06-multiple-reg-algebra-slide.knit

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# 计量经济学(Econometrics)

### 胡华平

### 西北农林科技大学

### 经济管理学院数量经济教研室

### huhuaping01@hotmail.com

### 2023-02-15

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---
class: center, middle, duke-orange,hide_logo
name: chapter06a

# 第06章 多元回归I：代数部分

.red[
[6.1 估计问题](#estimation)

[6.2 推断问题](#inference)

[6.3 受约束的最小二乘法](#restricted-reg)

[6.4 检验回归模型的结构或稳定性](#structure-break)
]

---
layout: false
class: center, middle, duke-softblue,hide_logo
name: estimation

# 6.1 多元回归分析：估计问题

---
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<div class="my-footer"><span>huhuaping@   <a href="#chapter06a">第06章 多元回归I：代数部分</a>
&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;
&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;
<a href="#inference">6.1 多元回归分析：估计问题</a></span></div>

---

## 三变量模型：符号与假定

三变量的PRM和PRF为:

`$$\begin {align} 
Y_{i}=\beta_{1}+\beta_{2} X_{2 i}+\beta_{3} X_{3 i}+u_{i} \\ 
E\left(Y_{i} | X_{2 i}, X_{3 i}\right)=\beta_{1}+\beta_{2} X_{2 i}+\beta_{3} X_{3 i}
\end {align}$$`

多元回归分析是以多个解释变量的固定值为条件的回归分析。

我们所获得的，是各个自变量X值固定时，Y的平均值或Y的平均响应（mean response）。

- 
`$\beta_1$`：截距项

- 
`$u_i$`：表示所有未包含到模型中来的变量对Y的平均影响。

- 
`$\beta_2,\beta_3$`：偏回归系数（partial regression coefficients）

- 
`$i$`：指第i次观测，当数据为时间序列时用t表示；

---

## 三变量模型：CLRM假设

**CLRM假设1-1**：模型是正确设置的。（这里大有学问，也是一切计量分析问题的根本来源）

**CLRM假设1-2**：模型应该是参数线性的。也即模型中**参数**必须线性，变量可以不是线性

`$$\begin {align} 
Y_{i}=\beta_1+ \beta_{2} X_{2i}+ \beta_{3} X_{3i}+u_{i}
\end {align}$$`

---

## 三变量模型：CLRM假设

**CLRM假设2-1**：X是固定的（给定的）或独立于误差项。也即自变量X**不是**随机变量。

`$$\begin{align}
Cov(X_{2i}, u_i)= Cov(X_{3i}, u_i)= 0, \quad i=1,2, \ldots, n \\ 
E(X_i, u_i)= 0
\end{align}$$`

**CLRM假设2-2**：X变量间不存在**完全共线性**

---

## 三变量模型：CLRM假设

**CLRM假设3-1**：假设随机干扰项均值为零。也即给定
`$X_i$`的情形下，假定随机干扰项
`$u_i$`的**条件期望**为零。

`$$\begin{align}
E(u|X_{2i},X_{3i})= 0
\end{align}$$`

**CLRM假设3-2**：随机干扰项的方差为同方差。也即给定
`$X_i$`的情形下，随机干扰项
`$u_i$`的方差，处处都是相等的。记为：

`$$\begin{align}
Var\left(u_i|(X_{2i},X_{3i}) \right) & = E \left[ \left( u_i -E(u_i) \right)^2|(X_{2i},X_{3i}) \right] \\
& = E(u_i^2|X_i)  = E(u_i^2)  \equiv \sigma^2
\end{align}$$`

---

## 三变量模型：CLRM假设

**CLRM假设3-3**：各个随机干扰之间无自相关。也即给定两个不同的自变量取值（
`$X_i,X_j;i \neq j$`）情形下，随机干扰项
`$u_i,u_j$`的相关系数为0。或者说
`$u_i,u_j$`最好是相互独立的。记为：

在
`$X_i$`为给定情形下，且
`$i,j \in (1, 2, \cdots, n); i \neq j$`，假定：

`$$\begin{align}
Cov(u_i, u_j|X_i,X_j) & = E \left[ \left( u_i -E(u_i) \right)\left( u_j -E(u_j) \right) \right]  \\
& = E(u_iu_j) \\
& \equiv 0
\end{align}$$`

