Journal of Applied Mathematics, Statistics and Informatics (JAMSI), 8 (2012), No. 2
SEQUENCES OF INEQUALITIES AMONG
DIFFERENCES OF GINI MEANS AND
DIVERGENCE MEASURES
INDER J. TANEJA
Abstract
In 1938, Gini [3] studied a mean having two parameters. Later, many authors studied properties of
this mean. In particular, it contains the famous means as harmonic, geometric, arithmetic, etc. Here
we considered a sequence of inequalities arising due to particular values of each parameter of Gini’s
mean. This sequence generates many nonnegative differences. Not all of them are convex. We have
established new sequences of inequalities of these differences. Some refinement inequalities are also
presented. Considering in terms of probability distributions these differences, we have made connections with some well known divergence measures.
Mathematics Subject Classification 2000: 94A17; 26A48; 26D07
Additional Key Words and Phrases: Arithmetic mean; Geometric Mean; Harmonic Mean; Gini
Mean; Power Mean; Differences of Means; Divergence measures
1.
GINI MEAN OF ORDER R AND S
The Gini [3] mean of order r and s is given by

1

ar +br r−s


s
 as +b
Er,s (a, b) = exp ar ln ra+brr ln b

a +b


√ab
r 6= s
r = s 6= 0 .
(1)
r=s=0
In particular when s = 0 in (1), we have
(
Er,0 (a, b) := Br (a, b) =
ar +br
√ 2
ab,
r1
,
r=
6 0
.
r=0
(2)
Again, when s = r − 1 in (1), we have
Er,r−1 (a, b) := Ks (a, b) =
ar + br
, r ∈ R.
+ br−1
ar−1
(3)
The expression (2) is famous as mean of order r or power mean. The expression (3)
is known as Lehmer mean[4]. Both these means are monotonically increasing in r.
Moreover, these two have the following inequality [2] among each other:
(
< Lr (a, b), r > 1
Br (a, b)
.
(4)
> Lr (a, b), r < 1
DOI 10.2478/v10294-012-0014-2
©University of SS. Cyril and Methodius in Trnava
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INDER J. TANEJA
Since Er,s = Es,r , the Gini-mean Er,s (a, b) given by (1) is an increasing function in r or s. Using the monotonicity property [1], [7], [5] we have the following
inequalities:
E−3,−2 ≤E−2,−1 ≤ E−3/2,−1/2 ≤ E−1,0 ≤ E−1/2,0 ≤
≤ E−1/2,1/2 ≤ E0,1/2 ≤ E0,1 ≤ E0,2 ≤ E1,2 ,
(5)
E−3,−2 ≤E−2,−1 ≤ E−3/2,−1/2 ≤ E−1,0 ≤ E−1/2,0 ≤
≤ E−1/2,1/2 ≤ E0,1/2 ≤ E0,1 ≤ E1/2,1 ≤ E1,2 .
(6)
or
Equivalently,
K−2 ≤K−1 ≤ K−1/2 ≤ (K0 = B−1 = H) ≤ B−1/2 ≤
≤ K1/2 = B0 = G ≤ B1/2 ≤ (K1 = B1 = A) ≤ (B2 = S) ≤ K2 ,
(7)
K−2 ≤K−1 ≤ K−1/2 ≤ (K0 = B−1 = H) ≤ B−1/2 ≤
≤ K1/2 = B0 = G ≤ B1/2 ≤ (K1 = B1 = A) ≤ E1/2,1 ≤ K2 .
(8)
or
where H, G, A and S are respectively , the harmonic, geometric, arithmetic and the
square-root means. We observe from the expression (7) that the means considered
either are the particular cases of (2) or of (3), while in the second case we have we
have a mean E1/2,1 , which is neither a particular case of (2) nor of (3). We can easily
find values proving that there is no relation between S and E1/2,1 . Summarizing
the expressions (7) and (8), we can write them in joint form as
P1 ≤ P2 ≤ P3 ≤ H ≤ P4 ≤ G ≤ N1 ≤ A ≤ (P5 or S) ≤ P6 ,
(9)
where P1 = K−2 , P2 = K−1 , P3 = K−1/2 , P4 = B−1/2 , N1 = B1/2 , P5 = E1/2,1 and
P5 = K2 .
In [9; ?], the author studied the following inequalities:
H ≤ G ≤ N1 ≤ N3 ≤ N2 ≤ A ≤ S,
(10)
where
N2 (a, b) =
!
√ ! r
√
a+ b
a+b
2
2
and
N3 (a, b) =
a+
√
ab + b
.
3
The expression N3 (a, b) is famous as Heron’s mean. The inequalities (10) admit
many non-negative differences. Bases on these difference the author [9; ?] proved
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SEQ. OF INEQ. AMONG DIFFERENCES OF GINI MEANS AND DIVERG. MEASURES
the following result:


