2. Preliminaries

We first give a brief overview of some preliminaries, namely, stochastic orders and majorization orders.

Let $U\subseteq \mathbb{R}$ denote a subset of the real line. Further, let, for any vector ${\bf \vartheta}=(\vartheta_1,\vartheta_2,\ldots,\vartheta_n)\in \mathbb{R}^n, \vartheta_{(1)}\leq \vartheta_{(2)}\leq \ldots \leq \vartheta_{(n)}$ denote the increasing arrangement of components in ϑ.

It is to be noted that ${\bf \vartheta}\;\stackrel{w}{\succeq}\;{\bf \eta}~ ({\bf \vartheta}\;\succeq_{w} \;{\bf \eta}) \implies {\bf \vartheta}\stackrel{m}{\succeq}{\bf \eta}$, but the reverse is not true. Later, we give some definitions related to multivariate majorization (Marshall et al. [Reference Marshall, Olkin and Arnold21], Chap. 15).

Next we give the definitions of some stochastic orders (see [Reference Shaked and Shanthikumar26]).

In what follows, we introduce a notation. Let

\begin{eqnarray*} \mathcal{U}_n&=&\left\{({\bf \vartheta},{\bf \eta})=\left( \begin{array}{ccc} \vartheta_1 & \cdots & \vartheta_n \\ \eta_1 & \cdots & \eta_n \\ \end{array} \right): \vartheta_i\gt0, \eta_i\gt0,\quad \text{and} \right. \\ & & \left. (\vartheta_i-\vartheta_j)(\eta_i-\eta_j)\geq 0, \ \forall~ i,j=1,\ldots,n \right\}. \end{eqnarray*}

Lemma 2.4. [Reference Kundu, Chowdhury, Nanda and Hazra17, Reference Marshall, Olkin and Arnold21]

Let $\varphi:\mathcal{E}\rightarrow \mathbb{R}$ is continuously differentiable on the interior of $\mathcal{E}$. Then, for ${\bf \vartheta},{\bf \eta}\in \mathcal{E}$,

\begin{eqnarray*} {\bf \vartheta}\stackrel{m}\succeq{\bf \eta}\;\implies\;\varphi({\bf \vartheta})\geq (\text{resp.}\;\leq)\;\varphi({\bf \eta}) \end{eqnarray*}

iff $\varphi_{(k)}(\mathbf{z})\;\text{is increasing (respectively, decreasing) in}\;k=1,\dots,n,$ where $\varphi_{(k)}= \partial\varphi({\mathbf{z}})/\partial z_k$ denotes the partial derivative of φ with respect to its kth argument.

Lemma 2.5. [Reference Marshall, Olkin and Arnold21]

Let $\varphi: S\rightarrow \mathbb{R}$ be a function, $S \subseteq \mathbb{R}^n$. Then, for ${\bf \vartheta},{\bf \eta}\in S,$

\begin{equation*} {\bf \vartheta}\succeq_{w} {\bf \eta}\; \implies\;\varphi({\bf \vartheta})\geq (\text{resp. }\leq)\; \varphi({\bf \eta})\end{equation*}

iff φ is increasing (respectively, decreasing) and Schur-convex (respectively, Schur-concave) on S. Similarly,

\begin{equation*}{\bf \vartheta}\stackrel{w}\succeq {\bf \eta}\; \implies\;\varphi({\bf \vartheta})\geq (\text{resp. }\leq)\; \varphi({\bf \eta})\end{equation*}

iff φ is decreasing (respectively, increasing) and Schur-convex (respectively, Schur-concave) on S. $\Box$

Lemma 2.6. [Reference Haidari, Najafabadi and Balakrishnan14, Reference Marshall, Olkin and Arnold21]

Let $\varphi:\mathcal{D}(\mathcal{E})\rightarrow \mathbb{R}$ be a continuous function and continuously differentiable on the interior of $\mathcal{D} (\mathcal{E})$. Then

\begin{equation*}\varphi({\bf \vartheta})\geq \varphi({\bf \eta})~ whenever~ {\bf \vartheta}\succeq_{w} {\bf \eta} ~ on~\mathcal{D}~(\mathcal{E})\end{equation*}

iff $\varphi_{(k)}(z)$ is a nonnegative decreasing (increasing) function in k for all z in the interior of $\mathcal{D} (\mathcal{E})$. Similarly,

\begin{equation*}\varphi({\bf \vartheta})\geq \varphi({\bf \eta})~ whenever~ {\bf \vartheta}\stackrel{w}{\succeq} {\bf \eta} ~ on~\mathcal{D}~(\mathcal{E})\end{equation*}

iff $\varphi_{(k)}(z)$ is a nonpositive decreasing (increasing) function of k for all z in the interior of $\mathcal{D} (\mathcal{E})$.

We conclude this section by giving the following useful definition (for details, see [Reference Nelsen22]).

3. Stochastic comparisons of largest claim amounts

In this section, we derive some stochastic comparison results for largest claim amounts of two different portfolios of risks. Unless otherwise specified, we assume that $X=(X_1,\ldots,X_n)$ and $Y=(Y_1,\ldots,Y_n)$ are two sets of independent r.v.’s.

Assume that $I_{p_i}$, $i=1,\ldots,n$, are independent Bernoulli r.v.’s, independent of X _i’s, with $E(I_{p_i})=p_i$. Denote multivariate Bernoulli random vector $\bf I=\left(I_{p_1},\ldots,I_{p_n}\right)$. Let $X_i^{*}=X_i I_{p_i}$, $i=1,\ldots,n$, and denote $X_{n:n}^*=\max(X_1^{*},\ldots,X_n^{*})$. Then $X_{n:n}^*$ corresponds to the largest claim amount in a portfolio of risks, where X _i’s represent random claim amount that can be made by a policy in an insurance period, and $I_{p_i}$’s indicate the occurrence of these claims. Further, suppose odds function of each X _i in X is proportional to that of a baseline r.v. with proportionality constant (odds ratio) $\alpha_i\gt0$, that is, $X_i\sim {\rm PO}(\bar{F},\alpha_i)$, $i=1,\ldots,n$, where $\bar{F}$ denotes the survival function of the baseline r.v. Let us denote $X_{n:n}^{\circ}=\max(X_1^{\circ},\ldots,X_n^{\circ})$, where $X_i^{\circ}=X_i I_{q_i}$ and $I_{q_i}$, $i=1,\ldots,n$, are independent Bernoulli r.v.’s, independent of X _i’s, with $E(I_{q_i})=q_i$.

Similarly suppose $Y_i\sim {\rm PO}(\bar{F},\beta_i)$, $\beta_i\gt0$, $i=1,\ldots,n$. Denote $Y_{n:n}^*=\max(Y_1^{*},\ldots,Y_n^{*})$, where $Y_i^{*}=Y_i I_{p_i}$, and $Y_{n:n}^{\circ}=\max(Y_1^{\circ},\ldots,Y_n^{\circ})$, where $Y_i^{\circ}=Y_i I_{q_i}$, $i=1,\ldots,n$.

Theorem 3.1 established that more heterogeneity among the odds of claim in terms of the weakly supermajorization order results in less largest claim amount in the sense of the usual stochastic orders when both portfolios have common occurrence of claim $\textbf{p}.$ By the symbol $a\stackrel{sign}{=} b$, we mean that a and b have the same sign.

Theorem 3.1. Let $\kappa:[0,1]\rightarrow R_{+}$ be a differentiable and strictly decreasing function. Then, for $(\kappa(\textbf{p}),\bf\alpha) \in \mathcal{U}_n$ and $(\kappa(\textbf{p}),\bf\beta) \in \mathcal{U}_n$,

\begin{equation*}\bf\alpha \stackrel{w}{\succeq} \bf\beta \implies X^{*}_{n:n} \leq_{st} Y^{*}_{n:n}.\end{equation*}

Proof. We have $F_{X^{*}_{n:n}}(x)= \prod_{i=1}^{n} (1-\kappa^{-1}(u_i)\bar{F}_{X_i}(x)) $, where $\bar{F}_{X_i}(x)=\frac{\alpha_i \bar{F}(x)}{1-\bar{\alpha}_i\bar{F}(x)}$, and $u_i=\kappa(p_i)$, $i=1,2,\ldots,n$. Note that $\bar{F}_{X_i}$ is increasing and concave in α _i. Now,

\begin{eqnarray*} \frac{\partial F_{X^{*}_{n:n}}(x)}{\partial \alpha_i}=\frac{-\kappa^{-1}(u_i)\frac{\partial\bar{F}_{X_i}}{\partial\alpha_i}}{1-\kappa^{-1}(u_i)\bar{F}_{X_i}(x)}F_{X^*_{n:n}}(x) &=& -\frac{\kappa^{-1}(u_i)\frac{\bar{F}(x)F(x)}{(1-\bar{\alpha}_i\bar{F}(x))^2}}{1-\kappa^{-1}(u_i)\frac{\alpha_i \bar{F}(x)}{1-\bar{\alpha}_i\bar{F}(x)}}F_{X^*_{n:n}}(x) \\ &=& -\frac{\kappa^{-1}(u_i) \bar{F}(x)F(x)}{(1-\bar{\alpha}_i\bar{F}(x))^2-\kappa^{-1}(u_i)\alpha_i(1-\bar{\alpha}_i\bar{F}(x))\bar{F}(x)}F_{X^*_{n:n}}(x)\\ &=& -g(z_i,\alpha_i)\bar{F}(x)F(x)F_{X^*_{n:n}}(x)~(say), \end{eqnarray*}

where $z_i=\kappa^{-1}(u_i)$. Again,

\begin{eqnarray*} \frac{\partial g}{\partial \alpha_i}&\stackrel{\rm sign}{=}& -\kappa^{-1}(u_i)\bar{F}(x)\left[(2-\kappa^{-1}(u_i))(1-\bar{F}(x))+2(1-\kappa^{-1}(u_i))\alpha_i \bar{F}(x)\right]\leq 0. \end{eqnarray*}

