请输入您要查询的字词:

 

单词 ProofOfBennettInequality
释义

proof of Bennett inequality


By Chernoff-Cramèr inequalityMathworldPlanetmath (http://planetmath.org/ChernoffCramerBound), we have:

Pr{i=1n(Xi-E[Xi])>ε}exp[-supt0(tε-ψ(t))]

where

ψ(t)=i=1n(lnE[etXi]-tE[Xi]).

Keeping in mind that the condition

Pr{|Xi|M}=1 i

implies that, for all i,

E[|Xi|k]Mk k0

(see here (http://planetmath.org/RelationBetweenAlmostSurelyAbsolutelyBoundedRandomVariablesAndTheirAbsoluteMoments) for a proof) and since lnxx-1 x>0, and

E[|X|k]Mk  E[|X|k]E[X2]Mk-2 k2,kN

(see here (http://planetmath.org/AbsoluteMomentsBoundingNecessaryAndSufficientCondition) for a proof), one has:

ψ(t)=i=1n(lnE[etXi]-tE[Xi])
i=1nE[etXi]-tE[Xi]-1
=i=1nE[k=0(tXi)kk!]-tE[Xi]-1
=i=1n(k=0tkE[Xik]k!)-tE[Xi]-1
=i=1n(k=2tkE[Xik]k!)
i=1n(k=2tkE[|Xi|k]k!)
i=1n(k=2tkE[Xi2]Mk-2k!)
=k=2tkMk-2i=1nE[Xi2]k!
=v2M2k=2(tM)kk!
=v2M2[exp(tM)-tM-1]

One can now write

supt0(tε-ψ(t))supt0(tε-v2M2(etM-tM-1))=supt>0[v2M2(M2εv2t-(etM-tM-1))].

By elementary calculus, we obtain the value of t that maximizes theexpression in round brackets:

topt=1Mln(1+Mεv2)

which, once plugged into the bound, yields

Pr{i=1n(Xi-E[Xi])>ε}exp[-v2M2((1+Mεv2)ln(1+Mεv2)-Mεv2)].

Observing that (1+x)ln(1+x)-xx2ln(1+x) x0 (see here (http://planetmath.org/ASimpleMethodForComparingRealFunctions)), one gets thesub-optimal yet more easily manageable formula:

Pr{i=1n(Xi-E[Xi])>ε}exp[-ε2Mln(1+εMv2)].
随便看

 

数学辞典收录了18232条数学词条,基本涵盖了常用数学知识及数学英语单词词组的翻译及用法,是数学学习的有利工具。

 

Copyright © 2000-2023 Newdu.com.com All Rights Reserved
更新时间:2025/5/4 5:45:58