4DN N ≤ 34 DN2 G ≤ DAG ≤ 4DAN2


1


 2 DAH ≤ 1 2 1



2 DSG ≤ DAG

 13 DSH ≤




2
1
DSA ≤
3 DSN1 ≤ 2 DSG





4

DSN2 ≤ 4DAN2


 35
2
4 DSN3 ≤ 3 DSN1
,
(11)
where DSA = S − A, DSH = S − H, etc. Some applications of the inequalities (11)
can be seen in [6], [7].
Combining (9) and (10), we have the following sequence of inequalities:
P1 6 P2 6 P3 6 H 6 P4 6 G 6 N1 6 N3 6 N2 6 A 6 (P5 or S) 6 P6 .
The inequalities (12) admits many nonnegative differences such as
h a
a i
a
= b ft
− fp
,
Dtp (a, b) = bgtp
b
b
b
where
(12)
(13)
gtp (x) = ft (x) − fp (x), ft (x) > fp (x), ∀x > 0.
More precisely, the expression (12) and the function f : (0, ∞) → R given in (13)
lead us to the following result:
fP1 (x) 6 fP2 (x) 6 fP3 (x) 6 fH (x) 6 fP4 (x) 6 fG (x) 6 fN1 (x) 6
6 fN3 (x) 6 fN2 (x) 6 fA (x) 6 (fP5 (x) or fS (x)) 6 fP6 (x).
(14)
Equivalently,
√
x(x + 1)
x ( x + 1)
4x
x(x2 + 1)
2x
6
6
6 √
6
2 6
x3 + 1
x2 + 1
1+x
x3/2 + 1
( x + 1)
!
√
√
2
r
√
√
x+1
x+1
x+1
x+ x+1
6 x6
6
6
6
2
3
2
2
!
r
2
x+1
x2 + 1
x+1
x2 + 1
√
6
.
6
or
6
2
2
x+1
x+1
(15)
The inequalities appearing in (12) produces many nonnegative differences. In
total, we have 77 differences. Some of them are not convex and some of them are
convex. Some of them are equal to each other with some multiplicative constants
such as
(i) DP6 A = DP6 H = 21 DAH .
(ii) DP5 P4 = DAN3 = 23 DAN1 = 13 DAG = 2DN3 N1 = 12 DN3 G = 23 DN1 G .
We leave to the readers to verify the convexity of the measures DP6 N2 , DP6 N3 ,
DP6 N1 , DP6 G , DP6 P4 , DP6 P2 , DP6 P1 , DP5 A , DP5 N2 , DP5 N3 , DP5 N1 , DP5 G , DP5 H ,
DAP4 , DP6 S , DSP4 , DP6 S and DSP4 . While, the convexity of the difference of means
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INDER J. TANEJA
DSA , DSN2 , DSN3 , DSN1 , DSG , DSH , DAN2 , DAH , DAG , DN2 N1 and DN2 G is already proved in Taneja [9; ?]. We observe that not all difference of means are
convex [12].
In this paper our aim is to produce new inequalities for the difference of means
arising due to inequalities given in (12). In another words we shall improve considerably the inequalities given in (11). For this we need first to know the convexity of the difference of means. In total, we have 77 differences. Some of them
are equal to each other with some multiplicative constants. Some of them are not
convex and some of them are convex.
2.
SEQUENCES OF INEQUALITIES
In this section we shall bring sequence of inequalities based on the differences
arising due to (12). The results given in this section are based on the applications
of the following lemma [11]:
Lemma 2.1. Let f1 , f2 : I ⊂ R+ → R be two convex functions satisfying the assumptions:
(i) f1 (1) = f10 (1) = 0, f2 (1) = f20 (1) = 0;
(ii) f1 and f2 are twice differentiable in R+ ;
(iii) there exists the real constants α, β such that 0 6 α < β and
α6
f100 (x)
6 β, f200 (x) > 0,
f200 (x)
for all x > 0 then we have the inequalities:
α φf2 (a, b) 6 φf1 (a, b) 6 β φf2 (a, b),
for all a, b ∈ (0, ∞), where the function ϕ(·) (a, b) is given by
b
φf (a, b) = a f
, a, b > 0.
a
Theorem 2.1. The following sequences of inequalities hold:
1
1
1
1
DP6 P1 6 DP6 P2 6 DSA 6 DSH 6 DAH 6
8
3
4 6
 2
D


