Random variable inequalities involving (k,s)-integration

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https://doi.org/10.26637/MJM0504/0005

Random variable inequalities involving

(k,s)-integration

M. Houas

1

* and Z. Dahmani

2

Abstract

In this paper, we use the(k,s)−Riemann-Liouville operator to establish new results on integral inequalities by

using fractional moments of continuous random variables. Keywords

(k,s)−Riemann-Liouville integral, random variable, (k,s)−fractional expectation, (k,s)−fractional variance, (k,s)−fractional moment.

AMS Subject Classification

26D15, 26A33, 60E15.

1_{Laboratory FIMA, UDBKM, University of Khemis Miliana, Algeria.} 2_{Laboratory LPAM, UMAB, University of Mostaganem, Algeria.}

*Corresponding author:ı[email protected];ı2_{[email protected]}

Article History: Received6June2017; Accepted7August2017 c2017 MJM.

1. Introduction

The study of integral inequalities is an important research subject in mathematical analysis. These inequalities have many applications in differential equations, probability theory and statistical problems, but one of the most useful appli-cations is to establish uniqueness of solutions in fractional boundary value problems. For detailed applications on the subject, one may refer to [9–12], and the references cited therein. Moreover, the integral inequalities involving frac-tional integration are also of great importance. For some earlier work on the topic, we refer to [2–5,7,8,14,16,17]. In [9], P. Kumar presented new results involving higher moments for continuous random variables. Also the author established some estimations for the central moments. Other results based on Gruss inequality and some applications of the truncated exponential distribution have been also discussed by the au-thor. In [10], Ostrowski type integral inequalities involving moments of a continuous random variable defined on a finite interval, is established. In [3], the author established several inequalities for the fractional dispersion and the fractional

variance functions of continuous random variables. Recently, A. Akkurt et al. [1] proposed new generalizations of the re-sults in [3]. Very recently, Z. Dahmani [4,6] presented new fractional integral results for the fractional moments of contin-uous random variables by correcting some results in [3]. In a very recent work, M. Tomar et al. [17] proposed new integral inequalities for the(k,s)−fractional expectation and variance functions of a continuous random variable.

Motivated by the results presented in [3,4,6,14,17], in this paper, we present some random variable integral inequalities for the(k,s)−fractional operator.

2. Preliminaries

We recall the notations and definitions of the(k,s)−fractional integration theory [13,14,17].

Definition 2.1. The Riemann–Liouville fractional integral of orderα≥0, for a continuous function f on[a,b]is defined by

Jα

a[f(t)] =Γ(1α)

Rt

a(t−τ) α−1

f(τ)dτ;α≥0,a<t≤b ,

(2.1)

whereΓ(α) =R∞

0 e

−u_uα−1_du_.

(2)

by

kJaα[f(t)] =kΓk1(α) Rt

a(t−τ)

α

k−1_f_(τ)dτ_; _α_>₀_, _a_<_t≤_b _,

(2.2)

whereΓk(α) =

R∞

0 e −uk

k_uα−1_du_.

Definition 2.3. The(k,s)−Riemann-Liouville fractional in-tegral of orderα>0, for a continuous function f on[a,b]is defined as

k

sJaα[f(t)] =

(s+1)1−αk kΓk(α)

Rt

a ts+1−τs+1 α_k−1

τsf(τ)dτ;α>0,a<t≤b ,

(2.3)

where k>0,s∈_R\ {−1}.

Theorem 2.4. Let f be continuous on[a,b],k>0,and s∈ R\ {−1}.Then,

k sJaα

h k sJ

β a[f(t)]

i

= k_sJaα+β[f(t)] = ksJ β a

_k

sJaα[f(t)]

,k>0,s∈R\ {−1} ,

(2.4)

for allα>0,β>0,a<t≤b.

Theorem 2.5. Let α >0,β >0,k>0 and s∈R\ {−1}. Then, we have

k sJaα

ts+1−as+1β_k−1

= Γk(β) (s+1)αkΓk(α+β) t

s+1₋_as+1β+α_k −1

,

(2.5)

Remark 2.6. (i):Taking s=0,k>0in(5),we obtain

kJaα

(t−a)βk−1

= Γk(β)

Γk(α+β)(t−a)

β+α

k −1_,_α_,_β_>₀_,_a_>₀_,_k_>₀ _.

