2. Preliminaries

We recall here the basic definitions and representations for tail-weight iterated distributions. These iterated distributions were introduced by Averous and Meste [Reference Averous and Meste6] and initially studied by Fagiuoli and Pellerey [Reference Fagiuoli and Pellerey21]. Let X be a nonnegative random variable with density function $f_X$ , distribution function $F_X$ , and tail function $\overline{F}_X=1-F_X$ .

Definition 2.1. For each $x\geq 0$ , define

(2.1) \begin{equation}\overline{T}_{X,0}(x)=f_{X}(x)\quad \mbox{and}\quad \widetilde{\mu}_{X,0}=\int_0^\infty \overline{T}_{X,0}(t)\,dt=1.\end{equation}

For each $s\geq 1$ , define the $s-\text{iterated}$ distribution $T_{X,s}$ by its tail $\overline{T}_{X,s}=1-T_{X,s}$ as follows:

(2.2) \begin{equation}\overline{T}_{X,s}(x)=\frac{1}{\widetilde{\mu}_{X,s-1}}\int_{x}^\infty\overline{T}_{X,s-1}(t)\,dt,\quad \mbox{where}\quad \widetilde{\mu}_{X,s}=\int_0^\infty \overline{T}_{X,s}(t)\,dt.\end{equation}

Moreover, we extend the domain of definition of each $\overline{T}_{X,s}$ by defining $\overline{T}_{X,s}(x)=1$ for $x<0$ .

The distribution $\overline{T}_{X,2}$ is also known as the equilibrium distribution of X, and plays an important role in ageing relations (see, for example, Chatterjee and Mukherjee [Reference Chatterjee, Mukherjee, Balakrishnan, Rao and North-Holland13]) and in renewal theory (see Cox [Reference Cox17]). Hence, the iteration process above defines, for each $s\geq 1$ , $\overline{T}_{X,s}$ as the equilibrium distribution of a random variable with tail $\overline{T}_{X,s-1}$ . Although the definitions are introduced in a recursive way, a closed form representation for the iterated distributions is available.

Lemma 2.1. (Lemma 2 and Remark 3 in Arab and Oliveira [Reference Arab and Oliveira3]) The tails $\overline{T}_{X,s}$ may be represented as

(2.3) \begin{equation}\overline{T}_{X,s}(x)=\frac{1}{\mathbb{E} X^{s-1}}\int_x^\infty f_{X}(t)(t-x)^{s-1}\,dt.\end{equation}

The $s-$ iterated distribution moments are given by

(2.4) \begin{equation}\widetilde{\mu}_{X,s}=\frac{1}{s}\frac{\mathbb{E} X^{s}}{\mathbb{E} X^{s-1}}.\end{equation}

Note that (2.3) may be rewritten as

(2.5) \begin{equation}\overline{T}_{X,s}(x)=\frac{1}{\mathbb{E} X^{s-1}}\mathbb{E}(X-x)_+^{s-1},\end{equation}

where $(X-x)_+=\max(0,X-x)$ is the residual lifetime at age x. Therefore, the $s-$ iterated distribution may be interpreted as the normalised survival moment of order $s-1$ .

One of the simplest and most common notions of ageing is defined through the monotonicity of the failure rate function of a distribution,

\[\frac{f_X(x)}{1-F_X(x)}=\frac{\overline{T}_{X,0}(x)}{\overline{T}_{X,1}(x)}.\]

The direct verification of this monotonicity is, in general, not a simple task, as for many distributions the tail does not have an explicit closed representation, or at least not a manageable one. Having defined iterated distributions, it becomes natural to proceed likewise with respect to the failure rate functions, as defined in Nanda et al. [Reference Nanda, Hazra, Al-Mutairi and Ghitany32] and also studied in Arab and Oliveira [Reference Arab and Oliveira3].

Definition 2.2. For each $s\geq 1$ and $x\geq 0$ , define the $s-$ iterated failure rate function as

\begin{equation*}r_{X,s}(x)=\frac{\overline{T}_{X,s-1}(x)}{\int_x^\infty\overline{T}_{X,s-1}(t)\,dt}=\frac{\overline{T}_{X,s-1}(x)}{\widetilde{\mu}_{X,s-1}\overline{T}_{X,s}(x)}.\end{equation*}

It is obvious that for $s=1$ we find the failure rate of X to be $r_{X,1}(x)=\frac{f_{X}(x)}{\overline{F}_{X}(x)}$ ; hence the monotonicity of the failure rate is expressed as the monotonicity of $r_{X,1}$ . We may extend this monotonicity notion by considering the $s-$ iterated distribution, as done in Averous and Meste [Reference Averous and Meste6], Fagiuoli and Pellerey [Reference Fagiuoli and Pellerey21], and Nanda et al. [Reference Nanda, Hazra, Al-Mutairi and Ghitany32], among many other references.

The above-mentioned references introduce a few other monotonicity notions, but we refer only to the ones to be addressed in the present paper. Note that it follows easily from the definition above that the $s{-}{\rm IFR}$ monotonicity of a random variable X implies that the variable is also $s{-}{\rm IFRA}$ .

We introduce next the order relations to be addressed.

Fagiuoli and Pellerey [Reference Fagiuoli and Pellerey21] and Nanda et al. [Reference Nanda, Hazra, Al-Mutairi and Ghitany32] concentrated on establishing relations between the ordering notions defined. It is useful to note that these order relations define partial order relations in the equivalence classes of $\mathcal{F}$ corresponding to the equivalence relation $F\sim G$ defined by $F(x)=G(kx)$ for some $k>0$ . In case of families of distributions that have a scale parameter, this allows one to choose the parameter in the most convenient way.

The exponential distribution plays an important role when dealing with ageing notions. Besides being a fixed point with respect to the iteration procedure, comparability with the exponential, either in the $s{-}{\rm IFR}$ or $s{-}{\rm IFRA}$ sense, is equivalent to $s{-}{\rm IFR}$ or $s{-}{\rm IFRA}$ monotonicity, as proved by Nanda et al. [Reference Nanda, Hazra, Al-Mutairi and Ghitany32] (see Theorems 3.2 and 4.3).

As an immediate consequence of the above, we have the following comparison results.

A general characterisation of the above order relations is given below (see Propositions 3.1 and 4.1 in Nanda et al. [Reference Nanda, Hazra, Al-Mutairi and Ghitany32]).

The above characterisation requires explicit expressions for the tails of the iterated distributions, which are often not available. Computationally tractable characterisations for deciding about the actual comparison of general distributions were studied in Arab and Oliveira [Reference Arab and Oliveira3]. We quote the characterisation proved in Theorem 27 and Corollary 29 of [Reference Arab and Oliveira3].

Theorem 2.3. Let X and Y be random variables with absolutely continuous distributions with densities $f_X$ and $f_Y$ and distribution functions $F_X,F_Y\in\mathcal{F}$ , respectively. Suppose that for every pair of constants $a>0$ and $b\in\mathbb{R}$ , either the function

\begin{equation*}H_s(x)=\frac{1}{\mathbb{E} Y^{s-1}}f_Y(x)-\frac{a^{s}}{\mathbb{E} X^{s-1}}f_X(ax+b)\end{equation*}

or the function

\begin{equation*}H_{s-1}(x)=\frac{1}{\mathbb{E} Y^{s-1}}\overline{F}_Y(x)-\frac{a^{s-1}}{\mathbb{E} X^{s-1}}\overline{F}_X(ax+b)\end{equation*}

changes sign at most twice as x traverses from 0 to $+\infty$ , and if the change of sign occurs twice, it is in the order ‘ $+,-,+$ ’. Then $F_X\leq_{s{-}{\rm IFR}}F_Y$ .

