随机过程英语讲义-15

合集下载

随机过程

随机过程

《随机过程》课程教学大纲课程编号:02200021课程名称:随机过程英文名称:Stochastic Processes课程类别:选修课总学时:72 讲课学时:68 习题课学时:4 学分: 4适用对象:数学与应用数学、信息与计算科学专业先修课程:数学分析、高等代数、概率论与数理统计一、课程简介随机过程是研究客观世界中随机演变过程规律性的学科,它的基本知识和方法不仅为数学、概率统计专业所必需,也为工程技术、生物信息及经济领域的应用和研究所需要。

本课程介绍随机过程研究领域的一些基础而重要的知识和技能。

二、课程性质、目的和任务随机过程是概率论的后续课程,具有比概率理论更加实用的应用方面,处理问题也更加贴近实际情况。

通过这门课程的学习,使学生了解随机过程的基本概念,掌握最常见而又有重要应用价值的诸如Poisson过程、更新过程、Markov过程、Brown运动的基本性质,能够处理基本的随机算法。

提高学生利用概率理论数学模型解决随机问题的能力。

通过本课程的学习,可以让数学专业的学生很方便地转向在金融管理、电子通讯等应用领域的研究。

三、课程基本要求通过本课程的学习,要求学生掌握随机过程的一般概念,知道常见的几类随机过程的定义、背景和性质;掌握泊松过程的定义与基本性质,了解它的实际背景,熟悉它的若干推广;掌握更新过程的定义与基本性质、更新函数、更新方程,了解更新定理及其应用,知道更新过程的若干推广;掌握离散时间的马尔可夫链的基本概念,熟练掌握转移概率、状态分类与性质,熟悉极限分布、平稳分布与状态空间的分解,了解分枝过程;掌握连续时间的马尔可夫链的定义、柯尔莫哥洛夫方程;掌握布朗运动的定义与基本性质,熟悉随机积分的定义与基本性质,了解扩散过程与伊藤公式,会求解一些简单的随机微分方程。

四、教学内容及要求第一章预备知识§1.概率空间;§2.随机变量和分布函数;§3.数字特征、矩母函数和特征函数;§4.条件概率、条件期望和独立性;§5.收敛性教学要求:本章主要是对概率论课程的复习和巩固,为后续学习做准备。

随机过程

随机过程

随机过程
1. 基本概念
1、随机过程0{}()t t X t T ≥∈是一个从T ⨯Ω到R 的映射其中T 是一个时间指标集,如果它是连续的正实数,随机过程0{}()t t X t T ≥∈就是连续时间随机过程,例如布朗运动;如果T
是一个离散时间指标集,随机过程
0{}()t t X t T ≥∈就是一个离散时间随机过程,例如随机游走。

随机过程可表达为:(,)(,)t X t ωω→
固定t ,可得随机过程(,)X t ω在时间t 的分布样本;固定ω,可得随机过程 (,)X t ω的轨迹。

2、研究随机过程的方法多种多样,主要可以分为两大类:一类是概率方法,其中用到轨道性质、停时和随机微分方程等;另一类是分析的方法,其中用到测度论、微分方程、半群理论、函数堆和希尔伯特空间等。

实际研究中常常两种方法并用。

B-S 模型
期权的定价方法
(1)Black —Scholes 公式
(2)二项式定价方法
(3)风险中性定价方法
(4)鞅定价方法等。

随机过程讲义(第二章)(PDF)

随机过程讲义(第二章)(PDF)

第二章 随机过程的一般概念2.1 随机过程的基本概念和例子定义2.1.1:设(P ,,F )Ω为概率空间,T 是某参数集,若对每一个,是该概率空间上的随机变量,则称为随机过程(Stochastic Process)。

T t ∈),(w t X ),w t (X 随机过程就是定义在同一概率空间上的一族随机变量。

随机过程可以看成定义在),(w t X Ω×T 上的二元函数,固定Ω∈0w ,即对于一个特定的随机试验,称为样本路径(Sample Path),或实现(realization),这是通常所观测到的过程;另一方面,固定,是一个随机变量,按某个概率分布随机取值。

),(0w t X T t ∈0),(0w t X抽象一点:令,即∏∈=Tt T R R T R 中的元素为),(T t x X t t ∈=,为其Borel域(插乘)(T R B σ域),随机过程实质上是()F ,Ω到())(,T T R R B 上的一个可测映射,在())(,T TR RB 上诱导出一个概率测度:T P ()B X P B P R B T T T ∈=∈∀)(),(B 。

