We introduce a partial order, variance majorization, onRn, which is analogous to the majorization order

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Volume 7, Issue 3, Article 79, 2006

A VARIANCE ANALOG OF MAJORIZATION AND SOME ASSOCIATED INEQUALITIES

MICHAEL G. NEUBAUER AND WILLIAM WATKINS DEPARTMENT OFMATHEMATICS

CALIFORNIASTATEUNIVERSITYNORTHRIDGE

NORTHRIDGE, 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

ABSTRACT. We introduce a partial order, variance majorization, onRⁿ, 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. INTRODUCTION

Let(x₁, . . . , x_n)and(y₁, . . . , y_n)be two sequences of real numbers in nonincreasing order.

The sequencexmajorizesyif

i

X

k=1

x_k≥

i

X

k=1

y_k,

fori= 1, . . . , nwith equality fori=n. Majorization is a partial order on the set of nonincreas- ing 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 functionF(x₁, . . . , x_n)ofn real variables is said to be Schur convex ifF(x)≥F(y)whenever the sequencexmajorizesy. Marshall and Olkin [6]

catalog many functions and results of this type with particular emphasis on statistical inequal- ities. As a simple example, take the product function F(x) = Qn

k=1x_k. If x, y are n-tuples of nonnegative real numbers and ifxmajorizesy, thenF(x) ≤ F(y). That is,−F is a Schur convex function. In particular, ify= (¯x, . . . ,x)¯ thenxmajorizesyand 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

¯

x. Another way to state this well-known elementary result is that the product of a sequencex of nonnegative reals with fixed meanx¯attains a maximum when the variance ofxis 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 sequencesxhaving 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 non- negative reals with fixed mean and fixed variance is attained when the sequence takes on only two valuesα < β and the multiplicity ofβis 1. This simple consequence of the main theorem is known as Cohn’s Inequality [1]: ifx1, . . . , xnare nonnegative reals then

n

Y

k=1

x_k≤αⁿ⁻¹β,

whereα and β are chosen so that the sequences x = (x₁, . . . , x_n)and (α, . . . , α, β) have the same means and the same variances.

2. MAJORIZATION AND VARIANCE MAJORIZATION

LetI(I^st) be the set of nondecreasing (strictly increasing) sequences inRⁿ: I={x∈Rⁿ :x₁ ≤x₂ ≤ · · · ≤x_n}

I^st ={x∈Rⁿ :x1 < x2 <· · ·< xn}.

The variance majorization order involves the variances of leading subsequences ofx ∈ I. So letx[i] = (x₁, . . . , x_i)be the leading subsequence ofxfori= 1, . . . , n. Note thatx[i]consists of theismallest components ofx. We denote the mean ofx[i]by

x[i] = (1/i)

i

X

k=1

x_k, and the variance ofx[i]by

Var(x[i]) = (1/i)

i

X

k=1

(x_k−x[i])². 2.1. Definitions of Variance Majorization and Majorization.

Definition 2.1 (Variance Majorization). Let x = (x1, . . . , xn) and y = (y1, . . . , yn) be se- quences of real numbers inIsuch thatx¯ = ¯yand Var(x) = Var(y). We say thatxis variance majorized byy(oryvariance majorizesx), if

Var(x[i])≤Var(y[i]), fori= 2, . . . n. We writex^vm≺ yory ^vm x.

For fixed meanmand variancev ≥0, variance majorization is a partial order on the set S(m, v) ={x∈I: ¯x=m,Var[x] =v},

which is the intersection ofIwith the sphere inRⁿcentered atm(1,1, . . . ,1)with radius√ nv, and the hyperplane throughm(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 ≥ · · · ≥z_n}.

Definition 2.2 (Majorization). Letx = (x₁, . . . , x_n) andy = (y₁, . . . , y_n)be sequences in D such thatx¯= ¯y. We say thatxis majorized byy(orymajorizesx), if

x[i]≤y[i], fori= 1, . . . , n. In this case we writex^maj≺ y.

The definition of majorization is usually given in this equivalent form:

i

X

k=1

xk≤

i

X

k=1

yk, fori= 1, . . . , nwith equality fori=n.

2.2. Least and Greatest Sequences with Respect to the Variance Majorization Order. Re- turning now to the variance majorization order, there is a least element and a greatest element inS(m, v)for eachmandv ≥0.

Lemma 2.1. Letmandv ≥0be real numbers and let x_min = (α₁, . . . , α₁, β₁) x_max = (α₂, β₂, . . . , β₂), where

α₁ =m−p

v/(n−1) β1 =m+p

(n−1)v α₂ =m−p

(n−1)v β₂ =m+p

v/(n−1).

Thenx_min, x_max∈S(m, v)and

x_min ^vm≺ x^vm≺ x_max, for allx∈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, x_min = (−1,−1,−1,−1,−1,5)andx_max = (−5,1,1,1,1,1).

By contrast, the least and greatest elements inD∩ {x: ¯x=m}with respect to the majoriza- tion order are(¯x, . . . ,x)¯ and(n¯x,0, . . . ,0).

3. VARIANCE MONOTONEFUNCTIONS ANDSCHUR CONVEX FUNCTIONS

Let I be a closed interval in R and let F(x₁, . . . , x_n) be a real-valued function defined on I∩Iⁿ.

