2. Preliminaries

Throughout this paper, the term ‘increasing’ is used for ‘monotone non-decreasing’ and ‘decreasing’ is used for ‘monotone non-increasing’. Expectations are assumed to be finite whenever they appear, and inverse functions are assumed to be right continuous. Let $\mathbb{R}=(-\infty,+\infty)$, $\mathbb{R}_{+}=[0,\infty)$, and $\mathbb{R}_{++}=(0,\infty)$. Denote $\mathcal{I}^n_{+}=\{{{x}}\in\mathbb{R}_{+}^n\;:\;0\leq x_1\leq x_2\leq\cdots\leq x_n\}$ and $\mathcal{D}^n_{+}=\{\textbf{{x}}\in\mathbb{R}_{+}^n\;:\;x_1\geq x_2\geq\cdots\geq x_n\geq 0\}$. We shall use $\stackrel{\rm sgn}{=}$ to express that both sides of an equality have the same sign.

1. (i) usual stochastic order (denoted by $X\geq_{\rm st} Y$) if $\bar{F}(x)\geq \bar{G}(x)$ for all $x\in\mathbb{R}_{+}$;

2. (ii) convex transform order (denoted by $X\geq_{\rm c}Y$) if $F^{-1}[G(x)]$ is convex in $x\in\mathbb{R}_{+}$, or equivalently, $G^{-1}[F(x)]$ is concave in $x\in\mathbb{R}_{+}$;

3. (iii) star order (denoted by $X\geq_{\star}Y$) if $F^{-1}[G(x)]$ is star-shaped in the sense that $F^{-1}[G(x)]/x$ is increasing in $x\in\mathbb{R}_{+}$;

4. (iv) superadditive order (denoted by $X\geq_{\rm su}Y$) if $F^{-1}[G(x)]$ is superadditive in $x\in\mathbb{R}_{+}$, that is, $F^{-1}[G(x+y)]\geq F^{-1}[G(x)]+F^{-1}[G(y)]$ for all $x,y\in\mathbb{R}_{+}$;

5. (v) new better than used in expectation (NBUE) order (denoted by $X\geq_{\rm NBUE}Y$) if

   \begin{equation*} \frac{1}{\mathbb{E}[X]}\int_{F^{-1}(u)}^{\infty}\bar{F}(x) \, \mathrm{d} x\geq \frac{1}{\mathbb{E}[Y]}\int_{G^{-1}(u)}^{\infty}\bar{G}(x) \, \mathrm{d} x \qquad\mbox{for all $u\in[0,1]$}; \end{equation*}

6. (vi) Lorenz order (denoted by $X\geq_{\rm Lorenz}Y$) if $L_X(p)\leq L_Y(p)$ for all $p\in[0,1]$, where the Lorenz curve $L_X$ corresponding to X is defined as $L_X(p)=\int_0^{p}F^{-1}(u) \, \mathrm{d} u/\mathbb{E}[X]$;

7. (vii) dispersive order (denoted by $X\geq_{\rm disp}Y$) if $F^{-1}(v)-F^{-1}(u)\geq G^{-1}(v)-G^{-1}(u)$, for $0< u\leq v<1$;

8. (viii) right-spread order (denoted by $X\geq_{\rm RS}Y$) if

   \begin{equation*} \int_{F^{-1}(u)}^{\infty}\bar{F}(t) \, \mathrm{d} t\geq \int_{G^{-1}(u)}^{\infty}\bar{G}(t) \, \mathrm{d} t \qquad\mbox{for all $u\in(0,1)$.} \end{equation*}

The class of transform orders contains the convex transform order, the star order, and the superadditive order, which are commonly used in actuarial science to compare the severities of different risks/losses, and in reliability theory to compare the lifetimes of different coherent systems. These three types of stochastic orders are scale invariant, and usually adopted to compare the skewness of the distributions of interested random variables. The skewness of the distribution of the extreme claim amount plays a key role in many practical scenarios. For instance, in the context of actuarial science, the claims are usually right and positively skewed since they are bounded on the left but unbounded on the right, and the tail on the right side of the density function is longer/fatter than the left side. In order to compare the skewness of the distributions of the largest claim amounts, it is natural to establish sufficient conditions for some transform orders between them to analyze the effects of the heterogeneity among occurrence probabilities and claim severity parameters on the skewness of their distributions.

The convex transform order was proposed in [Reference Van Zwet36] to compare the skewness of probability distributions. It is also called the increasing failure rate order in reliability theory since, when f and g exist, the convexity of $G^{-1}[F(x)]$ means that

\begin{equation*}\frac{f(F^{-1}(u))}{g(G^{-1}(u))}=\frac{\tilde{r}_F(F^{-1}(u))}{\tilde{r}_G(G^{-1}(u))}\end{equation*}

is increasing in $u\in[0, 1]$, where $\tilde{r}_F$ and $\tilde{r}_G$ stand for the reversed hazard rate functions of F and G, respectively. Thus, $X\leq_{\rm c}Y$ can be understood as X aging faster than Y in some stochastic sense. The star order is called increasing failure rate in average in reliability theory. The Lorenz order, which is called harmonic new better than used in expectation in reliability theory, can be used in economics to measure the inequality of incomes. It is known that $X\geq_{\star} Y$ implies $X\geq_{\rm su} Y$, and

\begin{equation*} X\geq_{\rm c} Y\Longrightarrow X\geq_{\star} Y\Longrightarrow X\geq_{\rm NBUE}Y\Longrightarrow X\geq_{\rm Lorenz}Y \Longrightarrow {\rm CV}(X)\geq {\rm CV}(Y),\end{equation*}

where ${\rm CV}(X)=\sqrt{{\rm Var}(X)}/\mathbb{E}(X)$ and ${\rm CV}(Y)=\sqrt{{\rm Var}(Y)}/\mathbb{E}(Y)$ denote the coefficients of variation of X and Y, respectively. The dispersive order, which is stronger than the right-spread order, is a kind of partial order for comparing the variabilities of two probability distributions. It is known that both the usual stochastic order and the right-spread order (also called excess wealth order in economics) imply increasing convex order. In the following we shall sometimes use $F\geq_{\rm sym} G$ to denote that the two independent random variables X and Y with respective distributions F and G are such that $X\geq_{\rm sym} Y$, where ‘$\geq_{\rm sym}$ indicates some specified stochastic order. For comprehensive discussions on these useful orders, we refer interested readers to the monographs [Reference Denuit, Dhaene, Goovaerts and Kaas10, Reference Marshall and Olkin30, Reference Shaked and Shanthikumar35].

One useful tool in deriving various inequalities in statistics and probability is the notion of majorization, which characterizes the heterogeneity and dispersiveness among the coordinates of real-valued vectors.

1. (i) A vector ${\textbf{{x}}}\in\mathbb{R}^n$ is said to majorize another vector ${\textbf{{y}}}\in\mathbb{R}^n$ (written as ${\textbf{{x}}}\stackrel{\rm m}{\succeq} {\textbf{{y}}}$) if $\sum_{i=1}^jx_{i\;:\;n}\leq \sum_{i=1}^j y_{i\;:\;n}$ for $j=1,\ldots,n-1$, and $\sum_{i=1}^nx_{i\;:\;n}= \sum_{i=1}^n y_{i\;:\;n}$;

2. (ii) A vector ${\textbf{{x}}}\in\mathbb{R}^n$ is said to weakly supermajorize another vector ${\textbf{{y}}}\in\mathbb{R}^n$ (written as ${\textbf{{x}}}\stackrel{\rm w}{\succeq} {\textbf{{y}}}$) if $\sum_{i=1}^jx_{i\;:\;n}\leq \sum_{i=1}^j y_{i\;:\;n}$ for $j=1,\ldots,n$;

3. (iii) A vector ${\textbf{{x}}}\in\mathbb{R}^n$ is said to weakly submajorize another vector ${\textbf{{y}}}\in\mathbb{R}^n$ (written as ${\textbf{{x}}}\succeq_{\rm w} {\textbf{{y}}}$) if $\sum_{j=i}^nx_{j\;:\;n}\geq \sum_{j=i}^n y_{j\;:\;n}$ for $i=1,\ldots,n$;

4. (iv) A vector ${\textbf{{x}}}\in\mathbb{R}^{n}_+$ is said to be p-larger than another vector ${\textbf{{y}}}\in\mathbb{R}^{n}_+$ (written as ${\textbf{{x}}}\stackrel{\rm p}{\succeq} {\textbf{{y}}}$) if $\prod_{i=1}^j x_{i\;:\;n}\leq \prod_{i=1}^j y_{i\;:\;n}$ for $j=1,\ldots,n$.

It is worth mentioning that the weak supermajorization order implies the p-larger order, and the majorization order implies both the weak supermajorization and submajorization orders. However, the reverse statement is not true in general. Interested readers are referred to [Reference Marshall, Olkin and Arnold31] for more detailed discussions on their properties and applications.

Next, we introduce the notion of Schur function and an important lemma that is useful to establish inequalities with respect to the majorization order.

Lemma 1. ([Reference Marshall, Olkin and Arnold31]) Consider the real-valued continuously differentiable function $\phi$ on $J^{n}$, where $J\subseteq\mathbb{R}$ is an open interval. Then, $\phi$ is Schur-convex (Schur-concave) on $J^{n}$ if and only if $\phi$ is symmetric on $J^{n}$, and for all $i\neq j$ and all ${{\textbf{{u}}}}\in J^{n}$,

\begin{equation*}(u_i-u_j)\left(\frac{\displaystyle{\partial\phi({{\textbf{{u}}}})}}{\displaystyle{\partial u_{i}}}-\frac{\displaystyle{\partial\phi({{\textbf{{u}}}})}}{\displaystyle{\partial u_{j}}}\right)\geq0 \ (\leq0),\end{equation*}

where $\frac{\displaystyle{\partial\phi({{\textbf{{u}}}})}}{\displaystyle{\partial u_{i}}}$ denotes the partial derivative of $\phi$ with respect to its ith argument, $i=1,\ldots,n$.

