## 数学代写|概率论代写Probability theory代考|Math561

statistics-lab™ 为您的留学生涯保驾护航 在代写概率论Probability theory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论Probability theory代写方面经验极为丰富，各种代写概率论Probability theory相关的作业也就用不着说。

## 数学代写|概率论代写Probability theory代考|PROPERTIES OF DISTRIBUTION FUNCTIONS

We shall establish some general properties of the distribution function of an arbitrary random variable. We need two facts about probability measures.
Theorem 1. Let $(\Omega, \mathscr{F}, P)$ be a probability space.
(a) If $A_1, A_2, \ldots$ is an expanding sequence of sets in $\mathscr{F}$, that is, $A_n \subset A_{n+1}$ for all $n$, and $A=\bigcup_{n=1}^{\infty} A_n$, then $P(A)=\lim _{n \rightarrow \infty} P\left(A_n\right)$.

(b) If $A_1, A_2, \ldots$ is a contracting sequence of sets in $\mathscr{F}$, that is, $A_{n+1} \subset A_n$ for all $n$, and $A=\bigcap_{n=1}^{\infty} A_n$, then $P(A)=\lim {n \rightarrow \infty} P\left(A_n\right)$. Proof. (a) We can write $$A=A_1 \cup\left(A_2-A_1\right) \cup\left(A_3-A_2\right) \cup \cdots \cup\left(A_n-A{n-1}\right) \cdots$$
(see Figure 2.5.1; note this is the expansion (1.3.11) in the special case of an expanding sequence). Since this is a disjoint union,
\begin{aligned} P(A) & =P\left(A_1\right)+P\left(A_2-A_1\right)+P\left(A_3-A_2\right)+\cdots \ & =P\left(A_1\right)+P\left(A_2\right)-P\left(A_1\right)+P\left(A_3\right)-P\left(A_2\right)+\cdots \quad \text { since } A_n \subset A_{n+1} \ & =\lim {n \rightarrow \infty} P\left(A_n\right) \end{aligned} (b) If $A=\bigcap{n=1}^{\infty} A_n$, then, by the DeMorgan laws, $A^c=\bigcup_{n=1}^{\infty} A_n{ }^c$. Now $A_{n+1} \subset A_n$; hence $A_n{ }^c \subset A_{n+1}^c$. Thus the sets $A_n{ }^c$ form an expanding sequence, so, by (a), $P\left(A_n^c\right) \rightarrow P\left(A^c\right)$; that is; $1-P\left(A_n\right) \rightarrow 1-P(A)$. The result follows.

## 数学代写|概率论代写Probability theory代考|JOINT DENSITY FUNCTIONS

We are going to investigate situations in which we deal simultaneously with several random variables defined on the same sample space. As an introductory example, suppose that a person is selected at random from a certain population, and his age and weight recorded. We may take as the sample space the set of all pairs $(x, y)$ of real numbers, that is, the Euclidean plane $E^2$, where we interpret $x$ as the age and $y$ as the weight. Let $R_1$ be the age of the person selected, and $R_2$ the weight; that is, $R_1(x, y)=x, R_2(x, y)=y$. We wish to assign probabilities to events that involve $R_1$ and $R_2$ simultaneously. A cross-section of the available data might appear as shown in Figure 2.6.1. Thus there are 4 million people whose age is between 20 and 25 and (simultaneously) whose weight is between 150 and 160 pounds, and so on. Now suppose that we wish to estimate the number of people between 22 and 23 years, and 154 and 156 pounds. There are 4 million people spread over 5 years and 10 pounds, or 4/50 million per year-pound. We are interested in a range of 1 year and 2 pounds, and so our estimate is $4 / 50 \times 1 \times 2=8 / 50$ million (see Figure 2.6.2). If the total population is 200 million, then
$$P\left{22 \leq R_1 \leq 23,154 \leq R_2 \leq 156\right}$$
should be approximately
$$\frac{8 / 50}{200}=.0008$$
Notation. $\left{22 \leq R_1 \leq 23,154 \leq R_2 \leq 156\right}$ means $\left{22 \leq R_1 \leq 23\right.$ and $\left.154 \leq R_2 \leq 156\right}$.

# 概率论代考

## 数学代写|概率论代写Probability theory代考|PROPERTIES OF DISTRIBUTION FUNCTIONS

(a)如果$A_1, A_2, \ldots$是$\mathscr{F}$中集合的展开序列，即$A_n \subset A_{n+1}$适用于所有$n$，而$A=\bigcup_{n=1}^{\infty} A_n$则$P(A)=\lim _{n \rightarrow \infty} P\left(A_n\right)$。

(b)如果$A_1, A_2, \ldots$是$\mathscr{F}$中的集合的收缩序列，即$A_{n+1} \subset A_n$适用于所有$n$，而$A=\bigcap_{n=1}^{\infty} A_n$，则$P(A)=\lim {n \rightarrow \infty} P\left(A_n\right)$。证明。我们可以写$$A=A_1 \cup\left(A_2-A_1\right) \cup\left(A_3-A_2\right) \cup \cdots \cup\left(A_n-A{n-1}\right) \cdots$$
(见图2.5.1;注意，这是在展开序列的特殊情况下的展开(1.3.11)。由于这是一个分裂的联盟，
\begin{aligned} P(A) & =P\left(A_1\right)+P\left(A_2-A_1\right)+P\left(A_3-A_2\right)+\cdots \ & =P\left(A_1\right)+P\left(A_2\right)-P\left(A_1\right)+P\left(A_3\right)-P\left(A_2\right)+\cdots \quad \text { since } A_n \subset A_{n+1} \ & =\lim {n \rightarrow \infty} P\left(A_n\right) \end{aligned} (b)如果$A=\bigcap{n=1}^{\infty} A_n$，则根据民主党法律，$A^c=\bigcup_{n=1}^{\infty} A_n{ }^c$。现在$A_{n+1} \subset A_n$;因此，$A_n{ }^c \subset A_{n+1}^c$。因此集合$A_n{ }^c$形成一个展开式序列，由(a)， $P\left(A_n^c\right) \rightarrow P\left(A^c\right)$;那就是;$1-P\left(A_n\right) \rightarrow 1-P(A)$。结果如下。

