Burkholder s inequality for multiindex martingales

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1 Annales Mathematicae et Informaticae 32 (25) pp Burkholder s inequality for multiindex martingales István Fazekas Faculty of Informatics, University of Debrecen fazekasi@inf.unideb.hu Dedicated to the memory of my teacher Péter Kiss Abstract Multiindex versions of Khintchine s and Burkholder s inequalities are presented. Key Words: Khintchine s inequality, Burkholder s inequality, random field, martingale. AMS Classification Number: 6G42, 6G6, 6E Introduction and notation Burkholder s inequality is a powerful tool of martingale theory. Let (Z n, F n ), n = 1, 2,..., be a martingale with difference X n = Z n Z n 1. Let p > 1. There exist finite and positive constants C p and D p depending only on p such that ( n ) ] p/2 1/p [ ( C p [E k=1 X2 k (E Z n p ) 1/p n ) ] p/2 1/p D p E k=1 X2 k, (1.1) see Burkholder s classical paper [1] and the textbook [2]. When the random variables X 1, X 2,... are independent (1.1) is called the Marcinkiewicz-Zygmund inequality (and in this particular case it is valid also for p = 1). Let ε i (t), i = 1, 2,..., be the Rademacher system on [, 1]. If X k = ε k a k, then we obtain Khintchine s inequality. There exist finite and positive constants A p and B p depending only on p such that for any real sequence a k, k = 1, 2,..., ( n ) 1/2 [ A p k=1 a2 k n ] 1/p ε p ( n k(t)a k dt B p k=1 k=1 a2 k ) 1/2. (1.2) Supported by the Hungarian Foundation of Scientific Researches under Grant No. OTKA T4767/24 and Grant No. OTKA T48544/25. 45

2 46 I. Fazekas This inequality is valid for p >. Actually, the standard proof of (1.1) is based on (1.2), see [1]). The two-index version of (1.1) is obtained in [8], see also [7]. The aim of this paper is to prove a multiindex version of Burkholder s inequality. The proof is based on the transform of a single parameter martingale. We also use the multiindex version of Khintchine s inequality (for the sake of completeness, we prove it). In [9] the second inequality of (3.2) was presented (without proof) for p > 2. It was applied to obtain a Brunk-Prokhorov type strong law of large numbers for martingale fields (see [9], Proposition 14). For a recent overview of multiindex random processes see [6]. In [6] a certain version of the Burkholder inequality was presented for continuous parameter random fields without the details of the proof (p. 257, Theorem 4.1.2). We do not use that theorem, we give a simple proof based on well-known one-parameter results. Our Burkholder type inequality can be used to prove convergence results for multiindex autoregressive type martingales (see [5], for the two-index case see [4]). We use the following notation. Let d be a fixed positive integer. Let N denote the set of positive integers, N the set of non-negative integers. The multidimensional indices will be denoted by k = (k 1,..., k d ), n = (n 1,..., n d ), N d. Relations, min are defined coordinatewise. I.e. k n means k 1 n 1,..., k d n d. Relation k < n means k n but k n. Let X p = (E X p ) 1/p for p >. Then X p1 X p2 for < p 1 p Khintchine s inequality Theorem 2.1. Let ε i (t), i = 1, 2,..., be the Rademacher system on [, 1]. Let p >. There exist finite and positive constants A p,d and B p,d depending only on p and d such that for any d-index sequence a k, k N d, 1/2 [ A p,d k) ( a2 ] 1/p ε p k 1 (t 1 ) ε kd (t d )a k dt1... dt d B p,d ( a2 k) 1/2. (2.1) Proof. First we remark that for d = 1 inequality (2.1) is the original Khintchine s inequality. Denote by ε i,ni, n i = 1, 2,..., i = 1, 2,..., d, independent sequences of independent Bernoulli random variables with P(ε i,ni = 1) = P(ε i,ni = 1) = 1/2 for each i and n i. Let s n = ( a2 k) 1/2 and Sn = ε 1,k 1 ε d,kd a k. Then, by the Fubini theorem, inequality (2.1) is equivalent to A p,d s n S n p B p,d s n. (2.2) Now we prove that the products ε 1,k1 ε d,kd, (k 1,..., k d ) N d, are pairwise independent Bernoulli variables. By induction, it is enough to prove that ε 1,k1 ε 2,k2,

