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

Posted on 2023年8月22日2023年8月22日 by statistics-lab

如果你也在怎样代写概率论Probability Theory 这个学科遇到相关的难题，请随时右上角联系我们的24/7代写客服。概率论Probability Theory是研究与随机现象有关的概率的数学分支。一个随机现象可能有几种结果。概率论用一定的形式概念描述某一特定结果发生的几率。

概率论Probability Theory某些随机变量在概率论中经常出现，因为它们很好地描述了许多自然或物理过程。因此，它们的分布在概率论中具有特殊的重要性。一些基本的离散分布有离散均匀分布、伯努利分布、二项式分布、负二项式分布、泊松分布和几何分布。重要的连续分布包括连续均匀分布、正态分布、指数分布、分布和分布。

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

我们将建立任意随机变量的分布函数的一些一般性质。我们需要两个关于概率度量的事实。
定理1。设$(\Omega, \mathscr{F}, P)$为概率空间。
(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

我们将研究同时处理在同一样本空间上定义的几个随机变量的情况。作为一个介绍性的例子，假设从一定的人群中随机选择一个人，并记录他的年龄和体重。我们可以取实数对的集合$(x, y)$作为样本空间，即欧几里得平面$E^2$，其中$x$表示年龄，$y$表示权重。设$R_1$为入选人的年龄，$R_2$为权重;也就是$R_1(x, y)=x, R_2(x, y)=y$。我们希望为同时涉及$R_1$和$R_2$的事件分配概率。可用数据的横截面可能如图2.6.1所示。因此，有400万人的年龄在20到25岁之间，(同时)体重在150到160磅之间，以此类推。现在假设我们希望估计年龄在22到23岁之间，体重在154到156磅之间的人的数量。有400万人分布在5年和10英镑，或每年4/ 5000万英镑。我们对1年2磅的范围感兴趣，因此我们的估计是$4 / 50 \times 1 \times 2=8 / 50$百万(参见图2.6.2)。如果总人口是2亿，那么
$$
P\left{22 \leq R_1 \leq 23,154 \leq R_2 \leq 156\right}
$$
应该近似
$$
\frac{8 / 50}{200}=.0008
$$
符号。$\left{22 \leq R_1 \leq 23,154 \leq R_2 \leq 156\right}$分别代表$\left{22 \leq R_1 \leq 23\right.$和$\left.154 \leq R_2 \leq 156\right}$。

数学代写|概率论代写Probability theory代考请认准statistics-lab™

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

金融工程代写

金融工程是使用数学技术来解决金融问题。金融工程使用计算机科学、统计学、经济学和应用数学领域的工具和知识来解决当前的金融问题，以及设计新的和创新的金融产品。

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

术语广义线性模型（GLM）通常是指给定连续和/或分类预测因素的连续响应变量的常规线性回归模型。它包括多元线性回归，以及方差分析和方差分析（仅含固定效应）。

有限元方法代写

有限元方法（FEM）是一种流行的方法，用于数值解决工程和数学建模中出现的微分方程。典型的问题领域包括结构分析、传热、流体流动、质量运输和电磁势等传统领域。

有限元是一种通用的数值方法，用于解决两个或三个空间变量的偏微分方程（即一些边界值问题）。为了解决一个问题，有限元将一个大系统细分为更小、更简单的部分，称为有限元。这是通过在空间维度上的特定空间离散化来实现的，它是通过构建对象的网格来实现的：用于求解的数值域，它有有限数量的点。边界值问题的有限元方法表述最终导致一个代数方程组。该方法在域上对未知函数进行逼近。[1] 然后将模拟这些有限元的简单方程组合成一个更大的方程系统，以模拟整个问题。然后，有限元通过变化微积分使相关的误差函数最小化来逼近一个解决方案。

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随机分析代写

随机微积分是数学的一个分支，对随机过程进行操作。它允许为随机过程的积分定义一个关于随机过程的一致的积分理论。这个领域是由日本数学家伊藤清在第二次世界大战期间创建并开始的。

时间序列分析代写

随机过程，是依赖于参数的一组随机变量的全体，参数通常是时间。随机变量是随机现象的数量表现，其时间序列是一组按照时间发生先后顺序进行排列的数据点序列。通常一组时间序列的时间间隔为一恒定值（如1秒，5分钟，12小时，7天，1年），因此时间序列可以作为离散时间数据进行分析处理。研究时间序列数据的意义在于现实中，往往需要研究某个事物其随时间发展变化的规律。这就需要通过研究该事物过去发展的历史记录，以得到其自身发展的规律。

回归分析代写

多元回归分析渐进（Multiple Regression Analysis Asymptotics）属于计量经济学领域，主要是一种数学上的统计分析方法，可以分析复杂情况下各影响因素的数学关系，在自然科学、社会和经济学等多个领域内应用广泛。

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，工具箱允许您学习和应用专业技术。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。

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

数学代写|概率论代写Probability theory代考|SF2940

Posted on 2023年8月22日2023年8月22日 by statistics-lab

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

在数值计算和理论分析中都很重要的对$n !$的估计是斯特林公式
$$
n ! \sim n^n e^{-n} \sqrt{2 \pi n}
$$
从某种意义上说
$$
\lim {n \rightarrow \infty} \frac{n !}{\left(n^n e^{-n} \sqrt{2 \pi n}\right)}=1 $$证明。定义$(2 n) !$ !(读取$2 n$半阶乘)为$2 n(2 n-2)(2 n-4) \cdots$$6(4)(2)$, $(2 n+1) ! !$为$(2 n+1)(2 n-1) \cdots(5)(3)(1)$。我们首先表明(a) $$ \frac{(2 n) ! !}{(2 n+1) ! !}<\frac{\pi}{2} \frac{(2 n-1) ! !}{(2 n) ! !}<\frac{(2 n-2) ! !}{(2 n-1) ! !} $$让$I_k=\int_0^{\pi / 2}(\cos x)^k d x, k=0,1,2, \ldots$然后$I_0=\pi / 2, I_1=1$。分部积分，得到$I_k=\int_0^{\pi / 2}(\cos x)^{k-1} d(\sin x)=\int_0^{\pi / 2}(k-1)(\cos x)^{k-2}$$\sin ^2 x d x$。因为$\sin ^2 x=1-\cos ^2 x$，我们有$I_k=(k-1) I{k-2}-(k-1) I_k$或$I_k=[(k-1) / k] I_{k-2}$。通过迭代得到$I_{2 n}=(\pi / 2)[(2 n-1) ! ! /$$(2 n) ! !]$和$I_{2 n+1}=[(2 n) ! ! /(2 n+1) ! !]$。由于$(\cos x)^k$随$k$减小，$I_k$也减小，因此$I_{2 n+1}<I_{2 n}<I_{2 n-1}$得到证明。
(b)让$Q_n=\left(\begin{array}{c}2^n \ n\end{array}\right) / 2^{2 n}$。然后
$$
\lim _{n \rightarrow \infty} Q_n \sqrt{n \pi}=1
$$
为了证明这一点，写下来
$$
\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}
$$

因此，通过(a)，
$$
\frac{(2 n) ! !}{(2 n+1) ! !}<\frac{\pi}{2} Q_n<\frac{(2 n-2) ! !}{(2 n-1) ! !}
$$
将这个不等式乘以
$$
\frac{(2 n-1) ! !}{(2 n-2) ! !}=\frac{(2 n-1) ! !}{(2 n) ! !} \frac{(2 n) ! !}{(2 n-2) ! !}=Q_n(2 n)
$$
获取
$$
\frac{2 n}{2 n+1}<n \pi Q_n{ }^2<1
$$
令$n \rightarrow \infty$，得到$n \pi Q_n{ }^2 \rightarrow 1$，证明(b)。
(c)斯特林公式的证明。让$c_n=n ! / n^n e^{-n} \sqrt{2 \pi n}$。我们必须将$c_n \rightarrow 1$表示为$n \rightarrow \infty$。考虑$(n+1) ! / n !=n+1$。我们有
$$
\begin{aligned}
\frac{(n+1) !}{n !} & =\frac{c_{n+1}(n+1)^{n+1} e^{-(n+1)} \sqrt{2 \pi(n+1)}}{c_n n^n e^{-n} \sqrt{2 \pi n}} \
& =\left(\frac{c_{n+1}}{c_n}\right) e^{-1}\left(\frac{n+1}{n}\right)^n \frac{(n+1)^{3 / 2}}{\sqrt{n}}
\end{aligned}
$$

