标签： STAT3061

统计代写|随机过程代写stochastic process代考|STANDARDIZED RANDOM VARIABLES

Posted on 2023年7月11日2023年7月11日 by statistics-lab

如果你也在怎样代写随机过程Stochastic Porcesses 这个学科遇到相关的难题，请随时右上角联系我们的24/7代写客服。随机过程Stochastic Porcesses在概率论和相关领域，是一个数学对象，通常被定义为一个随机变量系列。随机过程被广泛用作系统和现象的数学模型，这些系统和现象似乎以随机的方式变化。这方面的例子包括细菌种群的生长，由于热噪声而波动的电流，或气体分子的运动。随机过程在许多学科中都有应用，如生物学、化学、生态学、神经科学、物理学、图像处理、信号处理、控制理论、信息理论、计算机科学、密码学和电信。此外，金融市场中看似随机的变化也促使人们在金融领域广泛使用随机过程。

随机过程Stochastic Porcesses应用和对现象的研究反过来又激发了新的随机过程的提出。这类随机过程的例子包括维纳过程或布朗运动过程，路易-巴舍利耶用来研究巴黎证券交易所的价格变化，以及A.K.埃朗用来研究一定时期内发生的电话数量的泊松过程。这两个随机过程被认为是随机过程理论中最重要和最核心的，并且在巴切莱特和埃朗之前和之后，在不同的环境和国家中被反复和独立地发现了。

statistics-lab™ 为您的留学生涯保驾护航在代写随机过程stochastic process方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写随机过程stochastic process代写方面经验极为丰富，各种代写随机过程stochastic process相关的作业也就用不着说。

统计代写|随机过程代写stochastic process代考|STANDARDIZED RANDOM VARIABLES

统计代写|随机过程代写stochastic process代考|BERNOULLI AND BINOMIAL RANDOM VARIABLES

Bernoulli trials, named after the Swiss mathematician James Bernoulli, are perhaps the simplest type of random variable. They have only two possible outcomes. One outcome is usually called a success, denoted by $s$. The other outcome is called a failure, denoted by $f$. The experiment of flipping a coin is a Bernoulli trial. Its only outcomes are “heads” and “tails.” If we are interested in heads, we may call it a success; tails is then a failure. The experiment of tossing a die is a Bernoulli trial if, for example, we are interested in knowing whether the outcome is odd or even. An even outcome may be called a success, and hence an odd outcome a failure, or vice versa. If a fuse is inspected, it is either “defective” or it is “good.” So the experiment of inspecting fuses is a Bernoulli trial. A good fuse may be called a success, a defective fuse a failure.

统计代写|随机过程代写stochastic process代考|Expectations and Variances of Binomial Random Variables

Let $X$ be a binomial random variable with parameters $(n, p)$. Intuitively, we expect that the expected value of $X$ will be $n p$. For example, if we toss a fair coin 100 times, we expect that the average number of heads will be 50 , which is $100 \cdot 1 / 2=n p$. Also, if we choose 10 fuses from a set with $30 \%$ defective fuses, we expect the average number of defective fuses to be $n p=10(0.30)=3$. The formula $E(X)=n p$ can be verified directly from the definition of mathematical expectation as follows:
$$
\begin{aligned}
E(X) & =\sum_{x=0}^n x\left(\begin{array}{l}
n \
x
\end{array}\right) p^x(1-p)^{n-x}=\sum_{x=1}^n x \frac{n !}{x !(n-x) !} p^x(1-p)^{n-x} \
& =\sum_{x=1}^n \frac{n !}{(x-1) !(n-x) !} p^x(1-p)^{n-x} \
& =n p \sum_{x=1}^n \frac{(n-1) !}{(x-1) !(n-x) !} p^{x-1}(1-p)^{n-x} \
& =n p \sum_{x=1}^n\left(\begin{array}{c}
n-1 \
x-1
\end{array}\right) p^{x-1}(1-p)^{n-x} .
\end{aligned}
$$
Letting $i=x-1$ (by reindexing this sum), we obtain
$$
E(X)=n p \sum_{i=0}^{n-1}\left(\begin{array}{c}
n-1 \
i
\end{array}\right) p^i(1-p)^{(n-1)-i}=n p[p+(1-p)]^{n-1}=n p,
$$
where the next-to-last equality follows from binomial expansion (Theorem 2.5). To calculate the variance of $X$, from a procedure similar to the one we used to compute $E(X)$, we obtain (see Exercise 30)
$$
E\left(X^2\right)=\sum_{x=1}^n x^2\left(\begin{array}{l}
n \
x
\end{array}\right) p^x(1-p)^{n-x}=n^2 p^2-n p^2+n p .
$$

随机过程代考

统计代写|随机过程代写stochastic process代考|BERNOULLI AND BINOMIAL RANDOM VARIABLES

伯努利试验以瑞士数学家詹姆斯·伯努利(James Bernoulli)命名，它可能是最简单的随机变量类型。它们只有两种可能的结果。一个结果通常被称为成功，用$s$表示。另一个结果称为失败，用$f$表示。抛硬币的实验是伯努利实验。它唯一的结果是“正面”和“反面”。如果我们对头像感兴趣，我们可以称之为成功;反面就是失败。掷骰子的实验是伯努利试验，例如，如果我们感兴趣的是知道结果是奇数还是偶数。偶数的结果可能被称为成功，因此奇数的结果可能被称为失败，反之亦然。如果检查保险丝，要么是“有缺陷的”，要么是“好的”。所以检查保险丝的实验是伯努利试验。好的保险丝可以说是成功，有缺陷的保险丝可以说是失败。

伯努利试验的样本空间包含两个点$s$和$f$。由$X(s)=1$和$X(f)=0$定义的随机变量称为伯努利随机变量。因此，当伯努利试验结果成功时，伯努利随机变量的值为1，当伯努利试验失败时，伯努利随机变量的值为0。如果$p$是成功的概率，那么$1-p$(有时表示为$q$)就是失败的概率。因此$X$的概率质量函数为
$$
p(x)= \begin{cases}1-p \equiv q & \text { if } x=0 \ p & \text { if } x=1 \ 0 & \text { otherwise }\end{cases}
$$

统计代写|随机过程代写stochastic process代考|Expectations and Variances of Binomial Random Variables

设$X$为二项随机变量，参数为$(n, p)$。直观地，我们期望$X$的期望值为$n p$。例如，如果我们投掷一枚均匀硬币100次，我们预计正面的平均次数是50，也就是$100 \cdot 1 / 2=n p$。同样，如果我们从一组有$30 \%$缺陷的保险丝中选择10个保险丝，我们期望有缺陷的保险丝的平均数目为$n p=10(0.30)=3$。公式$E(X)=n p$可由数学期望的定义直接验证如下:
$$
\begin{aligned}
E(X) & =\sum_{x=0}^n x\left(\begin{array}{l}
n \
x
\end{array}\right) p^x(1-p)^{n-x}=\sum_{x=1}^n x \frac{n !}{x !(n-x) !} p^x(1-p)^{n-x} \
& =\sum_{x=1}^n \frac{n !}{(x-1) !(n-x) !} p^x(1-p)^{n-x} \
& =n p \sum_{x=1}^n \frac{(n-1) !}{(x-1) !(n-x) !} p^{x-1}(1-p)^{n-x} \
& =n p \sum_{x=1}^n\left(\begin{array}{c}
n-1 \
x-1
\end{array}\right) p^{x-1}(1-p)^{n-x} .
\end{aligned}
$$
让$i=x-1$(通过重新索引这个和)，我们得到
$$
E(X)=n p \sum_{i=0}^{n-1}\left(\begin{array}{c}
n-1 \
i
\end{array}\right) p^i(1-p)^{(n-1)-i}=n p[p+(1-p)]^{n-1}=n p,
$$
其中倒数第二的等式由二项展开(定理2.5)推导而来。要计算$X$的方差，可以使用与计算$E(X)$类似的过程，得到(参见练习30)
$$
E\left(X^2\right)=\sum_{x=1}^n x^2\left(\begin{array}{l}
n \
x
\end{array}\right) p^x(1-p)^{n-x}=n^2 p^2-n p^2+n p .
$$

数学代写|随机过程统计代写Stochastic process statistics代考请认准statistics-lab™

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

金融工程代写

金融工程是使用数学技术来解决金融问题。金融工程使用计算机科学、统计学、经济学和应用数学领域的工具和知识来解决当前的金融问题，以及设计新的和创新的金融产品。

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

术语广义线性模型（GLM）通常是指给定连续和/或分类预测因素的连续响应变量的常规线性回归模型。它包括多元线性回归，以及方差分析和方差分析（仅含固定效应）。

有限元方法代写

有限元方法（FEM）是一种流行的方法，用于数值解决工程和数学建模中出现的微分方程。典型的问题领域包括结构分析、传热、流体流动、质量运输和电磁势等传统领域。

有限元是一种通用的数值方法，用于解决两个或三个空间变量的偏微分方程（即一些边界值问题）。为了解决一个问题，有限元将一个大系统细分为更小、更简单的部分，称为有限元。这是通过在空间维度上的特定空间离散化来实现的，它是通过构建对象的网格来实现的：用于求解的数值域，它有有限数量的点。边界值问题的有限元方法表述最终导致一个代数方程组。该方法在域上对未知函数进行逼近。[1] 然后将模拟这些有限元的简单方程组合成一个更大的方程系统，以模拟整个问题。然后，有限元通过变化微积分使相关的误差函数最小化来逼近一个解决方案。

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

随机分析代写

随机微积分是数学的一个分支，对随机过程进行操作。它允许为随机过程的积分定义一个关于随机过程的一致的积分理论。这个领域是由日本数学家伊藤清在第二次世界大战期间创建并开始的。

时间序列分析代写

随机过程，是依赖于参数的一组随机变量的全体，参数通常是时间。随机变量是随机现象的数量表现，其时间序列是一组按照时间发生先后顺序进行排列的数据点序列。通常一组时间序列的时间间隔为一恒定值（如1秒，5分钟，12小时，7天，1年），因此时间序列可以作为离散时间数据进行分析处理。研究时间序列数据的意义在于现实中，往往需要研究某个事物其随时间发展变化的规律。这就需要通过研究该事物过去发展的历史记录，以得到其自身发展的规律。

回归分析代写

多元回归分析渐进（Multiple Regression Analysis Asymptotics）属于计量经济学领域，主要是一种数学上的统计分析方法，可以分析复杂情况下各影响因素的数学关系，在自然科学、社会和经济学等多个领域内应用广泛。

MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习和应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

统计代写|随机过程代写stochastic process代考|DISTRIBUTION FUNCTIONS

Posted on 2023年7月11日2023年7月11日 by statistics-lab

统计代写|随机过程代写stochastic process代考|DISTRIBUTION FUNCTIONS

Random variables are often used for the calculation of the probabilities of events. For example, in the experiment of throwing two dice, if we are interested in a sum of at least 8 , we define $X$ to be the sum and calculate $P(X>8)$. Other examples are the following:

If a bus arrives at a random time between 10:00 A.M. and 10:30 A.M. at a station, and $X$ is the arrival time, then $X<10 \frac{1}{6}$ is the event that the bus arrives before 10:10 A.M.
If $X$ is the price of gold per troy ounce on a random day, then $X \leq 400$ is the event that the price of gold remains at or below $\$ 400$ per troy ounce.
If $X$ is the number of votes that the next Democratic presidential candidate will get, then $X \geq 5 \times 10^7$ is the event that he or she will get at least 50 million votes.
If $X$ is the number of heads in 100 tosses of a coin, then $40<X \leq 60$ is the event that the number of heads is at least 41 and at most 60 .

