The best constant in the Khintchine inequality for...

The best constant in the Khintchine inequality for

slightly dependent random variables

Orli Herscovici 1

Joint work with Susanna Spektor 2

1Department of MathematicsUniversity of Haifa

2Department of Mathematics and Statistics SciencesSheridan College Institute of Technology and Advanced Learning

Online Asymptotic Geometric Analysis Seminar

June 20, 2020

Khintchine inequality (1923)

∀p ∈ (0,∞) ∃Ap,Bp s.t. for arbitrary N ∈ N

(∣∣∣N∑

∣∣∣p) 1

≤ Bp

where for i = 1, . . . ,N

ai ∈ R,

εi – mutually independent Rademacher r.v.s.,

P(εi = 1) = P(εi = −1) =1

Orli Herscovici (University of Haifa) Khintchine inequality Online AGA Seminar 2 / 31

Some related researches

1930 Littlewood; Paley and Zigmund – more systematic study of theinequality

1961 Steckin: B2n = ((2n − 1)!!)1

1964 Kahane – generalization to normed spaces

1970s Young, Szarek, Haagerup – best constants

Maurey and Pisier – study of geometric properties ofBanach spaces

Tomczak-Jaegermann – geometric properties, convexity

1980s Ball, Milman, Garling

1990s – Kahane, Latała, Oleszkiewicz, Tomczak-Jaegermann, Litvak,

Milman, König, Peškir, Eskenazis, Nayar, Tkocz, Spektor

convex bodieslogconcave random variables

Kahane-Khintchine inequality

Steinhaus random variables...

and many many others...

Khintchine inequality

D.J.H. Garling “Inequalities. A journey into linear analysis” (2007)

Theorem 12.3.1

There exist positive constants Ap, Bp, for 0 < p < ∞, such that if

a1, . . . ,aN are real numbers and ε1, . . . , εN are Rademacher random

variables, then

∣∣∣N∑

∣∣∣p) 1

≤ Ap

∣∣∣N∑

∣∣∣p) 1

If 0 < p < 2, we can take Bp = 1 and Ap ≤ 31p− 1

2 . If 2 ≤ p ≤ ∞ we can

take Bp ∼√

ep as p → ∞, and Ap = 1.

Khintchine inequality: proof

Consider the case 2 < p < ∞.

If 2k − 2 < p < 2k , then

∣∣∣N∑

∣∣∣2k−2

) 12k−2

∣∣∣N∑

∣∣∣p) 1

∣∣∣N∑

∣∣∣2k) 1

Thus it is sufficient to establish the existence and asymptotic

properties of B2k .(E

∣∣∣N∑

∣∣∣2k)

( N∑

k1+···+kN=2k

k1! · · · kN !a

1 · · · akN

N E(εk1

1 · · · εkN

k1+···+kN=2k

k1! · · · kN !a

1 · · · akN

N E(εk1

1 ) · · ·E(εkN

Khintchine inequality: proof (cont.)

E(εki

{1, if ki is even

0, if ki is odd

Many of therms in the sum are 0, and

∣∣∣N∑

∣∣∣2k

k1+···+kN=k

(2k1)! · · · (2kN)!a

1 · · · a2kN

But (2k1)! · · · (2kN)! ≥ 2k1k1! · · · 2kN kN ! = 2kk1! · · · kN !, and so

∣∣∣N∑

∣∣∣2k

≤ (2k)!

k1+···+kN=k

k1! · · · kN !a

1 · · · a2kN

=(2k)!

2kk!||a||2k

Slightly dependent Rademacher r.v.s.

Based on the O. Herscovici, S. Spektor, “The best constant in the Khintchine

inequality for slightly dependent random variables”, arXiv:1806.03562v3.

Our assumption:N∑

εi = M

M = 0 → we’ll consider,

M > 0 generalizes the previous case,

M < 0 similar to the case M > 0.

