Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 1, Issue 2, Article 21, 2000
SOME INEQUALITIES FOR THE EXPECTATION AND VARIANCE OF A
RANDOM VARIABLE WHOSE PDF IS n-TIME DIFFERENTIABLE
N.S. BARNETT, P. CERONE, S.S. DRAGOMIR, AND J. ROUMELIOTIS
S CHOOL OF C OMMUNICATIONS AND I NFORMATICS , V ICTORIA U NIVERSITY OF T ECHNOLOGY, PO B OX
14428, M ELBOURNE C ITY MC, V ICTORIA , 8001. AUSTRALIA
Neil.Barnett@vu.edu.au
URL: http://cams.vu.edu.au/staff/neilb.html
Peter.Cerone@vu.edu.au
URL: http://sci.vu.edu.au/staff/peterc.html
Sever.Dragomir@vu.edu.au
URL: http://rgmia.vu.edu.au/SSDragomirWeb.html
John.Roumeliotis@vu.edu.au
URL: http://www.staff.vu.edu.au/johnr/
Received 25 February, 2000; accepted 29 May, 2000
Communicated by S.P. Singh
A BSTRACT. Some inequalities for the expectation and variance of a random variable whose
p.d.f. is n-time differentiable are given.
Key words and phrases: Random Variable, Expectation, Variance, Dispertion.
2000 Mathematics Subject Classification. 60E15, 26D15.
1. I NTRODUCTION
Let f : [a, b] → R+ be the p.d.f. of the random variable X and
Z b
E (X) :=
tf (t) dt
a
its expectation and
Z
σ (X) =
b
12
12 Z b
2
2
t f (t) dt − [E (X)]
(t − E (X)) f (t) dt =
a
its dispersion or standard deviation.
ISSN (electronic): 1443-5756
c 2000 Victoria University. All rights reserved.
005-00
2
a
2
N.S. BARNETT , P. C ERONE , S.S. D RAGOMIR ,
AND
J. ROUMELIOTIS
In [1], using the identity
2
2
Z
[x − E (X)] + σ (X) =
(1.1)
b
(x − t)2 f (t) dt
a
and applying a variety of inequalities such as: Hölder’s inequality, pre-Grüss, pre-Chebychev,
pre-Lupaş, or Ostrowski type inequalities, a number of results concerning the expectation and
variance of the random variable X were obtained.
For example,
(1.2) σ 2 (X) + [x − E (X)]2

h
i
(b−a)2
a+b 2

(b
−
a)
+
x
−
kf k∞ , if f ∈ L∞ [a, b] ;


12
2




 h
i1
(b−x)2q+1 +(x−a)2q+1 q
≤
kf kp ,
if f ∈ Lp [a, b] ,

2q+1



p > 1, p1 + 1q = 1;



 b−a 2,
+ x − a+b
2
2
for all x ∈ [a, b], which imply, amongst other things, that
0 ≤ σ (X)

h
i1
1
1

(b−a)2
a+b 2 2
2
2

(b
−
a)
+
E
(X)
−
kf
k

∞ , if f ∈ L∞ [a, b] ;
12
2





n
o1
1
≤
(1.3)
[b−E(X)]2q+1 +[E(X)−a]2q+1 2q
2 ,
kf
k
if f ∈ Lp [a, b] ,

p
2q+1




p > 1, p1 + 1q = 1;