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

直观地说，随机变量是在随机实验中测量的量。如果$\Omega$是一个样本空间，实验结果为$\omega$，则进行测量过程，得到一个数字$R(\omega)$。因此，随机变量是样本空间上的实值函数。(正式的定义将推迟到本节后面的部分，它在某种程度上更具限制性。)

例1。投掷硬币10次，设$R$为正面的次数。我们取$\Omega=$所有长度为10的序列，其组成部分为$H$和$T ; 2^{10}$。一个典型的样本点是$\omega=H H T H T T H H T H$。对于这一点$R(\omega)=6$。另一个随机变量$R_1$是正面紧接着反面的次数。对于上面的$\omega$点，$R_1(\omega)=3$。
例2。从一定的人群中随机挑选一个人，测量他的身高和体重。我们可以将样本空间作为平面$E^2$，即所有实数对$(x, y)$的集合，第一个坐标$x$表示高度，第二个坐标$y$表示权重(我们可以通过为第一象限的补赋概率0来满足高度和权重非负的要求)。设$R_1$为被选者的身高，$R_2$为体重。然后$R_1(x, y)=x, R_2(x, y)=y$。另一个例子，设$R_3$等于两倍的高度加上重量的立方根;也就是$R_3=2 R_1+\sqrt[3]{R_2}$。然后是$R_3(x, y)=$$2 R_1(x, y)+\sqrt[3]{R_2(x, y)}=2 x+\sqrt[3]{y}$。

例3。掷两个骰子。我们可以取样本空间为所有整数对的集合$(x, y), x, y=1,2, \ldots, 6$(总共36个点)。
让$R_1=$公布第一次掷硬币的结果。然后$R_1(x, y)=x$。
设$R_2=$为两个面之和。然后$R_2(x, y)=x+y$。
设$R_3=1$，如果至少有一个面是偶数;$R_3=0$否则。
然后是$R_3(6,5)=1 ; R_3(3,6)=1 ; R_3(1,3)=0$，等等。

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

金融工程代写

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

有限元方法代写

随机分析代写

时间序列分析代写

回归分析代写

MATLAB代写

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

数学代写|概率论代写Probability theory代考|MAP4102

Posted on 2023年8月22日2023年8月22日 by statistics-lab

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

我们考虑一类问题，其中概率的分配可以用一种自然的方式进行。

设$\ ω $是一个有限或可数无限集，且设$\mathscr{F}$由$\ ω $的所有子集组成。

对于每个点$\omega_i \in \omega_i, i=1,2， \ldots$，赋一个非负数$p_i$，其中$\sum_i p_i=1$。如果$A$是$\Omega$的任意子集，令$P(A)=\sum_{\omega_i \in A} p_i$。然后可以验证$P$是一个概率测度;$P\left{\omega_i\right}=p_i$，任意事件$A$的概率通过将$A$各点的概率相加得到。这种类型的$(\Omega， \mathscr{F}， P)$称为离散概率空间。
例1。投掷一枚(有偏的)硬币两次(见图1.4.1)。
设$E_1=${至少一个头部$}$。然后

和
$$
P\left(E_2\right)=P\left(A_3\right)+P\left(A_4\right)=.4
$$
在$\Omega=\left{\omega_1, \ldots, \omega_n\right}$和$p_i=1 / n, i=1,2, \ldots, n$的特殊情况下，我们有
$$
P(A)=\frac{\text { number of points of } A}{\text { total number of points in } \Omega}=\frac{\text { favorable outcomes }}{\text { total outcomes }}
$$
对应于概率的经典定义。

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

有序序列的数目 $\left(a_{i_1}, \ldots, a_{i_r}\right)$，其中 $a_{i_k}$ 属于 $\left{a_1, \ldots, a_n\right}$是吗? $n \times n \times \cdots \times n$ （ $r$ 次数)，或
$$
n^r
$$
(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。)

例如，如果投掷三个骰子，可能的结果数是$6 \times 6 \times 6=216$。
订购样品尺寸$r$，不得更换
有序序列$\left(a_{i_1}, \ldots, a_{i_r}\right)$的个数为，其中$a_{i_k}$属于$\left{a_1, \ldots, a_n\right}$，但不允许重复(即$a_i$不能在序列中出现多次)
$$
n(n-1) \cdots(n-r+1)=\frac{n !}{(n-r) !}, \quad r=1,2, \ldots, n
$$
(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$(n)_r$。

例如，$1,2, \ldots, 9$可以组成的3位数个数，如果不能重复，则为$9(8)(7)=504$。

未订购的样品尺寸$r$，无更换
无序集合$\left{a_{i_1}, \ldots, a_{i_r}\right}$的数量，其中$a_{i_k}, k=1, \ldots, r$是$\left{a_1, \ldots, a_n\right}$的不同元素(即，从$n$中选择$r$不同对象的方法的数量)，如果不考虑顺序，则为
$$
\left(\begin{array}{l}
n \
r
\end{array}\right)=\frac{n !}{r !(n-r) !}
$$
要了解这一点，请考虑以下过程。
(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)。

执行(a)和(b)的结果是$n$中$r$对象的排列;因此
$$
\left(\begin{array}{l}
n \
r
\end{array}\right) r !=(n)_r=\frac{n !}{(n-r) !}
$$
或
$$
\left(\begin{array}{l}
n \
r
\end{array}\right)=\frac{n !}{r !(n-r) !}, \quad r=1,2, \ldots, n
$$
我们将$\left(\begin{array}{l}n \ 0\end{array}\right)$定义为$n ! / 0 ! n !=1$，以使$\left(\begin{array}{l}n \ r\end{array}\right)$的公式对$r=0,1, \ldots, n$有效。注意$\left(\begin{array}{c}n \ k\end{array}\right)=\left(\begin{array}{c}n \ n-k\end{array}\right)$。

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

金融工程代写

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

有限元方法代写

随机分析代写

时间序列分析代写

回归分析代写

MATLAB代写

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

数学代写|概率论代写Probability theory代考|DTSA5001

Posted on 2023年8月17日2023年8月17日 by statistics-lab

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。

为了确定思路，我们将设想给定一个无限的概率分布序列，对于每一个$n$，我们都有$n$相互独立的变量$\mathbf{X}_1, \ldots, \mathbf{X}_n$和规定的分布。我们假设均值和方差存在，然后
$$
\mu_k=\mathbf{E}\left(\mathbf{X}_k\right), \quad \sigma_k^2=\operatorname{Var}\left(\mathbf{X}_k\right)
$$
和$\mathbf{S}_n=\mathbf{X}_1+\cdots+\mathbf{X}_n$的均值$m_n$和方差$s_n^2$为
$$
m_n=\mu_1+\cdots+\mu_n, \quad s_n^2=\sigma_1^2+\cdots+\sigma_n^2
$$