The functions $H_s$ and $H_{s-1}$ may, respectively, be replaced by

\begin{equation*}P_s(x)=\log f_Y(x)-\log f_X(ax+b)+\log\frac{\mathbb{E} X^{s-1}}{a^{s}\mathbb{E} Y^{s-1}}\end{equation*}

and

\begin{equation*}P_{s-1}(x)=\log \overline{F}_Y(x)-\log\overline{F}_X(ax+b)+\log\frac{\mathbb{E} X^{s-1}}{a^{s-1}\mathbb{E} Y^{s-1}}.\end{equation*}

The next statement provides a criterion to verify the $s{-}{\rm IFRA}$ order relation. We do not include a proof, as this follows from reproducing the arguments presented in Arab and Oliveira [Reference Arab and Oliveira3] for the proof of the previous result.

As indicated in Theorem 2.2, the control of the sign variation of $\overline{T}_{Y,s}(x)-\overline{T}_{X,s}(ax+b)$ is crucial in characterising the $s{-}{\rm IFR}$ and $s{-}{\rm IFRA}$ orderings. This function is obtained after integration of the function $H_s$ or $H_{s-1}$ defined in Theorems 2.3 and 2.4. Below, we quote a result about sign variation after integration which was used in Arab and Oliveira [Reference Arab and Oliveira3] for the proof of Theorem 2.3, and which is an important tool for our results below.

Lemma 2.2. (Lemma 26 in Arab and Oliveira [Reference Arab and Oliveira3]) Let f and g be two real-valued functions defined on $[0,\infty)$ such that

\begin{equation*} g(x)=\int_{x}^\infty f(t)\,dt.\end{equation*}

Assume that, as x traverses from 0 to $+\infty$ , f(x) changes sign in one of the following orders: ‘ $-,+$ ’ or ‘ $+,-$ ’ or ‘ $+,-,+$ ’ or ‘ $-,+,-,+$ ’. Then g(x), as x traverses from 0 to $\infty$ , has sign variation equal to every possible final part of the sign variation of f(x).

The lifetime of a parallel system is expressed as the maximum of the lifetimes of its components. When these components have exponentially distributed lifetimes, the distribution function of the system’s lifetime is expressed as a linear combination of exponential terms. Later, it will be important to be able to count and localise the roots of such expressions. The following result will play an important role in this aspect.

3. Hereditary monotonicity properties

A common feature of iterated monotonicity properties is that of being hereditary with respect to the iteration parameter. However, the hereditary property does not hold for all the order relations defined, as we will be showing below by an example. We quote first the hereditary property for monotonicity of the iterated failure rate.

This result is included in Theorem 2 in Navarro and Hernandez [Reference Navarro and Hernandez33]. It implies that, for most distributions, it is enough to verify the ${1-}{\rm IFR}$ or the ${1-}{\rm DFR}$ property. Exhibiting distributions that do not have lower iterated monotonicity but satisfy it after a few iteration steps usually requires a suitable modification of known families of distributions. Such an example, using fattened-tail Pareto distributions, was given in Example 9 in Arab and Oliveira [Reference Arab and Oliveira3]. This suggests a way to construct distributions with failure rates that become monotone only after a few iteration steps. The example below shows the same effect for a distribution that is IFR, instead.

Example 3.1. Let X be a nonnegative random variable with density function

\[f(x)= \frac{(x^2+c)e^{-x}}{c+2}.\]

It is easily verified that

\begin{equation*}\overline{T}_{X,1}(x)=\frac{(x^2+2x+2+c)e^{-x}}{c+2}, \qquad \overline{T}_{X,2}(x)=\frac{(x^2+4x+6+c)e^{-x}}{6+c},\end{equation*}

and

\begin{equation*} r_{X,1}(x)=\frac{f(x)}{\overline{T}_{X,1}(x)}=\frac{x^2+c}{x^2+2x+2+c}, \qquad r_{X,2}(x)=\frac{6+c}{2+c}\!\times\!\frac{x^2+2x+2+c}{x^2+4x+6+c}.\end{equation*}

Differentiating, we find

\begin{equation*} r_{X,1}^{\prime}(x)=\frac{2x^2+4x-2c}{\left(x^2+2x+2+c\right)^2}, \qquad r_{X,2}^{\prime}(x)=\frac{6+c}{2+c}\!\times\!\frac{2x^2+8x+4-2c}{\left(x^2+4x+6+c\right)^2}.\end{equation*}

By choosing $c\in(0,2)$ we obtain $r_{X,1}$ that starts decreasing and eventually becomes increasing, while $r_{X,2}$ is increasing. That is, X is not ${1-}{\rm IFR}$ but is ${2-}{\rm IFR}$ . Moreover, it is also easy to verify that X is not ${1-}{\rm IFRA}$ .

Let us now look at the hereditary property for $s{-}{\rm IFRA}$ monotonicity, to show that the situation is quite different from what happens with $s{-}{\rm IFR}$ monotonicity.

Proof. Observe first that

\begin{equation*}\overline{T}_{Y,s}(x)=\frac{1}{c(s,\lambda)}\left(e^{-x}+\frac{e^{-\lambda x}}{\lambda^{s-1}}-\frac{e^{-(\lambda+1) x}}{(\lambda+1)^{s-1}}\right),\end{equation*}

where

\[c(s,\lambda)=1+\frac{1}{\lambda^{s-1}}-\frac{1}{(\lambda+1)^{s-1}}.\]

To prove that Y is ${1-}{\rm IFRA}$ , we need to verify that

\begin{equation*}-t_1(x)=\frac{\log(\overline{T}_{Y,1}(x))}{x}\end{equation*}

is decreasing. Taking into account Theorem 2.2, we need to prove that $H(x)=e^{-x}+e^{-\lambda x}-e^{-(1+\lambda) x}-e^{-ax}$ changes sign at most once, in the order ‘ $+,-$ ’, for every $a>0$ (note that we are here interested in proving the function is decreasing, while Theorem 2.2 characterised increasingness). Moreover, note that, for every $x\geq 0$ , $\overline{T}_{Y,1}(x)\geq e^{-x}$ , so it is enough to consider $a<1$ . Hence the sign pattern of the coefficients means that, according to Theorem 2.5, H has at most two real roots. Moreover, we have that $\lim_{x\rightarrow+\infty}H(x)=0^-$ , $H(0)=0$ , and $H^\prime(0)=a>0$ , so it follows that the second root does exist and is positive; consequently the sign variation of H(x) is ‘ $+,-$ ’. Therefore, we have proved that Y is ${1-}{\rm IFRA}$ .

To prove the second statement, we verify that

\begin{equation*}t_2(x)=\frac{-\log(\overline{T}_{Y,2}(x))}{x}\end{equation*}

is not monotone. Indeed, we have

\begin{equation*}\lim_{x\rightarrow0}t_2(x)=\frac{1}{1+\frac{1}{\lambda}-\frac{1}{1+\lambda}}<1,\qquad\mbox{and}\qquad\lim_{x\rightarrow+\infty}t_2(x)=1.\end{equation*}

We verify now that the equation $t_2(x)=1$ has one positive solution. Rewrite this as

\begin{equation*}t_2(x)=1\quad\Leftrightarrow\quad P(x)=\left(\frac{1}{\lambda+1}-\frac{1}{\lambda}\right)e^{-x}+\frac{e^{-\lambda x}}{\lambda}-\frac{e^{-(1+\lambda)x}}{1+\lambda}=0.\end{equation*}

Again from Theorem 2.5, P has at most two real roots. As $P(0)=0$ , $P^\prime(0)>\frac{1}{\lambda}-\frac{1}{\lambda+1}>0$ , and $\lim_{x\rightarrow+\infty} P(x)=0^-$ , there exists a strictly positive second root.

