2009
DOI: 10.1016/j.spa.2008.10.006
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On the exactness of the Wu–Woodroofe approximation

Abstract: Let (X i ) be a stationary process adapted to a filtration (F i ), E(X i ) = 0, E(X 2 i ) < ∞; by S n = n−1 i=0 X i we denote the partial sums and σ 2 n = S n 2 2 . Wu and Woodroofe [Wei Biao Wu, M. Woodroofe, Martingale approximation for sums of stationary processes, Ann. Probab. 32 (2004) 1674-1690] have shown that if E(S n | F 0 ) 2 = o(σ n ) then there exists an array of row-wise stationary martingale difference sequences approximating the partial sums S n . If ∞ n=1 E(S n |F 0 ) 2 n 3/2 < ∞ then by [M. Ma… Show more

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Cited by 5 publications
(4 citation statements)
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“…The role of these conditions is to control the antisymmetric ("non-reversible") part of the generator by the symmetric one. Another approach goes by imposing decay-rate conditions on time-correlations (e.g., Maxwell and Woodroofe [103], Derriennic and Lin [47], Peligrad and Utev [113], Klicnarová and Volný [88], Volný [133], etc.). However, unlike the reversible situations, it does not seem likely that a single condition will eventually cover all cases of interest.…”
Section: Martingale Approximations and Other Tricksmentioning
confidence: 99%
“…The role of these conditions is to control the antisymmetric ("non-reversible") part of the generator by the symmetric one. Another approach goes by imposing decay-rate conditions on time-correlations (e.g., Maxwell and Woodroofe [103], Derriennic and Lin [47], Peligrad and Utev [113], Klicnarová and Volný [88], Volný [133], etc.). However, unlike the reversible situations, it does not seem likely that a single condition will eventually cover all cases of interest.…”
Section: Martingale Approximations and Other Tricksmentioning
confidence: 99%
“…I would like to thank the referee, who also indicated to me Lemma 2.1 below, and the references [8] and [13].…”
Section: Introductionmentioning
confidence: 99%
“…In the paper [13], there is an example of a stationary ergodic sequence such that E(S n |F 0 ) 2 = o( √ n/ ln(n)) and the CLT fails, but again the variance does not grow linearly. In [8], there is an example for which E(S n |F 0 ) 2 = o( n/ ln(n)) and n −1 E(S 2 n ) → 1 and the CLT fails. In Remark 2 of [8], the authors ask the following question: does E(S n |F 0 ) 2 = o( √ n/ ln(n)) and n −1 E(S 2 n ) → 1 imply the CLT?…”
mentioning
confidence: 99%
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