一般代表的是时间。

根据参数集T 的性质,随机过程可以分为两大类: t 1)为可数集,如T {}L ,2,1,0=T 或{}L L ,1,0,1,−=T ,称为离散参数随机过程,也称为随机序列;2)为不可数集,如T {}0≥=t t T 或{}∞<<∞−=t t T ,称为连续参数随机过程。

随机过程的取值称为过程所处的状态(State),所有状态的全体称为状态空间(State Space)。

通常以表示随机过程的状态空间。

根据状态空间的特征,一般把随机过程分为两大类:T t t X ∈),(S 1) 离散状态,即取一些离散的值; )(t X 2)连续状态,即的取值范围是连续的。

)(t X离散参数离散状态随机过程: Markov 链 连续参数离散状态随机过程: Poisson 过程 离散参数连续状态随机过程: *Markov 序列连续参数连续状态随机过程: Gauss 过程,Brown 运动例2.1.1:一醉汉在路上行走,以的概率向前迈一步,以q 的概率向后迈一步,以p r 的概率在原地不动,1=++r q p ,选定某个初始时刻,若以记它在时刻的位置,则就是直线上的随机游动(Random Walk)。

随机过程英语讲义

随机过程英语讲义

随机过程英语讲义2.2 Properties of Poisson processesExample Suppose that people immigrate into a territory at a Poisson rateλ=1 per day. (a) What is the expected time until the tenth immigrant arrives? (b) What is the probability that the elapsed time between the tenth and the eleventh arrival exceeds two days? Solution: (a) E[S10]=10/λ= 10 days (b) P{X112}= e -2λ= e-2≈ 0.13332.2 Properties of Poisson processesArrival time distribution Proposition 2.2.2: The arrival time of the nth event Sn follows aΓ distribution with parameter (n,λ). (λt ) n 1 f (t )=λe λ t Proof: (n 1)!{Sn≤ t} {N(t)≥n} P{Sn≤ t}= P{N(t)≥n}=∑k=n∞e λ t(λ t ) k k!differentiating the two sides of equation with respect to t:2.2 Properties of Poisson processes2.2 Properties of Poisson processesSuppose we are told that exactly one event of a Poisson process has taken place by time t, and we are asked to determine the distribution of the time atwhich the event occurred. Since a Poisson process possesses stationary and independent increments, it seems reasonable that each interval in[0,t] of equal length should have the same probability of containing the event. In other words, the time of the event should be uniformly distributed over[0,t].2.2 Properties of Poisson processesThis result may be generalized, but before doing so we need to introduce the concept of order statistics.52.2 Properties of Poisson processesOrder statistics Order statistics Let Y1, Y2……Yn are n random variables, if we arrange these random variables from small to big, note Y(1)= y1 is the smallest in the sequence, Y(2)= y2 is the second smallest,…. Y(n)= yn is the biggest in the sequence. Y(1) Y(2)…… Y(n), Y(1)……Y(n) or y1……yn are the order statistics of Y1…Yn.2.2 Properties of Poisson processes,Let,f is density of distribution of Yi, if f follows the uniform density over (0,t), the joint density of{Y(i)} is:Y(1) .....Y( n )f( y1,..., yn )= n !Ci=1nn! f ( yi )= n, t0 y1 ... yn t2.2 Properties of Poisson processesPast arrival times givenC Joint density of past arrival times Proposition 2.2.3: Given that N(t)=n, the n arrival times S1… Sn have the same distribution as the order statistics corresponding to the n i.i.d. samples from U(0,t). that is,n! f S1 .....S n N ( t )(t1,..., t n n)= n, tProof:0 t1 ... t n t2.2 Properties of Poisson processesP{ti≤ Si≤ t+hi, i=1,…n|N(t)= n}P{one event in[ti, ti+ hi], 1≤ i≤ n, no events elsewhere in[0,t]}= P{N (t )= n}=λh1eλh1n!= n h1 ...hn t...λhn e e e λt (λ t ) n n!λhn1λ ( t h1 ...hn )P{N (t )= n}= eλt(λ t ) n, n= 0,1, 2,...... n!2.2 Properties of Poisson processesn! P{ti≤ Si≤ t+hi, i=1,…n|N(t)= n}= n h1 ...hn t P{ti≤ S i≤ ti+ hi, i= 1...n N (t )= n} n!= n h1....hn ttaking the limits as hi→ 0 for all i, we obtain n! f S1 ..... S n N ( t )(t1,..., t n n)= n t2.2 Properties of Poisson processesExample: A cable TV company collects$1/unit time from each subscriber. Subscribers sign up in ac cordance with a Poisson process with rateλ. What is the expected total revenue received in (0,t]? Solution: (Depends on the total number of subscribers and their arriving time)2.2 Properties of Poisson processesLet N(t) denote the number of subscribers, and Si denote the收益arrival time of the ith customer. The revenue generated by this customer in (0,t] is t-Si. Adding the revenues generated by all arrivals in (0,t] N (t ) ∑ (t Si ), E ∑ (t S i ) i=1 i=1 find the previous expectation by conditioning on N(t)N (t )N (t ) n n E ∑ (t S i ) N (t )= n = E ∑ (t S i ) N (t )= n = nt E ∑ S i N (t )= n i=1 i=1 i=12.2 Properties of Poisson processesLet U1,…Un be iid random variables which follow U(0,t). sot n n n n E ∑ Si N (t )= n= E ∑ U (i ) = E ∑U i =∑ E[U i]= n 2 i=1 i=1 i=1 i=1soN (t ) t t E ∑ (t Si ) N (t )= n = nt n= n 2 2 i=1Calculate the expectation by conditional expectation: N (t ) E[ N (t )]t 1 2=λt E ∑ (t S i ) = 2 2 i=12.2 Properties of Poisson processesDecomposition of Poisson process (an important application of Proposition 2.2.3) A Poisson process N={N(t),t≥0} with rateλ. We consider the case in which if an arrival occurs at time s, it is a type-1 arrival with probability P(s) and a type-2 arrival with probability 1-P(s). The type of arrival depends on the epoch of arrival. By using Proposition 2.2.3 we can prove the following propositon.2.2 Properties of Poisson processesProposition 2.2.4 Let Ni={Ni(t), t≥0}, i=1 and 2, where Ni(t) denotes the number of type-i arrivals in (0,t]. N1(t) and N2(t) are two independent Poisson random variables with meansλpt andλqt, where1 t p=∫ P ( s )ds and q= 1 p t 0 λ pt (λpt ) n λ qt (λqt ) m P{N1 (t )= n, N 2 (t )= m}= e e n! m!2.2 Properties of Poisson processes2.2 Properties of Poisson processesThe importance of the above proposition is illustrated by thefollowing example.17。

