Maximal inequalities for centered norms of sums of independent random vectors

0.12 MB·English

by  Rafał Latała

Maximal inequalities for centered norms of sums
of independent random vectors
5
1
0
Rafal Latal a∗
2

n
a
J
4 Abstract
k
] LetX1,X2,...,XnbeindependentrandomvariablesandSk =Pi=1Xi.
R We show that for any constants ak,
P P( max ||Sk|−ak|>11t)≤30 max P(||Sk|−ak|>t).
. 1≤k≤n 1≤k≤n
h
t Wealso discusssimilar inequalities forsumsofHilbertandBanach space
a
valued random vectors.
m
[
1 Introduction and Main Results

1
v
8 Let X1,X2,... be independent random vectors in a separable Banach space F.
9 TheL´evy-Ottavianimaximalinequality(seee.g. Proposition1.1.1in[2])states
6 that for any t>0,
0
0 P max S >3t 3 max P( S >t), (1.1)
1. (cid:16)1≤k≤nk kk (cid:17)≤ 1≤k≤n k kk
0
where here and in the rest of this note,
5
1
k
:
v S = X for k=1,2,....
k i
Xi Xi=1
r If, additionally, variables X are symmetric then the classical L´evy inequality
i
a
gives the sharper bound
P max S >t 2P( S >t).
k n
(cid:16)1≤k≤nk k (cid:17)≤ k k
Montgomery-Smith[4] showedthat if we replace symmetry assumptions by the
identical distribution then
P max S >C t C P( S >t), (1.2)
k 1 2 n
(cid:16)1≤k≤nk k (cid:17)≤ k k
where one may take C =30 and C =9.
1 2
∗ResearchpartiallysupportedbyMNiSWGrantno. NN201397437.
1
Maximal inequalities are fundamental tools in the study of convergence of
random series and limit theorems for sums of independent random vectors (see
e.g. [2] and [3]).
In some applications one needs to investigate asymptotic behaviour of cen-
terednorms ofsums, i.e. randomvariables ofthe form( S a )/b (cf. [1]).
n n n
k k−
For such purpose it is natural to ask whether in (1.1) or (1.2) one may replace
variables S by S a . The answer turns out to be positive in the real
k k k
k k |k k− |
case.
Theorem 1.1. Let X ,X ,...,X be independent real r.v.’s. Then for any
1 2 n
numbers a ,a ,...,a and t>0,
1 2 n
P max S a >11t 30 max P( S a >t). (1.3)
k k k k
(cid:16)1≤k≤n|| |− | (cid:17)≤ 1≤k≤n || |− |
Example. Let Y ,Y ,... be i.i.d. r.v.’s such that EY2 =1 and Var(Y2)< .
1 2 i i ∞
Let S = k X , where X = e Y and (e ) is an orthonormal system in a
k i=1 i i i i i
HilbertspaPce ;alsolet x denotethe normofavectorx . Thenfort>0,
H | | ∈H
Var(S 2) Var(Y2)
P( S √k t) P( S 2 k t√k) | k| = 1 .
|| k|− |≥ ≤ || k| − |≥ ≤ t2k t2
On the other hand if we choose j such that 2j0/2 t, then for n 2j0,
0
≥ ≥
p :=P max S √k t P max (S 2 k) 3t√k
n k k
(cid:16)1≤k≤n|| |− |≥ (cid:17)≥ (cid:16)2j0≤k≤n | | − ≥ (cid:17)
P S 2 2j 3 2j/2t
2j
≥ (cid:18)j0≤j[≤log2nn| | − ≥ · o(cid:19)
2j
P 2−j/2 (Y2 1) 6t
≥ (cid:18)j0+1≤[j≤log2nn i=2Xj−1+1 i − ≥ o(cid:19)
and lim p = 1 for any t > 0 by the CLT. It is not hard to modify this
n→∞ n
example in such a way that X be an i.i.d. sequence.
i
HenceTheorem1.1doesnotholdininfinitedimensionalHilbertspaceseven
if we assume that X are symmetric and identically distributed. However a
i
modification of (1.3) is satisfied in Hilbert spaces.
