Ordering results between extreme order statistics in models with dependence defined by Archimedean [survival] copulas

Abstract

Motivated by recent works about stochastic comparisons between extreme order statistics arising from heterogeneous and dependent random variables, where the dependency structure is defined by the family of Archimedean copulas and the marginal distributions follow some specific parametric distributions, in this work, we investigate the case in which the marginal distributions can have arbitrary distribution functions depending on some parameter. Such parameter can be a shape, scale or location parameter, but other kinds of parameters, as frailty, resilience or tilt parameters can be also considered. Hence, the modified proportional hazard rate scale (MPHRS) and the modified proportional reversed hazard rate scale (MPRHRS) models, among others, belong to the wide parametric model studied here. Under this setup, we provide some general results for the usual stochastic order, when the parameter vectors verify the p-larger order or the reciprocally majorization order, generalizing some of the existing results in the literature. Besides this, extreme order statistics arising from the dependent MPHRS and MPRHRS models are compared in the sense of the reversed hazard rate order and the hazard rate order as well.

Appendix

Proof of Corollary 3.6

(i) To obtain the desired result, from (3.11) and Theorem 3.1, it is sufficient to show that

$$\begin{aligned} {H_{G}(x;e^{a_i})}=\frac{1-[1-G(x e^{a_i})]^{\lambda }}{1-{{\bar{\alpha }}}[1- G(x e^{a_i})]^{\lambda }} \end{aligned}$$

is increasing and log-concave with respect to $a_i$, where $a_i=\log \delta _i$, for $i=1,\ldots ,n$. After some computations, the partial derivative of the above function with respect to $a_i$ is

$$\begin{aligned} \frac{\partial {H_{G}(x;e^{a_i})}}{\partial a_i}= & {} \alpha \lambda x e^{a_i} r_{G}(x e^{a_i}) \frac{[1- G(x e^{a_i})]^{\lambda }}{(1-{{\bar{\alpha }}}[1- G(x e^{a_i})]^{\lambda })^{2}}\nonumber \\= & {} {\frac{\lambda x e^{a_i} r_{G}(x e^{a_i})}{1-{{\bar{\alpha }}}[1- G(x e^{a_i})]^{\lambda }}{{{{\bar{H}}}}_{G}(x;e^{a_i})}\ge 0,} \end{aligned}$$

(5.1)

which means that ${ H_{G}(x;e^{a_i})}$ is increasing in $a_i$, for $i=1,\ldots ,n$. On the other hand,

$$\begin{aligned} \frac{\partial \log { H_{G}(x;e^{a_i})}}{\partial a_i}= & {} {\frac{\lambda x e^{a_i} r_{G}(x e^{a_i})}{1-[1- G(x e^{a_i})]^{\lambda }}{{{{\bar{H}}}}_{G}(x;e^{a_i})}.} \end{aligned}$$

(5.2)

Now, observe that ${{{\bar{H}}}_{G}(x;e^{a_i})}$ is decreasing in $a_i$ and, it is easy to verify that if $ur_{G}(u)$ is decreasing in u. Then, the function

$$\begin{aligned} \frac{\lambda x e^{a_i} r_{G}(x e^{a_i})}{1-[1- G(x e^{a_i})]^{\lambda }} \end{aligned}$$

is also decreasing in $a_i$. Therefore, (5.2) is decreasing in $a_i$, that is, ${H}_{G}(x;e^{a_i})$ is log-concave with respect to $a_i$.

(ii) Again, from (3.11) and Theorem 3.1, it is enough to verify that

$$\begin{aligned} {H_{G}(x;e^{a_i})}=\frac{1-[1- G(t)]^{e^{a_i}}}{1-{{\bar{\alpha }}}[1- G(t)]^{e^{a_i}}} \end{aligned}$$

is increasing and log-concave with respect to $a_i$, where $a_i=\log \mu _i$, for $i=1,\ldots ,n$, and $t=x \theta$. The partial derivative of the above function with respect to $a_i$ yields

$$\begin{aligned} \frac{\partial {H_{G}(x;e^{a_i})}}{\partial a_i}= -\frac{\alpha e^{a_i} [1- G(t)]^{e^{a_i}} \log \left( 1-{G}(t)\right) }{(1-{{\bar{\alpha }}}[1- G(t)]^{e^{a_i}})^{2}}\ge 0, \end{aligned}$$