P
N


2
9 6 2
 1
D
2
6 3
6 DP6 G 6 25 P5 H 6
DP N
DP N  3

3 DAP4

 72 6 3 6 25 6 1 

D
D
SP
P
P
4
6 4
5
7
4
2
6 4DN2 N1 6 DN2 G 6 DAG 6 4DAN2 6 DP5 G 6
3
3
6
4
6 DP5 N1 6 DP5 N3 6 DP5 N2 6 2DP5 A ,
5
3
DSA 6
4
5 DSN2
3
4 DSN3
6
2
DSN1 6
3
1
3 DP6 G
1
2 DSG
52
6
2
DP H
5 5
(16)
(17)
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SEQ. OF INEQ. AMONG DIFFERENCES OF GINI MEANS AND DIVERG. MEASURES
and
1
8 DP6 P1
2
13 DP5 P1
6
1
6 DP6 P2
2
9 DP5 P2
6
2
4
DP P 6 DP6 N2 6 DP6 S 6 DAG .
7 5 3
9
(18)
P ROOF. We shall prove each part separately.
(1) For DP6 P1 ≤ 43 DP6 P2 : Let us consider a function gP6 P1 P6 P2 (x) = fP006 P1 (x) fP006 P2 (x),
where fP006 P1 (x) and fP006 P2 (x) are as given by
!
2
2
(x − 1) − x + x2 +
2
2 x +1
+ 8x3
4
2
+ (x − 1) + x (x − 1)
fP006 P1 (x) =
(x3 + 1)3
and
fP006 P2 (x)
=
2 x6 + 15x4 + 16x3 + 15x2 + 1
3
(x + 1) (x2 + 1)
3
,
respectively. This gives β = gP6 P1 P6 P2 (1) = fP006 P1 (1) fP006 P2 (1) = 43 . Now
we shall show that 34 DP6 P2 − DP6 P1 ≥ 0. We can write 43 DP6 P2 − DP6 P1 =
b fP6 P2 P6 P1 (a/b), where
4
fP6 P2
P6 P1 (x) =
4
(x + 1) (x − 1)
fP P (x) − fP6 P1 (x) = 2
.
3 6 2
(x − x + 1) (x2 + 1)
(19)
Since fP6 P2 P6 P1 (x) > 0, ∀x > 0, x 6= 1, this proves the required result.
00
(x),
(2) For DP6 P2 ≤ 6DSA : Let us consider a function gP6 P2 SA (x) = fP006 P2 (x) fSA
00
(x) are as given by
where fP006 P2 (x) and fSA
2 x6 + 15x4 + 16x3 + 15x2 + 1
00
fP6 P2 (x) =
3
3
(x + 1) (x2 + 1)
and
00
fSA
(x) =
(x2
1
√
,
+ 1) 2x2 + 2
00
(1) = 6. Now
respectively. This gives β = gP6 P1 SA (1) = fP006 P2 (1) fSA
we shall show that 6DSA − DP6 P2 ≥ 0. We can write 6DSA − DP6 P2 =
b fSA P6 P2 (a/b), where
fSA
with
P6 P2 (x)
= 6fSA (x) − fP6 P2 (x) =
(x2
k1 (x)
,
+ 1) (x + 1)
(20)
√
k1 (x) = 3 2x + 2 (x + 1) x2 + 1 − 4x4 + 5x3 + 6x2 + 5x + 4 .
Now we shall show that k1 (x) > 0, ∀x > 0, x 6= 1. Let us consider
√
2
2
h1 (x) = 3 2x + 2 (x + 1) x2 + 1
− 4x4 + 5x3 + 6x2 + 5x + 4 .
4
= (x − 1) 2x4 + 4x3 + 3x2 + 4x + 2 .
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INDER J. TANEJA
Since h2 (x) > 0 giving k2 (x) > 0, ∀x > 0, x 6= 1. This proves the required
result.
Argument: Let a and b two positive numbers, i.e., a > 0 and b > 0. If a2 − b2 ≥ 0,
then we can conclude that a ≥ b because a − b = (a2 − b2 ) (a + b). We have used
this argument to prove k1 (x) ≥ 0, ∀x > 0. We shall use frequently this argument to
prove the other parts.
Remark 2.1. According the parts 1 and 2, we observe that it is sufficient to write
and show the nonnegativity of the expressions (19) and (20) for the other parts. The
rest part of the proof follows on similar lines. Throughout, it is understood that ∀x >
0, x 6= 1.
(3) For DAH ≤ 98 DP6 N2 : In this case we have
fP6 N2
with
AH (x)
=
k2 (x)
8
fP6 N2 (x) − fAH (x) =
,
9
18 (x + 1)
√
√
k2 (x) = 7x2 + 18x + 7 − 4 2x + 2 x + 1 (x + 1) .
Let us consider
2 √
2
√
h2 (x) = 7x2 + 18x + 7 − 4 2x + 2 x + 1 (x + 1) .
4 √
√
=
x−1
17x2 + 4x3/2 + 38x + 4 x + 17 .
(4) For DAH ≤ 67 DP6 N3 : In this case we have
fP6 N3
√
2
6
( x − 1)
.
AH (x) = fP6 N3 (x) − fAH (x) =
7
18 (x + 1)
(5) For DAH ≤ 45 DSP4 : In this case we have
fSP4
AH (x)
where
=
4
k3 (x)
fSP4 (x) − fAH (x) =
√
2,
5
10 (x + 1) ( x + 1)