[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
$$
\mathbf{P}\left{\frac{\left|\mathbf{S}_n-m_n\right|}{n}>\epsilon\right} \rightarrow 0 .
$$
序列$\left{\mathbf{X}_k\right}$被认为服从中心极限定理，如果对于每一个固定的$\alpha<\beta$
$$
\mathbf{P}\left{\alpha<\frac{\mathbf{S}_n-m_n}{s_n}<\beta\right} \rightarrow \mathfrak{N}(\beta)-\mathfrak{N}(\alpha) .
$$
大数定律和中心极限定理都适用于大量惊人的数列$\left{\mathbf{X}_k\right}$，这是概率论的显著特征之一。大数定律尤其适用。只要$\mathbf{X}_k$是一致有界的，也就是说，只要存在一个常数$A$，使得$\left.\rfloor \mathbf{X}_k\right\rfloor<A$对所有$k$。更一般地说，大数定律成立的一个充分条件是
$$
\frac{s_n}{n} \rightarrow 0 .
$$

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

我们将举出两个例子，说明中心极限定理在与概率论没有直接联系的问题上的应用。两者都与$n$ !$n$元素的排列$a_1, a_2, \ldots, a_n$，每个元素的概率都是$1 / n$ !
(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$$ s_n^2=\frac{1}{12} \sum{k=1}^n\left(k^2-1\right)=\frac{2 n^3+3 n^2-5 n}{72} \sim \frac{n^3}{36} .
$$
对于较大的$n$，我们有$\epsilon S_n>n \geq \mathbf{U}_k$，因此林德堡条件的变量$\mathbf{U}_k$与$\mathbf{X}_k$相同。因此，中心极限定理适用，并且我们得出在极限$\frac{n^2}{4} \pm \frac{\alpha}{6} \sqrt{n^3}$之间的反转数的排列数$\mathbf{N}_n$渐近地由$n !{\mathfrak{N}(\alpha)-\mathfrak{N}(-\alpha)}$给出。特别地，对于大约一半的排列，反转的数量位于极限$\frac{1}{4} n^2 \pm 0.11 \sqrt{n^3}$之间。

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广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

有限元方法代写

随机分析代写

时间序列分析代写

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数学代写|概率论代写Probability theory代考|Stat410

Posted on 2023年8月17日2023年8月17日 by statistics-lab

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

当我们考虑到成功次数$\mathbf{S}_n$与试验次数$n$的相关性时，伯努利试验与随机变量理论之间的联系就变得更加清晰了。每次试验$\mathbf{S}_n$增加1或0，我们可以写
$$
\mathbf{S}_n=\mathbf{X}_1+\cdots+\mathbf{X}_n,
$$
其中，如果$k$次试验成功，随机变量$\mathbf{X}_k$等于1，否则为零。因此$\mathbf{S}_n$是$n$相互独立的随机变量的和，每个随机变量都假设值1和0的概率分别为$p$和$q$。由此，只需要一步就可以考虑(1.1)式的和，其中$\mathbf{X}_k$是任意分布的相互独立变量。(弱)大数定律(VI,4)指出，对于大的$n$，成功的平均比例$\mathbf{S}_n / n$可能位于$p$附近。这是下列情况中的一个特例

大数定律。设$\left{\mathbf{X}_k\right}$为具有共同分布的相互独立的随机变量序列。如果期望$\mu=\mathbf{E}\left(\mathbf{X}_k\right)$存在，那么对于每个$\epsilon>0$都是$n>\infty$
$$
\mathbf{P}\left{\left|\frac{\mathbf{X}_1+\cdots+\mathbf{X}_n}{n}-\mu\right|>\epsilon\right} \rightarrow 0 ;
$$
换句话说，平均值$\mathbf{S}_n / n$与期望的差异小于任意规定的$\epsilon$的概率趋向于1。

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

假设$\mu=\mathbf{E}\left(\mathbf{X}_k\right)=0$并没有丧失一般性，否则我们就会用$\mathbf{X}_k-\mu$来代替$\mathbf{X}_k$，而这仅仅涉及到符号的改变。在$\sigma^2=\operatorname{Var}\left(\mathbf{X}_k\right)$存在的特殊情况下，大数定律是切比雪夫不等式IX，(6.2)的一个平凡结果
$$
\mathbf{P}\left{\left|\mathbf{S}_n\right|>t\right} \leq \frac{n \sigma^2}{t^2} .
$$
对于$t=\epsilon n$，右侧趋向于0，因此(1.2)为真。
第二时刻不存在的情况就比较困难。证明依赖于截断的通用方法，它是推导各种极限定理的标准工具。设$\delta$为稍后确定的正常数。对于每个$n$，我们定义$n$对随机变量，如下所示。
$$
\begin{aligned}
& \mathbf{U}_k=\mathbf{X}_k, \quad \mathbf{V}_k=0 \quad \text { if } \quad\left|\mathbf{X}_k\right| \leq \delta n, \
& \mathbf{U}_k=0, \quad \mathbf{V}_k=\mathbf{X}_k \quad \text { if } \quad\left|\mathbf{X}_k\right|>\delta n . \
&
\end{aligned}
$$
这里必须记住$k=1, \ldots, n$以及$\mathbf{U}_k$和$\mathbf{V}_k$对$n$的依赖性。根据这个定义
$$
\mathbf{X}_k=\mathbf{U}_k+\mathbf{V}_k
$$
为了证明大数定律，只要证明对于给定的$\epsilon>0$常数$\delta$可以选择为$n \rightarrow \infty$就足够了
$$
\mathbf{P}\left{\left|\mathbf{U}_1+\cdots+\mathbf{U}_n\right|>\frac{1}{2} \epsilon n\right} \rightarrow 0
$$
和
$$
\mathbf{P}\left{\left|\mathbf{V}_1+\cdots+\mathbf{V}_n\right|>\frac{1}{2} \epsilon n\right} \rightarrow 0 . \quad\left(\because \leq\left{\left|\frac{\sum U_k}{n}+\frac{\sum V_k}{n}\right|>\right.\right.
$$

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

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非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

有限元方法代写

随机分析代写

时间序列分析代写

回归分析代写

MATLAB代写

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

数学代写|概率论代写Probability theory代考|STAT414

Posted on 2023年8月17日2023年8月28日 by statistics-lab

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

设$\mathbf{X}$为分布为$\left{f\left(x_j\right)\right}$的随机变量，设$r \geq 0$为整数。如果随机变量$\mathbf{X}^r$的期望，即
$$
\mathbf{E}\left(\mathbf{X}^r\right)=\sum x_j^r f\left(x_j\right),
$$
存在，则称为$\mathbf{X}$关于原点的RTH矩。如果级数不绝对收敛，我们说$r$力矩不存在。既然$|\mathbf{X}|^{r-1} \leq|\mathbf{X}|^r+1$，那么只要第n个矩存在，那么$(r-1)$ st也存在，因此所有之前的矩都存在。

矩在一般理论中起着重要的作用，但在本卷中，我们将只使用第二个矩。如果它存在，均值也存在(4.2)
$$
\mu=\mathbf{E}(\mathbf{X}) .
$$

然后很自然地用随机变量与均值的偏差$\mathbf{X}-\mu$来代替随机变量$\mathbf{X}$。因为$(x-\mu)^2 \leq 2\left(x^2+\mu^2\right)$只要$\mathbf{E}\left(\mathbf{X}^2\right)$存在，$\mathbf{X}-\mu$的第二个矩就存在。它是由
$$
\mathbf{E}\left((\mathbf{X}-\mu)^2\right)=\sum_j\left(x_j^2-2 \mu x_j+\mu^2\right) f\left(x_j\right) .
$$
把右边分成三个单独的和，我们发现它等于$\mathbf{E}\left(\mathbf{X}^2\right)-2 \mu \mathbf{E}(\mathbf{X})+\mu^2=\mathbf{E}\left(\mathbf{X}^2\right)-\mu^2$。

定义。设$\mathbf{X}$为二阶矩为$\mathbf{E}\left(\mathbf{X}^2\right)$的随机变量，$\mu=\mathbf{E}(\mathbf{X})$为其均值。我们定义了一个叫做$\mathbf{X}$的方差
$$
\operatorname{Var}(\mathbf{X})=\mathbf{E}\left((\mathbf{X}-\mu)^2\right)=\mathbf{E}\left(\mathbf{X}^2\right)-\mu^2 .
$$
它的正平方根(或零)称为$\mathbf{X}$的标准差。
为简单起见，我们经常说分布的方差而不提及随机变量。“分散”是现在普遍接受的术语“方差”的同义词。

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

金融工程代写

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

有限元方法代写

随机分析代写

时间序列分析代写

回归分析代写

MATLAB代写

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

数学代写|概率论代写Probability theory代考|RELATION TO THE POISSON APPROXIMATION

Posted on 2023年7月5日2023年7月5日 by statistics-lab

如果你也在怎样代写概率论Probability Theory 这个学科遇到相关的难题，请随时右上角联系我们的24/7代写客服。概率论Probability Theory作为统计学的数学基础，对许多涉及数据定量分析的人类活动至关重要。概率论的方法也适用于对复杂系统的描述，只对其状态有部分了解，如在统计力学或顺序估计。二十世纪物理学的一个伟大发现是量子力学中描述的原子尺度的物理现象的概率性质。

概率论Probability Theory Math37500的核心课题包括离散和连续随机变量、概率分布和随机过程（为非决定性或不确定的过程或测量量提供数学抽象，这些过程或测量量可能是单一发生的，或以随机方式随时间演变）。尽管不可能完美地预测随机事件，但对它们的行为可以有很多说法。概率论中描述这种行为的两个主要结果是大数法则和中心极限定理。概率论是与概率有关的数学分支。虽然有几种不同的概率解释，但概率论以严格的数学方式处理这一概念，通过一组公理来表达它。

数学代写|概率论代写Probability theory代考|RELATION TO THE POISSON APPROXIMATION

The error of the normal approximation will be small if $n p q$ is large. On the other hand, if $n$ is large and $p$ small, the terms $b(k ; n, p)$ will be found to be near the Poisson probabilities $p(k ; \lambda)$ with $\lambda=n p$. For small $\lambda$ only the Poisson approximation can be used, but for large $\lambda$ we can use either the normal or the Poisson approximation. This implies that for large values of $\lambda$ it must be possible to approximate the Poisson distribution by the normal distribution, and in example X, (1.c) we shall see that this is indeed so (cf. also problem 9). Here we shall be content to illustrate the point by a numerical and a practical example.

Examples. (a) The Poisson distribution with $\lambda=100$ attributes to the set of integers $a, a+1, \ldots, b$ the probability
$$
P(a, b)=p(a ; 100)+p(a+1 ; 100)+\cdots+p(b ; 100) .
$$
This Poisson distribution may be considered as an approximation to the binomial distribution with $n=100,000,000$ and $p=10^{-6}$. Then $n p q \approx 100$ and so it is not far-fetched to approximate this binomial distribution by the normal, at least for values close to the central term 100 . But this means that $P(a, b)$ is being approximated by
$$
\mathfrak{N}((b-99.5) / 10)-\mathfrak{N}((a-100.5) / 10) .
$$

The following sample gives an idea of the degree of approximation.
$\begin{array}{lcc} & \text { Correct values } & \text { Normal approximation } \ P(85,90) & 0.11384 & 0.11049 \ P(90,95) & 0.18485 & 0.17950 \ P(95,105) & 0.41763 & 0.41768 \ P(90,110) & 0.70652 & 0.70628 \ P(110,115) & 0.10738 & 0.11049 \ P(115,120) & 0.05323 & 0.05335\end{array}$
(b) A telephone trunking problem. The following problem is, with some simplifications, taken from actual practice. ${ }^8$ A telephone exchange $A$ is to serve 2000 subscribers in a nearby exchange $B$. It would be too expensive and extravagant to install 2000 trunklines from $A$ to $B$. It suffices to make the number $N$ of lines so large that, under ordinary conditions, only one out of every hundred calls will fail to find an idle trunkline immediately at its disposal. Suppose that during the busy hour of the day each subscriber requires a trunkline to $B$ for an average of 2 minutes. At a fixed moment of the busy hour we compare the situation to a set of 2000 trials with a probability $p=\frac{1}{30}$ in each that a line will be required. Under ordinary conditions these trials can be assumed to be independent (although this is not true when events like unexpected showers or earthquakes cause many people to call for taxicabs or the local newspaper; the theory no longer applies, and the trunks will be “jammed”). We have, then, 2000 Bernoulli trials with $p=\frac{1}{30}$, and the smallest number $N$ is required such that the probability of more than $N$ “successes” will be smaller than 0.01 ; in symbols $\mathbf{P}\left{\mathbf{S}_{2000} \geq N\right}<0.01$.

数学代写|概率论代写Probability theory代考|LARGE DEVIATIONS

The DeMoivre-Laplace theorem describes the asymptotic behavior of $\mathbf{P}\left{z_1<\mathbf{S}_n^>z_1$. This is justified by the following lemma, which shows that when $z_1 \rightarrow \infty$ the upper limit $z_2$ plays no role. Lemma. If $x_n \rightarrow \infty$ then for every fixed $\eta>0$ $$ \frac{\mathbf{P}\left{\mathbf{S}_n^>x_n+\eta\right}}{\mathbf{P}\left{\mathbf{S}_n^>x_n\right}} \rightarrow 0 $$ that is, $$ \mathbf{P}\left{x_n<\mathbf{S}_n^ \leq x_n+\eta\right} \sim \mathbf{P}\left{\mathbf{S}_n^>x_n\right} . $$ In other words: When $\mathbf{S}_n^$ exceeds $x_n$ it is likely to be very close to $x_n$, and larger values play no role in the limit.

Proof. With the notation (3.2) for the binomial distribution we have
$$
\mathbf{P}\left{\mathbf{S}n^>x_n\right}=\sum{v=0}^{\infty} a_{r_n+v}, \quad \mathbf{P}\left{\mathbf{S}n^>x_n+\eta\right}=\sum{v=0}^{\infty} a_{s_n+v},
$$
where $r_n$ and $s_n$ are integers that differ at most by one unit from $x_n \sqrt{n p q}$ and $\left(x_n+\eta\right) \sqrt{n p q}$, respectively. Now it is obvious from (3.4)

that for large $n$
$$
\frac{a_{k+1}}{a_k}<1-p t_k<1-\frac{k}{n}<e^{k / n},
$$
and hence
$$
\frac{a_{s_n+v}}{a_{r_n+v}}<e^{-\left(s_n-r_n\right) r_n / n}<e^{-\frac{1}{2} \eta x_n p q} .
$$
By assumption $x_n \rightarrow \infty$, and so the terms of the second series in (6.3) tend to become negligible in comparison with the corresponding terms of the first series.