Usually, when dealing with a random variable $X$, for constants $a$ and $b(bb)$, $P(X \geq b), P(b \leq X \leq a), P(b<X \leq a), P(b \leq X<a)$, and $P(b<X<a)$ is our ultimate goal. For this reason we calculate $P(X \leq t)$ for all $t \in(-\infty,+\infty)$. As we will show shortly, if $P(X \leq t)$ is known for all $t \in \mathbf{R}$, then for any $a$ and $b$, all of the probabilities that are mentioned above can be calculated. In fact, since the real-valued function $P(X \leq t)$ characterizes $X$, it tells us almost everything about $X$. This function is called the distribution function of $X$.

统计代写|随机过程代写stochastic process代考|DISCRETE RANDOM VARIABLES

In Section 4.1 we observed that the set of possible values of a random variable might be finite, infinite but countable, or uncountable. For example, let $X, Y$, and $Z$ be three random variables representing the respective number of tails in flipping a coin twice, the number of flips until the first heads, and the amount of next year’s rainfall. Then the sets of possible values for $X, Y$, and $Z$ are the finite set ${0,1,2}$, the countable set ${1,2,3,4, \ldots}$, and the uncountable set ${x: x \geq 0}$, respectively. Whenever the set of possible values that a random variable $X$ can assume is at most countable, $X$ is called discrete. Therefore, $X$ is discrete if either the set of its possible values is finite or it is countably infinite. To each discrete random variable, a real-valued function $p: \mathbf{R} \rightarrow \mathbf{R}$, defined by $p(x)=P(X=x)$, is assigned and is called the probability mass function of $X$. (It is also called the probability function of $X$ or the discrete probability function of $X$.) Since the set of values of $X$ is countable, $p(x)$ is positive at most for a countable set. It is zero elsewhere; that is, if possible values of $X$ are $x_1, x_2, x_3, \ldots$, then $p\left(x_i\right) \geq 0(i=1,2,3, \ldots)$ and $p(x)=0$ if $x \notin\left{x_1, x_2, x_3, \ldots\right}$. Now, clearly, the occurrence of the event $\left{x_1, x_2, x_3, \ldots\right}$ is certain. Therefore, we have that $\sum_{i=1}^{\infty} P\left(X=x_i\right)=1$ or, equivalently, $\sum_{i=1}^{\infty} p\left(x_i\right)=1$.

Definition 4.3 The probability mass function $p$ of a random variable $X$ whose set of possible values is $\left{x_1, x_2, x_3, \ldots\right}$ is a function from $\mathbf{R}$ to $\mathbf{R}$ that satisfies the following properties.
(a) $p(x)=0$ if $x \notin\left{x_1, x_2, x_3, \ldots\right}$.
(b) $p\left(x_i\right)=P\left(X=x_i\right)$ and hence $p\left(x_i\right) \geq 0(i=1,2,3, \ldots)$.
(c) $\sum_{i=1}^{\infty} p\left(x_i\right)=1$.
Because of this definition, if, for a set $\left{x_1, x_2, x_3, \ldots\right}$, there exists a function $p: \mathbf{R} \rightarrow \mathbf{R}$ such that $p\left(x_i\right) \geq 0(i=1,2,3, \ldots), p(x)=0, x \notin\left{x_1, x_2, x_3, \ldots\right}$, and $\sum_{i=1}^{\infty} p\left(x_i\right)=1$, then $p$ is called a probability mass function.

随机过程代考

统计代写|随机过程代写stochastic process代考|DISTRIBUTION FUNCTIONS

随机变量常用于计算事件的概率。例如，在掷两个骰子的实验中，如果我们对至少为8的和感兴趣，我们定义$X$为和并计算$P(X>8)$。其他例子如下:

如果一辆公共汽车在上午10点到10点30分之间的随机时间到达一个车站，$X$是到达时间，那么$X<10 \frac{1}{6}$是公共汽车在上午10点10分之前到达的事件

如果$X$是任意一天每金衡盎司黄金的价格，那么$X \leq 400$是黄金价格保持在或低于$\$ 400$每金衡盎司的事件。

如果$X$是下届民主党总统候选人的得票数，那么$X \geq 5 \times 10^7$是至少获得5000万票的事件。

如果$X$是投掷100次硬币中正面出现的次数，那么$40<X \leq 60$是正面出现的次数最少为41次，最多为60次的事件。

通常，在处理随机变量$X$时，对于常量$a$和$b(bb)$, $P(X \geq b), P(b \leq X \leq a), P(b<X \leq a), P(b \leq X<a)$和$P(b<X<a)$是我们的最终目标。因此，我们计算所有$t \in(-\infty,+\infty)$的$P(X \leq t)$。我们将很快说明，如果所有$t \in \mathbf{R}$都知道$P(X \leq t)$，那么对于任何$a$和$b$，都可以计算出上面提到的所有概率。事实上，由于实值函数$P(X \leq t)$表征了$X$，它几乎告诉了我们关于$X$的一切。这个函数称为$X$的分布函数。

统计代写|随机过程代写stochastic process代考|DISCRETE RANDOM VARIABLES

在第4.1节中，我们观察到随机变量的可能值的集合可能是有限的、无限的但可数的或不可数的。例如，设$X, Y$和$Z$为三个随机变量，分别表示两次抛硬币时出现反面的次数、第一次出现正面的次数以及下一年的降雨量。那么$X, Y$和$Z$的可能值的集合分别是有限集${0,1,2}$、可数集${1,2,3,4, \ldots}$和不可数集${x: x \geq 0}$。当一个随机变量$X$所能假设的可能值的集合最多是可数的时候，$X$就被称为离散的。因此，$X$是离散的，如果它的可能值的集合是有限的或者它是可数无限的。对于每个离散随机变量，分配一个由$p(x)=P(X=x)$定义的实值函数$p: \mathbf{R} \rightarrow \mathbf{R}$，称为$X$的概率质量函数。(也称为$X$的概率函数或$X$的离散概率函数)因为$X$的值集合是可数的，所以对于可数集合，$p(x)$最多是正的。其他地方是零;也就是说，如果可能，$X$的值是$x_1, x_2, x_3, \ldots$，那么$p\left(x_i\right) \geq 0(i=1,2,3, \ldots)$和$p(x)=0$如果$x \notin\left{x_1, x_2, x_3, \ldots\right}$。现在，很明显，事件$\left{x_1, x_2, x_3, \ldots\right}$的发生是确定的。因此，我们得到$\sum_{i=1}^{\infty} P\left(X=x_i\right)=1$或$\sum_{i=1}^{\infty} p\left(x_i\right)=1$。

4.3随机变量$X$的可能值集为$\left{x_1, x_2, x_3, \ldots\right}$，其概率质量函数$p$是一个从$\mathbf{R}$到$\mathbf{R}$的函数，满足以下性质。
(a) $p(x)=0$如果$x \notin\left{x_1, x_2, x_3, \ldots\right}$。
(b) $p\left(x_i\right)=P\left(X=x_i\right)$，因此$p\left(x_i\right) \geq 0(i=1,2,3, \ldots)$。
(c) $\sum_{i=1}^{\infty} p\left(x_i\right)=1$。
由于这个定义，如果对于一个集合$\left{x_1, x_2, x_3, \ldots\right}$，存在一个函数$p: \mathbf{R} \rightarrow \mathbf{R}$使得$p\left(x_i\right) \geq 0(i=1,2,3, \ldots), p(x)=0, x \notin\left{x_1, x_2, x_3, \ldots\right}$和$\sum_{i=1}^{\infty} p\left(x_i\right)=1$，那么$p$被称为概率质量函数。

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

金融工程代写

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

有限元方法代写

随机分析代写

时间序列分析代写

回归分析代写

MATLAB代写

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

统计代写|随机过程代写stochastic process代考|CONDITIONAL PROBABILITY

Posted on 2023年7月11日2023年7月11日 by statistics-lab

统计代写|随机过程代写stochastic process代考|CONDITIONAL PROBABILITY

To introduce the notion of conditional probability, let us first examine the following question: Suppose that all of the freshmen of an engineering college took calculus and discrete math last semester. Suppose that $70 \%$ of the students passed calculus, $55 \%$ passed discrete math, and $45 \%$ passed both. If a randomly selected freshman is found to have passed calculus last semester, what is the probability that he or she also passed discrete math last semester? To answer this question, let $A$ and $B$ be the events that the randomly selected freshman passed discrete math and calculus last semester, respectively. Note that the quantity we are asked to find is not $P(A)$, which is 0.55 ; it would have been if we were not aware that $B$ has occurred. Knowing that $B$ has occurred changes the chances of the occurrence of $A$. To find the desired probability, denoted by the symbol $P(A \mid B)$ [read: probability of $A$ given $B$ ] and called the conditional probability of $A$ given $B$, let $n$ be the number of all the freshmen in the engineering college. Then $(0.7) n$ is the number of freshmen who passed calculus, and (0.45)n is the number of those who passed both calculus and discrete math. Therefore, of the (0.7)n freshmen who passed calculus, $(0.45) n$ of them passed discrete math as well. It is given that the randomly selected student is one of the (0.7) $n$ who passed calculus; we also want to find the probability that he or she is one of the (0.45) $n$ who passed discrete math. This is obviously equal to $(0.45) n /(0.7) n=0.45 / 0.7$. Hence
$$
P(A \mid B)=\frac{0.45}{0.7}
$$
But 0.45 is $P(A B)$ and 0.7 is $P(B)$. Therefore, this example suggests that
$$
P(A \mid B)=\frac{P(A B)}{P(B)}
$$

统计代写|随机过程代写stochastic process代考|THE MULTIPLICATION RULE

The relation
$$
P(A \mid B)=\frac{P(A B)}{P(B)}
$$
is also useful for calculating $P(A B)$. If we multiply both sides of this relation by $P(B)$ [note that $P(B)>0$ ], we get
$$
P(A B)=P(B) P(A \mid B),
$$
which means that the probability of the joint occurrence of $A$ and $B$ is the product of the probability of $B$ and the conditional probability of $A$ given that $B$ has occurred. If $P(A)>0$, then by letting $A=B$ and $B=A$ in (3.4), we obtain
$$
P(B A)=P(A) P(B \mid A) .
$$
Since $P(B A)=P(A B)$, this relation gives
$$
P(A B)=P(A) P(B \mid A) .
$$
Thus, to calculate $P(A B)$, depending on which of the quantities $P(A \mid B)$ and $P(B \mid A)$ is known, we may use (3.4) or (3.5), respectively. The following example clarifies the usefulness of these relations.