Definition 3.1 (Variance Monotone). The functionF is variance monotone increasing onI∩Iⁿ if

x^vm≺ y =⇒ F(x)≤F(y),

for allx, y ∈I∩Iⁿ. If−F is variance monotone increasing, we say thatF is variance monotone decreasing.

4 MICHAELG. NEUBAUER ANDWILLIAMWATKINS

8-1,-1,-1,-1,-1, 5<

8-2,-2,-1,-1, 2, 4< 8-2,-2,-2, 1, 1, 4<

8-3,-2, 0, 0, 1, 4<

8-2,-2,-2, 0, 3, 3<

8-3,-1,-1,-1, 3, 3<

8-4,-1, 0, 0, 2, 3<

8-3,-3, 1, 1, 1, 3< 8-3,-3, 0, 2, 2, 2<

8-4,-1,-1, 2, 2, 2< 8-4,-2, 1, 1, 2, 2<

8-5, 1, 1, 1, 1, 1<

Figure 1: Variance majorization partial order for integral sequences inS(0,6) Figure 2.1: Variance majorization partial order for integral sequences inS(0,30).

Definition 3.2 (Schur Convex). The functionF is Schur convex onD∩Iⁿif x^maj≺ y =⇒ F(x)≤F(y),

for allx, y ∈D∩Iⁿ.

3.1. The Main Result. The next theorem is the main result.

Theorem 3.1. Let I be a closed interval in Rⁿ, and letF(z1, . . . , zn) be a continuous, real- valued function onI∩Iⁿthat is differentiable on the interior ofI∩Iⁿwith gradient∇F(z) = (F₁(z), . . . , F_n(z)). Suppose that

(3.1) F₂(z)−F₁(z) z2−z1

≥ F₃(z)−F₂(z) z3−z2

≥ · · · ≥ F_n(z)−Fn−1(z) zn−zn−1

, for allz ∈I^st∩Iⁿ. ThenF is variance monotone increasing onI∩Iⁿ, that is

x^vm≺ y =⇒ F(x)≤F(y).

for allx, y ∈I∩Iⁿ.

For functions of the form F(z) = φ(x₁) + · · · +φ(x_n), Theorem 3.1 specializes to the following corollary:

Corollary 3.2. Letφ(t)be a continuous, real-valued function on a closed interval I such that φis twice-differentiable on the interior ofIandφ⁰⁰is nonincreasing. Then the function

F(x₁, . . . , x_n) =φ(x₁) +· · ·+φ(x_n)

is variance monotone increasing on the set of nondecreasing sequences inIⁿ. That is, x^vm≺ y =⇒ φ(x₁) +· · ·+φ(x_n)≤φ(y₁) +· · ·+φ(y_n),

for allx, y ∈I∩Iⁿ.

It turns out thatS(m, v) ⊂ Iⁿ when the intervalI = h

m−p

(n−1)v, m+p

(n−1)v)i (see Corollary 4.8). Thus the sequencesx_maxandx_min (described in Lemma 2.1) are inIⁿ. So, ifF is variance monotone increasing onI∩Iⁿ, then the maximum and minimum values ofF are attained atx_max andx_min. This means we can bound F(x)by expressions involving only the mean and variance ofx:

Corollary 3.3. Letmandv ≥0be real numbers andI =h

m−p

(n−1)v, m+p

(n−1)vi . LetF be a variance monotone increasing function onI∩Iⁿ. Then

F(α1, . . . , α1, β1)≤F(x)≤F(α2, β2, . . . , β2), for allx∈S(m, v), whereα1, 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]). LetF(z)be a continuous, real-valued function on Dthat is differentiable on in the interior ofD. Then

x^maj≺ y =⇒ F(x)≤F(y), for allx, y ∈Dif and only if

(3.2) F₁(z)≥F₂(z)≥ · · · ≥F_n(z), for allzin the interior ofD.

The result analogous to Corollary 3.2 for majorization is known as Karamata’s Theorem:

Corollary 3.5 ([4]). Letφbe a continuous, real-valued function on a closed intervalIsuch that φis twice differentiable on the interior ofI andφ⁰⁰is nonnegative. Then

x^maj≺ y =⇒ φ(x₁) +· · ·+φ(x_n)≤φ(y₁) +· · ·+φ(y_n), for allx, y ∈D∩Iⁿ.

4. SOMEVARIANCE MONOTONE AND SCHURCONVEX 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. LetE_k(z)denote thekth elementary symmetric func- tion of the sequencez = (z₁, . . . , z_n)∈Rⁿ:

E_k(z) = X

z_i1z_i2· · ·z_ik,

where the sum is taken over all sets ofkindices with1≤i₁ <· · ·< i_k ≤n.

Theorem 4.1. LetEkbe thekth elementary symmetric function. Then x^vm≺ y =⇒ E_k(y)≤E_k(x),

x^vm≺ y =⇒ E_k+1(x)

E_k(x) ≤ E_k+1(y) E_k(y) , for allx, y ∈I∩[0,∞)ⁿ.

The next corollary is obtained from Corollary 3.3 by evaluating the elementary symmetric functionE_katx_minandx_max.

