The best constant in the Khintchine inequality for...
Transcript of 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
Ap
(N∑
i=1
a 2i
) 12
≤ E
(∣∣∣N∑
i=1
aiεi
∣∣∣p) 1
p
≤ Bp
(N∑
i=1
a 2i
) 12
,
where for i = 1, . . . ,N
ai ∈ R,
εi – mutually independent Rademacher r.v.s.,
P(εi = 1) = P(εi = −1) =1
2
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
2n
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...
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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
Bp
(E
∣∣∣N∑
i=1
εiai
∣∣∣p) 1
p
≤(
N∑
i=1
a 2i
) 12
≤ Ap
(E
∣∣∣N∑
i=1
εiai
∣∣∣p) 1
p
.
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.
Orli Herscovici (University of Haifa) Khintchine inequality Online AGA Seminar 4 / 31
Khintchine inequality: proof
Consider the case 2 < p < ∞.
If 2k − 2 < p < 2k , then
(E
∣∣∣N∑
i=1
εiai
∣∣∣2k−2
) 12k−2
≤(E
∣∣∣N∑
i=1
εiai
∣∣∣p) 1
p
≤(E
∣∣∣N∑
i=1
εiai
∣∣∣2k) 1
2k
Thus it is sufficient to establish the existence and asymptotic
properties of B2k .(E
∣∣∣N∑
i=1
εiai
∣∣∣2k)
= E
( N∑
i=1
εiai
)2k
=∑
k1+···+kN=2k
(2k)!
k1! · · · kN !a
k1
1 · · · akN
N E(εk1
1 · · · εkN
N )
=∑
k1+···+kN=2k
(2k)!
k1! · · · kN !a
k1
1 · · · akN
N E(εk1
1 ) · · ·E(εkN
N )
Orli Herscovici (University of Haifa) Khintchine inequality Online AGA Seminar 5 / 31
Khintchine inequality: proof (cont.)
E(εki
i ) =
{1, if ki is even
0, if ki is odd
Many of therms in the sum are 0, and
E
∣∣∣N∑
i=1
εiai
∣∣∣2k
=∑
k1+···+kN=k
(2k)!
(2k1)! · · · (2kN)!a
2k1
1 · · · a2kN
N
But (2k1)! · · · (2kN)! ≥ 2k1k1! · · · 2kN kN ! = 2kk1! · · · kN !, and so
E
∣∣∣N∑
i=1
εiai
∣∣∣2k
≤ (2k)!
2kk!
∑
k1+···+kN=k
k!
k1! · · · kN !a
2k1
1 · · · a2kN
N
=(2k)!
2kk!||a||2k
2
Orli Herscovici (University of Haifa) Khintchine inequality Online AGA Seminar 6 / 31
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∑
i=1
εi = M
M = 0 → we’ll consider,
M > 0 generalizes the previous case,
M < 0 similar to the case M > 0.
Note
EM
∣∣∣N∑
i=1
εiai
∣∣∣2p
=∑
p1+...+pN=2ppi∈{0,...,2p}
(2p)!
p1! . . . pN !a
p1
1 . . . apN
N EM
( N∏
i=1
εpi
i
)
= EM
( N∏
i=1
εpi
i
)· (a1 + . . . + aN)
2p.
Orli Herscovici (University of Haifa) Khintchine inequality Online AGA Seminar 7 / 31
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,
EM
∣∣∣N∑
i=1
εiai
∣∣∣2p
≤ C2p2p ||a||
2p2 ,
where
C2p2p =
(N
2
)p+1
·√π Γ(
N2
)
Γ(p + N2 + 1
2)· (2p)!
2pp!.
Orli Herscovici (University of Haifa) Khintchine inequality Online AGA Seminar 8 / 31
Case M = 0: proof
Note that∏N
i=1 εpi
i = ±1. In our case
M = 0 =⇒∣∣ {i | εi = 1}
∣∣ =∣∣ {i | εi = −1}
∣∣ = ℓ
Let
i1, . . . , iℓ be the indexes of εij = −1,
iℓ+1, . . . , i2ℓ be the indexes of εij = 1,
then
N∏
i=1
εpi
i =ℓ∏
j=1
εpij
ij·
2ℓ∏
j=ℓ+1
εpij
ij=
ℓ∏
j=1
(−1)pij ·
2ℓ∏
j=ℓ+1
1pij = (−1)pi1
+...+piℓ
From here on we re enumerate the indexes ij → j , s.t. we have
N∏
i=1
εpi
i =
{1, if p1 + . . .+ pℓ is even,
−1, if p1 + . . .+ pℓ is odd.
