Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 7, Issue 3, Article 79, 2006
A VARIANCE ANALOG OF MAJORIZATION AND SOME ASSOCIATED
INEQUALITIES
MICHAEL G. NEUBAUER AND WILLIAM WATKINS
D EPARTMENT OF M ATHEMATICS
C ALIFORNIA S TATE U NIVERSITY N ORTHRIDGE
N ORTHRIDGE , CA 91330
michael.neubauer@csun.edu
bill.watkins@csun.edu
URL: www.csun.edu/˜vcmth006
Received 03 March, 2006; accepted 21 March, 2006
Communicated by I. Olkin
A BSTRACT. 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.
Key words and phrases: Inequality, Symmetric polynomial, Majorization, Schur convex, Variance.
2000 Mathematics Subject Classification. Primary: 26D05, 26D07; Secondary: 15A42.
1. I NTRODUCTION
Let (x1 , . . . , xn ) and (y1 , . . . , yn ) be two sequences of real numbers in nonincreasing order.
The sequence x majorizes y if
i
i
X
X
xk ≥
yk ,
k=1
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 inequalQ
ities. 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
ISSN (electronic): 1443-5756
c 2006 Victoria University. All rights reserved.
064-06
2
M ICHAEL G. N EUBAUER AND W ILLIAM WATKINS
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 variancemajorized by the other. In particular the maximum value of the product of a sequence of nonnegative reals with 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.
2. M AJORIZATION AND VARIANCE M AJORIZATION
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
x[i] = (1/i)
i
X
xk ,
k=1
and the variance of x[i] by
Var(x[i]) = (1/i)
i
X
(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
vm
for i = 2, . . . n. We write x ≺ y or y x.
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
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
√
nv,
D = {z ∈ R : z1 ≥ z2 ≥ · · · ≥ zn }.
J. Inequal. Pure and Appl. Math., 7(3) Art. 79, 2006
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VARIANCE - MAJORIZATION
3
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],
maj
for i = 1, . . . , n. In this case we write x ≺ y.
The definition of majorization is usually given in this equivalent form:
i
X
i
X
xk ≤
yk ,
k=1
k=1
for i = 1, . . . , n with equality for i = n.
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
(n − 1)v
p
α2 = m − (n − 1)v
p
β2 = m + v/(n − 1).
β1 = m +
Then xmin , xmax ∈ S(m, v) and
vm
vm
xmin ≺ x ≺ xmax ,
for all x ∈ S(m, v).
Figure 2.1 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).
3. VARIANCE M ONOTONE F UNCTIONS AND S CHUR C ONVEX F UNCTIONS
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.
J. Inequal. Pure and Appl. Math., 7(3) Art. 79, 2006
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4
M ICHAEL G. N EUBAUER AND W ILLIAM WATKINS
8-5, 1, 1, 1, 1, 1<
8-4, -1, 0, 0, 2, 3<
8-3, -2, 0, 0, 1, 4<
8-4, -1, -1, 2, 2, 2<
8-3, -1, -1, -1, 3, 3<
8-2, -2, -1, -1, 2, 4<
8-4, -2, 1, 1, 2, 2<
8-3, -3, 1, 1, 1, 3<
8-2, -2, -2, 1, 1, 4<
8-3, -3, 0, 2, 2, 2<
8-2, -2, -2, 0, 3, 3<
8-1, -1, -1, -1, -1, 5<
Figure 2.1: Variance majorization partial order for integral sequences in S(0, 30).
Definition 3.2
(Schur
Convex).
The function
F is
Schur
D ∩ I ninifS(0, 6)
Figure
1: Variance
majorization
partial
order
for convex
integral on
sequences
maj
x ≺ y =⇒ F (x) ≤ F (y),
for all x, y ∈ D ∩ I n .
3.1. The Main Result. The next theorem is the main result.
Theorem 3.1. Let I be a closed interval in Rn , and let F (z1 , . . . , zn ) be a continuous, realvalued 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)
F2 (z) − F1 (z)
F3 (z) − F2 (z)
Fn (z) − Fn−1 (z)
≥
≥ ··· ≥
,
z2 − z1
z3 − z2
zn − zn−1
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 .
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VARIANCE - MAJORIZATION
5
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
F (x1 , . . . , xn ) = φ(x1 ) + · · · + φ(xn )
is variance monotone increasing on the set of nondecreasing sequences in I n . That is,
vm
x ≺ y =⇒ φ(x1 ) + · · · + φ(xn ) ≤ φ(y1 ) + · · · + φ(yn ),
for all x, y ∈ I ∩ I n .
h
i
p
p
It turns out that S(m, v) ⊂ I n when the interval 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:
h
i
p
p
Corollary 3.3. Let m and v ≥ 0 be real numbers and 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
F1 (z) ≥ F2 (z) ≥ · · · ≥ Fn (z),
(3.2)
for all z in the interior of D.
The result analogous to Corollary 3.2 for majorization is known as Karamata’s Theorem:
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 .