Corollary 4.2. Letx∈I∩[0,∞)ⁿ. ThenA(v, m, k)≤Ek(x)≤B(v, m, k), where A(v, m, k) = E_k(x_max) =

n

k m+

r v n−1

^k−1

m−(k−1) r v

n−1

B(v, m, k) =E_k(x_min) = n

k m−

r v n−1

^k−1

m+ (k−1) r v

n−1

The inequality analogous to Theorem 4.1 for (regular) majorization is given next:

Theorem 4.3 ([6, p. 80]).

x^maj≺ y =⇒ Ek(y)≤Ek(x) x^maj≺ y =⇒ E_k+1(y)

E_k(y) ≤ E_k+1(x) E_k(x) , for allx, y ∈D∩[0,∞)ⁿ.

4.2. Moment Functions. Letpbe a positive real number and letφ(t) = t^p. Thepth moment function ofz ∈[0,∞)ⁿis given by

M_p(z) =z^p₁ +· · ·+z_n^p. The following results are applications of Corollaries 3.5 and 3.2:

Theorem 4.4. LetM_pbe thepth moment function. Then

x^maj≺ y =⇒ M_p(x)≤M_p(y), forp∈(−∞,0]∪[1,∞) x^maj≺ y =⇒ M_p(y)≤M_p(x), forp∈[0,1],

for allx, y ∈D∩[0,∞)ⁿand

x^vm≺ y =⇒ M_p(x)≤M_p(y), forp∈(−∞,0]∪[1,2]

x^vm≺ y =⇒ M_p(y)≤M_p(x), forp∈[0,1]∪[2,∞), for allx, y ∈I∩[0,∞)ⁿ.

Again we obtain bounds onM_p(x), which depend only on the mean and variance ofx, from Corollary 3.3:

Corollary 4.5. Letx∈I∩[0,∞)ⁿwith meanmare variancev. Let A(m, v, p)=M_p(x_min) =(n−1)

m−p v n−1

p

+

m+p

(n−1)vp

B(m, v, p)=M_p(x_max)=

m−p

(n−1)vp

+ (n−1)

m+p _v

n−1

p

. Then

A(m, v, p)≤M_p(x)≤B(m, v, p), forp∈(−∞,0)∪[1,2]

and

B(m, v, p)≤M_p(x)≤A(m, v, p), forp∈[0,1]∪[2,∞).

4.3. Entropy Function. The entropy function is defined forx∈[0,∞)ⁿby H(x) =−(x₁logx₁+· · ·+x_nlogx_n).

Lettingφ(t) =−tlogt, we haveφ⁰⁰(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 in- equalities:

Theorem 4.6. LetHbe the entropy function. Then

x^vm≺ y =⇒ H(y)≤H(x), for allx, y ∈I∩[0,∞)ⁿ, and

x^maj≺ y =⇒ H(y)≤H(x), for allx, 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. Letx, y ∈I. Then

x^vm≺ y =⇒ x1 ≥y1 andxn ≥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 inS(m, v)in terms ofmandv:

Corollary 4.8. Letx= (x₁, . . . , x_n)be a sequence inS(m, v). Then m−p

(n−1)v ≤ x₁ ≤ m−p

v/(n−1) m+p

v/(n−1) ≤ x_n ≤ m+p

(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. LetGbe a symmetric matrix with eigenvaluesλ₁ ≤ · · · ≤λ_n. Then tr(G)

n + _n^√1_n−1p

n||G||²−(tr(G))² ≤λ_n≤ tr(G)

n +

√n−1 n

pn||G||²−(tr(G))² tr(G)

n −

√n−1 n

pn||G||²−(tr(G))² ≤λ₁ ≤ tr(G)

n − _n^√1_n−1p

n||G||²−(tr(G))².

Corollary 4.9 follows from the fact that the meanmand variancev of the eigenvalues can be expressed in terms of the trace and Euclidean norm||G||ofGas follows:

m= tr(G) n v = 1

n X

i

(λ_i−m)²

= 1 n

X

i

λ²_i −2mλ_i+m²

!

= 1

n(tr(G²)−nm²)

= 1

n² n||G||²−tr(G)² .

5. RESTRICTINGS(m, v)TO AN INTERVAL

Lemma 2.1 guarantees that there is a least and a greatest element inS(m, v)with respect to the variance majorization order. Now we restrict the setS(m, v)to an intervalI = [m−δ, m+ β], withδ, β ≥0, containingm. From Corollary 4.8, we have

h

m−p

v/(n−1), m+p

v/(n−1)in

⊂S(m, v)⊂h

m−p

(n−1)v, m+p

(n−1)vin

. So if eitherδ, β <p

v/(n−1), thenS(m, v)∩Iⁿ=∅. However, ifS(m, v)∩Iⁿis not empty, then it contains a least element, but it may not contain a greatest element.

Lemma 5.1. Letmandv ≥0be real numbers. LetI be the intervalI = [m−δ, m+β]such thatS(m, v)∩Iⁿis not empty. Then there exist unique real numbersm−δ ≤α ≤γ < m+β and an integer1≤j ≤n−1such that the sequence

xmin = (

j

z }| { α, . . . , α, γ,

n−j−1

z }| {

m+β, . . . , m+β)∈S(m, v)∩Iⁿ. Moreoverx_min ^vm≺ xfor allx∈S(m, v)∩Iⁿ.

Example 5.1. Letn = 5. The least element inS(0,1944/5)∩[−36,24]⁵is(−18,−18,−12,24,24).