Example 3.10 Suppose that five good fuses and two defective ones have been mixed up. To find the defective fuses, we test them one-by-one, at random and without replacement. What is the probability that we are lucky and find both of the defective fuses in the first two tests?
Solution: Let $D_1$ and $D_2$ be the events of finding a defective fuse in the first and second tests, respectively. We are interested in $P\left(D_1 D_2\right)$. Using (3.5), we get
$$
P\left(D_1 D_2\right)=P\left(D_1\right) P\left(D_2 \mid D_1\right)=\frac{2}{7} \times \frac{1}{6}=\frac{1}{21} .
$$

随机过程代考

统计代写|随机过程代写stochastic process代考|CONDITIONAL PROBABILITY

为了介绍条件概率的概念，让我们首先检查以下问题:假设一所工程学院的所有大一新生上学期都选修了微积分和离散数学。假设学生中$70 \%$通过了微积分，$55 \%$通过了离散数学，$45 \%$两门都通过了。如果一个随机选择的大一新生上学期通过了微积分，那么他或她上学期也通过离散数学的概率是多少?为了回答这个问题，设$A$和$B$分别为随机选择的大一新生上学期通过离散数学和微积分考试的事件。请注意，我们要求查找的数量不是$P(A)$，而是0.55;如果我们不知道$B$已经发生了，它就会发生。知道$B$已经发生改变了$A$发生的几率。为了找到期望的概率，用符号$P(A \mid B)$[阅读:$A$给定$B$的概率]表示，并称为$A$给定$B$的条件概率，设$n$为工程学院所有新生的人数。那么$(0.7) n$是通过微积分的新生人数，(0.45)n是微积分和离散数学都通过的新生人数。因此，在通过微积分考试的(0.7)名新生中，有$(0.45) n$人通过了离散数学考试。假设随机选择的学生是通过微积分考试的(0.7)$n$中的一个;我们还想找出他或她是通过离散数学考试的(0.45)$n$之一的概率。这显然等于$(0.45) n /(0.7) n=0.45 / 0.7$。因此
$$
P(A \mid B)=\frac{0.45}{0.7}
$$
0.45是$P(A B)$, 0.7是$P(B)$。因此，这个例子表明
$$
P(A \mid B)=\frac{P(A B)}{P(B)}
$$

统计代写|随机过程代写stochastic process代考|THE MULTIPLICATION RULE

关系
$$
P(A \mid B)=\frac{P(A B)}{P(B)}
$$
也可用于计算$P(A B)$。如果我们在等式两边同时乘以$P(B)$[注意$P(B)>0$]，我们得到
$$
P(A B)=P(B) P(A \mid B),
$$
这意味着$A$和$B$同时出现的概率是$B$的概率和$A$的条件概率的乘积，假设$B$已经出现。如果$P(A)>0$，则在(3.4)中取$A=B$和$B=A$，我们得到
$$
P(B A)=P(A) P(B \mid A) .
$$
由于$P(B A)=P(A B)$，这个关系给出
$$
P(A B)=P(A) P(B \mid A) .
$$
因此，要计算$P(A B)$，取决于已知的量$P(A \mid B)$和$P(B \mid A)$，我们可以分别使用(3.4)或(3.5)。下面的示例说明了这些关系的有用性。

例3.10假设五个好的保险丝和两个有缺陷的保险丝混在一起。为了找到有缺陷的保险丝，我们逐个测试，随机，不更换。我们幸运地在前两次测试中找到两个有缺陷的保险丝的概率是多少?
解决方案:设$D_1$和$D_2$分别为在第一次和第二次测试中发现有缺陷的保险丝的事件。我们对$P\left(D_1 D_2\right)$感兴趣。使用(3.5)，我们得到
$$
P\left(D_1 D_2\right)=P\left(D_1\right) P\left(D_2 \mid D_1\right)=\frac{2}{7} \times \frac{1}{6}=\frac{1}{21} .
$$

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

金融工程代写

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

有限元方法代写

随机分析代写

时间序列分析代写

回归分析代写

MATLAB代写

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

统计代写|随机过程代写stochastic process代考|VARIANCES AND MOMENTS OF DISCRETE RANDOM VARIABLES

Posted on 2023年6月9日2023年6月9日 by statistics-lab

统计代写|随机过程代写stochastic process代考|VARIANCES AND MOMENTS OF DISCRETE RANDOM VARIABLES

Thus far, through many examples, we have explained the importance of mathematical expectation in detail. For instance, in Example 4.16, we have shown how expectation is applied in decision making. Also, in Example 4.17, concerning the lottery, we showed that the expected value of the winning amount per game gives an excellent estimation for the total amount a player will win if he or she plays a large number of times. In these and many other situations, mathematical expectation is the only quantity one needs to calculate. However, very frequently we face situations in which the expected value by itself does not say much. In such cases more information should be extracted from the probability mass function. As an example, suppose that we are interested in measuring a certain quantity. Let $X$ be the true value ${ }^{\dagger}$ of the quantity minus the value obtained by measurement. Then $X$ is the error of measurement. It is a random variable with expected value zero, the reason being that in measuring a quantity a very large number of times, positive and negative errors of the same magnitudes occur with equal probabilities. Now consider an experiment in which a quantity is measured several times, and the average of the errors is obtained to be a number close to zero. Can we conclude that the measurements are very close to the true value and thus are accurate? The answer is no because they might differ from the true value by relatively large quantities but be scattered both in positive and negative directions, resulting in zero expectation. Thus in this and similar cases, expectation by itself does not give adequate information, so additional measures for decision making are needed. One such quantity is the variance of a random variable.

Variance measures the average magnitude of the fluctuations of a random variable from its expected value. This is particularly important because random variables fluctuate from their expected values. To mathematically define the variance of a random variable $X$, the first temptation is to consider the expectation of the difference of $X$ from its expected value, that is, $E[X-E(X)]$. But the difficulty with this quantity is that the positive and negative deviations of $X$ from $E(X)$ cancel each other, and we always get 0 . This can be seen mathematically from the corollary of Theorem 4.2: Let $E(X)=\mu$; then
$$
E[X-E(X)]=E(X-\mu)=E(X)-\mu=E(X)-E(X)=0
$$

统计代写|随机过程代写stochastic process代考|BERNOULLI AND BINOMIAL RANDOM VARIABLES

The sample space of a Bernoulli trial contains two points, $s$ and $f$. The random variable defined by $X(s)=1$ and $X(f)=0$ is called a Bernoulli random variable. Therefore, a Bernoulli random variable takes on the value 1 when the outcome of the Bernoulli trial is a success and 0 when it is a failure. If $p$ is the probability of a success, then $1-p$ (sometimes denoted $q$ ) is the probability of a failure. Hence the probability mass function of $X$ is
$$
p(x)= \begin{cases}1-p \equiv q & \text { if } x=0 \ p & \text { if } x=1 \ 0 & \text { otherwise. }\end{cases}
$$
Note that the same symbol $p$ is used for the probability mass function and the Bernoulli parameter. This duplication should not be confusing since the $p$ ‘s used for the probability mass function often appear in the form $p(x)$.
An accurate mathematical definition for Bernoulli random variables is as follows.

随机过程代考

统计代写|随机过程代写stochastic process代考|VARIANCES AND MOMENTS OF DISCRETE RANDOM VARIABLES

到目前为止，通过许多例子，我们已经详细解释了数学期望的重要性。例如，在例4.16中，我们展示了期望如何应用于决策制定。同样，在例4.17中，关于彩票，我们展示了每次游戏中奖金额的期望值，可以很好地估计玩家在玩很多次时将赢得的总金额。在这些和许多其他情况下，数学期望是唯一需要计算的量。然而，我们经常面临的情况是期望值本身并不能说明什么。在这种情况下，应从概率质量函数中提取更多的信息。举个例子，假设我们对测量某个量感兴趣。设$X$为该量的真实值${ }^{\dagger}$减去测量得到的值。那么$X$是测量误差。这是一个期望值为零的随机变量，因为在对一个量进行非常多次的测量时，相同量级的正负误差出现的概率是相等的。现在考虑一个实验，其中一个量被测量了几次，误差的平均值接近于零。我们能不能得出这样的结论:测量值非常接近真实值，因此是准确的?答案是否定的，因为它们可能与真实值相差较大，但在正负方向上分散，导致零期望。因此，在这种和类似的情况下，期望本身并不能提供充分的信息，因此需要额外的决策措施。其中一个量就是随机变量的方差。

方差度量随机变量相对于其期望值的波动的平均幅度。这一点尤其重要，因为随机变量会从它们的期望值上下波动。要从数学上定义随机变量$X$的方差，首先要考虑的是$X$与其期望值(即$E[X-E(X)]$)之差的期望。但这个量的难点在于$X$和$E(X)$的正、负偏差相互抵消，我们总是得到0。这可以从数学上从定理4.2的推论中看出:设$E(X)=\mu$;然后
$$
E[X-E(X)]=E(X-\mu)=E(X)-\mu=E(X)-E(X)=0
$$

统计代写|随机过程代写stochastic process代考|BERNOULLI AND BINOMIAL RANDOM VARIABLES

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

金融工程代写

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

有限元方法代写

随机分析代写

时间序列分析代写

回归分析代写

MATLAB代写

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

统计代写|随机过程代写stochastic process代考|BAYES’ FORMULA

Posted on 2023年6月9日2023年6月9日 by statistics-lab

统计代写|随机过程代写stochastic process代考|BAYES’ FORMULA

To introduce Bayes’ formula, let us first examine the following problem. In a bolt factory, 30 , 50 , and $20 \%$ of production is manufactured by machines I, II, and III, respectively. If 4,5 , and $3 \%$ of the output of these respective machines is defective, what is the probability that a randomly selected bolt that is found to be defective is manufactured by machine III? To solve this problem, let $A$ be the event that a random bolt is defective and $B_3$ be the event that it is manufactured by machine III. We are asked to find $P\left(B_3 \mid A\right)$. Now
$$
P\left(B_3 \mid A\right)=\frac{P\left(B_3 A\right)}{P(A)} .
$$
So, to calculate $P\left(B_3 \mid A\right)$, we need to know the quantities $P\left(B_3 A\right)$ and $P(A)$. But neither of these is given. However, since $P\left(A \mid B_3\right)$ and $P\left(B_3\right)$ are known we use the relation
$$
P\left(B_3 A\right)=P\left(A \mid B_3\right) P\left(B_3\right)
$$
to find $P\left(B_3 A\right)$. To calculate $P(A)$, we use the law of total probability. Let $B_1$ and $B_2$ be the events that the bolt is manufactured by machines I and II, respectively. Then $\left{B_1, B_2, B_3\right}$ is a partition of the sample space; hence
$$
P(A)=P\left(A \mid B_1\right) P\left(B_1\right)+P\left(A \mid B_2\right) P\left(B_2\right)+P\left(A \mid B_3\right) P\left(B_3\right)
$$

