Math 632 - 统计代写答疑辅导

标签： Math 632

数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|Math 632

Posted on 2022年6月18日2022年6月18日 by statistics-lab

如果你也在怎样代写概率模型和随机过程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。

随机过程是随机量在时间或空间上演变的概率模型。演变受不同时间或地点的随机量之间的某种依赖关系所支配。

statistics-lab™ 为您的留学生涯保驾护航在代写概率模型和随机过程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。我们的专家在代写概率模型和随机过程代写方面经验极为丰富，各种代写概率模型和随机过程相关的作业也就用不着说。

我们提供的概率模型和随机过程及其相关学科的代写，服务范围广, 其中包括但不限于:

Statistical Inference 统计推断
Statistical Computing 统计计算
Advanced Probability Theory 高等概率论
Advanced Mathematical Statistics 高等数理统计学
(Generalized) Linear Models 广义线性模型
Statistical Machine Learning 统计机器学习
Longitudinal Data Analysis 纵向数据分析
Foundations of Data Science 数据科学基础

数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|Math 632

数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|Proof of the Strong Law of Large Numbers

In this section we give a proof of the strong law of large numbers. Our proof of the strong law makes use of the Borel-Cantelli lemma.

Borel-Cantelli Lemma. For a sequence of events $A_{i}, i \geq 1$, let $N$ denote the number of these events that occur. If $\sum_{i=1}^{\infty} P\left(A_{i}\right)<\infty$, then $P(N=\infty)=0$.

Proof. Suppose that $\sum_{i=1}^{\infty} P\left(A_{i}\right)<\infty$. Now, if $N=\infty$, then for every $n<\infty$ at least one of the events $A_{n}, A_{n+1}, \ldots$ will occur. That is, $N=\infty$ implies that $\cup_{i=n}^{\infty} A_{i}$ occurs for every $n$. Thus, for every $n$
$$
\begin{aligned}
P(N=\infty) & \leq P\left(\cup_{i=n}^{\infty} A_{i}\right) \
& \leq \sum_{i=n}^{\infty} P\left(A_{i}\right)
\end{aligned}
$$

where the final inequality follows from Boole’s inequality. Because $\sum_{i=1}^{\infty} P\left(A_{i}\right)<\infty$ implies that $\sum_{i=n}^{\infty} P\left(A_{i}\right) \rightarrow 0$ as $n \rightarrow \infty$, we obtain from the preceding upon letting $n \rightarrow \infty$ that $P(N=\infty)=0$, which proves the result.

Remark. The Borel-Cantelli lemma is actually quite intuitive, for if we define the indicator variable $I_{i}$ to equal 1 if $A_{i}$ occurs and to equal 0 otherwise, then $N=\sum_{i=1}^{\infty} I_{i}$, implying that
$$
E[N]=\sum_{i=1}^{\infty} E\left[I_{i}\right]=\sum_{i=1}^{\infty} P\left(A_{i}\right)
$$
Consequently, the Borel-Cantelli theorem states that if the expected number of events that occur is finite then the probability that an infinite number of them occur is 0 , which is intuitive because if there were a positive probability that an infinite number of events could occur then $E[N]$ would be infinite.

数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|Stochastic Processes

A stochastic process ${X(t), t \in T}$ is a collection of random variables. That is, for each $t \in T, X(t)$ is a random variable. The index $t$ is often interpreted as time and, as a result, we refer to $X(t)$ as the state of the process at time $t$. For example, $X(t)$ might equal the total number of customers that have entered a supermarket by time $t$; or the number of customers in the supermarket at time $t$; or the total amount of sales that have been recorded in the market by time $t$; etc.

The set $T$ is called the index set of the process. When $T$ is a countable set the stochastic process is said to be a discrete-time process. If $T$ is an interval of the real line, the stochastic process is said to be a continuous-time process. For instance, $\left{X_{n}, n=0,1, \ldots\right}$ is a discrete-time stochastic process indexed by the nonnegative integers; while ${X(t), t \geq 0}$ is a continuous-time stochastic process indexed by the nonnegative real numbers.

The state space of a stochastic process is defined as the set of all possible values that the random variables $X(t)$ can assume.

Thus, a stochastic process is a family of random variables that describes the evolution through time of some (physical) process. We shall see much of stochastic processes in the following chapters of this text.

Example 2.54. Consider a particle that moves along a set of $m+1$ nodes, labeled $0,1, \ldots, m$, that are arranged around a circle (see Fig. 2.3). At each step the particle is equally likely to move one position in either the clockwise or counterclockwise direction. That is, if $X_{n}$ is the position of the particle after its $n$th step then
$$
P\left{X_{n+1}=i+1 \mid X_{n}=i\right}=P\left{X_{n+1}=i-1 \mid X_{n}=i\right}=\frac{1}{2}
$$ where $i+1 \equiv 0$ when $i=m$, and $i-1 \equiv m$ when $i=0$. Suppose now that the particle starts at 0 and continues to move around according to the preceding rules until all the nodes $1,2, \ldots, m$ have been visited. What is the probability that node $i, i=1, \ldots, m$, is the last one visited?

数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|The Discrete Case

Recall that for any two events $E$ and $F$, the conditional probability of $E$ given $F$ is defined, as long as $P(F)>0$, by
$$
P(E \mid F)=\frac{P(E F)}{P(F)}
$$
Hence, if $X$ and $Y$ are discrete random variables, then it is natural to define the conditional probability mass function of $X$ given that $Y=y$, by
$$
\begin{aligned}
p_{X \mid Y}(x \mid y) &=P{X=x \mid Y=y} \
&=\frac{P{X=x, Y=y}}{P{Y=y}} \
&=\frac{p(x, y)}{p_{Y}(y)}
\end{aligned}
$$
for all values of $y$ such that $P{Y=y}>0$. Similarly, the conditional probability distribution function of $X$ given that $Y=y$ is defined, for all $y$ such that $P{Y=y}>0$, by
$$
\begin{aligned}
F_{X \mid Y}(x \mid y) &=P{X \leq x \mid Y=y} \
&=\sum_{a \leq x} p_{X \mid Y}(a \mid y)
\end{aligned}
$$
Finally, the conditional expectation of $X$ given that $Y=y$ is defined by
$$
E[X \mid Y=y]=\sum_{x} x P{X=x \mid Y=y}
$$ $$
=\sum_{x} x p_{X \mid Y}(x \mid y)
$$
In other words, the definitions are exactly as before with the exception that everything is now conditional on the event that $Y=y$. If $X$ is independent of $Y$, then the conditional mass function, distribution, and expectation are the same as the unconditional ones. This follows, since if $X$ is independent of $Y$, then
$$
\begin{aligned}
p_{X \mid Y}(x \mid y) &=P{X=x \mid Y=y} \
&=P{X=x}
\end{aligned}
$$

概率模型和随机过程代考

数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|Proof of the Strong Law of Large Numbers

在本节中，我们将证明大数的强定律。我们对强定律的证明使用了 Borel-Cantelli 引理。

Borel-Cantelli引理。对于一系列事件一个一世,一世≥1，让ñ表示发生的这些事件的数量。如果∑一世=1∞磷(一个一世)<∞，然后磷(ñ=∞)=0.

证明。假设∑一世=1∞磷(一个一世)<∞. 现在，如果ñ=∞，那么对于每个n<∞至少其中一项事件一个n,一个n+1,…会发生。那是，ñ=∞暗示∪一世=n∞一个一世发生在每个n. 因此，对于每个n

磷(ñ=∞)≤磷(∪一世=n∞一个一世) ≤∑一世=n∞磷(一个一世)

其中最终的不等式来自布尔不等式。因为∑一世=1∞磷(一个一世)<∞暗示∑一世=n∞磷(一个一世)→0作为n→∞，我们在让n→∞那磷(ñ=∞)=0，这证明了结果。

评论。Borel-Cantelli 引理实际上非常直观，因为如果我们定义指标变量我一世等于 1 如果一个一世发生并等于0，然后ñ=∑一世=1∞我一世, 意味着

和[ñ]=∑一世=1∞和[我一世]=∑一世=1∞磷(一个一世)
因此，Borel-Cantelli 定理指出，如果发生的事件的预期数量是有限的，那么发生无限数量的事件的概率为 0，这是直观的，因为如果存在无限数量的事件可能发生的正概率然后和[ñ]将是无限的。

数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|Stochastic Processes

随机过程X(吨),吨∈吨是随机变量的集合。也就是说，对于每个吨∈吨,X(吨)是一个随机变量。指数吨通常被解释为时间，因此，我们指的是X(吨)作为当时的过程状态吨. 例如，X(吨)可能等于按时间进入超市的顾客总数吨; 或当时超市的顾客数量吨; 或按时间记录在市场上的总销售额吨; 等等

套装吨称为过程的索引集。什么时候吨是一个可数集，随机过程被称为离散时间过程。如果吨是实线的区间，随机过程称为连续时间过程。例如，\left{X_{n}, n=0,1, \ldots\right}\left{X_{n}, n=0,1, \ldots\right}是由非负整数索引的离散时间随机过程；尽管X(吨),吨≥0是由非负实数索引的连续时间随机过程。

随机过程的状态空间定义为随机变量的所有可能值的集合X(吨)可以假设。

因此，随机过程是一组随机变量，描述了某些（物理）过程随时间的演变。我们将在本文的后续章节中看到许多随机过程。

例 2.54。考虑一个粒子沿着一组米+1节点，标记0,1,…,米，这些排列在一个圆圈周围（见图2.3）。在每个步骤中，粒子同样有可能在顺时针或逆时针方向上移动一个位置。也就是说，如果Xn是粒子在它之后的位置n然后步骤

P\left{X_{n+1}=i+1 \mid X_{n}=i\right}=P\left{X_{n+1}=i-1 \mid X_{n}=i\right }=\frac{1}{2}P\left{X_{n+1}=i+1 \mid X_{n}=i\right}=P\left{X_{n+1}=i-1 \mid X_{n}=i\right }=\frac{1}{2}在哪里一世+1≡0什么时候一世=米，和一世−1≡米什么时候一世=0. 现在假设粒子从0开始，并继续根据前面的规则继续四处移动，直到所有节点1,2,…,米被访问过。节点的概率是多少一世,一世=1,…,米, 是最后一个访问吗？

数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|The Discrete Case

回想一下任何两个事件和和F，条件概率和给定F被定义，只要磷(F)>0，经过

磷(和∣F)=磷(和F)磷(F)
因此，如果X和是是离散随机变量，那么定义条件概率质量函数是很自然的X鉴于是=是，经过

pX∣是(X∣是)=磷X=X∣是=是 =磷X=X,是=是磷是=是 =p(X,是)p是(是)
对于所有值是这样磷是=是>0. 类似地，条件概率分布函数为X鉴于是=是定义，所有是这样磷是=是>0，经过

FX∣是(X∣是)=磷X≤X∣是=是 =∑一个≤XpX∣是(一个∣是)
最后，条件期望X鉴于是=是定义为

和[X∣是=是]=∑XX磷X=X∣是=是

=∑XXpX∣是(X∣是)
换句话说，定义与以前完全相同，只是现在一切都取决于以下事件是=是. 如果X独立于是，那么有条件的质量函数、分布和期望与无条件的相同。随之而来的是X独立于是，然后

pX∣是(X∣是)=磷X=X∣是=是 =磷X=X

数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes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

数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|Jointly Distributed Random Variables

Posted on 2022年6月18日2022年6月18日 by statistics-lab

如果你也在怎样代写概率模型和随机过程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。

随机过程是随机量在时间或空间上演变的概率模型。演变受不同时间或地点的随机量之间的某种依赖关系所支配。

我们提供的概率模型和随机过程及其相关学科的代写，服务范围广, 其中包括但不限于:

Statistical Inference 统计推断
Statistical Computing 统计计算
Advanced Probability Theory 高等概率论
Advanced Mathematical Statistics 高等数理统计学
(Generalized) Linear Models 广义线性模型
Statistical Machine Learning 统计机器学习
Longitudinal Data Analysis 纵向数据分析
Foundations of Data Science 数据科学基础

数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|Jointly Distributed Random Variables

数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|Joint Distribution Functions

Thus far, we have concerned ourselves with the probability distribution of a single random variable. However, we are often interested in probability statements concerning two or more random variables. To deal with such probabilities, we define, for any

two random variables $X$ and $Y$, the joint cumulative probability distribution function of $X$ and $Y$ by
$$
F(a, b)=P{X \leq a, Y \leq b}, \quad-\infty0} p(x, y)
$$
Similarly,
$$
p_{Y}(y)=\sum_{x: p(x, y)>0} p(x, y)
$$

The probability mass function of $X$ may be obtained from $p(x, y)$ by
$$
p_{X}(x)=\sum_{y: p(x, y)>0} p(x, y)
$$
Similarly,
$$
p_{Y}(y)=\sum_{x: p(x, y)>0} p(x, y)
$$
We say that $X$ and $Y$ are jointly continuous if there exists a function $f(x, y)$, defined for all real $x$ and $y$, having the property that for all sets $A$ and $B$ of real numbers
$$
P{X \in A, Y \in B}=\int_{B} \int_{A} f(x, y) d x d y
$$
The function $f(x, y)$ is called the joint probability density function of $X$ and $Y$. The probability density of $X$ can be obtained from a knowledge of $f(x, y)$ by the following reasoning:
$$
\begin{aligned}
P{X \in A} &=P{X \in A, Y \in(-\infty, \infty)} \
&=\int_{-\infty}^{\infty} \int_{A} f(x, y) d x d y \
&=\int_{A} f_{X}(x) d x
\end{aligned}
$$

数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|Independent Random Variables

The random variables $X$ and $Y$ are said to be independent if, for all $a, b$,
$$
P{X \leq a, Y \leq b}=P{X \leq a} P{Y \leq b}
$$
In other words, $X$ and $Y$ are independent if, for all $a$ and $b$, the events $E_{a}={X \leq a}$ and $F_{b}={Y \leq b}$ are independent.

