Journal of Inequalities in Pure and
Applied Mathematics
A VARIANCE ANALOG OF MAJORIZATION AND SOME
ASSOCIATED INEQUALITIES
volume 7, issue 3, article 79,
2006.
MICHAEL G. NEUBAUER AND WILLIAM WATKINS
Department of Mathematics
California State University Northridge
Northridge, CA 91330.
EMail: michael.neubauer@csun.edu
EMail: bill.watkins@csun.edu
URL: http://www.csun.edu/˜vcmth006
Received 03 March, 2006;
accepted 21 March, 2006.
Communicated by: I. Olkin
Abstract
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2000
Victoria University
ISSN (electronic): 1443-5756
064-06
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Abstract
We introduce a partial order, variance majorization, on Rn , which is analogous
to the majorization order. A new class of monotonicity inequalities, based on
variance majorization and analogous to Schur convexity, is developed.
2000 Mathematics Subject Classification: Primary: 26D05, 26D07; Secondary:
15A42.
Key words: Inequality, Symmetric polynomial, Majorization, Schur convex, Variance.
A Variance Analog of
Majorization and Some
Associated Inequalities
Michael G. Neubauer and
William Watkins
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Contents
1
2
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Majorization and Variance Majorization . . . . . . . . . . . . . . . . . .
2.1
Definitions of Variance Majorization and Majorization
2.2
Least and Greatest Sequences with Respect to the
Variance Majorization Order . . . . . . . . . . . . . . . . . . . .
3
Variance Monotone Functions and Schur Convex Functions .
3.1
The Main Result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.2
Schur Convex Functions . . . . . . . . . . . . . . . . . . . . . . . .
4
Some Variance Monotone and Schur Convex Functions . . . . .
4.1
Elementary Symmetric Functions . . . . . . . . . . . . . . . .
4.2
Moment Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.3
Entropy Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.4
Coordinates of x . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5
Restricting S(m, v) to an Interval . . . . . . . . . . . . . . . . . . . . . . .
6
Proofs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.1
Helmert basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.2
Proof of Theorem 3.1 and Lemma 4.7 . . . . . . . . . . . . .
6.3
Proof of Corollary 3.2 . . . . . . . . . . . . . . . . . . . . . . . . . .
6.4
Proof of Theorem 4.1 . . . . . . . . . . . . . . . . . . . . . . . . . .
6.5
Proof of Lemma 5.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.6
Proof of Lemma 2.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.7
Example 5.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
References
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A Variance Analog of
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Associated Inequalities
Michael G. Neubauer and
William Watkins
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1.
Introduction
Let (x1 , . . . , xn ) and (y1 , . . . , yn ) be two sequences of real numbers in nonincreasing order. The sequence x majorizes y if
i
X
k=1
xk ≥
i
X
yk ,
k=1
for i = 1, . . . , n with equality for i = n. Majorization is a partial order on
the set of nonincreasing sequences having the same sum and it plays a large
role in the theory of inequalities dating back to the work of I. Schur [7]. Indeed a function F (x1 , . . . , xn ) of n real variables is said to be Schur convex
if F (x) ≥ F (y) whenever the sequence x majorizes y. Marshall and Olkin
[6] catalog many functions and results of this type with particular emphasis
on statistical
Q inequalities. As a simple example, take the product function
F (x) = nk=1 xk . If x, y are n-tuples of nonnegative real numbers and if x
majorizes y, then F (x) ≤ F (y). That is, −F is a Schur convex function. In
particular, if y = (x̄, . . . , x̄) then x majorizes y and therefore the product of n
nonnegative numbers with fixed mean is maximized when all of them are equal
to the mean x̄. Another way to state this well-known elementary result is that
the product of a sequence x of nonnegative reals with fixed mean x̄ attains a
maximum when the variance of x is zero.
Now suppose that the variance of x is also fixed. In this paper, we define a
partial order (variance majorization) on the set of sequences x having a fixed
mean and a fixed variance. We obtain a monotonicity result similar to the one
above for sequences in which one is variance-majorized by the other. In particular the maximum value of the product of a sequence of nonnegative reals with
A Variance Analog of
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Associated Inequalities
Michael G. Neubauer and
William Watkins
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fixed mean and fixed variance is attained when the sequence takes on only two
values α < β and the multiplicity of β is 1. This simple consequence of the
main theorem is known as Cohn’s Inequality [1]: if x1 , . . . , xn are nonnegative
reals then
n
Y
xk ≤ αn−1 β,
k=1
where α and β are chosen so that the sequences x = (x1 , . . . , xn ) and (α, . . . , α, β)
have the same means and the same variances.
A Variance Analog of
Majorization and Some
Associated Inequalities
Michael G. Neubauer and
William Watkins
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2.
Majorization and Variance Majorization
Let I (Ist ) be the set of nondecreasing (strictly increasing) sequences in Rn :
I = {x ∈ Rn : x1 ≤ x2 ≤ · · · ≤ xn }
Ist = {x ∈ Rn : x1 < x2 < · · · < xn }.
The variance majorization order involves the variances of leading subsequences
of x ∈ I. So let x[i] = (x1 , . . . , xi ) be the leading subsequence of x for i =
1, . . . , n. Note that x[i] consists of the i smallest components of x. We denote
the mean of x[i] by
i
X
x[i] = (1/i)
xk ,
k=1
A Variance Analog of
Majorization and Some
Associated Inequalities
