# Stochastic ordering of sums of exponential random variables. A geometric approach

François Castella, Bruno Sericola

► To cite this version:

François Castella, Bruno Sericola. Stochastic ordering of sums of exponential random variables. A geometric approach. Univ Rennes, Inria, CNRS, IRISA. 2025, pp.1-38. hal-05125505

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https://inria.hal.science/hal-05125505v1

Submitted on 23 Jun 2025

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# Stochastic ordering of sums of exponential random variables.
## A geometric approach

François Castella *

Bruno Sericola †

June 23, 2025

## Abstract

The stochastic ordering is useful to compare random variables in terms of their probability distribution: if $X$ and $Y$ are two real random variables, we say $X$ is less than $Y$ in the usual stochastic ordering whenever $\mathbb{P}(X > x) \le \mathbb{P}(Y > x)$ for all $x \in \mathbb{R}$. We then write $X \le_{\text{st}} Y$. For instance if $X$ and $Y$ are two durations then $X \le_{\text{st}} Y$ means that for all values of $x$, the duration $X$ has less chances to exceed $x$ than duration $Y$. In this paper, we consider the case of sums of independent and exponentially distributed random variables: if $X_i$, resp. $Y_i$ ($i = 1, \dots, n$), are independent random variables that are exponentially distributed with rates $\lambda_i$, resp. $\mu_i$, we examine under which conditions on the $\lambda_i$'s and $\mu_i$'s one can ensure $\sum_{i=1}^n X_i \le_{\text{st}} \sum_{i=1}^n Y_i$. Adopting a geometric point of view in the space of the parameters $\lambda_i$ and $\mu_i$, we come up with close to optimal conditions, improving previous results on this question, as well as shedding a geometric light on known results.

**Keywords.** Exponential distribution; Stochastic ordering; Convolutions.

## 1 Introduction

The exponential distribution is a fundamental probability distribution which is used in several application domains. In queueing theory, the exponential distribution arises in the interarrival times when the arrival process is modeled by a Poisson process. It is also the basic tool of the phase-type distributions, see [8] or [11], which can approximate any non negative probability distribution used for instance to model the durations of the service times in queueing systems. Some well-known phase-type distributions are Erlang distributions, hypo-exponential and hyper-exponential distributions. They are also used in fluid queues, see for instance [4], to represent the fact that information flows through a network as a fluid. In such queues, the packet nature of traffic is ignored: a packet stream actually appears as a fluid flow modeled by on-off sources in which the on and off periods are exponentially distributed. More generally, the performance evaluation of computing systems or networks is often analyzed using continuous-time Markov chains which are based on exponential distributions, see for instance [11]. Other examples of domains in which the exponential distribution is used are the reliability and dependability theory, see for instance [10], the gas physics and the hydrology.

A lot of research has been done on the stochastic orderings of such distributions. The stochastic ordering of order statistics has been analyzed in [9], where it is shown that if $X_i$ ($i = 1, \dots, n$) are independent exponential random variables with rates $\lambda_1, \dots, \lambda_n$, and if $Y_i$ ($i = 1, \dots, n$) are i.i.d. exponential random variables with common rate $(\lambda_1 + \dots + \lambda_n)/n$, then there exists a monotone coupling of the order statistics $X_{(1)}, \dots, X_{(n)}$ and $Y_{(1)}, \dots, Y_{(n)}$. This result has been extended in [1] to a more general class of distributions for which the failure rate function $r(x)$ is decreasing while $xr(x)$ is increasing. Other extensions are proposed in [13] and in [7] for Gamma distributions.

<footnote>*François Castella is with Univ. Rennes, Inria, CNRS, IRISA</footnote>

<footnote>†Bruno Sericola is with Inria, Univ. Rennes, CNRS, IRISA</footnote>
In the case of sums of exponentially distributed random variables, which is our main concern in this paper, the following result is obtained in [2], by using Schur properties of convolutions. Let $X_i$, resp. $Y_i$ ($i = 1, \dots, n$), be independent exponential random variables with rates $\lambda_i$, resp. $\mu_i$, and assume without loss of generality that $\lambda_1 \le \lambda_2 \le \dots \le \lambda_n$ and $\mu_1 \le \mu_2 \le \dots \le \mu_n$ (upon re-ordering). The authors of [2] prove the following sufficient condition

$$ \left( \sum_{i=1}^{k} \lambda_i \ge \sum_{i=1}^{k} \mu_i \text{ for any } k = 1, \dots, n, \quad \text{and} \quad \sum_{i=1}^{n} \lambda_i = \sum_{i=1}^{n} \mu_i \right) \quad \implies \quad \sum_{i=1}^{n} X_i \le_{\text{lr}} \sum_{i=1}^{n} Y_i, \quad (1) $$

where $\le_{\text{lr}}$ is the likelihood ratio ordering of random variables. Note that the definitions and properties of the orderings for random variables we use in this paper can be found in [12]. In other words, the authors of [2] exhibit an explicit sufficient criterion on the rates $\lambda_i$'s and $\mu_i$'s to ensure the ordering of the associated sums of independent exponential variables. This result is extended in [6] for Gamma distributed variables, namely variables following a $\Gamma$ distribution sharing a common shape parameter $a \ge 1$, and their result is valid for both the $\le_{\text{lr}}$ and $\le_{\text{disp}}$ stochastic orders, where $\le_{\text{disp}}$ is the dispersive ordering of random variables. Note in passing that, as is well known, if $X$ and $Y$ are two real random variables, then $X \le_{\text{lr}} Y \implies X \le_{\text{disp}} Y$ as well as $X \le_{\text{lr}} Y \implies X \le_{\text{st}} Y$, where $\le_{\text{st}}$ denotes stochastic ordering that is our main concern in this paper. It is defined as $X \le_{\text{st}} Y$ if and only if for any $x \in \mathbb{R}$ we have $\mathbb{P}(X > x) \le \mathbb{P}(Y > x)$. Note finally that if $X$ and $Y$ are non-negative random variables, we have the complete chain $X \le_{\text{lr}} Y \implies X \le_{\text{disp}} Y \implies X \le_{\text{st}} Y$.

In [3], the result obtained in [2] is extended in the following sense. Take $X_i$, resp. $Y_i$ ($i = 1, \dots, n$), independent exponential variables with rates $\lambda_i$, resp. $\mu_i$, with the convention that $\lambda_1 \le \lambda_2 \le \dots \le \lambda_n$ and $\mu_1 \le \mu_2 \le \dots \le \mu_n$ as before. The authors of [3] prove the following refined version of (1), namely

$$ \left( \prod_{i=1}^{k} \lambda_i \ge \prod_{i=1}^{k} \mu_i, \quad \text{for any } k = 1, \dots, n \right) \quad \implies \quad \sum_{i=1}^{n} X_i \le_{\text{st}} \sum_{i=1}^{n} Y_i. \quad (2) $$

Note that the assumption made in [2], namely $\sum_{i=1}^{k} \lambda_i \ge \sum_{i=1}^{k} \mu_i$ for all $k = 1, \dots, n$, with equality when $k = n$, implies the weaker assumption $\prod_{i=1}^{k} \lambda_i \ge \prod_{i=1}^{k} \mu_i$ for all $k = 1, \dots, n$, which is made in [3].

Besides, a weak form of the converse of (2) is proved in [3], namely

$$ \sum_{i=1}^{n} X_i \le_{\text{st}} \sum_{i=1}^{n} Y_i, \quad \implies \quad \left( \lambda_1 \ge \mu_1 \quad \text{and} \quad \prod_{i=1}^{n} \lambda_i \ge \prod_{i=1}^{n} \mu_i \right). \quad (3) $$

In other words, [3] proves that the condition $\prod_{i=1}^{k} \lambda_i \ge \prod_{i=1}^{k} \mu_i$ for any $k = 1, \dots, n$ is a sufficient condition to ensure $\sum_{i=1}^{n} X_i \le_{\text{st}} \sum_{i=1}^{n} Y_i$, while $\prod_{i=1}^{k} \lambda_i \ge \prod_{i=1}^{k} \mu_i$ in the two cases $k = 1$ and $k = n$ is a necessary condition to ensure $\sum_{i=1}^{n} X_i \le_{\text{st}} \sum_{i=1}^{n} Y_i$.

In [5], the authors generalize this result for Gamma distributed random variables, by showing, using the same notation, that if $\prod_{i=1}^{k} \lambda_i \ge \prod_{i=1}^{k} \mu_i$ for any $k = 1, \dots, n$, then $\sum_{i=1}^{n} X_i \le_{\text{disp}} \sum_{i=1}^{n} Y_i$, where $X_i$, resp. $Y_i$, have a Gamma distribution with scale parameter $\lambda_i$, resp. $\mu_i$, both having the same shape parameter $a \ge 1$.

In this paper, we generalize the result established in [3] concerning the stochastic ordering of sums of exponentially distributed variables. Let us briefly describe our results. To do so, from now on we fix $(\lambda_1, \dots, \lambda_n)$ with $\lambda_1 \le \dots \le \lambda_n$ and ask for an optimal criterion on the parameters $(\mu_1, \dots, \mu_n)$ with $\mu_1 \le \dots \le \mu_n$ which ensures $\sum_{i=1}^{n} X_i \le_{\text{st}} \sum_{i=1}^{n} Y_i$. For later convenience we set for any $k$, the notation

$$ P_k(\lambda) := \prod_{i=1}^{k} \lambda_i, \quad P_k(\mu) := \prod_{i=1}^{k} \mu_i, $$
so that with this notation the result obtained in [3] reads

$$ (P_k(\lambda) \ge P_k(\mu) \text{ for any } k = 1, \dots, n) \quad \Rightarrow \quad \sum_{i=1}^{n} X_i \le \text{st} \sum_{i=1}^{n} Y_i, \\ \text{and conversely} \quad \sum_{i=1}^{n} X_i \le \text{st} \sum_{i=1}^{n} Y_i, \quad \Rightarrow \quad P_1(\lambda) \ge P_1(\mu) \text{ and } P_n(\lambda) \ge P_n(\mu) $$

Our contribution is threefolds.

First we give a geometric approach to the analysis of the problem, and transform the question into a geometric question in the space of parameters $(\mu_1, \dots, \mu_n)$. This turns out to considerably simplify the analysis. Technically, our first contribution states that $\sum_{i=1}^n X_i \le \text{st} \sum_{i=1}^n Y_i$ provided we can draw a curve $(\lambda_1(t), \dots, \lambda_n(t))$ joining $(\lambda_1, \dots, \lambda_n)$ (when $t=0$) and $(\mu_1, \dots, \mu_n)$ (when $t=1$) in the space of parameters, in such a way that some explicit linear combination of the form

$$ Q_t(X) := \sum_{i=1}^{n} \alpha_i X^{\lambda_i(t)}, \qquad (4) $$

(with explicit values of the $\alpha_i$'s and $\lambda_i(t)$'s) remains non-negative whenever $t \in [0, 1]$ and $X \in [0, 1]$. Conversely we prove that when $Q_t(X)$ changes sign close to $X=1$, we may exhibit values of $x$ close to $x=0$ such that $\mathbb{P}(\sum_{i=1}^n X_i > x) > \mathbb{P}(\sum_{i=1}^n Y_i > x)$. This establishes a form of optimality of our geometric criterion to ensure $\sum_{i=1}^n X_i \le \text{st} \sum_{i=1}^n Y_i$.

Second, and as a consequence, we establish a criterion on $(\mu_1, \dots, \mu_n)$, which generalizes the criterion $P_k(\lambda) \ge P_k(\mu)$ for any $k=1, \dots, n$ established in [3], to ensure $\sum_{i=1}^n X_i \le \text{st} \sum_{i=1}^n Y_i$. Roughly speaking our criterion reads $P_k(\lambda) \ge \beta_k P_k(\mu)$ for any $k=1, \dots, n$, where the non-negative coefficients $\beta_k$ are allowed to satisfy $\beta_k < 1$ for some $k$'s. Our criterion is global in the sense that it allows to exhibit a whole set of values of the $\mu_i$'s, which may be far from the associated $\lambda_i$'s. More technically, our result roughly asserts the following (see below for the more precise statements). For any integer $k_0$ such that $1 < k_0 < n$, we compute an explicit exponent $\theta_0(\lambda) > 0$, only depending on the $\lambda_i$'s such that

$$ \left( P_k(\lambda) \ge P_k(\mu) \text{ for } k=1, \dots, n \text{ with } k \ne k_0 \quad \text{and} \quad P_{k_0}(\lambda) \ge P_{k_0}(\mu) \underbrace{\left( \frac{P_{k_0-1}(\mu)}{P_{k_0-1}(\lambda)} \right)^{\theta_0(\lambda)}}_{\le 1} \right) \\ \Rightarrow \sum_{i=1}^{n} X_i \le \text{st} \sum_{i=1}^{n} Y_i. \qquad (5) $$

With the notation used in this paragraph, the reader may recognize the factor $P_{k_0-1}(\mu)/P_{k_0-1}(\lambda) = \beta_{k_0}$ (and $\beta_k = 1$ for all other values of $k \ne k_0$). This factor always satisfies $\beta_{k_0} \le 1$, due to the assumption $P_{k_0}(\lambda) \ge P_{k_0}(\mu)$. It also satisfies $\beta_{k_0} < 1$ in general, unless we are in the very particular case $\lambda_{k_0-1} = \mu_{k_0-1}$.

Third, we establish that our criterion (5) is optimal in the following local sense: we prove that there exists a vicinity of the $n$-tuple $(\lambda_1, \dots, \lambda_n)$ in the space of parameters $(\mu_1, \dots, \mu_n)$, for which any $(\mu_1, \dots, \mu_n)$ in this vicinity which violates our criterion (5), necessarily satisfies $\sum_{i=1}^n X_i \le \text{st} \sum_{i=1}^n Y_i$ as well, in that we have $\mathbb{P}(\sum_{i=1}^n X_i > x) > \mathbb{P}(\sum_{i=1}^n Y_i > x)$ for all values of $x$ close to $x=0$. This proves that our criterion (5) is close to a necessary and sufficient condition in a strong but local sense. More technically, we establish there exists a small $\varepsilon_0 > 0$ and a $\varepsilon_0$-vicinity of $(\lambda_1, \dots, \lambda_n)$, such that for any $(\mu_1, \dots, \mu_n)$ in this $\varepsilon_0$-vicinity we have

$$ \left( P_k(\lambda) \ge P_k(\mu) \text{ for } k=1, \dots, n \text{ with } k \ne k_0 \quad \text{and} \quad P_{k_0}(\lambda) \le P_{k_0}(\mu) \left( \frac{P_{i_0}(\mu)}{P_{i_0}(\lambda)} \right)^{\alpha} \text{ where } \alpha > \theta_0(\lambda) \right) \\ \Rightarrow \sum_{i=1}^{n} X_i \le \text{st} \sum_{i=1}^{n} Y_i, \quad \text{in that } \mathbb{P}\left(\sum_{i=1}^{n} X_i > x\right) > \mathbb{P}\left(\sum_{i=1}^{n} Y_i > x\right), \quad \forall x \in [\varepsilon_0/2, \varepsilon_0]. \qquad (6) $$
All in all, our second and third contributions exhibit, in the particular circumstance where all $P_k(\mu)$'s but the one corresponding to $k = k_0$ satisfy $P_k(\lambda) \ge P_k(\mu)$, a somehow necessary and (locally) sufficient criterion for $P_{k_0}(\mu)$, to ensure $\sum_{i=1}^n X_i \le \sum_{i=1}^n Y_i$. This criterion reads, with our notation,

$$P_{k_0}(\lambda) \ge P_{k_0}(\mu) \times \left( \frac{P_{k_0-1}(\mu)}{P_{k_0-1}(\lambda)} \right)^{\theta_0(\lambda)},$$

with the additional explicit prefactor $(P_{k_0-1}(\mu)/P_{k_0-1}(\lambda))^{\theta_0(\lambda)} \le 1$ which is the key part of the result.

This paper is organized as follows. In section 2 we review some basic facts about sums of exponentially distributed variables and recover the necessary condition (3) already obtained in [3]. In section 3, we set up the key tool that allows to transform the question into a geometric problem (Theorem 6) and establish our first main contribution (4) in the form of Theorem 9. This enables us to recover the sufficient condition established in [3] as a particular case. In sections 4 and 5 we establish our main result (5), first in the case $n = 3$ (Theorem 14), next in the general case (Theorem 17). In sections 6 and 7 we prove the optimality property (6) in the form of Theorem 23 in the case $n = 3$ and Theorem 25 in the general case. Last, section 8 is devoted to numerical illustrations and graphs.

## 2 Notation and basic results - recovering the necessary condition of [3]

Let $X_i$ ($i = 1, \dots, n$) be independent exponential random variables with rates $\lambda_i$. On the other hand, let $Y_i$, ($i = 1, \dots, n$) be independent exponential random variables with rates $\mu_i$. Throughout the paper, we investigate the following question :

which optimal condition on the $\mu_i$'s ensures $\sum_{i=1}^n X_i \le \sum_{i=1}^n Y_i$ ?

In other words, introducing the complementary cumulative distribution functions

$$F(\lambda_1, \dots, \lambda_n)(x) = \mathbb{P}(X_1 + \dots + X_n > x), \text{ and } F(\mu_1, \dots, \mu_n)(x) = \mathbb{P}(Y_1 + \dots + Y_n > x),$$

our question reads

which optimal condition on the $\mu_i$'s ensures $F(\lambda_1, \dots, \lambda_n)(x) \le F(\mu_1, \dots, \mu_n)(x)$, for any $x \ge 0$ ?

This preliminary section is devoted to some basic facts on sums of exponential variables, and more precisely on the functions $F(\lambda_1, \dots, \lambda_n)(x)$ as functions of the $\lambda_i$'s and/or as functions of $x$. Note that in the sequel we will sometimes use the notation $F(\lambda_1, \dots, \lambda_n)$, removing the $x$ dependence of these various quantities (or $y$ or $z$ dependence...) whenever the context makes it clear.

We know that, up to defining for all $\lambda \ge 0$

$$e_{\lambda}(x) = 1_{\{x>0\}} \exp(-\lambda x), \qquad (7)$$

where $1_E$ denotes the indicator function of set $E$, we have

$$F(\lambda_1, \dots, \lambda_n)(x) = \lambda_1 \cdots \lambda_n \int_x^\infty (e_{\lambda_1} * \cdots * e_{\lambda_n}) (z) dz, \qquad (8)$$

where $*$ is the standard convolution operation. We shall make use of the following alternative formulation, namely

$$F(\lambda_1, \dots, \lambda_n)(x) = 1 - \lambda_1 \cdots \lambda_n \int_0^x (e_{\lambda_1} * \cdots * e_{\lambda_n}) (z) dz.$$
With this observation at hand, we may therefore write

$$F(\lambda_1, \dots, \lambda_n)(x) = 1 - (\lambda_1 \cdots \lambda_n) (e_0 * e_{\lambda_1} * \cdots * e_{\lambda_n}) (x). \quad (9)$$

In order to further simplify the notation in the sequel, it turns out to be useful to introduce

$$e_{\lambda_1, \dots, \lambda_n}(x) := (e_{\lambda_1} * \cdots * e_{\lambda_n}) (x). \quad (10)$$

This notation provides the equality

$$F(\lambda_1, \dots, \lambda_n)(x) = 1 - (\lambda_1 \cdots \lambda_n) e_{0, \lambda_1, \dots, \lambda_n}(x), \quad (11)$$

where we stress that in the above formulae, one may always remember that

$$e_{0, \lambda_1, \dots, \lambda_n}(x) = \int_0^x e_{\lambda_1, \dots, \lambda_n}(z) dz. \quad (12)$$

It is clear that the function $F(\lambda_1, \dots, \lambda_n)$ is symmetric in the parameters $\lambda_1, \dots, \lambda_n$, so that in the sequel we shall systematically limit our attention to the situation where

$$\lambda_1 \le \lambda_2 \le \cdots \le \lambda_n,$$

without any loss of generality.