## 数学代写|概率论代写Probability theory代考|JOINT DENSITY FUNCTIONS

$$P\left{22 \leq R_1 \leq 23,154 \leq R_2 \leq 156\right}$$

$$\frac{8 / 50}{200}=.0008$$

## 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|概率论代写Probability theory代考|SF2940

statistics-lab™ 为您的留学生涯保驾护航 在代写概率论Probability theory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论Probability theory代写方面经验极为丰富，各种代写概率论Probability theory相关的作业也就用不着说。

## 数学代写|概率论代写Probability theory代考|APPENDIX: STIRLING’S FORMULA

An estimate of $n !$ that is of importance both in numerical calculations and theoretical analysis is Stirling’s formula
$$n ! \sim n^n e^{-n} \sqrt{2 \pi n}$$
in the sense that
$$\lim {n \rightarrow \infty} \frac{n !}{\left(n^n e^{-n} \sqrt{2 \pi n}\right)}=1$$ Proof. Define $(2 n) !$ ! (read $2 n$ semifactorial) as $2 n(2 n-2)(2 n-4) \cdots$ $6(4)(2)$, and $(2 n+1) ! !$ as $(2 n+1)(2 n-1) \cdots(5)(3)(1)$. We first show that (a) $$\frac{(2 n) ! !}{(2 n+1) ! !}<\frac{\pi}{2} \frac{(2 n-1) ! !}{(2 n) ! !}<\frac{(2 n-2) ! !}{(2 n-1) ! !}$$ Let $I_k=\int_0^{\pi / 2}(\cos x)^k d x, k=0,1,2, \ldots$ Then $I_0=\pi / 2, I_1=1$. Integrating by parts, we obtain $I_k=\int_0^{\pi / 2}(\cos x)^{k-1} d(\sin x)=\int_0^{\pi / 2}(k-1)(\cos x)^{k-2}$ $\sin ^2 x d x$. Since $\sin ^2 x=1-\cos ^2 x$, we have $I_k=(k-1) I{k-2}-(k-1) I_k$ or $I_k=[(k-1) / k] I_{k-2}$. By iteration, we obtain $I_{2 n}=(\pi / 2)[(2 n-1) ! ! /$ $(2 n) ! !]$ and $I_{2 n+1}=[(2 n) ! ! /(2 n+1) ! !]$. Since $(\cos x)^k$ decreases with $k$, so does $I_k$, and hence $I_{2 n+1}<I_{2 n}<I_{2 n-1}$, and (a) is proved.
(b) Let $Q_n=\left(\begin{array}{c}2^n \ n\end{array}\right) / 2^{2 n}$. Then
$$\lim _{n \rightarrow \infty} Q_n \sqrt{n \pi}=1$$
To prove this, write
\begin{aligned} Q_n & =\frac{(2 n) !}{n ! n ! 2^{2 n}}=\frac{(2 n) !}{\left(2^n n !\right)^2} \ & =\frac{(2 n) !}{((2 n)(2 n-2) \cdots(4)(2))^2}=\frac{(2 n-1) ! !}{(2 n) ! !} \end{aligned}

Thus, by (a),
$$\frac{(2 n) ! !}{(2 n+1) ! !}<\frac{\pi}{2} Q_n<\frac{(2 n-2) ! !}{(2 n-1) ! !}$$
Multiply this inequality by
$$\frac{(2 n-1) ! !}{(2 n-2) ! !}=\frac{(2 n-1) ! !}{(2 n) ! !} \frac{(2 n) ! !}{(2 n-2) ! !}=Q_n(2 n)$$
to obtain
$$\frac{2 n}{2 n+1}<n \pi Q_n{ }^2<1$$
If we let $n \rightarrow \infty$, we obtain $n \pi Q_n{ }^2 \rightarrow 1$, proving (b).
(c) Proof of Stirling’s formula. Let $c_n=n ! / n^n e^{-n} \sqrt{2 \pi n}$. We must show that $c_n \rightarrow 1$ as $n \rightarrow \infty$. Consider $(n+1) ! / n !=n+1$. We have
\begin{aligned} \frac{(n+1) !}{n !} & =\frac{c_{n+1}(n+1)^{n+1} e^{-(n+1)} \sqrt{2 \pi(n+1)}}{c_n n^n e^{-n} \sqrt{2 \pi n}} \ & =\left(\frac{c_{n+1}}{c_n}\right) e^{-1}\left(\frac{n+1}{n}\right)^n \frac{(n+1)^{3 / 2}}{\sqrt{n}} \end{aligned}

## 数学代写|概率论代写Probability theory代考|DEFINITION OF A RANDOM VARIABLE

Intuitively, a random variable is a quantity that is measured in connection with a random experiment. If $\Omega$ is a sample space, and the outcome of the experiment is $\omega$, a measuring process is carried out to obtain a number $R(\omega)$. Thus a random variable is a real-valued function on a sample space. (The formal definition, which is postponed until later in the section, is somewhat more restrictive.)

Example 1. Throw a coin 10 times, and let $R$ be the number of heads. We take $\Omega=$ all sequences of length 10 with components $H$ and $T ; 2^{10}$ points altogether. A typical sample point is $\omega=H H T H T T H H T H$. For this point $R(\omega)=6$. Another random variable, $R_1$, is the number of times a head is followed immediately by a tail. For the point $\omega$ above, $R_1(\omega)=3$.
Example 2. Pick a person at random from a certain population and measure his height and weight. We may take the sample space to be the plane $E^2$, that is, the set of all pairs $(x, y)$ of real numbers, with the first coordinate $x$ representing the height and the second coordinate $y$ the weight (we can take care of the requirement that height and weight be nonnegative by assigning probability 0 to the complement of the first quadrant). Let $R_1$ be the height of the person selected, and let $R_2$ be the weight. Then $R_1(x, y)=x, R_2(x, y)=y$. As another example, let $R_3$ be twice the height plus the cube root of the weight; that is, $R_3=2 R_1+\sqrt[3]{R_2}$. Then $R_3(x, y)=$ $2 R_1(x, y)+\sqrt[3]{R_2(x, y)}=2 x+\sqrt[3]{y}$.

