Maximal inequalities for centered norms of sums of independent random vectors
aa r X i v : . [ m a t h . P R ] J a n Maximal inequalities for centered norms of sumsof independent random vectors
Rafa l Lata la ∗ Abstract
Let X , X , . . . , X n be independent random variables and S k = P ki =1 X i .We show that for any constants a k , P ( max ≤ k ≤ n || S k | − a k | > t ) ≤
30 max ≤ k ≤ n P ( || S k | − a k | > t ) . We also discuss similar inequalities for sums of Hilbert and Banach spacevalued random vectors.
Let X , X , . . . be independent random vectors in a separable Banach space F .The L´evy-Ottaviani maximal inequality (see e.g. Proposition 1.1.1 in [2]) statesthat for any t > P (cid:16) max ≤ k ≤ n k S k k > t (cid:17) ≤ ≤ k ≤ n P ( k S k k > t ) , (1.1)where here and in the rest of this note, S k = k X i =1 X i for k = 1 , , . . . . If, additionally, variables X i are symmetric then the classical L´evy inequalitygives the sharper bound P (cid:16) max ≤ k ≤ n k S k k > t (cid:17) ≤ P ( k S n k > t ) . Montgomery-Smith [4] showed that if we replace symmetry assumptions by theidentical distribution then P (cid:16) max ≤ k ≤ n k S k k > C t (cid:17) ≤ C P ( k S n k > t ) , (1.2)where one may take C = 30 and C = 9. ∗ Research partially supported by MNiSW Grant no. N N201 397437. k S n k − a n ) /b n (cf. [1]).For such purpose it is natural to ask whether in (1.1) or (1.2) one may replacevariables k S k k by |k S k k − a k | . The answer turns out to be positive in the realcase. Theorem 1.1.
Let X , X , . . . , X n be independent real r.v.’s. Then for anynumbers a , a , . . . , a n and t > , P (cid:16) max ≤ k ≤ n || S k | − a k | > t (cid:17) ≤
30 max ≤ k ≤ n P ( || S k | − a k | > t ) . (1.3) Example.
Let Y , Y , . . . be i.i.d. r.v.’s such that E Y i = 1 and Var( Y i ) < ∞ .Let S k = P ki =1 X i , where X i = e i Y i and ( e i ) is an orthonormal system in aHilbert space H ; also let | x | denote the norm of a vector x ∈ H . Then for t > P ( || S k | − √ k | ≥ t ) ≤ P ( || S k | − k | ≥ t √ k ) ≤ Var( | S k | ) t k = Var( Y ) t . On the other hand if we choose j such that 2 j / ≥ t , then for n ≥ j , p n := P (cid:16) max ≤ k ≤ n || S k | − √ k | ≥ t (cid:17) ≥ P (cid:16) max j ≤ k ≤ n ( | S k | − k ) ≥ t √ k (cid:17) ≥ P (cid:18) [ j ≤ j ≤ log n n | S j | − j ≥ · j/ t o(cid:19) ≥ P (cid:18) [ j +1 ≤ j ≤ log n n − j/ j X i =2 j − +1 ( Y i − ≥ t o(cid:19) and lim n →∞ p n = 1 for any t > X i be an i.i.d. sequence.Hence Theorem 1.1 does not hold in infinite dimensional Hilbert spaces evenif we assume that X i are symmetric and identically distributed. However amodification of (1.3) is satisfied in Hilbert spaces. Proposition 1.2.
Let X , . . . , X n be independent symmetric r.v.’s with valuesin a separable Hilbert space ( H , | | ) . Then for any sequence of real numbers a , . . . , a n and t ≥ , P (cid:16) max ≤ k ≤ n (cid:12)(cid:12) | S k | − a k (cid:12)(cid:12) ≥ t (cid:17) ≤ ≤ k ≤ n P (cid:0)(cid:12)(cid:12) | S k | − a k (cid:12)(cid:12) ≥ t (cid:1) . A first consequence of Proposition 1.2 is the following Hilbert-space versionof (1.3) under a regularity assumption on coefficients ( a k ).2 orollary 1.3. Let X , . . . , X n be as in Proposition 1.2, ≤ i ≤ n and non-negative real numbers a i , . . . , a n , α, β and t satisfy the condition a k ≤ αa l + βt for all i ≤ k, l ≤ n. (1.4) Then P (cid:16) max i ≤ k ≤ n || S k | − a k | ≥ (6 α + 2 β + 1) t (cid:17) ≤ i ≤ k ≤ n P (cid:0) || S k | − a k | ≥ t (cid:1) . In proofs of limit theorems one typically applies maximal inequalities touniformly estimate k S k k for cn ≤ k ≤ n , where c is some constant. Next twocorollaries show that if we restrict k to such a group of indices then, under i.i.d.and symmetry assumptions, (1.3) holds in Hilbert spaces. Corollary 1.4.