There is no greatest element inS(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 ma- jorization. There is always a least element and a greatest in D∩ [m, M]ⁿ. The sequence (¯x, . . . ,x)¯ ∈D∩[m, M]ⁿ is the least element. The greatest element with respect to majoriza- tion 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 inD∩[m, M]ⁿis 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. PROOFS

The main technique used in the proofs is to express the sequences inIas linear combinations of a special basis forRⁿ, the so-called Helmert basis.

6.1. Helmert basis. The purpose of this section is to describe the relationship between the coordinates of ann-tuplex∈Rⁿand the coordinates ofxwith respect to the so-called Helmert basis forRⁿ(see [6, p. 47] for a discussion of the Helmert basis). The Helmert basis forRⁿis defined as follows:

w₀^T = 1

√n(1,1, . . . ,1) w₁^T = 1

√2(−1,1,0, . . . ,0) w₂^T = 1

√6(−1,−1,2,0, . . . ,0) ...

w_i^T = 1 pi(i+ 1)(

i

z }| {

−1, . . . ,−1, i,0, . . . ,0) ...

w^T_n−1 = 1

p(n−1)n(−1, . . . ,−1, n−1).

It is clear that{w₀, w₁, . . . , wn−1}is an orthonormal basis forRⁿ. Thus every vectorx∈Rⁿis a linear combination x = Pn−1

k=0a_kw_k. The Helmert coefficient a₀ is determined by the mean

¯

xof the sequencex. Specifically,a₀ = √

n¯x. The other Helmert coefficients,a₁, . . . , a_n−1 are related to the variance, partial variances, and order of the coordinates ofx.

Lemma 6.1. Letxbe a sequence of real numbers withx=Pn−1

k=0akwk. Then Var(x[i]) = 1

i(a²₁+· · ·+a²_i−1), fori= 2, . . . , n. In particular,

Var(x) = 1

n(a²₁+· · ·+a²_n−1).

Proof. Sincea_k =x·w_k,

i−1

X

k=1

a²_k =x

i−1

X

k=1

w^T_kw_k

! x^T

=x((I_i−(1/i)J_i)⊕0n−i)x^T

=

i

X

k=1

x²_k−(1/i)(

i

X

k=1

x_k)²

!

=iVar(x[i]).

Thei×iidentity matrix is denoted byIiandJidenotes thei×imatrix all of whose entries are one. The fact thatPi−1

k=1w_k^Twk =Ii−(1/i)Ji follows from a simple inductive argument.

Let x = P

a_iw_i. In the next definition and lemma, we give necessary and sufficient con- ditions on the sequence (a₁, . . . , an−1) for the sequence xto be nondecreasing. (Clearly, the coefficienta₀ does not influence the relative ordering of the coordinates ofx.)

Definition 6.1 (Admissible Sequence). Let α = (α₁, . . . , αn−1)be a sequence of nonnegative real numbers. Thenαis admissible if

(6.1) (i−1)iα_i−1 ≤i(i+ 1)α_i,

for2≤i≤n−1.

Lemma 6.2. Let x = (x₁, . . . , x_n) be a vector in Rⁿ and a₀, . . . , an−1 be scalars such that x=Pn−1

i=0 a_iw_i. The following conditions are equivalent:

(1) xis nondecreasing.

(2) a_i ≥0, fori= 1, . . . , n−1, and the sequencea⁽²⁾ = (a²₁, . . . , a²_n−1)is admissible.

(3) thekth component ofai−1w_i−a_iwi−1is nonnegative for all k 6=iand nonpositive for k =i.

Proof. Let1≤i≤n−1. Thenx₂−x₁ =√

2a₁, and x_i+1−x_i = ia_i

pi(i+ 1) − (i−1)ai−1

p(i−1)i − a_i pi(i+ 1)

!

= (i+ 1)a_i

pi(i+ 1) −(i−1)ai−1

p(i−1)i

= 1 i

pi(i+ 1)ai −p

(i−1)iai−1

, (6.2)

for i ≥ 2. Thus x is nondecreasing if and only if a_i ≥ 0, for i = 1, . . . , n−1 and a⁽²⁾ is admissible. So Conditions 1 and 2 are equivalent.

Now let(ai−1w_i−a_iwi−1)_kbe thekcomponent ofai−1w_i−a_iwi−1. Then

(6.3) (a_i−1w_i−a_iw_i−1)_k =



















−√^ai−1

i(i+1) + √^ai

(i−1)i , ifk < i,

−√^ai−1

i(i+1) − ^(i−1)a√ ⁱ⁻¹

(i−1)i , ifk =i,

iai−1

√

i(i+1) , ifk =i+ 1,

0 , ifk > i+ 1.

To prove that Condition 2 implies Condition 3, suppose thatai ≥ 0fori = 1, . . . , n−1and thata⁽²⁾is admissible. It is clear from Equation (6.3) that(ai−1wi−aiwi−1)k ≥0for allk 6=i and that(a_i−1w_i−a_iw_i−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< iwe get that a⁽²⁾ is admissible anda_i ≥ 0for alli. Thus Condition 2

holds.