3 Burkholder s inequality for multiindex martingales 47 (k 1, k 2 ) N 2, are pairwise independent Bernoulli variables if ε 1,k1, k 1 N, and ε 2,k2, k 2 N, are independent sequences of pairwise independent Bernoulli variables. Indeed, if ε 1 and ε 2 are independent Bernoulli variables then their product is Bernoulli: P(ε 1 ε 2 = ±1) = 1/2. Now turn to the independence. It is obvious that the independence of ε 1, ε 2, ε 3, and ε 4 implies the independence of ε 1 ε 2 and ε 3 ε 4. Moreover, the independence of ε 1, ε 2 and ε 3 implies the independence of ε 1 ε 3 and ε 2 ε 3 : P (ε 1 ε 3 = ±1, ε 2 ε 3 = ±1) = 1 4 = P (ε 1ε 3 = ±1) P (ε 2 ε 3 = ±1). Therefore S n 2 2 is the variance of the sum of pairwise indepenent random variables, so we have s n = S n 2. In particular, (2.2) is true for p = 2. Now we show that the products ε 1,k1 ε d,kd, (k 1,..., k d ) N d, are not (completely) independent. Indeed, if ε 1, ε 2, ε 3, and ε 4 are independent Bernoulli variables, then ε 1 ε 3, ε 2 ε 3, ε 1 ε 4, and ε 2 ε 4 are not independent: P (ε 1 ε 3 = 1, ε 2 ε 3 = 1, ε 1 ε 4 = 1, ε 2 ε 4 = 1) = 1/8 1/16 = P (ε 1 ε 3 = 1) P (ε 2 ε 3 = 1) P (ε 1 ε 4 = 1) P (ε 2 ε 4 = 1). So relation (2.2) is really different from its one-index version. Now we prove the second part of (2.2). We start with the case of p 2. We use induction. For d = 1 it is the original Khintchine s inequality. Assume (2.2) for d 1. Let I k1,n 2,...,n d (t 2,..., t d ) = n 2 k 2 =1 n d Then, by the original Khintchine s inequality, k d =1 ε k 2 (t 2 ) ε kd (t d )a k1,k 2,...,k d. n 1 ε k 1 (t 1 )I k1,n 2,...,n k d (t 2,..., t d ) p dt 1 1=1 From here B p p,1 ( n1 ) p/2 k 1 =1 I2 k 1,n 2,...,n d (t 2,..., t d ). S n p p B p p,1 B p p,1 { n1 k 1=1 ( n1 [ k 1 =1 ( n 1 n2 B p p,1 B p,d 1 k 1=1 I 2 k 1,n 2,...,n d (t 2,..., t d )) p/2 dt 2... dt d ( I 2 k1,n 2,...,n d (t 2,..., t d ) ) p/2 dt2... dt d ] 2/p } p/2 k 2=1 n d k d =1 ) 1/2 2 a 2 k 1,k 2,...,k d p/2

4 48 I. Fazekas = (B p,d s n ) p, where we used the triangle inequality in L p/2 and (2.2) for d 1. So we proved the second part of (2.2) for p 2. As S n p S n 2 for < p 2, the second part of (2.2) is true for < p. Now turn to the first part of (2.2). We see that s n = S n 2 S n p for p 2. Therefore it is enough to prove the inequality for < p < 2. We follow the lines of [2], p Let < p < 2. Choose r 1, r 2 >, r 1 + r 2 = 1, pr 1 + 4r 2 = 2. By Holder s inequality and the second part of (2.2), we have s 2 n = S n 2 2 S n pr 1 p S n 4r2 4 S n pr 1 p Bs 4r 2 n. From here S n pr1 p (1/B)s 2 4r2 n Therefore the first part of (2.2) is true for < p < 2. = (1/B)s pr1 n. 3. Burkholder s inequality Let (X n, F n ), n N d, be a martingale difference. It means that F n, n N d, is an increasing sequence of σ-algebras, i.e. F k F n if k n; X n is F n -measurable and integrable; E(X n F k ) = if k < n. To obtain Burkholder s inequality, we shall assume the so called condition (F4). I. e. E {E(η F m ) F n } = E { } η F min{m,n (3.1) for each integrable random variable η and for each m, n N d (see, e.g., [6] and [3]). Denote by (Z n, F n ), n N d, the martingale corresponding to the difference (X n, F n ), n N d. More precisely, let Z n = and F n = {, Ω} if n N d \ N d and Z n = X k, n N d. Theorem 3.1. Let (Z n, F n ), n N d, be a martingale and (X n, F n ), n N d, the martingale difference corresponding to it. Assume that (3.1) is satisfied. Let p > 1. There exist finite and positive constants C p,d and D p,d depending only on p and d such that ( ) ] p/2 1/p [ ( ) ] p/2 1/p C p,d [E X2 k (E Z n p ) 1/p D p,d E X2 k. (3.2) Proof. We follow the lines of [8]. Let u i,ni {, 1}, n i = 1, 2,..., i = 1, 2,..., d. Let T n = u 1,k 1 u d,kd X k = n 1 k 1=1 u 1,k 1 Y k1,