We start with the following well-known lemma.

**Lemma 1 (Asymptotic expansion of $F(\lambda_1, \dots, \lambda_n)(x)$ as $x \to +\infty$)**

Assuming that $0 < \lambda_1 < \dots < \lambda_n$, we have the explicit formula

$$F(\lambda_1, \dots, \lambda_n)(x) = \sum_{i=1}^{n} \left( \prod_{j \neq i} \frac{\lambda_j}{\lambda_j - \lambda_i} \right) e^{-\lambda_i x}$$

and the explicit asymptotic expansion as $x \to +\infty$,

$$F(\lambda_1, \dots, \lambda_n)(x) \underset{x \to \infty}{\sim} \left( \prod_{j=2}^{n} \frac{\lambda_j}{\lambda_j - \lambda_1} \right) e^{-\lambda_1 x}.$$

**Remark 2** With our ordering of the $\lambda_i$'s we have of course $\prod_{j=2}^n \lambda_j / (\lambda_j - \lambda_1) > 0$, so that $F$ is positive at infinity, in $x$, as it should. More importantly, note that the lowest value $\lambda_1$ dominates the behaviour of $F$ close to $x = +\infty$. This is one important reason for our ordering of the $\lambda_i$'s.

Note also that the above formulae may be extended to the case when two or more of the $\lambda_i$'s coincide, by using the continuity of $F$ and by appropriately passing to the limit. For instance we may observe that when $0 < \lambda_1 < \lambda_2$ we have

$$F(\lambda_1, \lambda_2)(x) = \frac{\lambda_2}{\lambda_2 - \lambda_1} e^{-\lambda_1 x} + \frac{\lambda_1}{\lambda_1 - \lambda_2} e^{-\lambda_2 x},$$

so that an easy asymptotic expansion provides, passing to the limit $\lambda_2 \to \lambda_1$ (take $\lambda_2 = \lambda_1 + \varepsilon$, with $\varepsilon \to 0$) in the above relation,

$$F(\lambda_1, \lambda_1)(x) = (1 + \lambda_1 x) e^{-\lambda_1 x} \underset{x \to \infty}{\sim} \lambda_1 x e^{-\lambda_1 x}.$$

Similarly one may write when $0 < \lambda_1 < \lambda_2 < \lambda_3$

$$F(\lambda_1, \lambda_2, \lambda_3)(x) = \frac{\lambda_2 \lambda_3}{(\lambda_2 - \lambda_1)(\lambda_3 - \lambda_1)} e^{-\lambda_1 x} + \frac{\lambda_1 \lambda_3}{(\lambda_1 - \lambda_2)(\lambda_3 - \lambda_2)} e^{-\lambda_2 x} + \frac{\lambda_1 \lambda_2}{(\lambda_1 - \lambda_3)(\lambda_2 - \lambda_3)} e^{-\lambda_3 x}$$
and, passing to the limit $\lambda_2 \to \lambda_1$ as well as $\lambda_3 \to \lambda_1$ (take $\lambda_2 = \lambda_1 + \varepsilon$ and $\lambda_3 = \lambda_1 + 2\varepsilon$ with $\varepsilon \to 0$),
obtain

$$F(\lambda_1, \lambda_1, \lambda_1)(x) = e^{-\lambda_1 x} \left( 1 + \lambda_1 x + \frac{\lambda_1^2 x^2}{2} \right) \underset{x \to \infty}{\sim} \frac{\lambda_1^2 x^2}{2} e^{-\lambda_1 x}.$$

<figure><img src="image_2.png" /></figure>

The following theorem completes the picture provided by the previous lemma, in that it provides a fully explicit asymptotic expansion of $F(\lambda_1, \dots, \lambda_n)(x)$ as $x \to 0$ at any order.

**Theorem 3 (Asymptotic expansion of $F(\lambda_1, \dots, \lambda_n)(x)$ as $x \to 0$)**

Assuming that

$$0 < \lambda_1 \le \lambda_2 \le \dots \le \lambda_n$$

we have, for any integer $p$, the following $p$-terms asymptotic expansion of $F$ close to $x = 0$, namely

$$F(\lambda_1, \dots, \lambda_n)(x) = 1 - P_n(\lambda) \left[ \frac{x^n}{n!} - S_1^{(n)}(\lambda) \frac{x^{n+1}}{(n+1)!} + S_2^{(n)}(\lambda) \frac{x^{n+2}}{(n+2)!} + \dots + (-1)^p S_p^{(n)}(\lambda) \frac{x^{n+p}}{(n+p)!} \right] + O(x^{n+p+1}),$$

where $P_n(\lambda) = \lambda_1 \cdots \lambda_n$ as before and for any integer $k \ge 1$ we define the quantities

$$S_1^{(n)}(\lambda) = \sum_{1 \le i \le n} \lambda_i, \quad S_2^{(n)}(\lambda) = \sum_{1 \le i \le j \le n} \lambda_i \lambda_j, \quad \dots, \quad S_k^{(n)}(\lambda) = \sum_{1 \le i_1 \le i_2 \le \dots \le i_k \le n} \lambda_{i_1} \lambda_{i_2} \cdots \lambda_{i_k}.$$

Here the remainder term $O(x^{n+p+1})$ stands for a function of $x$ which satisfies, for any $0 \le x \le 1$, for any $M > 0$, and for any $\lambda_i$'s such that $0 < \lambda_1 \le \dots \le \lambda_n \le M$, the upper bound

$$|O(x^{n+p+1})| \le C_{M,n,p} x^{n+p+1},$$

for some constant $C_{M,n,p}$ which depends on $M$, $n$ and $p$, but it does not depend neither on $x$ nor on the choice of the $\lambda_i$'s.

**Proof.** This theorem is easily proved by recursion on $n$. Indeed, starting from the established formula (11)

$$F(\lambda_1, \dots, \lambda_n)(x) = 1 - P_n(\lambda) e_{0, \lambda_1, \dots, \lambda_n}(x),$$

it is enough to prove by recursion that for any $n$ we have

$$(e_0 * e_{\lambda_1} * \dots * e_{\lambda_n})(x) = \frac{x^n}{n!} - S_1^{(n)}(\lambda) \frac{x^{n+1}}{(n+1)!} + \dots + (-1)^p S_p^{(n)}(\lambda) \frac{x^{n+p}}{(n+p)!} + O(x^{n+p+1}). \quad (13)$$

For $n = 1$, formula(13) is obvious by writing

$$e_0 * e_{\lambda_1}(x) = \int_0^x e^{-\lambda_1 y} dy = x - \lambda_1 \frac{x^2}{2} + \lambda_1^2 \frac{x^3}{3!} + \dots + (-1)^p \lambda_1^p \frac{x^{1+p}}{(1+p)!} + \mathcal{O}(x^{2+p}).$$

Now, assuming (13) holds true for some $n$, by taking the convolution product with $e_{\lambda_{n+1}}(x)$ on both sides
of this equality we deduce

$$
\begin{align*}
(e_0 * e_{\lambda_1} * \cdots * e_{\lambda_n} * e_{\lambda_{n+1}})(x) &= \int_0^x (e_0 * e_{\lambda_1} * \cdots * e_{\lambda_n})(y) \exp(-\lambda_{n+1}(x-y)) dy \\
&= \int_0^x \left( \frac{y^n}{n!} - S_1^{(n)}(\lambda) \frac{y^{n+1}}{(n+1)!} + \cdots + (-1)^p S_p^{(n)}(\lambda) \frac{y^{n+p}}{(n+p)!} + O(x^{n+p+1}) \right) \\
&\qquad \times \left( 1 - \lambda_{n+1}(x-y) + \cdots + (-1)^p \lambda_{n+1}^p \frac{(x-y)^p}{p!} + O((x-y)^{p+1}) \right) dy \\
&= \int_0^x \frac{y^n}{n!} dy \\
&\quad - \int_0^x \left[ S_1^{(n)}(\lambda) \frac{y^{n+1}}{(n+1)!} + \lambda_{n+1} \frac{y^n(x-y)}{n!} \right] dy \\
&\quad + \int_0^x \left[ S_2^{(n)}(\lambda) \frac{y^{n+2}}{(n+2)!} + S_1^{(n)}(\lambda) \lambda_{n+1} \frac{y^{n+1}(x-y)}{(n+1)!} + \lambda_{n+1}^2 \frac{y^n(x-y)^2}{n!2!} \right] dy \\
&\quad + \cdots \\
&\quad + (-1)^p \int_0^x \left[ S_p^{(n)}(\lambda) \frac{y^{n+p}}{(n+p)!} + S_{p-1}^{(n)}(\lambda) \lambda_{n+1} \frac{y^{n+p-1}(x-y)}{(n+p-1)!} + \cdots \right. \\
&\qquad \left. + S_1^{(n)}(\lambda) \lambda_{n+1}^{p-1} \frac{y^{n+1}(x-y)^{p-1}}{(n+1)!(p-1)!} + \lambda_{n+1}^p \frac{y^n(x-y)^p}{n!p!} \right] dy \\
&\quad + O(x^{n+p+2}).
\end{align*}
$$

Noting that

$$
\begin{align*}
S_1^{(n)}(\lambda) + \lambda_{n+1} &= S_1^{(n+1)}(\lambda), \\
S_2^{(n)}(\lambda) + S_1^{(n)}(\lambda)\lambda_{n+1} + \lambda_{n+1}^2 &= S_2^{(n+1)}(\lambda), \\
\dots, \\
S_p^{(n)}(\lambda) + S_{p-1}^{(n)}(\lambda)\lambda_{n+1} + \dots + S_1^{(n)}(\lambda)\lambda_{n+1}^{p-1} + \lambda_{n+1}^p &= S_p^{(n+1)}(\lambda),
\end{align*}
$$

and observing as well that

$$
\int_0^x \frac{y^{n+p}}{(n+p)!} dy = \int_0^x \frac{y^{n+p-1}(x-y)}{(n+p-1)!} dy = \dots = \int_0^x \frac{y^{n+1}(x-y)^{p-1}}{(n+1)!(p-1)!} dy = \int_0^x \frac{y^n(x-y)^p}{n!p!} dy = \frac{x^{n+p+1}}{(n+p+1)!},
$$

which rests on a simple integration by parts in the variable $y$, we recover

$$
\begin{align*}
& (e_0 * e_{\lambda_1} * \cdots * e_{\lambda_n} * e_{\lambda_{n+1}})(x) \\
&= \frac{x^{n+1}}{(n+1)!} - S_1^{(n+1)}(\lambda) \frac{x^{n+2}}{(n+2)!} + S_2^{(n+1)}(\lambda) \frac{x^{n+3}}{(n+3)!} + \cdots + (-1)^p S_p^{(n+1)}(\lambda) \frac{x^{n+p+1}}{(n+p+1)!} + O(x^{n+p+2}).
\end{align*}
$$

Formula (13) is established for any $n$. The theorem follows.

<figure><img src="image_3.png" /></figure>

The above two asymptotic expansions of $F$ close to $x = 0$ and $x = +\infty$ imply the following important
yet basic property. This result is a refined analysis of the result obtained in [3].

**Corollary 4 (Necessary conditions to ensure**

Assuming that

$0 < \lambda_1 \le \dots \le \lambda_n \text{ and } 0 < \mu_1 \le \dots \le \mu_n$

and that for any $x \ge 0$ we have

$F(\lambda_1, \dots, \lambda_n)(x) \leq F(\mu_1, \dots, \mu_n)(x),$
then, the following holds true:

(i) (recovering the necessary condition of [3]) The two upper bounds

$$ \lambda_1 \ge \mu_1 \quad \text{and} \quad \prod_{i=1}^{n} \lambda_i \ge \prod_{i=1}^{n} \mu_i $$

are necessarily satisfied.

(ii) (extending the necessary condition of [3]) On top of that, in the case when we have the equality $\lambda_1 \cdots \lambda_n = \mu_1 \cdots \mu_n$ in the second upper bound above, then the following additional upper bound is necessary, namely

$$ \sum_{i=1}^{n} \lambda_i \le \sum_{i=1}^{n} \mu_i. $$

**Remark 5** One may push the discussion further and assert that if $\lambda_1 \cdots \lambda_n = \mu_1 \cdots \mu_n$ and $\lambda_1 + \cdots + \lambda_n = \mu_1 + \cdots + \mu_n$, then we have the additional necessary condition

$$ \sum_{1 \le i \le j \le n} \lambda_i \lambda_j \ge \sum_{1 \le i \le j \le n} \mu_i \mu_j. $$

This discussion may be pushed further at any order, i.e. if

$$ P_n(\lambda) = P_n(\mu) \quad \text{and} \quad S_{\ell}^{(n)}(\lambda) = S_{\ell}^{(n)}(\mu), \quad \text{for } \ell = 1, \dots, k $$

then we have the additional necessary condition

$$ (-1)^k S_{k+1}^{(n)}(\lambda) \le (-1)^k S_{k+1}^{(n)}(\mu), $$

This shows that the exact characterization of those parameters $\mu_i$'s for which we are able to guarantee $F(\lambda_1, \dots, \lambda_n)(x) \le F(\mu_1, \dots, \mu_n)(x)$ for all $x \ge 0$ is by nature a complex task. ■

**Proof.** The proof simply relies on the basic fact that the assumption $F(\lambda_1, \dots, \lambda_n)(x) \le F(\mu_1, \dots, \mu_n)(x)$ implies that the respective asymptotic expansions of these two functions meet the same inequality close to $x = 0$ and close to $x = \infty$.

The constraint $\lambda_1 \ge \mu_1$ then comes from the asymptotic behaviour at infinity which is given by Lemma 1.

The other conditions come from the asymptotic expansion of $F(\lambda_1, \dots, \lambda_n)(x)$ close to $x = 0$ given by Theorem 3. Part (i) (resp. part (ii)) is obtained by taking $p = 0$ (resp. $p = 1$) in this theorem. This can be easily extended if necessary to any value of $p$ as explained in Remark 5. ■

The remainder part of this text inspects the question of the converse statement, namely: which minimal conditions on the $\mu_i$'s ensure that $F(\lambda_1, \dots, \lambda_n)(x) \le F(\mu_1, \dots, \mu_n)(x)$ for any $x$?

This question turns out to be delicate.

## 3 Variations of $F(\lambda_1, \dots, \lambda_n)(x)$ with respect to $(\lambda_1, \dots, \lambda_n)$ - a geometric approach

In order to study the variations of $F(\lambda_1, \dots, \lambda_n)(x)$ with respect to $(\lambda_1, \dots, \lambda_n)$, we start from the formula

$$ F(\lambda_1, \dots, \lambda_n)(x) = 1 - (\lambda_1 \cdots \lambda_n) e_{0, \lambda_1, \dots, \lambda_n}(x), $$

where we recall that

$$ e_{0, \lambda_1, \dots, \lambda_n}(x) = (e_0 * e_{\lambda_1} * \dots * e_{\lambda_n})(x). $$

The key to our analysis is the following.
**Theorem 6 (Differentiation of F with respect to $\lambda_i$)**

(i) For any $\lambda > 0$ we have the relation

$$ \lambda \partial_{\lambda} (e_{0} * [\lambda e_{\lambda}]) (x) = (e_{\lambda} * [\lambda e_{\lambda}]) (x). $$

In other words the action of the differential operator $\lambda \partial_{\lambda}$ on the convolution product $e_0 * [\lambda e_{\lambda}]$ transforms the first factor $e_0$ into $e_{\lambda}$, leaving the other factor unchanged.

(ii) As a consequence, for any index $i = 1, \dots, n$ we have

$$ \lambda_i \partial_{\lambda_i} F(\lambda_1, \dots, \lambda_n)(x) = -(\lambda_1 \cdots \lambda_n) (e_{\lambda_i} * e_{\lambda_1, \dots, \lambda_n}) (x). $$

**Remark 7** At first glance the $\lambda$-derivative of $e_0 * e_{\lambda}$, when computed directly using differentiation inside an integral, should involve the convolution of products of exponential functions and polynomial functions (in the $x$ variable), whose sign is extremely difficult to analyze. The above theorem somehow circumvents this difficulty, in that it reformulates the $\lambda$-derivative of $e_0 * e_{\lambda}$ as a singular convolution of $e_{\lambda}$ with itself, a convolution whose sign is easily seen (see the proof below). This remark is essential. ■

**Remark 8** The reader should be careful about the homogeneity of the above formulae. In particular, note that the above theorem is easily formulated in terms of the differential operator $\lambda \partial_{\lambda}$, which is homogeneous of degree zero in $\lambda$, contrary to the more usual operator $\partial_{\lambda}$, which is homogeneous of degree $-1$. This is also a key point in the later analysis. ■

**Proof.** The proof is based on an easy reformulation of the involved convolution products. Namely, using an integration by parts, we write

$$ \begin{aligned} \lambda \partial_{\lambda} (e_0 * [\lambda e_{\lambda}]) (x) &= \lambda \partial_{\lambda} \left( \lambda \int_{0}^{x} e^{-\lambda y} dy \right) = \int_{0}^{x} \lambda(1 - \lambda y)e^{-\lambda y} dy \\ &= [(\lambda y - 1)e^{-\lambda y}]_{y=0}^{x} - \int_{0}^{x} \lambda e^{-\lambda y} dy = \lambda x e^{-\lambda x}, \end{aligned} $$

and we observe, on the other hand, that

$$ e_{\lambda} * [\lambda e_{\lambda}](x) = \int_{0}^{x} \lambda e^{-\lambda y} e^{-\lambda(x-y)} dy = \lambda e^{-\lambda x} \int_{0}^{x} dy = \lambda x e^{-\lambda x}. $$

Point (i) is proved.

To prove (ii) we write, say in the case $i = n$ to keep formulae simple,

$$ \begin{aligned} \lambda_n \partial_{\lambda_n} F(\lambda_1, \dots, \lambda_n)(x) &= -\lambda_n \partial_{\lambda_n} [\lambda_1 \cdots \lambda_n (e_0 * e_{\lambda_1} * \dots * e_{\lambda_{n-1}} * e_{\lambda_n}) (x)] \\ &= -\lambda_1 \cdots \lambda_{n-1} (e_{\lambda_1, \dots, \lambda_{n-1}} * [\lambda_n \partial_{\lambda_n} (e_0 * [\lambda_n e_{\lambda_n}])]) (x) \\ &= -\lambda_1 \cdots \lambda_{n-1} (e_{\lambda_1, \dots, \lambda_{n-1}} * [(e_{\lambda_n} * [\lambda_n e_{\lambda_n}]))] (x) \\ &= -\lambda_1 \cdots \lambda_n (e_{\lambda_n} * e_{\lambda_1, \dots, \lambda_{n-1}}) (x), \end{aligned} $$

where point (i) is used in the third equality. Relation (ii) is established. ■

Once the above theorem is proved, the following result is a geometric consequence which is a key tool in the sequel.

**Theorem 9 (Variations of $F(\lambda_1, \dots, \lambda_n)$ along specific curves)**

Fix $n$ positive numbers $\lambda_1, \dots, \lambda_n$ (not necessarily ordered). Fix $n$ real numbers $\alpha_1, \dots, \alpha_n$. Set, for $i = 1, \dots, n$,

$$ \lambda_i(t) = \lambda_i e^{-\alpha_i t}. $$
(i) We have the explicit formula

$$ \frac{d}{dt} F(\lambda_1(t), \dots, \lambda_n(t))(x) = \left( \prod_{i=1}^{n} \lambda_i(t) \right) \left[ e_{\lambda_1(t), \dots, \lambda_n(t)} * \{ \alpha_1 e_{\lambda_1(t)} + \dots + \alpha_n e_{\lambda_n(t)} \} \right] (x). \quad (14) $$

(ii) Assume that there is a $T > 0$ such that for any $t \in [0, T]$ and for any $x \ge 0$ we have

$$ q_t(x) := \alpha_1 e_{\lambda_1(t)}(x) + \dots + \alpha_n e_{\lambda_n(t)}(x) \ge 0. \quad (15) $$

Equivalently, assume that for any $t \in [0, T]$ and for any $X \in [0, 1]$ we have

$$ Q_t(X) := \alpha_1 X^{\lambda_1(t)} + \dots + \alpha_n X^{\lambda_n(t)} \ge 0, \quad (16) $$

where the reader may note the relation $q_t(x) = Q_t(e^{-x})$.