(2.6)

(ii):The formula(5),for s=0and k=1becomes Jα

a h

(t−a)β−1i= Γ(β)

Γ(α+β)(t−a) β+α−1

. (2.7)

Corollary 2.7. Let k>0and s∈_R\ {−1}.Then, for any α>0,we have

k

sJaα[1] = 1

(s+1)αkΓk(α+k) t

s+1₋_as+1α_k−2

. (2.8)

Remark 2.8. (i):For s=0,k>0in(8),we get

kJaα[1] =Γk(α1+k)(t−a)

α

k−2 _. _(2.9) (ii):For s=0,k=1in(8),we have

Jα

a [1] =Γ(α1+1)(t−a)

β−2 _. _(2.10) For more details on(k,s)−fractional integral, we refer the reader to [14,17].

We recall also the following definitions [3,4,17]

Definition 2.9. The(k,s)−fractional expectation function of orderα>0, for a random variable X with a positive p.d.f.

f defined on[a,b]is defined as

k

sEX,α(t):=

t

Z

a

ts+1−τs+1

α

k−1

τs+1f(τ)dτ,

(2.11)

α>0,k>0,s∈R\ {−1}and a<t≤b.

Definition 2.10. The(k,s)−fractional expectation function of orderα >0for the random variable X−E(X) with a positive probability density function f defined on [a,b] is defined as

k

sEX−E(X),α(t):=

t

Z

a

ts+1−τs+1αk−1

τs(τ−E(X))f(τ)dτ,

(2.12)

α>0,k>0,s∈_R\ {−1}and a<t≤b.

Definition 2.11. The(k,s)−fractional variance function of orderα>0for a random variable X having a positive p.d.f.

f on[a,b]is defined as

k

sσX2,α(t):=

t

Z

a

ts+1−τs+1

α

k−1_τs_(τ₋_E_(X₎₎2_f_(τ)_dτ_,

(2.13)

α>0,k>0,s∈R\ {−1}and a<t≤b.

We introduce also the following definition.

Definition 2.12. The(k,s)−fractional moment function of orders r>0,α >0 for a continuous random variable X having a p.d.f. f defined on[a,b]is defined as

k

sMr,α(t):=

t

Z

a

ts+1−τs+1

α

k−1

τs+rf(τ)dτ,

(2.14)

α>0,k>0,s∈R\ {−1}and a<t≤b.

Remark 2.13. If we take s=0,α=k=1in Definition 12, we obtain the classical moment of order r>0given by Mr:= b

R

a

τrf(τ)dτ.

We define the quantities that will be used later:

H(τ,ρ):= (g(τ)−g(ρ)) (h(τ)−h(ρ)), τ,ρ∈(a,t),a<t≤b,

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and

ϕ_kα_,_s(t,τ):=(s+1) 1−α

k

Γk(α)

ts+1−τs+1

α

k−1

τsp(τ),k>0,s∈R/{−1} τ∈(a,t),

(2.16) wherep:[a,b]→R+is a continuous function.

3. Inequalities for

(k

,

s)

−

operator

We prove:

Theorem 3.1. Let X be a continuous random variable having a p.d.f. f:[a,b]→_R+_.

(i):If f ∈L∞[a,b], then for all a<t≤b andα>0,k>

0,s∈R\ {−1}, k

sJaα[f(t)] ksEXr−1_(X−E(X))_,α(t)− k_sE_X−E(X)_,α(t)k_sMr−1,α(t)

≤ kfk2_∞hk

sJaα[1] ksJaα[tr]−ksJaα[t)]ksJaα

tr−1i, (3.1)

(ii):for all a<t≤b,the inequality k

sJaα[f(t)] ksEXr−1_(X−E(X))_,α(t)− k_sE_X−E(X)_,α(t)k_sMr−1,α(t)

≤ 1

2(t−a) t

r−1_−ar−1_k

sJaα[f(t)] 2

, (3.2)

is also valid forα>0,k>0,s∈_R/{−1}.

Proof. Using (2.15) and (2.16), we can write

t

Z

a

ϕ_kα_,_s(t,τ)H(τ,ρ)dτ= t

Z

a

ϕ_kα_,_s(t,τ) (g(τ)−g(ρ)) (h(τ)−h(ρ))dτ.

(3.3) Then Z t a Z t a

ϕ_kα_,_s(t,τ)ϕ_kα_,_s(t,ρ)H(τ,ρ)dτdρ (3.4) =

Z t

a

Z t

a

ϕ_kα_,_s(t,τ)ϕ_kα_,_s(t,τ) (g(τ)−g(ρ)) (h(τ)−h(ρ))dτdρ.

