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https://doi.org/10.26637/MJM0703/0011

Further results and applications on continuous

random variables

Mohamed DOUBBI BOUNOUA

1

, Zoubir DAHMANI

2

* and Zakaria BEKKOUCHE

3

Abstract

New results and new applications of fractional calculus for continuous random variables are presented. New expectation and variance identities of orderα≥1are established. Under a new fractional normalisation technique, other weighted random variable inequalities are generated and some classical results are deduced as special cases.

Keywords

Integral inequalities, Riemann-Liouville integral, random variable, fractional expectation, fractional variance, fractional moment.

AMS Subject Classification 26D15, 26A33, 60E15.

1Department of Mathematics, Shanghai China, China.

2Laboratory LPAM, UMAB University, Algeria.

3School of Mathematics and Statistics, Central China Normal University, China.

*Corresponding author:1doubbibounoua.mohamed@yahoo.fr; 2zzdahmani@yahoo.fr;3bezack@mails.ccnu.edu.cn

Article History: Received24March2019; Accepted09May2019 c2019 MJM.

Contents

1 Introduction. . . 429

2 Preliminaries. . . 430

3 Main Results. . . 430

References. . . 435

1. Introduction

The integral inequalities are very important in physics and applied sciences. For some applications of this theory, we refer the reader [2,4,5,8–11]. Let us now introduce some papers that have motivate the present work. We begin by [1] where, T. Cacoullos and V. Papathanasiou established the following nice covariance identity

Cov(h(X),g(X)) =E(z(X)g0(X)), (1.1) whereX is a continuous random variable ( CRV, for short) with support an interval(a,b),−∞≤a<b≤∞, f is

proba-bility density function ( pdf, for short) ofX,gandhare two continuous functions withσ2=V(X),|E(z(X)g0(X))|<∞

and

z(x):= 1

f(x)

x

Z

a

(E[h(X)]−h(t))f(t)dt. (1.2)

Under some other conditions, the same authors [2] proved that

Var[g(X)]≥E

2[z(X)g0(X)]

E[z(X)h(X)] , (1.3)

They also established the following theorem [3]:

Theorem 1.1. For every absolutely continuous function h with h0>0,the inequality:

Var[g(X)]≤E[z(X)

h0(X)(g

0(

X))2] (1.4)

holds, with equality iff g=c1h+c2, where

z f =

x

Z

a

(E(h)−h(t))f(t)dt. (1.5)

Very recently, based on the notions of [4], Z. Dahmani [5] proved the following theorem:

Theorem 1.2. Let X be a random variable with a p.d.f.

defined on[a,b],such thatµ=E(X),σ2=Var(X)and w∈ C([a,b]);Rx

a(b−t)α−1(µ−t)f(t)dt= (b−x)α−1σ2w(x)f(x).

Then, we have

Varg(X),α≥ σ4(X) VarX,αE

2

(2)

Motivated by the papers in [2,3,5,8], in this work, we fo-cus our attention on the application of the Riemann-Liouville fractional integration on random variables. We use some re-cently introduced concepts on continuous random variables to establish new identities on continuous random variables. We prove new lower bounds for the variances as well as for the expectations. For our results, some classical and excellent results of [1–3], that correspond to the standard integration of orderα=1,are generalised for anyα≥1.

2. Preliminaries

Definition 2.1. [7]The Riemann-Liouville fractional integral operator of orderα>0,for a continuous function f on[a,b]

is defined as

a[f(t)] =

1

Γ(α)

t

Z

a

(t−τ)α−1f(τ)dτ,α>0,a<t≤b. (2.1)

For anyα >0 and any positive continuous functionw defined on[a,b],we recall the definitions [4]:

Definition 2.2. The fractional w−weighted expectation func-tion of orderα,for a CRV X with a pdf f defined on[a,b]is given a by:

EX,α,w(t) =

1

Γ(α)

t

Z

a

(t−τ)α−1τw(τ)f(τ)dτ,a≤t<b,

(2.2)

Definition 2.3. The fractional w−weighted expectation func-tion of orderα,for X−E(X)is given by

EX−E(X),α,w(t) =

1

Γ(α)

t

Z

a

(t−τ)α−1(τ−E(X))

×w(τ)f(τ)dτ. (2.3)

Definition 2.4. The fractional w−weighted variance function of orderα,for X is defined as

σX2,α,w(t) = 1 Γ(α)

t

Z

a

(t−τ)α−1(τ−E(X))2w(τ)f(τ)dτ.

