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arXiv:0706.2679v4 [math.PR] 20 Nov 2007

Bounds on the concentration function

in terms of Diophantine approximation

Omer Friedland and Sasha Sodin

February 1, 2008

Abstract

We demonstrate a simple analytic argument that may be used to

bound the L´ evy concentration function of a sum of independent ran-

dom variables. The main application is a version of a recent inequality

due to Rudelson and Vershynin, and its multidimensional generalisa-

tion.

Des bornes pour la fonction de concentration en mati` ere

d’approximation Diophantienne. Nous montrons un simple raison-

nement analytique qui peut ˆ etre utile pour borner la fonction de

concentration d’une somme des variables al´ eatoires ind´ ependantes.

L’application principale est une version de l’in´ egalit´ e r´ ecente de Rudel-

son et Vershynin, et sa g´ en´ eralisation au cadre multidimensionel.

1Introduction

The P. L´ evy concentration function of a random variable S is defined as

Q(S) = sup

x∈R

P?|S − x| ≤ 1?.

Since the work of L´ evy, Littlewood–Offord, Erd˝ os, Esseen, Kolmogorov and

others, numerous results in probability theory concern upper bounds on the

1[omerfrie; sodinale]@post.tau.ac.il; address: School of Mathematical Sciences, Tel Aviv

University, Ramat Aviv, Tel Aviv 69978, Israel

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concentration function of the sum of independent random variables; a par-

ticularly powerful approach was introduced in the 1970-s by Hal´ asz [2].

This note was motivated by the recent work of Rudelson and Vershynin

[4]. Let X be a random variable; let X1,··· ,Xnbe independent copies of

X, and let a = (a1,··· ,an) be an n-tuple of real numbers.

In the Gaussian case X ∼ N(0,1), we have:

(where | · | stands for Euclidean norm), and consequently

?

k=1

?n

k=1akXk ∼ N(0,|a|2)

Q

n

?

akXk

?

=

?

2

π|a|(1 + o(1)) ,|a| → ∞ .(1)

On the other hand, if X has atoms, the left-hand side of (1) does not tend

to 0 as |a| → ∞. Therefore one may ask, for which a ∈ Rnis it true that

?

k=1

Q

n

?

akXk

?

≤ C/|a| ?(2)

Rudelson and Vershynin gave a bound in terms of Diophantine approxi-

mation of the vector a. Their approach makes use of a deep measure-theoretic

lemma from [2]. Our goal is to show a simpler analytic method that may

be of use in such problems. The following theorem is a (slightly improved)

version of [4, Theorem 1.3].

Theorem 1.1 Let X1,··· ,Xnbe independent copies of a random variable

X such that Q(X) ≤ 1 − p, and let a = (a1,··· ,an) ∈ Rn. If, for some

0 < D < 1 and α > 0,

|ηa − m| ≥ α

for

m ∈ Zn, η ∈?1/(2?a?∞), D?,

?

(3)

then

Q(

n

?

k=1

Xkak) ≤ Cexp(−cp2α2) +

1

pD

1

|a|

?

.(4)

Here and further C,c,C′,c1,··· > 0 denote numerical constants.

We also extend this result to the multidimensional case. The concentration

function of an Rd-valued random vector− →

Q(− →

S is defined as

S ) = sup

x∈RdP?|− →S −− →x | ≤ 1?.

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Theorem 1.2 Let X1,··· ,Xnbe independent copies of a random variable

X such that Q(X) ≤ 1−p, and let− →a1,··· ,− →an∈ Rdbe such that, for some

0 < D < d and α > 0,

n

?

k=1

(− →η ·− →ak− mk)2≥ α2for m1,··· ,mn∈ Z,− →η ∈ Rd

such that max

k

|− →η ·− →ak| ≥ 1/2, |− →η | ≤ D.

(5)

Then

Q(

n

?

k=1

Xk− →ak) ≤ Cd?

exp(−cp2α2)

?√d

pD

+

?d

det−1/2

?

n

?

k=1

− →ak⊗− →ak

??

