随机过程英文论文
广义平稳随机过程英语
广义平稳随机过程英语Generalized stationary stochastic processesA stochastic process is said to be stationary when the statistical properties of the process do not change over time. In other words, the mean, variance, and correlation structure of the process remain constant over time.However, in many real-world situations, the assumption of strict stationarity is not appropriate. For example, the dynamics of a financial market may change over time due to changes in regulations, technological advancements, or global economic conditions.Generalized stationary stochastic processes offer a more flexible framework that relaxes the assumption of strict stationarity. Instead, these processes assume that the statistical properties of the process may change over time, but they change slowly enough that they can be considered approximately stationary over a short period of time.One example of a generalized stationary stochastic process is the autoregressive integrated moving average (ARIMA) process. In an ARIMA process, the statistical properties of the process may change over time, but the changes are modeled as a function of past values of the process.Another example of a generalized stationary stochastic process is the autoregressive conditional heteroscedasticity (ARCH) process. In an ARCH process, the variance of the process may change over time, but the changes are modeled as a function of past squared residuals of the process.Overall, generalized stationary stochastic processes provide a more realistic framework for modeling real-world time series data that exhibit non-stationary behavior.。
随机过程在金融市场中的应用
随机过程在金融市场中的应用随机过程(Stochastic process)是一类描述不确定性的数学模型,它可以被用于描绘各种自然现象,例如气象、地震、生物学以及金融市场等等。
在金融市场中,对风险和不确定性的精确度量是非常重要的。
因此,随机过程成为了金融建模和风险管理中重要工具之一。
现在我们将探讨随机过程在金融市场中的应用。
1. 随机过程的定义在介绍随机过程在金融市场中的应用之前,我们先来了解一下随机过程的相关知识。
随机过程是指一个表示时间演变的随机变量族,它可以被看做是若干个随机变量的集合。
随机过程可以用一个或多个自变量来描述,例如时间或空间等。
这些自变量通常被称为“时空索引”,它们对应着各个时间或空间的状态。
随机过程通常有三个构成要素:状态空间(state space)、时空索引集(index set)和概率测度(probability measure)。
2. 在金融市场中,随机过程广泛应用于风险管理、金融衍生品定价和股票价格预测等领域。
下面我们来分别介绍一下这些应用。
2.1 风险管理随机过程在风险管理中的应用很广泛。
例如,一个公司可能需要计算其未来收入的概率分布,以便确定对冲或保险策略。
这通常需要建立一个代表公司未来收入的随机过程模型。
2.2 金融衍生品定价衍生品是一种由金融市场上的其他金融资产衍生出来的金融工具。
通俗的讲,衍生品就是一种基于其他金融资产的投资工具。
许多金融衍生品的定价是建立在随机过程模型的基础上完成的。
例如,期权和衍生品的定价公式中通常都涉及到随机过程。
2.3 股票价格预测随机过程在股票价格预测方面的应用也很广泛。
许多投资者会使用随机过程来建立股票价格预测模型。
这些模型通常会使用历史股价数据作为输入来计算出未来的股价走势。
3. 随机过程的种类在金融市场中,有以下几种随机过程被广泛应用。
3.1 随机游走过程随机游走过程是一种最简单的随机过程,它可以被认为是一种随机变量的序列。
markov随机过程 英文
0.9 0.1 P= 0.2 0.8
coke
0.2
pepsi
8
Markov Process
Coke vs. Pepsi Example (cont) Given that a person is currently a Pepsi purchaser, what is the probability that he will purchase Coke two purchases from now?
Heads .
P(H)
Tails -
1-P(H)
Markov Process
• Markov Property: The state of the system at time t+1 depends only
on the state of the system at time t
Pr[X t +1 = xt +1 | X 1 L X t = x1 L xt ] = Pr[X t +1 = xt +1 | X t = xt
0.9 0.1 P= 0.2 0.8
0.781 0.219 P = 0.438 0.562
3
Pr[X3=Coke] = 0.6 * 0.781 + 0.4 * 0.438 = 0.6438 Qi - the distribution in week i Q0=(0.6,0.4) - initial distribution Q3= Q0 * P3 =(0.6438,0.3562)
5
Markov Process
Gambler’s Example – Gambler starts with $10 - At each play we have one of the following: • Gambler wins $1 with probability p
课件——应用随机过程(英文版—G.A. PAVLIOTIS)
StochProc
January 16, 2011 2 / 367
This is a basic graduate course on stochastic processes, aimed towards PhD students in applied mathematics and theoretical physics.