(5.3)

i.e, ${H_{G}(x;e^{a_i})}$ is increasing in $a_i$, for $i=1,\ldots ,n$, and further,

$$\begin{aligned} \frac{\partial \log {H_{G}(x;e^{a_i})}}{\partial a_i}= & {} -\frac{\alpha e^{a_i} [1- G(t)]^{e^{a_i}} \log \left( 1-{G}(t)\right) }{(1-{{\bar{\alpha }}}[1- G(t)]^{e^{a_i}})(1-[1- G(t)]^{e^{a_i}})}\nonumber \\= & {} {-\alpha \log \left( 1-{G}(t)\right) \frac{ e^{a_i} [1- G(t)]^{e^{a_i}}}{1-[1- G(t)]^{e^{a_i}}}\cdot \frac{1}{1-{{{\bar{\alpha }}}}[1- G(t)]^{e^{a_i}}}}.\nonumber \\ \end{aligned}$$

(5.4)

Now, from Lemma A.12 in [8], it is known that the function

$$\begin{aligned} \frac{ e^{a_i} [1- G(t)]^{e^{a_i}}}{1-[1- G(t)]^{e^{a_i}}}\end{aligned}$$

is decreasing in $a_i$. Moreover, it can be easily verified that the function $1/(1-{{{\bar{\alpha }}}}[1- G(t)]^{e^{a_i}})$ is also decreasing in $a_i$ when $0<\alpha <1$. Therefore, (5.4) is a decreasing function in $a_i$, i.e., $H_{G}(x;e^{a_i})$ is log-concave in $a_i$. $\square$

Proof of Corollary 3.9

From (3.11) and Remark 3.5, it is required to show that

$$\begin{aligned} { H_{G}(x;e^{a_i})}=\frac{1-[{{\bar{G}}}(t)]^{\lambda }}{1-(1-e^{a_i})[{{{\bar{G}}}}(t)]^{\lambda }} \end{aligned}$$

is monotone and log-concave with respect to $a_i$, where $a_i=\log \beta _i$, for $i=1,\ldots ,n$, and $t=x \theta$. Differentiating the above function with respect to $a_i$, we have

$$\begin{aligned} \frac{\partial { H_{ G}(x;e^{a_i})}}{\partial a_i}=- \frac{e^{a_i} [{{{\bar{G}}}}(t)]^{\lambda } (1-[{{\bar{G}}}(t)]^{\lambda })}{(1-(1-e^{a_i})[{{{\bar{G}}}}(t)]^{\lambda })^{2}}=- { H_{ G}(x;e^{a_i})}{{{{\bar{H}}}}_{{{\bar{G}}}}(x;e^{a_i})} \le 0, \end{aligned}$$

i.e., ${H_{G}(x;e^{a_i})}$ is decreasing with respect to $a_{i}$, for $i=1,\ldots ,n$. Therefore, it is monotone. On the other hand,

$$\begin{aligned}\frac{\partial \log {H_{G}(x;e^{a_i})}}{\partial a_i}=- {{{{\bar{H}}}}_{{{{\bar{G}}}}}(x;e^{a_i})}, \end{aligned}$$

which is decreasing since the function ${{\bar{H}}}_{G}(x;e^{a_i})$ is increasing in $a_i$. Hence, ${H_{G}(x;e^{a_i})}$ is log-concave with respect to $a_i$. $\square$

Proof of Corollary 3.13

(i) From (3.11) and Theorem 3.10, it is required to prove that

$$\begin{aligned} { H_{ G}(x;1/{a_i})}=\frac{1-[1- G(x /{a_i})]^{\lambda }}{1-{{\bar{\alpha }}}[1- G(x /{a_i})]^{\lambda }} \end{aligned}$$

is decreasing and log-concave with respect to $a_i$, where $a_i=1/{\delta _i}$, for $i=1,\ldots ,n$. After some standard calculations, the partial derivative of the above function with respect to $a_i$ is