Proposition 1.2. Let X ,...,X be independent symmetric r.v.’s with values
1 n
in a separable Hilbert space ( , ). Then for any sequence of real numbers
H | |
a ,...,a and t 0,
1 n
≥
P max S 2 a 3t 6 max P S 2 a t .
k k k k
(cid:16)1≤k≤n(cid:12)| | − (cid:12)≥ (cid:17)≤ 1≤k≤n (cid:0)(cid:12)| | − (cid:12)≥ (cid:1)
(cid:12) (cid:12) (cid:12) (cid:12)
A first consequence of Proposition1.2 is the following Hilbert-space version
of (1.3) under a regularity assumption on coefficients (a ).
k
2
Corollary 1.3. Let X ,...,X be as in Proposition 1.2, 1 i n and non-
1 n
≤ ≤
negative real numbers a ,...,a , α, β and t satisfy the condition
i n
a αa +βt for all i k,l n. (1.4)
k l
≤ ≤ ≤
Then
P max S a (6α+2β+1)t 6 max P S a t .
k k k k
(cid:16)i≤k≤n|| |− |≥ (cid:17)≤ i≤k≤n || |− |≥
(cid:0) (cid:1)
In proofs of limit theorems one typically applies maximal inequalities to
uniformly estimate S for cn k n, where c is some constant. Next two
k
k k ≤ ≤
corollariesshowthatifwerestrictk tosuchagroupofindices then, under i.i.d.
and symmetry assumptions, (1.3) holds in Hilbert spaces.
Corollary 1.4. Let X ,X ,...,X be symmetric i.i.d. r.v.’s with values in a
1 2 n
separable Hilbert space ( , ). Then for any integer i such that n i n and
H | | 2 ≤ ≤
any sequence of positive numbers a ,...,a and t 0 we have
i n
≥
P max S a 19t 6 max P( S a t).
k k k k
(cid:16)i≤k≤n|| |− |≥ (cid:17)≤ i≤k≤n || |− |≥
Proof. We may obviously assume that
1
max P( S a t) .
k k
i≤k≤n || |− |≥ ≤ 6
Observe that for any k <l, the random variable S := l X has the same
k,l i=k i
distribution as Sl−k+1. P
Take k,l i,...,n , then
∈{ }
P(S 2a +2t) P(S a +t)+P(S a +t)
2k k k k k+1,2k k
| |≥ ≤ | |≥ | |≥
1
=2P(S a +t) .
k k
| |≥ ≤ 3
Therefore
P(a t S 2a +2t)
l l k
− ≤| |≤
P(a t S , S +S 2a +2t, S S 2a +2t)
l l l l+1,2k k l l+1,2k k
≥ − ≤| | | |≤ | − |≤
1 2
1 P(S <a t) 2P(S >2a +2t) 1 >0,
l l 2k k
≥ − | | − − | | ≥ − 6 − 3
where in the second inequality we used the symmetry of X . Hence we get
i
a 2a +3t and we may apply Corollary 1.3 with α=2 and β =3.
l k
≤
Corollary 1.5. Let X ,X ,...,X be as before. Then for any n i n and
1 2 n 2j ≤ ≤
any sequence of positive numbers a ,...,a and t 0 we have
i n
≥
P max S a 19t 6j max P( S a t).
k k k k
(cid:16)i≤k≤n|| |− |≥ (cid:17)≤ i≤k≤n || |− |≥
3
Corollary 1.4 naturally leads to the formulation of the following open ques-
tion.
Question. CharacterizeallseparableBanachspaces(E, )withthefollowing
kk
property. There exist constants C ,C < such that for any symmetric i.i.d.