√
2
√
k3 (x) =4 (x + 1) 2x + 2 x + 1 −
3
5x + 10x5/2 + 17x2 + 17x+
√
√
−
.
2
+10x ( x − 1) + 10 x + 5
Let us consider
h
√
2 i2
√
−
h3 (x) = 4 (x + 1) 2x + 2 x + 1
3
2
5x + 10x5/2 + 17x2 + 17x+
√
√
−
.
2
+10x ( x − 1) + 10 x + 5
6 7x3 + 70x5/2 + 201x2 +
√
√
=
x−1
.
+340x3/2 + 201x + 70 x + 7
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SEQ. OF INEQ. AMONG DIFFERENCES OF GINI MEANS AND DIVERG. MEASURES
(6) For DP6 N3 ≤
14
15 DP6 N1 :
fP6 N1
In this case we have
P6 N3 (x) =
√
2
14
( x − 1)
fP6 N1 (x) − fP6 N3 (x) =
.
15
30 (x + 1)
(7) For DP6 N3 ≤ 23 DP6 P4 : In this case we have
fP6 P4
P6 N3 (x) =
√ √
4
2
x ( x − 1)
fP6 P4 (x) − fP6 N3 (x) = √
.
2
3
3 ( x + 1) (x + 1)
(8) For DSP4 ≤ DP6 N1 : In this case we have
fP6 N1
SP4 (x)
= fP6 N1 (x) − fSP4 (x) =
k4 (x)
√
2,
4 (x + 1) ( x + 1)
where
k4 (x) = 3x3 + 4x5/2 + 9x2 + 9x + 4x
2
√
√
x−1 +4 x+3
p
2
√
−2 2x2 + 2 (x + 1) x + 1 .
Let us consider
h
i2
2
√
√
h4 (x) = 3x3 + 4x5/2 + 9x2 + 4x x − 1 + 9x + 4 x + 3 −
h p
2 i2
√
.
− 2 2x2 + 2 (x + 1) x + 1
!
√
4 x3 ( x − 2)2 + 4x3 + 20x5/2 +
√
=
x−1
.
√
2
+78x2 + 20x3/2 + 4x + (2 x − 1)
(9) For DSP4 ≤ 57 DP6 P4 : In this case we have
fP6 P4
SP4 (x)
=
5
k5 (x)
fP P (x) − fSP4 (x) =
√
2,
7 6 4
14 (x + 1) ( x + 1)
where
√
k5 (x) =10x3 + 20x5/2 + 26x2 + 26x + 20 x + 10
p
2
√
− 7 2x2 + 2 x + 1 (x + 1) .
Let us consider
2
√
h5 (x) = 10x3 + 20x5/2 + 26x2 + 26x + 20 x + 10 −
h p
i2
2
√
− 7 2x2 + 2 x + 1 (x + 1)
4 x4 + 8x7/2 + 94x3 + 264x5/2 +
√
√
=2 x−1
.
+386x2 + 264x3/2 + 94x + 8 x + 1
(10) For DP6 N2 ≤ 34 DP6 G : In this case we have
fP6 G
P6 N2 (x)
=
3
k6 (x)
fP G (x) − fP6 N2 (x) =
,
4 6
4 (x + 1)
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INDER J. TANEJA
where
k6 (x) =
√
√
√ 2x + 2 x + 1 (x + 1) − x2 + 1 + 3x3/2 + 3 x .
Let us consider
2 √
√ 2
√
2x + 2 x + 1 (x + 1) − x2 + 1 + 3x3/2 + 3 x
h6 (x) =
4 2
√
√
=
x−1
x + 2x3/2 + x + 2 x + 1 .
(11) For DP6 N1 ≤ 65 DP6 G : In this case we have
fP6 G
√
2
5
( x − 1)
.
P6 N1 (x) = fP6 G (x) − fP6 N1 (x) =
6
12 (x + 1)
(12) For DP6 P4 ≤ 76 DP6 G : In this case we have
fP6 G
h√
√ i
√
4
2
( x − 1) ( x − 1) + x
7
.
P6 P4 (x) = fP6 G (x) − fP6 P4 (x) =
6
12 (x + 1)
(13) For DP6 G ≤ 65 DP5 H : In this case we have
fP5 H
P6 G (x)
=
6
fP H (x) − fP6 G (x) =
5 5
h√
√ i
√
4
2
( x − 1) ( x − 1) + x
.
5 (x + 1)
(14) For DP6 G ≤ 2DAP4 : In this case we have
fAP4
√ √
4
x ( x − 1)
.
P6 G (x) = 2fAP4 (x) − fP6 G (x) = √
2
( x + 1) (x + 1)
(15) For DP5 H ≤ 10DN2 N1 : In this case we have
fN2 N1
where
P5 H (x)
= 10fN2 N1 (x) − fP5 H (x) =
k7 (x)
,
√
2
2 ( x + 1) (x + 1)
√
3
√
k7 (x) =5 2x + 2 x + 1 (x + 1) −
√
− 7x3 + 20x5/2 + 37x2 + 32x3/2 + 37x + 20 x + 7 .
Let us consider
i2
h √
3
√
h7 (x) = 5 2x + 2 x + 1 (x + 1) −
2
√
− 7x3 + 20x5/2 + 37x2 + 32x3/2 + 37x + 20 x + 7
4 x4 + 24x7/2 + 72x3 + 120x5/2 +
√
√
=
x−1
.
+126x2 + 120x3/2 + 72x + 24 x + 1
(16) For DAP4 ≤ 6DN2 N1 : In this case we have
fN2 N1
AP4 (x)
= 6fN2 N1 (x) − fAP4 (x) =
56
k8 (x)
√
2,
2 ( x + 1)
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SEQ. OF INEQ. AMONG DIFFERENCES OF GINI MEANS AND DIVERG. MEASURES
where
√
3
√
√
k8 (x) = 3 2x + 2 x + 1 − 2 2x2 + 7x3/2 + 6x + 7 x + 2 .
Let us consider
h √
2
3 i2 2
√
√
h8 (x) = 3 2x + 2 x + 1