Example 3. Throw two dice. We may take the sample space to be the set of all pairs of integers $(x, y), x, y=1,2, \ldots, 6$ (36 points in all).
Let $R_1=$ the result of the first toss. Then $R_1(x, y)=x$.
Let $R_2=$ the sum of the two faces. Then $R_2(x, y)=x+y$.
Let $R_3=1$ if at least one face is an even number; $R_3=0$ otherwise.
Then $R_3(6,5)=1 ; R_3(3,6)=1 ; R_3(1,3)=0$, and so on.

# 概率论代考

## 数学代写|概率论代写Probability theory代考|APPENDIX: STIRLING’S FORMULA

$$n ! \sim n^n e^{-n} \sqrt{2 \pi n}$$

## 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|概率论代写Probability theory代考|MAP4102

statistics-lab™ 为您的留学生涯保驾护航 在代写概率论Probability theory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论Probability theory代写方面经验极为丰富，各种代写概率论Probability theory相关的作业也就用不着说。

## 数学代写|概率论代写Probability theory代考|COMBINATORIAL PROBLEMS

We consider a class of problems in which the assignment of probabilities can be made in a natural way.

Let $\Omega$ be a finite or countably infinite set, and let $\mathscr{F}$ consist of all subsets of $\Omega$.

For each point $\omega_i \in \Omega, i=1,2, \ldots$, assign a nonnegative number $p_i$, with $\sum_i p_i=1$. If $A$ is any subset of $\Omega$, let $P(A)=\sum_{\omega_i \in A} p_i$. Then it may be verified that $P$ is a probability measure; $P\left{\omega_i\right}=p_i$, and the probability of any event $A$ is found by adding the probabilities of the points of $A$. An $(\Omega, \mathscr{F}, P)$ of this type is called a discrete probability space.
Example 1. Throw a (biased) coin twice (see Figure 1.4.1).
Let $E_1=$ {at least one head $}$. Then
$$E_1=A_1 \cup A_2 \cup A_3$$
Hence
\begin{aligned} P\left(E_1\right) & =P\left(A_1\right)+P\left(A_2\right)+P\left(A_3\right) \ & =.36+.24+.24=.84 \end{aligned}
Let $E_2=$ {tail on first toss $}$; then
$$E_2=A_3 \cup A_4$$

and
$$P\left(E_2\right)=P\left(A_3\right)+P\left(A_4\right)=.4$$
In the special case when $\Omega=\left{\omega_1, \ldots, \omega_n\right}$ and $p_i=1 / n, i=1,2, \ldots, n$, we have
$$P(A)=\frac{\text { number of points of } A}{\text { total number of points in } \Omega}=\frac{\text { favorable outcomes }}{\text { total outcomes }}$$
corresponding to the classical definition of probability.

## 数学代写|概率论代写Probability theory代考|Ordered samples of size $r$, with replacement

The number of ordered sequences $\left(a_{i_1}, \ldots, a_{i_r}\right)$, where the $a_{i_k}$ belong to $\left{a_1, \ldots, a_n\right}$, is $n \times n \times \cdots \times n$ ( $r$ times), or
$$n^r$$
(The term “with replacement” refers to the fact that if the symbol $a_{i_k}$ is selected at step $k$ it may be selected again at any future time.)

For example, the number of possible outcomes if three dice are thrown is $6 \times 6 \times 6=216$.
Ordered Samples of Size $r$, without Replacement
The number of ordered sequences $\left(a_{i_1}, \ldots, a_{i_r}\right)$, where the $a_{i_k}$ belong to $\left{a_1, \ldots, a_n\right}$, but repetition is not allowed (i.e., no $a_i$ can appear more than once in the sequence), is
$$n(n-1) \cdots(n-r+1)=\frac{n !}{(n-r) !}, \quad r=1,2, \ldots, n$$
(The first symbol may be chosen in $n$ ways, and the second in $n-1$ ways, since the first symbol may not be used again, and so on.) The above number is sometimes called the number of permutations of $r$ objects out of $n$, written $(n)_r$.

For example, the number of 3-digit numbers that can be formed from $1,2, \ldots, 9$, if no digit can be repeated, is $9(8)(7)=504$.

Unordered Samples of Size $r$, without Replacement
The number of unordered sets $\left{a_{i_1}, \ldots, a_{i_r}\right}$, where the $a_{i_k}, k=1, \ldots, r$, are distinct elements of $\left{a_1, \ldots, a_n\right}$ (i.e., the number of ways of selecting $r$ distinct objects out of $n$ ), if order does not count, is
$$\left(\begin{array}{l} n \ r \end{array}\right)=\frac{n !}{r !(n-r) !}$$
To see this, consider the following process.
(a) Select $r$ distinct objects out of $n$ without regard to order; this can be done in $\left(\begin{array}{l}n \ r\end{array}\right)$ ways, where $\left(\begin{array}{l}n \ r\end{array}\right)$ is to be determined.
(b) For each set selected in (a), say $\left{a_{i_1}, \ldots, a_{i_r}\right}$, select an ordering of $a_{i_1}, \ldots, a_{i_r}$. This can be done in $(r)_r=r$ ! ways (see Figure 1.4.2 for $n=3$, $r=2)$.

The result of performing (a) and (b) is a permutation of $r$ objects out of $n$; hence
$$\left(\begin{array}{l} n \ r \end{array}\right) r !=(n)_r=\frac{n !}{(n-r) !}$$
or
$$\left(\begin{array}{l} n \ r \end{array}\right)=\frac{n !}{r !(n-r) !}, \quad r=1,2, \ldots, n$$
We define $\left(\begin{array}{l}n \ 0\end{array}\right)$ to be $n ! / 0 ! n !=1$, to make the formula for $\left(\begin{array}{l}n \ r\end{array}\right)$ valid for $r=0,1, \ldots, n$. Notice that $\left(\begin{array}{c}n \ k\end{array}\right)=\left(\begin{array}{c}n \ n-k\end{array}\right)$.

