Abstract (X (1) i k. The reverse bound holds if in addition, the following symmetry condition holds almost surely

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1 Decouping Inequaities for the Tai Probabiities of Mutivariate U-statistics by Victor H. de a Peña 1 and S. J. Montgomery-Smith 2 Coumbia University and University of Missouri, Coumbia Abstract In this paper we present a decouping inequaity that shows that mutivariate U- statistics can be studied as sums of (conditionay independent random variabes. This resut has important impications in severa areas of probabiity and statistics incuding the study random graphs and mutipe stochastic integration. More precisey, we get the foowing resut: Theorem 1. Let {X j } be a sequence of independent random variabes in a measurabe space (S, S, and et {X (j i }, j = 1,..., k be k independent copies of {X i }. Let f i1... be famiies of functions of k variabes taking (S... S into a Banach space (B,. Then, for a n k 2, t > 0, there exist numerica constants C k depending on k ony so that, P ( C k P (C k 1... n 1... n f i1...,... t f i1...,..., X (k t. The reverse bound hods if in addition, the foowing symmetry condition hods amost surey f i1... (X i1, X i2,..., X ik = f iπ(1 i π(2...i π(k (X iπ(1, X iπ(2,..., X iπ(k, for a permutations π of (1,..., k. 1. Introduction In this paper we provide the mutivariate extension of the tai probabiity decouping inequaity for generaized U-statistics of order two and quadratic forms presented in de a Peña and Montgomery-Smith (1993. This type of inequaity permits the transfer of some resuts for sums of independent random variabes to the case of U-statistics. Our work buids mainy on recent work of Kwapien and Woyczynski (1992 as we as on resuts for U-statistics from Giné and Zinn (1992 and papers deaing with inequaities for mutiinear forms of symmetric and hypercontractive random variabes in de a Peña, Montgomery- Smith and Szuga (1992, and de a Peña (1992. It is to be remarked that the decouping inequaities for mutiinear forms introduced in McConne and Taqqu (1986 provided us with our first exposure to this decouping probem. For a more expanded ist of references on the subject see, for exampe, Kwapien and Woyczynski ( ,2 Supported in part by NSF grants. 2 Supported by the University of Missouri Research Board. AMS 1991 subject cassifications: Primary 60E15. Secondary 60D05. Key words and phrases: U-statistics, Decouping. 1

2 2. Main Resut Theorem 1. Let {X i } be a sequence of independent random variabes in a measurabe space (S, S, and et {X (j i }, j = 1,..., k be k independent copies of {X i }. Let f i1... be famiies of functions of k variabes taking (S... S into a Banach space (B,. Then, for a n k 2, t > 0, there exist numerica constants C k, Ck depending on k ony so that, P ( C k P (C k 1... n 1... n f i1...,... t f i1...,..., X (k t. If in addition, the foowing symmetry condition hods amost surey for a permutations π of (1,..., k, then P ( C k P ( C k f i1... (X i1, X i2,..., X ik = f iπ(1 i π(2...i π(k (X iπ(1, X iπ(2,..., X iπ(k 1... n 1... n f i1...,..., X (k t f i1...,... t. Note: In this paper we use the notation {... } to denote that a of,..., are different. 3. Preiminary Resuts Throughout this paper we wi be using two resuts found in earier work. The first one comes from de a Peña and Montgomery-Smith (1993. For competeness we reproduce the proof here. Lemma 1. Let X, Y be two i.i.d. random variabes. Then (1 P ( X t 3P ( X + Y 2t 3. Proof: Let X, Y, Z be i.i.d. random variabes. Then P ( X t = P ( (X + Y + (X + Z (Y + Z 2t P ( X + Y 2t/3 + P ( X + Z 2t/3 + P ( Y + Z 2t/3 = 3P ( X + Y 2t/3. The second resut comes from Kwapien and Woyczynski (1992 and can aso be found in de a Peña and Montgomery-Smith (