$$\begin{aligned} \frac{\partial { H_{ G}\left( x;1/{a_i}\right) }}{\partial a_i}= & {} -\frac{\alpha \lambda x}{a^{2}_i} r_{G}\left( {x}/{a_i}\right) \frac{[1- G(x /{a_i})]^{\lambda }}{(1-{{\bar{\alpha }}}[1- G(x/{a_i})]^{\lambda })^{2}}\\= & {} {-\frac{\lambda x r_{G}(x/a_i)}{a_i^2} \frac{{{{\bar{H}}}_{ {{{\bar{G}}}}}(x;1/{a_i})}}{1-{{\bar{\alpha }}}[1- G(x /{a_i})]^{\lambda }}} \le 0, \end{aligned}$$

i.e., ${ H_{ G}\left( x;1/{a_i}\right) }$ is decreasing in $a_i$, for $i=1,\ldots ,n$. Further,

$$\begin{aligned} \frac{\partial \log { H_{ G} (x;1/{a_i})}}{\partial a_i}= & {} - \frac{\lambda x r_{G}(x /{a_i})}{a^2_i} \frac{{{\bar{H}}}_{{{{\bar{G}}}}}(x;1/{a_i})}{1-[1- G(x/{a_i})]^{\lambda }}, \end{aligned}$$

which is decreasing in $a_i$, whenever $u^2r_{G}(u)$ is decreasing in u, since both functions ${{\bar{H}}}_{G}(x;1/{a_i})$ and $1/(1-[1- G(x/{a_i})]^{\lambda })$ are increasing in $a_i$.

(ii) Now, from (3.11) and Remark 3.12, it is sufficient to prove that

$$\begin{aligned} { H_{ G}(x;1/{a_i})}=\frac{1-[ 1- G(t)]^{\lambda }}{1-(1-1/{a_i})[ 1- G(t)]^{\lambda }} \end{aligned}$$

(5.5)

is monotone and log-concave in $a_i$, where $a_i=1/\beta _i$, for $i=1,\ldots ,n$, and $t=x \theta$. Differentiating the above function with respect to $a_i$ partially, we get

$$\begin{aligned} \frac{\partial { H_{ G}(x;1/{a_i})}}{\partial a_i}= & {} \frac{ [ 1- G(t)]^{\lambda } (1-[ 1- G(t)]^{\lambda })}{{a^2_i}(1-(1-1/{a_i})[ 1- G(t)]^{\lambda })^{2}}\\= & {} \frac{1}{a_i}{{{\bar{H}}}_{{{\bar{G}}}}(x;1/{a_i})}{ H_{ G}(x;1/{a_i})} \ge 0, \end{aligned}$$

i.e., ${ H_{ G}(x;1/{a_i})}$ is increasing in $a_i$, for $i=1,\ldots ,n$, i.e., it is monotone. On the other hand,

$$\begin{aligned} \frac{\partial \log { H_{ G}(x;1/{a_i})}}{\partial a_i}= \frac{1}{a_i}{{{\bar{H}}}_{{{{\bar{G}}}}}(x;1/{a_i})}, \end{aligned}$$

is decreasing in $a_i$, since

$$\begin{aligned}\frac{\partial ^2 \log { H_{ G}(x;1/{a_i})}}{\partial a_i^2}= -\frac{1}{a_{i}^2}{ {{{\bar{H}}}}_{ {{\bar{G}}}}(x;1/{a_i})}\left( 1+{ H_{ G}(x;1/{a_i})}\right) \le 0. \end{aligned}$$

Therefore, ${ H_{ G}(x;1/{a_i})}$ is log-concave in $a_i$. $\square$

Proof of Corollary 3.18

(i) From (3.17) and Theorem 3.14, it is required to establish that

$$\begin{aligned} {{{{\bar{H}}}}_{{{{\bar{G}}}}}(x;e^{a_i})}=\frac{e^{a_i}[{{\bar{G}}}(t)]^{\lambda }}{1-(1-e^{a_i})[{{{\bar{G}}}}(t)]^{\lambda }} \end{aligned}$$

is increasing and log-concave with respect to $a_i$, where $a_i=\log \beta _i$, for $i=1,\ldots ,n$, and $t=x \theta$. Differentiating the above function with respect to $a_i$, we have