统计代写|随机过程代写stochastic process代考|INDEPENDENCE

Let $A$ and $B$ be two events of a sample space $S$, and assume that $P(A)>0$ and $P(B)>0$. We have seen that, in general, the conditional probability of $A$ given $B$ is not equal to the probability of $A$. However, if it is, that is, if $P(A \mid B)=P(A)$, we say that $A$ is independent of $B$. This means that if $A$ is independent of $B$, knowledge regarding the occurrence of $B$ does not change the chance of the occurrence of $A$. The relation $P(A \mid B)=P(A)$ is equivalent to the relations $P(A B) / P(B)=P(A), P(A B)=P(A) P(B), P(B A) / P(A)=P(B)$, and $P(B \mid A)=P(B)$. The equivalence of the first and last of these relations implies that if $A$ is independent of $B$, then $B$ is independent of $A$. In other words, if knowledge regarding the occurrence of $B$ does not change the chance of occurrence of $A$, then knowledge regarding the occurrence of $A$ does not change the chance of occurrence of $B$. Hence independence is a symmetric relation on the set of all events of a sample space. As a result of this property, instead of making the definitions ” $A$ is independent of $B$ ” and ” $B$ is independent of $A$,” we simply define the concept of the “independence of $A$ and $B$.” To do so, we take $P(A B)=$ $P(A) P(B)$ as the definition. We do this because a symmetrical definition relating $A$ and $B$ does not readily follow from either of the other relations given [i.e., $P(A \mid B)=P(A)$ or $P(B \mid A)=P(B)]$. Moreover, these relations require either that $P(B)>0$ or $P(A)>0$, whereas our definition does not.
Definition 3.3 Two events $A$ and $B$ are called independent if
$$
P(A B)=P(A) P(B) .
$$
If two events are not independent, they are called dependent. If $A$ and $B$ are independent, we say that ${A, B}$ is an independent set of events.

随机过程代考

统计代写|随机过程代写stochastic process代考|BAYES’ FORMULA

为了介绍贝叶斯公式，让我们首先检查下面的问题。在螺栓工厂中，30％、50％和$20 \%$的产品分别由机器I、II和III生产。如果这些机器的4、5和$3 \%$的输出是有缺陷的，那么随机选择的被发现有缺陷的螺栓是由机器III生产的概率是多少?为了解决这个问题，设$A$为随机螺栓有缺陷的事件，$B_3$为机器III制造的事件。我们被要求找到$P\left(B_3 \mid A\right)$。现在
$$
P\left(B_3 \mid A\right)=\frac{P\left(B_3 A\right)}{P(A)} .
$$
为了计算$P\left(B_3 \mid A\right)$，我们需要知道$P\left(B_3 A\right)$和$P(A)$。但是这两个都没有给出。然而，由于$P\left(A \mid B_3\right)$和$P\left(B_3\right)$是已知的，我们使用关系
$$
P\left(B_3 A\right)=P\left(A \mid B_3\right) P\left(B_3\right)
$$
找到$P\left(B_3 A\right)$。为了计算$P(A)$，我们使用总概率定律。设$B_1$和$B_2$分别为螺栓由机器I和机器II制造的事件。则$\left{B_1, B_2, B_3\right}$是样本空间的一个分区;因此
$$
P(A)=P\left(A \mid B_1\right) P\left(B_1\right)+P\left(A \mid B_2\right) P\left(B_2\right)+P\left(A \mid B_3\right) P\left(B_3\right)
$$

统计代写|随机过程代写stochastic process代考|INDEPENDENCE

设$A$和$B$为样本空间$S$的两个事件，并假设$P(A)>0$和$P(B)>0$。我们已经看到，一般情况下，给定$B$的$A$的条件概率不等于$A$的概率。然而，如果是，也就是说，如果$P(A \mid B)=P(A)$，我们说$A$独立于$B$。这意味着，如果$A$独立于$B$，那么关于$B$发生的知识不会改变$A$发生的几率。关系$P(A \mid B)=P(A)$等价于关系$P(A B) / P(B)=P(A), P(A B)=P(A) P(B), P(B A) / P(A)=P(B)$和$P(B \mid A)=P(B)$。第一个和最后一个关系的等价意味着，如果$A$独立于$B$，那么$B$独立于$A$。换句话说，如果关于$B$发生的知识不会改变$A$发生的机会，那么关于$A$发生的知识不会改变$B$发生的机会。因此独立性是样本空间中所有事件集合上的对称关系。由于这个特性，我们没有定义“$A$独立于$B$”和“$B$独立于$A$”，而是简单地定义了“$A$和$B$的独立性”的概念，为此，我们将$P(A B)=$$P(A) P(B)$作为定义。我们这样做是因为关于$A$和$B$的对称定义不容易从给出的其他关系中得到[即$P(A \mid B)=P(A)$或$P(B \mid A)=P(B)]$]。此外，这些关系需要$P(B)>0$或$P(A)>0$，而我们的定义不需要。
3.3两个事件$A$和$B$称为独立if
$$
P(A B)=P(A) P(B) .
$$
如果两个事件不是独立的，则称为依赖事件。如果$A$和$B$是独立的，我们说${A, B}$是一组独立的事件。

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

金融工程代写

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

有限元方法代写

随机分析代写

时间序列分析代写

回归分析代写

MATLAB代写

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

统计代写|随机过程代写stochastic process代考|RANDOM SELECTION OF POINTS FROM INTERVALS

Posted on 2023年6月9日2023年6月9日 by statistics-lab

统计代写|随机过程代写stochastic process代考|RANDOM SELECTION OF POINTS FROM INTERVALS

In Section 1.6, we showed that the probability of the occurrence of any particular point in a random selection of points from an interval $(a, b)$ is 0 . This implies immediately that if $[\alpha, \beta] \subseteq(a, b)$, then the events that the point falls in $[\alpha, \beta],(\alpha, \beta),[\alpha, \beta)$, and $(\alpha, \beta]$ are all equiprobable. Now consider the intervals $\left(a, \frac{a+b}{2}\right)$ and $\left(\frac{a+b}{2}, b\right) ;$ since $\frac{a+b}{2}$ is the midpoint of $(a, b)$, it is reasonable to assume that
$$
p_1=p_2
$$
where $p_1$ is the probability that the point belongs to $\left(a, \frac{a+b}{2}\right)$ and $p_2$ is the probability that it belongs to $\left(\frac{a+b}{2}, b\right)$. The events that the random point belongs to $\left(a, \frac{a+b}{2}\right)$ and $\left(\frac{a+b}{2}, b\right)$ are mutually exclusive and
$$
\left(a, \frac{a+b}{2}\right) \cup\left[\frac{a+b}{2}, b\right)=(a, b) ;
$$
therefore,
$$
p_1+p_2=1
$$
This relation and (1.5) imply that
$$
p_1=p_2=1 / 2
$$

统计代写|随机过程代写stochastic process代考|WHAT IS SIMULATION?

Solving a scientific or an industrial problem usually involves mathematical analysis and/or simulation. To perform a simulation, we repeat an experiment a large number of times to assess the probability of an event or condition occurring. For example, to estimate the probability of at least one 6 occurring within four rolls of a die, we may do a large number of experiments rolling a die four times and calculate the number of times that at least one 6 is obtained. Similarly, to estimate the fraction of time that, in a certain bank all the tellers are busy, we may measure the lengths of such time intervals over a long period $X$, add them, and then divide by $X$. Clearly, in simulations, the key to reliable answers is to perform the experiment a large number of times or over a long period of time, whichever is applicable. Since manually this is almost impossible, simulations are carried out by computers. Only computers can handle millions of operations in short periods of time.

To simulate a problem that involves random phenomena, generating random numbers from the interval $(0,1)$ is essential. In almost every simulation of a probabilistic model,we will need to select random points from the interval $(0,1)$. For example, to simulate the experiment of tossing a fair coin, we draw a random number from $(0,1)$. If it is in $(0,1 / 2)$, we say that the outcome is heads, and if it is in $[1 / 2,1)$, we say that it is tails. Similarly, in the simulation of die tossing, the outcomes $1,2,3,4,5$, and 6 , respectively, correspond to the events that the random point from $(0,1)$ is in $(0,1 / 6),[1 / 6,1 / 3),[1 / 3,1 / 2),[1 / 2,2 / 3),[2 / 3,5 / 6)$, and $[5 / 6,1)$.

As discussed in Section 1.7, choosing a random number from a given interval is, in practice, impossible. In real-world problems, to perform simulation we use pseudorandom numbers instead. To generate $n$ pseudorandom numbers from a uniform distribution on an interval $(a, b)$, we take an initial value $x_0 \in(a, b)$, called the seed, and construct a function $\psi$ so that the sequence $\left{x_1, x_2, \ldots, x_n\right} \subset(a, b)$ obtained recursively from
$$
x_{i+1}=\psi\left(x_i\right), \quad 0 \leq i \leq n-1,
$$
satisfies certain statistical tests for randomness. (Choosing the tests and constructing the function $\psi$ are complicated matters beyond the scope of this book.) The function $\psi$ takes a seed and generates a sequence of pseudorandom numbers in the interval $(a, b)$. Clearly, in any pseudorandom number generating process, the numbers generated are rounded to a certain number of decimal places. Therefore, $\psi$ can only generate a finite number of pseudorandom numbers, which implies that, eventually, some $x_j$ will be generated a second time. From that point on, by (1.7), a pitfall is that the same sequence of numbers that appeared after $x_j$ ‘s first appearance will reappear. Beyond that point, numbers are not effectively random. One important aspect of the construction of $\psi$ is that the second appearance of any of the $x_j$ ‘s is postponed as long as possible.

随机过程代考

统计代写|随机过程代写stochastic process代考|RANDOM SELECTION OF POINTS FROM INTERVALS

在1.6节中，我们证明了在区间$(a, b)$中随机选择的点中出现任何特定点的概率为0。这意味着如果$[\alpha, \beta] \subseteq(a, b)$，那么点落在$[\alpha, \beta],(\alpha, \beta),[\alpha, \beta)$和$(\alpha, \beta]$中的事件都是等概率的。现在考虑区间$\left(a, \frac{a+b}{2}\right)$和$\left(\frac{a+b}{2}, b\right) ;$，因为$\frac{a+b}{2}$是$(a, b)$的中点，所以可以合理地假设
$$
p_1=p_2
$$
其中$p_1$是该点属于$\left(a, \frac{a+b}{2}\right)$的概率，$p_2$是该点属于$\left(\frac{a+b}{2}, b\right)$的概率。随机点所属的事件$\left(a, \frac{a+b}{2}\right)$和$\left(\frac{a+b}{2}, b\right)$互斥，且
$$
\left(a, \frac{a+b}{2}\right) \cup\left[\frac{a+b}{2}, b\right)=(a, b) ;
$$
因此，
$$
p_1+p_2=1
$$
这个关系式和式(1.5)表明
$$
p_1=p_2=1 / 2
$$

统计代写|随机过程代写stochastic process代考|WHAT IS SIMULATION?