# 概率论代考

## 数学代写|概率论代写Probability theory代考|COMBINATORIAL PROBLEMS

$$P\left(E_2\right)=P\left(A_3\right)+P\left(A_4\right)=.4$$

$$P(A)=\frac{\text { number of points of } A}{\text { total number of points in } \Omega}=\frac{\text { favorable outcomes }}{\text { total outcomes }}$$

## 数学代写|概率论代写Probability theory代考|Ordered samples of size $r$, with replacement

$$n^r$$
(术语“与替换”是指如果符号 $a_{i_k}$ 在第一步被选中 $k$ 可在将来任何时候再次选择。)

$$n(n-1) \cdots(n-r+1)=\frac{n !}{(n-r) !}, \quad r=1,2, \ldots, n$$
(第一个符号可以以$n$的方式选择，第二个符号可以以$n-1$的方式选择，因为第一个符号可能不会再次使用，以此类推。)上述数字有时称为$r$对象对$n$的排列次数，写为$(n)_r$。

$$\left(\begin{array}{l} n \ r \end{array}\right)=\frac{n !}{r !(n-r) !}$$

(a)从$n$中不顾顺序选择$r$不同的对象;这可以通过$\left(\begin{array}{l}n \ r\end{array}\right)$方式完成，其中$\left(\begin{array}{l}n \ r\end{array}\right)$是要确定的。
(b)对于(a)中选择的每个集合，例如$\left{a_{i_1}, \ldots, a_{i_r}\right}$，选择一个$a_{i_1}, \ldots, a_{i_r}$的排序。这可以在$(r)_r=r$完成!方式($n=3$, $r=2)$见图1.4.2)。

$$\left(\begin{array}{l} n \ r \end{array}\right) r !=(n)_r=\frac{n !}{(n-r) !}$$

$$\left(\begin{array}{l} n \ r \end{array}\right)=\frac{n !}{r !(n-r) !}, \quad r=1,2, \ldots, n$$

## 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|概率论代写Probability theory代考|DTSA5001

statistics-lab™ 为您的留学生涯保驾护航 在代写概率论Probability theory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论Probability theory代写方面经验极为丰富，各种代写概率论Probability theory相关的作业也就用不着说。

## 数学代写|概率论代写Probability theory代考|VARIABLE DISTRIBUTIONS

$\mathrm{Up}$ to now we have considered only variables $\mathbf{X}_k$ having the same distribution. This situation corresponds to a repetition of the same game of chance, but it is more interesting to see what happens if the type of game changes at each step. It is not necessary to think of gambling places; the statistician who applies statistical tests is engaged in a dignified sort of gambling, and in his case the distribution of the random variables changes from occasion to occasion.

To fix ideas we shall imagine that an infinite sequence of probability distributions is given so that for each $n$ we have $n$ mutually independent variables $\mathbf{X}_1, \ldots, \mathbf{X}_n$ with the prescribed distributions. We assume that the means and variances exist and put
$$\mu_k=\mathbf{E}\left(\mathbf{X}_k\right), \quad \sigma_k^2=\operatorname{Var}\left(\mathbf{X}_k\right)$$
The sum $\mathbf{S}_n=\mathbf{X}_1+\cdots+\mathbf{X}_n$ has mean $m_n$ and variance $s_n^2$ given by
$$m_n=\mu_1+\cdots+\mu_n, \quad s_n^2=\sigma_1^2+\cdots+\sigma_n^2$$

[cf. IX, (2.4) and IX,(5.6)]. In the special case of identical distributions we had $m_n=n \mu, s_n^2=n \sigma^2$.

The (weak) law of large numbers is said to hold for the sequence $\left{\mathbf{X}_k\right}$ if for every $\epsilon>0$
$$\mathbf{P}\left{\frac{\left|\mathbf{S}_n-m_n\right|}{n}>\epsilon\right} \rightarrow 0 .$$
The sequence $\left{\mathbf{X}_k\right}$ is said to obey the central limit theorem if for every fixed $\alpha<\beta$
$$\mathbf{P}\left{\alpha<\frac{\mathbf{S}_n-m_n}{s_n}<\beta\right} \rightarrow \mathfrak{N}(\beta)-\mathfrak{N}(\alpha) .$$
It is one of the salient features of probability theory that both the law of large numbers and the central limit theorem hold for a surprisingly large class of sequences $\left{\mathbf{X}_k\right}$. In particular, the law of large numbers halds. whenever the $\mathbf{X}_k$ are uniformly bounded, that is, whenever there cxists a constant $A$ such that $\left.\rfloor \mathbf{X}_k\right\rfloor<A$ for all $k$. More generally, a sufficient condition for the law of large numbers to hold is that
$$\frac{s_n}{n} \rightarrow 0 .$$

## 数学代写|概率论代写Probability theory代考|APPLICATIONS TO COMBINATORIAL ANALYSIS

We shall give two examples of applications of the central limit theorem to problems not directly connected with probability theory. Both relate to the $n$ ! permutations of the $n$ elements $a_1, a_2, \ldots, a_n$, to each of which we attribute probability $1 / n$ !.
(a) Inversions. In a given permutation the element $a_k$ is said to induce $r$ inversions if it precedes exactly $r$ elements with smaller index (i.e., elements which precede $a_k$ in the natural order). For example, in $\left(a_3 a_6 a_1 a_5 a_2 a_4\right)$ the elements $a_1$ and $a_2$ induce no inversion, $a_3$ induces two, $a_4$ none, $a_5$ two, and $a_6$ four. In $\left(a_6 a_5 a_4 a_3 a_2 a_1\right)$ the element $a_k$ induces $k-1$ inversions and there are fifteen inversions in all. The number $\mathbf{X}k$ of inversions induced by $a_k$ is a random variable, and $\mathbf{S}_n=\mathbf{X}_1+\cdots+\mathbf{X}_n$ is the total number of inversions. Here $\mathbf{X}_k$ assumes the values $0,1, \ldots, k-1$, each with probability $1 / k$, and therefore $$\mu_k=\frac{k-1}{2},$$ $$\sigma_k^2=\frac{1+2^2+\cdots+(k-1)^2}{k}-\left(\frac{k-1}{2}\right)^2=\frac{k^2-1}{12} .$$ The number of inversions produced by $a_k$ does not depend on the relative order of $a_1, a_2, \ldots, a{k-1}$, and the $\mathbf{X}k$ are therefore mutually independent. From (6.1) we get $$m_n=\frac{1+2+\cdots+(n-1)}{2}=\frac{n(n-1)}{4} \sim \frac{n^2}{4}$$ and $$s_n^2=\frac{1}{12} \sum{k=1}^n\left(k^2-1\right)=\frac{2 n^3+3 n^2-5 n}{72} \sim \frac{n^3}{36} .$$
For large $n$ we have $\epsilon S_n>n \geq \mathbf{U}_k$, and hence the variables $\mathbf{U}_k$ of the Lindeberg condition are identical with $\mathbf{X}_k$. Therefore the central limit theorem applies, and we conclude that the number $\mathbf{N}_n$ of permutations for which the number of inversions lies between the limits $\frac{n^2}{4} \pm \frac{\alpha}{6} \sqrt{n^3}$ is, asymptotically, given by $n !{\mathfrak{N}(\alpha)-\mathfrak{N}(-\alpha)}$. In particular, for about onehalf of all permutations the number of inversions lies between the limits $\frac{1}{4} n^2 \pm 0.11 \sqrt{n^3}$.

