Optimal comparison of weak and strong moments of random vectors with - - PowerPoint PPT Presentation

Optimal comparison of weak and strong moments

• f random vectors with applications

(joint work with Rafa l Lata la) Piotr Nayar

Institute of Mathematics University of Warsaw

16/09/2019, Jena

Theorem (Lata la, N., 2019) Let X be a random vector in Rn and ∅ = T ⊆ Rn. Then for p ≥ 2

• E sup

t∈T

|t, X|p 1/p ≤ 2√e n + p p sup

t∈T

(E|t, X|p)1/p .

Theorem (Lata la, N., 2019) Let X be a random vector in Rn and ∅ = T ⊆ Rn. Then for p ≥ 2

• E sup

t∈T

|t, X|p 1/p ≤ 2√e n + p p sup

t∈T

(E|t, X|p)1/p .

• Remark. This result is optimal (up to a universal constant) and

equality is achieved for any rotationally invariant random vector X and T = Bn

2 .

Theorem (Lata la, N., 2019) Let X be a random vector in Rn and ∅ = T ⊆ Rn. Then for p ≥ 2

• E sup

t∈T

|t, X|p 1/p ≤ 2√e n + p p sup

t∈T

(E|t, X|p)1/p .

• Remark. This result is optimal (up to a universal constant) and

equality is achieved for any rotationally invariant random vector X and T = Bn

2 .

Proof inspired by the Welch bound proof of Datta, Stephen and Douglas (2012).

Application I - p-summing constant. Theorem (Lata la, N., 2019) Let X be a random vector in Rn and ∅ = T ⊆ Rn. Then for p ≥ 2

• E sup

t∈T

|t, X|p 1/p ≤ 2√e n + p p sup

t∈T

(E|t, X|p)1/p .

Application I - p-summing constant. Theorem (Lata la, N., 2019) (strong vs. weak moments) Let X be a random vector in (Rn, · ). Then for p ≥ 2 (EXp)1/p ≤ 2√e n + p p sup

t∗≤1

(E|t, X|p)1/p .

Application I - p-summing constant. Corollary (Lata la, N., 2019) Let (F, · ) be a Banach space of dimension n. Then for any vectors x1, . . . , xl ∈ F we have  

l

• j=1

xjp  

1/p

≤ 2√e n + p p sup

t∗≤1

 

l

• j=1

| t, xj |p  

1/p

.

Application I - p-summing constant. Corollary (Lata la, N., 2019) Let (F, · ) be a Banach space of dimension n. Then for any vectors x1, . . . , xl ∈ F we have  

l

• j=1

xjp  

1/p

≤ 2√e n + p p sup

t∗≤1

 

l

• j=1

| t, xj |p  

1/p

. The best constant πp(F) in this inequality is called the p-summing constant of F. We have πp(ln

2 ) = (E|U1|p)−1/p ≈

• n+p

p .

Therefore πp(F) ≤ cπp(ldim F

2

).

Application I - p-summing constant. Corollary (Lata la, N., 2019) Let (F, · ) be a Banach space of dimension n. Then for any vectors x1, . . . , xl ∈ F we have  

l

• j=1

xjp  

1/p

≤ 2√e n + p p sup

t∗≤1

 

l

• j=1

| t, xj |p  

1/p

. The best constant πp(F) in this inequality is called the p-summing constant of F. We have πp(ln

2 ) = (E|U1|p)−1/p ≈

• n+p

p .

Therefore πp(F) ≤ cπp(ldim F

2

).

• Question. Is it true that πp(F) ≤ πp(ldim F

2

)?

Application II - concentration of measure theory. Theorem (Lata la, N., 2019 ) Every centered log-concave probability measure on Rn satisfies the

• ptimal concentration inequality in the sense of Lata

la and Wojtaszczyk with a constant ∼ n5/12.

Application II - concentration of measure theory. Theorem (Lata la, N., 2019 ) Every centered log-concave probability measure on Rn satisfies the

• ptimal concentration inequality in the sense of Lata

la and Wojtaszczyk with a constant ∼ n5/12.

• Remark. Previous best bound ∼ n1/2 was due to Lata

la.

Basic linear algebra facts.

Basic linear algebra facts. Lemma 1 (rank factorization) Suppose A is a k × l matrix of rank at most n. Then A can be written as a product A = TX, where T is k × n and X is n × l: A = TX =    t1 . . . tk   

• n

·   x1 · · · xl     

n = (ti, xj)i≤k,j≤l

Basic linear algebra facts. Lemma 1 (rank factorization) Suppose A is a k × l matrix of rank at most n. Then A can be written as a product A = TX, where T is k × n and X is n × l: A = TX =    t1 . . . tk   

• n

·   x1 · · · xl     

n = (ti, xj)i≤k,j≤l

• Proof. There exist vectors v(1), . . . , v(n) such that every column

a of A can be written as a =

n

• s=1

v(s)λs.

Basic linear algebra facts. Lemma 1 (rank factorization) Suppose A is a k × l matrix of rank at most n. Then A can be written as a product A = TX, where T is k × n and X is n × l: A = TX =    t1 . . . tk   

• n

·   x1 · · · xl     

n = (ti, xj)i≤k,j≤l

• Proof. There exist vectors v(1), . . . , v(n) such that every column

a(j) of A can be written as a(j) =

n

• s=1

v(s)λ(j)

s .

Basic linear algebra facts. Lemma 1 (rank factorization) Suppose A is a k × l matrix of rank at most n. Then A can be written as a product A = TX, where T is k × n and X is n × l: A = TX =    t1 . . . tk   

• n

·   x1 · · · xl     

n = (ti, xj)i≤k,j≤l

• Proof. There exist vectors v(1), . . . , v(n) such that every column

a(j) of A can be written as a(j)

i

=

n

• s=1

v(s)

i

λ(j)

s .

Basic linear algebra facts.

Basic linear algebra facts. Lemma 2 (rank of Hadamard product), Peng-Waldron, 2002 Suppose A = (aij) is a k × l matrix of rank at most n. Let m be a positive integer. Then A◦m := (am

ij ) has rank at most

n+m−1

m

• .

Basic linear algebra facts. Lemma 2 (rank of Hadamard product), Peng-Waldron, 2002 Suppose A = (aij) is a k × l matrix of rank at most n. Let m be a positive integer. Then A◦m := (am

ij ) has rank at most

n+m−1

m

• .
• Proof. There exist vectors v(1), . . . , v(n) such that every column

a = (a1, . . . , ak) of A can be written as a =

n

• s=1

v(s)λs, that is ai =

n

• s=1

v(s)

i

λs.

Basic linear algebra facts. Lemma 2 (rank of Hadamard product), Peng-Waldron, 2002 Suppose A = (aij) is a k × l matrix of rank at most n. Let m be a positive integer. Then A◦m := (am

ij ) has rank at most

n+m−1

m

• .
• Proof. There exist vectors v(1), . . . , v(n) such that every column

a = (a1, . . . , ak) of A can be written as a =

n

• s=1

v(s)λs, that is ai =

n

• s=1

v(s)

i

λs. am

i = n

• s1,s2,...,sm=1

v(s1)

i

v(s2)

i

· . . . · v(sm)

i

λs1λs2 · . . . · λsm.

Basic linear algebra facts. Lemma 2 (rank of Hadamard product), Peng-Waldron, 2002 Suppose A = (aij) is a k × l matrix of rank at most n. Let m be a positive integer. Then A◦m := (am

ij ) has rank at most

n+m−1

m

• .
• Proof. There exist vectors v(1), . . . , v(n) such that every column

a = (a1, . . . , ak) of A can be written as a =

n

• s=1

v(s)λs, that is ai =

n

• s=1

v(s)

i

λs. am

i = n

• s1,s2,...,sm=1

v(s1)

i

v(s2)

i

· . . . · v(sm)

i

λs1λs2 · . . . · λsm. We conclude that am ∈ span

• (v(s1)

i

· . . . · v(sm)

i

)i=1,...,k, 1 ≤ s1 ≤ . . . ≤ sm ≤ n

• .

Lemma 3 (case p = 2) Let X be a random vector in Rn and let us take ∅ = T ⊆ Rn. Then E sup

t∈T

|t, X|2 ≤ n sup

t∈T

E|t, X|2.

Lemma 3 (case p = 2) Let X be a random vector in Rn and let us take ∅ = T ⊆ Rn. Then E sup

t∈T

|t, X|2 ≤ n sup

t∈T

E|t, X|2. Let C be the covariance matrix of a symmetric, bounded and non-degenerate X.

Lemma 3 (case p = 2) Let X be a random vector in Rn and let us take ∅ = T ⊆ Rn. Then E sup

t∈T

|t, X|2 ≤ n sup

t∈T

E|t, X|2. Let C be the covariance matrix of a symmetric, bounded and non-degenerate X. We can assume supt∈T E|t, X|2 = 1 and T = {t ∈ Rn : E|t, X|2 ≤ 1} = {t ∈ Rn : Ct, t ≤ 1} = {t ∈ Rn : |C 1/2t| ≤ 1}.

Lemma 3 (case p = 2) Let X be a random vector in Rn and let us take ∅ = T ⊆ Rn. Then E sup

t∈T

|t, X|2 ≤ n sup

t∈T

E|t, X|2. Let C be the covariance matrix of a symmetric, bounded and non-degenerate X. We can assume supt∈T E|t, X|2 = 1 and T = {t ∈ Rn : E|t, X|2 ≤ 1} = {t ∈ Rn : Ct, t ≤ 1} = {t ∈ Rn : |C 1/2t| ≤ 1}. Then we have E sup

t∈T

|t, X|2 = E sup

|C 1/2t|≤1

|t, X|2 = E sup

|C 1/2t|≤1

|C 1/2t, C −1/2X|2 = E|C −1/2X|2 = n.

Proposition (case p = 2m) Let X be a random vector in Rn and let us take T ⊆ Rn. Suppose m is a positive integer. Then E sup

t∈T

|t, X|2m ≤ n + m − 1 m

• sup

t∈T

E|t, X|2m.

Proposition (case p = 2m) Let X be a random vector in Rn and let us take T ⊆ Rn. Suppose m is a positive integer. Then E sup

t∈T

|t, X|2m ≤ n + m − 1 m

• sup

t∈T

E|t, X|2m.

• Proof. Enough: for any k, l ≥ 1 and any vectors t1, . . . , tk and

x1, . . . , xl in Rn

l

• j=1

sup

1≤i≤k

|ti, xj|2m ≤ n + m − 1 m

• sup

1≤i≤k l

• j=1

|ti, xj|2m.

Proposition (case p = 2m) Let X be a random vector in Rn and let us take T ⊆ Rn. Suppose m is a positive integer. Then E sup

t∈T

|t, X|2m ≤ n + m − 1 m

• sup

t∈T

E|t, X|2m.

• Proof. Enough: for any k, l ≥ 1 and any vectors t1, . . . , tk and

x1, . . . , xl in Rn

l

• j=1

sup

1≤i≤k

|ti, xj|2m ≤ n + m − 1 m

• sup

1≤i≤k l

• j=1

|ti, xj|2m. From Lemma 2 the matrix A◦m = (ti, xjm) has rank at most N := n+m−1

m

• .

Proposition (case p = 2m) Let X be a random vector in Rn and let us take T ⊆ Rn. Suppose m is a positive integer. Then E sup

t∈T

|t, X|2m ≤ n + m − 1 m

• sup

t∈T

E|t, X|2m.

• Proof. Enough: for any k, l ≥ 1 and any vectors t1, . . . , tk and

x1, . . . , xl in Rn

l

• j=1

sup

1≤i≤k

|ti, xj|2m ≤ n + m − 1 m

• sup

1≤i≤k l

• j=1

|ti, xj|2m. From Lemma 2 the matrix A◦m = (ti, xjm) has rank at most N := n+m−1

m

• . From Lemma 1 there exist vectors ˜

t1, . . . , ˜ tk and ˜ x1, . . . , ˜ xl in RN such that ti, xjm = ˜ ti, ˜ xj.

Proposition (case p = 2m) Let X be a random vector in Rn and let us take T ⊆ Rn. Suppose m is a positive integer. Then E sup

t∈T

|t, X|2m ≤ n + m − 1 m

• sup

t∈T

E|t, X|2m.

• Proof. Enough: for any k, l ≥ 1 and any vectors t1, . . . , tk and

x1, . . . , xl in Rn

l

• j=1

sup

1≤i≤k

|ti, xj|2m ≤ n + m − 1 m

• sup

1≤i≤k l

• j=1

|ti, xj|2m. From Lemma 2 the matrix A◦m = (ti, xjm) has rank at most N := n+m−1

m

• . From Lemma 1 there exist vectors ˜

t1, . . . , ˜ tk and ˜ x1, . . . , ˜ xl in RN such that ti, xjm = ˜ ti, ˜

• xj. But from Lemma 3

l

• j=1

sup

1≤i≤k

|˜ ti, ˜ xj|2 ≤ N sup

1≤i≤k l

• j=1

|˜ ti, ˜ xj|2.

Cn,q ≤ Cn,p if p < q. Let pj = supi≤k |ti, xj|q−p and assume l

j=1 pj = 1. Let X be a

random vector such that P(X = xj) = pj.

Cn,q ≤ Cn,p if p < q. Let pj = supi≤k |ti, xj|q−p and assume l

j=1 pj = 1. Let X be a

random vector such that P(X = xj) = pj. Then

l

• j=1

sup

1≤i≤k

|ti, xj|q = E sup

1≤i≤k

|ti, X|p ≤ C p

n,p sup 1≤i≤k

E|ti, X|p = C p

n,p sup 1≤i≤k l

• j=1

|ti, xj|ppj ≤ C p

n,p sup 1≤i≤k

 

l

• j=1

|ti, xj|q  

p/q 



l

• j=1

pq/(q−p)

j

 

(q−p)/q

= C p

n,p sup 1≤i≤k

 

l

• j=1

|ti, xj|q  

p/q 



l

• j=1

sup

1≤i≤k

|ti, xj|q  

(q−p)/q

.

Cn,q ≤ Cn,p if p < q. Let pj = supi≤k |ti, xj|q−p and assume l

j=1 pj = 1. Let X be a

random vector such that P(X = xj) = pj. Then

l

• j=1

sup

1≤i≤k

|ti, xj|q = E sup

1≤i≤k

|ti, X|p ≤ C p

n,p sup 1≤i≤k

E|ti, X|p = C p

n,p sup 1≤i≤k l

• j=1

|ti, xj|ppj ≤ C p

n,p sup 1≤i≤k

 

l

• j=1

|ti, xj|q  

p/q 



l

• j=1

pq/(q−p)

j

 

(q−p)/q

= C p

n,p sup 1≤i≤k

 

l

• j=1

|ti, xj|q  

p/q 



l

• j=1

sup

1≤i≤k

|ti, xj|q  

(q−p)/q

. After rearranging we get Cn,q ≤ Cn,p.

Thank you!