6.2. Proof of Theorem 3.1 and Lemma 4.7. LetF be a differentiable, real-valued function on I∩Iⁿsatisfying Inequality (3.1). Letx, y be nondecreasing sequences inIⁿ such thatx¯ = ¯y, Var(x) =Var(y)andx^vm≺ y. Let

x=

n−1

X

k=0

a_kw_k

y=

n−1

X

k=0

b_kw_k, for scalarsa_k, b_k,k= 0, . . . , n−1, and let

a⁽²⁾ = (a²₁, . . . , a²_n−1), b⁽²⁾ = (b²₁, . . . , b²_n−1).

Sincex¯ = ¯y, we havea₀ = b₀. By Lemma 6.2,a_k, b_k ≥ 0fork = 1, . . . , n−1, a⁽²⁾ andb⁽²⁾ are admissible, and by Lemma 6.1

(6.4)

i

X

k=1

a²_k ≤

i

X

k=1

b²_k,

for1≤i≤n−1with equality in Inequality (6.4) fori=n−1since Var(x) = Var(y).

Next define a pathc(t) = (c₀, c(t)₁, . . . , c(t)n−1)fromatobbyc₀ =a₀ =b₀, and c(t)_k=

q

(1−t)a²_k+tb²_k,

fort ∈ [0,1]andk = 1, . . . , n−1. Thenc(t)⁽²⁾ = (1−t)a⁽²⁾ +tb⁽²⁾, from which it follows thatc⁽²⁾is admissible and that

i

X

k=1

a²_k ≤

i

X

k=1

c(t)²_k ≤

i

X

k=1

b²_k, fori= 1, . . . , n−1.

Now define a pathz(t)fromxtoyby

z(t) =a₀w₀+

n−1

X

k=1

c(t)_kw_k.

Sincec(t)_k ≥0andc(t)⁽²⁾is admissible,z(t)is a nondecreasing sequence.

Letj be the smallest index for whicha_j 6=b_j. Thenc(t)_k = a_k fork < j andc(t)_j > 0for t > 0. It is easy to verify that c⁰(t)_k = (b²_k−a²_k)/c(t)_k (unlessc(t)_k = 0). Thus the tangent vectorz⁰(t)is given by

z⁰(t) =

n−1

X

k=j

b²_k−a²_k c(t)_k wk

(6.5)

= (b²_j −a²_j) w_j

c_j − w_j+1 c_j+1

+ (b²_j+1+b²_j+2−a²_j+1−a²_j+2)

w_j+2

c_j+2 −w_j+3 c_j+3

+· · ·+ b²₁+b²₂+· · ·+b²_n−1−a²₁−a²₂− · · · −a²_n−1

w_n−2 cn−2

− w_n−1 cn−1

, fort >0. It follows from Inequality (6.4) thatz⁰(t)is a nonnegative linear combination of the vectors ^w_cⁱ⁻¹

i−1 −^w_cⁱ

i, fori=j+ 1, . . . , n−1.

We now show thatz(t)∈ Iⁿ for allt ∈ [0,1]. Indeed, both the first and the last coordinates ofz(t)are nonincreasing functions oft. Thus

y₁ =z(1)₁ ≤z(t)₁ ≤z(t)₂ ≤ · · · ≤z(t)_n ≤z(0)_n=x_n.

Sincey₁, x_n ∈ I, z(t) ∈Iⁿand thusF(z(t))is defined for allt ∈ [0,1]. To see thatz(t)₁ and z(t)_n are decreasing, we examine the first and last coordinates of the vectors ^w_cⁱ⁻¹

i−1 − ^w_cⁱ

i. The first coordinate is

−1

p(i−1)ic_i−1 + 1 pi(i+ 1)c_i,

which is nonpositive sincec⁽²⁾ is an admissible sequence. Thusz⁰(t)₁ ≤ 0andz(t)₁ is nonin- creasing int. This proves the first part of Lemma 4.7.

The last coordinate of ^w_cⁱ⁻¹

i−1 − ^w_cⁱ

i is zero unlessi=n−1and in that case it is

−(n−1) p(n−1)ncn−1

,

which is also nonpositive. Thusz(t)n is nonincreasing. This proves the other part of Lemma 4.7.

Finally to prove thatF(z(t))is an increasing function int, we show that dF

dt =∇F ·z⁰(t)≥0.

In view of Equation (6.5), it suffices to show that

∇F ·

w_i−1 ci−1

− w_i c_i

≥0, fori=j+ 1, . . . , n−1.

Sincew_kis orthogonal to the all-ones vectore,

∇F ·

wi−1

ci−1

− w_i ci

= (∇F −F_ie)·

wi−1

ci−1

− w_i ci

. For eachi= 1, . . . , n−1define a functionK_ionI^st∩Iⁿby

K_i(z) = F_i+1(z)−F_i(z) z_i+1−z_i . Now leti < j. Then ^Fj(z)−F_z ^i(z)

j−z_i is a convex combination ofKk(z)fork =i, . . . , j−1:

F_j(z)−F_i(z) z_j −z_i =

j−1

X

k=i

z_k+1−z_k

z_j −z_i K_k(z).

Thus by Condition (3.1),

F_j(z)−F_i(z)

z_j −z_i ≤K_i(z) and so

F_j(z)−F_i(z)≤(z_j−z_i)K_i(z) for alli < jandz∈I^st∩Iⁿ.