1 2
∞
r.v.’s X ,X ,...,X with values in E, any n i n, any positive constants
1 2 n 2 ≤ ≤
a ,...,a and t>0,
i n
P max S a C t) C max P( S a t). (1.5)
k k 1 2 k k
(cid:16)i≤k≤n|k k− |≥ ≤ i≤k≤n |k k− |≥
In particular does the above inequality hold in Lp with 1<p< ?
∞
In the last section of the paper we present an example showing that in a
general separable Banach space estimate (1.5) does not hold.
2 Proofs
Below we will use the following notation. By X˜ ,X˜ ,... we will denote the
1 2
independent copy of the random sequence X ,X ,.... We put
1 2
k n
S˜ := X˜ , S :=S S = X .
k i k,n n k−1 i
−
Xi=1 Xi=k
We start with the following simple lemma.
Lemma 2.1. Suppose that real numbers x, y, a, b and u satisfy the conditions
x a u, y a u, x+s b u, y+s b u and x y >2u.
|| |− |≤ || |− |≤ || |− |≤ || |− |≤ | − |
Then a b 2u and s 4u.
| − |≤ | |≤
Proof. Ifa<0 then x, y <uand x y <2u. So a 0andinthe sameway
| | | | | − | ≥
we show that b 0. Without loss of generality we may assume x < y, hence
≥
x ( a u, a+u),y (a u,a+u),x+s ( b u, b+u),y+s (b u,b+u).
∈ − − − ∈ − ∈ − − − ∈ −
Thus 2a 2u y x 2a+2u and 2b 2u (y+s) (x+s) 2b+2u and
− ≤ − ≤ − ≤ − ≤
therefore a b 2u. Moreover, b+a 2u s b+a+2u and we get
| − | ≤ − − ≤ ≤ −
s a b +2u 4u.
| |≤| − | ≤
Proof of Theorem 1.1. We may and will assume that
p:= max P( S a >t) (0,1/30).
k k
1≤k≤n || |− | ∈
Let
I := k: a 2t , I := k: P(S S˜ >2t)>5p
1 k 2 k k
{ ≤ } { | − | }
and
I := 1,...,n (I I ).
3 1 2
{ }\ ∪
First we show that
P max S a >11t 3p. (2.1)
k k
(cid:16)k∈I1|| |− | (cid:17)≤
4
Indeed, notice that for all k, a > t (otherwise p = 1). Therefore by the
k
−
L´evy-Ottavianiinequality (1.1),
P max S a >11t P(max S >9t) 3maxP(S >3t)
k k k k
(cid:16)k∈I1|| |− | (cid:17)≤ k∈I1| | ≤ k∈I1 | |
3maxP( S a >t) 3p.
k k
≤ k∈I1 || |− | ≤
Next we prove that
P max S a >11t 5p. (2.2)
k k
(cid:16)k∈I2|| |− | (cid:17)≤
Let us take k I and define the following events
2
∈
A := S S˜ >2t , A :=A S >4t
1 k k 2 1 k+1,n
{| − | } ∩{| | }
and
B := S a t, S˜ a t, S a t, S˜ +S a t .
k k k k n n k k+1,n n
{|| |− |≤ || |− |≤ || |− |≤ || |− |≤ }
We have P(A )+P(B) > 5p+1 4p > 1, hence A B = and by Lemma
1 1
2.1, a a 2t. Also by Lemm−a 2.1, A B = , h∩ence6 P(∅A )+P(B) 1.
k n 2 2
Ther|efor−e 5p|P≤(S >4t) P(A ) 4p.∩Thus f∅or all k I , a a ≤2t
k+1,n 2 2 k n
and P(S | 4t) | 1/5. L≤et ≤ ∈ | − |≤
k+1,n
| |≤ ≥
τ :=inf k I : S a >11t .
