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随机过程好书推荐

随机过程好书推荐

随机过程好书推荐随机过程实在太重要了,⽤当年林元烈上课给我们说的⼀句话,"随机数学充满了魅⼒与威⼒"来形容随机过程再合适不过了.当然,随机过程⽐较难,故有"随机过程随机过"⼀说.这⾥,我就把美国各个好⼀些的⼤学的随机过程课程以及使⽤的课本和书籍汇总⼀下,希望能对⼤家有⽤.随机过程主要有应⽤随机过程和理论概率两种课程.前者⽐较常见与⼯科,后者主要集中在数学系或者统计系.我就美国⼀些好⼀些的学校⾥使⽤的教材作⼀汇总1. Stochastic Processes by Sheldon Ross应⽤随机过程的经典书籍, 在美国IE系使⽤的⽐较多.Berkeley的IE系⼀直在⽤, 可能因为Ross以前就是Berkeley. 另外Cornell 好像也⽤. 因为⽐较应⽤, ⼀些学校统计系的硕⼠⽣也在⽤. 原来Stanford的⼀个⽼师上课的时候说, 这本书有⼀些cheating, 主要就是说使⽤了很多trick, ⽽对于随机过程的概率本质没有揭⽰出来, 我觉得还是挺中肯的. 当然作为研究⽣阶段⼊门的随机过程书籍, 还是挺不错的. ⼀些排名稍微低⼀些的⼯科系也喜欢使⽤这本书.2. A first course in stochastic processes and A second course in stochastic processes经典的硕⼠⽔平的随机过程教材, 主要特点就是在没有measure theoretical probability的前提下, 仍旧能够保持相当的严格性. 有些问题解决的⼗分巧妙. 主要原因还是因为作者对于随机过程有着很深的理解. ⽽且两册书介绍的⽐较全⾯, 适合以后毕业想去搞⾦融⼯程的⼈学习. 现在各个学校使⽤它们当课本的不多了, 但基本上都是重要的参考书.3. Adventures in stochastic processes最好的应⽤随机过程的书, 所谓最好, 主要原因是Resnick 很成功地在揭⽰概率本质的前提下来讨论随机过程, 美国⼀些顶尖的IE系(⽐如Gatech 和 Stanford)都使⽤它作为课本. 重要的内容都讲的⽐较好, ⽽且有⼀定的概率本质描述, 是必须推荐的⼀本书. 如果有些⼈学完了还觉得想再深⼊⼀部, 可以看看Resnick 的那本probability path.4. Introduction to Stochastic Processes, Second Edition by Gregory F. Lawler⾮常好的⼀本⼩书. 虽然很薄, 但⼏乎没有缺失什么重要的东西, ⽐较适合想速成的⼈学. 我第⼀次学随机过程就是⽤这本书, 仔细读过之后, 感觉写的⾮常清楚⽽且精炼. 这⾥推荐给那些想了解⼀下随机过程的朋友.5. Introduction to Probability Models, by Sheldon Ross6. An Introduction to Stochastic Modeling, by Karlin and Taylor这两本书都⽐较简单, 主要是给⼯科硕⼠⽣⽤的. ⼊门的时候可以稍微翻⼀翻.7. Probability and Random Processes, by Geoffrey R. Grimmett and David R. Stirzaker绝对的好书. 是⼀本不难但的使⽤measure theoretical probability讲随机过程的书. Stanford的概率课就把它列为参考书之⼀. MIT给⼯科和商学院学⽣学习的基础概率课也把他列为教材, 对于那些想学习随机过程, ⼜想了解严格数学构造的朋友们, 这的确是⼀本好书.本⽂来⾃: ⼈⼤经济论坛计量经济学与统计版,详细出处参考:。

金融随机分析 II Stochastic process

金融随机分析 II Stochastic process

n
i
bn 0
P 1 (ii) bn E X | F ni i1 0, and 2 n
(iii ) b
EX
n i 1
2 ni
E E X ni | Fi 1
2
0
12
Ref: Hall . P, Heyde C., Martingale limit theory and its applicatiom.1980, Academic Press. Inc
M k M a.s and
M
k
M dP 0 as k
10
Corollary 2.2.9 Let X L P ,N k k 1 be an
1

increasing family of σ-algebras, N k F and define N to be the σ-algebra generated by
h X k 1 k h X k 1 X k (b) for every bounded Borel-measurable function h :
(c) (d)
euX k1 k euX k1 X k , u



