Posted on Categories:Probability and Statistics, 概率与统计, 统计代写, 统计代考

## avatest™帮您通过考试

avatest™的各个学科专家已帮了学生顺利通过达上千场考试。我们保证您快速准时完成各时长和类型的考试，包括in class、take home、online、proctor。写手整理各样的资源来或按照您学校的资料教您，创造模拟试题，提供所有的问题例子，以保证您在真实考试中取得的通过率是85%以上。如果您有即将到来的每周、季考、期中或期末考试，我们都能帮助您！

•最快12小时交付

•200+ 英语母语导师

•70分以下全额退款

## 统计代写|概率与统计代考Probability and Statistics代写|INTRODUCTION

Consider a situation where we want to evaluate the probability $P(A)$ of some event $A$. Suppose that after finding $P(A)$, we learn that some other event, $B$, occurred. In many cases, such an information leads to a change in the assessment of the probability of the event $A$. The symbol used for this new probability will be $P(A \mid B)$, to be read “conditional probability of $A$, given $B$,” or “probability of event $A$, given that $B$ occurred.”
EXAMPLE $4.1$
Conditional probabilities are most easily interpreted as probabilities in subpopulations. Consider an attribute such as color blindness, known to occur much more often among men than among women. If $D$ is the event “a randomly selected person is color blind,” then $P(D)$ refers to the chance of color blindness in the whole population. Suppose now that the person selected is known to be a woman (event $W$ ). This information changes the assessment of probability of color blindness to $P(D \mid W)$, which is now the probability of color blindness in the subpopulation of women.
Questions that might arise here are the following:

How to use data on probabilities of color blindness separately among men and women to find the overall chance of color blindness, that is, to find $P(D)$ if we know $P(D \mid W)$ and $P(D \mid M)$ ?

How to find the probability that a randomly selected color blind person is a woman, that is, $P(W \mid D)$ ?

The first of these questions requires using the weighted average, usually referred to as the formula for total probability (Section 4.3). To answer the second question, one has to use the Bayes’ formula (Section 4.4).
The examples and exercises in this chapter are designed to provide practice in recognizing which probabilities are conditional and which are not.

## 统计代写|概率与统计代考Probability and Statistics代写|PROBLEMS

A computer file contains data on households in a certain city. Each entry line in this file contains various information about one household: income, socioeconomic status, number of children, their ages, and so on. One data line can then be selected at random (each line has the same probability of being selected). Consequently, probabilities of various events are interpretable as relative frequencies of entries in the data file with the corresponding property.

Let $X, Y$, and $Z$ be, respectively, the numbers of boys, girls, and cars in the households sampled. Let $A$ be the event that a household has a TV set, and let $B$ be the event that it has a swimming pool.
(i) Interpret the probabilities below as relative frequencies of occurrence of some attributes in certain subpopulations. (a) $P(A)$. (b) $P(A \mid Z>0)$. (c) $P(Z>0 \mid A)$. (d) $P(X=0 \mid X+Y=3)$. (e) $P\left(B \mid A^c\right)$. (f) $P\left[(X+Y=0)^c \mid A \cap B\right]$. (g) $P(X Y=$ $0 \mid X+Y>1)$
(ii) Use symbols to express probabilities corresponding to the following relative frequencies: (a) Relative frequency of households with two cars. (b) Relative frequency of households with no children among households with at least one car. (c) Relative frequency of households that have both a swimming pool and a TV set, among those who have either a swimming pool or a TV set and have at least one car.

# 概率与统计代写

## 统计代写|概率与统计代考概率统计代写|简介

.

EXAMPLE $4.1$

## 统计代写|概率与统计代考概率与统计代写|PROBLEMS

.

(ii)用符号表示对应于下列相对频率的概率:(a)拥有两辆汽车的家庭的相对频率。(b)至少拥有一辆汽车的家庭中没有子女的家庭的相对频率。(c)同时拥有游泳池和电视机的家庭的相对频率，在那些既拥有游泳池又拥有电视机并且至少拥有一辆汽车的家庭中

avatest.org 为您提供可靠及专业的论文代写服务以便帮助您完成您学术上的需求，让您重新掌握您的人生。我们将尽力给您提供完美的论文，并且保证质量以及准时交稿。除了承诺的奉献精神，我们的专业写手、研究人员和校对员都经过非常严格的招聘流程。所有写手都必须证明自己的分析和沟通能力以及英文水平，并通过由我们的资深研究人员和校对员组织的面试。