The course will consist of three parts: Fundamentals of the theory of stochastic processes, applications (reaction rate theory, surface diffusion....) and non-equilibrium statistical mechanics.
Pavliotis (IC)
StochProc
January 16, 2011 5 / 367
3. PART III: NON-EQUILIBRIUM STATISTICAL MECHANICS. Derivation of stochastic differential equations from deterministic dynamics (heat bath models, projection operator techniques etc.). The fluctuation-dissipation theorem. Linear response theory. Derivation of macroscopic equations (hydrodynamics) and calculation of transport coefficients.
The emphasis of the course will be on the presentation of analytical tools that are useful in the study of stochastic models that appear in various problems in applied mathematics, physics, chemistry and biology.
随机过程英文练习题_连续时间马尔科夫链南京大学
随机过程英文练习题_连续时间马尔科夫链南京大学6.2 Suppose that a one-celled organism can be in one of two states-either A or B. An individual in state A will change to state B at an exponential rate α; an individual in state B divides into two new individuals of type A at an exponential rate β. Define an appropriate continuous-time Markov chain for a population of such organisms and determine the appropriate parameters for this model.6.3 Consider two machines that are maintained by a single repairman. Machine i functions for an exponential time with rate μbefore breaking down, i = 1,2. The repair times (for either i machine) are exponential with rate μ. Can we analyze this as a birth and death process? If so, what are the parameters? If not, how can we analyze it?6.4 Potential customers arrive at a single-server station in accordance with Poisson process with rate λ. However, if the arrival finds n customers already in the station, then he will enter the sy stem with probability αn. Assuming an exponential service rate μ, set this up as a birth and death process and determine the birth and death rates.6.8 Consider two machines, both of which have an exponential1. There is a single repairman that can lifetime with meanλservice machines at an exponential rate μ. Set up the Kolmogorov backward equations; you need not solve them.6.9 The birth and death process with parameters λ=μμ=,0,n>0 is called a pure death process. Find P ij(t).nn6.13 A small barbershop, operated by a single barber, has room for at most two customers. Potential customers arrive at a Poisson rate of three per hour, and the successive service times are independent exponential random variables with mean 1/4 hour. What is(a) the average number of customers in the shop?(b) the proportion of potential customers that enter the shop?(c) If the barber could work twice as fast, how much more business would he do?6.14Potential customers arrive at a full-service, one-pump gasstation at a Poisson rate of 20 cars per hour. However,customers will only enter the station for gas if there areno more than two cars(including the one currently beingattended to) at the pump. Suppose the amount of timerequired to service a car is exponentially distributed witha mean of five minutes.(how about 2-pumps?)(a)What fraction of the attendant’s time will be spentservicing cars?(b)What fraction of potential customers are lost?6.19 A single repairperson looks after both machines 1 and 2. Each time it is repaired, machine i stays up for an exponential time with rate 2,1,=i i λ. When machine i fails, it requires an exponential distributed amount of time with rate i μ to complete its repair. The repairperson will always service machine 1 when it is down. For instance, if machine 1 fails when 2 is being repaired,then the repairperson will immediately stop work on machine 2 and start on 1. What proportion of time is machine 2 down?6.23 A job shop consists of three machines and two repairmen. The amount of time a machine works before