2 k k
{ ∈ || |− | }
Then
1
P(τ =k) P(τ =k, S 4t)
k+1,n
5 ≤ | |≤
P(τ =k, S a >11t 4t a a )
n n k n
≤ || |− | − −| − |
P(τ =k, S a >t)
n n
≤ || |− |
and
P max S a >11t = P(τ =k) 5 P(τ =k, S a >t)
k k n n
(cid:16)k∈I2|| |− | (cid:17) kX∈I2 ≤ kX∈I2 || |− |
5P( S a >t) 5p.
n n
≤ || |− | ≤
Finally we show
P max S a >11t 21p. (2.3)
k k
(cid:16)k∈I3|| |− | (cid:17)≤
To this end take any k I and notice that
3
∈
2max P(S a t),P(S +a t)
k k k k
{ | − |≤ | |≤ }
P(S a t)+P(S +a t)
k k k k
≥ | − |≤ | |≤
29
P( S a t) 1 p .
k k
≥ || |− |≤ ≥ − ≥ 30
5
If x a t and y+a t then x y 2a 2t>2t. Therefore
k k k
| − |≤ | |≤ | − |≥ −
5p P(S S˜ >2t)
k k
≥ | − |
P(S a t, S˜ +a t)+P(S +a t, S˜ a t)
k k k k k k k k
≥ | − |≤ | |≤ | |≤ | − |≤
=2P(S a t)P(S +a t).
k k k k
| − |≤ | |≤
So for any k I we may choose b = a such that
3 k k
∈ ±
30
P(S b t) 5p 6p.
k k
| − |≤ ≤ 29 ≤
Therefore
P(S +b >t) P( S a >t)+P(S b t) 7p
k k k k k k
| | ≤ || |− | | − |≤ ≤
and by the L´evy-Ottavianiinequality (1.1),
P(max S a >11t) P(max S +b >11t)
k k k k
k∈I3|| |− | ≤ k∈I3| |
11
3maxP(S +b > t) 21p.
k k
≤ k∈I3 | | 3 ≤
This shows (2.3).
Inequalities (2.1), (2.2) and (2.3) imply (1.3).
Proof of Proposition 1.2. It is enough to consider the case when
1
p:= max P S 2 a t < .
k k
1≤k≤n | | − ≥ 6
(cid:0)(cid:12) (cid:12) (cid:1)
(cid:12) (cid:12)
Notice that
P S 2 S 2 (a a ) 2t P S 2 a t +P S 2 a t
n k n k n n k k
| | −| | − − ≥ ≤ | | − ≥ | | − ≥
(cid:0)(cid:12) (cid:12) (cid:1) (cid:0)(cid:12) (cid:12) (cid:1) (cid:0)(cid:12) (cid:12) (cid:1)
Th(cid:12)erefore (cid:12) (cid:12) (cid:12) (cid:12) (cid:12)
P S 2+2 S ,S (a a ) 2t) 2p
k+1,n k k+1,n n k
| | h i− − ≥ ≤
(cid:0)(cid:12) (cid:12)
and by the symm(cid:12) etry (cid:12)
P S 2 2 S ,S (a a ) 2t 2p.
k+1,n k k+1,n n k
| | − h i− − ≥ ≤
(cid:0)(cid:12) (cid:12) (cid:1)
Thus by the tria(cid:12)ngle inequality (cid:12)
P S 2 (a a ) 2t 4p.
k+1,n n k
| | − − ≥ ≤
(cid:0)(cid:12) (cid:12) (cid:1)
Now let x be su(cid:12)ch that x2 a (cid:12)3t then by the triangle inequality
k
∈H || | − |≥
and symmetry
1 4p P x2+ S 2 a t
k+1,n n
− ≤ | | | | − ≥
P(cid:0)(cid:12)(cid:12)x2+ Sk+1,n 2+2 x(cid:12)(cid:12),Sk+(cid:1)1,n an t
≤ | | | | h i− ≥
+(cid:0)P(cid:12)(cid:12) x2+ Sk+1,n 2 2 x,Sk+1,n a(cid:12)(cid:12)n (cid:1)t
| | | | − h i− ≥
=2P(cid:0)(cid:12)(cid:12)x2+ Sk+1,n 2+2 x,Sk+1,n an(cid:12)(cid:12) t(cid:1)
| | | | h i− ≥
=2P(cid:0)(cid:12)(cid:12)x+Sk+1,n 2 an t . (cid:12)(cid:12) (cid:1)
| | − ≥
(cid:0)(cid:12) (cid:12) (cid:1)
(cid:12) (cid:12)
6
So for any x and k =1,2,...,n,
∈H
1 1
x2 a 3t P x+S 2 a t (1 4p) . (2.4)
k k+1,n n
| | − ≥ ⇒ | | − ≥ ≥ 2 − ≥ 6
(cid:12) (cid:12) (cid:0)(cid:12) (cid:12) (cid:1)
(cid:12) (cid:12) (cid:12) (cid:12)
Now let
τ :=inf k n: S 2 a 3t ,
k k
≤ | | − ≥
(cid:8) (cid:12) (cid:12) (cid:9)
then since τ =k σ(X1,...,Xk) we(cid:12)get by (2.4(cid:12)),
{ }∈
1
P τ =k, S 2 a t P(τ =k).