In terms of the joint distribution function $F$ of $X$ and $Y$, we have that $X$ and $Y$ are independent if
$$
F(a, b)=F_{X}(a) F_{Y}(b) \quad \text { for all } a, b
$$
When $X$ and $Y$ are discrete, the condition of independence reduces to
$$
p(x, y)=p_{X}(x) p_{Y}(y)
$$
while if $X$ and $Y$ are jointly continuous, independence reduces to
$$
f(x, y)=f_{X}(x) f_{Y}(y)
$$

To prove this statement, consider first the discrete version, and suppose that the joint probability mass function $p(x, y)$ satisfies Eq. (2.13). Then
$$
\begin{aligned}
P{X \leq a, Y \leq b} &=\sum_{y \leq b} \sum_{x \leq a} p(x, y) \
&=\sum_{y \leq b} \sum_{x \leq a} p_{X}(x) p_{Y}(y) \
&=\sum_{y \leq b} p_{Y}(y) \sum_{x \leq a} p_{X}(x) \
&=P{Y \leq b} P{X \leq a}
\end{aligned}
$$
and so $X$ and $Y$ are independent. That Eq. (2.14) implies independence in the continuous case is proven in the same manner and is left as an exercise.
An important result concerning independence is the following.

数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|Covariance and Variance of Sums of Random Variables

The covariance of any two random variables $X$ and $Y$, denoted by $\operatorname{Cov}(X, Y)$, is defined by
$$
\begin{aligned}
\operatorname{Cov}(X, Y) &=E[(X-E[X])(Y-E[Y])] \
&=E[X Y-Y E[X]-X E[Y]+E[X] E[Y]] \
&=E[X Y]-E[Y] E[X]-E[X] E[Y]+E[X] E[Y] \
&=E[X Y]-E[X] E[Y]
\end{aligned}
$$
Note that if $X$ and $Y$ are independent, then by Proposition $2.3$ it follows that $\operatorname{Cov}(X, Y)=0$

Let us consider now the special case where $X$ and $Y$ are indicator variables for whether or not the events $A$ and $B$ occur. That is, for events $A$ and $B$, define
$$
X=\left{\begin{array}{ll}
1, & \text { if } A \text { occurs } \
0, & \text { otherwise, }
\end{array} \quad Y= \begin{cases}1, & \text { if } B \text { occurs } \
0, & \text { otherwise }\end{cases}\right.
$$
Then,
$$
\operatorname{Cov}(X, Y)=E[X Y]-E[X] E[Y]
$$
and, because $X Y$ will equal 1 or 0 depending on whether or not both $X$ and $Y$ equal 1 , we see that
$$
\operatorname{Cov}(X, Y)=P{X=1, Y=1}-P{X=1} P{Y=1}
$$
From this we see that
$$
\begin{aligned}
\operatorname{Cov}(X, Y)>0 & \Leftrightarrow P{X=1, Y=1}>P{X=1} P{Y=1} \
& \Leftrightarrow \frac{P{X=1, Y=1}}{P{X=1}}>P{Y=1} \
& \Leftrightarrow P{Y=1 \mid X=1}>P{Y=1}
\end{aligned}
$$
That is, the covariance of $X$ and $Y$ is positive if the outcome $X=1$ makes it more likely that $Y=1$ (which, as is easily seen by symmetry, also implies the reverse).
In general it can be shown that a positive value of $\operatorname{Cov}(X, Y)$ is an indication that $Y$ tends to increase as $X$ does, whereas a negative value indicates that $Y$ tends to decrease as $X$ increases.

概率模型和随机过程代考

数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|Joint Distribution Functions

到目前为止，我们关注的是单个随机变量的概率分布。然而，我们经常对涉及两个或多个随机变量的概率陈述感兴趣。为了处理这样的概率，我们定义，对于任何

两个随机变量X和是, 的联合累积概率分布函数X和是经过

F(a, b)=P{X \leq a, Y \leq b}, \quad-\infty0} p(x, y)F(a, b)=P{X \leq a, Y \leq b}, \quad-\infty0} p(x, y)
相似地，

p是(是)=∑X:p(X,是)>0p(X,是)

的概率质量函数X可以从p(X,是)经过

pX(X)=∑是:p(X,是)>0p(X,是)
相似地，

p是(是)=∑X:p(X,是)>0p(X,是)
我们说X和是如果存在函数，则联合连续F(X,是), 为所有实数定义X和是, 具有对所有集合的属性一个和乙实数

磷X∈一个,是∈乙=∫乙∫一个F(X,是)dXd是
功能F(X,是)称为联合概率密度函数X和是. 的概率密度X可以从知识中获得F(X,是)通过以下推理：

磷X∈一个=磷X∈一个,是∈(−∞,∞) =∫−∞∞∫一个F(X,是)dXd是 =∫一个FX(X)dX

数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|Independent Random Variables

随机变量X和是据说是独立的，如果，对于所有一个,b,

磷X≤一个,是≤b=磷X≤一个磷是≤b
换句话说，X和是是独立的，如果，对于所有一个和b，事件和一个=X≤一个和Fb=是≤b是独立的。

就联合分布函数而言F的X和是, 我们有X和是是独立的，如果

F(一个,b)=FX(一个)F是(b) 对所有人一个,b
什么时候X和是是离散的，独立的条件简化为

p(X,是)=pX(X)p是(是)
而如果X和是是共同连续的，独立性简化为

F(X,是)=FX(X)F是(是)

为了证明这个陈述，首先考虑离散版本，并假设联合概率质量函数p(X,是)满足方程。(2.13)。然后

磷X≤一个,是≤b=∑是≤b∑X≤一个p(X,是) =∑是≤b∑X≤一个pX(X)p是(是) =∑是≤bp是(是)∑X≤一个pX(X) =磷是≤b磷X≤一个
所以X和是是独立的。那个方程。(2.14) 意味着连续情况下的独立性以相同的方式证明，并留作练习。
关于独立性的一个重要结果如下。

数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|Covariance and Variance of Sums of Random Variables

任意两个随机变量的协方差X和是，表示为这⁡(X,是), 定义为

这⁡(X,是)=和[(X−和[X])(是−和[是])] =和[X是−是和[X]−X和[是]+和[X]和[是]] =和[X是]−和[是]和[X]−和[X]和[是]+和[X]和[是] =和[X是]−和[X]和[是]
请注意，如果X和是是独立的，然后由命题2.3它遵循这⁡(X,是)=0

现在让我们考虑一下特殊情况X和是是事件是否发生的指示变量一个和乙发生。也就是说，对于事件一个和乙, 定义
$$
X=\left{

1, 如果一个发生 0, 否则， \四Y=

{1, 如果乙发生 0, 否则 \正确的。

吨H和n,
\operatorname{Cov}(X, Y)=E[XY]-E[X] E[Y]

一个nd,b和C一个在s和$X是$在一世ll和q在一个l1○r0d和p和nd一世nG○n在H和吨H和r○rn○吨b○吨H$X$一个nd$是$和q在一个l1,在和s和和吨H一个吨
\ 操作员名称 {Cov} (X, Y) = P {X = 1, Y = 1} -P {X = 1} P {Y = 1}

Fr○米吨H一世s在和s和和吨H一个吨

这⁡(X,是)>0⇔磷X=1,是=1>磷X=1磷是=1 ⇔磷X=1,是=1磷X=1>磷是=1 ⇔磷是=1∣X=1>磷是=1
$$
即协方差X和是如果结果是肯定的X=1更有可能是=1（从对称性很容易看出，这也意味着相反）。
一般来说，可以证明一个正值这⁡(X,是)是一个迹象，表明是趋于增加X确实如此，而负值表示是趋于减少X增加。

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

数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|E 2204

Posted on 2022年6月18日2022年6月18日 by statistics-lab

如果你也在怎样代写概率模型和随机过程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。

随机过程是随机量在时间或空间上演变的概率模型。演变受不同时间或地点的随机量之间的某种依赖关系所支配。

我们提供的概率模型和随机过程及其相关学科的代写，服务范围广, 其中包括但不限于:

Statistical Inference 统计推断
Statistical Computing 统计计算
Advanced Probability Theory 高等概率论
Advanced Mathematical Statistics 高等数理统计学
(Generalized) Linear Models 广义线性模型
Statistical Machine Learning 统计机器学习
Longitudinal Data Analysis 纵向数据分析
Foundations of Data Science 数据科学基础

数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|E 2204

数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|The Discrete Case

If $X$ is a discrete random variable having a probability mass function $p(x)$, then the expected value of $X$ is defined by
$$
E[X]=\sum_{x: p(x)>0} x p(x)
$$
In other words, the expected value of $X$ is a weighted average of the possible values that $X$ can take on, each value being weighted by the probability that $X$ assumes that value. For example, if the probability mass function of $X$ is given by
$$
p(1)=\frac{1}{2}=p(2)
$$
then
$$
E[X]=1\left(\frac{1}{2}\right)+2\left(\frac{1}{2}\right)=\frac{3}{2}
$$
is just an ordinary average of the two possible values 1 and 2 that $X$ can assume. On the other hand, if
$$
p(1)=\frac{1}{3}, \quad p(2)=\frac{2}{3}
$$
then
$$
E[X]=1\left(\frac{1}{3}\right)+2\left(\frac{2}{3}\right)=\frac{5}{3}
$$
is a weighted average of the two possible values 1 and 2 where the value 2 is given twice as much weight as the value 1 since $p(2)=2 p(1)$.
Example 2.15. Find $E[X]$ where $X$ is the outcome when we roll a fair die.
Solution: $\quad$ Since $p(1)=p(2)=p(3)=p(4)=p(5)=p(6)=\frac{1}{6}$, we obtain
$$
E[X]=1\left(\frac{1}{6}\right)+2\left(\frac{1}{6}\right)+3\left(\frac{1}{6}\right)+4\left(\frac{1}{6}\right)+5\left(\frac{1}{6}\right)+6\left(\frac{1}{6}\right)=\frac{7}{2}
$$
Example 2.16 (Expectation of a Bernoulli Random Variable). Calculate $E[X]$ when $X$ is a Bernoulli random variable with parameter $p$.
Solution: Since $p(0)=1-p, p(1)=p$, we have
$$
E[X]=0(1-p)+1(p)=p
$$
Thus, the expected number of successes in a single trial is just the probability that the trial will be a success.

数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|The Continuous Case

We may also define the expected value of a continuous random variable. This is done as follows. If $X$ is a continuous random variable having a probability density function $f(x)$, then the expected value of $X$ is defined by
$$
E[X]=\int_{-\infty}^{\infty} x f(x) d x
$$
Example 2.20 (Expectation of a Uniform Random Variable). Calculate the expectation of a random variable uniformly distributed over $(\alpha, \beta)$.

Solution: From Eq. (2.8) we have
$$
\begin{aligned}
E[X] &=\int_{\alpha}^{\beta} \frac{x}{\beta-\alpha} d x \
&=\frac{\beta^{2}-\alpha^{2}}{2(\beta-\alpha)} \
&=\frac{\beta+\alpha}{2}
\end{aligned}
$$
In other words, the expected value of a random variable uniformly distributed over the interval $(\alpha, \beta)$ is just the midpoint of the interval.

