### 统计代写|随机过程代写stochastic process代考|In a box

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。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (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

## 统计代写|随机过程代写stochastic process代考|One ball is drawn

EXERCISE 1.5. (a) In a box, there are two red (R), four blue (B), and eight green (G) balls. One ball is drawn at a time without replacement and its color is noted. The stochastic process $\left{X_{n}, n=1,2, \ldots\right}$ with the state space $S={R, B, G}$ doesn’t satisfy the Markovian property. It can be proved, for example, as $P\left(X_{3}=G \mid X_{1}=R, X_{2}=B\right)=\frac{6}{12}$, whereas $P\left(X_{3}=G \mid X_{1}=G, X_{2}=B\right)=$ $\frac{7}{12}$, thus, the color of the ball drawn at the third step depends on the colors of all previously drawn balls, not just the one drawn at step two.

(b) If the drawing is done with replacement, the process is a Markov chain. Since the balls are put back into the box, the colors of drawn balls are independent of each other. Let $C$ stand for any of the three colors: red, blue, or green. Then we can write
$$P\left(X_{3}=C \mid X_{1}=C, X_{2}=C\right)=P\left(X_{3}=C\right)=P\left(X_{3}=C \mid X_{2}=C\right),$$
and thus, the Markov property always holds. Note that a sequence of independent trials is a special case of a Markov chain.

EXERCISE 1.6. Let $O$ denote any outcome of a coin flip. The flips are considered independent, therefore, we obtain
$$P\left(X_{3}=O \mid X_{1}=0, X_{2}=O\right)=P\left(X_{3}=0\right)=P\left(X_{3}=O \mid X_{2}=0\right),$$
that is, the Markovian property always holds. The coin is fair, hence, the transition probability matrix is
where $\pi_{H}+\pi_{T}=1$. The solution is $\pi_{H}=\pi_{T}=0.5$. Note that a sequence of independent trials is a special case of a Markov chain.

## 统计代写|随机过程代写stochastic process代考|Below we find the transition

EXERCISE 1.7. (a) Below we find the transition probability matrix for Chapter 1 of “Moby Dick” by Herman Melville.
library (tidyverse)
chapter $1<-$ read_file (“./Loomings. txt”)

###### cleaning the text

lowercase<- tolower (chapterl)
no.blanks<- gsub (” “,” , lowercase)
no. line.breaks<- gsub (” $\backslash I \backslash \cap “, \quad n “$, no.blanks)

###### removing all punctuation

clean.string<- gsub(” [ [:punct: $]]^{\prime \prime}, “$, no.line.breaks)

###### splitting the string into characters

x $2<-$ strsplit (clean.string, “*)

###### shifting the text by one place

no.last<- substr(clean.string, 1 , nchar(clean.string)-1)
first.blank<- str_c(” “, no.llast)
$x 1<-$ strsplit (first.blank, “*)
vowels<-c (“a”, “e”, “in, “o”, “u”)
$\left.{ }^{n} \mathrm{~V} “, ” w^{\prime \prime}, ” \mathrm{x}^{n}, ” \mathrm{y}^{\prime \prime}, ” \mathrm{z}^{\prime \prime}\right)$
for (counter in $1: \operatorname{nchar}(x 2)){$
$v<-$ ifelse (x2 [[counter]] 8 ins vowels, 1,0$)$
cx- ifelae $(x 2[[$ cumber $]$ ) ing cunsonanLs, 1,0$)$

vv<- ifelse(x1[[counter]] \&ins vowels \& x2[[counter]] sin\& vowels, 1,0$)$
vc<- ifelse(x1 [[counter]] 8 in vowels \& x2[[counter]] \&ins consonants, 1, 0)
$c v<-$ ifelse $(x 1[[c o u n t e r]]$ \&ins consonants \& $x 2$ [[counter]] 8ins vowels, 1, 0)
cc<- ifelse $(x 1[$ [counter]] \&ins consonants \& $x 2[[$ counter]] sin\& consonants, 1,0$)$
1
sum (v)
3647
sum (c)
5871
sum (vv)
572
sum (vc)
3075
sum (cv)
3075
sum (cc)
2795
We check quickly that these numbers add up properly. Since the first chapter of “Moby Dick” starts and ends with consonants, all vowels are transitioned into and transitioned from. Thus, sum $(\mathrm{v})=3647=$ sum $(\mathrm{vV})+\operatorname{sum}(\mathrm{vc})=572+3075=\operatorname{sum}(\mathrm{vV})+\operatorname{sum}(\mathrm{cv}) .$ Also, all but the last consonant are transitioned from, therefore, sum (c) $-1=5870=$ sum (cv) $+$ sum $(\mathrm{cc})=3075$ $+2795$, and all but the first consonant are transitioned to, so sum (c) $-1=5870=s u m$ (vc) $+$ sum $(c c)=3075+2795$. The transition probability matrix is