3 Proposition 1. Let Y be any mean zero random variabe with vaues in a Banach space (B,. Then, for a aɛb, (2 P ( a + Y a κ 4, where κ = inf x ɛb (E x (Y 2 E(x (Y. (Here B denotes the famiy of inear functionas on B. 2 Proof: Note first that if ξ is a random variabe for which Eξ = 0, then P (ξ 0 1 (E ξ 2 4 E(ξ 2. From this, we deduce that P (x (Y 0 1 (E X (Y 2 4 E(X (Y The resut then foows, 2 because if x ɛb is such that x = 1 and x (a = a, then { a + Y a } contains {x (a + Y x (a} = {x (Y 0}. Lemma 2. Let x, a i1, a i1,..., a i1... beong to a Banach space (B,. Let {ɛ i } be a sequence of symmetric Bernoui random variabes. Then, P ( x + k r=1 1 i i... i r n a i1...i r ɛ i1...ɛ ir x c 1 k, for a universa constant 1 < c k < depending on k ony. Proof: Suppose that x, a i1, a i1,..., a i1... are in R, then since the ɛ s are hypercontractive, by equation (1.4 of Kwapien and Szuga (1991 and the easy argument of the proof of Lemma 3 in de a Peña and Montgomery-Smith (1993, for some σ > 0, we get (E k r= i r n a...i r ɛ i1...ɛ ir = (E k r=1 1 <...<i r n b...i r ɛ i1...ɛ ir σ k (E k r=1 1 <...< n b...i r ɛ i1...ɛ ir = σ k (E k r= n a...i r ɛ i1...ɛ ir 2 1 2, where b i1...i r = π S r a iπ(1...i π(r, and S r denotes the set of a permutations of {1,..., r}. Next, observe that ξ 4 σ 2 ξ 2 impies that ξ 2 σ 4 ξ 1. Take x ɛb so that x = 1 and x (x = x, then P ( x + k r=1 1 i i... i r n a...i r ɛ i1...ɛ ir x P (x (x + k r=1 1 i i... i r n x (a i1...i r ɛ i1...ɛ ir x (x = P ( k r=1 1 i i... i r n x (a i1...i r ɛ i1...ɛ ir 0 c 1 k Note: Throughout this paper we wi use c k and C k to denote numerica constants that depend on k ony and may change from appication to appication. 4. Proof of the Upper Bound: Our proof of this resut is obtained by appying the argument used in the proof of the upper bound in the bivariate case pus an inductive argument. Let {σ i } be a sequence of independent symmetric Bernoui random variabes, P (σ i = 1 = 1 2 and P (σ i = 1 = 1 2. Consider random variabes (Z (1 i, Z (2 i such that (Z (1 i, Z (2 i = i i if σ i = 1 and (Z (1 i, Z (2 i = (X (2 i i if σ i = 1. Then (1 + σ i and (1 σ i are either 0 or 2 and 3

4 these random variabes can be used to transform the probem from one invoving X s to one invoving Z s. Let us first iustrate the argument in the case that k = f i1 (Z (1, Z (1, Z (2 = {(1 + σ i1 (1 + σ i2 (1 + σ i3 f i1 +(1 + σ i1 (1 + σ i2 (1 σ i3 f i1 +(1 + σ i1 (1 σ i2 (1 + σ i3 f i1 (3 +(1 σ i1 (1 + σ i2 (1 + σ i3 f i1 (X (2 +(1 + σ i1 (1 σ i2 (1 σ i3 f i1 +(1 σ i1 (1 + σ i2 (1 σ i3 f i1 (X (2 +(1 σ i1 (1 σ i2 (1 + σ i3 f i1 (X (2 +(1 σ i1 (1 σ i2 (1 σ i3 f i1 (X (2 }, where the sign + is chosen ifthe superscript of X i agrees with that of Z i, and otherwise. Next, set T n,3 = 1 n {f i1 + f i1 +f i1 + f i1 (X (2 +f i1 + f i1 (X (2 +f i1 (X (2 + f i1 (X (2 }. Letting G 2 = σ i i, i = 1,..., n we get T n,3 = n E(f i1 (Z (1, Z (1, Z (2 G 2. More generay, for any 1 1,..., k 2, one can obtain the expansion 4

