Hermite–Hadamard type inequalities for r-convex positive stochastic processes

ABSTRACT

In the present work, we aim to find Hermite–Hadamard type inequality for r-convex positive stochastic processes. The case of the product of an r-convex and s-convex stochastic process is also investigated.

KEYWORDS:

• Hermite–Hadamard inequalities
• convex stochastic processes
• root-mean-square integral

2010 AMS SUBJECT CLASSIFICATIONS:

• 26D15
• 26A51
• 33B20
• 52A40
• 52A41

1. Introduction

Convex stochastic processes are introduced by Nikodem [1] in 1980. Skowroński [2] generalized certain well-known properties of convex functions to convex stochastic processes. In [3], Kotrys investigated the Hermite–Hadamard inequality involving convex stochastic processes. In recent works on convex stochastic processes, many authors studied different integral inequalities, see [3–9].

Motivated from the above works, we study the Hermite–Hadamard type inequalities for r-convex stochastic processes. We also consider the case of a product of an r-convex and s-convex stochastic process.

2. Preliminaries

Let $(\mathrm{\Omega },\mathcal{F},P)$ be an arbitrary probability space. A $\mathcal{F}$-measurable function $X:\mathrm{\Omega }⟶\mathbb{R}$ is said to be a random variable. Let $I\subset \mathbb{R}$ be an interval. Then a function $X:I\times \mathrm{\Omega }⟶\mathbb{R}$ is said to be the stochastic process if the function $X(t,.)$ is a random variable for all t in I.

Let $P-lim$ and $E\left[X\right]$ denote the limit in probability and the expected value of X, respectively. A stochastic process $X:I\times \mathrm{\Omega }⟶\mathbb{R}$ is said to be continuous in probability if $P-\underset{t\to t_0}{lim}$ $X(t,.)=X(t_0,.)$ for all $t_0\in I$ while it is said to mean-square continuous in I if $\underset{t\to t_0}{lim}E\left[\right(X\left(t\right)-X\left(t_0\right){)}^2]=0,\mathrm{\forall }{t}_0\in I.$

It is worthy to note that if $X:I\times \mathrm{\Omega }⟶\mathbb{R}$ is mean-square continuous, then it is continuous in I but the converse does not hold.

The mean-square integral is defined as: A random variable Y : $\mathrm{\Omega }⟶\mathbb{R}$ is said to be the mean-square integral of the stochastic process $X:I\times \mathrm{\Omega }⟶\mathbb{R}$ on $[a,b]\subset I$ with $E\left[\right(X\left(t\right){)}^2]<\mathrm{\infty }$ ∀$t\in I$, if for every normal sequence of partitions of $[a,b]$, the following relation holds $\underset{n\to \mathrm{\infty }}{lim}E\left[{\left(\sum _{k=1}^{n}X\right({\mathrm{\Theta }}_k\left)\right(t_k-t_{k-1})-Y)}^2\right]=0,$ where ${\mathrm{\Theta }}_k\in [t_{k-1},t_k],k=1,2,3,\dots ,n$ and $a=t_0<t_1<t_2<\dots <t_n=b$ is the partition of $[a,b]$. Thus, we can write $Y(.)={\int }_a^{b}X(t,.)\mathrm{d}t(a.e).$ The assumption of the mean-square continuity of the stochastic process is enough for the mean-square integral to exist.

From the definition of mean-square integral, we immediately have the following relation. $\begin{array}{rl}& X(t,.)\le Y(t,.)(\mathrm{a}.\mathrm{e}),\mathrm{for}t\in [a,b]==>{\int }_a^{b}X(t,.)\mathrm{d}t\\ & \le {\int }_a^{b}Y(t,.)\mathrm{d}t(\mathrm{a}.\mathrm{e}.).\end{array}$

That is a mean-square integral is monotonic. Throughout the entire paper, the monotonicity of mean-square integral and positivity of the stochastic process will be frequently used. Now, we define the following.