3. Convex transform order

Assume that the nonnegative random variable $X_{i}$ has survival function $\bar{G}(t,\lambda_{i})$, where $\lambda_i\in\mathbb{R}_{++}$ is an index parameter, for $i=1,2,\ldots,n$. Let $I_{i}$ be a Bernoulli random variable such that $\mathbb{E}[I_{i}]=p_{i}$ for $i=1,2,\ldots,n$. Note that the random variable $Y_i\;:\!=\;I_{p_{i}}X_{i}$ is discrete-continuous, which admits zero with probability $1-p_i$, and $X_{i}$ with probability $p_i$, for $i=1,2,\ldots,n$. Then, the distribution function of $Y_{i}$ is given by $F_{Y_{i}}(t)=1-p_i\bar{G}(t,\lambda_{i})$ for $t\in\mathbb{R}_{+}$. Let $Y_{i}^{\ast}\;:\!=\;I_{i}^{\ast}X_{i}^{\ast}$ be the claim amounts arising from another insurance portfolio for $i=1,2,\ldots,n$. Denote by $Y_{n\;:\;n}$ and $Y_{n\;:\;n}^{\ast}$ the largest claim amounts from $Y_{1},\ldots,Y_{n}$ and $Y_{1}^{\ast},\ldots,Y_{n}^{\ast}$, respectively.

The well-known Cauchy–Schwarz inequality introduced in the following lemma plays a critical role in obtaining the main results.

Lemma 2. ([Reference Mitrinović32]) Let $(a_{1},a_{2},\ldots,a_{n})$ and $(b_{1},b_{2},\ldots,b_{n})$ be two sequences of real numbers. Then,

\begin{equation*}\left(\sum_{i=1}^{n}a_{i}^{2}\right)\left(\sum_{i=1}^{n}b_{i}^{2}\right)\geq \left(\sum_{i=1}^{n}a_{i}b_{i}\right)^{2},\end{equation*}

where the equality holds if and only if the sequences $(a_{1},a_{2},\ldots,a_{n})$ and $(b_{1},b_{2},\ldots,b_{n})$ are proportional, i.e. there is a constant $\lambda$ such that $a_{k}=\lambda b_{k}$ for each $k\in\{1,2,\ldots,n\}$.

Now, we present a sufficient condition for the convex transform order between the largest claim amounts when one set of claims is heterogeneous and another set of claims is homogeneous.

Theorem 1. Let $X_{i}$ ($X_{i}^{\ast}$) be independent exponential random variables with hazard rate $\lambda_{i}$ ($\lambda_{i}^{\ast}=\lambda$), $i=1,2,\ldots,n$. Let $I_{1},\ldots,I_{n}$ ($I_{1}^{\ast},\ldots,I_{n}^{\ast}$) be a set of independent Bernoulli random variables such that $\mathbb{E}[I_{i}]=p_{i}$ ($\mathbb{E}[I_{i}^{\ast}]=p$), $i=1,2,\ldots,n$. Then,

\begin{equation*} p\geq1-\prod_{i=1}^{n}(1-p_{i})^{1/n}\Longrightarrow Y_{n\;:\;n}\geq_{\rm c}Y_{n\;:\;n}^{\ast}.\end{equation*}

Proof. Let $F_{n}$ ($f_{n}$) and $G_{n}$ ($g_{n}$) be the distribution (density) functions of $Y_{n\;:\;n}$ and $Y_{n\;:\;n}^{\ast}$, respectively. Then, for $x\in\mathbb{R}_{+}$, we have $F_{n}(x)=\prod_{i=1}^{n}(1-p_{i}\mathrm{e}^{-\lambda_{i}x})$ and $G_{n}(x)=(1-p\mathrm{e}^{-\lambda x})^{n}$. In light of [Reference Marshall and Olkin30, Proposition 21.A.7], it is sufficient to show that $G_{n}^{-1}F_{n}(x)$ is concave on $x\in\mathbb{R}_{+}$. Since $F_{n}(x)\geq F_{n}(0)=\prod_{i=1}^{n}(1-p_{i})$ and $G_{n}(x)\geq G_{n}(0)=(1-p)^{n}$, we know that the inverse of $G_{n}$ exists by the assumption that $p\geq1-\prod_{i=1}^{n}(1-p_{i})^{1/n}$. Note that, for $x\in\mathbb{R}_{+}$,

\begin{equation*}G_{n}^{-1}F_{n}(x)=-\frac{1}{\lambda}\ln\left(\frac{1-\prod_{i=1}^{n}(1-p_{i}\mathrm{e}^{-\lambda_{i}x})^{1/n}}{p}\right).\end{equation*}

Then, we have

\begin{equation*}g_{n}(G_{n}^{-1}F_{n}(x))=n\lambda\left[1-\prod_{i=1}^{n}\big(1-p_{i}\mathrm{e}^{-\lambda_{i}x}\big)^{1/n}\right]\prod_{i=1}^{n}\big(1-p_{i}\mathrm{e}^{-\lambda_{i}x}\big)^{(n-1)/n}.\end{equation*}

Upon differentiating $G_{n}^{-1}F_{n}(x)$ with respect to x, we obtain

\begin{eqnarray*}\frac{\mathrm{d} G_{n}^{-1}F_{n}(x)}{\mathrm{d} x}&=& \frac{f_{n}(x)}{g_{n}(G_{n}^{-1}F_{n}(x))}\\[5pt] &=& \frac{\sum_{i=1}^{n}\frac{p_{i}\lambda_{i}\mathrm{e}^{-\lambda_{i}x}}{1-p_{i}\mathrm{e}^{-\lambda_{i}x}}\prod_{i=1}^{n}(1-p_{i}\mathrm{e}^{-\lambda_{i}x})}{n\lambda\left[1-\prod_{i=1}^{n}(1-p_{i}\mathrm{e}^{-\lambda_{i}x})^{1/n}\right]\prod_{i=1}^{n}(1-p_{i}\mathrm{e}^{-\lambda_{i}x})^{(n-1)/n}}\\[5pt] &=& \frac{\sum_{i=1}^{n}\frac{p_{i}\lambda_{i}\mathrm{e}^{-\lambda_{i}x}}{1-p_{i}\mathrm{e}^{-\lambda_{i}x}}\prod_{i=1}^{n}(1-p_{i}\mathrm{e}^{-\lambda_{i}x})^{1/n}}{n\lambda\left[1-\prod_{i=1}^{n}(1-p_{i}\mathrm{e}^{-\lambda_{i}x})^{1/n}\right]}.\end{eqnarray*}

Thus, it is enough to show that

\begin{equation*} h(x)=\frac{\sum_{i=1}^{n}\frac{p_{i}\lambda_{i}}{\mathrm{e}^{\lambda_{i}x}-p_{i}}}{\prod_{i=1}^{n}(1-p_{i}\mathrm{e}^{-\lambda_{i}x})^{-1/n}-1}\end{equation*}

is decreasing in $x\in\mathbb{R}_{+}$. By taking the derivative of h(x), this is equivalent to proving that

\begin{multline*} {\sum_{i=1}^{n}\frac{p_{i}\lambda_{i}^2\mathrm{e}^{\lambda_{i}x}}{(\mathrm{e}^{\lambda_{i}x}-p_{i})^2}\left[\prod_{i=1}^{n}(1-p_{i}\mathrm{e}^{-\lambda_{i}x})^{-1/n}-1\right]} \\ \geq \frac{1}{n}\sum_{i=1}^{n}\frac{p_{i}\lambda_{i}}{\mathrm{e}^{\lambda_{i}x}-p_{i}}\sum_{i=1}^{n}\frac{p_{i}\lambda_{i}\mathrm{e}^{-\lambda_{i}x}}{1-p_{i}\mathrm{e}^{-\lambda_{i}x}}\prod_{i=1}^{n}(1-p_{i}\mathrm{e}^{-\lambda_{i}x})^{-1/n},\end{multline*}

i.e.

\begin{equation*} \sum_{i=1}^{n}\frac{p_{i}\lambda_{i}^2\mathrm{e}^{-\lambda_{i}x}}{(1-p_{i}\mathrm{e}^{-\lambda_{i}x})^2}\left[1-\prod_{i=1}^{n}(1-p_{i}\mathrm{e}^{-\lambda_{i}x})^{1/n}\right]\geq\frac{1}{n}\left(\sum_{i=1}^{n}\frac{p_{i}\lambda_{i}\mathrm{e}^{-\lambda_{i}x}}{1-p_{i}\mathrm{e}^{-\lambda_{i}x}}\right)^2.\end{equation*}

By applying the Cauchy–Schwarz inequality in Lemma 3, it can be seen that

\begin{equation*} \sum_{i=1}^{n}\frac{p_{i}\lambda_{i}^2\mathrm{e}^{-\lambda_{i}x}}{(1-p_{i}\mathrm{e}^{-\lambda_{i}x})^2}\sum_{i=1}^{n}p_{i}\mathrm{e}^{-\lambda_{i}x}\geq\left(\sum_{i=1}^{n}\frac{p_{i}\lambda_{i}\mathrm{e}^{-\lambda_{i}x}}{1-p_{i}\mathrm{e}^{-\lambda_{i}x}}\right)^2.\end{equation*}

Thus, we only need to show that

\begin{equation*} \sum_{i=1}^{n}\frac{p_{i}\lambda_{i}^2\mathrm{e}^{-\lambda_{i}x}}{(1-p_{i}\mathrm{e}^{-\lambda_{i}x})^2}\left[1-\prod_{i=1}^{n}(1-p_{i}\mathrm{e}^{-\lambda_{i}x})^{1/n}\right]\geq\frac{1}{n}\sum_{i=1}^{n}\frac{p_{i}\lambda_{i}^2\mathrm{e}^{-\lambda_{i}x}}{(1-p_{i}\mathrm{e}^{-\lambda_{i}x})^2}\sum_{i=1}^{n}p_{i}\mathrm{e}^{-\lambda_{i}x} ,\end{equation*}

i.e.

\begin{equation*} \frac{1}{n}\sum_{i=1}^{n}(1-p_{i}\mathrm{e}^{-\lambda_{i}x})\geq\prod_{i=1}^{n}(1-p_{i}\mathrm{e}^{-\lambda_{i}x})^{1/n},\end{equation*}

which holds naturally by applying the arithmetic–geometric mean inequality.