---

## 对多元回归方程的解释

三变量的PRM和PRF为:

`$$\begin {align} 
Y_{i}=\beta_{1}+\beta_{2} X_{2 i}+\beta_{3} X_{3 i}+u_{i} \\ 
E\left(Y_{i} | X_{2 i}, X_{3 i}\right)=\beta_{1}+\beta_{2} X_{2 i}+\beta_{3} X_{3 i}
\end {align}$$`

多元回归分析是以多个解释变量的固定值为条件的回归分析。

我们所获得的，是各个自变量X值固定时，Y的平均值或Y的平均响应（mean response）。偏回归系数的含义：

- 
`$\beta_2$`度量着在保持
`$X_{3i}$`不变的情况下，
`$X_{2i}$`每变化1个单位时，Y的均值的变化。换一句话说，   给出
`$X_{2i}$`的单位变化对Y均值的“直接”或“净”影响（净在不染有
`$X_{3i}$`的影响）。

- 
`$\beta_3$`则给出了
`$X_{3i}$`的单位变化对Y均值
`$E(Y)$`的“直接”或“净”影响，净在不沾有
`$X_{2i}$`的影响。

---

### 儿童死亡率案例：数据

研究关注儿童死亡率（`CM`，千分数）与人均GNP（`PGNP`，1980年的人均GNP）和妇女识字率（`FLR`，百分数）的关系，并构建如下PRM：

`$$\begin{equation} \begin{alignedat}{999} &CM=&& + \beta_{1} && + \beta_{2} PGNP&& + \beta_{3} FLR&&+u_i\\ \end{alignedat} \end{equation}$$`

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---

### 儿童死亡率案例：散点图

绘制散点图1：

---

### 儿童死亡率案例：散点图

绘制散点图2：

---

### 儿童死亡率案例：二元模型

儿童死亡率的二元回归模型如下：

`$$\begin{equation} \begin{alignedat}{999} &CM=&& + \beta_{1} && + \beta_{2} PGNP&& + \beta_{3} FLR&&+u_i\\ \end{alignedat} \end{equation}$$`

以上二元回归模型的OLS估计结果如下：

`$$\begin{equation} \begin{alignedat}{999} &\widehat{CM}=&&+263.64&&-0.01PGNP&&-2.23FLR\\ &\text{(t)}&&(22.7411)&&(-2.8187)&&(-10.6293)\\&\text{(se)}&&(11.5932)&&(0.0020)&&(0.2099)\\&\text{(fitness)}&& R^2=0.7077;&& \bar{R^2}=0.6981\\& && F^{\ast}=73.83;&& p=0.0000 \end{alignedat} \end{equation}$$`

- 是如何分离出人均国民收入PNGP对CM的“真实”或净影响呢？

- 是如何分离出妇女识字率FLR对CM的“真实”或净影响呢？

---

### 儿童死亡率案例：一元回归重现（PGNP纯影响）

.pull-left[
**步骤1**：妇女识字率FLR对儿童死亡率CM的回归**模型1**：

`$$\begin{equation} \begin{alignedat}{999} &CM=&& + \hat{\beta}_{1} && + \hat{\beta}_{2} FLR&&+e_i\\ \end{alignedat} \end{equation}$$`

`$$\begin{equation} \begin{alignedat}{999} &\widehat{CM}=&&+263.86&&-2.39FLR\\ &\text{(t)}&&(21.5840)&&(-11.2092)\\&\text{(se)}&&(12.2250)&&(0.2133)\\&\text{(fitness)}&& R^2=0.6696;&& \bar{R^2}=0.6643\\& && F^{\ast}=125.65;&& p=0.0000 \end{alignedat} \end{equation}$$`
]

