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精编资料,供您参考切比雪夫不等式证明精编5篇【前言导读】由三一刀客最美丽的网友为您分享整理的“切比雪夫不等式证明精编5篇”文档资料,以供您学习参考,希望这篇文档对您有所帮助,喜欢就分享给朋友们呢!切比雪夫不等式及其应用(摘要1天津理工大学2014届本科毕业论文切比雪夫不等式及其应用摘要切比雪夫不等式是概率论中重要的不等式之一。尤其在分布未知时,估计某些事件的概率的上下界时,常用到切比雪夫不等式。另外,大数定律是概率论极限理论的基础,而切比雪夫不等式又是证明大数定律的重要途径。如今,在切比雪夫不等式的基础上发展起来的一系列不等式都是研究中心极限定理的有力工具。作为一个理论工具,切比雪夫不等式的地位是很高的。本文首先介绍了切比雪夫不等式的一些基本理论,引出其概率形式,用现代概率方法证明了切比雪夫不等式并给出了其等号成立的充要条件。其次,从三大方面阐述了其在概率论中的应用,并且给出了切比雪夫大数定律和伯努利大数定律的证明。在充分了解切比雪夫不等式后,最后探索了其在生活中的应用,并且用切比雪夫不等式评价了irr的概率风险分析。关键词:切比雪夫不等式大数定律irrthechebyster’sinequalityanditsapplicationsabstractinprobabilitytheory,thechebyshev’sinequalityisoneoftheimportantinequalities.inparticularthedistributionisunknown,thechebyshev’sinequalityisusuallyusedwhenestimatingtheboundaryfromaboveorbelowofprobability.inaddition,thelawoflargenumbersisthebasisofthelimittheoryofprobability.thechebyshev’sinequalityisanimportantwaytoproveit.now,aseriesofinequalitiesthataredevelopedonthebasisofthechebyshev’sinequalityareapowerfultoolforthecentrallimittheorem.asatheoreticaltool,itsstatusisveryhigh.first,thisarticleintroducessomebasictheoryofthechebyshev’sinequality,itraises精编资料,供您参考thechebyshev’sinequality’sformofprobabilityandmakesaproveforthechebyshev’sinequalitywiththemethodofmodernprobability.furthermore,itgivesthenecessaryandsufficientconditionoftheestablishmentoftheequalsign.天津理工大学2014届本科毕业论文secondly,weintroducesitsfiveapplicationinprobabilitytheoryandgivestheproveofthechebyshevandbernoullilawoflargenumbers.afterthefullunderstandingofthechebyshev’sinequality,finally,weexploreitsapplicationinthelifeandgivetheprobabilisticriskassessmentoftheirrwiththechebyshev’sinequality.keywords:chebyshev’sinequalitylawoflargenumbersirr经典不等式证明-柯西不等式-排序不等式-切比雪夫不等式-均值不等式2mathwang几个经典不等式的关系一几个经典不等式(1)均值不等式设a1,a2,?an?0是实数a?a???a12n???111n?+??a1a2an其中ai?0,i?1,2,?n.当且仅当a1?a2???an时,等号成立。n(2)柯西不等式设a1,a2,?an,b1,b2,?bn是实数,则?a2122?a2???an??b12?b22???bn2???a1b1?a2b2???anbn?2当且仅当bi?0(i?1,2,?,n)或存在实数k,使得ai?kbi(i?1,2,?,n)时,等号成立。(3)排序不等式设a1?a2???an,b1?b2???bn为两个数组,c1,c2,?,cn是b1,b2,?,bn的任一排列,则a1b1?a2b2???anbn?a1c1?a2c2???ancn?a1bn?a2bn?1???anb1当且仅当a1?a2???an或b1?b2???bn时,精编资料,供您参考等号成立。(4)切比晓夫不等式对于两个数组:a1?a2???an,b1?b2???bn,有a1b1?a2b2???anbn?a1?a2???an??b1?b2???bn?a1bn?a2bn?1???anb1??????nnnn????当且仅当a1?a2???an或b1?b2???bn时,等号成立。二相关证明(1)用排序不等式证明切比晓夫不等式证明:由a1b1?a2b2???anbn?a1?a2???an??b1?b2???bn??????nnn?????n?a1b1?a2b2???anbn???a1?a2???an??b1?b2???bn?而?a1?a2???an??b1?b2???bn??a1b1?a2b2???anbn?a1b2?a2b3???anb1?a1b3?a2b4???anb2?a1b4?a2b5???anb3???a1bn?1?a2bn???anbn?2?a1bn?a2b1???anbn?1根据“顺序和?