(2.4)

Definition 2.5. The fractional w−weighted expectation of orderα for X is defined as

EX,α,w=

1

Γ(α)

b

Z

a

(b−τ)α−1τw(τ)f(τ)dτ. (2.5)

Definition 2.6. The fractional w−weighted variance of order αfor X is given by:

σX2,α,w= 1

Γ(α)

b

Z

a

(b−τ)α−1(τ−E(X))2w(τ)f(τ)dτ.(2.6)

3. Main Results

We begin this section by proving the following theorem. We have:

Theorem 3.1. Let X be a CRV with support an interval

[a,b],−∞<a<b<∞.Suppose that X has a pdf f and

let w be a positive continuous function on[a,b].Then, the following equality holds for anyα≥1 :

Ezg0,α,ω=Egh,α,w−Eh,α,wEg,α,w, (3.1)

where g is an absolutely continuous function withEzg0, α,ω

<

∞,Jaαw f(b) =1, h is a given function and z satisfies:

z(t) = 1

(b−t)α−1w(t)f(t)

× t

Z

a

(b−u)α−1w(u)f(u)(Eh,α,w−h(u))du. (3.2)

Proof.

We have

Ezg0, α,ω=

1

Γ(α)

b

Z

a

(b−t)α−1g0(t) 1

(b−t)α−1w(t)f(t)

× t

Z

a

(b−u)α−1w(u)f(u)(Eh,α,w−h(u))du

×w(t)f(t)dt

= 1 Γ(α)

b

Z

a

g0(t)

t

Z

a

(b−u)α−1w(u)f(u)

(3)

Therefore,

Ezg0, α,ω=

1

Γ(α)

g(t)

t

Z

a

(b−u)α−1w(u)f(u)

×(Eh,α,w−h(u))du|tt==ba

− 1

Γ(α) Zb

a

g(t)

×(b−t)α−1w(t)f(t)(Eh,α,w−h(t))dt

= 1 Γ(α)

g(b)

b

Z

a

(b−u)α−1w(u)f(u)

×(Eh,α,w−h(u))du

− 1

Γ(α) Zb

a

g(t)

×(b−t)α−1w(t)f(t)(Eh,α,w−h(t))dt

=g(b)

Eh,α,w

Γ(α)

b

Z

a

(b−u)α−1w(u)f(u)du

− 1

Γ(α)

b

Z

a

(b−u)α−1w(u)f(u)h(u)du

E

h,α,w

Γ(α)

b

Z

a

g(t)(b−t)α−1w(t)f(t)dt

− 1

Γ(α)

b

Z

a

g(t)(b−t)α−1w(t)f(t)h(t)dt

=g(b)

Eh,α,wJaαw f(b)−Eh,α,w

−Eh,α,wEg,α,w+Egh,α,w.

Thanks to the hypothesisJα

aw f(b) =1, we obtain

Ezg0,α,ω=Egh,α,w−Eh,α,wEg,α,w. The proof is thus achieved.

Remark 3.2. If we takeα=1,w(x) =1in(3.1), we obtain

(1.1).

Lemma 3.3. Let X be a CRV with support an interval[a,b]

,−∞≤a<b≤∞,and a pdf f and suppose that w:[a,b]→

R+is a continuous function. Then, for anyα≥1,we have σg2,α,w=Ezg0,α,ω, (3.3) where g is an absolutely continuous function such that

Ezg0,α,ω<∞,g(b) =E(g)and z(t) = 1

(b−t)α−1w(t)f(t)

× t

Z

a

(b−u)α−1(E(g)−g(u))w(u)f(u)du.

Proof:

We have

Ezg0,α,ω= 1

Γ(α)

b

Z

a

(b−t)α−1g0(t) 1

(b−t)α−1w(t)f(t)

× t

Z

a

(b−u)α−1(E(g)−g(u))

×w(u)f(u)duw(t)f(t)dt

= 1 Γ(α)

b

Z

a

g0(t)

t

Z

a

(b−u)α−1(E(g)−g(u))

×w(u)f(u)dudt.