Let X , X , . . . , X n be symmetric i.i.d. r.v.’s with values in aseparable Hilbert space ( H , | | ) . Then for any integer i such that n ≤ i ≤ n andany sequence of positive numbers a i , . . . , a n and t ≥ we have P (cid:16) max i ≤ k ≤ n || S k | − a k | ≥ t (cid:17) ≤ i ≤ k ≤ n P ( || S k | − a k | ≥ t ) . Proof.
We may obviously assume thatmax i ≤ k ≤ n P ( || S k | − a k | ≥ t ) ≤ . Observe that for any k < l , the random variable S k,l := P li = k X i has the samedistribution as S l − k +1 .Take k, l ∈ { i, . . . , n } , then P ( | S k | ≥ a k + 2 t ) ≤ P ( | S k | ≥ a k + t ) + P ( | S k +1 , k | ≥ a k + t )= 2 P ( | S k | ≥ a k + t ) ≤ . Therefore P ( a l − t ≤ | S l | ≤ a k + 2 t ) ≥ P ( a l − t ≤ | S l | , | S l + S l +1 , k | ≤ a k + 2 t, | S l − S l +1 , k | ≤ a k + 2 t ) ≥ − P ( | S l | < a l − t ) − P ( | S k | > a k + 2 t ) ≥ − − > , where in the second inequality we used the symmetry of X i . Hence we get a l ≤ a k + 3 t and we may apply Corollary 1.3 with α = 2 and β = 3. Corollary 1.5.
Let X , X , . . . , X n be as before. Then for any n j ≤ i ≤ n andany sequence of positive numbers a i , . . . , a n and t ≥ we have P (cid:16) max i ≤ k ≤ n || S k | − a k | ≥ t (cid:17) ≤ j max i ≤ k ≤ n P ( || S k | − a k | ≥ t ) . Question.
Characterize all separable Banach spaces ( E, k k ) with the followingproperty. There exist constants C , C < ∞ such that for any symmetric i.i.d.r.v.’s X , X , . . . , X n with values in E , any n ≤ i ≤ n , any positive constants a i , . . . , a n and t > P (cid:16) max i ≤ k ≤ n |k S k k − a k | ≥ C t ) ≤ C max i ≤ k ≤ n P ( |k S k k − a k | ≥ t ) . (1.5)In particular does the above inequality hold in L p with 1 < p < ∞ ?In the last section of the paper we present an example showing that in ageneral separable Banach space estimate (1.5) does not hold. Below we will use the following notation. By ˜ X , ˜ X , . . . we will denote theindependent copy of the random sequence X , X , . . . . We put˜ S k := k X i =1 ˜ X i , S k,n := S n − S k − = n X i = k X i . 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 | > u .Then | a − b | ≤ u and | s | ≤ u .Proof. If a < | x | , | y | < u and | x − y | < u . So a ≥ 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 2 a − u ≤ y − x ≤ a + 2 u and 2 b − u ≤ ( y + s ) − ( x + s ) ≤ b + 2 u andtherefore | a − b | ≤ u . Moreover, − b + a − u ≤ s ≤ − b + a + 2 u and we get | s | ≤ | a − b | + 2 u ≤ u . Proof of Theorem 1.1.