## 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。

Posted on Categories:Probability and Statistics, 概率与统计, 统计代写, 统计代考

## avatest™帮您通过考试

avatest™的各个学科专家已帮了学生顺利通过达上千场考试。我们保证您快速准时完成各时长和类型的考试，包括in class、take home、online、proctor。写手整理各样的资源来或按照您学校的资料教您，创造模拟试题，提供所有的问题例子，以保证您在真实考试中取得的通过率是85%以上。如果您有即将到来的每周、季考、期中或期末考试，我们都能帮助您！

•最快12小时交付

•200+ 英语母语导师

•70分以下全额退款

## 统计代写|概率与统计代考Probability and Statistics代写|NECESSITY OF THE AXIOMS

Looking at Axiom 3, one may wonder why do we need it for the case of countable (and not just finite) sequences of events. Indeed, the necessity of all three axioms, with only finite additivity in Axiom 3, can be easily justified simply by using probability to represent the limiting relative frequency of occurrences of events. Recall the symbol $N(A)$ from Section $2.1$ for the number of occurrences of the event $A$ in the first $N$ experiments. The nonnegativity axiom is simply a reflection of the fact that the count $N(A)$ cannot be negative. The norming axiom reflects the fact that event $\mathcal{S}$ is certain and must occur in every experiment so that $N(\mathcal{S})=N$, and hence, $N(\mathcal{S}) / N=1$. Finally, (taking the case of two disjoint events $A$ and $B$ ), we have $N(A \cup B)=N(A)+N(B)$, since whenever $A$ occurs, $B$ does not, and conversely. Thus, if probability is to reflect the limiting relative frequency, then $P(A \cup B)$ should be equal to $P(A)+P(B)$, since the frequencies satisfy the analogous condition $N(A \cup B) / N=N(A) / N+N(B) / N$.

The need for countable additivity, however, cannot be explained so simply. This need is related to the fact that to build a sufficiently powerful theory, one needs to take limits. If $A_1, A_2, \ldots$ is a monotone sequence of events (increasing or decreasing, i.e., $A_1 \subset A_2 \subset \cdots$ or $\left.A_1 \supset A_2 \cdots\right)$ then $\lim P\left(A_n\right)=P\left(\lim A_n\right)$, where the event $\lim A_n$ has been defined in Section 1.4. Upon a little reflection, one can see that such continuity is a very natural requirement. In fact, the same requirement has been taken for granted for over 2,000 years in a somewhat different context: in computing the area of a circle, one uses a sequence of polygons with an increasing number of sides, all inscribed in the circle. This leads to an increasing sequence of sets “converging” to the circle, and therefore the area of the circle is taken to be the limit of the areas of approximating polygons. The validity of this idea (i.e., the assumption of the continuity of the function $f(A)=$ area of $A$ ) was not really questioned until the beginning of the twentieth century. Research on the subject culminated with the results of Lebesgue.

## 统计代写|概率与统计代考Probability and Statistics代写|SUBJECTIVE PROBABILITY

Let us finally consider briefly the third interpretation of probability, namely as a degree of certainty, or belief, about the occurrence of an event. Most often, this probability is associated not so much with an event as with the truth of a proposition asserting the occurrence of this event.

The material of this section assumes some degree of familiarity with the concept of expectation, formally defined only in later chapters. For the sake of completeness, in the simple form needed here, this concept is defined below. In the presentation, we follow more or less the historical development, refining gradually the conceptual structures introduced. The basic concept here is that of a lottery, defined by an event, say $A$, and two objects, say $a$ and $b$. Such a lottery, written simply $a A b$, will mean that the participant $(\mathrm{X})$ in the lottery receives object $a$ if the event $A$ occurs, and receives object $b$ if the event $A^c$ occurs.
The second concept is that of expectation associated with the lottery $a A b$, defined as
$$u(a) P(A)+u(b) P\left(A^c\right),$$
where $u(a)$ and $u(b)$ are measures of how much the objects $a$ and $b$ are “worth” to the participant. When $a$ and $b$ are sums of money (or prices of objects $a$ and $b$ ), and we put $u(x)=x$, the quantity (2.13) is sometimes called expected value. In cases where $u(a)$ and $u(b)$ are values that person $\mathrm{X}$ attaches to $a$ and $b$ (at a given moment), these values do not necessarily coincide with prices. We then refer to $u(a)$ and $u(b)$ as utilities of $a$ and $b$, and the quantity (2.13) is called expected utility $(E U)$. Finally, when in the latter case, the probability $P(A)$ is the subjective assessment of likelihood of the event $A$ by $\mathrm{X}$, the quantity (2.13) is called subjective expected utility $(S E U)$.

