Table Of ContentReal zeroes of random polynomials, II
Descartes’ rule of signs and anti-concentration on the
symmetric group
Ken S¨oze
∗
6
1
0 Abstract
2
In this sequel to [8], we present a different approach to bounding the expected number of real zeroes of
n
randompolynomialswithrealindependentidenticallydistributedcoefficientsormoregenerally,exchange-
a
J ablecoefficients. Weshowthatthemeannumberofrealzeroesdoesnotgrowfasterthanthelogarithmof
9 thedegree. ThemainingredientsofourapproachareDescartes’ruleofsignsandanewanti-concentration
1 inequality for the symmetric group. This paper can be read independently of part I in this series.
]
R
P 1 The result on the expected number of real zeroes
.
h
t In this part, which can be read independently from the first part [8] (see the latter for a brief
a
m history of the problem), we will bound the expected number of real zeroes of random polyno-
[ mials with real independent identically distributed coefficients or, more generally, exchangeable
coefficients (the definition of exchangeability is recalled a few lines later).
1
v For a non-zero polynomial P and a subset A R, let N(A,P) denote the number of zeroes
⊆
8 of P, counted with multiplicity, that fall in A. We write N(P) for N(R,P) and N (P) for
∗
5 N(R 0 ,P). Everywhere in the paper, C,c,C ,c etc., denote positive numerical constants (not
8 \{ } ′ ′
4 depending on any parameters). However, the values of these constants may change from line to
0 line. With this notation, we are ready to state our main theorem and the key lemmas.
.
1
0 Theorem 1. Let u ,...,u , n 2, be real numbers, not all equal to zero. Let π be a uniform
0 n
6 random permutation of 0,1,...≥,n . Let P(x) = n u xk. Then
1 { } k=0 π(k)
:
v E[N (P)] 6PClogn.
∗
i
X
As an almost immediate corollary, we get a bound on the expected number of zeros for
r
a
random polynomials with i.i.d.or, more generally, exchangeable coefficients. Recall that random
variables λ ,...,λ are said to be exchangeable if the distribution of (λ ,...,λ ) is the
1 n π(1) π(n)
same as the distribution of (λ ,...,λ ) for any π S , where S is the group of permutations
1 n n n
∈
of 1,...,n . Note that if λ are i.i.d., then they are exchangeable.
k
{ }
Corollary 2. Let P (x) = λ +λ x+...+λ xn, n 2.
n 0 1 n
≥
1. If λ ,...,λ are exchangeable random variables, then E N (P )1l 6 Clogn.
0 n ∗ n Pn=0
{ 6 }
2. If λ ,...,λ are i.i.d. with p := P λ = 0 < 1, then E(cid:2)N(P )1l (cid:3)6 Clogn+ p0 .
0 n 0 { 0 } n {Pn6=0} 1−p0
(cid:2) (cid:3)
∗290W 232nd Str,Apt 4b, Bronx, NY 10463, USA;[email protected]
1
1 The result on the expected number of real zeroes 2
The reason for the indicator function 1l is that there may be a positive probability for
Pn=0
{ 6 }
all coefficients to vanish (in which case N(P ) is not defined). If the coefficients are i.i.d., then
n
P P = 0 = 1 pn+1, showing that restricting to this event leaves out only a tiny part of the
{ n 6 } − 0
probability space, provided that n is large and p is not too close to 1.
0
Proof. Condition on the multi-set of values λ ,...,λ (multi-set means that λ need not be
0 n k
{ }
distinct). Conditional on this multi-set being equal to u ,u ,...,u , by exchangeability, the
0 1 n
{ }
random vector (λ ,...,λ ) has the same distribution as (u ,...,u ), where π is a uniform
0 n π(0) π(n)
random permutation of 0,1,...,n . On the event P = 0, not all u can equal zero, and hence
n i
{ } 6
Theorem 1 applies to give the first part of the corollary.
For a non-zero polynomial P, we have N(P) = N (P) + N( 0 ,P). By the first part,
∗
{ }
E N (P )1l 6 Clogn. If the coefficients are i.i.d., the probability that N( 0 ,P )= k is
pk(1∗ pn) a{nPdn6=h0e}nce { } n
0(cid:2) − 0 (cid:3)
n 1
E N( 0 ,P )1l = (1 p ) − kpk 6 p0 ,
{ } n {Pn6=0} − 0 0 1 p0
k=0 −
(cid:2) (cid:3) X
where the last inequality follows by extending the sum up to infinity.
Our approach in this paper is based on Descartes’ rule of signs and on the following “relative
anti-concentration bound”.
Lemma 3. Let (ξ ,...,ξ ) be a vector of exchangeable random variables having a distribution
1 k
that is absolutely continuous with respect to Lebesgue measure on Rk. Then
k k
C
P jξ 6 ξ 6 , (1)
j j
k
n(cid:12)Xj=1 (cid:12) (cid:12)Xj=1 (cid:12)o
(cid:12) (cid:12) (cid:12) (cid:12)
and (cid:12) (cid:12) (cid:12) (cid:12)
k k
C
P ( 1)jjξ 6 ( 1)jξ 6 . (2)
j j
− − k
n(cid:12)Xj=1 (cid:12) (cid:12)Xj=1 (cid:12)o
(cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
In turn, Lemma 3 will be deduced from an anti-concentration bound for linear forms on the
symmetric group S . The following lemma may be considered the main technical result of this
n
paper and potentially of interest beyond its application to proving our main theorem.
Lemma4. Letn> 2andletw ,...,w bereal numberssuchthat n w = 0and n w2 = 1.
1 n i=1 i i=1 i
Let π be a random permutation uniformly distributed on S . Then, for every L R, we have
n
P ∈P
n
C
P w π(i) Ln 6 1 6 e cL. (3)
i − | |
− n
n(cid:12)Xi=1 (cid:12) o
(cid:12) (cid:12)
We remark that it is possible(cid:12) to strengthen th(cid:12)e statement to have C e cL2 on the right hand
n −
side (see a brief discussion in Section 5), but we shall not need that improvement in this paper.
Although the main application of this lemma is to prove Lemma 3, we shall also use it in several
other smaller ways. For this purpose, we record an easy corollary of Lemma 4.
Corollary 5. Suppose u ,...,u are real numbers, not all equal. Let π be a random permutation
1 n
uniformly distributed on S . Then, for every x R, we have
n
∈
n
C
P u π(i) = x 6 .
i
n
nXi=1 o
2 Descartes’ rule of signs and deduction of Theorem 1 from Lemma 3 3
Proof. Set w = (u u¯)/σ where u¯ = 1 n u and σ2 = n (u u¯)2. The assumption that
u ,...,u arie not ail−l equal ensures thant σi>=10 iand hence w ia=r1e wie−ll-defined, n w = 0 and
1n w2n= 1. Now apply Lemma 4 withPLn = (x 1n(n+P1i)u¯)/σ. The corollaryi=f1olloiws.
i=1 i − 2 P
P
2 Descartes’ rule of signs and deduction of Theorem 1 from Lemma 3
Recall that the number of sign-changes of a finite or infinite sequence b = (b ,b ,...) of real
0 1
numbers is defined as the supremum of all k for which there exist indices 0 6 i < i < ... < i
0 1 k
such that b b < 0 for each j = 1,2,...,k. Let S(b) denote the number of sign changes of b.
ij ij 1
−
Lemma 6 (Descartes’ rule of signs [6, Chapter 5, Problem 38]). Let f(x)= b +b x+b x2+...
0 1 2
be a non-zero power series with real coefficients and convergent in ( R,R). Let N+(f) be the
−
number of zeroes (counted with multiplicity) of f in the interval (0,R). Then, N+(f) 6 S(b).
In the remaining part of this section, we prove Theorem 1 assuming that Lemma 3 and
Lemma 4 are true.
2.1 Proof of Theorem 1 assuming Lemma 3 and Corollary 5
Set P(x) = n λ xj with λ = u where π is a uniform random permutation. Observe that
j=0 j j π(j)
xnP (1/x) = λ +λ x+...+λ xn is a random polynomial with the same distribution as P .
n n n 1 0 n
P −
Therefore, taking I = (0,1) and I = ( 1,0), we can write
− −
E[N (P )] = 2E[N(I,P )]+2E[N( I,P )]+E[N( 1 ,P )]+E[N( 1 ,P )], (4)
∗ n n n n n
− { } {− }
where N( 1 ,P ) are the multiplicities of zeroes at 1.
n
{± } ±
Bound for E[N(I,P )]: Consider the Taylor series with radius of convergence at least 1:
n
P (x) P (x)
F(x) = n = S xk and G(x) = n = T xk.
1 x k (1 x)2 k
− k>0 − k>0
X X
Then
λ +λ +...+λ if k 6 n,
0 1 k
S =
k
(Sn if k > n,
and
S +S +...+S = (k+1)λ +kλ +...+λ if k 6 n,
0 1 k 0 1 k
T =
k
(Tn+(k n)Sn if k > n.
−
Firstly, N(I,P ) = N(I,G) 6 S((T ) ), by Descartes’ rule. Secondly, S((T ) ) 6 1 +
n k k 0 k k 0
≥ ≥
S(T ,T ,...,T ) since, beyond n, there could be at most one change of sign in the sequence
0 1 n
(T ) . Thus,
k k n
≥ E[N(I,P )] 6 1+E[S(T ,...,T )]. (5)
n 0 n
Recall that λ = u where π is a uniform random permutation. Assume without loss of
j π(j)
generality that on the same probability space, we have standard Gaussian random variables Z ,
j
06 j 6 n, that are independent among themselves and independent of π. Set λε = λ +εZ for
j j j
ε > 0. Let Skε and Tkε be defined using (λεj)06j6n exactly as Sk and Tk are defined in terms of
(λj)j6n. Then,
n
S(T0ε,...,Tnε) 6 1l TεTε 60 .
k=1 { k k−1 }
X
2 Descartes’ rule of signs and deduction of Theorem 1 from Lemma 3 4
Since Tε = Tε + Sε, to have TεTε 6 0, it is necessary that Tε 6 Sε . Now, for any
k k 1 k k k 1 | k| | k|
fixed ε > 0, th−e random variables λε−, 0 6 j 6 n, are exchangeable, and have an absolutely
j
continuous distribution on Rn. By conclusion (1) in Lemma 3, it immediately follows that
P Tε 6 Sε 6 C/k and hence E[S(Tε,...,Tε)] 6 Clogn. Observe that C does not depend
{| k| | k|} 0 n
on ε (or anything else).
Since sign changes are defined by strict inequalities, we see that almost surely,
S(T ,...,T ) 6 liminfS(Tε,...,Tε).
0 n 0 n
ε 0
→
and hence, by Fatou’s lemma E[S(T ,...,T )] 6 Clogn. Plugging back this conclusion into (5),
0 n
we get E[N(I,P )] 6 Clogn.
n
Bound for E[N( I,P )]: Next we bound E[N( I,P )]. Replacing x by x, we have the
n n
− − −
analogue of (5):
N( I,P ) 6 1+S(T ,...,T ) (6)
− n 0′ n′
where S = k ( 1)jλ and T = S +...+S = k (k+1 j)( 1)jλ .
k′ j=0 − j k′ 0′ k′ j=0 − − j
Exactly as before, we define λε = λ +εZ , where Z are independent standard Gaussians
P j j j P j
that are also independent of π. Define Sj′ε and Tj′ε in terms of (λεj)j6n just as Sj′ and Tj′ are
defined in terms of (λj)j6n. By the lower semi-continuity of sign changes, by letting ε decrease
to zero, we may deduce that E[S(T ,...,T )] is bounded by Clogn provided we prove the same
0′ n′
bound for E[S(T ε,...,T ε)]. To do that, we write
0′ n′
n
E[S(T ε,...,T ε)] 6 P T εT ε 6 0
0′ n′ { k′ k′−1 }
k=1
X
n
6 P T ε 6 Sε .
{| k′ | | k′ |}
k=1
X
Now, use the bound (2) in Lemma 3 to get P T ε 6 Sε 6 C/k. Using this bound in (6), we
{| k′ | | k′ |}
get the inequality E[N( I,P )] 6 Clogn.
n
−
Bound for E[N( 1 ,P )]: If N( 1 ,P ) > 2, we must have P (1) = P (1) = 0, and therefore
n (k +1)u {=± 0}. Onbviously,{th}is cnannot happen if all u nare equan′l and not zero. Then,
k=0 π(k) k
Corollary 5 shows that this event has probability at most C/n. Therefore, E[N( 1 ,P )] 6 C
n
P { }
since the root at 1 has multiplicity at most n.
Now we turn to the root at 1. If N( 1 ,P ) 2, then P ( 1) = P ( 1) = 0 and hence,
n+1( 1)kkλ = 0. Using e−xchangeab{i−lity}, thne p≥robability onf−this evenn′t−is the same as the
k=1 − k−1
probability of
P
π(k)λ = σ(k)λ (7)
k 1 k 1
− −
kX∈En kX∈On
where π and σ are uniform random permutations of E = 2Z 1,2,...,n + 1 and O =
n n
∩ { }
(2Z+1) 1,2...,n+1 respectively, and π,σ are independent of each other and of λ ,...,λ .
0 n
∩{ }
Fix the values of λ ,...,λ and consider three cases.
0 n
Case 1: Suppose λ , k E are not all equal. In this case, fix σ (i.e., condition on σ) so that
k 1 n
− ∈
the right hand side of (7) is not random anymore. We may also write π(k) = 2π (k/2) where
′
π is a uniform random permutation of 1E = 1,2,..., n+1 . Apply Corollary 5 to π and
′ 2 n { ⌊ 2 ⌋} ′
conclude that the probability of the event in (7) is at most C/n.
Case 2: Suppose λ , k E are all equal but λ , k O are not all equal. Then
k 1 n k 1 n
− ∈ − ∈
we fix π and write σ(k) = 2σ ((k 1)/2) + 1 where σ is a uniform random permutation of
′ ′
−
1(O 1) = 0,1,..., n 1 . Apply Corollary 5 to σ and conclude that the probability of (7)
2 n− { ⌈ −2 ⌉} ′
is at most C/n.
3 Lemma 4 yields Lemma 3 5
Case 3: Suppose λ = A for all k E and λ = B for all k O . If A = B, then
k 1 n k 1 n
P is a non-zero mul−tiple of 1 + t + t2∈+ ... + tn (r−ecall that, by ass∈umption, all λ do not
n k
vanish simultaneously) and N( 1 ,P ) 6 1. Hence, we assume that A = B. In this case, let
n
{− } 6
τ be a uniform random permutation in 0,1,...,n and let λ = λ so that λ has the same
{ } ′k τ(k) ′
distribution as λ. The probability that λ are equal for all k E and equal for all k O is
′k ∈ n ∈ n
smaller than e cn for some c > 0. Outside this event of negligible probability, λ will fall into
− ′
one of the two cases considered above.
Thus, in all cases, P N( 1 ,P ) 2 6 C/n and hence E[N( 1 ,P )] 6C.
n n
{ {− } ≥ } {− }
Insummary,wehave shownthatthefirsttwotermsontherighthandsideof (4)arebounded
by Clogn and that the last two terms are bounded by C. Thus, E[N (P )] 6 Clogn. ✷
∗ n
Remark: The idea of employing the sign-changes of the Taylor series of the function (1
−
x) 1P (x) was used already in the pioneering paper of Bloch-P´olya [1] and then discussed by
− n
Kac [4]. Combining this idea with the classical Kolmogorov-Rogozin inequality for the concen-
tration function (see, for instance, [2]), one can get a cruder form of Theorem 1 with √n in place
of logn.
3 Lemma 4 yields Lemma 3
The proof of Lemma 3 is based on randomization over permutations acting on (ξ , ... ,ξ )
1 k
combined with estimate (3) in Lemma 4. Throughout, we say that π S acts on the tuple
k
∈
ξ = (ξ ,...,ξ ) by setting (πξ) = ξ ; we define similarly the action of π on functions of ξ.
1 k j π(j)
The proof of the first estimate in Lemma 3 employs the full permutation group S which keeps
k
invariant the joint distribution of the sums k ξ and k jξ . This is no longer possible
j=1 j j=1 j
when dealing with the sums k ( 1)jξ , k ( 1)jjξ , and we are forced to use subgroups of
j=1 − j Pj=1 − j P
the permutation group S . Which subgroup to use depends on whether k is odd or even, and
k
P P
we distinguish between these cases in what follows.
Theproof of the firstestimate (1) in Lemma 3 is significantly simpler than that of the second
estimate (2). The reader interested only in the case of symmetrically distributed i.i.d.s may skip
the proof of (2), which is contained in Sections 3.3 and 3.4.
3.1 A corollary to Lemma 4
We start with a straightforward corollary to Lemma 4, which may be interesting on its own.
Lemma 7. Let n > 2 and let u ,...,u be real numbers, not all equal to zero. Let π be a uniform
1 n
random permutation in S . Then
n
n n
C
P u π(i) 6 u 6 .
i i
n
n(cid:12)Xi=1 (cid:12) (cid:12)Xi=1 (cid:12)o
(cid:12) (cid:12) (cid:12) (cid:12)
Proof of Lemma 7: Ifu sareallequ(cid:12)al(andhen(cid:12)cen(cid:12)on-zer(cid:12)o),thentheprobabilityinthestatement
i
is zero and there is nothing to prove. Otherwise, write u = u¯+σw , where u¯ is the mean of
i i
u ,...,u , and σ2 = n (u u¯)2. Then, w ,...,w satisfy the hypotheses of Lemma 4.
1 n i=1 i − 1 n
Assume without loss of generality that u¯ > 0. We want to get a bound on
P
n n n
n(n+1)u¯ u¯
P u π(i) 6 u = P w π(i)+ 6 n
i i i
2 σ σ
n(cid:12)Xi=1 (cid:12) (cid:12)Xi=1 (cid:12)o n(cid:12)Xi=1 (cid:12) o
(cid:12) (cid:12) (cid:12) (cid:12) (cid:12)n (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12) (cid:12) n(n 1)u¯(cid:12)n(n+3)u¯
= P w π(i) − , .
i
∈ 2 σ 2 σ
nXi=1 h io
3 Lemma 4 yields Lemma 3 6
We cover the interval
n(n 1) u¯ n(n+3) u¯
− ,
2 σ 2 σ
(cid:20) (cid:21)
by nu¯ intervals of length 2, and apply (3) to each subinterval (the value of L is different for
⌈ σ ⌉
different intervals, but, in any case, L > c nu¯). We get
| | σ
n n
C nu¯
P uiπ(i) 6 ui 6 e−cnσu¯ . (8)
n σ
n(cid:12)Xi=1 (cid:12) (cid:12)Xi=1 (cid:12)o l m
(cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
Note that x e cx is bounded by 2 1 on [0, ) (for x 6 1 the bound is 1 while for x > 1 we
⌈ ⌉ − c ∨ ∞
bound x by 2x and use that max te t 6 1). Thus
−
⌈ ⌉ t>0
C nu¯ C 2 C
e−cnσu¯ 6 1 6 ′ .
n σ n c ∨ n
(cid:18) (cid:19)
l m
This completes the proof of Lemma 7. ✷
3.2 The first estimate in Lemma 3
Here, we use Lemma 7 to deduce (1). Let A be the set ξ ,...,ξ (note that ξ are distinct
1 k k
{ }
and non-zero with probability 1). Conditional on A= u ,...,u , the tuple (ξ ,...,ξ ) has the
1 k 1 k
{ }
same distribution as (u ,...,u ), where π is uniformly distributed in S . Lemma 7 applies
π(1) π(k) k
(as u i has the same distribution as u π(i)) to show that
i π(i) i i
P P
k k
C
P jξ 6 ξ A= u ,...,u 6 .
j j 1 k
{ } k
n(cid:12)Xj=1 (cid:12) (cid:12)Xj=1 (cid:12)(cid:12) o
(cid:12) (cid:12) (cid:12) (cid:12)(cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)(cid:12)
Since this holds for every realization of A, we get
k k
C
P jξ 6 ξ 6 ,
j j
k
n(cid:12)Xj=1 (cid:12) (cid:12)Xj=1 (cid:12)o
(cid:12) (cid:12) (cid:12) (cid:12)
completing the proof. (cid:12) (cid:12) (cid:12) (cid:12) ✷
3.3 The second estimate in Lemma 3, the odd case
Let k = 2m 1 and define
−
m 1 m
−
S = ξ , S = ξ , S = S S .