Sincec(t)⁽²⁾ is an admissible sequence, Lemma 6.2 guarantees that all components of ^w_cⁱ⁻¹

i−1 −

wi

ci are nonpositive except theith component. Of course theith component of∇F −Fieis zero.

It follows that

(∇F −F_ie)·

wi−1

ci−1

−w_i c_i

≥K_i(z)(z−z_ke)·

wi−1

ci−1

−w_i c_i

=K_k(z)z·

wi−1

c_i−1 − w_i c_i

= 0.

The last equality holds becausez =Pn−1

k=j ckwkis clearly orthogonal to ^w_cⁱ⁻¹

i−1 −^w_cⁱ

i.

We have shown that ^dF_dt ≥0, for z ∈ I^st∩Iⁿandt ∈ (0,1). ThusF(z(t))is an increasing function oft. SoF(x) = F(z(0))≤F(z(1)) =F(y).

By the continuity ofF,F is variance monotone increasing onI∩Iⁿ.

6.3. Proof of Corollary 3.2. Letφ(t)be a continuous, real-valued function on a closed interval Isuch thatφ(t)is twice-differentiable on the interior ofIandφ⁰⁰(t)is nonincreasing onI. Letz be an increasing sequence inIⁿ. Letibe an integer satisfyng1≤i≤n−1. By the mean-value theorem, there existsξ_iin the intervalz_i < ξ_i < z_i+1such that

φ⁰(z_i+1)−φ⁰(z_i)

z_i+1−z_i =φ⁰⁰(ξ_i).

Inequality (3.1) follows sinceξ₁ ≤ξ₂ ≤ · · · ≤ξn−1,φ⁰⁰is nonincreasing, andF_i(z) =φ⁰(z_i).

6.4. Proof of Theorem 4.1. SinceE₂(z)andE₁(z)are constant on S(m, v), we assume that k ≥ 3. Let i be an integer satisfying 1 ≤ i ≤ n−1 and let E_k^(i,i+1) be the kth elementary symmetric polynomial of then−2variablesz₁, z₂, . . . , z_i−1, z_i+2, . . . , z_n. Then

E_k(z) = E_k^(i,i+1)+ (z_i+z_i+1)E_k−1^(i,i+1)+z_iz_i+1E_k−2^(i,i+1). Thus

∂E_k(z)

∂z_i+1 − ∂E_k(z)

∂z_i =−(z_i+1−z_i)E_k−2^(i,i+1), and so

1 zi+1−zi

∂E_k(z)

∂zi+1

− ∂E_k(z)

∂zi

=−E_k−2^(i,i+1),

for all z ∈ I∩[0,∞]ⁿ. But the sequence z is nondecreasing so E_k−2^(i−1,i) ≥ E_k−2^(i,i+1) for i = 1, . . . , n−1. It follows from Theorem 3.1 that−E_k(z)is variance monotone increasing. Thus E_k(z)is variance monotone decreasing. This proves the first inequality in Theorem 4.1.

To prove that the functionF(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)

z₂−z₁ ≥ F₃(z)−F₂(z) z₃ −z₂ .

We write E_k for thekth elementary symmetric function of z₁, . . . , z_n andE_k⁰ for thekth ele- mentary symmetric function ofz₄, . . . , z_n. Then

(6.6) E_k =E_k⁰ + (z₁+z₂ +z₃)E_k−1⁰ + (z₁z₂+z₁z₃+z₂z₃)E_k−2⁰ +z₁z₂z₃E_k−3⁰ . It follows that

F₁ = ∂

∂z₁

E_k+1 E_k

= 1

E_k²[E_kE_k⁰ −E_k+1E_k−1⁰ + (z₂+z₃)(E_kE_k−1⁰ −E_k+1E_k−2⁰ ) +z₂z₃(E_kE_k−2⁰ −E_k+1E_k−3⁰ )].

Thus

F₂ −F₁ z2 −z1

=− 1 E_k²

E_kE_k−1⁰ −E_k+1E_k−2⁰ +z₃(E_k⁰E_k−2⁰ −E_k+1E_k−3⁰ ) . Similarly,

F₃ −F₂

z₃ −z₂ =− 1 E_k²

E_kE_k−1⁰ −E_k+1E_k−2⁰ +z₁(E_k⁰E_k−2⁰ −E_k+1E_k−3⁰ ) , so that

F₂−F₁

z₂−z₁ − F₃−F₂

z₃−z₂ = z₃−z₁

E_k² (E_kE_k−2⁰ −E_k+1E_k−3⁰ ).

Sincez₃−z₁andE_k²are positive, it remains to show thatE_kE_k−2⁰ −E_k+1E_k−3⁰ is nonnegative, which can be rewritten using Equation (6.6) as

E_kE_k−2⁰ −E_k+1E_k−3⁰ = (E_k⁰E_k−2⁰ −E_k+1⁰ E_k−3⁰ ) + (z₁ +z₂+z₃)(E_k−1⁰ E_k−2⁰ −E_k⁰E_k−3⁰ ) + (z₁z₂+z₁z₃+z₂z₂)(E_k−2⁰ E_k−2⁰ −E_k−1⁰ E_k−3⁰ ).

Each of the expressions inE_r⁰ 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,∞)ⁿands > r =⇒ Es−1E_r > E_sEr−1, withr =k−2ands=k+ 1, k, k−1.