∣∣∣N∑

∣∣∣2p

p1+...+pN=2ppi∈{0,...,2p}

p1! . . . pN !a

1 . . . apN

( N∏

)· (a1 + . . . + aN)

Slightly dependent Rademacher r.v.s.: case M = 0

Theorem (HS, 2020)

Let εi , 1 ≤ i ≤ N, be Rademacher random variables satisfying

condition∑N

i=1 εi = 0. Let a = (a1, . . . ,aN) ∈ RN . Then for any p ∈ N,

∣∣∣N∑

∣∣∣2p

≤ C2p2p ||a||

C2p2p =

·√π Γ(

Γ(p + N2 + 1

2)· (2p)!

Case M = 0: proof

Note that∏N

i=1 εpi

i = ±1. In our case

M = 0 =⇒∣∣ {i | εi = 1}

∣∣ =∣∣ {i | εi = −1}

∣∣ = ℓ

i1, . . . , iℓ be the indexes of εij = −1,

iℓ+1, . . . , i2ℓ be the indexes of εij = 1,

i =ℓ∏

2ℓ∏

j=ℓ+1

ℓ∏

(−1)pij ·

2ℓ∏

j=ℓ+1

1pij = (−1)pi1

+...+piℓ

From here on we re enumerate the indexes ij → j , s.t. we have

{1, if p1 + . . .+ pℓ is even,

−1, if p1 + . . .+ pℓ is odd.

Case M = 0: proof

Lemma (HS, 2020)

Let εi , i ≤ N, be Rademacher random variables satisfying condition∑Ni=1 εi = 0 and let p1 + . . . + pN = 2p, pi ∈ {0, . . . ,2p}. Then,

PDif := P+ − P− =

2p+N−12p

P+ = P

({N∏

εi = 0

P− = P

({N∏

i = −1

εi = 0

Proof of the Lemma about PDif

We have that

{1, if p1 + . . .+ pℓ is even,

−1, if p1 + . . .+ pℓ is odd.

Notation

Teven the number of all solutions of p1 + . . .+ pN = 2p, where

p1 + . . .+ pℓ is even

Todd the number of all solutions of p1 + . . .+ pN = 2p, where

p1 + . . .+ pℓ is odd

T the number of all solutions of p1 + . . .+ pN = 2p

=⇒ PDif =Teven − Todd

2p+N−12p

)is the number of weak compositions of 2p into N parts.

To find Teven − Todd , we divide the sequences summing to 2p into

classes and sum over each class separately.

2p = p1 + p2 + . . .+ pℓ + pℓ+1 + pℓ+2 + . . .+ p2ℓ

Now we consider the difference Teven − Todd for each class

c = (c1, c2, . . . , cℓ) := (p1 + pℓ+1,p2 + pℓ+2, . . . ,pℓ + p2ℓ).

if p1 + . . .+ pℓ = even =⇒ (−1)p1+...+pℓ = 1

if p1 + . . .+ pℓ = odd =⇒ (−1)p1+...+pℓ = −1

(Teven − Todd)c =∑

(−1)p1+...+pℓ

(Teven − Todd)c =∑

(−1)p1+...+pℓ

= ((−1)p1,1 + (−1)p1,2 + . . .) · · · ((−1)pℓ,1 + (−1)pℓ,2 + . . .)

(Teven − Todd)c =ℓ∏

pj+pj+ℓ=cj

(−1)pj

For fixed class c and any cj in this class, we have

pj 0 1 2 · · · cj

pj+ℓ cj cj−1 cj−2 · · · 0

=⇒∑

pj+pj+ℓ=cj

(−1)pj =

{0, if cj is odd,1, if cj is even.

Thus for any class c = (c1, . . . , cℓ) we have

=⇒ (Teven − Todd)c =

{0, if at least one of cj is odd,1, if all cj is even,

which means any cj = 2zj and

2p = c1 + . . .+ cℓ =⇒ p = z1 + . . .+ zℓ

Teven − Todd = # weak compositions of p into ℓ parts,

(p + ℓ− 1

PDif =Teven − Todd

(p+ℓ−1

2p+N−12p

Case M = 0: proof (cont.)