随机过程讲义 第一章

随机过程讲义 第一章

第一章 随机过程及其分类在概率论中,我们研究了随机变量,n 维随机向量。

在极限定理中我们研究了无穷多个随机变量,但只局限在它们之间相互独立的情形。

将上述情形加以推广,即研究一族无穷多个、相互有关的随机变量,这就是随机过程。

1. 随机过程的概念定义:设),,(P ∑Ω是一概率空间,对每一个参数T t ∈,),(ωt X 是一定义在概率空间),,(P ∑Ω上的随机变量,则称随机变量族});,({T t t X X T ∈=ω为该概率空间上的一随机过程。

其中R T ⊂是一实数集,称为指标集或参数集。

随机过程的两种描述方法: 用映射表示T X ,R T t X →Ω⨯:),(ω即),(⋅⋅X 是一定义在Ω⨯T 上的二元单值函数,固定T t ∈,),(⋅t X 是一定义在样本空间Ω上的函数,即为一随机变量;对于固定的Ω∈ω,),(ω⋅X 是一个关于参数T t ∈的函数,通常称为样本函数,或称随机过程的一次实现,所有样本函数的集合确定一随机过程。

记号),(ωt X 有时记为)(ωt X 或简记为)(t X 。

参数T 一般表示时间或空间。

常用的参数一般有:(1)},2,1,0{0 ==N T ;(2)},2,1,0{ ±±=T ;(3)],[b a T =,其中a 可以取0或∞-,b 可以取∞+。

当参数取可列集时,一般称随机过程为随机序列。

随机过程});({T t t X ∈可能取值的全体所构成的集合称为此随机过程的状态空间,记作S 。

S 中的元素称为状态。

状态空间可以由复数、实数或更一般的抽象空间构成。

实际应用中,随机过程的状态一般都具有特定的物理意义。

例1:抛掷一枚硬币,样本空间为},{T H =Ω,借此定义:⎩⎨⎧=时当出现,时当出现T 2H ,cos )(t t t X π ),(∞+-∞∈t 其中2/1}{}{==T P H P ,则)},(,)({∞+-∞∈t t X 是一随机过程。

