An Ostrowski's Type Inequality for a Random Variable whose Probability Density Function Belongs to L∞ [A,B]

AN OSTROWSKI'S TYPE INEQUALITY FOR A

RANDOM VARIABLE WHOSE PROBABILITY

DENSITY FUNCTION BELONGS TO L

1 AB]

N.S. BARNETT AND S.S. DRAGOMIR

Abstract. An inequality of Ostrowski's type for a random variable whose probability density function is inL1ab] in terms of the cumulative distribution function and expectation is given. An application for a Beta random variable is also given.

1 Introduction

The following theorem contains the integral inequality which is known in the literature as Ostrowski's inequality 1, p. 469]

Theorem 1.1.

Letf :I

R

!

R

be a dierentiable mapping in I (

I is the interior of I), and let ab 2 I

with

a < b: If f 0 : (

ab)!

R

is bounded on (ab), i.e., kf

0 k

1:= sup t2(ab)

jf 0(

t)j<1 then we have the inequality:

f(x); 1 b;a

b Z

a

f(t)dt "

1 4 +

; x;

a+b

2

2 (b;a)

2

#

(b;a)kf 0

k 1 (1.1)

for allx2ab]:The constant 14 is sharp in the sense that it can not be replaced by a smaller one.

In 2], S.S. Dragomir and S. Wang applied Ostrowski's inequality in Numeri-cal Analysis obtaining an estimation of the error bound for the quadrature rules of Riemann type in terms of the innity norm kk

1

: Application for special means : logarithmic mean, identric mean, p-logarithmic mean etc... were also given.

The main aim of this paper is to give an Ostrowski's type inequality for random variables whose probability density functions are inL

1

ab]: An ap-plication for a Beta Random Variable is also given.

2 The Results

LetX be a random variable with the probability density function f : ab]

R

!

R

+ and with cumulative distribution functionF(x) = Pr(Xx):

The following theorem holds

Date.November, 1998.

1991 Mathematics Subject Classication. Primary 26D15Secondary 65 x xx.

26 Barnett and Dragomir

Theorem 2.1.

Let f 2 L 1

ab] and put kfk

1 = sup t2ab]

f(t) < 1: Then we have the inequality:

Pr(X x);

b;E(X) b;a

"

1 4 +

; x;

a+b

2

2 (b;a)

2

#

(b;a)kfk 1 (2.1)

or equivalently,

Pr(X x);

E(X);a b;a

"

1 4 +

; x;

a+b

2

2 (b;a)

2

#

(b;a)kfk 1 (2.2)

for allx2ab]:The constant

1

4 in (2:1) and (2:2) is sharp. Proof. Letxy2ab]:Then

jF(x);F(y)j= y Z

x

f(t)dt jx;yjkfk 1

which shows thatF iskfk 1

;Lipschitzian on ab]: Consider the kernel p: ab]

2

!

R

given by

p(xt) := 8 <

:

t;a ift2ax] t;b ift2(xb] :

Then the Riemann-Stieltjes integral b R

a

p(xt)dF(t) exists for anyx 2ab] and the formula of integration by parts for Riemann-Stieltjes integral gives:

b Z

a

p(xt)dF(t) = x Z

a

(t;a)dF(t) + b Z

x

(t;b)dF(t) (2.3)

= (t;a)F(t)j b a

; x Z

a

F(t)dt+ (t;b)F(t)j b x

; b Z

x

F(t)dt

= (b;a)F(x); b Z

a

F(t)dt:

The integration by parts formula for Riemann-Stieltjes integral also gives

E(X) = b Z

a

tdF(t) =tF(t)j b a

; b Z

a

F(t)dt (2.4)

=bF(b);aF(a); b Z

a

F(t)dt=b; b Z

a

F(t)dt: Now, using (2:3) and (2:4), we get the equality

(b;a)F(x) +E(X);b= b Z

a

p(xt)dF(t) (2.5)

for allx2ab]: Now, assume that

n := a=x

(n)

0 <x

(n)

1 <:::<x

(n) n;1

<x

(n)

n =

b is a sequence of divisions with(

n)

!0 asn!1where

(

n) := max n

x

(n) i+1

;x

(n) i :

i= 0:::n;1 o

:

Ifp: ab]!

R

is Riemann integrable on ab] and: ab]!