(Agreement of Laplace transforms)
2.3 Markov Processes
Definition 2.3.1 Let (, , ) be a filtration under . Let {Xk,k=0,1,…}be a stochastic process on (, , ) . This process is said to be Markov if:

随机过程2016期末考试及答案

随机过程2016期末考试及答案

Stochastic Processes
Final Exam: Gauss Award - Page 3 of 9
26/12/16
2. (15 points) When three fair six-sided dice are rolled, what is the probability that the sum of the total numbers will be 12?
Stochastic Processes
Final Exam: Gauss Award - Page 6 of 9
26/12/16
5. (20 points) A Yule process starts at time 0 with one organism. This organism splits into two organisms after a time Y1 with the density fY1 (y ) = λe−λy , y ≥ 0. Each of these two organisms splits into two more organisms after independent exponentially distributed delays, each with the same density λe−λy . In general, each old and new organism continues to split forever after a delay y with the same density λe−λy . (a) (5 points) Let T1 be the time at which the first organism splits, and for each i > 1, let Ti be the interval from (i − 1)st splitting until the ith. Show that Ti is exponential with parameter iλ and explain why the Ti are independent. (b) (5 points) For each n ≥ 1, let the continuous random variable Sn be the time at which the n th splitting occurs, i.e. Sn = T1 + . . . + Tn . Find a simple expression for the distribution function of Sn . (c) (5 points) Let X (t) be the number of organisms at time t > 0. Express the distribution function of X (t) for each t > 0 in terms of Sn for each n. Show that X (t) is a random variable for each t > 0 (i.e., show that X (t) is finite with probability 1). (d) (5 points) Find E [X (t)] for each t > 0.