For the final statement, we want to prove the monotonicity of

\begin{equation*}r_{Y,s}(x)=\frac{\overline{T}_{Y,s-1}(x)}{\widetilde{\mu}_{Y,s-1}\overline{T}_{Y,s}(x)},\end{equation*}

which coincides with the monotonicity of

\begin{equation*}N(x)=\frac{e^{-x}+\frac{e^{-\lambda x}}{\lambda^{s-2}}-\frac{e^{-(\lambda+1) x}}{(\lambda+1)^{s-2}}}{e^{-x}+\frac{e^{-\lambda x}}{\lambda^{s-1}}-\frac{e^{-(\lambda+1) x}}{(\lambda+1)^{s-1}}}.\end{equation*}

We look at the numerator of $N^\prime(x)$ , which after some algebraic manipulation may be written as

\begin{equation*}Q(x)=-\frac{(\lambda-1)^2}{\lambda^{s-1}}e^{-(\lambda+1)x}+\frac{\lambda^2}{(\lambda+1)^{s-1}}e^{-(\lambda+2)x}+\frac{1}{(\lambda^2+\lambda)^{s-1}}e^{-(2\lambda+1)x}.\end{equation*}

Of course, the sign of $N^\prime(x)$ coincides with the sign of Q(x). Notice that if $\lambda>1$ , we have $\lambda+1<\lambda+2<2\lambda+1$ , while the two last terms interchange when $\lambda<1$ . Hence it follows from Theorem 2.5 that Q has at most one real root. Moreover, $\lim_{x\rightarrow+\infty}Q(x)=0^-$ . Therefore, if $Q(0)>0$ the sign variation of Q in $(0,+\infty)$ is ‘ $+,-$ ’, and if $Q(0)<0$ the sign variation is ‘ $-$ ’. We have that

\begin{equation*}Q(0)=\frac{\lambda^{s+1}+1-(\lambda-1)^2(1+\lambda)^{s-1}}{(\lambda^2+\lambda)^{s-1}},\end{equation*}

and this, as a function of s, will eventually become negative, as the numerator has a negative coefficient for $\lambda^s$ , the largest power in that expression.

The following is an immediate consequence of Example 3.1 and Proposition 3.1.

4. Simple failure rate monotonicity properties

This section presents simple properties of IFR or DFR distributions that are not of hereditary nature. We first highlight some improvements on classical moment bounds that may be derived from the iterated failure rate monotonicity.

Proof. A direct application of the Hölder inequality justifies the upper bound in the $s{-}{\rm IFR}$ case. Both the lower bound in the $s{-}{\rm IFR}$ and the upper bound in the $s{-}{\rm DFR}$ case follow by requiring the appropriate sign in the numerator of $r_{X,s}^\prime$ and taking into account (2.4) and (2.5).

We now have a look at the iterated failure rate properties of parallel systems. The lifetime of such a system is expressed mathematically as the maximum of the lifetimes of each one of the components; this has already been used to provide an example illustrating the nonhereditary nature of IFRA monotonicity (see Proposition 3.1). We recall here a well-known property of the monotonicity of parallel systems and derive a few simple consequences.

Proof. Taking into account Lemma 3.1, it is enough to verify that $X_{(n)}$ is ${1-}{\rm IFR}$ . Writing

\[r_{X_{(n)},1}(x)=\frac{nF^{n-1}(x)}{1+F(x)+\cdots+F^{n-1}(x)}\frac{f(x)}{1-F(x)},\]

the conclusion is immediate.

An easy consequence follows if we form the parallel system with components after a few iteration steps.

A related result was proved in Theorem 2.2 in Abouammoh and El-Neweihi [Reference Abouammoh and El-Neweihi1], which we quote here presented in a slightly more general wording.

The original statement by Abouammoh and El-Neweihi [Reference Abouammoh and El-Neweihi1] considers only the case where $s=2$ . The above version follows immediately by the hereditary property of $s{-}{\rm IFR}$ monotonicity. Both results prove the iterated monotonicity of maxima based on distributions constructed after some iteration steps. The statement in Proposition 4.3 has a more straightforward practical interpretation.

5. Nonhereditary property of the $s{-}{\rm IFR}$ ordering

We now have a look at hereditary properties of the $s{-}{\rm IFR}$ ordering. We shall prove that, unlike what happens with $s{-}{\rm IFR}$ monotonicity, the ordering relation is not an hereditary property. For the discussion, we need to recall one more stochastic order relation (see Section 4.B.2 in Shaked and Shanthikumar [Reference Shaked and Shanthikumar40]).

Definition 5.1. Let X and Y be random variables with distribution functions $F_X,F_Y\in\mathcal{F}$ . The random variable X is said to be more DMRL than Y, and we write $X\leq_{{\rm DMRL}}Y$ , if

\begin{equation*}d(x)=\frac{\overline{T}_{Y,2}(\overline{T}_{Y,1}^{-1}(x))}{\overline{T}_{X,2}(\overline{T}_{X,1}^{-1}(x))}\end{equation*}

is decreasing.

The following relation with failure rate order holds.

Nanda et al. [Reference Nanda, Hazra, Al-Mutairi and Ghitany32] mention in their Remark 3.1, without proof, that the DMRL order is equivalent to the ${2-}{\rm IFR}$ order. An immediate consequence of this remark is that if $X\leq_{{1-}{\rm IFR}}Y$ then $X\leq_{s{-}{\rm IFR}}Y$ for any $s\geq 2$ . Indeed, once it is proved that $X\leq_{{1-}{\rm IFR}}Y$ , Theorem 4.B.20 in Shaked and Shanthikumar [Reference Shaked and Shanthikumar40] implies that $X\leq_{{\rm DMRL}}Y$ ; hence, according to the remark of Nanda et al. [Reference Nanda, Hazra, Al-Mutairi and Ghitany32], $X\leq_{{2-}{\rm IFR}}Y$ . If we define now $X^\ast_2$ with tail function $\overline{T}_{X,2}$ , and $Y^\ast_2$ with tail function $\overline{T}_{Y,2}$ , the previous order relation means that $X^\ast_2\leq_{{1-}{\rm IFR}}Y^\ast_2$ . Therefore, iterating once again, and applying Theorem 4.B.20 from Shaked and Shanthikumar [Reference Shaked and Shanthikumar40] and the above-mentioned remark, it follows that $X^\ast_2\leq_{{2-}{\rm IFR}}Y^\ast_2$ , which is just a rewriting of $X\leq_{{3-}{\rm IFR}}Y$ . Repeating the above construction, it would follow that $X\leq_{s{-}{\rm IFR}}Y$ for every $s\geq 1$ . However, the equivalence mentioned in Remark 3.1 of Nanda et al. [Reference Nanda, Hazra, Al-Mutairi and Ghitany32] is, in general, not true. We can prove the stated equivalence only when one of the random variables is exponentially distributed. The construction of a counterexample for the general result requires a very careful choice of distribution functions, as described below in Proposition 5.2.

Proof. Taking into account the comments after Definition 2.4, it is enough to consider the case where Y has mean 1. Then we have that $\overline{T}_{Y,1}(x)=\overline{T}_{Y,2}=e^{-x}$ . Therefore $X\leq_{{\rm DMRL}}Y$ is equivalent to

\begin{equation*}d(x)=\frac{\overline{T}_{Y,2}(\overline{T}_{Y,1}^{-1}(x))}{\overline{T}_{X,2}(\overline{T}_{X,1}^{-1}(x))}=\frac{x}{\overline{T}_{X,2}(\overline{T}_{X,1}^{-1}(x))} \end{equation*}

being decreasing. On the other hand, $X\leq_{{2-}{\rm IFR}}Y$ is equivalent to $c_2(x)=\overline{T}_{Y,2}^{-1}(\overline{T}_{X,2}(x))$ being convex, or, alternatively, to $c_2^\prime$ being increasing. Differentiating,

\begin{equation*}c_2^\prime(x)=\frac{\overline{T}_{X,1}(x)}{\overline{T}_{Y,2}(\overline{T}_{Y,1}^{-1}(\overline{T}_{X,2}(x)))}=\frac{\overline{T}_{X,1}(x)}{\overline{T}_{X,2}(x)}.\end{equation*}

Hence $X\leq_{{2-}{\rm IFR}}Y$ is equivalent to $c_2^\prime(\overline{T}_{X,1}^{-1}(x))=d(x)$ being decreasing, which proves the equivalence.