6.5. Proof of Lemma 5.1. We may assume thatm = 0so thatI = [−δ, β].

Let z ∈ Rⁿ with mean z¯ = 0. There exist real numbers a = (a₁, . . . , an−1) such that z =Pn−1

i=1 a_iw_i Thenz ∈S(0, v)if and only ifasatisfies the following conditions:

(6.7)

a_i ≥0, for alli a⁽²⁾ is admissible Pn−1

i=1 a²_i =nv.

Let z ∈ S(0, v). The only vector among w₁, . . . , w_n−1 having a nonzeronth coordinate is wn−1. Thus

zn =

rn−1 n an−1.

To computez₁ in terms ofa_k, notice that the first coordinate of eachw_kis−1/p

k(k+ 1). So z₁ =−

n−1

X

k=1

a_k pk(k+ 1). Thusz ∈Iⁿif and only if

(6.8) (n−1)a²_n−1 ≤nβ²,

and

(6.9) −δ ≤ −

n−1

X

k=1

a_k pk(k+ 1).

Next, we establish another inequality for the sequencesafor whichz =P

a_kw_k ∈S(0, v)∩ Iⁿ. Sincea⁽²⁾ is admissible and Inequality (6.8) holds, we have

2a²₁ ≤6a²₂ ≤ · · · ≤k(k+ 1)a²_k ≤ · · · ≤(n−1)na²_n−1 ≤n²β². Thus

(6.10) a²_k≤n²β²

1

k − 1

k+ 1

, fork = 1, . . . , n−1.

We now defineb= (b₁, . . . , bn−1)so that it satisfies Conditions (6.7) and so that for all other asatisfying Conditions (6.7), we have

(6.11)

i

X

k=1

b²_k ≤

i

X

k=1

a²_k,

fori = 1, . . . , n−1. In view of the fact thatPn−1

k=1a²_k =nv =Pn−1

k=1b²_k, the inequalities above are equivalent to

(6.12)

n−1

X

k=i+1

a²_k ≤

n−1

X

k=i+1

b²_k, fori= 0, . . . , n−2.

We begin by specifying the integerj. The function f(j) :=nβ²

1 j − 1

n

, is decreasing inj with

f(1) =nβ²

1− 1 n

f(n) = 0.

We also have from Inequality (6.10) that v = 1

n

n−1

X

k=1

a²_k ≤nβ²

1− 1 n

=f(1).

Thus there exists1≤j ≤nsuch that

(6.13) nβ²

1 j+ 1 − 1

n

≤v < nβ² 1

j − 1 n

. Define the sequence of nonnegative realsb= (b₁, . . . , bn−1)as follows:

b²_i =n²β² 1

i − 1 i+ 1

, fori=j+ 1, . . . , n−1 (6.14)

b²_j =nv−n²β² 1

j+ 1 − 1 n

, (6.15)

b_i = 0, fori= 1, . . . , j−1.

It is clear that

n−1

X

k=1

b²_k =nv.

To check thatb⁽²⁾ is admissible, notice that

i(i+ 1)b²_i =n²β² = (i+ 1)(i+ 2)b²_i+1, fori=j+ 1, . . . , n−2. Also

j(j+ 1)b²_j =j(j+ 1)

nv−n²β² 1

j+ 1 − 1 n

≤j(j + 1)

n²β² 1

j − 1 n

−n²β² 1

j+ 1 − 1 n

=n²β²

= (j+ 1)(j+ 2)b²_j+1.

The inequality follows from the choice ofj in Inequality (6.13). Of course0 = i(i+ 1)b²_i ≤ (i+ 1)(i + 2)b²_i+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

a²_k ≤n²β² 1

i+ 1 − 1 n

. Fori≥j we have

n−1

X

k=i+1

b²_k =n²β²

n−1

X

k=i+1

1

k − 1

k+ 1

=n²β² 1

i+ 1 − 1 n

. So Inequality (6.12) holds fori≥j. Fori < j,

n−1

X

k=i+1

a²_k≤

n−1

X

k=1

a²_k =nv=

n−1

X

k=i+1

b²_k. So Inequality (6.12) holds fori < jtoo.

Now let

x_min =

n−1

X

k=1

b_kw_k.

We now show that x = x_min = (

j

z }| { α, . . . , α, γ,

n−j−1

z }| {

β, . . . , β) for some −δ ≤ α ≤ γ < β. From the proof of Lemma 6.2 we have x_i+1 = x_i if and only if i(i− 1)b²_i−1 = i(i + 1)b²_i. For i = 1, . . . , j −1we have bi = 0. Sox1 = · · · = xj = α. Andi(i−1)b²_i−1 = i(i+ 1)b²_i for i = j + 2, . . . , n−1. Sox_j+2 = · · · = x_n = β. The last equality follows from the choice of b_n−1.

We now show that −δ ≤ α. SinceS(0, v)∩Iⁿ is nonempty, letz ∈ S(0, v)∩Iⁿ and let z =Pn−1

i=1 a_iw_i for some nonnegative realsa= (a₁, . . . , an−1). Then

−δ ≤z₁ =−X a_i pi(i+ 1). We will show that

(6.16) 0≤

n−1

X

i=1

a_i−b_i pi(i+ 1), from which it follows that−δ ≤αand hence thatxmin ∈Iⁿ.