 b−a ,
+ E (X) − a+b
2
2
and
0 ≤ σ 2 (X) ≤ [b − E (X)] [E (X) − a] ≤
(1.4)
1
(b − a)2 .
4
In this paper more accurate inequalities are obtained by assuming that the p.d.f. of X is ntime differentiable and that f (n) is absolutely continuous on [a, b]. For other recent results on
the application of Ostrowski type inequalities in Probability Theory, see [2]-[4].
2. S OME P RELIMINARY I NTEGRAL I DENTITIES
The following lemma, which is interesting in itself, holds.
Lemma 2.1. Let X be a random variable whose probability distribution function f : [a, b] →
R+ is n-time differentiable and f (n) is absolutely continuous on [a, b]. Then
2
2
(2.1) σ (X) + [E (X) − x] =
n
X
(b − x)k+3 + (−1)k (x − a)k+3
k=0
f (k) (x)
(k + 3) k!
Z t
Z
1 b
2
n (n+1)
+
(t − x)
(t − s) f
(s) ds dt
n! a
x
for all x ∈ [a, b].
J. Inequal. Pure and Appl. Math., 1(2) Art. 21, 2000
http://jipam.vu.edu.au/
S OME I NEQUALITIES FOR THE E XPECTATION AND VARIANCE
3
Proof. Is by Taylor’s formula with integral remainder. Recall that
Z
n
X
(t − x)k (k)
1 t
(2.2)
f (t) =
f (x) +
(t − s)n f (n+1) (s) ds
k!
n!
x
k=0
for all t, x ∈ [a, b].
Together with
2
2
Z
σ (X) + [E (X) − x] =
(2.3)
b
(t − x)2 f (t) dt,
a
where f is the p.d.f. of the random variable X, we obtain
σ 2 (X) + [E (X) − x]2
" n
#
Z b
Z t
X (t − x)k
1
2
n
f (k) (x) +
(t − s) f (n+1) (s) ds dt
=
(t − x)
k!
n!
(2.4)
x
a
k=0
Z t
Z
Z
n
k+2
b
b
X
(t − x)
1
2
n (n+1)
(k)
=
f (x)
dt +
(t − x)
(t − s) f
(s) ds dt
k!
n! a
a
x
k=0
and since
Z
a
b
(b − x)k+3 + (−1)k (x − a)k+3
(t − x)k+2
dt =
,
k!
(k + 3) k!
the identity (2.4) readily produces (2.1)
Corollary 2.2. Under the above assumptions, we have
h
i
k
2 X
n
1 + (−1) (b − a)k+3
a
+
b
a+b
2
(k)
(2.5) σ (X) + E (X) −
=
f
2
2k+3 (k + 3) k!
2
k=0
!
2 Z t
Z 1 b
a+b
+
t−
(t − s)n f (n+1) (s) ds dt.
a+b
n! a
2
2
The proof follows by using (2.4) with x =
a+b
.
2
Corollary 2.3. Under the above assumptions,
(2.6) σ 2 (X) +
1
(E (X) − a)2 + (E (X) − b)2
2
"
#
n
X
(b − a)k+3 f (k) (a) + (−1)k f (k) (b)
=
(k + 3) k!
2
k=0
Z bZ b
1
+
K (t, s) (t − s)n f (n+1) (s) dsdt,
n! a a
where
K (t, s) :=



(t−a)2
2
if a ≤ s ≤ t ≤ b,

 − (t−b)2 if a ≤ t < s ≤ b.
2
J. Inequal. Pure and Appl. Math., 1(2) Art. 21, 2000
http://jipam.vu.edu.au/
4
N.S. BARNETT , P. C ERONE , S.S. D RAGOMIR ,
J. ROUMELIOTIS
AND
Proof. In (2.1), choose x = a and x = b, giving
(2.7) σ 2 (X) + [E (X) − a]2
Z t
Z
n
X
1 b
(b − a)k+3 (k)
2
n (n+1)
f (a) +
(t − a)
(t − s) f
(s) ds dt
=
(k
+
3)
k!
n!
a
a
k=0
and
(2.8) σ 2 (X) + [E (X) − b]2
Z t
Z
n
X
1 b
(−1)k (b − a)k+3 (k)
2
n (n+1)
=
f (b) +
(t − b)
(t − s) f
(s) ds dt.
(k + 3) k!
n! a
b
k=0
Adding these and dividing by 2 gives (2.6).
Taking into account that µ = E (X) ∈ [a, b], then we also obtain the following.
Corollary 2.4. With the above assumptions,
2
(2.9) σ (X) =
n
X
(b − µ)k+3 + (−1)k (µ − a)k+3
k=0
(k + 3) k!
1
+
n!
Z
f (k) (µ)
b
2
Z
(t − µ)
a
t
n
(t − s) f
(n+1)
(s) ds dt.
µ
Proof. The proof follows from (2.1) with x = µ ∈ [a, b].
Lemma 2.5. Let the conditions of Lemma 2.1 relating to f hold. Then the following identity is
valid.
(2.10) σ 2 (X) + [E (X) − x]2
Z
n
X
(b − x)k+3 + (−1)k (x − a)k+3 f (k) (x)
1 b
·
+
Kn (x, s) f (n+1) (s) ds,
=
k
+
3
k!
n!
a
k=0
where
(2.11)
K (x, s) =