breaking down is exponentially distributed with mean 10. If the amount of time it takes a single repairman to fix a machine is exponentially distributed with mean 8, then(a) What is the average number of machines not in use? (b) What proportion of time are both repairmen busy?6.24 Consider a taxi station where taxis and customers arrive in accordance with Poisson processes with respective rates of one and two per minute. A taxi will wait no matter how many other taxis are present. However, an arriving customer that does not find a taxi waiting leaves. Find(a) the average number of taxis waiting(b) the proportion of arriving customers that get taxis.。
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.。
随机过程论文——PageRank算法的马尔科夫过程分析
PageRank算法的马尔科夫过程分析一、PageRank简介大名鼎鼎的PageRank算法是Google排名运算法则(排名公式)的一个非常重要的组成部分,其用于衡量一个网站好坏的标准。
在揉合了诸如Title、Keywords标识等所有其它因素之后,Google利用PageRank来调整网页的排名,使得“等级/重要性”的网页会相对排在前面。
简单来说,Google通过下述几个步骤来实现网页在其搜索结果页面中排名:(1)找到所有与搜索关键词匹配的网页(2)根据页面因素如标题、关键词密度等排列等级(3)计算导入链接的锚文本中关键词(4)通过PageRank得分调整网站排名结果PageRank于2001年9月被授予美国专利,专利人是Google创始人之一的拉里.佩奇(Larry Page)。
所以,PageRank里面的Page并不是指网页,而是指佩奇~PageRank对于网页重要性的级别分为1~10级,10级为满级。
PR值越高说明该网页越受欢迎,也即越重要。
一个PR值为1的网站表明该网站不具备流行度,而PR值为7~10的网站则表明该网站是非常受欢迎的,或者说极其重要。
一般PR值达到4,就算是一个相当不错的网站了。
Google把自己网站的PR值设置为10~类似里氏震级,PageRank级别并不是线性增长的,而是按照一种指数刻度,打个比方PageRank4比PageRank3虽然只是高了一级,但却在影响力上高上6~7倍,因此,一个PageRank5的网页和一个PageRank8的网页之间差距会比你可能认为的要大的多。
PageRank的思路很简单,打个比方:如何判断一篇论文的价值,即被其他论文引述的次数越多就越重要,如果被权威的论文引用,那么该论文也很重要。
PageRank就是借鉴于这一思路,根据网站的外部链接和内部链接的数量和质量来衡量这个网站的价值,相当于每个到该页面的链接都是对该页面的一次投票,被链接的越多,就意味着被其他网站投票越多。
随机过程--Chapter 2
customer in (0,t] is t-Si. Adding the revenues generated by all
arrivals in (0,t]
N (t)
N (t )
(t Si ) ,
i 1
E (t Si )
i1
14
2.2 Properties of Poisson processes
Solution:
(a) E[S10]=10/= 10 days
(b) P{X11>2} = e -2 = e-2 0.1333
9
2.2 Properties of Poisson processes
Arrival time distribution
Proposition 2.2 :
The arrival time of the nth event Sn follows a Γ distribution
f (t) e t
each interarrival time {Xn, n1} follows an exponential
distribution with parameter .
8
2.2 Properties of Poisson processes
Example 1 Suppose that people immigrate into a territory at a Poisson
and X1 and X2 are independent
7
2.2 Properties of Poisson processes
Similarly, we obtain: P{ Xn>tXn-1=s} = e-t P{ Xn tXn-1= s} = 1- e-t
随机过程(双语)
第 1 页 共 3 页Ⅰ.Decide whether the following statements are true or false (52=10''⨯)1.If 1,,n X X are independent random variables, then11[][]n nk k k k E X E X ===∏∏. ( ) 2.A generating function ()P s uniquely determines its sequence. ( )3.Gamblers ruin on {0,1,2,3}, then {1,2} is a class and is closed. ( )4.Let {,0}n X n ≥ be a Markov chain with finite state space. Fix a state i , define the successive return times to state i by (0)inf{:}i n n X i τ==, and, for 0n ≥, (1)inf{():}i i m n m n X i ττ+=≥=, then {(),0}i n n τ≥ is a renewalsequence. ( )5.The Compound Poisson process has independent increments. ( ) Ⅱ.Fill in the blanks (103=30''⨯)1.If ~(;)X p k λ, then the generating function ()P s of X is .2.If ~;(),X b k n p and ~;(),Y b k m p , and X and Y are independent , then ~X Y + .3.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 ()1{(),0}N t i i X t X t ==≥∑, where ()X t is the quantity ordered by t . Then the range of ()X t is .4.Suppose a Markov chain on states {1,2,3} with transition matrix 010********P ⎛⎫ ⎪= ⎪ ⎪⎝⎭, then 12(2)p = .5.The state i is recurrent iff ()0n ii n p∞==∑ .6.A set of states C is closed iff for all ,:c ij i C j C p ∈∈= .第 2 页 共 3 页7.Suppose that if it rains today, then it will rain tomorrow with probability 0.5, and if it does not rain today, then it will rain tomorrow with probability 0.4. Then the probability that it rains today and it will still rain next three days is .8.If X is uniform on (0,1), then X has the Laplace transform X Ee λ-= .9.Suppose {(),0}N t t ≥ is a homogeneous Poisson process on [0,)∞ with rate 2α=, then {(1)1,(3)2}=P N N == .10. Suppose {((0,]),0}N t t Λ> is a mixed Poisson process, then ((0,])E N t Λ= .