Some New Results Related to the Likelihood Ratio and the Laplace Transform Ordering

Abstract

The purpose of this paper is twofold. First, we aim to study some new properties and relations for the likelihood ratio order. In particular, we present new integral inequalities involving the ratio of two probability density functions and the ratio of two corresponding cumulative distribution functions. Specifically, we provide alternative proofs for certain relationships between the likelihood ratio order and other stochastic orders. Second, we consider the Laplace ratio order and present new characterizations for the Laplace ratio ordering of two random variables. These results contribute to the literature on stochastic orders by providing new characterizations and inequalities that can be useful in probability theory, risk analysis and reliability theory.

1. Introduction

Stochastic order relations between random variables and distributions play a crucial role in statistical analysis by providing insights into the distribution and behavior of data. They are particularly important in estimating percentiles and medians, which are fundamental in descriptive statistics. Additionally, order statistics are widely used in reliability analysis, survival studies and risk assessment, where the minimum and maximum values help model extreme events. Their applications extend to non-parametric methods, ranking procedures and hypothesis testing, making them an essential tool in both theoretical and applied statistics. One of the earliest definitions of stochastic ordering, introduced by Lehmann [1], states that a random variable X with cumulative distribution function F is stochastically greater than a random variable Y with distribution function G if

$$F\left(x\right)\le G\left(x\right)\mathrm{for}\mathrm{all}x\in (-\infty ,\infty ).$$

(1)

We call this ordering stochastic ordering, and it is denoted by $X{\le }_{\mathrm{st}}Y.$ It is important to note that the inequality (1) can be rewritten as follows:

$$P(X\le x)\le P(Y\le x)\mathrm{for}\mathrm{all}x\in (-\infty ,\infty ).$$

(2)

Moreover, the inequalities (1) and (2) can be reformulated in a more general but actually equivalent manner as follows:

$$P(X\in U)\le P(Y\in U),$$

for all upper sets $U\subseteq \mathbb{R};$ that is, for all measurable sets U such that if $x\in U$, then every $y\ge x$ also belongs to $U.$

In certain cases, a pair of distributions may verify a stronger condition called likelihood ratio ordering. Assume that the cumulative distribution functions F and G possess the positive probability density functions f and g, respectively. We say that X is smaller than Y according to likelihood ratio ordering if the function ([2], Equation (1.C.1)).

$$t\mapsto {\displaystyle \frac{g\left(t\right)}{f\left(t\right)}}\mathrm{is}\mathrm{increasing}\mathrm{over}\mathrm{the}\mathrm{union}\mathrm{of}\mathrm{the}\mathrm{supports}\mathrm{of}X\mathrm{and}Y,$$

or, equivalently, that ([2], Equation (1.C.2)):

$$f\left(y\right)g\left(x\right)\le f\left(x\right)g\left(y\right)\mathrm{for}\mathrm{all}x\le y.$$

(3)

This ordering is denoted by $X{\le }_{lr}Y.$ It is important to note that by integrating (3) over $x\in A$ and $y\in B,$ where A and B are measurable sets in $\mathbb{R},$ it is seen that (3) is equivalent to ([2], Equation (1.C.3)):

$$P(X\in B)P(Y\in A)\le P(X\in A)P(Y\in B)\mathrm{such}\mathrm{that}A\le B,$$

(4)

where $A\le B$ means that $x\in A$ and $y\in B$ imply that $x\le y.$ It is worth mentioning that the inequality (4) does not directly involve the underlying densities, and thus, it applies equally to continuous distributions or to discrete distributions or even to mixed distributions.

A useful relationship between the likelihood ratio and the stochastic order is described in the following theorem (see [2], Theorem 1.C.1):

$$X{\le }_{\mathrm{lr}}Y⟹X{\le }_{\mathrm{st}}Y,$$

(5)

where X and Y are two continuous or discrete random variables. It is worth mentioning that in Remark 2, we present a simple proof for the above relation for two continuous random variables supported on $(0,\infty ).$

Before we proceed, we need to recall some other basic definitions and some required further notation in what follows.

Definition 1

([2], Section 5.A.1). Let X and Y be two non-negative random variables such that

$$E\left[e^{-sY}\right]\le E\left[e^{-sX}\right]\mathrm{holds}\mathrm{for}\mathrm{all}s>0.$$

Then X is said to be smaller than Y in the Laplace transform order. This ordering is denoted by $X{\le }_{Lt}Y.$ Moreover, if the function

$$s\mapsto {\displaystyle \frac{E\left[e^{-sY}\right]}{E\left[e^{-sX}\right]}}\mathrm{is}\mathrm{decreasing}\mathrm{on}(0,\infty ),$$

then X is said to be smaller than Y in the Laplace transform ratio order and denoted by $X{\le }_{Lt-r}Y.$ For some background on Laplace transform ratio order, the reader is referred to ([2], Section 5.B.1).

