assignmentutor™您的专属作业导师

assignmentutor-lab™ 为您的留学生涯保驾护航 在代写线性回归分析linear regression analysis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。我们的专家在代写线性回归分析linear regression analysis代写方面经验极为丰富，各种代写线性回归分析linear regression analysis相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

统计代写|线性回归分析代写linear regression analysis代考|STATISTICAL INDEPENDENCE

For any pair of random variables, independence implies that the pair are uncorrelated. For the normal distribution the converse is also true, as we now show.

THEOREM 2.4 Let $\mathbf{Y} \sim N_n(\boldsymbol{\mu}, \mathbf{\Sigma})$ and partition $\mathbf{Y}, \boldsymbol{\mu}$ and $\mathbf{\Sigma}$ as in Example 2.7. Then $\mathrm{Y}1$ and $\mathbf{Y}_2$ are independent if and only if $\mathbf{\Sigma}{12}=0$.

Proof. The m.g.f. of $\mathbf{Y}$ is $\exp \left(\mathbf{t}^{\prime} \boldsymbol{\mu}+\frac{1}{2} \mathbf{t}^{\prime} \mathbf{\Sigma} \mathbf{t}\right)$. Partition $\mathbf{t}$ conformably with $\mathbf{Y}$. Then the exponent in the m.g.f. above is
$$\mathbf{t}1^{\prime} \mu_1+\mathbf{t}_2^{\prime} \mu_2+\frac{1}{2} \mathbf{t}_1^{\prime} \Sigma{11} \mathbf{t}1+\frac{1}{2} \mathbf{t}_2^{\prime} \Sigma{22} \mathbf{t}2+\mathbf{t}_1^{\prime} \Sigma{12} \mathbf{t}_2 .$$

If $\Sigma_{12}=0$, the exponent can be written as a function of just $t_1$ plus a function of just $t_2$, so the m.g.f. factorizes into a term in $t_1$ alone times a term in $t_2$ alone. This implies that $\mathbf{Y}1$ and $\mathbf{Y}_2$ are independent. Conversely, if $\mathbf{Y}_1$ and $\mathbf{Y}_2$ are independent, then $$M\left(\mathrm{t}_1, \mathbf{0}\right) M\left(\mathbf{0}, \mathrm{t}_2\right)=M\left(\mathrm{t}_1, \mathrm{t}_2\right),$$ where $M$ is the m.g.f. of $\mathbf{Y}$. By (2.7) this implies that $\mathbf{t}_1^{\prime} \boldsymbol{\Sigma}{12} \mathbf{t}2=0$ for all $t_1$ and $t_2$, which in turn implies that $\Sigma{12}=0$. [This follows by setting $\mathbf{t}_1=(1,0, \ldots, 0)^{\prime}$, etc. $]$
We use this theorem to prove our next result.
THEOREM $2.5$ Let $\mathbf{Y} \sim N_n(\boldsymbol{\mu}, \mathbf{\Sigma})$ and define $\mathbf{U}=\mathbf{A Y}, \mathbf{V}=\mathbf{B Y}$. Then $\mathrm{U}$ and $\mathbf{V}$ are independent if and only if $\operatorname{Cov}[\mathbf{U}, \mathbf{V}]=\mathbf{A} \mathbf{\Sigma} \mathbf{B}^{\prime}=\mathbf{0}$.
Proof. Consider
$$\mathbf{W}=\left(\begin{array}{l} \mathbf{U} \ \mathbf{V} \end{array}\right)=\left(\begin{array}{l} \mathbf{A} \ \mathbf{B} \end{array}\right) \mathbf{Y}$$
Then, by Theorem $2.2$, the random vector $\mathbf{W}$ is multivariate normal with variancc-covariance matrix
$$\operatorname{Var}[\mathbf{W}]=\left(\begin{array}{c} \mathbf{A} \ \mathbf{B} \end{array}\right) \operatorname{Var}[\mathbf{Y}]\left(\mathbf{A}^{\prime}, \mathbf{B}^{\prime}\right)=\left(\begin{array}{ll} \mathbf{A} \mathbf{\Sigma} \mathbf{A}^{\prime} & \mathbf{A} \mathbf{\Sigma} \mathbf{B}^{\prime} \ \mathbf{B} \mathbf{\Sigma} \mathbf{A}^{\prime} & \mathbf{B} \mathbf{\Sigma} \mathbf{B}^{\prime} \end{array}\right)$$
Thus, by Theorem $2.4, \mathbf{U}$ and $\mathbf{V}$ are independent if and only if $\mathbf{A} \mathbf{\Sigma} \mathbf{B}^{\prime}=\mathbf{0}$.