Definition 2.1

A stochastic process $X:I\times \mathrm{\Omega }⟶[0,\mathrm{\infty })$ is said to be r-convex, if for each $u,v\in I$ and $\lambda \in [0,1]$ (1) $\begin{array}{rl}& X(\lambda u+(1-\lambda )v,.)\\ & \le \{\begin{array}{l}\left(\lambda X^r\right(u,.)+(1-\lambda \left)X^r\right(v,.){)}^{1/r},r\ne 0\\ \left(X\right(u,.\left){)}^{\lambda }\right(X(v,.){)}^{1-\lambda },r=0\end{array}(\mathrm{a}.\mathrm{e}.).\end{array}$(1) Note that 0-convex stochastic processes are logarithmic convex (see [9]) and 1-convex stochastic processes are the classical convex stochastic process. Note that if X is r-convex, then $X^r$ is convex stochastic process (r>0).

The above definition is analogue of the r-convex functions in the classical convex functions, see [10, 11].

Kotrys [3] studied the following well-known Hermite–Hadamard type inequality $\begin{array}{rl}X\left(\frac{u+v}{2}\right)& \le \frac{1}{v-u}{\int }_u^{v}X(t,.)\mathrm{d}x\\ & \le \frac{X(u,.)+X(v,.)}{2}(\mathrm{a}.\mathrm{e}),\end{array}$ where $X:I\times \mathrm{\Omega }⟶\mathbb{R}$ is Jensen convex and mean-square continuous stochastic process.

Hermite–Hadamard type inequalities for log-convex functions was investigated by Dragomir and Mond [12].

Pachpatte [13, 14] also gave some other refinements of these inequalities related with differentiable log-convex functions.

Tomar et al. [9] proved the following inequalities:

Let $X:I\times \mathrm{\Omega }⟶(0,\mathrm{\infty })$ be a log-convex stochastic process. Then for $u,v\in I$ with u<v, the following inequalities hold: $\begin{array}{rl}X(\frac{u+v}{2},.)& \le \frac{1}{v-u}{\int }_u^{v}X(t,.)\mathrm{d}t\\ & \le \frac{X(u,.)-X(v,.)}{\log X(u,.)-\log X(v,.)}\\ & =L\left(X\right(u,.),X(v,.\left)\right)(\mathrm{a}.\mathrm{e}),\end{array}$ where $L(p,q)$ $(p\ne q)$ is the logarithmic mean of real numbers p,q>0.

We also need the Jensen inequality for convex stochastic process proved by Sarikaya et al. [7].

For a convex stochastic process $X:I\times \mathrm{\Omega }⟶\mathbb{R}$, we have $\begin{array}{rl}& X\left(\frac{1}{v-u}{\int }_u^v\varphi \right(t,.)\mathrm{d}t,.)\\ & \le \frac{1}{v-u}{\int }_u^{v}X\circ \varphi (t,.)\mathrm{d}t(\mathrm{a}.\mathrm{e}.),\end{array}$ where $\varphi :I\times \mathrm{\Omega }⟶I\subset \mathbb{R}$ is an arbitrary non-negative integrable stochastic process. For a brief history and studies, we refer to [8, 15–20].

3. Main results

Theorem 3.1

Let $X:I\times \mathrm{\Omega }⟶[0,\mathrm{\infty })$ be a r-convex stochastic process with mean-square continuity in I.  Then for $u,v\in I$ with u<v, the below inequality holds (2) $\begin{array}{rl}\frac{1}{v-u}{\int }_u^{v}X(t,.)\mathrm{d}t& \le {\left[\frac{X^r(u,.)+X^r(v,.)}{2}\right]}^{1/r},\\ & r\ge 1,(\mathrm{a}.\mathrm{e}.).\end{array}$(2)

Proof.

From Jensen inequality, we obtain ${\left(\frac{1}{v-u}{\int }_u^{v}X\right(t,.\left)\mathrm{d}t\right)}^r\le \frac{1}{v-u}{\int }_u^v{X}^r(t,.)\mathrm{d}t(\mathrm{a}.\mathrm{e}.).$ Since $X^r$ is convex, then Hemite-Hadamard type inequality for convex stochastic processes yields us (see [3]) $\frac{1}{v-u}{\int }_u^v{X}^r(t,.)\mathrm{d}t.\le \frac{X^r(u,.)+X^r(v,.)}{2}(\mathrm{a}.\mathrm{e}.).$ Hence, $\frac{1}{v-u}{\int }_u^{v}X(t,.)\mathrm{d}t\le {\left[\frac{X^r(u,.)+X^r(v,.)}{2}\right]}^{1/r}(\mathrm{a}.\mathrm{e}.).$ This completes the proof.