For more details regarding the Laplace transform order, including its properties, we refer the interested reader to ([2], Chap. 5). In this regards, let us also mention that several notable contributions can be found in [3,4,5,6,7,8,9].

Definition 2

([2], Section 5.C.2). Consider now two non-negative random variables X and Y such that

$$E\left[X^n\right]\le E\left[Y^n\right]\mathrm{for}\mathrm{all}n\in \mathbb{N}.$$

Then X is said to be smaller than Y in the moments order, denoted by $X{\le }_{\mathrm{mom}}Y$. In addition, if

$${\displaystyle \frac{E\left[Y^n\right]}{E\left[X^n\right]}}\mathrm{is}\mathrm{increasing}\mathrm{in}n\in \mathbb{N},$$

then X is said to be smaller than Y in the moments ratio order and denoted as $X{\le }_{\mathrm{mom}-\mathrm{r}}Y.$

The moments order and the moments ratio order are reviewed in ([2], Section 5.C.2).

Definition 3.

Let X and Y be two non-negative random variables such that $E\left[e^{s_{0}Y}\right]<\infty$ for some $s_0>0,$ and

$$E\left[e^{sX}\right]\le E\left[e^{sY}\right],\mathrm{for}\mathrm{all}s>0.$$

Then X is said to be smaller than Y in the moment generating function order (or the exponential order), denoted as $X{\le }_{\mathrm{mgf}}Y$. For more details, see [2] (Section 5.C.3). See also ([10], Section 4.3, p. 45) and ([11], Section 3, p. 53).

In probability theory, the moment generating function essentially reduces to requiring that the moment generating functions of the non-negative random variables X and Y are uniformly ordered. Moreover, this ordering reflects the shared risk preferences of all decision makers whose utility functions take the following form

$$\phi \left(x\right)=1-exp(-sx),s>0,$$

or equivalently, whose pain functions take the following form:

$$\tilde{\phi }\left(x\right)=exp\left(sx\right)-1,s>0.$$

From the series expansion of the exponential function, it is clear that the moment generating function order is weaker than the moment order, i.e., we have ([2], Theorem 5.C.15):

$$X{\le }_{\mathrm{mom}}Y⟹X{\le }_{\mathrm{mgf}}Y.$$

Definition 4.

A function $f:I\subseteq \mathbb{R}⟶(0,\infty )$ is log-convex on the interval $I,$ if $logf$ is convex, i.e., if for all $x_1,x_2\in I$ and $\lambda \in [0,1],$ we have

$$f(\lambda x_1+(1-\lambda )x_2))\le {\left[f\left(x_1\right)\right]}^{\lambda }·{\left[f\left(x_2\right)\right]}^{1-\lambda }.$$

Similarly, a function $g:I\subseteq \mathbb{R}⟶(0,\infty )$ is said to be geometrically (or multiplicatively) convex if g is convex with respect to the geometric mean, i.e., if for all $x_1,x_2\in I$ and $\lambda \in [0,1]$, we have

$$g\left(x^{\lambda }{x}_2^{1-\lambda }\right)\le {\left[g\left(x_1\right)\right]}^{\lambda }·{\left[g\left(x_2\right)\right]}^{1-\lambda }.$$

Remark 1.

We note that if f and g are differentiable, then f is log-convex if and only if $x\mapsto f^{\prime }\left(x\right)/f\left(x\right)$ is increasing on I, while g is geometrically convex if and only if $x\mapsto x{g}^{\prime }\left(x\right)g\left(x\right)$ is increasing on $I.$ A similar definition and characterization of differentiable log-concave and geometrically concave functions also holds.

Motivated by the above definitions, our aim of the paper is to introduce the following new definitions.

Definition 5

(Geometrically convex order–Geometrically concave order). Let X and Y be two random variables such that

$$E\left[\phi \right(X\left)\right]\le E\left[\phi \right(Y\left)\right],$$

(6)

for all geometrically convex functions $\phi :(0,\infty )⟶(0,\infty ),$ provided the expectations exist. Then X is said to be smaller than Y in the geometrically convex order and denoted as $X{\le }_{\mathrm{Gcx}}Y.$

One can also define a geometrically concave order by requiring (6) to hold for all geometrically concave functions $h:[0,\infty )⟶(0,\infty ),$ denoted as $X{\le }_{\mathrm{Gcv}}Y.$

Moreover, if

$${\displaystyle \frac{E\left[\phi \right(sY\left)\right]}{E\left[\phi \right(sX\left)\right]}}\mathrm{is}\mathrm{increasing}\left(\mathrm{decreasing}\right)\mathrm{for}s>0,$$

where $h:[0,\infty )⟶(0,\infty )$ is geometrically convex (concave) function, then X is said to be smaller than Y in the geometrically convex (concave) ratio order and denoted as $X{\le }_{\mathrm{Gcx}-\mathrm{r}}Y\left(X{\le }_{\mathrm{Gcv}-\mathrm{r}}Y\right).$

