Multivariate stochastic comparisons of sequential order statistics with non-identical components

Abstract

Sequential order statistics (SOS) are useful tools for modeling the lifetimes of systems wherein the failure of a component has a significant impact on the lifetimes of the remaining surviving components. The SOS model is a general model that contains most of the existing models for ordered random variables. In this paper, we consider the SOS model with non-identical components and then discuss various univariate and multivariate stochastic comparison results in both one-and two-sample scenarios.

Appendix

Before presenting the proofs, we first introduce some acronyms that are exclusively used here: CIS - conditionally increasing in sequence; ILR - increasing likelihood ratio; IFR - increasing failure rate; DFR - decreasing failure rate; DRFR - decreasing reverse failure rate.

Proof of Part(a) of Theorem 4.1

We have

$$\begin{aligned} \bar{F}_{X_{1:n+1}^{\star }}\left( t\right) =\prod _{j=1}^{n+1} \bar{F}_j^{(1)}\left( t\right) \text { and } \bar{F}_{X_{1:n}^{\star }}\left( t\right) = \prod _{j=1}^{n} \bar{F}_j^{(1)}\left( t\right) , \quad t>0. \end{aligned}$$

Consequently, we have $X^{\star }_{1:n+1} \le _{st} X^{\star }_{1:n}$. Again, from Remark 3.1, we have

$$\begin{aligned} P\left( X^{\star }_{k:n} > x_k |X^{\star }_{1:n}= x_1, \ldots , X^{\star }_{k-1:n}= x_{k-1}\right) = {\left\{ \begin{array}{ll} \prod \limits _{j=1}^{n-k+1} \frac{ \bar{F}_j^{(k)} \left( x_k\right) }{ \bar{F}_j^{(k)}\left( x_{k-1}\right) } &{} \text {if }x_k \ge x_{k-1},\\ 1 &{} \text {if }x_k < x_{k-1}, \end{array}\right. } \end{aligned}$$

(A.1)

for $k = 2, \ldots , n$. As $\bar{F}\left( \cdot \right)$ is a decreasing function, we get $P(X^{\star }_{k:n} > x_k |X^{\star }_{1:n}= x_1,$ $\ldots , X^{\star }_{k-1:n}= x_{k-1})$ to be continuous and increasing in $x_{k-1} >0$. Then, $\left( X^{\star }_{1:n}, \ldots ,X^{\star }_{n:n} \right)$ is CIS. Thus, in view of Theorem 6.B.4 of Shaked and Shanthikumar (2007), we only need to show that

$$\begin{aligned} P\left( X^{\star }_{i:n+1}> t | X^{\star }_{i-1:n+1}= s\right)\le & {} P\left( X^{\star }_{i:n} > t | X^{\star }_{i-1:n}= s\right) \end{aligned}$$

(A.2)

for all $t>0$, and $i=2, \ldots , n$. Now, from (A.1), we get

$$\begin{aligned}{} & {} P\left( X^{\star }_{i:n+1}> t | X^{\star }_{i-1:n+1}= s\right) = \prod \limits _{j=1}^{n-i+2} \frac{ \bar{F}_j^{(i)}\left( t\right) }{ \bar{F}_j^{(i)}\left( s\right) } \le \prod \limits _{j=1}^{n-i+1} \frac{ \bar{F}_j^{(i)}\left( t\right) }{ \bar{F}_j^{(i)}\left( s\right) }\\{} & {} \quad = P\left( X^{\star }_{i:n} > t | X^{\star }_{i-1:n}= s\right) \end{aligned}$$

for all $t \ge s>0,$ where the inequality follows from the fact that $\bar{F}\left( \cdot \right)$ is a decreasing function. Further, for $t < s$, (A.2) is trivially true. Hence, the required result. $\square$

Proof of Part(b) of Theorem 4.1

Let $r_{j}^{(i)}(\cdot )$ be the hazard rate function of $F_{j}^{(i)}$, $j=1,\ldots , n-i+2$, $i=1,\ldots , n+1$. Further, let $\eta _{\cdot | \cdot } \left( \cdot \right)$ and $\lambda _{\cdot | \cdot } \left( \cdot \right)$ be the multivariate conditional hazard rate functions of $\left( X^{\star }_{1:n+1}, \ldots ,X^{\star }_{n:n+1} \right)$ and $\left( X^{\star }_{1:n}, \ldots ,X^{\star }_{n:n} \right)$, respectively. Note that $X^{\star }_{1:n} \le \ldots \le X^{\star }_{n:n}$. Then, from Remark 3.1 and (6.C.2) of Shaked and Shanthikumar (2007), we have

$$\begin{aligned} \lambda _{k | I}\left( u | \varvec{t}_I\right)= & {} {\left\{ \begin{array}{ll} \sum \limits _{j=1}^{n-i} r_{j}^{(i+1)} \left( u\right) &{} \text { for } k=i+1,\\ 0 &{} \text { for } k >i+1, \end{array}\right. } \end{aligned}$$

(A.3)

where $I = \left\{ 1, \ldots , i \right\}$ for some i. Similarly, we have

$$\begin{aligned} \eta _{k | I \cup J}\left( u | \varvec{s}_{I \cup J}\right)= & {} {\left\{ \begin{array}{ll} \sum \limits _{j=1}^{n-m+1} r_{j}^{(m+1)} \left( u\right) &{} \text { for } k=m+1, \\ 0 &{} \text { for } k >m+1, \end{array}\right. } \end{aligned}$$

(A.4)

where $I = \left\{ 1, \ldots , i \right\}$ and $J = \left\{ i+1, \ldots , m \right\}$, for some $m\ge i$ and $i=0, 1, \ldots , n-1$. In view of (6.D.13) of Shaked and Shanthikumar (2007), to prove the required result, it suffices to show that

$$\begin{aligned} \eta _{k | I \cup J}\left( u | \varvec{s}_{I \cup J}\right)\ge & {} \lambda _{k | I}\left( u | \varvec{t}_I\right) \text { for all } k \in \overline{I \cup J}, \end{aligned}$$

(A.5)

where $I = \left\{ 1, \ldots , i \right\}$ and $J = \left\{ i+1, \ldots , m \right\}$, for $0 \le i \le n-1$, $\varvec{s}_I \le \varvec{t}_I \le u\varvec{e}$ and $\varvec{s}_J \le u\varvec{e}$. Let us consider the following two cases.

Case I: Let $m>i$. As $\lambda _{k | I}\left( u | \varvec{t}_I\right) = 0, \text { for } k \ge m+1$, (A.5) holds readily.

Case II: Let $m=i$. Then,

$$\begin{aligned} \eta _{k | I \cup J}\left( u | \varvec{s}_{I \cup J}\right) = \sum \limits _{j=1}^{n-m+1} r_{j}^{(m+1)} \left( u\right) \ge \sum \limits _{j=1}^{n-m} r_{j}^{(m+1)} \left( u\right) = \lambda _{k | I}\left( u | \varvec{t}_I\right) , \text { for } k=m+1, \end{aligned}$$

where the inequality follows from the fact that $r_{n-m+1}^{(m+1)}\left( u\right)$ is positive in $u>0$. Further,

$\eta _{k | I \cup J}\left( u | \varvec{s}_{I \cup J}\right) = 0= \lambda _{k | I}\left( u | \varvec{t}_I\right) , \text { for } k>m+1$.

Thus, (A.5) holds, and hence the required result. $\square$

Proof of Part(c) of Theorem 4.1

In view of Lemma 3.1, the result holds if

$$\begin{aligned}{} & {} \prod _{i=1}^{n-1} \Bigg \{ \left( \frac{\bar{F}_{i} \left( x_i\right) }{\bar{F}_{i+1} \left( x_i\right) }\right) ^{ \sum \limits _{j=1}^{n-i+2} \alpha _{j}^{(i)} -1 } \left( {\bar{F}_{i+1} \left( x_i\right) } \right) ^{ \sum \limits _{j=1}^{n-i+2} \alpha _{j}^{(i)} - \sum \limits _{j=1}^{n-i+1} \alpha _{j}^{(i+1)} -1 } f_{i}\left( x_{i} \right) \Bigg \}\nonumber \\{} & {} \quad \left( {\bar{F}_{n} \left( x_n\right) } \right) ^{ \sum \limits _{j=1}^{2} \alpha _{j}^{(n)} -1 } f_{n}\left( x_{n} \right) \nonumber \\{} & {} \quad \times \prod _{i=1}^{n-1} \Bigg \{ \left( \frac{\bar{F}_{i} \left( y_i\right) }{\bar{F}_{i+1} \left( y_i\right) }\right) ^{ \sum \limits _{j=1}^{n-i+1} \alpha _{j}^{(i)} -1 } \left( {\bar{F}_{i+1} \left( y_i\right) } \right) ^{ \sum \limits _{j=1}^{n-i+1} \alpha _{j}^{(i)} - \sum \limits _{j=1}^{n-i} \alpha _{j}^{(i+1)} -1 } f_{i}\left( y_{i} \right) \Bigg \}\nonumber \\{} & {} \quad \left( {\bar{F}_{n} \left( y_n\right) } \right) ^{ \alpha _{1}^{(n)} -1 } f_{n}\left( y_{n} \right) \nonumber \\ {}{} & {} \le \prod _{i=1}^{n-1} \Bigg \{ \left( \frac{\bar{F}_{i} \left( x_i \wedge y_i \right) }{\bar{F}_{i+1} \left( x_i \wedge y_i\right) }\right) ^{ \sum \limits _{j=1}^{n-i+2} \alpha _{j}^{(i)} -1 } \left( {\bar{F}_{i+1} \left( x_i \wedge y_i \right) } \right) ^{ \sum \limits _{j=1}^{n-i+2} \alpha _{j}^{(i)} - \sum \limits _{j=1}^{n-i+1} \alpha _{j}^{(i+1)} -1 } f_{i}\left( x_i \wedge y_i \right) \Bigg \} \nonumber \\{} & {} \quad \times \left( {\bar{F}_{n} \left( x_n \wedge y_n \right) } \right) ^{ \sum \limits _{j=1}^{2} \alpha _{j}^{(n)} -1 } f_{n}\left( x_n \wedge y_n \right) \nonumber \\ {}{} & {} \quad \times \prod _{i=1}^{n-1} \Bigg \{ \left( \frac{\bar{F}_{i} \left( x_i \vee y_i \right) }{\bar{F}_{i+1} \left( x_i \vee y_i \right) }\right) ^{ \sum \limits _{j=1}^{n-i+1} \alpha _{j}^{(i)} -1 } \left( {\bar{F}_{i+1} \left( x_i \vee y_i \right) } \right) ^{ \sum \limits _{j=1}^{n-i+1} \alpha _{j}^{(i)} - \sum \limits _{j=1}^{n-i} \alpha _{j}^{(i+1)} -1 } f_{i}\left( x_i \vee y_i \right) \Bigg \} \nonumber \\{} & {} \quad \times \left( {\bar{F}_{n} \left( x_n \vee y_n\right) } \right) ^{ \alpha _{1}^{(n)} -1 } f_{n}\left( x_n \vee y_{n} \right) \end{aligned}$$