Then the function $t \in [0, T] \mapsto F(\lambda_1(t), \dots, \lambda_n(t))(x)$ is non-decreasing for any $x \ge 0$. In particular we have, for any $x \ge 0$ and $t \in [0, T]$,

$$ F(\lambda_1, \dots, \lambda_n)(x) \le F(\lambda_1(t), \dots, \lambda_n(t))(x). \quad (17) $$

**Remark 10** The key point here is that we have reduced the question of the sign of the variations of $F(\lambda_1, \dots, \lambda_n)$ with respect to $(\lambda_1, \dots, \lambda_n)$ (condition (17)), to the study of the mere sign of a single linear combination of exponentials (condition (15)), or, alternatively, to the study of the sign of linear combinations of powers of $X$ (condition (16)).

Eventually this theorem provides the sign of the variations of function $F(\lambda_1, \dots, \lambda_n)$ along the curve $t \mapsto (\lambda_1(t), \dots, \lambda_n(t))$ in the space of parameters $(\lambda_1, \dots, \lambda_n)$.

In that respect, the key point is formula (14), which in turn is a consequence of Theorem 6: it shows that, up to convolutions which preserve the sign, the whole phenomenon is governed by the sign of the linear combination $\alpha_1 e_{\lambda_1(t)} + \dots + \alpha_n e_{\lambda_n(t)}$. The reader is invited to compute the derivative of $F(\lambda_1(t), \dots, \lambda_n(t))$ directly, without Theorem 6. The sign of the obtained result is extremely difficult to analyze directly. ■

**Proof.** Set

$$ G(t)(x) = F(\lambda_1(t), \dots, \lambda_n(t))(x), $$

and observe in passing that $G(0)(x) = F(\lambda_1, \dots, \lambda_n)(x)$. Using Theorem 6 together with the chain rule and the fact that $\lambda'_i(t) = -\alpha_i \lambda_i(t)$ for any $i$, we have

$$ \begin{aligned} \frac{dG}{dt}(t)(x) &= \sum_{i=1}^{n} \lambda'_{i}(t) \left( \frac{\partial}{\partial \lambda_{i}} F \right) (\lambda_{1}(t), \dots, \lambda_{n}(t))(x) \\ &= -\sum_{i=1}^{n} \alpha_{i} \lambda_{i}(t) \left( \frac{\partial}{\partial \lambda_{i}} F \right) (\lambda_{1}(t), \dots, \lambda_{n}(t))(x) \\ &= -\sum_{i=1}^{n} \alpha_{i} \left( \lambda_{i} \frac{\partial}{\partial \lambda_{i}} F \right) (\lambda_{1}(t), \dots, \lambda_{n}(t))(x) \\ &= \sum_{i=1}^{n} \alpha_{i} \lambda_{1}(t) \cdots \lambda_{n}(t) \left[ e_{\lambda_{i}(t)} * e_{\lambda_{1}(t), \dots, \lambda_{n}(t)} \right] (x) \quad (\text{by Theorem 6}) \\ &= \lambda_{1}(t) \cdots \lambda_{n}(t) \left[ e_{\lambda_{1}(t), \dots, \lambda_{n}(t)} * \left\{ \sum_{i=1}^{n} \alpha_{i} e_{\lambda_{i}(t)} \right\} \right] (x). \end{aligned} $$

This proves

$$ \frac{dG}{dt}(t)(x) = \left( \prod_{i=1}^{n} \lambda_i(t) \right) \left[ e_{\lambda_1(t), \dots, \lambda_n(t)} * \{ \alpha_1 e_{\lambda_1(t)} + \dots + \alpha_n e_{\lambda_n(t)} \} \right] (x). \quad (18) $$
Point (i) is established.

Concerning point (ii), formula (18) establishes that the derivative of G inherits the sign of the linear combination of exponentials $q_t(x) := \alpha_1 e_{\lambda_1(t)} + \cdots + \alpha_n e_{\lambda_n(t)}$ whenever the latter has constant sign for $x \ge 0$, due to

$$ \frac{dG}{dt}(t)(x) = \left( \prod_{i=1}^{n} \lambda_i(t) \right) [e_{\lambda_1(t), \dots, \lambda_n(t)} * q_t](x). $$

There remains to observe that, setting

$$ X = e^{-x} $$

we have

$$ q_t(x) = \alpha_1 e_{\lambda_1(t)}(x) + \cdots + \alpha_n e_{\lambda_n(t)}(x) = Q_t(X). $$

This ends the proof of point (ii).

<figure><img src="image_4.png" /></figure>

The remainder part of this section is now devoted to recovering the result of [3] as an immediate consequence of Theorem 9.

**Corollary 11 (Recovering the sufficient condition established in [3])**

Take $2n$ ordered positive numbers $0 < \lambda_1 \le \lambda_2 \le \dots \le \lambda_n$ and $0 < \mu_1 \le \dots \le \mu_n$. For any $k = 1, \dots, n$, set as before the quantities

$$ P_k(\lambda) = \prod_{i=1}^{k} \lambda_i, \quad P_k(\mu) = \prod_{i=1}^{k} \mu_i. $$

Assume

$$ \forall k = 1, \dots, n, \quad P_k(\lambda) \ge P_k(\mu). $$

Then we have

$$ \forall x \ge 0, \quad F(\lambda_1, \dots, \lambda_n)(x) \le F(\mu_1, \dots, \mu_n)(x). $$

**Remark 12** In particular, in the situation of the above corollary, whenever $X_i$ and $Y_i$ are independent exponential random variables with rates $\lambda_i$ and $\mu_i$, we have

$$ \sum_{i=1}^{n} X_i \le \text{st} \sum_{i=1}^{n} Y_i. $$

<figure><img src="image_5.png" /></figure>

**Remark 13** An important but trivial particular case of Corollary 11 is the following simple decay property: whenever $0 < \lambda_1 \le \dots \le \lambda_n$ and $0 < \mu_1 \le \dots \le \mu_n$ satisfy $\lambda_i \ge \mu_i$ for $i = 1, \dots, n$, we have $F(\lambda_1, \dots, \lambda_n)(x) \le F(\mu_1, \dots, \mu_n)(x)$ for all $x$. ■

**Proof Corollary 11.** Since for any $k$ we have $P_k(\lambda) \ge P_k(\mu)$ we may find non-negative numbers $\alpha_k \ge 0$ such that

$$ P_k(\mu) = P_k(\lambda) e^{-\alpha_k} \quad \forall k = 1, \dots, n. $$

On the other hand, we may recover the $\lambda_i$'s and $\mu_i$'s from the $P_k(\lambda)$'s and $P_k(\mu)$'s through formulae

$$ \lambda_1 = P_1(\lambda), \quad \lambda_i = \frac{P_i(\lambda)}{P_{i-1}(\lambda)} \quad \forall i = 2, \dots, n, $$

and similarly for the $\mu_i$'s. As a consequence we have

$$ \mu_1 = \lambda_1 e^{-\alpha_1}, \quad \mu_i = \lambda_i e^{-(\alpha_i - \alpha_{i-1})} \quad \forall i = 2, \dots, n. $$
Let us now introduce the curve

$$ (\lambda_1(t), \lambda_2(t), \dots, \lambda_n(t)) = (\lambda_1 e^{-\alpha_1 t}, \lambda_2 e^{-(\alpha_2-\alpha_1)t}, \lambda_3 e^{-(\alpha_3-\alpha_2)t}, \dots, \lambda_n e^{-(\alpha_n-\alpha_{n-1})t}) $$

in the space of parameters $(\mu_1, \dots, \mu_n)$, where $t \in [0, 1]$. This curve satisfies

$$ (\lambda_1(0), \lambda_2(0), \dots, \lambda_n(0)) = (\lambda_1, \lambda_2, \dots, \lambda_n), \quad (\lambda_1(1), \lambda_2(1), \dots, \lambda_n(1)) = (\mu_1, \mu_2, \dots, \mu_n). $$

Besides, we claim that we have

$$ 0 \le \lambda_1(t) \le \lambda_2(t) \le \dots \le \lambda_n(t) \quad \forall t \in [0, 1]. $$

Let us prove this easy but important technical statement. First, this ordering holds true at $t = 0$ and $t = 1$. Second, upon setting $\beta_1 = \alpha_1$ and $\beta_i = \alpha_i - \alpha_{i-1}$ whenever $i = 2, \dots, n$, we have $\lambda_i(t) = \lambda_i e^{-\beta_i t}$ for all $i$'s. Therefore, for any given $i = 1, \dots, n-1$, two situations may occur. If $\beta_i \ge \beta_{i+1}$, we argue that the function $f_i(t) = \lambda_i(t)/\lambda_{i+1}(t)$ satisfies $df_i(t)/dt = -(\beta_i - \beta_{i+1})(\lambda_i/\lambda_{i+1})e^{-(\beta_i-\beta_{i+1})t} \le 0$, together with $f_i(0) \le 1$, from which we deduce $f_i(t) \le 1$ for all $t \in [0, 1]$, hence $\lambda_i(t) \le \lambda_{i+1}(t)$ for all $t \in [0, 1]$. If at variance $\beta_i < \beta_{i+1}$, we argue that $f_i(t)$ satisfies $df_i(t)/dt = -(\beta_i - \beta_{i+1})(\lambda_i/\lambda_{i+1})e^{-(\beta_i-\beta_{i+1})t} \ge 0$, together with $f_i(1) \le 1$, from which we deduce $f_i(t) \le 1$ for all $t \in [0, 1]$, hence $\lambda_i(t) \le \lambda_{i+1}(t)$ for all $t \in [0, 1]$. Our claim is proved.

Now, in the present case, Theorem 9 asserts that the variations of $F(\lambda_1(t), \dots, \lambda_n(t))(x)$ are governed by the sign of

$$ Q_t(X) = \alpha_1 X^{\lambda_1(t)} + (\alpha_2 - \alpha_1) X^{\lambda_2(t)} + \dots + (\alpha_n - \alpha_{n-1}) X^{\lambda_n(t)} $$

for $t \in [0, 1]$ and $X \in [0, 1]$. Therefore, we write

$$ \begin{aligned} Q_t(X) &= \alpha_1 (X^{\lambda_1(t)} - X^{\lambda_2(t)}) + \alpha_2 (X^{\lambda_2(t)} - X^{\lambda_3(t)}) + \dots + \alpha_{n-1} (X^{\lambda_{n-1}(t)} - X^{\lambda_n(t)}) + \alpha_n X^{\lambda_n(t)} \\ &= \alpha_1 X^{\lambda_1(t)} (1 - X^{\lambda_2(t)-\lambda_1(t)}) + \alpha_2 X^{\lambda_2(t)} (1 - X^{\lambda_3(t)-\lambda_2(t)}) + \dots \\ &\quad + \alpha_{n-1} X^{\lambda_{n-1}(t)} (1 - X^{\lambda_n(t)-\lambda_{n-1}(t)}) + \alpha_n X^{\lambda_n(t)}. \end{aligned} $$

In the above sum, for each $i = 1, \dots, n-1$ we may write, using $\lambda_{i+1}(t) \ge \lambda_i(t)$ for $t \in [0, 1]$, together with $X \in [0, 1]$, the lower bound

$$ \alpha_i X^{\lambda_i(t)} (1 - X^{\lambda_{i+1}(t)-\lambda_i(t)}) \ge 0. $$

This provides

$$ Q_t(X) \ge 0, \quad \forall t \in [0, 1], \quad \forall X \in [0, 1]. $$

From this we deduce, using Theorem 9, that

$$ \frac{d}{dt} F(\lambda_1(t), \dots, \lambda_n(t))(x) \ge 0, $$

for any $t \in [0, 1]$ and any $x \ge 0$. In particular

$$ F(\lambda_1, \dots, \lambda_n)(x) \le F(\mu_1, \dots, \mu_n)(x) $$

for all $x \ge 0$. The Corollary is proved. ■
## 4 A geometric approach to a non trivial extension of the sufficient condition in [3] - case $n = 3$

In the technically simpler case where $n = 3$, the sufficient condition in [3] reads: if $0 < \lambda_1 \le \lambda_2 \le \lambda_3$ and $0 < \mu_1 \le \mu_2 \le \mu_3$ satisfy

$$P_1(\lambda) \ge P_1(\mu), \quad P_2(\lambda) \ge P_2(\mu), \quad P_3(\lambda) \ge P_3(\mu),$$

then we have, for any $x \ge 0$, the relation

$$F(\lambda_1, \lambda_2, \lambda_3)(x) \le F(\mu_1, \mu_2, \mu_3)(x).$$

Besides, a necessary condition to ensure $F(\lambda_1, \lambda_2, \lambda_3)(x) \le F(\mu_1, \mu_2, \mu_3)(x)$ is

$$P_1(\lambda) \ge P_1(\mu), \quad P_3(\lambda) \ge P_3(\mu).$$

As a consequence, the only way to extend the condition in [3] is to investigate the situation when we have $P_2(\lambda) < P_2(\mu)$, while $P_1(\lambda) \ge P_1(\mu)$, and, $P_3(\lambda) \ge P_3(\mu)$. This is what we do here and below. We exhibit an optimal range of values of the $P_k(\mu)$'s, for which we have both

$$P_1(\lambda) \ge P_1(\mu), \quad P_2(\lambda) < P_2(\mu), \quad P_3(\lambda) \ge P_3(\mu),$$

and, for any $x \ge 0$, the bound

$$F(\lambda_1, \lambda_2, \lambda_3)(x) \le F(\mu_1, \mu_2, \mu_3)(x).$$

Our key result and major extension in that direction is the following. We show later in this text that our criterion is in some sense optimal.

**Theorem 14 (extended criterion in the case $n = 3$)**

Take $0 < \lambda_1 \le \lambda_2 \le \lambda_3$ and $0 < \mu_1 \le \mu_2 \le \mu_3$. Define

$$\theta_0(\lambda) = \frac{\lambda_2 - \lambda_1}{\lambda_3 - \lambda_2} \ge 0,$$

with $\theta_0(\lambda) = +\infty$ in the degenerate case $\lambda_2 = \lambda_3$. Assume

$$P_1(\lambda) > P_1(\mu), \quad P_3(\lambda) \ge P_3(\mu), \quad \text{but } P_2(\lambda) < P_2(\mu),$$

with the additional lower bound

$$P_2(\lambda) \ge P_2(\mu) \left( \frac{P_1(\mu)}{P_1(\lambda)} \right)^{\theta_0(\lambda)}.$$

Then we have, for any $x \ge 0$, the relation

$$F(\lambda_1, \lambda_2, \lambda_3)(x) \le F(\mu_1, \mu_2, \mu_3)(x).$$

**Remark 15** Since $P_1(\lambda) > P_1(\mu)$, we obviously have

$$\left( \frac{P_1(\mu)}{P_1(\lambda)} \right)^{\theta_0(\lambda)} \le 1.$$

Hence the above result is a strict extension of the condition in [3]. In other words, the present criterion extends the set of $\mu$'s for which we can guarantee $F(\lambda_1, \lambda_2, \lambda_3)(x) \le F(\mu_1, \mu_2, \mu_3)(x)$ for all $x \ge 0$, from the set obtained in [3]

$$A := \{ 0 < \mu_1 \le \mu_2 \le \mu_3 \text{ such that } P_1(\lambda) \ge P_1(\mu), \quad P_2(\lambda) \ge P_2(\mu), \quad P_3(\lambda) \ge P_3(\mu) \}, \quad (19)$$
to the extended set

$$ \mathcal{A}^{\text{ext}} := \left\{ 0 < \mu_1 \le \mu_2 \le \mu_3 \text{ s.t. } P_1(\lambda) \ge P_1(\mu), P_2(\lambda) \ge P_2(\mu) \left( \frac{P_1(\mu)}{P_1(\lambda)} \right)^{\theta_0(\lambda)}, P_3(\lambda) \ge P_3(\mu) \right\}. \quad (20) $$

Naturally, we have the inclusion $\mathcal{A} \subset \mathcal{A}^{\text{ext}}$, due to the fact that $P_1(\mu)/P_1(\lambda) \le 1$.

<figure><img src="image_6.png" /></figure>

**Remark 16** Note that choosing $\mu = (\mu_1, \mu_2, \mu_3)$ with positive coefficients is equivalent to choosing a triple $(P_1(\mu), P_2(\mu), P_3(\mu))$, since the knowledge of the $P_k(\mu)$'s allows to recover the $\mu_i$'s through formulae $\mu_1 = P_1(\mu)$, $\mu_2 = P_2(\mu)/P_1(\mu)$, $\mu_3 = P_3(\mu)/P_2(\mu)$. Even more, and in order to manipulate normalized coefficients, for a given $(\lambda_1, \lambda_2, \lambda_3)$, choosing $\mu = (\mu_1, \mu_2, \mu_3)$ is equivalent to choosing a triple

$$ \left( x = \frac{P_1(\mu)}{P_1(\lambda)}, y = \frac{P_2(\mu)}{P_2(\lambda)}, z = \frac{P_3(\mu)}{P_3(\lambda)} \right). \qquad (21) $$

With this remark in mind, the sets $\mathcal{A}$ and $\mathcal{A}^{\text{ext}}$ are best described in terms of the above variables $(x, y, z)$. In these variables indeed, the set $\mathcal{A}$ is roughly the cube

$$ \mathcal{C} = \{(x, y, z) \text{ s.t. } x \le 1, y \le 1, z \le 1\}, \quad \text{where } (x, y, z) \text{ are given by (21)}, $$

while $\mathcal{A}^{\text{ext}}$ is roughly the union of the above cube $\mathcal{C}$ with the subgraph of the hyperboloid $y = 1/x^{\theta_0(\lambda)}$, namely, in these variables, $\mathcal{A}^{\text{ext}}$ is roughly the set

$$ \mathcal{C}^{\text{ext}} = \{(x, y, z) \text{ s.t. } x \le 1, y \le \frac{1}{x^{\theta_0(\lambda)}}, z \le 1\}, \quad \text{where } (x, y, z) \text{ are given by (21)}. $$

Naturally we have $\mathcal{C} \subset \mathcal{C}^{\text{ext}}$ due to the bound $x \le 1$. Besides, the larger the exponent $\theta_0(\lambda)$ is, the larger the set $\mathcal{C}^{\text{ext}}$ is. In other words, if $\lambda_3 > \lambda_2$ is close to $\lambda_2$, the set $\mathcal{C}^{\text{ext}}$ is large because $\theta_0(\lambda)$ has a large value, while if $\lambda_3$ is far from $\lambda_2$ this set is narrow.

More technically, the reader's attention is drawn to the fact that in the variables $(x, y, z)$ the set $\mathcal{A}^{\text{ext}}$ is in fact a subset of $\mathcal{C}^{\text{ext}}$ (and not the whole of $\mathcal{C}^{\text{ext}}$), because the constraint $\mu_1 \le \mu_2 \le \mu_3$ imposes $P_1(\mu) \le (P_2(\mu)/P_1(\mu)) \le (P_3(\mu)/P_2(\mu))$, and therefore $y \ge ax^2$ (dashed black line in figures 1, 2, 3) as well as $z \ge by^2$, or equivalently $y \le \sqrt{z/b}$ (dashed red or green lines in figures 1, 2, 3), for some constants $a$ and $b$. We do not dwell on this technical point, since it is anyhow a common point of all relevant descriptions of the sets of $\mu_i$'s such that $F(\lambda_1, \lambda_2, \lambda_3) \ge F(\mu_1, \mu_2, \mu_3)$ once we reduce, as it is natural, the analysis to $\mu_i$'s and $\lambda_i$'s that are ordered according to $\lambda_1 \le \lambda_2 \le \lambda_3$ and $\mu_1 \le \mu_2 \le \mu_3$.