4. S OME VARIANCE M ONOTONE AND S CHUR C ONVEX F UNCTIONS
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.
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6
M ICHAEL G. N EUBAUER AND W ILLIAM WATKINS
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 ,
where the sum is taken over all sets of k indices with 1 ≤ i1 < · · · < ik ≤ n.
Theorem 4.1. Let Ek be the kth elementary symmetric function. Then
vm
x ≺ y =⇒ Ek (y) ≤ Ek (x),
vm
x ≺ y =⇒
Ek+1 (x)
Ek+1 (y)
≤
,
Ek (x)
Ek (y)
for all x, y ∈ I ∩ [0, ∞)n .
The next corollary is obtained from Corollary 3.3 by evaluating the elementary symmetric
function Ek at xmin and xmax .
Corollary 4.2. Let x ∈ I ∩ [0, ∞)n . Then A(v, m, k) ≤ Ek (x) ≤ B(v, m, k), where
r
r
k−1 n
v
v
A(v, m, k) = Ek (xmax ) =
m+
m − (k − 1)
k
n−1
n−1
r
r
k−1 n
v
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:
Theorem 4.3 ([6, p. 80]).
maj
x ≺ y =⇒ Ek (y) ≤ Ek (x)
maj
x ≺ y =⇒
Ek+1 (x)
Ek+1 (y)
≤
,
Ek (y)
Ek (x)
for all x, y ∈ D ∩ [0, ∞)n .
4.2. Moment Functions. Let p be a positive real number and let φ(t) = tp . The pth moment
function of z ∈ [0, ∞)n is given by
Mp (z) = z1p + · · · + znp .
The following results are applications of Corollaries 3.5 and 3.2:
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 .
Again we obtain bounds on Mp (x), which depend only on the mean and variance of x, from
Corollary 3.3:
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VARIANCE - MAJORIZATION
7
Corollary 4.5. Let x ∈ I ∩ [0, ∞)n with mean m are variance v. Let
p
p
p v p + m + (n − 1)v
A(m, v, p)=Mp (xmin ) =(n − 1) m − n−1
p
p
p v p
B(m, v, p)=Mp (xmax )= m − (n − 1)v + (n − 1) m + n−1
.
Then
A(m, v, p) ≤ Mp (x) ≤ B(m, v, p), for p ∈ (−∞, 0) ∪ [1, 2]
and
B(m, v, p) ≤ Mp (x) ≤ A(m, v, p), for p ∈ [0, 1] ∪ [2, ∞).
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
vm
x ≺ y =⇒ H(y) ≤ H(x),
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.
Lemma 4.7. Let x, y ∈ I. Then
vm
x ≺ y =⇒ x1 ≥ y1 and xn ≥ yn .
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 −p (n − 1)v ≤ x1 ≤ m − pv/(n − 1)
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−1 pn||G||2 − (tr(G))2 ≤ λ ≤ tr(G) − √1 pn||G||2 − (tr(G))2 .
1
n
n
n
n n−1
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8
M ICHAEL G. N EUBAUER AND W ILLIAM WATKINS
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 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
=
5. R ESTRICTING S(m, v) TO
AN I NTERVAL
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
h
in
p
p
p
p
m − v/(n − 1), m + v/(n − 1) ⊂ 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 .
vm
Moreover xmin ≺ x for all x ∈ S(m, v) ∩ I n .
Example 5.1. Let n = 5. The least element in S(0, 1944/5)∩[−36, 24]5 is (−18, −18, −12, 24, 24).
There is no greatest element in S(0, 1944/5) ∩ [−36, 24]5 . See Section 6.7.
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].
6. P ROOFS
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.
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VARIANCE - MAJORIZATION
9
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
i
1
=p
i(i + 1)
z }| {
(−1, . . . , −1, i, 0, . . . , 0)
..
.
1
T
wn−1
=p
(n − 1)n
(−1, . . . , −1, n − 1).
It is clear that {w0 , w1 , . . . ,P
wn−1 } is an orthonormal basis for Rn . Thus every vector x ∈ Rn is
a linear combination x = n−1
The Helmert coefficient a0 is determined by the mean
k=0 ak wk . √
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
a2k = x
k=1
i−1
X
!
wkT wk
xT
k=1
= x ((Ii − (1/i)Ji ) ⊕ 0n−i ) xT
!
i
i
X
X
=
x2k − (1/i)(
xk )2
k=1
k=1
= iVar(x[i]).
The i × i identity matrix
Pi−1 isTdenoted by Ii and Ji denotes the i × i matrix all of whose entries are
one. The fact that k=1 wk wk = Ii − (1/i)Ji follows from a simple inductive argument.
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.)
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10
M ICHAEL G. N EUBAUER AND W ILLIAM WATKINS
Definition 6.1 (Admissible Sequence). Let α = (α1 , . . . , αn−1 ) be a sequence of nonnegative
real numbers. Then α is admissible if
(6.1)
(i − 1)iαi−1 ≤ i(i + 1)αi ,
for 2 ≤ i ≤ n − 1.
Lemma
6.2. Let x = (x1 , . . . , xn ) be a vector in Rn and a0 , . . . , an−1 be scalars such that
Pn−1
x = 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)
(i + 1)ai
(i − 1)ai−1
=p
− p
i(i + 1)
(i − 1)i
p
1 p
(6.2)
i(i + 1)ai − (i − 1)iai−1 ,
=
i
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