文档:随机过程(雷斯尼克,英文)-Chapter1-2作业题提示

文档:随机过程(雷斯尼克,英文)-Chapter1-2作业题提示

Adventures in Stochastic ProcessesChapter 1 Preliminaries1.1. (a) Let X be the outcome of tossing a fair die. What is the gf of X? Use the gf to find EX.(b) Toss a die repeatedly. Let n μ be the number of ways to throw die until the sum of the faces is n. (So 11μ= (first throw equals 1), 22μ= (either the first throw equals 2 or the first 2 throws give 1 each), and so on. Find the generating function of{,1n 6}n μ≤≤ .解:(a) X 的概率分布为 1[],1,2,3,4,5,66P X k k ===,X 的生成函数为 66611111()[]66kk kk k k P s P X k s s s ======⋅=∑∑∑,X 的期望为 6611111117()||662k s s k k EX P s k s k -===='==⋅==∑∑.(b) n μ:点数之和为(1)n n ≥的投掷方法数,则 点数之和为1的投掷方法:第一次投掷点数为1,即0112μ==,点数之和为2的投掷方法: 情形1,第一次投掷点数为2, 情形2,前两次投掷点数均为1,即1222μ==,点数之和为3的投掷方法: 情形1,第一次投掷点数为3,情形2,前两次投掷点数为(1,2),(2,1), 情形3,前三次投掷点数均为1,即012232222C C Cμ=++=,点数之和为6的投掷方法: 情形1,第一次投掷点数为6,情形2,前两次投掷点数为下列组合之一:1和5,2和4,3和3,情形3,前三次投掷点数为下列组合之一:1,1和4,1,2和3,2,2和2, 情形4,前四次投掷点数为下列组合之一:1,1,1和3,1,1,2和2, 情形5,前五次投掷点数为下列组合之一:1,1,1,1和2, 情形6,前六次投掷点数均为1,即015565552C C C μ=+++=,于是,n μ(6)n ≤的生成函数为66111()2nn n n n n P s s s μ-===⋅=⋅∑∑1.2. Let {},1n X n ≥ be iid Bernoulli random variables with 11[1]1[0]P X p P X ===-=and let 1nn i i S X ==∑ be the number of successes in n trials. Show n S has a binomial distribution by the following method: (1) Prove for 0,11n k n ≥≤≤+1[][][1 ] n n n P S k pP S k qP S k +===-+=.(2) Solve the recursion using generating functions. 解:(1) 由全概率公式,得1111111[][1][|1][0][|0]n n n n n n n P S k P X P S k X P X P S k X +++++++=====+===[1][]n n pP S k qP S k ==-+=(2) 1110()[]n k n n k P s P S k s +++===∑10([1][])n k n n k pP S k qP S k s +===-+=∑1110[1][]n nk kn n k k ps P S k sq P S k s +-====-+=∑∑11[][]n nlkn n l k ps P S l s q P S k s ====+=∑∑211()()()()()n n n ps q P s ps q P s ps q +-=+=+=+所以 1~(;1,)n S b k n p ++1.3 Let {,1}n X n ≥ be iid non-negative integer valued random variables independent of the non-negative integer valued random variable N and suppose()()11(), Var , , Var E X X EN N <∞<∞<∞<∞.Set 1nn i i S X ==∑. Use generating functions to check211Var()Var()()Var()N S EN X EX N =+ 证明:由1()(())N S N X P s P P s =所以 11111()()|(())()|()()N N S s N X X s E S P s P Ps P s E N E X =='''===,1111211()|[(())(())(())()]|N S s N X X N X X s P s P Ps P s P P s P s ==''''''''=+ 11112((1))((1))((1))(1)NX X N X X P P P P P P ''''''=+ (1(1)1X P =) 222111()()()()EN EN EX E N EX EX =-+- 22111Var()()EN X EN EX ENEX =+-又 2211()|()()N S s N N N P s E S ES E S ENEX =''=-=- 所以 22211()Var()()N E S EN X EN EX =+ 因此 22Var()()()N N N S E S ES =-2222111Var()()-()()EN X EN EX EN EX =+211Var()()Var()EN X EX N =+.1.4. What are the range and index set for the following stochastic processes : (a) Let i X be the quantity of beer ordered by the th i customer at Happy Harry's and let ()N t be the number of customers to arrive by time t . The process is(){}()10,N t i i X t X t ==≥∑ where ()X t is the quantity ordered by time t .(b) Thirty-six points are chosen randomly in Alaska according to some probability distribution. A circle of random radius is drawn about each point yielding a random set S . Let ()X A be the value of the oil in the ground under region A S ⋂. The process is () {,}X B B Alaska ⊂.(c) Sleeping Beauty sleeps in one of three positions: (1) On her back looking radiant. (2) Curled up in the fetal position.(3) In the fetal position, sucking her thumb and looking radiant only to an orthodontist.Let ()X t be Sleeping Beauty's position at time t. The process is (){} ,0X t t ≥. (d) For 0,1,n =, let n X be the value in dollars of property damage to West PalmBeach, Florida and Charleston, South Carolina by the th n hurricane to hit the coast of the United States.解:(a) The range is {0,1,2,,}S =∞,the index is {|0}T t t =≥;(b) The range is [0,)S =∞,the index is {1,2,,36}T =;(c) The range is {1,2,3}S =,the index is {|0}T t t =≥; (d) The range is [0,)S =∞,the index is {0,1,2,}T =.1.5. If X is a non-negative integer valued random variable with~{},()X k X p P s Es =express the generating functions if possible, in terms of () P s , of (a) []P X n ≤, (b)[]P X n <, (c) []P X n ≥. 解:0()[]k k P s P X k s ∞===∑1000()[]k kki k k i P s P X k s p s ∞∞===⎛⎫=≤= ⎪⎝⎭∑∑∑001i k i i i k i i s s p p s ∞∞∞===⎛⎫== ⎪-⎝⎭∑∑∑ 011()11i i i s p P s s s ∞===--∑; 12000()[]k kki k k i P s P X k s p s ∞∞-===⎛⎫=<= ⎪⎝⎭∑∑∑10101i k i i i k i i s s p p s +∞∞∞==+=⎛⎫== ⎪-⎝⎭∑∑∑0()11i i i s ss p P s s s∞===--∑; 300()[]kki k k i k P s P X k s p s ∞∞∞===⎛⎫=≥= ⎪⎝⎭∑∑∑100011i i k i i i k i s s p p s +∞∞===-⎛⎫== ⎪-⎝⎭∑∑∑ 0011()111ii ii i s sP s p p s s s s ∞∞==-=-=---∑∑. 1.8 In a branching process 2()P s as bs c =++, where 0,0,0,(1)1a b c P >>>=. Compuct π. Give a condition for sure extinction. 解:由(1)1P a b c =++=,可得 1()b a c -=-+,2()s P s as bs c ==++ 2(1)0as b s c +-+=2(+)0as a c s c -+=,1cs s a== (1)21m P a b '==+≤.1.10. Harry lets his health habits slip during a depressed period and discovers spots growing between his toes according to a branching process with generating function23456()0.150 .050.030.070.40.250.05P s s s s s s s =++++++Will the spots survive? With what probability?解:由 2345()0 .050.060.21 1.6 1.250.3P s s s s s s '=+++++, 可得 (1)0 .050.060.21 1.6 1.250.3 3.471m P '==+++++=>, 又由 23456()0.150 .050.030.070.40.250.05s P s s s s s s s ==++++++, 依据1π<,可得=0.16π.1.23. For a branching process with offspring distribution,0,1,01,n n p pq n p q p =≥+=<<解: ()1pP s qs=- ()1ps P s qs==- 210qs s q -+-=1s = 或 p s q=1(1)1k k qm P p kq p∞='===≤∑, 112p p p -≤⇒≥.Chapter 2 Markov Chains2.1. Consider a Markov chain on states {0, 1, 2} with transition matrix0.30.30.4=0.20.70.10.20.30.5P ⎛⎫⎪⎪ ⎪⎝⎭.Compute 20[2|0]P X X == and 210[2,2|0]P X X X ===.解:由题意得 20.230.420.350.220.580.20.220.420.36P ⎛⎫⎪= ⎪ ⎪⎝⎭,(2)202[2|0]0.35P X X p ====, 120[2,2|0]P X X X === 2110[2|2][2|0]P X X P X X =====(1)(1)22020.50.40.2p p =⋅=⨯=2.8. Consider a Markov chain on {1, 2, 3} with transition matrix1001112631313515P ⎛⎫ ⎪ ⎪⎪= ⎪ ⎪ ⎪⎝⎭. Find ()3n i f for 1,2,3,n =.解:当1i =时,对任意1n ≥,()1313[(1)]0n f P n τ===;当2i =时,对于1n ≥,()112323222311[(1)]()63n n n f P n p p τ--====⋅; 当3i =时,对于1n =,(1)3333331[(1)1]15f P p τ====, 对于2n ≥,()222333332222331111[(1)]()()56356n n n n f P n p p p τ---===⋅⋅=⋅⋅=⋅. Exercise. Consider a Markov chain on states {1,2,3,4,5} with transition matrix1000001000120012000120120120120P ⎛⎫ ⎪ ⎪ ⎪= ⎪ ⎪ ⎪⎝⎭,(1) What are the equivalence classes ?(2) Which states are transient and which states are recurrent ?(3) What are the periods of each state? (详细过程自己完成!)解:(1) 分为三类:{1},{2}和{3,4,5}.(2) 1,2为正常返状态,3,4,5为瞬过状态.(3) 状态1,2的周期为1,状态3,4,5的周期为2.。

