# 统计代写|线性回归分析代写linear regression analysis代考|STAT5110

## 统计代写|线性回归分析代写linear regression analysis代考|Introducing Further Explanatory Variables

If we wish to introduce further explanatory variables into a less-than-full-rank model, we can, once again, reduce the model to one of full rank. As in Section 3.7, we see what happens when we add $\mathbf{Z} \gamma$ to our model $\mathbf{X} \boldsymbol{\beta}$. It makes sense to assume that $\mathbf{Z}$ has full column rank and that the columns of $\mathbf{Z}$ are linearly independent of the columns of $\mathbf{X}$. Using the full-rank model
$$\mathbf{Y}=\mathbf{X}_1 \alpha+\mathbf{Z} \gamma+\varepsilon$$
where $\mathbf{X}_1$ is $n \times r$ of rank $r$, we find that Theorem $3.6$ (ii), (iii), and (iv) of Section 3.7.1 still hold. To see this, one simply works through the same steps of the theorem, but replacing $\mathbf{X}$ by $\mathbf{X}_1, \boldsymbol{\beta}$ by $\alpha$, and $\mathbf{R}$ by $\mathbf{I}_n-\mathbf{P}$, where $\mathbf{P}=\mathbf{X}_1\left(\mathbf{X}_1^{\prime} \mathbf{X}_1\right)^{-1} \mathbf{X}_1$ is the unique projection matrix projecting onto $\mathcal{C}(\mathbf{X})$.

Referring to Section $3.8$, suppose that we have a set of linear restrictions $\mathrm{a}_i^{\prime} \beta=$ $0(i=1,2 \ldots, q)$, or in matrix form, $\mathbf{A} \beta=0$. Then a realistic assumption is that these constraints are all estimable. This implies that $\mathbf{a}_i^{\prime}=\mathbf{m}_i^{\prime} \mathbf{X}$ for some $\mathbf{m}_i$, or $\mathbf{A}=\mathbf{M X}$, where $\mathbf{M}$ is $q \times n$ of rank $q$ [as $q=\operatorname{rank}(\mathbf{A}) \leq \operatorname{rank}(\mathbf{M})$ by A.2.1]. Since $\mathbf{A} \beta=\mathbf{M X} \beta=\mathbf{M} \theta$, we therefore find the restricted least squares estimate of $\theta$ by minimizing $|\mathbf{Y}-\theta|^2$ subject to $\theta \in \mathcal{C}(\mathbf{X})=\Omega$ and $\mathbf{M} \theta=0$, that is, subject to
$$\boldsymbol{\theta} \in \mathcal{N}(\mathbf{M}) \cap \Omega \quad(=\omega, \text { say }) .$$
If $\mathbf{P}{\Omega}$ and $\mathbf{P}\omega$ are the projection matrices projecting onto $\Omega$ and $\omega$, respectively, then we want to find $\hat{\theta}\omega=\mathbf{P}\omega \mathbf{Y}$. Now, from B.3.2 and B.3.3,
$$\mathbf{P}{\Omega}-\mathbf{P}\omega=\mathbf{P}{\omega^{\perp} \cap \Omega},$$ where $\omega^{\perp} \cap \Omega=\mathcal{C}(\mathbf{B})$ and $\mathbf{B}=\mathbf{P}{\Omega} \mathbf{M}^{\prime}$. Thus
$\hat{\boldsymbol{\theta}}\omega=\mathbf{P}\omega \mathbf{Y}$
$=\mathbf{P}{\Omega} \mathbf{Y}-\mathbf{P}{\omega^{\perp} \cap \Omega} \mathbf{Y}$
$=\hat{\theta}_{\Omega}-\mathbf{B}\left(\mathbf{B}^{\prime} \mathbf{B}\right)^{-} \mathbf{B}^{\prime} \mathbf{Y}$

## 统计代写|线性回归分析代写linear regression analysis代考|GENERALIZED LEAST SQUARES

Having developed a least squares theory for the full-rank model $\mathbf{Y}=\mathbf{X} \boldsymbol{\beta}+\varepsilon$, where $E[\varepsilon]=0$ and $\operatorname{Var}[\varepsilon]=\sigma^2 \mathbf{I}_n$, we now consider what modifications are necessary if we allow the $\varepsilon_i$ to be correlated. In particular, we assume that $\operatorname{Var}[\varepsilon]=\sigma^2 \mathbf{V}$, where $\mathbf{V}$ is a known $n \times n$ positive-definite matrix.