− 4x + 14x3/2 + 12x + 14 x + 4
4 2
√
√
=2 x−1
x + 2x3/2 + 2 x + 1 .
(17) For DAN2 ≤ 16 DP5 G : In this case we have
P5 G (x)
fAN2
=
1
k9 (x)
√
fP5 G (x) − fAN2 (x) =
,
6
12 ( x + 1)
where k9 (x) = k8 (x) > 0.
(18) For DP5 G ≤ 23 DP5 N1 : In this case we have
√
2
3
( x − 1)
fP5 G (x) − fP5 N1 (x) =
.
2
8 (x + 1)
P5 N1 (x) =
fP5 G
(19) For DP5 N1 ≤ 65 DP5 N3 : In this case we have
(20) For DP5 N3 ≤
10
9 DP5 N2
fP5 N2
√
2
6
( x − 1)
fP5 N3 (x) − fP5 N1 (x) =
.
5
20 (x + 1)
P5 N1 (x) =
fP5 N3
P5 N3 (x)
: In this case we have
=
10
k10 (x)
fP N (x) − fP5 N3 (x) =
√
2,
9 5 2
18 ( x + 1)
where
√
3
√
√
k10 (x) = 2 4x2 + 9x3/2 + 14x + 9 x + 4 − 5 2x + 2 x + 1 .
Let us consider
h i2 h √
3 i2
√
√
h10 (x) = 2 4x2 + 9x3/2 + 14x + 9 x + 4
− 5 2x + 2 x + 1
4 2
√
√
=2 x−1
7x + 22x3/2 + 32x + 22 x + 7 .
(21) For DP5 N2 ≤ 32 DP5 A : In this case we have
fP5 A
P5 N2 (x)
where
k11 (x) =
√
=
3
k11 (x)
fP A (x) − fP5 N2 (x) = √
2,
2 5
4 ( x + 1)
2x + 2
√
x+1
3
√
− (x + 1) x + 6 x + 1 .
Let us consider
h√
3 i2 2
√
√
2x + 2 x + 1
− (x + 1) x + 6 x + 1
4
√
√
=
x − 1 (x + 1) x + 4 x + 1 .
h11 (x) =
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INDER J. TANEJA
Parts 1-21 and the expression (11) together proves the sequences of inequalities
appearing in (16).
(22) For DSN2 ≤ 56 DSN1 : In this case we have
fSN1
SN2 (x)
where
k12 (x) = 6
=
5
fSN1 (x) − fSN2 (x) =
6
1
24
× k12 (x),
h p
√
2 i
√
√
x + 1 2x + 2 − 2 2x2 + 2 + 5 x + 1
.
In order to prove k12 (x) > 0, ∀x > 0, x 6= 1, here we shall apply twice the
argument given in part 2. Let us consider
hp
√
2
2 i2
√
√
2x2 + 2 + 5 x + 1
h12 (x) = 6 x + 1 2x + 2 − 2
2
√ √
√
= 39x2 + 41x3/2 + 3 x x − 1 + 41 x + 39
2 p
√
2x2 + 2.
− 20 x + 1
Applying again the same process over h12 (x), we get
h
i2
2
√ √
√
h12a (x) = 39x2 + 41x3/2 + 3 x x − 1 + 41 x + 39
h
i2
2 p
√
− 20 x + 1
2x2 + 2
2
4 √
√
=
x−1
721x2 + 3116x3/2 + 4806x + 3116 x + 721 .
Since h12a (x) > 0, giving h12 (x) > 0. This proves k12 (x) > 0.
(23) For DSN1 ≤ 12 DP6 G : In this case we have
fP6 G
SN1 (x)
=
k13 (x)
1
fP G (x) − fSN1 (x) =
,
2 6
4 (x + 1)
where
k13 (x) = 3x2 + 2x + 3 − 2 (x + 1)
p
2x2 + 2.
Let us consider
h13 (x) = 3x2 + 2x + 3
2
h
i2
p
− 2 (x + 1) 2x2 + 2
4
= (x − 1) .
(24) For DSG ≤ 54 DP5 H : In this case, we have
fP5 H
SG (x)
=
4
k14 (x)
fP5 H (x) − fSG (x) =
,
5
4 (x + 1)
where
2
√
√
k14 (x) = 2 4x3 + 5x5/2 + 11x2 + 3x x − 1 + 11x + 5 x + 4 −
2 p
√
− 5 (x + 1) x + 1
2x2 + 2.
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SEQ. OF INEQ. AMONG DIFFERENCES OF GINI MEANS AND DIVERG. MEASURES
Let us consider
i2
h 2
√
√
h14 (x) = 2 4x3 + 5x5/2 + 11x2 + 3x x − 1 + 11x + 5 x + 4
−
h
i
2
2 p
√
− 5 (x + 1) x + 1
2x2 + 2
4 7x4 + 8x7/2 + 64x3 + 120x5/2 +
√
√
.
=2 x−1
+242x2 + 120x3/2 + 64x + 8 x + 7
(25) For DP6 G ≤ 65 DP5 H : In this case we have
fP5 N3
h√
√ i √
2
4
x
−
1)
+
x ( x − 1)
(
6
.
P5 N1 (x) = fP5 H (x) − fP6 G (x) =
√
2
5
5 ( x + 1) (x + 1)
Parts 22-25 and the expression (11) together proves the sequences of inequalities appearing in (17).
(26) For DP6 P1 ≤
fP5 P2
P6 P1 (x)
(27) For DP5 P1 ≤
fP6 P2
(29) For DP6 P2 ≤
fP5 P3
16
9 fP5 P2 (x)
13
12 fP6 P2 (x)
: In this case, we have
13
9 fP5 P2 (x)
− fP5 P1 (x) =
 5

