The best constant in the Khintchine inequality for...
Transcript of The best constant in the Khintchine inequality for...
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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
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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
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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.
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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 )
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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
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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.
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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!.
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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.
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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
}).
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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
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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
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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.
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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
)
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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
)
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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
).
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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
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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!
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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
)
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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
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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!
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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
).
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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}.
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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
).
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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 → ∞.
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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!
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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 → ∞.
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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!
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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
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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
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Thank you for your attention
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