So $g(z_i,\alpha_i)$ is decreasing in α _i. Further,

\begin{eqnarray*}\frac{\partial g}{\partial z_i} \stackrel{\rm sign}{=}(1-\bar{\alpha_i}\bar{F}(x))^2 \geq 0, \end{eqnarray*}

so $g(z_i,\alpha_i)$ is increasing in $z_i = \kappa^{-1}(u_i)$, and so it is decreasing in u _i as $z_i = \kappa^{-1}(u_i)$ is decreasing in u _i. Without loss of generality, we assume that $\alpha_1 \geq \alpha_2 \geq \cdots \geq \alpha_n$ so that $(\bf\kappa(\textbf{p}),\bf\alpha) \in \mathcal{U}_n$ implies $\kappa(p_1)\geq \kappa(p_2) \geq \cdots \geq \kappa(p_n)$. Now for any pair $i,j$ such that $1\leq i\lt j \leq n$, we have $\alpha_i \geq \alpha_j$ and $u_i \geq u_j $. Thus, we have

(3)\begin{eqnarray} g(z_i,\alpha_i)\leq g(z_j,\alpha_i)\leq g(z_j,\alpha_j) \implies \frac{\partial F_{X^{*}_{n:n}}(x)}{\partial \alpha_j}\leq \frac{\partial F_{X^{*}_{n:n}}(x)}{\partial \alpha_i}\leq 0. \end{eqnarray}

So from Lemma 2.6, we get

\begin{equation*}\bf\alpha\stackrel{w}{\succeq} \bf\beta \implies X^{*}_{n:n} \leq_{st} Y^{*}_{n:n}. \end{equation*}

The following example demonstrates Theorem 3.1.

Example 3.2. Suppose that $\{X_1,X_2,X_3,X_4\}$ and $\{Y_1,Y_2,Y_3,Y_4\}$ are two sets of independent nonnegative r.v.’s with $X_i \sim {\rm PO}(\bar{F}(x),\alpha_i)$ and $Y_i \sim {\rm PO}(\bar{F}(x),\beta_i),\;i=1,2,3,4$, where $\bar{F}(x)={\rm e}^{-(0.5x)^{2}},\;x\gt0.$ Further, suppose that $\{I_{p_1},I_{p_2}, I_{p_3},I_{p_4}\}$ is a set of Bernoulli r.v.’s, independent of X _i’s and Y _i’s. Set $(\alpha_1,\alpha_2,\alpha_3,\alpha_4)=(0.2,0.6,1.5,2.8)$, $(\beta_1,\beta_2,\beta_3,\beta_4)=(0.5,0.8,2.5,4.8)$, $(p_1,p_2,p_3,p_4)=(0.95,0.65,0.5,0.35)$, and $\kappa(p)=1/p^2$, satisfying all the conditions of Theorem 3.1. We consider the transformation $x=t/(1-t)$ so that, for $t\in[0,1)$, we have $x\in[0,\infty)$. Then denote the distribution functions of $X^*_{n:n}$ and $Y^*_{n:n}$ by $F_{X^*_{n:n}}(t/(1-t))=\zeta_1(t)$ and $F_{Y^{*}_{n:n}}(t/(1-t))=\zeta_2(t)$. Figure 1 shows that $\zeta_1(t)\geq \zeta_2(t),\text{for all } t\in[0,1)$, and hence, $X^*_{n:n}\leq_{st} Y^*_{n:n}$.

Figure 1.

Plots of $\zeta_1(t)$ and $\zeta_2(t)$, $t\in[0,1]$.

Next we provide a counterexample to show that the stochastic ordering result in Theorem 3.1 may not hold if we relax the weakly supermajorize condition under a strictly decreasing function.

Counterexample 3.3.

In Example 3.2, let us take $(\alpha_1,\alpha_2,\alpha_3,\alpha_4)=(0.2,0.9,1.5,4.5)$, $(\beta_1,\beta_2,\beta_3,\beta_4)=(0.35,0.4,2.9,3.8)$ so that $\bf\alpha \stackrel{w}{\nsucceq} \bf\beta$. In Figure 2, we have plotted $\zeta_1(t)-\zeta_2(t)$ for all $t\in[0,1)$, from which it is clear that stochastic ordering result of Theorem 3.1 does not hold.

Figure 2.

Plot of $\zeta_1(t)-\zeta_2(t)$, $t\in[0,1]$.

Theorem 3.4 establishes that largest claim amounts of two portfolios of risks might be increased in terms of the usual stochastic order with the increased heterogeneity among the probabilities of occurrence of claims when both the portfolio of risks have common odds of claim vector $\bf\alpha.$

Theorem 3.4. Let $\kappa:[0,1]\rightarrow R_{+}$ be a differentiable and strictly increasing concave function. Then, for $(\bf\kappa(\textbf{p}),\bf\alpha) \in \mathcal{U}_n$ and $(\bf\kappa(\textbf{q}),\bf\alpha) \in \mathcal{U}_n$,

\begin{equation*}(\kappa(p_1),\kappa(p_2),\ldots,\kappa(p_n))\succeq _{w} (\kappa(q_1),\kappa(q_2),\ldots,\kappa(q_n)) \implies X^{*}_{n:n} \geq_{st} X^{\circ}_{n:n}.\end{equation*}

Proof. Here $F_{X^{*}_{n:n}}(x)= \prod_{i=1}^{n} (1-\kappa^{-1}(u_i)\bar{F}_{X_i}(x)) $, where $\bar{F}_{X_i}(x)=\frac{\alpha_i \bar{F}(x)}{1-\bar{\alpha}_i\bar{F}(x)}$. It is to be noted that $\bar{F}_{X_i}(x)$ is increasing in α _i. Also, κ ⁻¹ is strictly increasing and convex. Let $\phi(\textbf{u})=-F_{X^{*}_{n:n}}(x)$. We have

\begin{equation*}\frac{\partial \phi(\textbf{u})}{\partial u_i} = \frac{\bar{F}_{X_i}(x)\frac{{\rm d}\kappa^{-1}(u_i)}{{\rm d}u_i}}{1-\kappa^{-1}(u_i)\bar{F}_{X_i}(x)}F_{X^*_{n:n}}(x)\geq 0\end{equation*}

according to $\kappa^{-1}(.)$ is increasing.

Let $\ell(\tau_i,u_i)= \frac{\tau_i\frac{{\rm d}\kappa^{-1}(u_i)}{{\rm d}u_i}}{1-\kappa^{-1}(u_i)\tau_i}$, where $\tau_i=\bar{F}_{X_i}(x)$. Then

\begin{equation*}\frac{\partial \ell}{\partial u_i}\stackrel{\rm sign}{=}\tau_i (1-\tau_i \kappa^{-1}(u_i)) \frac{{\rm d}^2\kappa^{-1}(u_i)}{{\rm d}u_{i}^2}+\tau_i^2 \left(\frac{{\rm d}\kappa^{-1}(u_i)}{{\rm d}u_i}\right)^2 \geq 0,\end{equation*}

which holds as $\kappa^{-1}(u_i)$ is convex. So, $\ell(\tau_i,u_i)$ is increasing in u _i. Further,

\begin{equation*}\frac{\partial \ell}{\partial \tau_i}\stackrel{\rm sign}{=}(1-\tau_i \kappa^{-1}(u_i)) \frac{{\rm d}\kappa^{-1}(u_i)}{{\rm d}u_i}-\tau_i \frac{{\rm d}\kappa^{-1}(u_i)}{{\rm d}u_i}(-\kappa^{-1}(u_i)) = \frac{{\rm d}\kappa^{-1}(u_i)}{{\rm d}u_i}\geq 0, \end{equation*}

since $\kappa^{-1}(u_i)$ is increasing in u _i. Then $\ell(\tau_i,u_i)$ is increasing in τ _i (i.e., in $\bar{F}_{X_i}(x)$), so that it is increasing in α _i as $\bar{F}_{X_i}(x)$ is increasing in α _i.

Without loss of generality, we assume that $\alpha_1 \geq \alpha_2 \geq \cdots \geq \alpha_n$, so that $(\bf\kappa(\textbf{p}),\bf\alpha) \in \mathcal{U}_n$ implies $\kappa(p_1)\geq \kappa(p_2) \geq \cdots \geq \kappa(p_n)$. Now for any pair $i,j$ with $1\leq i\lt j \leq n$, we have $\alpha_i \geq \alpha_j$ and $u_i \geq u_j $. Then, if $\kappa^{-1}(u_i)$ is increasing and convex in u _i, we have

(4)\begin{eqnarray} &&\nonumber \ell(\tau_i,u_i) \geq \ell(\tau_j,u_i) \geq \ell(\tau_j,u_j), \end{eqnarray}

\begin{eqnarray} && \text{i.e.} ~ \frac{\partial \phi(\textbf{u})}{\partial u_i}\geq \frac{\partial \phi(\textbf{u})}{\partial u_j}\geq 0. \end{eqnarray}

So from Lemma 2.6, we have

\begin{equation*}(\kappa(p_1),\kappa(p_2),\ldots,\kappa(p_n))\succeq _{w} (\kappa(q_1),\kappa(q_2),\ldots,\kappa(q_n)) \implies X^{*}_{n:n} \geq_{\rm st} X^{\circ}_{n:n}.\end{equation*}

We illustrate Theorem 3.4 with the following example.

Example 3.5. Suppose that $\{X_1,X_2,X_3,X_4\}$ is a set of independent nonnegative r.v.’s with $X_i\sim {\rm PO}(\bar{F}(x),\alpha_i),i=1,2,3,4$, where $\bar{F}(x)={\rm e}^{-(x/2)^{1.5}},\;x\gt0.$ Set $\bf\alpha=(\alpha_1,\alpha_2,\alpha_3,\alpha_4)=(0.9,1.36,2.55,3.5)$, $(p_1,p_2,p_3,p_4)=(0.35,0.5,0.8,0.9),$ $(q_1,q_2,q_3,q_4)=(0.2,0.4,0.65,0.8) $, and $\kappa(p) = p/(1+p)$, satisfying all the conditions of Theorem 3.4. We consider the transformation $x=t/(1-t)$ so that, for $t\in[0,1)$, we have $x\in[0,\infty).$ After this substitution, let us denote the respective distribution functions of $X^*_{n:n}$ and $X^{\circ}_{n:n}$ by $F_{X^*_{n:n}}(t/(1-t))=\xi_1(t)$ and $F_{X^{\circ}_{n:n}}(t/(1-t))=\xi_2(t)$. From Figure 3, it is clear that $\xi_1(t)\leq \xi_2(t),~\forall t\in[0,1)$, and hence, $X^*_{n:n}\geq_{st}X^{\circ}_{n:n}$.