First, it has been shown by Ramsey (1926) that the degree of certainty about the occurrence of an event (of a given person) can be measured. Consider an event $A$, and the following choice suggested to $\mathrm{X}$ (whose subjective probability we want to determine). $\mathrm{X}$ is namely given a choice between the following two options:

1. Sure option: receive some fixed amount $\$ u$, which is the same as lottery$(\$u) B(\$ u)$, for any event$B$. 2. A lottery option. Receive some fixed amount, say$\$100$, if $A$ occurs, and receive nothing if $A$ does not occur, which is lottery $(\$ 100) A(\$0)$. One should expect that if $u$ is very small, $\mathrm{X}$ will probably prefer the lottery. On the other hand, if $u$ is close to $\$ 100, \mathrm{X}$may prefer the sure option. Therefore, there should exist an amount$u^$such that$\mathrm{X}$will be indifferent between the sure option with$u^$and the lottery option. With the amount of money as a representation of its value (or utility), the expected return from the lottery equals $$0(1-P(A))+100 P(A)=100 P(A),$$ which, in turn, equals$u^$. Consequently, we have$P(A)=u^ / 100$. Obviously, under the stated assumption that utility of money is proportional to the dollar amount, the choice of$\$100$ is not relevant here, and the same value for $P(A)$ would be obtained if we choose another “base value” in the lottery option (this can be tested empirically).

# 概率与统计代写

## 统计代写|概率与统计代考概率与统计代写|主观概率

$$u(a) P(A)+u(b) P\left(A^c\right),$$
，其中$u(a)$和$u(b)$是衡量对象$a$和$b$对参与者的“价值”有多少。当$a$和$b$是钱的总和(或物品的价格$a$和$b$)，我们写上$u(x)=x$，数量(2.13)有时被称为期望值。如果$u(a)$和$u(b)$是$\mathrm{X}$附加在$a$和$b$上的值(在给定时刻)，这些值不一定与价格一致。然后我们将$u(a)$和$u(b)$称为$a$和$b$的效用，数量(2.13)称为期望效用$(E U)$。最后，当在后一种情况下，概率$P(A)$是$\mathrm{X}$对事件$A$的可能性的主观评价时，这个量(2.13)称为主观期望效用$(S E U)$。

1. 确定选项:接收一定的固定金额$\$ u$，相当于彩票$(\$u) B(\$ u)$，任何事件$B$2. 抽奖选项。如果$A$出现，就收到固定的金额，比如$\$100$，如果$A$没有出现，就什么也得不到，这就是彩票$(\$ 100) A(\$0)$。人们应该预料到，如果$u$非常小，$\mathrm{X}$可能更喜欢抽签。另一方面，如果$u$接近$\$ 100, \mathrm{X}$可能更喜欢确定的选项。 因此，应该存在一个数量$u^$，使得$\mathrm{X}$在带有$u^$的确定选项和抽签选项之间是无所谓的。用钱的数量来表示它的价值(或效用)，彩票的预期收益等于 $$0(1-P(A))+100 P(A)=100 P(A),$$ ，反过来等于$u^$。因此，我们有$P(A)=u^ / 100$。显然，在假定货币效用与美元金额成正比的前提下，选择$\$100$在这里是无关的，如果我们在抽签选项中选择另一个“基础值”(这可以通过经验检验)，将获得相同的$P(A)$值。

avatest.org 为您提供可靠及专业的论文代写服务以便帮助您完成您学术上的需求，让您重新掌握您的人生。我们将尽力给您提供完美的论文，并且保证质量以及准时交稿。除了承诺的奉献精神，我们的专业写手、研究人员和校对员都经过非常严格的招聘流程。所有写手都必须证明自己的分析和沟通能力以及英文水平，并通过由我们的资深研究人员和校对员组织的面试。

## 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。

Posted on Categories:Probability and Statistics, 概率与统计, 统计代写, 统计代考

## avatest™帮您通过考试

avatest™的各个学科专家已帮了学生顺利通过达上千场考试。我们保证您快速准时完成各时长和类型的考试，包括in class、take home、online、proctor。写手整理各样的资源来或按照您学校的资料教您，创造模拟试题，提供所有的问题例子，以保证您在真实考试中取得的通过率是85%以上。如果您有即将到来的每周、季考、期中或期末考试，我们都能帮助您！