解决科学或工业问题通常涉及数学分析和/或模拟。为了进行模拟，我们将实验重复很多次，以评估事件或情况发生的概率。例如，为了估计在掷骰子四次中至少出现一个6的概率，我们可以做大量的实验，将骰子掷四次，并计算得到至少一个6的次数。类似地，为了估计某家银行所有柜员忙碌的时间比例，我们可以测量一段很长的时间间隔的长度$X$，将它们相加，然后除以$X$。显然，在模拟中，获得可靠答案的关键是进行大量的实验或在很长一段时间内进行实验，无论哪种情况都适用。由于人工模拟几乎是不可能的，所以模拟是由计算机进行的。只有计算机才能在短时间内处理数百万次操作。

要模拟涉及随机现象的问题，必须从间隔$(0,1)$生成随机数。在几乎每个概率模型的模拟中，我们都需要从$(0,1)$区间中选择随机点。例如，为了模拟抛一枚均匀硬币的实验，我们从$(0,1)$中随机抽取一个数字。如果是在$(0,1 / 2)$，我们说结果是正面，如果是在$[1 / 2,1)$，我们说结果是反面。类似地，在掷骰子模拟中，结果$1,2,3,4,5$和6分别对应于$(0,1)$中的随机点在$(0,1 / 6),[1 / 6,1 / 3),[1 / 3,1 / 2),[1 / 2,2 / 3),[2 / 3,5 / 6)$和$[5 / 6,1)$中的事件。

正如第1.7节所讨论的，从给定的区间中选择一个随机数实际上是不可能的。在现实世界的问题中，我们使用伪随机数来进行模拟。为了从区间$(a, b)$上的均匀分布生成$n$伪随机数，我们取一个初始值$x_0 \in(a, b)$，称为种子，并构造一个函数$\psi$，以便从递归地获得序列$\left{x_1, x_2, \ldots, x_n\right} \subset(a, b)$
$$
x_{i+1}=\psi\left(x_i\right), \quad 0 \leq i \leq n-1,
$$
满足一定的随机性统计检验。(选择测试和构造函数$\psi$是复杂的事情，超出了本书的范围。)函数$\psi$接受一个种子，并在$(a, b)$区间内生成一系列伪随机数。显然，在任何伪随机数生成过程中，生成的数字被四舍五入到小数点后若干位。因此，$\psi$只能生成有限数量的伪随机数，这意味着最终会生成一些$x_j$。从这一点开始，通过(1.7)，一个陷阱是在$x_j$第一次出现之后出现的相同的数字序列将重新出现。超过这个点，数字就不是随机的了。构建$\psi$的一个重要方面是，任何$x_j$的第二次出现都尽可能地推迟。

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

金融工程代写

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

有限元方法代写

随机分析代写

时间序列分析代写

回归分析代写

MATLAB代写

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

统计代写|随机过程代写stochastic process代考|Processes with independent increments in a separable Banach space

Posted on 2023年5月26日2023年5月26日 by statistics-lab

统计代写|随机过程代写stochastic process代考|Processes with independent increments in a separable Banach space

Processes with independent increments in a separable Banach space. Let $\mathscr{X}$ be a separable Banach space, $\mathfrak{B}$ be the $\sigma$-algebra of its Borel subsets. Then ${\mathscr{X}, \mathfrak{B}}$ satisfies conditions a) and b) stated in the beginning of this Section. Thus we may consider processes with independent increments with values in $\mathscr{X}$.

Let $\xi(t)$ be a process defined on $[a, b)$. Consider the process $\overline{\xi(t)}$ independent of $\xi(t)$ with the same finite-dimensional distributions. Then $\xi(t)-\bar{\xi}(t)=\xi^(t)$ will be a symmetric process with independent increments. We shall assume that $\xi^(t)$ is a separable process. We now show that in such a case, $\xi^(t)$ has no discontinuities of the second kind. First note that for any closed convex set $S$ $$ \mathrm{P}\left{\xi^(s) \in S, a \leqslant s \leqslant t\right} \geqslant 1-2 \mathrm{P}\left{\xi^(t) \notin S\right} . $$ Indeed for an arbitrary partition $a=s_0l\left(s_v\right)\right}$ has no common points with $S$ (the existence of such a functional follows from the existence of a “supporting” hyperplane which separates points outside the convex set from the convex set), then denoting by $V$ the half-space ${x: l(x)<0}$ we obtain $$ \begin{aligned} \mathrm{P}{v \leqslant n}= & 1-\mathrm{P}\left{\xi^\left(s_k\right) \in S, k=0, \ldots, n\right} \
& \leqslant \mathrm{P}{v \leqslant n} 2 \mathrm{P}\left{\xi^(t)-\xi^\left(s_v\right) \in V \mid v \leqslant n\right} \leqslant 2 \mathrm{P}\left{\xi^(t) \notin S\right} . \end{aligned} $$ In view of the separability of the process $\xi^(t)$ equation (18) follows from (19). We now construct closed convex compact sets $K_{m, n}$ such that
$$
\mathrm{P}\left{\xi^\left(b-\frac{1}{n}\right) \notin K_{m, n}\right} \leqslant \frac{1}{m} . $$ Then $$ \mathrm{P}\left{\xi^(s) \in K_{m, n}, a \leqslant s \leqslant b-\frac{1}{n}\right} \geqslant 1-\frac{2}{m} .
$$

统计代写|随机过程代写stochastic process代考|Some properties of sample functions

Some properties of sample functions. The process $\xi(t)$ is called a step process if it has only a finite number of jumps on each interval $[a, b-\varepsilon]$ and if $\xi(t)$ is constant between the jumps.

Theorem 6. In order for a stochastically continuous process $\xi(t)$ defined on $[a, b)$ and taking values in a separable Banach space $\mathscr{X}$ to be a step process it is necessary and sufficient that its characteristic functional $\varphi_t(l)$ be of the form
$$
\varphi_t(l)=\varphi_a(l) \exp \left{\int\left(e^{i l(x)}-1\right) \Pi_t(d x)\right}
$$
where $\Pi_t(A)$ for all $t \in[a, b)$ is a finite measure and is a continuous and monotone function of $t$ for $A \in \mathfrak{B}$.

Proof. Let $\xi(t)$ be a step process. Let $\Delta_{\varepsilon}={x:|x|>\varepsilon}$. Then $\xi(t)-\xi(a)=\lim {\varepsilon \rightarrow 0} \xi{\Delta_{\varepsilon}}(t)$ and moreover for $t<b$ there exists the $\lim {\varepsilon \rightarrow 0} v_t\left(\Delta{\varepsilon}\right)=v_t$, where $v_t$ is the total number of jumps of the process $\xi(s)$ up to the moment $t$. Being the limit of Poisson variables, the variable $v_t$ will also have a Poisson distribution. Moreover
$$
\mathrm{E} v_t=\lim {\varepsilon \rightarrow 0} \mathrm{E} v_t\left(\Delta{\varepsilon}\right)=\lim {\varepsilon \rightarrow 0} \Pi_t\left(\Delta{\varepsilon}\right)
$$
Hence, $\lim {\varepsilon \rightarrow 0} \Pi_t\left(\Delta{\varepsilon}\right)<\infty$. Set $\lim {\varepsilon \rightarrow 0} \Pi_t\left(\Delta{\varepsilon}\right)=\Pi_t(\mathscr{X})$. Then $\Pi_t(A)$ is defined and is finite for all $A \in \mathfrak{B}$. Furthermore
$$
\mathrm{E} e^{i l\left(\zeta_{\Delta_{\varepsilon}}(t)\right)}=\exp \left{\int_{\Delta_{\varepsilon}}\left(e^{i l(x)}-1\right) \Pi_t(d x)\right} .
$$

随机过程代考

统计代写|随机过程代写stochastic process代考|Processes with independent increments in a separable Banach space

可分离巴拿赫空间中具有独立增量的过程。设$\mathscr{X}$为可分巴拿赫空间，$\mathfrak{B}$为其Borel子集的$\sigma$ -代数。那么${\mathscr{X}, \mathfrak{B}}$满足本节开头所述的a)和b)条件。因此，我们可以考虑具有$\mathscr{X}$值的独立增量的过程。

假设$\xi(t)$是在$[a, b)$上定义的进程。考虑具有相同有限维分布的独立于$\xi(t)$的过程$\overline{\xi(t)}$。那么$\xi(t)-\bar{\xi}(t)=\xi^(t)$将是一个具有独立增量的对称过程。我们假定$\xi^(t)$是一个可分离过程。现在我们证明，在这种情况下，$\xi^(t)$没有第二类不连续。首先注意，对于任何封闭凸集$S$$$ \mathrm{P}\left{\xi^(s) \in S, a \leqslant s \leqslant t\right} \geqslant 1-2 \mathrm{P}\left{\xi^(t) \notin S\right} . $$的确，对于任意划分$a=s_0l\left(s_v\right)\right}$与$S$没有共同点(这样一个泛函的存在遵循一个“支持”超平面的存在，该平面将凸集外的点与凸集分开)，然后用$V$表示半空间${x: l(x)<0}$，得到$$ \begin{aligned} \mathrm{P}{v \leqslant n}= & 1-\mathrm{P}\left{\xi^\left(s_k\right) \in S, k=0, \ldots, n\right} \
& \leqslant \mathrm{P}{v \leqslant n} 2 \mathrm{P}\left{\xi^(t)-\xi^\left(s_v\right) \in V \mid v \leqslant n\right} \leqslant 2 \mathrm{P}\left{\xi^(t) \notin S\right} . \end{aligned} $$考虑到过程的可分性$\xi^(t)$式(18)由式(19)推导而来。我们现在构造闭凸紧集$K_{m, n}$，使得
$$
\mathrm{P}\left{\xi^\left(b-\frac{1}{n}\right) \notin K_{m, n}\right} \leqslant \frac{1}{m} . $$然后 $$ \mathrm{P}\left{\xi^(s) \in K_{m, n}, a \leqslant s \leqslant b-\frac{1}{n}\right} \geqslant 1-\frac{2}{m} .
$$

统计代写|随机过程代写stochastic process代考|Some properties of sample functions

样例函数的一些性质。如果过程$\xi(t)$在每个间隔$[a, b-\varepsilon]$上只有有限数量的跳跃，并且$\xi(t)$在跳跃之间是恒定的，则称为阶跃过程。