The upcoming sections are organized as follows: In Section 2, we recollect some lemmas that will be helpful to establish some of the main results of this study. In Section 3, we present bilateral integral functional inequalities involving the ratio of the probability density functions and the ratio of the corresponding cumulative distribution functions under the assumption that they satisfy the likelihood ratio ordering. Moreover, relationships between the likelihood ratio orders and the Gcx-orders are given. In particular, we propose alternative proofs for some existing results in order statistics. In the final section, we present new characterizations of the Laplace ratio ordering of two random variables. More precisely, we derive a lower bound for the Laplace transform under the likelihood ratio ordering. As a direct consequence, we establish sufficient conditions under which this ordering implies the Laplace transform ordering. Furthermore, we give two-sided exponential bounds for $E\left[e^{-sX}\right]$ on bounded support and present a sufficient condition that ensures Laplace ordering between two random variables.

2. Two Useful Lemmas

Our aim in this section is to present two useful lemmas that will be helpful to establish some of the main results.

The following lemma is called the Chebyshev integral inequality ([12], p. 40).

Lemma 1.

If $\varphi ,\psi :[a,b]⟶\mathbb{R}$ are integrable functions and synchronous (both increasing or both decreasing), and $\rho :[a,b]⟶\mathbb{R}$ is a positive integrable function, then

$${\int }_a^b\rho \left(t\right)\varphi \left(t\right)dt{\int }_a^b\rho \left(t\right)\psi \left(t\right)dt\le {\int }_a^b\rho \left(t\right)dt{\int }_a^b\rho \left(t\right)\varphi \left(t\right)\psi \left(t\right)dt.$$

(7)

Note that if one of the functions ϕ or ψ is asynchronous (one is decreasing and the other is increasing), then (7) is reversed.

The next lemma is the so-called monotone form of l’Hospital’s rule; see [13] for a proof.

Lemma 2.

Let $f,g:[a,b]\to \mathbb{R}$ be two continuous functions that are differentiable on $(a,b).$ Furthermore, let $g^{\prime }\left(x\right)\ne 0$ on $(a,b).$ If $f^{\prime }/g^{\prime }$ is increasing (decreasing) on $(a,b),$ then the functions

$$x⟼{\displaystyle \frac{f\left(x\right)-f\left(a\right)}{g\left(x\right)-g\left(a\right)}}\mathrm{and}x⟼{\displaystyle \frac{f\left(x\right)-f\left(b\right)}{g\left(x\right)-g\left(b\right)}},$$

are also increasing (decreasing) on $(a,b).$

Remark 2.

We note that by means of the monotone form of l’Hospital’s rule given in the above lemma, we re-obtain the relation (5) for two continuous random variables supported on $(0,\infty ).$ Indeed, let F and G be the cumulative distribution functions corresponding to the random variables X and Y supported on $(0,\infty ).$ Assume the function $t\mapsto {\displaystyle \frac{g\left(t\right)}{f\left(t\right)}}$ is increasing on $(0,\infty );$ here, the functions f and g are the probability density functions associated with F and $G.$ Then, according to Lemma 2, we conclude that the function

$$t\mapsto {\displaystyle \frac{G\left(t\right)}{F\left(t\right)}}={\displaystyle \frac{G\left(t\right)-G\left(0\right)}{F\left(t\right)-F\left(0\right)}},$$

is also increasing on $(0,\infty ).$

3. First Set of Main Results

In the first result of this section, we report on the bilateral integral functional inequalities related to the ratio of the probability density functions, and the ratio of their cumulative distributions satisfy the likelihood ratio ordering. Moreover, under some conditions, we show that the likelihood ratio implies the geometric convex order and the moment generating function order.

Theorem 1.

Let X and Y be positive continuous random variables with probability density functions f and g, respectively. Assume that $X{\le }_{lr}Y$, then the following assertions are valid:

(a).

If the functions ${\displaystyle \frac{G\left(t\right)}{F\left(t\right)}}$ and ${\displaystyle \frac{g\left(t\right)}{f\left(t\right)}}$ are integrable on $(-\infty ,\infty )$, then

$${\int }_{-\infty }^{\infty }{\displaystyle \frac{G\left(t\right)}{F\left(t\right)}}dt\le {\int }_{-\infty }^{\infty }{\displaystyle \frac{g\left(t\right)}{f\left(t\right)}}dt.$$

(8)

(b).

Suppose that the functions ${\displaystyle \frac{G\left(t\right)}{F\left(t\right)}}$ and ${\displaystyle \frac{g\left(t\right)}{f\left(t\right)}}$ are integrable on $[a,b]$ where $-\infty <a<b<\infty .$ If the function $t\mapsto f\left(t\right)$ is increasing, then the following inequality

$${\displaystyle \frac{1}{b-a}}{\int }_a^b{\displaystyle \frac{g\left(t\right)}{f\left(t\right)}}dt\le {\displaystyle \frac{G\left(b\right)}{F\left(b\right)}},$$

(9)

holds true. Furthermore, if the function $t\mapsto f\left(t\right)$ is decreasing on $[a,b],$ then the inequality (9) is reversed.