Illustrations are provided in figures 1, 2, 3.

<figure><img src="image_7.png" /></figure>
<figure><img src="image_8.png" /><figcaption>Figure 1: Case $(\lambda_1, \lambda_2, \lambda_3) = (1, 2, 3)$ - Representation of the sets $\mathcal{A}$ and $\mathcal{A}^{\text{ext}}$ in the variables $(x, y, z)$ given by the change of variables (21), inside the particular plane $z = 1$.</figcaption></figure>

The region inside the black cube (and above the dashed black line due to the constraint $\mu_1 \le \mu_1$) represents the set $\mathcal{A}$ that is covered by the analysis in [3] (recall we fixed $z = 1$).

The region inside the red lines (and below the dashed red line due to the constraint $\mu_2 \le \mu_3$) represents the set $\mathcal{A}^{\text{ext}} \setminus \mathcal{A}$ that is covered by our analysis.

<figure><img src="image_9.png" /><figcaption>Figure 2: Similar representation of $\mathcal{A}$ and $\mathcal{A}^{\text{ext}}$, in the plane $z = 0.9$ (red) and $z = 0.5$ (green), when $(\lambda_1, \lambda_2, \lambda_3) = (1, 2, 3)$.</figcaption></figure>

<figure><img src="image_10.png" /><figcaption>Figure 3: Similar representation of $\mathcal{A}$ and $\mathcal{A}^{\text{ext}}$, in the plane $z = 1$, when $(\lambda_1, \lambda_2, \lambda_3) = (1, 2, 2.3)$ (red), $(\lambda_1, \lambda_2, \lambda_3) = (1, 2, 4)$ (green)</figcaption></figure>

**Proof of Theorem 14.** Our assumptions may be summarized as

$$ \lambda_1 \le \lambda_2 \le \lambda_3, \quad \mu_1 \le \mu_2 \le \mu_3 $$
and

$$P_1(\lambda) > P_1(\mu), \quad P_3(\lambda) \geq P_3(\mu), \quad P_2(\lambda) < P_2(\mu) \leq P_2(\lambda) \left( \frac{P_1(\lambda)}{P_1(\mu)} \right)^{\theta_0(\lambda)}.$$

By continuity of all involved functions, it is enough to prove the Theorem in the case when

$$0 < \lambda_1 < \lambda_2 < \lambda_3, \quad 0 < \mu_1 < \mu_2 < \mu_3.$$

(If $\lambda_1 = \lambda_2$, use the Theorem for the values $\lambda_1$ and $\lambda_2 + \varepsilon$ and let $\varepsilon \to 0$, and similarly if $\lambda_3 = \lambda_2$). Finally, and due to the fact that the function $F$ decays in the variable $P_3(\mu)$ in the domain $\mu_1 \le \mu_2 \le \mu_3$ by Corollary 11, we may always assume, without loss of generality, that

$$P_3(\mu) = P_3(\lambda).$$

Indeed, if $P_3(\mu) \le P_3(\lambda)$, there is an $\alpha_3 \ge 0$ such that $P_3(\mu) = P_3(\lambda) e^{-\alpha_3}$, and we may always define $(\tilde{\mu}_1, \tilde{\mu}_2, \tilde{\mu}_3)$ such that $P_1(\tilde{\mu}) = P_1(\mu)$, $P_2(\tilde{\mu}) = P_2(\mu)$, and $P_3(\tilde{\mu}) = P_3(\lambda) = P_3(\mu) e^{+\alpha_3}$. This corresponds to setting $(\tilde{\mu}_1, \tilde{\mu}_2, \tilde{\mu}_3) = (\mu_1, \mu_2, \mu_3 e^{+\alpha_3})$ and we indeed have the ordering $\tilde{\mu}_1 \le \tilde{\mu}_2 \le \tilde{\mu}_3$ as necessary to apply Corollary 11. This gives $F(\mu_1, \mu_2, \mu_3)(x) \ge F(\tilde{\mu}_1, \tilde{\mu}_2, \tilde{\mu}_3)(x)$ for all $x$, and it is enough to prove the Theorem when the $\mu_i$'s are replaced by the $\tilde{\mu}_i$'s.

All in all, we thus restrict ourselves to proving the Theorem under the assumptions

$$0 < \lambda_1 < \lambda_2 < \lambda_3, \quad 0 < \mu_1 < \mu_2 < \mu_3$$

and

$$P_1(\lambda) > P_1(\mu), \quad P_3(\lambda) = P_3(\mu), \quad P_2(\lambda) < P_2(\mu) \le P_2(\lambda) \left( \frac{P_1(\lambda)}{P_1(\mu)} \right)^{\theta_0(\lambda)}.$$

Let us come to the proof.

Under our assumptions, we may define $\alpha > 0$ and $\beta > 0$ such that

$$P_1(\mu) = P_1(\lambda) e^{-\alpha}, \quad P_2(\mu) = P_2(\lambda) e^{+\beta}, \quad \text{and, of course, } P_3(\mu) = P_3(\lambda).$$

With these notations in mind, our assumption $P_2(\mu) \le P_2(\lambda) (P_1(\lambda)/P_1(\mu))^{\theta_0(\lambda)}$ reads

$$\beta \le \alpha \theta_0(\lambda) = \alpha \frac{\lambda_2 - \lambda_1}{\lambda_3 - \lambda_2}. \qquad (22)$$

On top of that, knowing the value of the $P_k(\mu)$'s allows to recover the $\mu_i$'s, namely $\mu_1 = \lambda_1 e^{-\alpha}$, $\mu_2 = \lambda_2 e^{\alpha+\beta}$, $\mu_3 = \lambda_3 e^{-\beta}$. This is due to the general fact that $\mu_k = \lambda_k (P_k(\mu)/P_k(\lambda)) (P_{k-1}(\lambda)/P_{k-1}(\mu))$.

This being set we now introduce the functions

$$\lambda_1(t) = \lambda_1 e^{-\alpha t}, \quad \lambda_2(t) = \lambda_2 e^{(\alpha+\beta)t}, \quad \text{and} \quad \lambda_3(t) = \lambda_3 e^{-\beta t},$$

so that $(\lambda_1(0), \lambda_2(0), \lambda_3(0)) = (\lambda_1, \lambda_2, \lambda_3)$ and $(\lambda_1(1), \lambda_2(1), \lambda_3(1)) = (\mu_1, \mu_2, \mu_3)$. Note in passing that we also have $\lambda_1(t) < \lambda_2(t) < \lambda_3(t)$ for any $t \in [0, 1]$. Indeed, the function $\lambda_2(t) - \lambda_1(t)$ clearly increases with $t$, while satisfying $\lambda_2(0) - \lambda_1(0) = \lambda_2 - \lambda_1 > 0$, and the function $\lambda_3(t) - \lambda_2(t)$ clearly decreases with $t$, while satisfying $\lambda_3(1) - \lambda_2(1) = \mu_3 - \mu_2 > 0$.

With these definitions at hand, we define accordingly

$$F(t)(x) = F(\lambda_1(t), \lambda_2(t), \lambda_3(t))(x),$$

so that $F(0)(x) = F(\lambda_1, \lambda_2, \lambda_3)(x)$ and $F(1)(x) = F(\mu_1, \mu_2, \mu_3)(x)$. Therefore, in order to prove that $F(\lambda_1, \lambda_2, \lambda_3)(x) \le F(\mu_1, \mu_2, \mu_3)(x)$ for all $x \ge 0$, it suffices to prove that for any $t \in [0, 1]$ and $x \ge 0$ we have

$$\partial_t F(t)(x) \ge 0.$$
We now use Theorem 9 to assert that proving $\partial_t F(t)(x) \ge 0$ boils down to proving

$$Q_t(X) \ge 0, \quad \forall t \in [0, 1], \quad \forall X \in [0, 1],$$

upon setting

$$Q_t(X) = \alpha X^{\lambda_1(t)} - (\alpha + \beta) X^{\lambda_2(t)} + \beta X^{\lambda_3(t)}.$$

We are now bound to study the sign of $Q_t(X)$. To do so we observe that

$$Q_t(X) = X^{\lambda_1(t)} P_t(X)$$

where

$$P_t(X) = \alpha - (\alpha + \beta) X^{\lambda_2(t) - \lambda_1(t)} + \beta X^{\lambda_3(t) - \lambda_1(t)},$$

so that we need proving $P_t(X) \ge 0$ whenever $X \in [0, 1]$ and $t \in [0, 1]$.

We have for any $t \in [0, 1]$

$$P_t(0) = \alpha > 0, \quad \text{and} \quad P_t(1) = 0.$$

Besides, studying the variations of $P_t(X)$ as $X \in [0, 1]$, we observe

$$P'_{t}(X) = \left[ -(\alpha + \beta)(\lambda_{2}(t) - \lambda_{1}(t)) + \beta(\lambda_{3}(t) - \lambda_{1}(t))X^{\lambda_{3}(t) - \lambda_{2}(t)} \right] X^{\lambda_{2}(t) - \lambda_{1}(t) - 1} \\ = R_{t}(X) X^{\lambda_{2}(t) - \lambda_{1}(t) - 1}$$

upon setting

$$R_t(X) := -(\alpha + \beta)(\lambda_2(t) - \lambda_1(t)) + \beta(\lambda_3(t) - \lambda_1(t)) X^{\lambda_3(t) - \lambda_2(t)}.$$

Now, for any $t \in [0, 1]$, the function $R_t(X)$ satisfies $R_t(0) = -(\alpha + \beta)(\lambda_2(t) - \lambda_1(t)) < 0$, and $R_t$ increases as $X \in [0, 1]$. Hence for any $t \in [0, 1]$, the function $R_t$ changes sign at most once as $X \in [0, 1]$. As a consequence, for any $t \in [0, 1]$ the function $P'_t(X)$ changes sign at most once on the interval $X \in [0, 1]$, and $P'_t(X) < 0$ in some $(t$-dependent) vicinity of $X = 0$. This is the key step where we benefit the fact that we work in the case $n = 3$, i.e. our parameters are the three positive numbers $(\lambda_1, \lambda_2, \lambda_3)$. Under these circumstances indeed, $P_t(X)$ is a sum of three terms, so that $P'_t(X)$ is the sum of two terms, and this latter function changes sign at most once.

As a consequence, in our situation, we have eventually proved, for any fixed $t \in [0, 1]$, that

$$Q_t(X) \ge 0 \quad \forall X \in [0, 1] \iff P'_t(1) \le 0. \qquad (23)$$

There remains to prove that $P'_t(1) \le 0$.

To do so, we compute

$$P'_{t}(1) = -(\alpha + \beta) \left[ \lambda_{2} e^{(\alpha+\beta)t} - \lambda_{1} e^{-\alpha t} \right] + \beta \left[ \lambda_{3} e^{-\beta t} - \lambda_{1} e^{-\alpha t} \right].$$

Hence, when $t = 0$, we obtain

$$\begin{aligned} P'_{t}(1)|_{t=0} &= -(\alpha + \beta)(\lambda_{2} - \lambda_{1}) + \beta(\lambda_{3} - \lambda_{1}) \\ &= -\alpha(\lambda_{2} - \lambda_{1}) + \beta(\lambda_{3} - \lambda_{2}) \\ &\le -\alpha(\lambda_{2} - \lambda_{1}) + \alpha \frac{\lambda_{2} - \lambda_{1}}{\lambda_{3} - \lambda_{2}}(\lambda_{3} - \lambda_{2}) = 0, \end{aligned}$$

due to our assumption on the ratio $P_2(\mu)/P_2(\lambda)$, used in the form of estimate (22). This proves $P'_t(1) \le 0$ when $t = 0$.

To treat the other values of $t$, we may write, on the other hand

$$P'_{t}(1) = \lambda_{3} \beta e^{-\beta t} - \lambda_{2} (\alpha + \beta) e^{(\alpha+\beta)t} + \lambda_{1} \alpha e^{-\alpha t}.$$
Hence, in the case $\alpha \ge \beta$, we may observe that

$$P'_{t}(1) = e^{-\beta t} S(t)$$

upon defining

$$S(t) := \lambda_3 \beta - \lambda_2 (\alpha + \beta) e^{(\alpha+2\beta)t} + \lambda_1 \alpha e^{-(\alpha-\beta)t}.$$

There remains to observe that $S(0) \le 0$ (since $P'_{t}(1) \le 0$ when $t = 0$) and that $t \mapsto S(t)$ is clearly decaying (due to our assumption $\alpha \ge \beta$), to deduce that $S(t) \le 0$ for all $t \ge 0$, hence that $P'_{t}(1) \le 0$ for any $t \ge 0$. Similarly, in the case $\alpha < \beta$, we may observe that

$$P'_{t}(1) = e^{-\alpha t} \tilde{S}(t)$$

upon defining

$$\tilde{S}(t) := \lambda_3 \beta e^{-(\beta-\alpha)t} - \lambda_2 (\alpha + \beta) e^{(\alpha+2\beta)t} + \lambda_1 \alpha.$$

There remains to observe that $\tilde{S}(0) \le 0$ (since $P'_{t}(1) \le 0$ when $t = 0$) and that $t \mapsto \tilde{S}(t)$ is clearly decaying, (due to our assumption $\alpha < \beta$), to deduce as before $P'_{t}(1) \le 0$ for any $t \ge 0$.

Summarizing, we have proved that our assumption $\beta \le \theta_0(\lambda) \alpha$ implies in any circumstance

$$P'_{t}(1) \le 0, \quad \forall t \in [0, 1].$$

which in turn implies (see (23))

$$Q_t(X) \le 0, \quad \forall t \in [0, 1], \quad \forall X \in [0, 1].$$

We deduce that $F(\lambda_1, \lambda_2, \lambda_3)(x) \le F(\mu_1, \mu_2, \mu_3)(x)$ for any $x \ge 0$ as claimed.

The Theorem is proved.

<figure><img src="image_11.png" /></figure>

# 5 A geometric approach to a non trivial extension of the sufficient condition in [3] - general case

In the general case when $n \ge 3$, extending the sufficient condition in [3] means examining the situation when

$$P_1(\lambda) \ge P_1(\mu), \quad P_n(\lambda) \ge P_n(\mu)$$

yet

$$P_k(\lambda) < P_k(\mu) \quad \text{for some indices } 1 < k < n.$$

This is what we do now. Our key result and major extension is the following. We show later in this text that our criterion is in some sense optimal.

## Theorem 17 (An extended quantitative criterion when $n \ge 4$)

Take $0 < \lambda_1 \le \dots \le \lambda_n$ and $0 < \mu_1 \le \dots \le \mu_n$. Assume there is an index $k_0$ with $1 < k_0 < n$ such that

$$P_k(\lambda) \ge P_k(\mu), \quad \forall k = 1, \dots, n \quad \text{with } k \ne k_0$$

but

$$P_{k_0-1}(\lambda) > P_{k_0-1}(\mu), \quad P_{k_0}(\lambda) < P_{k_0}(\mu).$$

Assume that, setting

$$\theta_0(\lambda) = \frac{\lambda_{k_0} - \lambda_{k_0-1}}{\lambda_{k_0+1} - \lambda_{k_0}} \ge 0,$$
(with $\theta_0(\lambda) = +\infty$ if $\lambda_{k_0+1} = \lambda_{k_0}$) we have the additional lower bound

$$P_{k_0}(\lambda) \geq P_{k_0}(\mu) \left( \frac{P_{k_0-1}(\mu)}{P_{k_0-1}(\lambda)} \right)^{\theta_0(\lambda)}.$$

Last, assume the technical requirement

$$\lambda_{k_0-2} \leq \frac{P_{k_0-2}(\mu)}{P_{k_0-2}(\lambda)} \mu_{k_0-1}, \quad \frac{P_{k_0+1}(\lambda)}{P_{k_0+1}(\mu)} \mu_{k_0+1} \leq \lambda_{k_0+2}. \qquad (24)$$

Then we have, for any $x \geq 0$, the relation

$$F(\lambda_1, \dots, \lambda_n)(x) \leq F(\mu_1, \dots, \mu_n)(x).$$

In fact, the above Theorem (as well as its version in the case $n = 3$) is basically a consequence of the following more general but more abstract result.

**Theorem 18 (An extended abstract criterion)**

Take $0 < \lambda_1 \leq \dots \leq \lambda_n$ and $0 < \mu_1 \leq \dots \leq \mu_n$. Define the real numbers $\alpha_k$ as

$$P_k(\mu) = P_k(\lambda) e^{-\alpha_k}, \quad \text{for } k = 1, \dots, n.$$

(here the $\alpha_k$'s may be $\leq 0$ or $\geq 0$). Define

$$\lambda_1(t) = \lambda_1 e^{-\alpha_1 t}, \quad \lambda_i(t) = \lambda_i e^{-(\alpha_i - \alpha_{i-1})t}, \quad \text{for } i = 2, \dots, n$$

so that for all $i$ we have $\lambda_1(0) = \lambda_i$ and $\lambda_i(1) = \mu_i$.

Assume that for any $X \in [0, 1]$ and $t \in [0, 1]$ we have

$$\alpha_1 X^{\lambda_1(t)} + (\alpha_2 - \alpha_1) X^{\lambda_2(t)} + \dots + (\alpha_n - \alpha_{n-1}) X^{\lambda_n(t)} \geq 0.$$

Then we have, for any $x \geq 0$, the relation

$$F(\lambda_1, \dots, \lambda_n)(x) \leq F(\mu_1, \dots, \mu_n)(x).$$

**Remark 19 (on Theorem 17)**

Since $P_{k_0-1}(\lambda) \geq P_{k_0-1}(\mu)$, we obviously have

$$\left( \frac{P_{k_0-1}(\mu)}{P_{k_0-1}(\lambda)} \right)^{\theta_0(\lambda)} \leq 1,$$

Hence Theorem 17 is a strict extension of the condition in [3], in the generic case when $\lambda_{k_0+1} > \lambda_{k_0}$ and $P_{k_0-1}(\lambda) > P_{k_0-1}(\mu)$, for which we have

$$\left( \frac{P_{k_0-1}(\mu)}{P_{k_0-1}(\lambda)} \right)^{\theta_0(\lambda)} < 1.$$

Second, concerning the technical requirement (24), it may be rephrased

$$\frac{P_{k_0-2}(\mu)}{P_{k_0-2}(\lambda)} \geq \frac{\lambda_{k_0-2}}{\mu_{k_0-1}}, \quad \frac{P_{k_0+1}(\mu)}{P_{k_0+1}(\lambda)} \geq \frac{\mu_{k_0+1}}{\lambda_{k_0+2}}.$$

In other words, the two ratios $P_{k_0-2}(\mu)/P_{k_0-2}(\lambda)$ and $P_{k_0+1}(\mu)/P_{k_0+1}(\lambda)$, which by assumption are $\leq 1$, need to meet the above additional constraints $P_{k_0-2}(\mu)/P_{k_0-2}(\lambda) \geq \lambda_{k_0-2}/\mu_{k_0-1}$ resp. $P_{k_0+1}(\mu)/P_{k_0+1}(\lambda) \geq \mu_{k_0+1}/\lambda_{k_0+2}$. It may also be rephrased

$$P_{k_0-1}(\mu) \geq \lambda_{k_0-2} P_{k_0-2}(\lambda), \quad P_{k_0}(\mu) \geq \frac{P_{k_0+1}(\lambda)}{\lambda_{k_0+2}},$$
so that, all in all, the present Theorem deals with situations where we have

$$P_{k_0-1}(\lambda) > P_{k_0-1}(\mu) \geq \lambda_{k_0-2} P_{k_0-2}(\lambda), \quad P_{k_0}(\mu) \geq \frac{P_{k_0+1}(\lambda)}{\lambda_{k_0+2}}, \quad P_{k_0}(\mu) > P_{k_0}(\lambda).$$

Last, the reader may note that the additional technical requirement (24) does not appear in the case $n = 3$, see the statement of Theorem 14, as it is natural from the indices involved. In the similar spirit, note that in the special situation when $k_0 = 2$, the requirement (24) becomes simply $\mu_3 P_3(\lambda)/P_3(\mu) \leq \lambda_4$ (only the second constraint in (24) needs to be satisfied), while at the other end of the spectrum, when $k_0 = n - 1$, it becomes $\lambda_{n-3} \leq \mu_{n-2} P_{n-3}(\mu)/P_{n-3}(\lambda)$ (only the first constraint in (24) needs to be satisfied). ■

**Remark 20 (possible variants of Theorem 17)**

Once there is a $k_0$ such that $P_{k_0}(\lambda) < P_{k_0}(\mu)$ and the sufficient criterion in [3] is violated, a whole variety of assumptions on the $P_k(\mu)$'s can clearly be made, other than $P_{k_0}(\lambda) \geq P_{k_0}(\mu) [P_{k_0-1}(\mu)/P_{k_0-1}(\lambda)]^{\theta_0(\lambda)}$, to ensure the ordering $F(\lambda_1, \dots, \lambda_n)(x) \leq F(\mu_1, \dots, \mu_n)(x)$. This is transparent from the actual proof of Theorem 17 we provide below. The point is, our approach allows to consider various situations where the indices $k = 1, \dots, n$ can be split into two subsets A and B for which:

* $P_k(\lambda) \geq P_k(\mu)$ whenever $k \in A$ (in the spirit of the criterion in [3]),

* $P_k(\lambda) < P_k(\mu)$ whenever $k \in B$ (in violation of the criterion in [3]),

* the ratios $P_k(\lambda)/P_k(\mu)$ whenever $k \in B$ (all being $\leq 1$) are appropriately lower bounded by the ratios $P_k(\mu)/P_k(\lambda)$ whenever $k \in A$ (all being $\leq 1$), according to a criterion that needs to be carefully designed by hand.