 (−1)n+1 ψn (s − a, x − s) , a ≤ s ≤ x

ψn (b − s, s − x) ,
x<s≤b
with
(2.12) ψn (u, v) =
un+1
· (n + 2) (n + 1) u2
(n + 3) (n + 2) (n + 1)
+2 (n + 3) (n + 1) uv + (n + 3) (n + 2) v 2 .
J. Inequal. Pure and Appl. Math., 1(2) Art. 21, 2000
http://jipam.vu.edu.au/
S OME I NEQUALITIES FOR THE E XPECTATION AND VARIANCE
5
Proof. From (2.1), an interchange of the order of integration gives
Z
Z t
1 b
2
(t − x) dt
(t − s)n f (n+1) (s) ds
n! a
x
Z xZ s
1
=
−
(t − x)2 (t − s)n f (n+1) (s) dtds
n!
a
a
Z bZ b
2
n (n+1)
+
(t − x) (t − s) f
(s) dtds
x
=
Z
1
n!
s
b
K̃n (x, s) f (n+1) (s) ds,
a
where
K̃n (x, s) =

Rs
2
n
 pn (x, s) = − a (t − x) (t − s) dt, a ≤ s ≤ x

qn (x, s) =
Rb
s
(t − x)2 (t − s)n dt,
x < s < b.
To prove the lemma it is sufficient to show that K ≡ K̃.
Now,
Z s−a
Z s
2
n
n+1
p̃n (x, s) = −
(t − x) (t − s) dt = (−1)
(u + x − s)2 un du
a
0
Z s−a
u2 + 2 (x − s) u + (x − s)2 un du
= (−1)n+1
0
n+1
= (−1)
ψn (s − a, x − s) ,
where ψ (·, ·) is as given by (2.12). Further,
Z b
Z
2
n
q̃n (x, s) =
(t − x) (t − s) dt =
s
b−s
[u + (s − x)]2 un du = ψn (b − s, s − x) ,
0
where, again, ψ (·, ·) is as given by (2.12). Hence K ≡ K̃ and the lemma is proved.
3. S OME I NEQUALITIES
We are now able to obtain the following inequalities.
Theorem 3.1. Let X be a random variable whose probability density function f : [a, b] → R+
is n-time differentiable and f (n) is absolutely continuous on [a, b], then
n
X
(b − x)k+3 + (−1)k (x − a)k+3 (k)
2
2
(3.1) σ (X) + [E (X) − x] −
f (x)
(k + 3) k!
k=0

kf (n+1) k∞ n+4
n+4 

(x
−
a)
+
(b
−
x)
, if f (n+1) ∈ L∞ [a, b] ;

(n+1)!(n+4)






1

n+3+ 1
q +(b−x)n+3+ q
(n+1)
(x−a)
f
k
kp
≤
,
if f (n+1) ∈ Lp [a, b] ,
1
1

n!
n+3+
(
(nq+1) q
q)




p > 1, p1 + 1q = 1;



(n+1)