n n
| | − ≥ ≥ 6
(cid:0) (cid:12) (cid:12) (cid:1)
(cid:12) (cid:12)
Hence
n
1 1
P S 2 a t P(τ =k)= P max S 2 a 3t
n n k k
(cid:0)(cid:12)| | − (cid:12)≥ (cid:1)≥ 6Xk=1 6 (cid:16)1≤k≤n(cid:12)| | − (cid:12)≥ (cid:17)
(cid:12) (cid:12) (cid:12) (cid:12)
and the proposition follows.
Proof of Corollary 1.3. We may consider variables S ,X ,...,X instead of
i i+1 n
X ,...,X and assume that i=1. Let a:=min a . We will analyze two
1 n 1≤k≤n k
cases.
Case 1. a 3t. Then by (1.4) we get a (3α+β)t for all k. Thus by the
k
≤ ≤
L´evy inequality,
P max S a (6α+2β+1)t P max S (3α+β+1)t
k k k
(cid:16) k || |− |≥ (cid:17)≤ (cid:16) k | |≥ (cid:17)
2P(S (3α+β+1)t)
n
≤ | |≥
2P( S a t)
n n
≤ || |− |≥
2maxP( S a t).
k k
≤ k || |− |≥
Case 2. a 3t. Notice first that for any s>0 we have
≥
S 2 a2 s(2a +s) S a s S 2 a2 sa . (2.5)
{|| k| − k|≥ k }⊂{|| k|− k|≥ }⊂{|| k| − k|≥ k}
Indeed, the last inclusion follows since S 2 a2 = (S +a ) S a
|| k| − k| | k| k || k|− k| ≥
a S a . To see the first inclusion in (2.5) observe that
k k k
|| |− |
S 2 a2 s(2a +s) S a s S +a 2a +s
{|| k| − k|≥ k }⊂{|| k|− k|≥ }∪{| k| k ≥ k }
S a s .
k k
⊂{|| |− |≥ }
Now by (2.5) we get
P max S a (6α+2β+1)t
k k
(cid:16) k || |− |≥ (cid:17)
P max S 2 a2 (6α+2β+1)at .
≤ (cid:16) k || k| − k|≥ (cid:17)
7
Hence by Proposition 1.2,
1
P max S a (6α+2β+1)t 6maxP S 2 a2 (6α+2β+1)at .
(cid:16) k || k|− k|≥ (cid:17)≤ k (cid:0)|| k| − k|≥ 3 (cid:1)
But 1(6α+2β+1)a 2(αa+βt)+t 2a +t for all k by (1.4). Thereforeby
3 ≥ ≥ k
(2.5),
P max S a (6α+2β+1)t 6maxP( S 2 a2 t(2a +t))
(cid:16) k || k|− k|≥ (cid:17)≤ k || k| − k|≥ k
6maxP( S a t).
k k
≤ k || |− |≥
3 Example
Let us fix a positive integer n and put
n
I = j Z: j n .
n
n ∈ 2 ≤ ≤ o
Let t = n2+j for j =1,2,...,n, then
j j
jt =n2+j and (j 1)t n2 for j I . (3.1)
j j n
− ≤ ∈
Let N be a large integer (to be fixed later) and let F be the space of all
double-indexed sequences a=(a ) with the norm
i,j 0≤i≤N,j∈In
j
(a ) =max a +t a .