As an immediate consequence, we have the hereditary property of the $s{-}{\rm IFR}$ order with respect to exponentially distributed random variables. This proves the remark of Nanda et al. [Reference Nanda, Hazra, Al-Mutairi and Ghitany32] for the particular choice of the exponential as the reference distribution.

However, the same hereditary property does not hold when comparing general random variables with respect to the $s{-}{\rm IFR}$ ordering. That is, Remark 3.1 in Nanda et al [Reference Nanda, Hazra, Al-Mutairi and Ghitany32] is, in general, not true, as is proven in the proposition that follows.

Proof. Given $c_1,c_2>0$ , we say that a random variable X has branched Pareto distribution with parameters $c_1,c_2$ , $X\sim{\rm BP}(c_1,c_2)$ , if its survival function is

\begin{equation*}\overline{T}_{X,1}(x)=\frac{c_1^2}{(x+c_1)^2}{\mathbb{I}}_{[0,c_1]}(x)+\frac{(c_1+c_2)^2}{4(x+c_2)^2}{\mathbb{I}}_{(c_1,+\infty)}(x).\end{equation*}

Explicit expressions for the 2-iterated distribution and for the corresponding inverse functions are

\begin{equation*}\begin{array}{l}\displaystyle\overline{T}_{X,2}(x)=\frac{4}{3c_1+c_2}\left(\frac{c_1^2}{x+c_1}+\frac{c_2-c_1}{4}\right){\mathbb{I}}_{[0,c_1]}(x)+\frac{(c_1+c_2)^2}{(3c_1+c_2)(x+c_2)}{\mathbb{I}}_{(c_1,+\infty)}(x), \\ \\\displaystyle\overline{T}_{X,1}^{-1}(x)=\left(\frac{c_1+c_2}{2\sqrt{x}}-c_2\right){\mathbb{I}}_{[0,\frac14]}(x)+ \left(\frac{c_1}{\sqrt{x}}-c_1\right){\mathbb{I}}_{(\frac14,+\infty)}(x), \\ \\\displaystyle\overline{T}_{X,2}^{-1}(x)=\left(\!\frac{(c_1+c_2)^2}{(3c_1+c_2)x}-c_2\!\right){\mathbb{I}}_{[0,\frac{c_1+c_2}{3c_1+c_2}]}(x)+\left(\!\frac{4c_1^2}{(3c_1+c_2)x-(c_2-c_1)}-c_1\!\right){\mathbb{I}}_{(\frac{c_1+c_2}{3c_1+c_2},+\infty)}(x).\end{array}\end{equation*}

Moreover,

\begin{equation*}\overline{T}_{X,2}(\overline{T}_{X,1}^{-1})(x)=\frac{2(c_1+c_2)\sqrt{x}}{3c_1+c_2}{\mathbb{I}}_{[0,\frac{1}{4}]}(x)+\frac{4}{3c_1+c_2}\left( c_1\sqrt{x}+\frac{c_2-c_1}{4}\right){\mathbb{I}}_{(\frac{1}{4},+\infty)}(x),\end{equation*}

and

\begin{equation*}\overline{T}_{X,1}(\overline{T}_{X,2}^{-1})(x)=\frac{(3c_1+c_2)^2}{4(c_1+c_2)^2} x^2{\mathbb{I}}_{[0,\frac{c_1+c_2}{3c_1+c_2}]}(x)+\frac{((3c_1+c_2)x-(c_2-c_1))^2}{16c_1^2}{\mathbb{I}}_{(\frac{c_1+c_2}{3c_1+c_2},+\infty)}(x).\end{equation*}

Choosing suitably the parameters $c_1$ and $c_2$ , we obtain the counterexample. A possible choice is $X\sim{\rm BP}(5,10)$ and $Y\sim{\rm BP}(2,6)$ . For these parameters, we find

\begin{equation*}d(x)=\frac{10}{9}{\mathbb{I}}_{[0,\frac14)}(x)+\frac{25\left( 2\sqrt{x}+1\right)}{12\left( 5\sqrt{x}+\frac{5}{4}\right)}{\mathbb{I}}_{[\frac14,+\infty)}(x),\end{equation*}

which is decreasing,

\begin{equation*}c_2^\prime(x)=\frac{81}{100}{\mathbb{I}}_{[0,\frac35]}(x)+\frac{9x^2}{(5x-1)^2}{\mathbb{I}}_{(\frac35,\frac23]}(x)+\frac{4(3x-1)^2}{(5x-1)^2}{\mathbb{I}}_{(\frac23,1]}(x),\end{equation*}

which is not monotone, and

\begin{equation*}c_1(x)=\overline{T}_{Y,1}^{-1}(\overline{T}_{X,1})(x)=\left(\frac{2(x+5)}{5}-2\right){\mathbb{I}}_{[0,5]}(x)+\left(\frac{8(x+10)}{15}-6\right){\mathbb{I}}_{(5,+\infty)}(x),\end{equation*}

which is convex.

6. A criterion for $s{-}{\rm IFR}$ ordering and a first application

We have recalled (see Theorem 2.3) the criterion introduced by Arab and Oliveira [Reference Arab and Oliveira3] to establish the $s{-}{\rm IFR}$ order between two different random variables, and we have mentioned, in Theorem 2.4, the straightforward extension to establish the $s{-}{\rm IFRA}$ order. The criterion introduced in Theorem 2.3 was used in Arab and Oliveira [Reference Arab and Oliveira3] to establish the iterated order within the families of gamma or Weibull distributions. The proofs given in Arab and Oliveira [Reference Arab and Oliveira3] required a careful analysis of the sign variation of

\begin{equation*}P_s(x)=\log f_Y(x)-\log f_X(ax+b)+\log\frac{\mathbb{E} X^{s-1}}{a^{s}\mathbb{E} Y^{s-1}}\end{equation*}

(or of $P_{s-1}$ , defined in Theorem 2.3). A close look at those proofs shows that the difficult cases to handle always correspond to $b<0$ , where a correct positioning of the roots is needed. So it would be quite useful if we could reduce the need to verify the criterion of Theorem 2.3 to only the case $a>0$ and $b\geq 0$ . We can obtain such a simplification with the help of the $s{-}{\rm IFRA}$ ordering.

Proof. To prove that $X\leq_{s{-}{\rm IFR}}Y$ , we need to verify that $c_s(x)=\overline{T}_{Y,s}^{-1}(\overline{T}_{X,s}(x))$ is convex, or, equivalently, that $\overline{T}_{X,s}^{-1}(\overline{T}_{Y,s}(x))$ is concave. Taking into account Theorem 20 in Arab and Oliveira [Reference Arab and Oliveira3], this is equivalent to verifying that $V(x)=\overline{T}_{X,s}^{-1}(\overline{T}_{Y,s}(x))-(ax+b)$ has, for every pair of real numbers a and b, at most the sign variation ‘ $-,+,-$ ’. The assumption $X\leq_{s{-}{\rm IFRA}}Y$ means that $\frac{c_s(x)}{x}$ is increasing, or, equivalently, that

\begin{equation*}\frac{\overline{T}_{X,s}^{-1}(\overline{T}_{Y,s}(x))}{x}\end{equation*}

is decreasing. For $x>0$ , the sign variation of V(x) is the same as the sign variation of

\begin{equation*}\frac{V(x)}{x}=\left(\frac{\overline{T}_{X,s}^{-1}(\overline{T}_{Y,s}(x))}{x}-a\right)-\frac{b}{x}.\end{equation*}

The expression in the parentheses is decreasing, and for $b<0$ , $\frac{b}{x}$ is increasing. Therefore $\frac{V(x)}{x}$ has at most one root, so the proof is concluded.

We may now prove a comparison result ordering two distributions, one from the Weibull family and the other from the gamma family.