定理6。为了使定义在$[a, b)$上并在可分离的Banach空间$\mathscr{X}$上取值的随机连续过程$\xi(t)$是阶跃过程，其特征泛函$\varphi_t(l)$具有以下形式是充分必要的
$$
\varphi_t(l)=\varphi_a(l) \exp \left{\int\left(e^{i l(x)}-1\right) \Pi_t(d x)\right}
$$
其中$\Pi_t(A)$对于所有$t \in[a, b)$是有限测度，对于$A \in \mathfrak{B}$是$t$的连续单调函数。

证明。让$\xi(t)$成为一个循序渐进的过程。让$\Delta_{\varepsilon}={x:|x|>\varepsilon}$。然后是$\xi(t)-\xi(a)=\lim {\varepsilon \rightarrow 0} \xi{\Delta_{\varepsilon}}(t)$，对于$t<b$，还有$\lim {\varepsilon \rightarrow 0} v_t\left(\Delta{\varepsilon}\right)=v_t$，其中$v_t$是进程跳转的总数$\xi(s)$，到目前为止$t$。作为泊松变量的极限，变量$v_t$也将具有泊松分布。而且
$$
\mathrm{E} v_t=\lim {\varepsilon \rightarrow 0} \mathrm{E} v_t\left(\Delta{\varepsilon}\right)=\lim {\varepsilon \rightarrow 0} \Pi_t\left(\Delta{\varepsilon}\right)
$$
因此，$\lim {\varepsilon \rightarrow 0} \Pi_t\left(\Delta{\varepsilon}\right)<\infty$。设置$\lim {\varepsilon \rightarrow 0} \Pi_t\left(\Delta{\varepsilon}\right)=\Pi_t(\mathscr{X})$。然后定义$\Pi_t(A)$，并且对于所有$A \in \mathfrak{B}$都是有限的。此外
$$
\mathrm{E} e^{i l\left(\zeta_{\Delta_{\varepsilon}}(t)\right)}=\exp \left{\int_{\Delta_{\varepsilon}}\left(e^{i l(x)}-1\right) \Pi_t(d x)\right} .
$$

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

金融工程代写

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

有限元方法代写

随机分析代写

时间序列分析代写

回归分析代写

MATLAB代写

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

统计代写|随机过程代写stochastic process代考|Processes with a semi-Markov chance interference

Posted on 2023年5月26日2023年5月26日 by statistics-lab

统计代写|随机过程代写stochastic process代考|Processes with a semi-Markov chance interference

We construct a measure which corresponds to this process on $\left{\mathscr{F}^, \mathscr{N}^\right}$, where $\mathscr{F}^$ is the set of functions $x(t)$ defined on the sum of the intervals $\cup\left[t_k, t_{k+1}\right)$, $0=t_0$ is the $\sigma$-algebra generated by the cylinders on $\mathscr{F}^*$. Such a measure will depend on $x$ and $y$ which are initial values of the Markov and semi-Markov processes, respectively.

Let $\mathscr{F}y$ be the set of piecewise constant functions $y(t)$ with values in $\mathscr{Y}$. Let $y(t)=y_k$ if $\zeta_k \leqslant t<\zeta{k+1}, 0=\zeta_0<\zeta_1<\cdots<\zeta_n<\cdots, \zeta_n \uparrow \infty$. We associate a measure $\mathrm{P}x^{y(\cdot)}$ on $\left{\mathscr{F}^, \mathscr{N}^\right}$ with the function $y(\cdot)$ in the following manner: if $$ A=\bigcap{k=0}^n \theta_{\zeta_k} A_k
$$
where $A_k$ is a cylindrical set on $\mathscr{F}^*$ defined by the values of $x(t)$ for $0 \leqslant t<\zeta_{k+1}-\zeta_k$, then
$$
\begin{aligned}
\mathrm{P}x^{y(\cdot)}(A)= & \int \mathrm{P}_x^{y_0}\left{A_0 \cap C{\zeta_1}\left(d x_1\right)\right} \int \mathrm{P}{x_1}^{y_1}\left{A_1 \cap C{\zeta_2-\zeta_1}\left(d x_2\right)\right} \times \cdots \
& \times \int \mathrm{P}{x{n-1}}^{y_{n-1}}\left{A_{n-1} \cap C_{\zeta_n-\zeta_{n-1}}\left(d x_n\right)\right} \mathrm{P}{x_n}^{y_n}\left(A_n\right), \quad C_t(B)={x(\cdot): x(t) \in B} . \end{aligned} $$ It is easy to verify that $\mathrm{P}_x^{y(\cdot)}(A)$ is an $\mathscr{N}{y y}$-measurable function on $\mathscr{F}y$, where $\mathscr{N}{\mathscr{X}}$ is a $\sigma$-algebra generated by the cylincers in $\mathscr{F}{a y}$. Therefore the integral $$ \mathrm{P}{y, x}(A)=\int \mathrm{P}x^{y(\cdot)}(A) \mu{y y}(d y(\cdot)),
$$
where $\mu_{x y}(\cdot)$ is a measure on $\left{\mathscr{F}{y y}, \mathscr{N}{q y}\right}$ corresponding to the component of the semi-Markov process $(y(t) ; \xi(t))$, is meaningful.

统计代写|随机过程代写stochastic process代考|The ergodic theorem for processes with a discrete chance interference

The ergodic theorem for processes with a discrete chance interference. Let $(x(t) ; \xi(t))$ be a process with a discrete chance interference in the phase space ${\mathscr{X}, \mathfrak{B}}$. We assume that the process is regular, i.e. that $\xi(t)$ possesses a finite number of jumps on each finite interval. We shall be concerned with the behavior of
$$
\frac{1}{T} \int_0^T f(x(t), \xi(t)) d t
$$

as $T \rightarrow \infty$, where $f(x, s)$ is a $\mathfrak{B} \times \mathfrak{B}{+}$-measurable function. Below we shall derive sufficient conditions for expression (42) to have with probability 1 the limit of the form $$ S(f)=\iiint_0^{\infty} Q{x, 0}(t, d y, \mathscr{X}) f(y, t) d t \pi(d x) / \iint_0^{\infty} Q_{x, 0}(t, \mathscr{X}, \mathscr{X}) d t \pi(d x)
$$
as $T \rightarrow \infty$, where $Q_{x, 0}\left(t, B_1, B_2\right)$ is the function associated with a process with discrete chance interference by relation (38) and $\pi(d x)$ is the stationary distribution of the Markov chain in $(\mathscr{X}, \mathfrak{B})$ with the transition probability
$$
\pi(x, B)=Q_{x, 0}(0, \mathscr{X}, B),
$$
i. e. for all $B \in \mathfrak{B}$
$$
\pi(B)=\int \pi(d x) \pi(x, B)
$$
In order that expression (43) be meaningful it is necessary that there exist a measure $\pi$ which satisfies (45) and the function $f(y, t)$ be such that
$$
\iiint_0^{\infty} Q_{x, 0}(t, d y, \mathscr{X})|f(y, t)| d t \pi(d x)<\infty
$$
Recall that the Markov chain $\left{x_n\right}$ in ${\mathscr{X}, \mathfrak{B}}$ is called ergodic if there exists for this chain a stationary distribution $\pi(d x)$ and for all $x \in \mathscr{X}$ and $\mathfrak{B}$-measurable function $f$ such that $\int|f(x)| \pi(d x)<\infty$ we have, with probability $\mathrm{P}{x, 0}=1$, $$ \lim {n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^n f\left(x_k\right)=\int f(x) \pi(d x) .
$$

随机过程代考

统计代写|随机过程代写stochastic process代考|Processes with a semi-Markov chance interference

我们在$\left{\mathscr{F}^, \mathscr{N}^\right}$上构造了一个与此过程相对应的度量，其中$\mathscr{F}^$是在区间$\cup\left[t_k, t_{k+1}\right)$和上定义的函数集$x(t)$, $0=t_0$是在$\mathscr{F}^*$上由圆柱体生成的$\sigma$ -代数。这样的度量将取决于$x$和$y$，它们分别是马尔可夫过程和半马尔可夫过程的初始值。

设$\mathscr{F}y$为分段常数函数$y(t)$的集合，其值在$\mathscr{Y}$中。让$y(t)=y_k$如果$\zeta_k \leqslant t<\zeta{k+1}, 0=\zeta_0<\zeta_1<\cdots<\zeta_n<\cdots, \zeta_n \uparrow \infty$。我们以以下方式将$\left{\mathscr{F}^, \mathscr{N}^\right}$上的度量$\mathrm{P}x^{y(\cdot)}$与函数$y(\cdot)$关联起来:如果$$ A=\bigcap{k=0}^n \theta_{\zeta_k} A_k
$$
其中$A_k$是$\mathscr{F}^*$上的一个圆柱形集，由$x(t)$的值定义为$0 \leqslant t<\zeta_{k+1}-\zeta_k$，那么
$$
\begin{aligned}
\mathrm{P}x^{y(\cdot)}(A)= & \int \mathrm{P}x^{y_0}\left{A_0 \cap C{\zeta_1}\left(d x_1\right)\right} \int \mathrm{P}{x_1}^{y_1}\left{A_1 \cap C{\zeta_2-\zeta_1}\left(d x_2\right)\right} \times \cdots \ & \times \int \mathrm{P}{x{n-1}}^{y{n-1}}\left{A_{n-1} \cap C_{\zeta_n-\zeta_{n-1}}\left(d x_n\right)\right} \mathrm{P}{x_n}^{y_n}\left(A_n\right), \quad C_t(B)={x(\cdot): x(t) \in B} . \end{aligned} $$很容易验证$\mathrm{P}x^{y(\cdot)}(A)$是$\mathscr{F}y$上的一个$\mathscr{N}{y y}$ -可测量函数，其中$\mathscr{N}{\mathscr{X}}$是$\mathscr{F}{a y}$中的圆柱体生成的一个$\sigma$ -代数。因此积分$$ \mathrm{P}{y, x}(A)=\int \mathrm{P}x^{y(\cdot)}(A) \mu{y y}(d y(\cdot)), $$ 其中$\mu{x y}(\cdot)$是$\left{\mathscr{F}{y y}, \mathscr{N}{q y}\right}$上对应于半马尔可夫过程$(y(t) ; \xi(t))$分量的测度，是有意义的。

统计代写|随机过程代写stochastic process代考|The ergodic theorem for processes with a discrete chance interference

具有离散偶然性干扰过程的遍历定理。设$(x(t) ; \xi(t))$为相空间${\mathscr{X}, \mathfrak{B}}$中具有离散机会干涉的过程。我们假设这个过程是规则的，即$\xi(t)$在每个有限区间上具有有限数量的跳跃。我们将关注……的行为
$$
\frac{1}{T} \int_0^T f(x(t), \xi(t)) d t
$$

