### 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。