# 概率论代考

## 数学代写|概率论代写Probability theory代考|VARIABLE DISTRIBUTIONS

$\mathrm{Up}$ 到目前为止，我们只考虑了具有相同分布的变量$\mathbf{X}_k$。这种情况对应于相同的机会游戏的重复，但更有趣的是，如果游戏类型在每一步发生变化，会发生什么。没有必要想到赌博的地方;应用统计检验的统计学家从事的是一种体面的赌博，在他的情况下，随机变量的分布随场合而变化。

$$\mu_k=\mathbf{E}\left(\mathbf{X}_k\right), \quad \sigma_k^2=\operatorname{Var}\left(\mathbf{X}_k\right)$$

$$m_n=\mu_1+\cdots+\mu_n, \quad s_n^2=\sigma_1^2+\cdots+\sigma_n^2$$

[参见IX，(2.4)和IX，(5.6)]。在相同分布的特殊情况下我们有$m_n=n \mu, s_n^2=n \sigma^2$。

(弱)大数定律对于数列$\left{\mathbf{X}_k\right}$ if对于每一个$\epsilon>0$都成立
$$\mathbf{P}\left{\frac{\left|\mathbf{S}_n-m_n\right|}{n}>\epsilon\right} \rightarrow 0 .$$

$$\mathbf{P}\left{\alpha<\frac{\mathbf{S}_n-m_n}{s_n}<\beta\right} \rightarrow \mathfrak{N}(\beta)-\mathfrak{N}(\alpha) .$$

$$\frac{s_n}{n} \rightarrow 0 .$$

## 数学代写|概率论代写Probability theory代考|APPLICATIONS TO COMBINATORIAL ANALYSIS

(a)倒置。在给定的排列中，如果元素$a_k$恰好位于具有较小索引的$r$元素之前(即，以自然顺序位于$a_k$之前的元素)，则会导致$r$倒排。例如，在$\left(a_3 a_6 a_1 a_5 a_2 a_4\right)$中，元素$a_1$和$a_2$不诱导反转，$a_3$诱导2,$a_4$ none, $a_5$ two和$a_6$ four。在$\left(a_6 a_5 a_4 a_3 a_2 a_1\right)$中，元素$a_k$引起$k-1$反转，总共有15个反转。由$a_k$引起的反转数$\mathbf{X}k$为随机变量，$\mathbf{S}_n=\mathbf{X}_1+\cdots+\mathbf{X}_n$为反转总数。这里$\mathbf{X}_k$假设值$0,1, \ldots, k-1$，每个值的概率为$1 / k$，因此为$$\mu_k=\frac{k-1}{2},$$$$\sigma_k^2=\frac{1+2^2+\cdots+(k-1)^2}{k}-\left(\frac{k-1}{2}\right)^2=\frac{k^2-1}{12} .$$。$a_k$产生的倒排数量不依赖于$a_1, a_2, \ldots, a{k-1}$的相对顺序，因此$\mathbf{X}k$是相互独立的。从(6.1)我们得到$$m_n=\frac{1+2+\cdots+(n-1)}{2}=\frac{n(n-1)}{4} \sim \frac{n^2}{4}$$和$$s_n^2=\frac{1}{12} \sum{k=1}^n\left(k^2-1\right)=\frac{2 n^3+3 n^2-5 n}{72} \sim \frac{n^3}{36} .$$

## 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|概率论代写Probability theory代考|Stat410

statistics-lab™ 为您的留学生涯保驾护航 在代写概率论Probability theory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论Probability theory代写方面经验极为丰富，各种代写概率论Probability theory相关的作业也就用不着说。

## 数学代写|概率论代写Probability theory代考|IDENTICALLY DISTRIBUTED VARIABLES

The connection between Bernoulli trials and the theory of random variables becomes clearer when we consider the dependence of the number $\mathbf{S}_n$ of successes on the number $n$ of trials. With each trial $\mathbf{S}_n$ increases by 1 or 0 , and we can write
$$\mathbf{S}_n=\mathbf{X}_1+\cdots+\mathbf{X}_n,$$
where the random variable $\mathbf{X}_k$ equals 1 if the $k$ th trial results in success and zero otherwise. Thus $\mathbf{S}_n$ is a sum of $n$ mutually independent random variables, each of which assumes the values 1 and 0 with probabilities $p$ and $q$. From this it is only one step to consider sums of the form (1.1) where the $\mathbf{X}_k$ are mutually independent variables with an arbitrary distribution. The (weak) law of large numbers of VI,4, states that for large $n$ the average proportion of successes $\mathbf{S}_n / n$ is likely to lie near $p$. This is a special case of the following

Law of Large Numbers. Let $\left{\mathbf{X}_k\right}$ be a sequence of mutually independent random variables with a common distribution. If the expectation $\mu=\mathbf{E}\left(\mathbf{X}_k\right)$ exists, then for every $\epsilon>0$ as $n>\infty$
$$\mathbf{P}\left{\left|\frac{\mathbf{X}_1+\cdots+\mathbf{X}_n}{n}-\mu\right|>\epsilon\right} \rightarrow 0 ;$$
in words, the probability that the average $\mathbf{S}_n / n$ will differ from the expectation by less than an arbitrarily prescribed $\epsilon$ tends to one.