Corollary 3.1

Let $X:I\times \mathrm{\Omega }⟶[0,\mathrm{\infty })$ be 1-convex stochastic process with mean-square continuity in I. Then for $u,v\in I$ with u<v, the following inequality holds. Then $\frac{1}{v-u}{\int }_u^{v}X(t,.)\mathrm{d}t\le \frac{X(u,.)+X(v,.)}{2}.$

Theorem 3.2

Let $X:I\times \mathrm{\Omega }⟶(0,\mathrm{\infty })$ be r-convex stochastic process ($r\ge 0$) with mean-square continuity in I. Then for $u,v\in I$  with u<v, $X(u,.)\ne X(v,.)$ the following inequalities hold (3) $\begin{array}{rl}& \frac{1}{v-u}{\int }_u^{v}X(t,.)\mathrm{d}t\\ & \le \{\begin{array}{l}\frac{r}{r+1}\left[\frac{X^{r+1}(v,.)-X^{r+1}(u,.)}{X^r(v,.)-X^r(u,.)}\right],r\ne 0\\ \frac{X(v,.)-X(u,.)}{\log X(v,.)-\log X(u,.)},r=0\end{array}(\mathrm{a}.\mathrm{e}.).\end{array}$(3)

Proof.

For $r=0,$ Tomar et al. [9] proved this result. We proceed for the case r>0. Since X is r-convex stochastic process, for all $\lambda \in [0,1],$ we have $\begin{array}{rl}& X(\lambda u+(1-\lambda )v,.)\\ & \le \left(\lambda X^r\right(u,.)+(1-\lambda \left)X^r\right(v,.){)}^{1/r}(\mathrm{a}.\mathrm{e}.).\end{array}$ Therefore by using the same method as of [3], we have $\begin{array}{rl}& \frac{1}{v-u}{\int }_u^{v}X(t,.)\mathrm{d}t\\ & \le {\int }_0^1{\left(X^r\right(v,.)+\lambda (X^r(u,.)-X^r(v,.)\left)\right)}^{1/r}\mathrm{d}\lambda (\mathrm{a}.\mathrm{e}.).\end{array}$ Putting $\tau =X^r(v,.)+\lambda \left(X^r\right(u,.)-X^r(v,.\left)\right)$, we have $\begin{array}{rl}& \frac{1}{v-u}{\int }_u^{v}X(t,.)\mathrm{d}t\le \frac{1}{X^r(v,.)-X^r(u,.)}{\int }_{X^r(u,.)}^{X^r(v,.)}{\left(\tau \right)}^{1/r}\mathrm{d}\tau \\ & =\frac{r}{r+1}\left[\frac{X^{r+1}(v,.)-X^{r+1}(u,.)}{X^r(v,.)-X^r(u,.)}\right],(\mathrm{a}.\mathrm{e}.)\end{array}$ which completes the proof.

Note that for $r=1,$ in the above theorem, we have the same inequality again as in Corollary 3.1.

Theorem 3.3

Let $X:I\times \mathrm{\Omega }⟶(0,\mathrm{\infty })$ be r-convex ($0\le r\le s$) stochastic process with mean-square continuity in I. Then X is s-convex stochastic process.

Proof.

To prove this, we need the following inequality for non-negative real numbers x, y (4) $\begin{array}{rl}{x}^{1-\lambda }{y}^{\lambda }& \le \left(\right(1-\lambda )x^r+\lambda y^r{)}^{1/r}\\ & \le \left(\right(1-\lambda )x^s+\lambda y^s{)}^{1/s},\text{(see [11])}\end{array}$(4) where $0\le \lambda \le 1,$ $0\le r\le s$. Since X is r-convex stochastic process, by inequality (4) for all $u,v\in I$,$\lambda \in [0,1],$ we obtain $\begin{array}{rl}& X(\lambda u+(1-\lambda )v,.)\\ & \le \{\begin{array}{l}\left({\lambda X}^r{(u,.)+(1-\lambda )X}^r(v,.){)}^{1/r}\right({\le \lambda X}^s(u,.)\\ +(1-\lambda )X^s(v,.){)}^{1/s},0<r\le s\\ {\left(X\right(u,.\left)\right)}^{\lambda }{\left(X\right(v,.\left)\right)}^{1-\lambda }\le \left(\lambda X^s\right(u,.)\\ +(1-\lambda )X^s(v,.){)}^{1/s},0=r\le s.\end{array}(\mathrm{a}.\mathrm{e}.).\end{array}$ Hence, X is a s-convex stochastic process.

