The best constant in the Khinchine inequality for slightly dependent random variables

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1 arxiv: v1 [math.pr] 10 Jun 018 The best constant in the Khinchine inequality for slightly dependent random variables Susanna Spektor Abstract We compute the best constant in the Khintchine inequality under assumption that the sum of Rademacher random variables is zero. 010 Classification: 46B06, 60C05 Keywords: Khintchine inequality, Rademacher random variables. 1 Introduction The classical Khintchine inequality states that for any p 0, there exists constants A p and B p, such that 1/ p 1/p 1/ A p ai E a i ε i B p a i, for arbitrary. Here a i, i 1,..., R and ε i, i 1,..., is a sequence of Rademacher random variables, i.e. mutually independent random variables with distribution Pε i 1 Pε i 1 1. The computation of the best possible constants has attracted a lot of interest. For the classical case, Haagerup found the best constants for general p 1, in [1]. Also Khintchine inequalities for different kinds of random variables were investigated, for example, rotationally invariant random vectors in [] or k-wise independent random variables in [3]. In our work we are dealing with the Khinchine inequality for slightly dependent Rademacher random variables, established in [4]: p 1/p E S a i ε i C p a, 1 1

2 where p and by E S we denote an expectation with condition that S ε i 0. Let us note here that condition requires even number of elements, i.e. l. In the present paper, we introduce a combinatorial method which enables us to compute the best constant C p in 1. Our main result is the following theorem. Theorem 1.1. Let ε i,i, be Rademacher random variables satisfying condition. Let a a 1,...,a R. Then for any q, q E S a i ε i! q + 1! q!! 1! q q! a q. 3 The paper is organized as following. In the next section we provide the necessary combinatorial results. In Section 3, we will establish the best constant of the Khinchine inequality for slightly dependent random variables. Some combinatorial results Lemma.1. Let ε i,i, be Rademacher random variables satisfying condition and let q q q, q i 0,...,q}. Then, P Dif P q + 1! q! 1! i 1} ε i 0} P i 1} ε i 0} Proof. Let us note here that εq i i 1 if and only if the sum of those q i for which ε i 1 is even, while ε i 0 holds whenever half of ε i 1. Thus, to calculate given probability is sufficient to calculate all combinations that satisfy both conditions. 4

3 Since l ε i 0 holds whenever half of ε i 1, then renumerating i, we get l i l 1 q i l il+1 ow, if q q l is odd, then l εq i i 1 and if q q l is even, then l εq i i 1. Thus, we obtained the following question: For given numbers l and q, find number of ways of choosing q 1,...,q l such that q q l is odd and q q l q, for q i 0,...,l}, i 1,...,l. Thenumberofallsolutionstogetq q q l forq i 0,...,l}, i 1,...,l is q +l 1 F. q Denote now by S the number of all solutions for which the sum of the first l integers q i is odd and by T - the number of all solutions for which the sum of the first l integers q i is even so, F S +T. We have then 1 q i. S We can write now, that T +S T S and T T +S+T S. P Dif T S. 5 It is left to find T S. For this we divide the sequences summing to q into classes and sum over each class separately. We know that q q q l +...+q l. 6 Fix class c c 1,...,c l, where c j 0,...,q} and c c l q. For j 1,...,l we take partition of 6 such that c j q j + q l j+1. For each such class c we consider the difference T S c. Then T S T S c. over all c 3

4 We associating the even numbers q i to 1 and odd numbers q i to 1. We have now, T S c l j1 q 1,...,q l :q j +q l j+1 c j } q j +q l j+1 c j 1 qj. Solutions of the sum under the product are just pairs Therefore, q j +q l j+1 c j 1 qj We have that q j q l j+1 0 c j 1 c j 1.. c j 0. q j +q l j+1 c j 1 q1+...+ql 1, all c j are even, j 1,...,l 0, not all c j are even j 1,...,l. T S c 1, all c j are even, j 1,...,l 0, not all c j are even j 1,...,l. ote, the classes in which not all of c j are even would not change the number T S. Therefore, T S would be equal to the number of classes c c 1,...,c l, where all c j z j are even, which is the number of all possible ways of choosing z j, such that q z z l. Therefore, q +l 1 T S. 7 q Combining 5 and 7, we obtain the bound on P Dif. 4

5 Lemma.. Let ε i,i, be Rademacher random variables satisfying condition and let q q q, q i 0,...,q}. Then, E S i!! q + 1! q! 1! C,q. 8 Proof. Denote D i : ε i 1} and D c i : ε i 1}. ote, the cardinalities cardd cardd c. We have, E S i 1 P S i 1 1 P S i 1 P P Dif. ε i 0 The P Dif have been calculated in 4. Let us find P ε i 0. We have that event } ε i 0 cardd ε i 1, i D & ε 1, i D c }. Thus, P ε i Relations 4 and 9 gives the desired result. 3 Proof of the main Theorem Using multinomial theorem, due to linearity of conditional expectation, we obtain q q! E S ε i a i q 1!...q! aq a q E S i. 10 q q q q i 0,...,q} The conditional expectation in 10 have been computed in 8. 5

6 Letting q i k i,i 1,...,, and, since k i! q k i!, we have q E S ε i a i C,q C,q C,q q! q q! k k q k k q C,q q! q q! a q. q q q q i 0,...,q} q! q 1!...q! ak a k q! q 1!...q! ak a k q! k 1!...k! ak a k Remark 3.1. Using Stirling s formula and the fact that e /, for, expression! q + 1! q!! 1 in 3 can be approximated as! q q! Thus, letting p q, one would get References! q + 1! q!! 1! q q! qq. p 1/p E S ε i a i p a. [1] U. Haagerup, The best constants in the Khintchine inequality, Studia Math., 70, [] H. König and S. Kwapień, Best Khintchine type inequalities for sums of independent, rotationally invariant random vectors, Positivity 5 001, [3] B. Pass, S. Spektor, Khinchine type inequality for k-dependent Rademacher random variables. Statistics and Probability Letters ,

7 [4] S. Spektor, Khinchine inequality for dependent random variables. Canad. Math. Bull ,