随机过程精品课件 (15)

随机过程精品课件 (15)


C(t1, tn ) C(t2 , tn ) C(tn , tn )
m(ti ) E[ X (ti )]
C(ti , t j ) E{[ X (ti ) m(ti )][ X (t j ) m(t j )]} C(t j , ti )
E[ X ( t ) X ( s )] m( s )m( t )
E{[ X (t ) X ( s) X ( s)]X ( s)} m( s)m(t ) E {[ X ( t ) X ( s )]E[ X ( s )]} E[ X ( s )] m st
2 2
X(t) - X(s) 与X(s)相互 独立.
m ( s t ) m ( s ) m ( t ),
D( s t ) D( s ) D( t )
命题:若 y ( s t ) y ( s ) y ( t ), 则对任意实数 t, 有 y ( t ) ty (1). 可证得1)和2). 证3) C ( s , t ) E {[ X ( t ) m ( t )][ X ( s ) m ( s )]}
2
2
2
(ti ti 1 )
2
4.具有马氏性 证
因 X (t ) 是维纳过程
所以 增量 X (t 因此
s ) X ( s ) 与时刻 s 以前的状态 X ( ) ( 0 s )独立,
P{X(t s) a | X(s) x ,X() ,0 s }
0, s t E X ( t ) 0, R ( s , t ) δ ( s t ) , s t
2
称其为高斯白噪声过程,它是独立过程. 高斯白噪声是典型的随机干扰数学模型, 普遍存在于电流的波动,通信设备各部分的 波动,电子发射的波动等各种波动现象中. 如金融、电子工程中常用的线性模型— 自回归模型(AR(p))

《随机过程》课件

《随机过程》课件

f1(x1, t1)
F1(x1, t1) x1
4
● 随机过程 (t) 的二维分布函数:
F2 (x1, x2 ;t1,t2 , ) P (t1) x1, (t2 ) x2
● 随机过程 (t)的二维概率密度函数:
f2
(x1,
x2 ; t1, t2
)
2F2 (x1, x2;t1,t2 ) x1 x2
Dξ t Eξ 2 t 2atξ t a2 t
E[ξ 2 (t)] 2at Eξ t a2 (t)
E[ξ 2 (t)] a2 (t)



所以 a(t
,) 的方偏差离等程于x度2均f。1方(
x值,
t与)d均x值平[a方(t之)]差2











t

均方值
均值平方
8
● 相关函数
在通信系统中所遇到的信号及噪声,大多数可视为平稳的随机过程。 因此,研究平稳随机过程有着很大的实际意义。
13
● 2.2 各态历经性 ● 问题的提出:我们知道,随机过程的数字特征(均值、相关函数)是对随 机过程的所有样本函数的统计平均,但在实际中常常很难测得大量的样本, 这样,我们自然会提出这样一个问题:能否从一次试验而得到的一个样本 函数x(t)来决定平稳过程的数字特征呢? ● 回答是肯定的。平稳过程在满足一定的条件下具有一个有趣而又非常有用 的特性,称为“各态历经性”(又称“遍历性”)。具有各态历经性的过 程,其数字特征(均为统计平均)完全可由随机过程中的任一实现的时间 平均值来代替。 ● 下面,我们来讨论各态历经性的条件。
R(t1,t2 ) E[ (t1) (t2 )]

随机过程英语讲义-1

随机过程英语讲义-1

18
执行器
控制对象
传感器
y(k+2)
乱 序

络 y(k-1)
y(k+1)
控制器
y(k)
Solution: Denote sample space by S then, S = { 1, 2, 3, …, }.
19
Problem 2: Measure the time between two message arrivals at a message center. Solution: Denote sample space by S then, S = { t: t ≥ 0 } = [ 0, where t denotes time. ),
backslash, sometimes escape 反斜线转义符,有时表 示转义符或续行符 tilde 波浪符
'
apostrophe hyphen
撇号[ə'pɔstrəfi]
-...
‘’
连字号
dash 破折号 dots/ ellipsis 省略号 single quotation marks 单引号 double quotation marks 双引号 parallel 双线号 arrow 箭号;参见号
20
We shall suppose that for each event E of the sample space S a number P(E) is defined and satisfies the following three axioms: Axiom (1) 0 ≤ P(E) ≤ 1. Axiom (2) P(S)=1. Axiom (3) For any sequence of events E1 , E2 ,…that are mutually exclusive, that is, events for which Ei∩Ej=Φ when i≠ j (where Φ is the null set) ,