Figure 3.

Plots of $\xi_1(t)$ and $\xi_2(t)$, $t\in[0,1].$

Next we provide a counterexample to show that the stochastic ordering result in Theorem 3.4 may not hold if we relax the weakly submajorize condition under an increasing concave function.

Counterexample 3.6.

In Example 3.5, let us take $(p_1,p_2,p_3,p_4)=(0.1,0.2,0.85,0.95)$ and $(q_1,q_2,q_3,q_4)=(0.5,0.55,0.75,0.8)$ so that $(\kappa(p_1),\kappa(p_2),\kappa(p_3),\kappa(p_4))\nsucceq _{w} (\kappa(q_1),\kappa(q_2),\kappa(q_3),\kappa(q_4))$. In Figure 4, we have plotted $\xi_1(t)-\xi_2(t)~\forall t\in[0,1)$, from which it is clear that none of these distribution functions dominate each other.

Figure 4.

Plot of $\xi_1(t)-\xi_2(t)$, $t\in[0,1]$

Cai and Wei [Reference Cai and Wei10] proposed some multivariate dependence notions based on stochastic arrangement increasing (SAI) notion, including weakly SAI through left tail probability (LWSAI), to model multivariate dependent risks. Since then it has been applied in finance and actuarial science to model dependent stochastic returns and risks [Reference Cai and Wei10, Reference Zhang, Li and Cheung29, Reference Zhang, Zhao and Cheung30]. For a random vector $\bf X=(X_1,X_2,\ldots,X_n)$, one of the ways to define or describe its dependence notion is to characterize the expectations of the transformations of the random vector [Reference Cai and Wei9]. For any (i, j) with $1\leq i\lt j\leq n$, denote

\begin{equation*}\mathcal{G}^{i,j}_{\text{LWSAI}}(n)= \left\{g(\bf x): g(\bf x)- g(\pi_{ij}(\bf x)) ~\text{is}~ \text{decreasing}~ \text{in} ~x_i\leq x_j\right\},\end{equation*}

where $\pi_{i,j}$ is the special permutation of transposition defined as $\pi_{i,j}(\bf x)=(x_1,\ldots,x_j,\ldots,x_i,\ldots,x_n)$. A random vector $\bf X=(X_1,X_2,\ldots,X_n)$ or its distribution is said to be the LWSAI [Reference Cai and Wei10] if $\mathbb{E}[g(\bf X)]\geq \mathbb{E}[g(\tau_{i,j}(\bf X))]\text{for any } g(\bf x)\in \mathcal{G}^{i,j}_{LWSAI}(n)$ and any $1\leq i \lt j\leq n$.

Next, we present a stochastic ordering result when the occurrence probabilities are interdependent in terms of LWSAI. Let us denote $S_k = \{\bf\chi| \chi_i =0~ \text{or}~ 1,~ i=1,2,\ldots,n,~ \chi_1+\cdots+\chi_n = k\}$, $k=0,\ldots,n$, and $S_{k}^{i,j}(\eta_i,\eta_j) =\{\bf\chi \in S_k| \chi_i = \eta_i, \chi_j = \eta_j,~\eta_i,\eta_j\in \{0,1\}\}$, for any $1\leq i \neq j \leq n$, $k=1,\ldots,n-1$. Then $S_k =\bigcup_{\eta_i,\eta_j\in \{0,1\}}S_{k}^{i,j}(\eta_i,\eta_j)$. Also denote $p(\bf\chi)=\mathbb{P}(\bf I=\bf\chi)=\mathbb{P}\left(I_{p_1}=\chi_1,\ldots,I_{p_n}=\chi_n\right)$.

Theorem 3.8. Suppose that $\bf I=\left(I_{p_1},\ldots,I_{p_n}\right)$ is LWSAI. If

\begin{equation*}\bf\alpha \stackrel{m}{\succeq} \bf\beta ~\text{such that}~ \alpha_1\leq \alpha_2\leq \cdots \leq \alpha_n ~\text{and}~ \beta_1\leq \beta_2\leq \cdots \leq \beta_n,\end{equation*}

then $X^*_{n:n}\leq_{st} Y^*_{n:n}$.

Proof. From Theorem 4.1 of Kundu et al. [Reference Kundu, Hazra and Nanda18], it follows that $\bf\alpha \stackrel{m}{\succeq} \bf\beta \implies X_{n:n}\leq_{st} Y_{n:n}$, that is, $F_{X_{n:n}}(t)\geq F_{Y_{n:n}}(t)$ for all t, where $X_{n:n} = \max(X_1,X_2,\ldots,X_n)$. We desire to show that $F_{X^*_{n:n}}(t)\geq F_{Y^*_{n:n}}(t)$ for all $t\in \Re_{+}$. By the nature of majorization order, it suffices to prove the result when $(\alpha_i,\alpha_j)\stackrel{m}\succeq (\beta_i,\beta_j)$ for some pair $1\leq i\lt j\leq n$ and $\alpha_r=\beta_r$ for all $r\neq i,j$. Now, we have

\begin{eqnarray*} F_{X^*_{n:n}}(t)&=& \mathbb{P}\left(\max\{I_{p_1}X_1,\ldots,I_{p_n}X_n\}\leq t\right)\\ &=& \sum_{k=0}^{n} \sum_{\chi\in S_k} \mathbb{P}\left(\max\{I_{p_1}X_1,\ldots,I_{p_n}X_n\}\leq t|\bf I=\bf \chi\right)p(\bf \chi)\\ &=& p(\bf 0)+p(\bf 1)\mathbb{P}(\max\{X_1,\ldots,X_n\}\leq t)+ \sum_{k=1}^{n-1}\sum_{\bf \chi\in S_k} p(\bf\chi) \mathbb{P}(\max\{\chi_1 X_1,...,\chi_nX_n\}\leq t)\\ &=& p(\bf 0)+p(\bf 1)F_{X_{n:n}}(t)+ \sum_{k=1}^{n-1}\left\{\sum_{\bf \chi\in S^{i,j}_k(0,0)} p(\bf\chi)\prod_{r\neq i,j}^{n}\mathbb{P}(\chi_rX_r\leq t)\right.\\ & &\left. +\sum_{\bf \chi\in S^{i,j}_k(0,1)} p(\bf\chi)F_{X_j}(t)\prod_{r\neq i,j}^{n}\mathbb{P}(\chi_rX_r\leq t)+\sum_{\bf \chi\in S^{i,j}_k(0,1)}p(\tau_{i,j}(\bf\chi))F_{X_i}(t)\prod_{r\neq i,j}^{n}\mathbb{P}(\chi_rX_r\leq t)\right.\\ &+&\left. \sum_{\bf \chi\in S^{i,j}_k(1,1)} p(\bf\chi)F_{X_i}(t)F_{X_j}(t)\prod_{r\neq i,j}^{n}\mathbb{P}(\chi_rX_r\leq t)\right\}. \end{eqnarray*}

Similarly,

\begin{eqnarray*} F_{Y^*_{n:n}}(t)&=& p(\bf 0)+p(\bf 1)F_{Y_{n:n}}(t)+ \sum_{k=1}^{n-1}\left\{\sum_{\bf \chi\in S^{i,j}_k(0,0)} p(\bf\chi)\prod_{r\neq i,j}^{n}\mathbb{P}(\chi_rY_r\leq t)\right.\\ & &\left. +\sum_{\bf \chi\in S^{i,j}_k(0,1)} p(\bf\chi)F_{Y_j}(t)\prod_{r\neq i,j}^{n}\mathbb{P}(\chi_rY_r\leq t)+\sum_{\bf \chi\in S^{i,j}_k(0,1)}p(\tau_{i,j}(\bf\chi))F_{Y_i}(t)\prod_{r\neq i,j}^{n}\mathbb{P}(\chi_rY_r\leq t)\right.\\ &+&\left. \sum_{\bf \chi\in S^{i,j}_k(1,1)} p(\bf\chi)F_{Y_i}(t)F_{Y_j}(t)\prod_{r\neq i,j}^{n}\mathbb{P}(\chi_rY_r\leq t)\right\}. \end{eqnarray*}