(A.6)

for all $x_1 \le \ldots \le x_n$ and $y_1 \le \ldots \le y_n$. So, to prove the required result, it suffices to show that the above inequality holds on the set $E_1 = \{ 1 \le i \le n-1:x_i \ge y_i \}$. Now, consider the set $E_1$. Then, (A.6) reduces to

$$\begin{aligned}{} & {} \prod _{i \in E_1} \Bigg \{ \left( \frac{\bar{F}_{i} \left( x_i\right) }{\bar{F}_{i+1} \left( x_i\right) }\right) ^{ \alpha _{n-i+2}^{(i)} } \left( {\bar{F}_{i+1} \left( x_i\right) } \right) ^{ \alpha _{n-i+2}^{(i)} - \alpha _{n-i+1}^{(i+1)} } \Bigg \} \nonumber \\{} & {} \quad \times \left( {\bar{F}_{n} \left( x_n\right) } \right) ^{ \sum \limits _{j=1}^{2} \alpha _{j}^{(n)} -1 } \left( {\bar{F}_{n} \left( y_n\right) } \right) ^{ \alpha _{1}^{(n)} -1 } f_{n}\left( x_{n} \right) f_{n}\left( y_{n} \right) \nonumber \\ {}{} & {} \quad \le \prod _{i \in E_1} \Bigg \{ \left( \frac{\bar{F}_{i} \left( y_i\right) }{\bar{F}_{i+1} \left( y_i\right) }\right) ^{ \alpha _{n-i+2}^{(i)} }\left( {\bar{F}_{i+1} \left( y_i\right) } \right) ^{ \alpha _{n-i+2}^{(i)} - \alpha _{n-i+1}^{(i+1)} } \Bigg \} \nonumber \\ {}{} & {} \quad \times \left( {\bar{F}_{n} \left( x_n \wedge y_n \right) } \right) ^{ \sum \limits _{j=1}^{2} \alpha _{j}^{(n)} -1 } \left( {\bar{F}_{n} \left( x_n \vee y_n\right) } \right) ^{ \alpha _{1}^{(n)} -1 } f_{n}\left( x_n \wedge y_n \right) f_{n}\left( x_n \vee y_{n} \right) .\nonumber \\ \end{aligned}$$

(A.7)

Now, from the condition that “$F_i \le _{hr} F_{i+1}$, $i=1,\ldots , n-1$”, we get

$$\begin{aligned} \frac{\bar{F}_{i} \left( u\right) }{\bar{F}_{i+1} \left( u\right) } \le \frac{\bar{F}_{i} \left( v\right) }{\bar{F}_{i+1} \left( v\right) }, \text { for all } u \ge v > 0. \end{aligned}$$

(A.8)

Again, from the condition that “$\alpha _{n-i+2}^{(i)} \ge \alpha _{n-i+1}^{(i+1)}$, for $i=1, \ldots , n-1$”, we get

$$\begin{aligned} \left( {\bar{F}_{i+1} \left( u\right) } \right) ^{ \alpha _{n-i+2}^{(i)} - \alpha _{n-i+1}^{(i+1)} } \le \left( {\bar{F}_{i+1} \left( v\right) } \right) ^{ \alpha _{n-i+2}^{(i)} - \alpha _{n-i+1}^{(i+1)} }, \text { for all } u \ge v > 0. \end{aligned}$$

(A.9)

Upon using (A.8) and (A.9), we obtain (A.7). Hence, the required result. $\square$

Proof of Part(a) of Theorem 4.2

From the definition of SOS, we have $X^{\star }_{1:n} \le _{st} X^{\star }_{2:n}$. Again, from Remark 3.1, we have

$$\begin{aligned} P\left( X^{\star }_{k:n} > x_k |X^{\star }_{1:n}= x_1, \ldots , X^{\star }_{k-1:n}= x_{k-1}\right) = {\left\{ \begin{array}{ll} \prod \limits _{j=1}^{n-k+1} \frac{ \bar{F}_j^{(k)}\left( x_k\right) }{ \bar{F}_j^{(k)}\left( x_{k-1}\right) } &{} \text {for }x_k \ge x_{k-1},\\ 1 &{} \text {for }x_k < x_{k-1}, \end{array}\right. }\nonumber \\ \end{aligned}$$

(A.10)

for $k = 2, \ldots , n$. As $\bar{F} \left( \cdot \right)$ is a decreasing function, we have $P(X^{\star }_{k:n} > x_k |X^{\star }_{1:n}= x_1,$ $\ldots , X^{\star }_{k-1:n}= x_{k-1})$ to be continuous and increasing in $x_{k-1} >0$. Then, $\left( X^{\star }_{1:n}, \ldots ,X^{\star }_{n:n} \right)$ is CIS. Thus, in view of Theorem 6.B.4 of Shaked and Shanthikumar (2007), we only need to show that

$$\begin{aligned} P\left( X^{\star }_{i:n}> t | X^{\star }_{i-1:n}= s\right)\le & {} P\left( X^{\star }_{i+1:n} > t | X^{\star }_{i:n}= s\right) \end{aligned}$$

(A.11)

for all $t>0$, and $i=2, 3, \ldots , n-1$. Now, from (A.10), we get

$$\begin{aligned}{} & {} P\left( X^{\star }_{i:n}> t | X^{\star }_{i-1:n}= s\right) = \prod \limits _{j=1}^{n-i+1} \frac{ \bar{F}_j^{(i)}\left( t\right) }{ \bar{F}_j^{(i)}\left( s\right) } \le \prod \limits _{j=1}^{n-i} \frac{ \bar{F}_j^{(i+1)}\left( t\right) }{ \bar{F}_j^{(i+1)}\left( s\right) }\\{} & {} \quad = P\left( X^{\star }_{i+1:n} > t | X^{\star }_{i:n}= s\right) \end{aligned}$$

for all $t \ge s>0,$ where the inequality follows from the facts that $\bar{F} \left( \cdot \right)$ is a decreasing function and $F_j^{(i)} \le _{hr} F_{j}^{(i+1)}$, for $j=1, \ldots , n-i$, $i=2, \ldots , n-1$. Further, for $t < s$, (A.11) holds readily, and hence the required result. $\square$

Proof of Part(b) of Theorem 4.2

Let $r_{j}^{(i)}(\cdot )$ be the hazard rate function of $F_{j}^{(i)}$, for $j=1,\ldots , n-i+1$, $i=1,\ldots , n$. Further, let $\eta _{\cdot | \cdot } \left( \cdot \right)$ and $\lambda _{\cdot | \cdot } \left( \cdot \right)$ be the multivariate conditional hazard rate functions of $\left( X^{\star }_{1:n}, \ldots ,X^{\star }_{n-1:n} \right)$ and $\left( X^{\star }_{2:n}, \ldots ,X^{\star }_{n:n} \right)$, respectively. Note that $X^{\star }_{1:n} \le \ldots \le X^{\star }_{n:n}$. Then, from Remark 3.1 and (6.C.2) of Shaked and Shanthikumar (2007), we have

$$\begin{aligned} \lambda _{k | I}\left( u | \varvec{t}_I\right)= & {} {\left\{ \begin{array}{ll} \sum \limits _{j=1}^{n-i-1} r_{j}^{(i+2)} \left( u\right) &{} \text { for } k=i+1,\\ 0 &{} \text { for } k >i+1, \end{array}\right. } \end{aligned}$$

(A.12)

where $I = \left\{ 1, \ldots , i \right\}$ for some i. Similarly, we have

$$\begin{aligned} \eta _{k | I \cup J}\left( u | \varvec{s}_{I \cup J}\right)= & {} {\left\{ \begin{array}{ll} \sum \limits _{j=1}^{n-m} r_{j}^{(m+1)} \left( u\right) &{} \text { for } k=m+1,\\ 0 &{} \text { for } k >m+1, \end{array}\right. } \end{aligned}$$

(A.13)

where $I = \left\{ 1, \ldots , i \right\}$ and $J = \left\{ i+1, \ldots , m \right\}$, for some $m\ge i$ and $i=0, 1, \ldots , n-2$. In view of (6.D.13) of Shaked and Shanthikumar (2007), to prove the required result, it suffices to show that

$$\begin{aligned} \eta _{k | I \cup J}\left( u | \varvec{s}_{I \cup J}\right)\ge & {} \lambda _{k | I}\left( u | \varvec{t}_I\right) \text { for all } k \in \overline{I \cup J}, \end{aligned}$$

(A.14)

where $I = \left\{ 1, \ldots , i \right\}$ and $J = \left\{ i+1, \ldots , m \right\}$, for $0 \le i \le n-2$, $\varvec{s}_I \le \varvec{t}_I \le u\varvec{e}$ and $\varvec{s}_J \le u\varvec{e}$. Let us consider the following three cases.

Case I: Let $i\ge 1$ and $m>i$. As $\lambda _{k | I}\left( u | \varvec{t}_I\right) = 0, \text { for } k \ge m+1$, (A.14) holds readily.

Case II: Let $i\ge 1$ and $m=i$. Then,

$$\begin{aligned}{} & {} \eta _{k | I \cup J}\left( u | \varvec{s}_{I \cup J}\right) = \sum \limits _{j=1}^{n-m} r_{j}^{(m+1)} \left( u\right) \ge \sum \limits _{j=1}^{n-m-1} r_{j}^{(m+2)} \left( u\right) \nonumber \\{} & {} \quad = \lambda _{k | I}\left( u | \varvec{t}_I\right) , \text { for } k=m+1, \end{aligned}$$

where the first inequality follows from the assumption that “$F_{j}^{(m+1)} \le _{hr} F_{j}^{(m+2)}$, $j=1, \ldots , n-m-1$, $m=1, \ldots , n-2$”. Further,

$\eta _{k | I \cup J}\left( u | \varvec{s}_{I \cup J}\right) = 0= \lambda _{k | I}\left( u | \varvec{t}_I\right) , \text { for } k>m+1.$ Thus, (A.14) holds.

Case III: Let $i = 0$. Then, (A.14) can be written as $\eta _{k | J}\left( u | \varvec{s}_{ J}\right) \ge \lambda _{k | \emptyset }\left( u | \varvec{t}_{ \emptyset }\right) .$ Note that $\lambda _{k | \emptyset }\left( u | \varvec{t}_{ \emptyset }\right) = 0$, for $k\ge 2$. Thus, we only need to show that $\eta _{1 | \emptyset }\left( u | \varvec{s}_{ \emptyset }\right) \ge \lambda _{1 | \emptyset }\left( u | \varvec{t}_{ \emptyset }\right) ,$ or equivalently, $X^{\star }_{1:n} \le _{hr} X^{\star }_{2:n}$. From Remark 3.1, we have

$$\begin{aligned} \frac{{r}_{X^{\star }_{1:n}}\left( t\right) }{{r}_{X^{\star }_{2:n}}\left( t\right) }= & {} \frac{{f}_{X^{\star }_{1:n}}\left( t\right) }{ \sum _{j=1}^{n-1} r_{j}^{(2)} \left( t\right) \int _{0}^{t} \prod _{j=1}^{n-1} \frac{\bar{F}_j^{(2)} \left( t\right) }{\bar{F}_j^{(2)} \left( z\right) } {f}_{X^{\star }_{1:n}}\left( z\right) dz } + \frac{{r}_{X^{\star }_{1:n}}\left( t\right) }{ \sum _{j=1}^{n-1} r_{j}^{(2)} \left( t\right) }\nonumber \\ \end{aligned}$$

(A.15)

for all $t>0$. Now, from the given condition that “$F_{j}^{(1)} \;\le _{hr } \; F_{j}^{(2)}$, $j=1,\ldots , n-1$”, we get ${{r}_{X^{\star }_{1:n}}\left( t\right) }/{ \sum _{j=1}^{n-1} r_{j}^{(2)} \left( t\right) } \ge 1 \text { for all } t>0.$ Furthermore, we have

$$\begin{aligned} \frac{{f}_{X^{\star }_{1:n}}\left( t\right) }{ \sum _{j=1}^{n-1} r_{j}^{(2)} \left( t\right) \int _{0}^{t} \prod _{j=1}^{n-1} \frac{\bar{F}_j^{(2)} \left( t\right) }{\bar{F}_j^{(2)} \left( z\right) } {f}_{X^{\star }_{1:n}}\left( z\right) dz } \ge 0 \text { for all }t>0. \end{aligned}$$

Consequently, from (A.15), we get ${r}_{X^{\star }_{1:n}}\left( t\right) \; \ge \; {r}_{X^{\star }_{2:n}}\left( t\right)$ for all $t>0$, and thus $X^{\star }_{1:n} \le _{hr} X^{\star }_{2:n}$, so that (A.14) follows. Hence, the required result. $\square$