.pull-right[
**步骤2**：妇女识字率FLR对人均国民收入PGNP的回归**模型2**：

`$$\begin{equation} \begin{alignedat}{999} &PGNP=&& + \hat{\beta}_{1} && + \hat{\beta}_{2} FLR&&+e_i\\ \end{alignedat} \end{equation}$$`

`$$\begin{equation} \begin{alignedat}{999} &\widehat{PGNP}=&&-39.30&&+28.14FLR\\ &\text{(t)}&&(-0.0535)&&(2.1950)\\&\text{(se)}&&(734.9526)&&(12.8211)\\&\text{(fitness)}&& R^2=0.0721;&& \bar{R^2}=0.0571\\& && F^{\ast}=4.82;&& p=0.0319 \end{alignedat} \end{equation}$$`
]

---

### 儿童死亡率案例：一元回归重现（PGNP纯影响）

**步骤3**：分别得到两次一元线性回归的残差
`$e_{1i}$`（`resid1`）和
`$e_{2i}$`（`resid2`），然后进行**无截距**回归分析：

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对两个残差序列进一步构造如下的**无截距**回归模型：

`$$\begin{equation} \begin{alignedat}{999} &\widehat{resid1}=&&+\hat{\beta}_{1}resid2\\ \end{alignedat} \end{equation}$$`

---

### 儿童死亡率案例：一元回归重现（PGNP纯影响）

<img src="../pic/chpt6-children-mod-demo.png" width="713" style="display: block; margin: auto;" />
.pull-left[
残差模型将得到如下回归分析结果：

`$$\begin{equation} \begin{alignedat}{999} &\widehat{resid1}=&&-0.01resid2\\ &\text{(t)}&&(-2.8645)\\&\text{(se)}&&(0.0020)\\&\text{(fitness)}&& R^2=0.1152;&& \bar{R^2}=0.1012\\& && F^{\ast}=8.21;&& p=0.0057 \end{alignedat} \end{equation}$$`
]

.pull-right[
对比原来的二元回归模型结果：

---

### 儿童死亡率案例：一元回归重现（FLR纯影响）

.pull-left[
**步骤1**：人均国民收入PGNP对儿童死亡率CM的回归**模型3**：

`$$\begin{equation} \begin{alignedat}{999} &CM=&& + \hat{\beta}_{1} && + \hat{\beta}_{2} PGNP&&+e_i\\ \end{alignedat} \end{equation}$$`

`$$\begin{equation} \begin{alignedat}{999} &\widehat{CM}=&&+157.42&&-0.01PGNP\\ &\text{(t)}&&(15.9893)&&(-3.5157)\\&\text{(se)}&&(9.8456)&&(0.0032)\\&\text{(fitness)}&& R^2=0.1662;&& \bar{R^2}=0.1528\\& && F^{\ast}=12.36;&& p=0.0008 \end{alignedat} \end{equation}$$`
]

.pull-right[
**步骤2**：人均国民收入PGNP对妇女识字率FLR的回归**模型4**：

`$$\begin{equation} \begin{alignedat}{999} &FLR=&& + \hat{\beta}_{1} && + \hat{\beta}_{2} PGNP&&+e_i\\ \end{alignedat} \end{equation}$$`

`$$\begin{equation} \begin{alignedat}{999} &\widehat{FLR}=&&+47.60&&+0.00PGNP\\ &\text{(t)}&&(13.3876)&&(2.1950)\\&\text{(se)}&&(3.5553)&&(0.0012)\\&\text{(fitness)}&& R^2=0.0721;&& \bar{R^2}=0.0571\\& && F^{\ast}=4.82;&& p=0.0319 \end{alignedat} \end{equation}$$`
]