Since $\mathbf{V}$ is positive-definite, there exists an $n \times n$ nonsingular matrix $\mathbf{K}$ such that $\mathbf{V}=\mathbf{K K}^{\prime}$ (A.4.2). Therefore, setting $\mathbf{Z}=\mathbf{K}^{-1} \mathbf{Y}, \mathbf{B}=\mathbf{K}^{-1} \mathbf{X}$, and a $\eta=\mathbf{K}^{-1} \varepsilon$, we have the model $\mathbf{Z}=\mathbf{B} \beta+\eta$, where $\mathbf{B}$ is $n \times p$ of $\operatorname{rank} p$ (A.2.2).
Also, $E[\eta]=0$ and
$\operatorname{Var}[\eta]=\operatorname{Var}\left[\mathbf{K}^{-1} \varepsilon\right]=\mathbf{K}^{-1} \operatorname{Var}[\varepsilon] \mathbf{K}^{-1^{\prime}}=\sigma^2 \mathbf{K}^{-1} \mathbf{K K}^{\prime} \mathbf{K}^{\prime-1}=\sigma^2 \mathbf{I}_n$.
Minimizing $\boldsymbol{\eta}^{\prime} \boldsymbol{\eta}$ with respect to $\beta$, and using the theory of Section 3.1, the least squares estimate of $\beta$ for this transformed model is
\begin{aligned} \boldsymbol{\beta}^* &=\left(\mathbf{B}^{\prime} \mathbf{B}\right)^{-1} \mathbf{B}^{\prime} \mathbf{Z} \ &=\left(\mathbf{X}^{\prime}\left(\mathbf{K K}^{\prime}\right)^{-1} \mathbf{X}\right)^{-1} \mathbf{X}^{\prime}\left(\mathbf{K K}^{\prime}\right)^{-1} \mathbf{Y} \ &=\left(\mathbf{X}^{\prime} \mathbf{V}^{-1} \mathbf{X}\right)^{-1} \mathbf{X}^{\prime} \mathbf{V}^{-1} \mathbf{Y} \end{aligned}
with expected value
$$E\left[\boldsymbol{\beta}^\right]=\left(\mathbf{X}^{\prime} \mathbf{V}^{-1} \mathbf{X}\right)^{-1} \mathbf{X}^{\prime} \mathbf{V}^{-1} \mathbf{X} \boldsymbol{\beta}=\boldsymbol{\beta},$$ dispersion matrix \begin{aligned} \operatorname{Var}\left[\beta^\right] &=\sigma^2\left(\mathbf{B}^{\prime} \mathbf{B}\right)^{-1} \ &=\sigma^2\left(\mathbf{X}^{\prime} \mathbf{V}^{-1} \mathbf{X}\right)^{-1}, \end{aligned}
and residual sum of squares
\begin{aligned} \mathbf{f}^{\prime} \mathbf{f} &=\left(\mathbf{Z}-\mathbf{B} \boldsymbol{\beta}^\right)^{\prime}\left(\mathbf{Z}-\mathbf{B} \boldsymbol{\beta}^\right) \ &=\left(\mathbf{Y}-\mathbf{X} \boldsymbol{\beta}^\right)^{\prime}\left(\mathbf{K} \mathbf{K}^{\prime}\right)^{-1}\left(\mathbf{Y}-\mathbf{X} \boldsymbol{\beta}^\right) \ &=\left(\mathbf{Y}-\mathbf{X} \boldsymbol{\beta}^\right)^{\prime} \mathbf{V}^{-1}\left(\mathbf{Y}-\mathbf{X} \boldsymbol{\beta}^\right) \end{aligned}