Michael G. Neubauer and
William Watkins
and the variance of x[i] by
Var(x[i]) = (1/i)
i
X
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(xk − x[i])2 .
k=1
2.1.
Definitions of Variance Majorization and Majorization
Definition 2.1 (Variance Majorization). Let x = (x1 , . . . , xn ) and y =
(y1 , . . . , yn ) be sequences of real numbers in I such that x̄ = ȳ and Var(x) =
Var(y). We say that x is variance majorized by y (or y variance majorizes x), if
Var(x[i]) ≤ Var(y[i]),
vm
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vm
for i = 2, . . . n. We write x ≺ y or y x.
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For fixed mean m and variance v ≥ 0, variance majorization is a partial
order on the set
S(m, v) = {x ∈ I : x̄ = m, Var[x] = v},
which is the√intersection of I with the sphere in Rn centered at m(1, 1, . . . , 1)
with radius nv, and the hyperplane through m(1, 1, . . . , 1) orthogonal to the
vector (1, 1, . . . , 1).
By contrast, majorization is a partial order of the set of nonincreasing sequences
D = {z ∈ R : z1 ≥ z2 ≥ · · · ≥ zn }.
Definition 2.2 (Majorization). Let x = (x1 , . . . , xn ) and y = (y1 , . . . , yn ) be
sequences in D such that x̄ = ȳ. We say that x is majorized by y (or y majorizes
x), if
x[i] ≤ y[i],
The definition of majorization is usually given in this equivalent form:
xk ≤
k=1
for i = 1, . . . , n with equality for i = n.
i
X
k=1
Michael G. Neubauer and
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maj
for i = 1, . . . , n. In this case we write x ≺ y.
i
X
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2.2.
Least and Greatest Sequences with Respect to the
Variance Majorization Order
Returning now to the variance majorization order, there is a least element and a
greatest element in S(m, v) for each m and v ≥ 0.
Lemma 2.1. Let m and v ≥ 0 be real numbers and let
xmin = (α1 , . . . , α1 , β1 )
xmax = (α2 , β2 , . . . , β2 ),
where
α1 = m −
p
v/(n − 1),
p
α2 = m − (n − 1)v,
p
β1 = m + (n − 1)v,
p
β2 = m + v/(n − 1).
Then xmin , xmax ∈ S(m, v) and
A Variance Analog of
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vm
vm
xmin ≺ x ≺ xmax ,
for all x ∈ S(m, v).
Figure 1 shows the Hasse diagram for the variance majorization partial order
for all integral sequences of length six with sum 0 and sum of squares equal to
30. In this case, xmin = (−1, −1, −1, −1, −1, 5) and xmax = (−5, 1, 1, 1, 1, 1).
By contrast, the least and greatest elements in D ∩ {x : x̄ = m} with respect
to the majorization order are (x̄, . . . , x̄) and (nx̄, 0, . . . , 0).
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8-5, 1, 1, 1, 1, 1<
8-4, -1, 0, 0, 2, 3<
8-4, -1, -1, 2, 2, 2<
8-4, -2, 1, 1, 2, 2<
A Variance Analog of
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Michael G. Neubauer and
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8-3, -2, 0, 0, 1, 4<
8-3, -1, -1, -1, 3, 3<
8-3, -3, 1, 1, 1, 3<
8-3, -3, 0, 2, 2, 2<
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8-2, -2, -1, -1, 2, 4<
8-2, -2, -2, 1, 1, 4<
8-2, -2, -2, 0, 3, 3<
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8-1, -1, -1, -1, -1, 5<
Figure 1: Variance majorization partial order for integral sequences in S(0, 30).
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3.
Variance Monotone Functions and Schur
Convex Functions
Let I be a closed interval in R and let F (x1 , . . . , xn ) be a real-valued function
defined on I ∩ I n .
Definition 3.1 (Variance Monotone). The function F is variance monotone
increasing on I ∩ I n if
vm
x ≺ y =⇒ F (x) ≤ F (y),
for all x, y ∈ I ∩ I n . If −F is variance monotone increasing, we say that F is
variance monotone decreasing.
Definition 3.2 (Schur Convex). The function F is Schur convex on D ∩ I n if
maj
x ≺ y =⇒ F (x) ≤ F (y),
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The Main Result
The next theorem is the main result.
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Theorem 3.1. Let I be a closed interval in R , and let F (z1 , . . . , zn ) be a
continuous, real-valued function on I ∩ I n that is differentiable on the interior
of I ∩ I n with gradient ∇F (z) = (F1 (z), . . . , Fn (z)). Suppose that
(3.1)
Michael G. Neubauer and
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for all x, y ∈ D ∩ I n .
3.1.
A Variance Analog of
Majorization and Some
Associated Inequalities
F3 (z) − F2 (z)
Fn (z) − Fn−1 (z)
F2 (z) − F1 (z)
≥
≥ ··· ≥
,
z2 − z1
z3 − z2
zn − zn−1
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for all z ∈ Ist ∩ I n . Then F is variance monotone increasing on I ∩ I n , that is
vm
x ≺ y =⇒ F (x) ≤ F (y).
for all x, y ∈ I ∩ I n .
For functions of the form F (z) = φ(x1 ) + · · · + φ(xn ), Theorem 3.1 specializes to the following corollary:
Corollary 3.2. Let φ(t) be a continuous, real-valued function on a closed interval I such that φ is twice-differentiable on the interior of I and φ00 is nonincreasing. Then the function
A Variance Analog of
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F (x1 , . . . , xn ) = φ(x1 ) + · · · + φ(xn )
Michael G. Neubauer and
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is variance monotone increasing on the set of nondecreasing sequences in I n .
That is,
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vm
x ≺ y =⇒ φ(x1 ) + · · · + φ(xn ) ≤ φ(y1 ) + · · · + φ(yn ),
for all x, y ∈ I ∩ I n .
It turns out that S(m, v) ⊂ I n when the interval
h
i
p
p
I = m − (n − 1)v, m + (n − 1)v)
(see Corollary 4.8). Thus the sequences xmax and xmin (described in Lemma
2.1) are in I n . So, if F is variance monotone increasing on I ∩ I n , then the
maximum and minimum values of F are attained at xmax and xmin . This means
we can bound F (x) by expressions involving only the mean and variance of x:
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Corollary 3.3. Let m and v ≥ 0 be real numbers and
h
i
p
p
I = m − (n − 1)v, m + (n − 1)v .