---

### 儿童死亡率案例：一元回归重现（FLR纯影响）

**步骤3**：分别得到两次一元线性回归的残差
`$e_{3i}$`（`resid3`）和
`$e_{4i}$`（`resid4`），然后进行**无截距**回归分析：

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对两个残差序列进一步构造如下的**无截距**回归模型：

`$$\begin{equation} \begin{alignedat}{999} &\widehat{resid3}=&&+\hat{\beta}_{1}resid4\\ \end{alignedat} \end{equation}$$`

---

### 儿童死亡率案例：一元回归重现（FLR纯影响）

.pull-left[
残差模型将得到如下回归分析结果：

`$$\begin{equation} \begin{alignedat}{999} &\widehat{resid3}=&&-2.23resid4\\ &\text{(t)}&&(-10.8021)\\&\text{(se)}&&(0.2066)\\&\text{(fitness)}&& R^2=0.6494;&& \bar{R^2}=0.6438\\& && F^{\ast}=116.69;&& p=0.0000 \end{alignedat} \end{equation}$$`
]

.pull-right[
对比原来的二元回归模型结果：

---

## 偏回归系数的OLS估计：模型

样本回归函数SRF、样本回归模型SRM、样本回归函数的离差形式：
`$$\begin {align} 
 \hat{Y}_{i} &=\hat{\beta}_{1}+\hat{\beta}_{2} X_{2 i}+\hat{\beta}_{3} X_{3 i} \\ 
 Y_{i} &=\hat{\beta}_{1}+\hat{\beta}_{2} X_{2 i}+\hat{\beta}_{3} X_{3 i}+e_{i} \\ 
 \hat{y}_{i} &=\hat{\beta}_{2} x_{2 i}+\hat{\beta}_{3} x_{3 i}
\end {align}$$`

总体回归函数PRF和总体回归模型PRM：

`$$\begin {align} 
\mathrm{E}\left( Y_{i} | (X_{2 i}, X_{3 i} ) \right) & =\beta_{1}+\beta_{2} X_{2 i}+\beta_{3} X_{3 i} \\ 
Y_{i} &=\beta_{1}+\beta_{2} X_{2 i}+\beta_{3} X_{3 i}+u_{i}
\end {align}$$`

---

## 偏回归系数的OLS估计：OLS方法

`$$\begin {align} 
\begin{array}{c}{\mathrm{E}\left(Y_{i}\left|\left(X_{2 i}, X_{3 i}\right)=\beta_{1}+\beta_{2} X_{2 i}+\beta_{3} X_{3 i}\right.\right.} \\ {Y_{i}=\beta_{1}+\beta_{2} X_{2 i}+\beta_{3} X_{3 i}+u_{i}}\end{array}
\end {align}$$`

`$$\begin {align} 
\frac{\partial \sum{e_i^2}} {\partial \hat{\beta}_{1}} & =2 \sum\left(Y_{i}-\hat{\beta}_{1}-\hat{\beta}_{2} X_{2 i}-\hat{\beta}_{3} X_{3 i}\right)(-1)  =0 \\ 
\frac{\partial \sum{e_i^2}} {\partial \hat{\beta}_{2}} & =2 \sum\left(Y_{i}-\hat{\beta}_{1}-\hat{\beta}_{2} X_{2 i}-\hat{\beta}_{3} X_{3 i}\right)\left(-X_{2 i}\right)  =0 \\ 
\frac{\partial \sum e_i^2} {\partial \hat{\beta}_{3}} & =2 \sum\left(Y_{i}-\hat{\beta}_{1}-\hat{\beta}_{2} X_{2 i}-\hat{\beta}_{3} X_{3 i}\right)\left(-X_{3 i}\right)  =0
\end {align}$$`

`$$\begin {align} 
\overline{Y} &=\hat{\beta}_{1}+\hat{\beta}_{2} \overline{X}_{2}+\hat{\beta}_{3} \overline{X}_{3} \\ 
\sum_{i} Y_{2 i} &=\hat{\beta}_{1} \sum X_{2 i}+\hat{\beta}_{2} \sum X_{2 i}^{2}+\hat{\beta}_{3} \sum X_{2 i} X_{3 i} \\ 
\sum Y_{i} X_{3 i} &=\hat{\beta}_{1} \sum X_{3 i}+\hat{\beta}_{2} \sum X_{2 i} X_{3 i}+\hat{\beta}_{3} \sum X_{3 i}^{2}
\end {align}$$`