## 统计代写|随机过程代写stochastic process代考|The code below computes

The code below computes the limiting probabilities and the proportions of vowels and consonants in the text.

###### specifying the transition probability matrix

tm<- matrix (c (sum(vv)/sum $(v), \operatorname{sum}(v c) / \operatorname{sum}(v), \operatorname{sum}(\mathrm{cv}) /(\operatorname{sum}(c)-1)$,
sum $(c c) /(s u m(c)-1))$, nrow=2, ncol=2, byrow=TRUE)

###### creating Markov chain object

library (markovchain)
mc<- new (“markovchain”, transitionMatrix=tm, states=c $\left.\left({ }^{n} v^{\prime \prime}, ” c^{*}\right)\right)$

###### computing limiting probabilities

$$\begin{array}{rr} v & c \ 0.383209 & 0.616791 \end{array}$$

###### computing proportions of vowels and consonants

print (prop. vowels<- $\operatorname{sum}(v) /(\operatorname{sum}(v)+\operatorname{sum}(c)))$
$0.3831687$
print (prop. cons<- sum (c) / (sum (v) $+\operatorname{sum}(c)))$
$0.6168313$
From the output, the limiting probabilities are equal to the actual proportions of vowels and consonants.

## 统计代写|随机过程代写stochastic process代考|One ball is drawn

(b) 如果绘图是用替换完成的，则该过程是马尔可夫链。由于球被放回盒子里，所以抽出的球的颜色是相互独立的。让C代表三种颜色中的任何一种：红色、蓝色或绿色。然后我们可以写

## 统计代写|随机过程代写stochastic process代考|Below we find the transition

###### 清理文本

no.blanks<- gsub (” “,” , 小写)

###### 删除所有标点符号

clean.string <- gsub (”[[: dot:]]′′,“, no.line.breaks)

###### 将字符串拆分为字符

X2<−strsplit (clean.string, “*)

###### 将文本移动一个位置

no.last<- substr(clean.string, 1 , nchar(clean.string)-1)
first.blank<- str_c(” “, no.llast)
X1<−strsplit (first.blank, “*)

n 在“,”在′′,”Xn,”是′′,”和′′)
for ($1 中的计数器：\operatorname{nchar}(x 2)){v <-一世F和ls和(X2[[C这在n吨和r]]8一世ns在这在和ls,1,0)CX−一世F和l一种和（×2[[C在米b和r])一世nGC在ns这n一种n大号s,1,0)$

vv<- ifelse(x1[[counter]] \&ins 元音 \& x2[[counter]] sin\& 元音, 1,0)
vc<- ifelse(x1 [[counter]] 8 in vowels \& x2[[counter]] \&ins consonants, 1, 0)
C在<−如果别的(X1[[C这在n吨和r]]\&ins 辅音\&X2[[counter]] 8ins 元音，1, 0)
cc<- ifelse(X1[[计数器]] \&ins 辅音 \&X2[[计数器]] sin\& 辅音，1,0)
1
sum (v)
3647
sum (c)
5871
sum (vv)
572
sum (vc)
3075
sum (cv)
3075
sum (cc)
2795

## 统计代写|随机过程代写stochastic process代考|The code below computes

###### 指定转移概率矩阵

tm<- 矩阵 (c (sum(vv)/sum(在),和⁡(在C)/和⁡(在),和⁡(C在)/(和⁡(C)−1),

###### 创建马尔可夫链对象

mc<- new (“markovchain”, transitionMatrix=tm, states=c(n在′′,”C∗))

###### 计算元音和辅音的比​​例

print (prop. vowels<-和⁡(在)/(和⁡(在)+和⁡(C)))
0.3831687
print (prop. cons<- sum (c) / (sum (v)+和⁡(C)))
0.6168313

## 有限元方法代写

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。

## 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。