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On stochastic comparisons of minimum order statistics from the location–scale family of distributions

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Abstract

We consider stochastic comparisons of minimum order statistics from the location–scale family of distributions that contain most of the popular lifetime distributions. Under certain assumptions, we show that the minimum order statistic of one set of random variables dominates that of another set of random variables with respect to different stochastic orders. Furthermore, we illustrate our results using some well-known specific distributions.

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Acknowledgements

The authors are thankful to the Editor-in-Chief, the Associate Editor and the anonymous Reviewers for their valuable constructive comments which lead to an improved version of the manuscript. The first author sincerely acknowledges the financial support from the Claude Leon Foundation, South Africa. The work of the second author was supported by Indian Institute of Management Calcutta, India. The third author was supported by the NFR (national Research Foundation of South Africa) Grant No. 103613. The financial support from National Board of Higher Mathematics, Govt. of India (No. 2/48(25)/2014/NBHM (R.P.) R & D II/1393 dt. 03.02.2015) is duly acknowledged by Asok K. Nanda.

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Correspondence to Maxim Finkelstein.

Appendix

Appendix

Proof of Theorem 2.1

We only prove the result under the assumption \(\{{\varvec{\lambda }}\in \mathcal {E},\; {\varvec{\mu }}\in \mathcal {E},\;{\varvec{\theta }}\in \mathcal {E}\}\). It could be proven in the same line under the second assumption. Note that

$$\begin{aligned} \bar{F}_{X_{1:n}}(x;{{\varvec{\lambda }}},{{\varvec{\theta }}})\;\text { is increasing in each }\lambda _i,\;\text {for }i=1,2,\ldots ,n. \end{aligned}$$
(4.1)

Further,

$$\begin{aligned} \frac{\partial {\bar{F}}_{X_{1:n}}(x;{{\varvec{\lambda }}},{{\varvec{\theta }}})}{\partial \lambda _k}\text { is decreasing in }k=1,2,\ldots ,n, \end{aligned}$$

if, for \(1\le p\le q\le n\),

$$\begin{aligned} \frac{1}{\theta _p}r\left( \frac{x-\lambda _p}{\theta _p}\right) \ge \frac{1}{\theta _q}r\left( \frac{x-\lambda _q}{\theta _q}\right) . \end{aligned}$$
(4.2)

Consider the following two cases.

  • Case I Let \(x\le \lambda _p\le \lambda _q\) or \(\lambda _p\le x\le \lambda _q\). Then, (4.2) is trivially true.

  • Case II Let \(\lambda _p\le \lambda _q\le x\). Then, (4.2) holds because \(r(\cdot )\) is an increasing function.

Thus, (4.2) holds for both the cases, and hence, by Lemma 3.3 of Kundu et al. (2016), we get that

$$\begin{aligned} {\bar{F}}_{X_{1:n}}(x;{{\varvec{\lambda }}},{{\varvec{\theta }}})\text { is Schur-concave in } {{\varvec{\lambda }}}\in \mathcal {E}. \end{aligned}$$
(4.3)

Finally, on using (4.1) and (4.3) in Lemma 4.3 of Kundu et al. (2016), we get the required result. \(\square \)

Proof of Theorem 2.6

We only prove the result under the assumption \(\{{\varvec{\lambda }}\in \mathcal {E},\; {\varvec{\mu }}\in \mathcal {E},\;{\varvec{\theta }}\in \mathcal {E}\}\). It could be proven in the same line under the second assumption. Note that

$$\begin{aligned} r_{X_{1:n}}(x;{{\varvec{\lambda }}},{{\varvec{\theta }}})\;\text { is decreasing in each }\lambda _i,\;\text {for }i=1,2,\ldots ,n. \end{aligned}$$
(4.4)

Further,

$$\begin{aligned} \frac{\partial r_{X_{1:n}}(x;{{\varvec{\lambda }}},{{\varvec{\theta }}})}{\partial \lambda _k}\text { is increasing in }k=1,2,\ldots ,n, \end{aligned}$$

if, for \(1\le p\le q\le n\),

$$\begin{aligned} \frac{1}{\theta ^2_p} r'\left( \frac{x-\lambda _p}{\theta _p}\right) \ge \;\frac{1}{\theta _q^2} r'\left( \frac{x-\lambda _q}{\theta _q}\right) . \end{aligned}$$
(4.5)

Since, \(\lambda _q\ge \lambda _p\) and \(\theta _q\ge \theta _p\), we have that

$$\begin{aligned} \frac{1}{\theta _p^2}\ge \frac{1}{\theta _q^2}\; \text{ and } \;\frac{x-\lambda _p}{\theta _p}\ge \frac{x-\lambda _q}{\theta _q}. \end{aligned}$$
(4.6)

Consider the following three cases:

  • Case I Let \(\lambda _q\ge \lambda _p\ge x\). Then (4.5) holds trivially.