Proof. 

(a).

The assumption $X{\le }_{lr}Y$ means that the function $t\mapsto {\displaystyle \frac{g\left(t\right)}{f\left(t\right)}}$ is increasing on $(-\infty ,\infty ).$ Therefore, we obtain

$$\begin{array}{cc}\hfill {\int }_a^b{\displaystyle \frac{G\left(x\right)}{F\left(x\right)}}dx& ={\int }_a^b{\displaystyle \frac{1}{F\left(x\right)}}\left({\int }_{-\infty }^{x}g\left(t\right)dt\right)dx\hfill \\ \hfill & ={\int }_a^b{\displaystyle \frac{1}{F\left(x\right)}}\left({\int }_{-\infty }^x\left({\displaystyle \frac{g\left(t\right)}{f\left(t\right)}}\right)·f\left(t\right)dt\right)dx\hfill \\ \hfill & \le {\int }_a^b{\displaystyle \frac{g\left(x\right)}{f\left(x\right)F\left(x\right)}}\left({\int }_{-\infty }^{x}f\left(t\right)dt\right)dx\hfill \\ \hfill & ={\int }_a^b{\displaystyle \frac{g\left(x\right)}{f\left(x\right)}}dx.\hfill \end{array}$$

Therefore, we conclude the asserted inequality (8).

(b).

We consider the function $\rho ,\varphi ,\psi :[a,b]⟶\mathbb{R}$, defined by

$$\rho \left(t\right)=1,\varphi \left(t\right)={\displaystyle \frac{g\left(t\right)}{f\left(t\right)}}\mathrm{and}\psi \left(t\right)=f\left(t\right).$$

Under our assumptions, the functions $\varphi$ and $\psi$ are synchoronous. From the Chebyshev inequality (7), it follows that

$$\left({\int }_a^{b}f\left(t\right)dt\right)\left({\int }_a^b{\displaystyle \frac{g\left(t\right)}{f\left(t\right)}}dt\right)\le \left({\int }_a^b\rho \left(t\right)dt\right)\left({\int }_a^{b}g\left(t\right)dt\right),$$

which is equivalent to

$$F\left(b\right)·\left({\int }_a^b{\displaystyle \frac{g\left(t\right)}{f\left(t\right)}}dt\right)\le (b-a)G\left(b\right).$$

Hence, the inequality (9) holds true if f is increasing and reversed if f is decreasing.

□

Theorem 2.

Let X and Y be non-negative random variables with probability density functions $f\left(t\right)$ and $g\left(t\right).$ Let

$$h:[0,\infty )⟶(0,\infty )$$

be a differentiable and geometrically convex function. Assume the following conditions hold:

(i). 

For all $s>0,$ the expectations $E\left[h\right(sX\left)\right]$ and $E\left[h\right(sY\left)\right]$ exist and are finite.

(ii). 

For each fixed $s>0$, the functions $t\mapsto t{h}^{\prime }\left(st\right)f\left(t\right)$ and $t\mapsto t{h}^{\prime }\left(st\right)g\left(t\right)$ are integrable on $(0,\infty ),$ and there exist integrable functions ${\varphi }_X\left(t\right)$ and ${\varphi }_Y\left(t\right)$, independent of s, such that

$$|t{h}^{\prime }\left(st\right)f\left(t\right)|\le {\varphi }_X\left(t\right),\left|t{h}^{\prime }\left(st\right)g\left(t\right)\right|\le {\varphi }_Y\left(t\right),$$

for all $t>0.$ Then, the following implications hold:

$$X{\le }_{lr}Y⟹X{\le }_{Gcx-r}Y⟹X{\le }_{Gcx}Y.$$

Proof. 

We consider the functions $\rho ,\psi ,\varphi$ defined on $(0,\infty )$ by

$$\rho \left(t\right)=h\left(st\right)f\left(t\right),\psi \left(t\right)={\displaystyle \frac{g\left(t\right)}{f\left(t\right)}}\mathrm{and}\varphi \left(t\right)={\displaystyle \frac{t{h}^{\prime }\left(st\right)}{h\left(st\right)}}.$$

We observe that the functions $\psi$ and $\phi$ are increasing on $(0,\infty )$ under the given conditions. On the other hand, we have

$${\int }_0^{\infty }\rho \left(t\right)dt=E\left[h\left(sX\right)\right],{\int }_0^{\infty }\rho \left(t\right)\psi \left(t\right)dt=E\left[h\left(sY\right)\right],$$

and

$${\int }_0^{\infty }\rho \left(t\right)\phi \left(t\right)dt={\displaystyle \frac{\partial E\left[h\right(sX\left)\right]}{\partial s}},{\int }_0^{\infty }\rho \left(t\right)\phi \left(t\right)\psi \left(t\right)dt={\displaystyle \frac{\partial E\left[h\right(sY\left)\right]}{\partial s}}.$$