统计代写|线性回归分析代写linear regression analysis代考|DISTRIBUTION OF QUADRATIC FORMS

Quadratic forms in normal variables arise frequently in the theory of regression in connection with various tests of hypotheses. In this section we prove some simple results concerning the distribution of such quadratic forms.

Let $\mathbf{Y} \sim N_n(\mu, \boldsymbol{\Sigma})$, where $\boldsymbol{\Sigma}$ is positive-definite. We are interested in the distribution of random variables of the form $\mathbf{Y}^{\prime} \mathbf{A Y}=\sum_{i=1}^n \sum_{j=1}^n a_{i j} Y_i Y_j$. Note that we can always assume that the matrix $\mathbf{A}$ is symmetric, since if not we can replace $a_{i j}$ with $\frac{1}{2}\left(a_{i j}+a_{j i}\right)$ without changing the value of the quadratic form. Since $\mathbf{A}$ is symmetric, we can diagonalize it with an orthogonal transformation; that is, there is an orthogonal matrix $\mathbf{T}$ and a diagonal matrix D with
$$\mathbf{T}^{\prime} \mathbf{A T}=\mathbf{D}=\operatorname{diag}\left(d_1, \ldots, d_n\right) .$$
The diagonal elements $d_i$ are the eigenvalues of $\mathbf{A}$ and can be any real numbers.

We begin by assuming that the random vector in the quadratic form has a $N_n\left(\mathbf{0}, \mathbf{I}n\right)$ distribution. The general case can be reduced to this through the usual transformations. By Example 2.3, if $\mathbf{T}$ is an orthogonal matrix and $\mathbf{Y}$ has an $N_n\left(\mathbf{0}, \mathbf{I}_n\right)$ distribution, so does $\mathbf{Z}=\mathbf{T}^{\prime} \mathbf{Y}$. Thus we can write $$\mathbf{Y}^{\prime} \mathbf{A Y}=\mathbf{Y}^{\prime} \mathbf{T} \mathbf{D} \mathbf{T}^{\prime} \mathbf{Y}=\mathbf{Z}^{\prime} \mathbf{D} \mathbf{Z}=\sum{i=1}^n d_i Z_i^2,$$
so the distribution of $\mathbf{Y}^{\prime} \mathbf{A Y}$ is a linear combination of independent $\chi_1^2$ random variables. Given the values of $d_i$, it is possible to calculate the distribution, at least numerically. Farebrother [1990] describes algorithms for this.

There is an important special case that allows us to derive the distribution of the quadratic form exactly, without recourse to numerical methods. If $r$ of the eigenvalues $d_i$ are 1 and the remaining $n-r$ zero, then the distribution is the sum of $r$ independent $\chi_1^2$ ‘s, which is $\chi_r^2$. We can recognize when the eigenvalues are zero or 1 using the following theorem.

THEOREM 2.6 Let $\mathbf{A}$ be a symmetric matrix. Then $\mathbf{A}$ has $r$ eigenvalues equal to 1 and the rest zero if and only if $\mathbf{A}^2=\mathbf{A}$ and rank $\mathbf{A}=r$.

线性回归分析代写

统计代写|线性回归分析代写linear regression analysis代考|STATISTICAL INDEPENDENCE

$$M\left(\mathrm{t}_1, \mathbf{0}\right) M\left(\mathbf{0}, \mathrm{t}_2\right)=M\left(\mathrm{t}_1, \mathrm{t}_2\right),$$

$$\mathbf{W}=(\mathbf{U} \mathbf{V})=(\mathbf{A} \mathbf{B}) \mathbf{Y}$$

统计代写|线性回归分析代写linear regression analysis代考|DISTRIBUTION OF QUADRATIC FORMS

$$\mathbf{T}^{\prime} \mathbf{A T}=\mathbf{D}=\operatorname{diag}\left(d_1, \ldots, d_n\right)$$

$$\mathbf{Y}^{\prime} \mathbf{A} \mathbf{Y}=\mathbf{Y}^{\prime} \mathbf{T D} \mathbf{T}^{\prime} \mathbf{Y}=\mathbf{Z}^{\prime} \mathbf{D Z}=\sum i=1^n d_i Z_i^2,$$

有限元方法代写

assignmentutor™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。

assignmentutor™您的专属作业导师
assignmentutor™您的专属作业导师