  • Case II Let \(\lambda _q\ge x\ge \lambda _p\). Then (4.5) follows from (4.6) and the fact that \(r(\cdot )\) is an increasing function.

  • Case III Let \(x\ge \lambda _q\ge \lambda _p\). Then (4.5) follows from (4.6) and the fact that r(u) is both increasing and convex in u. Thus, for all the cases, (4.5) holds. So, by Lemma 3.3 of Kundu et al. (2016), we have that

    $$\begin{aligned} r_{X_{1:n}}(x;{{\varvec{\lambda }}},{{\varvec{\theta }}})\text { is Schur-convex in } {{\varvec{\lambda }}}\in \mathcal {E}. \end{aligned}$$
    (4.7)

Finally, on using (4.4) and (4.7) in Lemma 4.3 of Kundu et al. (2016), we get the required result. \(\square \)

Proof of Theorem 2.11

Let

$$\begin{aligned}&\mathcal {F}=\{{\varvec{\mu }}=(\mu _1,\mu _2,\ldots ,\mu _n):\;\mu _i=\theta _1^{-1},\\&\qquad \text { for }1\le i\le n_1 \text { and } \mu _i=\theta _2^{-1}, \text { for }n_1+1\le i\le n \}. \end{aligned}$$

Further, let \(u(x)=xr(x)\), \(v(x)=xr'(x)/ r(x)\), and \(\omega (x)=u(x)v'(x)\), for all \(x>0\). Note that \(r_{X_{1:n}}(x)/r_{Y_{1:n}}(x)\) is decreasing in x if

$$\begin{aligned} \Lambda (\mu _1,\mu _2,\ldots ,\mu _n){\mathop {=}\limits ^\mathrm{def.}}\frac{\sum \limits _{i=1}^n u(\mu _i (x-\lambda ))v(\mu _i (x-\lambda ))}{\sum \limits _{i=1}^n u(\mu _i (x-\lambda ))} \end{aligned}$$

is Schur-concave in \((\mu _1,\mu _2,\ldots ,\mu _n)\in \mathcal {F}\). Consider the following two cases:

  • Case I Let \(1\le i,j\le n_1\) or \(n_1+1\le i,j\le n\). Then

    $$\begin{aligned} \frac{\partial \Lambda }{\partial \mu _i}- \frac{\partial \Lambda }{\partial \mu _j}=0 \end{aligned}$$
  • Case II Let \(1\le i\le n_1\) and \(n_1+1\le j\le n\). Then

    $$\begin{aligned}&\frac{\partial \Lambda }{\partial \mu _i}- \frac{\partial \Lambda }{\partial \mu _j}{\mathop {=}\limits ^\mathrm{sgn}}n_1n_2\{v(a_1 (x-\lambda ))-v(a_2 (x-\lambda ))\}\{ u'(a_1 (x-\lambda ))u(a_2 (x-\lambda ))\\&\qquad +\, u'(a_2 (x-\lambda ))u(a_1 (x-\lambda ))\}\\&\qquad +\, \{n_1 u(a_1 (x-\lambda ))+n_2 u(a_2 (x-\lambda ))\}\{n_1 \omega (a_1 (x-\lambda ))- n_2\omega (a_2 (x-\lambda ))\}, \end{aligned}$$

    where \(a_1=\theta _1^{-1}\) and \(a_2=\theta _2^{-1}\). From the hypothesis, we have that u(x) is increasing in x, and v(x) is decreasing and concave in x. This implies that w(x) is non-positive and decreasing in x. Thus, it follows that \(\frac{\partial \Lambda }{\partial \mu _i}-\frac{\partial \Lambda }{\partial \mu _j}\le 0\) whenever the set of conditions \(\{(\theta _1,\theta _2)\in ~\mathcal {E}, (\delta _1,\delta _2)\in \mathcal {E},(n_1,n_2)\in \mathcal {D}\}\) holds. Further, \(\frac{\partial \Lambda }{\partial \mu _i}-\frac{\partial \Lambda }{\partial \mu _j}\ge 0\) whenever the set of conditions \(\{(\theta _1,\theta _2)\in \mathcal {D}, (\delta _1,\delta _2)\in \mathcal {D}, (n_1,n_2)\in \mathcal {E}\}\) holds. Hence the result follows from Lemma 3.3 of Kundu et al. (2016) and Theorem A.3 of Marshall et al. (2011, p. 83).\(\square \)

