2018
DOI: 10.1016/j.spl.2017.09.013
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A large deviation inequality forβ-mixing time series and its applications to the functional kernel regression model

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Cited by 5 publications
(5 citation statements)
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“…random variables [52], the exponent of the tail bound has a sublinear dependence on sample size due to the presence of n. n is close to n when γ is large corresponding to a faster decaying β-mixing coefficient (Definition 4). This sublinear relation with respect to n is also reported by [49] and [48] as well under exponential α and β-mixing conditions with γ = 1. They provided an O(exp(−nϵ/(log n log log n))) tail bound, which is a faster rate in n compared to Proposition 5 with γ = 1.…”
Section: Resultssupporting
confidence: 79%
See 1 more Smart Citation
“…random variables [52], the exponent of the tail bound has a sublinear dependence on sample size due to the presence of n. n is close to n when γ is large corresponding to a faster decaying β-mixing coefficient (Definition 4). This sublinear relation with respect to n is also reported by [49] and [48] as well under exponential α and β-mixing conditions with γ = 1. They provided an O(exp(−nϵ/(log n log log n))) tail bound, which is a faster rate in n compared to Proposition 5 with γ = 1.…”
Section: Resultssupporting
confidence: 79%
“…The α-mixing concentration inequality in [47,Theorem 2] gives the tail bound on the relative deviation (scaled by variance) instead of the absolute deviation. The β-mixing results in [48] and the α-mixing results in [49] provide a subexponential bound of O(exp(−ϵ)) which is a weaker dependency on ϵ than we desired. The detailed discussion of the concentration inequalities we derived is postponed until the main results are introduced.…”
Section: Resultsmentioning
confidence: 72%
“…Indeed, this follows with some calculations (see the proof of Theorem 11.2 [22]). Furthermore, we need a result, which follows using the β-mixing property and a lemma in [38], s,t∈I(l,u)…”
Section: Proofs Of the Results In Section 2 And Sectionmentioning
confidence: 99%
“…For (4.23), use that A n,t = Ψ(X t−1 ) − Ψ(X t−1 ) − ε. We apply Proposition 2 in Krebs (2018b) which allows us to exchange the first probability in (4.23) by the expectation w.r.t. the product measure, i.e.,…”
Section: Technical Resultsmentioning
confidence: 99%