# 线性回归分析代考

## 统计代写|线性回归分析代写linear regression analysis代考|Introducing Further Explanatory Variables

$$\mathbf{Y}=\mathbf{X}1 \alpha+\mathbf{Z} \gamma+\varepsilon$$ 在哪里 $\mathbf{X}_1$ 是 $n \times r$ 等级 $r$ ，我们发现定理3.6第 $3.7 .1$ 节的 (ii)、 (iii) 和 (iv) 仍然有效。要看到这一点，只需 完成定理的相同步骤，但替换 $\mathbf{X}$ 经过 $\mathbf{X}_1, \boldsymbol{\beta}$ 经过 $\alpha$ ，和 $\mathbf{R}$ 经过 $\mathbf{I}_n-\mathbf{P}$ ，在哪里 $\mathbf{P}=\mathbf{X}_1\left(\mathbf{X}_1^{\prime} \mathbf{X}_1\right)^{-1} \mathbf{X}_1$ 是投影到的唯一投影矩阵 $\mathcal{C}(\mathbf{X})$. 参考部分 $3.8$ ，假设我们有一组线性限制 $\mathrm{a}_i^{\prime} \beta=0(i=1,2 \ldots, q)$ ，或以矩阵形式， $\mathbf{A} \beta=0$. 然后一个 现实的假设是这些约束都是可估计的。这意味着 $\mathbf{a}_i^{\prime}=\mathbf{m}_i^{\prime} \mathbf{X}$ 对于一些 $\mathbf{m}_i$ ，或者 $\mathbf{A}=\mathbf{M X}$ ，在哪里 $\mathbf{M}$ 是 $q \times n$ 等级 $q$ [作为 $q=\operatorname{rank}(\mathbf{A}) \leq \operatorname{rank}(\mathbf{M})$ 由 A.2.1]。自从 $\mathbf{A} \beta=\mathbf{M X} \beta=\mathbf{M} \theta$ ，因此我们找到限 制最小二乘估计 $\theta$ 通过最小化 $|\mathbf{Y}-\theta|^2$ 受制于 $\theta \in \mathcal{C}(\mathbf{X})=\Omega$ 和 $\mathbf{M} \theta=0$ ，也就是说，服从 $\boldsymbol{\theta} \in \mathcal{N}(\mathbf{M}) \cap \Omega \quad(=\omega$, say $)$ 如果 $\mathbf{P} \Omega$ 和 $\mathbf{P} \omega$ 是投影到的投影矩阵 $\Omega$ 和 $\omega$ ，分别，那么我们要找到 $\hat{\theta} \omega=\mathbf{P} \omega \mathbf{Y}$. 现在，从 B.3.2 和 B.3.3, $$\mathbf{P} \Omega-\mathbf{P} \omega=\mathbf{P} \omega^{\perp} \cap \Omega,$$ 在哪里 $\omega^{\perp} \cap \Omega=\mathcal{C}(\mathbf{B})$ 和 $\mathbf{B}=\mathbf{P} \Omega \mathbf{M}^{\prime}$. 因此 $$\hat{\boldsymbol{\theta}} \omega=\mathbf{P} \omega \mathbf{Y}$$ $=\mathbf{P} \Omega \mathbf{Y}-\mathbf{P} \omega^{\perp} \cap \Omega \mathbf{Y}$ $$=\hat{\theta}{\Omega}-\mathbf{B}\left(\mathbf{B}^{\prime} \mathbf{B}\right)^{-} \mathbf{B}^{\prime} \mathbf{Y}$$

## 统计代写|线性回归分析代写linear regression analysis代考|GENERALIZED LEAST SQUARES

$\operatorname{Var}[\eta]=\operatorname{Var}\left[\mathbf{K}^{-1} \varepsilon\right]=\mathbf{K}^{-1} \operatorname{Var}[\varepsilon] \mathbf{K}^{-1^{\prime}}=\sigma^2 \mathbf{K}^{-1} \mathbf{K}^{\prime} \mathbf{K}^{\prime-1}=\sigma^2 \mathbf{I}_n$.

$$\boldsymbol{\beta}^*=\left(\mathbf{B}^{\prime} \mathbf{B}\right)^{-1} \mathbf{B}^{\prime} \mathbf{Z} \quad=\left(\mathbf{X}^{\prime}\left(\mathbf{K} \mathbf{K}^{\prime}\right)^{-1} \mathbf{X}\right)^{-1} \mathbf{X}^{\prime}\left(\mathbf{K} \mathbf{K}^{\prime}\right)^{-1} \mathbf{Y}=\left(\mathbf{X}^{\prime} \mathbf{V}^{-1} \mathbf{X}\right)^{-1} \mathbf{X}^{\prime} \mathbf{V}^{-1} \mathbf{Y}$$

myassignments-help数学代考价格说明

1、客户需提供物理代考的网址，相关账户，以及课程名称，Textbook等相关资料~客服会根据作业数量和持续时间给您定价~使收费透明，让您清楚的知道您的钱花在什么地方。

2、数学代写一般每篇报价约为600—1000rmb，费用根据持续时间、周作业量、成绩要求有所浮动(持续时间越长约便宜、周作业量越多约贵、成绩要求越高越贵)，报价后价格觉得合适，可以先付一周的款，我们帮你试做，满意后再继续，遇到Fail全额退款。

3、myassignments-help公司所有MATH作业代写服务支持付半款，全款，周付款，周付款一方面方便大家查阅自己的分数，一方面也方便大家资金周转，注意:每周固定周一时先预付下周的定金，不付定金不予继续做。物理代写一次性付清打9.5折。

Math作业代写、数学代写常见问题

myassignments-help擅长领域包含但不是全部: