aa r X i v : . [ m a t h . P R ] J un Esseen type bounds of the remainder in acombinatorial CLT
Andrei N. FrolovDept. of Mathematics and MechanicsSt. Petersburg State UniversitySt. Petersburg, RussiaE-mail address: [email protected] 9, 2018
Abstract
We derive Esseen type bounds of the remainder in a combinatorial central limittheorem for independent random variables without third moments.
AMS 2000 subject classification:
Key words: combinatorial central limit theorem, Berry-Esseen inequality, Esseen inequality
Let k c ij k be a n × n matrix of real numbers such that n X i =1 c ij = n X j =1 c ij = 0 . (1)Let k Y ij k be a n × n matrix of independent random variables with EY ij = 0 and EY ij = σ ij .Assume that π = ( π (1) , π (2) , . . . π ( n )) is a random permutation of , , . . . , n with theuniform distribution on the set of all such permutations. Assume that π and k Y ij k areindependent.Denote S n = n X i =1 c iπ ( i ) + n X i =1 Y iπ ( i ) . n > . Then ES n = 0 , B n = DS n = 1 n − n X i,j =1 c ij + 1 n n X i,j =1 σ ij . To avoid triviality, we assume in the sequel that B n > .Put ∆ n = sup x ∈ R (cid:12)(cid:12)(cid:12) P ( S n < x p B n ) − Φ( x ) (cid:12)(cid:12)(cid:12) , where Φ( x ) is the standard normal distribution function.First results on asympotical normality of S n were obtained in the case P ( Y ij = 0) = 1 for i, j n . Motivated by statistical applications, Wald and Wolfowits (1944)stated conditions, sufficient for ∆ n → as n → ∞ when c ij = a i b j . Noether (1949)proved that these conditions maybe replaced by weaker ones. Hoeffding (1951) consideredgeneral case of c ij and obtained a combinatorial central limit theorem (CLT). Furtherresults on combinatorial CLT were obtained by Motoo (1957) and Kolchin and Chistyakov(1973). Von Bahr (1976) and Ho and Chen (1978) derived bounds for the remainder incombinatorial CLT in the case on non-degenerated Y ij . Botlthausen (1984) obtainedEsseen type inequality for the remainder in the case of degenerated Y ij . The constant wasnot be specified in the latter paper. Further results of this type were proved by Goldstein(2005) (see also Chen, Goldstein and Shao (2011)). They contain explicit constants in theinequalities. For non-degenerated Y ij , Esseen type inequalities were stated by Neammaneeand Suntornchost (2005), Neammanee and Rattanawong (2009) and Chen and Fang (2012)(see also comments on p.2 of Chen and Fang (2012)).At the moment, best results on bounds in combinatorial CLT are obtained by Stein’smethod. The detailed discussion of this approach may be found in Chen and Fang (2012)and references therein.We start with the following known result. Theorem A. If E | Y ij | < ∞ for i, j n then there exists an absolute positiveconstant A such that ∆ n A B / n n n X i,j =1 | c ij + Y ij | . (2)Chen and Fang (2012) have proved that inequality (2) holds with A = 447 .In this paper, we derive generalizations of (2) to the case of random variables Y ij without third moments. We apply the classical technique of truncation in a similar wayas in Petrov (1995) where generalizations of the Esseen’s inequality may be found.Our first result is as follows. 2 heorem 1. There exists an absolute positive constant A such that ∆ n A B / n n n X i,j =1 | c ij | + 1 B n n n X i,j =1 α ij + 1 B / n n n X i,j =1 β ij ! , (3) where α ij = EY ij I {| Y ij | > √ B n } and β ij = E | Y ij | I {| Y ij | < √ B n } for i, j n , and I {·} denotes the indicator of the event in brackets. Remark 1.
One can put A = max { , A + 5 } in Theorem 1. Theorem 1 implies the next result.
Theorem 2.
Let g ( x ) be a positive, even function such that g ( x ) and x/g ( x ) are non-decreasing for x > . Suppose that g ij = EY ij g ( Y ij ) < ∞ for i, j n . Then ∆ n A B / n n n X i,j =1 | c ij | + 1 B n g ( √ B n ) n n X i,j =1 g ij ! . (4)It is clear that α ij g ij /g ( √ B n ) and β ij √ B n g ij /g ( √ B n ) for i, j n . Hence (3)implies (4). At the same time, putting g ( x ) = min {| x | , √ B n } in Theorem 2, we concludethat (4) yields (3) with A instead of A .One can obtain variants of Theorem 2 as follows. Writing α ij = EY ij I { Y ij − p B n } + EY ij I { Y ij > p B n } allows estimate α ij by a sum of different moments of random variables Y − ij and Y + ij , where Y ± ij = max {± Y ij , } . The same maybe done also for β ij . So, if g − ( x ) and g + ( x ) , x > ,are positive, non-decreasing functions such that x/g − ( x ) and x/g + ( x ) are non-decreasing,then α ij g − ij g − ( √ B n ) + g + ij g + ( √ B n ) and β ij √ B n g − ij g − ( √ B n ) + √ B n g + ij g + ( √ B n ) , where g − ij = E ( Y − ij ) g − ( Y − ij ) and g + ij = E ( Y + ij ) g + ( Y + ij ) . Further generalizations maybederived in the same way for g − ( x ) and g + ( x ) depending on i and j .One of the most important case of g ( x ) is g ( x ) = | x | δ , δ ∈ (0 , , in which Theorem 2turns to the following result. Theorem 3.
Assume that E | Y ij | δ < ∞ for i, j n , where δ ∈ (0 , . Then ∆ n A B / n n n X i,j =1 | c ij | + 1 B δ/ n n n X i,j =1 E | Y ij | δ ! . (5)3eneralizations of Theorem 3 maybe obtained in the same way as it was mentionedfor Theorem 2 above.Inequality (2) is better than (5) for δ = 1 . But if P ( Y ij = 0) = 1 for i, j n , thenright-hand sides of these bounds coincide up to constants.The second term in righthand side of (5) is the Lyapunov type ratio for Y ij . Theorem 3yields combinatorial CLT under Lyapunov type condition on Y ij . Theorem 1 also impliescombinatorial CLT under Lindeberg type condition on Y ij . The latter follows from (3)and β ij p B n (cid:16) EY ij I {| Y ij | > ε p B n } + εσ ij (cid:17) for i, j n and all ε ∈ (0 , . Proof of Theorem 1.
Assume first that B n = 1 .Put ¯ Y ij = Y ij I {| Y ij | < } , ¯ a ij = E ¯ Y ij , ¯ σ ij = D ¯ Y ij , ¯ a i. = 1 n n X j =1 ¯ a ij , ¯ a .j = 1 n n X i =1 ¯ a ij , ¯ a .. = 1 n n X i,j =1 ¯ a ij , for i, j n . Denote ¯ S n = n X i =1 c iπ ( i ) + n X i =1 ¯ Y iπ ( i ) . Hence ¯ e n = E ¯ S n = n ¯ a .. , ¯ B n = D ¯ S n = 1 n − n X i,j =1 ( c ij + ¯ a ij − ¯ a i. − ¯ a .j + ¯ a .. ) + 1 n n X i,j =1 ¯ σ ij . Put F n ( x ) = P ( S n < x ) , ¯ F n ( x ) = P (cid:0) ¯ S n < x (cid:1) . Denote ∆ n = sup x ∈ R | F n ( x ) − ¯ F n ( x ) | , ∆ n = sup x ∈ R | ¯ F n ( x ) − Φ(( x − ¯ e n ) / p ¯ B n ) | , ∆ n = sup x ∈ R | Φ(( x − ¯ e n ) / p ¯ B n ) − Φ( x/ p ¯ B n ) | , ∆ n = sup x ∈ R | Φ( x/ p ¯ B n ) − Φ( x ) | . It is clear that ∆ n = sup x ∈ R | F n ( x ) − Φ( x ) | X k =1 ∆ nk . C n = 1 n n X i,j =1 | c ij | , L n = 1 n n X i,j =1 EY ij I {| Y ij | > } . We have − ¯ B n = B n − ¯ B n = 1 n n X i,j =1 ( σ ij − ¯ σ ij ) + 1 n − n X i,j =1 ( c ij − ( c ij + ¯ a ij − ¯ a i. − ¯ a .j + ¯ a .. ) )= 1 n n X i,j =1 ( σ ij − ¯ σ ij ) − n − n X i,j =1 (¯ a ij − ¯ a i. − ¯ a .j + ¯ a .. ) − n − n X i,j =1 c ij (¯ a ij − ¯ a i. − ¯ a .j + ¯ a .. )= 1 n n X i,j =1 ( σ ij − ¯ σ ij ) − n − n X i,j =1 (¯ a ij − ¯ a i. − ¯ a .j + ¯ a .. ) − n − n X i,j =1 c ij ¯ a ij . In the last equality, we have used (1).In the sequel, we will repeatedly use that | ¯ a ij | < for i, j n .We have σ ij − ¯ σ ij = EY ij I {| Y ij | > } + (¯ a ij ) EY ij I {| Y ij | > } + | ¯ a ij | (6)for i, j n .Taking into account that ( x + y ) x + y ) for all real x and y , we get (¯ a ij − ¯ a i. − ¯ a .j + ¯ a .. ) a ij + ¯ a i. + ¯ a .j + ¯ a .. ) | ¯ a ij | + | ¯ a i. | + | ¯ a .j | + | ¯ a .. | ) (7)for i, j n . Moreover, n X i,j =1 | ¯ a i. | n X i,j =1 n n X j =1 | ¯ a ij | ! = n X i,j =1 | ¯ a ij | (8)and, in the same way, n X i,j =1 | ¯ a .j | n X i,j =1 | ¯ a ij | . (9)Further, n X i,j =1 | ¯ a .. | n X i,j =1 n n X i,j =1 | ¯ a ij | ! = n X i,j =1 | ¯ a ij | . (10)5oting that xy x / y / / for all non-negative x and y , we get n X i,j =1 | c ij ¯ a ij | n X i,j =1 | c ij | + 23 n X i,j =1 | ¯ a ij | / n X i,j =1 | c ij | + 23 n X i,j =1 | ¯ a ij | . (11)Making use of EY ij = 0 , | ¯ a ij | = | EY ij I {| Y ij | < }| = | EY ij I {| Y ij | > }| E | Y ij | I {| Y ij | > } EY ij I {| Y ij | > } (12)for i, j n .It follows by (6)–(12) that | B n − ¯ B n | (cid:18) nn − n n − (cid:19) L n + n n − C n . This yields that for n > , | B n − ¯ B n | L n + C n . (13)Assume that ¯ B n ̺B n where ̺ ∈ (0 , . Then L n + C n − ̺ > > ∆ n (14)and the conclusion of Theorem 1 holds.In the sequel, we assume that ¯ B n > ̺B n .Now we will estimate ∆ nk , k = 1 , . Since { S n < x } ⊂ (cid:8) ¯ S n < x (cid:9) ∪ n [ i =1 (cid:8) | Y iπ ( i ) | > (cid:9) , we have F n ( x ) ¯ F n ( x ) + n X i =1 P (cid:0) | Y iπ ( i ) | > (cid:1) = ¯ F n ( x ) + 1 n n X i,j =1 P ( | Y ij | > ¯ F n ( x ) + 1 n n X i,j =1 EY ij I {| Y ij | > } = ¯ F n ( x ) + L n . From the other hand (cid:8) ¯ S n < x (cid:9) ⊂ { S n < x } ∪ n [ i =1 (cid:8) | Y iπ ( i ) | > (cid:9) , which yields that ¯ F n ( x ) F n ( x ) + L n .
6t follows that ∆ n L n . (15)Applying Theorem A with c ij + ¯ a ij − ¯ a i. − ¯ a .j + ¯ a .. and ¯ Y ij − ¯ a ij instead of c ij and Y ij correspondingly, we get inequality ∆ n A n ¯ B / n n X i,j =1 E (cid:12)(cid:12) c ij + ¯ Y ij − ¯ a i. − ¯ a .j + ¯ a .. (cid:12)(cid:12) . Using that | x + y | | x | + | y | ) for all real x and y , we obtain ∆ n A n̺ / n X i,j =1 ( | c ij | + E (cid:12)(cid:12) ¯ Y ij (cid:12)(cid:12) + | ¯ a i. | + | ¯ a .j | + | ¯ a .. | ) A n̺ / n X i,j =1 ( | c ij | + E (cid:12)(cid:12) ¯ Y ij (cid:12)(cid:12) + | ¯ a i. | + | ¯ a .j | + | ¯ a .. | ) A ̺ / ( C n + 3 L n + D n ) , (16)where D n = 1 n n X i,j =1 E (cid:12)(cid:12) ¯ Y ij (cid:12)(cid:12) = 1 n n X i,j =1 E | Y ij | I {| Y ij | < } . The inequality sup x ∈ R | Φ( x + y ) − Φ( x ) | | y |√ π holds for all real y . By the latter inequality and (12), we have ∆ n | ¯ e n |√ π ¯ B / n | ¯ e n |√ π̺ / √ π̺ / n n X i,j =1 | ¯ a ij | L n √ π̺ / . (17)Note that sup x ∈ R | Φ( xy ) − Φ( x ) | y − √ πe for all y > . If ¯ B n < B n = 1 , then ∆ n √ πe p ¯ B n − ! and we conclude by (13) that p ¯ B n − B n − ¯ B n p ¯ B n ( √ B n + p ¯ B n ) L n + C n ̺ / (1 + ̺ / ) . y ∈ (0 , , then sup x ∈ R | Φ( xy ) − Φ( x ) | − yy √ πe . So, for ¯ B n > B n = 1 , inequality ∆ n √ πe (cid:16)p ¯ B n − (cid:17) holds and we have p ¯ B n − B n − ¯ B n √ B n + p ¯ B n . It follows that ∆ n L n + C n √ πe̺ / (1 + ̺ / ) . (18)Relations (15)–(18) imply ∆ n A ( L n + C n + D n ) , when B n = 1 . If B n = 1 , we replace c ij and Y ij by c ij / √ B n and Y ij / √ B n in the previouspart of the proof. Then we obtain from the last inequality that ∆ n A n n X i,j =1 | c ij | B / n + 1 n n X i,j =1 E Y ij B n I (cid:26) | Y ij |√ B n > (cid:27) + 1 n n X i,j =1 E | Y ij | B / n I (cid:26) | Y ij |√ B n < (cid:27)! , which coincides with (3). ✷ Proof of Remark 1.
From (14)–(18) we get A = max (cid:26) − ̺ , A ̺ / + 1 √ π̺ / + 36 √ πe̺ / (1 + ̺ / ) (cid:27) . This yields that A < max (cid:26) − ̺ , A ̺ / + 0 . ̺ / + 5 . ̺ / (1 + ̺ / ) (cid:27) . Taking ̺ / = 0 . , we get A < max { , A + 5 } . ✷ References
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