Ⅲ.Computation (410=40''⨯)1.In a branching process, 2()0.20.20.6P s s s =++. Compute the extinction probability π.2.If X is a non-negative integer valued random variable with ~{}k X p ,()X P s Es =, express the generating functions, in terms of ()P s , of []P X n ≤. 3.Suppose {(),0}N t t ≥ is a homogeneous Poisson process on [0,)∞ with rate 2α=. Compact(1) {(1)(3)}E N N ;(2) {(1)1,(3)2}P N N ==.4.Three occupational levels were identified:(1) upper level (executive, managerial, high administrative, professional)(2) middle level (high grade supervisor, non-manual, skilled manual)(3) lower level (semi-skilled or unskilled).Transition probabilities from generation to generation were estimated to be10.40.50.120.20.60.230.10.50.4P ⎛⎫ ⎪= ⎪ ⎪⎝⎭第 3 页 共 3 页 Compute the probability (2)1,1,2,3i f i =.Ⅳ.Analysis (10')Consider a Markov chain on states {1,2,3,4} with transition matrix1212001000013230120120P ⎛⎫ ⎪ ⎪= ⎪ ⎪⎝⎭, (1) What are the equivalence classes ? (2) Which states are transient and which states are recurrent ?(3) What are the periods of each state.Ⅴ.Application (10')A small shop opens at 8 A.M. From 8 until 11 A.M. customers seem to arrive, on the average, at a steadily increasing rate that starts with an initial rate of 5 customers per hour at 8 A.M. and reaches a maximum of 20 customers per hour at 11 A.M. From 11 A.M. until 1 P.M. the (average) rate seems to remain constant at 20 customers per hour. However, the (average) arrival rate then drops steadily from 1 P.M. until closing time at 5 P.M. at which time it has the value of 12 customers per hour. If we assume that the numbers of customers arriving at the shop during disjoint time periods are independent, then(1) what is the probability that no customers arrive between 8:30 AM and 9:30AM ?(2) what is the expected number of arrivals in the period?。
简明随机过程导论(英文版-Niels O.Nggaard)
Example 1.2 Consider the closed interval [0, 1], this is a Borel set. We
remove
the
open
interval
(
1 3
,
2 3
).
We
are
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with
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1 3
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Introduction to Stochastic Processes
Niels O. Nygaard
1 Measure and Integration Theory
1.1 σ-algebras
Consider a set Ω. We consider a system of subsets F of Ω. We assume that F satisfies the following axioms
Basically you should think of a σ-algebra as a system of subsets in which you can perform any of the usual set-theoretic operations (union, intersection, difference) on countably many sets. We get an obvious example by taking F to be the system of all subsets of Ω, P(Ω). At the other extreme the system consisting only of Ω itself and ∅ is a σ-algebra.
国外研究生随机过程introduction; the coxrossrubinstein model
p
+
01..81 (1
−
p)
=
1.25 1.1
0.8
+
0.8 1.1
0.2
=
1.16 1.1
>
1.
The difference 1.16/1.1 − 1 can be thought of as the
price of risk.
In other words, the expected return is 0.25 × 0.8 − 0.2 × 0.2 = 0.16 > R = 0.1;
1
Binomial branch
• Two time points: t = 0 (“now”) and t = T = 1 (“tomorrow”).
• Two assets:
– Cash bond / bank account, with value Bt, earning an interest rate R;
Examples:
• Forward contract on a stock: a contract written at time t = 0, to buy a stock at time t = T at the strike price K.
• European stock option: a contract written at time t = 0, giving the right but not the obligation to buy a stock at time t = T at strike price K. The option costs C now.
Remarks:
1. R measures the time value of money (determined by market preferences): 1$ now is worth (1 + R)$ at time t = 1. Assumption: limitless lending and borrowing at the same rate!