---

## 回归系数的OLS估计：回归系数

`$$\begin {align} 
\hat{\beta}_{1} &=\overline{Y}-\hat{\beta}_{2} \overline{X}_2-\hat{\beta}_{3} \overline{X}_{3} \\ 
\hat{\beta}_{2} &=\frac{\left(\sum y_{i} x_{2 i}\right)\left(\sum x_{3 i}^{2}\right)-\left(\sum y_{i} x_{3 i}\right)\left(\sum x_{2 i} x_{3 i}\right)}{\left(\sum x_{2 i}^{2}\right)\left(\sum x_{3 i}^{2}\right)-\left(\sum x_{2 i} x_{3 i}\right)^{2}} \\ 
\hat{\beta}_{3} &=\frac{\left(\sum y_{i} x_{3 i}\right)\left(\sum x_{2 i}^{2}\right)-\left(\sum y_{i} x_{2 i}\right)\left(\sum x_{2 i} x_{3 i}\right)}{\left(\sum x_{2 i}^{2}\right)\left(\sum x_{3 i}^{2}\right)-\left(\sum x_{2 i} x_{3 i}\right)^{2}} 
\end {align}$$`

**偏斜率系数**
`$(\hat{\beta}_2,\hat{\beta}_3)$`OLS估计量公式的特点：

- 公式是**对称的**。通过对调
`$x_{2i},x_{3i}$`而得到另一个。

- 两个方程的分母完全相同。

- 三变量情形是双变量情形的自然而然的推广。

---

## 回归系数的OLS估计：随机干扰项的方差

对于二元回归模型：

`$$\begin {align} 
Y_{i} &=\beta_{1}+\beta_{2} X_{2 i}+\beta_{3} X_{3 i}+u_{i}
\end {align}$$`

在CLRM假设下，
`$var(u_i) \equiv \sigma^2$`，可以证明其OLS估计量（证明略）：

.pull-left[
`$$\begin {align} 
\hat{\sigma}^{2}=\frac{\sum \mathrm{e}_{i}^{2}}{n-3}
\end {align}$$`
]

.pull-right[
`$$\begin {align} 
(n-3) \frac{\hat{\sigma}^{2}}{\sigma^{2}} \sim \chi^{2}(n-3)
\end {align}$$`
]

`$$\begin {align} 
\begin{aligned} \sum e_{i}^{2} &=\sum\left(e_{i} e_{i}\right) =\sum e_{i}\left(y_{i}-\hat{\beta}_{2} x_{2 i}-\hat{\beta}_{3} x_{3 i}\right) \\ 
&=\sum e_{i} y_{i}  && \leftarrow \left[ \sum e_{i} x_{2 i}=\sum e_{i} x_{3 i}=0 \right] \\
&=\sum y_{i}\left(y_{i}-\hat{\beta}_{2} x_{2 i}-\hat{\beta}_{3} x_{3 i}\right) \\ 
&=\sum y_{i}^{2}-\hat{\beta}_{2} \sum y_{i} x_{2 i}-\hat{\beta}_{3} \sum y_{i} x_{3 i} \end{aligned}
\end {align}$$`

---

## 回归系数的OLS估计：方差和标准差

`$\hat{\beta}_1$`的**真实方差**：

`$$\begin {align} 
\operatorname{var}\left(\hat{\beta}_{1}\right) \equiv \sigma_{\hat{\beta}_1}^{2}
=\left[\frac{1}{n}+\frac{\overline{X}_{2}^{2} \sum x_{3 i}^{2}+\overline{X}_{3}^{2} \sum x_{2 i}^{2}-2 \overline{X}_{2} \overline{X}_{3} \sum x_{2 i} x_{3 i}}{\sum x_{2 i}^{2} \sum x_{3 i}^{2}-\left(\sum x_{2 i} x_{3 i}\right)^{2}}\right] \cdot \sigma^{2}
\end {align}$$`