Example 2.21 (Expectation of an Exponential Random Variable). Let $X$ be exponentially distributed with parameter $\lambda$. Calculate $E[X]$.
Solution:
$$
E[X]=\int_{0}^{\infty} x \lambda e^{-\lambda x} d x
$$
Integrating by parts $\left(d v=\lambda e^{-\lambda x} d x, u=x\right)$ yields
$$
\begin{aligned}
E[X] &=-\left.x e^{-\lambda x}\right|{0} ^{\infty}+\int{0}^{\infty} e^{-\lambda x} d x \
&=0-\left.\frac{e^{-\lambda x}}{\lambda}\right|_{0} ^{\infty} \
&=\frac{1}{\lambda}
\end{aligned}
$$

数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|Expectation of a Function of a Random Variable

Suppose now that we are given a random variable $X$ and its probability distribution (that is, its probability mass function in the discrete case or its probability density function in the continuous case). Suppose also that we are interested in calculating not the expected value of $X$, but the expected value of some function of $X$, say, $g(X)$. How do we go about doing this? One way is as follows. Since $g(X)$ is itself a random variable, it must have a probability distribution, which should be computable from a knowledge of the distribution of $X$. Once we have obtained the distribution of $g(X)$, we can then compute $E[g(X)]$ by the definition of the expectation.
Example 2.23. Suppose $X$ has the following probability mass function:
$$
p(0)=0.2, \quad p(1)=0.5, \quad p(2)=0.3
$$
Calculate $E\left[X^{2}\right]$
Solution: Letting $Y=X^{2}$, we have that $Y$ is a random variable that can take on one of the values $0^{2}, 1^{2}, 2^{2}$ with respective probabilities
$$
\begin{aligned}
&p_{Y}(0)=P\left{Y=0^{2}\right}=0.2 \
&p_{Y}(1)=P\left{Y=1^{2}\right}=0.5 \
&p_{Y}(4)=P\left{Y=2^{2}\right}=0.3
\end{aligned}
$$
Hence,
$$
E\left[X^{2}\right]=E[Y]=0(0.2)+1(0.5)+4(0.3)=1.7
$$
Note that
$$
1.7=E\left[X^{2}\right] \neq(E[X])^{2}=1.21
$$

概率模型和随机过程代考

数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|The Discrete Case

如果X是具有概率质量函数的离散随机变量p(X), 那么期望值X定义为

和[X]=∑X:p(X)>0Xp(X)
换句话说，期望值X是可能值的加权平均值X可以取，每个值都由以下概率加权X假定该值。例如，如果概率质量函数X是（谁）给的

p(1)=12=p(2)
然后

和[X]=1(12)+2(12)=32
只是两个可能值 1 和 2 的普通平均值X可以假设。另一方面，如果

p(1)=13,p(2)=23
然后

和[X]=1(13)+2(23)=53
是两个可能值 1 和 2 的加权平均值，其中值 2 的权重是值 1 的两倍，因为p(2)=2p(1).
例 2.15。寻找和[X]在哪里X是我们掷出公平骰子时的结果。
解决方案：自从p(1)=p(2)=p(3)=p(4)=p(5)=p(6)=16，我们获得

和[X]=1(16)+2(16)+3(16)+4(16)+5(16)+6(16)=72
示例 2.16（伯努利随机变量的期望）。计算和[X]什么时候X是带参数的伯努利随机变量p.
解决方案：由于p(0)=1−p,p(1)=p，我们有

和[X]=0(1−p)+1(p)=p
因此，单次试验的预期成功次数就是试验成功的概率。

数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|The Continuous Case

我们也可以定义一个连续随机变量的期望值。这是按如下方式完成的。如果X是具有概率密度函数的连续随机变量F(X), 那么期望值X定义为

和[X]=∫−∞∞XF(X)dX
示例 2.20（均匀随机变量的期望）。计算均匀分布的随机变量的期望(一个,b).

解决方案：从方程式。(2.8) 我们有

和[X]=∫一个bXb−一个dX =b2−一个22(b−一个) =b+一个2
换句话说，随机变量的期望值均匀分布在区间上(一个,b)只是区间的中点。

示例 2.21（指数随机变量的期望）。让X随参数呈指数分布λ. 计算和[X].
解决方案：

和[X]=∫0∞Xλ和−λXdX
分部整合(d在=λ和−λXdX,在=X)产生
$$
\begin{aligned}
E[X] &=-\left.xe^{-\lambda x}\right| {0} ^{\infty}+\int {0}^{\infty} e^{-\lambda x} dx \
&=0-\left.\frac{e^{-\lambda x}}{\ lambda}\right|_{0} ^{\infty} \
&=\frac{1}{\lambda}
\end{aligned}
$$

数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|Expectation of a Function of a Random Variable

假设现在给定一个随机变量X及其概率分布（即离散情况下的概率质量函数或连续情况下的概率密度函数）。还假设我们感兴趣的不是计算的期望值X, 但某些函数的期望值为X，说，G(X). 我们该如何做呢？一种方法如下。自从G(X)它本身是一个随机变量，它必须有一个概率分布，它应该可以根据分布的知识来计算X. 一旦我们得到分布G(X)，然后我们可以计算和[G(X)]根据期望的定义。
例 2.23。认为X具有以下概率质量函数：

p(0)=0.2,p(1)=0.5,p(2)=0.3
计算和[X2]
解决方案：让是=X2, 我们有是是一个随机变量，可以取其中一个值02,12,22有各自的概率

\begin{对齐} &p_{Y}(0)=P\left{Y=0^{2}\right}=0.2 \ &p_{Y}(1)=P\left{Y=1^{2}\右}=0.5 \ &p_{Y}(4)=P\left{Y=2^{2}\right}=0.3 \end{对齐}\begin{对齐} &p_{Y}(0)=P\left{Y=0^{2}\right}=0.2 \ &p_{Y}(1)=P\left{Y=1^{2}\右}=0.5 \ &p_{Y}(4)=P\left{Y=2^{2}\right}=0.3 \end{对齐}
因此，

和[X2]=和[是]=0(0.2)+1(0.5)+4(0.3)=1.7
注意

1.7=和[X2]≠(和[X])2=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

数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|MAP 4102

Posted on 2022年6月18日2022年6月18日 by statistics-lab

如果你也在怎样代写概率模型和随机过程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。

随机过程是随机量在时间或空间上演变的概率模型。演变受不同时间或地点的随机量之间的某种依赖关系所支配。

我们提供的概率模型和随机过程及其相关学科的代写，服务范围广, 其中包括但不限于:

Statistical Inference 统计推断
Statistical Computing 统计计算
Advanced Probability Theory 高等概率论
Advanced Mathematical Statistics 高等数理统计学
(Generalized) Linear Models 广义线性模型
Statistical Machine Learning 统计机器学习
Longitudinal Data Analysis 纵向数据分析
Foundations of Data Science 数据科学基础

数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|MAP 4102

数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|The Geometric Random Variable

Suppose that independent trials, each having probability $p$ of being a success, are performed until a success occurs. If we let $X$ be the number of trials required until the first success, then $X$ is said to be a geometric random variable with parameter $p$. Its probability mass function is given by
$$
p(n)=P{X=n}=(1-p)^{n-1} p, \quad n=1,2, \ldots
$$

Eq. (2.4) follows since in order for $X$ to equal $n$ it is necessary and sufficient that the first $n-1$ trials be failures and the $n$th trial a success. Eq. (2.4) follows since the outcomes of the successive trials are assumed to be independent.
To check that $p(n)$ is a probability mass function, we note that
$$
\sum_{n=1}^{\infty} p(n)=p \sum_{n=1}^{\infty}(1-p)^{n-1}=1
$$

数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|The Poisson Random Variable

A random variable $X$, taking on one of the values $0,1,2, \ldots$, is said to be a Poisson random variable with parameter $\lambda$, if for some $\lambda>0$,
$$
p(i)=P{X=i}=e^{-\lambda} \frac{\lambda^{i}}{i !}, \quad i=0,1, \ldots
$$
Eq. (2.5) defines a probability mass function since
$$
\sum_{i=0}^{\infty} p(i)=e^{-\lambda} \sum_{i=0}^{\infty} \frac{\lambda^{i}}{i !}=e^{-\lambda} e^{\lambda}=1
$$
The Poisson random variable has a wide range of applications in a diverse number of areas, as will be seen in Chapter $5 .$

An important property of the Poisson random variable is that it may be used to approximate a binomial random variable when the binomial parameter $n$ is large and $p$ is small. To see this, suppose that $X$ is a binomial random variable with parameters $(n, p)$, and let $\lambda=n p$. Then
$$
\begin{aligned}
P{X=i} &=\frac{n !}{(n-i) ! i !} p^{i}(1-p)^{n-i} \
&=\frac{n !}{(n-i) ! i !}\left(\frac{\lambda}{n}\right)^{i}\left(1-\frac{\lambda}{n}\right)^{n-i} \
&=\frac{n(n-1) \cdots(n-i+1)}{n^{i}} \frac{\lambda^{i}}{i !} \frac{(1-\lambda / n)^{n}}{(1-\lambda / n)^{i}}
\end{aligned}
$$
Now, for $n$ large and $p$ small
$$
\left(1-\frac{\lambda}{n}\right)^{n} \approx e^{-\lambda}, \quad \frac{n(n-1) \cdots(n-i+1)}{n^{i}} \approx 1, \quad\left(1-\frac{\lambda}{n}\right)^{i} \approx 1
$$
Hence, for $n$ large and $p$ small,
$$
P{X=i} \approx e^{-\lambda} \frac{\lambda^{i}}{i !}
$$

数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|Continuous Random Variables

In this section, we shall concern ourselves with random variables whose set of possible values is uncountable. Let $X$ be such a random variable. We say that $X$ is a continuous random variable if there exists a nonnegative function $f(x)$, defined for all real $x \in$ $(-\infty, \infty)$, having the property that for any set $B$ of real numbers
$$
P{X \in B}=\int_{B} f(x) d x
$$
The function $f(x)$ is called the probability density function of the random variable $X$. In words, Eq. (2.6) states that the probability that $X$ will be in $B$ may be obtained by integrating the probability density function over the set $B$. Since $X$ must assume some value, $f(x)$ must satisfy
$$
1=P{X \in(-\infty, \infty)}=\int_{-\infty}^{\infty} f(x) d x
$$

All probability statements about $X$ can be answered in terms of $f(x)$. For instance, letting $B=[a, b]$, we obtain from Eq. (2.6) that
$$
P{a \leq X \leq b}=\int_{a}^{b} f(x) d x
$$
If we let $a=b$ in the preceding, then
$$
P{X=a}=\int_{a}^{a} f(x) d x=0
$$
In words, this equation states that the probability that a continuous random variable will assume any particular value is zero.

概率模型和随机过程代考

数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|The Geometric Random Variable

假设独立的试验，每个试验都有概率p是一个成功，执行，直到成功发生。如果我们让X是直到第一次成功所需的试验次数，然后X据说是一个带参数的几何随机变量p. 其概率质量函数由下式给出

p(n)=磷X=n=(1−p)n−1p,n=1,2,…

方程。(2.4) 如下，因为为了X等于n第一个是必要且充分的n−1试炼是失败的，而n试验成功。方程。（2.4）如下，因为连续试验的结果被假定为独立的。
检查p(n)是概率质量函数，我们注意到

∑n=1∞p(n)=p∑n=1∞(1−p)n−1=1

数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|The Poisson Random Variable

随机变量X，取其中一个值0,1,2,…, 称为泊松随机变量，带参数λ, 如果对于某些λ>0,

p(一世)=磷X=一世=和−λλ一世一世!,一世=0,1,…
方程。(2.5) 定义了一个概率质量函数，因为

∑一世=0∞p(一世)=和−λ∑一世=0∞λ一世一世!=和−λ和λ=1
泊松随机变量在许多领域都有广泛的应用，我们将在本章中看到5.

泊松随机变量的一个重要性质是，当二项式参数n很大并且p是小。要看到这一点，假设X是带参数的二项式随机变量(n,p)，然后让λ=np. 然后

磷X=一世=n!(n−一世)!一世!p一世(1−p)n−一世 =n!(n−一世)!一世!(λn)一世(1−λn)n−一世 =n(n−1)⋯(n−一世+1)n一世λ一世一世!(1−λ/n)n(1−λ/n)一世
现在，对于n大而p小的

(1−λn)n≈和−λ,n(n−1)⋯(n−一世+1)n一世≈1,(1−λn)一世≈1
因此，对于n大而p小的，

磷X=一世≈和−λλ一世一世!

数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|Continuous Random Variables

在本节中，我们将关注其可能值的集合是不可数的随机变量。让X成为这样一个随机变量。我们说X如果存在非负函数，则为连续随机变量F(X), 为所有实数定义X∈ (−∞,∞), 具有对任何集合的属性乙实数

磷X∈乙=∫乙F(X)dX
功能F(X)称为随机变量的概率密度函数X. 换句话说，方程式。(2.6) 表明概率X将在乙可以通过对集合上的概率密度函数进行积分来获得乙. 自从X必须假设一些价值，F(X)必须满足

1=磷X∈(−∞,∞)=∫−∞∞F(X)dX

所有关于的概率陈述X可以从以下方面回答F(X). 例如，让乙=[一个,b]，我们从方程式获得。(2.6) 那

磷一个≤X≤b=∫一个bF(X)dX
如果我们让一个=b在前面，那么

磷X=一个=∫一个一个F(X)dX=0
换句话说，这个等式表明连续随机变量假设任何特定值的概率为零。

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

数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|STAT3004

Posted on 2022年6月18日2022年6月18日 by statistics-lab

如果你也在怎样代写概率模型和随机过程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。

随机过程是随机量在时间或空间上演变的概率模型。演变受不同时间或地点的随机量之间的某种依赖关系所支配。

我们提供的概率模型和随机过程及其相关学科的代写，服务范围广, 其中包括但不限于:

Statistical Inference 统计推断
Statistical Computing 统计计算
Advanced Probability Theory 高等概率论
Advanced Mathematical Statistics 高等数理统计学
(Generalized) Linear Models 广义线性模型
Statistical Machine Learning 统计机器学习
Longitudinal Data Analysis 纵向数据分析
Foundations of Data Science 数据科学基础

数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|Random Variables

It frequently occurs that in performing an experiment we are mainly interested in some functions of the outcome as opposed to the outcome itself. For instance, in tossing dice we are often interested in the sum of the two dice and are not really concerned about the actual outcome. That is, we may be interested in knowing that the sum is seven and not be concerned over whether the actual outcome was $(1,6)$ or $(2,5)$ or $(3,4)$ or $(4,3)$ or $(5,2)$ or $(6,1)$. These quantities of interest, or more formally, these real-valued functions defined on the sample space, are known as random variables.

Since the value of a random variable is determined by the outcome of the experiment, we may assign probabilities to the possible values of the random variable.