For instance, in Theorem 17, we have $A = \{1, 2, \dots, k_0 - 1, k_0 + 1, \dots, n\}$, while $B = \{k_0\}$ and the constraint is $P_{k_0}(\lambda)/P_{k_0}(\mu) \geq [P_{k_0-1}(\mu)/P_{k_0-1}(\lambda)]^{\theta_0(\lambda)}$.

Technically, the to-be-designed criterion, depending on the subsets A and B, is to be deduced from the analysis of the sign the linear combination $\alpha_1 X^{\lambda_1(t)} + (\alpha_2 - \alpha_1) X^{\lambda_2(t)} + \dots + (\alpha_n - \alpha_{n-1}) X^{\lambda_n(t)}$ appearing in the general Theorem 18, for $X \in [0, 1]$ and $t \in [0, 1]$. We are therefore bound to analyze the sign of such general linear combinations of powers of X for $X \in [0, 1]$. Even for a fixed value of t, this task may be extremely tedious, if not impossible, in general.

The fact that Theorem 17 is easily formulated, and proved, basically relies on the observation that in the present situation, the relevant linear combination only contains three terms. We refer to the proof made in the case $n = 3$ on this technical point. ■

**Proof of Theorem 18.**

With the notation of the Theorem, we set $F(t)(x) = F(\lambda_1(t), \dots, \lambda_n(t))(x)$. By Theorem 9 the variations of $F$ as $t \in [0, 1]$ are governed by the sign of the linear combination $\alpha_1 X^{\lambda_1(t)} + (\alpha_2 - \alpha_1) X^{\lambda_2(t)} + \dots + (\alpha_n - \alpha_{n-1}) X^{\lambda_n(t)}$. The Theorem is proved. ■

**Proof of Theorem 17.**

The proof is a direct consequence of Theorem 18 and follows the lines of the proof of Theorem 14.

First, we need to clean up the situation. In fact, the main point lies in establishing that it is enough to prove the Theorem in the reference situation when

$$P_k(\lambda) = P_k(\mu) \quad \forall k \neq k_0 - 1, k \neq k_0, \text{ while } P_{k_0-1}(\lambda) > P_{k_0-1}(\mu) \text{ and } P_{k_0}(\lambda) < P_{k_0}(\mu).$$

This is where the technical assumption (24) is involved. Indeed, our assumption $P_k(\lambda) \geq P_k(\mu)$ for all $k \neq k_0 - 1$ and $k \neq k_0$ implies the existence of $\alpha_k \geq 0$ ($i = 1, \dots, n, i \neq k_0 - 1, i \neq k_0$) such that

$$P_k(\mu) = P_k(\lambda) e^{-\alpha_k} \quad \forall k \neq k_0 - 1, k \neq k_0,$$
from which we deduce, expressing the $\mu_i$'s and $\lambda_i$'s ($i \neq k_0 - 1, i \neq k_0$) in terms of the $P_k(\mu)$'s and $P_k(\lambda)$'s ($k \neq k_0 - 1, k \neq k_0$), the relations

$$ 
\begin{aligned}
\mu_1 &= \lambda_1 e^{-\alpha_1}, & \mu_2 &= \lambda_2 e^{-\alpha_2+\alpha_1}, & \dots, & \mu_{k_0-2} &= \lambda_{k_0-2} e^{-\alpha_{k_0-2}+\alpha_{k_0-3}}, \\
\mu_{k_0+1} &= \frac{P_{k_0+1}(\lambda)}{P_{k_0}(\lambda)} e^{-\alpha_{k_0+1}}, & \mu_{k_0+2} &= \lambda_{k_0+2} e^{-\alpha_{k_0+2}+\alpha_{k_0+1}}, & \dots, & \mu_n &= \lambda_n e^{-\alpha_n+\alpha_{n-1}},
\end{aligned}
$$

while

$$ \mu_{k_0-1} = \frac{P_{k_0-1}(\mu)}{P_{k_0-2}(\lambda)} e^{+\alpha_{k_0-2}}, \quad \mu_{k_0} = \frac{P_{k_0}(\mu)}{P_{k_0-1}(\mu)}. $$

Replacing now the $\alpha_k$'s ($k \neq k_0 - 1, k \neq k_0$) by 0 in the above formula leads us to defining the intermediate point

$$ (\tilde{\mu}_1, \dots, \tilde{\mu}_n) = (\lambda_1, \dots, \lambda_{k_0-2}, \frac{P_{k_0-1}(\mu)}{P_{k_0-2}(\lambda)}, \mu_{k_0}, \frac{P_{k_0+1}(\lambda)}{P_{k_0}(\mu)}, \lambda_{k_0+2}, \dots, \lambda_n). $$

This point satisfies by construction $P_k(\tilde{\mu}) \ge P_k(\mu)$ for all $k$ since $P_k(\tilde{\mu}) = P_k(\lambda)$ whenever $k \le k_0 - 2$, and $P_{k_0-1}(\tilde{\mu}) = P_{k_0-1}(\mu)$, together with $P_{k_0}(\tilde{\mu}) = P_{k_0}(\mu)$, and finally $P_k(\tilde{\mu}) = P_k(\lambda)$ whenever $k \ge k_0 + 1$. Now, the technical assumption (24) ensures that we have the ordering

$$ \tilde{\mu}_1 \le \dots \le \tilde{\mu}_{k_0-2} \le \tilde{\mu}_{k_0-1} \le \tilde{\mu}_{k_0} \le \tilde{\mu}_{k_0+1} \le \tilde{\mu}_{k_0+2} \le \dots \le \tilde{\mu}_n. $$

Indeed, all these inequalities, when written in the form $\tilde{\mu}_i \le \tilde{\mu}_{i+1}$, are obvious except in the cases $i = k_0 - 2$, $i = k_0 - 1$, $i = k_0$, and $i = k_0 + 1$. In the cases $i = k_0 - 2$, $i = k_0 - 1$ we argue that

$$ \lambda_{k_0-2} \le \frac{P_{k_0-1}(\mu)}{P_{k_0-2}(\lambda)} = \underbrace{\frac{P_{k_0-2}(\mu)}{P_{k_0-2}(\lambda)}}_{\le 1} \mu_{k_0-1} \le \mu_{k_0}, $$

where the first inequality uses the first assumption in (24). In the cases $i = k_0$, $i = k_0 + 1$ we argue that

$$ \mu_{k_0} \le \frac{P_{k_0+1}(\lambda)}{P_{k_0}(\mu)} = \underbrace{\frac{P_{k_0+1}(\lambda)}{P_{k_0+1}(\mu)}}_{\ge 1} \mu_{k_0+1} \le \lambda_{k_0+2} $$

where the second inequality uses the second assumption in (24). Thus, due to the correct ordering of the $\tilde{\mu}_i$'s, we are in position to use that $P_k(\tilde{\mu}) \ge P_k(\mu)$ for all $k$, to deduce $F(\mu_1, \dots, \mu_n)(x) \ge F(\tilde{\mu}_1, \dots, \tilde{\mu}_n)(x)$ for all $x$ and it is indeed enough to prove the Theorem with the $\mu_i$'s replaced by the $\tilde{\mu}_i$'s, i.e. in the situation when $P_k(\tilde{\mu}) = P_k(\lambda)$ for all $k \neq k_0 - 1$ and $k \neq k_0$.

With these reductions at hand, and dropping the tildas, we may assume there is an $\alpha > 0$ and a $\beta > 0$ such that

$$ P_{k_0-1}(\mu) = P_{k_0-1}(\lambda) e^{-\alpha}, \quad P_{k_0}(\mu) = P_{k_0}(\lambda) e^{+\beta}, \quad P_k(\mu) = P_k(\lambda) \text{ for all other values of } k. $$

In that situation we recover the relations

$$ 
\begin{aligned}
\mu_1 &= \lambda_1, \dots, \mu_{k_0-2} = \lambda_{k_0-2}, \\
\mu_{k_0-1} &= \lambda_{k_0-1} e^{-\alpha}, \quad \mu_{k_0} = \lambda_{k_0} e^{+(\alpha+\beta)}, \quad \mu_{k_0+1} = \lambda_{k_0+1} e^{-\beta}, \\
\mu_{k_0+2} &= \lambda_{k_0+2}, \dots, \mu_n = \lambda_n.
\end{aligned}
$$

As in the case $n = 3$, and as in Theorem 18, we now draw the curve $(\lambda_1(t), \dots, \lambda_n(t))$ which joins $(\lambda_1, \dots, \lambda_n)$ at $t = 0$ and $(\mu_1, \dots, \mu_n)$ at $t = 1$ through the formula

$$ 
\begin{aligned}
\lambda_i(t) &= \lambda_i \text{ for } i \neq k_0 - 1, k_0, k_0 + 1, \\
\lambda_{k_0-1}(t) &= \lambda_{k_0-1} e^{-\alpha t}, \quad \lambda_{k_0}(t) = \lambda_{k_0} e^{+(\alpha+\beta)t}, \quad \lambda_{k_0+1}(t) = \lambda_{k_0+1} e^{-\beta t}.
\end{aligned}
$$
The rest of the proof is now as in the case $n = 3$, namely the assumption

$$P_{k_0}(\lambda) \geq P_{k_0}(\mu) \left( \frac{P_{k_0-1}(\mu)}{P_{k_0-1}(\lambda)} \right)^{\theta_0(\lambda)},$$

which also reads

$$\beta \leq \theta_0(\lambda) \alpha = \frac{\lambda_{k_0} - \lambda_{k_0-1}}{\lambda_{k_0+1} - \lambda_{k_0}} \alpha$$

ensures that

$$\alpha X^{\lambda_{k_0-1}(t)} - (\alpha + \beta) X^{\lambda_{k_0}(t)} + \beta X^{\lambda_{k_0+1}(t)} \geq 0, \quad \forall X \in [0, 1], \quad \forall t \in [0, 1].$$

We deduce that $t \mapsto F(\lambda_1(t), \dots, \lambda_n(t))(x)$ decays as $t \in [0, 1]$, and the result follows. ■

We end up this section by illustrating a possible variant of Theorem 17, in the spirit of Remark 20.

**Theorem 21 (An example of a variant of Theorem 17)**

Take $0 < \lambda_1 \le \dots \le \lambda_n$ and $0 < \mu_1 \le \dots \le \mu_n$. Assume there are three indices $i_0, j_0, k_0$ with $1 < i_0 < j_0 < k_0 < n$ such that

$$\begin{align*}
P_k(\lambda) &= P_k(\mu) \quad \text{whenever } 1 \le k \le i_0 - 1, \\
P_k(\lambda) &< P_k(\mu) \quad \text{whenever } i_0 \le k \le j_0 - 1, \\
P_k(\lambda) &> P_k(\mu) \quad \text{whenever } j_0 \le k \le k_0 - 1, \\
P_k(\lambda) &= P_k(\mu) \quad \text{whenever } k_0 \le k \le n.
\end{align*}$$

Assume moreover that

$$\begin{align*}
\frac{P_k(\mu)}{P_k(\lambda)} &= \frac{P_{i_0}(\mu)}{P_{i_0}(\lambda)} \quad \text{whenever } i_0 \le k \le j_0 - 1, \\
\frac{P_k(\mu)}{P_k(\lambda)} &= \frac{P_{j_0}(\mu)}{P_{j_0}(\lambda)} \quad \text{whenever } j_0 \le k \le k_0 - 1.
\end{align*}$$

Setting

$$\theta_0(\lambda) = \frac{\lambda_{j_0} - \lambda_{i_0}}{\lambda_{k_0} - \lambda_{j_0}} \geq 0,$$

(with $\theta_0(\lambda) = +\infty$ if $\lambda_{k_0} = \lambda_{j_0}$), assume we have the additional lower bound

$$P_{j_0}(\lambda) \geq P_{j_0}(\mu) \left( \frac{P_{i_0}(\mu)}{P_{i_0}(\lambda)} \right)^{\theta_0(\lambda)}.$$

Then we have, for any $x \ge 0$, the relation

$$F(\lambda_1, \dots, \lambda_n)(x) \le F(\mu_1, \dots, \mu_n)(x).$$

**Remark 22** Obviously, many variants of this Theorem may be formulated, allowing for instance $P_k(\lambda) \ge P_k(\mu)$ for $1 \le k \le i_0 - 1$ and for $k_0 \le k \le n$. We do not dwell on this very technical aspect. ■

**Proof of Theorem 21.**

The proof is a simplified version of that of Theorem 17. Our assumptions imply there is an $\alpha > 0$ and a $\beta > 0$ such that

$$\begin{align*}
P_k(\mu) &= P_k(\lambda) e^{-\alpha} \quad \text{whenever } i_0 \le k \le j_0 - 1, \\
P_k(\mu) &= P_k(\lambda) e^{+\beta} \quad \text{whenever } j_0 \le k \le k_0 - 1, \\
P_k(\mu) &= P_k(\lambda) \quad \text{for all other values of } k.
\end{align*}$$
Hence we recover the relations

$$ 
\begin{aligned}
\mu_{i_0} &= \lambda_{i_0} e^{-\alpha}, \quad \mu_{j_0} = \lambda_{j_0} e^{+(\alpha+\beta)}, \quad \mu_{k_0} = \lambda_{k_0} e^{-\beta}, \\
\mu_i &= \lambda_i \text{ for all other values of } i.
\end{aligned}
$$

As in the case $n = 3$, and as in Theorem 17, we now draw the curve $(\lambda_1(t), \dots, \lambda_n(t))$ which joins $(\lambda_1, \dots, \lambda_n)$ at $t = 0$ and $(\mu_1, \dots, \mu_n)$ at $t = 1$ through the formula

$$ 
\begin{aligned}
\lambda_i(t) &= \lambda_i \text{ for } i \neq i_0, j_0, k_0, \\
\lambda_{i_0}(t) &= \lambda_{i_0} e^{-\alpha t}, \quad \lambda_{j_0}(t) = \lambda_{j_0} e^{+(\alpha+\beta)t}, \quad \lambda_{k_0}(t) = \lambda_{k_0} e^{-\beta t}.
\end{aligned}
$$

The rest of the proof is now as in the case $n = 3$, namely the assumption

$$ \frac{P_{j_0}(\lambda)}{P_{j_0}(\mu)} \geq \left( \frac{P_{i_0}(\mu)}{P_{i_0}(\lambda)} \right)^{\theta_0(\lambda)} $$

also reads

$$ \beta \leq \theta_0(\lambda) \alpha \text{ with } \theta_0(\lambda) = \frac{\lambda_{j_0} - \lambda_{i_0}}{\lambda_{k_0} - \lambda_{j_0}}. $$

(since, by definition, $e^{\alpha} = P_{i_0}(\lambda)/P_{i_0}(\mu)$ and $e^{\beta} = P_{j_0}(\mu)/P_{j_0}(\lambda)$). This ensures that

$$ \alpha X^{\lambda_{i_0}(t)} - (\alpha + \beta) X^{\lambda_{j_0}(t)} + \beta X^{\lambda_{k_0}(t)} \geq 0, \quad \forall X \in [0, 1], \quad \forall t \in [0, 1]. $$

We deduce that $t \mapsto F(\lambda_1(t), \dots, \lambda_n(t))(x)$ decays as $t \in [0, 1]$, and the result follows. ■

# 6 Optimality of our conditions - case $n = 3$

In this section we prove that Theorem 14 is optimal in the following sense

**Theorem 23 (optimality of the extended criterion derived in Theorem 14)**

*Fix $0 < \lambda_1 < \lambda_2 < \lambda_3$. Take some*

$$ \theta > \theta_0(\lambda) = \frac{\lambda_2 - \lambda_1}{\lambda_3 - \lambda_2}. $$

*Define the set*

$$ \Omega_{\theta}(\lambda) = \left\{ \begin{array}{l} (\mu_1, \mu_2, \mu_3) \text{ such that } 0 < \mu_1 < \mu_2 < \mu_3, P_1(\lambda) > P_1(\mu), P_3(\lambda) \geq P_3(\mu), \text{ but } P_2(\lambda) < P_2(\mu), \\ \text{together with } P_2(\lambda) \leq P_2(\mu) \left(\frac{P_1(\mu)}{P_1(\lambda)}\right)^{\theta} \end{array} \right\} $$

Then, there is a small parameter $\varepsilon_0$, and an associated vicinity $V_{\varepsilon_0}$, of size $\varepsilon_0$ of the point $(\lambda_1, \lambda_2, \lambda_3)$ such that

$$ \forall (\mu_1, \mu_2, \mu_3) \in \Omega_{\theta}(\lambda) \cap V_{\varepsilon_0}, \quad \forall x \in \left[\frac{\varepsilon_0}{2}, \varepsilon_0\right], \quad \text{we have } F(\lambda_1, \lambda_2, \lambda_3)(x) > F(\mu_1, \mu_2, \mu_3)(x). $$

Technically, the vicinity $V_{\varepsilon_0}$, may be defined, for instance, as follows. Pick up a parameter $\delta > 0$ such that $\delta \leq (\lambda_3 - \lambda_2)(\theta - \theta_0(\lambda))/20$. Pick up an arbitrary small parameter $0 < \eta < 1/2$. Then we may set

$$ V_{\varepsilon_0} = \left\{ (\mu_1, \mu_2, \mu_3) \text{ such that } |\mu_i - \lambda_i| \leq \varepsilon_0 \ (i = 1, 2, 3) \text{ and } |\mu_1 - \lambda_1| \geq \eta \varepsilon_0 \\ \text{and } \frac{P_3(\mu)}{P_3(\lambda)} \geq \left( \frac{P_1(\mu)}{P_1(\lambda)} \right)^{\delta \varepsilon_0} \right\}. $$
**Remark 24** Note that the set $\Omega_{\theta}(\lambda)$ is (almost) exactly the set of parameters $\mu_i$'s which violate our criterion $P_2(\lambda) \ge P_2(\mu)$ ($P_1(\mu)/P_1(\lambda)^{\theta_0(\lambda)}$ (the "almost" part of the statement stemming from the constraint $\theta > \theta_0(\lambda)$, where $\theta$ may be arbitrary close to $\theta_0(\lambda)$).