According to Lemma 1, we establish that the following inequality

$$E\left[h\left(sX\right)\right]·{\displaystyle \frac{\partial E\left[h\right(sY\left)\right]}{\partial s}}-E\left[h\left(sY\right)\right]·{\displaystyle \frac{\partial E\left[h\right(sX\left)\right]}{\partial s}}\ge 0,$$

(10)

holds true for all $s>0.$ Now, we define the function $T:(0,\infty )⟶(0,\infty )$ by

$$\begin{array}{cc}\hfill T\left(s\right)& ={\displaystyle \frac{E\left[h\right(sY\left)\right]}{E\left[h\right(sX\left)\right]}}\hfill \\ \hfill & ={\displaystyle \frac{{\int }_0^{\infty }h\left(st\right)g\left(t\right)dt}{{\int }_0^{\infty }h\left(st\right)f\left(t\right)dt}}.\hfill \end{array}$$

Therefore, in view of (10), we obtain

$$\begin{array}{cc}\hfill & {\left({\int }_0^{\infty }h\left(st\right)f\left(t\right)dt\right)}^2{T}^{\prime }\left(s\right)\hfill \\ & =\left({\int }_0^{\infty }t{h}^{\prime }\left(st\right)g\left(t\right)dt\right)·\left({\int }_0^{\infty }h\left(st\right)f\left(t\right)dt\right)\hfill \\ \hfill & -\left({\int }_0^{\infty }h\left(st\right)g\left(t\right)dt\right)·\left({\int }_0^{\infty }t{h}^{\prime }\left(st\right)f\left(t\right)dt\right)\hfill \\ \hfill & =E\left[h\left(sX\right)\right]·{\displaystyle \frac{\partial E\left[h\right(sY\left)\right]}{\partial s}}-E\left[h\left(sY\right)\right]·{\displaystyle \frac{\partial E\left[h\right(sX\left)\right]}{\partial s}}\hfill \\ \hfill & \ge 0.\hfill \end{array}$$

(11)

Consequently, the function $s\mapsto E\left[h\right(sY\left)\right]/E\left[h\right(sX\left)\right]$ is increasing on $(0,\infty ).$ Consequently, we deduce that if $X{\le }_{lr}Y$, then $X{\le }_{\mathrm{Gcx}-\mathrm{r}}Y.$ Furthermore, we have

$$\begin{array}{cc}\hfill {\displaystyle \frac{E\left[h\right(sY\left)\right]}{E\left[h\right(sX\left)\right]}}& \ge \underset{s⟶0}{lim}{\displaystyle \frac{{\int }_0^{\infty }h\left(st\right)g\left(t\right)dt}{{\int }_0^{\infty }h\left(st\right)f\left(t\right)dt}}\hfill \\ \hfill & ={\displaystyle \frac{{\int }_0^{\infty }g\left(t\right)dt}{{\int }_0^{\infty }f\left(t\right)dt}}=1.\hfill \end{array}$$

The proof of Theorem 2 is completed. □

Remark 3.

It is worth mentioning that the inequality (11) is reversed when the function h is geometrically concave and the conditions (i) and (ii) of Theorem 2 are satisfied. Therefore, the following relations

$$Y{\le }_{lr}X⟹X{\le }_{Gcv-r}Y⟹X{\le }_{Gcv}Y$$

hold true.

The following result is well known (see, for instance, ([2], Theorem 5.B.10)); we provide an alternative proof.

Corollary 1.

Let X and Y be two non-negative random variables supported on $(0,\infty )$ with probability density functions $f\left(t\right)$ and $g\left(t\right),$ and assume that both have finite first moments. Then, the following implications hold:

$$X{\le }_{lr}Y⟹X{\le }_{Lt-r}Y⟹X{\le }_{Lt}Y.$$

Proof. 

Assume that $X{\le }_{lr}Y,$ then the function $t\mapsto {\displaystyle \frac{g\left(t\right)}{f\left(t\right)}}$ is increasing on $(0,\infty ).$ Now, we set $h_1\left(t\right)=e^{-t}$, then the function $t\mapsto t{h}_1^{\prime }\left(t\right)/h_1\left(t\right)$ is decreasing, i.e., the function $t\mapsto h_1\left(t\right)$ is geometrically concave on $[0,\infty ).$ Therefore, by applying Theorem 2, we establish that the function $s\mapsto {\displaystyle \frac{E\left[e^{-sY}\right]}{E\left[e^{-sX}\right]}}$ is decreasing on $(0,\infty ).$ This, in turn, implies that $X{\le }_{Lt-r}.$ Moreover, we obtain that

$$\begin{array}{cc}\hfill {\displaystyle \frac{E\left[e^{-sY}\right]}{E\left[e^{-sX}\right]}}& \le \underset{s\to 0}{lim}{\displaystyle \frac{E\left[e^{-sY}\right]}{E\left[e^{-sX}\right]}}\hfill \\ & =\underset{s\to 0}{lim}{\displaystyle \frac{{\int }_0^{\infty }{e}^{-sx}f\left(x\right)dx}{{\int }_0^{\infty }{e}^{-sx}g\left(x\right)dx}}\hfill \\ \hfill & ={\displaystyle \frac{{\int }_0^{\infty }f\left(x\right)dx}{{\int }_0^{\infty }g\left(x\right)dx}}=1.\hfill \end{array}$$

(12)

This, in turn, implies that the following inequality,

$$E\left[e^{-sY}\right]\le E\left[e^{-sX}\right],$$

(13)

is valid for all $s>0.$ This completes the proof. □

Corollary 2.