Upon using the condition that $\mathbb{P}(\chi_rX_r\leq t)=\mathbb{P}(\chi_rY_r\leq t) $ for all $r\neq i,j$, we have

\begin{eqnarray*} F_{X^{\star}_{n:n}}(t)-F_{Y^*_{n:n}}(t)&=& p(\bf 1)[F_{X_{n:n}}(t)-F_{Y_{n:n}}(t)] \\ & & \sum_{k=1}^{n-1}\left\{\sum_{\bf \chi\in S^{i,j}_k(0,1)}p(\bf \chi)[F_{X_j}(t)-F_{Y_j}(t)]\prod_{r\neq i,j}^{n}\mathbb{P}(\chi_rX_r\leq t)+\right.\\ & &\left. \sum_{\bf \chi\in S^{i,j}_{k}(0,1)}p(\tau_{i,j}(\bf \chi))[F_{X_i}(t)-F_{Y_i}(t)]\prod_{r\neq i,j}^{n}\mathbb{P}(\chi_rX_r\leq t)+\right.\\ & &\left. \sum_{\bf \chi\in S^{i,j}_k(1,1)}p(\bf \chi)[F_{X_i}(t)F_{X_j}(t)-F_{Y_i}(t)F_{Y_j}(t)]\prod_{r\neq i,j}^{n}\mathbb{P}(\chi_rX_r\leq t)\right\} \\ &\geq & \sum_{k=1}^{n-1}\left\{\sum_{\bf \chi\in S^{i,j}_k(0,1)}p(\tau_{i,j}(\bf \chi))[F_{X_j}(t)-F_{Y_j}(t)]\prod_{r\neq i,j}^{n}\mathbb{P}(\chi_rX_r\leq t)+\right.\\ & &\left. \sum_{\bf \chi\in S^{i,j}_{k}(0,1)}p(\tau_{i,j}(\bf \chi))[F_{X_i}(t)-F_{Y_i}(t)]\prod_{r\neq i,j}^{n}\mathbb{P}(\chi_rX_r\leq t)+\right.\\ & &\left. \sum_{\bf \chi\in S^{i,j}_k(1,1)}p(\bf \chi)[F_{X_i}(t)F_{X_j}(t)-F_{Y_i}(t)F_{Y_j}(t)]\prod_{r\neq i,j}^{n}\mathbb{P}(\chi_rX_r\leq t)\right\} \\ &=& \sum_{k=1}^{n-1}\left\{\sum_{\bf \chi\in S^{i,j}_k(0,1)}p(\tau_{i,j}(\bf \chi))[F_{X_i}(t)+F_{X_j}(t)-F_{Y_i}(t)-F_{Y_j}(t)]\prod_{r\neq i,j}^{n}\mathbb{P}(\chi_rX_r\leq t)+\right.\\ & &\left. \sum_{\bf \chi\in S^{i,j}_k(1,1)}p(\bf \chi)[F_{X_i}(t)F_{X_j}(t)-F_{Y_i}(t)F_{Y_j}(t)]\prod_{r\neq i,j}^{n}\mathbb{P}(\chi_rX_r\leq t) \right\} \\ &\geq& 0, \end{eqnarray*}

where the first inequality follows from the fact that $F_{X_{n:n}}(t)\geq F_{Y_{n:n}}(t)$, Lemma 3.7, and the fact that $F_{X_i}(t)=\frac{F(t)}{1-\bar{\alpha}_i\bar{F}(t)}$ is decreasing in α _i. For the last inequality, we have the following explanation. Since $F_{X_i}(t)$ is convex in α _i, it follows that $F_{X_i}(t)+F_{X_j}(t)\geq F_{Y_i}(t)+F_{Y_j}(t).$ Let

\begin{equation*}\phi(\alpha_i,\alpha_j)=F_{X_i}(t)F_{X_j}(t)=\frac{F^2(t)}{(1-\bar{\alpha}_i\bar{F}(t))(1-\bar{\alpha}_j\bar{F}(t))}.\end{equation*}

Then, for $1\leq i\lt j\leq n$,

\begin{equation*}\frac{\partial \phi}{\partial \alpha_i}-\frac{\partial \phi}{\partial \alpha_j}=\frac{(\alpha_i-\alpha_j)\bar{F}(x)}{(1-\bar{\alpha}_i\bar{F}(t))^2(1-\bar{\alpha}_j\bar{F}(t))^2}\leq 0.\end{equation*}

So, from Lemma 2.4, we get that $(\alpha_i,\alpha_j)\stackrel{m}\succeq (\beta_i,\beta_j)\Rightarrow \phi(\alpha_i,\alpha_j)\geq \phi(\beta_i,\beta_j)$, and thus the proof is completed.

We illustrate Theorem 3.8 with the following example.

Example 3.10. Suppose that $\{X_1,X_2\}$ and $\{Y_1,Y_2\}$ are two sets of independent nonnegative r.v.’s with $X_i \sim {\rm PO}(\bar{F}(x),\alpha_i),i=1,2,$ and $Y_i \sim {\rm PO}(\bar{F}(x),\beta_i),i=1,2,$ where $\bar{F}(x)={\rm e}^{-(0.08x)^{0.08}},\;x\gt0.$ Set $(\alpha_1,\alpha_2)=(0.55,1.45),\ (\beta_1,\beta_2)=(0.65,1.35),\ p(0,0)=P(I_{p_1}=0, I_{p_2}=0)=0.14,\ p(0,1)=0.47,\ p(1,0)=0.25,\ {\rm and} p(1,1)=0.14.$ Then $I=\{I_{p_1},I_{p_2}\}$ is LWSAI. We consider the transformation $x=t/(1-t)$ so that for $t\in[0,1)$, we have $x\in[0,\infty).$ After this substitution, let us denote the respective distribution functions of $X^*_{n:n}$ and $Y^*_{n:n}$ by $F_{X^*_{n:n}}(t/(1-t))=\varphi_1(t)$ and $F_{Y^{*}_{n:n}}(t/(1-t))=\varphi_2(t)$. Figure 5 shows that $\varphi_1(t)\geq \varphi_2(t)\ \text{for all } t\in[0,1)$. Hence, $X^*_{n:n}\leq_{st} Y^*_{n:n}$. $\Box$

Figure 5.

Plots of $\varphi_1(t)$ and $\varphi_2(t)$, $t\in[0,1]$.

Theorems 3.11–3.13 compare the largest claim amounts of two interdependent heterogeneous portfolios of risks where the joint distribution functions are modeled using copulas.

Theorem 3.11. Let $X_{1},X_{2},\ldots,X_{n}\; (Y_{1},Y_{2},\ldots,Y_{n})~$ be nonnegative r.v.’s with $X_{i}\sim {\rm PO}(\bar{F},\alpha_i)\;(Y_{i}\sim {\rm PO}(\bar{F},\beta_i)),\;i=1,2,\ldots,n$, and let the associated copula be $\mathit{C}.$ Further, suppose that $\{\mathit{I}_{p_1},\mathit{I}_{p_2},\ldots,\mathit{I}_{p_n}\}$ is a set of independent Bernoulli r.v.’s, independent of X _i’s $(Y_{i}$’s $)$. Then

\begin{equation*}\alpha_i\leq \beta_i,\forall i=1,2,\ldots,n\implies X^*_{n:n}\leq_{st} Y^*_{n:n}.\end{equation*}

Proof. The distribution function of $X^*_{n:n}$ can be written as

\begin{eqnarray*} G_{X^*_{n:n}}(t)&=& \mathbb{P}(X^*_{1}\leq t,X^*_{2}\leq t,\ldots,X^*_{n}\leq t)\\ &=& \mathbb{P}(\mathit{I}_{p_1}X_{1}\leq t,\ldots,\mathit{I}_{p_n}X_{n}\leq t)\\ &=& \sum_{\bf \chi\in \{0,1\}^n}\mathbb{P}\left(\mathit{I}_{p_1} X_{1}\leq t,\ldots,\mathit{I}_{p_n} X_{n}\leq t| \bf I=\bf \chi \right)p(\bf \chi)\\ &=& \sum_{\bf \chi\in \{0,1\}^n}p(\bf \chi)\mathbb{P}\left(\chi_1 X_{1}\leq t,\ldots,\chi_n X_{n}\leq t\right)\\ &=& \sum_{\bf \chi\in \{0,1\} ^n}p(\bf \chi)\mathit{C}\left([F_{X_1})]^{\chi_1},\ldots,[F_{X_n}]^{\chi_n} \right). \end{eqnarray*}

Since $F_{X_i}(x)=\frac{F(x)}{1-\bar{\alpha}_i \bar{F}(x)}$ is decreasing in α _i and the copula is component-wise increasing, we have that $G_{X^*_{n:n}}(x)$ is decreasing in $\alpha_i,$ for $i=1,2,\ldots,n.$ Hence, the desired result follows.

Theorem 3.12. Let $X_{1},X_{2},\ldots,X_{n}$ be nonnegative r.v.’s with $X_{i}\sim PO(\bar{F},\alpha_i),~i=1,2,\ldots,n$, and let the associated copula be $\mathit{C}\;(\mathit{C}^{\prime}).$ Further, suppose that $\{\mathit{I}_{p_1},\mathit{I}_{p_2},\ldots,\mathit{I}_{p_n}\}$ is a set of independent Bernoulli r.v.’s, independent of X _i’s. Then

\begin{equation*}\mathit{C^\prime}\prec \mathit{C} \implies X^*_{n:n}\leq_{st} X^{*^{\prime}}_{n:n},\end{equation*}

where the r.v.’s $X^*_{n:n}$ and $ X^{*^{\prime}}_{n:n}$ represent the largest claim amount with the associated copula $\mathit{C}\;(\mathit{C}^{\prime}).$

For the next theorem, proof follows from Theorems 3.11 and 3.12 and, hence, omitted.

Theorem 3.13. Let $X_{1},X_{2},\ldots,X_{n} \;(Y_{1},Y_{2},\ldots,Y_{n})~$ be nonnegative r.v.’s with $X_{i}\sim {\rm PO}(\bar{F},\alpha_i)\;(Y_{i}\sim {\rm PO}(\bar{F},\beta_i)),\;i=1,2,\ldots,n$, and let the associated copula be $\mathit{C}\;(\mathit{C}^{\prime}).$ Further, suppose that $\{\mathit{I}_{p_1},\mathit{I}_{p_2},\ldots,\mathit{I}_{p_n}\}$ is a set of independent Bernoulli r.v.’s, independent of X _i’s $(Y_{i}$’s $)$. Then

\begin{equation*}\alpha_i\leq \beta_i,\ \forall i=1,2,\ldots,n ,\ \text{and }\ \alpha_i\leq \beta_i,\ \forall i=1,2,\ldots,n,\ \text{and }\ \mathit{C^\prime}\prec \mathit{C} \implies X^*_{n:n}\leq_{st} Y^*_{n:n}.\end{equation*}

The following example demonstrates the result given in the above theorem.