As a special case of the above theorem, we deduce the following results.

Corollary 3.2

Let $X:I\times \mathrm{\Omega }⟶(0,\mathrm{\infty })$ be r-convex ($0\le r\le 1$) stochastic process with mean-square continuity in the interval I. Then X  is a convex stochastic process.

Corollary 3.3

Let $X:I\times \mathrm{\Omega }⟶(0,\mathrm{\infty })$ be r-convex ($0\le r\le s$) stochastic process with mean-square continuity in the interval I. Then the following inequalities holds: $\begin{array}{rl}& \frac{1}{v-u}{\int }_u^{v}X(t,.)\mathrm{d}t\\ & \le \{\begin{array}{l}\frac{r}{r+1}\left[\frac{X^{r+1}(v,.)-X^{r+1}(u,.)}{X^r(v,.)-X^r(u,.)}\right]\\ \le \frac{s}{s+1}\left[\frac{X^{s+1}(v,.)-X^{s+1}(u,.)}{X^s(v,.)-X^s(u,.)}\right],0<r\le s\\ \frac{X(v,.)-X(u,.)}{\log X(v,.)-\log X(u,.)}\\ \le \frac{s}{s+1}\left[\frac{X^{s+1}(v,.)-X^{s+1}(u,.)}{X^s(v,.)-X^s(u,.)}\right],0=r\le s.\end{array}\\ & (\mathrm{a}.\mathrm{e}.).\end{array}$

Proof.

The proof follows at once by using Theorem 3.3 and proceeding on similar lines as Theorem 3.1.

Theorem 3.4

Let $X,Y:I\times \mathrm{\Omega }⟶(0,\mathrm{\infty })$ be r-convex and s-convex stochastic process with mean-square continuity respectively. Then for $u,v\in I$  with u<v, the following inequalities hold: $\begin{array}{rl}& \frac{1}{v-u}{\int }_u^{v}X(t,.)Y(t,.)\mathrm{d}t\\ & \le \frac{1}{2}\frac{r}{r+2}\left[\frac{X^{r+2}(v,.)-X^{r+2}(u,.)}{X^r(v,.)-X^r(u,.)}\right]\\ & +\frac{1}{2}\frac{s}{s+2}\left[\frac{Y^{s+2}(v,.)-Y^{s+2}(u,.)}{Y^s(v,.)-Y^s(u,.)}\right]\\ & \left(X\right(u,.)\ne X(v,.),Y(u,.)\ne Y(v,.\left)\right),(\mathrm{a}.\mathrm{e}.).\end{array}$

Proof.