概率论代考

数学代写|概率论代写Probability theory代考|RELATION TO THE POISSON APPROXIMATION

如果$n p q$较大，则正态近似的误差较小。另一方面，如果$n$较大，$p$较小，则发现$b(k ; n, p)$项与$\lambda=n p$接近泊松概率$p(k ; \lambda)$。对于小的$\lambda$，只能使用泊松近似，但是对于大的$\lambda$，我们可以使用正态或泊松近似。这意味着，对于较大的$\lambda$值，一定可以用正态分布来近似泊松分布，在例X， (1.c)中，我们将看到确实是这样(同样参见问题9)。在这里，我们将满足于用一个数值和一个实际的例子来说明这一点。

例子。(a)具有$\lambda=100$属性的泊松分布为整数集$a, a+1, \ldots, b$的概率
$$
P(a, b)=p(a ; 100)+p(a+1 ; 100)+\cdots+p(b ; 100) .
$$
这种泊松分布可以看作是$n=100,000,000$和$p=10^{-6}$的二项分布的近似。然后是$n p q \approx 100$所以用正态分布近似这个二项分布并不牵强，至少对于接近中心项100的值。但这意味着$P(a, b)$被近似为
$$
\mathfrak{N}((b-99.5) / 10)-\mathfrak{N}((a-100.5) / 10) .
$$

下面的示例给出了近似程度的概念。
$\begin{array}{lcc} & \text { Correct values } & \text { Normal approximation } \ P(85,90) & 0.11384 & 0.11049 \ P(90,95) & 0.18485 & 0.17950 \ P(95,105) & 0.41763 & 0.41768 \ P(90,110) & 0.70652 & 0.70628 \ P(110,115) & 0.10738 & 0.11049 \ P(115,120) & 0.05323 & 0.05335\end{array}$
(b)电话干线问题。下面的问题经过一些简化，是根据实际情况提出的。${ }^8$电话交换机$A$将为附近交换机$B$的2000名用户提供服务。安装2000条从$A$到$B$的干线太昂贵和奢侈了。它足以使线路数量$N$如此之大，以至于在通常情况下，每100个呼叫中只有一个不能立即找到空闲的干线供其使用。假设在一天的繁忙时段，每个用户平均需要2分钟的主干线路到$B$。在繁忙时段的一个固定时刻，我们将这种情况与一组2000次试验进行比较，每次试验需要一条线路的概率为$p=\frac{1}{30}$。在一般情况下，这些试验可以被认为是独立的(尽管当诸如意外的阵雨或地震等事件导致许多人要求出租车或当地报纸时，情况并非如此;这个理论不再适用，后备箱将被“卡住”)。然后，我们有2000次伯努利试验$p=\frac{1}{30}$，并且需要最小的数字$N$，以便超过$N$的“成功”概率将小于0.01;用符号$\mathbf{P}\left{\mathbf{S}_{2000} \geq N\right}<0.01$表示。

数学代写|概率论代写Probability theory代考|LARGE DEVIATIONS

DeMoivre-Laplace定理描述了$\mathbf{P}\left{z_1<\mathbf{S}_n^>z_1$的渐近行为。下面的引理证明了这一点，即当$z_1 \rightarrow \infty$上限$z_2$不起作用时。引理。如果$x_n \rightarrow \infty$，那么对于每个固定的$\eta>0$$$ \frac{\mathbf{P}\left{\mathbf{S}_n^>x_n+\eta\right}}{\mathbf{P}\left{\mathbf{S}_n^>x_n\right}} \rightarrow 0 $$，即$$ \mathbf{P}\left{x_n<\mathbf{S}_n^ \leq x_n+\eta\right} \sim \mathbf{P}\left{\mathbf{S}_n^>x_n\right} . $$，换句话说:当$\mathbf{S}_n^$超过$x_n$时，它很可能非常接近$x_n$，较大的值在限制中不起作用。

证明。用(3.2)表示二项分布
$$
\mathbf{P}\left{\mathbf{S}n^>x_n\right}=\sum{v=0}^{\infty} a_{r_n+v}, \quad \mathbf{P}\left{\mathbf{S}n^>x_n+\eta\right}=\sum{v=0}^{\infty} a_{s_n+v},
$$
其中$r_n$和$s_n$是整数，它们与$x_n \sqrt{n p q}$和$\left(x_n+\eta\right) \sqrt{n p q}$的差别不超过1个单位。从(3.4)中可以明显看出

对于大的$n$
$$
\frac{a_{k+1}}{a_k}<1-p t_k<1-\frac{k}{n}<e^{k / n},
$$
因此
$$
\frac{a_{s_n+v}}{a_{r_n+v}}<e^{-\left(s_n-r_n\right) r_n / n}<e^{-\frac{1}{2} \eta x_n p q} .
$$
通过假设$x_n \rightarrow \infty$，因此(6.3)中第二个系列的项与第一个系列的相应项相比，往往变得可以忽略不计。

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

金融工程代写

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

有限元方法代写

随机分析代写

时间序列分析代写

回归分析代写

MATLAB代写

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

数学代写|概率论代写Probability theory代考|OBSERVATIONS FITTING THE POISSON DISTRIBUTI0N

Posted on 2023年7月5日2023年7月5日 by statistics-lab

数学代写|概率论代写Probability theory代考|OBSERVATIONS FITTING THE POISSON DISTRIBUTI0N

(a) Radioactive disintegrations. A radioactive substance emits $\alpha$ particles; the number of particles reaching a given portion of space during time $t$ is the best-known example of random events obeying the Poisson law. Of course, the substance continues to decay, and in the long run the density of $\alpha$-particles will decline. However, with radium it takes years before a decrease of matter can be detected; for relatively short periods the conditions may be considered constant, and we have an ideal realization of the hypotheses which led to the Poisson distribution.

In a famous experiment ${ }^{13}$ a radioactive substance was observed during $N=2608$ time intervals of 7.5 seconds each; the number of particles reaching a counter was obtained for each period. Table 3 records the number $N_k$ of periods with exactly $k$ particles. The total number of particles is $T=\sum k N_k=10,094$, the average $T / N=3.870$. The theoretical values $N p(k ; 3.870)$ are seen to be rather close to the observed numbers $N_k$. To judge the closeness of fit, an estimatc of the probable magnitude of chance fluctuations is required. Statisticians judge the closeness of fit by the $\chi^2$-criterion. Measuring by this standard, we should expect that under ideal conditions about 17 out of 100 comparable cases would show worse agreement than exhibited in table 3.
(b) Flying-bomb hits on London. As an example of a spatial distribution of random points consider the statistics of flying-bomb hits in the south of London during World War II. The entire area is divided into $N=576$ small areas of $t=\frac{1}{4}$ square kilometers each, and table 4 records the number $N_k$ of areas with exactly $k$ hits. ${ }^{14}$ The total number of hits is $T=\sum k N_k=537$, the average $\lambda t=T / N=0.9323 \ldots$ The fit of the Poisson distribution is surprisingly good; as judged by the $\chi^2$-criterion, under ideal conditions some 88 per cent of comparable observations should show a worse agreement. It is interesting to notc that most people believed in a tendency of the points of impact to cluster. If this were true, there would be a higher frequency of areas with either many hits or no hit and a deficiency in the intermediate classes. Table 4 indicates perfect randomness and homogeneity of the area; we have here an instructive illustration of the established fact that to the untrained eye randomness appears as regularity or tendency to cluster.

数学代写|概率论代写Probability theory代考|WAITING TIMES. THE NEGATIVE BINOMIAL DISTRIBUTION

Consider a succession of $n$ Bernoulli trials and let us inquire how long it will take for the $r$ th success to turn up. Here $r$ is a fixed positive integer. The total number of successes in $n$ trials may, of course, fall short of $r$, but the probability that the $r$ th success occurs at the trial number $v \leq n$ is clearly independent of $n$ and depends only on $v, r$, and $p$. Since necessarily $v \geq r$, it is preferable to write $v=k+r$. The probability that the rth success occurs at the trial number $r+k$ (where $k=0,1, \ldots)$ will be denoted by $f(k ; r, p)$. It equals the probability that exactly $k$ failures precede the rth success. This event occurs if, and only if, among the $r+k-1$ trials there are exactly $k$ failures and the following, or $(r+k)$ th, trial results in success; the corresponding probabilities are $\left(\begin{array}{c}r+k-1 \ k\end{array}\right) \cdot p^{r-1} q^k$ and $p$, whence
$$
f(k ; r, p)=\left(\begin{array}{c}
r+k-1 \
k
\end{array}\right) \cdot p^r q^k
$$
Rewriting the binomial coefficient in accordance with II,(12.4), we find the alternative form
$$
f(k ; r, p)=\left(\begin{array}{c}
-r \
k
\end{array}\right) p^r(-q)^k, \quad k=0,1,2, \ldots
$$
Suppose now that Bernoulli trials are continued as long as necessary for $r$ successes to turn up. A typical sample point is represented by a sequence containing an arbitrary number, $k$, of letters $F$ and exactly $r$ letters $S$, the sequence terminating by an $S$; the probability of such a point is, by definition, $p^r q^k$. We must ask, however, whether it is possible that the trials never end, that is, whether an infinite sequence of trials may produce fewer than $r$ successes. Now $\sum_{k=0}^{\infty} f(k ; r, p)$ is the probability that the $r$ th success occurs after finitely many trials; accordingly, the possibility of an infinite sequence with fewer than $r$ successes can be discounted if, and only if,
$$
\sum_{k=0}^{\infty} f(k ; r, p)=1
$$