4x + 16x9/2 + 44x4 + 88x7/2 +
√
4

( x − 1)
+139x3 + 162x5/2 + 139x2 +
√
3/2
+88x + 16 x + 44x + 4
.
=
√
2
2
9 (x + 1) (x3 + 1) ( x + 1)
=
12
7 DP5 P3
P6 P2 (x)
: In this case, we have
− fP5 P1 (x) =

 5
x + 30x9/2 + 89x4 + 152x7/2 +
√
4

( x − 1)  +181x3 + 190x5/2 + 181x2 +
√
+152x3/2 + 89x + 30 x + 1
.
=
√
2
12 (x2 + 1) (x3 + 1) ( x + 1)
=
13
9 DP5 P2
P5 P1 (x)
: In this case, we have
− fP6 P1 (x) =

 5
7x + 10x9/2 + 23x4 + 64x7/2 +
√
4

( x − 1)
+142x3 + 180x5/2 + 142x2 +
√
3/2
+64x + 23x + 10 x + 7
.
=
√
2
2
9 (x + 1) (x3 + 1) ( x + 1)
=
13
12 DP6 P2
P5 P1 (x)
(28) For DP5 P1 ≤
fP5 P2
16
9 DP5 P2
: In this case, we have
12
7 fP5 P3 (x)
− fP6 P2 (x) =
4
√
5x + x7/2 + 17x3 + 22x5/2 +
4
√
( x − 1)
+38x2 + 22x3/2 + 17x + x + 5
=
.
√
2
7 (x2 + 1) (x3 + 1) ( x + 1)
=
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INDER J. TANEJA
(30) For DP5 P2 ≤ 97 DP5 P3 : In this case, we have
fP5 P3
P5 P2 (x)
= 97 fP5 P3 (x) − fP5 P2 (x) =
3
√
2x + 6x5/2 + 16x2 +
4
√
( x − 1)
+21x3/2 + 16x + 6 x + 2
=
.
√
√
2
7 (x2 + 1) (x − x + 1) ( x + 1)
14
9 DP6 N2 : In this case, we have
14
P6 N2 (x) = 9 fP6 N2 (x) − fP5 P3 (x) =
 6
11/2
(31) For DP5 P3 ≤
fP5 P3

x + 268x
+ 1143x5 + 782x9/2 +
4
7/2

√
+ 4478x3 +
4  +2597x + 2014x

( x − 1) 
5/2
2
3/2
 +2014x + 2597x + 782x +

√
+1143x + 268 x + 1
.
=
√
2
18 (x3 + 1) ( x + 1)
(32) For DP6 N2 ≤ 94 DP6 S : In this case, we have
fP6 S
P6 N2 (x)
=
9
k15 (x)
fP S (x) − fP6 N2 (x) =
,
4 6
8 (x + 1)
where
p
√
√
k15 (x) = 2 2x + 2 x + 1 (x + 1) + 10x2 + 10 − 9 (x + 1) 2x2 + 2.
In order to prove k15 (x) > 0, we shall apply twice the argument given in part
2. Let us consider
√
2
√
h15 (x) = 2 2x + 2 x + 1 (x + 1) + 10x2 + 10 −
i2
h
p
− 9 (x + 1) 2x2 + 2
√
√
= 40 2x + 2 x + 1 x2 + 1 (x + 1) −
2
3 √
46x4 + 8 (x + 1) ( x − 1) +
−
.
+260x3 + 28x2 + 260x + 46
Applying again over h15 (x) the argument given in part 2, we get
√
2
√
h15a (x) = 40 2x + 2 x + 1 x2 + 1 (x + 1) −
2
2
3 √
46x4 + 8 (x + 1) ( x − 1) +
−
+260x3 + 28x2 + 260x + 46


√
71 + 2316 x + 11960x5/2 + 4090x+
4  +11960x7/2 + 11180x3/2 + 2316x11/2 + 
√
.
=4 x−1 

 +11180x9/2 + 12021x4 + 12021x2 +
3
5
6
+6004x + 4090x + 71x
Since h15a (x) > 0 giving h15 (x) > 0, and consequently, we have k15 (x) > 0.
(33) For DP6 S ≤ DAG : In this case, we have
fAG
P6 S (x)
= fAG (x) − fP6 S (x) =
60
k16 (x)
,
2 (x + 1)
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SEQ. OF INEQ. AMONG DIFFERENCES OF GINI MEANS AND DIVERG. MEASURES
with
k16 (x) =
h
p
√ i
2
2x2 + 2 (x + 1) − (x − 1) + 2 (x + 1) x .
Now we shall show that k16 (x) > 0, ∀x > 0, x 6= 1. Let us consider
hp
i2 h
√ i2
2
h16 (x) =
2x2 + 2 (x + 1) − (x − 1) + 2 (x + 1) x
6 √
2
√
x−1
x+1 .
=
Parts 26-33 proves the sequences of inequalities appearing in (18).
Remark 2.2. In view of Theorem 2.1, we have the following improvement over the inequalities (11):
1
DSH 6 12 DAH 6 4DN2 N1 6 43 DN2 G 6 DAG 6 4DAN2