Proof. Choose X with ${\rm Weibull}(\alpha,1)$ distribution with density $f_X(x)=\alpha x^{\alpha-1}e^{-x^\alpha}$ , and Y with $\Gamma(\alpha,1)$ distribution with density $f_Y(x)=\frac{1}{\Gamma(\alpha)}x^{\alpha-1}e^{-x}$ . We are taking $\theta_1=\theta_2=1$ , as these are scale parameters, so their value does not affect the order relation between the random variables. We want to prove that $X\leq_{s{-}{\rm IFR}}Y$ . Put $V_s(x)=\overline{T}_{Y,s}(x)-\overline{T}_{X,s}(ax+b)$ . We will now analyse the sign variation of $V_s$ for $x\geq 0$ .

1. Step 1. The $s{-}{\rm IFRA}$ ordering. In the definition of $V_s$ , take $b=0$ . Then we have

   \begin{equation*}P_s(x)=-(\alpha-1)\log(a)-x+a^\alpha x^\alpha-\log(\alpha\Gamma(\alpha))+\log\frac{\mathbb{E} X^{s-1}}{a^s\mathbb{E} Y^{s-1}},\end{equation*}
   implying that $\lim_{x\rightarrow+\infty}P_s(x)=+\infty$ , and $P_s^\prime(x)=-1+a^\alpha\alpha x^{\alpha-1}$ , so the sign variation of $P_s^\prime$ is ‘ $-,+$ ’, and the monotonicity of $P_s$ is ‘ $\searrow\nearrow$ ’. If $P_s(0)<0$ , the sign variation of $P_s$ is ‘ $-,+$ ’, so, using Lemma 2.2, the sign variation of $V_s$ is at most ‘ $-,+$ ’. If $P_s(0)\geq 0$ , the sign variation of $P_s$ may be ‘ $+,-,+$ ’. The function $V_s$ is obtained by integrating the function $H_s$ given in Theorem 2.3, so, again based on Lemma 2.2, and taking into account that $V_s(0)=0$ , the sign variation of $V_s$ is at most ‘ $-,+$ ’. Therefore, we have proved that $X\leq_{s{-}{\rm IFRA}}Y$ .

2. Step 2. The $s{-}{\rm IFR}$ ordering. We consider now $V_s$ with $a>0$ and $b>0$ . Then we have

   \begin{equation*}P_s(x)=(\alpha-1)\left(\log x-\log(ax+b)\right)-x+(ax+b)^\alpha-\log(\alpha\Gamma(\alpha))+\log\frac{\mathbb{E} X^{s-1}}{a^s\mathbb{E} Y^{s-1}}.\end{equation*}

It is obvious that $\lim_{x\rightarrow+\infty}P_s(x)=+\infty$ . Differentiating, we have that

\begin{equation*}P_s^\prime(x)=\frac{\alpha-1}{x}-\frac{a(\alpha-1)}{ax+b}-1+a\alpha(ax+b)^{\alpha-1}=\frac{N_s(x)}{x(ax+b)},\end{equation*}

where $N_s(x)=a\alpha x(ax+b)^\alpha-ax^2-bx+b(\alpha-1)$ . Hence, as we will be considering x such that the denominator is positive, the sign of $P_s^\prime$ is determined by the sign of $N_s$ . Differentiating $N_s$ , we obtain $N_s^{\prime\prime\prime}(x)=a^3\alpha^2(\alpha-1)(ax+b)^{\alpha-3}\bigl(a(\alpha+1)x+3b\bigr)$ . Therefore, ${\rm sgn}(N_s^{\prime\prime\prime})={\rm sgn}\bigl(a(\alpha+1)x+3b\bigr)$ . As $a(\alpha+1)x+3b\geq 0$ , it follows that $N_s^{\prime\prime\prime}(x)\geq 0$ ; hence $N_s^{\prime\prime}$ is increasing. We have that $\lim_{x\rightarrow+\infty}N_s^{\prime\prime}(x)=+\infty$ , and $N_s^{\prime\prime}(0)=2a(a\alpha^2b^{\alpha-1}-1)$ ; this last quantity may be either positive or negative. Looking now at $N_s^\prime$ , we have $N_s^\prime(+\infty)=+\infty$ , and $N_s^\prime(0)=b(a\alpha b^{\alpha-1}-1)$ , which may be either positive or negative, irrespective of the sign of $N_s^{\prime\prime}$ at the origin. Finally, we have $N_s(0)=b(\alpha-1)>0$ and $\lim_{x\rightarrow+\infty}N_s(x)=+\infty$ . The table below summarises the maximum sign variation possibilities, taking into account the behaviour just described.

As ${\rm sgn}(P_s^\prime)={\rm sgn}(N_s)$ , it follows that the possible monotonicities for $P_s$ are ‘ $\nearrow$ ’ or ‘ $\nearrow\searrow\nearrow$ ’. Going back to the expression for $P_s$ , we verify that $\lim_{x\rightarrow0^+}P_s(x)=-\infty$ and $\lim_{x\rightarrow+\infty}P_s(x)=+\infty$ ; therefore the possible sign variations of $P_s$ are ‘ $-,+$ ’ or ‘ $-,+,-,+$ ’. Based again on Lemma 2.2, for the first case it follows that the possible sign variations for $V_s$ are ‘ $-,+$ ’ or ‘ $+$ ’, while in the second case, the possible sign variations for $V_s$ are ‘ $-,+,-,+$ ’, ‘ $+,-,+$ ’, ‘ $-,+$ ’, or ‘ $+$ ’. Taking into account that $V_s(0)=1-\overline{T}_{X,s}(b)\geq0$ , actually only the sign variations starting at positive values are possible; that is, the possibilities are ‘ $+$ ’ or ‘ $+,-,+$ ’, so the proof is concluded.

Using the criterion proved in Theorem 6.1, we may complete the comparison within the gamma or the Weibull families of distributions which was partially given in Propositions 30–33 in Arab and Oliveira [Reference Arab and Oliveira3]. We state here the complete result.

Proof. Given Propositions 30–33 in Arab and Oliveira [Reference Arab and Oliveira3], we only need to consider the case where $1>\alpha^\prime>\alpha>0$ . The result follows from repeating the steps for the proof of Proposition 6.1, with the arguments used in Propositions 30 and 32 in Arab and Oliveira [Reference Arab and Oliveira3].

7. Failure rate ordering of exponentially distributed parallel systems

We now apply our results to prove extended ordering relations among parallel systems with components that have exponentially distributed lifetimes. We will be extending Theorem 3.1 by Kochar and Xu [Reference Kochar and Xu26], where these authors prove that a parallel system where the components have the same exponential distribution ages faster than a same-sized system where the components have exponential lifetimes with different mean values. We will be using the criterion introduced in Theorem 6.1 to extend, for general $s\geq 1$ , this ageing characterisation of parallel systems proved by Kochar and Xu [Reference Kochar and Xu26].

Throughout this section we take

(7.1) \begin{align}\begin{array}{ll@{\qquad}l}X=\max(X_1,X_2),& \; & \text{where}\, X_1\, \text{and}\, X_2\, \text{are\, independent\, mean}\, 1\, \\ & & \text{exponentially\, distributed};\\Y=\max(Y_1,Y_2),& &{\text{where}\, Y_1\, \text{is\, mean\, 1\, exponentially\, distributed},\, Y_2\, \text{is}\,}\\ & & {\text{mean}\, 1/}\lambda < {1\, \text{exponentially\, distributed,\, and}\, Y_1,\, Y_2\,}\\ &&\text{are\, independent.}\end{array}\end{align}

The choice made for the mean values of the components’ lifetimes is not really essential, but makes our proofs easier to explain. The only important fact is that $X_1$ and $X_2$ have the same mean. Indeed, taking into account the comments after Definition 2.4, we may always renormalise the variables to reduce to the present case. This section studies the $s-$ iterated failure rate order between X and Y. The main tool for the analysis is the result about roots of polynomials of exponentials recalled in Theorem 2.5.