Example 2.1. Letting $X$ denote the random variable that is defined as the sum of two fair dice; then
$$
\begin{aligned}
&P{X=2}=P{(1,1)}=\frac{1}{36}, \
&P{X=3}=P{(1,2),(2,1)}=\frac{2}{36} \
&P{X=4}=P{(1,3),(2,2),(3,1)}=\frac{3}{36} \
&P{X=5}=P{(1,4),(2,3),(3,2),(4,1)}=\frac{4}{36} \
&P{X=6}=P{(1,5),(2,4),(3,3),(4,2),(5,1)}=\frac{5}{36} \
&P{X=7}=P{(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)}=\frac{6}{36} \
&P{X=8}=P{(2,6),(3,5),(4,4),(5,3),(6,2)}=\frac{5}{36} \
&P{X=9}=P{(3,6),(4,5),(5,4),(6,3)}=\frac{4}{36} \
&P{X=10}=P{(4,6),(5,5),(6,4)}=\frac{3}{36}, \
&P{X=11}=P{(5,6),(6,5)}=\frac{2}{36}, \
&P{X=12}=P{(6,6)}=\frac{1}{36}
\end{aligned}
$$
In other words, the random variable $X$ can take on any integral value between two and twelve, and the probability that it takes on each value is given by Eq. (2.1). Since $X$ must take on one of the values two through twelve, we must have
$$
1=P\left{\bigcup_{n=2}^{12}{X=n}\right}=\sum_{n=2}^{12} P{X=n}
$$
which may be checked from Eq. (2.1).

数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|Discrete Random Variables

As was previously mentioned, a random variable that can take on at most a countable number of possible values is said to be discrete. For a discrete random variable $X$, we define the probability mass function $p(a)$ of $X$ by
$$
p(a)=P{X=a}
$$
The probability mass function $p(a)$ is positive for at most a countable number of values of $a$. That is, if $X$ must assume one of the values $x_{1}, x_{2}, \ldots$, then
$$
\begin{array}{ll}
p\left(x_{i}\right)>0, & i=1,2, \ldots \
p(x)=0, & \text { all other values of } x
\end{array}
$$
Since $X$ must take on one of the values $x_{i}$, we have
$$
\sum_{i=1}^{\infty} p\left(x_{i}\right)=1
$$
The cumulative distribution function $F$ can be expressed in terms of $p(a)$ by
$$
F(a)=\sum_{\text {all }} p\left(x_{i} \leq a\right.
$$
For instance, suppose $X$ has a probability mass function given by
$$
p(1)=\frac{1}{2}, \quad p(2)=\frac{1}{3}, \quad p(3)=\frac{1}{6}
$$
then, the cumulative distribution function $F$ of $X$ is given by
$$
F(a)= \begin{cases}0, & a<1 \ \frac{1}{2}, & 1 \leq a<2 \ \frac{5}{6}, & 2 \leq a<3 \ 1, & 3 \leq a\end{cases}
$$
This is graphically presented in Fig. 2.1.

数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|The Binomial Random Variable

Suppose that $n$ independent trials, each of which results in a “success” with probability $p$ and in a “failure” with probability $1-p$, are to be performed. If $X$ represents the number of successes that occur in the $n$ trials, then $X$ is said to be a binomial random variable with parameters $(n, p)$.

The probability mass function of a binomial random variable having parameters $(n, p)$ is given by
$$
p(i)=\left(\begin{array}{c}
n \
i
\end{array}\right) p^{i}(1-p)^{n-i}, \quad i=0,1, \ldots, n
$$
where
$$
\left(\begin{array}{l}
n \
i
\end{array}\right)=\frac{n !}{(n-i) ! i !}
$$
equals the number of different groups of $i$ objects that can be chosen from a set of $n$ objects. The validity of Eq. (2.3) may be verified by first noting that the probability of any particular sequence of the $n$ outcomes containing $i$ successes and $n-i$ failures is, by the assumed independence of trials, $p^{i}(1-p)^{n-i}$. Eq. (2.3) then follows since there are $\left(\begin{array}{l}n \ i\end{array}\right)$ different sequences of the $n$ outcomes leading to $i$ successes and $n-i$ failures. For instance, if $n=3, i=2$, then there are $\left(\begin{array}{l}3 \ 2\end{array}\right)=3$ ways in which the three trials can result in two successes. Namely, any one of the three outcomes $(s, s, f),(s, f, s),(f, s, s)$, where the outcome $(s, s, f)$ means that the first two trials are successes and the third a failure. Since each of the three outcomes $(s, s, f),(s, f, s),(f, s, s)$ has a probability $p^{2}(1-p)$ of occurring the desired probability is thus $\left(\begin{array}{l}3 \ 2\end{array}\right) p^{2}(1-p)$.

概率模型和随机过程代考

数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|Random Variables

经常发生的是，在进行实验时，我们主要对结果的某些功能感兴趣，而不是结果本身。例如，在掷骰子时，我们通常对两个骰子的总和感兴趣，而并不真正关心实际结果。也就是说，我们可能有兴趣知道总和是 7，而不关心实际结果是否为(1,6)或者(2,5)或者(3,4)或者(4,3)或者(5,2)或者(6,1). 这些感兴趣的量，或者更正式地说，在样本空间上定义的这些实值函数，被称为随机变量。

由于随机变量的值是由实验结果决定的，我们可以为随机变量的可能值分配概率。

例 2.1。让X表示定义为两个公平骰子之和的随机变量；然后

磷X=2=磷(1,1)=136, 磷X=3=磷(1,2),(2,1)=236 磷X=4=磷(1,3),(2,2),(3,1)=336 磷X=5=磷(1,4),(2,3),(3,2),(4,1)=436 磷X=6=磷(1,5),(2,4),(3,3),(4,2),(5,1)=536 磷X=7=磷(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)=636 磷X=8=磷(2,6),(3,5),(4,4),(5,3),(6,2)=536 磷X=9=磷(3,6),(4,5),(5,4),(6,3)=436 磷X=10=磷(4,6),(5,5),(6,4)=336, 磷X=11=磷(5,6),(6,5)=236, 磷X=12=磷(6,6)=136
换句话说，随机变量X可以取 2 到 12 之间的任何整数值，它取每个值的概率由方程式给出。(2.1)。自从X必须取值 2 到 12 之一，我们必须有

1=P\left{\bigcup_{n=2}^{12}{X=n}\right}=\sum_{n=2}^{12} P{X=n}1=P\left{\bigcup_{n=2}^{12}{X=n}\right}=\sum_{n=2}^{12} P{X=n}
这可以从方程式检查。(2.1)。

数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|Discrete Random Variables

如前所述，一个最多可以取可数个可能值的随机变量被称为离散变量。对于离散随机变量X，我们定义概率质量函数p(一个)的X经过

p(一个)=磷X=一个
概率质量函数p(一个)对于最多可数个值是正的一个. 也就是说，如果X必须假设其中一个值X1,X2,…，然后

p(X一世)>0,一世=1,2,… p(X)=0, 所有其他值 X
自从X必须采用其中一个值X一世，我们有

∑一世=1∞p(X一世)=1
累积分布函数F可以表示为p(一个)经过

F(一个)=∑全部 p(X一世≤一个
例如，假设X具有由下式给出的概率质量函数

p(1)=12,p(2)=13,p(3)=16
那么，累积分布函数F的X是（谁）给的

F(一个)={0,一个<1 12,1≤一个<2 56,2≤一个<3 1,3≤一个
这在图 2.1 中以图形方式呈现。

数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|The Binomial Random Variable

假设n独立的试验，每一次试验都有一定的概率导致“成功”p并在“失败”的概率1−p, 将被执行。如果X表示在该过程中发生的成功次数n试炼，然后X被称为具有参数的二项式随机变量(n,p).

具有参数的二项式随机变量的概率质量函数(n,p)是（谁）给的

p(一世)=(n 一世)p一世(1−p)n−一世,一世=0,1,…,n
在哪里

(n 一世)=n!(n−一世)!一世!
等于不同组的数量一世可以从一组中选择的对象n对象。等式的有效性。（2.3）可以通过首先注意到任何特定序列的概率来验证n结果包含一世成功和n−一世失败是，通过假定试验的独立性，p一世(1−p)n−一世. 方程。(2.3) 然后如下，因为有(n 一世)不同的序列n结果导致一世成功和n−一世失败。例如，如果n=3,一世=2, 那么有(3 2)=3三项试验可以导致两次成功的方式。即，三种结果中的任何一种(s,s,F),(s,F,s),(F,s,s), 结果在哪里(s,s,F)表示前两次试验成功，第三次失败。由于三个结果中的每一个(s,s,F),(s,F,s),(F,s,s)有概率p2(1−p)因此，发生期望的概率是(3 2)p2(1−p).

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

数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|Introduction to Probability Theory

Posted on 2022年6月18日2022年6月18日 by statistics-lab

如果你也在怎样代写概率模型和随机过程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。

随机过程是随机量在时间或空间上演变的概率模型。演变受不同时间或地点的随机量之间的某种依赖关系所支配。

我们提供的概率模型和随机过程及其相关学科的代写，服务范围广, 其中包括但不限于:

Statistical Inference 统计推断
Statistical Computing 统计计算
Advanced Probability Theory 高等概率论
Advanced Mathematical Statistics 高等数理统计学
(Generalized) Linear Models 广义线性模型
Statistical Machine Learning 统计机器学习
Longitudinal Data Analysis 纵向数据分析
Foundations of Data Science 数据科学基础

数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|Introduction to Probability Theory

数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|Sample Space and Events

Suppose that we are about to perform an experiment whose outcome is not predictable in advance. However, while the outcome of the experiment will not be known in advance, let us suppose that the set of all possible outcomes is known. This set of all possible outcomes of an experiment is known as the sample space of the experiment and is denoted by $S$.
Some examples are the following.

If the experiment consists of the flipping of a coin, then
$$
S={H, T}
$$
where $H$ means that the outcome of the toss is a head and $T$ that it is a tail.
If the experiment consists of rolling a die, then the sample space is
$$
S={1,2,3,4,5,6}
$$
where the outcome $i$ means that $i$ appeared on the die, $i=1,2,3,4,5,6$.
If the experiment consists of flipping two coins, then the sample space consists of the following four points:
$$
S={(H, H),(H, T),(T, H),(T, T)}
$$
The outcome will be $(H, H)$ if both coins come up heads; it will be $(H, T)$ if the first coin comes up heads and the second comes up tails; it will be $(T, H)$ if the

first comes up tails and the second heads; and it will be $(T, T)$ if both coins come up tails.

If the experiment consists of rolling two dice, then the sample space consists of the following 36 points:
$$
S=\left{\begin{array}{l}
(1,1),(1,2),(1,3),(1,4),(1,5),(1,6) \
(2,1),(2,2),(2,3),(2,4),(2,5),(2,6) \
(3,1),(3,2),(3,3),(3,4),(3,5),(3,6) \
(4,1),(4,2),(4,3),(4,4),(4,5),(4,6) \
(5,1),(5,2),(5,3),(5,4),(5,5),(5,6) \
(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)
\end{array}\right}
$$
where the outcome $(i, j)$ is said to occur if $i$ appears on the first die and $j$ on the second die.

数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|Probabilities Defined on Events

Consider an experiment whose sample space is $S$. For each event $E$ of the sample space $S$, we assume that a number $P(E)$ is defined and satisfies the following three conditions:
(i) $0 \leqslant P(E) \leqslant 1$.
(ii) $P(S)=1$.
(iii) For any sequence of events $E_{1}, E_{2}, \ldots$ that are mutually exclusive, that is, events for which $E_{n} E_{m}=\emptyset$ when $n \neq m$, then
$$
P\left(\bigcup_{n=1}^{\infty} E_{n}\right)=\sum_{n=1}^{\infty} P\left(E_{n}\right)
$$
We refer to $P(E)$ as the probability of the event $E$.

Example 1.1. In the coin tossing example, if we assume that a head is equally likely to appear as a tail, then we would have
$$
P({H})=P({T})=\frac{1}{2}
$$
On the other hand, if we had a biased coin and felt that a head was twice as likely to appear as a tail, then we would have
$$
P({H})=\frac{2}{3}, \quad P({T})=\frac{1}{3}
$$
Example 1.2. In the die tossing example, if we supposed that all six numbers were equally likely to appear, then we would have
$$
P({1})=P({2})=P({3})=P({4})=P({5})=P({6})=\frac{1}{6}
$$
From (iii) it would follow that the probability of getting an even number would equal
$$
\begin{aligned}
P({2,4,6}) &=P({2})+P({4})+P({6}) \
&=\frac{1}{2}
\end{aligned}
$$

数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|Conditional Probabilities

Suppose that we toss two dice and that each of the 36 possible outcomes is equally likely to occur and hence has probability $\frac{1}{36}$. Suppose that we observe that the first die is a four. Then, given this information, what is the probability that the sum of the two dice equals six? To calculate this probability we reason as follows: Given that the initial die is a four, it follows that there can be at most six possible outcomes of our experiment, namely, $(4,1),(4,2),(4,3),(4,4),(4,5)$, and $(4,6)$. Since each of these outcomes originally had the same probability of occurring, they should still have equal probabilities. That is, given that the first die is a four, then the (conditional) probability of each of the outcomes $(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)$ is $\frac{1}{6}$ while the (conditional) probability of the other 30 points in the sample space is 0 . Hence, the desired probability will be $\frac{1}{6}$.

If we let $E$ and $F$ denote, respectively, the event that the sum of the dice is six and the event that the first die is a four, then the probability just obtained is called the conditional probability that $E$ occurs given that $F$ has occurred and is denoted by
$$
P(E \mid F)
$$
A general formula for $P(E \mid F)$ that is valid for all events $E$ and $F$ is derived in the same manner as the preceding. Namely, if the event $F$ occurs, then in order for $E$ to occur it is necessary for the actual occurrence to be a point in both $E$ and in $F$, that is, it must be in $E F$. Now, because we know that $F$ has occurred, it follows that $F$ becomes our new sample space and hence the probability that the event $E F$ occurs will equal the probability of $E F$ relative to the probability of $F$. That is,
$$
P(E \mid F)=\frac{P(E F)}{P(F)}
$$
Note that Eq. (1.5) is only well defined when $P(F)>0$ and hence $P(E \mid F)$ is only defined when $P(F)>0$.