Proof of Theorem 2.14

Suppose that the first set of conditions holds. Then, the weak submajorization order gives that \(\theta _2^{-1}\ge \delta _2^{-1}\) and \(b\theta _1^{-1}+n_2\theta _2^{-1}\ge b\delta _1^{-1}+n_2\delta _2^{-1}\), for \(b=n_1, n_1-1,\ldots , 2, 1\). If \(n_1\theta _1^{-1}+n_2\theta _2^{-1}= n_1\delta _1^{-1}+n_2\delta _2^{-1}\) holds, then the result follows from Theorem 2.11. Suppose that \(n_1\theta _1^{-1}+n_2\theta _2^{-1}> n_1\delta _1^{-1}+n_2\delta _2^{-1}\). Then, there exists some \(\theta \) such that \(n_1\theta _1^{-1}+n_2\theta ^{-1}= n_1\delta _1^{-1}+n_2\delta _2^{-1}\) and \(\theta _2^{-1} >\theta ^{-1} \ge \delta _2^{-1}\). Let \(X^*_{1:n}\) be the minimum order statistic formed from the set of random variables (\(X_1^*,X_2^*,\ldots ,X_n^*\)), where \(X_i^*\sim LS(\lambda ,\theta _1)\), for \(i=1,2,\ldots ,n_1\) and \(X_j^*\sim LS(\lambda ,\theta )\), for \(j=n_1+1,n_1+2,\ldots , n_1+n_2\;(=n)\). Then, on using Lemma 4.4 of Kundu et al. (2016), \(X^*_{1:n}\ge _{R-HR}Y_{1:n}\) follows from Theorem 2.11. Further, note that \(\theta _2^{-1} >\theta ^{-1} \ge \theta _1^{-1}\) and

$$\begin{aligned}&\left( \underbrace{{\theta _1}^{-1},{\theta _1}^{-1},\ldots ,{\theta _1}^{-1},}_{n_1} \underbrace{{\theta _2}^{-1},{\theta _2}^{-1},\ldots ,{\theta _2}^{-1}}_{n_2}\right) \\&\quad \succeq _{w}\left( \underbrace{{\theta _1}^{-1},{\theta _1}^{-1},\ldots ,{\theta _1}^{-1},}_{n_1} \underbrace{{\theta }^{-1},{\theta }^{-1},\ldots ,{\theta }^{-1}}_{n_2}\right) . \end{aligned}$$

Thus, \(X_{1:n}\ge _{R-HR}X^*_{1:n}\) follows from Theorem 2.13. Hence \(X_{1:n}\ge _{R-HR}Y_{1:n}\). The proof follows in a similar way under the second set of conditions.\(\square \)

Proof of Theorem 2.15

Writing \(D_2(x)= r_{X_{1:n}}(x;\lambda ,{{\varvec{\theta }}})/ r_{Y_{1:n}}(x;\lambda ,\theta )\), we have

where the first inequality follows from (C4), (C7) and equation (1.5) of Mitrinović et al. (1993, p. 240) whereas the second inequality follows from (C7) and (C8). \(\square \)

Proof of Lemma 3.1(i)

Writing \(\Delta _1(u)=ur'_Z(u)/r_Z(u)\), we have

$$\begin{aligned} \Delta _1(u)=-\left( 1+\frac{\alpha }{\gamma }\right) +\left( \frac{\alpha +\beta }{\gamma }\right) \left( \frac{1}{1+u^{\frac{1}{\gamma }}}\right) +ur_Z(u). \end{aligned}$$
(4.8)

Differentiating the above expression with respect to u, we get

$$\begin{aligned} \Delta _1'(u)=\frac{1}{u}\left[ -\left( \frac{\alpha +\beta }{\gamma ^2}\right) \left\{ \frac{u^{\frac{1}{\gamma }}}{\left( 1+u^{\frac{1}{\gamma }}\right) ^2}\right\} +ur_Z(u)\left( 1+\Delta _1(u)\right) \right] . \end{aligned}$$
(4.9)

From Lemma 3.1, we have that \(\lim _{u \rightarrow 0}\Delta _1(u)=-\,1+\beta /\gamma \) and \(\lim _{u \rightarrow \infty } \Delta _1(u)=-\,1\). If possible, let \(\min _{u>0}\Delta _1(u)<-\,1\). Then, \(\Delta _1(u)\) has to cross the level \(u=-\,1\) at least once. Let \(v\in (0,\infty )\) be any point such that \(\Delta _1(v)=-\,1\). Then, from (4.9), we have that \(\Delta _1'(v)<0.\) This means that, if \(\Delta _1(u)\) crosses the level \(u=-\,1\), then it will cross exactly once and from above. Assume that \(u_0\) (\(>0\)) is the point where \(\Delta _1(u)\) crosses the level \(u=-\,1\). Then, for \(u_0<u< \infty \), we have \(\Delta _1(u)<-\,1\). On using this in (4.9), we get that \(\Delta _1(u)\) is decreasing in \(u\in (u_0,\infty )\). This gives that \(\Delta _1(u)> \lim u_{u\rightarrow \infty }\Delta _1(u)=-\,1\), which is a contradiction to the fact that \(\Delta _1(u)<-\,1\). Thus, the point \(u_0\) does not exist, and hence the minimum of \(\Delta _1(u)\) is always larger or equal to \(-\,1\). Therefore, the result is proved.\(\square \)