随机过程讲义英文版

随机过程讲义英文版

1
Answer: That X and Y are independent and have densities X and Y means the following. For any intervals I and J, possibly unbounded,
P{X ∈ I, Y ∈ J} = P{X ∈ I}P{Y ∈ J} =
3. Sum of independent random variables. Convolution Suppose that X and Y are independent random variables defined on some probability space and have densities f and g respectively. Calculate the density of the random variable Z X + Y .
/itprnn/book.html).
Answer: Page 60 - 61 of [1]. Remark: If you remember, I wrote P{H = i|D = 3} for the first question and P{H = i|D = 2, H = 2} for the second. Actually we should also write P{H = i|D = 3, H = 3} for the first one. However, for the first set of conditional probabilities, the information {H = 3} is already contained in {D = 3} because we are sure that the M.C. opens an empty door (but not for the second set, why?), so the results are the same. But for clarity, we should have written down everything.

随机过程英语讲义-5

随机过程英语讲义-5

m X (t ) := E[X (t )]
t ∈T

Similarly, we can define autocorrelation function
R X (s, t ) := E[X ( s) X (t )] s, t ∈ T
Properties:
Autocovariance function
21
Orthogonal increment process(正交增量过程) Let {X (t ), t ≥ 0} be a random process. If for arbitrary 0 ≤ t1 ≤ t 2 ≤ t 3 ,
E [( X (t 3 ) − X (t 2 ) )( X (t 2 ) − X (t1 ) )] = 0
( cos(a) cos(b) = (cos(a + b ) + cos(a − b )) is used.)
1 2
Multiple random processes:
Given two random processes {X (t ), t ∈ T1 } and {Y (t ), t ∈ T2 } . Let
Solution: Recall that the probability density function of ξ is
⎧1 ⎪ , 0 < x < 2π , p ( x) = ⎨ 2π ⎪ 0, otherwise ⎩
thus
m X (t ) = E[X (t )] = E[a cos(ωt + ξ )]
11
Example:Find m X (t ) , C X (s, t ) and R X ( s, t ) of a random process

随机过程英语讲义-9

随机过程英语讲义-9

Setting t=0,
[m( s )]n P{N ( s ) = n} = e − m ( s ) , n!
E[N(s)] = m(s)
n = 0 ,1,2 ,......
--- mean value function
16
2.3 Non-homogenous Poisson processes
Proof:
λ (t) = arrival rate at time t
12
2.3 Non-homogenous Poisson processes
Definition: The counting process N={N(t), t≥0} is called a non-homogeneous Poisson process with intensity function {λ(t), t≥0} if it possesses the following properties: (a) N(0)=0 (b) It satisfies the independent-increment property (c) P{N(t+h)-N(t)=1}=λ(t)h+o(h) (d) P{N(t+h)-N(t) ≥2}=o(h)
20
2.3 Non-homogenous Poisson processes
Solution: (1) arrivals constitute a nonhomogeneous Poisson process with intensity function λ(t) given by
13
2.3 Non-homogenous Poisson processes
Definition of Poisson processes: A counting process N ={N(t), t ≥0} is a Poisson process with rate λ >0, if it possesses the following properties: (a) N(0) = 0, (b) It satisfies the stationary and independent increment properties (c) P{N(h)=1}=λh+o(h) (d) P{N(h) ≥ 2}=o(h)

随机过程讲义

随机过程讲义
为事件A出现的情况下,事件B的条件概率,或 简称事件B关于事件A的条件概率。
2.基本公式
定理1(乘法公式)
假设 若 则
A1,A2, ,An为任意n个事件( n 2 ),
P(A1 A2 An) 0
P(A1 A2 An) P( A1 ) P( A2 | A1 ) P( A3 | A1 A2 )