Stochastic Processes SolutionbookII 随机过程习题解答

Stochastic Processes SolutionbookII  随机过程习题解答
k i−1 ,ti ]
,
Xti − Xti−1
i=1
2
=
i=1 n k=1 ti
k σs
k dWs
+
ti−1
µs ds
n ti ti−1 ti 2 k k σs dWs ti ti−1 l l σs dWs
m
=
i=1
+2
k,l=1,k=l k k σs dWs
+2
ti−1
µs ds
k=1
+
ti−1
µs ds
As thus
Π
→ 0, we have Var Wtik − Wtik−1 E [W i , W j ]t
Wtik − Wtik−1
Wtjk − Wtjk−1
= 0 and
n k=1 i j [W , W ]t =
Wtjk − Wtjk−1 = 0.
converges to its expectation, and thus
t
f (t, x) = eαx− 2 α t ,
1
2
α ∈ R.
f (t, Bt ) = f (0, B0 ) −
0 t
1 2 α f (s, Bs ) ds + 2
t
αf (s, Bs ) dBs +
0
1 2
t
α2 f (s, Bs ) ds
0
=1+
0
αf (s, Bs ) dBs
t 0 αf (s, Bs ) dBs t≥0
S 1 2 E(YS YS ) = E 0 S 2 1 Yu dYu + 0 S S 1 2 Yu dYu + 0 S

第6章通信理论 principles of communications systems,modulation,and noise 课后答案第六版

第6章通信理论 principles of communications systems,modulation,and noise 课后答案第六版

=2
#
The statistical average variance is this expression minus E 2 [X (t)]. b. The time average autocorrelation function is hx (t) x (t + )i = A2 cos (2 f0 ) 2
Problem 6.3 a. The sketches would consist of squarewaves of random delay with respect to t = 0. 1
2 b. X (t) takes on y two values A and fX (x) =
CHAPTER 6. STOCHASTIC PROCESSES A, and these are equally likely. Thus A) + 1 (x + A) 2
2 a sin (! 0 t) sin [! 0 (t + )] !2 0 2 a fcos (! 0 ) cos [! 0 (2t 2! 2 0
2 a
da
+ )]g
c. No, it is not wide sense stationary because the ensemble-average autocorrelation function is time dependent (not a function only of time di¤erence).
Chapter 6
Stochastic Processes
6.1 Problem Solutions
Problem 6.1 The various sample functions are as follows. Sample functions for case (a) are horizontal lines at levels 2A; 0; 2A, each case of which occurs equally often (with probability 1/3). Sample functions for case (b) are horizontal lines at levels 3A; 2A; A; A; 2A; 3A, each case of which occurs equally often (with probability 1/6). Sample functions for case (c) are horizontal lines at levels 4A; 2A; 2A; 4A; or oblique straight lines of slope A or A, each case of which occurs equally often (with probability 1/6). Problem 6.2 a. For case (a) of problem 6.1, since the sample functions are constant with time and are less than or equal 2A, the probability is one. For case (b) of problem 6.1, since the sample functions are constant with time and 5 out of 6 are less than or equal to 2A, the probability is 5/6. For case (c) of problem 6.1, the probability is 5/6 because 5 out of 6 of the sample functions will be less than 2A at t = 2. b. The probabilities are now 2/3, 1/2, and 1/2, respectively. c. The probabilities are 1, 5/6, and 5/6, respectively.

信息与计算科学专业课程简介

信息与计算科学专业课程简介

信息与计算科学专业课程简介课程代码:3112001131.课程名称:解析几何 Analytic Geometry总学时: 64 周学时: 4学分: 3 开课学期:一修读对象:必修预修课程:无内容简介:《解析几何》是学科基础课程,是所有数学专业及应用数学专业的主要的基础课。

它是用代数的方法来研究几何图形性质的一门学科。

《解析几何》包括向量与坐标,轨迹与方程,平面与空间直线,柱面、锥面、旋转曲面与二次曲面,二次曲线的一般理论与二次曲面的一般理论等。

选用教材:吕林根,许子道,《解析几何》(第四版),高等教育出版社,2006年。

参考书目:周建伟,《解析几何》,高等教育出版社,2005年。

课程代码:311200214、311200314、311200616、3112007152.课程名称:数学分析Ⅰ-Ⅳ Mathematical AnalysisⅠ-Ⅳ总学时:334 周学时:4,4,6,5学分: 18 开课学期:一,二,三,四修读对象:必修预修课程:无内容简介:《数学分析》是学科基础课程,是所有数学专业及应用数学专业第一基础课。

它提供了利用函数性质分析和解决实际问题的方法, 培养学生严谨的抽象思维能力,为学习其他学科奠定基础。

主要内容有:实数、函数、极限论,函数的连续性。

一元函数微分学,微分学基本定理。

一元微分学应用,实数完备性基本定理,闭区间上连续函数性质的证明,不定积分,定积分及应用,非正常积分。

数项级数,函数列与函数项级数,幂级数,付里叶级数,多元函数的极限与连续,多元函数微分学。

隐函数定理及其应用,重积分,含参量非正常积分,曲线积分与曲面积分。

选用教材:华东师范大学数学系,《数学分析》(第三版)(上、下册),高等教育出版社,2001年。

参考书目:① 陈纪修,《数学分析》(第二版),高等教育出版社2004年。

② 刘玉琏,傅沛仁,《数学分析讲义》(第三版),高等教育出版社,1992年。

课程代码:311200416、3112005153.课程名称:高等代数Ⅰ-Ⅱ Advanced AlgebraⅠ-Ⅱ总学时:198 周学时:6,5学分: 11 开课学期:二,三修读对象:必修预修课程:无内容简介:《高等代数》是学科基础课程。

StochasticProcessesinPhysicsandChemistry

StochasticProcessesinPhysicsandChemistry

PY542INFORMATION Fall2008Instructor:Sidney Redner(321SCI,x2618)Office Hours:Tues.&Fri.9-10:30am,and by appointment.General:This course treats non-equilibrium statistical mechanics and transport phenom-ena.Because of the rapid developments in thefield,the breadth of topics,and the lack of an established formalism,most of the classic texts no longer seem appropriate for this course.For this reason,the“unofficial”course text is a book that I am currently writing with2co-authors.It is continuously being updated and individual chapters are posted on the course website.Other books that should be helpful during the semester include:(i)N.G.Van Kampen,Stochastic Processes in Physics and Chemistry(North-Holland). This gives an excellent treatment of stochastic processes.Buy it used if you can.I would have assigned this as the text if the price was a factor2smaller.(ii)S.Redner,A Guide to First-Passage Processes(Cambridge University Press).This book gives background on random walks and diffusion processes,as well as a reference for the portion of the course onfirst-passage phenomena.If you purchase the hardcover version,I will refund you my royalty(approximately$5.50per book),but the paperback version is much cheaper.I will also post relevant excerpts on the course website. (iii)F.Reif,Statistical and Thermal Physics(McGraw-Hill).A standard advanced un-dergraduate text for statistical mechanics.The last few chapters provide a particularly useful introduction to various aspects of non-equilibrium processes.(iv)K.Huang,Statistical Mechanics2nd edition(Wiley).Relevant chapters are3and 5that deal with kinetic theory and transport phenomena.(i)N.Wax(editor),Selected Papers on Noise and Stochastic Processes(Dover).This book contains reprints of some of the most important classic research articles on stochas-tic processes.Although out of print,it may be possible to obtain used somewhere.How-ever,the book contains reprints of articles that are generally available on the web.The most useful is“Stochastic Problems in Physics in Astronomy”by S.Chandrasekhar, Rev.Mod.Phys.15,1–89(1943).Other useful articles include“On the Theory of the Brownian Motion”,by G.E.Uhlenbeck and L.S.Orenstein,Phys.Rev.26,823–41(1930)&“On the Theory of the Brownian Motion II”by M.C.Wang and G.E. Uhlenbeck,Rev.Mod.Phys.17,323–42(1945).(v)R.Kubo,M.Toda and N.Hashitsume,Statistical Physics II(Springer-Verlag). Contains a particularly good discussion of linear response theory and thefluctuation-dissipation theorem.(vi)J.A.McLennan,Introduction to Non-Equilibrium Statistical Mechanicsi(Prentice-Hall).This book contains a thorough discussion of the Boltzmann transport equation. (vii)H.J.Kreuzer,Non-Equilibrium Thermodynamics and its Statistical Foundations (Oxford University Press).Comprehensively treats transport theory from the macro-scopic viewpoint and has an excellent discussion of the Rayleigh-B´e nard instability.Course organization:Lectures:Lectures will be held on Tuesdays and Thursdays from2:00—3:30in SCI B58.The accompanying outline represents a rough approximation to the material that will be covered this semester.Discussion:Sections will be held weekly starting Wed.Sept.3at2:00pm in PRB365.Homework:Approximately10assignments will be handed out.While some collab-oration on homework is acceptable,what is turned in should represent your personal effort.Exams and Grading:The average of the homeworks will count approximately30±5% of the total class grade.I will give one midterm exam(exact format to be determined) that will count approximately30±5%of the total class grade.For thefinal,I am currently planning a take-home but time-limitedfinal exam that will count for the approximately remaining40%of the total course grade.。