k1  kf
(x − a)n+3 + (b − x)n+3 ,
n!(n+3)
J. Inequal. Pure and Appl. Math., 1(2) Art. 21, 2000
http://jipam.vu.edu.au/
6
N.S. BARNETT , P. C ERONE , S.S. D RAGOMIR ,
AND
J. ROUMELIOTIS
for all x ∈ [a, b], where k·kp (1 ≤ p ≤ ∞) are the usual Lebesque norms on [a, b], i.e.,
Z
kgk∞ := ess sup |g (t)|
kgkp :=
and
t∈[a,b]
b
p1
|g (t)| dt , p ≥ 1.
p
a
Proof. By Lemma 2.1,
2
2
n
X
(b − x)k+3 + (−1)k (x − a)k+3
f (k) (x)
k!
(k
+
3)
k=0
Z t
Z b
1
2
n (n+1)
=
(t − x)
(t − s) f
(s) ds dt
n! a
x
:= M (a, b; x) .
σ (X) + [E (X) − x] −
(3.2)
Clearly,
Z t
n
(n+1)
(t − x) (t − s) f
(s) ds dt
a
"x
Z
#
Z b
(n+1) t
1
2
n
(t − x) sup f
(s) |t − s| ds dt
n! a
s∈[x,t]
x
(n+1) Z
2
n+1
b
f
(t − x) |t − x|
∞
dt
n!
n+1
(n+1) Za
b
f
∞
|t − x|n+3 dt
(n + 1)! a
(n+1) Z
Z b
x
f
n+3
n+3
∞
(x − t)
dt +
(t − x)
dt
(n + 1)!
a
x
(n+1) f
(x − a)n+4 + (b − x)n+4
1
|M (a, b; x)| ≤
n!
≤
≤
=
=
=
Z
b
2
∞
(n + 1)! (n + 4)
and the first inequality in (3.1) is obtained.
For the second, we use Hölder’s integral inequality to obtain
p1
Z t
1q Z t
p
(t − x)2 |t − s|nq ds f (n+1) (s) ds dt
a
x
x
Z b
p1 Z b
(n+1) p
nq+1
1
f
≤
(s) ds
(t − x)2 |t − x| q dt
n!
a
a
(n+1)
Z
b
1
1 f
p
|t − x|n+2+ q dt
=
1
n! (nq + 1) q a
(n+1) "
#
(b − x)n+3+ 1q + (x − a)n+3+ 1q
1 f
p
=
.
n! (nq + 1) 1q
n + 3 + 1q
1
|M (a, b; x)| ≤
n!
Z
b
J. Inequal. Pure and Appl. Math., 1(2) Art. 21, 2000
http://jipam.vu.edu.au/
S OME I NEQUALITIES FOR THE E XPECTATION AND VARIANCE
7
Finally, note that
Z t
Z
(n+1) 1 b
2
n
|M (a, b; x)| ≤
(t − x) |t − x| f
(s) ds dt
n! a
x
(n+1) Z
b
f
1
|t − x|n+2 dt
≤
n!
#
(n+1) "a
f
(x − a)n+3 + (b − x)n+3
1
=
n!
n+3
and the third part of (3.1) is obtained.
It is obvious that the best inequality in (3.1) is when x =
a+b
,
2
giving Corollary 3.2.
Corollary 3.2. With the above assumptions on X and f ,
(3.3)
h
i
k
k+3
2 X
n
1
+
(−1)
(b
−
a)
2
a + b σ (X) + E (X) − a + b −
f (k)
k+3
2
2 (k + 3) k!
2
k=0

kf (n+1) k∞

(b − a)n+4 , if f (n+1) ∈ L∞ [a, b] ;

n+3

2
(n+1)!(n+4)






n+3+ 1
kf (n+1) kp
q
(b−a)
, if f (n+1) ∈ Lp [a, b] , p > 1,
≤
1
n+2+ 1
1
q
q
2
n!
n+3+
(nq+1)

(
q)


1

+ 1q = 1;

p



 kf (n+1) k1
(b − a)n+3 .
2n+2 n!(n+3)
The following corollary is interesting as it provides the opportunity to approximate the variance when the values of f (k) (µ) are known, k = 0, ..., n.
Corollary 3.3. With the above assumptions and µ =
(3.4)
a+b
,
2
we have
n
X
(b − µ)k+3 + (−1)k (µ − a)k+3 (k)
2
f (µ)
σ (X) −
(k + 3) k!
k=0

kf (n+1) k∞ 

(µ − a)n+4 + (b − µ)n+4 , if f (n+1) ∈ L∞ [a, b] ;

(n+1)!(n+4)