Proof of Part(c) of Theorem 4.2

Note that the multivariate likelihood ratio order is closed under marginalization. So, to prove the required result, it suffices to show that $\left( 0, X^{\star }_{1:n}, \ldots ,X^{\star }_{n-1:n} \right) \le _{lr} \left( X^{\star }_{1:n}, \ldots ,X^{\star }_{n:n} \right)$, which holds, in view of Lemma 3.1, if

$$\begin{aligned}{} & {} \prod _{i=1}^{n-2} \Bigg \{ \left( \frac{\bar{F}_{i} \left( x_{i+1}\right) }{\bar{F}_{i+1} \left( x_{i+1}\right) }\right) ^{ \sum \limits _{j=1}^{n-i+1} \alpha _{j}^{(i)} -1 } \left( {\bar{F}_{i+1} \left( x_{i+1}\right) } \right) ^{ \sum \limits _{j=1}^{n-i+1} \alpha _{j}^{(i)} - \sum \limits _{j=1}^{n-i} \alpha _{j}^{(i+1)} -1 } f_{i}\left( x_{i+1} \right) \Bigg \} \nonumber \\{} & {} \times \left( {\bar{F}_{n-1} \left( x_n\right) } \right) ^{ \sum \limits _{j=1}^{2} \alpha _{j}^{(n-1)} -1 } f_{n-1}\left( x_{n} \right) \times \prod _{i=2}^{n-1} \Bigg \{ \left( \frac{\bar{F}_{i} \left( y_i\right) }{\bar{F}_{i+1} \left( y_i\right) }\right) ^{ \sum \limits _{j=1}^{n-i+1} \alpha _{j}^{(i)} -1 } \nonumber \\{} & {} \left( {\bar{F}_{i+1} \left( y_i\right) } \right) ^{ \sum \limits _{j=1}^{n-i+1} \alpha _{j}^{(i)} - \sum \limits _{j=1}^{n-i} \alpha _{j}^{(i+1)} -1 } f_{i}\left( y_{i} \right) \Bigg \} \left( {\bar{F}_{n} \left( y_n\right) } \right) ^{ \alpha _{1}^{(n)} -1 } f_{n}\left( y_{n} \right) \nonumber \\{} & {} \le \prod _{i=1}^{n-2} \Bigg \{ \left( \frac{\bar{F}_{i} \left( x_{i+1} \wedge y_{i+1} \right) }{\bar{F}_{i+1} \left( x_{i+1} \wedge y_{i+1}\right) }\right) ^{ \sum \limits _{j=1}^{n-i+1} \alpha _{j}^{(i)} -1 } \nonumber \\{} & {} \left( {\bar{F}_{i+1} \left( x_{i+1} \wedge y_{i+1} \right) } \right) ^{ \sum \limits _{j=1}^{n-i+1} \alpha _{j}^{(i)} - \sum \limits _{j=1}^{n-i} \alpha _{j}^{(i+1)} -1 } f_{i}\left( x_{i+1} \wedge y_{i+1} \right) \Bigg \} \nonumber \\{} & {} \times \left( {\bar{F}_{n-1} \left( x_n \wedge y_n \right) } \right) ^{ \sum \limits _{j=1}^{2} \alpha _{j}^{(n-1)} -1 } f_{n-1}\left( x_n \wedge y_n \right) \nonumber \\{} & {} \times \prod _{i=2}^{n-1} \Bigg \{ \left( \frac{\bar{F}_{i} \left( x_i \vee y_i \right) }{\bar{F}_{i+1} \left( x_i \vee y_i \right) }\right) ^{ \sum \limits _{j=1}^{n-i+1} \alpha _{j}^{(i)} -1 } \nonumber \\{} & {} \left( {\bar{F}_{i+1} \left( x_i \vee y_i \right) } \right) ^{ \sum \limits _{j=1}^{n-i+1} \alpha _{j}^{(i)} - \sum \limits _{j=1}^{n-i} \alpha _{j}^{(i+1)} -1 } f_{i}\left( x_i \vee y_i \right) \Bigg \} \nonumber \\{} & {} \times \left( {\bar{F}_{n} \left( x_n \vee y_n\right) } \right) ^{ \alpha _{1}^{(n)} -1 } f_{n}\left( x_n \vee y_{n} \right) \end{aligned}$$

(A.16)

for all $x_2 \le \ldots \le x_n$ and $y_1 \le \ldots \le y_n$. Let $E_2 = \{ 1 \le i \le n-2:x_{i+1} \ge y_{i+1} \}$. Note that the above inequality follows if it holds on $E_2$. Given $E_2$, (A.16) reduces to

$$\begin{aligned}{} & {} \prod _{i \in E_2} \Bigg \{ \left( \frac{\bar{F}_{i} \left( x_{i+1}\right) }{\bar{F}_{i+1} \left( x_{i+1}\right) }\right) ^{ \sum \limits _{j=1}^{n-i+1} \alpha _{j}^{(i)} -1 } \left( {\bar{F}_{i+1} \left( x_{i+1}\right) } \right) ^{ \sum \limits _{j=1}^{n-i+1} \alpha _{j}^{(i)} - \sum \limits _{j=1}^{n-i} \alpha _{j}^{(i+1)} -1 } f_{i}\left( x_{i+1} \right) \Bigg \} \nonumber \\{} & {} \times \left( {\bar{F}_{n-1} \left( x_n\right) } \right) ^{ \sum \limits _{j=1}^{2} \alpha _{j}^{(n-1)} -1 } f_{n-1}\left( x_{n} \right) \nonumber \\ {}{} & {} \times \prod _{i \in E_2} \Bigg \{ \left( \frac{\bar{F}_{i+1} \left( y_{i+1}\right) }{\bar{F}_{i+2} \left( y_{i+2}\right) }\right) ^{ \sum \limits _{j=1}^{n-i} \alpha _{j}^{(i+1)} -1 } \left( {\bar{F}_{i+2} \left( y_{i+1}\right) } \right) ^{ \sum \limits _{j=1}^{n-i} \alpha _{j}^{(i+1)} - \sum \limits _{j=1}^{n-i-1} \alpha _{j}^{(i+2)} -1 } f_{i+1}\left( y_{i+1} \right) \Bigg \} \nonumber \\ {}{} & {} \times \left( {\bar{F}_{n} \left( y_n\right) } \right) ^{ \alpha _{1}^{(n)} -1 } f_{n}\left( y_{n} \right) \nonumber \\ {}{} & {} \le \prod _{i \in E_2} \Bigg \{ \left( \frac{\bar{F}_{i} \left( y_{i+1} \right) }{\bar{F}_{i+1} \left( y_{i+1}\right) }\right) ^{ \sum \limits _{j=1}^{n-i+1} \alpha _{j}^{(i)} -1 } \left( {\bar{F}_{i+1} \left( y_{i+1} \right) } \right) ^{ \sum \limits _{j=1}^{n-i+1} \alpha _{j}^{(i)} - \sum \limits _{j=1}^{n-i} \alpha _{j}^{(i+1)} -1 } f_{i}\left( y_{i+1} \right) \Bigg \} \nonumber \\ {}{} & {} \times \left( {\bar{F}_{n-1} \left( x_n \wedge y_n \right) } \right) ^{ \sum \limits _{j=1}^{2} \alpha _{j}^{(n-1)} -1 } f_{n-1}\left( x_n \wedge y_n \right) \nonumber \\ {}{} & {} \times \prod _{i \in E_2} \Bigg \{ \left( \frac{\bar{F}_{i+1} \left( x_{i+1} \right) }{\bar{F}_{i+2} \left( x_{i+1}\right) }\right) ^{ \sum \limits _{j=1}^{n-i} \alpha _{j}^{(i+1)} -1 } \left( {\bar{F}_{i+2} \left( x_{i+1} \right) } \right) ^{ \sum \limits _{j=1}^{n-i} \alpha _{j}^{(i+1)} - \sum \limits _{j=1}^{n-i-1} \alpha _{j}^{(i+2)} -1 } f_{i+1}\left( x_{i+1} \right) \Bigg \} \nonumber \\ {}{} & {} \times \left( {\bar{F}_{n} \left( x_n \vee y_n\right) } \right) ^{ \alpha _{1}^{(n)} -1 } f_{n}\left( x_n \vee y_{n} \right) . \end{aligned}$$

(A.17)

Now, from the condition that “$\left( { \sum \limits _{j=1}^{n-i+1} \alpha _{j}^{(i)} - \sum \limits _{j=1}^{n-i} \alpha _{j}^{(i+1)} }-1 \right)$ is positive and decreasing in $i \in \{1,\ldots , n-1\}$”, we get ${ \sum \limits _{j=1}^{n-i+1} \alpha _{j}^{(i)} - \sum \limits _{j=1}^{n-i} \alpha _{j}^{(i+1)} } \ge \sum \limits _{j=1}^{n-i} \alpha _{j}^{(i+1)} - \sum \limits _{j=1}^{n-i-1} \alpha _{j}^{(i+2)} \ge 1$ and $\sum \limits _{j=1}^{n-i} \alpha _{j}^{(i+1)} \ge 1$, for $i=1, \ldots , n-2$. Further, by using these along with the condition that “$F_{i} \le _{hr} F_{i+1}$”, we get

$$\begin{aligned}{} & {} \left( \frac{\bar{F}_{i} \left( x_{i+1}\right) }{\bar{F}_{i+1} \left( x_{i+1}\right) }\right) ^{ \sum \limits _{j=1}^{n-i+1} \alpha _{j}^{(i)} - \sum \limits _{j=1}^{n-i} \alpha _{j}^{(i+1)} } \le \; \left( \frac{\bar{F}_{i} \left( y_{i+1}\right) }{\bar{F}_{i+1} \left( y_{i+1}\right) }\right) ^{ \sum \limits _{j=1}^{n-i+1} \alpha _{j}^{(i)} - \sum \limits _{j=1}^{n-i} \alpha _{j}^{(i+1)} }, \end{aligned}$$

(A.18)

$$\begin{aligned}{} & {} \bar{F}^{ \sum \limits _{j=1}^{n-i+1} \alpha _{j}^{(i)} - \sum \limits _{j=1}^{n-i} \alpha _{j}^{(i+1)} -1}_{i+1} \le _{hr} \bar{F}^{\sum \limits _{j=1}^{n-i} \alpha _{j}^{(i+1)} - \sum \limits _{j=1}^{n-i-1} \alpha _{j}^{(i+2)} -1}_{i+2} \nonumber \\{} & {} \quad \text { and } \bar{F}^{ \sum \limits _{j=1}^{2} \alpha _{j}^{(n-1)} -1}_{n-1} \le _{hr} \bar{F}^{ \alpha _{1}^{(n)} -1}_{n}, \end{aligned}$$

(A.19)

for all $x_{i+1} \ge y_{i+1} > 0$, for $i=1, \ldots , n-2$. Again, from the condition that “$\left( \bar{F}_{i+1}(u)\right) ^2/\bar{F}_{i}(u) \bar{F}_{i+2}(u)$ is increasing in $u>0$”, we get

$$\begin{aligned} \left( \frac{\left( \bar{F}_{i+1}(y_{i+ 1})\right) ^2}{\bar{F}_{i}(y_{i+1}) \bar{F}_{i+2}(y_{i+1})} \right) ^{ \sum \limits _{j=1}^{n-i} \alpha _{j}^{(i+1)} -1 } \le \; \left( \frac{\left( \bar{F}_{i+1}(x_{i+1})\right) ^2}{\bar{F}_{i}(x_{i+1}) \bar{F}_{i+2}(x_{i+1})} \right) ^{ \sum \limits _{j=1}^{n-i} \alpha _{j}^{(i+1)} -1 } \end{aligned}$$

(A.20)

for all $x_{i+1} \ge y_{i+1} > 0$. Further, we have the assumption that “$F_{i} \le _{lr} F_{i+1}$, for $i=1, \ldots , n-1$”. Finally, upon combining this with (A.18), (A.19) and (A.20), we get (A.17), and hence the required result. $\square$

Proof of Part(a) of Theorem 4.3

We have

$$\begin{aligned} \bar{F}_{X^{\star }_{2:n+1} }\left( t\right)= & {} P\left( X^{\star }_{1:n+1}> t\right) + \int _{o}^{t} P\left( X^{\star }_{2:n+1}> t | X^{\star }_{1:n+1} =z\right) {f}_{X^{\star }_{1:n+1}}\left( z\right) \, dz \nonumber \\= & {} \int _{0}^{\infty } \bar{G}_2\left( z|t\right) \; dF_{X_{1:n+1}^{\star }}(z), \quad t>0, \end{aligned}$$