pi P( X xi ) pij

(i 1,2,
j 1,2,)
p j P(Y y j ) pij
i 1
j 1
分别称为( X , Y )关于 X 和 Y 的边缘分布律。
X和Y相互独立的充要条件是
pij pi p j
连续型
若随机变量(X,Y)的概率密度为
P(Ai1 Ai2 Ais) P(Ai1)P(Ai2) P(Ais)
则称事件
A1,A2, ,An 相互独立。
美国有一对夫妻连续生了8个儿子。他们原本只想要4 个小孩,但是当前面4个小孩都是男孩时,他们想再生一 个女孩,直到连续生了7个男孩。后来他们的医生都保证 说,按照平均数定律,下次生女孩的概率是99%。不幸的 是,第8次还是男孩。因为生孩子和扔硬币一样,连续8个 男孩的概率固然很小,但是在已经生了7个男孩之后,下 一个是女孩的概率仍然是50%。
2
2
3.性质
(1)
E (C ) C
n n
D(C ) 0
2
E(CX ) CE ( X ) D(CX ) C D( X )
(2)
E ( X i ) E ( X i )
i 1 i 1
(3) 若X和Y相互独立,则
E( XY ) E ( X ) E(Y )

随机过程英语讲义-14

随机过程英语讲义-14

14
6.5 Limiting Probabilities
Hence, in the long run the rate at which transitions into state j occur must equal the rate at which transitions out of state j occur. Now when the process is in state j it leaves at rate vj , and, since Pj is the proportion of time it is in state j, it thus follows that vjPj = rate at which the process leaves state j. Similarly, when the process is in state i it departs to j at rate qij , and, since Pi is the proportion of time in state i, we see that the rate at which transitions from i to j occur is equal to qij Pi . Hence, Piqij = rate at which the process enters state j .
5
6.4 Kolmogorov Differential Equations
Computing the Transition Probabilities
If we define rij by Where qij=viPij is the transtion rate from i to j then the Kolmogorov backward equations can be written as and the forward equations as

随机过程Ch3-possion过程

随机过程Ch3-possion过程

由均值函数知,单位时间内事件A发生的平均数为。 称为过程的速率或强度。
3.2 泊松过程的性质
RX ( s, t ) E[ X ( s ) X (t )] E[ X ( s )( X (t ) X ( s ) X ( s ))] E[ X ( s )( X (t ) X ( s ))] E[( X ( s )) 2 ] E[( X ( s ))] E[( X (t ) X ( s ))] D[ X ( s )] E[ X ( s )]) (
3.2 泊松过程的性质

P W k(1) W1( 2 )
D
y
y=x D x


f ( x , y )dxdy
(1) k
f(x, y)为W 与 W1( 2 ) 的联合概率密度 由于X1(t)与X2(t)独立,故
f ( x , y ) fW (1 ) ( x ) fW ( 2 ) ( y )
Wn-2 Wn-1
Wn
FTn (t ) P Tn t 1 P Tn t 1 e t
3.2 泊松过程的性质
•等待时间Wn的分布 定理3.3设{X(t), t 0}是参数为的泊松过程, {Wn, n 1}是相应等待时间序列, 则Wn服从参数为n与的分布, 概率密度为
t (t ) n 1 ,t 0 e fW n ( t ) (n 1)! 0 , t 0
3.2 泊松过程的性质
证 Wn Ti (n 1) ,Ti为时间间隔
i 1
n
T1 0
T2
Tn W2 Wn-1 Wn t
j
Wn t X (t ) n
• 定义3.2:称计数过程{X(t),t 0 }是泊 松过程,如果X(t)满足 (1) X(0)=0; (2) X(t)是独立增量过程; (3)在任一长度为t的区间中,事件A发生的 次数服从参数t> 0(参数> 0)的泊松分 布,即对任意s, t 0,有 n t ( t ) P X (t s ) X ( s ) n e , n! n 0,1, 2,

随机过程英语讲义-2

随机过程英语讲义-2
Some useful properties:

Two jointly distributed random variables X and Y are said to be uncorrelated if their covariance , defined by Cov(X , Y )=E[( X − EX )(Y − EY )] =E[ XY ] − E[ X ]E[Y ] is zero.
33
Example:
34
35
1.4 Moments and Central Moments
Definition : The kth moment is The first moment = Average = Expectation:
μ X = E[ X ] = ∫ xf X ( x)dx
−∞
5)
P (a < X < b) = FX (b) − FX (a )Leabharlann 2Example:13
Joint PDF
Joint CDF
14
Properties:
1) FX ,Y ( x, y ) is non-decreasing. 2)
∫ ∫


-∞ -∞
f
X ,Y
( x, y )dxdy = 1
17
PMF of g(X) Suppose what are the PMF of Y ?
18
Expected Value of g(X)