Let X and Y be two non-negative random variables supported on $(0,\infty )$ with probability density functions $f\left(t\right)$ and $g\left(t\right)$ such that $E\left[e^{s_{0}Y}\right]<\infty$ for some $s_0>0.$ Then

$$X{\le }_{\mathrm{lr}}Y⟹X{\le }_{\mathrm{mgf}}Y.$$

Proof. 

In our case, we set $h_2\left(t\right)=e^t$ then the function $t\mapsto t{h}_2^{\prime }\left(t\right)/h_2\left(t\right)=t$ is increasing on $(0,\infty )$. Therefore, thanks to Theorem 2, we conclude that the function $s\mapsto E\left[e^{sY}\right]/E\left[e^{sX}\right]$ is increasing on $(0,\infty )$. Hence, we obtain

$$\begin{array}{cc}\hfill {\displaystyle \frac{E\left[e^{sY}\right]}{E\left[e^{sX}\right]}}& \ge \underset{s⟶0}{lim}{\displaystyle \frac{E\left[e^{sY}\right]}{E\left[e^{sX}\right]}}\hfill \\ & =\underset{s⟶0}{lim}{\displaystyle \frac{{\int }_0^{\infty }{e}^{st}g\left(t\right)dt}{{\int }_0^{\infty }{e}^{st}f\left(t\right)dt}}\hfill \\ \hfill & =\underset{s⟶0}{lim}{\displaystyle \frac{{\int }_0^{\infty }g\left(t\right)dt}{{\int }_0^{\infty }f\left(t\right)dt}}=1.\hfill \end{array}$$

(14)

Hence, X is smaller than Y in the moment generating function order. □

The relationship between the orders ${\le }_{\mathrm{lr}},{\le }_{\mathrm{mom}-\mathrm{r}}$ and ${\le }_{\mathrm{mom}}$ are described in the next Proposition.

Proposition 1.

Let X and Y be two non-negative random variables supported on $(0,\infty )$ with probability density functions $f\left(t\right)$ and $g\left(t\right)$ such that $E\left[Y^n\right]<\infty$ for all $n\in \mathbb{N}.$ Then,

$$X{\le }_{\mathrm{lr}}Y⟹X{\le }_{\mathrm{mom}-\mathrm{r}}Y⟹X{\le }_{\mathrm{mom}}Y.$$

Proof. 

Under the given conditions, the function $t\mapsto g\left(t\right)/f\left(t\right)$ is increasing on $(0,\infty ).$ Now, we define the function $\mathcal{E}:(0,\infty )⟶(0,\infty )$ by

$$\mathcal{E}\left(s\right)={\displaystyle \frac{E\left[Y^s\right]}{E\left[X^s\right]}},s>0.$$

Therefore, we obtain

$$\begin{array}{cc}\hfill E^2\left[X^s\right]{\mathcal{E}}^{\prime }\left(s\right)& =E\left[X^s\right]·{\displaystyle \frac{\partial E\left[Y^s\right]}{\partial s}}-E\left[Y^s\right]·{\displaystyle \frac{\partial E\left[X^s\right]}{\partial s}}\hfill \\ \hfill & =\left({\int }_0^{\infty }{t}^{s}f\left(t\right)dt\right)·\left({\int }_0^{\infty }log\left(t\right)t^{s}g\left(t\right)dt\right)\hfill \\ & -\left({\int }_0^{\infty }{t}^{s}g\left(t\right)dt\right)·\left({\int }_0^{\infty }log\left(t\right)t^{s}f\left(t\right)dt\right).\hfill \end{array}$$

(15)

Another use of the Chebyshev integral inequality (7) is that $\rho ,\varphi ,\psi :(0,\infty )⟶\mathbb{R}$, defined by

$$\rho \left(t\right)=t^{s}f\left(t\right),\varphi \left(t\right)=log\left(t\right),\mathrm{and}\psi \left(t\right)={\displaystyle \frac{g\left(t\right)}{f\left(t\right)}}.$$

We observe that the functions $\varphi$ and $\psi$ are increasing on $(0,\infty )$ under the given conditions. Then, from (7), it follows that

$$\left({\int }_0^{\infty }{t}^{s}f\left(t\right)dt\right)·\left({\int }_0^{\infty }{t}^{s}log\left(t\right)g\left(t\right)dt\right)\ge \left({\int }_0^{\infty }{t}^{s}g\left(t\right)dt\right)·\left({\int }_0^{\infty }{t}^{s}log\left(t\right)f\left(t\right)dt\right).$$