Lemma (HS, 2020)

Let εi , i ≤ N, be Rademacher random variables satisfying condition

εi = 0, (1)

and let p1 + . . .+ pN = 2p, pi ∈ {0, . . . ,2p}. Denote by EM an

expectation with condition (1). Then,

p+N2−1

2p+N−12p

Proof of the Lemma about EM

)= 1 × PM

)− 1 × PM

i = −1

(∑Ni=1 εi = 0

PDif – calculated

Let us find P

(∑Ni=1 εi = 0

). Denote D = {i : εi = 1} and

Dc = {i : εi = −1}. Note, the cardinalities card(D) = card(Dc) = N2 .

The event{N∑

εi = 0

}= {εi = 1 | ∀i ∈ D}

⊎{εi = −1 | ∀i ∈ Dc}.

=⇒ P

εi = 0

∣∣∣N∑

∣∣∣2p

)· (a1 + . . . + aN)

For any a = (a1, . . . ,aN)

a1 + . . . + aN ≤ |a1|+ . . .+ |aN | ≡ ||a||1||a||2 ≤ ||a||1 ≤

√N||a||2

(a1 + . . .+ aN)2p ≤ Np||a||2p

∣∣∣N∑

∣∣∣2p

≤2N(

p+N2−1

)(2p+N−1

)Np||a||2p2

The constant

C2p2p =

p+N2−1

)(2p+N−1

) · Np

=2N−1

)!(p + N

2− 1)!

(2p + N − 1)!· Np · (2p)!

Since x! = Γ(x + 1) = xΓ(x), we have

(p + N2− 1)! = Γ(p + N

(2p + N − 1)! = Γ(2p + N).

Applying duplication formula Γ(2x) = π− 12 22x−1Γ(x)Γ(x + 1

Γ(2p + N), we obtain

C2p2p =

·√π Γ(

Γ(p + N2 + 1

2)· (2p)!

Case M = 0: asymptotic approximation

Proposition (HS, 2020)

The constant C2p2p has the following upper bound.

C2p2p ≤ 2NNp

(N + 1)p·(

N2 !)2

N!· (2p)!

2pp!∼ e− p

√πN

2· (2p)!

as N → ∞.

PROOF.

C2p2p =

·√π Γ(

Γ(p + N2+ 1

2)· (2p)!

From Γ(x + 1) = xΓ(x) we obtain

2+ p +

p−1∏

(N + 1

It is easy to see that

p−1∏

(N + 1

Duplication formula with integer n gives Γ(

n + 12

(2n)!√π

22nn!.

Therefore

C2p2p ≤

)p+1· 2N(N

2 )!Γ(

· (2p)!

(N + 1)p· 2N

(N2 !)2

N!· (2p)!

Let us consider now an asymptotic behaviour of the constant C2p2p as

N → ∞.Orli Herscovici (University of Haifa) Khintchine inequality Online AGA Seminar 20 / 31

We have to considerNp

(N+1)p

For the first term we have

(N + 1)p=

)N·(− pN )

∼ e− pN ,

while the second term can be approximated by applying(

)∼ 22n√

(see Elezovic, “Asymptotic expansions of central binomial coefficients

and Catalan numbers” (2014)).

Finally,

C2p2p ∼ e− p

√πN

2· (2p)!