The following result can be readily obtained from Theorem 1 by applying the fact that the convex transform order implies the star order and the Lorenz order.

The next result provides a lower bound on the coefficient of variation of the largest claim amount from a set of heterogeneous and independent insurance portfolios.

Corollary 2. Under the setup of Theorem 1, if $p\geq1-\prod_{i=1}^{n}(1-p_{i})^{1/n}$,

\begin{equation*} {\rm CV}(Y_{n\;:\;n})\geq \sqrt{\frac{\sum_{k=1}^{n}\binom{n}{k}p^{k}(1-p)^{n-k}\Big(\sum_{j=1}^{k}\frac{1}{j^2}+\big(\sum_{j=1}^{k}\frac{1}{j}\big)^2\Big)}{\Big(\sum_{k=1}^{n}\binom{n}{k}p^{k}(1-p)^{n-k}\sum_{j=1}^{k}\frac{1}{j}\Big)^2}-1}.\end{equation*}

Proof. Let $X_{k\;:\;k}^{\ast}$ be the largest order statistics from indpendent and identically distributed random variables $X_{i_{1}}^{\ast},\ldots,X_{i_{k}}^{\ast}$ with common hazard rate $\lambda$, for any $1\leq i_{1}<\cdots<i_{k}\leq n$. From [Reference Barlow and Proschan4, p. 60], it follows that

\begin{equation*} \mathbb{E}[X_{k\;:\;k}^{\ast}]=\frac{1}{\lambda}\sum_{j=1}^{k}\frac{1}{j} , \qquad \mbox{Var}[X_{k\;:\;k}^{\ast}]=\frac{1}{\lambda^2}\sum_{j=1}^{k}\frac{1}{j^2}.\end{equation*}

Note that the probability that the largest claim amount $Y_{n\;:\;n}^{\ast}$ takes value 0 is equal to $(1-p)^n$. Then, we have

\begin{equation*} \mathbb{E}[Y_{n\;:\;n}^{\ast}]=\sum_{k=1}^{n}\binom{n}{k}p^{k}(1-p)^{n-k}\mathbb{E}[X_{k\;:\;k}^{\ast}]=\frac{1}{\lambda}\sum_{k=1}^{n}\binom{n}{k}p^{k}(1-p)^{n-k}\sum_{j=1}^{k}\frac{1}{j}\end{equation*}

and

\begin{eqnarray*}\mathbb{E}[(Y_{n\;:\;n}^{\ast})^2]&=&\sum_{k=1}^{n}\binom{n}{k}p^{k}(1-p)^{n-k}\mathbb{E}[(X_{k\;:\;k}^{\ast})^2]\\&=&\frac{1}{\lambda^2}\sum_{k=1}^{n}\binom{n}{k}p^{k}(1-p)^{n-k}\left(\sum_{j=1}^{k}\frac{1}{j^2}+\left(\sum_{j=1}^{k}\frac{1}{j}\right)^2\right).\end{eqnarray*}

Then, by substituting $\mathbb{E}[Y_{n\;:\;n}^{\ast}]$ and $\mathbb{E}[(Y_{n\;:\;n}^{\ast})^2]$ into

\begin{equation*}{\rm CV}(Y_{n\;:\;n}^{\ast})=\frac{\sqrt{{\rm Var}[Y_{n\;:\;n}^{\ast}]}}{\mathbb{E}[Y_{n\;:\;n}^{\ast}]}=\sqrt{\frac{\mathbb{E}[(Y_{n\;:\;n}^{\ast})^2]}{\mathbb{E}^2[Y_{n\;:\;n}^{\ast}]}-1},\end{equation*}

the desired result can be obtained.

The next result gives equivalent characterizations of the dispersive order and the right-spread order between the largest claim amounts from two sets of independent heterogeneous and homogeneous insurance portfolios.

1. (i) $Y_{n\;:\;n}\geq_{\rm disp}Y_{n\;:\;n}^{\ast}$ if and only if $Y_{n\;:\;n}\geq_{\rm st}Y_{n\;:\;n}^{\ast}$;

2. (ii) $Y_{n\;:\;n}\geq_{\rm RS}Y_{n\;:\;n}^{\ast}$ if and only if $\mathbb{E}[Y_{n\;:\;n}]\geq\mathbb{E}[Y_{n\;:\;n}^{\ast}]$.

Proof.

1. (i) By using [Reference Ahmed, Alzaid, Bartoszewicz and Kochar1, Theorem 3], for two random variables X and Y, $X\geq_{\rm su}Y$ and $X\geq_{\rm st}Y$ imply that $X\geq_{\rm disp}Y$. According to Theorem 1, we have $Y_{n\;:\;n}\geq_{\rm c}Y_{n\;:\;n}^{\ast}$ and thus $Y_{n\;:\;n}\geq_{\rm su}Y_{n\;:\;n}^{\ast}$. Hence, the condition $Y_{n\;:\;n}\geq_{\rm st}Y_{n\;:\;n}^{\ast}$ implies that $Y_{n\;:\;n}\geq_{\rm disp}Y_{n\;:\;n}^{\ast}$. Conversely, observing that $Y_{n\;:\;n}\geq_{\rm disp}Y_{n\;:\;n}^{\ast}$ implies that $Y_{n\;:\;n}\geq_{\rm st}Y_{n\;:\;n}^{\ast}$, since their distributions have a common left-hand point of the support. Then, the required result follows immediately.

2. (ii) It was shown in [Reference Fernández-Ponce, Kochar and Muñoz-Perez15, Theorem 4.3] that $X\geq_{\rm NBUE}Y$ and $\mathbb{E}[X]\geq\mathbb{E}[Y]$ imply $X\geq_{\rm RS}Y$. Note that $Y_{n\;:\;n}\geq_{\rm c}Y_{n\;:\;n}^{\ast}$ implies that $Y_{n\;:\;n}\geq_{\rm NBUE}Y_{n\;:\;n}^{\ast}$, and $Y_{n\;:\;n}\geq_{\rm RS}Y_{n\;:\;n}^{\ast}$ implies that $\mathbb{E}[Y_{n\;:\;n}]\geq\mathbb{E}[Y_{n\;:\;n}^{\ast}]$. Then, the required result can be obtained by applying Theorem 1.

The next result presents sufficient conditions for the usual stochastic order and the dispersive order between the largest claim amounts arising from two sets of independent heterogeneous and homogeneous claims.

Proof. It suffices to show that $Y_{n\;:\;n}\geq_{\rm st}Y_{n\;:\;n}^{\ast}$ since the usual stochastic order implies the dispersive order according to the proof of Theorem 2(i). Note that the distribution function of $Y_{n\;:\;n}$ is given by $F_{n}(x)=\prod_{i=1}^{n}(1-p_{i}\mathrm{e}^{-\lambda_{i}x})$. Let $\psi(\boldsymbol{\lambda})=-\prod_{i=1}^{n}(1-p_{i}\mathrm{e}^{-\lambda_{i}x})$. Since $\bar{\lambda}=\frac{1}{n}\sum_{i=1}^{n}\lambda_i$, it holds that $(\lambda_1,\ldots,\lambda_n)\stackrel{\rm m}{\succeq}(\bar{\lambda},\ldots,\bar{\lambda})$. Observe that, for any $1\leq i<j\leq n$,

\begin{eqnarray*} \frac{\partial\psi({\boldsymbol\lambda})}{\partial\lambda_i}-\frac{\partial\psi({\boldsymbol\lambda})}{\partial\lambda_j}&=&\frac{p_ix\mathrm{e}^{-\lambda_ix}}{1-p_{i}\mathrm{e}^{-\lambda_{i}x}}\psi({\boldsymbol\lambda})-\frac{p_jx\mathrm{e}^{-\lambda_jx}}{1-p_{j}\mathrm{e}^{-\lambda_{j}x}}\psi({\boldsymbol\lambda})\\&\stackrel{\rm sgn}{=}&\frac{p_j\mathrm{e}^{-\lambda_jx}}{1-p_{j}\mathrm{e}^{-\lambda_{j}x}}-\frac{p_i\mathrm{e}^{-\lambda_ix}}{1-p_{i}\mathrm{e}^{-\lambda_{i}x}}\leq0,\end{eqnarray*}

where the inequality is due to $p_i\geq p_j$ and $\lambda_i\leq\lambda_j$. Then, in light of Lemma 1, we have $\psi({\boldsymbol\lambda})\geq\psi((\bar{\lambda},\ldots,\bar{\lambda}))$, which in turn implies that