Hence

(s+1)2(1−αk) k2_Γ2

k(α) t Z a t Z a

ts+1−τs+1

α

k−1 _ts+1₋_ρs+1α_k−1 τsρs

(3.5)

×p(τ)p(ρ) (g(τ)−g(ρ)) (h(τ)−h(ρ))dτdρ = 2k_sJα

a[p(t)] ksJ α

a[pgh(t)]−2ksJ α

a[pg(t)]ksJ α

a [ph(t)].

In (3.5), we choosep(t) =f(t),g(t) =t−E(X)andh(t) = tr−1,t∈(a,b), we obtain

(s+1)2(1−αk) k2_Γ2

k(α) t Z a t Z a

ts+1−τs+1

α

k−1 _ts+1₋ ρs+1

α

k−1

τsρs (3.6)

×(τ−ρ) τr−1−ρr−1f(τ)f(ρ)dτdρ

= 2k_sJα

a[f(t)]ksJaα

tr−1(t−E(X))f(t) −2k_sJα

a[(t−E(X))f(t)]ksJaα

tr−1f(t)

= 2k_sJα

a[f(t)]ksEXr−1_(X−E(X))_,α(t)−2

k

sEX−E(X),α(t)

k

sMr−1,α(t).

Using the factf ∈L∞([a,b]), we can write

(s+1)2(1−αk) k2_Γ2

k(α) t Z a t Z a

ts+1−τs+1α_k−1

ts+1−ρs+1α_k−1

τsρs (3.7)

×(τ−ρ) τr−1−ρr−1f(τ)f(ρ)dτdρ

≤ kfk2_∞(s+1) 2(α

k−1)

k2_Γ2_(α₎ t Z a t Z a

ts+1−τs+1αk−1 _ts+1_−ρs+1αk−1 τsρs

×(τ−ρ) τr−1−ρr−1dτdρ

= kfk2_∞h2k_sJα

a [1] ksJaα[tr]−2ksJaα[t] ksJaα

tr−1i

.

By (3.6) and (3.7), we have

k

sJaα[f(t)]skEXr−1_(X−E(X))_,α(t)−

k

sEX−E(X),α(t)

k

sMr−1,α(t)

≤ kfk2_∞hk

sJaα[1] ksJaα[tr]−ksJaα[t] ksJaα

tr−1 i

. (3.8)

Since sup

τ,ρ∈[a,t]

|τ−ρ|

τr−1−ρr−1

= (t−a) tr−1−ar−1

, then we observe that

(s+1)2(1−αk) k2_Γ2_(α₎

t Z a t Z a

ts+1−τs+1αk−1 _ts+1_−ρs+1αk−1 τsρs

×(τ−ρ) τr−1−ρr−1f(τ)f(ρ)dτdρ (3.9)

≤ sup

τ,ρ∈[a,t]

|τ−ρ|

τr−1−ρr−1

(s+1) 2(α

k−1)

k2_Γ2_(α₎ t Z a t Z a

× ts+1−ρs+1

α

k−1_τs_ρs_f_(τ)_f_(ρ)dτdρ

= (t−a) tr−1−ar−1 (k_sJα

a[f(t)])2.

Thanks to (3.6) and (3.9), we obtain

k

sJaα[f(t)]skEXr−1_(X−E(X))_,α(t)−

k

sEX−E(X),α(t)

k

sMr−1,α(t)

≤ (t−a) tr−1−ar−1 (k_sJα

a[f(t)])2. (3.10)

We prove also the following result:

Theorem 3.2. Let X be a continuous random variable having a p.d.f. f :[a,b]→R+. Then we have:

(i∗)For any k>0,s∈R/{−1}andα>0,β>0, k

sJaα[f(t)]ksEXr−1_(X−E(X))_,β(t) + _skJ_aβ[f(t)]_skE_Xr−1_(X−E(X))_,α(t)

−k

sEX,α(t)ksMr−1,β(t)−skEX,β(t) ksMr−1,α(t) (3.11) ≤ kfk2_∞hk

sJaα[1]ksJaβ[tr] +ksJaβ[1]ksJaα[tr] −k_sJα

a [t]ksJaβ

tr−1−k_sJβ a[t] skJ

α a

tr−1i,a<t≤b,

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(ii∗)The inequality

k

sJaα[f(t)]ksEXr−1_(X−E(X))_,β(t) + k_sJ_aβ[f(t)]k_sE_Xr−1_(X−E(X))_,α(t)