5 Burkholder s inequality for multiindex martingales 49 where Y k1 = Y k1,n 2,...,n d = n 2 First we show that k 2 =1 n d k d =1 u 2,k 2 u d,kd X k1,k 2,...,k d. E Z n p M d E T n p. (3.3) We use induction. For d = 1 (3.3) is included in [1], p. 152 (because T n is a transform of the martingale Z n and vice versa). Now we assume that (3.3) is true for d 1. Let n 2,..., n d be fixed, F k1 = F k1,n 2,...,n d. Then, using (3.1), we can show that (Y k1, F k1 ), k 1 = 1, 2,..., is a martingale difference. As the martingale n 1 k Y 1=1 k 1 = n 1 k u 1=1 1,k 1 (u 1,k1 Y k1 ) is a transform of the martingale n1 k (u 1=1 1,k 1 Y k1 ), by [1], p. 152, E n 1 Y p k 1 M1 E n 1 (u 1,k 1 Y k1 ) p. (3.4) k 1 =1 k 1 =1 { Now, using (3.1), we can show that for any fixed n 1 the (d 1)-index sequence n1 k 1 =1 X } k 1,k 2,...,k d, F n1,k 2,...,k d, (k2,..., k d ) N d 1, is a martingale difference. Therefore, using (3.3) for d 1, we obtain E Z n p = E n 2 n d [ n1 ] X p k 1,...,k k 2=1 k d =1 k d 1=1 M d 1 E n 2 n d [ u n1 ] 2,k 2 u X p d,kd k 1,...,k k 2=1 k d =1 k d = 1=1 = M d 1 E n 1 Y p k 1 Md 1 M 1 E n 1 (u 1,k 1 Y k1 ) p = (3.5) k 1=1 k 1=1 = M d E T n p. In (3.5) we applied (3.4). So we proved (3.3). Because Z n and T n are each other s transforms, (3.3) implies N d E T n p E Z n p M d E T n p. (3.6) Now we prove the first part of (3.2). By (2.1), E X 2 k p/2 = = p 1 1 E ε k1 (t 1 ) ε kd (t d )X k dt 1... dt d 1 1 p E ε k1 (t 1 ) ε kd (t d )X k dt 1... dt d p E N d X k dt 1... dt d 1 1 N d E Z n p.

6 5 I. Fazekas In the third step we applied (3.6). We turn to the second part of (3.2). By (3.6), [ E Z n p p] M d E ε k 1 (t 1 ) ε kd (t d )X k. From here, using (2.1), E Z n p M d M d B p p,d E ( X2 k) p/2. [ p] E ε k 1 (t 1 ) ε kd (t d )X k dt 1... dt d The proof is complete. 4. Final comments Burkholder s inequality is valid for martingales with values in R t (t is a fixed ( t ) 1/p. positive integer). For p > and x = (x 1,..., x t ) R t let x p = i=1 x i p Let (X n, F n ), n N d, be a martingale difference with values in R t. Assume that condition (F4) is satisfied. Let (Z n, F n ), n N d, be the martingale corresponding to the difference (X n, F n ), n N d. Theorem 4.1. Let (Z n, F n ), n N d, be a martingale with values in R t and (X n, F n ), n N d, the martingale difference corresponding to it. Assume that (3.1) is satisfied. Let p > 1. There exist finite and positive constants C and D depending only on t, p and d such that C E p/2 1/p X k 2 2 (E Z n p 2 )1/p D E p/2 1/p X k 2 2. (4.1) Proof. It is known that for any p, q > there exist < c, d < such that c x p x q d x p for all x R t. Applying this observation and (3.2) we obtain (4.1). Using this theorem we can prove limit theorems for autoregressive type martingale fields. For details see [5] and [4] including the d-index case and the two-index case, respectively. References [1] Burkholder, D. L., Martingale transforms. Ann. Math. Stat. Vol. 37 (1966),

7 Burkholder s inequality for multiindex martingales 51 [2] Chow, Y. S., Teicher, H., Probability Theory. Springer, New York, [3] Fazekas, I., Convergence of vector valued martingales with multidimensional indices. Publ. Math. Debrecen, Vol. 3 (1983), [4] Fazekas, I., On the convergence of regression type martingal fields. Proc. of the 5th Pannonian Symp., 43 52, Reidel, Dordrecht, [5] Karácsony, Zs., Autoregressive type martingale fields. Accepted by Acta Mathematica Academiae Paedagogicae Nyiregyhaziensis. [6] Khosnevisan, D., Multiparameter Processes. Springer, New York, 22. [7] Ledoux, M., Inégalités de Burkholder pour martingales indexées par N N. Twoindex random processes, Lecture Notes in Math. Vol. 863, , Springer, Berlin, [8] Métraux, Ch., Quelques inégalités pour martingales à paramètre bidimensionnel. Séminaire de Probabilités XII, Université de Strasbourg, Lecture Notes in Math. Vol. 649, , Springer, Berlin, [9] Noszály, Cs., Tómács, T., A general approach to strong laws of large numbers for fields of random variables. Annales Univ. Sci. Budapest Eötvös Sect. Math. Vol. 43 (2), István Fazekas Faculty of Informatics University of Debrecen P.O. Box Debrecen Hungary