(1) \begin{equation}\prod_{i=1}^{n}(1-p_{i}\mathrm{e}^{-\lambda_{i}x})\leq\prod_{i=1}^{n}(1-p_{i}\mathrm{e}^{-\bar{\lambda}x})\leq\prod_{i=1}^{n}(1-p_{i}\mathrm{e}^{-\lambda x}).\end{equation}

Next, we need to show that $\prod_{i=1}^{n}(1-p_{i}\mathrm{e}^{-\lambda x})\leq(1-p\mathrm{e}^{-\lambda x})^n$. It is easy to see that $(p_1,\ldots,p_n)\stackrel{\rm m}{\succeq}(\bar{p},\ldots,\bar{p})$. Let $\phi({\textbf{{p}}})=-\prod_{i=1}^{n}(1-p_i\mathrm{e}^{-\lambda x})$. Observe that, for any $1\leq i<j\leq n$,

\begin{equation*} \frac{\partial\phi({\textbf{{p}}})}{\partial p_i}-\frac{\partial\phi({\textbf{{p}}})}{\partial p_j}=\frac{\mathrm{e}^{-\lambda x}}{1-p_j\mathrm{e}^{-\lambda x}}\phi({\textbf{{p}}})-\frac{\mathrm{e}^{-\lambda x}}{1-p_i\mathrm{e}^{-\lambda x}}\phi({\textbf{{p}}})\stackrel{\rm sgn}{=} p_i-p_j\geq0.\end{equation*}

Then, by applying Lemma 1, it follows that $\phi({\textbf{{p}}})\geq\phi((\bar{p},\ldots,\bar{p}))$, which further implies that $\prod_{i=1}^{n}(1-p_{i}\mathrm{e}^{-\lambda x})\leq(1-\bar{p}\mathrm{e}^{-\lambda x})^n\leq(1-p\mathrm{e}^{-\lambda x})^n$. Then, the desired result is proved by combining this with (1).

Under certain conditions restricted on the occurrence probabilities, Theorem 3 states that the largest claim amount from the heterogenous insurance portfolio not only has a greater tail function than the largest claim amount from the homogeneous insurance portfolio, but also is more dispersive in the distance between the corresponding quantiles for any two fixed confidence levels. It is of interest to investigate whether the condition $\lambda\geq\bar{\lambda}$ in Theorem 3 could be relaxed to $\lambda\geq\tilde{\lambda}\;:\!=\;\left(\prod_{i=1}^n\lambda_i\right)^{1/n}$; this is left as an open problem.

The next theorem establishes an equivalent characterization for the right-spread ordering of the largest claim amounts.

Theorem 4. Under the setup of Theorem 1, if $p\geq1-\prod_{i=1}^{n}(1-p_{i})^{1/n}$ then $Y_{n\;:\;n}\geq_{\rm RS}Y_{n\;:\;n}^{\ast}$ if and only if $\lambda\geq\lambda^{\ast}$, where

(2) \begin{equation} \lambda^{\ast}=\frac{\sum_{k=1}^{n}\binom{n}{k}p^{k}(1-p)^{n-k}\sum_{j=1}^{k}\frac{1}{j}}{\sum_{k=1}^{n}(-1)^{k+1}\sum\limits_{1\leq i_1\leq\cdots\leq i_k\leq n}\left(\frac{\prod_{j=1}^{k}p_{i_j}}{\sum_{j=1}^{k}\lambda_{i_j}}\right)}. \end{equation}

Proof. According to the proof of Corollary 2, we have

\begin{equation*} \mathbb{E}[Y_{n\;:\;n}^{\ast}]=\frac{1}{\lambda}\sum_{k=1}^{n}\binom{n}{k}p^{k}(1-p)^{n-k}\sum_{j=1}^{k}\frac{1}{j}.\end{equation*}

On the other hand, we can calculate that

\begin{eqnarray*}\mathbb{E}[Y_{n\;:\;n}]&=&\int_{0}^{\infty}\left(1-\prod_{i=1}^{n}\big(1-p_{i}\mathrm{e}^{-\lambda_{i}x}\big)\right)\mathrm{d} x\\[5pt] &=& \int_{0}^{\infty}\left(\sum_{k=1}^{n}(-1)^{k+1}\sum\limits_{1\leq i_1\leq\cdots\leq i_k\leq n}\prod_{j=1}^{k}p_{i_j}\mathrm{e}^{-\sum_{j=1}^{k}\lambda_{i_j}x}\right)\mathrm{d} x\\[5pt] &=&\sum_{k=1}^{n}(-1)^{k+1}\sum\limits_{1\leq i_1\leq\cdots\leq i_k\leq n}\left(\frac{\prod_{j=1}^{k}p_{i_j}}{\sum_{j=1}^{k}\lambda_{i_j}}\right).\end{eqnarray*}

Then the desired result is proved by applying Theorem 2(ii).

Since the right-spread ordering between two random variables implies the ordering of their variances, the following corollary is a direct result of Theorem 4.

Corollary 3. Under the setup of Theorem 1, suppose that $p\geq1-\prod_{i=1}^{n}(1-p_{i})^{1/n}$. Then, $\lambda\geq\lambda^{\ast}$, where $\lambda^{\ast}$ is defined in (2), implies

(3) \begin{align} {\rm Var}[Y_{n\;:\;n}] & \geq \frac{1}{\lambda^2}\left[\sum_{k=1}^{n}\binom{n}{k}p^{k}(1-p)^{n-k}\left(\sum_{j=1}^{k}\frac{1}{j^2}+\left(\sum_{j=1}^{k}\frac{1}{j}\right)^2\right)\right. \notag\\ & \qquad\qquad\qquad\qquad\qquad\qquad\left.-\left(\sum_{k=1}^{n}\binom{n}{k}p^{k}(1-p)^{n-k}\sum_{j=1}^{k}\frac{1}{j}\right)^2\right], \end{align}

The following example illustrates the lower bound given in Corollary 3.

Figure 1. Plot of the lower bound of ${\rm Var}[Y_{3\;:\;3}]$ in Example 1 with respect to $p\in[0.7116,1]$.

To conclude this section, we establish the convex transform and star orderings for the largest claim amounts from two sets of homogeneous insurance portfolios when the baseline distributions of claim severities are different.

Proof. Let $F_{n}$ and $G_{n}$ be the distribution functions of $Y_{n\;:\;n}$ and $Y_{n\;:\;n}^{\ast}$, respectively. Then, for $x\in\mathbb{R}_{+}$, we have $F_{n}(x)=(1-p\bar{F}_1(x))^n$ and $G_{n}(x)=(1-p\bar{F}_2(x))^n$. First, it is easy to see that $F_{n}^{-1}(G_{n}(x))$ is well defined on $x\in\mathbb{R}_{+}$. Then, the convex transform ordering result is equivalent to showing that $F_{n}^{-1}(G_{n}(x))=\bar{F}_1^{-1}(\bar{F}_2(x))$ is convex in $x\in\mathbb{R}_{+}$, and the star ordering result is equivalent to showing that $F_{n}^{-1}(G_{n}(x))/x=\bar{F}_1^{-1}(\bar{F}_2(x))/x$ is increasing in $x\in\mathbb{R}_{+}$. Note that $\bar{F}_1^{-1}(\bar{F}_2(x))=F_1^{-1}(1-\bar{F}_2(x))=F_1^{-1}(F_2(x))$ for all $x\in\mathbb{R}_{+}$. Thus, the convex transform order holds if and only if $F_1^{-1}(F_2(x))$ is convex in $x\in\mathbb{R}_{+}$, and star order holds if and only if $F_1^{-1}(F_2(x))$ is star-shaped in $x\in\mathbb{R}_{+}$, which are implied by (ii) and (iii) in Definition 1, respectively.

A natural interesting question is to study the effects of the occurrence probabilities and the number of claims on the variability of largest claim amounts from homogeneous insurance portfolios by means of the convex transform order and the star order; this remains open and needs further investigation.

4. Star order

In the context of insurance practice, some insureds may have higher occurrence probabilities/claim sizes than others (in the sense of some stochastic orders between their claim severity distributions) in an insurance portfolio. Therefore, a natural way for describing this phenomenon falls in the multiple-outlier claims model. This section carries out stochastic comparisons on the largest claim amounts arising from multiple-outlier scaled or PHR claims in the sense of the star and dispersive orderings.

4.1. Multiple-outlier scaled claims

In this subsection we investigate the effect of heterogeneity among scale parameters on the skewness of the largest claim amounts from independent multiple-outlier scaled claims. The following lemma is originally due to [Reference Saunders and Moran34] and plays a key role in proving the main results.

Lemma 3. Let $\{G_{\lambda} \mid \lambda \in \mathbb{R}_{+}\}$ be a class of distribution functions such that $G_{\lambda}$ is supported on some open interval $I \subseteq \mathbb{R}_{+}$ and has density $g_{\lambda}$ which does not vanish on any subinterval of I. Then, $G_{\lambda} \leq_{\star} G_{\lambda^{\ast}}$, for $\lambda \leq \lambda^{\ast}$ if and only if $\big({\frac{\partial G_{\lambda}(x)}{\partial\lambda}}\big)/({xg_{\lambda}(x)})$ is decreasing in x, where $\frac{\partial G_{\lambda}(\cdot)}{\partial\lambda}$ is the derivative of $G_{\lambda}$ with respect to $\lambda$.

To begin with, let $G(\cdot)$ be the baseline distribution function with its reversed hazard rate function given by $\tilde{r}(\cdot)$.