e 2j o 2j 1 e o
− −
j=1 j=1
X X
Set ξ = ξ 1 S , ξ = ξ 1S ,
2′j 2j − m−1 e 2′j−1 2j−1− m o
m 1 m
−
T = jξ , T = jξ , T = T T .
e 2′j o 2′j−1 e− o
j=1 j=1
X X
Then
2m 1
−
( 1)jξ = S S = S,
j e o
− −
j=1
X
3 Lemma 4 yields Lemma 3 7
and
2m 1 m 1 m m
− −
( 1)jjξ = 2 jξ 2 jξ + ξ
j 2j 2j 1 2j 1
− − − −
j=1 j=1 j=1 j=1
X X X X
m 1 m
− 1 m(m 1) 1 m(m+1)
= 2 jξ2′j +2 m 1Se· 2− −2 jξ2′j−1−2m So· 2 +So
j=1 − j=1
X X
= 2T 2T +mS mS
e o e o
− −
m
= 2 T + S .
2
h i
Thus, we aim at proving the existence of a numerical constant C so that
m 1 C
P T + S 6 S 6 . (9)
| 2 | 2 | | m
n o
3.3.1 The subgroups Se and So of S
k k k
Let Se denote the subgroup of S that includes those permutations that involve only the even
k k
indices 2j m 1. Similarly, let So denote the subgroup of S that includes those permutations
{ }j=−1 k k
that involve only the odd indices 2j 1 m . Let S denote the subgroup of S consisting of
{ − }j=1 k k
permutations π = π π where π Se and π So (S is a subgroup because the subgroups
Se and So commutee◦). oNote thaet ∈S aknd S ,oa∈ndbhkencek S, are invariant under the action of
k k e o
any π S on ξ; in particular, π = π π S acts obn ξ = ξ k again as a permutation.
∈ k e◦ o ∈ k ′ { j′}j=1
Instead of considering π drawn uniformly from S , we consider π and π drawn uniformly from
k e o
Se and Sbo, respectively, and take π = π π , wbritingTπ = π T , Tπ = π T and Tπ = Tπ Tπ.
k k e◦ o o o o e e e e − o
Proving (9) then reduces to proving that for any fixed tuple ξ with distinct entries,
m 1 C
Pξ Tπ + S 6 S 6 , (10)
| 2 | 2| | m
n o
where Pξ denotes averaging with respect to the product of uniform measures on Se and So,
k k
with ξ fixed1. Henceforth, we assume that S = 0, which happens almost surely because of the
6
assumption that (ξ ,...,ξ ) has an absolutely continuous distribution.
1 k
3.3.2 Proof of estimate (2) in the case k = 2m 1
−
Introduce the partition of R determined by
I = [(j 1/2)S,(j +1/2)S), j Z.
j
− ∈
1 Here,aswellasinthecasek=2m,weusethefollowing observation. Letf: Rk →RbeaBorel function(in
thecurrentinstance, f = 1|S|−|T + mS|) andlet ξ=(ξ ,...,ξ ) beavectorof exchangeablerandom variables.
2 2 1 k
Then, for any subset S⊂S and any probability distribution P on S, one has
e k e e
P(cid:8)f(ξ)≥0(cid:9)6 sup Pe(cid:8)σ∈Se: f(xσ1,...,xσk)≥0(cid:9).
x1,...,xk
3 Lemma 4 yields Lemma 3 8
We have then
m 1 m 1
Pξ Tπ + S 6 S = Pξ Tπ Tπ + S 6 S
2 2| | e − o 2 2 | |
n(cid:12) (cid:12) o n(cid:12) (cid:12) o
(cid:12) (cid:12) (cid:12) (cid:12)
6 Pξ Tπ I ,Tπ I
e ∈ je o ∈ jo
|je−jjeXo,j+o∈m2Z|64 (cid:8) (cid:9)
= Pξ Tπ I Pξ Tπ I .