3
DSA 6 4
.
DSN2


6 32 DSN1 6 12 DSG
 53
4 DSN3
2.1
Refinement Inequalities
In this subsection, we shall present improvement over the inequalities given (12).
Results obtained in Theorem 2.1, can be written in an individual form. Let us divide them in two two groups. The first one is with three measures and the second
group is with four measures.
Group 1
(1) P2 ≤
P6 +3P1
;
4
2S+H
≤ A;
3
P6 +14N1
(3)
≤ N3 ;
15
P6 +2P4
≤ N3 ;
(4)
3
5P6 +2P4
;
(5) S ≤
7
P6 +3G
(6)
≤ N2 ;
4
P6 +5G
(7)
≤ N1 ;
6
4
(8) G ≤ P6 +6P
;
7
2N2 +G
(9)
≤ N1 ;
3
3A+G
(10) N2 ≤ 4 ;
(11) N1 ≤ P5 +2G
;
3
(2)
P5 +5N1
;
6
3
N2 ≤ P5 +9N
;
10
P5 +2N2
A≤
;
3
S+4N2
≤ A;
5
S+3N3
≤ A;
4
S+5N1
≤ N2 ;
6
S+8N1
≤ N3 ;
9
S+3G
≤
N1 ;
4
9P1 +4P5
P2 ≤
;
13
2P5 +7P2
P3 ≤
;
9
5P6 +4N2
S≤
.
9
(12) N3 ≤
(13)
(14)
(15)
(16)
(17)
(18)
(19)
(20)
(21)
(22)
Group 2
(1) 6A + P6 ≤ 6S + P2 ;
(5) S + N1 ≤ P4 + P6 ;
(2) 9A + 8N2 ≤ 9H + 8P6 ;
(6) 6H + 5P6 ≤ 6P5 + 5G;
(3) 9A + 6N3 ≤ 7H + 6P6 ;
(7) 2P4 + P6 ≤ 2A + G;
(4) 5A + 4P4 ≤ 5H + 4S;
(8) 10N1 + P5 ≤ 10N2 + H;
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INDER J. TANEJA
(9) A + 6N1 ≤ P4 + 6N2 ;
(10) G + 6A ≤ P5 + 6N2 ;
(11) G + 2S ≤ P6 + 2N1 ;
(12) 4H + 5S ≤ 4P5 + 5G;
(13) 16P2 + 9P6 ≤ 16P5 + 9P1 ;
(14) 13P2 + 12P5 ≤ 13P6 + 12P1 ;
(15) 12P3 + 7P6 ≤ 12P5 + 7P2 ;
(16) 14N2 + 9P5 ≤ 14P6 + 9P3 ;
(17) P6 + G ≤ A + S.
Based on the inequalities appearing in Groups 1 and 2, we have the theorem
giving the refinement over the inequalities appearing in (12).
Theorem 2.2. The following inequalities hold:
P6 +5G 10N2 +H−P5 2N2 +G
4
10
6
≤
≤
N
≤
G ≤ P6 +6P
≤
≤
1
P4 +6N2 −A
S+3G
7
3
4
6