R

isL-lipschitzian (lipschitzian with the constantL), then :

b Z

a

p(x)d(x) = lim ( n) !0 n;1 X i=0 p

(n) i

h

x

(n) i+1

; x

(n) i i (2.6) lim ( n) !0 n;1 X i=0 p

(n) i

x

(n) i+1

;x

(n) i

x

(n) i+1

; x

(n) i

x

(n) i+1

;x

(n) i L lim ( n) !0 n;1 X i=0 p

(n) i

x

(n) i+1

;x

(n) i =L b Z a

jp(x)jdx:

Applying the inequality (2:6) for the mappingsp(x) andF()we get :

b Z

a

p(xt)dF(t) kfk 1

b Z

a

j p(xt)jdt

=kfk 1 2 4 x Z a

(t;a)dt+ b Z

x

(b;t)dt 3

5= kfk

1 "

(x;a)

2_{+ (}

b;x)

2 2 # = " 1 4 (b;a)

2₊ x;

a+b 2

2 #

kfk 1

28 Barnett and Dragomir Finally, by the identity (2:5) we deduce:

F(x);

b;E(X) b;a

"

1 4 +

; x;

a+b

2

2 (b;a)

2

#

(b;a)kfk 1 for allx2ab] which is (2:1):

Now, taking into account the fact that

Pr(X x) = 1;Pr(Xx) the inequality (2:2) is also obtained.

To prove the sharpness of the constant k =

1

4 assume that the inequality (2:1) holds with a constantc>0i.e.,

Pr(Xx);

b;E(X) b;a

"

c+ ;

x; a+b

2

2 (b;a)

2

#

(b;a)kfk 1 (2.7)

for allx2ab]:

Assume thatX0is a random variable having the probability density function f0: 01]!

R

,f0(t) = 1:Then

Pr(X0 x) =x x201]

E(X0) = 12 and

kf0k 1= 1

: Consequently, (2:7) becomes

x; 1 2 c+

x; 1 2

2

for allx201]:

Choosing x = 0 we get c 14 and the sharpness of the constant is thus proved.

The above theorem has some interesting corollaries for the expectation ofX as follows:

Corollary 2.2.

Under the above assumptions, we have the double inequality:

b; 1 2 (b;a)

2

kfk 1

E(X)a+ 12 (b;a)

2

kfk 1

: (2.8)

Proof. We know that

aE(X)b: Now, choosex=ain (2:1) to obtain

b;E(X) b;a

1

2 (b;a)kfk 1

i.e.,

b;E(X) 1 2 (b;a)

2

kfk 1 which is equivalent to the rst inequality in (2:8):

Also, choose x=bin (2:1) to get 1;

b;E(X) b;a

1

2 (b;a)kfk 1 i.e.,

E(X);a 1 2 (b;a)

2

kfk 1 and the inequality (2:8) is proved.

Remark 2.1.

We know that

1 = b Z

a

f(x)dx(b;a)kfk 1

which gives us

kfk 1

1 b;a

:

Now, if we assume thatkfk

1 is not too large, i.e., kfk

1

2 b;a (2.9)

then

a+ 12 (b;a)

2

kfk 1

b and

b; 1 2 (b;a)

2

kfk 1

a

which shows that the inequality (2:8) is a tighter inequality thanaE(X)b when (2:9) holds.

Another equivalent inequality to (2:8) which can be more useful in practice is the following one:

Corollary 2.3.

With the above assumptions, we have the inequality

E(X); a+b

2

(b;a)

2 2

kfk 1

; 1 b;a

30 Barnett and Dragomir Proof. From the inequality (2:8) we have :

b; a;b

2 ;

1 2 (b;a)

2

kfk 1

E(X); a+b

2

a; a+b

2 +12 (b;a)

2

kfk 1 i.e.,

; (b;a)

2 2

kfk 1

; 1 b;a

E(X); a+b

2

(b;a)

2 2

kfk 1

; 1 b;a

which is exactly (2:10):

This corollary provides the possibility of nding a sucient condition in terms ofkfk

1for the expectation

E(X) to be close to the mean value a+b

2 :

Corollary 2.4.

LetX andf be as above and">0:If

kfk 1

1 b;a

+ 2" (b;a)

2 (2.11)

then

E(X); a+b

2 ": The proof is obvious and we shall omit the details. The following corollary of Theorem 2.1 also holds:

Corollary 2.5.