Theorem 6. Let $X_{1},\ldots,X_{n_{1}}$ ($X_{1}^{\ast},\ldots,X_{n_{1}}^{\ast}$) be independent scaled random variables with distributions $G(\lambda_{1}x)$ ($G(\mu_{1}x)$), and $X_{n_{1}+1},\ldots,X_{n}$ ($X_{n_{1}+1}^{\ast},\ldots,X_{n}^{\ast}$) be another set of independent scaled random variables with distributions $G(\lambda_{2}x)$ ($G(\mu_{2}x)$) for $x\in\mathbb{R}_+$. Let $I_{1},\ldots,I_{n_{1}}$ be independent Bernoulli random variables such that $\mathbb{E}[I_{i}]=p_{1}$ for $i=1,2,\ldots,n_{1}$, and let $I_{n_{1}+1},\ldots,I_{n}$ be another set of independent Bernoulli random variables such that $\mathbb{E}[I_{j}]=p_{2}$ for $j=n_{1}+1,\ldots,n$. Suppose that $(p_{1}-p_{2})(\lambda_{1}-\lambda_{2})\geq0$ and $(\lambda_{1}-\lambda_{2})(\mu_{1}-\mu_{2})\geq0$. If both $x\tilde{r}(x)$ and $x\tilde{r}'(x)/\tilde{r}(x)$ are decreasing in $x\in\mathbb{R}_{+}$, we have

\begin{equation*} \frac{\lambda_{2\;:\;2}}{\lambda_{1\;:\;2}}\geq\frac{\mu_{2\;:\;2}}{\mu_{1\;:\;2}}\Longrightarrow Y_{n\;:\;n}\geq_{\rm \star}Y_{n\;:\;n}^{\ast}.\end{equation*}

Proof. Let $n_{2}=n-n_{1}$. Without loss of generality, it is assumed that $p_{1}\geq p_{2}$, $\lambda_{1}\geq\lambda_{2}$, and $\mu_{1}\geq\mu_{2}$. Note that the the distributions of $Y_{n\;:\;n}$ and $Y_{n\;:\;n}^{\ast}$ have the same probability at the value 0. The proof is completed by using the following two steps.

Case 1: $\lambda_{1}+\lambda_{2}=\mu_{1}+\mu_{2}=c$. Let $\lambda_{1}=\lambda\geq\lambda_{2}$ and $\mu_{1}=\mu\geq\mu_{2}$. Based on Lemma 3, it is enough to show that $\big({\frac{\partial G_{n,\lambda}(x)}{\partial \lambda}}\big)({xg_{n,\lambda}(x)})$ is decreasing in $x\in\mathbb{R}_{+}$ for $\lambda\in[\frac{c}{2},c]$. Note that the distribution of $Y_{n\;:\;n}$ can be denoted as $G_{n,\lambda}(x)=[1-p_{1}\bar{G}(\lambda x)]^{n_{1}}[1-p_{2}\bar{G}((c-\lambda)x)]^{n_{2}}$, with its density function defined as

\begin{align*} g_{n,\lambda}(x) & = [1-p_{1}\bar{G}(\lambda x)]^{n_{1}-1}[1-p_{2}\bar{G}((c-\lambda)x)]^{n_{2}-1} \\ & \quad \times \left\{n_{1}p_{1}\lambda g(\lambda x)[1-p_{2}\bar{G}((c-\lambda)x)]+n_{2}p_{2}(c-\lambda) g((c-\lambda)x)[1-p_{1}\bar{G}(\lambda x)]\right\}\end{align*}

and

\begin{align*} \frac{\partial G_{n,\lambda}(x)}{\partial \lambda} & = x[1-p_{1}\bar{G}(\lambda x)]^{n_{1}-1}[1-p_{2}\bar{G}((c-\lambda)x)]^{n_{2}-1} \\ & \quad \times \left\{n_{1}p_{1}g(\lambda x)[1-p_{2}\bar{G}((c-\lambda)x)]-n_{2}p_{2}g((c-\lambda)x)[1-p_{1}\bar{G}(\lambda x)]\right\}.\end{align*}

Then, it suffices to prove that

\begin{align*} \frac{\frac{\partial G_{n,\lambda}(x)}{\partial \lambda}}{xg_{n,\lambda}(x)} & = \frac{n_{1}p_{1}g(\lambda x)[1-p_{2}\bar{G}((c-\lambda)x)]-n_{2}p_{2}g((c-\lambda)x)[1-p_{1}\bar{G}(\lambda x)]}{n_{1}p_{1}\lambda g(\lambda x)[1-p_{2}\bar{G}((c-\lambda)x)]+n_{2}p_{2}(c-\lambda) g((c-\lambda)x)[1-p_{1}\bar{G}(\lambda x)]} \\ & = \left\{n_{1}p_{1}g(\lambda x)[1-p_{2}\bar{G}((c-\lambda)x)]-n_{2}p_{2}g((c-\lambda)x)[1-p_{1}\bar{G}(\lambda x)]\right\} \\ & \quad \times \left\{\lambda \{n_{1}p_{1}g(\lambda x)[1-p_{2}\bar{G}((c-\lambda)x)]-n_{2}p_{2}g((c-\lambda)x)[1-p_{1}\bar{G}(\lambda x)]\}\right. \\ & \qquad \quad \left. + \, c n_{2}p_{2}g((c-\lambda)x)[1-p_{1}\bar{G}(\lambda x)]\right\}^{-1} \\ & = \left(\lambda+\frac{c n_{2}p_{2}g((c-\lambda)x)[1-p_{1}\bar{G}(\lambda x)]}{n_{1}p_{1}g(\lambda x)[1-p_{2}\bar{G}((c-\lambda)x)]-n_{2}p_{2}g((c-\lambda)x)[1-p_{1}\bar{G}(\lambda x)]}\right)^{-1} \\ & = \left[\lambda+c\left(\frac{n_{1}}{n_{2}}\times\frac{p_{1}g(\lambda x)[1-p_{2}\bar{G}((c-\lambda)x)]}{p_{2}g((c-\lambda)x)[1-p_{1}\bar{G}(\lambda x)]}-1\right)^{-1}\right]^{-1}\end{align*}

is decreasing in $x\in\mathbb{R}_{+}$ for $\lambda\in[\frac{c}{2},c]$. Thus, it is sufficient to prove that

\begin{eqnarray*}\Delta(x)&=&\frac{p_{1}g(\lambda x)[1-p_{2}\bar{G}((c-\lambda)x)]}{p_{2}g((c-\lambda)x)[1-p_{1}\bar{G}(\lambda x)]}\\&=& \frac{\tilde{r}(\lambda x)}{\tilde{r}((c-\lambda)x)}\times\frac{p_{1}G(\lambda x)[1-p_{2}\bar{G}((c-\lambda)x)]}{p_{2}G((c-\lambda)x)[1-p_{1}\bar{G}(\lambda x)]}\\&=:&\Delta_{1}(x)\times\Delta_{2}(x)\end{eqnarray*}

is decreasing in $x\in\mathbb{R}_{+}$ for $\lambda\in[\frac{c}{2},c]$, where

\begin{equation*}\Delta_{1}(x)=\frac{\tilde{r}(\lambda x)}{\tilde{r}((c-\lambda)x)} , \qquad \Delta_{2}(x)=\frac{p_{1}G(\lambda x)[1-p_{2}\bar{G}((c-\lambda)x)]}{p_{2}G((c-\lambda)x)[1-p_{1}\bar{G}(\lambda x)]}.\end{equation*}

On one hand, by using the assumption that $x\tilde{r}'(x)/\tilde{r}(x)$ is decreasing in $x\in\mathbb{R}_{+}$, we have

\begin{equation*}\left[\frac{\tilde{r}(\lambda x)}{\tilde{r}((c-\lambda)x)}\right]'\stackrel{\rm sgn}{=}\frac{\lambda x\tilde{r}'(\lambda x)}{\tilde{r}(\lambda x)}-\frac{(c-\lambda)x\tilde{r}'((c-\lambda)x)}{\tilde{r}((c-\lambda)x)}\leq0.\end{equation*}

Next, we will prove that $\Delta_{2}(x)$ is also decreasing in $x\in\mathbb{R}_{+}$. Observe that