Example 3.14. Suppose $\{X_1,X_2\}$ and $\{Y_1,Y_2\}$ are two sets of independent nonnegative r.v.’s with $X_i \sim {\rm PO}(\bar{F}(x),\alpha_i),\;i=1,2,\ \text{and }\ Y_i \sim {\rm PO}(\bar{F}(x),\beta_i),\;i=1,2,$ where $\bar{F}(x)={\rm e}^{-(0.05x)^{0.5}},\;x\gt0.$ Set $(\alpha_1,\alpha_2)=(0.5,1.25)$, $(\beta_1,\beta_2)=(0.75,1.55)$, $p(0,0)=0.89,\;p(0,1)=0.06,\;p(1,0)$ $\left.=0.04,\;p(1,1)=0.01\right.$. Here we take $C(x_1,x_2) = {\rm e}^{-\left\{(\log(x_1))^{\theta_1}+(\log(x_2))^{\theta_1}\right\}^{1/\theta_1}}$ and $C^{\prime}(x_1,x_2)={\rm e}^{-\left\{(\log(x_1))^{\theta_2}+(\log(x_2))^{\theta_2}\right\}^{1/\theta_2}}$, where $\theta_1=2$ and $\theta_2=5$. We consider the transformation $x=t/(1-t)$, so that for $t\in[0,1)$, we have $x\in[0,\infty)$. After this substitution, we denote the distribution functions of $X^*_{n:n}$ and $Y^*_{n:n}$ by $F_{X^*_{n:n}}(t/(1-t))=\nu_1(t)$ and $F_{Y^{*}_{n:n}}(t/(1-t))=\nu_2(t)$, respectively. Figure 6 shows that $\nu_1(t)\geq \nu_2(t)\ \text{for all } t\in[0,1)$. Hence, $X^*_{n:n}\leq_{st} Y^*_{n:n}$. $\Box$

Figure 6.

Plots of $\nu_1(t)$ and $\nu_2(t)$, $t\in[0,1]$.

The following theorem compares the largest claim amounts of two sets of heterogeneous portfolios of risks in terms of the reversed hazard rate order. Here we assume that the odds ratios are the same but the probabilities of occurrences of claims are different.

Proof. We have $F_{X^*_{n:n}}(t)= \prod_{i=1}^{n}(1-\kappa^{-1}(u_i)\bar{F}_{\alpha}(t))$, where $\kappa(p_i)=u_i$, $i=1,\ldots,n$. Since $X_i \sim {\rm PO}(\bar F, \alpha)$ for $i=1,\ldots,n.$ We have $\bar F_{\alpha}(x) = \frac{\alpha \bar F(x)}{1- \bar \alpha \bar F(x)}$.

Now $ f_{{X^*_{n:n}}}(t)= \sum_{i=1}^{n}\left(\frac{\kappa^{-1}(u_i)f_{\alpha}(t)}{1-\kappa^{-1}(u_i)\bar{F}_{\alpha}(t)}\right)F_{X^*_{n:n}}(t)$ and therefore

\begin{equation*}\tilde{r}_{X^*_{n:n}}(t)=\sum_{i=1}^{n}\frac{\kappa^{-1}(u_i)f_{\alpha}(t)}{1-\kappa^{-1}(u_i)\bar{F}_{\alpha}(t)}.\end{equation*}

So, we have

\begin{equation*}\frac{\partial \tilde{r}_{X^*_{n:n}}(t)}{\partial u_i}= \frac{\frac{{\rm d} \kappa^{-1}(u_i)}{{\rm d} u_i}f_{\alpha}(t)}{(1-\kappa^{-1}(u_i)\bar{F}_{\alpha}(t))} + \frac{\frac{{\rm d} \kappa^{-1}(u_i)}{{\rm d} u_i}\kappa^{-1}(u_i)f_{\alpha}(t)\bar{F}_{\alpha}(t)}{(1-\kappa^{-1}(u_i)\bar{F}_{\alpha}(t))^2} \end{equation*}

and

\begin{equation*}\frac{\partial \tilde{r}_{X^*_{n:n}}(t)}{\partial u_j}= \frac{\frac{{\rm d} \kappa^{-1}(u_j)}{{\rm d} u_j}f_{\alpha}(t)}{(1-\kappa^{-1}(u_j)\bar{F}_{\alpha}(t))} + \frac{\frac{{\rm d} \kappa^{-1}(u_j)}{{\rm d} u_j}\kappa^{-1}(u_j)f_{\alpha}(t)\bar{F}_{\alpha}(t)}{(1-\kappa^{-1}(u_j)\bar{F}_{\alpha}(t))^2}.\end{equation*}

Now, consider the following two cases.

Case I: Let κ be strictly decreasing and convex. Then κ ⁻¹ is strictly decreasing and convex. Consequently, $\tilde{r}_{X^*_{n:n}}(t)$ is decreasing in u _i. Further,

\begin{eqnarray*} (u_i-u_j)\left(\frac{\partial\tilde{r}_{X^*_{n:n}}(t)}{\partial u_i}-\frac{\partial \tilde{r}_{X^*_{n:n}}(t)}{\partial u_j}\right)&\stackrel{\rm sign}{=}&(u_i-u_j)\left[\left(\frac{\frac{{\rm d} \kappa^{-1}(u_i)}{{\rm d} u_i}}{1-\kappa^{-1}(u_i)\bar{F}_{\alpha}(t)}-\frac{\frac{{\rm d} \kappa^{-1}(u_j)}{{\rm d} u_j}}{1-\kappa^{-1}(u_j)\bar{F}_{\alpha}(t)}\right)\right.+\\ & &\left. \left(\frac{\kappa^{-1}(u_i)\frac{{\rm d} \kappa^{-1}(u_i)}{{|rm d} u_i}}{1-\kappa^{-1}(u_i)\bar{F}_{\alpha}(t)}-\frac{\kappa^{-1}(u_j)\frac{{|rm d} \kappa^{-1}(u_j)}{{\rm d} u_j}}{1-\kappa^{-1}(u_j)\bar{F}_{\alpha}(t)}\right) \right]\geq 0, \end{eqnarray*}

which follows from the fact that κ ⁻¹ is decreasing and convex. Consequently, we have that $\tilde{r}_{X^*_{n:n}}(t)$ is decreasing and Schur-convex in u _i. So, from Lemma 2.5, the first part of the theorem follows.

Case II: Let κ be strictly increasing and concave. Then κ ⁻¹ is strictly increasing and convex. Consequently, $\tilde{r}_{X^*_{n:n}}(t)$ is increasing in u _i. Further,

\begin{eqnarray*} (u_i-u_j)\left(\frac{\partial\tilde{r}_{X^*_{n:n}}(t)}{\partial u_i}-\frac{\partial \tilde{r}_{X^*_{n:n}}(t)}{\partial u_j}\right)&\stackrel{\rm sign}{=}& (u_i-u_j)\left[\left(\frac{\frac{{\rm d} \kappa^{-1}(u_i)}{{\rm d} u_i}}{1-\kappa^{-1}(u_i)\bar{F}_{\alpha}(t)}-\frac{\frac{{\rm d} \kappa^{-1}(u_j)}{{\rm d} u_j}}{1-\kappa^{-1}(u_j)\bar{F}_{\alpha}(t)}\right)\right.+\\ & &\left. \left(\frac{\kappa^{-1}(u_i)\frac{{\rm d} \kappa^{-1}(u_i)}{{\rm d} u_i}}{1-\kappa^{-1}(u_i)\bar{F}_{\alpha}(t)}-\frac{\kappa^{-1}(u_j)\frac{{\rm d} \kappa^{-1}(u_j)}{{\rm d} u_j}}{1-\kappa^{-1}(u_j)\bar{F}_{\alpha}(t)}\right) \right]\geq 0, \end{eqnarray*}

which follows from the fact that κ ⁻¹ is increasing and convex. Consequently, we have that $\tilde{r}_{X^*_{n:n}}(t)$ is increasing and Schur-convex in u _i. So, from Lemma 2.5, the second part of the theorem follows.

We illustrate Theorem 3.15 with the following example.

Example 3.16. Suppose that $\{X_1,X_2,X_3,X_4\}$ is a set of independent nonnegative r.v.’s with $X_i \sim {\rm PO}(\bar{F}(x),\alpha),\ i=1,2,3,4$, where $\bar{F}(x)={\rm e}^{-(0.5x)^{1.5}}, \;x\gt0,\ \text{and } \alpha=0.75.$ Further, suppose that $\{I_{p_1},I_{p_2}, I_{p_3},I_{p_4}\}$ and $\{I_{q_1},I_{q_2}, I_{q_3},I_{q_4}\}$ are two sets of Bernoulli r.v.’s, independent of X _i’s, $i = 1,2,3,4.$ Set $(p_1,p_2,p_3,p_4)=(0.35,0.65,0.85,0.96),\ (q_1,q_2,q_3,q_4)=(0.15,0.35,0.55,0.82)$. Let $\kappa(x)=\log(1+x)$, which is strictly increasing and concave. We consider the transformation $x=t/(1-t)$ so that for $t\in[0,1)$, we have $x\in[0,\infty).$ After this substitution, let us denote the respective reverse hazard functions by $\tilde{r}_{X^*_{n:n}}(t/(1-t)=\tilde{r}_1(t)$ and $\tilde{r}_{X^{\circ}_{n:n}}(t/(1-t))=\tilde{r}_2(t)$. From Figure 7, we see that $\tilde{r}_1(t)\geq \tilde{r}_2(t)\text{ for all } t\in[0,1)$. Hence, $X^*_{n:n}\geq_{rh} X^{\circ}_{n:n} $. $\Box$

Figure 7.

Plots of $\tilde{r}_1(t)$ and $\tilde{r}_2(t)$, $t\in[0,1]$.

Next, we provide a counterexample to show that the ordering result in Theorem 3.15 may not hold if we relax the stated majorization conditions.

Counterexample 3.17.

In Example 3.16, let us take $(p_1,p_2,p_3,p_4)=(0.1,0.2,0.85,0.95)$ and $(q_1,q_2,q_3,q_4)=(0.5,0.65,0.8,0.85)$ so that $(\kappa(p_1),\kappa(p_2),\kappa(p_3),\kappa(p_4))\nsucceq _{w} (\kappa(q_1),\kappa(q_2),\kappa(q_3),\kappa(q_4))$. In Figure 8, we have plotted $\tilde{r}_1(t)-\tilde{r}_2(t)$ for all $t\in[0,1)$, which shows that the hazard rate ordering result of Theorem 3.15(ii) does not hold in this case.

Figure 8.