Let F be a variance monotone increasing function on I ∩ I n . Then
F (α1 , . . . , α1 , β1 ) ≤ F (x) ≤ F (α2 , β2 , . . . , β2 ),
for all x ∈ S(m, v), where α1 , β1 , α2 , β2 are defined as in Lemma 2.1.
3.2.
Schur Convex Functions
By comparison, the following theorem by Schur is the result analogous to Theorem 3.1 for majorization. It plays the central role in the theory of majorization
inequalities:
Theorem 3.4 ([7]). Let F (z) be a continuous, real-valued function on D that
is differentiable on in the interior of D. Then
maj
x ≺ y =⇒ F (x) ≤ F (y),
for all x, y ∈ D if and only if
(3.2)
F1 (z) ≥ F2 (z) ≥ · · · ≥ Fn (z),
for all z in the interior of D.
The result analogous to Corollary 3.2 for majorization is known as Karamata’s Theorem:
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Corollary 3.5 ([4]). Let φ be a continuous, real-valued function on a closed
interval I such that φ is twice differentiable on the interior of I and φ00 is nonnegative. Then
maj
x ≺ y =⇒ φ(x1 ) + · · · + φ(xn ) ≤ φ(y1 ) + · · · + φ(yn ),
for all x, y ∈ D ∩ I n .
A Variance Analog of
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Michael G. Neubauer and
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4.
Some Variance Monotone and Schur Convex
Functions
In this section, we give a short list of some common functions that are monotone
in both the regular majorization and variance majorization orders. And we give
as corollaries some samples of the kinds of inequalities one can obtain from
Corollary 3.3.
4.1.
Elementary Symmetric Functions
Let Ek (z) denote the kth elementary symmetric function of the sequence z =
(z1 , . . . , zn ) ∈ Rn :
X
Ek (z) =
zi1 zi2 · · · zik ,
A Variance Analog of
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where the sum is taken over all sets of k indices with 1 ≤ i1 < · · · < ik ≤ n.
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Theorem 4.1. Let Ek be the kth elementary symmetric function. Then
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vm
x ≺ y =⇒ Ek (y) ≤ Ek (x),
vm
Ek+1 (x)
Ek+1 (y)
x ≺ y =⇒
≤
,
Ek (x)
Ek (y)
for all x, y ∈ I ∩ [0, ∞)n .
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The next corollary is obtained from Corollary 3.3 by evaluating the elementary symmetric function Ek at xmin and xmax .
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Corollary 4.2. Let x ∈ I ∩ [0, ∞)n . Then A(v, m, k) ≤ Ek (x) ≤ B(v, m, k),
where
r
r
k−1 v
n
v
A(v, m, k) = Ek (xmax ) =
m+
m − (k − 1)
k
n−1
n−1
r
r
k−1 v
n
v
B(v, m, k) = Ek (xmin ) =
m−
m + (k − 1)
n−1
k
n−1
The inequality analogous to Theorem 4.1 for (regular) majorization is given
next:
A Variance Analog of
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Associated Inequalities
Theorem 4.3 ([6, p. 80]).
Michael G. Neubauer and
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maj
x ≺ y =⇒ Ek (y) ≤ Ek (x)
maj
Ek+1 (y)
Ek+1 (x)
x ≺ y =⇒
≤
,
Ek (y)
Ek (x)
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for all x, y ∈ D ∩ [0, ∞)n .
4.2.
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Moment Functions
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p
Let p be a positive real number and let φ(t) = t . The pth moment function of
z ∈ [0, ∞)n is given by
Mp (z) =
II
I
z1p
+ · · · + znp .
The following results are applications of Corollaries 3.5 and 3.2:
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Theorem 4.4. Let Mp be the pth moment function. Then
maj
x ≺ y =⇒ Mp (x) ≤ Mp (y), for p ∈ (−∞, 0] ∪ [1, ∞)
maj
x ≺ y =⇒ Mp (y) ≤ Mp (x), for p ∈ [0, 1],
for all x, y ∈ D ∩ [0, ∞)n and
vm
x ≺ y =⇒ Mp (x) ≤ Mp (y), for p ∈ (−∞, 0] ∪ [1, 2]
vm
x ≺ y =⇒ Mp (y) ≤ Mp (x), for p ∈ [0, 1] ∪ [2, ∞),
for all x, y ∈ I ∩ [0, ∞)n .
A Variance Analog of
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Michael G. Neubauer and
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Again we obtain bounds on Mp (x), which depend only on the mean and
variance of x, from Corollary 3.3:
n
Corollary 4.5. Let x ∈ I ∩ [0, ∞) with mean m are variance v. Let
p
p
p v p A(m, v, p)=Mp (xmin ) =(n − 1) m − n−1
+ m + (n − 1)v
p
p
p v p
.
B(m, v, p)=Mp (xmax )= m − (n − 1)v + (n − 1) m + n−1
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A(m, v, p) ≤ Mp (x) ≤ B(m, v, p), for p ∈ (−∞, 0) ∪ [1, 2]
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and
B(m, v, p) ≤ Mp (x) ≤ A(m, v, p), for p ∈ [0, 1] ∪ [2, ∞).
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4.3.
Entropy Function
The entropy function is defined for x ∈ [0, ∞)n by
H(x) = −(x1 log x1 + · · · + xn log xn ).
Letting φ(t) = −t log t, we have φ00 (t) = −1/t, which is nonpositive and
increasing on [0, ∞). Thus −φ satisfies the conditions of Corollaries 3.2 and
3.5. Thus we have the following inequalities:
Theorem 4.6. Let H be the entropy function. Then
A Variance Analog of
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vm
x ≺ y =⇒ H(y) ≤ H(x),
Michael G. Neubauer and
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for all x, y ∈ I ∩ [0, ∞)n , and
maj
x ≺ y =⇒ H(y) ≤ H(x),
for all x, y ∈ D ∩ [0, ∞)n .
4.4.
Coordinates of x
The smallest and the largest coordinates of a sequence are variance monotone
decreasing.
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Lemma 4.7. Let x, y ∈ I. Then
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x ≺ y =⇒ x1 ≥ y1 and xn ≥ yn .
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We call this result a lemma because it is a part of the proof of Theorem 3.1
rather than a consequence of it.
When combined with Corollary 3.3, Lemma 4.7 gives bounds for the smallest and largest coordinates of a sequence in S(m, v) in terms of m and v:
Corollary 4.8. Let x = (x1 , . . . , xn ) be a sequence in S(m, v). Then
p
p
m − (n − 1)v ≤ x1 ≤ m − v/(n − 1)
p
p
m + v/(n − 1) ≤ xn ≤ m + (n − 1)v.
Applying Corollary 4.8 to the eigenvalues of a symmetric matrix, we recover
an equivalent form of an inequality of Wolkowicz and Styan [8, Theorem 2.1]