(16)

Owing to the inequalities (15) and (16), we conclude that the function $s\mapsto \mathcal{E}\left(s\right)$ is increasing on $(0\infty ).$ Therefore, for all $s>0$, we obtain $\mathcal{E}\left(s\right)\ge {lim}_{s⟶0}\mathcal{E}\left(0\right)=1.$ In particular, for all $n\in \mathbb{N}$, we conclude that $E\left(X^n\right)\le E\left(Y^n\right).$ This ends the proof of Proposition 1. □

4. Second Set of Main Results

In this section, we derive new inequalities involving the Laplace transform of non-negative random variables with bounded support. Under likelihood ratio ordering, we establish a lower bound for the Laplace transform and identify some conditions under which this ordering implies Laplace transform ordering. In addition, two-sided exponential bounds for $E\left[e^{-sX}\right]$ are obtained. Finally, we establish a sufficient criterion that ensures the Laplace transform ordering between two random variables.

Our first main result in this section reads as follows.

Proposition 2.

Suppose that the function $f:[0,\infty )\to (0,\infty )$ is of the class $C^1\left([0,\infty )\right)$ such that $f^{\prime }\left(0\right)\le 0.$ Let X and $X^{\prime }$ be non-negative random variables supported on $[0,\infty )$ with probability density functions $f\left(t\right)$ and $f^{\prime }\left(t\right)$, respectively. If $X{<}_{lr}{X}^{\prime },$ then the following estimation,

$$E\left[e^{-sX}\right]\ge {\displaystyle \frac{f^2\left(0\right)}{f\left(0\right)s-f^{\prime }\left(0\right)}},$$

(17)

holds true for all $s>0.$

Proof. 

Assume that $X{<}_{lr}{X}^{\prime },$ then the function $t\mapsto {\displaystyle \frac{f^{\prime }\left(t\right)}{f\left(t\right)}}$ is increasing. From Lemma 2, it follows that the function $\Xi$ defined on $[0,\infty )$ by

$$\Xi \left(t\right)={\displaystyle \frac{log\left(f\right(t\left)\right)-log\left(f\right(0\left)\right)}{t}}$$

(18)

is increasing on $[0,\infty ).$ From this fact and from the l’Hospital rule, we obtain

$$\begin{array}{cc}\hfill \Xi \left(t\right)\ge & \underset{t⟶0}{lim}{\displaystyle \frac{log\left(f\right(t\left)\right)-log\left(f\right(0\left)\right)}{t}}\hfill \\ \hfill & ={\displaystyle \frac{f^{\prime }\left(0\right)}{f\left(0\right)}}\hfill \end{array}$$

(19)

for all $t>0.$ This, in turn, implies that

$$f\left(t\right)\ge f\left(0\right){\mathrm{e}}^{{\displaystyle \frac{t{f}^{\prime }\left(0\right)}{f\left(0\right)}}},t>0.$$

(20)

Owing to the above inequality, we establish that

$$\begin{array}{cc}\hfill E\left[e^{-tX}\right]& ={\int }_0^{\infty }{e}^{-tx}f\left(x\right)dx\hfill \\ \hfill & \ge f\left(0\right){\int }_0^{\infty }{e}^{-{\displaystyle \frac{x(f\left(0\right)t-f^{\prime }\left(0\right))}{f\left(0\right)}}}dx\hfill \\ \hfill & ={\displaystyle \frac{f^2\left(0\right)}{f\left(0\right)t-f^{\prime }\left(0\right)}},\hfill \end{array}$$

(21)

which completes the proof. □

Remark 4.

We note that if $X^{\prime }{\le }_{\mathrm{lr}}X$, then the inequality (17) is reversed. Indeed, since $X^{\prime }{\le }_{\mathrm{lr}}X,$ the function $t\mapsto {\displaystyle \frac{f^{\prime }\left(t\right)}{f\left(t\right)}}$ is decreasing. According to Lemma 2, we conclude that the function $t\mapsto \Xi \left(t\right)$, defined by (18), is also decreasing on $[0,\infty ).$ This observation, together with l’Hospital’s rule, implies that the inequality in (20) is reversed. The rest follows immediately.

Corollary 3.

Suppose that the functions $f,g:[0,\infty )\to (0,\infty )$ are of the class $C^1\left([0,\infty )\right)$ such that $max(f^{\prime }\left(0\right),g^{\prime }\left(0\right))\le 0$, satisfying the following inequality

$${\displaystyle \frac{g^2\left(0\right)}{f\left(0\right)t-f^{\prime }\left(0\right)}}\le {\displaystyle \frac{f^2\left(0\right)}{f\left(0\right)t-f^{\prime }\left(0\right)}}.$$

(22)

Then, the following assertion

$$X{\le }_{\mathrm{lr}}{X}^{\prime }\mathrm{and}{Y}^{\prime }{\le }_{\mathrm{lr}}Y⟹X{\le }_{\mathrm{Lt}}Y$$

holds true.