Since g ( X ) is itself a random variable , it follows from (∗) that E[ g ( X )] = ∫ x dFg ( x),

随机过程导论Introduction

随机过程导论Introduction

8
Syllabus
Introduction to probability theory Poisson processes Renewal processes Discrete-time Markov chains Continuous-time Markov chains Brownian motion

For Random events Never be able to characterize them with absolute certainty; Be analyzed in a probabilistic fashion viห้องสมุดไป่ตู้ probability models
6
Preface
Examples of stochastic processes
14
Outline
Chapter 6: Brownian Motion
Brownian motion is a Markov process with a continuous state space and a continuous index set Brownian motion process was studied originally by English botanist Robert Brown for modeling the motion of a small particle immersed in a liquid subject to molecular collisions. A mathematical model of the process was first derived by Einstein.
15
12
Outline

随机过程英语讲义-11

随机过程英语讲义-11
STOCHASTIC PROCESSES AND ITS APPLICATIONS
Prof. Xia Yuanqing School of Automation Beijing Institute of Technology E-mail:yuanqing.xia@ Assistant Dai Li School of Automation Beijing Institute of Technology E-mail:daili1887@
Given that it rained on Monday and Tuesday, what is the probability that it will rain on Thursday?
4.2 Chapman-Kolmogorov Equations
23
4.1 Introduction and Examples
The following proposition formalizes the Markov property, which says that the future and the past are conditionally independent given the present state of the chain.
1 2
Two Simple Examples to Show the Formula: two state
20
4.1 Introduction and Examples
Proof:
21
4.1 Introduction and Examples
22
4.1 Introduction and Examples
P[Yn = k ] = e k!

随机过程1-5(2)

随机过程1-5(2)
m 2
0
f (u
k 1
n
k
) X ( uk )( t k t k 1 ) f (v l ) X (v l )( sl sl 1 ) 0
l 1
0

0
/
lim [ f ( uk ) f ( u j ) RX ( uk , u j )( t k t k 1 )( t j t j 1 )

b
X ( t )dt
/
a
例(补充):设随机过程 X(t) 的均值函数为
m X (t ) t 1
2
试求 Y ( s )

s 0
s
X ( t )d t
0
的均值函数。
解: 由均方积分的性质知
EY (s)

E X ( t )d t

s 0
( t 1 )d t
2
s
2
s.
3
第1章 随机过程基本概念 例(补充):设随机过程 X(t) 的协方差函数为
0 k 1
n
下面看均方积分存在的一个充分条件。
第1章 随机过程基本概念
第9页
定理4 f ( t ) X ( t ) 在区间 [a, b] 上均方可积的充分条件是二 重积分 b b f ( s ) f ( t ) RX ( s, t )dsdt
a a
存在;且有
E | f ( t ) X ( t )dt |
第7页
四、均方积分
定义 设 { X ( t ), t [ a , b ]} 是随机过程, f ( t ) 把区间 [a, b] 分成 n 个子区间,分点为
a t0 t1 ... t n b
  1. 1、下载文档前请自行甄别文档内容的完整性,平台不提供额外的编辑、内容补充、找答案等附加服务。
  2. 2、"仅部分预览"的文档,不可在线预览部分如存在完整性等问题,可反馈申请退款(可完整预览的文档不适用该条件!)。
  3. 3、如文档侵犯您的权益,请联系客服反馈,我们会尽快为您处理(人工客服工作时间:9:00-18:30)。
36
5.3 Azuma’s Inequality for Martingales
Theorem 5.3.3 Azuma's Inequality Let Zn, n ≥ 1 be a martingale with mean μ = E[ Z n ]. Let Z 0 = μ and suppose that for nonnegative constants α i , βi , i ≥ 1,
E[ f ( X )] ≤
β α +β
f (−α ) +
α α +β
f (β )
30
5.3 Azuma’s Inequality for Martingales
Proof: Since f is convex it follows that, in the region −α ≤ x ≤ β , it is never above the line segment connecting the points (−α , f (−α )) and ( β , f ( β )) . (See Figure 5.3.1.)
f (α , β ) = e β + e − β + α (e β − e − β ) − 2 exp{αβ + β 2 / 2}
would assume a strictly positive maximum in the interior of the Region R = {(α , β ) :| α |≤ 1,| β |≤ 100}. Setting the partial derivatives of f equal to 0,évy
5
Joseph Doob
6
5.0 Martingales origins
7
5.1 Martingale
Definiton:
8
5.1 Martingale
Remark:
9
5.1 Martingale ,
10
5.1 Martingale
11
5.1 Martingale ,
STOCHASTIC PROCESSES AND ITS APPLICATIONS
Prof. Xia Yuanqing School of Automation Beijing Institute of Technology E-mail:yuanqing.xia@ Assistant Dai Li School of Automation Beijing Institute of Technology E-mail:daili1887@
33
5.3 Azuma’s Inequality for Martingales
Now the preceding inequality is true when α = −1 or +1 and when β is large (say when | β |≥ 100 ). Thus, if Lemma 5.3.2 were false, then the function
Proposition:
18
5.2 Stopping Times
19
5.2 Stopping Times
20
5.2 Stopping Times
21
5.2 Stopping Times
22
5.2 Stopping Times
Theorem:
23
5.2 Stopping Times
24
5.2 Stopping Times
or, expanding in a Taylor series,
β 2i +1 /(2i )! =∑ β 2i +1 /(2i + 1)! ∑
i =0 i =0


