On Z p -norms of random vectors

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1 On Z -norms of random vectors Rafa l Lata la Abstract To any n-dimensional random vector X we may associate its L -centroid body Z X and the corresonding norm. We formulate a conjecture concerning the bound on the Z X-norm of X and show that it holds under some additional symmetry assumtions. We also relate our conjecture with estimates of covering numbers and Sudakov-tye minorization bounds. 1 Introduction. Formulation of the Problem. Let 2 and X = X 1,..., X n be a random vector in R n such that E X <. We define the following two norms on R n : t MX := E t, X 1/ and t ZX := su{ t, s : s MX 1}. By M X and Z X we will also denote unit balls in these norms, i.e. M X := {t R n : t MX 1} and Z X := {t R n : t ZX 1}. The set Z X is called the L -centroid body of X or rather of the distribution of X. It was introduced under a different normalization for uniform distributions on convex bodies in [9]. Investigation of L -centroid bodies layed a crucial role in the Paouris roof of large deviations bounds for Euclidean norms of log-concave vectors [10]. Such bodies also aears in questions related to the otimal concentration of log-concave vectors [7]. Let us introduce a bit of useful notation. We set t := t 2 = t, t and B2 n = {t R n : t 1}. By Y = E Y 1/ we denote the L -norm of a random variable Y. Letter C denotes universal constants that may differ at each occurence, we write f g if 1 C f g Cf. Let us begin with a simle case, when a random vector X is rotationally invariant. Then X = RU, where U has a uniform distribution on S n 1 and R = X is a nonnegative random variable, indeendent of U. We have for any vector t R n and 2, t, U = t U 1 n + t, Suorted by the National Science Centre, Poland grant 2015/18/A/ST1/

2 where U 1 is the first coordinate of U. Therefore So t MX = U 1 R t and t ZX = U 1 1 E X Z X 1/ = U1 1 This motivates the following roblem. R 1 R 1 t. n + E X 1/ = U Problem 1. Is it true that for at least a large class of centered n-dimensional random vectors X, 1/2 n + E X 2 Z X C for 2, or maybe even 1/ E X n + Z X C Notice that the roblem is linearly-invariant, since for 2? AX ZAX = X ZX for any A GLn. 2 For any centered random vector X with nondegenerate covariance matrix, random vector Y = CovX 1/2 X is isotroic i.e. centered with identity covariance matrix. We have M 2 Y = Z 2 Y = B n 2, hence E X 2 Z 2 X = E Y 2 Z 2 Y = E Y 2 = n. Next remark shows that the answer to our roblem is ositive in the case n. Remark 1. For n and any n-dimensional random vector X we have E X Z X 1/ 10. Proof. Let S be a 1/2-net in the unit ball of M X such that S 5 n such net exists by the volume-based argument, cf. [1, Corollary ]. Then 1/ E X Z X 1/ 2 E su t, X 2 E t S t S 2 S 1/ sue t, X 1/ 2 5 n/. t S t, X 1/ 2

3 L -centroid bodies lay an imortant role in the study of vectors uniformly distributed on convex bodies and a more general class of log-concave vectors. A random vector with a nondenerate covariance matrix is called log-concave if its density has the form e h, where h: R n, ] is convex. If X is centered and log-concave then t, X λ q t, X q for q 2, 3 where λ = 2 λ = 1 if X is symmetric and log-concave and λ = 3 for arbitrary log-concave vectors. One of oen roblems for log-concave vectors [7] states that for such vectors, arbitrary norm and q 1, E X q 1/q C E X + su t, X q. t 1 In articular one may exect that for log-concave vectors E X q Z X 1/q C E X ZX + su t M X t, X q C E X ZX + As a result it is natural to state the following variant of Problem 1. max{, q}. Problem 2. Let X be a centered log-concave n-dimensional random vector. Is it true that E X q n Z X 1/q C for 2 n, 1 q n. In Section 2 we show that Problems 1 and 2 have affirmative solutions in the class of unconditional vectors. In Section 3 we relate our roblems to estimates of covering numbers. We also show that the first estimate in Problem 1 holds if the random vector X satisfies the Sudakov-tye minorization bound. 2 Bounds for unconditional random vectors In this section we consider the class of unconditional random vectors in R n, i.e. vectors X having the same distribution as ε 1 X 1, ε 2 X 2,..., ε n X n, where ε i is a sequence of indeendent symmetric ±1 random variables Rademacher sequence, indeendent of X. Our first result shows that formula 1 may be extended to the unconditional case for even. We use the standard notation for a multiindex α = α 1,..., α n, x R n and m = α i, x α := i xα i i and m α := m!/ i α i!. 3