## 数学代写|概率论代写Probability theory代考|PROOF OF THE LAW OF LARGE NUMBERS

There is no loss of generality in assuming that $\mu=\mathbf{E}\left(\mathbf{X}_k\right)=0$, for otherwise we would replace $\mathbf{X}_k$ by $\mathbf{X}_k-\mu$, and this involves merely a change of notation. In the special case where $\sigma^2=\operatorname{Var}\left(\mathbf{X}_k\right)$ exists the law of large numbers is a trivial consequence of Chebyshev’s inequality IX,(6.2) according to which
$$\mathbf{P}\left{\left|\mathbf{S}_n\right|>t\right} \leq \frac{n \sigma^2}{t^2} .$$
For $t=\epsilon n$ the right side tends to 0 , and so (1.2) is true.
The case where the second moment does not exist is more difficult. The proof depends on the versatile method of truncation which is a standard tool in deriving various limit theorems. Let $\delta$ be a positive constant to be determined later. For each $n$ we define $n$ pairs of random variables as follows.
\begin{aligned} & \mathbf{U}_k=\mathbf{X}_k, \quad \mathbf{V}_k=0 \quad \text { if } \quad\left|\mathbf{X}_k\right| \leq \delta n, \ & \mathbf{U}_k=0, \quad \mathbf{V}_k=\mathbf{X}_k \quad \text { if } \quad\left|\mathbf{X}_k\right|>\delta n . \ & \end{aligned}
Here $k=1, \ldots, n$ and the dependence of the $\mathbf{U}_k$ and $\mathbf{V}_k$ on $n$ must be borne in mind. By this definition
$$\mathbf{X}_k=\mathbf{U}_k+\mathbf{V}_k$$
and to prove the law of large numbers it suffices to show that for given $\epsilon>0$ the constant $\delta$ can be chosen so that as $n \rightarrow \infty$
$$\mathbf{P}\left{\left|\mathbf{U}_1+\cdots+\mathbf{U}_n\right|>\frac{1}{2} \epsilon n\right} \rightarrow 0$$
and
$$\mathbf{P}\left{\left|\mathbf{V}_1+\cdots+\mathbf{V}_n\right|>\frac{1}{2} \epsilon n\right} \rightarrow 0 . \quad\left(\because \leq\left{\left|\frac{\sum U_k}{n}+\frac{\sum V_k}{n}\right|>\right.\right.$$

# 概率论代考

## 数学代写|概率论代写Probability theory代考|IDENTICALLY DISTRIBUTED VARIABLES

$$\mathbf{S}_n=\mathbf{X}_1+\cdots+\mathbf{X}_n,$$

$$\mathbf{P}\left{\left|\frac{\mathbf{X}_1+\cdots+\mathbf{X}_n}{n}-\mu\right|>\epsilon\right} \rightarrow 0 ;$$

## 数学代写|概率论代写Probability theory代考|PROOF OF THE LAW OF LARGE NUMBERS

$$\mathbf{P}\left{\left|\mathbf{S}_n\right|>t\right} \leq \frac{n \sigma^2}{t^2} .$$

\begin{aligned} & \mathbf{U}_k=\mathbf{X}_k, \quad \mathbf{V}_k=0 \quad \text { if } \quad\left|\mathbf{X}_k\right| \leq \delta n, \ & \mathbf{U}_k=0, \quad \mathbf{V}_k=\mathbf{X}_k \quad \text { if } \quad\left|\mathbf{X}_k\right|>\delta n . \ & \end{aligned}

$$\mathbf{X}_k=\mathbf{U}_k+\mathbf{V}_k$$

$$\mathbf{P}\left{\left|\mathbf{U}_1+\cdots+\mathbf{U}_n\right|>\frac{1}{2} \epsilon n\right} \rightarrow 0$$

$$\mathbf{P}\left{\left|\mathbf{V}_1+\cdots+\mathbf{V}_n\right|>\frac{1}{2} \epsilon n\right} \rightarrow 0 . \quad\left(\because \leq\left{\left|\frac{\sum U_k}{n}+\frac{\sum V_k}{n}\right|>\right.\right.$$

## 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|概率论代写Probability theory代考|STAT414

statistics-lab™ 为您的留学生涯保驾护航 在代写概率论Probability theory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论Probability theory代写方面经验极为丰富，各种代写概率论Probability theory相关的作业也就用不着说。

## 数学代写|概率论代写Probability theory代考|EXAMPLES AND APPLICATIONS

(a) Binomial distribution. Let $\mathbf{S}_n$ be the number of successes in $n$ Bernoulli trials with probability $p$ for success. We know that $\mathbf{S}_n$ has the binomial distribution ${b(k ; n, p)}$, whence $\mathbf{E}\left(\mathbf{S}_n\right)=\sum k b(k ; n, p)=$ $=n p \sum b(k-1 ; n-1, p)$. The last sum includes all terms of the binomial distribution for $n-1$ and hence equals 1 . Therefore the mean of the binomial distribution is
$$\mathbf{E}\left(\mathbf{S}_n\right)=n p .$$
The same result could have been obtained without calculation by a method which is often expedient. Let $\mathbf{X}_k$ be the number of successes scored at the $k$ th trial. This random variable assumes only the values 0 and 1 with corresponding probabilities $q$ and $p$. Hence
$$\mathbf{E}\left(\mathbf{X}_k\right)=0 \cdot q+1 \cdot p=p$$
and since
$$\mathrm{S}_n=\mathbf{X}_1+\mathbf{X}_2+\cdots+\mathbf{X}_n$$
we get (3.1) directly from (2.4).
(b) Poisson distribution. If $\mathbf{X}$ has the Poisson distribution $p(k ; \lambda)=$ $=e^{-\lambda} \lambda^k / k$ ! (where $\left.k=0,1, \ldots\right)$ then
$$\mathbf{E}(\mathbf{X})=\sum k p(k ; \lambda)=\lambda \sum p(k-1 ; \lambda) .$$

## 数学代写|概率论代写Probability theory代考|THE VARIANCE

Let $\mathbf{X}$ be a random variable with distribution $\left{f\left(x_j\right)\right}$, and let $r \geq 0$ be an integer. If the expectation of the random variable $\mathbf{X}^r$, that is,
$$\mathbf{E}\left(\mathbf{X}^r\right)=\sum x_j^r f\left(x_j\right),$$
exists, then it is called the rth moment of $\mathbf{X}$ about the origin. If the series does not converge absolutely, we say that the $r$ th moment does not exist. Since $|\mathbf{X}|^{r-1} \leq|\mathbf{X}|^r+1$, it follows that whenever the rth moment exists so does the $(r-1)$ st, and hence all preceding moments.