However, the reader's attention is drawn to the fact that the vicinity $V_{\varepsilon_0}$ is not a simple ball of radius $\varepsilon_0$ around $(\lambda_1, \lambda_2, \lambda_3)$. The requirement is more precise. First, the additional constraint $|\mu_1 - \lambda_1| \ge \eta\varepsilon_0$ requires that the distance between $\mu_1$ and $\lambda_1$ is not only $\le \varepsilon_0$, but actually of order $\varepsilon_0$ in that $\eta\varepsilon_0 \le |\mu_1 - \lambda_1| \le \varepsilon_0$, so $V_{\varepsilon_0}$ is an annulus in the variable $\mu_1$. Second, and more importantly, the vicinity $V_{\varepsilon_0}$ is "flat" in the direction of the parameter $P_3(\mu)$. Let us make this comment clear. All the above sets are in fact best expressed in the variable $(x = P_1(\mu)/P_1(\lambda), y = P_2(\mu)/P_2(\lambda), z = P_3(\mu)/P_3(\lambda))$ introduced in Remark 16. In these variables, the set of "authorized" variables, for which we can ensure $F(\lambda_1, \lambda_2, \lambda_3) < F(\mu_1, \mu_2, \mu_3)$, reads roughly

$$ \{0 \le x \le 1, \quad y \le \frac{1}{x^{\theta_0(\lambda)}}, \quad 0 \le z \le 1\}, $$

while the set $\Omega_{\theta}(\lambda)$ is essentially

$$ \{0 \le x \le 1, \quad y \ge \frac{1}{x^{\theta}}, \quad 0 \le z \le 1\}, $$

with $\theta > \theta_0(\lambda)$. Lastly, the vicinity $V_{\varepsilon_0}$ is a vicinity of $(x = 1, y = 1, z = 1)$ inside (a subset of) the set

$$ \{x^{\delta\varepsilon_0} \le z \le 1\}. $$

The point we wish to stress is that the vicinity $V_{\varepsilon_0}$ is "flat" in the direction $z$, being an open angle with aperture $\delta\varepsilon_0$ in the $(x, z)$ variables. (thanks to the expansion $x^{\delta\varepsilon_0} = 1 + \delta\varepsilon_0(1-x) + o(1-x)$ as $x \to 1$). Figure 24 below shows the shape of $\Omega_{\theta}(\lambda) \cap V_{\varepsilon_0}$ in the $(x, y, z)$ variables, a vicinity of $(1, 1, 1)$.

<figure><img src="image_12.png" /><figcaption>Figure 4: Shape of the set $\Omega_{\theta}(\lambda) \cap V_{\varepsilon_0}$ in the $(x, y, z)$ variables.</figcaption></figure>

This set is roughly a dihedron limited by the three curves $(x, y, z) = (1, y, 1)$ inside the plane $z = 1$ (green), $(x, y, z) = (x, 1/x^{\theta}, 1)$ inside the plane $z = 1$ (blue), and $(x, y, z) = (x, 1/x^{\theta}, x^{\delta\varepsilon_0})$ inside the plane $y = 1/x^{\theta}$ (dashed red).

The angle between the blue and dashed red curves is roughly $\delta\varepsilon_0$ and the angle between the blue and green curves is roughly $\theta$.

The whole dihedron limited by the two thick black lines $(x, y, z) = (x, 1, 1)$, and $(x, y, z) = (1, 1, z)$, and the blue line is the extended "authorized" values of $(x, y, z)$ studied in Theorem 14, while the dihedron limited by the two thick black lines and the fine black line $(x, y, z) = (1, y, 1)$ is the domain studied in [3].
**Proof.**

**First Step: preparation of the proof.**

Fix a $\theta$ and a $\delta$ as in the Theorem, as well as an $\varepsilon_0 > 0$ whose value is to be made precise later.

Pick up a $(\mu_1, \mu_2, \mu_3) \in \Omega_{\theta}(\lambda) \cap V_{\varepsilon_0}$.

We first exploit the information $(\mu_1, \mu_2, \mu_3) \in \Omega_{\theta}(\lambda)$.

Due to the assumptions on the $P_k(\mu)$'s that characterize the set $\Omega_{\theta}(\lambda)$, we can find $\alpha > 0$, $\beta > 0$ and $\gamma \ge 0$ such that

$$P_1(\mu) = P_1(\lambda) e^{-\alpha}, \quad P_2(\mu) = P_2(\lambda) e^{+\beta}, \quad P_3(\mu) = P_3(\lambda) e^{-\gamma}.$$

Besides, we also have the constraint

$$\beta \ge \alpha \theta, \tag{25}$$

which comes as a consequence of the requirement $P_2(\lambda) \le P_2(\mu) (P_1(\mu)/P_1(\lambda))^{\theta}$. Coming back to the $\mu_i$'s, we have

$$\mu_1 = \lambda_1 e^{-\alpha}, \quad \mu_2 = \lambda_2 e^{\alpha+\beta}, \quad \mu_3 = \lambda_3 e^{-(\beta+\gamma)},$$

and the constraint $0 < \mu_1 < \mu_2 < \mu_3$ imposes $\alpha + 2\beta + \gamma < \log(\lambda_3/\lambda_2)$ because $\mu_2 < \mu_3$ (the constraint $\mu_1 < \mu_2$ is automatically satisfied).

Second, we exploit the additional constraint $(\mu_1, \mu_2, \mu_3) \in V_{\varepsilon_0}$ as in the statement of the Theorem.

Since $|\mu_1 - \lambda_1| = \lambda_1 (1 - e^{-\alpha})$ and similarly for $|\mu_2 - \lambda_2|$ and $|\mu_3 - \lambda_3|$, using the first order Taylor expansion of the exponential in the particular form $|e^y - (1 + y)| \le 2y^2$ whenever $|y| \le 1$ (say), we obtain from the assumption $|\mu_i - \lambda_i| \le \varepsilon_0$ ($i = 1, 2, 3$) the existence of a universal constant $C_0 > 0$ such that

$$\alpha, \beta, \gamma \le C_0 \varepsilon_0.$$

The additional constraint $|\mu_1 - \lambda_1| \ge \eta \varepsilon_0$ also gives a constant $c_0$ such that

$$\alpha \ge c_0 \eta \varepsilon_0.$$

On top of that, the constraint $P_3(\lambda)/P_3(\mu) \le (P_1(\lambda)/P_1(\mu))^{\delta\varepsilon_0}$ imposes

$$\gamma \le \delta \varepsilon_0 \alpha, \tag{26}$$

a technical constraint whose usefulness will become clear later. In particular (26) together with the information $\alpha \le C_0 \varepsilon_0$ imposes that $\gamma$ not only is of order $\varepsilon_0$ but in fact of order $\varepsilon_0^2$, in that

$$\gamma \le C_1(\lambda) \varepsilon_0^2$$

for some $\lambda$ dependent constant $C_1(\lambda) > 0$. Note in passing that the above mentioned constraint $\alpha + 2\beta + \gamma < \log(\lambda_3/\lambda_2)$ is now automatically satisfied provided $\varepsilon_0$ is chosen so as to fulfill the requirement $4C_0\varepsilon_0 < \log(\lambda_3/\lambda_2)$.

With these elements in mind, introduce the path

$$\lambda_1(t) = \lambda_1 e^{-\alpha t}, \quad \lambda_2(t) = \lambda_2 e^{(\alpha+\beta)t}, \quad \lambda_3(t) = \lambda_3 e^{-(\beta+\gamma)t},$$

where $t \in [0, 1]$, as we did in various proofs before. This path joins $(\lambda_1, \lambda_2, \lambda_3)$ (at $t = 0$) and $(\mu_1, \mu_2, \mu_3)$ (at $t = 1$). Introduce as well

$$F(t)(x) = F(\lambda_1(t), \lambda_2(t), \lambda_3(t))(x).$$

We know from Theorem 9 that

$$\frac{d}{dt} F(t)(x) = \left( \prod_{i=1}^{3} \lambda_i(t) \right) \left[ e_{\lambda_1(t), \lambda_2(t), \lambda_3(t)} * \{ \alpha e_{\lambda_1(t)} - (\alpha + \beta) e_{\lambda_2(t)} + (\beta + \gamma) e_{\lambda_3(t)} \} \right] (x),$$
or, in other words

$$ \frac{d}{dt} F(t)(x) = \left( \prod_{i=1}^{3} \lambda_i(t) \right) [e_{\lambda_1(t), \lambda_2(t), \lambda_3(t)} * q_t](x), \quad (27) $$

up to defining

$$ q_t(x) := [\alpha e_{\lambda_1(t)} - (\alpha + \beta) e_{\lambda_2(t)} + (\beta + \gamma) e_{\lambda_3(t)}](x). \quad (28) $$

Therefore, for any given $x > 0$, proving that $F(\lambda_1, \lambda_2, \lambda_3)(x) > F(\mu_1, \mu_2, \mu_3)(x)$ boils down to proving that we have $q_t(x) < 0$ in some strong enough sense.

The reader's attention is drawn here to the following fact. Our Theorem only deals with values of $x$ that are in some $\varepsilon_0$-vicinity of $x = 0$ so our subsequent analysis only deals with such $x$'s. In fact, the bounds presented in the Theorem become false as $x$ leaves some $\varepsilon_0$-vicinity of $x = 0$.

Let us come to more quantitative bounds.

## Second step. Estimating $q_t(x)$ for small values of $x$.

By the standard first order expansion of the exponential function we can find a constant $C_1(\lambda) > 0$ (still denoted $C_1(\lambda)$), only depending on the $\lambda_i$'s, and whose exact value may change from line to line in the computation below), such that, for any $t \in [0, 1]$, $x \in [0, \varepsilon_0]$, and $0 < \alpha, \beta, \gamma < C_0 \varepsilon_0$ (where $\varepsilon_0$ is the smallness parameter we picked up above), we have

$$ \begin{aligned} & [\alpha e_{\lambda_1(t)} - (\alpha + \beta) e_{\lambda_2(t)} + (\beta + \gamma) e_{\lambda_3(t)}](x) \\ &= \alpha e^{-\lambda_1(t)x} - (\alpha + \beta) e^{-\lambda_2(t)x} + (\beta + \gamma) e^{-\lambda_3(t)x} \\ &= \alpha (1 - \lambda_1(t)x + R_1(t, x)) - (\alpha + \beta) (1 - \lambda_2(t)x + R_2(t, x)) + (\beta + \gamma) (1 - \lambda_3(t)x + R_3(t, x)) \\ &= \gamma - x [\alpha \lambda_1(t) - (\alpha + \beta) \lambda_2(t) + (\beta + \gamma) \lambda_3(t)] + R_4(t, x) \\ &= \gamma - x [-\alpha (\lambda_2(t) - \lambda_1(t)) + \beta (\lambda_3(t) - \lambda_2(t)) + \gamma \lambda_3(t)] + R_4(t, x), \end{aligned} $$

for some $(t, x)$ dependent remainders $R_i(t, x)$ for which we can ensure

$$ |R_i(t, x)| \le C_1(\lambda) \varepsilon_0^2 \quad (i = 1, 2, 3), \quad |R_4(t, x)| \le C_1(\lambda) \varepsilon_0^3, $$

for any values of $(t, x, \alpha, \beta, \gamma)$ as before. This is due to the fact that the various arguments $\lambda_i(t)x$ ($i = 1, 2, 3$) involved in the exponentials all belong to some fixed vicinity of the origin (of size $O(\varepsilon_0)$) thanks to the assumptions $|x| \le \varepsilon_0$, $t \in [0, 1]$ and $\alpha, \beta, \gamma \le C_0 \varepsilon_0$. On top of that, we may write the zero-th order expansion

$$ \lambda_1(t) = \lambda_1 e^{-\alpha t} = \lambda_1(1 + R_5(t)) $$

for some remainder satisfying $|R_5(t)| \le C_0 \varepsilon_0$, and similarly for $\lambda_2(t)$ and $\lambda_3(t)$. This gives

$$ \begin{aligned} & [\alpha e_{\lambda_1(t)} - (\alpha + \beta) e_{\lambda_2(t)} + (\beta + \gamma) e_{\lambda_3(t)}](x) \\ &= \gamma - x \underbrace{-[\alpha(\lambda_2 - \lambda_1) + \beta(\lambda_3 - \lambda_2) + \gamma \lambda_3]}_{=: A_{\alpha, \beta, \gamma}} + R_6(t, x), \end{aligned} \quad (29) $$

with $|R_6(t, x)| \le C_1(\lambda) \varepsilon_0^3$. Our assumption on the ratio $P_2(\mu)/P_2(\lambda)$, which imposes $\beta \ge \alpha \theta$ as observed in (25), provides the following useful observation, recalling $\theta_0(\lambda) = (\lambda_2 - \lambda_1)/(\lambda_3 - \lambda_2)$, namely

$$ \begin{aligned} A_{\alpha, \beta, \gamma} &= -\alpha (\lambda_2 - \lambda_1) + \alpha \theta (\lambda_3 - \lambda_2) + (\beta - \alpha \theta) (\lambda_3 - \lambda_2) + \gamma \lambda_3 \\ &= \alpha (\lambda_3 - \lambda_2) [\theta - \theta_0(\lambda)] + (\beta - \alpha \theta) (\lambda_3 - \lambda_2) + \gamma \lambda_3, \end{aligned} $$

so that

$$ A_{\alpha, \beta, \gamma} \ge \alpha (\lambda_3 - \lambda_2) [\theta - \theta_0(\lambda)] > 0. \quad (30) $$
At this level of the analysis we have replaced $\alpha e_{\lambda_1(t)} - (\alpha + \beta)e_{\lambda_2(t)} + (\beta + \gamma)e_{\lambda_3(t)}$ by an asymptotic expansion that is suitable to a sign analysis.

### Third Step. Estimating the variations of F

We start from

$$ \begin{aligned} \frac{d}{dt} F(t)(x) &= \left( \prod_{i=1}^{3} \lambda_i(t) \right) \left[ e_{\lambda_1(t), \lambda_2(t), \lambda_3(t)} * \{ \alpha e_{\lambda_1(t)} - (\alpha + \beta) e_{\lambda_2(t)} + (\beta + \gamma) e_{\lambda_3(t)} \} \right] (x) \\ &= \left( \prod_{i=1}^{3} \lambda_i(t) \right) \int_{0}^{x} \int_{0}^{y_1} \int_{0}^{y_2} \int_{0}^{y_3} e^{-\lambda_1(t)(x-y_1)} e^{-\lambda_2(t)(y_1-y_2)} e^{-\lambda_3(t)(y_2-y_3)} \\ & \qquad [\gamma - y_4 A_{\alpha, \beta, \gamma} + R_6(t, y_4)] dy_1 dy_2 dy_3 dy_4, \end{aligned} $$

where the second line uses (29). Using the first order expansion of the exponential we may use above the fact that $\alpha, \beta, \gamma \le C_0 \varepsilon_0$ to deduce that, whenever $t \in [0, 1]$ and $x \in [0, \varepsilon_0]$, we have

$$ e^{-\lambda_1(t)(x-y_1)} = 1 + R_7(t, x, y_1) $$

for a remainder term which satisfies $|R_7(t, x, y_1)| \le C_1(\lambda)\varepsilon_0$ whenever $y_1 \in [0, x]$, and similarly for the terms involving $\lambda_2$ and $\lambda_3$. This provides, whenever $0 \le y_4 \le y_3 \le y_2 \le y_1 \le x \le \varepsilon_0$, and using the reinforced information $\gamma \le C_1(\lambda)\varepsilon_0^2$, the relation

$$ \begin{aligned} & e^{-\lambda_1(t)(x-y_1)} e^{-\lambda_2(t)(y_1-y_2)} e^{-\lambda_3(t)(y_2-y_3)} [\gamma - y_4 A_{\alpha, \beta, \gamma} + R_6(t, y_4)] \\ &= \gamma - y_4 A_{\alpha, \beta, \gamma} + R_8(t, y_4, y_3, y_2, y_1, x), \end{aligned} $$

for a remainder term which satisfies $|R_8(t, y_4, y_3, y_2, y_1, x)| \le C_1(\lambda)\varepsilon_0^3$. By explicit integration we obtain

$$ \begin{aligned} \left( \prod_{i=1}^{3} \lambda_i(t) \right)^{-1} \frac{d}{dt} F(t)(x) &= \int_{0}^{x} \int_{0}^{y_1} \int_{0}^{y_2} \int_{0}^{y_3} [\gamma - y_4 A_{\alpha, \beta, \gamma}] dy_1 dy_2 dy_3 dy_4 + R_9(t, x) \\ &= \frac{x^4}{24} \gamma - \frac{x^5}{120} A_{\alpha, \beta, \gamma} + R_9(t, x), \end{aligned} $$

for a remainder term which satisfies $|R_9(t, x)| \le C_1(\lambda)\varepsilon_0^7$. Therefore, we come up with

$$ \left( \prod_{i=1}^{3} \lambda_i(t) \right)^{-1} \frac{d}{dt} F(t)(x) \le \frac{x^4}{24} \left[ \gamma - A_{\alpha, \beta, \gamma} \frac{x}{5} \right] + C_1(\lambda) \varepsilon_0^7, \quad \forall x \in [0, \varepsilon_0]. \quad (31) $$

There remains to prove that the right-hand-side of (31) is negative whenever $\varepsilon_0/2 \le x \le \varepsilon_0$. To do so we write, whenever $x \ge \varepsilon_0/2$, that

$$ \gamma - A_{\alpha, \beta, \gamma} \frac{x}{5} \le \gamma - A_{\alpha, \beta, \gamma} \frac{\varepsilon_0}{10}. $$

Hence, using the technical information $\gamma \le \delta\varepsilon_0\alpha$ derived in (26), together with the upper bound (30) on $A_{\alpha, \beta, \gamma}$ we obtain, whenever $x \ge \varepsilon_0/2$, the upper bound

$$ \begin{aligned} \gamma - A_{\alpha, \beta, \gamma} \frac{x}{5} &\le \delta \varepsilon_0 \alpha - \alpha(\lambda_3 - \lambda_2) [\theta - \theta_0(\lambda)] \frac{\varepsilon_0}{10} \\ &\le \alpha \varepsilon_0 \left[ \delta - \frac{(\lambda_3 - \lambda_2) [\theta - \theta_0(\lambda)]}{10} \right] \\ &\le -\alpha \varepsilon_0 \frac{(\lambda_3 - \lambda_2) [\theta - \theta_0(\lambda)]}{20} \end{aligned} $$
thanks to our technical assumption $0 < \delta \le (\lambda_3 - \lambda_2) [\theta - \theta_0(\lambda)]/20$ on $\delta$.

All in all we have established

$$x \ge \frac{\varepsilon_0}{2} \implies \gamma - A_{\alpha,\beta,\gamma} \frac{x}{5} \le -K \alpha \varepsilon_0 \quad \text{with } K = \frac{(\lambda_3 - \lambda_2) [\theta - \theta_0(\lambda)]}{20}.$$

We are in position to conclude.