\begin{align*}\Delta^{\prime}_{2}(x) & \stackrel{\rm sgn}{=} \left\{p_{1}\lambda g(\lambda x)[1-p_{2}\bar{G}((c-\lambda)x)]+p_{1}p_{2}(c-\lambda)g((c-\lambda)x)G(\lambda x)\right\} \\&\quad\ \times p_{2}G((c-\lambda)x)[1-p_{1}\bar{G}(\lambda x)]\\&\quad\ -\left\{p_{2}(c-\lambda)g((c-\lambda)x)[1-p_{1}\bar{G}(\lambda x)]+p_{1}p_{2}\lambda g(\lambda x)G((c-\lambda)x)\right\} \\&\quad\ \times p_{1}G(\lambda x)[1-p_{2}\bar{G}((c-\lambda)x)]\\&=p_{1}p_{2}[1-p_{1}\bar{G}(\lambda x)][1-p_{2}\bar{G}((c-\lambda)x)]\\&\quad\times\left[\lambda g(\lambda x)G((c-\lambda)x)-(c-\lambda)g((c-\lambda)x)G(\lambda x)\right]\\&\quad+p_{1}p_{2}(G\lambda x)G((c-\lambda)x)\left\{p_{2}(c-\lambda)g((c-\lambda)x)[1-p_{1}\bar{G}(\lambda x)]\right.\\&\qquad\left.-p_{1}\lambda g(\lambda x)[1-p_{2}\bar{G}((c-\lambda)x)]\right\}\\&\stackrel{\rm sgn}{=} \lambda x\tilde{r}(\lambda x)-(c-\lambda)x\tilde{r}((c-\lambda)x)+\frac{p_{2}(c-\lambda)x g((c-\lambda)x)}{1-p_{2}\bar{G}((c-\lambda)x)}-\frac{p_{1}\lambda x g(\lambda x)}{1-p_{1}\bar{G}(\lambda x)}\\&= \lambda x\tilde{r}(\lambda x)-(c-\lambda)x\tilde{r}((c-\lambda)x)\\&\quad+(c-\lambda)x\tilde{r}((c-\lambda)x)\times\frac{p_{2} G((c-\lambda)x)}{1-p_{2}\bar{G}((c-\lambda)x)}-\lambda x\tilde{r}(\lambda x)\times\frac{p_{1}G(\lambda x)}{1-p_{1}\bar{G}(\lambda x)}\\&= \lambda x\tilde{r}(\lambda x)-(c-\lambda)x\tilde{r}((c-\lambda)x)\\&\quad+(c-\lambda)x\tilde{r}((c-\lambda)x)\times\left(1+\frac{\frac{1}{p_{2}}-1}{G((c-\lambda)x)}\right)^{-1}-\lambda x\tilde{r}(\lambda x)\times\left(1+\frac{\frac{1}{p_{1}}-1}{G(\lambda x)}\right)^{-1}\\&\leq\lambda x\tilde{r}(\lambda x)-(c-\lambda)x\tilde{r}((c-\lambda)x)\\&\quad+(c-\lambda)x\tilde{r}((c-\lambda)x)\times\left(1+\frac{\frac{1}{p_{1}}-1}{G(\lambda x)}\right)^{-1}-\lambda x\tilde{r}(\lambda x)\times\left(1+\frac{\frac{1}{p_{1}}-1}{G(\lambda x)}\right)^{-1}\\&=\left[\lambda x\tilde{r}(\lambda x)-(c-\lambda)x\tilde{r}((c-\lambda)x)\right]\times\left[1-\left(1+\frac{\frac{1}{p_{1}}-1}{G(\lambda x)}\right)^{-1}\right]\\&\leq0,\end{align*}

where the first inequality is based on $1\geq p_{1}\geq p_{2}\geq0$ and $G((c-\lambda)x)\leq G(\lambda x)$, and the last inequality is due to the decreasing property of $x\tilde{r}(x)$ in $x\in\mathbb{R}_+$. Thus, the proof is finished.

Case 2: $\lambda_{1}+\lambda_{2}\neq\mu_{1}+\mu_{2}$. In this case, there exists some $\alpha>0$ such that $\lambda_{1}+\lambda_{2}=\alpha(\mu_{1}+\mu_{2})$. 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 distribution $G(\alpha\mu_{1}x)$ and $Z_{n_{1}+1},\ldots,Z_{n}$ have distribution $G(\alpha\mu_{2}x)$. According to Case 1, $Y_{n\;:\;n}\geq_{\rm \star}Z_{n\;:\;n}$. Since the star order is scale invariant, it follows that $Y_{n\;:\;n}\geq_{\rm \star}Y_{n\;:\;n}^{\ast}$.

Remark 2. By considering the claim severity as the scale model $G(\lambda_i x)$ in Theorem 6, $i=1,2$, it holds that $G(\lambda_1 x)\leq G(\lambda_2 x)$ for $\lambda_1\leq\lambda_2$ and $x>0$. This means that the policyholders with claim severity distribution $G(\lambda_1 x)$ have larger claims than the policyholders with claim severity distribution $G(\lambda_2 x)$ in the sense of the usual stochastic order. The result of Theorem 6 implies that, under suitable conditions imposed on the baseline distribution, the occurrence probabilities, and the scale parameters, more heterogeneity among the scale parameters leads to a more skewed distribution function of the largest claim amount. For the case of $p_1=p_2=1$, the result of Theorem 6 reduces to [Reference Ding, Yang and Ling14, Theorem 8(a)], for which it is only required that $x\tilde{r}'(x)/\tilde{r}(x)$ is decreasing in $x\in\mathbb{R}_{+}$.

There are many distribution families that satisfy the conditions of Theorem 6. For example, a random variable X is said to have a generalized gamma distribution if its density function can be written as $g_{\alpha,\beta}(x)=\frac{\alpha}{\Gamma(\beta/\alpha)}x^{\beta-1}\mathrm{e}^{-x^{\alpha}}$, $x\in\mathbb{R}_{+}$, where $\alpha>0$ and $\beta>0$ are the shape parameters. It includes the widely used exponential $(\alpha=1,\beta=1)$, Weibull $(\alpha=\beta)$, and gamma $(\alpha=1)$ distributions as special cases. If we denote by $\tilde{r}(x;\alpha,\beta)$ the reversed hazard rate function of the generalized random variable, [Reference Khaledi, Farsinezhad and Kochar21] proved that $x\tilde{r}(x;\alpha,\beta)$ is decreasing in $x\in\mathbb{R}_{+}$ for $\alpha,\beta>0$, and [Reference Ding, Yang and Ling14] showed that $x\tilde{r}'(x;\alpha,\beta)/\tilde{r}(x;\alpha,\beta)$ is decreasing in $x\in\mathbb{R}_{+}$ for $\alpha,\beta>0$. Thus, the conditions of Theorem 6 always hold for the generalized gamma distribution.

Next, we give a numerical example to illustrate Theorem 6.

Figure 2. Plots of coefficients of variation of $Y_{n\;:\;n}$ and $Y_{n\;:\;n}^{\ast}$ under different settings.

Example 2. Suppose the baseline distribution is exponential with parameter 1. Figure 2 displays the plots of the coefficients of variation of $Y_{n\;:\;n}$ and $Y_{n\;:\;n}^{\ast}$ under the following six settings:

1. (a) $(n_1,n_2)=(8,2)$, $(\lambda_1,\lambda_2)=(0.9,0.3)$, $(\mu_1,\mu_2)=(0.25,0.1)$;

2. (b) $(n_1,n_2)=(8,2)$, $(\lambda_1,\lambda_2)=(0.9,0.3)$, $(\mu_1,\mu_2)=(1.6,0.6)$;

3. (c) $(n_1,n_2)=(2,10)$, $(\lambda_1,\lambda_2)=(0.9,0.3)$, $(\mu_1,\mu_2)=(0.25,0.1)$;

4. (d) $(n_1,n_2)=(2,10)$, $(\lambda_1,\lambda_2)=(0.9,0.3)$, $(\mu_1,\mu_2)=(1.6,0.6)$;

5. (e) $(n_1,n_2)=(2,4)$, $(\lambda_1,\lambda_2)=(0.9,0.3)$, $(\mu_1,\mu_2)=(0.6,1.6)$;

6. (f) $(n_1,n_2)=(4,2)$, $(\lambda_1,\lambda_2)=(0.9,0.3)$, $(\mu_1,\mu_2)=(0.6,1.6)$.

Under the settings of (a)–(d), these assumptions satisfy the conditions of Theorem 6 and thus the coefficient of variation of $Y_{n\;:\;n}$ is always larger than that of $Y_{n\;:\;n}^{\ast}$ for $p_1\geq p_2$. On the other hand, as observed in Figures 2(e) and 2(f), their coefficients of variation will cross each other if the condition $(\lambda_1-\lambda_2)(\mu_1-\mu_2)\geq0$ is not fulfilled.

4.2. Multiple-outlier PHR claims

Independent random variables $X_1,\ldots, X_n$ are said to follow a proportional hazard rates model if the survival function of $X_i$ can be written as $\bar{F}_i(x)=[\bar{F}(x)]^{\lambda_{i}}$ for $i = 1,\ldots,n$, where $\bar{F}(x)$ is the baseline survival function. Let $h(\cdot)$ be the hazard rate function of F. Then the survival function of $X_i$ can be written as $\bar{F}_i(x) = \mathrm{e}^{-\lambda_{i}R(x)}$ for $i = 1,\ldots,n$, where $R(x) =\int_{0}^{x}h(t) \, \mathrm{d} t$ is the cumulative hazard rate function of X. Many well-known distributions are special cases of the PHR model such as the exponential, Weibull, Pareto, and Lomax distributions.

This subsection studies the effects of heterogeneity among hazard rate parameters on the skewness of the largest claim amount from independent multiple-outlier PHR claims.

Theorem 7. Let $X_{1},\ldots,X_{n}$ ($X_{1}^{\ast},\ldots,X_{n}^{\ast}$) be independent random variables with survival functions $(\bar{F}^{\lambda_{1}}{\textbf{1}}_{n_{1}},\bar{F}^{\lambda_{2}}{\textbf{1}}_{n_{2}})$ ($(\bar{F}^{\mu_{1}}{\textbf{1}}_{n_{1}},\bar{F}^{\mu_{2}}{\textbf{1}}_{n_{2}})$), where $n_{1}+n_{2}=n$. Let $I_{1},\ldots,I_{n}$ be a set of independent Bernoulli random variables such that $\mathbb{E}[I_{i}]=p$ for $i=1,2,\ldots,n$. Suppose that $(\lambda_{1}-\lambda_{2})(\mu_{1}-\mu_{2})\geq0$. If $\frac{R(x)}{xh(x)}$ is increasing in $x\in\mathbb{R}_{+}$, we have $(\lambda_{1}{\textbf{1}}_{n_{1}},\lambda_{2}{\textbf{1}}_{n_{2}})\stackrel{\rm m}{\succeq}(\mu_{1}{\textbf{1}}_{n_{1}},\mu_{2}{\textbf{1}}_{n_{2}})\Longrightarrow Y_{n\;:\;n}\geq_{\rm \star}Y_{n\;:\;n}^{\ast}$,