4 Proosition 2. We have for any k = 1, 2,... and any n-dimensional unconditional random vector X such that E X 2k <, 1/2k E X 2k Z 2k X c2k := α 1 =k k 2 α 2k 2α 1/2k n + k, k where the summation runs over all multiindices α = α 1,..., α n with nonnegative integer coefficients such that α 1 = n α i = k. Proof. Observe first that For any t, s R n we have E t, X 2k 2k = E t i ε i X i = t, s k = So by the Cauchy-Schwarz inequality, α 1 =k α 1 =k k t α s α. α s k Z 2k X = su{ t, s k : E t, X 2k 1} To see that c 2k n + k/k observe that k α 2 2k = 2α Therefore, since 1 2l l 2 2l, we get n + k 1 4 k k 2k k 1 n 2αi α i 2k t 2α EX 2α. 2α α 1 =k. n + k 1 c 2k 2k 4k. k k 2 α 2k s2α EX 2α 2α 1/2. Corollary 3. Let X be an unconditional n-dimensional random vector. Then 1/2k n + E X 2k Z X C for any ositive integer k 2. 4

5 Proof. By the monotonicity of L 2k -norms we may and will assume that k = /2. Then by Proosition 2, 1/2k 1/2k n + k E X 2k Z X E X 2k Z 2k X C k n + C. In the unconditional log-concave case we may bound higher moments of X ZX. Theorem 4. Let X be an unconditional log-concave n-dimensional random vector. Then for, q 2, n E X q + n + Z X 1/q C + su t, X q C + q. t M X In order to show this result we will need the following lemma. Lemma 5. Let 2 n, X be an unconditional random vector in R n such that E X < and E X i = 1. Then s ZX su su t i s i + C 1 su t i s i. 4 I [n], I t MX 1 i I t MX 1, t 2 1/2 Proof. We have by the unconditionality of X and Jensen s inequality, t MX = t i ε i X i t i ε i E X i. By the result of Hitczenko [5], for numbers a 1,..., a n, a i ε i a i + a i 2 i i> where a i i n denotes the nonincreasing rearrangement of a i i n. Thus i> t i 2 1/2 C 1 t MX 1/2, 5 and 4 easily follows. 5

6 Proof of Theorem 4. The last bound in the assertion follows by 3. It is easy to see that increasing q if necessary it is enough to consider the case q n. If q n then the similar argument as in the roof of Remark 1 shows that 1/q E X q Z X 2 5 n/q su t M X t, X q 10 su t M X t, X q. Finally, consider the remaining case n q n. By 2 we may assume that E X i = 1 for all i. By the log-concavity t, X q1 C q 1 q 2 t, X q2 for q 1 q 2 1, in articular σ i := X i 2 C. Let E 1,..., E n be i.i.d. symmetric exonential random variables with variance 1. By [6, Theorem 3.1] we have su t t MX 1, t 2 i X i 1/2 q C su t t MX 1, t 2 i σ i E i + su t, X q. 1/2 t MX 1, t 2 1/2 1 We have and su t MX 1, t 2 1/2 su t, X q su t, X q t MX 1, t 2 1/2 t MX 1 t i σ i E i 1 n 1 σ 2 i E2 i 1 n n σ 2 i C. 1 Thus su t MX 1, t 2 1/2 n t i X i C + q su t MX 1 t, X q. Let for each I [n], P I X = X i i I and S I be a 1/2-net in M P I X of cardinality at 6

7 most 5 I. We have su su I [n], I t MX 1 t i X i 2 q i I 2 where the last estimate follows from q n. Hence the assertion follows by Lemma 5. su I [n], I t S I E I [n], I t S I su t i X i i I q q t i X i i I 2 5 /q {I [n], I } 1/q su I en /q 10 su t t M X i X i i I q C su t i X i, t M X i I q 1/q su t i X i t S I i I q Corollary 6. Let X be an unconditional log-concave n-dimensional random vector and 2 n. Then 1 n C E X n 1/ n n Z X E X Z X C 6 and P X ZX 1 n 1C n C, P X ZX Ct e tn for t 1. Proof. The uer bound in 6 easily follows by Theorem 4. In fact we have for t 1, E X t n Z X 1/t n Ct n, hence the Chebyshev inequality yields the uer tail bound for X ZX. To establish lower bounds we may assume that X is additionally isotroic. Then by the result of Bobkov and Nazarov [3] we have t, X C t 2 + t. This easily gives E X ZX 1 n C EX n/2 1 n C, where the last inequality follows by Lemma 7 below. 7