1

n+3+ 1
q +(b−µ)n+3+ q

 kf (n+1) kp (µ−a)
,
if f (n+1) ∈ Lp [a, b] ,
1
≤
n!(n+3+ 1q )
q
(nq+1)



p > 1, p1 + 1q = 1;



(n+1)

k1 n+3
n+3  kf
(µ
−
a)
+
(b
−
µ)
.
n!(n+3)
The following result also holds.
J. Inequal. Pure and Appl. Math., 1(2) Art. 21, 2000
http://jipam.vu.edu.au/
8
N.S. BARNETT , P. C ERONE , S.S. D RAGOMIR ,
AND
J. ROUMELIOTIS
Theorem 3.4. Let X be a random variable whose probability density function f : [a, b] → R+
is n-time differentiable and f (n) is absolutely continuous on [a, b], then
(3.5)
"
#
n
k+3
k (k)
(k)
X
2
1
(b
−
a)
f
(a)
+
(−1)
f
(b)
σ (X) + (E (X) − a)2 + (E (X) − b)2 −
2
(k
+
3)
k!
2
k=0

1
f (n+1) (b − a)n+4 ,
if f (n+1) ∈ L∞ [a, b] ;

(n+4)(n+1)!
∞


1

(n+1) (b−a)n+3+ q

21/q−1
f
, if f (n+1) ∈ Lp [a, b] ,
1
1
p (nq+1) 1q
≤
n!(qn+1) q [(n+2)q+2] q


p > 1, p1 + 1q = 1;


 1 (n+1) f
(b − a)n+3 ,
2n!
1
where k·kp (1 ≤ p ≤ ∞) are the usual Lebesque p−norms.
Proof. Using Corollary 2.3,
"
#
n
k+3
k (k)
(k)
X
(b
−
a)
f
(a)
+
(−1)
f
(b)
1
2
2
2
σ (X) + (E (X) − a) + (E (X) − b) −
2
(k
+
3)
k!
2
k=0
Z Z
1 b b
≤
|K (t, s)| |t − s|n f (n+1) (s) dsdt
n! a a
=: N (a, b) .
It is obvious that
Z Z
(n+1) 1 b b
N (a, b) ≤ f
|K (t, s)| |t − s|n dsdt
∞ n!
a
a
Z Z t
Z b
(n+1) 1 b
n
n
|K (t, s)| |t − s| ds +
|K (t, s)| |t − s| ds dt
= f
∞ n!
a
a
t
#
Z b"
2
n+1
2
n+1
1 (t
−
a)
(t
−
a)
(t
−
b)
(b
−
t)
f (n+1) =
·
+
·
dt
∞
n!
2
n+1
2
n+1
a
Z b
(n+1) 1
=
f
(t − a)n+3 + (b − t)n+3 dt
∞
2 (n + 1)!
"a
#
n+4
n+4
(n+1) 1
(b
−
a)
(b
−
a)
f
=
+
∞
2 (n + 1)!
n+4
n+4
(n+1) f
∞
=
(b − a)n+4
(n + 4) (n + 1)!
so the first part of (3.5) is proved.
J. Inequal. Pure and Appl. Math., 1(2) Art. 21, 2000
http://jipam.vu.edu.au/
S OME I NEQUALITIES FOR THE E XPECTATION AND VARIANCE
9
Using Hölder’s integral inequality for double integrals,
N (a, b)
1q
Z b Z b
p1 Z b Z b
(n+1) p
1
q
qn
≤
f
(s) dsdt
×
|K (t, s)| |t − s| dsdt
n!
a
a
a
a
1 1q
Z b
(b − a) p f (n+1) p Z b Z t
|K (t, s)|q |t − s|qn ds +
|K (t, s)|q |t − s|qn ds dt
=
n!
a
a
t
"
"
# # 1q
1 Z
Z
2q Z b
b
(b − a) p f (n+1) p
(t − a)2q t
(t
−
b)
=
|t − s|qn ds +
|t − s|qn ds dt
q
q
n!
2
2
a
a
t
"
"
# # 1q
1 (b − a) p f (n+1) p Z b (t − a)2q (t − a)qn+1 (t − b)2q (b − t)qn+1
=
+
dt
n!
2q (qn + 1)
2q (qn + 1)
a
1 1q Z b
1q
Z b
(b − a) p f (n+1) p 1
(n+2)q+1
(n+2)q+1
=
· q
(t − a)
dt +
(b − t)
dt
n!
2 (qn + 1)
a
a
# 1q
1 1q "
(b − a) p f (n+1) p 1
(b − a)(n+2)q+2 (b − a)(n+2)q+2
=
· q
+
n!
2 (qn + 1)
(n + 2) q + 2
(n + 2) q + 2
=
=
1
2
21/q f (n+1) p (b − a)n+2+ p + q
1
1
n!2 (qn + 1) q ((n + 2) q + 2) q
i
h
1
21/q−1 f (n+1) p (b − a)n+3+ q
1
1
n! (qn + 1) q [(n + 2) q + 2] q
and the second part of (3.5) is proved.
Finally, we observe that
1
N (a, b) ≤
sup |K (t, s)| |t − s|n
n! (t,s)∈[a,b]2
2
1 (b − a)
· (b − a)n (b − a)
n!
2
1
=
(b − a)n+3 f (n+1) 1 ,
2n!
Z bZ
a
Z
=
which is the final result of (3.5).
J. Inequal. Pure and Appl. Math., 1(2) Art. 21, 2000
b
(n+1) f
(s) dsdt
a
b
(n+1) f
(s) ds
a
http://jipam.vu.edu.au/
10
N.S. BARNETT , P. C ERONE , S.S. D RAGOMIR , AND J. ROUMELIOTIS
The following particular case can be useful in practical applications. For n = 0, (3.1) becomes
"
#
2
2
(b
−
a)
a
+
b
(3.6) σ 2 (X) + [E (X) − x]2 − (b − a) x −
+
f (x)
2
12
 kf 0 k ∞