xi ex = ∑ i =0 i !

which is clearly not possible when β ≠ 0 . Hence, if the lemma is not true, we can conclude that the strictly positive maximal value of f (α , β ) occurs when β = 0 . However, f (α , 0) = 0 and thus the lemma is proven.
3
5.0 Martingales origins
The concept of martingale in probability theory was introduced by Paul Pierre Levy, and much of the original development of the theory was done by Joseph Leo Doob among others. Part of the motivation for that work was to show the impossibility of successful betting strategies. This is the essence of the Martingale.
E[ f ( X )] ≤
E [X] = 0
f ( β ).
32
β α +β
f (−α ) +
α α +β
5.3 Azuma’s Inequality for Martingales
Lemma 5.3.2 For 0 ≤ θ ≤ 1
θe
(1−θ ) x
+ (1 − θ )e
−θ x
≤e
x2 / 8
Proof: Letting θ = (1 + α ) / 2 and x = 2 β , we must show that for −1 ≤ α ≤ 1 ,
(1 + α )e
β (1−α )
+ (1 − α )e
− β (1+α )
≤ 2e
β2 /2
or, equivalently,
e β + e − β + α (e β − e− β ) ≤ 2 exp{αβ + β 2 / 2}.
α α +β
f (β ) +
1 [ f ( β ) − f (−α )]x α +β
it follows, since −α ≤ X ≤ β , that
f (X ) ≤
β α +β
f (−α ) +
α α +β
f (β ) +
1 [ f ( β ) − f (−α )] X . α +β
Taking expectations gives the result, that is,
12
5.1 Martingale ,
Why?
= E[ Z n | Z1 , L , Z n ] = Zn .
13
5.1 Martingale
Why?
14
5.1 Martingale
15
5.1 Martingale
16
5.2 Stopping Times
Definiton:
17
5.2 Stopping Times
35
5.3 Azuma’s Inequality for Martingales
We just consider that α ≠ 0 when β ≠ 0 . We can see that
e β + e− β α 1+ α β = 1 + . ⇒ β (e β + e − β ) = e β − e − β e − e− β β
∂f (α , β ) = 0 ⇒ e β − e − β + α (e β + e − β ) = 2(α + β ) exp{αβ + β 2 / 2} [1] ∂β ∂f (α , β ) = 0 ⇒ e β − e − β = 2 β exp{αβ + β 2 / 2} [2] ∂α
34
5.3 Azuma’s Inequality for Martingales
Figure. 5.3.1
31
5.3 Azuma’s Inequality for Martingales
As the formula for this line segment is
y − f ( β ) y − f (−α ) = x−β x − (−α ) ⇔ y=
β α +β
f (−α ) +
1
Chapter 5
Martingales
2
5.0 Martingales origins
Originally, martingale referred to a class of betting strategies that was popular in 18th century France. The simplest of these strategies was designed for a game in which the gambler wins his stake if a coin comes up heads and loses it if the coin comes up tails. The strategy had the gambler double his bet after every loss so that the first win would recover all previous losses plus win a profit equal to the original stake. As the gambler's wealth and available time jointly approach infinity, his probability of eventually flipping heads approaches 1, which makes the martingale betting strategy seem like a sure thing. However, the exponential growth of the bets eventually bankrupts its users.
相关文档
最新文档