**Last step. Conclusion.**

Summarizing, the previous step establishes

$$\left(\prod_{i=1}^{3} \lambda_{i}(t)\right)^{-1} \frac{d}{dt} F(t)(x) \leq -K \alpha \varepsilon_{0} \frac{x^{4}}{24} + C_{1}(\lambda) \varepsilon_{0}^{7}, \quad \text{whenever } \frac{\varepsilon_{0}}{2} \leq x \leq \varepsilon_{0}, \text{ and } t \in [0, 1].$$

Hence

$$\left(\prod_{i=1}^{3} \lambda_{i}(t)\right)^{-1} \frac{d}{dt} F(t)(x) \leq -\varepsilon_{0}^{5}\left(\frac{K}{24} \alpha-C_{1}(\lambda) \varepsilon_{0}^{2}\right) \quad \forall x \in\left[\frac{\varepsilon_{0}}{2}, \varepsilon_{0}\right], \quad \forall t \in[0,1].$$

There remains to choose $\varepsilon_0$ so small as to satisfy

$$\varepsilon_0^2 \le \frac{K \alpha}{48 C_1(\lambda)},$$

which is possible since we have $c_0 \eta \varepsilon_0 \le \alpha \le C_0 \varepsilon_0$ (see the first step of our proof). We recover

$$\left(\prod_{i=1}^{3} \lambda_{i}(t)\right)^{-1} \frac{d}{dt} F(t)(x) \leq -\varepsilon_{0}^{5} \frac{K \alpha}{48} < 0 \quad \forall x \in\left[\frac{\varepsilon_{0}}{2}, \varepsilon_{0}\right], \quad \forall t \in[0,1].$$

Our Theorem is proved.

<figure><img src="image_13.png" /></figure>

# 7 Optimality of our conditions - general case

In this section we prove that Theorems 17 and 18 are optimal in the following sense

**Theorem 25 (optimality of the extended quantitative criterion derived in Theorem 17)**

Fix $0 < \lambda_1 < \dots < \lambda_n$. Assume there is an index $k_0$ with $1 < k_0 < n$ such that

$$P_k(\lambda) \geq P_k(\mu), \quad \forall k = 1, \dots, n \quad \text{with } k \neq k_0$$

but

$$P_{k_0}(\lambda) < P_{k_0}(\mu).$$

Take some

$$\theta > \theta_0(\lambda) = \frac{\lambda_{k_0} - \lambda_{k_0-1}}{\lambda_{k_0+1} - \lambda_{k_0}} > 0.$$

Define the set

$$\Omega_{\theta}(\lambda) = \left\{ \begin{array}{ll} (\mu_1, \dots, \mu_n) & \text{such that } 0 < \mu_1 < \dots < \mu_n, P_k(\lambda) \ge P_k(\mu), \quad \text{for } k \ne k_0, k_0-1, \\ P_{k_0-1}(\lambda) > P_{k_0-1}(\mu), & \text{but } P_{k_0}(\lambda) < P_{k_0}(\mu), \\ \text{together with } P_{k_0}(\lambda) \le P_{k_0}(\mu) \left(\frac{P_{k_0-1}(\mu)}{P_{k_0-1}(\lambda)}\right)^{\theta} & \end{array} \right\}.$$
Then, there is a small parameter $\epsilon_0$, and an associated a vicinity $V_{\epsilon_0}$, of size $\epsilon_0$ of the point $(\lambda_1, \dots, \lambda_n)$ such that

$$ \forall(\mu_1, \dots, \mu_n) \in \Omega_{\theta}(\lambda) \cap V_{\epsilon_0}, \quad \forall x \in \left[\frac{\epsilon_0}{2}, \epsilon_0\right], \quad \text{we have } F(\lambda_1, \dots, \lambda_n)(x) > F(\mu_1, \dots, \mu_n)(x). $$

Technically, the vicinity $V_{\epsilon_0}$, may be defined, for instance, as follows. Pick up a parameter $\delta > 0$ such that $\delta \le (\lambda_{k_0} - \lambda_{k_0-1})(\theta - \theta_0(\lambda))/(4(n+2))$. Pick up an arbitrary small parameter $0 < \eta < 1/2$. Then we may set

$$ V_{\epsilon_0} = \left\{ \begin{array}{ll} (\mu_1, \dots, \mu_n) \text{ such that } |\mu_i - \lambda_i| \le \epsilon_0 \quad (i = 1, \dots, n), & |\mu_{k_0-1} - \lambda_{k_0-1}| \ge \eta \epsilon_0, \\ \\ \text{and } \frac{P_k(\mu)}{P_k(\lambda)} \ge \left( \frac{P_{k_0-1}(\mu)}{P_{k_0-1}(\lambda)} \right)^{\delta \epsilon_0} & \text{for all } k \ne k_0, k_0-1 \end{array} \right\}. $$

**Remark 26** As in the case $n=3$, and up to disregarding the technical requirement (24) that appears in the statement of Theorem 17 (and not here), note that the set $\Omega_{\theta}(\lambda)$ is exactly the set of parameters $(\mu_1, \dots, \mu_n)$ which violate our optimal criterion $P_{k_0}(\lambda) \ge P_{k_0}(\mu) (P_{k_0-1}(\mu)/P_{k_0-1}(\lambda))^{\theta_0(\lambda)}$.

Note also that, as in the case $n=3$, the vicinity $V_{\epsilon_0}$ is not a simple ball of radius $\epsilon_0$ around $(\lambda_1, \dots, \lambda_n)$. Firstly, the difference $|\mu_{k_0-1} - \lambda_{k_0-1}|$ needs to be both $\le \epsilon_0$ and $\ge \eta \epsilon_0$. Secondly, when rephrased in terms of the variables $P_k(\mu)$, this vicinity is somehow "flat" in the directions $k \ne k_0, k_0-1$, due to the constraints $P_k(\mu)/P_k(\lambda) \ge (P_{k_0-1}(\mu)/P_{k_0-1}(\lambda))^{\delta \epsilon_0}$ for $k \ne k_0, k_0-1$.

**Theorem 27 (Optimality of our general abstract criterion)**

Take $0 < \lambda_1 \le \dots \le \lambda_n$. Take three constants $K_1 > 0$, $K_3 > K_2 > 0$, and $K_4 > 0$ with $K_1/K_2 < 1$. Then, there exists an $\epsilon_0 > 0$ such that for any $\epsilon \le \epsilon_0$, the following holds.

Pick up any $0 < \mu_1 \le \dots \le \mu_n$ such that $|\mu_i - \lambda_i| \le \epsilon$. Define the real numbers $\alpha_k$ as

$$ P_k(\mu) = P_k(\lambda) e^{-\alpha_k}, \quad \text{and} \quad k = 1, \dots, n. $$

(here the $\alpha_k$'s may be $\le 0$ or $\ge 0$). The $\alpha_k$'s necessarily are of order $\epsilon$ at most. Define

$$ \lambda_1(t) = \lambda_1 e^{-\alpha_1 t}, \quad \lambda_i(t) = \lambda_i e^{-(\alpha_i - \alpha_{i-1})t} \quad i = 2, \dots, n, $$

so that for all $i$ we have $\lambda_i(0) = \lambda_i$ and $\lambda_i(1) = \mu_i$. Define for any $X \in [0, 1]$ and $t \in [0, 1]$

$$ Q_t(X) = \alpha_1 X^{\lambda_1(t)} + (\alpha_2 - \alpha_1) X^{\lambda_2(t)} + \dots + (\alpha_n - \alpha_{n-1}) X^{\lambda_n(t)}. $$

Assume that

$$ \begin{gathered} Q_t(X) \le a - b(1-X) + R, \quad \text{whenever } X \in [1-\epsilon, 1], \quad t \in [0, 1], \tag{32} \\ \text{with } 0 \le a \le K_1 \epsilon^2, \quad K_2 \epsilon \le b \le K_3 \epsilon, \quad |R| \le K_4 \epsilon^3. \end{gathered} $$

Then, whenever $\epsilon/2 \le x \le \epsilon$, the relation

$$ F(\lambda_1, \dots, \lambda_n)(x) > F(\mu_1, \dots, \mu_n)(x) $$

holds true.

**Remark 28** Theorem 27 proves in some sense that Theorem 18 is locally optimal, in that, if the criterion $Q_t(X) \ge 0$ is violated close to $X=1$, then the relation $F(\lambda_1, \dots, \lambda_n)(x) \le F(\mu_1, \dots, \mu_n)(x)$ is violated close to $x=0$.

Note that our assumption (32) means roughly that $Q_t(X) \approx a - b(1-X)$ close to $X=1$, a function that is $\ge 0$ whenever $X \in [1-(a/b), 1]$ but this function is $< 0$ whenever $X \in [1-(a/b), 1-\epsilon]$, hence in particular whenever $X \in [1-(K_1/K_2)\epsilon, 1-\epsilon]$.

In the more general case when the relation $Q_t(X) \ge 0$ is violated far from $X=1$, it is difficult if not impossible to conclude concerning the sign of $F(\lambda_1, \dots, \lambda_n)(x) - F(\mu_1, \dots, \mu_n)(x)$. This is due to the fact that (see the proof of the present Theorem) the sign of $F(\lambda_1, \dots, \lambda_n)(x) - F(\mu_1, \dots, \mu_n)(x)$ is given by the sign of $e^{\lambda_1(t), \dots, \lambda_n(t)}(x) * Q_t(e^{-x})$, so that the convolution makes it difficult to follow the signs.
**Proof of Theorem 25**

The proof is very similar to the one made in the case $n = 3$. We give here a short version of the proof, insisting only on the points that are specific to the case $n \ge 3$.

**First Step: preparation of the proof.**

Fix a $\theta$ as in the Theorem and pick up a $(\mu_1, \dots, \mu_n)$ in the set $\Omega_\theta(\lambda)$. Due to the assumptions on the $P_k(\mu)$'s that characterize the set $\Omega_\theta(\lambda)$, we can find $\alpha_k \ge 0$ ($k = 1, \dots, n$) such that

$$P_k(\mu) = P_k(\lambda) e^{-\alpha_k} \quad \text{for } k \neq k_0, \quad P_{k_0}(\mu) = P_{k_0}(\lambda) e^{+\alpha_{k_0}}.$$

Besides, we also have

$$\alpha_{k_0-1} > 0, \quad \alpha_{k_0} > 0.$$

On top of that, we have the constraint

$$\alpha_{k_0} \ge \alpha_{k_0-1} \theta, \quad (33)$$

which comes as a consequence of the requirement $P_{k_0}(\lambda) \le P_{k_0}(\mu) (P_{k_0-1}(\mu)/P_{k_0-1}(\lambda))^\theta$. Coming back to the $\mu_i$'s, we have

$$\begin{align*}
\mu_1 &= \lambda_1 e^{-\alpha_1}, & \mu_2 &= \lambda_2 e^{-(\alpha_2-\alpha_1)}, & \dots & \mu_{k_0-1} &= \lambda_{k_0-1} e^{-(\alpha_{k_0-1}-\alpha_{k_0-2})}, \\
\mu_{k_0} &= \lambda_{k_0} e^{+(\alpha_{k_0}+\alpha_{k_0-1})}, & \mu_{k_0+1} &= \lambda_{k_0+1} e^{-(\alpha_{k_0+1}+\alpha_{k_0})}, & \dots & \mu_{k_0+2} &= \lambda_{k_0+2} e^{-(\alpha_{k_0+2}-\alpha_{k_0+1})}, \\
\dots, & \mu_n &= \lambda_n e^{-(\alpha_n-\alpha_{n-1})}, & & & &
\end{align*}$$

and the constraint $0 < \mu_1 < \dots < \mu_n$ is automatically satisfied provided the $\alpha_k$'s are small enough (by continuity, using $\lambda_1 < \dots < \lambda_n$ together with the fact that the $\mu_i$'s are close to the $\lambda_i$'s), which is the case since we establish below that all $\alpha_k$'s are of order $\varepsilon_0$. On top of that we have $\alpha_{k_0-1} \ge c_0 \eta \varepsilon_0$ for some constant $c_0 > 0$.

Now, pick up an $\varepsilon_0 > 0$ whose value is to be made precise later, pick up a $\delta > 0$ as in the Theorem, and impose the additional constraint $(\mu_1, \dots, \mu_n) \in V_{\varepsilon_0}$ as in the statement of the Theorem. Since $|\mu_1 - \lambda_1| = \lambda_1 (1 - e^{-\alpha_1})$ and similarly for the other $\mu_i$'s, using the first order Taylor expansion of the exponential gives the existence of a universal constant $C > 0$, independent of $\varepsilon$, such that

$$\forall k, \quad \alpha_k \le C \varepsilon_0.$$

On top of that, the constraints $\frac{P_k(\mu)}{P_k(\lambda)} \ge \left(\frac{P_{k_0-1}(\mu)}{P_{k_0-1}(\lambda)}\right)^{\delta\varepsilon_0} (k \neq k_0, k_0-1)$ give

$$\alpha_k \le \delta \varepsilon_0 \alpha_{k_0-1} \quad \text{for } k \neq k_0, k_0-1. \quad (34)$$

a technical constraint whose usefulness will become clear later. In particular (34) together with the information $\alpha_{k_0-1} \le C \varepsilon_0$ imposes that the $\alpha_k$'s for $k \neq k_0, k_0-1$ are not only of order $\varepsilon_0$ but in fact of order $\varepsilon_0^2$, in that

$$\alpha_k \le C \varepsilon_0^2, \quad \text{for } k \neq k_0, k_0-1.$$

With these elements in mind, introduce the path

$$\begin{align*}
\lambda_1(t) &= \lambda_1 e^{-\alpha_1 t}, & \lambda_2(t) &= \lambda_2 e^{-(\alpha_2-\alpha_1)t}, & \dots, & \lambda_{k_0-1}(t) &= \lambda_{k_0-1} e^{-(\alpha_{k_0-1}-\alpha_{k_0-2})t}, \\
\lambda_{k_0}(t) &= \lambda_{k_0} e^{+(\alpha_{k_0}+\alpha_{k_0-1})t}, & \lambda_{k_0+1}(t) &= \lambda_{k_0+1} e^{-(\alpha_{k_0+1}+\alpha_{k_0})t}, & \dots, & \lambda_{k_0+2}(t) &= \lambda_{k_0+2} e^{-(\alpha_{k_0+2}-\alpha_{k_0+1})t}, \\
\dots, & \lambda_n(t) &= \lambda_n e^{-(\alpha_n-\alpha_{n-1})t}, & & & &
\end{align*}$$

where $t \in [0, 1]$. This path joins $(\lambda_1, \dots, \lambda_n)$ (at $t = 0$) and $(\mu_1, \dots, \mu_n)$ (at $t = 1$). Introduce as well

$$F(t)(x) = F(\lambda_1(t), \dots, \lambda_n(t))(x).$$
We know from Theorem 9 that

$$ \frac{d}{dt} F(t)(x) = \left( \prod_{i=1}^{n} \lambda_i(t) \right) \left[ e_{\lambda_1(t), \dots, \lambda_n(t)} * q_t \right] (x), \quad (35) $$

with

$$ q_t(x) = \alpha_1 e_{\lambda_1(t)} + (\alpha_2 - \alpha_1) e_{\lambda_2(t)} + \dots + (\alpha_{k_0-1} - \alpha_{k_0-2}) e_{\lambda_{k_0-1}(t)} \\ - (\alpha_{k_0} + \alpha_{k_0-1}) e_{\lambda_{k_0}(t)} + (\alpha_{k_0+1} + \alpha_{k_0}) e_{\lambda_{k_0+1}(t)} \\ + (\alpha_{k_0+2} - \alpha_{k_0+1}) e_{\lambda_{k_0+1}(t)} + \dots + (\alpha_n - \alpha_{n-1}) e_{\lambda_n(t)}. \quad (36) $$

Therefore, for any given $\varepsilon_0/2 \le x \le \varepsilon_0$, proving that $F(\lambda_1, \dots, \lambda_n)(x) > F(\mu_1, \dots, \mu_n)(x)$ boils down to proving that $q_t(x) < 0$ close to $x = 0$, in some strong enough sense.

Let us come to more quantitative upper bounds.

## Second step. Estimating $q_t(x)$.

As in the proof performed in the case $n = 3$, we make an asymptotic expansion of $q_t(x)$ when $x$ is of order $\varepsilon_0$ and all $\alpha_k$'s are of order $\varepsilon_0$ as well. This gives

$$ q_t(x) = \alpha_n - x \left[ \lambda_1 \alpha_1 + \lambda_2 (\alpha_2 - \alpha_1) + \dots + \lambda_{k_0-1} (\alpha_{k_0-1} - \alpha_{k_0-2}) \right. \\ \left. - \lambda_{k_0} (\alpha_{k_0} + \alpha_{k_0-1}) + \lambda_{k_0+1} (\alpha_{k_0+1} + \alpha_{k_0}) \right. \\ \left. + \lambda_{k_0+2} (\alpha_{k_0+2} - \alpha_{k_0+1}) + \dots + \lambda_n (\alpha_n - \alpha_{n-1}) \right] + R_t(x) $$

for a remainder term $R_t(x)$ which satisfies $|R_t(x)| \le C \varepsilon_0^3$, for all $x \le \varepsilon_0$ and all $t \in [0, 1]$, where $C > 0$ is, as before, some irrelevant constant that does not depend on $\varepsilon$, $t$, $x$ (but it does depend on the chosen $\lambda_i$'s). Using the reinforced information that the $\alpha_k$'s have order $\varepsilon_0^2$ whenever $k \ne k_0, k_0 + 1$ gives the simplified formula

$$ q_t(x) = \alpha_n - x \underbrace{\left[ \lambda_{k_0-1} \alpha_{k_0-1} - \lambda_{k_0} (\alpha_{k_0} + \alpha_{k_0-1}) + \lambda_{k_0+1} \alpha_{k_0} \right]}_{=:A_\alpha} + R_t(x) $$

whenever $x \le \varepsilon_0$, $t \in [0, 1]$, for some remainder $R_t$ (still denoted by the same letter) as before. We observe that

$$ \begin{aligned} A_\alpha &= \lambda_{k_0-1} \alpha_{k_0-1} - \lambda_{k_0} (\alpha_{k_0} + \alpha_{k_0-1}) + \lambda_{k_0+1} \alpha_{k_0} = -\alpha_{k_0-1} (\lambda_{k_0} - \lambda_{k_0-1}) + \alpha_{k_0} (\lambda_{k_0+1} - \lambda_{k_0}) \\ &\ge -\alpha_{k_0-1} (\lambda_{k_0} - \lambda_{k_0-1}) + \alpha_{k_0-1} \theta (\lambda_{k_0+1} - \lambda_{k_0}) \\ &= \alpha_{k_0-1} (\lambda_{k_0+1} - \lambda_{k_0}) [\theta - \theta_0(\lambda)] > 0, \end{aligned} $$

where we have used the key constraint $\alpha_{k_0} \ge \alpha_{k_0-1} \theta$, see (33), which comes from the definition of the set $\Omega_\theta(\lambda)$, together with $\theta > \theta_0(\lambda)$.