Case M > 0

Theorem (HS, 2020)

Let εi , i ≤ N, be Rademacher random variables satisfying condition∑ni=1 εi = M ≥ 0. Let a = (a1, . . . ,aN) ∈ R

N . Then for any p ∈ N,

∣∣∣∣∣

2p ≤ C

2p2p ||a||

C2p2p =

)(2p+N−1

(p − m + N−M

2 − 1

p − m

)(2m + M − 1

Case M > 0: sketch of the proof

∑Ni=1 εi = M =⇒ |{i | εi = −1}| = ℓ, and |{i | εi = 1}| = M + ℓ,

where ℓ = N−M2

a renumeration of variables:

i =ℓ∏

(−1)pi

2ℓ+M∏

j=ℓ+1

{1, p1 + . . .+ pℓ is even,

−1, p1 + . . .+ pℓ is odd.

a construction of classes c:

2p = p1 + . . .+ pℓ + . . . + p2ℓ + . . .+ p2ℓ+M

c = (c1, . . . , cℓ,p2ℓ+1, . . . ,p2ℓ+M),

where cj = pj + pℓ+j , and cj ,pi ∈ {0, . . . ,2p}.

Case M > 0: sketch of the proof

for each such class c consider the difference (Teven − Todd)c

2p = 2z1 + . . . + 2zℓ + p2ℓ+1 + . . .+ p2ℓ+M

{p2ℓ+1 + . . .+ p2ℓ+M = 2m,2z1 + . . .+ 2zℓ = 2p − 2m

for some m.

(Teven − Todd)c =

(p − m + ℓ− 1

p − m

)(2m + M − 1

Asymptotics of growing sample with given M

Theorem (HS, 2020)

Let ε1, . . . , εN be Rademacher random variables, i.e. such that

P(εi = 1) = P(εi = −1) = 1/2, with condition that∑N

i=1 εi = M, where

0 < M < N is fixed. Then, for any integer p ≥ 2

C2p2p ∼

√π(N2 − M2)

e−MpN

(M − 1)!

(2m + M − 1)!

(p − m)!(2m)!

N − M

when N → ∞.

Asymptotics of proportionally growing samples

Theorem (HS, 2020)

i=1 εi = M. Let a

number of negative εi is n and a number of positive εi is αn, for some

fixed real α > 1. Then, for any integer p ≥ 2

C2p2p ∼

√2πnα

ααn 2(α+1)n

(α+ 1)(α+1)n

(α+ 1)p·

(α− 1)2mnm

(p − m)!(2m)!,

when n → ∞.

If α → 1, then

C2p2p ∼

√πn

2p2p ∼ e− p

√πN

2· (2p)!

i=1 εi = M. Let a

number of negative εi is n and a number of positive εi is αn, for some

fixed real α > 1. Then for any p ≥ 2, the coefficient C2p2p has the

following upper bound.

C2p2p ≤

√2πnα

ααn 2(α+1)n

(α+ 1)(α+1)n

(α− 1)2pnp

(p + 1)p

(α+ 1)pp!,

where n → ∞.

i=1 εi = M, withM = βN, 0 < β < 1. Then, for any integer p ≥ 2,

C2p2p ∼

√πN

2(1 − β2)

(1 + β

1 − β

) βN2 (1 − β)p(2p)!

2mβ2mNm

(1 − β)m(p − m)!(2m)!

when N → ∞.

If β → 0, then

C2p2p ∼

√πN

C2p2p ∼ e− p

√πN

2· (2p)!

Applications

Moments =⇒ information about the tail of random variable

Possible applications:

Risk and portfolio management

Epidemiological studies

The next example is based on the article by A.B. Kashlak, S.

Myroshnychenko, S. Spektor, “Analytic permutation testing via

Kahane-Khintchine inequalities”, arXiv:2001.01130

Automatic Speech Recognition

One of the methods is a permutation test. It maps a sample of size n

onto the symmetric group with n! elements.

Disadvantages of the method:

long computation time

a conservative test – more likely to get a false negative and miss a

significant result

Applications

Proposed solution:

Applying Kahane-Khintchine’s type inequality, which effectively

consider the entire discrete distribution at once without the need to

simulate

Main idea:

the moment bounds from the

Kahane-Khintchine’s type

inequality directly imply bounds

on the tail probability of the test

statistic, which means

bounding the p-value for testing

for significance

Thank you for your attention

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