乱序和”(在n?1个部分同时使用),可得n?a1b1?a2b2???anbn???a1?a2???an??b1?b2???bn?即得a1b1?a2b2???anbn?a1?a2???an??b1?b2???bn??????nnn????同理,根据“乱序和?反序和”,可得?a1?a2???an??b1?b2???bn?a1bn?a2bn?1???anb1?????nnn????综合即证(2)用排序不等式证明“几何—算数平均不等式”?证明:构造两个数列:a1?a2???annaa?aa1aa,x2?122,?xn?12nn?1ccc1c1c21cny1??,y2??,?yn???1精编资料,供您参考x1a1x2a1a2xna1a2?anx1?其中c?。因为两个数列中相应项互为倒数,故无论大小如何,乘积的和:............................x1y1?x2y2??xnyn总是两数组的反序和。于是由“乱序和?反序和”,总有.........x1yn?x2y1??xnyn?1?x1y1?x2y2??xnyn于是aa1a2????n?1?1???1ccc即a1?a2???an?nc即证a1?a2???an?c?na1?a2???an(3)用切比晓夫不等式证明“算数—开方平均不等式”:?n证明:不妨设a1?a2???an,222a1?a2???an?a1?a2???an??a1?a2???an?a1?a2???an。???????nnnn????由切比晓夫不等式,右边不等式显然成立。即证。(4)用切比晓夫不等式证明“调和—算数平均不等式”n?+??a1a2an?a1?a2???ann证明:n111?+??a1a2an?a1?a2???ann1?11?+??a1a2an?a1?a2???an??????nn?????111?精编资料,供您参考a??a????a?12n?a1a2an??1?。n???不妨设a1?a2???an,则111????,由切比晓夫不等式,上式成立。即证。anan?1a1(5)用均值不等式和切比晓夫不等式证明柯西不等式证明:不妨设a1?a2???an,b1?b2???bn由切比晓夫不等式,有a1b1?a2b2???anbn?a1?a2???an??b1?b2???bn??????。nnn????由均值不等式,有a1?a2???an?nb1?b2???bn?n所以a1b1?a2b2???anbn?n两边平方,即得?a1b1?a2b2???anbn??a1?a2???anb22?b2???bn.即证。(6)补充“调和—几何平均不等式”的证明111????a?a2???ananaa21证明?1中的ai换成。?1nainn?两边取倒数,即得?+??a1a2an应用切比雪夫3应用切比雪夫不等式解题切比雪夫不等式是解决不等式问题的强力武器之一。本文对该不等式及其应用进行简单的介绍。一、切比雪夫不等式及其推论1?ai?bin1②若a1?a2?????an,b1?b2?????bn.则有?精编资料,供您参考aibi??ai?bi(切比雪夫不等式)n①若a1?a2?????an,b1?b2?????bn.则有?aibi?常见的方法是运用排序不等式,但最简单的证法是通过恒等变形。证明1:①式左边为顺序和,记为s,则s?a1b1?a2b2?????anbn,s?a1b2?a2b3?????anb1,s?a1b3?a2b4?????anb2,??????,s?a1bn?a2b1?????anbn?1.将上面n个式子相加,并按列求和即得结论。②证明同上(左边反序和不等号反向即可)。证明2:推论1设xi?r?(i?1,2,???,n),实数p,q均不为零。则⑴当p,q同号时,?xi?1nnp?qi1npnq??xi??xini?1i?11npnq??xi??xi.ni?1i?1⑵当p,q异号时,?xi?1p?qi该推论直接应用切比雪夫不等式即证。推论2设xi?r?(i?1,2,???,n),ns则x?1,r?s??x?i?i.?iri?1i?1i?1nnn1nnn1nr?sns1r?snss证明:事实上,?xi??xi?xi??n(?xi)??xi??xini?1ni?1i?1i?1i?1i?1r推论3设a1,a2,???,an,b1,b2,???,bn?r且a1?a2?????an,b1?b2?????bn或a1?a2?????an,b1?b2?????bn,mi?r?(i?1,2,???,n)则?m??mab??ma??mbiiiiiiiii?1i?1i?1i?1nnnn1nn证明:事实上,?mi??miaibi??miai??mibi???mimj(ai?aj)(bi?bj)(好范文网0.2i?1j?1i?1i?1i?1i?1推论3是切比雪夫不等式的加权形式。显然,当m1?m2?????mn时,就是切比雪夫不等式。nnnn注意:切比雪夫与推论3等号成立的条件均为a1?a2?????an,b1?b2?????bn中至少一组成立。二、切比雪夫不等式的应用1、构造两组数证明不等式。此类问题最关键、也是最难的步骤就是构造,选择两组数时往往需要很强的技巧。例1、已知0?a?b?c?d?e,例2、设xi?r?(i?1,2,???,n),精编资料,供您参考nn?(n?1)i?1ad?cd?cb?be?ea?。求证:。a?1?5?xi?1ni?1求证:i?1例3、设xi?r?(i?1,2,???,n),k?1.n1n1nxik?1求证:?(2014,女子数学奥林匹克)xi??k???1?xx1?xi?1i?1ii?1ii?1in2、去分母。能用切比雪夫不等式去分母的分式不等式,往往当变量排序后,分式的值也可以排序。一般的,当分母的值与分式的值都能排序时,可考虑用这种方法。ak3?(第四届中国东南)例4、设a,b,c?0,abc?1.求证:对整数k(k?2),?b?c2例5、设a,b,c?0,a?b?c
本文标题:切比雪夫不等式证明精编5篇
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