概率论代考

数学代写|概率论代写Probability theory代考|OBSERVATIONS FITTING THE POISSON DISTRIBUTI0N

(a)放射性衰变。放射性物质释放$\alpha$粒子;在一段时间内到达给定空间部分的粒子数量$t$是最著名的服从泊松定律的随机事件的例子。当然，这种物质会继续衰变，从长远来看，$\alpha$ -粒子的密度会下降。然而，镭需要数年才能检测到物质的减少;在相对较短的时期内，这些条件可以认为是恒定的，我们对导致泊松分布的假设有了理想的实现。

在一个著名的实验${ }^{13}$中，一种放射性物质在$N=2608$中每隔7.5秒被观察到;每个周期到达计数器的粒子数被计算出来。表3记录了恰好有$k$个粒子的周期个数$N_k$。粒子总数为$T=\sum k N_k=10,094$，平均值为$T / N=3.870$。理论值$N p(k ; 3.870)$被认为与观测到的数字$N_k$相当接近。为了判断拟合的密切程度，需要估计机会波动的可能幅度。统计学家通过$\chi^2$ -标准来判断拟合的接近程度。按照这个标准来衡量，我们应该预期，在理想条件下，100个可比较的案例中约有17个会显示出比表3所示更差的一致性。
(b)飞弹击中伦敦。作为随机点空间分布的一个例子，考虑一下二战期间伦敦南部的飞弹命中统计数据。整个区域被划分为$N=576$小区域，每个小区域$t=\frac{1}{4}$平方公里，表4记录了精确命中$k$的区域数量$N_k$。${ }^{14}$总命中数为$T=\sum k N_k=537$，平均值为$\lambda t=T / N=0.9323 \ldots$，泊松分布的拟合性出奇地好;根据$\chi^2$ -标准判断，在理想条件下，大约88％的可比观测结果应显示出较差的一致性。有趣的是，大多数人都相信撞击点有聚集的趋势。如果这是真的，那么就会有更高频率的区域出现许多命中或没有命中，并且在中级职业中存在缺陷。表4显示了区域的完全随机性和均匀性;我们在这里有一个有启发性的例子，说明了一个既定的事实，即对未经训练的眼睛来说，随机性表现为规律性或聚集的趋势。

数学代写|概率论代写Probability theory代考|WAITING TIMES. THE NEGATIVE BINOMIAL DISTRIBUTION

考虑一系列$n$伯努利试验，让我们询问需要多长时间才能取得$r$的成功。这里$r$是一个固定正整数。当然，$n$试验中成功的总数可能低于$r$，但$r$次成功发生在试验号$v \leq n$的概率显然与$n$无关，而仅取决于$v, r$和$p$。因为必须$v \geq r$，所以最好写$v=k+r$。第n次成功发生在试验号$r+k$处的概率(其中$k=0,1, \ldots)$用$f(k ; r, p)$表示)。它等于刚好$k$次失败先于第n次成功的概率。当且仅当在$r+k-1$试验中恰好有$k$失败且以下试验或$(r+k)$次试验成功时，发生此事件;对应的概率为$\left(\begin{array}{c}r+k-1 \ k\end{array}\right) \cdot p^{r-1} q^k$和$p$，其中
$$
f(k ; r, p)=\left(\begin{array}{c}
r+k-1 \
k
\end{array}\right) \cdot p^r q^k
$$
根据II，(12.4)重写二项式系数，我们得到另一种形式
$$
f(k ; r, p)=\left(\begin{array}{c}
-r \
k
\end{array}\right) p^r(-q)^k, \quad k=0,1,2, \ldots
$$
现在假设伯努利试验一直持续到$r$成功出现的必要时间。一个典型的样本点由一个序列表示，该序列包含一个任意数字$k$，由字母$F$和精确的$r$个字母$S$组成，该序列以$S$结束;根据定义，这个点出现的概率是$p^r q^k$。然而，我们必须问，试验是否可能永远不会结束，也就是说，无穷无尽的试验序列是否会产生少于$r$的成功。现在$\sum_{k=0}^{\infty} f(k ; r, p)$是在有限次试验后$r$次成功的概率;因此，一个成功数小于$r$的无穷序列的可能性，当且仅当:
$$
\sum_{k=0}^{\infty} f(k ; r, p)=1
$$

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

金融工程代写

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

有限元方法代写

随机分析代写

时间序列分析代写

回归分析代写

MATLAB代写

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

数学代写|概率论代写Probability theory代考|APPLICATIONS TO GENETICS

Posted on 2023年7月5日2023年7月5日 by statistics-lab

数学代写|概率论代写Probability theory代考|APPLICATIONS TO GENETICS

The theory of heredity, originated by G. Mendel (1822-1884), provides instructive illustrations for the applicability of simple probability models. We shall restrict ourselves to indications concerning the most elementary problems. In describing the biological background, we shall necessarily oversimplify and concentrate on such facts as are pertinent to the mathematical treatment.

Heritable characters depend on special carriers, called genes. All cells of the body, except the reproductive cells or gametes, carry exact replicas of the same gene structure. The salient fact is that genes appear in pairs. The reader may picture them as a vast collection of beads on short pieces of string, the chromosomes. These also appear in pairs, and paired genes occupy the same position on paired chromosomes. In the simplest case each gene of a particular pair can assume two forms (alleles), $A$ and $a$. Then three different pairs can be formed, and, with respect to this particular pair, the organism belongs to one of the three genotypes $A A, A a, a a$ (there is no distinction between $A a$ and $a A$ ). For example, peas carry a pair of genes such that $A$ causes red blossom color and $a$ causes white. The three genotypes are in this case distinguishable as red, pink, and white. Each pair of genes determines one heritable factor, but the majority of observable properties of organisms depend on several factors. For some characteristics (e.g., eye color and left-handedness) the influence of one particular pair of genes is predominant, and in such cases the effects of Mendelian laws are readily observable. Other characteristics, such as height, can be understood as the cumulative effect of a very large number of genes [cf. example X, (5.c)]. Here we shall study genotypes and inheritance for only one particular pair of genes with respect to which we have the three genotypes $A A, A a$, $a a$. Frequently there are $N$ different forms $A_1, \ldots, A_N$ for the two genes and, accordingly, $N(N+1) / 2$ genotypes $A_1 A_1, A_1 A_2, \ldots, A_N A_N$. The theory applies to this case with obvious modifications (cf. problem 27). The following calculations apply also to the case where $A$ is dominant and a recessive. By this is meant that $A a$-individuals have the same observable properties as $A A$, so that only the pure $a a$-type shows an observable influence of the $a$-gene. All shades of partial dominance appear in nature. Typical partially recessive properties are blue eyes, left-handedness, etc.