LetX andf be as above. Then we have the inequality: Pr

X

a+b 2

; 1 2 (2.12)

1

4 (b;a)kfk

1+ 1

b;a

E(X); a+b

2

3

4 (b;a)kfk 1

; 1 2: Proof. If we choose in (2:1)x=

a+b

2 we get Pr

X

a+b 2

;

b;E(X) b;a

1

4 (b;a)kfk 1

which is clearly equivalent to Pr

X

a+b 2

; 1

2 + b;1a

E(X); a+b

2

1

4 (b;a)kfk 1

: Now, using the triangle inequality, we get

Pr

X

a+b 2 ; 1 2 = Pr X

a+b 2

; 1

2 +b;1a

E(X); a+b

2

; 1 b;a

E(X); a+b

2

Pr

X

a+b 2

; 1

2 +b;1a

E(X); a+b

2

+ 1 b;a

E(X); a+b

2

1

4 (b;a)kfk

1+ 1

b;a

E(X); a+b

2

3

4 (b;a)kfk 1

; 1 2 and the desired inequality is proved.

Remark 2.2.

A similar result applies for Pr

X

a+b 2

: We shall omit the details.

Finally, the following result holds:

Corollary 2.6.

LetX andf be as above. Then we have the inequality:

E(X); a+b

2

1 4 (b;a)

2

kfk 1+ (

b;a) Pr

X

a+b 2 ; 1 2 : (2.13)

Proof. As in the above corollary, we have: 1

b;a

E(X); a+b

2

Pr

X

a+b 2

; 1

2 + b;1a

E(X); a+b

2

+ Pr

X

a+b 2 ; 1 2 1

4 (b;a)kfk

1+ Pr

X

a+b 2

32 Barnett and Dragomir

Remark 2.3.

If we assume thatf 2Cab] thenF is dierentiable on(ab) and by Ostrowski's inequality(1:1) applied for F we get

F(x); 1 b;a

b Z

a

F(t)dt "

1 4 +

; x;

a+b

2

2 (b;a)

2

#

(b;a)kfk 1

for allx2ab]:

Now, using the identity (2:4) we recapture the inequality (2:1) and (2:2) for random variables whose probability density functions are continuous on ab]:

3 Application for a Beta Random Variable

A Beta random variable Xwith parameters (pq) has the probability density function

f(x:pq) := x

p;1(1 ;x)

q;1 B(pq)

0<x<1 where

=f(pq) :pq>0g and

B(pq) :=

1

Z

0

t p;1(1

;t) q;1

dt:

We observe that for 0<p<1

kf(pq)k

1= sup x2(01)

" x

p;1(1 ;x)

q;1 B(pq)

# =1:

Assume thatpq 1:Then

df(xpq) dx

= 1 B(pq)

h

(p;1)x p;2(1

;x) q;1

;(q;1)x p;1(1

;x) q;2

i

= x p;2(1

;x) q;2 B(pq)

(p;1)(1;x);(q;1)x]

=x p;2(1

;x) q;2 B(pq)

;(p+q;2)x+ (p;1)]: We observe that

df(xpq) dx

= 0

i x0 = p;1 p+q;2

2 (01) (forpq>1) and then

df(xpq) dx

> 0 on (0x0) and df(xpq)

dx

<0 on (x01):Consequently, kf(pq)k

1=

f(x0pq) = ( p;1)

p;1 (q;1)

q;1 B(pq)(p+q;2)

p+q;2 : On the other hand we have

E(X) = 1 B(pq)

1

Z

0

xx p;1(1

;x) q;1

dx=

B(p+ 1q) B(pq) =

p p+q

:

Now, using Theorem 2.1 we can state the following proposition:

Proposition 3.1.

LetX be a Beta random variable with the parameters(pq), pq 1:Then we have the inequality :

Pr(X x); q p+q

" 1 4 +

x; 1 2

2 #

(p;1) p;1

(q;1) q;1 B(pq)(p+q;2)

p+q;2 and

Pr(X x); p p+q

" 1 4 +

x; 1 2

2 #

(p;1) p;1

(q;1) q;1 B(pq)(p+q;2)

p+q;2 wherex201]: Particularly, we have

Pr

X

1 2

; q p+q

1 4

(p;1) p;1

(q;1) q;1 B(pq)(p+q;2)

p+q;2 and

Pr

X 1 2

; p

p+q

1 4

(p;1) p;1

(q;1) q;1 B(pq)(p+q;2)

p+q;2

References

1] D.S. MITRINOVIC, J.E. PECARIC and A.M. FINK, Inequalities for Func-tions and Their Integrals and Derivatives, Kluwer Academic Publishers, Dor-drecht, 1994.

2] S.S. DRAGOMIR and S. WANG, Applications of Ostrowski's inequality to the estimation of error bounds for some special means and some numerical quadrature rules, Appl. Math. Lett.,

11

(1998), 105-109.

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