(A.21)

where

$$\begin{aligned} \bar{G}_2\left( z|t\right) = {\left\{ \begin{array}{ll} \prod \limits _{j=1}^{n} \frac{ \bar{F}_j^{(2)}\left( t\right) }{ \bar{F}_j^{(2)}\left( z\right) } &{} \text {if }z \le t,\\ 1 &{} \text {if }z > t. \end{array}\right. } \end{aligned}$$

Note that $\bar{F}_{j}^{(2)}(t) \le {\bar{F}_{j}^{(2)}(t)}/{\bar{F}_{j}^{(2)}(z)}$, for all $0 <z \le t$. This, together with the assumption that “$F_j^{(1)} \le _{st} F_{j}^{(2)}$, $j=1, \ldots , n$”, imply that

$$\begin{aligned} \bar{G}_2\left( z|t\right) \ge \prod \limits _{j=1}^{n} { \bar{F}_j^{(2)}\left( t\right) } \nonumber \ge \prod \limits _{j=1}^{n} { \bar{F}_j^{(1)}\left( t\right) } \text { for all } 0 < z \le t, \end{aligned}$$

which further implies that $X^{\star }_{1:n} \le _{st} X^{\star }_{2:n+1}$. Again, from Remark 3.1, we have

$$\begin{aligned} P\left( X^{\star }_{k:n} > x_k |X^{\star }_{1:n}= x_1, \ldots , X^{\star }_{k-1:n}= x_{k-1}\right) = {\left\{ \begin{array}{ll} \prod \limits _{j=1}^{n-k+1} \frac{ \bar{F}_j^{(k)}\left( x_k\right) }{ \bar{F}_j^{(k)}\left( x_{k-1}\right) } &{} \text { for }x_k \ge x_{k-1},\\ 1 &{} \text { for }x_k < x_{k-1}, \end{array}\right. } \nonumber \\ \end{aligned}$$

(A.22)

for $k = 2, \ldots , n$. As $\bar{F}\left( \cdot \right)$ is a decreasing function, we get $P(X^{\star }_{k:n} > x_k |X^{\star }_{1:n}= x_1,$ $\ldots , X^{\star }_{k-1:n}= x_{k-1})$ to be continuous and increasing in $x_{k-1} >0$. Then, $\left( X^{\star }_{1:n}, \ldots ,X^{\star }_{n:n} \right)$ is CIS. Thus, in view of Theorem 6.B.4 of Shaked and Shanthikumar (2007), we only need to show that

$$\begin{aligned} P\left( X^{\star }_{i:n}> t | X^{\star }_{i-1:n}= s\right)\le & {} P\left( X^{\star }_{i+1:n+1} > t | X^{\star }_{i:n+1}= s\right) \end{aligned}$$

(A.23)

for all $t>0$, and $i=2, \ldots , n$. Now, from (A.22), we get

$$\begin{aligned} P\left( X^{\star }_{i:n}> t | X^{\star }_{i-1:n}= s\right)= & {} \prod \limits _{j=1}^{n-i+1} \frac{ \bar{F}_j^{(i)}\left( t\right) }{ \bar{F}_j^{(i)}\left( s\right) } \le \prod \limits _{j=1}^{n-i+1} \frac{ \bar{F}_j^{(i+1)}\left( t\right) }{ \bar{F}_j^{(i+1)}\left( s\right) } \nonumber \\ {}= & {} P\left( X^{\star }_{i+1:n} > t | X^{\star }_{i:n}= s\right) \end{aligned}$$

$\text {for all } t \ge s>0,$ where the inequality follows from the facts that $\phi \left( \cdot \right)$ is a decreasing function and $F_j^{(i)} \le _{hr} F_{j}^{(i+1)}$, for $j=1, \ldots , n-i+1$, $i=2, \ldots , n$. Hence, (A.23) holds for $t \ge s>0$. Further, for $t < s$, (A.23) holds readily, and hence, the required result. $\square$

Proof of Part(b) of Theorem 4.3

Let $r_{j}^{(i)}(\cdot )$ be the hazard rate function of $F_{j}^{(i)}$, for $j=1,\ldots , n-i+2$, $i=1,\ldots , n+1$. Further, let $\eta _{\cdot | \cdot } \left( \cdot \right)$ and $\lambda _{\cdot | \cdot } \left( \cdot \right)$ be the multivariate conditional hazard rate functions of $\left( X^{\star }_{1:n}, \ldots ,X^{\star }_{n-1:n} \right)$ and $\left( X^{\star }_{2:n+1}, \ldots ,X^{\star }_{n+1:n+1} \right)$, respectively. Note that $X^{\star }_{1:n} \le \ldots \le X^{\star }_{n:n}$. Then, from Remark 3.1 and (6.C.2) of Shaked and Shanthikumar (2007), we have

$$\begin{aligned} \lambda _{k | I}\left( u | \varvec{t}_I\right)= & {} {\left\{ \begin{array}{ll} \sum \limits _{j=1}^{n-i} r_{j}^{(i+2)} \left( u\right) &{} \text { for } k=i+1,\\ 0 &{} \text { for } k >i+1, \end{array}\right. } \end{aligned}$$

(A.24)

where $I = \left\{ 1, \ldots , i \right\}$, for some i. Similarly, we have

$$\begin{aligned} \eta _{k | I \cup J}\left( u | \varvec{s}_{I \cup J}\right)= & {} {\left\{ \begin{array}{ll} \sum \limits _{j=1}^{n-m} r_{j}^{(m+1)} \left( u\right) &{} \text { for } k=m+1,\\ 0 &{} \text { for } k >m+1, \end{array}\right. } \end{aligned}$$

(A.25)

where $I = \left\{ 1, \ldots , i \right\}$ and $J = \left\{ i+1, \ldots , m \right\}$, for some $m\ge i$ and $i=0, 1, \ldots , n-1$. In view of (6.D.13) of Shaked and Shanthikumar (2007), to prove the required result, it suffices to show that

$$\begin{aligned} \eta _{k | I \cup J}\left( u | \varvec{s}_{I \cup J}\right)\ge & {} \lambda _{k | I}\left( u | \varvec{t}_I\right) \text { for all } k \in \overline{I \cup J}, \end{aligned}$$

(A.26)

where $I = \left\{ 1, \ldots , i \right\}$ and $J = \left\{ i+1, \ldots , m \right\}$, for $0 \le i \le n-1$, $\varvec{s}_I \le \varvec{t}_I \le u\varvec{e}$ and $\varvec{s}_J \le u\varvec{e}$. Now, consider the following three cases.

Case I: Let $i\ge 1$ and $m>i$. As $\lambda _{k | I}\left( u | \varvec{t}_I\right) = 0, \text { for } k \ge m+1$, (A.26) holds readily.

Case II: Let $i\ge 1$ and $m=i$. Then,

$$\begin{aligned} \eta _{k | I \cup J}\left( u | \varvec{s}_{I \cup J}\right) = \sum \limits _{j=1}^{n-m} r_{j}^{(m+1)} \left( u\right) \nonumber \ge \sum \limits _{j=1}^{n-m} r_{j}^{(m+2)} \left( u\right) \nonumber = \lambda _{k | I}\left( u | \varvec{t}_I\right) , \text { for } k=m+1, \end{aligned}$$

where the inequality follows from the assumption that “$F_{j}^{(m+1)} \le _{hr} F_{j}^{(m+2)}$, for $m=1, \ldots , n-1$”. Further,

$\eta _{k | I \cup J}\left( u | \varvec{s}_{I \cup J}\right) = 0= \lambda _{k | I}\left( u | \varvec{t}_I\right) , \text { for } k>m+1$. Thus, (A.26) holds.

Case III: Let $i = 0$. Then, (A.26) can be equivalently written as

$\eta _{k | J}\left( u | \varvec{s}_{ J}\right) \ge \lambda _{k | \emptyset }\left( u | \varvec{t}_{ \emptyset }\right)$. Note that $\lambda _{k | \emptyset }\left( u | \varvec{t}_{ \emptyset }\right) = 0$, for $k\ge 2$. Thus, we only need to show that $\eta _{1 | \emptyset }\left( u | \varvec{s}_{ \emptyset }\right) \ge \lambda _{1 | \emptyset }\left( u | \varvec{t}_{ \emptyset }\right) ,$ or equivalently, $X^{\star }_{1:n} \le _{hr} X^{\star }_{2:n+1}$. From Remark 3.1, we have

$$\begin{aligned} \frac{{r}_{X^{\star }_{1:n}}\left( t\right) }{{r}_{X^{\star }_{2:n+1}}\left( t\right) }= & {} \frac{{f}_{X^{\star }_{1:n}}\left( t\right) }{ \sum \limits _{j=1}^{n-1} r_{j}^{(2)} \left( t\right) \int _{0}^{t} \prod \limits _{j=1}^{n} \frac{\bar{F}_j^{(2)} \left( t\right) }{\bar{F}_j^{(2)} \left( z\right) } {f}_{X^{\star }_{1:n+1}}\left( z\right) dz } + \frac{{r}_{X^{\star }_{1:n}}\left( t\right) }{ \sum \limits _{j=1}^{n} r_{j}^{(2)} \left( t\right) } \end{aligned}$$

(A.27)

for all $t>0$. Now, from the given condition that “$F_{j}^{(1)} \;\le _{hr } \; F_{j}^{(2)}$, $j=1,2,\ldots , n$", we get ${{r}_{X^{\star }_{1:n}}\left( t\right) }\Big /{ \sum \limits _{j=1}^{n} r_{j}^{(2)} \left( t\right) } \ge 1 \text { for all } t>0$. Furthermore, we have

$$\begin{aligned} \frac{{f}_{X^{\star }_{1:n}}\left( t\right) }{ \sum \limits _{j=1}^{n} r_{j}^{(2)} \left( t\right) \int _{0}^{t} \prod \limits _{j=1}^{n} \frac{\bar{F}_j^{(2)} \left( t\right) }{\bar{F}_j^{(2)} \left( z\right) } {f}_{X^{\star }_{1:n+1}}\left( z\right) dz } \ge 0 \text { for all }t>0. \end{aligned}$$

Consequently, from (A.27), we get ${r}_{X^{\star }_{1:n}}\left( t\right) \; \ge \; {r}_{X^{\star }_{2:n}}\left( t\right)$, for all $t>0$, and so $X^{\star }_{1:n} \le _{hr} X^{\star }_{2:n+1}$. Thus, (A.26) follows, and hence, the required result. $\square$

