切比雪夫不等式练习题.docx
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切比雪夫不等式练习题.docx
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切比雪夫不等式练习题
切比雪夫不等式练习题
第一章
习题一1.4
证由切比雪夫不等式及E|?
|?
0
P?
1?
P?
1?
nE|?
|?
1
故P?
P?
limP?
1。
n?
1n
由切比雪夫不等式P?
E|?
|/n及E|?
|?
?
,得
P?
P与有相同的n阶自协方差矩阵。
故由平稳序列{Xt}的n阶自协方差矩阵退化知,对任给整数k?
1,存在非零实向量b?
使得var[Tn?
k?
1
i?
k?
{|?
|?
n})?
limP?
0。
n?
1nbi?
k?
1]?
0。
不妨假设bn?
0,则有对任给整数k?
1,Xn?
k可由Xk,Xk?
1,?
Xn?
k?
1线性表出。
对m?
n?
1,Xn可由X1,X2,?
Xn?
1线性表出,Xn?
1可由X2,X2,?
Xn线性
表出,故Xn?
1可由X1,X2,?
Xn?
1线性表出。
假设对所有n?
m?
n?
k,Xm可由X1,X2,?
Xn?
1线性表出。
则对
m?
n?
k?
1,由于Xn?
k?
1可由Xk?
1,Xk?
2,?
Xn?
k线性表出,由假设,Xn?
k?
1也可由X1,X2,?
Xn?
1线性表出。
根据,,对任何m?
n,Xm可由X1,X2,?
Xn?
1线性表出,即存在常数a0,a1,?
an?
1,使得Xm?
a0?
?
aiXn?
i,
i?
1n?
1a.s.。
习题四.3
解显然服从二维正态分布,且EXt?
EXs?
0。
记t?
12k?
l,s?
12m?
n,其中0?
l?
11,0?
n?
11,则Xt12i?
l,Xs12j?
n,这里?
0?
0。
i?
0j?
0km
由于{?
t}是正态白噪声WN,故
当l?
n,即t?
s时,?
t,s?
cov?
0;
当l?
n?
0,即t?
s,t?
12k时,?
t,s?
cov?
min?
2?
[min2]?
;12
12),t?
12k时,当l?
n?
0,即t?
s?
min?
所以
2?
]?
1)?
2。
12t,tt,s?
~N,其中?
?
T,Σ。
?
s,s?
?
t,s?
?
第二章
习题二
1X2.tt?
t?
1,Xt?
?
t?
a?
t?
1
习题三
3.提示:
当{Xt}与{Yt}的特征多项式满足A?
B时,是AR序列。
习题五
5.提示:
利用第一章7.4和第二章定理3.1。
Zt}仍然{
编号
毕业论文
题目:
切比雪夫不等式的推广及应用学院:
数学与统计学院专业:
数学与应用数学作者姓名:
指导教师:
职称:
完成日期:
2013年月24日
二○一三年五月
切比雪夫不等式的推广及应用
摘要本文给出切比雪夫不等式的三种形式的推广,并利用契比雪夫不等式研究随机变量落入某一区域的概率,求解证明概率方面的不等式,证明切比雪夫大数定理和特殊不等式等四个方面的应用.
关键词切比雪夫不等式;推广;应用;实例.中图分类号O211.1
Thepromotionandapplicationofchebyshevinequality
SongQiaoguoInstructorZhuFuguo
Abstract:
Chebyshevinequalityispresentedinthispaperthethreeformsofpromotion,andusethechebyshevinequalitystudyrandomvariablesintotheprobabilityofacertainarea,solvingtheprobabilityofinequality,provechebyshevtheoremoflargenumberandtheapplicationofthefouraspects,suchasspecialinequalities.
Abstract:
chebyshevinequality;Promotion;Applications;Theinstance
1引言
概率论是一门研究随机现象数量规律的科学,而切比雪夫不等式又是概率论中介绍的极少数的重要不等式之一,尤其是在分布未知时某些事件的概率上下界常用切比雪夫不等式.又如大数定理是概率论极限理论的基础,而切比雪夫不等式又是证明它的重要途径.作为一种理论工具,切比雪夫不等式不等式有很高的地位.虽然它的证明其理论成果相对比较完善,但一般的概率论与数理统计教材对大数定律的介绍篇幅较少,但不够广泛.我们知道,数学的各门分支之间都是有一定联系的,若我们在学习中能把这些联系点找出来并加以对比分析与应用,则既加深了对知识的理解,贯通了新旧知识的联系,又拓宽了知识的应用范围,同时也活跃了思维,无论从深度上还是从广度上都是一个飞跃.对切比雪夫不等式的应用问题的推广也是一项非常
有价值的研究方向,通过对这些问题的应用推广,不仅能加深对切比雪夫不等式的理解,而且能使之更为有效的应用其他知识领域中.
预备知识
定义1?
1?
若随机变量X有数学期望E?
X?
和方差D?
X?
则对于任意的正数?
?
0,总有:
P?
X?
E?
?
定义2
?
2?
?
?
D
?
2
.
如果函数f和g对于一切x1,x2均成立
f2)gg)f)与0g成似序;,与g成反序.
定义3?
3?
设连续型随机变量X的概率密度函数为f,若积分?
敛,则称?
?
?
?
?
xfdx收
xfdx为X的数学期望,则D?
E?
E2为X的方差.
主要结论及证明
定理1?
2?
切比雪夫不等式积分形式
如果连续函数f与g在区间?
a,b?
上成似序,则成立如下不等式
?
b
a
fdx?
gdx?
?
fgdx
a
a
bb
相反,如果f与g成反序,则不等号反向.
证明引入辅助函数
F?
?
fgdx?
?
fdx?
gdx,
a
a
a
t
t
t
F求导得
F’?
?
fgdx?
fg?
f?
gdx?
g?
fdx
a
a
a
ttt
fg?
fg?
fg?
gf?
dx
at
t
f?
f)?
?
g?
g?
dx.
a
由于f与与g在区间?
a,b?
上成似序,故有
?
f?
f?
?
g?
g?
?
0,
于是F’?
0,因此F在?
a,b?
上单调递增,又F?
0,?
F?
0,即
?
fgdx?
?
fdx?
gdx?
0,
a
a
a
b
b
b
?
?
fdx?
gdx?
?
fgdx.
a
a
a
bbb
同理反序成立.
定理2?
4?
切比雪夫不等式有限形式
若l?
和m?
是两个实序列,且满l1?
l2ln,
m1?
m2mn,或l1?
l2ln,m1?
m2mn,则成立如下不等式
1n1n1n
limi?
.?
ni?
1ni?
1ni?
1
证明设l1,l2,?
ln,m1,m2,?
mn为两个有相同次序的序列,有排序不等式得
l1m1?
l2m2?
?
lnmn?
l1m1?
l2m2lnmn,l1m1?
l2m2?
?
lnmn?
l1m2?
l2m3lnm1,l1m1?
l2m2?
?
lnmn?
l1m3?
l2m4lnm2,
?
?
l1m1?
l2m2?
?
lnmn?
l1mn?
l2m1lnmn?
1,
将这n个式子相加得到
n?
limi?
i?
1
i?
1
i?
1
n
n
n
不等式两边同时除以n2,得
1n1n1n
limi?
.?
ni?
1ni?
1ni?
1
定理3
?
4?
设a?
b?
?
R,?
i?
0,则当
a1?
a2an,b1?
b2bn或者a1?
a2an,b1?
b2bn时,有如下不等
式成立
i?
1
i?
1
i?
1
i?
1
当a1?
a2an,b1?
b2bn或者a1?
a2an,b1?
b2bn时,也有如下不等式成立
i?
1
i?
1
i?
1
i?
1
并且当?
i?
0,对于任意的i?
1,2,?
n时,则,中等式成立的条件是
a1?
a2an或b1?
b2bn.
证明先证明成立.
i?
1
i?
1
i?
1
i?
1
kkkk
用数学归纳法
k?
1时
1?
则不等式成立.
假设k?
n时
i?
1
i?
1
i?
1
i?
1
n
n
n
n
成立.
下证k?
n?
1时
i?
1
i?
1
i?
1
i?
1
n?
1n?
1nn
n?
1an?
1?
?
ibi?
?
n?
1bn?
1?
?
iai?
?
n2?
1an?
1bn?
1
i?
1
i?
1
i?
1
i?
1
nnnn
天津理工大学2011届本科毕业论文
第一章绪论··············································································································1
第二章切比雪夫不等式的基本理论·······································································
2.1切比雪夫不等式的有限形式和积分形式···········································································
2.切比雪夫不等式的概率形式·······························································································
第三章切比雪夫不等式在概率论中的应用····························································
3.1估计概率·······························································································································
3.1.1随机变量取值的离散程度··························································································
3.1.随机变量取值偏离E超过3?
的概率··································································
3.1.估计事件X?
E?
?
的概率·················································································
3.1.估计随机变量落入有限区间的概率··········································································
3.求解或证明一些有关概率不等式·······················································································
3.2.1求解相关不等式···························································································
3.2.证明相关不等式·························································································10
3.证明大数定律······················································································································11
3.3.1切比雪夫大数定律·····································································································11
3.3.伯努利大数定律········································································································12
第四章切比雪夫不等式在其他领域的应用··························································14
4.1生活中的小概率事件·········································································································14
4.切比雪夫不等式在经济评价风险中的应用·····································································15
4.2.1IRR的多元线性函数·······························································································15
4.2.IRR的概率分析·······································································································16
4.2.应用···························································································································17
4.前向神经网络容错性分析的切比雪夫不等式法·····························································0
4.3.1前向神经网络的随机故障模型···············································································0
4.3.连接故障对单个神经元容错性能的影响·······························································1
参考文献··················································································································致谢··························································································································5
天津理工大学2011届本科毕业论文
第一章绪论
概率论是一门研究随机现象数量规律的科学,是近代数学的重要组成部分。
随机现象在自然界和人类生活中无处不在,因而大多数的应用研究,无论是在工业、农业、经济、军事和科学技术中,其本质都是现实过程中的大量随机作用的影响。
这个观点强有力地推动了概率论的飞速发展,使其理论与方法被广泛地应用于各个行业。
而概率论极限理论的创立更使其锦上添花,以至在近代数学中异军突起。
历史上第一个极限定理是属于雅各布·伯努利,后人称之为“大数定律”。
因其遗著《猜度术》于171年出版,故概率史家称171年为伯努利大数定律创立年。
伯努利大数定律给出了频率估计概率的理论依据,同时开创了概率论中极限理论的先河,标志着概率论成为独立的数学分支。
1837年,泊松对大数定律提出一个较宽松的条件,进而得到泊松大数定律。
之后,由于有些数学家过分强调概率论在伦理学中的应用,又加上概率论自身基础不牢固,大多数数学家往往把概率论看作是有争议的课题,排除在精密科学之外。
切比雪夫正是在概率论门庭冷落的年代从事其研究的。
切比雪夫在1866年发表的论文《论均值》中,提出了著名的切比雪夫大数定律。
该论文给出如下三个定理[1]:
定理1.1:
若以a,b,c,?
表示x,y,z,?
的数学期望,用a1,b1,c1,?
表示相应的平方
x2,y2,z2,?
的数学期望,则对任何?
,x?
y?
z?
?
落在
a?
b?
ca1?
b1?
c1a2?
b2?
c2?
?
和
a?
b?
ca1?
b1?
c1a2?
b2?
c2?
?
之间的的概率总小于1?
1
?
2
定理1.2:
若以a,b,c,?
表示x,y,z,?
的数学期望,用a1,b1,c1,?
表示相应的平方
x2,y2,z2,?
的数学期望,则不论t取何值,N个量x,y,z,?
的算术平均值和他们相应的数学期望的算术平均值的差不超过
1a1?
b1?
c1?
?
a2?
b2?
c2?
?
?
tNN
t2
的概率对任何t都将大于1?
。
N
1
天津理工大学2011届本科毕业论文
定理1.3:
如果量u1,u2,u3?
和它们的平方u1,u2,u3?
的数学期望不超过一给定的值,则N个量的算术平均值和其数学期望的算术平均值之差不小于某一给定的概率,且当N趋于无穷时,其值趋于1。
.
这就是切比雪夫大数定律,用今天的符号可表示为:
定理1.4:
设X1,X2,X3,?
Xn,?
是两两不相关的随机变量序列,且其方差一致有界,则对任意的?