5 2 k f i1... (Z ( 1,..., Z ( k (4 = Again, 1 j 1,...,j k 2 The appropriate extension of T n,3 is T n,k = T n,k = 2 k From Lemma 1 we get, 3P (3 3P (3 T n,k + {3P (3 T n,k t +3P (3 P ( 1... n 1... n 1... n (1±σ i1...(1±σ ik f i1...,..., X (j k n 1 j 1,...,j k n 1... n {3P (3 T n,k t (5 + C k P (C k 1 j 1,...,j k 2, not a j s equa f i1...,..., X (j k. E(f i1... (Z ( 1,..., Z ( k G 2. f i1...,... t {f i1...,... + f i1... (X (2,... } 2t = f i1...,... + f i1... (X (2,... T n,k 2t f i1...,..., X (j k t} 1 j 1,...j k 2, not a j s equa 1... n f i1...,..., X (j k t}. (Reca that C k, c k are numerica constants that depend on k ony and may change from appication to appication. Observe aso that using (4 and the fact that the σ s are independent from the X s, Lemma 2 with x = T n,k gives for any fixed 1 1,..., k 2, (6 P (2 k 1... n f i1... (Z ( 1,..., Z ( k T n,k G 2 c 1 k, Integrating over { T n,k t} and using the fact that { i i : i = 1,..., n} has the same joint distribution as {(Z (1 i, Z (2 i : i = 1,..., n} we obtain that (7 P (2 k f i1... (X ( 1,..., X ( k t n 5

6 = P (2 k 1... n f i1... (Z ( 1,..., Z ( k t c 1 k P ( T n,k t It is obvious that the upper bound decouping inequaity hods for the case of U- statistics of order 1. Assume that it hods for U-statistics of orders 2,..., k 1. Putting (5 and (7 together with 1 1,..., k 2, not a s equa we get, P ( 1... n f...,... t {3P (3 T n,k t + C k P (C k 1 j 1,...,j k 2, not a j s equa C k P (C k 1 j 1,...,j k 2, not a j s equa C k P (C k 1... n f n 1... n,..., X (k t, f i1...,..., X (j k t} f i1...,..., X (j k t where again, the ast ine foows by the decouping resut for U-statistics of orders 2,..., k 1 of the inductive hypothesis. Since the statement not a j s equa means that there are ess than k j s equa, the variabes whose j s are equa can be decouped using (conditionay on the other variabes the decouping inequaities for U-statistics of order 2,..., k 1. Next we give the proof of the ower bound. 5. Proof of the Lower Bound In order to show the ower bound we require the foowing resut. Lemma 3. Let 1 k. Then there is a constant C k such that P ( 1... n C 1 k P ( f i1...,... t 1... n 1 j 1,...,j k f i1..., X (j 2,..., X (j k C k t. Proof: Let {δ r }, r = 1,...,, be a sequence of random variabes for which P (δ r = 1 = 1 and P (δ r = 0 = 1 1, and r=1 δ r = 1. Set ɛ r = δ r 1 for r = 1,...,. Then, it is easy to see that there exists σ > 0 depending ony upon such that for any rea number x 0 and any sequence of rea constants {a i } (8 x 0 + r=1 a r ɛ r 4 x 0 + σ 1 a r ɛ r 2. One can aso use the resuts of Section 6.9 of Kwapien and Woyczynski (1992 (Pg. 180, 181 to assert this since the ɛ s satisfy the conditions 1. through 3. stated there. 6 r=1

7 Let {(δ i1,..., δ i, i = 1,..., n} be n independent copies of (δ 1,..., δ. As before, we define (9 ɛ ij = δ ij 1. Since the vectors E i = (ɛ i1,..., ɛ i are independent, by an argument given in Kwapien and Szuga (1991, for i = 1,..., n, for a constants x 0, a ij in R, (10 x 0 + n i=1 r=1 and recentering, we obtain (11 x 0 + a ir ɛ ir 4 x 0 + σ 1 n i=1 r=1 n i=1 r=1 a ir δ ir 4 σ 1 x 0 + a ir ɛ ir 2 σ 1 x 0 + n i=1 r=1 a ir δ ir 2. n i=1 r=1 a ir ɛ ir 2, Next we use the sequence E i, i = 1,..., n in defining the anaogue of the Z s used in our proof of the upper bound. For each i, et Z i = X (j i if δ ij = 1. Then, {Z i, i = 1,..., n} has the same joint distribution as {X (1 i, i = 1,..., n} and f i1... (Z i1,.., Z ik = 1 j 1,j 2,...,j k The fact that Eδ ir j r = 1 for a i r, j r gives, where G = σ( i E(f i1... (Z i1,..., Z ik G = ( 1 k,..., X ( i, i = 1,..., n. Let U n = 1... n f... (Z i1,..., Z ik = 1... n 1 j 1,...,j k δ i1 j 1...δ ik j k f i1...,..., X (j k. 1 j 1,...,j k f i1...,..., X (j k, δ i1 j 1...δ ik j k f i1...,, X (j k. Let D i = (δ i1,..., δ i. Since the D s are independent of the X s, if we et g i1... (D i1,..., D ik = 1 j 1,...,j k δ i1 j 1...δ ik j k f i1...,, X (j k. then, since f i1... (X i1,..., X ik = f i(π(1...i (π(k (X iπ(1,..., X iπ(k, 7