Proof of Part(c) of Theorem 4.3

Note that the multivariate likelihood ratio order is closed under marginalization. So, to prove the required result, it suffices to show that $\left( 0, X^{\star }_{1:n}, \ldots ,X^{\star }_{n:n} \right) \le _{lr} \left( X^{\star }_{1:n+1}, \ldots ,X^{\star }_{n+1:n+1} \right)$, which holds, in view of Lemma 3.1, if

$$\begin{aligned}{} & {} \prod _{i=1}^{n-1} \Bigg \{ \left( \frac{\bar{F}_{i} \left( x_{i+1}\right) }{\bar{F}_{i+1} \left( x_{i+1}\right) }\right) ^{ \sum \limits _{j=1}^{n-i+1} \alpha _{j}^{(i)} -1 } \left( {\bar{F}_{i+1} \left( x_{i+1}\right) } \right) ^{ \sum \limits _{j=1}^{n-i+1} \alpha _{j}^{(i)} - \sum \limits _{j=1}^{n-i} \alpha _{j}^{(i+1)} -1 } f_{i}\left( x_{i+1} \right) \Bigg \} \nonumber \\ {}{} & {} \times \left( {\bar{F}_{n} \left( x_{n+1}\right) } \right) ^{ \alpha _{1}^{(n)} -1 } f_{n}\left( x_{n+1} \right) \nonumber \\{} & {} \times \prod _{i=2}^{n} \Bigg \{ \left( \frac{\bar{F}_{i} \left( y_i\right) }{\bar{F}_{i+1} \left( y_i\right) }\right) ^{ \sum \limits _{j=1}^{n-i+2} \alpha _{j}^{(i)} -1 } \left( {\bar{F}_{i+1} \left( y_i\right) } \right) ^{ \sum \limits _{j=1}^{n-i+2} \alpha _{j}^{(i)} - \sum \limits _{j=1}^{n-i+1} \alpha _{j}^{(i+1)} -1 } f_{i}\left( y_{i} \right) \Bigg \} \nonumber \\ {}{} & {} \times \left( {\bar{F}_{n+1} \left( y_{n+1}\right) } \right) ^{ \alpha _{1}^{(n+1)} -1 } f_{n+1}\left( y_{n+1} \right) \nonumber \\ {}{} & {} \le \prod _{i=1}^{n-1} \Bigg \{ \left( \frac{\bar{F}_{i} \left( x_{i+1} \wedge y_{i+1} \right) }{\bar{F}_{i+1} \left( x_{i+1} \wedge y_{i+1}\right) }\right) ^{ \sum \limits _{j=1}^{n-i+1} \alpha _{j}^{(i)} -1 } \nonumber \\{} & {} \left( {\bar{F}_{i+1} \left( x_{i+1} \wedge y_{i+1} \right) } \right) ^{ \sum \limits _{j=1}^{n-i+1} \alpha _{j}^{(i)} - \sum \limits _{j=1}^{n-i} \alpha _{j}^{(i+1)} -1 } f_{i}\left( x_{i+1} \wedge y_{i+1} \right) \Bigg \} \nonumber \\ {}{} & {} \times \left( {\bar{F}_{n} \left( x_{n+1} \wedge y_{n+1} \right) } \right) ^{ \alpha _{1}^{(n)} -1 } f_{n}\left( x_{n+1} \wedge y_{n+1} \right) \nonumber \\ {}{} & {} \times \prod _{i=2}^{n} \Bigg \{ \left( \frac{\bar{F}_{i} \left( x_i \vee y_i \right) }{\bar{F}_{i+1} \left( x_i \vee y_i \right) }\right) ^{ \sum \limits _{j=1}^{n-i+2} \alpha _{j}^{(i)} -1 } \left( {\bar{F}_{i+1} \left( x_i \vee y_i \right) } \right) ^{ \sum \limits _{j=1}^{n-i+2} \alpha _{j}^{(i)} - \sum \limits _{j=1}^{n-i+1} \alpha _{j}^{(i+1)} -1 } f_{i}\left( x_i \vee y_i \right) \Bigg \} \nonumber \\ {}{} & {} \times \left( {\bar{F}_{n+1} \left( x_{n+1} \vee y_{n+1}\right) } \right) ^{ \alpha _{1}^{(n+1)} -1 } f_{n+1}\left( x_{n+1} \vee y_{n+1} \right) \end{aligned}$$

(A.28)

for all $x_2 \le \ldots \le x_{n+1}$ and $y_1 \le \ldots \le y_{n+1}$. Let $E_3 = \{ 1 \le i \le n-1:x_{i+1} \ge y_{i+1} \}$. Then, the above inequality follows if it holds on $E_3$. Given $E_3$, (A.28) reduces to

$$\begin{aligned}{} & {} \prod _{i \in E_3} \Bigg \{ \left( \frac{\bar{F}_{i} \left( x_{i+1}\right) }{\bar{F}_{i+1} \left( x_{i+1}\right) }\right) ^{ \sum \limits _{j=1}^{n-i+1} \alpha _{j}^{(i)} -1 } \left( {\bar{F}_{i+1} \left( x_{i+1}\right) } \right) ^{ \sum \limits _{j=1}^{n-i+1} \alpha _{j}^{(i)} - \sum \limits _{j=1}^{n-i} \alpha _{j}^{(i+1)} -1 } f_{i}\left( x_{i+1} \right) \Bigg \} \nonumber \\ {}{} & {} \times \left( {\bar{F}_{n} \left( x_{n+1}\right) } \right) ^{ \alpha _{1}^{(n)} -1 } f_{n}\left( x_{n+1} \right) \nonumber \\ {}{} & {} \times \prod _{i \in E_3} \Bigg \{ \left( \frac{\bar{F}_{i+1} \left( y_{i+1}\right) }{\bar{F}_{i+2} \left( y_{i+2}\right) }\right) ^{ \sum \limits _{j=1}^{n-i+1} \alpha _{j}^{(i+1)} -1 } \nonumber \\{} & {} \left( {\bar{F}_{i+2} \left( y_{i+1}\right) } \right) ^{ \sum \limits _{j=1}^{n-i+1} \alpha _{j}^{(i+1)} - \sum \limits _{j=1}^{n-i} \alpha _{j}^{(i+2)} -1 } f_{i+1}\left( y_{i+1} \right) \Bigg \} \nonumber \\ {}{} & {} \times \left( {\bar{F}_{n+1} \left( y_{n+1}\right) } \right) ^{ \alpha _{1}^{(n+1)} -1 } f_{n+1}\left( y_{n+1} \right) \nonumber \\ {}{} & {} \le \prod _{i \in E_3} \Bigg \{ \left( \frac{\bar{F}_{i} \left( y_{i+1} \right) }{\bar{F}_{i+1} \left( y_{i+1}\right) }\right) ^{ \sum \limits _{j=1}^{n-i+1} \alpha _{j}^{(i)} -1 } \nonumber \\{} & {} \left( {\bar{F}_{i+1} \left( y_{i+1} \right) } \right) ^{ \sum \limits _{j=1}^{n-i+1} \alpha _{j}^{(i)} - \sum \limits _{j=1}^{n-i} \alpha _{j}^{(i+1)} -1 } f_{i}\left( y_{i+1} \right) \Bigg \} \nonumber \\ {}{} & {} \times \left( {\bar{F}_{n} \left( x_{n+1} \wedge y_{n+1} \right) } \right) ^{ \alpha _{1}^{(n)} -1 } f_{n}\left( x_{n+1} \wedge y_{n+1} \right) \nonumber \\ {}{} & {} \times \prod _{i \in E_3} \Bigg \{ \left( \frac{\bar{F}_{i+1} \left( x_{i+1} \right) }{\bar{F}_{i+2} \left( x_{i+1}\right) }\right) ^{ \sum \limits _{j=1}^{n-i+1} \alpha _{j}^{(i+1)} -1 } \nonumber \\{} & {} \left( {\bar{F}_{i+2} \left( x_{i+1} \right) } \right) ^{ \sum \limits _{j=1}^{n-i+1} \alpha _{j}^{(i+1)} - \sum \limits _{j=1}^{n-i} \alpha _{j}^{(i+2)} -1 } f_{i+1}\left( x_{i+1} \right) \Bigg \} \nonumber \\ {}{} & {} \times \left( {\bar{F}_{n+1} \left( x_{n+1} \vee y_{n+1}\right) } \right) ^{ \alpha _{1}^{(n+1)} -1 } f_{n+1}\left( x_{n+1} \vee y_{n+1} \right) . \end{aligned}$$

(A.29)

Now, from the condition that “$\big (\sum \limits _{j=1}^{n-i+1} \left( \alpha _{j}^{(i)} - \alpha _{j}^{(i+1)} \right) -1 \big )$ is positive and decreasing in $i \in \{1,\ldots , n\}$”, we get ${ \sum \limits _{j=1}^{n-i+1} \alpha _{j}^{(i)} - \sum \limits _{j=1}^{n-i} \alpha _{j}^{(i+1)} } \ge {\sum \limits _{j=1}^{n-i+1} \alpha _{j}^{(i+1)} - \sum \limits _{j=1}^{n-i} \alpha _{j}^{(i+2)} } \ge 1$, $\sum \limits _{j=1}^{n-i+1} \alpha _{j}^{(i+1)} \ge 1$, for $i=1, \ldots , n-1$, and ${ \alpha _{1}^{(n)} } \ge { \alpha _{1}^{(n+1)} } \ge 1$. These, together with the condition that “$F_{i} \le _{hr} F_{i+1}$”,

imply that

$$\begin{aligned} \left( \frac{\bar{F}_{i} \left( x_{i+1}\right) }{\bar{F}_{i+1} \left( x_{i+1}\right) }\right) ^{ \sum \limits _{j=1}^{n-i+1} \alpha _{j}^{(i)} - \sum \limits _{j=1}^{n-i+1} \alpha _{j}^{(i+1)} } \le \; \left( \frac{\bar{F}_{i} \left( y_{i+1}\right) }{\bar{F}_{i+1} \left( y_{i+1}\right) }\right) ^{ \sum \limits _{j=1}^{n-i+1} \alpha _{j}^{(i)} - \sum \limits _{j=1}^{n-i+1} \alpha _{j}^{(i+1)} },\nonumber \\ \end{aligned}$$

(A.30)

$$\begin{aligned} \bar{F}^{ \sum \limits _{j=1}^{n-i+1} \alpha _{j}^{(i)} - \sum \limits _{j=1}^{n-i} \alpha _{j}^{(i+1)} -1}_{i+1} \le _{hr} \bar{F}^{\sum \limits _{j=1}^{n-i+1} \alpha _{j}^{(i+1)} - \sum \limits _{j=1}^{n-i} \alpha _{j}^{(i+2)} -1}_{i+2} \text { and } \bar{F}^{ \alpha _{1}^{(n)} -1}_{n} \le _{hr} \bar{F}^{ \alpha _{1}^{(n+1)} -1}_{n+1},\nonumber \\ \end{aligned}$$

(A.31)

for all $x_{i+1} \ge y_{i+1} > 0$, for $i=1, \ldots , n-1$. Moreover, from the condition that “$\left( \bar{F}_{i+1}(u)\right) ^2/\bar{F}_{i}(u) \bar{F}_{i+2}(u)$ is increasing in $u>0$”, we get

$$\begin{aligned} \left( \frac{\left( \bar{F}_{i+1}(y_{i+ 1})\right) ^2}{\bar{F}_{i}(y_{i+1}) \bar{F}_{i+2}(y_{i+1})} \right) ^{ \sum \limits _{j=1}^{n-i+1} \alpha _{j}^{(i+1)} -1 } \le \; \left( \frac{\left( \bar{F}_{i+1}(x_{i+1})\right) ^2}{\bar{F}_{i}(x_{i+1}) \bar{F}_{i+2}(x_{i+1})} \right) ^{ \sum \limits _{j=1}^{n-i+1} \alpha _{j}^{(i+1)} -1 } \end{aligned}$$

(A.32)

for all $x_{i+1} \ge y_{i+1} > 0$. Additionally, we have the assumption that “$F_{i} \le _{lr} F_{i+1}$, for $i=1, \ldots , n$”. Finally, upon combining this with (A.30), (A.31) and (A.32), we get (A.29). Hence, the required result. $\square$

Proof of Part(a) of Theorem 4.4

From Lemma 3.4, we have

$$\begin{aligned} X^{\star }_{i:n} {\mathop {=}\limits ^{d}} D^{-1} \left( \sum _{j=1}^{i}W_{j,n} \right) \text { and } X^{\star }_{i+1:n} {\mathop {=}\limits ^{d}} D^{-1} \left( \sum _{j=1}^{i+1}W_{j,n} \right) . \end{aligned}$$

Note that the likelihood ratio order is closed under increasing transformations. As F is continuous, $D^{-1} \left( \cdot \right)$ is increasing and so the required result holds if and only if $\sum _{j=1}^{i}W_{j,n} \le _{lr} \sum _{j=1}^{i+1}W_{j,n}$, for all $i=1,\ldots , n-1$. Now, we have $W_{j, n}$, $j=1,\ldots ,i+1$, to be independent and non-negative random variables. Thus, in view of Theorem 1.C.9 of Shaked and Shanthikumar (2007), the result follows (i.e., the above inequality holds) provided $W_{j,n}$ is ILR, for all $j=1,\ldots , i+1$. From (3.9), we have

$$\begin{aligned} \frac{f'_{W_{j,n}} \left( t\right) }{f_{W_{j,n}} \left( t\right) }= & {} -\sum _{l=1}^{n-j+1} \alpha _{l}^{(j)}, \quad t>0, \end{aligned}$$

which implies that ${f'_{W_{j,n}} \left( t\right) } / {f_{W_{j,n}} \left( t\right) }$ is decreasing in $t>0$ and so $W_{j,n}$ is ILR, for all $j=1, \ldots , i+1$. Hence, the required result. $\square$