Orli Herscovici (University of Haifa) Khintchine inequality Online AGA Seminar 9 / 31
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− =
(p+N
2−1
p
)(
2p+N−12p
) ,
where
P+ = P
({N∏
i=1
εpi
i = 1
}⋂{
N∑
i=1
εi = 0
}),
P− = P
({N∏
i=1
εpi
i = −1
}⋂{
N∑
i=1
εi = 0
}).
Orli Herscovici (University of Haifa) Khintchine inequality Online AGA Seminar 10 / 31
Proof of the Lemma about PDif
We have that
N∏
i=1
εpi
i =
{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
T
Orli Herscovici (University of Haifa) Khintchine inequality Online AGA Seminar 11 / 31
Proof of the Lemma about PDif
T =(
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 =∑
c
(−1)p1+...+pℓ
Orli Herscovici (University of Haifa) Khintchine inequality Online AGA Seminar 12 / 31
Proof of the Lemma about PDif
(Teven − Todd)c =∑
c
(−1)p1+...+pℓ
= ((−1)p1,1 + (−1)p1,2 + . . .) · · · ((−1)pℓ,1 + (−1)pℓ,2 + . . .)
(Teven − Todd)c =ℓ∏
j=1
∑
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.
Orli Herscovici (University of Haifa) Khintchine inequality Online AGA Seminar 13 / 31
Proof of the Lemma about PDif
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
p
),
and
PDif =Teven − Todd
T=
(p+ℓ−1
p
)(
2p+N−12p
)
Orli Herscovici (University of Haifa) Khintchine inequality Online AGA Seminar 14 / 31
Case M = 0: proof (cont.)
Lemma (HS, 2020)
Let εi , i ≤ N, be Rademacher random variables satisfying condition
N∑
i=1
εi = 0, (1)
and let p1 + . . .+ pN = 2p, pi ∈ {0, . . . ,2p}. Denote by EM an
expectation with condition (1). Then,
EM
(N∏
i=1
εpi
i
)=
2N
(NN2
)(
p+N2−1
p
)(
2p+N−12p
)
Orli Herscovici (University of Haifa) Khintchine inequality Online AGA Seminar 15 / 31
Proof of the Lemma about EM
EM
(N∏
i=1
εpi
i
)= 1 × PM
(N∏
i=1
εpi
i = 1
)− 1 × PM
(N∏
i=1
εpi
i = −1
)
=PDif
P
(∑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∑
i=1
εi = 0
}= {εi = 1 | ∀i ∈ D}
⊎{εi = −1 | ∀i ∈ Dc}.
=⇒ P
(N∑
i=1
εi = 0
)=
1
2N
(NN2
).
Orli Herscovici (University of Haifa) Khintchine inequality Online AGA Seminar 16 / 31
Case M = 0: proof (cont.)
EM
∣∣∣N∑
i=1
εiai
∣∣∣2p
= EM
(N∏
i=1
ε pi
i
)· (a1 + . . . + aN)
2p
For any a = (a1, . . . ,aN)
a1 + . . . + aN ≤ |a1|+ . . .+ |aN | ≡ ||a||1||a||2 ≤ ||a||1 ≤
√N||a||2
(a1 + . . .+ aN)2p ≤ Np||a||2p
2
EM
∣∣∣N∑
i=1
εiai
∣∣∣2p
≤2N(
p+N2−1
p
)(
NN2
)(2p+N−1
2p
)Np||a||2p2
Orli Herscovici (University of Haifa) Khintchine inequality Online AGA Seminar 17 / 31
Case M = 0: proof (cont.)
The constant
C2p2p =
2N(
p+N2−1
p
)(
NN2
)(2p+N−1
2p
) · Np
=2N−1
(N2
)!(p + N
2− 1)!
(2p + N − 1)!· Np · (2p)!
p!.
Since x! = Γ(x + 1) = xΓ(x), we have
(p + N2− 1)! = Γ(p + N
2),
(2p + N − 1)! = Γ(2p + N).
Applying duplication formula Γ(2x) = π− 12 22x−1Γ(x)Γ(x + 1
2) to
Γ(2p + N), we obtain
C2p2p =
(N
2
)p+1
·√π Γ(
N2
)
Γ(p + N2 + 1
2)· (2p)!
2pp!
Orli Herscovici (University of Haifa) Khintchine inequality Online AGA Seminar 18 / 31
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
N
√πN
2· (2p)!
2pp!,
as N → ∞.
PROOF.
C2p2p =
(N
2
)p+1
·√π Γ(
N2
)
Γ(p + N2+ 1
2)· (2p)!