概率模型和随机过程代考

数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|Sample Space and Events

假设我们要进行一项实验，其结果无法提前预测。然而，虽然实验的结果不会提前知道，但让我们假设所有可能结果的集合都是已知的。实验的所有可能结果的集合称为实验的样本空间，并表示为小号.
一些例子如下。

如果实验包括掷硬币，那么
小号=H,吨
在哪里H表示掷骰子的结果是正面，并且吨它是一条尾巴。
如果实验由掷骰子组成，则样本空间为
小号=1,2,3,4,5,6
结果在哪里一世意思是一世出现在模具上，一世=1,2,3,4,5,6.
如果实验是抛两枚硬币，那么样本空间由以下四点组成：
小号=(H,H),(H,吨),(吨,H),(吨,吨)
结果将是(H,H)如果两个硬币都出现正面；这将是(H,吨)如果第一枚硬币正面朝上，第二枚硬币反面；这将是(吨,H)如果

第一个出现反面，第二个出现正面；它会是(吨,吨)如果两个硬币都出现反面。

如果实验由掷两个骰子组成，那么样本空间由以下 36 个点组成：
S=\left{\begin{array}{l} (1,1),(1,2),(1,3),(1,4),(1,5),(1,6) \ ( 2,1),(2,2),(2,3),(2,4),(2,5),(2,6) \ (3,1),(3,2),(3, 3),(3,4),(3,5),(3,6) \ (4,1),(4,2),(4,3),(4,4),(4,5) ,(4,6) \ (5,1),(5,2),(5,3),(5,4),(5,5),(5,6) \ (6,1),( 6,2),(6,3),(6,4),(6,5),(6,6) \end{array}\right}S=\left{\begin{array}{l} (1,1),(1,2),(1,3),(1,4),(1,5),(1,6) \ ( 2,1),(2,2),(2,3),(2,4),(2,5),(2,6) \ (3,1),(3,2),(3, 3),(3,4),(3,5),(3,6) \ (4,1),(4,2),(4,3),(4,4),(4,5) ,(4,6) \ (5,1),(5,2),(5,3),(5,4),(5,5),(5,6) \ (6,1),( 6,2),(6,3),(6,4),(6,5),(6,6) \end{array}\right}
结果在哪里(一世,j)据说如果发生一世出现在第一个骰子上并且j在第二个模具上。

数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|Probabilities Defined on Events

考虑一个样本空间为小号. 对于每个事件和样本空间的小号, 我们假设一个数磷(和)被定义并满足以下三个条件：
(i)0⩽磷(和)⩽1.
(二)磷(小号)=1.
(iii) 对于任何事件序列和1,和2,…是互斥的，也就是说，事件和n和米=∅什么时候n≠米，然后

磷(⋃n=1∞和n)=∑n=1∞磷(和n)
我们指磷(和)作为事件的概率和.

例 1.1。在抛硬币的例子中，如果我们假设一个正面和一个尾巴出现的可能性相同，那么我们将有

磷(H)=磷(吨)=12
另一方面，如果我们有一个有偏差的硬币，并且觉得正面出现的可能性是反面的两倍，那么我们会有

磷(H)=23,磷(吨)=13
例 1.2。在掷骰子的例子中，如果我们假设所有六个数字出现的可能性相同，那么我们将有

磷(1)=磷(2)=磷(3)=磷(4)=磷(5)=磷(6)=16
从 (iii) 可以得出，得到偶数的概率等于

磷(2,4,6)=磷(2)+磷(4)+磷(6) =12

数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|Conditional Probabilities

假设我们掷两个骰子，36 种可能的结果中的每一种发生的可能性相同，因此具有概率136. 假设我们观察到第一个骰子是 4。那么，给定这些信息，两个骰子之和等于 6 的概率是多少？为了计算这个概率，我们推理如下：鉴于初始骰子是 4，因此我们的实验最多可以有六个可能的结果，即，(4,1),(4,2),(4,3),(4,4),(4,5)，和(4,6). 由于这些结果中的每一个最初具有相同的发生概率，因此它们应该仍然具有相同的概率。也就是说，假设第一个骰子是 4，那么每个结果的（条件）概率(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)是16而样本空间中其他 30 个点的（条件）概率为 0 。因此，期望的概率将是16.

如果我们让和和F分别表示骰子和为6的事件和第一个骰子是4的事件，则刚刚得到的概率称为条件概率和鉴于发生F已经发生并表示为

磷(和∣F)
一个通用公式磷(和∣F)对所有事件都有效和和F以与前面相同的方式导出。也就是说，如果事件F发生，然后为了和要发生，实际发生必须是两者中的一个点和并且在F，也就是说，它必须在和F. 现在，因为我们知道F已经发生，因此F成为我们的新样本空间，因此事件发生的概率和F发生的概率等于和F相对于概率F. 那是，

磷(和∣F)=磷(和F)磷(F)
请注意，方程式。(1.5) 仅在以下情况下被明确定义磷(F)>0因此磷(和∣F)仅在以下情况下定义磷(F)>0.

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

数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|Random Walks

Posted on 2022年6月18日2022年6月18日 by statistics-lab

如果你也在怎样代写概率模型和随机过程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。

随机过程是随机量在时间或空间上演变的概率模型。演变受不同时间或地点的随机量之间的某种依赖关系所支配。

我们提供的概率模型和随机过程及其相关学科的代写，服务范围广, 其中包括但不限于:

Statistical Inference 统计推断
Statistical Computing 统计计算
Advanced Probability Theory 高等概率论
Advanced Mathematical Statistics 高等数理统计学
(Generalized) Linear Models 广义线性模型
Statistical Machine Learning 统计机器学习
Longitudinal Data Analysis 纵向数据分析
Foundations of Data Science 数据科学基础

数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|Random Walks in 1D

Consider a “random walker” on a one-dimensional lattice. Time will be discrete, and at every time step the walker will take a step in one of the possible two directions, with equal probability. What is the mean value of, say, coordinate $x$, after $N$ steps? From symmetry, it is clearly zero. But we certainly don’t expect the walker to be precisely at the origin after a large number of steps. Its typical magnitude away from the origin may be found by calculating the RMS (root-mean-square) of the trajectories, i.e., the second moment of the distance from the origin.

Let’s say the walker was at position $x$ after $N$ steps. After an additional step, it could be in one of two possible positions. Averaging the squared distance between these two possibilities, conditioned on being at position $x$ at time $N$, gives us
$$
\frac{1}{2}\left[(x+1)^{2}+(x-1)^{2}\right]=x^{2}+1 .
$$
Upon averaging over the value of $x^{2}$ at time $N$, we find that
$$
\left\langle x_{N+1}^{2}\right\rangle=\left\langle x_{N}^{2}\right\rangle+1
$$
therefore, repeating the argument $N-1$ times, we find that
$$
\left\langle x_{N}^{2}\right\rangle=N
$$

This implies that the typical distance from the origin scales like $\sqrt{N}$. What about the position distribution? If $N$ is even, then it is clear that the probability to be a distance $M$ from the origin after $N$ steps is zero for odd $M$, and for even $M$ it is
$$
p_{M}=\frac{1}{2^{N}}\left(\begin{array}{l}
N \
R
\end{array}\right)=\frac{1}{2^{N}} \frac{N !}{R !(N-R) !}
$$
where $R$ is the number of steps to the right, thus $R-(N-R)=2 R-N=M$. We can now evaluate this probability for $N \gg M$ using Stirling’s formula, which provides an (excellent) approximation for $N$ !, namely $N ! \approx \sqrt{2 \pi N}\left(\frac{N}{e}\right)^{N}$. This leads to
$$
\begin{aligned}
p_{M} & \approx \frac{1}{\sqrt{2 \pi}} \frac{1}{2^{N}} \frac{N^{N+1 / 2}}{R^{R+1 / 2}(N-R)^{N-R+1 / 2}} \
&=\frac{1}{\sqrt{2 \pi}} e^{-N \log (2)+(N+1 / 2) \log (N)-(R+1 / 2) \log (R)-(N-R+1 / 2) \log (N-R)}
\end{aligned}
$$
We can proceed by using our assumption that $N \gg M$, implying that $R$ is approximately equal to $N / 2$.

数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|Derivation of the Diffusion Equation for Random Walks

We shall now approach the problem with more generality, following nearly precisely the derivation by Einstein. We will work in $3 \mathrm{D}$ but the approach would be the same

in any dimension. The approach will have discrete time steps, but the step direction will be a continuous random variable, described by a probability distribution $g(\vec{\Delta})$ (here $\vec{\Delta}$ is a vector describing the step in the 3D space – not to be confused with the Laplacian operator!). We will not limit the random walker to a lattice, though it is possible to implement such a scenario by taking $g(\vec{\Delta})$ to be a sum of $\delta$-functions (can you see how?).

We will seek to obtain the probability distribution $p(\vec{r})$, i.e., $p(\vec{r}) d V$ will be the probability to find the particle in a volume $d V$ around the point $\vec{r}$ (at some point in time corresponding to a a given number of steps). If the original problem is cast on a lattice, this distribution will be relevant to describe the coarse grained problem, when we shall zoom-out far enough such that we will not care about the details at the level of the lattice constant.

If at time $t$ the probability distribution is described by $p(\vec{r}, t)$, let us consider what it will be a time $\tau$ later, where $\tau$ denotes the duration of each step. As you can guess, in a realistic scenario the time of a step is non-constant, and $\tau$ would be the mean step time. Thus, we haven’t lost too much in making time discrete – but we did make an assumption that a mean time exists. In Chapter 7 we will revisit this point, and show that in certain situations when the mean time diverges, we can get subdiffusion (slower spread of the probability distribution over time compared with diffusion).

To find $p(\vec{r}, t+\tau)$, we need to integrate over all space, and consider the probability to have the “right” jump size to bring us to $\vec{r}$. This leads to
$$
p(\vec{r}, t+\tau)=\int p(\vec{r}-\vec{\Delta}, t) d^{3} \vec{\Delta} g(\vec{\Delta}) .
$$
To proceed, we will Taylor expand $p$, assuming that the probability distribution is smooth on the scale of the typical jump. If we expand it to first order, we will find that
$$
p(\vec{r}-\vec{\Delta}) \approx p(\vec{r})-(\nabla p) \cdot \vec{\Delta} .
$$
If the diffusion process is isotropic in space, there is no difference between a jump in the $\pm \vec{\Delta}$ direction, so the integral associated with the second term trivially vanishes:
$$
\int((\nabla p) \cdot \vec{\Delta}) g(\vec{\Delta}) d^{3} \vec{\Delta}=0 .
$$
This means we have to expand to second order:
$$
p(\vec{r}-\vec{\Delta}) \approx p(\vec{r})-(\nabla p) \cdot \vec{\Delta}+\left.\frac{1}{2} \sum_{i, j} \frac{\partial^{2} p}{\partial x_{i} \partial x_{j}}\right|{\vec{r}} \Delta{i} \Delta_{j}
$$

数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|Markov Processes and Markov Chains

Let us approach our analysis of the random walker from a new perspective, which will be easier to generalize to other networks. Within our model for diffusion on a 1D lattice, the probability to go to site $j$ does not depend on the history of the random walker, but only on its current state – this is an example of a Markov process (named after Andrey Markov). A familiar childhood example of a Markov process is the game Chutes and Ladders (as well as many other board games – see also Problem 2.15). But we emphasize that whether a process is Markovian depends on the space: Consider, for example, a game where we throw a die at each round, and keep track of the running sum. This is clearly a Markov process – knowing the current sum determines the probabilities to go to the next states. But what about a game in which we reset the sum to zero each time we get two identical numbers in a row? In this case, the process is memory dependent, so it is non-Markovian. But, if we work in the space where a state is defined by a pair, the running sum, and the result of the last throw – then the process becomes Markovian again. It is also worth noting that in cases where time steps are discrete, the process is referred to as a Markov chain. In what follows we will deal with Markov chains, though in the next chapter we will study Markov processes with continuous time. For an extended discussion of Markov chains, see Feller (2008).