(cid:13)(cid:13) i,j 0≤i≤N,j∈In(cid:13)(cid:13) j∈In(cid:18)| 0,j| j1≤i1<i2X<...<ij≤NXs=1| is,j|(cid:19)
(cid:13) (cid:13)
Let (e ) be a standard basis of F, so that (a )= a e .
i,j i,j i,j i,j i,j
Define random vectors X1,X2,...,Xn by the forPmula
X = (Y e +R e ),
l l,j 0,j l,j Nl,j
jX∈In
where(Y ,R ) and(N ) areindependentr.v’s,P(R = 1)=1/2,
Y are sl,yjmml,ejtrl≤icn,Pj∈(IYn = 1l l)≤n= 1 P(Y = 0) = p (wl,jith ±p a small
l,j | l,j| 2n − k,j n n
positive number to be specified later) and N are uniformly sampled from the
l
set 1,...,N .
{ }
Obviously X ,X ,...,X are i.i.d. and symmetric. As usual we set S =
1 2 n k
X +X +...+X . Let
1 2 k
A= N ,N ,...,N are pairwise distinct .
1 2 n
{ }
Notice that P(Ac) 0 when N . On the set A we have for k n,
→ →∞ ≤
k
S =max Y +t min k,j .
k l,j j
k k j∈In(cid:16)(cid:12)(cid:12)Xl=1 (cid:12)(cid:12) { }(cid:17)
(cid:12) (cid:12)
8
For j >k we have by (3.1),
k
Y +t min k,j <1+t (j 1) n2+1,
l,j j j
(cid:12)(cid:12)Xl=1 (cid:12)(cid:12) { } − ≤
(cid:12) (cid:12)
hence on the set A, for k I we get
n
∈
k k
S = max Y +n2+j = Y +n2+k.
k l,j l,k
k k j∈In,j≤k(cid:16)(cid:12)(cid:12)Xl=1 (cid:12)(cid:12) (cid:17) (cid:12)(cid:12)Xl=1 (cid:12)(cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
Take 0<t< 1 then for k I ,
2nC1 ∈ n
k
P( S (n2+k) t) P(Ac)+P Y =0 P(Ac)+kp
k l,k n
|k k− |≥ ≤ (cid:16)Xl=1 6 (cid:17)≤
and
k
P max S (n2+k) tC P max Y =0 P(Ac).
k 1 l,k
(cid:16)k∈In|k k− |≥ (cid:17)≥ (cid:16)k∈In(cid:12)(cid:12)Xl=1 (cid:12)(cid:12)6 (cid:17)−
(cid:12) (cid:12)
Thelastnumberisofordern2p ifN islargeandp issmall. Thisshowsthatif
n n
(1.5) holdsfor i= n/2 in F thenC mustbe ofordern. So (1.5) cannothold
2
⌈ ⌉
with absolute constants C and C in (infinite dimensional) separable Banach
1 2
spaces.
References
[1] E. Gin´e and D. Mason, The law of the iterated logarithm for the integrated
squared deviation of a kernel density estimator, Bernoulli 10 (2004), 721–
752.
[2] S. Kwapien´ and W.A. Woyczyn´ski Random series and stochastic integrals:
single and multiple, Birkh¨auser, Boston, MA, 1992.
[3] M.LedouxandM.Talagrand,ProbabilityinBanachspaces,Springer-Verlag,
Berlin, 1991.
[4] S. Montgomery-Smith, Comparison of sums of independent identically dis-
tributed random vectors Probab. Math. Statist. 14 (1993), 281–285.
Institute of Mathematics
University of Warsaw
Banacha 2
02-097 Warszawa
Poland
[email protected]
9