Plot of $\tilde{r}_1(t)-\tilde{r}_2(t)$, $t\in[0,1]$.

4. Star order for multiple-outlier claim

Let X _i, $i=1,2,\ldots,r$, have a common distribution F, and X _j, $j=r+1,r+2,\ldots,n$, have a common distribution G, where $r=1,2,\ldots,n-1$. This type of model is known as outlier model, where F is called the original distribution, whereas the G is called the outlier distribution. For $r=1,2,\ldots,n-2$, it is called multiple-outlier model. In actuarial practice, even for a portfolio of risks consisting of similar kind of insureds, it may happen that some insureds have different (higher/lower) probabilities of occurrence of claims, claim sizes or odds of claims than the rest. Then this phenomena falls in the multiple-outlier claims model.

Star order is one of the most important transform order used to compare the skewness of probability distributions. Since in general the insurance claims follow positively skewed and heavy-tailed distributions, it is therefore of interest to establish sufficient conditions for star order between them to analyze the effects of the heterogeneity among occurrence probabilities and claim severity parameters (e.g., the odds ratio in our considered model) on the skewness of their distributions.

In Theorem 4.2, we derive stochastic comparisons on the largest claim amounts in case of multiple-outlier claims model with respect to star order.

The following lemma is derived from Saunders and Moran [Reference Saunders and Moran25], which will be used in proving Theorem 4.2.

Theorem 4.2. Let $X_i\sim {\rm PO}(\bar{F},\alpha_1) (Y_i\sim {\rm PO}(\bar{F},\beta_1))$ for $i=1,2,\ldots,n_1$, and let $X_j\sim {\rm PO}(\bar{F},\alpha_2) (Y_j\sim {\rm PO}(\bar{F},\beta_2))$ for $j=n_1+1,n_1+2,\ldots,n_1+n_2\;(=n)$. Assume that X _i’s are independent and that the Y _j’s are independent. Further, let $I_{p_i}$, $i=1,2,\ldots,n_1$, be independent Bernoulli r.v.’s such that $\mathbb{E}[I_{p_i}]=p_1$, and let $I_{p_j}$, $j=n_1+1,\ldots,n$, be another set of independent Bernoulli r.v.’s such that $\mathbb{E}[I_{p_j}]=p_2$. Then, for $n_1p_1\geq n_2p_2$, $p_1\geq p_2$, $\alpha_1\leq \alpha_2$, and $\beta_1\leq \beta_2$,

\begin{equation*}\frac{\alpha_1}{\alpha_2}\leq \frac{\beta_1}{\beta_2}\implies X^{*}_{n:n} \geq_{\star} X^{*}_{n:n}.\end{equation*}

Proof. Consider the following two cases.

Case I: Let $\alpha_1+\alpha_2=\beta_1+\beta_2=c$ (say). Further, let $\alpha_1=\alpha\leq \alpha_2$ and $\beta_1=\beta\leq\beta_2$, so that $\alpha\in[0,c/2]$. Then the distribution function of $X^{*}_{n:n}$ is given by

(5)\begin{equation} F_{n,\alpha}(x)=\left[1-p_1\bar{F}_\alpha(x)\right]^{n_1}\left[1-p_2\bar{F}_{c-\alpha}(x)\right]^{n_2}. \end{equation}

Here $\bar{F}_{\alpha}(x) = \frac{\alpha\bar{F}(x)}{1-\overline{\alpha}\bar{F}(x)}$ and $\bar{F}_{c-\alpha}(x) = \frac{(c-\alpha)\bar{F}(x)}{1-\overline{(c-\alpha)}\bar{F}(x)}$.

The density function corresponding to Eq. (5) is

\begin{eqnarray*} f_{n,\alpha}(x)&=&\left[1-p_1\bar{F}_\alpha(x)\right]^{n_1-1}\left[1-p_2\bar{F}_{c-\alpha}(x)\right]^{n_2-1}f(x) \\ & &\times\left[\frac{n_1p_1\alpha(1-p_2\bar{F}_{c-\alpha}(x))}{(1-\bar{\alpha}\bar{F}(x))^2}+ \frac{n_2p_2(c-\alpha)(1-p_1\bar{F}_{\alpha}(x))}{(1-\overline{(c-\alpha)}\bar{F}(x))^2}\right]. \end{eqnarray*}

Now,

\begin{eqnarray*} \frac{\partial F_{n,\alpha}(x)}{\partial\alpha} &=& F(x)\bar{F}(x)\left[1-p_1\bar{F}_\alpha(x)\right]^{n_1-1}\left[1-p_2\bar{F}_{c-\alpha}(x)\right]^{n_2-1} \\ & &\times\left[\frac{-n_1p_1(1-p_2\bar{F}_{c-\alpha}(x))}{(1-\bar{\alpha}\bar{F}(x))^2} + \frac{n_2p_2(1-p_1\bar{F}_{\alpha}(x))}{(1-\overline{(c-\alpha)}\bar{F}(x))^2} \right]. \end{eqnarray*}

Let $\Lambda_1(x)=\frac{1-p_2\bar{F}_{c-\alpha}(x)}{(1-\bar{\alpha}\bar{F}(x))^2}$ and $\Lambda_2(x)=\frac{1-p_1\bar{F}_{\alpha}(x)}{(1-\overline{(c-\alpha)}\bar{F}(x))^2}. $ Then, by using Lemma 4.1, it suffices to prove that

\begin{eqnarray*} \frac{\frac{\partial F_{n,\alpha}(x)}{\partial\alpha}}{xf_{n,\alpha}(x)} &=& \frac{F(x)\bar{F}(x)}{xf(x)}\left[\frac{-n_1p_1\Lambda_1(x)+n_2p_2\Lambda_2(x)}{n_1p_1\alpha\Lambda_1(x)+n_2p_2(c-\alpha)\Lambda_2(x)}\right] \\ &=& \frac{F(x)\bar{F}(x)}{xf(x)} \times \Lambda(x) \end{eqnarray*}

is increasing in $x\in \mathbb{R}_+$, for $\alpha \in[0,c/2]$, where

\begin{eqnarray*} \Lambda(x) &=& \left[\frac{n_1p_1\alpha\Lambda_1(x)+n_2p_2(c-\alpha)\Lambda_2(x)}{-n_1p_1\Lambda_1(x)+n_2p_2\Lambda_2(x)}\right]^{-1} \\ &=& \left[\frac{cn_2p_2\Lambda_2(x)}{n_2p_2\Lambda_2(x)-n_1p_1\Lambda_1(x)} -\alpha \right]^{-1} \\ &=& \left[c\left(1-\frac{n_1p_1}{n_2p_2}\frac{\Lambda_1(x)}{\Lambda_2(x)}\right)^{-1}-\alpha \right]^{-1}. \end{eqnarray*}

Further, let

\begin{eqnarray*} \Lambda_3(x)&=&\frac{\Lambda_1(x)}{\Lambda_2(x)}\\ &=&\frac{\left(\frac{1-p_2\bar{F}_{c-\alpha}(x)}{(1-\bar{\alpha}\bar{F}(x))^2}\right)}{\left(\frac{1-p_1\bar{F}_{\lambda}(x)}{(1-\overline{(c-\alpha)}\bar{F}(x))^2}\right)} \\ &=& \left(\frac{[1-\overline{(c-\alpha)}\bar{F}(x)-p_2(c-\alpha)\bar{F}(x)]}{(1-\bar{\alpha}\bar{F}(x))}\right)\left(\frac{(1-\bar{\alpha}\bar{F}(x))(1-\overline{(c-\alpha)}\bar{F}(x))^2}{[1-\bar{\alpha}\bar{F}(x)-p_1\alpha\bar{F}(x)]}\right) \\ &=& \left(\frac{[F(x)+(1-p_2)(c-\alpha)\bar{F}(x)]}{[F(x)+(1-p_1)\alpha\bar{F}(x)}]\right)\left(\frac{1-\overline{(c-\alpha)}\bar{F}(x)}{1-\bar{\alpha}\bar{F}(x)}\right) \\ &=& \Delta_1(x)\times\Delta_2(x). \end{eqnarray*}

It is clear that $\Delta_1(x)\geq 0$ and $\Delta_2(x)\geq 0\;\forall x\in \mathbb{R}_+$. Now we have

\begin{align*} \Delta^{\prime}_1(x)\stackrel{\rm sign}{=} &[1-(1-p_2)(c-\alpha)][F(x)+\alpha(1-p_1)\bar{F}(x)]\\&-[1-\alpha(1-p_1)][F(x)+(1-p_2)(c-\alpha)\bar{F}(x)] \\ \stackrel{\rm sign}{=}& [\alpha(1-p_1)-(1-p_2)(c-\alpha)](\bar{F}(x)+F(x)) \\ \stackrel{\rm sign}{=}& [\alpha(1-p_1)-(1-p_2)(c-\alpha)] \\ &\le 0, \end{align*}

which holds as $\alpha\leq c-\alpha$ and $p_1\geq p_2$. Further,

\begin{eqnarray*} \Delta^{\prime}_2(x)&\stackrel{\rm sign}{=}& (\overline{c-\alpha})(1-\bar{\alpha}\bar{F}(x))(1-\bar{c-\alpha}\bar{F}(x)) \\ &\stackrel{\rm sign}{=}& (\bar{c-\alpha}) - \bar{\alpha}\leq 0. \end{eqnarray*}

Hence, ultimately we have $\Lambda^{\prime}_3(x)\leq 0.$ Consequently, $\Lambda_3(x)$ is nonnegative and decreasing in $x.$

\begin{eqnarray*} \text{Now } &&\alpha\leq (c-\alpha) \\ & & \implies \frac{1-\overline{(c-\alpha)}\bar{F}(x)}{1-\bar{\alpha}\bar{F}(x)}\geq 1 \end{eqnarray*}

\begin{eqnarray*} \text{and }&& (c-\alpha)\geq \alpha ,p_1\geq p_2 \\ & &\implies F(x)+(1-p_2)(c-\alpha)\bar{F}(x)\geq F(x)+(1-p_1)\alpha\bar{F}(x), \\ & & \text{and hence, } \Lambda_3(x)\geq 1. \end{eqnarray*}

Now if $n_1\geq n_2$, then $n_1p_1\geq n_2p_2$. On combining all of these results, we have

\begin{eqnarray*} && \frac{n_1p_1}{n_2p_2}\Lambda_3(x)\geq 1 \\ & & \implies \left(1-\frac{n_1p_1}{n_2p_2}\Lambda_3(x)\right)\leq 0. \end{eqnarray*}

Hence, $\left(1-\frac{n_1p_1}{n_2p_2}\Lambda_3(x)\right)$ is increasing in x, which implies $\left[\left(1-\frac{n_1p_1}{n_2p_2}\Lambda_3(x)\right)^{-1}-\alpha\right]^{-1}$ is increasing in $x.$ So ultimately we have $\Lambda(x)$ is increasing in $x.$ This completes the proof.