Proof of Part(b) of Theorem 4.4

From Lemma 3.4, we have

$$\begin{aligned} X^{\star }_{i:n+1}{\mathop {=}\limits ^{d}} D^{-1} \left( \sum _{j=1}^{i}W_{j,n+1} \right) \text { and } X^{\star }_{i:n} {\mathop {=}\limits ^{d}} D^{-1} \left( \sum _{j=1}^{i}W_{j,n} \right) . \end{aligned}$$

Note that the likelihood ratio order is closed under increasing transformations. As F is continuous, $D^{-1} \left( \cdot \right)$ is increasing and so the required result holds if and only if $\sum _{j=1}^{i}W_{j, n+1} \le _{lr} \sum _{j=1}^{i}W_{j, n}$, for all $i=1,\ldots ,n$. Now, we have $W_{j, n}$ and $W_{j, n+1}$, $j=1,\ldots ,i$, to be independent and non-negative random variables. Thus, in view of Theorem 1.C.9 of Shaked and Shanthikumar (2007), the result follows (i.e., the above inequality holds) provided $W_{j,n}$ and $W_{j,n+1}$ are ILR, and $W_{j, n+1} \le _{lr} W_{j, n}$, for all $j=1,\ldots , i$. From (3.9), we have

$$\begin{aligned} \frac{f'_{W_{j,n}} \left( t\right) }{f_{W_{j,n}} \left( t\right) }= & {} -\sum _{l=1}^{n-j+1} \alpha _{l}^{(j)}, \quad t>0, \end{aligned}$$

which implies that ${f'_{W_{j,n}} \left( t\right) } / {f_{W_{j,n}} \left( t\right) }$ is decreasing in $t>0$ and so $W_{j,n}$ is ILR, for all $j=1, \ldots , i$. Similarly, we get $W_{j,n+1}$ to be ILR, for all $j=1, \ldots , i$. Again, the condition that “$\alpha _{n-j+2}^{(j)}>0$" implies $W_{j, n+1} \le _{lr} W_{j, n}$, for all $j=1,\ldots , i$. Hence, the required result. $\square$

Proof of Part(c) of Theorem 4.4

From Lemma 3.4, we have

$$\begin{aligned} X^{\star }_{i:n} {\mathop {=}\limits ^{d}} D^{-1} \left( \sum _{j=1}^{i}W_{j,n} \right) \text { and } X^{\star }_{i+1:n+1} {\mathop {=}\limits ^{d}} D^{-1} \left( \sum _{j=1}^{i+1}W_{j,n+1} \right) . \end{aligned}$$

Note that the likelihood ratio order is closed under increasing transformations. As F is continuous, $D^{-1} \left( \cdot \right)$ is increasing and so the required result holds if and only if $\sum _{j=1}^{i}W_{j, n} \le _{lr} \sum _{j=1}^{i+1}W_{j, n+1}$, $i=1,\ldots , n$. Now, we have $W_{j, n}$ and $W_{j, n+1}$, $j=1,\ldots ,i+1$, to be independent and non-negative random variables. Thus, in view of Theorem 1.C.9 of Shaked and Shanthikumar (2007), the result follows (i.e., the above inequality holds) provided $W_{j,n}$, $W_{1,n+1}$ and $W_{j+1,n+1}$ are ILR, and $W_{j, n} \le _{lr} W_{j+1, n+1}$, for all $j=1,\ldots , i$. From (3.9), we have

$$\begin{aligned} \frac{f'_{W_{j,n}} \left( t\right) }{f_{W_{j,n}} \left( t\right) }= & {} -\sum _{l=1}^{n-j+1} \alpha _{l}^{(j)}, \quad t>0, \end{aligned}$$

which implies that ${f'_{W_{j,n}} \left( t\right) } / {f_{W_{j,n}} \left( t\right) }$ is decreasing in $t>0$ and so $W_{j,n}$ are ILR, $j=1, \ldots , i$. Similarly, we get $W_{j,n+1}$ to be ILR, for all $j=1, \ldots , i+1$. Again, from the condition that “$\alpha _{l}^{(j)} \ge \alpha _{l}^{(j+1)}$, for $l=1, \ldots , n-j+1$”, we get $W_{j, n} \le _{lr} W_{j+1, n+1}$, for all $j=1,\ldots , i$. Hence, the required result. $\square$

Proof of Par(a) of Theorem 4.5

From Lemma 3.4, we have

$$\begin{aligned} X^{\star }_{i:n} {\mathop {=}\limits ^{d}} D^{-1} \left( \sum _{j=1}^{i}W_{j,n} \right) \text { and } X^{\star }_{i+1:n} {\mathop {=}\limits ^{d}} D^{-1} \left( \sum _{j=1}^{i+1}W_{j,n} \right) . \end{aligned}$$

As F is DFR, $D^{-1} \left( \cdot \right)$ is increasing and convex and so in view of Theorem 3.B.10(a) of Shaked and Shanthikumar (2007), the result holds if $\sum _{j=1}^{i}W_{j,n} \le _{disp} \sum _{j=1}^{i+1}W_{j,n}$ and $\sum _{j=1}^{i}W_{j,n} \le _{st} \sum _{j=1}^{i+1}W_{j,n}$, for all $i=1, \ldots , n-1$. Now, we have $W_{j, n}$, $j=1,\ldots ,i+1$, to be independent and non-negative random variables. Thus, in view of Theorem 1.1 of Khaledi and Kochar (2000), the result follows (i.e., the above inequality holds) provided $W_{j,n}$ is ILR, for all $j=1,\ldots ,i+1$. From (3.9), we have

$$\begin{aligned} \frac{f'_{W_{j,n}} \left( t\right) }{f_{W_{j,n}} \left( t\right) }= & {} -\sum _{l=1}^{n-j+1} \alpha _{l}^{(j)}, \quad t>0, \end{aligned}$$

which implies that ${f'_{W_{j,n}} \left( t\right) } / {f_{W_{j,n}} \left( t\right) }$ is decreasing in $t>0$ and so $W_{j,n}$ is ILR, for all $j=1, \ldots , i+1$. Hence, the required result. $\square$

Proof of Part(b) of Theorem 4.5

From Lemma 3.4, we have

$$\begin{aligned} X^{\star }_{i:n+1} {\mathop {=}\limits ^{d}} D^{-1} \left( \sum _{j=1}^{i}W_{j,n+1} \right) \text { and } X^{\star }_{i:n} {\mathop {=}\limits ^{d}} D^{-1} \left( \sum _{j=1}^{i}W_{j,n} \right) . \end{aligned}$$

As F is DFR, $D^{-1} \left( \cdot \right)$ is increasing and convex and so in view of Theorem 3.B.10(a) of Shaked and Shanthikumar (2007), the result holds if $\sum _{j=1}^{i}W_{j, n+1} \le _{disp} \sum _{j=1}^{i}W_{j, n}$ and $\sum _{j=1}^{i}W_{j, n+1} \le _{st} \sum _{j=1}^{i}W_{j, n}$, for all $i=1,\ldots ,n$. Now, we have $W_{j, n}$ and $W_{j, n+1}$, $j=1,\ldots ,i$, to be independent and non-negative random variables. Thus, in view of Theorem 1.1 of Khaledi and Kochar (2000), the result follows (i.e., the above inequality holds) provided $W_{j,n}$ and $W_{j,n+1}$ are ILR, $W_{j, n+1} \le _{disp} W_{j, n}$ and $W_{j, n+1} \le _{st} W_{j, n}$, for all $j=1,\ldots , i$. From (3.9), we have

$$\begin{aligned} \frac{f'_{W_{j,n}} \left( t\right) }{f_{W_{j,n}} \left( t\right) }= & {} -\sum _{l=1}^{n-j+1} \alpha _{l}^{(j)}, \quad t>0, \end{aligned}$$

which implies that ${f'_{W_{j,n}} \left( t\right) } / {f_{W_{j,n}} \left( t\right) }$ is decreasing in $t>0$ and so $W_{j,n}$ is ILR, for all $j=1, \ldots , i$. Similarly, we get $W_{j,n+1}$ to be ILR, for all $j=1, \ldots , i$. Again, from the condition that “$\alpha _{n-j+2}^{(j)} >0$”, we get

$$\begin{aligned} -\left\{ \frac{1}{ \sum _{l=1}^{n-j+1} \alpha _{l}^{(j)} } - \frac{1}{ \sum _{l=1}^{n-j+2} \alpha _{l}^{(j)} } \right\} \ln \left( 1-u \right) \text { to be increasing in } u \in \left( 0,1 \right) , \end{aligned}$$

which further implies that ${F}^{-1}_{ W_{j,n} } \left( u\right) - {F}^{-1}_{ W_{j,n+1} } \left( u\right)$ is increasing in $u \in (0,1)$. Thus, we get $W_{j, n+1} \le _{disp} W_{j, n}$, for all $j=1,\ldots , i$. Further, it can easily be verified that $W_{j, n+1} \le _{st} W_{j, n}$, for all $j=1,\ldots , i$. Hence, the required result. $\square$

Proof of Part(c) of Theorem 4.5

From Lemma 3.4, we have

$$\begin{aligned} X^{\star }_{i:n} {\mathop {=}\limits ^{d}} D^{-1} \left( \sum _{j=1}^{i}W_{j,n} \right) \text { and } X^{\star }_{i+1:n+1} {\mathop {=}\limits ^{d}} D^{-1} \left( \sum _{j=1}^{i+1}W_{j,n+1} \right) . \end{aligned}$$

As F is DFR, $D^{-1} \left( \cdot \right)$ is increasing and convex and so in view of Theorem 3.B.10(a) of Shaked and Shanthikumar (2007), the result holds if $\sum _{j=1}^{i}W_{j, n} \le _{disp} \sum _{j=1}^{i+1}W_{j, n+1}$ and $\sum _{j=1}^{i}W_{j, n} \le _{st} \sum _{j=1}^{i+1}W_{j, n+1}$, for all $i=1,\ldots , n$. Now, we have $W_{j, n}$, $j=1,\ldots ,i$, and $W_{j, n+1}$, $j=1,\ldots ,i+1$, to be independent and non-negative random variables. Thus, in view of Theorem 1.1 of Khaledi and Kochar (2000), the result follows (i.e., the above inequality holds) provided $W_{j,n}$, $W_{1,n+1}$ and $W_{j,n+1}$ are ILR, $W_{j, n} \le _{disp} W_{j+1, n+1}$ and $W_{j, n} \le _{st} W_{j+1, n+1}$, for all $j=1,\ldots , i$. From (3.9), we have

$$\begin{aligned} \frac{f'_{W_{j,n}} \left( t\right) }{f_{W_{j,n}} \left( t\right) }= & {} -\sum _{l=1}^{n-j+1} \alpha _{l}^{(j)}, \quad t>0, \end{aligned}$$

which implies that ${f'_{W_{j,n}} \left( t\right) } / {f_{W_{j,n}} \left( t\right) }$ is decreasing in $t>0$ and so $W_{j,n}$ is ILR, for all $j=1, \ldots , i$. Similarly, it can be shown that $W_{j,n+1}$ is ILR, for all $j=1, \ldots , i+1$. Again, from the condition that “$\alpha _{l}^{(j)} \ge \alpha _{l}^{(j+1)}$, for $l=1, \ldots , n-j+1$”, we get

$$\begin{aligned} -\left\{ \frac{1}{ \sum _{l=1}^{n-j+1} \alpha _{l}^{(j+1)} } - \frac{1}{ \sum _{l=1}^{n-j+1} \alpha _{l}^{(j)} } \right\} \ln \left( 1-u \right) \text { to be increasing in } u \in \left( 0,1 \right) , \end{aligned}$$

which further implies that ${F}^{-1}_{ W_{j+1,n+1} } \left( u\right) - {F}^{-1}_{ W_{j,n} } \left( u\right)$ is increasing in $u \in (0,1)$. Again, this implies that $W_{j, n} \le _{disp} W_{j+1, n+1}$, for all $j=1,\ldots , i$. Further, it can easily be verified that $W_{j, n} \le _{st} W_{j+1, n+1}$, for all $j=1,\ldots , i$. Hence, the required result. $\square$

Proof of Part(a) of Theorem 4.6

Note that the hazard rate order is closed under increasing transformations, and so $X^{\star }_{i:n} \le _{hr} X^{\star }_{i+1:n}$ holds if and only if $D_{i+1}\left( X^{\star }_{i:n}\right) \le _{hr} D_{i+1}\left( X^{\star }_{i+1:n}\right)$; here, $D_{i+1}$ is the cumulative hazard rate function of $F_{i+1}$. Further, from Lemma 3.3, we have $X^{\star }_{i+1:n} = D_{i+1}^{-1}\left( W_{ i+1,n } + D_{i+1}\left( X_{i:n}^{\star }\right) \right)$ and so the above inequality holds if and only if $D_{i+1}\left( X^{\star }_{i:n}\right) \le _{hr} W_{i+1, n } + D_{i+1}\left( X_{i:n}^{\star }\right)$. Now, we have $Y_l^{ (i+1) }$, $l=1, \ldots , n-i$, and $X^{\star }_{i:n}$ to be independent, which implies that $W_{i+1,n}$ and $D_{i+1}\left( X^{\star }_{i:n}\right)$ are independent. Moreover, $W_{ i+1,n }$ is a non-negative random variable.