 P +2G
5

4 
3


≤ P6 +2P


3
P4 + P6 − S
≤ N3 ≤
≤
P6 +14N1




15

 P4 + P6 − S S+8N
1
9
 8P6 +9H−9A 






8





 S+4N2




 3A+G

P5 +9N3
5
≤
A
≤
≤
S+3N
3
4
10
≤
≤ N2 ≤




4

 2S+H








3
 14P6 +9P3 −9P5

≤
14
5H+4S−4P4
5
P5 +6N2 −G
2
≤ P5 +2N
6
3
 4H+5S−5G
≤ P5

4

≤

S≤



≤
P6 +2N1 −G
2
≤
7H+6P6 −6N3
7




≤
12P5 +13P2 −12P1
13

A+S−G


6P5 +5G−6H
5

2A + G − 2P4
2P4 +5P6
7
4N2 +5P6
9
≤

≤
≤ P6 ≤



12P5 +7P2 −12P3
7
6S + P2 − 6A
≤
16P5 +9P1 −16P2
9
(21)
and
P2 ≤







P6 +3P1
4
≤ N1 ≤
4P5 +9P1
13
P3 ≤
P6 +3G
4
≤ N3 ≤
2P5 +7P2
9
≤
P5 +5N1
6
S+5N1
6
≤ N2
≤A≤
3I+2P4
2
≤
P5 ≤
S
T +2A
2
≤
3J+16G
16
.
≤ N1
(22)
The proof of the above theorem is based on the same techniques applied for the
Theorem 2.1. This shall be dealt [13] elsewhere.
3.
CONNECTIONS WITH INFORMATION MEASURES
Let
)
n
X
P = (p1 , p2 , ..., pn ) pi > 0,
pi = 1 , n > 2,
(
Γn =
i=1
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SEQ. OF INEQ. AMONG DIFFERENCES OF GINI MEANS AND DIVERG. MEASURES
be the set of all complete finite discrete probability distributions. For all P, Q ∈ Γn ,
the following inequalities are already proved by author [9]:
4
1
1
1
DAH 6 I 6 4DN2 N1 6 DN2 G 6 DAG 6 4 DAN2 6 J 6 T 6
Ψ,
2
3
8
16
(23)
where
" n
X
#
n
1 X
2pi
2qi
I(P ||Q) =
pi ln
+
,
qi ln
2 i=1
pi + qi
pi + qi
i=1
n
X
pi
,
J(P ||Q) =
(pi − qi ) ln
qi
i=1
n X
pi + qi
pi + qi
T (P ||Q) =
ln
√
2
2 pi qi
i=1
and
Ψ(P ||Q) =
n
X
(pi − qi )2 (pi + qi )
i=1
pi qi
.
The measures I(P ||Q), J(P ||Q) and T (P ||Q) are the respectively, the well-know
Jensen-Shannon divergence, J−divergence and arithmetic and geometric mean divergence. The measure Ψ(P ||Q) is symmetric chi-square divergence. For detailed study
on these measures refer to Taneja [8; ?; ?]. Moreover, DAH (P ||Q) = 12 ∆(P ||Q) and
DAG (P ||Q) = h(P ||Q), where ∆(P ||Q) and h(P ||Q) are the well-known triangular’s and Hellinger’s discriminations respectively.
In the theorem below we shall make connections of the classical divergence measures with the inequalities given in (16).
Theorem 3.1. The following inequalities hold:
2
4
5 DP5 H
6 4DN2 N1 6 DN2 G 6 DAG 6 4DAN2 6
2
3
3 DAP4 6 I
6
2
1
D
6 5 DP5 N3 6 43 DP5 N2 6 2DP5 A
6 DP5 G 6 1 P5 N1
6T 6
Ψ.
3
16
8J
(24)
P ROOF. In view of the inequalities given in (16), it sufficient to show the in3
equalities DAP4 6 32 I, DP5 G 6 16
J and DP5 A 6 12 T. In each part of these three
expressions we have logarithmic expressions. We shall apply a different approach
to show them.
(i) For DAP4 6 32 I: Let us consider
gAP4 _I (x) =
00
fAP
(x)
12x(x + 1)
4
=√ √
4 , x 6= 1, x > 0.
fI00 (x)
x ( x + 1)
Calculating the first order derivative of the function gAP4 _I (x) with respect to x,
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INDER J. TANEJA
one gets
0
gAP
(x)
4 _I
(
√
√
3
6 x3/2 − 3x + 3 x − 1
> 0,
6 ( x − 1)
=−
= −√ √
√ √
5
5
< 0,
x ( x + 1)
x ( x + 1)
x<1
x>1
(25)
In view of (25), we conclude that the function gAP4 _I (x) is increasing in x ∈ (0, 1)
and decreasing in x ∈ (1, ∞), and hence
βAP4 _I =
sup gAP4 _I (x) = gAP4 _I (1) =
x∈(0,∞)
3
.
2
(26)
By the application of Lemma 2.1 with (26) we get the required result.
3
16 J:
Let us consider
√
√
fP005 G (x)
3 x 4x3/2 + 4 x + x2 − 2x + 1
gP5 G_J (x) = 00
=
.
√
4
fJ (x)
4 (x + 1) ( x + 1)
(ii) For DP5 G 6
Calculating the first order derivative of the function gP5 G_J (x) with respect to x,
one gets
(
√
√
3
2
3/2
3
(
x
−
1)
x
+
8x
+
6x
+
8
x
+
1
> 0, x < 1
gP0 5 G_J (x) = −
(27)
√ √
5
2
< 0, x > 1
8 x ( x + 1) (x + 1)
In view of (27), we conclude that the function gP5 G_J (x) is increasing in x ∈ (0, 1)
and decreasing in x ∈ (1, ∞), and hence
βP5 G_J =
sup gP5 G_J (x) = gP5 G_J (1) =
x∈(0,∞)
3
.
2
(28)
By the application of Lemma 2.1 with (28) we get the required result.
(iii) For DP5 A 6 21 T: Let us consider
√
√
fP005 A (x)
2 x 4x3/2 + 4 x + x2 − 6x + 1 (x + 1)
=
.
gP5 A_T (x) = 00
√
4
fT (x)
(x2 + 1) ( x + 1)
Calculating the first order derivative of the function gP5 A_T (x) with respect to x,
one gets
√
√
2
√
3
8 x x2 + 1 ( x − 1) +
(
( x − 1)
+x4 + 14x3 + 10x2 + 14x + 1
> 0, x < 1
0
gP5 A_T (x) = −
(29)
√ √
5
2
2
< 0, x > 1
x ( x + 1) (x + 1)
In view of (29), we conclude that the function gP5 A_T (x) is increasing in x ∈ (0, 1)
and decreasing in x ∈ (1, ∞), and hence
βP5 A_T =
sup gP5 A_T (x) = gP5 A_T (1) =
x∈(0,∞)
1
.
2
(30)
By the application of Lemma 2.1 with (30) we get the required result.
Here we have referred Lemma 2.1, but its extension for the probability distributions is already proved in [9; ?].
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SEQ. OF INEQ. AMONG DIFFERENCES OF GINI MEANS AND DIVERG. MEASURES
REFERENCES
P. CZINDER and Z. PALES, Local monotonicity properties of two-variable Gini means and the comparison theorem revisited, J. Math. Anal. Appl., 301(2005), 427-438.
CHAO-PING CHEN, Asymptotic Representations for Stolarsky, Gini and the Generalized Muirhead
Means, RGMIA COLLECTION, 11(4)(2008), 1-13.
C. GINI, Di una formula compressiva delle medie, Metron, 13(1938) 3-22.
D. H. LEHMER, On the compounding of certain means, J. Math. Anal. Appl., 36 (1971), 183-200.
J. SÁNDER, A Note on Gini Mean, General Mathematics, 12(4)(2004), 17-21.
H. N. SHI, J. ZHANG and DA-MAO LI, Schur-Geometric Convexity for Difference of Means, Applied
Mathematics E-Notes, 10(2010), 275-284
S. SIMIĆ, A Simple Proof of Monotonicity for Stolarsky and Gini Means, Kragujevac J. Math, 32(2009),
75-79.
I.J. TANEJA, New Developments in Generalized Information Measures, Chapter in: Advances in
Imaging and Electron Physics, Ed. P.W. Hawkes, 91(1995), 37-136.
I.J. TANEJA, On Symmetric and Nonsymmeric Divergence Measures and Their Generalizations,
Chapter in: Advances in Imaging and Electron Physics, 138(2005), 177-250.
I.J. TANEJA, Refinement Inequalities Among Symmetric Divergence Measures, The Australian Journal of Mathematical Analysis and Applications, 2(1)(2005), Art. 8, pp. 1-23.
I.J. TANEJA, Refinement of Inequalities among Means, Journal of Combinatorics, Information and Systems Sciences, 31(2006), 357-378.
I.J. TANEJA, Inequalities among Differences of Gini Means and Divergence Measures, Available at:
http://arxiv.org/abs/1105.5802.
I.J. TANEJA, Refinement of Gini-Means Inequalities and Connections with Divergence Measures,
Available at: http://arxiv.org/abs/1111.5241.
Inder J. Taneja,
Departamento de Matemática
Universidade Federal de Santa Catarina
88.040-900 Florianópolis, SC, Brazil.
e-mail: [email protected]
http://www.mtm.ufsc.br/∼taneja
October May 2012
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