8 we have that, g i1... (D i1,..., D ik = g i(π(1...i (π(k (D iπ(1,..., D iπ(k. Therefore, the two sided decouping inequaity in de a Peña (1992 can be appied and, for every convex increasing function Φ, every G -measurabe function T, and k independent copies D (r i, r = 1,..., k of D i there exists numerica constants A k, B k so that E(Φ(A k T + E(Φ( T + E(Φ(B k T n 1... n 1... n This resut with (11 shows that conditionay on G g i1... (D i1,..., D ik G g i1... (D (1,..., D (k G g i1... (D i1,..., D ik G. (12 U n T n 4 σ k B k U n T n 2, A k where T n = E(U n G = ( 1 k 1... n 1 j 1,...,j k f i1..., X (j 2,..., X (j k. (See aso the proofs of Lemma 2 and Lemma of Kwapien and Woyczynski (1992. Thus we have that, (13 P ( U n T n G c 1 k. This foows from the use of (12 and Proposition 1 with a = T n and Y = U n T n. We aso use the fact that for any random variabe ξ and positive constant c, ξ 4 c ξ 2 impies that ξ 2 c 2 ξ 1 (See aso the proof of Lemma 2 for the approach to transfer the probem from one on Banach space vaued random variabes to one on rea vaued. Integrating (13 over the set { T n t} we get P ( 1... n f...,... t c 1 k P (C k = P ( and Lemma 3 is proved n 1... n 1 j 1,...,j k f i1... (Z i1, Z i2,..., Z ik t 8 f i1..., X (j 2,..., X (j k t,

9 The end of the proof of the ower bound foows by using induction and the iterative procedure introduced to obtain the proof of the ower bound mutivariate decouping inequaity in de a Peña (1992. We give a different expression of the same proof, motivated by ideas from de a Peña, Montgomery-Smith and Szuga (1992. We wi use S k to denote the set of permutations of {1,..., k}. The Mazur-Oricz formua tes us that for any 1 j 1,..., j k k that 1... n 0 δ 1,...,δ k 1 ( 1 k δ 1... δ k δ j1... δ jk is 0 uness j 1,..., j k is a permutation of 1,..., k, in which case it is 1. Hence f i1... (X (π(1,..., X (π(k π S k = 0 δ 1,...,δ k 1 By the symmetry properties on f, f i1... = 1 k! Therefore, Pr( 1... n =,..., X (k ( 1 k δ 1... δ k 1... n 0 δ 1,...,δ k 1 0 δ 1,...,δ k 1 k =1 ( k 1 j 1,...,j k k ( 1 k δ 1... δ k f i1...,..., X (k t Pr( Pr( 1... n 1 j 1,...,j k k 1... n 1 j 1,...,j k δ j1... δ jk f i1...,..., X (j k. 1 j 1,...,j k k δ j1... δ jk f i1...,..., X (j k. δ j1... δ jk f i1...,..., X (j k k!t/2 k f i1...,..., X (j k k!t/2 k, and this combined with Lemma 3 is sufficient to show the resut. 9

10 7. References 1. de a Peña, V. H. (1992. Decouping and Khintchine s inequaities for U-statistics. Ann. Probab. 20 4, de a Peña, V. H., Montgomery-Smith and Szuga, J. (1992. Contraction and decouping inequaities for mutiinear forms and U-statistics. Preprint. 3. de a Peña, V. H. and Montgomery-Smith (1993. Bounds on the tai probabiity of U-statistics and quadratic forms. Preprint. 4. Giné, E. and Zinn, J. (1992. A remark on convergence in distribution of U-statistics. Preprint. 5. Kwapien, S. Szuga, J. (1991. Hypercontraction methods in moment inequaities for series of independent random variabes in normed spaces. Ann. Probab. 19 (1, Kwapien, S. and Woyczynski, W. (1992. Random Series and Stochastic Integras: Singe and Mutipe. Birkhauser. 7. McConne, T. and Taqqu, M. (1986. Decouping inequaities for mutiinear forms in independent symmetric random variabes. Ann. Probab. 14(3,