− √ai−1 + √ ai
, if k < i,


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,
vm
Var(x) = Var(y) and x ≺ y. Let
x=
n−1
X
ak wk
k=0
y=
n−1
X
bk wk ,
k=0
for scalars ak , bk , k = 0, . . . , n − 1, and let
a(2) = (a21 , . . . , a2n−1 ),
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b(2) = (b21 , . . . , b2n−1 ).
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VARIANCE - MAJORIZATION
11
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
a2k
≤
k=1
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 +
n−1
X
c(t)k wk .
k=1
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) z 0 (t) =
n−1 2
X
b − a2
k
k
wk
c(t)
k
k=j
wj
wj+1
wj+2 wj+3
2
2
2
2
2
2
= (bj − aj )
−
+ (bj+1 + bj+2 − aj+1 − aj+2 )
−
cj
cj+1
cj+2
cj+3
wn−2 wn−1
2
2
2
2
2
2
+ · · · + b1 + b2 + · · · + bn−1 − a1 − a2 − · · · − an−1
−
,
cn−2
cn−1
for t > 0. It follows from Inequality (6.4) that z 0 (t) is a nonnegative linear combination of the
i−1
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 .
Since y1 , xn ∈ I, z(t) ∈ I n and thus F (z(t)) is defined for all t ∈ [0, 1]. To see that z(t)1 and
i−1
z(t)n are decreasing, we examine the first and last coordinates of the vectors wci−1
− wcii . The
first coordinate is
−1
1
p
+p
,
(i − 1)ici−1
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.
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12
M ICHAEL G. N EUBAUER AND W ILLIAM WATKINS
The last coordinate of
wi−1
ci−1
−
wi
ci
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
−
≥ 0,
∇F ·
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
For each i = 1, . . . , n − 1 define a function Ki on Ist ∩ I n by
Ki (z) =
Now let i < j. Then
Fj (z)−Fi (z)
zj −zi
Fi+1 (z) − Fi (z)
.
zi+1 − zi
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
z
j − 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 .
i−1
Since c(t)(2) is an admissible sequence, Lemma 6.2 guarantees that all components of wci−1
−
wi
are nonpositive except the ith component. Of course the ith component of ∇F − Fi e is zero.
ci
It follows that
wi−1 wi
wi−1 wi
(∇F − Fi e) ·
−
≥ Ki (z)(z − zk e) ·
−
ci−1
ci
ci−1
ci
wi−1 wi
= Kk (z)z ·
−
ci−1
ci
= 0.
P
wi−1
wi
The last equality holds because z = n−1
k=j ck wk is clearly orthogonal to ci−1 − ci .
≥ 0, for z ∈ Ist ∩ I n and t ∈ (0, 1). Thus F (z(t)) is an increasing
We have shown that dF
dt
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 .
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VARIANCE - MAJORIZATION
13
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. Since E2 (z) and E1 (z) are constant on S(m, v), we assume that
(i,i+1)
k ≥ 3. Let i 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
(i,i+1)
Ek (z) = Ek
(i,i+1)
+ (zi + zi+1 )Ek−1
(i,i+1)
+ zi zi+1 Ek−2 .
Thus
∂Ek (z) ∂Ek (z)
(i,i+1)
−
= −(zi+1 − zi )Ek−2 ,
∂zi+1
∂zi
and so
1
zi+1 − zi
∂Ek (z) ∂Ek (z)
−
∂zi+1
∂zi
(i,i+1)
= −Ek−2 ,
(i−1,i)
(i,i+1)
for all z ∈ I ∩ [0, ∞]n . But the sequence z is nondecreasing so Ek−2 ≥ 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
F2 (z) − F1 (z)
F3 (z) − F2 (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)
0
0
0
Ek = 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
= 2 [Ek Ek0 − Ek+1 Ek−1
+ (z2 + z3 )(Ek Ek−1
− Ek+1 Ek−2
)
Ek
0
0
+ z2 z3 (Ek Ek−2
− Ek+1 Ek−3
)].
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
F2 − F1 F3 − F2
z3 − z1
0
0
−
=
(Ek Ek−2
− Ek+1 Ek−3
).
z2 − z1
z3 − z2
Ek2
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14
M ICHAEL G. N EUBAUER AND W ILLIAM WATKINS
0
0
Since z3 −z1 and Ek2 are positive, it remains to show that Ek Ek−2
−Ek+1 Ek−3
is nonnegative,
which can be rewritten using Equation (6.6) as
0
0
0
0
0
0
0
0
)
− Ek0 Ek−3
Ek−2
) + (z1 + z2 + z3 )(Ek−1
Ek−3
− Ek+1
= (Ek0 Ek−2
− Ek+1 Ek−3
Ek Ek−2
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]:
z ∈ [0, ∞)n and s > r =⇒ Es−1 Er > Es Er−1 ,
with r = k − 2 and s = k + 1, k, k − 1.
6.5. Proof of Lemma 5.1. We may assume that m = 0 so that I = [−δ, β].
n
Let
Pzn−1∈ R with mean z̄ = 0. There exist real numbers a = (a1 , . . . , an−1 ) such that
z = i=1 ai wi Then z ∈ S(0, v) if and only if a satisfies the following conditions:
ai ≥ 0, for all i