We may and will assume that p := max ≤ k ≤ n P ( || S k | − a k | > t ) ∈ (0 , / . Let I := { k : a k ≤ t } , I := { k : P ( | S k − ˜ S k | > t ) > p } and I := { , . . . , n } \ ( I ∪ I ) . First we show that P (cid:16) max k ∈ I || S k | − a k | > t (cid:17) ≤ p. (2.1)4ndeed, notice that for all k , a k > − t (otherwise p = 1). Therefore by theL´evy-Ottaviani inequality (1.1), P (cid:16) max k ∈ I || S k | − a k | > t (cid:17) ≤ P (max k ∈ I | S k | > t ) ≤ k ∈ I P ( | S k | > t ) ≤ k ∈ I P ( || S k | − a k | > t ) ≤ p. Next we prove that P (cid:16) max k ∈ I || S k | − a k | > t (cid:17) ≤ p. (2.2)Let us take k ∈ I and define the following events A := {| S k − ˜ S k | > t } , A := A ∩ {| S k +1 ,n | > t } and B := {|| S k | − a k | ≤ t, || ˜ S k | − a k | ≤ t, || S n | − a n | ≤ t, || ˜ S k + S k +1 ,n | − a n | ≤ t } . We have P ( A ) + P ( B ) > p + 1 − p >
1, hence A ∩ B = ∅ and by Lemma2.1, | a k − a n | ≤ t . Also by Lemma 2.1, A ∩ B = ∅ , hence P ( A ) + P ( B ) ≤ p P ( | S k +1 ,n | > t ) ≤ P ( A ) ≤ p . Thus for all k ∈ I , | a k − a n | ≤ t and P ( | S k +1 ,n | ≤ t ) ≥ /
5. Let τ := inf { k ∈ I : || S k | − a k | > t } . Then 15 P ( τ = k ) ≤ P ( τ = k, | S k +1 ,n | ≤ t ) ≤ P ( τ = k, || S n | − a n | > t − t − | a k − a n | ) ≤ P ( τ = k, || S n | − a n | > t )and P (cid:16) max k ∈ I || S k | − a k | > t (cid:17) = X k ∈ I P ( τ = k ) ≤ X k ∈ I P ( τ = k, || S n | − a n | > t ) ≤ P ( || S n | − a n | > t ) ≤ p. Finally we show P (cid:16) max k ∈ I || S k | − a k | > t (cid:17) ≤ p. (2.3)To this end take any k ∈ I and notice that2 max { P ( | S k − a k | ≤ t ) , P ( | S k + a k | ≤ t ) }≥ P ( | S k − a k | ≤ t ) + P ( | S k + a k | ≤ t ) ≥ P ( || S k | − a k | ≤ t ) ≥ − p ≥ . | x − a k | ≤ t and | y + a k | ≤ t then | x − y | ≥ a k − t > t . Therefore5 p ≥ P ( | S k − ˜ S k | > t ) ≥ P ( | S k − a k | ≤ t, | ˜ S k + a k | ≤ t ) + P ( | S k + a k | ≤ t, | ˜ S k − a k | ≤ t )= 2 P ( | S k − a k | ≤ t ) P ( | S k + a k | ≤ t ) . So for any k ∈ I we may choose b k = ± a k such that P ( | S k − b k | ≤ t ) ≤ p ≤ p. Therefore P ( | S k + b k | > t ) ≤ P ( || S k | − a k | > t ) + P ( | S k − b k | ≤ t ) ≤ p and by the L´evy-Ottaviani inequality (1.1), P (max k ∈ I || S k | − a k | > t ) ≤ P (max k ∈ I | S k + b k | > t ) ≤ k ∈ I P ( | S k + b k | > t ) ≤ p. 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 p := max ≤ k ≤ n P (cid:0)(cid:12)(cid:12) | S k | − a k (cid:12)(cid:12) ≥ t (cid:1) < . Notice that P (cid:0)(cid:12)(cid:12) | S n | − | S k | − ( a n − a k ) (cid:12)(cid:12) ≥ t (cid:1) ≤ P (cid:0)(cid:12)(cid:12) | S n | − a n (cid:12)(cid:12) ≥ t (cid:1) + P (cid:0)(cid:12)(cid:12) | S k | − a k (cid:12)(cid:12) ≥ t (cid:1) Therefore P (cid:0)(cid:12)(cid:12) | S k +1 ,n | + 2 h S k , S k +1 ,n i − ( a n − a k ) (cid:12)(cid:12) ≥ t ) ≤ p and by the symmetry P (cid:0)(cid:12)(cid:12) | S k +1 ,n | − h S k , S k +1 ,n i − ( a n − a k ) (cid:12)(cid:12) ≥ t (cid:1) ≤ p. Thus by the triangle inequality P (cid:0)(cid:12)(cid:12) | S k +1 ,n | − ( a n − a k ) (cid:12)(cid:12) ≥ t (cid:1) ≤ p. Now let x ∈ H be such that || x | − a k | ≥ t then by the triangle inequalityand symmetry1 − p ≤ P (cid:0)(cid:12)(cid:12) | x | + | S k +1 ,n | − a n (cid:12)(cid:12) ≥ t (cid:1) ≤ P (cid:0)(cid:12)(cid:12) | x | + | S k +1 ,n | + 2 h x, S k +1 ,n i − a n (cid:12)(cid:12) ≥ t (cid:1) + P (cid:0)(cid:12)(cid:12) | x | + | S k +1 ,n | − h x, S k +1 ,n i − a n (cid:12)(cid:12) ≥ t (cid:1) = 2 P (cid:0)(cid:12)(cid:12) | x | + | S k +1 ,n | + 2 h x, S k +1 ,n i − a n (cid:12)(cid:12) ≥ t (cid:1) = 2 P (cid:0)(cid:12)(cid:12) | x + S k +1 ,n | − a n (cid:12)(cid:12) ≥ t (cid:1) .
6o for any x ∈ H and k = 1 , , . . . , n , (cid:12)(cid:12) | x | − a k (cid:12)(cid:12) ≥ t ⇒ P (cid:0)(cid:12)(cid:12) | x + S k +1 ,n | − a n (cid:12)(cid:12) ≥ t (cid:1) ≥
12 (1 − p ) ≥ . (2.4)Now let τ := inf (cid:8) k ≤ n : (cid:12)(cid:12) | S k | − a k (cid:12)(cid:12) ≥ t (cid:9) , then since { τ = k } ∈ σ ( X , . . . , X k ) we get by (2.4), P (cid:0) τ = k, (cid:12)(cid:12) | S n | − a n (cid:12)(cid:12) ≥ t (cid:1) ≥ P ( τ = k ) . Hence P (cid:0)(cid:12)(cid:12) | S n | − a n (cid:12)(cid:12) ≥ t (cid:1) ≥ n X k =1 P ( τ = k ) = 16 P (cid:16) max ≤ k ≤ n (cid:12)(cid:12) | S k | − a k (cid:12)(cid:12) ≥ t (cid:17) and the proposition follows. Proof of Corollary 1.3.
We may consider variables S i , X i +1 , . . . , X n instead of X , . . . , X n and assume that i = 1. Let a := min ≤ k ≤ n a k . We will analyze twocases. Case 1. a ≤ t . Then by (1.4) we get a k ≤ (3 α + β ) t for all k . Thus by theL´evy inequality, P (cid:16) max k || S k | − a k | ≥ (6 α + 2 β + 1) t (cid:17) ≤ P (cid:16) max k | S k | ≥ (3 α + β + 1) t (cid:17) ≤ P ( | S n | ≥ (3 α + β + 1) t ) ≤ P ( || S n | − a n | ≥ t ) ≤ k P ( || S k | − a k | ≥ t ) . Case 2. a ≥ t . Notice first that for any s > {|| S k | − a k | ≥ s (2 a k + s ) } ⊂ {|| S k | − a k | ≥ s } ⊂ {|| S k | − a k | ≥ sa k } . (2.5)Indeed, the last inclusion follows since || S k | − a k | = ( | S k | + a k ) || S k | − a k | ≥ a k || S k | − a k | . To see the first inclusion in (2.5) observe that {|| S k | − a k | ≥ s (2 a k + s ) } ⊂ {|| S k | − a k | ≥ s } ∪ {| S k | + a k ≥ a k + s }⊂ {|| S k | − a k | ≥ s } . Now by (2.5) we get P (cid:16) max k || S k | − a k | ≥ (6 α + 2 β + 1) t (cid:17) ≤ P (cid:16) max k || S k | − a k | ≥ (6 α + 2 β + 1) at (cid:17) . P (cid:16) max k || S k | − a k | ≥ (6 α + 2 β + 1) t (cid:17) ≤ k P (cid:0) || S k | − a k | ≥