Let us denote by $\boldsymbol{P}$ the matrix describing the transition probabilities, i.e., $\boldsymbol{P}{i j}$ will be the probability to go from $i$ to $j$. The key insight to note is that for a Markov chain if we know the current probabilities to be at every site, which we will denote by the vector $\vec{p}$, and we know the matrix $P$, we can easily find the vector of probabilities to be at every site after an additional move. By the definition of the matrix $P$, the probability to be in the $i$ th site after an additional move is given by $$ p{i}^{n+1}=\sum_{j} p_{j}^{n} \boldsymbol{P}_{j i}
$$

概率模型和随机过程代考

数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|Random Walks in 1D

考虑一维晶格上的“随机游走者”。时间将是离散的，并且在每个时间步，步行者将以相等的概率朝可能的两个方向之一迈出一步。比如坐标的平均值是多少X，后ñ脚步？从对称性来看，它显然是零。但是我们当然不期望步行者在经过大量的步骤后精确地在原点。其远离原点的典型幅度可以通过计算轨迹的RMS（均方根）来找到，即距离原点的距离的二阶矩。

假设步行者在位置X后ñ脚步。在一个额外的步骤之后，它可能处于两个可能的位置之一。平均这两种可能性之间的平方距离，以位置为条件X有时ñ，给我们

12[(X+1)2+(X−1)2]=X2+1.
在平均超过的值X2有时ñ, 我们发现

⟨Xñ+12⟩=⟨Xñ2⟩+1
因此，重复论证ñ−1次，我们发现

⟨Xñ2⟩=ñ

这意味着与原点的典型距离像ñ. 位置分布呢？如果ñ是偶数，那么很明显是距离的概率米从起源之后ñ奇数的步数为零米, 甚至对于米这是

p米=12ñ(ñ R)=12ññ!R!(ñ−R)!
在哪里R是向右的步数，因此R−(ñ−R)=2R−ñ=米. 我们现在可以评估这个概率ñ≫米使用斯特林公式，它提供了一个（极好的）近似值ñ!，即ñ!≈2圆周率ñ(ñ和)ñ. 这将导致

p米≈12圆周率12ñññ+1/2RR+1/2(ñ−R)ñ−R+1/2 =12圆周率和−ñ日志⁡(2)+(ñ+1/2)日志⁡(ñ)−(R+1/2)日志⁡(R)−(ñ−R+1/2)日志⁡(ñ−R)
我们可以继续使用我们的假设：ñ≫米, 意味着R大约等于ñ/2.

数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|Derivation of the Diffusion Equation for Random Walks

我们现在将更普遍地处理这个问题，几乎完全遵循爱因斯坦的推导。我们将在3D但方法是一样的

在任何维度。该方法将具有离散的时间步长，但步长方向将是一个连续的随机变量，由概率分布描述G(Δ→)（这里Δ→是描述 3D 空间中步骤的向量——不要与拉普拉斯算子混淆！）。我们不会将随机游走者限制在一个格上，尽管可以通过采用G(Δ→)是总和d-functions（你能看到怎么做吗？）。

我们将寻求获得概率分布p(r→)， IE，p(r→)d在将是在体积中找到粒子的概率d在围绕点r→（在某个时间点对应于给定数量的步骤）。如果将原始问题投射到格上，则该分布将与描述粗粒度问题相关，当我们将缩小到足够远以至于我们不会关心格常数级别的细节时。

如果当时吨概率分布描述为p(r→,吨), 让我们考虑一下这将是什么时候τ后来，在哪里τ表示每个步骤的持续时间。正如您可以猜到的，在现实场景中，步骤的时间是非恒定的，并且τ将是平均步长。因此，我们在使时间离散方面并没有损失太多——但我们确实假设存在平均时间。在第 7 章中，我们将重新讨论这一点，并表明在某些情况下，当平均时间发散时，我们可以得到次扩散（与扩散相比，概率分布随时间的扩散更慢）。

寻找p(r→,吨+τ)，我们需要对所有空间进行积分，并考虑具有“正确”跳跃大小的概率以将我们带到r→. 这将导致

p(r→,吨+τ)=∫p(r→−Δ→,吨)d3Δ→G(Δ→).
为了继续，我们将泰勒展开p，假设概率分布在典型跳跃的尺度上是平滑的。如果我们把它展开到一阶，我们会发现

p(r→−Δ→)≈p(r→)−(∇p)⋅Δ→.
如果扩散过程在空间中是各向同性的，则在空间中的跳跃之间没有区别±Δ→方向，因此与第二项相关的积分微不足道地消失：

∫((∇p)⋅Δ→)G(Δ→)d3Δ→=0.
这意味着我们必须扩展到二阶：

p(r→−Δ→)≈p(r→)−(∇p)⋅Δ→+12∑一世,j∂2p∂X一世∂Xj|r→Δ一世Δj

数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|Markov Processes and Markov Chains

让我们从一个新的角度来分析随机游走者，这将更容易推广到其他网络。在我们的一维晶格扩散模型中，到达现场的概率j不依赖于随机游走者的历史，而只依赖于它的当前状态——这是马尔科夫过程的一个例子（以安德烈马尔科夫命名）。马尔可夫过程的一个熟悉的童年例子是滑梯和梯子游戏（以及许多其他棋盘游戏——另见习题 2.15）。但我们强调，一个过程是否是马尔可夫取决于空间：例如，考虑一个游戏，我们在每一轮掷骰子，并跟踪运行总和。这显然是一个马尔科夫过程——知道当前的总和决定了进入下一个状态的概率。但是，如果每次我们连续获得两个相同的数字时，我们将总和重置为零，那又会怎样呢？在这种情况下，该过程是依赖于内存的，因此它是非马尔可夫的。但是，如果我们在一个状态由一对定义的空间中工作，运行总和，以及最后一次抛出的结果——然后该过程再次变为马尔可夫。还值得注意的是，在时间步长离散的情况下，该过程称为马尔可夫链。接下来我们将处理马尔可夫链，尽管在下一章中我们将研究具有连续时间的马尔可夫过程。有关马尔可夫链的详细讨论，请参阅 Feller (2008)。

让我们用磷描述转移概率的矩阵，即磷一世j将是从一世至j. 要注意的关键见解是，对于马尔可夫链，如果我们知道每个站点的当前概率，我们将用向量表示p→，我们知道矩阵磷，我们可以很容易地找到在额外移动后在每个站点的概率向量。通过矩阵的定义磷, 出现在一世额外移动后的第一个站点由

p一世n+1=∑jpjn磷j一世

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

数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|MATH 106

Posted on 2022年6月18日2022年6月18日 by statistics-lab

如果你也在怎样代写概率模型和随机过程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。

随机过程是随机量在时间或空间上演变的概率模型。演变受不同时间或地点的随机量之间的某种依赖关系所支配。

我们提供的概率模型和随机过程及其相关学科的代写，服务范围广, 其中包括但不限于:

Statistical Inference 统计推断
Statistical Computing 统计计算
Advanced Probability Theory 高等概率论
Advanced Mathematical Statistics 高等数理统计学
(Generalized) Linear Models 广义线性模型
Statistical Machine Learning 统计机器学习
Longitudinal Data Analysis 纵向数据分析
Foundations of Data Science 数据科学基础

数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|MATH 106

数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|Assignment 3

Suppose $N=\left(N_{t}, t \geqslant 0\right)$ is a Poisson process with rate $\lambda=2$. Let $n_{1}=7, n_{2}=10$, $t_{1}=2$, and $t_{2}=12$.
(a) Determine $\mathbb{P}\left(N_{t_{1}}=n_{1}, N_{t_{2}}=n_{2}\right)$.
$[0.5]$
(b) Determine $\mathbb{P}\left(N\left(1, t_{1}\right]=n_{1}, N\left(t_{1}-1, t_{2}\right]=n_{2}\right)$.
$[0.5]$
(c) Determine $\mathbb{E}\left[N\left(1, t_{1}\right] \mid N\left(t_{1}-1, t_{2}\right]=n_{2}\right]$.
$[0.5]$
(d) Determine $\mathbb{E}\left[N\left(t_{1}-1, t_{2}\right] \mid N\left(1, t_{1}\right]=n_{1}\right]$.
$[0.5]$
Let $\left(N_{t}, t \geqslant 0\right)$ be a (non-homogeneous) Poisson counting process with rate function $\lambda(t)=1-\mathrm{e}^{-t}$. For $0 \leqslant s \leqslant t, m \in{0,1, \ldots, n}$, and for $n \in{0,1,2, \ldots}$, determine:
$$
\mathbb{P}\left(N_{s}=m \mid N_{t}=n\right)
$$
$[2]$
The Sturt Stony Desert is a “gibber” desert partly located in south-western Queensland (Australia) with an estimated area of $29750 \mathrm{~km}^{2}$. Suppose that you happen to be camping with friends and want to take a picture of the fat-tailed dunnart (Sminthopsis crassicaudata). One of your friends claims that its appearance in this area follows a spatial Poisson process with constant rate $0.5$ per $\mathrm{km}^{2}$. You’d really like to capture a spectacular photo of this unusual desert creature, but before setting out you want to make sure you have a reasonable chance of success if your friend’s claim is correct. You and your friend can carry enough water and food as well as your photographic gear to scour a $0.25 \mathrm{~km}^{2}$ area before you have to return to camp.
(a) If your friend is correct, what is the probability that you would see at least one fat-tailed dunnart on a single trip?
[1]
(b) How many trips (visiting distinct areas) should you expect to take before you have 6 sightings of the fat-tailed dunnart?
(c) Suppose a recent survey of put the total population of the fat-tailed dunnart in the Sturt Stony Desert at $5 \times 10^{5}$. Given this information, what is the conditional probability that you would see at least one fat-tailed dunnart on a single trip?
(d) * Suppose now that your friend also claims that if you see a fat-tailed dunnart, you’re more likely to see another one nearby. Is this consistent with your friend’s earlier claim? Argue why or why not.
[1]

数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|Solutions to Assignment 3

Suppose $N=\left(N_{t}, t \geqslant 0\right)$ is a Poisson process with rate $\lambda=2$. Let $n_{1}=7, n_{2}=10$, $t_{1}=2$, and $t_{2}=12$.
(a) Determine $\mathbb{P}\left(N_{t_{1}}=n_{1}, N_{t_{2}}=n_{2}\right)$.
Solution: We may compute:
$[0.5]$
$$
\mathbb{P}\left(N_{2}=7, N_{12}=10\right)=\mathbb{P}\left(N_{2}=7, N(2,12]=3\right)=\mathbb{P}\left(N_{2}=7\right) \times \mathbb{P}(N(2,12]=3)
$$
$$
\begin{aligned}
&=\mathbb{P}\left(N_{2}=7\right) \times \mathbb{P}\left(N_{10}=3\right)=\mathrm{e}^{-4} \
&=\mathrm{e}^{-24} \frac{4^{7} \times 20^{3}}{7 ! \times 3 !}=\mathrm{e}^{-24} \frac{131072000}{30240} \
&=\mathrm{e}^{-24} \frac{819200}{189} \approx 1.6363 \times 10^{-7}
\end{aligned}
$$
(b) Determine $\mathbb{P}\left(N\left(1, t_{1}\right]=n_{1}, N\left(t_{1}-1, t_{2}\right]=n_{2}\right)$.
Solution: In this case, the intervals overlap only at one end. In general, one may need to split in to three non-overlapping intervals $\left(1, t_{1}-1\right],\left(t_{1}-1, t_{1}\right]$, and $\left(t_{1}, t_{2}\right]$ and sum up over all the possible cases. Here, we find:
$$
\begin{aligned}
\mathbb{P}(N(1,2]=7, N(1,12]=10) &=\mathbb{P}(N(1,2]=7, N(2,12]=3) \
&=\mathbb{P}(N(1,2]=7) \times \mathbb{P}(N(2,12]=3) \
&=\mathbb{P}\left(N_{1}=7\right) \times \mathbb{P}\left(N_{10}=3\right) \
&=\mathrm{e}^{-2} \frac{2^{7}}{7 !} \times \mathrm{e}^{-20} \frac{20^{3}}{3 !}=\mathrm{e}^{-22} \frac{2^{7} \times 20^{3}}{7 ! \times 3 !} \
&=\mathrm{e}^{-22} \frac{1024000}{30240}=\mathrm{e}^{-22} \frac{6400}{189} \
& \approx 9.4458 \times 10^{-9}
\end{aligned}
$$
(c) Determine $\mathbb{E}\left[N\left(1, t_{1}\right] \mid N\left(t_{1}-1, t_{2}\right]=n_{2}\right]$.
Solution: In this case, we may directly determine that, for $x \in{0,1, \ldots, 10}$, we have
$$
\begin{aligned}
\mathbb{P}(N(1,2]=x \mid N(1,12]=10) &=\cdots=\frac{\mathbb{P}(N(1,2]=x) \mathbb{P}(N(2,12]=10-x)}{\mathbb{P}(N(1,12]=10)} \
&=\cdots=\left(\begin{array}{c}
10 \
x
\end{array}\right)\left(\frac{1}{11}\right)^{x}\left(1-\frac{1}{11}\right)^{10-x}
\end{aligned}
$$
and zero otherwise. That is, $(N(1,2] \mid N(1,12]=10) \sim \operatorname{Bin}(10,1 / 11)$. Thus, $\mathbb{E}[N(1,2] \mid N(1,12]=10]=10 / 11 \approx 0.9091$
(d) Determine $\mathbb{E}\left[N\left(t_{1}-1, t_{2}\right] \mid N\left(1, t_{1}\right]=n_{1}\right]$.
Solution: In this case, we have $\mathbb{E}[N(1,12] \mid N(1,2]=7]=7+\mathbb{E} N(2,12]$. Moreover, $N(2,12] \sim \operatorname{Poi}(20)$ so we conclude that $\mathbb{E}[N(1,12] \mid N(1,2]=7]=$ $7+20=27$
$[0.5]$