a(2) is admissible
Pn−1 2
i=1 ai = nv.
(6.7)
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
p
To compute z1 in terms of ak , notice that the first coordinate of each wk is −1/ k(k + 1). So
z1 = −
n−1
X
ak
p
k(k + 1)
k=1
.
n
Thus z ∈ I if and only if
(n − 1)a2n−1 ≤ nβ 2 ,
(6.8)
and
(6.9)
−δ ≤ −
n−1
X
ak
p
k(k + 1)
k=1
.
Next, we establish another inequality for the sequences a for which z =
n
I . Since a(2) is admissible and Inequality (6.8) holds, we have
P
ak wk ∈ S(0, v) ∩
2a21 ≤ 6a22 ≤ · · · ≤ k(k + 1)a2k ≤ · · · ≤ (n − 1)na2n−1 ≤ n2 β 2 .
Thus
(6.10)
a2k
2 2
≤n β
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
k=1
J. Inequal. Pure and Appl. Math., 7(3) Art. 79, 2006
b2k
≤
i
X
a2k ,
k=1
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VARIANCE - MAJORIZATION
for i = 1, . . . , n − 1. In view of the fact that
are equivalent to
n−1
X
(6.12)
Pn−1
k=1
a2k ≤
a2k = nv =
n−1
X
15
Pn−1
k=1
b2k , the inequalities above
b2k ,
k=i+1
k=i+1
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
2
f (1) = nβ 1 −
n
f (n) = 0.
We also have from Inequality (6.10) that
n−1
1X 2
1
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
(6.13)
nβ
−
≤ v < 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)
bi = n β
−
, for i = j + 1, . . . , n − 1
i
i+1
1
1
2
2 2
(6.15)
bj = nv − n β
−
,
j+1 n
bi = 0, for i = 1, . . . , j − 1.
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
j(j +
1)b2j
1
1
2 2
= 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).
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16
M ICHAEL G. N EUBAUER AND W ILLIAM WATKINS
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
1
1
2 2
2
ak ≤ n β
−
.
i
+
1
n
k=i+1
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
a2k ≤
n−1
X
k=1
k=i+1
n−1
X
a2k = nv =
b2k .
k=i+1
So Inequality (6.12) holds for i < j too.
Now let
xmin =
n−1
X
bk wk .
k=1
n−j−1
j
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 .
WePnow show that −δ ≤ α. Since S(0, v) ∩ I n is nonempty, let z ∈ S(0, v) ∩ I n and let
z = n−1
i=1 ai wi for some nonnegative reals a = (a1 , . . . , an−1 ). Then
X
a
p i
−δ ≤ z1 = −
.
i(i + 1)
We will show that
n−1
X
a − bi
pi
,
0≤
i(i + 1)
i=1
(6.16)
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
Then
(6.17)
X
n−1
n−2 i
X
X
ai − b i
1
1
p
=
−
(a2k − b2k ),
c
c
i(i + 1)
i
i+1
i=1
i=1
k=1
where
ci =
J. Inequal. Pure and Appl. Math., 7(3) Art. 79, 2006
p
i(i + 1)(ai + bi ),
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VARIANCE - MAJORIZATION
17
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 .
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
l−1
X
(6.18)
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
(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 .
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
X
a2k ≤
k=j
n−1
X
b2k .
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
X
k=1
a2k
≤
j
X
b2j = 0,
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,
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18
M ICHAEL G. N EUBAUER AND W ILLIAM WATKINS
for k = 1, . . . , j − 1 and for k = j + 1, . . . , n − 1, and
so a = b. It follows that xmin = y.
Pn−1
k=1
a2k =
Pn−1
k=1
a2k . Thus aj = bj and
6.6. Proof of Lemma 2.1. Let xmin = (α1 , . . . , α1 , β1 ) be defined as in the statement of
vm
Lemma 2.1. 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 .
Now let x ∈ S(m, v) with
x = a0 w0 + a1 w1 + · · · + an−1 wn−1 .
We must show that
i
X
(6.19)
a2i
≤
k=1
i
X
b2i ,
k=1
for i = 1, . . . , n − 1. We need the next lemma to finish the proof.
Lemma 6.3. Let r = (r1 , . . . , rm ) be an admissible sequence of nonnegative real numbers with
P
rk = t. Let s = (s1 , . . . , sm ) be the unique sequence of nonnegative reals with
2s1 = 6s2 = 12s3 = · · · = m(m + 1)sm
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)
P The condition on 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
where +m−1 is a vector in Rm−1 with nonnegative components.
J. Inequal. Pure and Appl. Math., 7(3) Art. 79, 2006
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VARIANCE - MAJORIZATION
19
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 + r 2
.
Pr = 
..