Thus, in view of Lemma 1.B.3 of Shaked and Shanthikumar (2007), the result follows (i.e., the above inequality holds) provided $D_{i+1}\left( X_{i:n}^{\star }\right)$ is IFR. Now, we prove the statement that “$D_{i+1}\left( X_{i:n}^{\star }\right)$ is IFR" through induction. First, we show that this statement is true for $i=1$, i.e., $D_{2}\left( X_{1:n}^{\star }\right)$ is IFR. From Definition 3.2, the cumulative hazard rate function of $D_{2}\left( X^{\star }_{1:n}\right)$ is given by

$$\begin{aligned} \Delta _{D_{2}\left( X^{\star }_{1:n}\right) } \left( t\right) = \sum _{j=1}^{n} \alpha _{j}^{(1)} \left( D_1 \circ D_2^{-1}\right) \left( t \right) , \quad t>0. \end{aligned}$$

Now, from the condition that “$F_1 \le _c F_2$” and the increasing property of $D_2^{-1}\left( \cdot \right)$, we get

$$\begin{aligned} \frac{\partial ^2 u}{\partial t^2}= & {} \frac{\partial }{\partial t} \left( \frac{r_1\left( D_{2}^{-1}\left( t\right) \right) }{r_{2}\left( D_{2}^{-1}\left( t\right) \right) } \right) \ge 0 \text { for all } t>0, \end{aligned}$$

From this, we get $\frac{\partial ^2}{\partial t^2}\left( \Delta _{D_{2}(X^{\star }_{1:n})} \left( t\right) \right) \ge 0$, for all $t>0$, and so $D_{2}\left( X^{\star }_{1:n} \right)$ is IFR. Thus, the statement is true for $i=1$. Now, we assume the statement to be true for $i=j-1$, i.e., $D_{j}\left( X^{\star }_{j-1:n}\right)$ is IFR. Now, upon using this, we show that $D_{j+1}\left( X^{\star }_{j:n}\right)$ is IFR. From Lemma 3.3, we get

$D_{j+1}\left( X^{\star }_{j:n}\right) = \left( D_{j+1} \circ D_j^{-1}\right) \left( Q_{j,n}\right)$,

where $Q_{j,n} = W_{j,n} + D_{j}\left( X^{\star }_{j-1:n}\right)$. Then, the cumulative hazard rate function of $D_{j+1}\left( X^{\star }_{j:n}\right)$ is given by

$$\begin{aligned} \Delta _{D_{j+1}\left( X^{\star }_{j:n}\right) } \left( t\right) =\Delta _{ Q_{j,n} }\left( \left( D_{j} \circ D_{j+1}^{-1}\right) \left( t\right) \right) . \end{aligned}$$

(A.33)

Further, we have $Y_l^{ (j) }$, $l=1,\ldots , n-j+1$, and $X^{\star }_{j-1:n}$ to be independent. This implies that $W_{j,n}$ and $D_{j}\left( X^{\star }_{j-1:n}\right)$ are independent. Again, by using (3.7), we get

$$\begin{aligned} \frac{\partial ^2}{ \partial t^2} \Delta _{W_{j,n}} \left( t\right)= & {} \frac{\partial }{ \partial t} \left( \sum _{l=1}^{n-j+1} \alpha _{l}^{(j)} t \right) \\ {}\ge & {} 0 \text { for all } t>0, \end{aligned}$$

which implies that $W_{j,n}$ is IFR. Further, from the induction hypothesis, we have $D_{j}\left( X^{\star }_{j-1:n}\right)$ to be IFR. Upon combining these two facts, we get $Q_{j,n}$ to be IFR, which in turn implies that

$\Delta _{Q_{j,n}} \left( t\right) \text { is increasing and convex in } t>0$.

Again, by proceeding in a manner similar to the case when $i=1$, we can easily obtain

$D_j \circ D_{j+1}^{-1} \left( t\right) \text { to be convex in }t>0$.

Finally, upon using these two facts in (A.33), we get $\Delta _{D_{j+1}\left( X^{\star }_{j:n}\right) }$ to be convex in $t>0$. Consequently, $D_{j+1}\left( X^{\star }_{j:n}\right)$ is IFR and hence the statement gets proved for $i=j$. Thus, by induction, we get $D_{i+1}\left( X^{\star }_{i:n}\right)$ to be IFR, for all i. Hence, the required result. $\square$

Proof of Part(b) of Theorem 4.6

Note that the reverse hazard rate order is closed under increasing transformations and so $X^{\star }_{i:n} \le _{hr} X^{\star }_{i+1:n}$ holds if and only if $D_{i+1}\left( X^{\star }_{i:n}\right) \le _{rh} D_{i+1}\left( X^{\star }_{i+1:n}\right)$; here, $D_{i+1}$ is the cumulative hazard rate function of $F_{i+1}$. Further, from Lemma 3.3, we have $X^{\star }_{i+1:n} = D_{i+1}^{-1}\left( W_{ i+1,n } + D_{i+1}\left( X_{i:n}^{\star }\right) \right)$. Thus, the above inequality holds if and only if $D_{i+1}\left( X^{\star }_{i:n}\right) \le _{rh} W_{i+1, n } + D_{i+1}\left( X_{i:n}^{\star }\right)$. Now, we have $Y_l^{ (i+1) }$, $l=1, \ldots , n-i$, and $X^{\star }_{i:n}$ to be independent, which implies that $W_{i+1,n}$ and $D_{i+1}\left( X^{\star }_{i:n}\right)$ are independent. Moreover, $W_{ i+1,n }$ is a non-negative random variable.

Thus, in view of Lemma 1.B.44 of Shaked and Shanthikumar (2007), the result follows (i.e., the above inequality holds) provided $D_{i+1}\left( X_{i:n}^{\star }\right)$ is DRFR. Now, we prove the statement that “$D_{i+1}\left( X_{i:n}^{\star }\right)$ is DRFR”through induction. First, we show that this statement is true for $i=1$, i.e., $D_{2}\left( X_{1:n}^{\star }\right)$ is DRFR. From Definition 3.2, the cumulative reverse hazard rate function of $D_{2}\left( X^{\star }_{1:n}\right)$ is given by

$$\begin{aligned} \tilde{\Delta }_{D_{2}\left( X^{\star }_{1:n}\right) } \left( t\right) = -\ln \left( 1- \prod _{j=1}^{n}e^{-\alpha _{j}^{(1)} \left( D_1 \circ D_2^{-1}\right) \left( t \right) } \right) , \quad t>0, \end{aligned}$$

which gives

$$\begin{aligned} \frac{\partial ^2}{\partial t^2}\left( \tilde{ \Delta }_{D_{2}\left( X^{\star }_{1:n}\right) } \left( t \right) \right)= & {} -\left( \frac{\partial u}{\partial t} \right) ^2 \frac{\partial }{\partial u} \left( \frac{ {\sum \limits _{j=1}^{n}\alpha _{j}^{(1)}} e^{-\sum \limits _{j=1}^{n}\alpha _{j}^{(1)} u}}{1- e^{- \sum \limits _{j=1}^{n}\alpha _{j}^{(1)} u}}\right) \nonumber \\ {}{} & {} - \left( \frac{\sum \limits _{j=1}^{n}\alpha _{j}^{(1)} e^{- \sum \limits _{j=1}^{n}\alpha _{j}^{(1)} u}}{1- e^{- \sum \limits _{j=1}^{n}\alpha _{j}^{(1)} u}}\right) \frac{\partial ^2 u}{\partial t^2} \end{aligned}$$

(A.34)

for $t>0$, where $u= \left( D_1 \circ D_2^{-1}\right) \left( t\right)$. Now, from the fact that “${e^{-a u}}/ {(1- e^{-a u})}$ is positive and decreasing in $u>0$, for $a>0$", we get

${\sum \limits _{j=1}^{n}\alpha _{j}^{(1)} e^{- \sum \limits _{j=1}^{n}\alpha _{j}^{(1)} u}} \Big /{\Big (1- e^{- \sum \limits _{j=1}^{n}\alpha _{j}^{(1)} u}\Big )}$ to be positive and decreasing in $u>0.$

Again, from the condition that “$F_1 \ge _c F_2$” and the increasing property of $D_2^{-1}\left( \cdot \right)$, we get

$$\begin{aligned} \frac{\partial ^2 u}{\partial t^2}= & {} \frac{\partial }{\partial t} \left( \frac{r_1\left( D_{2}^{-1}\left( t\right) \right) }{r_{2}\left( D_{2}^{-1}\left( t\right) \right) } \right) \le 0 \text { for all } t>0, \end{aligned}$$

Upon using these two facts in (A.34), we get $\frac{\partial ^2}{\partial t^2}\left( \tilde{\Delta }_{D_{2}(X^{\star }_{1:n})} \left( t\right) \right) \ge 0$ for all $t>0$, and so $D_{2}\left( X^{\star }_{1:n} \right)$ is DRFR. Thus, the statement is true for $i=1$. Now, we assume the statement to be true for $i=j-1$, i.e., $D_{j}\left( X^{\star }_{j-1:n}\right)$ is DRFR. Now, upon using this, we show that $D_{j+1}\left( X^{\star }_{j:n}\right)$ is DRFR. From Lemma 3.3, we get

$D_{j+1}\left( X^{\star }_{j:n}\right) = \left( D_{j+1} \circ D_j^{-1}\right) \left( Q_{j,n}\right) ,$

where $Q_{j,n} = W_{j,n} + D_{j}\left( X^{\star }_{j-1:n}\right)$. Then, the cumulative reverse hazard rate function of $D_{j+1}\left( X^{\star }_{j:n}\right)$ is given by

$$\begin{aligned} \tilde{\Delta }_{D_{j+1}\left( X^{\star }_{j:n}\right) } \left( t\right) =\tilde{\Delta }_{ Q_{j,n} }\left( \left( D_{j} \circ D_{j+1}^{-1}\right) \left( t\right) \right) . \end{aligned}$$

(A.35)

Further, we have $Y_l^{ (j) }$, $l=1,\ldots , n-j+1$, and $X^{\star }_{j-1:n}$ to be independent. This implies that $W_{j,n}$ and $D_{j}\left( X^{\star }_{j-1:n}\right)$ are independent. Again, by using (3.7), we get

$$\begin{aligned} \frac{\partial ^2}{ \partial t^2} \Delta _{W_{j,n}} \left( t\right)= & {} -\frac{\partial }{ \partial t} \left( \frac{\sum \limits _{l=1}^{n-j+1}\alpha _{l}^{(j)} e^{- \sum \limits _{l=1}^{n-j+1}\alpha _{l}^{(j)} u}}{1- e^{- \sum \limits _{l=1}^{n-i+1}\alpha _{l}^{(j)} u}} \right) \ge 0 \text { for all } t>0, \end{aligned}$$

which implies that $W_{j,n}$ is DRFR. Further, from the induction hypothesis, we have $D_{j}\left( X^{\star }_{j-1:n}\right)$ to be DRFR. Upon combining these two facts, we get $Q_{j,n}$ to be DRFR. Further, this implies that

