedge l_{12})-(\bf{\lambda}\circ(\bf{X}\wedge\bf{l}_1))_{\{1,2\}}\right)\right]\\ & \quad -\mathbb{E}\left[\tilde{u}\left(\sum_{i=1}^{n}X_i-(X_2\wedge l_{22})-(\bf{\lambda}\circ(\bf{X}\wedge\bf{l}_1))_{\{1,2\}}\right)\right]\\ & \quad +\mathbb{E}\left[\tilde{u}\left(\sum_{i=1}^{n}X_i-(X_1\wedge l_{11})-(\bf{\lambda}\circ(\bf{X}\wedge\bf{l}_1))_{\{1,2\}}\right)\right]\\ & \quad -\mathbb{E}\left[\tilde{u}\left(\sum_{i=1}^{n}X_i-(X_1\wedge l_{21})-(\bf{\lambda}\circ(\bf{X}\wedge\bf{l}_1))_{\{1,2\}}\right)\right]. \end{align*}

Next, we need to show the non-negativity of $\Delta _{2}$. For any ${\bf \lambda }\in \Lambda ^{1,2}_{k}(0,1)$, if $x_1\geq l_{12}$, then

$$\sum_{i=1}^{n}x_i-(x_2\wedge l_{12})-b_{1,2}\geq\sum_{i=1}^{n}x_i-(x_1\wedge l_{21})-b_{1,2}$$

and

$$\sum_{i=1}^{n}x_i-(x_2\wedge l_{12})-b_{1,2}\geq\sum_{i=1}^{n}x_i-(x_2\wedge l_{22})-b_{1,2};$$

if $x_1\leq l_{12}$, then

\begin{align*} & \sum_{i=1}^{n}x_i-(x_1\wedge l_{11})-b_{1,2}\geq\sum_{i=1}^{n}x_i-(x_2\wedge l_{12})-b_{1,2},\\ & \sum_{i=1}^{n}x_i-(x_1\wedge l_{11})-b_{1,2}\geq\sum_{i=1}^{n}x_i-(x_1\wedge l_{21})-b_{1,2}, \end{align*}

and

$$\sum_{i=1}^{n}x_i-(x_1\wedge l_{11})-b_{1,2}\geq\sum_{i=1}^{n}x_i-(x_2\wedge l_{22})-b_{1,2}.$$

Moreover,

\begin{align*} & \left(\sum_{i=1}^{n}x_i-(x_1\wedge l_{11})-b_{1,2}\right)+\left(\sum_{i=1}^{n}x_i-(x_2\wedge l_{12})-b_{1,2}\right)\nonumber\\ & \quad \geq \left(\sum_{i=1}^{n}x_i-(x_1\wedge l_{21})-b_{1,2}\right)+\left(\sum_{i=1}^{n}x_i-(x_2\wedge l_{22})-b_{1,2}\right). \end{align*}

It then holds that $(\sum _{i=1}^{n}x_i-(x_1\wedge l_{11})-b_{1,2},\sum _{i=1}^{n}x_i-(x_2\wedge l_{12})-b_{1,2})\succeq _{\rm w}$ $(\sum _{i=1}^{n}x_i-(x_1\wedge l_{21})-b_{1,2},\sum _{i=1}^{n}x_i-(x_2\wedge l_{22})-b_{1,2})$, which implies $(\tilde {u}(\sum _{i=1}^{n}x_i-(x_1\wedge l_{11})-b_{1,2}),\tilde {u}(\sum _{i=1}^{n}x_i-(x_2\wedge l_{12})-b_{1,2}))\succeq _{\rm w}$ $(\tilde {u}(\sum _{i=1}^{n}x_i-(x_1\wedge l_{21})-b_{1,2}),\tilde {u}(\sum _{i=1}^{n}x_i-(x_2\wedge l_{22})-b_{1,2}))$ upon applying Lemma 2.2. Hence, we have

\begin{align*} & \tilde{u}\left(\sum_{i=1}^{n}x_i-(x_1\wedge l_{11})-b_{1,2}\right)+\tilde{u}\left(\sum_{i=1}^{n}x_i-(x_2\wedge l_{12})-b_{1,2}\right)\nonumber\\ & \quad \geq \tilde{u}\left(\sum_{i=1}^{n}x_i-(x_1\wedge l_{21})-b_{1,2}\right)+\tilde{u}\left(\sum_{i=1}^{n}x_i-(x_2\wedge l_{22})-b_{1,2}\right). \end{align*}

By using the double expectation formula, we have $\Delta _{2}\geq 0$, and this in turn implies $\eta (\bf {l}_1)\geq \eta (\bf {l}_2)$. Hence, the proof is finished.

Under the setting of Theorem 4.6, it is not hard to see that the worst allocation of policy limits can be determined as $(l,0,\ldots,0)$, as summarized in the following corollary.

The following example illustrates the results of Theorem 4.6 and Corollary 4.7.

Example 4.8. Under the setup of Example 4.1, we consider $(X_{1},X_{2})=(-\lambda _1^{-1}\log {U},-\lambda _2^{-1}\log {U})$, where $U$ is uniformly distributed on $(0,1)$ and $(\lambda _{1},\lambda _{2})=(0.6,0.2)$. It is easy to check that $(X_{1},X_{2})$ is comonotonic and $X_{1}\leq _{\rm st}X_{2}$. Consider the following four allocation polices $\bf {l}_{11}=(8,0)$, $\bf {l}_{12}=(7,1)$, $\bf {l}_{13}=(5,3)$, $\bf {l}_{14}=(4,4)$ within $\mathcal {D}^2_+$. Let $\bar {d}=6$. We denote

$$U^{I}_{\bf{l}_{r}}=X_{1}-I_{1}(X_{1}\wedge l_{r,1})+X_{2}-I_{2}(X_{2}\wedge l_{r,2}),$$

where $\bf {l}_{r}=(l_{r,1},l_{r,2})$ for $r=11,12,13,14$. It is plain that $\bf {l}_{11} \stackrel {{\rm m}}{\succeq } \bf {l}_{12}\stackrel {{\rm m}}{\succeq } \bf {l}_{13}\stackrel {{\rm m}}{\succeq } \bf {l}_{14}$. The values of $\hat {\Delta }^{I}_{\bf {l}_{11}}(\alpha _{j})$, $\hat {\Delta }^{I}_{\bf {l}_{12}}(\alpha _{j})$, $\hat {\Delta }^{I}_{\bf {l}_{13}}(\alpha _{j})$, and $\hat {\Delta }^{I}_{\bf {l}_{14}}(\alpha _{j})$ are plotted for different values of $\alpha _{j}$, for $j=1,2,\ldots,20$. As displayed in Figure 7, it holds that $\hat {\Delta }^{I}_{\bf {l}_{11}}(\alpha )\geq \hat {\Delta }^{I}_{\bf {l}_{12}}(\alpha )\geq \hat {\Delta }^{I}_{\bf {l}_{13}}(\alpha )\geq \hat {\Delta }^{I}_{\bf {l}_{14}}(\alpha )$ for $\alpha \in [0,1]$. Therefore, the results of Theorem 4.6 and Corollary 4.7 are validated by applying the strong law of large numbers.

Figure 7. Plots of $\hat {\Delta }^{I}_{\bf {l}_{r}}(\alpha _{j})$ for $j=1,2,\ldots,20$ and $r=11,12,13,14$.