Looking at the population as a whole, we conceive of the pairing of parents as the result of a second chance process. We shall investigate only the so-called random mating, which is defined by this condition: If $r$ descendants in the first filial generation are chosen at random, then their parents form a random sample of size $r$, with possible repetitions, from the aggregate of all possible parental pairs. In other words, each descendant is to be regarded as the product of a random selection of parents, and all selections are mutually independent. Random mating is an idealized model of the conditions prevailing in many natural populations and in field experiments. However, if red peas are sown in one corner of the field and white peas in another, parents of like color will unite more often than under random mating. Preferential selectivity (such as blondes preferring blondes) also violates the condition of random mating. Extreme non-random mating is represented by self-fertilizing plants and artificial inbreeding. Some such assortative mating systems will be analyzed mathematically, but for the most part we shall restrict our attention to random mating.

数学代写|概率论代写Probability theory代考|SEX-LINKED CHARACTERS

In the introduction to the preceding section it was mentioned that genes lie on chromosomes. These appear in pairs and are transmitted as units, so that all genes on a chromosome stick together. ${ }^9$ Our scheme for the inheritance of genes therefore applies also to chromosomes as units. Sex is determined by two chromosomes; females are $X X$, males $X Y$. The mother necessarily transmits an $X$-chromosome, and the sex of offspring depends on the chromosome transmitted by the father. Accordingly, male and female gametes are produced in equal numbers. The difference in birth rate for boys and girls is explained by variations in prenatal survival chances.

We said that both genes and chromosomes appear in pairs, but there is an exception inasmuch as the genes situated on the $X$-chromosome have no corresponding gene on $Y$. Females have two $X$-chromosomes, and hence two of such $X$-linked genes; however, in males the $X$-genes appear as singles. Typical are two sex-linked genes causing colorblindness and haemophilia. With respect to each of them, females can still be classified into the three genotypes, $A A, A a, a a$, but, having only one gene, males have only the two genotypes $A$ and $a$. Note that a son always has the father’s $Y$-chromosome so that a sex-linked character cannot be inherited from father to son. However, it can pass from father to daughter and from her to a grandson.

We now proceed to adapt the analysis of the preceding section to the present situation. Assume again random mating and let the frequencies of the genotypes $A A, A a, a a$ in the female population be $u, 2 v, w$, respectively. As before put $p=u+v, q=v+w$. The frequencies of the two male genotypes $A$ and $a$ will be denoted by $p^{\prime}$ and $q^{\prime}\left(p^{\prime}+q^{\prime}=1\right)$. Then $p$ and $p^{\prime}$ are the frequencies of the $A$-gene in the female and male populations, respectively. The probability for a female descendant to be of genotype $A A, A a, a a$ will be denoted by $u_1, 2 v_1, w_1$; the analogous probabilities for the male types $A$ and $a$ are $p_1^{\prime}, q_1^{\prime}$. Now a male offspring receives his $X$-chromosome from the female parent, and hence
$$
p_1^{\prime}=p, \quad q_1^{\prime}=q .
$$
For the three female genotypes we find, as in section 5 ,
$$
u_1=p p^{\prime}, \quad 2 v_1=p q^{\prime}+q p^{\prime}, \quad w_1=q q^{\prime} .
$$
Hence
$$
p_1=u_1+v_1=\frac{1}{2}\left(p+p^{\prime}\right), \quad q_1=v_1+w_1=\frac{1}{2}\left(q+q^{\prime}\right)
$$

概率论代考

数学代写|概率论代写Probability theory代考|APPLICATIONS TO GENETICS

孟德尔(G. Mendel, 1822-1884)提出的遗传理论为简单概率模型的适用性提供了有益的例证。我们将把自己限制在有关最基本问题的指示上。在描述生物学背景时，我们必然会过分简化，而把注意力集中在与数学处理有关的事实上。

可遗传的性状依赖于特殊的载体，称为基因。身体的所有细胞，除了生殖细胞或配子，都携带相同基因结构的精确副本。显著的事实是，基因是成对出现的。读者可能会把它们想象成一大堆串在短串上的珠子，也就是染色体。它们也成对出现，成对的基因在成对的染色体上占据相同的位置。在最简单的情况下，一对特定基因中的每个基因可以有两种形式(等位基因)，$a$和$a$。然后可以形成三对不同的组合，并且，就这对组合而言，生物体属于三种基因型中的一种:A A, A A, A A$(在A A$和A A$之间没有区别)。例如，豌豆携带一对基因，使得$a$导致红色开花，$a$导致白色开花。在这种情况下，三种基因型可区分为红色、粉红色和白色。每对基因决定一个遗传因素，但大多数可观察到的生物特性取决于几个因素。对于某些特征(例如，眼睛颜色和左撇子)，一对特定基因的影响占主导地位，在这种情况下，孟德尔定律的影响很容易观察到。其他特征，如身高，可以理解为大量基因的累积效应[参见例子X， (5.c)]。在这里，我们将只研究一对特定基因的基因型和遗传，我们有三种基因型:A A、A A、A A。这两个基因通常有$N$不同的形式$A_1， \ldots, A_N$，相应地，$N(N+1) / 2$基因型$A_1 A_1, A_1 A_2， \ldots, A_N A_N$。该理论适用于这种情况，但有明显的修改(参见问题27)。下面的计算也适用于$A$为显性和隐性的情况。这意味着A – A -个体具有与A – A – $相同的可观察属性，因此只有纯粹的A – A – $类型显示出A – $基因的可观察影响。自然界中存在着各种程度的部分支配。典型的部分隐性特征是蓝眼睛、左撇子等。

从整体上看，我们认为父母的配对是第二次机会过程的结果。我们只研究所谓的随机交配，它是由以下条件定义的:如果在第一代子代中随机选择$r$后代，则它们的父母从所有可能的父母对的总和中形成一个大小为$r$的随机样本，并可能重复。换句话说，每个后代都被看作是父母随机选择的产物，所有的选择都是相互独立的。随机交配是许多自然种群和野外实验中普遍存在的条件的理想模型。然而，如果在田地的一个角落播种红豌豆，在另一个角落播种白豌豆，相同颜色的亲本将比随机交配更容易结合。择优选择(比如金发女郎偏爱金发女郎)也违反了随机交配的条件。极端非随机交配以植物自交和人工近交为代表。一些这样的选型交配系统将被数学分析，但在大多数情况下，我们将把注意力限制在随机交配上。

数学代写|概率论代写Probability theory代考|SEX-LINKED CHARACTERS

在前一节的介绍中提到，基因位于染色体上。它们成对出现，并作为单位传播，因此染色体上的所有基因都粘在一起。${ }^9$因此，我们的基因遗传方案也适用于作为单位的染色体。性别是由两条染色体决定的;女性是$X X$，男性是$X Y$。母亲必然会遗传一条$X$ -染色体，后代的性别取决于父亲遗传的染色体。因此，产生的雄性配子和雌性配子数量相等。男孩和女孩出生率的差异可以用产前存活率的差异来解释。

我们说过，基因和染色体都是成对出现的，但也有例外，因为位于$X$ -染色体上的基因在$Y$上没有相应的基因。女性有两条$X$ -染色体，因此有两个与$X$相关的基因;然而，在男性中，$X$ -基因是单独出现的。典型的是导致色盲和血友病的两个性别连锁基因。对于它们中的每一个，雌性仍然可以分为三个基因型$A A, A a, a a$，但是，只有一个基因，雄性只有两个基因型$A$和$a$。请注意，儿子总是有父亲的$Y$ -染色体，因此性别连锁的特征不能从父亲遗传给儿子。然而，它可以从父亲传给女儿，从女儿传给孙子。