$$\begin{aligned} \frac{\partial {{{{\bar{H}}}}_{{{{\bar{G}}}}}(x;e^{a_i})}}{\partial a_i}= \frac{e^{a_i} [{{{\bar{G}}}}(t)]^{\lambda } (1-[{{\bar{G}}}(t)]^{\lambda })}{(1-(1-e^{a_i})[{{{\bar{G}}}}(t)]^{\lambda })^{2}}= { H_{ G}(x;e^{a_i})}{{{{\bar{H}}}}_{{{\bar{G}}}}(x;e^{a_i})} \ge 0, \end{aligned}$$

i.e., ${{{{\bar{H}}}}_{{{{\bar{G}}}}}(x;e^{a_i})}$ is increasing with respect to $a_{i}$, for $i=1,\ldots ,n$. On the other hand, we get

$$\begin{aligned} \frac{\partial \log {{{{\bar{H}}}}_{{{\bar{G}}}}(x;e^{a_i})}}{\partial a_i}= { H_{ G}(x;e^{a_i})}. \end{aligned}$$

Now, observe that ${ H_{ G}(x;e^{a_i})}$ is decreasing in $a_i$ since ${{{{\bar{H}}}}_{{{{\bar{G}}}}}(x;e^{a_i})}$ is increasing in $a_i$. Hence, ${{{{\bar{H}}}}_{{{\bar{G}}}}(x;e^{a_i})}$ is log-concave in $a_i$.

(ii) Again, from (3.17) and Theorem 3.14, it is sufficient to establish that

$$\begin{aligned} {{{{\bar{H}}}}_{{{{\bar{G}}}}}(x;e^{a_i})}=\frac{\alpha [{{{\bar{G}}}}(x e^{a_i})]^{\lambda }}{1-{{\bar{\alpha }}}[{{{\bar{G}}}}(x e^{a_i})]^{\lambda }} \end{aligned}$$

(5.6)

is decreasing and log-convex with respect to $a_i$, where $a_i=\log \delta _i$, for $i=1,\ldots ,n$. From (5.1), clearly ${{{{\bar{H}}}}_{{{{\bar{G}}}}}(x;e^{a_i})}$ is decreasing in $a_i$ and further,

$$\begin{aligned} \frac{\partial \log {{{{\bar{H}}}}_{{{{\bar{G}}}}}(x;e^{a_i})}}{\partial a_i}=- \frac{\lambda x e^{a_i} r_{G}(x e^{a_i})}{1-{{\bar{\alpha }}}[{{\bar{G}}}(x e^{a_i})]^{\lambda }}. \end{aligned}$$

Now, the second order partial derivative of $\log {{{\bar{H}}}_{{{{\bar{G}}}}}(x;e^{a_i})}$ with respect to $a_i$ is obtained as

$$\begin{aligned}&\frac{\partial ^2\log {{{{\bar{H}}}}_{{{\bar{G}}}}(x;e^{a_i})}}{\partial a_i^2} \overset{sign}{=}\nonumber \\&\quad -\left( (1-{{\bar{\alpha }}}[{{{\bar{G}}}}(x e^{a_i})]^{\lambda })\frac{\partial \,( x e^{a_i} r_{G}(x e^{a_i}))}{\partial a_i} -{{\bar{\alpha }}}\lambda \left( x e^{a_i} r_{G}(x e^{a_i})\right) ^2 [{{{\bar{G}}}}(x e^{a_i})]^{\lambda } \right) \nonumber \\ \end{aligned}$$

(5.7)

which is non-negative whenever the function $u r_{G}(u)$ is decreasing in u and $0<\alpha <1$. Hence, ${{{\bar{H}}}_{{{{\bar{G}}}}}(x;e^{a_i})}$ is log-convex in $a_i$ under these conditions. $\square$

Proof of Corollary 3.20

(i)From (3.17), the same idea than in Remark 3.5 and Theorem 3.14, it is required to show that (5.6) is monotone and log-concave with respect to $a_i$, where $a_i=\log \delta _i$, for $i=1,\ldots ,n$. From the proof of Corollary 3.6(ii), we know that ${{{{\bar{H}}}}_{{{\bar{G}}}}(x;e^{a_i})}$ is decreasing in $a_i$, so it is monotone. Finally, observe that (5.7) is non-positive when the function $xr_{G}(x)$ is increasing in x and $\alpha \ge 1$.