(x − a)4 + (b − x)4 ,
if f 0 ∈ L∞ [a, b] ;

4




i
 qkf 0 kp h
3+ 1q
3+ 1q
(x
−
a)
+
(b
−
x)
, if f 0 ∈ Lp [a, b] ,
≤
3q+1


p > 1, p1 + 1q = 1;


h
i

2

 kf 0 k (b−a) + x − a+b 2 ,
1
12
2
for all x ∈ [a, b]. In particular, for x = a+b
,
2
2
a+b
(b − a)3
a + b 2
(3.7) σ (X) + E (X) −
−
f
2
12
2
 0
kf k∞

(b − a)4 , if f 0 ∈ L∞ [a, b] ;

32






3+ 1
qkf 0 kp (b−a) q
≤
, if f 0 ∈ Lp [a, b] ,
2+ 1
q (3q+1)

2



p > 1, p1 + 1q = 1;



0
 kf k1 (b − a)3 ,
12
which is, in a sense, the best inequality that can be obtained from (3.6). If in (3.6) x = µ =
E (X), then
"
#
2
(b − a)2
a+b
2
(3.8) σ (X) − (b − a) E (X) −
+
f (E (X))
2
12
 kf 0 k ∞
(E (X) − a)4 + (b − E (X))4 , if f 0 ∈ L∞ [a, b] ;


4




 kf 0 kp 4
4
(E
(X)
−
a)
+
(b
−
E
(X))
, if f 0 ∈ Lp [a, b] , p > 1,
1
≤
3+ q )
(




h
i


 kf 0 k (b−a)2 + E (X) − a+b 2 .
1
12
2
In addition, from (3.5),
3 1
(b
−
a)
f
(a)
+
f
(b)
(3.9) σ 2 (X) + (E (X) − a)2 + (E (X) − b)2 −
2
3
2
 1 0
kf k∞ (b − a)4 ,
if f 0 ∈ L∞ [a, b] ;

4





3+ 1q
1
0
, if f 0 ∈ Lp [a, b] , p > 1,
1
1 kf kp (b − a)
≤
q
q
n!2 (q+1)