−k

sEX,α(t)ksMr−1,β(t)− ksEX,β(t)ksMr−1,α(t) (3.12) ≤ (t−a) tr−1−ar−1k

sJaα[f(t)]ksJaβ[f(t)],a<t≤b

is also valid for any k>0,s∈_R/{−1}andα>0,β >0. Proof. Multiplying both sides of (3.4) byϕ_kβ_,_s(t,ρ),where

ϕ_kβ_,_s(t,ρ):=(s+1) 1−α

k

Γk(α)

ts+1−ρs+1

α

k−1

τsp(τ),ρ∈(a,t), a<t≤b,

(3.13) we can obtain

Z t

a

Z t

a

ϕ_kα_,_s(t,τ)ϕ_kβ_,_s(t,ρ)H(τ,ρ)dτdρ (3.14) =

Z t

a

Z t

a

ϕ_kα_,_s(t,τ)ϕ_kβ_,_s(t,ρ) (g(τ)−g(ρ)) (h(τ)−h(ρ))dτdρ.

Hence,

(s+1)2

1−α+β_k

k2_Γ

k(α)Γk(β) t Z a t Z a

ts+1−ρs+1β_k−1

(3.15)

×τsρsp(τ)p(ρ) (g(τ)−g(ρ)) (h(τ)−h(ρ))dτdρ

= k_sJα

a [p(t)]ksJaβ[pgh(t)] +ksJaβ[p(t)] ksJaα[pgh(t)] −k

sJaα[ph(t)]ksJaβ[pg(t)]−ksJaβ[ph(t)]ksJaα[pg(t)].

In (3.15), we takep(t) = f(t),g(t) =t−E(X),h(t) =tr−1. So, we get

(s+1)2

1−α+β k

k2_Γ

ts+1−τs+1αk−1 _ts+1_−ρs+1βk−1 τsρs

×(τ−ρ) τr−1−ρr−1)f(τ)f(ρ)dτdρ (3.16)

= k_sJα

a [f(t)]ksJaβ

tr−1(t−E(X))f(t) +k_sJβ

a[f(t)] ksJaα

tr−1(t−E(X))f(t)

−k_sJα

a[(t−E(X))f(t)] ksJaβ

tr−1f(t)− k_sJβ

a[(t−E(X))f(t)]ksJaα

tr−1f(t)

= k_sJα

a [f(t)]ksEXr−1_(X−E(X))_,β(t) + k_sJ_aβ[f(t)]k_sE_Xr−1_(X−E(X))_,α(t)

−k

sEX,α(t)ksMr−1,β(t)− ksEX,β(t)ksMr−1,α(t).

We have also

(s+1)2

1−α+β k

k2_Γ

ts+1−τs+1αk−1 _ts+1_−ρs+1βk−1 τsρs

≤ kfk2_∞(s+1) 21−α+β

k

k2_Γ

ts+1−τs+1αk−1 _ts+1₋ ρs+1

β

k−1 τsρs

×(τ−ρ) τr−1−ρr−1dτdρ

= kfk2_∞hk sJ

α a[1]ksJ

β

a[tr] + ksJ β a[1]ksJ

α a [tr]

− k

sJaα[t]ksJaβ

tr−1−k

sJaβ[t] ksJaα

tr−1i.

By (3.16) and (3.17), we obtain (3.11). To prove (3.12), we remark that

(s+1)2

1−α+β_k

k2_Γ

ts+1−τs+1

α

k−1 _ts+1₋_ρs+1β_k−1 τsρs

≤ sup

τ,ρ∈[a,t]

|(τ−ρ|

τr−1−ρr−1 (s+1)2

1−α+β_k

k2_Γ

ts+1−ρs+1β_k−1

×τsρsf(τ)f(ρ)dτdρ

= (t−a) tr−1−ar−1_k

sJaα[f(t)] ksJaβ[f(t)].

Therefore, by (3.16) and (3.18), we get (3.12).

Remark 3.3. If we takeα =β in Theorem 15, we obtain Theorem 14.

We give also the following(k,s)−fractional integral in-equality:

Theorem 3.4. Let X be a continuous random variable hav-ing a p.d.f. f :[a,b]→R+. Assume that there exist con-stantsϕ,φ such thatϕ≤ f(t)≤φ.Then, for all k>0,s∈ R/{−1},α>0,we have

k

sJaα[f(t)] ksM2r,α(t)−ksMr2,α(t)≤

1 4(b

r₋_ar₎2k

sJaα[f(t)] 2

, a<t≤b.