Proof of Lemma 3.1(ii)

If possible, let \(\max _{u>0}\Delta _1(u)>-\,1+\beta /\gamma \). Then, \(\Delta _1(u)\) has to cross the level \(u=-\,1+\beta /\gamma \) at least once and from above because \(\lim _{u \rightarrow 0}\Delta _1(u)=-\,1+\beta /\gamma \) and \(\lim _{u \rightarrow \infty } \Delta _1(u)=-\,1\). Let \(w\in (0,\infty )\) be any point such that \(\Delta _1(w)=-\,1+\beta /\gamma \). Then, from (4.8), we have

$$\begin{aligned} wr_Z(w)=\left( \frac{\alpha +\beta }{\gamma }\right) \left( \frac{w^{\frac{1}{\gamma }}}{1+w^{\frac{1}{\gamma }}}\right) . \end{aligned}$$

On using this and the fact that \(g(u){\mathop {=}\limits ^{\text {def.}}}\beta -1/(1+u^{1/\gamma })\) is positive for all \(u>0\), we have, from (4.9), that \(\Delta _1'(w)>0\). This means that, if \(\Delta _1(u)\) crosses the level \(u=-\,1+\beta /\gamma \), then it will cross exactly once and from below. In other wards, \(\Delta _1(u)\) never crosses the level \(u=-\,1+\beta /\gamma \) from above, which is a contradiction. Thus, the maximum of \(\Delta _1(u)\) does not exceed the level \(u=-\,1+\beta /\gamma \). Hence, the result follows.\(\square \)

Proof of Proposition 3.1(iii)

From Proposition 3.1(i) and Lemma 3.1, we have that \(ur_Z(u)\) is increasing in u with \(\lim _{u \rightarrow 0}ur_Z(u)=0\) and \(\lim _{u \rightarrow \infty }u r_Z(u)=\alpha /\gamma \). Thus,

$$\begin{aligned} ur_Z(u)\le \frac{\alpha }{\gamma },\quad \text { for all }u>0. \end{aligned}$$

On using this and (4.9), we have

$$\begin{aligned} \left[ u\frac{\tilde{r}_Z'(u)}{\tilde{r}_Z(u)}\right] '\le & {} \left( \frac{\alpha +\beta }{\gamma }\right) \left( \frac{u^{-1}}{1+u^{\frac{1}{\gamma }}}\right) \xi _1(u), \end{aligned}$$
(4.10)

where

$$\begin{aligned} \xi _1(u)=u r_Z(u)-\frac{u^{\frac{1}{\gamma }}}{\gamma \left( 1+u^{\frac{1}{\gamma }}\right) }. \end{aligned}$$

Thus, to prove the result, it suffices to show that \(\xi _1(u)\le 0\). Now, on using Lemma 3.1, we have

$$\begin{aligned} \lim _{u\rightarrow 0} \xi _1(u)=0 \text { and } \lim _{u\rightarrow \infty } \xi _1(u)=\frac{\alpha -1}{\gamma }<0. \end{aligned}$$

Let us assume that \(\max _{0<u<\infty }\xi _1(u)>0\). Then, \(\xi _1(u)\) has to cross the level zero at least once and from above. This means that there must exist a point \(w\in (0,\infty )\) such that \(\xi _1(w)=0\) and \(\xi _1'(w)<0\). Let \(v\in (0,\infty )\) be any point such that \(\xi _1(v)=0\). On using this, we get, after some algebraic calculation, that \(\xi _1'(v)>0\). This shows that there does not exist any such point w where both \(\xi _1(w)=0\) and \(\xi _1'(w)<0\) hold. Thus, our assumption is not correct, and hence \(\xi _1(u)\le 0\), for all \(u> 0\). Therefore, the result is proved. \(\square \)

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Hazra, N.K., Kuiti, M.R., Finkelstein, M. et al. On stochastic comparisons of minimum order statistics from the location–scale family of distributions. Metrika 81, 105–123 (2018). https://doi.org/10.1007/s00184-017-0636-x

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