As already mentioned in course of the proof of Proposition 3.1, it is easily verified that

\begin{equation*}\overline{T}_{Y,s}(x)=\frac{1}{c(s,\lambda)}\left(e^{-x}+\frac{e^{-\lambda x}}{\lambda^{s-1}}-\frac{e^{-(\lambda+1) x}}{(\lambda+1)^{s-1}}\right),\end{equation*}

where

\begin{equation*}c(s,\lambda)=1+\frac{1}{\lambda^{s-1}}-\frac{1}{(\lambda+1)^{s-1}}.\end{equation*}

The tail of the distribution of X is obtained by replacing $\lambda$ by 1 in these expressions.

For sake of readability, we will present the various partial results leading to the comparison of X and Y in $s{-}{\rm IFR}$ order in a series of propositions.

Proof. Define

\begin{align*}U_s(x)&=\overline{T}_{X,s}(x)-\overline{T}_{Y,s}(x) \\[3pt] & = \frac{2^s e^{-x}-e^{-2x}}{2^s-1}-\frac{1}{c(s,\lambda)}\left(e^{-x}+\frac{e^{-\lambda x}}{\lambda^{s-1}}-\frac{e^{-(\lambda+1) x}}{(\lambda+1)^{s-1}}\right) \\[3pt] &=\left(\frac{2^s}{2^s-1}-\frac{1}{c(s,\lambda)}\right)e^{-x}-\frac{e^{-2x}}{2^s-1} -\frac{e^{-\lambda x}}{c(s,\lambda)\lambda^{s-1}} +\frac{e^{-(\lambda+1)x}}{c(s,\lambda)(\lambda+1)^{s-1}}.\end{align*}

We are considering $\lambda>1$ , so the signs of the coefficients of $U_s$ , after ordering decreasingly with respect to the exponents, are ‘ $+,-,-,+$ ’ (the signs of the coefficients of $e^{-\lambda x}$ and $e^{-2x}$ are the same, so we do not need to consider two cases). So, taking into account Theorem 2.5, $U_s$ has at most two real roots. One root is easily located, as $U_s(0)=0$ . Moreover, notice that $\lim_{x\rightarrow-\infty}U_s(x)=+\infty$ , the limit being governed by the sign of the coefficient of $e^{-x}$ , while $\lim_{x\rightarrow+\infty}U_s(x)=0^+$ , this limit being governed by the sign of the coefficient of the exponential with the smallest exponent. In order to locate the remaining root, we need to differentiate: for $k<s$ , we have

\begin{equation*}U_s^{(k)}(x)=(-1)^k\left[\frac{2^s e^{-x}-2^ke^{-2x}}{2^s-1}-\frac{1}{c(s,\lambda)}\left(e^{-x}+\frac{e^{-\lambda x}}{\lambda^{s-1-k}}-\frac{e^{-(\lambda+1) x}}{(\lambda+1)^{s-1-k}}\right)\right],\end{equation*}

and

\begin{equation*}U_s^{(s)}(x)=(-1)^s\left[\frac{2^s e^{-x}-2^se^{-2x}}{2^s-1}-\frac{1}{c(s,\lambda)}\left(e^{-x}+\lambda e^{-\lambda x}-(\lambda+1)e^{-(\lambda+1) x}\right)\right].\end{equation*}

Hence the signs of the coefficients of the exponentials alternate with each differentiation, and $U_s^{(s)}(0)=0$ . It is now convenient to separate into two cases.

1. s even: The signs of the coefficients in $U_s^{(s)}$ are ‘ $+,-,-,+$ ’, implying that $\lim_{x\rightarrow-\infty}U_s^{(s)}(x)=+\infty$ , $\lim_{x\rightarrow+\infty}U_s^{(s)}(x)=0^+$ , and $U_s^{(s)}$ has at most two real roots. As $U_s^{(s)}(0)=0$ , depending on the location of the second root, the sign variation in $(0,+\infty)$ of $U_s^{(s)}$ may be either ‘ $+$ ’ or ‘ $-,+$ ’. The sign of $U_s^{(s-1)}(0)$ is not determined, and the signs of the limits at $\pm\infty$ are reversed with respect to $U_s^{(s)}$ , so we need to consider the two possibilities leading to the following possible situations:

1. Therefore there are only two possible sign variations for $U_s^{(s-1)}$ when $x\in(0,+\infty)$ : ‘ $-$ ’ or ‘ $+,-$ ’. We may now proceed to characterise the possible sign variations in $(0,+\infty)$ of $U_s^{(s-2)}$ , repeating the above arguments. Again, notice that it is not possible to determine the sign of $U_s^{(s-2)}(0)$ , so the possibilities are as follows:

1. We find for the sign variation in $(0,+\infty)$ of $U_s^{(s-2)}$ exactly the same behaviour as for $U_s^{(s)}$ , so we may repeat the arguments above to find that $U_s^\prime$ has the same sign variation in $(0,+\infty)$ as $U_s^{(s-1)}$ ; that is, it is either ‘ $-$ ’ or ‘ $+,-$ ’. Going back to $U_s$ , recall that $\lim_{x\rightarrow-\infty}U_s(x)=+\infty$ , $\lim_{x\rightarrow+\infty}U_s(x)=0^+$ , and $U_s(0)=0$ . This behaviour does not allow for the case of $U_s^\prime$ being always negative, so the sign variation of $U_s^\prime$ is ‘ $+,-$ ’, which implies that $U_s(x)\geq 0$ for every $x\geq 0$ .

2. s odd: For this case, we have that the signs of the coefficients in $U_s^{(s)}$ are ‘ $-,+,+,-$ ’, $\lim_{x\rightarrow-\infty}U_s^{(s)}(x)=-\infty$ , $\lim_{x\rightarrow+\infty}U_s^{(s)}(x)=0^-$ , $U_s^{(s)}(0)=0$ , and $U_s^{(s)}$ has at most two real roots. Therefore the only possible sign variation in $(0,+\infty)$ for $U_s^{(s)}$ is ‘ $+,-$ ’, which is what we found for the $s-1$ derivative in the previous case. Hence, repeating the arguments, we find the same conclusion; that is, again, $U_s(x)\geq 0$ for every $x\geq 0$ .

Corollary 7.1. Let X and Y be defined as in (7.1). Then

\begin{equation*}\frac{\overline{T}_{Y,s}^{-1}\overline{T}_{X,s}(x)}{x}\leq 1\end{equation*}

for every $x>0$ .

Proof. We have just proved that $\overline{T}_{X,s}(x)\geq\overline{T}_{Y,s}(x)$ , which, as $\overline{T}_{Y,s}$ is a decreasing function, implies that $\overline{T}_{Y,s}^{-1}\overline{T}_{X,s}(x)\leq x$ ; hence the result is proved.

Proof. We need to prove that

\begin{equation*}t_s(x)=\frac{\overline{T}_{Y,s}^{-1}(\overline{T}_{X,s}(x))}{x} \end{equation*}

is increasing for $x\geq 0$ , or, equivalently, that the sign variation in $(0,+\infty)$ of $t_s(x)-a$ is at most ‘ $-,+$ ’. The previous corollary means that we need only to consider $0<a\leq 1$ . This is still equivalent to proving that, for the described choice of a, $\overline{T}_{X,s}(x)-\overline{T}_{Y,s}(ax)$ behaves, at most, as ‘ $+,-$ ’. Reversing this expression, this is equivalent to proving that $V_s(x)=\overline{T}_{Y,s}(x)-\overline{T}_{X,s}(ax)$ behaves, at most, as ‘ $-,+$ ’, now for $a>1$ . This is the same formulation as in Theorem 2.2 with a reduced scope for the choice of a. To write the expression explicitly, we have

\begin{equation*}V_s(x)=\frac{1}{c(s,\lambda)}\left(e^{-x}+\frac{e^{-\lambda x}}{\lambda^{s-1}}-\frac{e^{-(\lambda+1) x}}{(\lambda+1)^{s-1}}\right)-\frac{2^s e^{-ax}-e^{-2ax}}{2^s-1}.\end{equation*}

We will be using Theorem 2.5 to identify the maximum number the roots of $V_s$ , then proceed in a similar way as before to locate them and infer the sign variation of the function. As in the proof of Proposition 7.1, we start by differentiating to obtain

\begin{equation*}V_s^{(s)}(x)=(-1)^s\left[\frac{1}{c(s,\lambda)}\left(e^{-x}+\lambda e^{-\lambda x}-(\lambda+1)e^{-(\lambda+1) x}\right)-2^sa^s\frac{e^{-ax}-e^{-2ax}}{2^s-1}\right],\end{equation*}

so we have $V_s^{(s)}(0)=0$ . To apply Theorem 2.5, we need to order decreasingly with respect to the exponents the exponential terms in $V_s$ , which means we need to separate into several cases, depending on the location of a with respect to $\lambda$ , and verify in each case that the sign variation of $V_s$ is at most ‘ $-,+$ ’. Recall that both these parameters are larger than or equal to 1.