•最快12小时交付

•200+ 英语母语导师

•70分以下全额退款

## 统计代写|概率与统计代考Probability and Statistics代写|AXIOMS OF PROBABILITY

Let $\mathcal{S}$ be the sample space, namely the set of all outcomes of an experiment. Formally, probability, to be denoted by $P$, is a function defined on the class of all events, ${ }^1$ satisfying the following conditions (usually referred to as axioms):
Axiom 1 (Nonnegativity):
$P(A) \geq 0$ for every event $A$.
Axiom 2 (Norming):
$$P(\mathcal{S})=1 .$$
$$P\left(A_1 \cup A_2 \cup \cdots\right)=\sum_{i=1}^{\infty} P\left(A_i\right)$$
for every sequence of pairwise disjoint events $A_1, A_2, \ldots$, (such that $A_i \cap A_j=\emptyset$ for all $i \neq j$ ).

If the sample space $\mathcal{S}$ is finite or countable, one can define a probability function $P$ as follows: Let $f$ be a nonnegative function defined on $\mathcal{S}$, satisfying the condition $\sum_{s \in \mathcal{S}} f(s)=1$. Then, $P$ may be defined for every subset $A$ of $\mathcal{S}$ as $P(A)=\sum_{s \in A} f(s)$. One can easily check that $P$ satisfies all three axioms.

Indeed, $P(A) \geq 0$ because $f$ is nonnegative, and $P(S)=\sum_{s \in S} f(s)=1$. Finally, let $A_1, A_2, \ldots$ be a sequence of disjoint subsets of $S$. Then,
\begin{aligned} P\left(A_1\right)+P\left(A_2\right)+\cdots &=\sum_{s \in A_1} f(s)+\sum_{s \in A_2} f(s)+\cdots \ &=\sum_{s \in A_1 \cup A_2 \cup \cdots} f(s)=P\left(\bigcup_{i=1}^{\infty} A_i\right) \end{aligned}

## 统计代写|概率与统计代考Probability and Statistics代写|CLASSICAL PROBABILITY

For the so-called classical or logical interpretation of probability, we will assume that the sample space $\mathcal{S}$ contains a finite number $N$ of outcomes and all of these outcomes are equally probable.

Obviously, in this case, each of the outcomes has the same probability $1 / N$, and for every event $A$
$$P(A)=\frac{\text { number of outcomes in } A}{N} .$$
In many real situations, the outcomes in the sample space reveal a certain symmetry, derived from physical laws, from logical considerations, or simply from the sampling scheme used. In such cases, one can often assume that the outcomes are equiprobable and use (2.7) as a rule for computing probabilities. Obviously, the function $P$ in (2.7) satisfies the axioms of probability.

To use some very simple examples, in tossing a regular die, each face has the same probability $1 / 6$. Then the probability of the event $A=$ “outcome odd” is $P(A)=3 / 6=1 / 2$, since there are three odd outcomes among the possible six.

The case above is rather trivial, but considerations of symmetry can sometimes lead to unexpectedly simple solutions of various problems.

# 概率与统计代写

## 统计代写|概率与统计代考概率与统计代写|AXIOMS OF Probability

$P(A) \geq 0$对于每个事件$A$。

$$P(\mathcal{S})=1 .$$

$$P\left(A_1 \cup A_2 \cup \cdots\right)=\sum_{i=1}^{\infty} P\left(A_i\right)$$

.

\begin{aligned} P\left(A_1\right)+P\left(A_2\right)+\cdots &=\sum_{s \in A_1} f(s)+\sum_{s \in A_2} f(s)+\cdots \ &=\sum_{s \in A_1 \cup A_2 \cup \cdots} f(s)=P\left(\bigcup_{i=1}^{\infty} A_i\right) \end{aligned}

## 统计代写|概率与统计代考概率与统计代写|经典概率

.

$$P(A)=\frac{\text { number of outcomes in } A}{N} .$$
，在许多实际情况下，样本空间中的结果显示出某种对称性，这种对称性来自物理定律，来自逻辑考虑，或者仅仅来自所使用的抽样方案。在这种情况下，人们通常可以假设结果是等可能的，并使用(2.7)作为计算概率的规则。显然，(2.7)中的函数$P$满足概率公理

avatest.org 为您提供可靠及专业的论文代写服务以便帮助您完成您学术上的需求，让您重新掌握您的人生。我们将尽力给您提供完美的论文，并且保证质量以及准时交稿。除了承诺的奉献精神，我们的专业写手、研究人员和校对员都经过非常严格的招聘流程。所有写手都必须证明自己的分析和沟通能力以及英文水平，并通过由我们的资深研究人员和校对员组织的面试。

## 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。