`$\hat{\beta}_1$`的**样本方差**：

`$$\begin {align} 
S_{\hat{\beta}_1}^{2}
=\left[\frac{1}{n}+\frac{\overline{X}_{2}^{2} \sum x_{3 i}^{2}+\overline{X}_{3}^{2} \sum x_{2 i}^{2}-2 \overline{X}_{2} \overline{X}_{3} \sum x_{2 i} x_{3 i}}{\sum x_{2 i}^{2} \sum x_{3 i}^{2}-\left(\sum x_{2 i} x_{3 i}\right)^{2}}\right] \cdot \hat{\sigma}^{2}
\end {align}$$`

---

## 回归系数的OLS估计：方差和标准差

`$\hat{\beta}_2$`的**真实方差**：

`$$\begin {align} 
\operatorname{var}\left(\hat{\beta}_{2}\right) &\equiv
\sigma_{\hat{\beta}_{2}}^{2}  =\frac{\sum x_{3 i}^{2}}{\left(\sum x_{2 i}^{2}\right)\left(\sum x_{3 i}^{2}\right)-\left(\sum x_{2 i} x_{3 i}\right)^{2}} \sigma^{2} 
\\ 
& =\frac{\sigma^{2}}{\sum x_{2 i}^{2}\left(1-r_{23}^{2}\right)} && \leftarrow \left[ r_{23}^{2}=\frac{\left(\sum x_{2 i} x_{3 i}\right)^{2}}{\sum x_{2 i}^{2} \sum x_{3 i}^{2}} \right] 
\end {align}$$`

`$\hat{\beta}_2$`的**样本方差**：

`$$\begin {align} 
S_{\hat{\beta}_{2}}^{2} 
& =\frac{\sum x_{3 i}^{2}}{\left(\sum x_{2 i}^{2}\right)\left(\sum x_{3 i}^{2}\right)-\left(\sum x_{2 i} x_{3 i}\right)^{2}} \hat{\sigma}^{2} 
\\ 
& =\frac{\hat{\sigma}^{2}}{\sum x_{2 i}^{2}\left(1-r_{23}^{2}\right)} && \leftarrow \left[ r_{23}^{2}=\frac{\left(\sum x_{2 i} x_{3 i}\right)^{2}}{\sum x_{2 i}^{2} \sum x_{3 i}^{2}} \right] 
\end {align}$$`

---

## 回归系数的OLS估计：方差和标准差

`$\hat{\beta}_3$`的**真实方差**：

`$$\begin {align} 
\operatorname{var}\left(\hat{\beta}_{3}\right) &\equiv
\sigma_{\hat{\beta}_{3}}^{2}  =\frac{\sum x_{2 i}^{2}}{\left(\sum x_{2 i}^{2}\right)\left(\sum x_{3 i}^{2}\right)-\left(\sum x_{2 i} x_{3 i}\right)^{2}} \sigma^{2} 
\\ 
& =\frac{\sigma^{2}}{\sum x_{3 i}^{2}\left(1-r_{23}^{2}\right)} && \leftarrow \left[ r_{23}^{2}=\frac{\left(\sum x_{2 i} x_{3 i}\right)^{2}}{\sum x_{2 i}^{2} \sum x_{3 i}^{2}} \right] 
\end {align}$$`

`$\hat{\beta}_3$`的**样本方差**：

`$$\begin {align} 
S_{\hat{\beta}_{3}}^{2} 
& =\frac{\sum x_{2 i}^{2}}{\left(\sum x_{2 i}^{2}\right)\left(\sum x_{3 i}^{2}\right)-\left(\sum x_{2 i} x_{3 i}\right)^{2}} \hat{\sigma}^{2} 
\\ 
& =\frac{\hat{\sigma}^{2}}{\sum x_{3 i}^{2}\left(1-r_{23}^{2}\right)} && \leftarrow \left[ r_{23}^{2}=\frac{\left(\sum x_{2 i} x_{3 i}\right)^{2}}{\sum x_{2 i}^{2} \sum x_{3 i}^{2}} \right] 
\end {align}$$`