数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|Assignment 4

Consider an abstract probability space $(\Omega, \mathcal{F}, \mathbb{P})$. Answer the following questions.
(a) Suppose $A_{1}, A_{2}$, and $A_{3}$ form a partition of $\Omega$. Write down the smallest $\sigma$ algebra $\mathcal{F}{3}$ containing $A{1}, A_{2}$, and $A_{3}$. How many elements are in $\mathcal{F}{3}$ ? (b) Extending (a), suppose now $A{1}, A_{2}, \ldots, A_{n}$ form a partition of $\Omega$ for $n \in$ ${3,4,5, \ldots}$. What is the cardinality of the smallest $\sigma$-algebra, $\mathcal{F}{n}$, containing $A{1}, \ldots, A_{n} ?$
$[1]$
Show by way of example that if $\mathcal{F}$ and $\mathcal{G}$ are two $\sigma$-algebras of subsets of $\Omega$ then $\mathcal{H}=\mathcal{F} \cup \mathcal{G}$ is not in general also a $\sigma$-algebra of subsets of $\Omega$.
$[1]$
Consider an abstract space $\Omega$ and two events $A$ and $B$.
(a) In general, write down the smallest $\sigma$-algebra containing $A$ and $B$.
$[1]$
(b) If $A$ and $B$ are independent with $\mathbb{P}(A)=0.4$ and $\mathbb{P}(B)=0.5$, determine the prohahilities of all elements of the smallest $\sigma$-algehra found in (a).
[?]
Consider the Borel $\sigma$-algebra on $\mathbb{R}, \mathscr{B} \equiv \mathscr{B}(\mathbb{R})$; that is, the $\sigma$-algebra generated by all intervals of the form $(-\infty, b]$, for all $b \in \mathbb{R}$. Show that $\mathscr{B}$ contains sets of the form $[a, b)$ and $[a, b]$.
[1]
Using only the three Kolmogorov axioms of a probability measure, show that if $A$ and $B$ are events satisfying $A \subseteq B$ then $\mathbb{P}(A) \leqslant \mathbb{P}(B)$.
$[1]$
Let $(\Omega, \mathcal{F}, \mathbb{P})$ be a probability space, and $A_{1}, A_{2}, \ldots$ be an increasing sequence of events; that is, $A_{1} \subseteq A_{2} \subseteq \cdots$. Using only the Kolmogorov axioms, prove that $\mathbb{P}$ is continuous from below:
$$
\lim {n \rightarrow \infty} \mathbb{P}\left(A{n}\right)=\mathbb{P}\left(\cup_{n=1}^{\infty} A_{n}\right)
$$
Hint: Work with a new sequence of events $B_{1}:=A_{1}$ and $B_{n}:=A_{n} \backslash A_{n-1}$.
$[2]$
* Let $(\Omega, \mathcal{F}, \mathbb{P})$ be a probability space, and $A_{1}, A_{2}, \ldots$ be a decreasing sequence of events; that is, $A_{1} \supseteq A_{2} \supseteq \cdots$. Using the Kolmogorov axioms and/or the continuity from below property, prove that $\mathbb{P}$ is continuous from above:
$$
\lim {n \rightarrow \infty} \mathbb{P}\left(A{n}\right)=\mathbb{P}\left(\cap_{n=1}^{\infty} A_{n}\right)
$$
Let $(\Omega, \mathcal{F}, \mathbb{P})$ be a probability space and $X: \Omega \rightarrow \mathbb{R}$ be a random variable. Denote the Borel $\sigma$-algebra on $\mathbb{R}$ as $\mathscr{B}(\mathbb{R})$, and define the function $\mu_{X}(B)=\mathbb{P}\left(X^{-1}(B)\right)$ for all $B \in \mathscr{B}(\mathbb{R})$. Show that $\left(\mathbb{R}, \mathscr{B}(\mathbb{R}), \mu_{X}\right)$ is a probability space. Hint: Use the fact that $\mathbb{P}$ is a probability measure and the definition of a random variable.

概率模型和随机过程代考

数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|Assignment 3

认为ñ=(ñ吨,吨⩾0)是一个具有速率的 Poisson 过程λ=2. 让n1=7,n2=10, 吨1=2，和吨2=12.
(a) 确定磷(ñ吨1=n1,ñ吨2=n2).
[0.5]
(b) 确定磷(ñ(1,吨1]=n1,ñ(吨1−1,吨2]=n2).
[0.5]
(c) 确定和[ñ(1,吨1]∣ñ(吨1−1,吨2]=n2].
[0.5]
(d) 确定和[ñ(吨1−1,吨2]∣ñ(1,吨1]=n1].
[0.5]
让(ñ吨,吨⩾0)是具有速率函数的（非齐次）泊松计数过程λ(吨)=1−和−吨. 为了0⩽s⩽吨,米∈0,1,…,n，并且对于n∈0,1,2,…，决定：
磷(ñs=米∣ñ吨=n)
[2]
斯特石沙漠是部分位于昆士兰州西南部（澳大利亚）的“gibber”沙漠，估计面积为29750 ķ米2. 假设你碰巧和朋友一起露营，想拍一张肥尾邓纳特（Sminthopsis crassicaudata）的照片。你的一个朋友声称它在这个区域的出现遵循一个恒定速率的空间泊松过程0.5每ķ米2. 您真的很想为这种不寻常的沙漠生物拍摄一张壮观的照片，但在出发之前，您要确保如果您朋友的说法正确，您有合理的成功机会。您和您的朋友可以携带足够的水和食物以及您的摄影器材来冲刷0.25 ķ米2必须返回营地之前的区域。
(a) 如果你的朋友是正确的，你在一次旅行中至少看到一个肥尾邓纳特的概率是多少？
[1]
(b) 在您看到 6 次肥尾杜纳特之前，您应该预计进行多少次旅行（访问不同的区域）？
(c) 假设最近的一项调查将斯图特石质沙漠中的肥尾邓纳特总种群数量设为5×105. 给定这些信息，您在一次旅行中至少看到一个肥尾邓纳特的条件概率是多少？
(d) * 现在假设你的朋友也声称如果你看到一只肥尾杜纳特，你更有可能在附近看到另一个。这与你朋友之前的说法一致吗？争论为什么或为什么不。
[1]

数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|Solutions to Assignment 3

认为ñ=(ñ吨,吨⩾0)是一个具有速率的 Poisson 过程λ=2. 让n1=7,n2=10, 吨1=2，和吨2=12.
(a) 确定磷(ñ吨1=n1,ñ吨2=n2).
解决方案：我们可以计算：
[0.5]
磷(ñ2=7,ñ12=10)=磷(ñ2=7,ñ(2,12]=3)=磷(ñ2=7)×磷(ñ(2,12]=3)
=磷(ñ2=7)×磷(ñ10=3)=和−4 =和−2447×2037!×3!=和−2413107200030240 =和−24819200189≈1.6363×10−7
(b) 确定磷(ñ(1,吨1]=n1,ñ(吨1−1,吨2]=n2).
解决方案：在这种情况下，间隔仅在一端重叠。一般来说，一个人可能需要分成三个不重叠的区间(1,吨1−1],(吨1−1,吨1]，和(吨1,吨2]并总结所有可能的情况。在这里，我们发现：
磷(ñ(1,2]=7,ñ(1,12]=10)=磷(ñ(1,2]=7,ñ(2,12]=3) =磷(ñ(1,2]=7)×磷(ñ(2,12]=3) =磷(ñ1=7)×磷(ñ10=3) =和−2277!×和−202033!=和−2227×2037!×3! =和−22102400030240=和−226400189 ≈9.4458×10−9
(c) 确定和[ñ(1,吨1]∣ñ(吨1−1,吨2]=n2].
解：在这种情况下，我们可以直接确定，对于X∈0,1,…,10，我们有
磷(ñ(1,2]=X∣ñ(1,12]=10)=⋯=磷(ñ(1,2]=X)磷(ñ(2,12]=10−X)磷(ñ(1,12]=10) =⋯=(10 X)(111)X(1−111)10−X
否则为零。那是，(ñ(1,2]∣ñ(1,12]=10)∼垃圾桶⁡(10,1/11). 因此，和[ñ(1,2]∣ñ(1,12]=10]=10/11≈0.9091
(d) 确定和[ñ(吨1−1,吨2]∣ñ(1,吨1]=n1].
解决方案：在这种情况下，我们有和[ñ(1,12]∣ñ(1,2]=7]=7+和ñ(2,12]. 而且，ñ(2,12]∼接着⁡(20)所以我们得出结论和[ñ(1,12]∣ñ(1,2]=7]= 7+20=27
[0.5]

数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|Assignment 4

考虑一个抽象的概率空间(Ω,F,磷). 回答以下问题。
(a) 假设一个1,一个2，和一个3形成一个分区Ω. 写下最小的σ代数F3包含一个1,一个2，和一个3. 有多少元素F3? (b) 扩展 (a)，假设现在一个1,一个2,…,一个n形成一个分区Ω为了n∈ 3,4,5,…. 最小的基数是多少σ-代数，Fn, 包含一个1,…,一个n?
[1]
举例说明如果F和G是两个σ- 子集的代数Ω然后H=F∪G也不是一般的σ- 子集的代数Ω.
[1]
考虑一个抽象空间Ω和两个事件一个和乙.
(a) 一般来说，写下最小的σ-代数包含一个和乙.
[1]
(b) 如果一个和乙独立于磷(一个)=0.4和磷(乙)=0.5, 确定所有最小元素的概率σ-algehra 在 (a) 中发现。
[?]
考虑博雷尔σ-代数是R,乙≡乙(R); 那就是σ- 由形式的所有区间生成的代数(−∞,b]，对所有人b∈R. 显示乙包含表单集[一个,b)和[一个,b].
[1]
仅使用概率测度的三个 Kolmogorov 公理，证明如果一个和乙事件是否令人满意一个⊆乙然后磷(一个)⩽磷(乙).
[1]
让(Ω,F,磷)是一个概率空间，并且一个1,一个2,…是一个不断增加的事件序列；那是，一个1⊆一个2⊆⋯. 仅使用 Kolmogorov 公理，证明磷从下面连续：
林n→∞磷(一个n)=磷(∪n=1∞一个n)
提示：使用新的事件序列乙1:=一个1和乙n:=一个n∖一个n−1.
[2]
* 让(Ω,F,磷)是一个概率空间，并且一个1,一个2,…是一个递减的事件序列；那是，一个1⊇一个2⊇⋯. 使用 Kolmogorov 公理和/或来自下属性的连续性，证明磷从上面是连续的：
林n→∞磷(一个n)=磷(∩n=1∞一个n)
让(Ω,F,磷)是一个概率空间并且X:Ω→R是一个随机变量。表示 Borelσ-代数是R作为乙(R), 并定义函数μX(乙)=磷(X−1(乙))对所有人乙∈乙(R). 显示(R,乙(R),μX)是一个概率空间。提示：使用以下事实磷是概率测度和随机变量的定义。

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
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数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|MATH 355

Posted on 2022年6月18日2022年6月18日 by statistics-lab

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随机过程是随机量在时间或空间上演变的概率模型。演变受不同时间或地点的随机量之间的某种依赖关系所支配。

我们提供的概率模型和随机过程及其相关学科的代写，服务范围广, 其中包括但不限于:

Statistical Inference 统计推断
Statistical Computing 统计计算
Advanced Probability Theory 高等概率论
Advanced Mathematical Statistics 高等数理统计学
(Generalized) Linear Models 广义线性模型
Statistical Machine Learning 统计机器学习
Longitudinal Data Analysis 纵向数据分析
Foundations of Data Science 数据科学基础

数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|MATH 355

数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|Solutions to Assignment 1

Let $X$ and $Y$ be random variables defined on a common probability space. Assuming $\operatorname{Var}(X)<\infty$, show that $\operatorname{Var}(X)=\mathbb{E} \operatorname{Var}(X \mid Y)+\operatorname{Var}(\mathbb{E}[X \mid Y])$. [Hint: Use defn. of $\operatorname{Var}(X)$ and conditional expectation tricks.]

Solution: Recall that provided $\mathbb{E}|Z|<\infty$, we have that $\mathbb{E} Z=\mathbb{E E}[Z \mid Y]$. Applying this to $X$ and $X^{2}$, we know that $\mathbb{E} X=\mathbb{E}[X \mid Y]$ and $\mathbb{E} X^{2}=\mathbb{E E}\left[X^{2} \mid Y\right]$, both of which exist and are finite since $\operatorname{Var}(X)<\infty$ by assumption. Now, $\operatorname{Var}(X)=$ $\mathbb{E} X^{2}-(\mathbb{E} X)^{2}$, and $\operatorname{Var}(X \mid Y)=\mathbb{E}\left[X^{2} \mid Y\right]-(\mathbb{E}[X \mid Y])^{2}$. Hence
$$
\begin{aligned}
\operatorname{Var}(X) &=\mathbb{E} E\left[X^{2} \mid Y\right]-(\mathbb{E} E[X \mid Y])^{2} \
&=\mathbb{E E}\left[X^{2} \mid Y\right]-\mathbb{E}\left[(\mathbb{E}[X \mid Y])^{2}\right]+\mathbb{E}\left[(\mathbb{E}[X \mid Y])^{2}\right]-(\mathbb{E} E[X \mid Y])^{2} \
&=\mathbb{E} \operatorname{Var}(X \mid Y)+\operatorname{Var}(\mathbb{E}[X \mid Y])
\end{aligned}
$$
$[1]$

Let $X$ be a non-negative random variable with probability density function (pdf) $f$
(a) Show that $\mathbb{E} X=\int_{0}^{\infty} \mathbb{P}(X \geqslant x) \mathrm{d} x$.
Solution: Note that $\mathbb{P}(X \geqslant x)=\int_{x}^{\infty} f(u)$ du. Hence
$$
\begin{aligned}
\int_{0}^{\infty} \mathbb{P}(X \geqslant x) \mathrm{d} x &=\int_{0}^{\infty} \int_{x}^{\infty} f(u) \mathrm{d} u \mathrm{~d} x \
&=\int_{0}^{\infty} f(u) \int_{0}^{u} 1 \mathrm{~d} x \mathrm{~d} u \
&=\int_{0}^{\infty} u f(u) \mathrm{d} u=\mathbb{E} X
\end{aligned}
$$
where the second line follows from swapping integration order.
$[1]$
(b) Using (a), show that $\mathbb{E}\left[X^{\alpha}\right]=\int_{0}^{\infty} \alpha x^{\alpha-1} \mathbb{P}(X \geqslant x) \mathrm{d} x$ for any $\alpha>0$.
Solution: Write $Y=X^{\alpha}$ which is still a non-negative random variable with some pdf. Then, from (a), we know that $\mathbb{E Y = \int _ { 0 } ^ { \infty } \mathbb { P } ( Y \geqslant y ) \mathrm { d } y \text { . Change }}$ variables via $y=x^{\alpha}$ so $\mathrm{d} y=\alpha x^{\alpha-1} \mathrm{~d} x$; note that $x=y^{1 / \alpha}$ has the same limits as $y$ for any $\alpha>0$. Hence $\mathbb{E}\left[X^{\alpha}\right]=\mathbb{E} Y=\int_{0}^{\infty} \mathbb{P}(X \geqslant x) \alpha x^{\alpha-1} \mathrm{~d} x$. A minor rearrangement yields the result.
Suppose $X_{1}, X_{2}, \ldots, X_{n}$ are independent random variables, with cdfs $F_{1}, F_{2}, \ldots$, $F_{n}$, respectively. Express the cdf of $M=\min \left(X_{1}, \ldots, X_{n}\right)$ in terms of the $\left{F_{i}\right}$.