(3.19)

Proof. Using Theorem 2.4 of [17], we can write

k

sJaα[p(t)] ksJaα[phg(t)]−ksJaα[ph(t)] ksJaα[pg(t)] ≤ 1 4 k

sJaα[p(t)] 2

(φ−ϕ)2.

(3.20) In (3.20), we replacehbyg,we will have

k

sJaα[p(t)]ksJaα

pg2(t)−(k_sJα

a[pg(t)])2 ≤ 1 4 k

sJaα[p(t)] 2

(φ−ϕ)2.

(3.21) Takingp(t) = f(t),g(t) =tr,a<t≤bin (3.21), we obtain

k

sJaα[f(t)]ksJaα

t2rf(t) −(k

sJaα[trf(t)])2 ≤ 1 4 k

sJaα[f(t)] 2

(φ−ϕ)2.

(3.22) In (3.22), we takeφ=brandϕ=ar, then we have

0≤k_sJα

a[f(t)]ksJaα

t2rf(t)−(k_sJα

a[trf(t)])2≤

1 4(b

r₋_ar₎2k

sJaα[f(t)] 2

.

(3.23) This implies that

k

sJaα[f(t)]ksM2r,α(t)−ksMr2,α(t)≤

1 4(b

r_−ar₎2k

sJaα[f(t)] 2

.

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Finally, we present the following result:

Theorem 3.5. Let X be a continuous random variable hav-ing a p.d.f. f :[a,b]→R+. Assume that there exist con-stantsϕ,φ such thatϕ≤ f(t)≤φ.Then, for all k>0,s∈ R/{−1},α>0,β>0,we have

k

sJaα[f(t)] ksM2r,β(t) + ksJaβ[f(t)]ksM2r,α(t) (3.25) +2arbr k_sJα

a[f(t)]ksJaβ[f(t)] ≤ (ar+br)k_sJα

a[f(t)]ksMr,β(t) + (ar+br)ksJaβ[f(t)] ksMr,α(t),a<t≤b.

Proof. We takep(t) =f(t),g(t) =tr,a<t≤band by The-orem 2.5 of [17], we can write

h k

sJaα[f(t)] ksJaβ

t2rf(t)+k_sJβ

a[f(t)]ksJaα

t2rf(t) (3.26)

−2k_sJα

a [trf(t)]ksJaβ[trf(t)] i2

≤ hφk_sJα

a[f(t)]−ksJaα[trf(t)] ksJaβ[trf(t)]−ϕksJaβ[f(t)]

+k_sJα

a [trf(t)]−ϕskJaα[f(t)] φskJaβ[f(t)]−ksJaβ[trf(t)] i2

.

Combining (3.21) and (3.26) and taking into account the fact that the left-hand side of (3.21) is positive, we can write

k

sJaα[f(t)] ksJaβ

t2rf(t) +k_sJβ

a[f(t)]skJaα

t2rf(t)

−2s_kJα

a [trf(t))]ksJ β

a[trf(t)] (3.27) ≤ φ k_sJ_aα[f(t)]− k_sJ_aα[trf(t)] k_sJ_aβ[trf(t)]−ϕk_sJ_aβ[f(t)]

+k_sJα

a[trf(t)]−ϕksJaα[f(t)] φskJaβ[f(t)]−ksJaβ[trf(t)]

.

Therefore

k

sJaα[f(t)] ksM2r,β(t) + ksJaβ[f(t)]ksM2r,α(t)−2ksMr,β(t)ksMr,α(t) ≤ φ k_sJα

a[f(t)]− ksMr,α(t) ksMr,β(t)−ϕksJaβ[f(t)]

(3.28)

+k_sMr,α(t)−ϕksJaα[f(t)] φ ksJaβ[f(t)]−ksMr,β(t)

.

This implies that

k

sJaα[f(t)] ksM2r,β(t) + ksJaβ[f(t)]ksM2r,α(t) +2ϕ φs_kJ_aα[f(t)]k_sJ_aβ[f(t)] ≤ (φ+ϕ)k_sJα

a[f(t)]ksMr,β(t) + ksJaβ[f(t)] ksMr,α(t)

. (3.29)

In (3.29), we takeφ=br,ϕ=ar, then we have

k

sJaα[f(t)]ksM2r,β(t) + ksJaβ[f(t)] ksM2r,α(t) +2arbr skJaα[f(t)] ksJaβ[f(t)] ≤ (ar+br)k_sJα

a[f(t)]ksMr,β(t) + skJaβ[f(t)] ksMr,α(t)

. (3.30)

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

ISSN(O):2321−5666

Random variable inequalities involving (k,s)-integration