Proof. Without loss of generality, it is assumed that $\lambda_{1}\leq\lambda_{2}$ and $\mu_{1}\leq\mu_{2}$. Let $\lambda=\lambda_{2}$, $\mu=\mu_{2}$, and $n_{1}\lambda_{1}+n_{2}\lambda_{2}=n_{1}\mu_{1}+n_{2}\mu_{2}=1$. Then we have $\lambda_{1}=(1-n_{2}\lambda_{2})/n_{1}$ and $\lambda\in[1/(n_{1}+n_{2}),1/n_{2})$. The distribution function of $Y_{n\;:\;n}$ is given by

\begin{equation*}F_{\lambda, n\;:\;n}(x)=\left[1-p\exp\left\{-\left(\frac{1-n_{2}\lambda}{n_{1}}\right)R(x)\right\}\right]^{n_{1}}\left[1-p\mathrm{e}^{-\lambda R(x)}\right]^{n_{2}}.\end{equation*}

Taking the derivative of $F_{Y_{n\;:\;n}}(x)$ with respective to $\lambda$ gives

\begin{multline*}\frac{\partial F_{\lambda, n\;:\;n}(x)}{\partial \lambda} = -R(x)\left[\frac{n_{2}p\exp\left\{-\left(\frac{1-n_{2}\lambda}{n_{1}}\right)R(x)\right\}}{1-p\exp\left\{-\left(\frac{1-n_{2}\lambda}{n_{1}}\right)R(x)\right\}}-\frac{n_{2}p\mathrm{e}^{-\lambda R(x)}}{1-p\mathrm{e}^{-\lambda R(x)}}\right]\\\times\left[1-p\exp\left\{-\left(\frac{1-n_{2}\lambda}{n_{1}}\right)R(x)\right\}\right]^{n_{1}}\left[1-p\mathrm{e}^{-\lambda R(x)}\right]^{n_{2}}.\end{multline*}

The density function of $Y_{n\;:\;n}$ is

\begin{multline*}f_{\lambda, n\;:\;n}(x) = h(x)\left[\frac{(1-n_{2}\lambda)p\exp\left\{-\left(\frac{1-n_{2}\lambda}{n_{1}}\right)R(x)\right\}}{1-p\exp\left\{-\left(\frac{1-n_{2}\lambda}{n_{1}}\right)R(x)\right\}}+\frac{\lambda n_{2}p\mathrm{e}^{-\lambda R(x)}}{1-p\mathrm{e}^{-\lambda R(x)}}\right] \\ \times\left[1-p\exp\left\{-\left(\frac{1-n_{2}\lambda}{n_{1}}\right)R(x)\right\}\right]^{n_{1}}\left[1-p\mathrm{e}^{-\lambda R(x)}\right]^{n_{2}}.\end{multline*}

By making use of Lemma 3, it is equivalent to showing that

\begin{equation*}-\frac{\frac{\partial F_{\lambda, n\;:\;n}(x)}{\partial \lambda}}{xf_{\lambda, n\;:\;n}(x)}=\frac{R(x)}{xh(x)}\cdot\frac{\frac{n_{2}p\exp\left\{-\left(\frac{1-n_{2}\lambda}{n_{1}}\right)R(x)\right\}}{1-p\exp\left\{-\left(\frac{1-n_{2}\lambda}{n_{1}}\right)R(x)\right\}}-\frac{n_{2}p\mathrm{e}^{-\lambda R(x)}}{1-p\mathrm{e}^{-\lambda R(x)}}}{\frac{(1-n_{2}\lambda)p\exp\left\{-\left(\frac{1-n_{2}\lambda}{n_{1}}\right)R(x)\right\}}{1-p\exp\left\{-\left(\frac{1-n_{2}\lambda}{n_{1}}\right)R(x)\right\}}+\frac{\lambda n_{2}p\mathrm{e}^{-\lambda R(x)}}{1-p\mathrm{e}^{-\lambda R(x)}}}=:\frac{R(x)}{xh(x)}\cdot\frac{\Omega_{1}(x)}{\Omega_{2}(x)}\end{equation*}

is increasing in $x\in\mathbb{R}_{+}$. Note that

\begin{eqnarray*}\Omega_{1}(x)&=&\frac{n_{2}p\exp\left\{-\left(\frac{1-n_{2}\lambda}{n_{1}}\right)R(x)\right\}}{1-p\exp\left\{-\left(\frac{1-n_{2}\lambda}{n_{1}}\right)R(x)\right\}}-\frac{n_{2}p\mathrm{e}^{-\lambda R(x)}}{1-p\mathrm{e}^{-\lambda R(x)}}\\&\stackrel{\rm sgn}{=}&\left[\frac{1}{p}\exp\left\{\left(\frac{1-n_{2}\lambda}{n_{1}}\right)R(x)\right\}-1\right]^{-1}-\left[\frac{1}{p}\mathrm{e}^{\lambda R(x)}-1\right]^{-1}\geq0,\end{eqnarray*}

where the inequality is due to $(1-n_{2}\lambda)/n_{1}\leq \lambda$. Based on the assumption that $\frac{R(x)}{xh(x)}$ is increasing in $x\in\mathbb{R}_{+}$, it suffices to show that $\Omega_{3}(x)=\Omega_{1}(x)/\Omega_{2}(x)$ is also increasing in $x\in\mathbb{R}_{+}$. Observe that

\begin{equation*}\Omega_{3}(x)=\left[n_{2}^{-1}\left(1-\frac{p\mathrm{e}^{-\lambda R(x)}}{1-p\mathrm{e}^{-\lambda R(x)}}\times\left(\frac{p\exp\left\{-\left(\frac{1-n_{2}\lambda}{n_{1}}\right)R(x)\right\}}{1-p\exp\left\{-\left(\frac{1-n_{2}\lambda}{n_{1}}\right)R(x)\right\}}\right)^{-1}\right)^{-1}-\lambda\right]^{-1}.\end{equation*}

It is enough to prove that

\begin{equation*}\Omega_{4}(x)=\frac{p\mathrm{e}^{-\lambda R(x)}}{1-p\mathrm{e}^{-\lambda R(x)}} \!\left(\frac{p\exp\left\{-\left(\frac{1-n_{2}\lambda}{n_{1}}\right)R(x)\right\}}{1-p\exp\left\{-\left(\frac{1-n_{2}\lambda}{n_{1}}\right)R(x)\right\}}\right)^{-1}\!\!=\frac{\exp\left\{\left(\frac{1-n_{2}\lambda}{n_{1}}\right)R(x)\right\}-p}{\mathrm{e}^{\lambda R(x)}-p}\end{equation*}

is decreasing in $x\in\mathbb{R}_{+}$, which can be obtained from the fact that $(1-n_{2}\lambda)/n_{1}\leq \lambda$ and $x/(1-p\mathrm{e}^{-x})$ is increasing in $x\in\mathbb{R}_{+}$ for any $p\in[0,1]$, and the observation that

\begin{align*} \Omega^{\prime}_{4}(x) & \stackrel{\rm sgn}{=} \frac{1-n_{2}\lambda}{n_{1}}h(x)\exp\left\{\left(\frac{1-n_{2}\lambda}{n_{1}}\right)R(x)\right\}\left[\mathrm{e}^{\lambda R(x)}-p\right] \\[5pt] & \quad \ - \lambda h(x)\mathrm{e}^{\lambda R(x)}\left[\exp\left\{\left(\frac{1-n_{2}\lambda}{n_{1}}\right)R(x)\right\}-p\right] \\[5pt] & \stackrel{\rm sgn}{=} \frac{\frac{1-n_{2}\lambda}{n_{1}}R(x)}{1-p\exp\left\{\left(-\frac{1-n_{2}\lambda}{n_{1}}\right)R(x)\right\}}-\frac{\lambda R(x)}{1-p\mathrm{e}^{-\lambda R(x)}}\leq0.\end{align*}

Hence, the desired result follows.

The next example provides an illustration for the condition $\frac{R(x)}{xh(x)}$ is increasing in $x\in\mathbb{R}_{+}$ employed in Theorem 7 when the baseline distribution is taken as the power-generalized Weibull distribution.

Example 3. A random variable X is said to have the power-generalized Weibull distribution, denoted by $X\sim PGW(\nu,\gamma)$, if its density function is given as $f(x;\nu,\gamma)=\frac{\nu}{\gamma}x^{\nu-1}(1+x^\nu)^{\frac{1}{\gamma}-1}\exp\big\{1-(1+x^\nu)^{\frac{1}{\gamma}}\big\}$, $x>0$, $\nu,\gamma>0$. Then its survival function can be derived as $\bar{F}(x;\nu,\gamma)=\exp\big\{1-(1+x^\nu)^{\frac{1}{\gamma}}\big\}$ for $x>0$ and $\nu,\gamma>0$. The hazard rate function is $h(x;\nu,\gamma)=\frac{\nu}{\gamma}x^{\nu-1}(1+x^\nu)^{\frac{1}{\gamma}-1}$ for $x>0$ and $\nu,\gamma>0$. Then, we have

\begin{equation*} \psi_1(x)\;:\!=\;\frac{R(x;\nu,\gamma)}{xh(x;\nu,\gamma)}=\frac{\int_0^x t^{\nu-1}(1+t^\nu)^{\frac{1}{\gamma}-1} \, \mathrm{d} t}{x^{\nu}(1+x^\nu)^{\frac{1}{\gamma}-1}},\qquad x>0,\quad \nu,\gamma>0.\end{equation*}