1. Case 1: $1<a<2a<\lambda<\lambda+1$ . As previously, we need to treat separately the cases where s is even and where s is odd.

   1. s even: The sign pattern of the coefficients in $V_s$ and in $V_s^{(s)}$ is now ‘ $+,-,+,+,-$ ’, indicating that each function has at most three real roots. Moreover, this sign pattern implies that $\lim_{x\rightarrow-\infty}V_s^{(s)}(x)=-\infty$ , $\lim_{x\rightarrow+\infty}V_s^{(s)}(x)=0^+$ , and $V_s^{(s)}(0)=0$ . Therefore the maximum possible sign variation for $V_s^{(s)}$ on $(0,+\infty)$ is ‘ $+,-,+$ ’. As before, the sign of $V_s^{(s-1)}(0)$ is not determined, so we need to analyse each possibility. Recall that the coefficient signs and the signs at each limit when $x\rightarrow\pm\infty$ are reversed with each differentiation, meaning that $\lim_{x\rightarrow-\infty}V_s^{(s-1)}(x)=+\infty$ and $\lim_{x\rightarrow+\infty}V_s^{(s-1)}(x)=0^-$ . Taking this into account, the possibilities are as follows:

1. To proceed to the analysis of $V_s^{(s-2)}$ , notice first that the sign of this function at the origin is not determined, and that $\lim_{x\rightarrow-\infty}V_s^{(s-2)}(x)=-\infty$ , $\lim_{x\rightarrow+\infty}V_s^{(s-2)}(x)=0^+$ . Therefore the possible sign variations are as follows:

1. This means that the maximum possible sign variation for $V_s^{(s-2)}$ is the same as for $V_s^{(s)}$ ; hence we may recurse on the argument to arrive at the conclusion that the maximum possible sign variation for $V_s^\prime$ on $(0,+\infty)$ is ‘ $-,+,-$ ’. Taking into account that $V_s(0)=0$ , $\lim_{x\rightarrow-\infty}V_s(x)=-\infty$ , $\lim_{x\rightarrow+\infty}V_s(x)=0^+$ , and $V_s$ has at most three real roots, its sign variation in $(0,+\infty)$ can be at most ‘ $-,+$ ’.

2. s odd: The sign pattern of the coefficients in $V_s^{(s)}$ is now ‘ $-,+,-,-,+$ ’, and $\lim_{x\rightarrow-\infty}V_s^{(s)}(x)=+\infty$ , $\lim_{x\rightarrow+\infty}V_s^{(s)}(x)=0^-$ , and $V_s^{(s)}(0)=0$ . This means that the maximum possible sign variation for $V_s^{(s)}$ on $(0,+\infty)$ is now ‘ $-,+,-$ ’, which corresponds to the behaviour of the $s-1$ derivative in the previous case, so the same conclusion still holds.

   Therefore, for this case we have verified that the sign variation in $(0,+\infty)$ of $V_s$ is at most ‘ $-,+$ ’.

1. Case 2: $1<a<\lambda<2a<\lambda+1$ . The sign pattern of the coefficients coincides with the one observed in the previous case, so the result also holds in this case.

2. Case 3: $1<a<\lambda<\lambda+1<2a$ . The sign pattern of the coefficients of the function $V_s$ is now ‘ $+,-,+,-,+$ ’; hence there could exist up to four real roots. Of course, we still have $V_s(0)=0$ , so we need to locate the remaining roots. Because of the number of possible roots, a direct application of the arguments in the previous cases with $V_s$ does not allow us to draw conclusions about a sign variation compatible with $s{-}{\rm IFRA}$ order. Note that in this case we have $a>\frac{\lambda+1}{2}$ , so for every fixed $x\geq 0$ , $\overline{T}_{X,s}(ax)<\overline{T}_{X,s}(\frac{\lambda+1}{2}x)$ ; therefore $V_s(x)=\overline{T}_{Y,s}(x)-\overline{T}_{X,s}(ax)>V_{\ast,s}(x)=\overline{T}_{Y,s}(x)-\overline{T}_{X,s}(\frac{\lambda+1}{2}x)$ . We shall prove that $V_{\ast,s}(x)\geq 0$ for $x\geq0$ , so the same holds for $V_s$ . Rewriting $V_{\ast,s}$ , with the exponentials already ordered decreasingly with respect to their exponents, we have

   \begin{align*}V_{\ast,s}(x)&=\frac{1}{c(s,\lambda)}e^{-x} -\frac{2^s}{2^s-1}e^{-\frac{(\lambda+1)}{2}x} +\frac{1}{c(s,\lambda)\lambda^{s-1}}e^{-\lambda x}\\ &\quad +\left(\frac{1}{2^s-1}-\frac{1}{c(s,\lambda)(\lambda+1)^{s-1}}\right)e^{-(\lambda+1)x}.\end{align*}
   The coefficient of the last exponential is easily seen to be positive, so the sign pattern of the coefficients in $V_{\ast,s}$ is ‘ $+,-,+,+$ ’; hence, besides having $V_{\ast,s}(0)=0$ , we have $\lim_{x\rightarrow-\infty}V_{\ast,s}(x)=+\infty$ and $\lim_{x\rightarrow+\infty}V_{\ast,s}(x)=0^+$ . Moreover, taking into account Theorem 2.5, $V_{\ast,s}$ has at most two real roots. To complete the study of the sign variation we need, as before, to separate the cases based on the value of s.

   1. s even. Repeating the arguments above, the sign pattern for the coefficients of $V_{\ast,s}^{(s)}$ is the same as for $V_{\ast,s}$ . Therefore we may repeat the arguments used in the course of the proof of Proposition 7.1 for the case where s is even, to derive that $V_{\ast,s}(x)\geq 0$ ; hence $V_s(x)\geq 0$ for every $x\geq 0$ .

   2. s odd. As in the proof of Proposition 7.1, this corresponds to the behaviour of the $s-1$ derivative when s is even, so the result holds in this case.

3. Case 4: $1<\lambda<a<2a<\lambda+1$ . The sign pattern of the coefficients, after ordering the exponentials, is ‘ $+,+,-,+,-$ ’, meaning that there are at most three real roots. This is exactly the same sign pattern we found in Case 1 above. So, repeating the arguments from Case 1, the same sign variation for $V_s$ follows.

4. Case 5: $1<\lambda<a<\lambda+1<2a$ . This is the simplest case to analyse. The sign pattern for the coefficients of $V_s$ is ‘ $+,+,-,-,+$ ’, implying that there are at most two real roots, $\lim_{x\rightarrow-\infty}V_s(x)=+\infty$ , and $\lim_{x\rightarrow+\infty}V_s(x)=0^+$ . This is easily seen to be compatible with two possible sign variations in $(0,+\infty)$ : ‘ $-,+$ ’ or ‘ $+$ ’.