Hence, we get

Ezg0,α,ω= 1

Γ(α)g(t)

t

Z

a

(b−u)α−1(E(g)−g(u))

×w(u)f(u)du|b a−

1

Γ(α)

b

Z

a

g(t)

×(b−t)α−1(E(g)−g(t))w(t)f(t)dt

= 1 Γ(α)g(b)

b

Z

a

(b−u)α−1(E(g)−g(u))

×w(u)f(u)du− 1

Γ(α)

b

Z

a

(b−t)α−1

×(g(t)(E(g)−g2(t))w(t)f(t)dt.

We use the fact thatg(b) =E(g),then it yields that

Ezg0,α,ω= 1

Γ(α)

b

Z

a

(b−u)α−1(E2(g)−E(g)g(u))

×w(u)f(u)du− 1

Γ(α)

b

Z

a

(b−t)α−1

×(g(t)E(g)−g2(t))w(t)f(t)dt

= 1 Γ(α)

b

Z

a

(b−t)α−1((g2(t) +E2(g)

−2g(t)E(g))w(t)f(t)dt

= 1 Γ(α)

b

Z

a

(b−t)α−1(g(t)−E(g))2w(t)f(t)dt

g2,α,w.

Remark 3.4. If we take g(x) =x in Lemma 3.3, we obtain E(X) =b,and

(4)

with,

z(t) = 1

(b−t)α−1w(t)f(t) t

Z

a

(b−u)α−1(E[X]−u)

×w(u)f(u)du

= 1

(b−t)α−1w(t)f(t) t

Z

a

(b−u)αw(u)f(u)du. (3.5)

Now, taking three continuous functionsU,V,Qdefined on

[a,b],we prove the following result:

Theorem 3.5. Let U,V and Q be three continuous functions on[a,b] and Jα

aQ(b) =1,then

aUV Q(b)−JaαU Q(b)JaαV Q(b) =Jaα[(U−JaαU Q(b)) ×(V−Jα

aV Q(b))Q(b)

(3.6)

is valid, for anyα≥1.

Proof

We have

a[(U−JaαU Q(b)) (V−JaαV Q(b))] (b)

= 1

Γ(α)

b

Z

a

(b−t)α−1(U(t)−Jα

aU Q(b))

×(V(t)−Jα

aV Q(b))Q(t)dt

= 1

Γ(α)

b

Z

a

(b−t)α−1U(t)V(t)Q(t)dt−J

α

aV Q(b)

Γ(α)

× b

Z

a

(b−t)α−1U(t)Q(t)dt−J

α

aU Q(b)

Γ(α)

× b

Z

a

(b−t)α−1V(t)Q(t)dt+J

α

aU Q(b)JaαV Q(b)

Γ(α)

× 1

Γ(α)

b

Z

a

(b−t)α−1Q(t)dt

= Jα

aUV Q(b)−2JaαU Q(b)JaαV Q(b)

+Jα

aU Q(b)JaαV Q(b)JaαQ(b).

SinceJα

aQ(b) =1,then we ca write

a [(U−JaαU Q(b)) (V−JaαV Q(b))Q] (b)

= Jα

aUV Q(b)−JaαU Q(b)JaαV Q(b).

Taking into account the above theorem, we prove the following theorem:

Theorem 3.6. Let X be a continuous random variable with support an interval[a,b],−∞<a<b<∞,having a pdf f

and let w:[a,b]→R+be a continuous function. Then, for anyα≥1,we have

Ezg20, α,w

Ezh0, α,w

≤E[gE

g,α,w]2,α,w,0<α, (3.7)

where g is an absolutely continuous function with Ezg0,α,ω<

∞and Jα

aw f(b) =1,h is a given function and z is given by

z(t) = 1

(b−t)α−1w(t)f(t) t

Z

a

(b−u)α−1w(u)f(u)

×(Eh,α,w−h(u))du. Proof

Thanks to (3.1), we observe that

Ezg0,

α,ω = Egh,α,w−Eh,α,wEg,α,w

= Jα

aghw f(b)−Jaαhw f(b)Jaαgw f(b).