.
r1 + r2 + · · · + rm
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
Next observe that T Q is the tridiagonal matrix:

−q2
t2
0 ···
0 0
 t2 −q3
t
·
·
·
0 0
3

...

0 0
0
t3 −q4

TQ = 
.
.
.

..
..
..


0
0 tm−1 −qm tm
0
0
0
0 1


 
U1 u

,
=
0m−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
for some w ∈ Rm−1 .
Next we show that −U1 is a symmetric M-matrix.
diagonal entries 1/2, 1/3, . . . , 1/(n + 1). Then

2 −1
0
 −1
2 −1

 0 −1
2
−DU1 D = 
.
.
 ..
..

 0
0
0
0
0
Let D be the n × n diagonal matrix with
···
···
···
..
.
0
0
0

0
0 

0 
,


−1
2 −1 
0
0 −1
2
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
w
U1
w +m−1
−1
Pr = PT Tr =
≤t
.
t
1
0m−1 1
J. Inequal. Pure and Appl. Math., 7(3) Art. 79, 2006
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20
M ICHAEL G. N EUBAUER AND W ILLIAM WATKINS
But
Ps = PT
−1
U1−1 w
Ts =
0m−1 1
0m−1
w
.
=t
1
t
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
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
−
= 1200,
b3 = 5 · 24
3 5
1
2
2
2 1
−
= 720.
b4 = 5 · 24
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. Suppose there is an
vm
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 S(0, 1944/5) ∩
vm
[−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 are the Helmert basis for R5 :
 1

√
− √12 − √16 − √112 − √120
5
 1

√1
 √5
− √16 − √112 − √120 
2



√2
√1
√1 
0
−
−
H =  √15
.
6
12
20 
 1

3
1
 √
√
0
0
− √20 
 5

12
√1
√4
0
0
0
5
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).
J. Inequal. Pure and Appl. Math., 7(3) Art. 79, 2006
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VARIANCE - MAJORIZATION
21
Thus
a(2) = (0, 648, 216, 675, 405)
b(2) = (0, 648, 384, 192, 720).
The partial sums of a(2) and b(2) are
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)
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 that v ≺ w for all v ∈ S(0, 1944/5). Express w
P
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 ,
k=1
for i = 1, . . . , n − 1 so that
i
X
k=1
c2k
≤
i
X
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 .
R EFERENCES
[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 AND G. PÓLYA, Inequalities, Cambridge, 1967.
[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
(1980) 471–506.
AND
G. STYAN, Bounds on eigenvalues using traces, Linear Algebra Appl., 29
J. Inequal. Pure and Appl. Math., 7(3) Art. 79, 2006
http://jipam.vu.edu.au/