We begin with the following identity:

a_i−b_i = 1 a_i+b_i

i

X

k=1

(a²_k−b²_k)−

i−1

X

k=1

(a²_k−b²_k)

! . Then

(6.17)

n−1

X

i=1

a_i−b_i pi(i+ 1) =

n−2

X

i=1

1 c_i − 1

c_i+1 i

X

k=1

(a²_k−b²_k), where

c_i =p

i(i+ 1)(a_i+b_i),

fori= 1, . . . , n−2. Both sequencesa⁽²⁾ andb⁽²⁾ are admissible. Thus c_i =p

i(i+ 1)(a_i+b_i)≤p

(i+ 1)(i+ 2)(a_i+1+b_i+1) = c_i+1,

for i = 1, . . . , n−2. It follows that1/c_i −1/c_i+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 thatx_min ∈Iⁿ.

Finally, we show thatx_min is unique. Let y= (

l

z }| { α₁, . . . , α₁, γ₁,

n−l−1

z }| {

β, . . . , β)∈S(0, v)∩Iⁿ,

where−δ ≤α₁ ≤γ₁ < β. We begin the proof thatx=x_min =yby showing thatl =j. From the definition ofbwe have

b₁ =· · ·=bj−1 = 0.

Sincey₁ =· · ·=y_l=α₁ we have

a₁ =· · ·=a_l−1 = 0.

But (6.18)

l−1

X

k=1

b²_k ≤

l−1

X

k=1

a²_k= 0, andbj+1 >0, sol−1≤j.

We now show thatj ≤l.

Sincey_l+1 =· · ·=y_n, we have

(l+ 1)(l+ 2)a²_l+1 = (l+ 1)(l+ 2)a²_l+2 =· · ·= (n−1)na²_n−1 =n²β². In addition,

(j + 1)(j+ 2)b²_j+1= (j+ 2)(j + 3)b²_j+2 =· · ·= (n−1)nbn−1 =n²β². Then

a²_k =n²β² 1

k − 1

k+ 1

, fork =l+ 1, . . . , n−1, and

b²_k =n²β² 1

k − 1

k+ 1

, fork =j+ 1, . . . , n−1.

Supposel < j. Thena_k =b_k,k= 1, . . . , j+ 1anda_j > b_j. This contradicts the fact that

n−1

X

k=j

a²_k ≤

n−1

X

k=j

b²_k. Thereforej ≤land soj =lorj =l−1.

Supposej =l−1.Then from Inequality (6.18) we haveb_j = 0andγ =α. Since

j

X

k=1

a²_k ≤

j

X

k=1

b²_j = 0,

we havea_j = 0and so γ₁ =α₁, which contradicts the assumption thatα₁ < γ₁. It follows that j =l.

To finish the proof of uniqueness, we show thata=b. This follows immediately since a_k =b_k= 0,

fork = 1, . . . , j−1and fork =j+ 1, . . . , n−1, andPn−1

k=1a²_k =Pn−1

k=1a²_k. Thusa_j =b_j and soa =b. It follows thatxmin =y.

6.6. Proof of Lemma 2.1. Let x_min = (α₁, . . . , α₁, β₁) be defined as in the statement of Lemma 2.1. Clearly, Var(x_min[i]) = 0, fori= 2, . . . , n−1. Thusx_min ^vm≺ x, for allx∈S(m, v).

The argument forx_max= (α₂, β₂, . . . , β₂)is more involved. Letb₀andb = (b₁, . . . , bn−1)be the coordinates ofx_maxin the Helmert basis:

x_max= (α₂, β₂, . . . , β₂) =b₀w₀+b₁w₁+· · ·+bn−1wn−1.

Since the second through nth coordiates of x_max are the same, the admissible sequence b⁽²⁾ satisfies:

2b²₁ = 6b²₂ = 12b²₃ =· · ·= (n−1)nb²_n−1. Now letx∈S(m, v)with

x=a₀w₀+a₁w₁+· · ·+a_n−1w_n−1. We must show that

(6.19)

i

X

k=1

a²_i ≤

i

X

k=1

b²_i,

fori= 1, . . . , n−1. We need the next lemma to finish the proof.

Lemma 6.3. Letr= (r₁, . . . , r_m)be an admissible sequence of nonnegative real numbers with Pr_k=t. Lets= (s₁, . . . , s_m)be the unique sequence of nonnegative reals with

2s₁ = 6s₂ = 12s₃ =· · ·=m(m+ 1)s_m andP

sk=t. Then

i

X

k=1

r_k ≤

i

X

k=1

s_k, fori= 1, . . . , m.

Equation (6.19) follows from Lemma 6.3 withm=n−1,r=a⁽²⁾,s=b⁽²⁾, andt=nv.

Proof. (Lemma 6.3) The condition on the sum and for admissibility can be expressed in matrix form as follows: P

r_k =tandr is admissible if and only if the firstm−1coordinates ofT r are nonnegative and the last coordinate ist, where

T =























−t1 t2 0 · · · 0 0 −t₂ t₃ 0 · · · 0 ... ... . .. ... ...

0 0 −t_i t_i+1 0

0 0 . .. . .. ...

0 0 · · · −t_m−1 t_m

1 1 · · · 1 1























andt_i =i(i+ 1). We can express this condition symbolically as T r =

+_m−1 t

,

where+m−1 is a vector inR^m−1 with nonnegative components.