e ∈ je o ∈ jo
|je−jjeXo,j+o∈m2Z|64 (cid:8) (cid:9) (cid:8) (cid:9)
Note that in the range of summation of the last expression we have that max(j , j ) > m/5 if
o e
| | | |
m > 40. Assuming this is the case, and using that
Pξ Tπ I = Pξ Tπ I = 1,
e ∈ je o ∈ jo
Xje (cid:8) (cid:9) Xjo (cid:8) (cid:9)
we obtain that
m 1
Pξ Tπ + S 6 S 6 sup Pξ Tπ I + sup Pξ Tπ I . (11)
| 2 | 2| | e ∈ j o ∈ j
j>m/5 j>m/5
n o | | | |
(cid:8) (cid:9) (cid:8) (cid:9)
We now apply estimate (3) in Lemma 4. The argument is the same for either T or T , so for
e o
concreteness set
m 1
−
β2 = (ξ )2
2′j
j=1
X
and consider the term in the right-hand side of (11) involving T . Set w = ξ /β so that
e j 2′j
w = 0 and w2 = 1. Then, for any j,
j j j j
P P
m 1
− jS S
Pξ Tπ I 6 P w π (j) 6 | | .
e ∈ j j e − β β
(cid:8) (cid:9) n(cid:12)(cid:12)Xj=1 (cid:12)(cid:12) o
(cid:12) (cid:12)
By Lemma 4, the RHS does not exceed C e−cβj|mS| |S| . Since we are interested only in js with
m β
|j| ≥ m/5, the latter expression is bounded by mC(cid:6)e−c|(cid:7)βS| |Sβ| . Since ⌈x⌉e−cx is bounded above,
we get the bound C/m.
(cid:6) (cid:7)
The same argument applies with Tπ replacing Tπ. Substituting in (11) completes the proof
o e
of the lemma when k is odd.
3.4 The second estimate in Lemma 3, the even case
Let k = 2m. We need to show that
2m 2m
C
P ( 1)jjξ 6 ( 1)jξ 6 .
j j
− − m
n(cid:12)Xj=1 (cid:12) (cid:12)Xj=1 (cid:12)o
(cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
Put
2m m
S = ( 1)jξ = (ξ ξ )
j 2j 2j 1
− − −
j=1 j=1
X X
3 Lemma 4 yields Lemma 3 9
andsetη = ξ ξ ,notingthatS = m η . Asintheoddcase,wecenterη byintroducing
j 2j− 2j−1 j=1 j j
η = η 1S. Then
j′ j − m P
2m m m m m
( 1)jjξ = (2j)ξ (2j 1)ξ = 2 jη + ξ
j 2j 2j 1 j 2j 1
− − − − −
j=1 j=1 j=1 j=1 j=1
X X X X X
m m m
= 2 jη +(m+1)S + ξ = 2 jη +mΛ
j′ 2j 1 j′
−
j=1 j=1 j=1
X X X
with
m
1 1
Λ= 1+ S + ξ . (12)
2j 1
m m −
(cid:16) (cid:17) Xj=1
Thus, we need to estimate
m
P 2 jη +mΛ 6 S .
j′
n(cid:12) Xj=1 (cid:12) (cid:12) (cid:12)o
(cid:12) (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
3.4.1 A randomization over local and global permutations
We introduce two subgroups of S . The first, which we refer to as local permutations, swaps the
k
entriesofthepairs(ξ ,ξ ). Thissubgroupisgeneratedbythetranspositionsτ ,j = 1,...,m,
2j 1 2j j
which map (2j 1,2j)− (2j,2j 1). Writing Sloc for the subgroup of local permutations, we
− → − k
note that Sloc = 2m.
| k |
The second subgroup of S that we employ, which we refer to as global permutations, swaps
k
the whole pairs. This subgroup is generated by the permutations θ , 1 6 j < j 6 m, which
jj ′
′
map
(1,2,...,2j 1,2j,...,2j 1,2j ,...,k 1,k) (1,2,...,2j 1,2j ,...,2j 1,2j,...,k 1,k).