Moments play an important role in the general theory, but in the present volume we shall use only the second moment. If it exists, so does the mean (4.2)
$$\mu=\mathbf{E}(\mathbf{X}) .$$

It is then natural to replace the random variable $\mathbf{X}$ by its deviation from the mean, $\mathbf{X}-\mu$. Since $(x-\mu)^2 \leq 2\left(x^2+\mu^2\right)$ the second moment of $\mathbf{X}-\mu$ exists whenever $\mathbf{E}\left(\mathbf{X}^2\right)$ exists. It is given by
$$\mathbf{E}\left((\mathbf{X}-\mu)^2\right)=\sum_j\left(x_j^2-2 \mu x_j+\mu^2\right) f\left(x_j\right) .$$
Splitting the right side into three individual sums, we find it equal to $\mathbf{E}\left(\mathbf{X}^2\right)-2 \mu \mathbf{E}(\mathbf{X})+\mu^2=\mathbf{E}\left(\mathbf{X}^2\right)-\mu^2$.

Definition. Let $\mathbf{X}$ be a random variable with second moment $\mathbf{E}\left(\mathbf{X}^2\right)$ and let $\mu=\mathbf{E}(\mathbf{X})$ be its mean. We define a number called the variance of $\mathbf{X}$ by
$$\operatorname{Var}(\mathbf{X})=\mathbf{E}\left((\mathbf{X}-\mu)^2\right)=\mathbf{E}\left(\mathbf{X}^2\right)-\mu^2 .$$
Its positive square root (or zero) is called the standard deviation of $\mathbf{X}$.
For simplicity we often speak of the variance of a distribution without mentioning the random variable. “Dispersion” is a synonym for the now generally accepted term “variance.”

# 概率论代考

## 数学代写|概率论代写Probability theory代考|EXAMPLES AND APPLICATIONS

(a)二项分布。设$\mathbf{S}_n$为$n$伯努利试验的成功次数，成功的概率为$p$。我们知道$\mathbf{S}_n$有二项分布${b(k ; n, p)}$，因此$\mathbf{E}\left(\mathbf{S}_n\right)=\sum k b(k ; n, p)=$$=n p \sum b(k-1 ; n-1, p)。最后一个和包含n-1二项分布的所有项，因此等于1。因此二项分布的均值是$$ \mathbf{E}\left(\mathbf{S}_n\right)=n p . $$用一种常常是权宜之计的方法，不用计算也可以得到同样的结果。设\mathbf{X}_k为k第1次试验的成功次数。此随机变量仅假设值0和1具有相应的概率q和p。因此$$ \mathbf{E}\left(\mathbf{X}_k\right)=0 \cdot q+1 \cdot p=p $$既然$$ \mathrm{S}_n=\mathbf{X}_1+\mathbf{X}_2+\cdots+\mathbf{X}_n $$我们直接从(2.4)得到(3.1)。 (b)泊松分布。如果\mathbf{X}有泊松分布p(k ; \lambda)=$$=e^{-\lambda} \lambda^k / k$ !(其中$\left.k=0,1, \ldots\right)$ then
$$\mathbf{E}(\mathbf{X})=\sum k p(k ; \lambda)=\lambda \sum p(k-1 ; \lambda) .$$

## 数学代写|概率论代写Probability theory代考|THE VARIANCE

$$\mathbf{E}\left(\mathbf{X}^r\right)=\sum x_j^r f\left(x_j\right),$$

$$\mu=\mathbf{E}(\mathbf{X}) .$$

$$\mathbf{E}\left((\mathbf{X}-\mu)^2\right)=\sum_j\left(x_j^2-2 \mu x_j+\mu^2\right) f\left(x_j\right) .$$

$$\operatorname{Var}(\mathbf{X})=\mathbf{E}\left((\mathbf{X}-\mu)^2\right)=\mathbf{E}\left(\mathbf{X}^2\right)-\mu^2 .$$

## 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|概率论代写Probability theory代考|MATH407

statistics-lab™ 为您的留学生涯保驾护航 在代写概率论Probability theory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论Probability theory代写方面经验极为丰富，各种代写概率论Probability theory相关的作业也就用不着说。

## 数学代写|概率论代写Probability theory代考|INTERPRETATION IN NUMBER THEORY LANGUAGE

Let $x$ be a real number in the interval $0 \leq x<1$, and let
$$x=. a_1 a_2 a_3 \cdots$$
be its decimal expansion (so that each $a_j$ stands for one of the digits $0,1, \ldots, 9)$. This expansion is unique except for numbers of the form $a / 10^n$ (where $a$ is an integer), which can be written either by means of an expansion containing infinitely many zeros or by means of an expansion containing infinitely many nines. To avoid ambiguities we now agree not to use the latter form.

The decimal expansions are connected with Bernoulli trials with $p=\frac{1}{10}$, the digit 0 representing success and all other digits failure. If we replace in (6.1) all zeros by the letter $S$ and all other digits by $F$, then (6.1) represents a possible outcome of an infinite sequence of Bernoulli trials with $p=\frac{1}{10}$. Conversely, an arbitrary sequence of letters $S$ and $F$ can be obtained in the described manner from the expansion of certain numbers $x$. In this way every event in the sample space of Bernoulli trials is represented by a certain aggregate of numbers $x$. For example, the event “success at the $n$th trial” is represented by all those $x$ whose $n$th decimal is zero. This is an aggregate of $10^{n-1}$ intervals each of length $10^{-n}$, and the total length of these intervals equals $\frac{1}{10}$, which is the probability of our event. Every particular finite sample sequence of length $n$ corresponds to an aggregate of certain intervals; for example, the sequence $S F S$ is represented by the nine intervals $0.01 \leq x<0.011,0.02 \leq x<0.021, \ldots$, $0.09 \leq x<0.091$. The probability of each such sample sequence equals the total length of the corresponding intervals on the $x$-axis. Probabilities of more complicated events are always expressed in terms of probabilities of finite sample sequences, and the calculation proceeds according to the same addition rule that is valid for the familiar Lebesgue measure on the $x$-axis. Accordingly, our probabilities will always coincide with the measure of the corresponding aggregate of points on the $x$-axis. We have thus a means of translating all limit theorems for Bernoulli trials with $p=\frac{1}{10}$ into theorems concerning decimal expansions. The phrase “with probability one” is equivalent to “for almost all $x$ ” or “almost everywhere.”