In (3.6), we takeU=g,V =h andQ=w f. Then, we observe that

Ezg0,α,ω = Jaα[(g−Jaαgw f(b)) (h−Jaαhw f(b))w f] (b)

= 1

Γ(α)

b

Z

a

(b−t)α−1(g(t)−Jα

agw f(b))

×(h(t)−Jα

ahw f(b))w f(t)dt.

Thanks to Cauchy-Shwarz inequality, we obtain

Ezg20, α,ω

 1

Γ(α)

b

Z

a

(b−t)α−1(g(t)−Jα

agw f(b)) 2

w f(t)dt 

×

 1

Γ(α)

b

Z

a

(b−t)α−1(h(t)−Jα

ahw f(b))w f(t)dt

= Jα

a

h

(g−Jα

agw f(b))

2w fi(b)Jα

a

h

(h−Jα

ahw f(b))

2w fi(b)

= E(gJα

agw f(b))2,α,wE(h−Jα

ahw f(b))2,α,w

= E

(g−Eg,α,w)

2

,α,wE(h−Eh,α,w)

2

,α,w. (3.8)

Again, takingg=hin (3.1), we get

Ezh0,

α,ω = Eh2,α,w− Eh,α,w 2

= Eh2,α,w−2 Eh,α,w 2

+ Eh,α,w 2

.

Thanks to the hypothesisJα

aw f(b) =1, it yields that

Ezh0,α,ω = Eh2,α,w−2 Eh,α,w

2

+ Eh,α,w

2 Jα

aw f(b)

= Jα

a h2w f

(b)−2Eh,α,wJaα(hw f) (b)

+ Eh,α,w 2

aw f(b)

= Jα

a

h

h2−2Eh,α,wh+ Eh,α,w

2i w f(b)

= Jα

a

h−Eh,α,w

2 w f

(b)

= E

(h−Eh,α,w)

2

(5)

Using (3.8) and (3.9), we deduce that

Ezg20,α,w

Ezh0, α,w

≤E[gE

g,α,w]2,α,w.

The proof is thus achieved.

Remark 3.7. If we take α=1and w(x) =1in(3.7),we obtain(1.3).

Lemma 3.8. Let X be a CRV with support an interval[a,b]

,−∞<a<b<∞,having a pdf f and let w:[a,b]→R+be

a continuous function. Then, for anyα≥1,we have E(gE

g,α,w)2,α,w (3.10)

= 1

[Γ(α)]2 Z b

a

Z y a

(b−x)α−1(b−y)α−1(g(x)−g(y))2

×f(x)f(y)w(x)w(y)dxdy,

where g is an absolutely continuous function and Jα

aw f(b) =1. Proof.

We have

1 [Γ(α)]2

Rb a

Ry a(b−x)

α−1(by)α−1(g(x)g(y))2

×f(x)f(y)w(x)w(y)dxdy= 1

[Γ(α)]2

Rb a(b−y)

α−1

×f(y)w(y)Ry a(b−x)

α−1g2(x)f(x)w(x)dxdy

+Rb a(b−y)

α−1g2(y)f(y)w(y)Ry a(b−x)

α−1

×f(x)w(x)dxdy−2Rb a(b−y)

α−1g(y)f(y)

×w(y)Ry a(b−x)

α−1g(x)f(x)w(x)dxdy

.

(3.11)

Then, we can write

Rb a(b−y)

α−1f(y)w(y)Ry a(b−x)

α−1g2(x)

×f(x)w(x)dxdy=Rb a(b−y)

α−1f(y)w(y)dy

Rb

a(b−x)

α−1g2(x)f(x)w(x)dxRb

a(b−y)

α−1

×g2(y)f(y)w(y)Ry a(b−x)

α−1f(x)w(x)dxdy

(3.12)

and

Z b

a

(b−y)α−1g(y)f(y)w(y)

Z y

a

(b−x)α−1g(x)

×f(x)w(x)dxdy=

Z b

a

(b−y)α−1g(y)f(y)w(y)dy 2

Z b

a

(b−y)α−1g(y)f(y)w(y)

Z y

a

(b−x)α−1

×g(x)f(x)w(x)dxdy. (3.13)