′ ′ ′ ′
− − − 7→ − − −
For instance, if we originally had the tuple (ξ ,ξ ,ξ ,ξ ,ξ ,ξ ), we can get something like
1 2 3 4 5 6
(ξ ,ξ ,ξ ,ξ ,ξ ,ξ ) after some local permutation, and then (ξ ,ξ ,ξ ,ξ ,ξ ,ξ ) after a global
2 1 3 4 6 5 3 4 6 5 2 1
permutation. Writing Sgl for the subgroup of global permutations, we note that Sgl = m!.
k | k |
We will consider in what follows permutations π from S that decompose as π = πg πl with
k
πl Sloc and πg Sgl and randomize over πg and πl with πg and πl independent◦and uni-
∈ k ∈ k
formly distributed over Sgl and Sloc. Note that unlike odd and even permutations considered
k k
in case k is odd, local and global permutations do not in general commute. We note that
The quantities S and Λ are invariant under the action of Sgl. In what follows, the explicit
• k
form (12) of Λ will be irrelevant, only the Sgl-invariance will matter.
k
Random choice of πl is equivalent to placing independent random signs in from of η , ...,
1
•
η .
m
We fix ξ (and hence, η), and denote by Pξ the law of (πg πl)ξ conditioned on ξ. We need
◦
to show that
m
C
Pξ 2 πg(j)(πlη) +mΛ 6 S 6 .
′j πl πl m
n(cid:12) Xj=1 (cid:12) (cid:12) (cid:12)o
(cid:12) (cid:12)
(cid:12) (cid:12)
Here, we put S = πlS, Λ(cid:12) = πlΛ, and, abusing no(cid:12)tation, we denote by πg(j) the image of
πl πl
j 1,2, ... ,m under πg viewed as a permutation on m letters.
∈{ }
3 Lemma 4 yields Lemma 3 10
3.4.2 Good and bad local permutations
Put
m
Λ
β2 = (πlη) 2, L = πl .
πl ′j πl 2β
πl
j=1
X(cid:0) (cid:1)
We need to estimate
m
1 S
Pξ β πg(j)(πlη)′j +mLπl 6 |2βπl|
πl πl
n(cid:12) Xj=1 (cid:12) o
(cid:12) (cid:12)
We wish to apply about S /(cid:12)β times Lemma 4 with w(cid:12)= (πlη) /β (this is a point where
| πl| πl j ′j πl
we use the randomization over Sgl). An obstacle is that, for some πl Sloc, the quantity β
k ∈ k πl
can be much smaller than S . The randomization over Sloc will help us to circumvent this
| πl| k
obstacle.
Put
m
B2 = η2,
j
j=1
X
and note that this quantity is both Sgl- and Sloc-invariant. The next two claims show that
k k
outside of a tiny part of all local permutations, β is comparable with B.
πl
Given a tuple η = (η ,...,η ) we call a permutation πl Sloc good if
1 m ∈ k
1 3
m 6 j :(πlη) >0 6 m
j
4 |{ }| 4
and bad otherwise. Let Sloc,good denote the subset of Sloc consisting of good permutations.
k k
Claim 8. For all πl Sloc,good, it holds that 1B2 6 β2 6 B2.
∈ k 5 πl
Proof of Claim 8. Since
m m m
1 1 1
β2 = (πlη) 2 = (πlη) S 2 = η2 S2 = B2 S2 ,
πl ′j j − m πl j − m πl − m πl
j=1 j=1 j=1
X X(cid:0) (cid:1) X
the upper bound β2 6 B2 is immediate. It remains to prove the claimed lower bound. Note
πl
however that
m
1 1
β2 = (πlη) S 2 > S 2.
πl j − m πl m πl
Xj=1(cid:0) (cid:1) j: sgn((πlηX)j)=−sgn(Sπl)(cid:0) (cid:1)
Since πl Sloc,good, the sum on the right-side is over at least 1m terms, and thus we obtain
∈ k 4
that
1 1 1
β2 > m S 2 = (B2 β2 ).
πl 4 m πl 4 − πl
The claim follows. (cid:0) (cid:1)
Thus, for good local permutations πl, the parameter β2 is controlled by B2. The next claim
πl
asserts that most of local permutations are good:
Claim 9. Let (ξ ,...,ξ ) be an exchangeable random vector having a distribution that is abso-
1 2m
lutely continuous to Lebesgue measure on Rk. Let η = ξ ξ , 16 j 6 m. Then
j 2j 2j 1
− −
P πl Sloc,good 6 e cm.
6∈ k −
(cid:8) (cid:9)