## 数学代写|概率论代写Probability theory代考|PROBLEMS FOR SOLUTION

1. Find an integer $\beta$ such that in rolling dicc therc are about even chances that a run of three consecutive aces appears before a non-ace run of length $\beta$.
2. Consider repeated independent trials with three possible outcomes $A, B$, $C$ and corresponding probabilities $p, q, r(p+q+r=1)$. Find the probability that a run of $\alpha$ consecutive $A$ ‘s will occur before a $B$-run of length $\beta$.
3. Continuation. Find the probability that an $A$-run of length $\propto$ will occur before either a $B$-run of length $\beta$ or a $C$-run of length. $\gamma$.
4. In a sequence of Bernoulli trials let $A_n$ be the event that a run of $n$ consecutive successes occurs between the $2^n$ th and the $2^{n+1}$ st trial. If $p \geq \frac{1}{2}$, there is probability one that infinitely many $A_n$ occur; if $p<\frac{1}{2}$, then with probability one only finitely many $A_n$ occur. $5 .{ }^7$ Denote by $\mathbf{N}_n$ the length of the success run beginning at the $n$th trial (i.e., $\mathbf{N}_n=0$ if the $n$th trial results in $F$, etc.). Prove that with probability one $$\lim \sup \frac{\mathbf{N}_n}{\log n}=1$$ where $\log$ denotes the logarithm to the basis $1 / p$. Hint: Consider the event $A_n$ that the $n$th trial is followed by a run of more than $a \log n$ successes. For $a>1$ the calculation is straightforward. For $a<1$ consider the subsequence of trials number $a_1, a_2, \ldots$ where $a_n$ is an integer very close to $n \log n$.
5. From the law of the iterated logarithm conclude: With probability one it will happen for infinitely many $n$ that all $\mathbf{S}_k^*$ with $n<k<17 n$ are positive. (Note: Considerably stronger statements can be proved using the results of chapter III.)
6. Let $\phi(t)$ be a positive monotonically increasing function, and let $n_r$ be the nearest integer to $e^{r / \log r}$. If
$$\sum \frac{1}{\phi\left(n_r\right)} e^{-\frac{1}{2} \phi^2\left(n_r\right)}$$
converges, then with probability one, the inequality
$$\mathbf{S}_n>n p+\sqrt{n p q} \phi(n)$$
takes place only for infinitely many $n$. Note that without loss of generality we may suppose that $\phi(n)<10 \sqrt{\log \log n}$; the law of the iterated logarithm takes care of the larger $\phi(n)$.
1. Prove $^8$ that the series (7.2) converges if, and only if,
$$\sum \frac{\phi(n)}{n} e^{-\frac{1}{2} \phi^2(n)}$$
converges. Hint: Collect the terms for which $n_{r-1}2 \log \log n$.
2. From the preceding problem conclude that with probability one
$$\lim \sup \left[\mathbf{S}_n^*-\sqrt{2 \log \log n}\right] \frac{\sqrt{2 \log \log n}}{\log \log \log n}=\frac{3}{2} .$$

# 概率论代考

## 数学代写|概率论代写Probability theory代考|INTERPRETATION IN NUMBER THEORY LANGUAGE


x =。A_1 a_2 a_3 \cdots


\mathbf{E}(Y \mid B)=\mathbf{E}\left(Y \mathbf{1}_B\right) / \mathbf{P}(B) .
$$没有出现严重的困难。 另一方面，如果\mathbf{P}(B)=0，那么这种方法根本不起作用。实际上，在这种情况下，如何定义\mathbf{P}(Y \in S \mid B)之类的东西是相当不清楚的。不幸的是，我们经常希望以概率为0的事件为条件。 ## 数学代写|概率论代写Probability theory代考|Conditioning on a random variable 例13.1.1。设(X, Y)均匀分布在三角形T=\left{(x, y) \in \mathbf{R}^2 ; 0 \leq y \leq 2, y \leq x \leq 2\right}上;如图13.1.2所示。(即对于Borel S \subseteq \mathbf{R}^2为\mathbf{P}((X, Y) \in S)=\frac{1}{2} \lambda_2(S \cap T)，其中\lambda_2为\mathbf{R}^2的勒贝格测度;网址:d \mathbf{P}=\frac{1}{2} \mathbf{1}_T d x d y。)那么\mathbf{P}\left(Y>\frac{3}{4} \mid X=1\right)是什么?\mathbf{E}(Y \mid X=1)是什么?由于\mathbf{P}(X=1)=0，目前尚不清楚如何进行。我们将在下面回到这个例子。 由于这个问题，我们采取了不同的方法。给定一个随机变量X，我们将认为条件概率(如\mathbf{P}(A \mid X))和条件期望值(如\mathbf{E}(Y \mid X))本身都是随机变量。我们将把它们看作是“随机”值X的函数。这是非常违反直觉的:我们习惯于将\mathbf{P}(\cdots)和\mathbf{E}(\cdots)视为数字，而不是随机变量。然而，我们将把它们视为随机变量，我们将看到这允许我们部分地解决在测度0集合上的条件反射困难(如上文{X=1})。 我们的想法是，一旦我们将这些量定义为随机变量，那么我们就可以要求它们满足某些性质。首先，我们需要这个$$
\mathbf{E}[\mathbf{P}(A \mid X)]=\mathbf{P}(A), \quad \mathbf{E}[\mathbf{E}(Y \mid X)]=\mathbf{E}(Y) .
$$换句话说，这些随机变量必须有正确的期望值。 不幸的是，这并没有完全指定随机变量\mathbf{P}(A \mid X)和\mathbf{E}(Y \mid X)的分布;实际上，有无穷多个不同的分布具有相同的均值。因此，我们将提出更严格的要求。为了说明这一点，回想一下，如果\mathcal{G}是一个子\sigma -代数(即\sigma -代数包含在主\sigma -代数\mathcal{F})，那么一个随机变量Z是\mathcal{G} -可测量的，如果{Z \leq z} \in \mathcal{G}对于所有z \in \mathbf{R}。(这也是{Z=z}=$${Z \leq z} \backslash \bigcup_n\left{Z \leq z-\frac{1}{n}\right} \in \mathcal{G}$。)另外，$\sigma(X)={{X \in B}: B \subseteq \mathbf{R}$Borel$}\$。

## 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。