我们现在着手使前一节的分析适应当前的情况。再次假设随机交配，让基因型$A A, A a, a a$在雌性种群中的频率分别为$u, 2 v, w$。如前所述，输入$p=u+v, q=v+w$。两种男性基因型$A$和$a$的频率将用$p^{\prime}$和$q^{\prime}\left(p^{\prime}+q^{\prime}=1\right)$表示。那么$p$和$p^{\prime}$分别是$A$ -基因在女性和男性人群中的频率。基因型为$A A, A a, a a$的女性后代的概率用$u_1, 2 v_1, w_1$表示;男性类型$A$和$a$的类似概率为$p_1^{\prime}, q_1^{\prime}$。现在雄性后代从雌性父母那里得到$X$ -染色体，因此
$$
p_1^{\prime}=p, \quad q_1^{\prime}=q .
$$
对于我们在第5节中发现的三种女性基因型，
$$
u_1=p p^{\prime}, \quad 2 v_1=p q^{\prime}+q p^{\prime}, \quad w_1=q q^{\prime} .
$$
因此
$$
p_1=u_1+v_1=\frac{1}{2}\left(p+p^{\prime}\right), \quad q_1=v_1+w_1=\frac{1}{2}\left(q+q^{\prime}\right)
$$

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金融工程代写

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

有限元方法代写

随机分析代写

时间序列分析代写

回归分析代写

MATLAB代写

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

数学代写|概率论代写Probability theory代考|THE MAIN LEMMA

Posted on 2023年6月27日2023年6月27日 by statistics-lab

数学代写|概率论代写Probability theory代考|THE MAIN LEMMA

As we saw, the probability of a return to the origin at epoch $2 v$ equals the quantity $u_{2 v}$ of (2.3). As the theory of fluctuations in random walks began to take shape it came as a surprise that almost all formulas involved this probability. One reason for this is furnished by the following simple lemma, which has a mild surprise value of its own and provides the key to the deeper theorems of the next section.

Lemma $1^{10}$ The probability that no return to the origin occurs up to and including epoch $2 n$ is the same as the probability that a return occurs at epoch $2 n$. In symbols,
$$
\mathbf{P}\left{\mathbf{S}1 \neq 0, \ldots, \mathbf{S}{2 n} \neq 0\right}=\mathbf{P}\left{\mathbf{S}{2 n}=0\right}=u{2 n}
$$
Here, of course, $n>0$. When the event on the left occurs either all the $\mathbf{S}j$ are positive, or all are negative. The two contingencies being equally probable we can restate (3.1) in the form $$ \mathbf{P}\left{\mathbf{S}_1>0, \ldots, \mathbf{S}{2 n}>0\right}=\frac{1}{2} u_{2 n}
$$
Proof. Considering all the possible values of $\mathbf{S}{2 n}$ it is clear that $$ \mathbf{P}\left{\mathbf{S}_1>0, \ldots, \mathbf{S}{2 n}>0\right}=\sum_{r=1}^{\infty} \mathbf{P}\left{\mathbf{S}1>0, \ldots, \mathbf{S}{2 n-1}>0, \mathbf{S}{2 n}=2 r\right} $$ (where all terms with $r>n$ vanish). By the ballot theorem the number of paths satisfying the condition indicated on the right side equals $N{2 n-1,2 r-1}-N_{2 n-1,2 r+1}$, and so the $r$ th term of the sum equals
$$
\frac{1}{2}\left(p_{2 n-1,2 r-1}-p_{2 n-1,2 r+1}\right)
$$

数学代写|概率论代写Probability theory代考|LAST VISIT AND LONG LEADS

We are now prepared for a closer analysis of the nature of chance fluctuations in random walks. The results are startling. According to widespread beliefs a so-called law of averages should cnsurc that in a long coin-tossing game each player will be on the winning side for about half the time, and that the lead will pass not infrequently from one player to the other. Imagine then a huge sample of records of ideal coin-tossing games, each consisting of exactly $2 n$ trials. We pick one at random and observe the epoch of the last tie (in other words, the number of the last trial at which the accumulated numbers of heads and tails were equal). This number is even, and we denote it by $2 k$ (so that $0 \leq k \leq n$ ). Frequent changes of the lead would imply that $k$ is likely to be relatively close to $n$, but this is not so. Indeed, the next theorem reveals the amazing fact that the distribution of $k$ is symmetric in the sense that any value $k$ has exactly the same probability as $n-k$. This symmetry implies in particular that the inequalities $k>n / 2$ and $k<n / 2$ are equally likely. 11 With probability $\frac{1}{2}$ no equalization occurred in the second half of the game, regardless of the length of the game. Furthermore, the probabilities near the end points are greatest; the most probable values for $k$ are the extremes 0 and $n$. These results show that intuition leads to an erroneous picture of the probable effects of chance fluctuations. A few numerical results may be illuminating.

概率论代考

数学代写|概率论代写Probability theory代考|THE MAIN LEMMA

如我们所见，在epoch $2 v$返回原点的概率等于(2.3)的量$u_{2 v}$。随着随机漫步波动理论的形成，几乎所有的公式都涉及到这种概率，这让人感到惊讶。下面的一个简单引理提供了一个原因，它本身有一个轻微的惊喜值，并为下一节的更深层次的定理提供了关键。

引理$1^{10}$在历元$2 n$之前(包括历元)没有返回原点的概率与历元$2 n$发生返回的概率相同。在符号中，
$$
\mathbf{P}\left{\mathbf{S}1 \neq 0, \ldots, \mathbf{S}{2 n} \neq 0\right}=\mathbf{P}\left{\mathbf{S}{2 n}=0\right}=u{2 n}
$$
当然，这里是$n>0$。当左边的事件发生时，要么所有的$\mathbf{S}j$都是正的，要么都是负的。这两种偶然性是同等可能的，我们可以用$$ \mathbf{P}\left{\mathbf{S}1>0, \ldots, \mathbf{S}{2 n}>0\right}=\frac{1}{2} u{2 n}
$$的形式重述(3.1)
证明。考虑$\mathbf{S}{2 n}$的所有可能值，显然是$$ \mathbf{P}\left{\mathbf{S}1>0, \ldots, \mathbf{S}{2 n}>0\right}=\sum{r=1}^{\infty} \mathbf{P}\left{\mathbf{S}1>0, \ldots, \mathbf{S}{2 n-1}>0, \mathbf{S}{2 n}=2 r\right} $$(其中包含$r>n$的所有项都消失了)。根据选票定理，满足右边条件的路径数等于$N{2 n-1,2 r-1}-N_{2 n-1,2 r+1}$，因此和的第$r$项等于
$$
\frac{1}{2}\left(p_{2 n-1,2 r-1}-p_{2 n-1,2 r+1}\right)
$$

数学代写|概率论代写Probability theory代考|LAST VISIT AND LONG LEADS

现在，我们准备对随机漫步中偶然性波动的本质进行更细致的分析。结果令人吃惊。根据普遍的信念，所谓的平均定律应该确保在一个长期的抛硬币游戏中，每个玩家在大约一半的时间里都是获胜的一方，并且领先将经常从一个玩家传递给另一个玩家。那么，想象一下理想抛硬币游戏的大量记录样本，每个游戏都包含$2 n$次试验。我们随机选择一个，并观察最后一次平局的时间(换句话说，最后一次试验中累积的正面和反面次数相等)。这个数是偶数，我们用$2 k$表示它(因此$0 \leq k \leq n$)。领先地位的频繁变化意味着$k$可能相对接近$n$，但事实并非如此。的确，下一个定理揭示了一个惊人的事实，即$k$的分布是对称的，因为任何值$k$与$n-k$的概率都完全相同。这种对称性特别意味着不等式$k>n / 2$和$k<n / 2$的可能性是相等的。11有可能$\frac{1}{2}$在比赛后半段没有出现扳平的情况，无论比赛时长如何。此外，端点附近的概率最大;$k$最可能的值是极端值0和$n$。这些结果表明，直觉会导致人们对偶然波动的可能影响产生错误的认识。一些数值结果可能具有启发性。

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写