8 By the Paley-Zygmund inequality we get P X ZX 1 n P X C ZX 1 2 E X Z X E X Z X 2 4E X 2 Z X c. Lemma 7. Let X by a symmetric isotroic n-dimensional log-concave vector. EX n/2 1 C. Then Proof. Let a i > 0 be such that PX i a i = 3/8. Then by the log-concavity of X i, P X i ta i = 2PX i ta i 3/4 t for t 1 and integration by arts yields X i 2 Ca i. Thus a i c 1 for a constant c 1 > 0. Let S = n I { X i c 1 }. Then ES = n P X i c 1 3n/4. On the other hand ES n 2 + npx n/2 c 1, so EX n/2 c 1PX n/2 c 1 c 1 /4. The next examle shows that the tail and moment bounds in Corollary 6 are otimal. Examle. Let X = X 1,..., X n be an isotroic random vector with i.i.d. symmetric exonential coordinates i.e. X has the density 2 n/2 ex 2 x 1. Then E X i 1/ /2, so 2 e i M X and P X ZX t n/ P X i t n/2 e t n/ 2 and for q = s n, s 1, 1/q E X q 2 Z X X i q cq/ = cs n/. 3 General case aroach via entroy numbers In this section we roose a method of deriving estimates for Z -norms via entroy estimates for M -balls and Euclidean distance. We use a standard notation for sets T, S R n, by NT, S we denote the minimal number of translates of S that are enough to cover T. If S is the ε-ball with resect to some translation-invariant metric d then NT, S is also denoted as NT, d, ε and is called the metric entroy of T with resect to d. We are mainly interested in log-concave vectors or random vectors which satisfy moment estimates t, X λ q t, X q for q 2. 7 Let us start with a simle bound. 8

9 Proosition 8. Suose that X is isotroic in R n and 7 holds. Then for any 2 and ε > 0 we have 1/2 E X 2 eλ Z X ε n + max {, log NM X, εb2 n }. Proof. Let N = NM X, εb2 n and choose t 1,..., t N M X such that M X N t i + εb2 n. Then x ZX ε x + su t i, x. i N Let r = max{, log N}. We have 1/2 E su t i, X 2 i N 1/r E su t i, X r i N N 1/r su i N 1/r E t i, X r t i, X r eλ r su t i, X eλ r i. Remark 9. The Paouris inequality [10] states that for isotroic log-concave vectors and q 2, E X q 1/q C n + q, so for such vectors and q 2, 1/q E X q 2e Z X Cε n + q + max{, q, log NM X, εb2 n }. Unfortunately, the known estimates for entroy numbers of M -balls are rather weak. Theorem 10 [4, Proosition 9.2.8]. Assume that X is isotroic log-concave and 2 n. Then log N t M X, B n 2 C n log2 log t t {, 1 for 1 t min C } n log 2. Corollary 11. Let X be isotroic log-concave, then 1/ E X n 3/4 Z X C log log n for 2 1 C n3/7 log 2/7 n. Proof. We aly Theorem 10 with t = n/ 1/4 log log 1/2 n and Proosition 8 with ε = t 1/2. Remark 12. Suose that X is centered and the following stronger bound than 7 satisfied for examle for Gaussian vectors holds t, X λ q t, X q for q

10 Then for any 2 n, 1 2n 1/2 1/n n E X λ 2 Z X E X n Z X 10λ. Proof. Without loss of generality we may assume that X is isotroic. We have so M X λ 1 2/B n 2 and t, X λ /2 t, X 2 = λ /2 t, E X 2 Z X 1/2 1 2 λ E X 2 1/2 1 2n = λ. On the other hand let S be a 1/2-net in M X of cardinality at most 5 n. Then 1/n 1/n 1/n E X n Z X 2 E su t, X n 2 E t, X n t S t S n n 2 S 1/n su t, X n 10λ t S su t, X 10λ t S. Recall that the Sudakov minoration rincile [11] states that if G is an isotroic Gaussian vector in R n then for any bounded T R n and ε > 0, E su t, G 1 log C ε NT, εb2 n. So we can say that a random vector X in R n satisfies the L 2 -Sudakov minoration with a constant C X if for any bounded T R n and ε > 0, E su t, X 1 ε log NT, εb2 n C. X Examle. Any unconditional n-dimensional random vector satisfies the L 2 -Sudakov minoration with constant C logn + 1/min i n E X i. Indeed, we have by the unconditionality, Jensen s inequality and the contraction rincile, E su t i X i = E su t i ε i X i E su t i ε i E X i min i n E X i E su t i ε i. 10