如$T \rightarrow \infty$，其中$f(x, s)$是一个$\mathfrak{B} \times \mathfrak{B}{+}$ -可测量函数。下面我们将推导出表达式(42)的充分条件，以概率1得到形式$$ S(f)=\iiint_0^{\infty} Q{x, 0}(t, d y, \mathscr{X}) f(y, t) d t \pi(d x) / \iint_0^{\infty} Q_{x, 0}(t, \mathscr{X}, \mathscr{X}) d t \pi(d x)
$$的极限
为$T \rightarrow \infty$，其中$Q_{x, 0}\left(t, B_1, B_2\right)$为通过关系式(38)与离散机会干扰过程相关的函数，$\pi(d x)$为$(\mathscr{X}, \mathfrak{B})$中马尔可夫链的平稳分布，具有转移概率
$$
\pi(x, B)=Q_{x, 0}(0, \mathscr{X}, B),
$$
即为所有$B \in \mathfrak{B}$
$$
\pi(B)=\int \pi(d x) \pi(x, B)
$$
为了使表达式(43)有意义，必须存在一个满足(45)的测度$\pi$，并且函数$f(y, t)$是这样的
$$
\iiint_0^{\infty} Q_{x, 0}(t, d y, \mathscr{X})|f(y, t)| d t \pi(d x)<\infty
$$
回想一下，${\mathscr{X}, \mathfrak{B}}$中的马尔可夫链$\left{x_n\right}$被称为遍历链，如果这个链存在一个平稳分布$\pi(d x)$，并且对于所有$x \in \mathscr{X}$和$\mathfrak{B}$ -可测量函数$f$，使得$\int|f(x)| \pi(d x)<\infty$我们有，概率为$\mathrm{P}{x, 0}=1$， $$ \lim {n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^n f\left(x_k\right)=\int f(x) \pi(d x) .
$$

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

金融工程代写

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

有限元方法代写

随机分析代写

时间序列分析代写

回归分析代写

MATLAB代写

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

统计代写|随机过程代写stochastic process代考|Multiplicative Additive Functionals, Excessive Functions

Posted on 2023年5月26日2023年5月26日 by statistics-lab

统计代写|随机过程代写stochastic process代考|Multiplicative Additive Functionals, Excessive Functions

Definition and basic properties of additive and multiplicative functionals. Let $\left{\mathscr{F}, \mathcal{N}, \mathrm{P}_x\right}$ be a homogeneous Markov process in a certain phase space ${\mathscr{X}, \mathfrak{B}}$. We shall consider special families of $\mathscr{N}_t$-measurable numerical variables. Since they are $\mathscr{N}_t$-measurable the shift-operator $\theta_h$ is applicable to them.

A family of variables $\alpha_t$ defined for $t \geqslant 0$ on $\mathscr{F}$ is called a homogeneous multiplicative functional if it satisfies the conditions:
M1) $\alpha_t$ is measurable with respect to $\mathscr{N}t$ for all $t \geqslant 0$, M2) for all $t \geqslant 0$ and $h>0$ relation $\alpha{t+h}=\alpha_h \theta_h \alpha_t$ is satisfied for all $x \in \mathscr{X}$ with probability $P_x=1$.

A family of variables $\varphi_t$ defined for $t \geqslant 0$ on $\mathscr{F}$ is called a homogeneous additive functional provided:
A 1) for all $t \geqslant 0$ the variable $\varphi_t$ is measurable with respect to $\mathcal{N}t$. A 2) for all $t \geqslant 0$ and $h>0$ relation $\varphi{t+h}=\varphi_h+\theta_h \varphi_t$ is satisfied for all $x \in \mathscr{X}$ with probability $P_x=1$.
We present examples of multiplicative functionals.
I. Let $\mathscr{X}$ be a topological space and let the sample function of the process $x(t)$ be continuous from the right, and let $g(x)$ be a bounded continuous function on $\mathscr{x}$. Then for all $t \geqslant 0$ the quantity
$$
\int_0^t g\left(x_s\right) d s
$$

is defined and is $\mathscr{N}t$-measurable. Set $$ \alpha_t=\exp \left{\int_0^t g\left(x_s\right) d s\right} $$ Since and $$ \theta_h \alpha_t=\exp \left{\int_0^t \theta_h g\left(x_s\right) d s\right}=\exp \left{\int_h^{t+h} g\left(x_s\right) d s\right} $$ $$ \alpha_h \theta_h \alpha_t=\exp \left{\int_0^h g\left(x_s\right) d s\right} \exp \left{\int_h^{t+h} g\left(x_s\right) d s\right}=\alpha{t+h},
$$
it follows that $\alpha_t$ is a multiplicative functional.

统计代写|随机过程代写stochastic process代考|Random time substitution

(This quantity is well defined if $\varphi(+\infty)>t$.) Thus we have a family of random processes: a random process $x\left(\tau_t\right)$ on the probability space $\left{\mathscr{F}, \mathcal{N}, \mathrm{P}_x\right}$ corresponds to each $x$. It turns out that this family is a Markov family of random processes. We construct a homogeneous Markov process with marginal distributions of the following structure: given that the initial value of the process is $x$, the marginal distributions of the process coincide with the marginal distributions of the process $x\left(\tau_t\right)$ on the probability space $\left{\mathscr{F}, \mathcal{N}, \mathrm{P}_x\right}$.

We introduce the set of functions $\mathscr{F}: x(t)=x(\tau)$ on $[0, \tilde{\zeta})$ where the variables $\tau_t$ are defined by relation (25) for all $x(\cdot) \in \mathscr{F}$ and $t \in[0, \tilde{\zeta}), \tilde{\zeta}=\varphi(+\infty)$. Denote by $\tilde{\mathcal{N}}$ the $\sigma$-algebra generated by the variables $\tilde{x}(t)$, by $\tilde{N}t$ the $\sigma$-algebra generated by $\tilde{x}(s)$ for $s \leqslant t$. Finally denote by $\tilde{\mathrm{P}}_x$ the measure on $\mathscr{N}$ defined by relation $$ \tilde{\mathrm{P}}_x(C)=\mathrm{P}_x\left{x\left(\tau_0\right) \in C\right} $$ for any cylinder $C$ in $\mathscr{N}$; since $x\left(\tau_t\right)$ is an $\mathscr{N}$-measurable variable it follows that the set $\left{x(\cdot): x\left(\tau_0\right) \in C\right}$ is $\mathcal{N}$-measurable for any cylinder $C$. To verify that $\left{\tilde{\mathscr{F}}, \tilde{\mathcal{N}}, \tilde{\mathrm{P}}_x\right}$ is a Markov process it is sufficient to show that $$ \tilde{\mathrm{P}}_x\left{\tilde{x}(t+h) \in B \mid \tilde{\mathscr{N}}_t\right}=\tilde{\mathrm{P}}{\tilde{x}(t)}{\tilde{x}(h) \in B}
$$
with probability $\mathrm{P}x=1$ for any $x \in \mathscr{X}, B \in \mathcal{B}, t>0$ and $h>0$. Relation (26) is equivalent to the following: $$ \mathrm{P}_x\left{x\left(\tau{t+h}\right) \in B \mid \mathcal{N}{\tau_t}\right}=\mathrm{P}{x\left(\tau_t\right)}\left{x\left(\tau_h\right) \in B\right}
$$
with probability $P_x=1$. It follows from the strong Markov property of the process $\left{\mathscr{F}, \mathcal{N}, P_x\right}$ that $(27)$ is satisfied provided
$$
\theta_{\tau_t}\left[x\left(\tau_h\right)\right]=x\left(\tau_{h+t}\right)
$$
with probability $P_x=1$.

随机过程代考

统计代写|随机过程代写stochastic process代考|Multiplicative Additive Functionals, Excessive Functions

加性泛函和乘法泛函的定义和基本性质。设$\left{\mathscr{F}, \mathcal{N}, \mathrm{P}_x\right}$为某相空间${\mathscr{X}, \mathfrak{B}}$中的齐次马尔可夫过程。我们将考虑$\mathscr{N}_t$可测量数值变量的特殊族。由于它们是$\mathscr{N}_t$可测量的，移位运算符$\theta_h$适用于它们。

在$\mathscr{F}$上为$t \geqslant 0$定义的变量族$\alpha_t$如果满足以下条件，则称为齐次乘法泛函:
对于所有$t \geqslant 0$, M1) $\alpha_t$相对于$\mathscr{N}t$是可测量的，对于所有$t \geqslant 0$和$h>0$, M2)关系$\alpha{t+h}=\alpha_h \theta_h \alpha_t$对于所有$x \in \mathscr{X}$是满足的，概率为$P_x=1$。

在$\mathscr{F}$上为$t \geqslant 0$定义的一组变量$\varphi_t$称为齐次加性泛函，提供:
1)对于所有$t \geqslant 0$，变量$\varphi_t$相对于$\mathcal{N}t$是可测量的。A 2)对于所有$t \geqslant 0$和$h>0$关系$\varphi{t+h}=\varphi_h+\theta_h \varphi_t$，对于所有$x \in \mathscr{X}$都以$P_x=1$的概率满足。
我们给出乘法函数的例子。
1、设$\mathscr{X}$为拓扑空间，设过程$x(t)$的样本函数从右连续，设$g(x)$为$\mathscr{x}$上的有界连续函数。然后对于所有$t \geqslant 0$数量
$$
\int_0^t g\left(x_s\right) d s
$$

是定义的并且是$\mathscr{N}t$ -可测量的。输入$$ \alpha_t=\exp \left{\int_0^t g\left(x_s\right) d s\right} $$ Since和$$ \theta_h \alpha_t=\exp \left{\int_0^t \theta_h g\left(x_s\right) d s\right}=\exp \left{\int_h^{t+h} g\left(x_s\right) d s\right} $$$$ \alpha_h \theta_h \alpha_t=\exp \left{\int_0^h g\left(x_s\right) d s\right} \exp \left{\int_h^{t+h} g\left(x_s\right) d s\right}=\alpha{t+h},
$$
由此可知$\alpha_t$是一个乘法泛函。

统计代写|随机过程代写stochastic process代考|Random time substitution

(这个数量定义得很好，如果$\varphi(+\infty)>t$。)这样我们就有了一组随机过程:概率空间$\left{\mathscr{F}, \mathcal{N}, \mathrm{P}_x\right}$上的一个随机过程$x\left(\tau_t\right)$对应于每个$x$。这个族是随机过程的马尔可夫族。我们构造了一个齐次马尔可夫过程，其边缘分布具有如下结构:给定过程的初值为$x$，过程的边缘分布与过程$x\left(\tau_t\right)$在概率空间$\left{\mathscr{F}, \mathcal{N}, \mathrm{P}_x\right}$上的边缘分布重合。