Eventually we have proved that, whenever $t \in [0, 1]$ and $x \le \varepsilon_0$, we have

$$ q_t(x) = \alpha_n - x A_\alpha + R_t(t, x), \quad (37) $$

with $|R_t(t, x)| \le C_1(\lambda) \varepsilon_0^3$, and

$$ A_\alpha \ge \alpha_{k_0-1} (\lambda_{k_0+1} - \lambda_{k_0}) [\theta - \theta_0(\lambda)] > 0. \quad (38) $$

## Third Step. Estimating the variations of F
We start from

$$
\begin{align*}
\frac{d}{dt} F(t)(x) &= \left( \prod_{i=1}^{n} \lambda_i(t) \right) [e_{\lambda_1(t), \dots, \lambda_n(t)} * q_t](x) \\
&= \left( \prod_{i=1}^{n} \lambda_i(t) \right) \int_0^x \int_0^{y_1} \dots \int_0^{y_n} e^{-\lambda_1(t)(x-y_1)} e^{-\lambda_2(t)(y_1-y_2)} \dots e^{-\lambda_n(t)(y_{n-1}-y_n)} \\
&\qquad \qquad \qquad [\alpha_n - y_n A_\alpha + R_t(t, y_n)] dy_1 \dots dy_n,
\end{align*}
$$

where the second line uses (37). Using the first order expansion of the exponential we may use that $\alpha_k \le C \varepsilon_0$
for any $k$ to deduce that, whenever $t \in [0, 1]$ and $x \in [0, \varepsilon_0]$, we have

$$
e^{-\lambda_1(t)(x-y_1)} = 1 + R(t, x, y_1),
$$

for a remainder term which satisfies $|R(t, x, y_1)| \le C \varepsilon_0$ whenever $y_1 \in [0, x]$, and similarly for the terms involving $\lambda_2, \dots, \lambda_n$. This provides, whenever $0 \le y_n \le \dots \le y_2 \le y_1 \le x \le \varepsilon_0$, and using the reinforced information $\alpha_n \le C_1(\lambda) \varepsilon_0^2$, the relation

$$
\begin{align*}
& e^{-\lambda_1(t)(x-y_1)} e^{-\lambda_2(t)(y_1-y_2)} \dots e^{-\lambda_n(t)(y_{n-1}-y_n)} [\alpha_n - y_n A_\alpha + R_t(t, y_n)] \\
&= \alpha_n - y_n A_\alpha + R_t(t, y_n, \dots, y_1, x),
\end{align*}
$$

for a remainder term which satisfies

$$
|R_t(t, y_n, \dots, y_1, x)| \le C \epsilon_0^3.
$$

By explicit integration we obtain

$$
\begin{align*}
\left( \prod_{i=1}^{n} \lambda_i(t) \right)^{-1} \frac{d}{dt} F(t)(x) &= \int_{0}^{x} \int_{0}^{y_1} \cdots \int_{0}^{y_n} [\alpha_n - y_n A_\alpha] dy_1 \cdots dy_n + R_t(x) \\
&= \frac{x^{n+1}}{(n+1)!} \alpha_n - \frac{x^{n+2}}{(n+2)!} A_\alpha + R_t(x),
\end{align*}
$$

for a remainder term which satisfies

$$
|R_t(x)| \le C \epsilon_0^{n+4}.
$$

Therefore, we come up with

$$
\left( \prod_{i=1}^{n} \lambda_i(t) \right)^{-1} \frac{d}{dt} F(t)(x) \le \frac{x^{n+1}}{(n+1)!} \left[ \alpha_n - A_\alpha \frac{x}{n+2} \right] + C \epsilon_0^{n+4}, \quad \forall x \in [0, \epsilon_0]. \quad (39)
$$

There remains to prove that the right-hand-side of (31) is negative whenever $\epsilon_0/2 \le x \le \epsilon_0$.

To do so we write, whenever $x \ge \epsilon_0/2$, the estimate

$$
\alpha_n - A_\alpha \frac{x}{n+2} \le \alpha_n - A_\alpha \frac{\epsilon_0}{2(n+2)}.
$$

Hence, using the technical information $\gamma \le \delta \epsilon_0 \alpha_{k_0-1}$ derived in (34), together with the upper bound (38)
on $A_\alpha$ we obtain, whenever $x \ge \epsilon_0/2$, that

$$
\begin{align*}
\alpha_n - A_\alpha \frac{x}{n+2} &\le \delta \epsilon_0 \alpha_{k_0-1} - \alpha_{k_0-1} (\lambda_{k_0+1} - \lambda_{k_0}) [\theta - \theta_0(\lambda)] \frac{\epsilon_0}{2(n+2)} \\
&\le \alpha_{k_0-1} \epsilon_0 \left[ \delta - \frac{(\lambda_{k_0+1} - \lambda_{k_0}) [\theta - \theta_0(\lambda)]}{2(n+2)} \right] \\
&\le -\alpha_{k_0-1} \epsilon_0 \frac{(\lambda_{k_0+1} - \lambda_{k_0}) [\theta - \theta_0(\lambda)]}{4(n+2)},
\end{align*}
$$
thanks to our technical assumption $0 < \delta \le (\lambda_{k_0+1} - \lambda_{k_0}) [\theta - \theta_0(\lambda)]/(4(n+2))$.

All in all we have established

$$ \frac{\varepsilon_0}{2} \le x \le \varepsilon_0 \quad \Rightarrow \quad \alpha_n - A_\alpha \frac{x}{n+2} \le -K \alpha_{k_0-1} \varepsilon_0 \quad \text{with } K = \frac{(\lambda_{k_0+1} - \lambda_{k_0}) [\theta - \theta_0(\lambda)]}{4(n+2)}. $$

We are in position to conclude.

**Last step. Conclusion.**

Summarizing, the previous step establishes

$$ \left( \prod_{i=1}^{n} \lambda_i(t) \right)^{-1} \frac{d}{dt} F(t)(x) \le -K \alpha_{k_0-1} \varepsilon_0 \frac{\varepsilon_0^{n+1}}{(n+1)!} + C \varepsilon_0^{n+4}, \quad \forall x \in \left[ \frac{\varepsilon_0}{2}, \varepsilon_0 \right], \quad \forall t \in [0, 1]. $$

Hence

$$ \left( \prod_{i=1}^{n} \lambda_i(t) \right)^{-1} \frac{d}{dt} F(t)(x) \le -\varepsilon_0^{n+2} \left( \frac{K}{(n+1)!} \alpha_{k_0-1} - C \varepsilon_0^2 \right) \quad \forall x \in \left[ \frac{\varepsilon_0}{2}, \varepsilon_0 \right], \quad \forall t \in [0, 1]. $$

There remains to choose $\varepsilon_0$ so small as to satisfy

$$ \varepsilon_0^2 \le \frac{K \alpha}{2 C (n+1)!}, $$

which is possible since we have $\alpha \ge c_0 \eta \varepsilon_0$. We recover

$$ \left( \prod_{i=1}^{n} \lambda_i(t) \right)^{-1} \frac{d}{dt} F(t)(x) \le -\varepsilon_0^{n+2} \frac{K \alpha}{4(n+1)!} < 0 \quad \forall x \in \left[ \frac{\varepsilon_0}{2}, \varepsilon_0 \right], \quad \forall t \in [0, 1]. $$

Our Theorem is proved.

**Proof of Theorem 27**

With the notation of the Theorem, we set

$$ F(t)(x) = F(\lambda_1(t), \dots, \lambda_n(t))(x), $$

as well as

$$ q_t(x) = Q_t(e^{-x}). $$

Our assumption on $Q_t$ provides, whenever $x \le \varepsilon \le \varepsilon_0$, the bound

$$ q_t(x) \le a - bx + \tilde{R}, $$

with $|\tilde{R}| \le C \varepsilon^3$ for some $\varepsilon$-independent constant $C$. We know that

$$ \begin{aligned} \frac{d}{dt} F(t)(x) &= \left( \prod_{i=1}^{n} \lambda_i(t) \right) [e_{\lambda_1(t), \dots, \lambda_n(t)} * q_t](x) \\ &= \left( \prod_{i=1}^{n} \lambda_i(t) \right) \int_0^x \int_0^{y_1} \dots \int_0^{y_n} e^{-\lambda_1(t)(x-y_1)} e^{-\lambda_2(t)(y_1-y_2)} \dots e^{-\lambda_n(t)(y_{n-1}-y_n)} q_t(y_n) dy_1 \dots dy_n, \end{aligned} $$

First order expansions as in the previous proof establish that, provided $\varepsilon$ is small enough, we have the relation

$$ \left( \prod_{i=1}^{n} \lambda_i(t) \right)^{-1} \frac{d}{dt} F(t)(x) = \int_{0}^{x} \int_{0}^{y_1} \dots \int_{0}^{y_n} q_t(y_n) dy_1 \dots dy_n + R_t(x), $$
for some remainder $R_t$ satisfying

$$|R_t(x)| \leq C\varepsilon^{n+4},$$

whenever $t \in [0, 1]$, $x \leq \varepsilon_0$. We deduce, for any $x \leq \varepsilon_0$, and $t \in [0, 1]$, the bound

$$ \left( \prod_{i=1}^{n} \lambda_i(t) \right)^{-1} \frac{d}{dt} F(t)(x) \leq a \frac{x^{n+1}}{(n+1)!} - b \frac{x^{n+2}}{(n+2)!} + C \varepsilon^{n+4} = \frac{x^{n+1}}{(n+1)!} \left[ a - b \frac{x}{n+2} \right] + C \varepsilon^{n+4}. $$

There remains to observe that, for $x \geq \varepsilon/2$, we have

$$ a - b \frac{x}{n+2} \leq a - b \frac{\varepsilon}{2(n+2)} \leq K_1 \varepsilon_0^2 - K_2 \frac{\varepsilon}{2(n+2)} \leq -K_2 \frac{\varepsilon}{4(n+2)} < 0, $$

up to choosing $\varepsilon$ small enough, so as to ensure $\varepsilon_0 \leq K_2/(4K_1(n+2))$.

The proof is complete.

<figure><img src="image_14.png" /></figure>

# 8 Numerical examples

In this section we give explicit numerical illustrations of the phenomena studied in this text.

**First illustration. We begin with an illustration in the case $(\lambda_1, \lambda_2, \lambda_3) = (1, 2, 4)$.

In that case we have

$$(P_1(\lambda), P_2(\lambda), P_3(\lambda)) = (1, 2, 8), \quad \theta_0(\lambda) = 1/2.$$

As a consequence the set of values for which the criterion in [3] guarantees $F(\lambda_1, \lambda_2, \lambda_3) \leq F(\mu_1, \mu_2, \mu_3)$ is the set

$$ \mathcal{A} = \{ (\mu_1, \mu_2, \mu_3) \text{ such that } 0 < \mu_1 \leq \mu_2 \leq \mu_3, \text{ and } P_1(\mu) \leq 1, P_2(\mu) \leq 2, P_3(\mu) \leq 8 \}, $$

while our analysis in Theorem 14 guarantees that there holds $F(\lambda_1, \lambda_2, \lambda_3)(x) \leq F(\mu_1, \mu_2, \mu_3)(x)$ for all $x$ whenever $(\mu_1, \mu_2, \mu_3)$ belongs to the larger set

$$ \mathcal{A}^{\text{ext}} = \left\{ (\mu_1, \mu_2, \mu_3) \text{ s.t. } 0 < \mu_1 \leq \mu_2 \leq \mu_3, \text{ and } P_1(\mu) \leq 1, P_2(\mu) \leq \frac{2}{P_1(\mu)^{1/2}}, P_3(\mu) \leq 8 \right\}. $$

In terms of the variables $\mu_i$'s, these sets are rephrased as

$$ \mathcal{A} = \left\{ (\mu_1, \mu_2, \mu_3) \text{ s.t. } 0 < \mu_1 \leq \mu_2 \leq \mu_3, \text{ and } \mu_1 \leq 1, \mu_2 \leq \frac{2}{\mu_1}, \mu_3 \leq \frac{8}{\mu_1 \mu_2} \right\}, $$

resp.

$$ \mathcal{A}^{\text{ext}} = \left\{ (\mu_1, \mu_2, \mu_3) \text{ s.t. } 0 < \mu_1 \leq \mu_2 \leq \mu_3, \text{ and } \mu_1 \leq 1, \mu_2 \leq \frac{2}{\mu_1^{3/2}}, \mu_3 \leq \frac{8}{\mu_1 \mu_2} \right\}. $$

To end up the present illustration, let us last mention that our analysis proves that the set $\mathcal{A}^{\text{ext}}$ is optimal in the sense that there exists a small vicinity of $(\lambda_1, \lambda_2, \lambda_3)$ in the complementary set $(\mathcal{A}^{\text{ext}})^c$ for which the relation $F(\lambda_1, \lambda_2, \lambda_3)(x) \leq F(\mu_1, \mu_2, \mu_3)(x)$ is violated close to $x = 0$ no matter which parameters $(\mu_1, \mu_2, \mu_3)$ are chosen in the above mentioned vicinity. Technically, the vicinity is of the form

$$ \Omega_{\theta}(\lambda) \cap V_{\varepsilon_0} = \left\{ (\mu_1, \mu_2, \mu_3) \text{ which are } \varepsilon_0\text{-close to } (1, 2, 8), \text{ s.t. } 0 < \mu_1 \leq \mu_2 \leq \mu_3, \text{ and } P_1(\mu) \leq 1, P_2(\mu) \geq \frac{2}{P_1(\mu)^{\theta}}, 8P_1(\mu)^{\beta} \leq P_3(\mu) \leq 8 \right\}, $$

where $\theta > 1/2$ is chosen close to $1/2$, and $\beta > 0$ is chosen close to 0.
<figure><img src="image_15.png" /></figure>

Figure 5: Representation of the various sets in the variables ($P_1(\mu)$, $P_2(\mu)$, $P_3(\mu)$).

Blue zone: the set $\mathcal{A}$ studied in [3].

Green zone: the extension of $\mathcal{A}^{\text{ext}} \setminus \mathcal{A}$ of $\mathcal{A}$ provided in the present paper.

Red zone: the vicinity $\Omega_{\theta}(\lambda) \cap V_{\varepsilon_0}$ proving a form of optimality of our criterion.

Yellow zone $(M, h, g, m) \setminus (M, G, H, I)$: zone for which our analysis is not able to determine the sign of $F(\lambda_1, \lambda_2, \lambda_3)(x) - F(\mu_1, \mu_2, \mu_3)(x)$ (if any).

Figure 8 illustrates the various sets.

We end up by illustrating the phenomena at stake in the two cases when $(\mu_1, \mu_2, \mu_3)$ is chosen in the red region $\Omega_{\theta}(\lambda) \cap V_{\varepsilon_0}$, where we are able to establish that our extended criterion is violated and the relation $F(\lambda_1, \lambda_2, \lambda_3)(x) \le F(\mu_1, \mu_2, \mu_3)(x)$ is violated (for $x$ close to 0), and the second, ambiguous case, when $(\mu_1, \mu_2, \mu_3)$ is chosen in the yellow region without lying in the red region, for which our extended criterion is violated but the sign of $F(\lambda_1, \lambda_2, \lambda_3)(x) - F(\mu_1, \mu_2, \mu_3)(x)$ is numerically seen to be difficult to decide, and various situations may occur.
<figure><img src="image_16.png" /><figcaption>Figure 6: Illustration of what happens when $(\mu_1, \mu_2, \mu_3)$ is chosen so that $(P_1(\mu), P_2(\mu), P_3(\mu))$ lies in the upper triangle (MHG) in figure 8. In that case $q_t(x)$ is negative for all small values of $x$ (see (28)) and the convolution that allows to rebuild the variations of $F$ with $\mu$ (see (27)) preserves this sign for small values of $x$ as well.</figcaption></figure>

<figure><img src="image_17.png" /><figcaption>Figure 7: Illustration of what happens when $(\mu_1, \mu_2, \mu_3)$ is chosen so that $(P_1(\mu), P_2(\mu), P_3(\mu))$ lies inside the red zone in figure 8. In that case $q_t(x)$ is positive for very small values of $x$, then negative, then becomes positive again when $x$ leaves a vicinity of $x = 0$, (see (28)) and the convolution that allows to rebuild the variations of $F$ with $\mu$ (see (27)) turns out to preserve this sign for small values of $x$ as well.</figcaption></figure>
<figure><img src="image_18.png" /><figcaption>Figure 8: Illustration of what may happen when $(\mu_1, \mu_2, \mu_3)$ is chosen in the yellow region in figure 8, without belonging to the red region. In that case $q_t(x)$ may indeed reach negative values for small values of $x$, but the convolution that allows to rebuild the variations of $F$ with $\mu$ does not preserve this sign.</figcaption></figure>

<figure><img src="image_19.png" /><figcaption>Figure 9: Illustration of another scenario when $(\mu_1, \mu_2, \mu_3)$ is chosen in the yellow region in figure 8, without belonging to the red region. In that case $q_t(x)$ may indeed reach negative values for small values of $x$, and the convolution that allows to rebuild the variations of $F$ with $\mu$ does preserve this change of sign.</figcaption></figure>

**Second illustration.** We continue with an illustration in the case $(\lambda_1, \lambda_2, \lambda_3, \lambda_4) = (1, 3, 4, 6)$.

$$ (P_1(\lambda), P_2(\lambda), P_3\lambda), P_4\lambda)) = (1, 3, 12, 72). $$

As a consequence the set of values for which the criterion in [3] guarantees $F(\lambda_1, \lambda_2, \lambda_3) \le F(\mu_1, \mu_2, \mu_3)$ is
the set

$$ \mathcal{A} = \left\{ (\mu_1, \mu_2, \mu_3, \mu_4) \text{ such that } 0 < \mu_1 \le \mu_2 \le \mu_3 \le \mu_4, \text{ and } P_1(\mu) \le 1, P_2(\mu) \le 3, P_3(\mu) \le 12, P_4(\mu) \le 72 \right\}. $$

while our analysis in Theorem 17 gurantees that there holds $F(\lambda_1, \lambda_2, \lambda_3, \lambda_4)(x) \le F(\mu_1, \mu_2, \mu_3, \mu_4)(x)$ for all $x$ whenever $(\mu_1, \mu_2, \mu_3, \mu_4)$ belongs to the larger set (obtained as the union of the two sets provided by Theorem 17 when choosing successively $k_0 = 2$ and $k_0 = 3$, and using the reformulation of the requirement (24) as observed in Remark 19)

$$ \begin{aligned} \mathcal{A}^{\text{ext}} = & \left\{ (\mu_1, \mu_2, \mu_3, \mu_4) \text{ such that } 0 < \mu_1 \le \mu_2 \le \mu_3 \le \mu_4, P_2(\mu) \ge \frac{12}{6} = 2, \right. \\ & \text{and } P_1(\mu) \le 1, P_2(\mu) \le \frac{3}{P_1(\mu)^2}, P_3(\mu) \le 12, P_4(\mu) \le 72 \right\} \\ & \cup \left\{ (\mu_1, \mu_2, \mu_3, \mu_4) \text{ such that } 0 < \mu_1 \le \mu_2 \le \mu_3 \le \mu_4, P_2(\mu) \ge 1, \right. \\ & \text{and } P_1(\mu) \le 1, P_2(\mu) \le 3, P_3(\mu) \le \frac{12}{P_2(\mu)^{1/5}}, P_4(\mu) \le 72 \right\}. \end{aligned} $$

## References

[1] A. D. Barbour, T. Lindvall, and L. C. G. Rogers. Stochastic ordering of order statistics. *Journal of Applied Probability*, 28(2), 1991.

[2] A. D. Boland, E. El-Neweihi, and F. Proschan. Schur properties of convolutions of exponential geometric random variables. *Journal of Multivariate Analysis*, 48(1), 1991.

[3] J-L. Bon and E. Păltănea. Ordering properties of convolutions of exponential random variables. *Lifetime Data Analysis*, 5, 1999.

[4] F. Guillemin and B. Sericola. Volume and duration of losses in finite buffer fluid queues. *Journal of Applied Probability*, 52(3), 2015.

[5] B-E. Khaledi and S. C. Kochar. Ordering convolutions of gamma random variables. *Sankhyā: The Indian Journal of Statistics*, 66(3), 2004.

[6] R. M. Korwar. On stochastic orders for sums of independent random variables. *Journal of Multivariate Analysis*, 80(2), 2002.

[7] S. Lihong and Z. Xinsheng. Stochastic comparisons of order statistics from gamma distributions. *Journal of Multivariate Analysis*, 93(1), 2005.

[8] M. F. Neuts. *Matrix-Geometric Solutions in Stochastic Models: An Algorithmic Approach*. Johns Hopkins University Press, Baltimore, MD, 1981.

[9] F. Proschan and J. Sethuraman. Stochastic comparison of order statistics from heterogeneous populations, with applications in reliability. *Journal of Multivariate Analysis*, 6(4), 1976.

[10] G. Rubino and B. Sericola. *Markov Chains and Dependability Theory*. Cambridge University Press, 2014.

[11] B. Sericola. *Markov Chains. Theory, Algorithms and Applications*. John Wiley & Sons, 2013.

[12] M. Shaked and J. G. Shantikumar. *Stochastic Orders*. Springer, New York, 2007.

[13] Y. Yu. Stochastic ordering of exponential family distributions and their mixtures. *Journal of Applied Probability*, 46(1), 2009.