随机过程英语讲义-15

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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) ,
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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.
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