$\tilde{\Delta }_{Q_{j,n}} \left( t\right) \text { is decreasing and convex in } t>0.$

Again, by proceeding in a manner similar to the case when $i=1$, we can easily obtain

$D_j \circ D_{j+1}^{-1} \left( t\right) \text { to be concave in }t>0.$

Finally, upon using these two facts in (A.35), we get $\tilde{\Delta }_{D_{j+1}\left( X^{\star }_{j:n}\right) }$ to be convex in $t>0$. Consequently, $D_{j+1}\left( X^{\star }_{j:n}\right)$ is DRFR and hence the statement gets proved for $i=j$. Thus, by induction, we get $D_{i+1}\left( X^{\star }_{i:n}\right)$ to be DRFR, for all i. Hence, the required result. $\square$

Proof of Part(a) of Theorem 4.7

In view of Theorem 3.11 of Belzunce et al. (2001), it suffices to prove that $X^{\star }_{1:n} \le _{st} Z^{\star }_{1:n}$ and $\left( X^{\star }_{i:n} | X^{\star }_{i-1:n} = x \right) \le _{hr} \left( Z^{\star }_{i:n} | Z^{\star }_{i-1:n} = x \right)$, for all $x>0$, for $i=2, \ldots , n$. Now, from the definition of SOS, the reliability functions of $X_{1:n}^{\star }$ and $Z_{1:n}^{\star }$ are given by

$$\begin{aligned} \bar{F}_{X_{1:n}^{\star }}\left( t\right) = \prod _{j=1}^{n} \bar{F}_{j}^{(1)}\left( t\right) \text { and } \bar{F}_{Z_{1:n}^{\star }}\left( t\right) =\prod _{j=1}^{n} \bar{G}_{j}^{(1)}\left( t\right) , \quad t>0, \end{aligned}$$

which, in view of the condition that “${F}_{j}^{(1)} \le _{st} {G}_{j}^{(1)}$, for $j=1,\ldots , n$”, implies that $\bar{F}_{X_{1:n}^{\star }}\left( t\right) \le \bar{F}_{Z_{1:n}^{\star }}\left( t\right)$ for $t>0$. Again, from Remark 3.1, we get, for $i=2,\ldots ,n$,

$$\begin{aligned} \bar{F}_{X^{\star }_{i:n} | X^{\star }_{i-1:n} =x } (t) = \prod _{j=1}^{n-i+1} \frac{ \bar{F}_{j}^{(i)}\left( t\right) }{ \bar{F}_{j}^{(i)}\left( x\right) } \text { and } \bar{F}_{Z^{\star }_{i:n} | Z^{\star }_{i-1:n} =x }(t) = \prod _{j=1}^{n-i+1} \frac{ \bar{G}_{j}^{(i)}\left( t\right) }{ \bar{G}_{j}^{(i)}\left( x\right) }, \quad t \ge x. \end{aligned}$$

These imply that

$$\begin{aligned} {r}_{X^{\star }_{i:n} | X^{\star }_{i-1:n} =x } (t) = \sum _{j=1}^{n-i+1} {r}_j^{(i)} \left( t \right) \text { and } {r}_{Z^{\star }_{i:n} | Z^{\star }_{i-1:n} =x } (t) = \sum _{j=1}^{n-i+1} {h}_j^{(i)} \left( t \right) , \quad t>0, \end{aligned}$$

where ${r}_j^{(i)}$ and $h_j^{(i)}$ are the hazard rate functions of $F_j^{(i)}$ and $G_j^{(i)}$, respectively. From the condition that “$F_j^{(i)} \le _{hr} G_j^{(i)}$, for $i=2,\ldots ,n$”, we have ${r}_j^{(i)} \left( t \right) \ge {h}_j^{(i)} \left( t \right)$ for all $t>0$, and so $\left( X^{\star }_{i:n} | X^{\star }_{i-1:n} = x \right) \le _{hr} \left( Z^{\star }_{i:n} | Z^{\star }_{i-1:n} = x \right)$ for all $x>0$, for $i=2, \ldots , n$. Hence, the required result. $\square$

Proof of Part(c) of Theorem 4.7

In view of Theorem 3.13 of Belzunce et al. (2001), it is enough to prove that $\left( X^{\star }_{i:n} | X^{\star }_{i-1:n} = x \right) \le _{hr} \left( Z^{\star }_{i:n} | Z^{\star }_{i-1:n} = x \right)$ and $\left( X^{\star }_{i:n} | X^{\star }_{i-1:n} = x \right) \ge _{c} \left( Z^{\star }_{i:n} | Z^{\star }_{i-1:n} = x \right)$ for all $x>0$, and ${r}_{Z^{\star }_{i+1:n} | Z^{\star }_{i:n} =x } (t) - {r}_{Z^{\star }_{i:n} | Z^{\star }_{i-1:n} =x } (t) \ge {r}_{Z^{\star }_{i+1:n} | Z^{\star }_{i:n} =x } (t) - {r}_{Z^{\star }_{i:n} | Z^{\star }_{i-1:n} =x } (t)$ for $i=1,2, \ldots , n-1$. Now, from the definition of SOS, the reliability functions of $X_{1:n}^{\star }$ and $Z_{1:n}^{\star }$ are given by

$$\begin{aligned} \bar{F}_{X_{1:n}^{\star }}\left( t\right) = \prod _{j=1}^{n} \bar{F}^{\alpha _{j}^{(1)}}\left( t\right) \text { and } \bar{F}_{Z_{1:n}^{\star }}\left( t\right) = \prod _{j=1}^{n} \bar{G}^{\beta _{j}^{(1)}}\left( t\right) , \quad t>0. \end{aligned}$$

These imply that

$$\begin{aligned} {r}_{X_{1:n}^{\star }}\left( t\right) =r(t) \sum _{j=1}^{n} {\alpha }_j^{(1)} \text { and } {r}_{Z_{1:n}^{\star }}\left( t\right) = h(t)\sum _{j=1}^{n} {\beta }_j^{(1)}, \quad t>0, \end{aligned}$$

where r and h are the hazard rate functions of F and G, respectively. Now, from the conditions that “${F} \le _{hr} {G}$”, “${F} \ge _{c} {G}$” and “$\sum _{j=1}^{n} {\alpha }_j^{(1)} \ge \sum _{j=1}^{n} {\beta }_j^{(1)}$”, we get ${X_{1:n}^{\star }} \le _{hr} {Z_{1:n}^{\star }}$ and ${X_{1:n}^{\star }} \ge _{c} {Z_{1:n}^{\star }}$. Again, from Remark 3.1, we get, for $i=2,\ldots ,n$,

$$\begin{aligned} \bar{F}_{X^{\star }_{i:n} | X^{\star }_{i-1:n} =x } (t) = \prod _{j=1}^{n-i+1} \left( \frac{ \bar{F}\left( t\right) }{ \bar{F}\left( x\right) } \right) ^{ {\alpha }_{j}^{(i)} } \text { and } \bar{F}_{Z^{\star }_{i:n} | Z^{\star }_{i-1:n} =x }(t) = \prod _{j=1}^{n-i+1} \left( \frac{ \bar{G}\left( t\right) }{ \bar{G}\left( x\right) } \right) ^{ {\beta }_{j}^{(i)} }, \quad t \ge x, \end{aligned}$$

which imply that

$$\begin{aligned} {r}_{X^{\star }_{i:n} | X^{\star }_{i-1:n} =x } (t) = r(t) \sum _{j=1}^{n-i+1} {\alpha }_j^{(i)} \text { and } {r}_{Z^{\star }_{i:n} | Z^{\star }_{i-1:n} =x } (t) = h(t) \sum _{j=1}^{n-i+1} {\beta }_j^{(i)}, \quad t>0. \end{aligned}$$

From the conditions that “${F} \le _{hr} {G}$”, “${F} \ge _{c} {G}$” and “$\sum _{j=1}^{n-i+1} {\alpha }_j^{(i)} \ge \sum _{j=1}^{n-i+1} {\beta }_j^{(i)}$, for $i=2,\ldots ,n$”, we get $\left( X^{\star }_{i:n} | X^{\star }_{i-1:n} = x \right) \le _{hr} \left( Z^{\star }_{i:n} | Z^{\star }_{i-1:n} = x \right)$ and $\left( X^{\star }_{i:n} | X^{\star }_{i-1:n} = x \right) \ge _{c} \left( Z^{\star }_{i:n} | Z^{\star }_{i-1:n} = x \right)$, for all $x>0$, for $i=2, \ldots , n$. Further, from the conditions that “${F} \le _{hr} {G}$” and “$\sum _{j=1}^{n-i+1} {\alpha }_j^{(i)} - \sum _{j=1}^{n-i} {\alpha }_j^{(i+1)} \ge \sum _{j=1}^{n-i+1} {\beta }_j^{(i)} - \sum _{j=1}^{n-i} {\beta }_j^{(i+1)} \ge 0$, for $i=1,\ldots ,n-1$”, we get ${r}_{Z^{\star }_{i+1:n} | Z^{\star }_{i:n} =x } (t) - {r}_{Z^{\star }_{i:n} | Z^{\star }_{i-1:n} =x } (t) \ge {r}_{Z^{\star }_{i+1:n} | Z^{\star }_{i:n} =x } (t) - {r}_{Z^{\star }_{i:n} | Z^{\star }_{i-1:n} =x } (t)$ for $i=1, \ldots , n-1$. Hence, the required result. $\square$

Proof of Theorem 4.8

Let

$$\begin{aligned} A_{l,n} = \min \left\{ - \frac{1}{\beta _{1}^{(l)}} \ln \left( 1-U^{(l)}_1\right) , \ldots , -\frac{1}{\beta _{n-l+1}^{(l)}} \ln \left( 1-U^{(l)}_{n-l+1}\right) \right\} , \quad l=1,2,\ldots , n, \end{aligned}$$

where $U_j^l,\; j=1,\ldots ,n-l+1,\; l=1,\ldots ,n$, are the same as in Lemma 3.2. Now, from Lemma 3.4, we can write

$$\begin{aligned} X^{\star }_{i:n}{\mathop {=}\limits ^{d}} D^{-1} \left( \sum _{j=1}^{i}W_{j,n} \right) \text { and } Z^{\star }_{i:n} {\mathop {=}\limits ^{d}} D^{-1} \left( \sum _{j=1}^{i}A_{j,n} \right) . \end{aligned}$$

Note that the likelihood ratio order is closed under increasing transformations. As F is continuous, $D^{-1} \left( \cdot \right)$ is increasing and so the required result holds if and only if $\sum _{j=1}^{i}W_{j, n} \le _{lr} \sum _{j=1}^{i}A_{j, n}$. Now, we have $W_{j, n}$ and $A_{j, n}$, $j=1,\ldots ,i$, to be independent and non-negative random variables. Thus, in view of Theorem 1.C.9 of Shaked and Shanthikumar (2007), the result follows (i.e., the above inequality holds) provided $W_{j,n}$ and $A_{j,n}$ are ILR, and $W_{j, n} \le _{lr} A_{j, n}$, for all $j=1,\ldots , i$. From (3.9), we have

${f'_{W_{j,n}} \left( t\right) } /{f_{W_{j,n}} \left( t\right) } = -\sum _{l=1}^{n-j+1} \alpha _{l}^{(j)}, \;t>0,$

which implies that ${f'_{W_{j,n}} \left( t\right) } / {f_{W_{j,n}} \left( t\right) }$ is decreasing in $t>0$ and so $W_{j,n}$ is ILR. Similarly, it can be shown that $A_{j,n}$ is also ILR, for all $j=1, \ldots , i$. Again, we have the condition that “$\varvec{\alpha }^{(j)} \overset{w}{\preceq } \varvec{\beta }^{(j)}$”. Then, from Theorem 3.1 of Li and Li (2015), we obtain $W_{j, n} \le _{lr} A_{j, n}$, for all $j=1,\ldots , i$. Hence, the required result. $\square$