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

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。

• 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

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}

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

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.

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

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

pX∣是(X∣是)=磷X=X∣是=是 =磷X=X,是=是磷是=是 =p(X,是)p是(是)

FX∣是(X∣是)=磷X≤X∣是=是 =∑一个≤XpX∣是(一个∣是)

=∑XXpX∣是(X∣是)

pX∣是(X∣是)=磷X=X∣是=是 =磷X=X

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。

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

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。

• 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

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.

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

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)

p是(是)=∑X:p(X,是)>0p(X,是)

pX(X)=∑是:p(X,是)>0p(X,是)

p是(是)=∑X:p(X,是)>0p(X,是)

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

F(一个,b)=FX(一个)F是(b) 对所有人 一个,b

p(X,是)=pX(X)p是(是)

F(X,是)=FX(X)F是(是)

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

$$X=\left{ 1, 如果 一个 发生 0, 否则， \四Y= {1, 如果 乙 发生 0, 否则 \正确的。 吨H和n, \operatorname{Cov}(X, Y)=E[XY]-E[X] E[Y] 一个nd,b和C一个在s和X是在一世ll和q在一个l1○r0d和p和nd一世nG○n在H和吨H和r○rn○吨b○吨HX一个nd是和q在一个l1,在和s和和吨H一个吨 \ 操作员名称 {Cov} (X, Y) = P {X = 1, Y = 1} -P {X = 1} P {Y = 1} Fr○米吨H一世s在和s和和吨H一个吨 这⁡(X,是)>0⇔磷X=1,是=1>磷X=1磷是=1 ⇔磷X=1,是=1磷X=1>磷是=1 ⇔磷是=1∣X=1>磷是=1$$

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。

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

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。

• 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

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
$$1.7=E\left[X^{2}\right] \neq(E[X])^{2}=1.21$$

p(1)=12=p(2)

p(1)=13,p(2)=23

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

\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

p(0)=0.2,p(1)=0.5,p(2)=0.3

\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}

1.7=和[X2]≠(和[X])2=1.21

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。

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

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。

• 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

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.

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

p(n)=磷X=n=(1−p)n−1p,n=1,2,…

∑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

p(一世)=磷X=一世=和−λλ一世一世!,一世=0,1,…

∑一世=0∞p(一世)=和−λ∑一世=0∞λ一世一世!=和−λ和λ=1

(1−λn)n≈和−λ,n(n−1)⋯(n−一世+1)n一世≈1,(1−λn)一世≈1

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

1=磷X∈(−∞,∞)=∫−∞∞F(X)dX

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。

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

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。

• 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

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 $\left(\begin{array}{l}3 \ 2\end{array}\right) p^{2}(1-p)$.

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

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}

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

p(一个)=磷X=一个

p(X一世)>0,一世=1,2,… p(X)=0, 所有其他值 X

∑一世=1∞p(X一世)=1

F(一个)=∑全部 p(X一世≤一个

p(1)=12,p(2)=13,p(3)=16

F(一个)={0,一个<1 12,1≤一个<2 56,2≤一个<3 1,3≤一个

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

p(一世)=(n 一世)p一世(1−p)n−一世,一世=0,1,…,n

(n 一世)=n!(n−一世)!一世!

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。

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

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。

• 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

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.

1. 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.
2. 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, $i=1,2,3,4,5,6$.
3. 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.

1. 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 $P(F)>0$.

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

1. 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。
2. 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.
3. 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

1. 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

(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

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。

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

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。

• 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

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 $$p{i}^{n+1}=\sum_{j} p_{j}^{n} \boldsymbol{P}_{j i}$$

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

12[(X+1)2+(X−1)2]=X2+1.

⟨Xñ+12⟩=⟨Xñ2⟩+1

⟨Xñ2⟩=ñ

p米=12ñ(ñ R)=12ññ!R!(ñ−R)!

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)

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

p(r→,吨+τ)=∫p(r→−Δ→,吨)d3Δ→G(Δ→).

p(r→−Δ→)≈p(r→)−(∇p)⋅Δ→.

∫((∇p)⋅Δ→)G(Δ→)d3Δ→=0.

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

p一世n+1=∑jpjn磷j一世

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。

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

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。

• 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

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

1. 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 $\mathbb{P}\left(N_{t_{1}}=n_{1}, N_{t_{2}}=n_{2}\right)$.
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(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 $\mathbb{E}\left[N\left(1, t_{1}\right] \mid N\left(t_{1}-1, t_{2}\right]=n_{2}\right]$.
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(d) Determine $\mathbb{E}\left[N\left(t_{1}-1, t_{2}\right] \mid N\left(1, t_{1}\right]=n_{1}\right]$.
$[0.5]$
2. 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:
$$\mathbb{P}\left(N_{s}=m \mid N_{t}=n\right)$$
$[2]$
3. 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

1. 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

1. 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 $A{1}, \ldots, A_{n} ?$
$[1]$
2. 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 $\Omega$.
$[1]$
3. 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).
[?]
4. 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]
5. Using only the three Kolmogorov axioms of a probability measure, show that if $A$ and $B$ are events satisfying $A \subseteq B$ then $\mathbb{P}(A) \leqslant \mathbb{P}(B)$.
$[1]$
6. 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 $B_{n}:=A_{n} \backslash A_{n-1}$.
$[2]$
7. * 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:
$$\lim {n \rightarrow \infty} \mathbb{P}\left(A{n}\right)=\mathbb{P}\left(\cap_{n=1}^{\infty} A_{n}\right)$$
8. 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.

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

1. 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].
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(d) 确定和[ñ(吨1−1,吨2]∣ñ(1,吨1]=n1].
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2. 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]
3. 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

1. 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

1. 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]
2. 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]
3. 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。
[?]
4. 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]
5. 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]
6. 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]
7. * 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)
8. 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。

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。

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

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。

• 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

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

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}
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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.
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(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.
2. 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

1. 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?
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(f) Continuing, as a function of $p, q, u$, and $d$, what is the expected number of hops required?
2. 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.
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(b) Calculate the probability that $X_{3}=1$.
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(c) Find the unique stationary (and limiting) distribution of the chain.
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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|Solutions to Assignment 2

1. 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}

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

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。]

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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。
2. 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

1. 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？
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(f) 继续，作为p,q,在， 和d，所需的预期跳数是多少？
2. 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。
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(b) 计算概率X3=1.
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(c) 找出链的唯一平稳（和限制）分布。
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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|Solutions to Assignment 2

1. 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？

rX在,−d=prX+1在,−d+qrX−1在,−d.

(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.

rX=rX−r−d=∑是=−(d−1)X在是,

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.

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。