that bounds the maximum and minimum eigenvalues by expressions involving
only the trace and Euclidean norm of the matrix:
Corollary 4.9. Let G be a symmetric matrix with eigenvalues λ1 ≤ · · · ≤ λn .
Then
tr(G) + √1 pn||G||2 − (tr(G))2 ≤ λ ≤ tr(G) + √n−1 pn||G||2 − (tr(G))2
n
n
n
n
n n−1
tr(G) −
n
√
n−1
n
n||G||2 − (tr(G))2 ≤ λ1 ≤ tr(G)
−
n
p
√1
n n−1
p
n||G||2 − (tr(G))2 .
Corollary 4.9 follows from the fact that the mean m and variance v of the
eigenvalues can be expressed in terms of the trace and Euclidean norm ||G|| of
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G as follows:
tr(G)
n
1X
v=
(λi − m)2
n i
m=
=
1
n
!
X
λ2i − 2mλi + m2
i
1
= (tr(G2 ) − nm2 )
n
1
= 2 n||G||2 − tr(G)2 .
n
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5.
Restricting S(m, v) to an Interval
Lemma 2.1 guarantees that there is a least and a greatest element in S(m, v)
with respect to the variance majorization order. Now we restrict the set S(m, v)
to an interval I = [m − δ, m + β], with δ, β ≥ 0, containing m. From Corollary
4.8, we have
h
in
p
p
m − v/(n − 1), m + v/(n − 1)
h
in
p
p
⊂ S(m, v) ⊂ m − (n − 1)v, m + (n − 1)v .
p
So if either δ, β < v/(n − 1), then S(m, v) ∩ I n = ∅. However, if S(m, v) ∩
I n is not empty, then it contains a least element, but it may not contain a greatest
element.
Lemma 5.1. Let m and v ≥ 0 be real numbers. Let I be the interval I =
[m − δ, m + β] such that S(m, v) ∩ I n is not empty. Then there exist unique real
numbers m − δ ≤ α ≤ γ < m + β and an integer 1 ≤ j ≤ n − 1 such that the
sequence
n−j−1
xmin
j
}|
{
z }| { z
= (α, . . . , α, γ, m + β, . . . , m + β) ∈ S(m, v) ∩ I n .
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vm
Moreover xmin ≺ x for all x ∈ S(m, v) ∩ I n .
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5
Example 5.1. Let n = 5. The least element in S(0, 1944/5) ∩ [−36, 24]
is (−18, −18, −12, 24, 24). There is no greatest element in S(0, 1944/5) ∩
[−36, 24]5 . See Section 6.7.
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The situation for restricting the sequences to a closed interval is a little
different for majorization. There is always a least element and a greatest in
D ∩ [m, M ]n . The sequence (x̄, . . . , x̄) ∈ D ∩ [m, M ]n is the least element. The
greatest element with respect to majorization takes at most three values for its
coordinates, two of which are the end points of the closed interval. That is, the
greatest element in for majorization in D ∩ [m, M ]n is of the form
(M, . . . , M, θ, m, . . . , m).
A discussion of restricting the majorization order to an interval is given in
[5] and [6, page 132].
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6.
Proofs
The main technique used in the proofs is to express the sequences in I as linear
combinations of a special basis for Rn , the so-called Helmert basis.
6.1.
Helmert basis
The purpose of this section is to describe the relationship between the coordinates of an n-tuple x ∈ Rn and the coordinates of x with respect to the so-called
Helmert basis for Rn (see [6, p. 47] for a discussion of the Helmert basis). The
Helmert basis for Rn is defined as follows:
1
w0T = √ (1, 1, . . . , 1)
n
1
w1T = √ (−1, 1, 0, . . . , 0)
2
1
w2T = √ (−1, −1, 2, 0, . . . , 0)
6
..
.
wiT
1
=p
i(i + 1)
i
z }| {
(−1, . . . , −1, i, 0, . . . , 0)
..
.
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1
T
wn−1
=p
(n − 1)n
(−1, . . . , −1, n − 1).
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It is clear that {w0 , w1 , . . . , wn−1 } is an orthonormal
basis for Rn . Thus every
P
n−1
vector x ∈ Rn is a linear combination x = k=0 ak wk . The Helmert coefficient
√
a0 is determined by the mean x̄ of the sequence x. Specifically, a0 = nx̄.
The other Helmert coefficients, a1 , . . . , an−1 are related to the variance, partial
variances, and order of the coordinates of x.
P
Lemma 6.1. Let x be a sequence of real numbers with x = n−1
k=0 ak wk . Then
1
Var(x[i]) = (a21 + · · · + a2i−1 ),
i
for i = 2, . . . , n. In particular,
Var(x) =
1 2
(a + · · · + a2n−1 ).
n 1
Proof. Since ak = x · wk ,
i−1
X
k=1
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a2k = x
i−1
X
!
wkT wk
xT
k=1
= x ((Ii − (1/i)Ji ) ⊕ 0n−i ) xT
!
i
i
X
X
2
2
=
xk − (1/i)(
xk )
k=1
k=1
= iVar(x[i]).
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The i × i identity matrix is denoted P
by Ii and Ji denotes the i × i matrix all of
T
whose entries are one. The fact that i−1
k=1 wk wk = Ii − (1/i)Ji follows from a
simple inductive argument.
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P
Let x =
ai wi . In the next definition and lemma, we give necessary and
sufficient conditions on the sequence (a1 , . . . , an−1 ) for the sequence x to be
nondecreasing. (Clearly, the coefficient a0 does not influence the relative ordering of the coordinates of x.)
Definition 6.1 (Admissible Sequence). Let α = (α1 , . . . , αn−1 ) be a sequence
of nonnegative real numbers. Then α is admissible if
(i − 1)iαi−1 ≤ i(i + 1)αi ,
(6.1)
for 2 ≤ i ≤ n − 1.
n
Lemma 6.2. Let
Px = (x1 , . . . , xn ) be a vector in R and a0 , . . . , an−1 be scalars
such that x = n−1
i=0 ai wi . The following conditions are equivalent:
1. x is nondecreasing.
2. ai ≥ 0, for i = 1, . . . , n − 1, and the sequence a(2) = (a21 , . . . , a2n−1 ) is
admissible.
3. the kth component of ai−1 wi − ai wi−1 is nonnegative for all k 6= i and
nonpositive for k = i.
√
Proof. Let 1 ≤ i ≤ n − 1. Then x2 − x1 = 2a1 , and
!
(i − 1)ai−1
ai
iai
xi+1 − xi = p
− p
−p
i(i + 1)
(i − 1)i
i(i + 1)
(6.2)
(i + 1)ai
(i − 1)ai−1
=p
− p
i(i + 1)
(i − 1)i
p
p
1
i(i + 1)ai − (i − 1)iai−1 ,
=
i
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for i ≥ 2. Thus x is nondecreasing if and only if ai ≥ 0, for i = 1, . . . , n − 1
and a(2) is admissible. So Conditions 1 and 2 are equivalent.
Now let (ai−1 wi − ai wi−1 )k be the k component of ai−1 wi − ai wi−1 . Then

, if k < i,
− √ai−1 + √ ai


i(i+1)
(i−1)i




 − √ai−1 − (i−1)a
√ i−1 , if k = i,
i(i+1)
(i−1)i
(6.3)
(ai−1 wi − ai wi−1 )k =
ia

i−1

√
, if k = i + 1,


i(i+1)



0
, if k > i + 1.
To prove that Condition 2 implies Condition 3, suppose that ai ≥ 0 for
i = 1, . . . , n − 1 and that a(2) is admissible. It is clear from Equation (6.3) that
(ai−1 wi − ai wi−1 )k ≥ 0 for all k 6= i and that (ai−1 wi − ai wi−1 )i ≤ 0.
Conversely, suppose that Condition 3 holds. With k = 3, i = 2 in Equation
(6.3), we get a1 ≥ 0. With k = 1 < i we get that a(2) is admissible and ai ≥ 0
for all i. Thus Condition 2 holds.
6.2.
Proof of Theorem 3.1 and Lemma 4.7
Let F be a differentiable, real-valued function on I ∩ I n satisfying Inequality
(3.1). Let x, y be nondecreasing sequences in I n such that x̄ = ȳ, Var(x) =
vm
Var(y) and x ≺ y. Let
x=
n−1
X
k=0
ak wk ,
y=
n−1
X
k=0
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bk wk ,
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for scalars ak , bk , k = 0, . . . , n − 1, and let
a(2) = (a21 , . . . , a2n−1 ),
b(2) = (b21 , . . . , b2n−1 ).
Since x̄ = ȳ, we have a0 = b0 . By Lemma 6.2, ak , bk ≥ 0 for k = 1, . . . , n − 1,
a(2) and b(2) are admissible, and by Lemma 6.1
i
X
(6.4)
a2k
≤
i
X
b2k ,
k=1
k=1
for 1 ≤ i ≤ n − 1 with equality in Inequality (6.4) for i = n − 1 since Var(x) =
Var(y).
Next define a path c(t) = (c0 , c(t)1 , . . . , c(t)n−1 ) from a to b by c0 = a0 =
b0 , and
q
c(t)k =
(1 − t)a2k + tb2k ,
for t ∈ [0, 1] and k = 1, . . . , n − 1. Then c(t)(2) = (1 − t)a(2) + tb(2) , from
which it follows that c(2) is admissible and that
i
X
k=1
a2k ≤
i
X
c(t)2k ≤
k=1
i
X
b2k ,
k=1
for i = 1, . . . , n − 1.
Now define a path z(t) from x to y by
z(t) = a0 w0 +
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n−1
X
k=1
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c(t)k wk .
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Since c(t)k ≥ 0 and c(t)(2) is admissible, z(t) is a nondecreasing sequence.
Let j be the smallest index for which aj 6= bj . Then c(t)k = ak for k < j
and c(t)j > 0 for t > 0. It is easy to verify that c0 (t)k = (b2k − a2k )/c(t)k (unless
c(t)k = 0). Thus the tangent vector z 0 (t) is given by
(6.5)
0
z (t) =
n−1 2
X
b − a2
k
k
wk
c(t)k
wj
wj+1
2
2
= (bj − aj )
−
cj
cj+1
k=j
+
(b2j+1
+
b2j+2
−
a2j+1
−
a2j+2 )
wj+2 wj+3
−
cj+2
cj+3
+ · · · + b21 + b22 + · · · + b2n−1
wn−2 wn−1
2
2
2
− a1 − a2 − · · · − an−1
−
,
cn−2
cn−1
0
for t > 0. It follows from Inequality (6.4) that z (t) is a nonnegative linear
i−1
combination of the vectors wci−1
− wcii , for i = j + 1, . . . , n − 1.
We now show that z(t) ∈ I n for all t ∈ [0, 1]. Indeed, both the first and the
last coordinates of z(t) are nonincreasing functions of t. Thus
y1 = z(1)1 ≤ z(t)1 ≤ z(t)2 ≤ · · · ≤ z(t)n ≤ z(0)n = xn .
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Since y1 , xn ∈ I, z(t) ∈ I and thus F (z(t)) is defined for all t ∈ [0, 1]. To see
that z(t)1 and z(t)n are decreasing, we examine the first and last coordinates of
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the vectors
wi−1
ci−1
−
wi
.
ci
The first coordinate is
−1
p
(i − 1)ici−1
1
+p
i(i + 1)ci
,
which is nonpositive since c(2) is an admissible sequence. Thus z 0 (t)1 ≤ 0 and
z(t)1 is nonincreasing in t. This proves the first part of Lemma 4.7.
i−1
The last coordinate of wci−1
− wcii is zero unless i = n − 1 and in that case it is
−(n − 1)
p
(n − 1)ncn−1
,
which is also nonpositive. Thus z(t)n is nonincreasing. This proves the other
part of Lemma 4.7.
Finally to prove that F (z(t)) is an increasing function in t, we show that
dF
= ∇F · z 0 (t) ≥ 0.
dt
In view of Equation (6.5), it suffices to show that
wi−1 wi
∇F ·
−
≥ 0,
ci−1
ci
for i = j + 1, . . . , n − 1.
Since wk is orthogonal to the all-ones vector e,
wi−1 wi
wi−1 wi
∇F ·
−
= (∇F − Fi e) ·
−
.
ci−1
ci
ci−1
ci
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For each i = 1, . . . , n − 1 define a function Ki on Ist ∩ I n by
Ki (z) =
Fi+1 (z) − Fi (z)
.
zi+1 − zi
F (z)−F (z)
Now let i < j. Then j zj −zii is a convex combination of Kk (z) for k =
i, . . . , j − 1:
j−1
Fj (z) − Fi (z) X zk+1 − zk
=
Kk (z).
zj − zi
zj − zi
k=i
Thus by Condition (3.1),
Fj (z) − Fi (z)
≤ Ki (z)
zj − zi
and so
Fj (z) − Fi (z) ≤ (zj − zi )Ki (z)
for all i < j and z ∈ Ist ∩ I n .
Since c(t)(2) is an admissible sequence, Lemma 6.2 guarantees that all comi−1
ponents of wci−1
− wcii are nonpositive except the ith component. Of course the
ith component of ∇F − Fi e is zero. It follows that
wi−1 wi
wi−1 wi
−
≥ Ki (z)(z − zk e) ·
−
(∇F − Fi e) ·
ci−1
ci
ci−1
ci
wi−1 wi
= Kk (z)z ·
−
ci−1
ci
= 0.
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P
wi−1
The last equality holds because z = n−1
k=j ck wk is clearly orthogonal to ci−1 −
wi
.
ci
We have shown that dF
≥ 0, for z ∈ Ist ∩ I n and t ∈ (0, 1). Thus F (z(t))
dt
is an increasing function of t. So F (x) = F (z(0)) ≤ F (z(1)) = F (y).
By the continuity of F , F is variance monotone increasing on I ∩ I n .
6.3.
Proof of Corollary 3.2
Let φ(t) be a continuous, real-valued function on a closed interval I such that
φ(t) is twice-differentiable on the interior of I and φ00 (t) is nonincreasing on I.
Let z be an increasing sequence in I n . Let i be an integer satisfyng 1 ≤ i ≤
n − 1. By the mean-value theorem, there exists ξi in the interval zi < ξi < zi+1
such that
φ0 (zi+1 ) − φ0 (zi )
= φ00 (ξi ).
zi+1 − zi
Inequality (3.1) follows since ξ1 ≤ ξ2 ≤ · · · ≤ ξn−1 , φ00 is nonincreasing, and
Fi (z) = φ0 (zi ).
6.4.
Proof of Theorem 4.1
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Since E2 (z) and E1 (z) are constant on S(m, v), we assume that k ≥ 3. Let i
(i,i+1)
be an integer satisfying 1 ≤ i ≤ n − 1 and let Ek
be the kth elementary
symmetric polynomial of the n − 2 variables z1 , z2 , . . . , zi−1 , zi+2 , . . . , zn . Then
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Ek (z) =
(i,i+1)
Ek
+ (zi +
(i,i+1)
zi+1 )Ek−1
+
(i,i+1)
zi zi+1 Ek−2 .
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Thus
∂Ek (z) ∂Ek (z)
(i,i+1)
−
= −(zi+1 − zi )Ek−2 ,
∂zi+1
∂zi
1
∂Ek (z) ∂Ek (z)
(i,i+1)
−
= −Ek−2 ,
zi+1 − zi
∂zi+1
∂zi
and so
(i−1,i)
for all z ∈ I ∩ [0, ∞]n . But the sequence z is nondecreasing so Ek−2 ≥
(i,i+1)
Ek−2 for i = 1, . . . , n − 1. It follows from Theorem 3.1 that −Ek (z) is
variance monotone increasing. Thus Ek (z) is variance monotone decreasing.
This proves the first inequality in Theorem 4.1.
To prove that the function F (z) = Ek+1 (z)/Ek (z) is variance monotone
increasing, we must show that Inequality (3.1) holds. It suffices to show that
F3 (z) − F2 (z)
F2 (z) − F1 (z)
≥
.
z2 − z1
z3 − z2
We write Ek for the kth elementary symmetric function of z1 , . . . , zn and Ek0 for
the kth elementary symmetric function of z4 , . . . , zn . Then
(6.6) Ek =
0
0
0
Ek0 + (z1 + z2 + z3 )Ek−1
+ (z1 z2 + z1 z3 + z2 z3 )Ek−2
+ z1 z2 z3 Ek−3
.
It follows that
∂
Ek+1
F1 =
∂z1
Ek
1
0
0
0
+ (z2 + z3 )(Ek Ek−1
− Ek+1 Ek−2
)
= 2 [Ek Ek0 − Ek+1 Ek−1
Ek
0
0
+ z2 z3 (Ek Ek−2
− Ek+1 Ek−3
)].
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Thus
F2 − F1
1 0
0
0
0
= − 2 Ek Ek−1
− Ek+1 Ek−2
+ z3 (Ek0 Ek−2
− Ek+1 Ek−3
) .
z2 − z1
Ek
Similarly,
F3 − F2
1 0
0
0
0
= − 2 Ek Ek−1
− Ek+1 Ek−2
+ z1 (Ek0 Ek−2
− Ek+1 Ek−3
) ,
z3 − z2
Ek
so that
z3 − z1
F2 − F1 F3 − F2
0
0
−
=
(Ek Ek−2
− Ek+1 Ek−3
).
z2 − z1
z3 − z2
Ek2
0
0
it remains to show that Ek Ek−2
−Ek+1 Ek−3
Since z3 −z1 and Ek2
are positive,
is nonnegative, which can be rewritten using Equation (6.6) as
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0
Ek Ek−2
=
0
− Ek+1 Ek−3
0
0
0
(Ek0 Ek−2
− Ek+1
Ek−3
)
+
0
0
0
+ (z1 + z2 + z3 )(Ek−1
Ek−2
− Ek0 Ek−3
)
0
0
0
0
(z1 z2 + z1 z3 + z2 z2 )(Ek−2 Ek−2 − Ek−1 Ek−3 ).
Each of the expressions in Er0 above are nonnegative. These weak inequalities follow from a simple counting argument. Or we can use an old result in
Hardy, Littlewood and Pólya [3, p. 52]:
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z ∈ [0, ∞) and s > r =⇒ Es−1 Er > Es Er−1 ,
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with r = k − 2 and s = k + 1, k, k − 1.
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6.5.
Proof of Lemma 5.1
We may assume that m = 0 so that I = [−δ, β].
Let z ∈ RnP
with mean z̄ = 0. There exist real numbers a = (a1 , . . . , an−1 )
such that z = n−1
i=1 ai wi Then z ∈ S(0, v) if and only if a satisfies the following conditions:
ai ≥ 0, for all i
(6.7)
a(2) is admissible
Pn−1 2
i=1 ai = nv.
Let z ∈ S(0, v). The only vector among w1 , . . . , wn−1 having a nonzero nth
coordinate is wn−1 . Thus
r
n−1
zn =
an−1 .
n
To compute
z1 in terms of ak , notice that the first coordinate of each wk is
p
−1/ k(k + 1). So
n−1
X
a
p k
z1 = −
.
k(k + 1)
k=1
Thus z ∈ I n if and only if
(6.8)
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(n − 1)a2n−1 ≤ nβ 2 ,
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and
(6.9)
−δ ≤ −
n−1
X
ak
p
k(k + 1)
k=1
.
for the sequences a for which z =
P Next, we establish nanother inequality
ak wk ∈ S(0, v) ∩ I . Since a(2) is admissible and Inequality (6.8) holds, we
have
2a21 ≤ 6a22 ≤ · · · ≤ k(k + 1)a2k ≤ · · · ≤ (n − 1)na2n−1 ≤ n2 β 2 .
A Variance Analog of
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Thus
(6.10)
a2k
2 2
≤n β
Michael G. Neubauer and
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1
1
−
,
k k+1
for k = 1, . . . , n − 1.
We now define b = (b1 , . . . , bn−1 ) so that it satisfies Conditions (6.7) and so
that for all other a satisfying Conditions (6.7), we have
(6.11)
i
X
b2k ≤
k=1
i
X
a2k ,
(6.12)
n−1
X
k=i+1
a2k ≤
n−1
X
k=i+1
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k=1
for i = 1, . . . , n − 1. In view of the fact that
inequalities above are equivalent to
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Pn−1
k=1
a2k = nv =
Pn−1
k=1
b2k , the
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b2k ,
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for i = 0, . . . , n − 2.
We begin by specifying the integer j. The function
1
2 1
f (j) := nβ
−
,
j n
is decreasing in j with
1
f (1) = nβ 1 −
n
f (n) = 0.
2
We also have from Inequality (6.10) that
n−1
X
1
1
2
2
v=
a ≤ nβ 1 −
= f (1).
n k=1 k
n
Thus there exists 1 ≤ j ≤ n such that
1
1
1
2 1
2
−
≤ v < nβ
−
.
(6.13)
nβ
j+1 n
j n
Define the sequence of nonnegative reals b = (b1 , . . . , bn−1 ) as follows:
1
2
2 2 1
(6.14)
−
, for i = j + 1, . . . , n − 1
bi = n β
i
i+1
1
1
2
2 2
(6.15)
−
,
bj = nv − n β
j+1 n
bi = 0, for i = 1, . . . , j − 1.
A Variance Analog of
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Associated Inequalities
Michael G. Neubauer and
William Watkins
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It is clear that
n−1
X
b2k = nv.
k=1
To check that b(2) is admissible, notice that
i(i + 1)b2i = n2 β 2 = (i + 1)(i + 2)b2i+1 ,
for i = j + 1, . . . , n − 2. Also
1
1
2 2
2
j(j + 1)bj = j(j + 1) nv − n β
−
j+1 n
1 1
1
1
2 2
2 2
≤ j(j + 1) n β
−
−n β
−
j n
j+1 n
= n2 β 2
= (j + 1)(j + 2)b2j+1 .
The inequality follows from the choice of j in Inequality (6.13). Of course
0 = i(i + 1)b2i ≤ (i + 1)(i + 2)b2i+1 for i = 1, . . . , j − 1. Thus b(2) is admissible.
It follows that b satisfies Conditions (6.7).
Next we show that Inequalities (6.12) hold. So suppose that a = (a1 , . . . , an−1 )
satisfies Conditions (6.7) and (6.8). Then from Inequality (6.10) we get
n−1
X
k=i+1
a2k
2 2
≤n β
1
1
−
.
i+1 n
A Variance Analog of
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Associated Inequalities
Michael G. Neubauer and
William Watkins
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For i ≥ j we have
n−1
X
b2k
k=i+1
n−1 X
1
1
=n β
−
k
k+1
k=i+1
1
1
= n2 β 2
−
.
i+1 n
2 2
So Inequality (6.12) holds for i ≥ j. For i < j,
n−1
X
k=i+1
a2k
≤
n−1
X
a2k
= nv =
k=1
n−1
X
b2k .
k=i+1
So Inequality (6.12) holds for i < j too.
Now let
n−1
X
xmin =
bk wk .
k=1
j
A Variance Analog of
Majorization and Some
Associated Inequalities
n−j−1
z }| { z }| {
We now show that x = xmin = (α, . . . , α, γ, β, . . . , β) for some −δ ≤ α ≤ γ <
β. From the proof of Lemma 6.2 we have xi+1 = xi if and only if i(i − 1)b2i−1 =
i(i + 1)b2i . For i = 1, . . . , j − 1 we have bi = 0. So x1 = · · · = xj = α. And
i(i − 1)b2i−1 = i(i + 1)b2i for i = j + 2, . . . , n − 1. So xj+2 = · · · = xn = β.
The last equality follows from the choice of bn−1 .
We now show
that −δ ≤ α. Since S(0, v)∩I n is nonempty, let z ∈ S(0, v)∩
P
n−1
I n and let z = i=1 ai wi for some nonnegative reals a = (a1 , . . . , an−1 ). Then
X
a
p i
−δ ≤ z1 = −
.
i(i + 1)
Michael G. Neubauer and
William Watkins
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We will show that
(6.16)
n−1
X
a − bi
pi
0≤
,
i(i + 1)
i=1
from which it follows that −δ ≤ α and hence that xmin ∈ I n .
We begin with the following identity:
1
ai − b i =
ai + b i
i
X
k=1
(a2k − b2k ) −
i−1
X
!
(a2k − b2k ) .
k=1
Michael G. Neubauer and
William Watkins
Then
(6.17)
A Variance Analog of
Majorization and Some
Associated Inequalities
X
i
n−1
n−2 X
X
ai − b i
1
1
p
=
−
(a2k − b2k ),
ci ci+1 k=1
i(i + 1)
i=1
i=1
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where
ci =
p
i(i + 1)(ai + bi ),
for i = 1, . . . , n − 2. Both sequences a(2) and b(2) are admissible. Thus
p
p
ci = i(i + 1)(ai + bi ) ≤ (i + 1)(i + 2)(ai+1 + bi+1 ) = ci+1 ,
for i = 1, . . . , n − 2. It follows that 1/ci − 1/ci+1 ≥ 0. Inequalities (6.11) hold
so that the expression on the right-hand side of Equation (6.17) is nonnegative.
Therefore Inequality (6.16) holds. This proves that −δ ≤ α. It follows that
xmin ∈ I n .
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Finally, we show that xmin is unique. Let
n−l−1
l
z }| {
z }| {
y = (α1 , . . . , α1 , γ1 , β, . . . , β) ∈ S(0, v) ∩ I n ,
where −δ ≤ α1 ≤ γ1 < β. We begin the proof that x = xmin = y by showing
that l = j.
From the definition of b we have
b1 = · · · = bj−1 = 0.
Since y1 = · · · = yl = α1 we have
a1 = · · · = al−1 = 0.
But
A Variance Analog of
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Associated Inequalities
Michael G. Neubauer and
William Watkins
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(6.18)
l−1
X
b2k ≤
k=1
l−1
X
a2k = 0,
k=1
and bj+1 > 0, so l − 1 ≤ j.
We now show that j ≤ l.
Since yl+1 = · · · = yn , we have
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(l + 1)(l + 2)a2l+1 = (l + 1)(l + 2)a2l+2 = · · · = (n − 1)na2n−1 = n2 β 2 .
In addition,
(j + 1)(j + 2)b2j+1 = (j + 2)(j + 3)b2j+2 = · · · = (n − 1)nbn−1 = n2 β 2 .
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Then
a2k
2 2
1
1
−
,
k k+1
1
1
−
,
k k+1
=n β
for k = l + 1, . . . , n − 1, and
b2k
2 2
=n β
for k = j + 1, . . . , n − 1.
Suppose l < j. Then ak = bk , k = 1, . . . , j + 1 and aj > bj . This contradicts
the fact that
n−1
n−1
X
X
a2k ≤
b2k .
k=j
Michael G. Neubauer and
William Watkins
k=j
Therefore j ≤ l and so j = l or j = l − 1.
Suppose j = l − 1.Then from Inequality (6.18) we have bj = 0 and γ = α.
Since
j
j
X
X
b2j = 0,
a2k ≤
k=1
A Variance Analog of
Majorization and Some
Associated Inequalities
k=1
we have aj = 0 and so γ1 = α1 , which contradicts the assumption that α1 < γ1 .
It follows that j = l.
To finish the proof of uniqueness, we show that a = b. This follows immediately since
ak = bk = 0,
P
Pn−1 2
2
for k = 1, . . . , j − 1 and for k = j + 1, . . . , n − 1, and n−1
k=1 ak =
k=1 ak .
Thus aj = bj and so a = b. It follows that xmin = y.
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6.6.
Proof of Lemma 2.1
Let xmin = (α1 , . . . , α1 , β1 ) be defined as in the statement of Lemma 2.1.
vm
Clearly, Var(xmin [i]) = 0, for i = 2, . . . , n − 1. Thus xmin ≺ x, for all
x ∈ S(m, v).
The argument for xmax = (α2 , β2 , . . . , β2 ) is more involved. Let b0 and
b = (b1 , . . . , bn−1 ) be the coordinates of xmax in the Helmert basis:
xmax = (α2 , β2 , . . . , β2 ) = b0 w0 + b1 w1 + · · · + bn−1 wn−1 .
Since the second through nth coordiates of xmax are the same, the admissible
sequence b(2) satisfies:
2b21 = 6b22 = 12b23 = · · · = (n − 1)nb2n−1 .
A Variance Analog of
Majorization and Some
Associated Inequalities
Michael G. Neubauer and
William Watkins
Now let x ∈ S(m, v) with
x = a0 w0 + a1 w1 + · · · + an−1 wn−1 .
We must show that
(6.19)
i
X
k=1
a2i ≤
i
X
b2i ,
k=1
for i = 1, . . . , n − 1. We need the next lemma to finish the proof.
Lemma 6.3. Let rP
= (r1 , . . . , rm ) be an admissible sequence of nonnegative
real numbers with
rk = t. Let s = (s1 , . . . , sm ) be the unique sequence of
nonnegative reals with
2s1 = 6s2 = 12s3 = · · · = m(m + 1)sm
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and
P
sk = t. Then
i
X
k=1
rk ≤
i
X
sk ,
k=1
for i = 1, . . . , m.
Equation (6.19) follows from Lemma 6.3 with m = n − 1, r = a(2) , s = b(2) ,
and t = nv.
Proof. (Lemma 6.3) The condition on
P the sum and for admissibility can be
expressed in matrix form as follows:
rk = t and r is admissible if and only
if the first m − 1 coordinates of T r are nonnegative and the last coordinate is t,
where


−t1
t2 0
···
0
 0 −t2 t3
0 ···
0 

..
.. . .
.. 
..


.
.
.
.
. 



0
−ti ti+1
0 
T = 0

.. 
..
..
 0
.
.
0
. 


 0
0
···
−tm−1 tm 
1
1
···
1 1
and ti = i(i + 1). We can express this condition symbolically as
+m−1
Tr =
,
t
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Michael G. Neubauer and
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where +m−1 is a vector in R
m−1
with nonnegative components.
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From the definition of s we have
0m−1
Ts =
,
t
where 0m−1 is the zero vector in Rm−1 . So T r ≥ T s. (The inequalities are
coordinate-wise.)
Now let P be the m × m lower triangular matrix with ones on and below the
main diagonal. Then


r1


r1 + r2


Pr = 
.
..


.
r1 + r2 + · · · + rm
A Variance Analog of
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Associated Inequalities
Michael G. Neubauer and
William Watkins
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We must show that P r ≤ P s.
Let Q = P −1 . Then Q is the lower triangular matrix with diagonal entries
equal to 1 and subdiagonal entries equal to −1. All other entries are zero:


1
0
0 ··· 0
 −1
1
0 ··· 0 




.
.
.
.
.
.
Q= 0
.
.
.
. 


 0

−1
1 0
0 ···
0 −1 1
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Next observe that T Q is the tridiagonal matrix:

−q2
t2
0 ···
0 0
 t2 −q3
t3 · · ·
0 0

...

0
t3 −q4
0 0

TQ = 
.
.
.

..
..
..


0
0 tm−1 −qm tm
0
0
0
0


 
U1 u

,
=
0m−1 1



1
where ti = i(i + 1), qi = ti−1 + ti = 2i2 , U1 is the (m − 1) × (m − 1) submatrix
of T Q in the upper left corner, and u = (0, . . . , 0, tm )T . Thus (T Q)−1 = P T −1
has the following form:
−1
U1
w
−1
PT =
,
0m−1 1
A Variance Analog of
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Associated Inequalities
Michael G. Neubauer and
William Watkins
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for some w ∈ Rm−1 .
Next we show that −U1 is a symmetric M-matrix. Let D be the n × n
diagonal matrix with diagonal entries 1/2, 1/3, . . . , 1/(n + 1). Then


2 −1
0 ···
0
0
 −1
2 −1 · · ·
0
0 


 0 −1

2
·
·
·
0
0


−DU1 D =  ..
,
... ...
 .



 0
0
0 −1
2 −1 
0
0
0
0 −1
2
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which is positive definite. It follows that −U1 is also positive definite and hence
an M-matrix. Thus all of the entries of U1−1 are negative. (See [2], Theorem
2.5.3.)
−1
U1
w +m−1
w
−1
Pr = PT Tr =
≤t
.
0m−1 1
t
1
But
Ps = PT
−1
U1−1 w
Ts =
0m−1 1
0m−1
w
=t
.
t
1
Thus, P r ≤ P s.
6.7.
Example 5.1
First we show that z = (−18, −18, −12, 24, 24) is the least element in
S(0, 1944/5) ∩ [−36, 24]5 with respect to the variance majorization order. We
will construct the coordinates of z = b0 w0 + b1 w1 + b2 w2 + b3 w3 + b4 w4 in the
Helmert basis, w0 , w1 , w2 , w3 , w4 .
The mean of z is 0 so b0 = 0. To compute the other coordinates we first
determine the index j, which is defined in Equation (6.13):
1
1
1944
1
2
2 1
5 · 24
−
≤
≤ 5 · 24
−
,
j+1 5
5
j 5
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Associated Inequalities
Michael G. Neubauer and
William Watkins
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so j = 2. Now from Equations (6.14) we have:
b1 = 0,
b22
2
2
1 1
−
= 24,
3 5
= 1944 − 5 · 24
1
2
2
2 1
b3 = 5 · 24
−
= 1200,
3 5
1
2
2
2 1
b4 = 5 · 24
−
= 720.
3 4
Then
√
√
√
z = 24w2 + 1200w3 + 720w4
= 2(−1, −1, 2, 0, 0) + 10(−1, −1, −1, 3, 0) + 6(−1, −1, −1, −1, 4)
= (−18, −18, −12, 24, 24).
Next we show that S(0, 1944/5) ∩ [−36, 24]5 has no greatest element. Supvm
pose there is an element w ∈ S(0, 1944/5) such that x ≺ w for all x ∈
S(0, 1944/5) ∩ [−36, 24]5 . We will prove that w 6∈ [−36, 24]5 .
The vectors x = (−36, 0, 0, 18, 18) and y = (−36, 0, 6, 6, 24) are both in
vm
S(0, 1944/5) ∩ [−36, 24]5 . So x, y ≺ w. To express x, y in terms of their
coordinates in the Helmert basis for R5 , let H be the 5×5 matrix whose columns
A Variance Analog of
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Associated Inequalities
Michael G. Neubauer and
William Watkins
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are the Helmert basis for R5 :
 1
√




H=



5
√1
5
√1
5
√1
5
1
√
5
− √12 − √16 − √112 − √120

√1
2

− √16 − √112 − √120 


√2
√1
√1  .
0
−
−
6
12
20 
√3
− √120 
0
0

12
4
√
0
0
0
20
Then the coordinate vectors of x, y with respect to the Helmert basis are
√ √
√ √
a = xH = (0, 18 2, 6 6, 15 3, 9 5)
√ √ √
√
b = yH = (0, 18 2, 8 6, 8 3, 12 5).
Thus
A Variance Analog of
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Michael G. Neubauer and
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a(2) = (0, 648, 216, 675, 405)
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b(2) = (0, 648, 384, 192, 720).
The partial sums of a(2) and b(2) are
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a(2) : (0, 648, 864, 1539, 1944)
b(2) : (0, 648, 1032, 1224, 1944)
The maximum of each of the five partial sums is
(0, 648, 1032, 1539, 1944)
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and the sequence with these partial sums is
c(2) = (0, 648, 384, 507, 405).
Thus the least upper bound of x, y is
cH T = (−37, −1, 5, 15, 18).
vm
Now let w be a sequence in S(0, 1944/5) such P
that v ≺ w for all v ∈ S(0, 1944/5).
Express w in terms of the Helmert basis, w = di wi , for nonnegative di . Then
i
X
a2k
k=1
≤
i
X
d2k
and
k=1
i
X
b2k
k=1
≤
i
X
d2k ,
A Variance Analog of
Majorization and Some
Associated Inequalities
Michael G. Neubauer and
William Watkins
k=1
for i = 1, . . . , n − 1 so that
Title Page
i
X
k=1
c2k ≤
i
X
Contents
d2k ,
k=1
vm
for i = 1, . . . , n − 1. Thus (−37, −1, 5, 15, 18) ≺ w and by Lemma 4.7,
w1 ≤ −37. So w 6∈ [−36, 24]5 .
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References
[1] J.H.E. COHN, Determinants with elements ±1, J. London Math. Soc., 42
(1967), 436–442.
[2] R.A. HORN AND C.R. JOHNSON, Topics in Matrix Analysis. Cambridge,
1991.
[3] G.H. HARDY, J.E. LITTLEWOOD
bridge, 1967.
AND
G. PÓLYA, Inequalities, Cam-
[4] J. KARAMATA, Sur une inégalité rélative aux fonctions convexes, Publ.
Math. Univ. Belgrade, 1 (1932) 145–148.
[5] J.H.B. KEMPERMAN, Momentproblems for sampling without replacement, I,II,III. Nederl. Akad. Wetensch. Proc. Ser. A, 75 (1973), 149–188.
[6] A.W. MARSHALL AND I. OLKIN, Inequalities: Theory of Majorization
and Its Applications, Academic Press, 1979.
[7] I. SCHUR, Über eine Klasse von Mittelbildungen mit Anwendungen auf die
Determinantentheorie, Sitzungsber. Berlin Math. Gesellschaft, 22 (1923),
9–20. (Issai Schur Collected Works (A. Brauer and H. Rohrbach, eds.) Vol.
II. Springer-Verlag, Berlin, 1973, 416–427.)
[8] H. WOLKOWICZ AND G. STYAN, Bounds on eigenvalues using traces,
Linear Algebra Appl., 29 (1980) 471–506.
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Associated Inequalities
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