By(3.11),(3.12)and(3.13),we obtain 1

[Γ(α)]2

Z b

a

Z y

a

(b−x)α−1(b−y)α−1(g(x)−g(y))2

×f(x)f(y)w(x)w(y)dxdy

= 1

[Γ(α)]2 Z b

a

(b−y)α−1f(y)w(y)dy Z b

a

(b−x)α−1g2(x)

×f(x)w(x)dx−

Z b

a

(b−y)α−1g(y)f(y)w(y)dy 2#

= Jα

a f w(b)Jaαg2f w(b)−(Jaαg f w(b)) 2

= Jα

ag2f w(b)−(Jaαg f w(b)) 2

= Eg2,α,w−Eg2,α,w

= Eg2,α,w−2Eg2,α,w+Eg2,α,w

= Eg2,α,w−2Eg2,α,w+Jaαf w(b)Eg2,α,w

= E(g−Eg,α,w)2,α,w.

Finally, based on Lemma 3.7, we present to the reader the following results:

Theorem 3.9. Let X be a CRV with support an interval

[a,b],−∞<a<b<∞,having a pdf f and let w:[a,b]→R+

be a continuous function. Then, for anyα≥1, the following inequality holds

E[gE

g,α,w]2,α,w≤Ez(g0)2

h0 ,α,w

, (3.14)

where g is an absolutely continuous function, withEzg0, α,ω

<

∞and Jaαw f(b) =1,h is continuous function on[a,b],with

0<h0and

z(t) = 1

(b−t)α−1w(t)f(t) (3.15)

× t

Z

a

(b−u)α−1w(u)f(u)(Eh,α,w−h(u))du.

Proof:By Lemma 3.7, we have

E[gEg,α,w]2,α,w

= 1

[Γ(α)]2

b

Z

a y

Z

a

(b−x)α−1(b−y)α−1[g(y)−g(x)]2

×f(x)w(x)f(y)w(y)dxdy

= 1

[Γ(α)]2

b

Z

a y

Z

a

(b−x)α−1(b−y)α−1

y

Z

x

g0(t)dt 

2

×f(x)w(x)f(y)w(y)dxdy.

(6)

Thanks to the condition onh0and by Cauchy-Schwarz inequality, we get

y

Z

x

g0(t)dt   2 =   y Z x p

h0(t) g 0(t)

p h0(t)

! dt

2

≤ (h(y)−h(x))

y

Z

x

[g0(t)]2

h0(t) dt.(3.17)

By(3.16)and(3.17), we can write

E[gE

g,α,w]2,α,w

≤ 1

[Γ(α)]2

b Z a y Z a

(b−x)α−1(b−y)α−1(h(y)−h(x))

× y

Z

x

[g0(t)]2

h0(t) dt f(x)w(x)f(y)w(y)dxdy.

On the other hand, since

 

x≤t≤y a≤x≤y a≤y≤b

 

a≤x≤t t≤y≤b a≤t≤b

,

then, by changing the order of integration, it yields that

E[gEg,α,w]2,α,w

≤ 1

[Γ(α)]2

b

Z

a

[g0(t)]2

h0(t)   b Z t t Z a

(b−x)α−1(b−y)α−1

×(h(y)−h(x))f(x)w(x)f(y)w(y)dxdy

dt.

(3.19)

Also, we have

b Z t t Z a

(b−x)α−1(b−y)α−1(h(y)−h(x))f(x)w(x)

×f(y)w(y)dxdy=

b Z a t Z a

(b−x)α−1(b−y)α−1

×(h(y)−h(x))f(x)w(x)f(y)w(y)dxdy

− t Z a t Z a

(b−x)α−1(b−y)α−1(h(y)−h(x))f(x)w(x)

×f(y)w(y)dxdy

=

b

Z

a

(b−y)α−1h(y)f(y)w(y)dy

t

Z

a

(b−x)α−1f(x)w(x)dx

− b

Z

a

(b−y)α−1f(y)w(y)dy

t

Z

a

(b−x)α−1h(x)f(x)w(x)dx

− t

Z

a

(b−y)α−1h(y)f(y)w(y)dy

t

Z

a

(b−x)α−1f(x)w(x)dx

+

t

Z

a

(b−y)α−1f(y)w(y)dy

t

Z

a

(b−x)α−1h(x)f(x)w(x)dx

=

b

Z

a

(b−y)α−1h(y)f(y)w(y)dy

t

Z

a

(b−x)α−1f(x)w(x)dx

− b

Z

a

(b−y)α−1f(y)w(y)dy

t

Z

a

(b−x)α−1h(x)f(x)w(x)dx.