我们在$[0, \tilde{\zeta})$上引入一组函数$\mathscr{F}: x(t)=x(\tau)$，其中变量$\tau_t$由关系(25)为所有$x(\cdot) \in \mathscr{F}$和$t \in[0, \tilde{\zeta}), \tilde{\zeta}=\varphi(+\infty)$定义。用$\tilde{\mathcal{N}}$表示由变量$\tilde{x}(t)$生成的$\sigma$ -代数，用$\tilde{N}t$表示由$\tilde{x}(s)$生成的$\sigma$ -代数表示$s \leqslant t$。最后用$\tilde{\mathrm{P}}x$表示由关系$$ \tilde{\mathrm{P}}_x(C)=\mathrm{P}_x\left{x\left(\tau_0\right) \in C\right} $$定义的对于$\mathscr{N}$中的任意圆柱体$C$的$\mathscr{N}$上的度量;由于$x\left(\tau_t\right)$是一个$\mathscr{N}$可测量的变量，因此，对于任何圆柱体$C$，集合$\left{x(\cdot): x\left(\tau_0\right) \in C\right}$都是$\mathcal{N}$可测量的。为了验证$\left{\tilde{\mathscr{F}}, \tilde{\mathcal{N}}, \tilde{\mathrm{P}}_x\right}$是一个马尔可夫过程，只要证明$$ \tilde{\mathrm{P}}_x\left{\tilde{x}(t+h) \in B \mid \tilde{\mathscr{N}}_t\right}=\tilde{\mathrm{P}}{\tilde{x}(t)}{\tilde{x}(h) \in B} $$就足够了对于任意$x \in \mathscr{X}, B \in \mathcal{B}, t>0$和$h>0$的概率为$\mathrm{P}x=1$。关系式(26)等价于如下:$$ \mathrm{P}_x\left{x\left(\tau{t+h}\right) \in B \mid \mathcal{N}{\tau_t}\right}=\mathrm{P}{x\left(\tau_t\right)}\left{x\left(\tau_h\right) \in B\right} $$ 概率是$P_x=1$。由过程$\left{\mathscr{F}, \mathcal{N}, P_x\right}$的强马尔可夫性质可知$(27)$是满足的 $$ \theta{\tau_t}\left[x\left(\tau_h\right)\right]=x\left(\tau_{h+t}\right)
$$
概率是$P_x=1$。

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

金融工程代写

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

有限元方法代写

随机分析代写

时间序列分析代写

回归分析代写

MATLAB代写

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

统计代写|随机过程代写stochastic process代考|Multiplicative Functionals

Posted on 2023年5月15日2023年5月15日 by statistics-lab

统计代写|随机过程代写stochastic process代考|Multiplicative Functionals

Multiplicative functionals and semi-stochastic kernels. Let $\left{\xi(t, \omega), \Im_t^s, \mathrm{P}_{s, x}\right}$ be a Markov process in the phase space ${\mathscr{X}, \mathfrak{B}}$.

Definition 1. A family of real random variables $\left{\mu_t^s, 0 \leqslant s \leqslant t<\infty\right}$ is called a multiplicative functional of a Markov process provided for all $s, t(0 \leqslant s \leqslant t<\infty)$ :
a) the random variable $\mu_t^s$ is $\tilde{\mathfrak{N}}t^s$-measurable; b) $\mu_t^s \mu_u^t=\mu_u^s\left(\bmod \mathrm{P}{s, x}\right)$ for any $(s, x) \in[0, \infty) \times \mathscr{X}$ and for arbitrary $t \in[s, u]$;
c) $0 \leqslant \mu_t^s \leqslant 1$.
Recall that $\mathfrak{N}t^s$ denotes the $\sigma$-algebra generated by the random elements $\xi(u), u \in[s, t]$ and $\tilde{\mathfrak{N}}_t^s$ is the completion of $\mathfrak{N}_t^s$ in $\tilde{\mathfrak{N}}^s$ with respect to the set of measures $\left{\mathrm{P}{s, q}, q \in Q\right}$ where $Q$ is the set of all probability measures on $\mathfrak{B}$.

It follows from the definition that $\mu_t^s$ is a monotonically non-decreasing function of $t$.

A multiplicative functional is called continuous from the right if for all $(s, x) \in[0, \infty) \times \mathscr{X}$ the function $\mu_t^s, t \geqslant s$, is continuous from the right for each $t$ almost surely with respect to $\mathrm{P}{s, x}$. A functional is called measurable if for any $s$ the random process $\left{\mu_t^s, t \geqslant s\right}$ is $\tilde{\mathfrak{N}}_t^s$-progressively measurable. Since $\mu_s^s=\mu_s^s \mu_s^s=\left(\mu_s^s\right)^2, \mu_s^s$ can take on only two values, 1 or 0 . It follows from the zero or one law that $\mathrm{P}{s, x}\left(\mu_s^s=1\right)=1$ or 0 . The point $(s, x)$ at which $\mathrm{P}_{s, x}\left(\mu_s^s=1\right)=1$ is called the permanency points of a multiplicative functional.

统计代写|随机过程代写stochastic process代考|An integral equation associated with multiplicative functionals

An integral equation associated with multiplicative functionals. In certain problems the analytic expression for the kernels $Q(s, x, t, B)$ generated by a given multiplicative functional is of interest. In the case of a multiplicative functional of the integral type one can obtain an integral equation satisfied by the function $Q(s, x, t, B)$.
Let
$$
\mu_t^s=\exp \left{-\int_s^t f[u, \xi(u)] d u\right}
$$
where $f(t, x)$ is non-negative $\mathscr{T} \times \mathfrak{B}$-measurable and the integral appearing in formula (12) converges for all $(s, t), 0 \leqslant s \leqslant t \leqslant \xi$. Since for almost all $s$, we have
$$
\frac{d \mu_t^s}{d s}=f[s, \xi(s)] \mu_t^s
$$
it follows that
$$
\mu_t^s=1-\int_s^t f[u, \xi(u)] \mu_t^u d u \quad(t<\zeta)
$$
Multiplying both sides of the obtained expression by $\chi[B, \xi(t)]$ and integrating both sides with respect to $\mathrm{P}{s, x}$ we obtain $$ Q(s, x, t, B)=P(s, x, t, B)-\int_s^t \mathrm{E}{s, x} f[u, \xi(u)] \mu_t^u \chi[B, \xi(t)] d u
$$
Utilizing the equalities
$$
\begin{aligned}
\mathrm{E}{s, x} f[u, \xi(u)] \mu_t^u \chi[B, \xi(t)] & =\mathrm{E}{s, x} f[u, \xi(u)] \mathrm{E}{u, \xi(u)} \chi[B, \xi(t)] \mu_t^u \ & =\mathrm{E}{s, x} f[u, \xi(u)] Q(u, \xi(u), t, B),
\end{aligned}
$$
we find that the kernel $Q(s, x, t, B)$ satisfies the following integral equation:
$$
Q(s, x, t, B)=P(s, x, t, B)-\int_s^t f(u, y) Q(u, y, t, B) P(s, x, u, d y) .
$$

随机过程代考

统计代写|随机过程代写stochastic process代考|Multiplicative Functionals

乘法泛函与半随机核。设$\left{\xi(t, \omega), \Im_t^s, \mathrm{P}_{s, x}\right}$为相空间${\mathscr{X}, \mathfrak{B}}$中的马尔可夫过程。

定义:一组实数随机变量$\left{\mu_t^s, 0 \leqslant s \leqslant t<\infty\right}$被称为马尔可夫过程的乘法泛函，提供给所有$s, t(0 \leqslant s \leqslant t<\infty)$:
A)随机变量$\mu_t^s$为$\tilde{\mathfrak{N}}t^s$可测;B) $\mu_t^s \mu_u^t=\mu_u^s\left(\bmod \mathrm{P}{s, x}\right)$适用于任何$(s, x) \in[0, \infty) \times \mathscr{X}$和任意$t \in[s, u]$;
C) $0 \leqslant \mu_t^s \leqslant 1$。
回想一下，$\mathfrak{N}t^s$表示$\sigma$——由随机元素$\xi(u), u \in[s, t]$和$\tilde{\mathfrak{N}}_t^s$生成的代数是$\tilde{\mathfrak{N}}^s$中$\mathfrak{N}_t^s$相对于测度集$\left{\mathrm{P}{s, q}, q \in Q\right}$的补全，其中$Q$是$\mathfrak{B}$上所有概率测度的集合。

由定义可知$\mu_t^s$是$t$的单调非递减函数。

如果对所有$(s, x) \in[0, \infty) \times \mathscr{X}$函数$\mu_t^s, t \geqslant s$从右连续到每个$t$几乎肯定是对$\mathrm{P}{s, x}$连续的，那么一个乘法泛函就称为从右连续的。如果对于任意$s$随机过程$\left{\mu_t^s, t \geqslant s\right}$是$\tilde{\mathfrak{N}}t^s$ -逐步可测量的，则称为可测量的函数。因为$\mu_s^s=\mu_s^s \mu_s^s=\left(\mu_s^s\right)^2, \mu_s^s$只能取两个值，1或0。它遵循0或1定律$\mathrm{P}{s, x}\left(\mu_s^s=1\right)=1$或0。在$\mathrm{P}{s, x}\left(\mu_s^s=1\right)=1$处的点$(s, x)$被称为乘法泛函的永久点。

统计代写|随机过程代写stochastic process代考|An integral equation associated with multiplicative functionals

与乘法函数有关的积分方程。在某些问题中，由一个给定的乘法泛函生成的核的解析表达式$Q(s, x, t, B)$是有意义的。对于积分型的乘法泛函，可以得到由$Q(s, x, t, B)$函数满足的积分方程。
让
$$
\mu_t^s=\exp \left{-\int_s^t f[u, \xi(u)] d u\right}
$$
其中$f(t, x)$是非负的$\mathscr{T} \times \mathfrak{B}$ -可测的，式(12)中的积分对所有$(s, t), 0 \leqslant s \leqslant t \leqslant \xi$收敛。因为几乎所有的$s$，我们已经
$$
\frac{d \mu_t^s}{d s}=f[s, \xi(s)] \mu_t^s
$$
由此得出
$$
\mu_t^s=1-\int_s^t f[u, \xi(u)] \mu_t^u d u \quad(t<\zeta)
$$
得到的表达式两边同时乘以$\chi[B, \xi(t)]$两边对$\mathrm{P}{s, x}$积分得到$$ Q(s, x, t, B)=P(s, x, t, B)-\int_s^t \mathrm{E}{s, x} f[u, \xi(u)] \mu_t^u \chi[B, \xi(t)] d u
$$
利用等式
$$
\begin{aligned}
\mathrm{E}{s, x} f[u, \xi(u)] \mu_t^u \chi[B, \xi(t)] & =\mathrm{E}{s, x} f[u, \xi(u)] \mathrm{E}{u, \xi(u)} \chi[B, \xi(t)] \mu_t^u \ & =\mathrm{E}{s, x} f[u, \xi(u)] Q(u, \xi(u), t, B),
\end{aligned}
$$
我们发现核$Q(s, x, t, B)$满足以下积分方程:
$$
Q(s, x, t, B)=P(s, x, t, B)-\int_s^t f(u, y) Q(u, y, t, B) P(s, x, u, d y) .
$$

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