?
0,皆有
limPsN?
EsN1n?
?
222
这里sN?
?
Xi。
若随机试验中的每次试验随机事件发生的概率相等,则为伯努利大数定律。
i?
1N
又因相互独立的随机变量列必定两两无关,故泊松大数定律也是切比雪夫大数定律的特例。
要证明定理1.4,我们需要用到切比雪夫不等式。
其实在上面三个定理中已经给出了切比雪夫不等式,定理1.2我们用今天的数学语言来描述就是:
定理1.5:
设X1,X2,X3,?
Xn,?
是两两不相关的独立随机变量序列,且其方差存在,若sN?
?
Xi,则对任意的?
?
0,皆有
i?
1N
PsN?
EsN1?
?
VarXi?
1Ni
?
2。
不难发现这就是切比雪夫不等式,以此我们也可以得出定理1.4的证明,关于其证明我们在下文会提到。
作为概率论极限理论中介绍的极少数重要不等式之一,它的应用是非常多的,它可以解决和说明很多关于分布的信息,尤其在估计某些事件的概率的上下界时我们常会用到切比雪夫不等式。
另外,切比雪夫不等式和切比雪夫大数定律是概率论极限理论的基础,其中切比雪夫不等式又是证明大数定律的重要工具和理论基础,而且以切比雪夫不等式的基础上发展起来了一系列不等式是研究中心极限定理的有力工具,切比雪夫不等式作为一个理论工具,它的地位是很高的。
事实上,马尔可夫不等式也是切比雪夫不等式的第一种推广形式。
在切比雪夫不等式的诞生至今,切比雪夫不等式的应用性质还没有条理性的给出,本文将在切比雪夫不等式的应用方面进行探究。
2
天津理工大学2011届本科毕业论文
第二章切比雪夫不等式的基本理论
2.1切比雪夫不等式的有限形式和积分形式
定理2.1[2]:
设x1,x2,...,xn,y1,y2,...,yn为任意两组实数,若
x1?
x2?
...?
xn且y1?
y2?
...?
yn或x1?
x2?
...?
xn且y1?
y2?
...?
yn,则
1n1n1n
xi)?
xiyi?
ni?
1ni?
1ni?
1
当且仅当x1?
x2?
...?
xn或y1?
y2?
...?
yn时,和中的不等式等号成立。
证明:
设x1,x2,...,xn,y1,y2,...,yn为两个有相同次序的序列,由排序不等式有x1y1?
x2y2?
...?
xnyn?
x1y1?
x2y2?
...?
xnynx1y1?
x2y2?
...?
xnyn?
x1y2?
x2y3?
...?
xny1x1y1?
x2y2?
...?
xnyn?
x1y3?
x2y4?
...?
xny…………
x1y1?
x2y2?
...?
xnyn?
x1yn?
x2y1?
...?
xnyn?
1
把上述n个式子相加,得
n?
xiyi?
i?
1i?
1i?
1nnn
上式两边同除以n2,得
1n1n1n
xi)?
xiyi?
和g在区间[a,b]上单调递增或递减且分段连续,则
?
b
afgdx?
b1bfdx?
?
agdxb?
a?
a
若f和g中一个单调递增,另一个单调递减,则
?
fgdx?
abb1bfdx?
gdx?
?
aab?
a
证明:
令A?
?
fdx,B?
?
gdx,C?
?
fgdx,下面我们只证且只aabbba
考虑单调递增,即证C?
1ABb?
a
由f和g在区间[a,b]上单调递增,我们可以得到对?
x,y?
[a,b]有
?
f)?
g)?
0
对上式两边关于[a,b]进行积分,得
x,y?
[a,b]f)?
g)dd?
0
即
?
[?
g?
fg?
fg?
fg)dx]dy?
0aabb
?
[C?
Ag?
Bf?
fg]dy?
0ab
C?
AB?
AB?
C?
0于是,C?
1AB。
b?
a
2.切比雪夫不等式的概率形式
定理2.3[4]:
设随机变量X的数学期望E与方差D存在,则对于任意正数?
,有不等式P[X?
E?
?
]?
或P[X?
E?
?
]?
1?
D?
D?
2
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- 不等式 练习题