Case II: Let $\alpha_1+\alpha_2\neq \beta_1+\beta_2.$ In this case, there exists some κ > 0 such that $\alpha_1+\alpha_2=\kappa(\beta_1+\beta_2).$ Now, let $Z_{n:n}$ be the largest claim amount from $I_{1}Z_1,\ldots,I_{n_1}Z_{n_1},I_{n_1+1}Z_{n_1+1},\ldots,I_{n}Z_{n},$ where $Z_1,\ldots,Z_{n_1}$ have the distribution $F_{\kappa \mu_1}$ and $Z_{n_1+1},\ldots,Z_{n}$ have the distribution $F_{\kappa\mu_2}$. Finally, on using the result of Case I and the scale invariant property of the star order, the desired result follows.

Example 4.3. Suppose that $\{X_1,X_2\}$ and $\{Y_1,Y_2\}$ are two sets of independent nonnegative r.v.’s with $X_i \sim {\rm PO}(\bar{F}(x),\alpha_i),\ i=1,2,\ \text{and }\ Y_i \sim {\rm PO}(\bar{F}(x),\beta_i),\ i=1,2,$ where $\bar{F}(x)={\rm e}^{-x},\;x\gt0.$ Set $(\alpha_1,\alpha_2)=(0.5,1.9),\ (\beta_1,\beta_2)=(0.8,1.2),\ (p_1,p_2)=(1/4,1/8),\ {\rm and}\ (n_1,n_2)=(3,2).$ Then all the conditions of Theorem 4.2 are satisfied. In Figure 9, we have plotted the derivative of ${F^{-1}_{X^{*}_{n:n}}(t)}/{F^{-1}_{Y^{*}_{n:n}}(t)}$ with respect to t from which it is clear that ${F^{-1}_{X^{*}_{n:n}}(t)}/{F^{-1}_{Y^{*}_{n:n}}(t)}$ is increasing for $t\in(0,1).$ Hence, $X^*_{n:n}\geq_{\star} Y^*_{n:n}$.\\[-37pt]

Figure 9.

Plots of derivative of ${F^{-1}_{X^{*}_{n:n}}(t)}/{F^{-1}_{Y^{*}_{n:n}}(t)}$ with respect to t for $t\in(0,1)$.

5. Aggregate claim amount

The aggregate claim of a portfolio is the sum of all amounts payable during the reference period. Our next theorem derives sufficient conditions that the aggregate claim amount increases on reducing the heterogeneity in the sense of majorization among the concerned parameters of a considered semiparametric family of distributions when they are in ascending order. Here we assume that occurrence probabilities are arranged according to LWSAI.

Proof. Let $A(\bf I,\alpha)= \sum_{i=1}^{n}I_{i}X_{\alpha_i}$ and $A(\bf I, \bf\beta)= \sum_{i=1}^{n}I_{i}X_{\beta_i}.$ We have to prove that $F_{A(\bf I,\bf \alpha)}(t)\geq F_{A(\bf I,\bf\beta)}(t)~\forall t\in \Re_{+}$. By the nature of majorization order, it suffices to prove it when $(\alpha_i,\alpha_j)\stackrel{m}{\succeq}(\beta_i,\beta_j)$ for some pair $1\leq i\lt j\leq n,$ and $\alpha_r=\beta_r$ for all $r\neq i,j.$ Note that

\begin{eqnarray*} F_{A(\bf I,\bf \alpha)}(t)&=& \mathbb{P}\left(\sum_{i=1}^{n}I_{p_i}X_{\alpha_i}\leq t\right)\\ &=& \sum_{k=0}^{n}\sum_{\bf \chi\in S_{k}}\mathbb{P}\left(\sum_{i=1}^{n}I_{p_i}X_{\alpha_i}\leq t \mid \bf I=\bf \chi\right)p(\bf \chi)\\ &=& p(\bf 0)+p(\bf 1)\mathbb{P}\left(\sum_{i=1}^{n}X_{\alpha_i}\leq t\right)+\sum_{k=1}^{n}\sum_{\bf \chi\in S_{k}}p(\bf \chi)\mathbb{P}\left(\sum_{i=1}^{n}\chi_{i}X_{\alpha_i}\leq t\right)\\ &=& p(\bf 0)+p(\bf 1)\mathbb{P}\left(\sum_{i=1}^{n}X_{\alpha_i}\leq t\right)+\sum_{k=1}^{n-1}\left\{\sum_{\chi\in S^{i,j}_k(0,0)}p(\bf \chi)\mathbb{P}\left(\sum_{r\neq i,j}^{n}\chi_r X_{\alpha_r}\leq t\right)\right.+\\ & &\left. \sum_{\chi\in S^{i,j}_k(0,1)}p(\bf \chi)\mathbb{P}\left(X_{\alpha_j}+\sum_{r\neq i,j}^{n}\chi_r X_{\alpha_r}\leq t\right)\right.+\\ & &\left.\sum_{\chi\in S^{i,j}_k(0,1)}p(\tau_{i,j}(\bf \chi))\mathbb{P}\left(X_{\alpha_i}+\sum_{r\neq i,j}^{n}\chi_r X_{\alpha_r}\leq t\right)\right.+\\ & &\left. \sum_{\chi\in S^{i,j}_k(1,1)}p(\bf \chi)\mathbb{P}\left(X_{\alpha_i}+X_{\alpha_j}+\sum_{r\neq i,j}^{n}\chi_r X_{\alpha_r}\leq t\right) \right\}. \end{eqnarray*}

Similarly,

\begin{eqnarray*} F_{A(\bf I,\bf \beta)}(t)&=& \mathbb{P}\left(\sum_{i=1}^{n}I_{i}X_{\beta_i}\leq t\right)\\ &=& p(\bf 0)+p(\bf 1)\mathbb{P}\left(\sum_{i=1}^{n}X_{\beta_i}\leq t\right)+\sum_{k=1}^{n-1}\left\{\sum_{\chi\in S^{i,j}_k(0,0)}p(\bf \chi)\mathbb{P}\left(\sum_{r\neq i,j}^{n}\chi_r X_{\beta_r}\leq t\right)\right.+\\ & &\left. \sum_{\chi\in S^{i,j}_k(0,1)}p(\bf \chi)\mathbb{P}\left(X_{\beta_j}+\sum_{r\neq i,j}^{n}\chi_r X_{\beta_r}\leq t\right)\right.+\\ & &\left.\sum_{\chi\in S^{i,j}_k(0,1)}p(\tau_{i,j}(\bf \chi))\mathbb{P}\left(X_{\beta_i}+\sum_{r\neq i,j}^{n}\chi_r X_{\beta_r}\leq t\right)\right.+\\ & &\left. \sum_{\chi\in S^{i,j}_k(1,1)}p(\bf \chi)\mathbb{P}\left(X_{\beta_i}+X_{\beta_j}+\sum_{r\neq i,j}^{n}\chi_r X_{\beta_r}\leq t\right) \right\}. \end{eqnarray*}

Under assumption (ii), it holds that

(6)\begin{equation}\mathbb{P}\left(\sum_{i=1}^{n}X_{\alpha_i}\leq t\right)\geq \mathbb{P}\left(\sum_{i=1}^{n}X_{\beta_i}\leq t\right)\end{equation}

and, for any $\bf \chi \in S^{i,j}_{k}(1,1),k=1,\ldots,n-1,$

(7)\begin{equation} \mathbb{P}\left(X_{\alpha_i}+X_{\alpha_j}+\sum_{r\neq i,j}^{n}\chi_r X_{\alpha_i}\leq t\right)\geq \mathbb{P}\left(X_{\beta_i}+X_{\beta_j}+\sum_{r\neq i,j}^{n}\chi_r X_{\beta_i}\leq t\right). \end{equation}