 1 0
kf k1 (b − a)3 ,
2
which provides an approximation for the variance in terms of the expectation and the values of
f at the end points a and b.
J. Inequal. Pure and Appl. Math., 1(2) Art. 21, 2000
http://jipam.vu.edu.au/
S OME I NEQUALITIES FOR
THE
E XPECTATION
AND
VARIANCE
11
Theorem 3.5. Let X be a random variable whose p.d.f. f : [a, b] → R+ is n−time differentiable
and f (n) is absolutely continuous on [a, b]. Then
n
k+3
k
k+3
(k)
X
(b − x)
+ (−1) (x − a)
f (x) 2
2
(3.10) σ (X) + (E (X) − x) −
·
k+3
k! k=0
 (n+1)
k∞
n+4
n+4 kf

(x
−
a)
+
(b
−
x)
,
if f (n+1) ∈ L∞ [a, b] ;


(n+1)!(n+4)





h
i 1 (n+1) k
1
(n+3)q+1
(n+3)q+1 q kf
p
q
≤
(x
−
a)
+
(b
−
x)
C
, if f (n+1) ∈ Lp [a, b] , p > 1;
n!






(n+1)

k1
 b−a a+b n+3 kf
+
x
−
·
,
2
2
n!(n+3)
where
Z
C=
(3.11)
0
1
un+3
un+2
un+1
+ 2 (1 − u)
+ (1 − u)2
n+3
n+2
n+1
q
du.
Proof. From (2.10),
n
k+3
k
k+3
(k)
X
f
(x)
(b
−
x)
+
(−1)
(x
−
a)
2
(3.12) σ (X) + (E (X) − x)2 −
·
k+3
k! k=0
Z b
1
(n+1)
= Kn (x, s) f
(s) ds .
n! a
Now, on using the fact that from (2.11), (2.12), ψn (u, v) ≥ 0 for u, v ≥ 0,
Z b
1
(n+1)
(3.13) Kn (x, s) f
(s) ds
n! a
(n+1) Z
Z b
x
f
∞
≤
ψn (s − a, x − s) ds +
ψn (b − s, s − x) ds .
n!
a
x
Further,
ψn (u, v) =
(3.14)
un+3
un+2
un+1
+ 2v
+ v2
n+3
n+2
n+1
and so
Z
(3.15)
a
x
ψn (s − a, x − s) ds
#
Z x"
n+1
(s − a)n+3
(s − a)n+2
(s
−
a)
=
+ 2 (x − s)
+ (x − s)2
ds
n+3
n+2
n+1
a
Z 1 n+3
n+1
λ
λn+2
n+4
2 λ
= (x − a)
+ 2 (1 − λ)
+ (1 − λ)
dλ,
n+3
n+2
n+1
0
s−a
where we have made the substitution λ = x−a
.
Collecting powers of λ gives
1
2
1
2λn+2
λn+1
n+3
λ
−
+
−
+
n+3 n+2 n+1
(n + 2) (n + 1) n + 1
J. Inequal. Pure and Appl. Math., 1(2) Art. 21, 2000
http://jipam.vu.edu.au/
12
N.S. BARNETT , P. C ERONE , S.S. D RAGOMIR , AND J. ROUMELIOTIS
and so, from (3.15),
Z x
(3.16)
ψn (s − a, x − s) ds
a
1
1
2
1
= (x − a)
−
+
n+4 n+3 n+2 n+1
1
2
−
+
(n + 3) (n + 2) (n + 1) (n + 2) (n + 1)
n+4
(x − a)n+4
=
.
(n + 4) (n + 1)
Similarly, on using (3.14),
#
Z b
Z b"
n+1
(b − s)n+2
(b − s)n+3
2 (b − s)
+ 2 (s − x)
+ (s − x)
ds
ψn (b − s, s − x) ds =
n+3
n+2
n+1
x
x
b−s
and making the substitution ν = b−x
gives
Z b
Z
n+4
ψn (b − s, s − x) ds = (b − x)
x
0
n+4
=
(3.17)
1
n+1
ν n+3
ν n+2
2 ν
+ 2 (1 − ν)
+ (1 − ν)
dν
n+3
n+2
n+1
(b − x)
,
(n + 4) (n + 1)
where we have used (3.15) and (3.16). Combining (3.16) and (3.17) gives the first inequality in
(3.10). For the second inequality in (3.10), we use Hölder’s integral inequality to obtain
Z b
1q
f (n+1) (s) Z b
1
p
q
(n+1)
(3.18)
Kn (x, s) f
(s) ds ≤
|Kn (x, s)| ds .
n!
n!
a
a
Now, from (2.11) and (3.14)
Z b
Z
q