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0
随 向 . 机 量
2 一维随机向量即随机变 . 量
0
3 对 n个 数 1, x2 ,L, xn , 于 实 x
0
{X1(ω) ≤ x1, X2 (ω) ≤ x2 ,L, Xn (ω) ≤ xn}
n i=1
= I{Xi (ω) ≤ xi }∈ℑ
2.分布函数 2.分布函数
F 称 (x1, x2 ,L, xn ) = P{X1(ω) ≤ x1, X2 (ω) ≤ x2 L, Xn (ω) ≤ xn} ˆ r 为 机 量 (ω) = ( X1(ω), X2 (ω),L, Xn (ω)) 随 向 X 的 布 数 分 函 .
j =1 ∞ ∞
p• j = P{ = y j } = ∑ pij , Y
i=1
j =1 2,L ,
对 一 pi• , p• j > 0, 有 于 切 在 = y j条 下 X的 件 布 为 件 , 条 分 列 Y P{X = xi | Y = y j } = P{X = xi ,Y = y j } P{Y = y j } = pij p• j
2 连续函数、单调函数是 一元Borel函数 .
0
30 Borel函数类关于代数运算及极限运算封闭 .
2. 定 : 设 是 Ω, ℑ, P)上 r.v., g(x)是 元 理 X ( 的 一 Borel 函 ,则 = g( X )是 Ω, ℑ, P)上 r.v.. 数 Y ( 的
y 3. 推广: 设 = g(x1,L, xn )是R 到R 的一个映射 ,
n