(ii) From (3.17), the same idea than in Remark 3.5 and Theorem 3.14, we need to show that

$$\begin{aligned} {{{{\bar{H}}}}_{{{\bar{G}}}}(x,e^{a_i})}=\frac{\alpha [{{\bar{G}}}(t)]^{e^{a_i}}}{1-{{\bar{\alpha }}}[{{{\bar{G}}}}(t)]^{e^{a_i}}} \end{aligned}$$

is monotone and log-concave with respect to $a_i$, where $a_i=\log \mu _i$, for $i=1,\ldots ,n$, and $t=x \theta$. From (5.3), it is known that ${{{{\bar{H}}}}_{{{\bar{G}}}}(x;e^{a_i})}$ is decreasing in $a_i$, i.e., monotone, and

$$\begin{aligned}&\frac{\partial \log {{{{\bar{H}}}}_{{{{\bar{G}}}}}(x;e^{a_i})}}{\partial a_i}\\&\quad = \frac{e^{a_i}\log {{{\bar{G}}}}(t)}{1-{{{\bar{\alpha }}}}[{{\bar{G}}}(t)]^{e^{a_i}}}. \end{aligned}$$

Now, the second order partial derivative of $\log {{{\bar{H}}}_{{{{\bar{G}}}}}(x;e^{a_i})}$ with respect to $a_i$ is obtained as

$$\begin{aligned} \frac{\partial ^2\log {{{{\bar{H}}}}_{{{\bar{G}}}}(x;e^{a_i})}}{\partial a_i^2}= e^{a_i}\log {{{\bar{G}}}}(t) \Big (1-{{{\bar{\alpha }}}} [{{{\bar{G}}}}(t)]^{e^{a_i}}+{{{\bar{\alpha }}}} [{{\bar{G}}}(t)]^{e^{a_i}} e^{a_i}\log {{{\bar{G}}}}(t) \Big )\le 0 \end{aligned}$$

when ${{{\bar{\alpha }}}}<0$, i.e., ${{{{\bar{H}}}}_{{{\bar{G}}}}(x;e^{a_i})}$ is log-concave when $\alpha \ge 1$. $\square$

Proof of Corollary 3.24

Now, from (3.17), the same idea than in Remark 3.5 and Theorem 3.22, we need to show that

$$\begin{aligned} { {{\bar{H}}}_{{{{\bar{G}}}}}(x,1/{a_i})}=\frac{(1/a_i)[ {{\bar{G}}}(t)]^{\lambda }}{1-(1-1/{a_i})[ {{{\bar{G}}}}(t)]^{\lambda }} \end{aligned}$$

(5.8)

is monotone and log-concave in $a_i$, where $a_i=1/\beta _i$, for $i=1,\ldots ,n$, and $t=x \theta$. From the proof of Corollary 3.13 (ii) above, we know that ${{{\bar{H}}}_{{{{\bar{G}}}}}(x;1/{a_i})}$ is decreasing with respect to $a_i$ since ${ H_{ G}(x;1/{a_i})}$ is increasing, i.e., it is monotone. On the other hand,

$$\begin{aligned} \frac{\partial \log {{{\bar{H}}}_{{{\bar{G}}}}(x;1/{a_i})}}{\partial a_i}=- \frac{1}{a_i}{{H}_{G}(x;1/{a_i})}, \end{aligned}$$

is increasing in $a_i$, since

$$\begin{aligned}\frac{\partial ^2 \log {{{\bar{H}}}_{{{{\bar{G}}}}}(x;1/{a_i})}}{\partial a_i^2}= \frac{1}{a_{i}^2}{{H}_{G}(x;1/{a_i})}\left( 1+{{{\bar{H}}}_{{{\bar{G}}}}(x;1/{a_i})}\right) \ge 0. \end{aligned}$$

Therefore, ${{{\bar{H}}}_{{{{\bar{G}}}}}(x;1/{a_i})}$ is log-convex with respect to $a_i$. $\square$