,

(6)

where C,c > 0 are numerical constants.

Of course, Theorem 1.1 follows formally from Theorem 1.2. For simplicity of

exposition we will prove Theorem 1.1 and indicate the adjustments that are

necessary for d > 1.

2Proof of Theorem 1.1

Step 1: By Chebyshev’s inequality and the identity

exp(−y2) =

?+∞

−∞

exp?2iyη − η2? dη

√π

it follows that

P

??????

n

?

k=1

Xkak− x

?????≤ 1

?

≤ eEexp

−

exp

?

n

?

?

k=1

Xkak− x

?2

= eE

?+∞

−∞

2i

?

n

?

k=1

Xkak− x

?

η − η2

?

dη

√π.

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Now we can swap the expectation with the integral and take absolute value:

P

??????

n

?

k=1

Xkak− x

?????≤ 1

?

≤ e

?+∞

?+∞

−∞

n

?

n

?

k=1

φ(2akη)exp?−2ixη − η2? dη

|φ(2akη)|exp?−η2? dη

√π

≤ e

−∞

k=1

√π,

where φ(η) = Eexp(iηX) is the characteristic function of every one of the

Xk. Therefore

Q(

n

?

k=1

Xkak) ≤ e

?+∞

−∞

n

?

k=1

|φ(2akη)|exp?−η2? dη

√π.

(7)

Step 2 (this step is analogous to [2, §3] and [4, 4.2]): First,

|φ(η)| ≤ exp

?

−1

2(1 − |φ(η)|2)

?

.

Let X′be an independent copy of X, X#= X − X′. Observe that

q = P?|X#| ≥ 2?≥ p2/2

and

|φ(η)|2= E exp?iηX#?

= Ecos?ηX#?≤ (1 − q) + qE?cos?ηX#???|X#| ≥ 2?

therefore

;

?+∞

−∞

n

?

?+∞

??+∞

k=1

|φ(2akη)|exp?−η2? dη

exp

2E

√π

≤

−∞

?

−q

?

n

?

n

?

k=1

?1 − cos?2akηX#?????|X#| ≥ 2

?1 − cos?2akηX#??− η2

?

− η2

?

dη

√π

≤ E

−∞

exp

?

−q

2

k=1

?

dη

√π

??? |X#| ≥ 2

?

.

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Replace the conditional expectation with supremum over the possible

values z ≥ 2 of |X#| and recall that

1 − cosθ ≥ c1min

m∈Z|θ − 2πm|2;

then

?+∞

−∞

n

?

k=1

|φ(2ηak)|exp?−η2? dη

≤ sup

z≥2

−∞

√π

?+∞

?+∞

?+∞

exp

?

?

−q

2

n

?

k=1

?1 − cos?2zηak

n

?

−c3p2

k=1

??− η2

?

dη

√π

≤ sup

z≥2

−∞

exp−c2p2

k=1

min

mk

|zηak− πmk|2− η2

?

dη

√π

= sup

z≥2/π

−∞

exp

?

n

?

min

mk|ηak− mk|2− (η/z)2

?

dη

z√π.

(8)

Step 3: Denote

A =

?

η ∈ R

???∀m ∈ Zn: |ηa − m| ≥ α/2

?

,B = R \ A .

Then the last integral in (8) can be split into

?+∞

−∞

=

?

A

+

?

B

,(9)

and

?

A

≤ exp?−c3p2α2?

. (10)

On the other hand, if η′,η′′∈ B, then |η′a − πm′|,|η′′a − πm′′| < α/2 for

some m′,m′′∈ Zn, and hence

|(η′− η′′)a − (m′− m′′)| < α .

Therefore by (3) either |η′− η′′| < 1/(2?a?∞) or |η′− η′′| > D. In other

words, B ⊂?

jBj, where Bjare intervals of length ≤ 1/?a?∞such that any

two points belonging to different Bjare at least D-apart.

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