(3.20)

Multiplying(3.20)by 1

Γ(α),we obtain 1 Γ(α) b Z t t Z a

(b−x)α−1(b−y)α−1(h(y)−h(x))f(x)w(x)

×f(y)w(y)dxdy=Jα

a (h f w) (b) t

Z

a

(b−x)α−1f(x)

×w(x)dx−Jα

a(f w) (b) t

Z

a

(b−x)α−1h(x)f(x)w(x)dx

= Jα

a (h f w) (b) t

Z

a

(b−x)α−1f(x)w(x)dx

− t

Z

a

(b−x)α−1h(x)f(x)w(x)dx

=

t

Z

a

(b−x)α−1[Jα

a(h f w) (b)−h(x)]f(x)w(x)dx

=

t

Z

a

(b−x)α−1

Eh,α,w−h(x)

f(x)w(x)dx. (3.21)

By (3.18) and (3.21), we get

E[g−Eg,α,w]2,α,w

≤ 1

Γ(α)

b

Z

a

[g0(t)]2

h0(t)

t

Z

a

(b−x)α−1Eh,α,w−h(x)

f(x)w(x)dx  dt = 1 Γ(α) b Z a

[g0(t)]2

h0(t)

(b−t)α−1w(t)f(t)z(t)dt

= Jα

a

z(g0)2

h0 w f

(7)

Theorem 3.9 is thus achieved.

Remark 3.10. If we take α=1,and w(x) =1in (3.14), we obtain (1.4).

References

[1] T. Cacoullos and V. Papathanasiou: On upper bounds

for the variance of functions of random varialbes, Statist. Probab. Lett., 7 (1985), pp. 175-184.

[2] T. Cacoullos and V. Papathanasiou: Caracterizations of

distributuons by variance bounds. Math. Statist., 4 (1989), pp. 351-356.

[3] T. Cacoullos and V. Papathanasiou: Characterizations of

distributions by generalisations of variance bounds and simple proofs of the CLT, J. Statist. Plann. Inference, 63 (1997), pp. 157-171.

[4] Z. Dahmani, A.E. Bouziane, M. Houas and M.Z.

Sarikaya: New w-weighted concepts for continuous ran-dom variables with applications, Note di Matematica, 37(1) (2017), pp. 23-40.

[5] Z. Dahmani: New applications of fractional calculus on

probabilistic random variables, Acta Math. Univ. Come-nianaeVol. LXXXVI, 2 (2017), pp. 299-307.

[6] F. Goodarzi, M. Amini and G.R. Mohtashami Borzadaran:

On upper bounds for the variance of functions of random variables with weighted distributions, Labachevskii Jour-nal of Mathematics, 37(4) (2016), pp. 422-435.

[7] R. Gorenflo, F. Mainardi: Fractional calculus: integral

and differential equations of fractional order, Springer Verlag, Wien, (1997), pp. 223-276.

[8] G.R. Mohtashami Borzadaran and D.N. Shanbhag:

Statist. Probab. Lett., 39 (1998), pp. 109-117.

[9] M.Z. Sarikaya, M.E. Kiris,N. Celik: Hermite-Hadamard

type inequality for h-convex stochastic processes. AIP Conference Proceedings 1726, 020076 (2016).

[10] M.Z. Sarikaya, T. Tunc, H. Budak: Simpson’s type

in-equality for F-convex function. FACTA Univ. FUMI, 1(10), (2018), pp. 747-753.

[11] M.Z. Sarikaya, H. Yildiz, H. Budak: On weighted

Iyengar-type inequalities for conformable fractional inte-grals. Mathematical Sciences, 11(4), (2017), pp. 327-331.

? ? ? ? ? ? ? ? ? ISSN(P):2319−3786 Malaya Journal of Matematik

References

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