11 On the other hand, the classical Sudakov minoration and the contraction rincile yields 1 C ε log NT, εb2 n E su C logn + 1E su t i g i E max g i E su i n t i ε i. t i ε i However the L 2 -Sudakov minoration constant may be quite large in the isotroic case even for unconditional vectors if we do not assume that L 1 and L 2 norms of X i are comarable. Indeed, let PX = ±n 1/2 e i = 1 2n for i = 1,..., n, where e 1,..., e n is the canonical basis of R n. Then X is isotroic and unconditional. Let T = {t R n : t n 1/2 }. Then However, by the volume-based estimate, hence E su t, X 1. NT, εb n 2 volt volεb n 2 1 n, Cε su ε log NT, εb2 n 1 n. ε>0 C Thus the L 2 -Sudakov constant C X n/c in this case. Next roosition shows that random vectors with uniformly log-convex density satisfy the L 2 -Sudakov minoration. Proosition 13. Suose that a symmetric random vector X in R n has the density of the form e h such that Hessh αid for some α > 0. Then X satisfies the L 2 -Sudakov minoration with constant C X C α. Proof. We will follow the method of the roof of the dual classical Sudakov inequality cf and its roof in [8]. Let T be a bounded symmetric set and A := E su t, X. By the duality of entroy numbers [2] we need to show that log 1/2 Nε 1 B n 2, T o Cε 1 α 1/2 A for ε > 0 or equivalently that NδB n 2, 6AT o excαδ 2 for δ >

12 To this end let N = NδB2 n, 6AT o. If N = 1 there is nothing to show, so assume that N 2. Then we may choose t 1,..., t N δb2 n such that the balls t i + 3AT 0 are disjoint. Let µ = µ X be the distribution of X. By the Chebyshev inequality, µ3at 0 = 1 P su t, X > 3A 2 3. Observe also that for any symmetric set K and t R n, µt + K = e hx t dx = e hx+t dx = K K e hx t+hx+t/2 dx. K By Taylor s exansion we have for some θ [0, 1], K 1 2 ehx t + e hx+t dx hx t + hx + t 2 Thus and = hx Hesshx + θtt, t + Hesshx θtt, t hx 1 2 α t 2. µt + K e hx α t 2 /2 = e α t 2 /2 µk K 1 N µt i + 3AT 0 N e α t i 2 /2 µ3at 0 2N /2 3 e αδ2 N 1/3 e αδ2 /2 and 9 easily follows. Proosition 14. Suose that X satisfies the L 2 -Sudakov minoration with constant C X. Then for any 2 N M X, ec X B n 2 e. In articular if X is isotroic we have for 2 n, 1/2 n E X 2 Z X e C X + 1. Proof. Suose that N = NM X, ec X 1/2 B2 n e. We can choose t 1,..., t N M X, such that t i t j 2 ec X 1/2. We have E su t i, X 1 ec X 1/2 log N > e. i N C X 12

13 However on the other hand, E su t i, X i N 1/ E su t i, X i N E t i, X i N 1/ N 1/ max t i, X e. i To show the second estimate we roceed in a similar way as in the roof of Proosition 8. We choose T M X such that T e and M X T + ec X 1/2 B2 n. We have so that X ZX ec X 1/2 X + su t, X 1/2 1/2 E X 2 Z X ecx 1/2 E X 2 1/2 + E su t, X 2. Vector X is isotroic, so E X 2 = n and since T M X and 2 we get 1/2 E su t, X 2 1/ E su t, X T 1/ max t, X e. 1/ E t, X References [1] S. Artstein-Avidan, A. Giannooulos and V. D. Milman, Asymtotic Geometric Analysis. Part I, Mathematical Surveys and Monograhs 202, American Mathematical Society, Providence, RI, [2] S. Artstein, V. D. Milman and S. J. Szarek, Duality of metric entroy, Ann. of Math , [3] S. Bobkov and F. L. Nazarov, On convex bodies and log-concave robability measures with unconditional basis, in: Geometric Asects of Functional Analysis, 53 69, Lecture Notes in Math. 1807, Sringer, Berlin, [4] S. Brazitikos, A. Giannooulos, P. Valettas and B. H. Vritsiou, Geometry of isotroic convex bodies, Mathematical Surveys and Monograhs 196, American Mathematical Society, Providence, RI, [5] P. Hitczenko, Domination inequality for martingale transforms of a Rademacher sequence, Israel J. Math

14 [6] R. Lata la, Weak and strong moments of random vectors, Marcinkiewicz Centenary Volume, Banach Center Publ , [7] R. Lata la, On some roblems concerning log-concave random vectors, Convexity and Concentration, , IMA Vol. Math. Al. 161, Sringer, [8] M. Ledoux and M. Talagrand, Probability in Banach Saces. Isoerimetry and Processes, Sringer, Berlin, [9] E. Lutvak and G. Zhang, Blaschke-Santaló inequalities, J. Differential Geom , [10] G. Paouris, Concentration of mass on convex bodies, Geom. Funct. Anal , [11] V. N. Sudakov, Gaussian measures, Cauchy measures and ε-entroy, Soviet Math. Dokl , Rafa l Lata la Institute of Mathematics University of Warsaw Banacha Warszawa Poland rlatala@mimuw.edu.l 14