⇔∀自 数 (2 ≤ m ≤ n)以 1≤ i1 < i2 <L< im ≤ n, 然 m 及 有 i1 ,L, Xim 相 独 X 互 立
i=1
例 . 设 X,Y) ~ N a, b,σ ,σ , r , 5 (
2 1 2 2
(
)
则 ,Y相 独 ⇔r = 0. X 互 立
1 例 . 设随机变量X ,Y相互独立 且皆以概率 取 6 , 2 值 与−1, 令Z = XY, 试证: X ,Y, Z两两独立 1 , 但不相互独立 .
n 1
若对R 中一切 Borel集B , 有 1
1
g (B ) ={( x1,L, xn ) : g(x1,L, xn ) ∈B }∈β , 1 ˆ 1 n 则称g(x1,L, xn )为n元Borel (可测函数 ) .
−1
设( X1,L, Xn )是概率空间 Ω, ℑ, P)上的随机向量 ( , , g(x1,L, xn )是n元Borel函数 则Y = g( X1,L, Xn ) 是(Ω, ℑ, P)上的随机变量 .
称 在 = x下 Y的 为 X , 条件 分布 数 函 . f (x, y) f (x, y) 记f X|Y (x | y) = , fY| X ( y | x) = fY ( y) f X (x)
, 分 称 在 = y下 X的 件 率 度 别 为 Y 条 概 密 及 X = x下 Y的 件 率 度 在 , 条 概 密 .
x, y ∈ R
其 a, b,σ1 > 0,σ2 > 0,−1< r <1 常 , 中 为 数 称 X ,Y)服 参 为a, b,σ1,σ2 , r的二 ( 从 数 维 正 分 , 为 X ,Y) ~ N(a, b,σ ,σ , r). 态 布 记 (
2 1 2 2
求 缘概 密 及 件概 密 . 边 率 度 条 率 度
例 . 设随机向量( X ,Y)的概率密度为 7 g(x)h( y), a ≤ x ≤ b, c ≤ y ≤ d f (x, y) = 0, 其它 其中a < b, c < d为常数 g(x)在 a, b]上非负可 , [ 积 h( y)在 c, d]上非负可积 证明: X ,Y独立 , [ , .
*
Borel函数与随机变量的函数 Borel函数与随机变量的函数
1 1
1.定义: 设y = g(x)是R 到R 的一个映射 若∀B ∈β1, , 1 g (B ) ={x : g(x) ∈B }∈β,则称g(x)是一 1 ˆ 1 1 元Borel (可测 函数 ) .
−1
注:0 一元Borel函数是(R1, β1)到 R1, β1)的可测函数 1 ( .
i =1 2,L , 在 = xi条 下 Y的 件 布 为 X 件 , 条 分 列
P{Y = y j | X = xi } =
P{X = xi ,Y = y j } P{X = xi }
=
pij pi•
j =1,2,L
称 (x | y j ) = P{X ≤ x | Y = y j } F =
i:xi ≤x
X |Y = y ~ N
σ1 a + r σ2 σ2 σ1
) (x − a),σ (1− r )).
2 1 2 2 2 2
( y −b),σ (1− r ) ,
3)联 分 决 边 分 ,反 不 立 合 布 定 缘 布 之 成 .
二、随机变量的独立性
1.定 义
1 设X1(ω),L, Xn (ω)是定义在概率空间 Ω, ℑ, P) (
§2 随机向量 随机变量的独立性
一、随机向量及其分布
1.定 : 若 X1(ω), X2 (ω),L, Xn (ω)定 在 一 率 义 r.v. 义 同 概 空 (Ω, ℑ, P)上 则 间 , 称 r X (ω) = ( X1(ω), X2 (ω),L, Xn (ω))为 个n维 一
r n 注 1 X (ω)是 个 (Ω, ℑ)到 R , βn )的 测 数 : 一 从 ( 可 函 . r −1 r ∀Bn ∈βn , X (Bn ) ={ω : X (ω) ∈Bn} ˆ ={ω : ( X1(ω), X2 (ω),L, Xn (ω)) ∈Bn}∈ℑ
序列独立⇔ 其中任意有限个随机变量相互独立 .
r 30 记随机向量Xi = ( Xi1,L, Xini )(i =1,2,L, k) 如果∀xi1, xi 2 ,L, xini ∈R, (i =1,2,L, k)有 P{I( Xi1 ≤ xi1,L, Xini ≤ xini )}
i=1 k
r r = ∏P{Xi1 ≤ xi1,L, Xini ≤ xini },则称X1,L, Xk
−∞ −∞ +∞ +∞
称FX|Y (x | y) = P{X ≤ x | Y = y} ˆ
= lim P{X ≤ x | y − ε < Y ≤ y} ˆ
P{X ≤ x, y − ε < Y ≤ y} = lim ε →+0 P{y − ε < Y ≤ y}
ε →+0
F(x, y) − F(x, y − ε ) = lim ε →+0 F ( y) − F ( y − ε ) Y Y [F(x, y) − F(x, y − ε )] / ε = lim ε →+0 [F ( y) − F ( y − ε )] / ε Y Y
3.分布函数的性质 3.分布函数的性质
1 对任一xi , i =1,2,L, n, F(x1, x2 ,L, xn )关于xi是单
0
调增加的 .
2 对任一xi , i =1,2,L, n, F(x1, x2 ,L, xn )关于xi右连续 .
0
30 F(x1,L, xi−1,−∞, xi+1,L, xn ) = lim F(x1,L, xn ) = 0 ˆ
例.设 维 3 二 r.v.(X, Y)的 率 度 概 密 为 f (x, y)
x − a y −b ( y −b)2 1 (x − a)2 + = exp− − 2r 2 2 σ σ 2 , 2 2(1− r ) σ1 2πσ1σ2 1− r σ1 2 2 1
结论: 1)正态随机向量的边缘分布仍是正态分布 但相 , 2 2 反结论不成立 若 X,Y) ~ N(a, b,σ1 ,σ2 , r), . (
则 ~ N(a,σ ),Y ~ N(b,σ ). X
2 1 2 2
2)正态随机向量的部分分 量对另外一部分分量 的条件分布仍是正态分 . 布
( Y | X = x ~ N(b + r
⇔∀实数x1,L, xn有F(x1,L, xn ) = F (x1)LFn (xn ) 1
⇔离 型 .v. ∀一 可 值 1,L, an , 散 r 组 能 a 有 {X1 = a1,L, Xn = an} = ∏P{Xi = ai } P
i=1 n
连 型 .v. ∀实 x1,L, xn , 有 (x1,L, xn ) = ∏ fi (xi ) 续 r 数 f
0 x1 xn
度函数 .
3 概率密度函数的性质
0
(1) f (x1,L, xn ) ≥ 0
(2) ∫ L f (x1,L, xn )dx1 Ldxn =1 ∫
−∞ −∞ +∞ +∞
∂F (3) = f (x1,L, xn )(几 处 成 ) 乎 处 立 ∂x1 L∂xn
n
(4) P{a1 < X1 ≤ b ,L, an < Xn ≤ bn} 1 ∫ L f (x1,L, xn )dxn Ldx1 = ∫
∂F(x, y) ∂y =
∂ y x x f (u, v)dudv ∫−∞ ∫−∞ ∂y ∫−∞ f (u, y)du = = fY ( y) fY ( y)
dF ( y) Y dy
=∫
x
−∞
f (u, y) du( fY ( y) > 0 设 ) fY ( y)
为 Y = y下 X的 件 布 数 在 , 条 分 函 . y f (x, v) 同 样的 F | X ( y | x) = ∫ , Y dv 设 X (x) > 0 ( f ) −∞ f (x) X
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