|Kn (x, s)| ds =
a
x
Z
q
b
ψ q (b − s, s − x) ds
a
x
h
i
(n+3)q+1
(n+3)q+1
= C (x − a)
+ (b − x)
,
ψ (s − a, x − s) ds +
where C is as defined in (3.11) and we have used (3.15) and (3.16). Substitution into (3.18)
gives the second inequality in (3.10).
Finally, for the third inequality in (3.10). From (3.12),
(3.19)
Z b
1
(n+1)
Kn (x, s) f
(s) ds
n!
a
Z x
Z b
(n+1) (n+1) 1
≤
ψn (s − a, x − s) f
(s) ds +
ψn (b − s, s − x) f
(s) ds
n!
a
x
Z x
Z b
(n+1) (n+1) 1
f
f
≤
ψn (x − a, 0)
(s) ds + ψn (b − x, 0)
(s) ds ,
n!
a
x
where, from (3.14),
(3.20)
J. Inequal. Pure and Appl. Math., 1(2) Art. 21, 2000
ψn (u, 0) =
un+3
.
n+3
http://jipam.vu.edu.au/
S OME I NEQUALITIES FOR
THE
E XPECTATION
AND
VARIANCE
13
Hence, from (3.19) and (3.20)
(
)
Z b
n+3
n+3
1
(n+1) 1
(x
−
a)
(b
−
x)
f
(·)1
Kn (x, s) f (n+1) (s) ds ≤
max
,
n!
n!
n+3
n+3
a
1
=
[max {x − a, b − x}]n+3 f (n+1) (·)1 ,
n! (n + 3)
which, on using the fact that for X, Y ∈ R
X − Y X +Y
+
max {X, Y } =
2
2 gives, from (3.12), the third inequality in (3.10). The theorem is now completely proved.
Remark 3.6. The results of Theorem 3.5 may be compared with those of Theorem 3.1. Theorem 3.5 is based on the single integral identity developed in Lemma 2.5, while Theorem 3.1 is
based on the double integral identity representation for the bound. It may be noticed from (3.1)
and (3.10) that the bounds are the same for f (n+1) ∈ L∞ [a, b], that for f (n+1) ∈ L1 [a, b] (which
is always true since f (n) is absolutely continuous) the bound obtained in (3.1) is better and for
f (n+1) ∈ Lp [a, b], p > 1, the result is inconclusive.
R EFERENCES
[1] N.S. BARNETT, P. CERONE, S.S. DRAGOMIR AND J. ROUMELIOTIS, Some inequalities for the
dispersion of a random variable whose p.d.f. is defined on a finite interval, J. Inequal. Pure and Appl.
Math., accepted for publication.
[2] N.S. BARNETT AND S.S. DRAGOMIR, An inequality of Ostrowski type for cumulative distribution
functions, Kyungpook Math. J., 39(2) (1999), 303–311.
[ONLINE] (Preprint available) http://rgmia.vu.edu.au/v1n1.html
[3] N.S. BARNETT AND S.S. DRAGOMIR, An Ostrowski type inequality for a random variable whose
probability density function belongs to L∞ [a, b], Nonlinear Anal. Forum, 5 (2000), 125–135.
[ONLINE] (Preprint available) http://rgmia.vu.edu.au/v1n1.html
[4] S.S. DRAGOMIR, N.S. BARNETT AND S. WANG, An Ostrowski type inequality for a random
variable whose probability density function belongs to Lp [a, b], p > 1, Math. Inequal. Appl., 2(4)
(1999), 501–508.
J. Inequal. Pure and Appl. Math., 1(2) Art. 21, 2000
http://jipam.vu.edu.au/