5. Case 6: $1<\lambda<\lambda+1<a<2a$ . This case produces the same sign pattern for the coefficients as Case 5, so the same conclusion about the sign variation of $V_s$ follows.

Thus we have verified that in all possible cases, the sign variation of $V_s$ is at most ‘ $-,+$ ’; hence, for $x\geq 0$ , $t_s(x)$ is increasing. The proposition is proved.

The previous result establishes the $s{-}{\rm IFRA}$ order, so we may now proceed to the proof of the $s{-}{\rm IFR}$ relation between these two random variables.

Proof. The plan for the proof is the same as for Proposition 7.2. The difference here is that we will be interested in proving the convexity of the relevant functions and we will not be able to automatically locate one of their roots. Taking into account Theorem 2.2, Remark 2.2, and Theorem 6.1, it is sufficient to verify that $V_s(x)=\overline{T}_{Y,s}(x)-\overline{T}_{X,s}(ax+b)$ changes sign at most twice, in the order ‘ $+,-,+$ ’, for every $a>0$ and $b\geq 0$ . The case $b=0$ was treated in Proposition 7.2, so we may assume in the sequel that $b>0$ . Although the function is similar to the one considered in Proposition 7.2, one should notice that now $V_s(0)=1-\overline{T}_{X,s}(b)>0$ . Of course, the sign patterns of the coefficients are similar, but we must take into account the extra terms $e^{-b}$ and $e^{-2b}$ .

We start by writing out explicitly the expressions for $V_s$ and its s-order derivative $V_s^{(s)}$ :

\begin{equation*}\begin{array}{l}\displaystyle V_s(x)=\frac{1}{c(s,\lambda)}\left(e^{-x}+\frac{e^{-\lambda x}}{\lambda^{s-1}}-\frac{e^{-(\lambda+1) x}}{(\lambda+1)^{s-1}}\right)-\frac{2^s e^{-(ax+b)}-e^{-2(ax+b)}}{2^s-1}, \\ \\\displaystyle V_s^{(s)}(x)=(-1)^s\left[\frac{1}{c(s,\lambda)}\left(e^{-x}+\lambda e^{-\lambda x}-(\lambda+1)e^{-(\lambda+1) x}\right)-2^sa^s\frac{e^{-(ax+b)}-e^{-2(ax+b)}}{2^s-1}\right].\end{array}\end{equation*}

Note that

\begin{equation*}V_s^{(s)}(0)=\frac{(-1)^{s+1}2^s a^se^{-b}(1-e^{-b})}{2^s-1},\end{equation*}

which has the same sign as $(-1)^{s+1}$ when $b>0$ .

1. Case 1: $1<a<2a<\lambda<\lambda+1$ . The sign pattern of the coefficients, after ordering the exponentials in decreasing order of their exponents, is ‘ $+,-,+,+,-$ ’, so $V_s$ has at most three real roots. Moreover, $\lim_{x\rightarrow-\infty}V_s(x)=-\infty$ and $\lim_{x\rightarrow+\infty}V_s(x)=0^+$ , so, recalling that $V_s(0)>0$ , it follows that the sign variation of $V_s$ on $(0,+\infty)$ is either ‘ $+$ ’ or ‘ $+,-,+$ ’.

2. Case 2: $1<a<\lambda<2a<\lambda+1$ . This case is treated exactly like the previous one.

3. Case 3: $1<a<\lambda<\lambda+1<2a$ . As before, this case requires a more careful analysis, as the number of possible roots is larger. The sign pattern of the coefficients of the function $V_s$ is now ‘ $+,-,+,-,+$ ’, so we may have up to four real roots for $V_s$ , and $\lim_{x\rightarrow-\infty}V_s(x)=+\infty$ , $\lim_{x\rightarrow+\infty}V_s(x)=0^+$ . We again separate according to s being even or odd.

   1. s even. In this case, we have $V_s^{(s)}(0)<0$ , so the sign variation in $(0,+\infty)$ of $V_s^{(s)}$ is either ‘ $-,+,-,+$ ’ or ‘ $-,+$ ’. Now, taking into account that each differentiation step reverses all signs, and that, with the exception of the s-order derivative, the signs of the derivatives at the origin are not determined, we have the following possibilities for the sign variations:

1. Hence the maximum possible sign variation for $V_s^{(s-2)}$ on $(0,+\infty)$ is the same as for $V_s^{(s)}$ , so we repeat the argument to obtain that the maximum possible sign variation for $V_s^\prime$ on $(0,+\infty)$ is ‘ $+,-,+,-$ ’. Therefore the monotonicity of $V_s$ is ‘ $\nearrow\searrow\nearrow\searrow$ ’, which, recalling that $V_s(0)>0$ , implies that the sign variation of $V_s$ may be ‘ $+$ ’ or ‘ $+,-,+$ ’.

2. s odd. Now we have $V_s^{(s)}(0)>0$ , which, taking into account the signs for $V_s^{(s)}$ at $\pm\infty$ , implies a sign variation, on $(0,+\infty)$ , of ‘ $+,-$ ’ or ‘ $+,-,+,-$ ’. That is, we find the same behaviour as for the $s-1$ derivative in the previous case, so the same conclusion about the sign variation of $V_s$ follows.

1. Case 4: $1<\lambda<a<2a<\lambda+1$ . This case exhibits the same behaviour as Case 1 above, so the same conclusion holds.

2. Case 5: $1<\lambda<a<\lambda+1<2a$ . The sign pattern of the coefficients of $V_s$ , after ordering the exponentials in the usual way, is ‘ $+,+,-,-,+$ ’, implying that there are at most two real roots, $\lim_{x\rightarrow-\infty}V_s(x)=+\infty$ , and $\lim_{x\rightarrow+\infty}V_s(x)=0^+$ . As $V_s(0)>0$ , the only possibilities for the sign variation for $V_s$ on $(0,+\infty)$ are ‘ $+$ ’ and ‘ $+,-,+$ ’.

3. Case 6: $1<\lambda<\lambda+1<a<2a$ . This case exhibits the same behaviour as the previous one.

4. Case 7: $0<a<1$ . In this case, regardless of the actual value of a, the sign pattern of the coefficients is ‘ $-,+,+,+,-$ ’, so $\lim_{x\rightarrow-\infty}V_s(x)=-\infty$ and $\lim_{x\rightarrow+\infty}V_s(x)=0^-$ . As $V_s(0)>0$ , the only possible sign variation on $(0,+\infty)$ is ‘ $+,-$ ’.

We have verified that, for all relevant choices of the parameters a and b, the assumptions of Theorem 2.2 are satisfied, so it follows that $X\leq_{s{-}{\rm IFR}}Y$ .

In the result that follows, we compare the ageing properties of a parallel system with n components with independent and identically distributed exponential lifetimes, to those of a parallel system with k components with independent and identically distributed exponential lifetimes, assuming that $k<n$ .

Proof. As the parameters $\lambda$ and $\beta$ are scale parameters, we may take $\lambda=\beta=1$ . The random variables $X_m$ and $Y_k$ have the following tail distributions: $\overline{F}_m(x)=1-(1-e^{-x})^m$ , $\overline{F}_k(x)=1-(1-e^{-x})^k$ , respectively. The proposition follows from proving that $c_1(x)=\overline{F}^{-1}_k(\overline{F}_m)=-\log(1-(1-e^{-x})^{\frac{m}{k}})$ is convex. As $c_1(x)$ is differentiable, the convexity is characterised by the nonnegativeness of the second derivative. Computing derivatives, and taking into account that the sign of $c_1^{\prime\prime}$ is determined by the sign of its numerator, it can be seen that the sign of $c_1^{\prime\prime}$ is the same as the sign of

\begin{equation*}Q(x)=\frac{m}{k}e^{-x}+(1-e^{-x})^{\frac{m}{k}}-1.\end{equation*}

This is positive for every $x\geq 0$ and $2\leq k<m$ , so the conclusion follows.

As an immediate consequence, we have the following ordering for the order statistics.