Proof. 

Since the functions $f,g$ is of the class $C^1\left([0,\infty )\right)$ such that $max(f^{\prime }\left(0\right),g^{\prime }\left(0\right))\le 0$, we deduce that $X{\le }_{\mathrm{lr}}{X}^{\prime }$ and $Y^{\prime }{\le }_{\mathrm{lr}}Y.$ According to Proposition 2, we conclude that

$$E\left[e^{-sX}\right]\ge {\displaystyle \frac{f^2\left(0\right)}{f\left(0\right)s-f^{\prime }\left(0\right)}},$$

(23)

and

$$E\left[e^{-sY}\right]\le {\displaystyle \frac{g^2\left(0\right)}{g\left(0\right)s-g^{\prime }\left(0\right)}}.$$

(24)

Now, combining (23) and (24) with (22), we establish that

$$E\left[e^{-sY}\right]\le E\left[e^{-sX}\right],\mathrm{for}\mathrm{all}s>0.$$

The proof of Corollary 3 is now complete. □

Theorem 3.

Let X be a real-valued continuous random variable supported on $[0,a],a>0$ with probability density function $f.$ For $s>0$, the following holds:

$$\begin{array}{cc}\hfill P(X<a)·exp\left(-{\displaystyle \frac{sE\left[X\right]}{P(X<a)}}\right)& \le E\left[e^{-sX}\right]\hfill \\ \hfill & \le P(X<a)+{\displaystyle \frac{(e^{-sa}-1)}{a}}E\left(X\right).\hfill \end{array}$$

(25)

Proof. 

Since the function $x\mapsto e^{-sx}$ is convex on $[0,a]$, then we take its arc one estimates by the secant on $[0,a];$ that is,

$$e^{-sx}\le 1+{\displaystyle \frac{x}{a}}(e^{-sx}-1),t\in [0,a).$$

Therefore, we obtain

$$\begin{array}{cc}\hfill E\left[e^{-sX}\right]& ={\int }_0^a{e}^{-sx}f\left(x\right)dx\hfill \\ \hfill & \le {\int }_0^{a}f\left(x\right)dx+{\displaystyle \frac{(e^{-xs}-1)}{a}}{\int }_0^{a}xf\left(x\right)dx\hfill \\ \hfill & =P(X<a)+{\displaystyle \frac{(e^{-sa}-1)}{a}}E\left(X\right).\hfill \end{array}$$

Now, focus on the left-hand side of (25). Let us recall Jensen’s integral inequality (see, for instance, ([14], Chap. I, Equation (7.15)), that is

$$\Psi \left(\mathcal{A}\left(\phi \right)\right)\le \mathcal{A}(\Psi \left(\phi \right)),$$

(26)

where $\Psi$ is a convex function and $\phi$ is integrable with respect to a probability measure $\sigma$, and $\mathcal{A}\left(\phi \right)$ is defined by

$$\mathcal{A}\left(\phi \right)=\left({\int }_c^d\phi \left(t\right)d\sigma \left(t\right)\right)/\left({\int }_c^{d}d\sigma \left(t\right)\right).$$

Upon setting

$$(c,d)=(0,a),\Psi \left(x\right)=e^{-sx},\phi \left(x\right)=x\mathrm{and}d\sigma \left(x\right)=f\left(x\right)dx$$

in the Jensen integral inequality and straightforward calculation, we establish the left-hand side of inequality (25). □

In view of the above Theorem, we compute the following result.

Corollary 4.

Let X and Y be a two real-valued continuous random variable supported on $[0,a],a>0$ with probability density functions f and g. If the following inequality

$$P(Y\le a)+{\displaystyle \frac{(e^{-sa}-1)}{a}}E\left(Y\right)\le P(X\le a)·exp\left(-{\displaystyle \frac{sE\left[X\right]}{P(X\le a)}}\right)$$

is valid for all $s>0,$ then $X{\le }_{\mathrm{Lt}}Y.$

5. Conclusions

In this investigation, we have established new properties of the likelihood ratio and Laplace transform orderings by identifying a gap in the application of the Chebyshev integral inequality and the monotone form of L’Hospital’s rule. Specifically, by addressing this gap, we derived new bilateral integral functional inequalities related to the ratio of probability density functions and the ratio of their cumulative distributions that satisfy the likelihood ratio ordering. Additionally, we presented alternative proofs for existing results in order statistics. Moreover, we introduced the geometrically convex order and derived its relationship with the likelihood ratio order. Finally, we provided new characterizations of the Laplace transform ordering.

Several potential directions for further research include extending the results to multivariate distributions and exploring applications in reliability theory and survival analysis. One could also generalize the integral inequalities to broader classes of functions and develop numerical methods to verify stochastic orderings in data. Additionally, the derived inequalities could be used to characterize more probability distributions and to explore connections with other stochastic orders. However, these goals will be addressed and presented in future work.