So,

\begin{eqnarray*} \psi^{\prime}_1(x) &\stackrel{\rm sgn}{=}& x^{\nu-1}(1+x^\nu)^{\frac{1}{\gamma}-1}\times x^{\nu}(1+x^\nu)^{\frac{1}{\gamma}-1}-\int_0^x t^{\nu-1}(1+t^\nu)^{\frac{1}{\gamma}-1} \, \mathrm{d} t \\ &&\quad\times\left[\nu x^{\nu-1}(1+x^\nu)^{\frac{1}{\gamma}-1}+\left(\frac{1}{\gamma}-1\right)\nu x^{2\nu-1}(1+x^\nu)^{\frac{1}{\gamma}-2}\right] \\ &\stackrel{\rm sgn}{=}& x^{\nu}(1+x^\nu)^{\frac{1}{\gamma}} -\nu\left(1+\frac{x^\nu}{\gamma}\right)\int_0^x t^{\nu-1}(1+t^\nu)^{\frac{1}{\gamma}-1} \, \mathrm{d} t.\end{eqnarray*}

Note that it is not easy to judge the sign of $\psi^{\prime}_1(x)$ on $x>0$ for arbitrary $\nu,\gamma>0$. Here, we consider the following two special cases:

Case 1: $\nu=1$ and $\gamma\geq1$. For this case, one can note that

\begin{equation*} \psi^{\prime}_1(x) \stackrel{\rm sgn}{=} x(1+x)^{\frac{1}{\gamma}}-\left(1+\frac{x}{\gamma}\right)\int_0^x (1+t^\nu)^{\frac{1}{\gamma}-1} \, \mathrm{d} t = x+r-r(1+x)^{\frac{1}{\gamma}}=:\psi_2(x).\end{equation*}

Since $\psi^{\prime}_2(x)=1-(1+x)^{\frac{1}{\gamma}-1}\geq0$ for all $x>0$ and $\gamma\geq1$, it follows that $\psi_2(x)$ is increasing in $x>0$. Thus, $\psi_2(x)\geq\lim_{x\to0_+}\psi_2(x)=0$, which implies that $\psi^{\prime}_1(x)\geq0$, i.e. $\psi_1(x)$ is increasing in $x>0$ when $\nu=1$ and $\gamma\geq1$.

Case 2: $\gamma=1$. It is easy to check that $\psi^{\prime}_1(x)=0$, which means that $\psi_1(x)$ is a constant for $x>0$. Simple calculations can be implemented to derive that $\psi_1(x)=\frac{1}{\nu}$. This case corresponds to the traditional Weibull distribution, and has been also verified in some recent works; see [Reference Amini-Seresht, Qiao, Zhang and Zhao2, Reference Kochar and Xu25], for example.

Remark 3. The result of Theorem 7 was proved in [Reference Ahmed, Alzaid, Bartoszewicz and Kochar2, Theorem 3.10] for the case of $p=1$. One may wonder whether the result of Theorem 7 could be generalized to the setting of different occurrence probabilities; however, we cannot prove it with a similar proof method to Theorem 7. We leave it as an open problem.

The next example states that the condition $(\lambda_1-\lambda_2)(\mu_1-\mu_2)\geq0$ in Theorem 7 cannot be removed.

Example 4. Let $\bar{F}(x)=\mathrm{e}^{-x}$ for $x\in\mathbb{R}_+$. Assume that $(n_1,n_2)=(3,4)$, $(\lambda_1,\lambda_2)=(1.9,0.3)$, and $(\mu_1,\lambda_2)=(13/30,1.4)$. It is plain that $(\lambda_1-\lambda_2)(\mu_1-\mu_2)<0$ and $(\lambda_{1}{\textbf{1}}_{n_{1}},\lambda_{2}{\textbf{1}}_{n_{2}})\stackrel{\rm m}{\succeq}(\mu_{1}{\textbf{1}}_{n_{1}},\mu_{2}{\textbf{1}}_{n_{2}})$. As shown in Figure 3, the coefficients of variation of $Y_{n\;:\;n}$ and $Y_{n\;:\;n}^{\ast}$ cross each other for $p\in[0,1]$. Hence, $(\lambda_1-\lambda_2)(\mu_1-\mu_2)\geq0$ is necessary in Theorem 7.

The following result can be proved by using a similar proof method to Theorem 7; the details are omitted for brevity.

Figure 3. Plots of coefficients of variation of $Y_{n\;:\;n}$ and $Y_{n\;:\;n}^{\ast}$.

Theorem 8. Let $X_{1},\ldots,X_{n}$ ($X_{1}^{\ast},\ldots,X_{n}^{\ast}$) be independent random variables with survival functions $(\bar{F}^{\lambda_{1}}{\textbf{1}}_{n_{1}},\bar{F}^{\lambda}{\textbf{1}}_{n_{2}})$ ($(\bar{F}^{\lambda_{2}}{\textbf{1}}_{n_{1}},\bar{F}^{\lambda}{\textbf{1}}_{n_{2}})$), where $n_{1}+n_{2}=n$. Let $I_{1},\ldots,I_{n}$ be a set of independent Bernoulli random variables such that $\mathbb{E}[I_{i}]=p$ for $i=1,\ldots,n$. Suppose that $\lambda_{1}\leq\lambda_{2}\leq\lambda$. If $\frac{R(x)}{xh(x)}$ is increasing in $x\in\mathbb{R}_{+}$, we have $Y_{n\;:\;n}\geq_{\rm \star}Y_{n\;:\;n}^{\ast}$.

A general result follows stating that the largest claim amount becomes more skewed if the hazard rate parameters have more heterogeneity.

Theorem 9. Under the same setup as Theorem 7, we assume that $\lambda_{1}\leq\mu_{1}\leq\mu_{2}\leq\lambda_{2}$. If $\frac{R(x)}{xh(x)}$ is increasing in $x\in\mathbb{R}_{+}$, we have $(\lambda_{1}{\textbf{1}}_{n_{1}},\lambda_{2}{\textbf{1}}_{n_{2}})\stackrel{\rm w}{\succeq}(\mu_{1}{\textbf{1}}_{n_{1}},\mu_{2}{\textbf{1}}_{n_{2}})\Longrightarrow Y_{n\;:\;n}\geq_{\rm \star}Y_{n\;:\;n}^{\ast}$.

Proof. There is some $\lambda^{\prime}_{1}$ such that $\lambda_{1}<\lambda^{\prime}_{1}\leq\mu_{1}$, $(\lambda^{\prime}_{1}{\textbf{1}}_{n_{1}},\lambda_{2}{\textbf{1}}_{n_{2}})\stackrel{\rm m}{\succeq}(\mu_{1}{\textbf{1}}_{n_{1}},\mu_{2}{\textbf{1}}_{n_{2}})$. Let $Z_{1},\ldots,Z_{n}$ be a set of independent random variables with tail functions $(\bar{F}^{\lambda^{\prime}_{1}}{\textbf{1}}_{n_{1}}$, $\bar{F}^{\lambda_{2}}{\textbf{1}}_{n_{2}})$. Denote $Z_{n\;:\;n}^{\ast}=\max\{I_{p}Z_{1},\ldots,I_{p}Z_{n_{1}},I_{p}Z_{n_{1}+1},\ldots,I_{p}Z_{n}\}$. From Theorem 8, we know that $Y_{n\;:\;n}\geq_{\rm \star}Z_{n\;:\;n}^{\ast}$. According to Theorem 7, $Z_{n\;:\;n}^{\ast}\geq_{\rm \star}Y_{n\;:\;n}^{\ast}$ holds. Hence, the theorem is proved.

It is interesting to note that the result of Theorem 9 is contained in [Reference Amini-Seresht, Qiao, Zhang and Zhao2, Theorem 3.11] under the setting of $p=1$. Indeed, it was proved in [Reference Amini-Seresht, Qiao, Zhang and Zhao2, Theorem 3.15] that the conclusion still holds if the weak supermajorization order is replaced by the p-larger order. We conjecture that the result of Theorem 9 might also hold under the p-larger order; this is left as an open problem.

The next result shows that more heterogeneity among the hazard rate parameters leads to more variability of the largest claim amount according to the dispersive order.

Theorem 10. Under the same setup as Theorem 7, we assume that $\lambda_{1}\leq\mu_{1}\leq\mu_{2}\leq\lambda_{2}$. If $\frac{R(x)}{xh(x)}$ is increasing in $x\in\mathbb{R}_{+}$, we have $(\lambda_{1}{\textbf{1}}_{n_{1}},\lambda_{2}{\textbf{1}}_{n_{2}})\stackrel{\rm w}{\succeq}(\mu_{1}{\textbf{1}}_{n_{1}},\mu_{2}{\textbf{1}}_{n_{2}})\Longrightarrow Y_{n\;:\;n}\geq_{\rm disp}Y_{n\;:\;n}^{\ast}$.

Proof. It was proved in [Reference Ahmed, Alzaid, Bartoszewicz and Kochar1] that, for two random variables X and Y, if $X \leq_{\star}Y$ then $X \leq_{\rm st} Y$ implies that $X \leq_{\rm disp}Y$. According to [Reference Balakrishnan, Zhang and Zhao3, Theorem 3.7], $Y_{n\;:\;n}\geq_{\rm st}Y_{n\;:\;n}^{\ast}$ holds. Then, the desired result can be obtained from Theorem 9.

Remark 4. Since both $Y_{n\;:\;n}$ and $Y_{n\;:\;n}^{\ast}$ are discrete-continuous random variables, the dispersive order in Theorem 10 should be understood in the sense that $F^{-1}_{Y_{n\;:\;n}}(0)=F^{-1}_{Y_{n\;:\;n}^{\ast}}(0)=(1-p)^{n}$ and $F^{-1}_{Y_{n\;:\;n}}(u)-F^{-1}_{Y_{n\;:\;n}}(v)\geq F^{-1}_{Y_{n\;:\;n}^{\ast}}(u)-F^{-1}_{Y_{n\;:\;n}^{\ast}}(v)$ for any $(1-p)^{n}<v\leq u<1$.