13 (6 α + 2 β + 1) at (cid:1) . But (6 α + 2 β + 1) a ≥ αa + βt ) + t ≥ a k + t for all k by (1.4). Therefore by(2.5), P (cid:16) max k || S k | − a k | ≥ (6 α + 2 β + 1) t (cid:17) ≤ k P ( || S k | − a k | ≥ t (2 a k + t )) ≤ k P ( || S k | − a k | ≥ t ) . Let us fix a positive integer n and put I n = n j ∈ Z : n ≤ j ≤ n o . Let t j = n + jj for j = 1 , , . . . , n , then jt j = n + j and ( j − t j ≤ n for j ∈ I n . (3.1)Let N be a large integer (to be fixed later) and let F be the space of alldouble-indexed sequences a = ( a i,j ) ≤ i ≤ N,j ∈ I n with the norm (cid:13)(cid:13)(cid:13) ( a i,j ) ≤ i ≤ N,j ∈ I n (cid:13)(cid:13)(cid:13) = max j ∈ I n (cid:18) | a ,j | + t j X ≤ i
1) = 1 / Y l,j are symmetric P ( | Y l,j | = n ) = 1 − P ( Y k,j = 0) = p n (with p n a smallpositive number to be specified later) and N l are uniformly sampled from theset { , . . . , N } .Obviously X , X , . . . , X n are i.i.d. and symmetric. As usual we set S k = X + X + . . . + X k . Let A = { N , N , . . . , N n are pairwise distinct } . Notice that P ( A c ) → N → ∞ . On the set A we have for k ≤ n , k S k k = max j ∈ I n (cid:16)(cid:12)(cid:12)(cid:12) k X l =1 Y l,j (cid:12)(cid:12)(cid:12) + t j min { k, j } (cid:17) . j > k we have by (3.1), (cid:12)(cid:12)(cid:12) k X l =1 Y l,j (cid:12)(cid:12)(cid:12) + t j min { k, j } < t j ( j − ≤ n + 1 , hence on the set A , for k ∈ I n we get k S k k = max j ∈ I n ,j ≤ k (cid:16)(cid:12)(cid:12)(cid:12) k X l =1 Y l,j (cid:12)(cid:12)(cid:12) + n + j (cid:17) = (cid:12)(cid:12)(cid:12) k X l =1 Y l,k (cid:12)(cid:12)(cid:12) + n + k. Take 0 < t < nC then for k ∈ I n , P ( |k S k k − ( n + k ) | ≥ t ) ≤ P ( A c ) + P (cid:16) k X l =1 Y l,k = 0 (cid:17) ≤ P ( A c ) + kp n and P (cid:16) max k ∈ I n |k S k k − ( n + k ) | ≥ tC (cid:17) ≥ P (cid:16) max k ∈ I n (cid:12)(cid:12)(cid:12) k X l =1 Y l,k (cid:12)(cid:12)(cid:12) = 0 (cid:17) − P ( A c ) . The last number is of order n p n if N is large and p n is small. This shows that if(1.5) holds for i = ⌈ n/ ⌉ in F then C must be of order n . So (1.5) cannot holdwith absolute constants C and C in (infinite dimensional) separable Banachspaces. References [1] E. Gin´e and D. Mason,
The law of the iterated logarithm for the integratedsquared deviation of a kernel density estimator , Bernoulli (2004), 721–752.[2] S. Kwapie´n and W.A. Woyczy´nski Random series and stochastic integrals:single and multiple , Birkh¨auser, Boston, MA, 1992.[3] M. Ledoux and M. Talagrand,
Probability in Banach spaces , Springer-Verlag,Berlin, 1991.[4] S. Montgomery-Smith,
Comparison of sums of independent identically dis-tributed random vectors
Probab. Math. Statist. (1993), 281–285.Institute of MathematicsUniversity of WarsawBanacha 202-097 WarszawaPoland [email protected]@mimuw.edu.pl