随机过程理论及应用(中英文0600006
随机过程理论及应用(中英文0600006)一、课程代码:0600006课内学时: 48 学分: 3二、适用范围(学科、专业、层次等)控制科学与工程、控制工程三、先修课程线性代数、微积分、概率论四、教学目标随机过程理论及应用是自动控制专业研究生所必修的一门基础课程,该课程覆盖了概率论和随机过程的基本知识,包括泊松过程、马尔可夫链、鞅和布朗运动等。
在这门课程中,我们旨在讲授随机过程的一些基本理论,并扩展到其在控制、通信、经济和金融等领域的一些应用。
通过学习这门课程可以让学生学会以概率的方式来思考问题、看待问题和解决问题。
五、考核与成绩评定:成绩以百分制衡量。
成绩评定依据:课堂成绩10%,课后作业20%,考试70%。
六、教学方式课堂讲授、课堂讨论、论文分析七、教学大纲(大纲撰写人:闫莉萍)1.预备知识 6学时1.1概率的公理化定义1.2随机变量与数字特征1.3矩母函数与特征函数1.4条件数学期望1.5随机过程的基本概念1.6随机过程的有限维分布和数字特征1.7随机过程的分类2.二阶矩过程与均分分析 6学时2.1基本概念2.2H空间与均方分析2.3宽平稳过程的概念和基本性质3.泊松过程 6学时3.1定义3.2与泊松过程相关的若干分布3.3泊松过程的推广3.4泊松过程的应用4. 离散时间马尔可夫过程 8学时4.1定义4.2转移概率矩阵4.3Chapman-Kolmogorov方程4.4状态的分类与状态空间分解4.5平稳分布4.6离散参数马尔科夫链的随机模拟与蒙特卡罗方法4.7应用5. 连续时间马尔可夫过程 6学时5.1定义与基本概念5.2转移概率矩阵5.3Kolmogorov微分方程5.4强马尔可夫性与嵌入马尔可夫链5.5连续马尔可夫过程的随机模拟5.6应用6. 鞅 6学时6.1基本概念6.2上(下)鞅及分解定理6.3停时和停时定理6.4鞅收敛定理6.5连续参数鞅7. 布朗运动 6学时7.1定义7.2布朗运动的性质7.3最大值与首中时7.4布朗运动的变形与推广8. 伊藤过程 4学时8.1伊藤积分8.2伊藤公式8.3伊藤微分8.4应用实例九、参考书及学生必读参考资料:1. 闫莉萍, 夏元清, 杨毅. 随机过程理论及其在自动控制中的应用[M]. 北京:国防工业出版社, 2012.2. Sheldon M. Rose. Stochastic Processes (Second Edition) [M]. John Wiley & SonsInc., 1996.3. 龚光鲁, 钱敏平. 应用随机过程[M]. 北京: 清华大学出版社, 2007.4. 林元烈. 应用随机过程[M]. 北京: 清华大学出版社, 2002.。
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
维纳随机过程范文
维纳随机过程范文维纳随机过程(Wiener process)又称布朗运动(Brownian motion),是一种随机过程,最早由罗伯特·维纳(Robert Wiener)在1923年引入,用来描述随机运动的数学模型。
维纳随机过程被广泛应用于金融、物理学、工程学等领域,具有重要的理论和实际意义。
1.W(0)=0,即在t=0时刻,随机过程的取值为0;2. 对于任意的t1<t2<...<tn,随机变量W(t2)-W(t1),W(t3)-W(t2),...,W(tn)-W(tn-1)是相互独立的;3.对于任意的t1<t2,随机变量W(t2)-W(t1)服从均值为0,方差为(t2-t1)的正态分布。
1. 增量独立性:对于任意的s<t1<t2<...<tn,W(t1+s)-W(s),W(t2+s)-W(t1+s),...,W(tn+s)-W(tn-1+s)是相互独立的;2. 高斯性:对于任意的t1<t2<...<tn,W(t1),W(t2),...,W(tn)是服从正态分布的随机变量;3.零均值:对于任意的t,E[W(t)]=0,即维纳随机过程的期望值为0。
维纳随机过程在金融领域的应用非常广泛,特别是在金融风险管理和衍生产品定价中起到重要作用。
它被用来模拟股票价格、汇率、利率等金融指标的随机波动。
维纳随机过程假设价格的波动是由无数个微小的随机因素叠加产生的,这些随机因素的大小和方向是随机的,并且无法预测。
维纳随机过程的一个重要特征是随机性,这使得它成为金融市场的波动和不确定性的有效描述工具。
在金融风险管理中,通过模拟维纳随机过程来计算股票、指数、期货等的价值变动,可以帮助投资者评估风险水平并制定相应的风险管理策略。
在衍生产品定价中,维纳随机过程被用来计算期权、期货、利率互换等衍生产品的价格。
维纳随机过程还被应用于金融工程学中的套利交易和对冲策略的设计。
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
随机过程英语讲义-2
吧
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)
吧
吧
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
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-7
∑
k =n
∞
e −λ t
(λ t ) k k!
differentiating the two sides of equation with respect to t :
28
2.2 Properties of Poisson processes
29
P{N (t ) = n} = e −λt
E[N(t)] = λt
(λt ) n , n!
n = 0 ,1,2 ,......