2pp!
From Γ(x + 1) = xΓ(x) we obtain
Γ
(N
2+ p +
1
2
)=
p−1∏
i=0
(N + 1
2+ i
)Γ
(N
2+
1
2
)
Orli Herscovici (University of Haifa) Khintchine inequality Online AGA Seminar 19 / 31
Case M = 0: asymptotic approximation
It is easy to see that
p−1∏
i=0
(N + 1
2+ i
)≥(
N + 1
2
)p
.
Duplication formula with integer n gives Γ(
n + 12
)=
(2n)!√π
22nn!.
Therefore
C2p2p ≤
(N
2
)p+1· 2N(N
2 )!Γ(
N2
)(
N+12
)pN!
· (2p)!
2pp!
=Np
(N + 1)p· 2N
(N2 !)2
N!· (2p)!
2pp!.
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
Case M = 0: asymptotic approximation
We have to considerNp
(N+1)p
(N2!)
2
N!
For the first term we have
Np
(N + 1)p=
(1 +
1
N
)−p
=
(1 +
1
N
)N·(− pN )
∼ e− pN ,
while the second term can be approximated by applying(
2nn
)∼ 22n√
πn
(see Elezovic, “Asymptotic expansions of central binomial coefficients
and Catalan numbers” (2014)).
Finally,
C2p2p ∼ e− p
N
√πN
2· (2p)!
2pp!
Orli Herscovici (University of Haifa) Khintchine inequality Online AGA Seminar 21 / 31
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,
EM
∣∣∣∣∣
N∑
i=1
εiai
∣∣∣∣∣
2p ≤ C
2p2p ||a||
2p2 ,
where
C2p2p =
2NNp
(N
N+M2
)(2p+N−1
2p
)p∑
m=0
(p − m + N−M
2 − 1
p − m
)(2m + M − 1
2m
).
Orli Herscovici (University of Haifa) Khintchine inequality Online AGA Seminar 22 / 31
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:
N∏
i=1
εpi
i =ℓ∏
i=1
(−1)pi
2ℓ+M∏
j=ℓ+1
1pj =
{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}.
Orli Herscovici (University of Haifa) Khintchine inequality Online AGA Seminar 23 / 31
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
2m
).
Orli Herscovici (University of Haifa) Khintchine inequality Online AGA Seminar 24 / 31
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)
2N
e−MpN
(M − 1)!
(2p)!
2p·
p∑
m=0
(2m + M − 1)!
(p − m)!(2m)!
(2
N − M
)m
,
when N → ∞.
Orli Herscovici (University of Haifa) Khintchine inequality Online AGA Seminar 25 / 31
Asymptotics of proportionally growing samples
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. 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
α+ 1
ααn 2(α+1)n
(α+ 1)(α+1)n
(2p)!
(α+ 1)p·
p∑
m=0
(α− 1)2mnm
(p − m)!(2m)!,
when n → ∞.
If α → 1, then
C2p2p ∼
√πn
(2p)!
2pp!C
2p2p ∼ e− p
N
√πN
2· (2p)!
2pp!
Orli Herscovici (University of Haifa) Khintchine inequality Online AGA Seminar 26 / 31
Asymptotics of proportionally growing samples
Proposition (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. 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
α+ 1
ααn 2(α+1)n
(α+ 1)(α+1)n
(α− 1)2pnp
(p + 1)p
(2p)!
(α+ 1)pp!,
where n → ∞.
Orli Herscovici (University of Haifa) Khintchine inequality Online AGA Seminar 27 / 31
Asymptotics of proportionally growing samples
Proposition (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, withM = βN, 0 < β < 1. Then, for any integer p ≥ 2,
C2p2p ∼
√πN
2(1 − β2)
N+12
(1 + β
1 − β
) βN2 (1 − β)p(2p)!
2p
p∑
m=0
2mβ2mNm
(1 − β)m(p − m)!(2m)!
when N → ∞.
If β → 0, then
C2p2p ∼
√πN
2
(2p)!
2pp!
C2p2p ∼ e− p
N
√πN
2· (2p)!
2pp!
Orli Herscovici (University of Haifa) Khintchine inequality Online AGA Seminar 28 / 31
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
Orli Herscovici (University of Haifa) Khintchine inequality Online AGA Seminar 29 / 31
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
Orli Herscovici (University of Haifa) Khintchine inequality Online AGA Seminar 30 / 31
Thank you for your attention
Orli Herscovici (University of Haifa) Khintchine inequality Online AGA Seminar 31 / 31