Then combining above two, we have

\begin{eqnarray*} &&F_{A(\bf I,\bf \alpha)}(t)-F_{A(\bf I,\bf \beta)}(t)\\ & &=p(\bf 1)\left[\mathbb{P}\left(\sum_{i=1}^{n}X_{\alpha_i}\leq t\right)-\mathbb{P}\left(\sum_{i=1}^{n}X_{\beta_i}\leq t\right)\right]\\ & &+ \sum_{k=1}^{n-1}\left\{\sum_{\bf\chi\in S^{i,j}_{k}(0,1)}p(\bf\chi)\left[\mathbb{P}\left(X_{\alpha_j}+\sum_{r\neq i,j}^{n}\chi_r X_{\alpha_r}\leq t\right)-\mathbb{P}\left(X_{\beta_j}+\sum_{r\neq i,j}^{n}\chi_r X_{\alpha_r}\leq t\right) \right]\right.\\ & &+\left.\sum_{\bf\chi\in S^{i,j}_{k}(0,1)}p(\tau_{i,j}\bf\chi)\left[\mathbb{P}\left(X_{\alpha_i}+\sum_{r\neq i,j}^{n}\chi_r X_{\alpha_r}\leq t\right)-\mathbb{P}\left(X_{\beta_i}+\sum_{r\neq i,j}^{n}\chi_r X_{\alpha_r}\leq t\right) \right] \right.\\ & &+\left. \sum_{\chi\in S^{i,j}_k(1,1)}p(\bf \chi) \left[\mathbb{P}\left(X_{\alpha_i}+X_{\alpha_j}+\sum_{r\neq i,j}^{n}\chi_r X_{\alpha_i}\leq t\right)\geq \mathbb{P}\left(X_{\beta_i}+X_{\beta_j}+\sum_{r\neq i,j}^{n}\chi_r X_{\beta_i}\leq t\right)\right] \right\}\\ & &\geq \sum_{k=1}^{n-1}\left\{\sum_{\bf\chi\in S^{i,j}_{k}(0,1)}p(\bf\chi)\left[\mathbb{P}\left(X_{\alpha_j}+\sum_{r\neq i,j}^{n}\chi_r X_{\alpha_r}\leq t\right)-\mathbb{P}\left(X_{\beta_j}+\sum_{r\neq i,j}^{n}\chi_r X_{\alpha_r}\leq t\right) \right]\right.\\ & &+\left. \sum_{\bf\chi\in S^{i,j}_{k}(0,1)}p(\tau_{i,j}\bf\chi)\left[\mathbb{P}\left(X_{\alpha_i}+\sum_{r\neq i,j}^{n}\chi_r X_{\alpha_r}\leq t\right)-\mathbb{P}\left(X_{\beta_i}+\sum_{r\neq i,j}^{n}\chi_r X_{\alpha_r}\leq t\right) \right]\right\}\\ & &\geq \sum_{k=1}^{n-1}\left\{\sum_{\bf\chi\in S^{i,j}_{k}(0,1)}p(\tau_{i,j}\bf\chi)\left[\mathbb{P}\left(X_{\alpha_j}+\sum_{r\neq i,j}^{n}\chi_r X_{\alpha_r}\leq t\right)-\mathbb{P}\left(X_{\beta_j}+\sum_{r\neq i,j}^{n}\chi_r X_{\alpha_r}\leq t\right) \right]\right.\\ & &+\left. \sum_{\bf\chi\in S^{i,j}_{k}(0,1)}p(\tau_{i,j}\bf\chi)\left[\mathbb{P}\left(X_{\alpha_i}+\sum_{r\neq i,j}^{n}\chi_r X_{\alpha_r}\leq t\right)-\mathbb{P}\left(X_{\beta_i}+\sum_{r\neq i,j}^{n}\chi_r X_{\alpha_r}\leq t\right) \right]\right\} \\ & & = \sum_{k=1}^{n-1}\left\{\sum_{\bf\chi\in S^{i,j}_{k}(0,1)}p(\tau_{i,j}\bf\chi)\idotsint_{\mathbb{R}^{n-2}_{+}}\left[\mathbb{P}\left(X_{\alpha_i}\leq t-\sum_{r\neq i,j}^{n}\chi_r x_{\alpha_r}\right) +\mathbb{P}\left(X_{\alpha_j}\leq t-\sum_{r\neq i,j}^{n}\chi_r x_{\alpha_r}\right)\right.\right.\\ & &-\left.\left. \mathbb{P}\left(X_{\beta_i}\leq t-\sum_{r\neq i,j}^{n}\chi_r x_{\alpha_r}\right)- \mathbb{P}\left(X_{\beta_j}\leq t-\sum_{r\neq i,j}^{n}\chi_r x_{\alpha_r}\right)\right]\prod_{r\neq i,j}^{n}g_{X_{\alpha_r}}(x_{\alpha_r})\,{\rm d}x_{\alpha_r} \right\} \\ &=& \sum_{k=1}^{n-1}\left\{\sum_{\bf\chi\in S^{i,j}_{k}(0,1)}p(\tau_{i,j}\bf\chi)\idotsint_{\mathbb{R}^{n-2}_{+}}\left[\bar{F}_{X_{\beta_i}}\left(t-\sum_{r\neq i,j}^{n}\chi_r x_{\alpha_r}\right) +\bar{F}_{X_{\beta_j}}\left(t-\sum_{r\neq i,j}^{n}\chi_r x_{\alpha_r}\right)\right.\right.\\ & &\left.\left. -\bar{F}_{X_{\alpha_i}}\left(t-\sum_{r\neq i,j}^{n}\chi_r x_{\alpha_r}\right) - \bar{F}_{X_{\alpha_j}}\left(t-\sum_{r\neq i,j}^{n}\chi_r x_{\alpha_r}\right) \right]\prod_{r\neq i,j}^{n}g_{X_{\alpha_r}}(x_{\alpha_r})\,{\rm d}x_{\alpha_r} \right\} \\ & & \geq 0, \end{eqnarray*}

where $g_{X_{\alpha_r}}(x)$ is the density function of $X_{\alpha_r}$. The first inequality follows from Eqs. (6) and (7), the second inequality from Lemma 3.7, and finally the last inequality is due to the fact that $\bar{F}_{X_{\alpha}}$ is concave in $\alpha\in \mathbb{R}_{+}$ as per assumption (i).

Remark 5.2. Theorem 5.1 hold true for the PO model, that is, for $X_{\alpha_i}\sim {\rm PO}(\bar{F},\alpha_i)\;(X_{\beta_i}\sim {\rm P}O(\bar{F},\beta_i)),\ i=1,\ldots,n$, with $\bar{F}(t)={\rm e}^{-\lambda t}$, λ > 0. This family of distribution is known as Marshall–Olkin extended exponential distribution. Note that $\bar{F}_{X_{\alpha}}(t)={\alpha \bar{F}(t)}/{1-\bar{\alpha}\bar{F}(t)}$ is increasing and concave in α. The condition (ii), that is, the survival function of $X_{\mu_1}+ X_{\mu_2}$ is Schur-concave in $(\mu_1,\mu_2)$ follows from the Corrollary F.12.a. (p. 235) of [Reference Marshall, Olkin and Arnold21] with the fact that both the survival function $\bar{F}_{X_{\mu}}(t)={\mu e^{-\lambda t}}/{1-\bar{\mu}e^{-\lambda t}}$ and p.d.f. $f_{X_{\mu}}(t)={\mu \cdot \lambda e^{-\lambda t} }/{(1-\bar{\mu}e^{-\lambda t})^2}$ are concave in µ.

It is to be noted that exponentiated Weibull distribution having the distribution function $F(t;\alpha,\beta))=\left(1-{\rm e}^{-t^{\beta}}\right)^{\alpha}$, $\alpha, \beta\gt0$, also satisfies both the conditions (i) and (ii) of Theorem 5.1 with respect to the parameter α.

Remark 5.3. It is worth to be mention that in Theorem 4.7 of Zhang et al. [Reference Zhang, Zhao and Cheung30], they compared aggregate claim amounts of two sets of heterogeneous portfolios under the assumption that the survival function $\bar{F}(x;\alpha)$ is decreasing and convex in α > 0.

The following example illustrates the result given in Theorem 5.1.

Example 5.4. Suppose that $\{X_1,X_2\}$ and $\{Y_1,Y_2\}$ are two sets of independent nonnegative r.v.’s with $X_i \sim {\rm PO}(\bar{F}(x),\alpha_i),i=1,2, \text{and } Y_i \sim {\rm PO}(\bar{F}(x),\beta_i),i=1,2,$ where $\bar{F}(x)={\rm e}^{-0.3x},\;x\gt0.$ Set $(\alpha_1,\alpha_2)=(0.4,2.6),\ (\beta_1,\beta_2)=(0.8,2.2),\ p(0,0)=P(I_{p_1}=0, I_{p_2}=0)=0.15,\ p(0,1)=0.46,\ p(1,0)=0.34,\ {\rm and}\ p(1,1)=0.05.$ Then, $I=\{I_{p_1},I_{p_2}\}$ is LWSAI. We consider the transformation $x=t/(1-t)$ so that, for $t\in[0,1)$, we have $x\in[0,\infty)$. After this substitution, we denote the distribution functions of $\sum_{i=1}^2 I_{p_i}X_{\alpha_i} \text{and } \sum_{i=1}^2 I_{p_i}X_{\beta_i}$ by $F_{A( I,\alpha)}$ and $F_{A(I,\beta)}$, respectively. $F_{A( I,\alpha)}(t/(1-t))=\psi_1(t)$ and $F_{A(I,\beta)}(t/(1-t))=\psi_2(t)$, respectively. From Figure 10, it is clear that $\psi_1(t)\geq \psi_2(t)\ \text{for all } t\in[0,1)$. Hence, $\sum_{i=1}^2 I_{p_i}X_{\alpha_i} \leq_{st} \sum_{i=1}^2 I_{p_i}X_{\beta_i}.$

Figure 10.

Plots of $\psi_1(t)$ and $\psi_2(t)$, $t\in[0,1]$.

6. Concluding remarks

In this paper, we study some stochastic comparison results for the largest and the aggregate claim amounts under the setup of PO model. We derive the results with respect to the usual stochastic order, the reversed hazard rate order, and the star order. Further, some numerical examples are given to illustrate the derived results.

As discussed, the PO model is one of the important semi-parametric models. It has a large number of applications in different disciplines including financial engineering and actuarial science. It is worth mentioning that the results obtained by us apply to the whole family of extended distributions, for example, extended–exponential, Weibull, gamma, Pareto, Lomax, and linear failure-rate distribution, which have been studied by many researchers in different areas of applications (see [Reference Barmalzan, Ayat, Balakrishnan and Roozegar5] and references therein). Further, the largest claim and the aggregate claim amounts contain useful information in determining the key factors in a given insurance policy. For instance, the largest claim amount is referred to as the probable maximum loss, which helps to determine the amount of funds required to pay future claims on policies. Thus, the proposed study may be helpful to the actuaries in determining which portfolio of risks is better (in some stochastic sense) among a list of portfolios with respect to the largest and aggregate claim amounts. Apart from this, our theoretical results can be used to compare the lifetimes of two parallel systems whose components are subject to random shocks instantaneously.

Even though we discussed a lot of results, there are still many open problems that may be explored. Here the results are mostly developed for the usual stochastic order. Thus, the study of the same problem, as in here, for other stochastic orders (namely, likelihood ratio order, convex order, etc.) can be considered as a potential problem.