数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|Assignment 2

Suppose a stochastic wallaby is hopping along-side the length of an infinitely long road in search of a tasty snack in the form of Livistona Rand. palm seeds. Each hop that our stochastic wallaby takes is of size 1. Our stochastic wallaby hops up the road with probability $p \in(0,1)$, and down the road with probability $q=1-p$.
(a) Suppose there is a single Livistona Rand. palm which has dropped its seeds $u$ hops up the road from where our stochastic wallaby currently is, and no such palm anywhere down the road. As a function of $p, q$, and $u$, what is the probability that the wallaby ever gets the opportunity to have its tasty snack of palm seeds?
(b) Continuing, suppose $p=q=\frac{1}{2}$. As a function of $u$, what is the expected number of hops required?
(c) * Continuing, as a function of $u$, what is the variance of the number of hops required?
(d) * Now for general $p \in(0,1)$, determine the expected number of hops and the variance of the number of hops required as a function of $p, q$, and $u$.
(e) Suppose now that is an additional Livistona Rand. palm which has dropped its seeds $d$ hops down the road from where our stochastic wallaby currently is. As a function of $p, q, u$, and $d$, what is the probability that the wallaby ever gets the opportunity to have its tasty snack of palm seeds?
$[1]$
(f) Continuing, as a function of $p, q, u$, and $d$, what is the expected number of hops required?
Let $X=\left(X_{n}, n=0,1, \ldots\right)$ be a Markov chain with state-space $E={0,1,2}$, initial distribution $\pi^{(0)}=(0,1,0)$, and one-step transition matrix
$$
\mathbf{P}=\frac{1}{15}\left(\begin{array}{lll}
8 & 1 & 6 \
3 & 5 & 7 \
4 & 9 & 2
\end{array}\right)
$$
(a) Draw the transition diagram for this Markov chain.
$[1]$
(b) Calculate the probability that $X_{3}=1$.
$[1]$
(c) Find the unique stationary (and limiting) distribution of the chain.
$[1]$

数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|Solutions to Assignment 2

Suppose a stochastic wallaby is hopping along-side the length of an infinitely long road in search of a tasty snack in the form of Livistona Rand. palm seeds. Each hop that our stochastic wallaby takes is of size 1. Our stochastic wallaby hops up the road with probability $p \in(0,1)$, and down the road with probability $q=1-p$.
(a) Suppose there is a single Livistona Rand. palm which has dropped its seeds $u$ hops up the road from where our stochastic wallaby currently is, and no such palm anywhere down the road. As a function of $p, q$, and $u$, what is the probability that the wallaby ever gets the opportunity to have its tasty snack of palm seeds?

Solution: Let $\left(S_{n}, n=0,1, \ldots\right)$ be the position of the stochastic wallaby, and denote its initial location as 0 ; that is, $S_{0}=0$. Then its position on the $(n+1)$-st hop can be written in terms of its position at the $n$-th step as $S_{n+1}=S_{n}+2 B_{n+1}-1$, for $n=0,1,2, \ldots$, where $B_{1}, B_{2}, \ldots$ iid $\operatorname{Ber}(p)$. Denote the (random) time at which the stochastic wallaby first visits position $x$ as $\tau_{x}=\inf \left{n \in \mathbb{N}: S_{n}=x\right}$. Then we seek $\mathbb{P}{0}\left(\tau{u}<\infty\right)$.

Let us introduce a new Livistona Rand. palm at position $-d$, and consider $r_{x}^{u,-d}=\mathbb{P}{x}\left(\tau{u}<\tau_{-d}\right)$. Then in particular we know that $\lim {d \rightarrow \infty} \tau{-d} \rightarrow \infty$, and so the original quantity of interest can be obtained as $\mathbb{P}{0}\left(\tau{u}<\infty\right)=$ $\lim {d \rightarrow \infty} r{0}^{u,-d}$.
Now, we know $r_{u}^{u,-d}=\mathbb{P}{u}\left(\tau{u}<\tau_{-d}\right)=1$ and $r_{-d}^{u,-d}=\mathbb{P}{-d}\left(\tau{u}<\tau_{-d}\right)=0$.
By one step analysis we also have that, for $-d<x<u$,
$$
r_{x}^{u,-d}=p r_{x+1}^{u,-d}+q r_{x-1}^{u,-d} .
$$
Recalling $1=p+q$, we may rearrange this to read
$$
(p+q) r_{x}^{u,-d}=p r_{x+1}^{u,-d}+q r_{x-1}^{u,-d} \Longleftrightarrow \underbrace{\left(r_{x+1}^{u,-d}-r_{x}^{u,-d}\right)}{v{x+1}}=\underbrace{\frac{q}{p}}{e} \cdot \underbrace{\left(r{x}^{u,-d}-r_{x-1}^{u,-d}\right)}{v{x}} .
$$
That is, $v_{x+1}=\varrho v_{x}$. Repeated application of this recursion yields $v_{x}=$ $\varrho^{x+d-1} v_{-(d-1)}$

Now, we also have (dropping the superscripts for notational convenience) that
$$
r_{x}=r_{x}-r_{-d}=\sum_{y=-(d-1)}^{x} v_{y},
$$
and so
$$
\begin{aligned}
r_{x} &=v_{-(d-1)} \sum_{y=-(d-1)}^{x} \varrho^{y+d-1}=\left(r_{-(d-1)}-r_{-d}\right) \sum_{z=0}^{x+d-1} \varrho^{z} \
&=r_{-(d-1)} \times \begin{cases}\frac{1-g^{z+d}}{1-\varrho}, & \varrho \neq 1, \
(x+d), & \varrho=1 .\end{cases}
\end{aligned}
$$

概率模型和随机过程代考

数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|Solutions to Assignment 1

让X和是是在公共概率空间上定义的随机变量。假设曾是⁡(X)<∞，显示曾是⁡(X)=和曾是⁡(X∣是)+曾是⁡(和[X∣是]). [提示：使用定义。的曾是⁡(X)和条件期望技巧。]

解决方案：回想一下提供的和|从|<∞, 我们有和从=和和[从∣是]. 将此应用于X和X2，我们知道和X=和[X∣是]和和X2=和和[X2∣是]，两者都存在并且是有限的，因为曾是⁡(X)<∞通过假设。现在，曾是⁡(X)= 和X2−(和X)2，和曾是⁡(X∣是)=和[X2∣是]−(和[X∣是])2. 因此

曾是⁡(X)=和和[X2∣是]−(和和[X∣是])2 =和和[X2∣是]−和[(和[X∣是])2]+和[(和[X∣是])2]−(和和[X∣是])2 =和曾是⁡(X∣是)+曾是⁡(和[X∣是])
[1]

让X是具有概率密度函数的非负随机变量 (pdf)F
(a) 证明和X=∫0∞磷(X⩾X)dX.
解决方案：注意磷(X⩾X)=∫X∞F(在)你。因此
∫0∞磷(X⩾X)dX=∫0∞∫X∞F(在)d在 dX =∫0∞F(在)∫0在1 dX d在 =∫0∞在F(在)d在=和X
其中第二行来自交换集成顺序。
[1]
(b) 使用 (a)，证明和[X一个]=∫0∞一个X一个−1磷(X⩾X)dX对于任何一个>0.
解决方案：写是=X一个这仍然是一个带有一些 pdf 的非负随机变量。那么，从（a），我们知道和是=∫0∞磷(是⩾是)d是 . 改变变量通过是=X一个所以d是=一个X一个−1 dX; 注意X=是1/一个具有相同的限制是对于任何一个>0. 因此和[X一个]=和是=∫0∞磷(X⩾X)一个X一个−1 dX. 一个小的重排产生结果。
认为X1,X2,…,Xn是独立的随机变量，用 cdfsF1,F2,…, Fn，分别。表达 cdf 的米=分钟(X1,…,Xn)方面\left{F_{i}\right}\left{F_{i}\right}.

数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|Assignment 2

假设一只随机的小袋鼠沿着一条无限长的道路跳跃，以寻找 Livistona Rand 形式的美味小吃。棕榈种子。我们的随机小袋鼠跳的每一跳的大小都是 1。我们的随机小袋鼠有概率在路上跳跃p∈(0,1), 并且有概率地走下去q=1−p.
(a) 假设有一个 Livistona Rand。落下种子的棕榈在从我们的随机小袋鼠目前所在的地方跳上马路，而且在路上的任何地方都没有这样的手掌。作为一个函数p,q，和在，小袋鼠有机会吃到美味的棕榈籽零食的概率是多少？
(b) 继续，假设p=q=12. 作为一个函数在，所需的预期跳数是多少？
(c) * 继续，作为以下函数的函数在，所需跳数的方差是多少？
(d) * 现在一般p∈(0,1), 确定预期的跳数和所需跳数的方差p,q，和在.
(e) 现在假设这是一个额外的 Livistona Rand。落下种子的棕榈d从我们的随机小袋鼠目前所在的地方跳下来。作为一个函数p,q,在，和d，小袋鼠有机会吃到美味的棕榈籽零食的概率是多少？
[1]
(f) 继续，作为p,q,在，和d，所需的预期跳数是多少？
让X=(Xn,n=0,1,…)是具有状态空间的马尔可夫链和=0,1,2, 初始分布圆周率(0)=(0,1,0), 和一步转移矩阵
磷=115(816 357 492)
(a) 画出这条马尔可夫链的转移图。
[1]
(b) 计算概率X3=1.
[1]
(c) 找出链的唯一平稳（和限制）分布。
[1]

数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|Solutions to Assignment 2

假设一只随机的小袋鼠沿着一条无限长的道路跳跃，以寻找 Livistona Rand 形式的美味小吃。棕榈种子。我们的随机小袋鼠跳的每一跳的大小都是 1。我们的随机小袋鼠有概率在路上跳跃p∈(0,1), 并且有概率地走下去q=1−p.
(a) 假设有一个 Livistona Rand。落下种子的棕榈在从我们的随机小袋鼠目前所在的地方跳上马路，而且在路上的任何地方都没有这样的手掌。作为一个函数p,q，和在，小袋鼠有机会吃到美味的棕榈籽零食的概率是多少？

解决方案：让(小号n,n=0,1,…)是随机小袋鼠的位置，并将其初始位置表示为 0 ；那是，小号0=0. 然后它的位置在(n+1)-st hop 可以根据它在n- 步骤为小号n+1=小号n+2乙n+1−1，为了n=0,1,2,…，在哪里乙1,乙2,…独立同居贝尔⁡(p). 表示随机小袋鼠首次访问位置的（随机）时间X作为\tau_{x}=\inf \left{n \in \mathbb{N}: S_{n}=x\right}\tau_{x}=\inf \left{n \in \mathbb{N}: S_{n}=x\right}. 然后我们寻求磷0(τ在<∞).

让我们介绍一个新的 Livistona Rand。掌心位置−d, 并考虑rX在,−d=磷X(τ在<τ−d). 然后我们特别知道林d→∞τ−d→∞，因此原始感兴趣的数量可以得到为磷0(τ在<∞)= 林d→∞r0在,−d.
现在，我们知道r在在,−d=磷在(τ在<τ−d)=1和r−d在,−d=磷−d(τ在<τ−d)=0.
通过一步分析，我们也有，因为−d<X<在,

rX在,−d=prX+1在,−d+qrX−1在,−d.
召回1=p+q，我们可以重新排列它来阅读

(p+q)rX在,−d=prX+1在,−d+qrX−1在,−d⟺(rX+1在,−d−rX在,−d)⏟在X+1=qp⏟和⋅(rX在,−d−rX−1在,−d)⏟在X.
那是，在X+1=ϱ在X. 重复应用此递归会产生在X= ϱX+d−1在−(d−1)

现在，我们也有（为了符号方便，去掉上标）

rX=rX−r−d=∑是=−(d−1)X在是,
所以

rX=在−(d−1)∑是=−(d−1)Xϱ是+d−1=(r−(d−1)−r−d)∑和=0X+d−1ϱ和 =r−(d−1)×{1−G和+d1−ϱ,ϱ≠1, (X+d),ϱ=1.

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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