Since X is r-convex stochastic process and Y is s-convex stochastic convex, for all $\lambda \in [0,1],$ we have $\begin{array}{rl}& X(\lambda u+(1-\lambda )v,.)\le \left(\lambda X^r\right(u,.)\\ & +(1-\lambda )X^r(v,.){)}^{1/r}(\mathrm{a}.\mathrm{e}.),\\ & Y(\lambda u+(1-\lambda )v,.))\le (\lambda Y^s(u,.)\\ & +(1-\lambda )Y^s(v,.){)}^{1/s}(\mathrm{a}.\mathrm{e}.).\end{array}$ Therefore, $\begin{array}{rl}& \frac{1}{v-u}{\int }_u^{v}X(t,.)Y(t,.)\mathrm{d}t\\ & ={\int }_0^{1}X(\lambda u+(1-\lambda )v,.)Y(\lambda u+(1-\lambda )v,.))\mathrm{d}\lambda \\ & \le {\int }_0^1\left(\lambda X^r\right(u,.)+(1-\lambda \left)X^r\right(v,.\left){)}^{1/r}\right(\lambda Y^s(u,.)\\ & +(1-\lambda )Y^s(v,.){)}^{1/s}\mathrm{d}\lambda \\ & ={\int }_0^1{\left(X^r\right(v,.)+\lambda (X^r(u,.)-X^r(v,.)\left)\right)}^{1/r}\\ & \times {\left(Y^u\right(v,.)+\lambda (Y^s(u,.)-Y^s(v,.)\left)\right)}^{1/s}\mathrm{d}\lambda (\mathrm{a}.\mathrm{e}.).\end{array}$ Now applying Cauchy's inequality, we obtain $\begin{array}{rl}& {\int }_0^1{\left(X^r\right(v,.)+\lambda (X^r(u,.)-X^r(v,.)\left)\right)}^{1/r}\\ & \times {\left(Y^s\right(v,.)+\lambda (Y^s(u,.)-Y^s(v,.)\left)\right)}^{1/s}\mathrm{d}\lambda \\ & \le \frac{1}{2}{\int }_0^1{\left[X^r\right(v,.)+\lambda (X^r(u,.)-X^r(v,.)\left)\right]}^{2/r}\mathrm{d}\lambda \\ & +\frac{1}{2}{\int }_0^1{\left[Y^s\right(v,.)+\lambda (Y^s(u,.)-Y^s(v,.)\left)\right]}^{2/s}\mathrm{d}\lambda (\mathrm{a}.\mathrm{e}.).\end{array}$ If we choose $\tau =X^r(v,.)+\lambda \left(X^r\right(u,.)-X^r(v,.\left)\right)$, $\eta =Y^s(v,.)+\lambda \left(Y^s\right(u,.)-Y^s(v,.\left)\right)$, then we obtain the following inequality $\begin{array}{rl}& {\int }_0^1{\left(X^r\right(v,.)+\lambda (X^r(u,.)-X^r(v,.)\left)\right)}^{1/r}\\ & \times {\left(Y^s\right(v,.)+\lambda (Y^s(u,.)-Y^s(v,.)\left)\right)}^{1/s}\mathrm{d}\lambda \\ & \le \frac{1}{2}\frac{1}{X^r(v,.)-X^r(u,.)}{\int }_{X^r(u,.)}^{X^r(v,.)}{\left(\tau \right)}^{2/r}\mathrm{d}\tau \\ & +\frac{1}{2}\frac{1}{Y^s(v,.)-Y^s(u,.)}{\int }_{Y^s(u,.)}^{Y^s(v,.)}{\left(\eta \right)}^{2/s}\mathrm{d}\eta \\ & =\frac{1}{2}\frac{r}{r+2}\left[\frac{X^{r+2}(v,.)-X^{r+2}(u,.)}{X^r(v,.)-X^r(u,.)}\right]\\ & +\frac{1}{2}\frac{s}{s+2}\left[\frac{Y^{s+2}(v,.)-Y^{s+2}(u,.)}{Y^s(v,.)-Y^s(u,.)}\right],(\mathrm{a}.\mathrm{e}.),\end{array}$ which leads us to the required result.

Note: By putting $X=Y,$ $r=s=2,$ in the above theorem, we have the following inequality $\frac{1}{v-u}{\int }_u^v{X}^2(t,.)\mathrm{d}t\le \frac{X^2(u,.)+X^2(v,.)}{2}(\mathrm{a}.\mathrm{e}.).$ The following result can be easily proved by proceeding on similar lines as in the above theorem.

Theorem 3.5

Let $X:I\times \mathrm{\Omega }⟶[0,\mathrm{\infty })$ be r-convex and $Y:I\times \mathrm{\Omega }⟶(0,\mathrm{\infty })$ be positive 0-convex stochastic processes with mean-square continuity in I respectively. Then for $u,v\in I$  with u<v, the following inequalities hold: $\begin{array}{rl}& \frac{1}{v-u}{\int }_u^{v}X(t,.)Y(t,.)\mathrm{d}t\\ & \le \frac{1}{2}\frac{r}{r+2}\left[\frac{X^{r+2}(v,.)-X^{r+2}(u,.)}{X^r(v,.)-X^r(u,.)}\right]\\ & +\frac{1}{4}\left[\frac{Y^2(v,.)-Y^2(u,.)}{\log Y(v,.)-\log Y(u,.)}\right]\\ & \left(X\right(u,.)\ne X(v,.)(\mathrm{a}.\mathrm{e}.).\end{array}$

Acknowledgments

The authors would like to thank the referees of this article for their insightful comments which greatly improves the entire presentation of the paper. W. Ul-Haq and Z. Al-Hussain thank Deanship of Scientific Research (DSR) for providing excellent research facilities.

Disclosure statement

No potential conflict of interest was reported by the authors.

ORCID

Wasim Ul-Haq  http://orcid.org/0000-0001-7148-1890

Nasir Rehman  http://orcid.org/0000-0001-9210-7953

Ziyad Ali Alhussain  http://orcid.org/0000-0001-8593-0239