11
2.1 Definition
12
2.1 Definition
Little o(t) results
Let o(t) denote any function of t that satisfies o(t)/t → 0 as t → 0. Examples include o(t) =tn, n>1 If the process {A(t), t≥0} is a Poisson process having arrival rate λ, then
23
2.2 Properties of Poisson processes
Proof: {X1>t} ⇔ {N(t)=0} P{X1>t} = P{N(t)=0} = e-λt P{ X1≤ t} = 1- e-λt differentiating the two sides of equation with respect to t, We obtain the density of distribution of X1:
f (t ) = λe − λ t
X1 follows an exponential distribution with parameter λ
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姓名:李范佩专业:031041202 学号:031041202Random Signal Analysischief contents1. Introduction of the random process2. Definition of the random process3. The digital characteristic of the random process4. Stationary random process and ergodic property5. The normal random process6. Markov chain7. Spectrum analysis of the stationary random process8. Analysis of the random signal through the linear system9. Analysis of the random signal through the nonlinear systemIntroductiona. Random process which is aim at the dynamical phenomenon that varies with the time, is the quantitative description for the relationship of the series of random events.b. Application: Atmosphere field, communication engineering,computer science and so on.c. Target: To find the inherent law from the events which is seeming external disorderThe definition of the random processWe suppose the sample space of the random expriment is S= {ξ},if there exists a corresponding function X(ξ,t),t ∈T for each ξ ,thus we can get gens function {X(ξ,t),t ∈T} about ‘t ’ for all the ξ,these function famlily about ‘t ’ are called random process,and recorded X(ξ,t).The random process can be redfined as follow:If X(ξ, ti) is random variable for each preset ti of the time (i = 1,2,3, …),then X(ξ,t) is called random process.Individual comprehension:Random process can be taken for the extension in the time –domain of the random variable. It is the combination of the random variable which is continuous and varying with timeProbability Distribution of Random ProcessThe definition of the probability distributionIf we suppose {X(t),t T } is random process,for arbitrary fixed t1,t2, …,tn ∈T,and real number x1, x2 , …,xn ∈ R,then we mark Fx(x1,x2 , …,xn ,t1,t2, …,tn ) = P{X(t1) ≤ x1, X(t2) ≤x2 , … X(tn) ≤ xn } as the n-dimensional distribution function of the random process {X(t),t ∈ T } .n-dimensional probability density functionfx(x1, x2 , …,xn , t1,t2, …,tn)= is called the n-dimensional distribution function of the random process {X(t),t 121212(,,,;,,,)X n n nF x x x t t t x x x ∂∂∂∂∈ T } .Finite dimensional distribution gens functions or n-dimensional probability density functions can fully determine the whole statistical property of the random process.The Digital Characteristics of The Random ProcessIn practical application we cannot fully determine the finte dimensional distribution gens functions to analyse it.Thus we just exploit the digitalcharacteristics to describe the random process.The digital characteristics includes mathematical expectation, variance, correlation function.a. Mathematical expectation The random process {X(t),t ∈ T }b. mx(t) = E[X(t)] = ,mx(t) shows the average of sample function value in the time t.b. Squared value is called squared value it shows the power of the random signalc. Variance The variance shows the rate of deviation of sample function value related to mx(t) Correlation FunctionIf x ,y obey the same distribution ,then it is called the autocorrelation function,otherwise called cross-correlation.(;)x xf x t dx ∞-∞⎰222()[()](;)x x t E X t x f x t dx ψ∞-∞==⎰22()[()][(()())]x x t D X t E X t m t σ==-12{(),}{(),} are the random process.For arbitray fixed t ,X t t T and Y t t T t T ∈∈∈121212(,)[()()](,;,)xy xy R t t E X t Y t x yf x y t t dxdy∞∞-∞-∞==⎰⎰Correlation function shows the degree of correlation of the sample value in the different time .For t1 = t2,Thus we can conclude that mx(t) and correlation function are the basic digital characteristics.Stationary Random Process and Ergodic PropertyIf the probability property of the random process is independent of the time shifting ,then we remember this random process as stationa ry random process. a. Stationnary random process is also classified into two types,the one is called sensu stricto random process, the other is called generalized random process. b. For the random process X(t),if it ’s n-dimensional probability density function is independent of the time start, just meet the following equation:this random process is called sensu stricto random process.c. For the random process X(t), if it meets these properties:⑴ (constant)2212(,)(,)[(()())][()]()x x x x K t t K t t E X t m t D X t t σ==-==12121212(,,,;,,,)(,,,;,,,)x n n x n n f x x x t t t f x x x t t t τττ=+++ (())()x E X t m t =。