---

## 回归系数的OLS估计：协方差

随机变量
`$\hat{\beta}_{2}$`和
`$\hat{\beta}_{3}$`之间的**协方差**为：

`$$\begin {align} 
\operatorname{cov}\left(\hat{\beta}_{2}, \hat{\beta}_{3}\right) & =\frac{-r_{23} \sigma^{2}}{\left(1-r_{23}^{2}\right) \sqrt{\sum x_{2 i}^{2}} \sqrt{\sum x_{3 i}^{2}}} && \leftarrow \left[ r_{23}^{2}=\frac{\left(\sum x_{2 i} x_{3 i}\right)^{2}}{\sum x_{2 i}^{2} \sum x_{3 i}^{2}} \right] 
\end {align}$$`

随机变量
`$\hat{\beta}_{2}$`和
`$\hat{\beta}_{3}$`之间的**样本协方差**为：

`$$\begin {align} 
S^2_{\hat{\beta}_{2} \hat{\beta}_{3}} & =\frac{-r_{23} \hat{\sigma}^{2}}{\left(1-r_{23}^{2}\right) \sqrt{\sum x_{2 i}^{2}} \sqrt{\sum x_{3 i}^{2}}} && \leftarrow \left[ r_{23}^{2}=\frac{\left(\sum x_{2 i} x_{3 i}\right)^{2}}{\sum x_{2 i}^{2} \sum x_{3 i}^{2}} \right] 
\end {align}$$`

---

## OLS估计量的特征1、2

**特征1**：三变量回归面通过均值点
`$(\bar{Y},\bar{X}_2, \bar{X}_3)$`

`$$\begin{align}
\bar{Y} = \hat{\beta}_1 +\hat{\beta}_2\bar{X}_2  +\hat{\beta}_3\bar{X}_3
\end{align}$$`

`$$\begin{align}
\bar{Y} = \hat{\beta}_1 +\hat{\beta}_2\bar{X}_2  +\hat{\beta}_3\bar{X}_3 + \cdots +\hat{\beta}_k\bar{X}_k
\end{align}$$`

**特征2**：
`$Y_i$`的**估计值**(
`$\hat{Y}_i$`)的均值(
`$\bar{\hat{Y_i}}$`)等于Y的样本均值(
`$\bar{Y}$`)

`$$\begin{align}
\hat{Y_i} &= \hat{\beta}_1 +\hat{\beta}_2{X}_{2i} +\hat{\beta}_3{X}_{3i} \\
& =(\bar{Y} - \hat{\beta}_2\bar{X}_2 -\hat{\beta}_3\bar{X}_3 ) + \hat{\beta_2}X_{2i} +\hat{\beta}_3 X_{3i} \\
& = \bar{Y} + \hat{\beta}_2(X_{2i} - \bar{X}_2) + \hat{\beta}_3(X_{3i} - \bar{X}_3) 
\end{align}$$`

`$$\begin{align}
&\Rightarrow  1/n\sum{\hat{Y_i}} =  1/n\sum{ \left( \bar{Y} - \hat{\beta}_2(X_{2i} - \bar{X}_2) - \hat{\beta}_3(X_{3i} - \bar{X}_3) \right) } \\
&\Rightarrow  \bar{\hat{Y_i}}  = \bar{Y}
\end{align}$$`

---

## OLS估计量的特征3

**特征3**：残差的均值(
`$\bar{e_i}$`)为零：

`$$\begin{align}
\frac{\partial \sum e_{i}^{2}}{\partial \hat{\beta}_{1}}=2 \sum\left(Y_{i}-\hat{\beta}_{1}-\hat{\beta}_{2} X_{2 i}-\hat{\beta}_{3} X_{3 i}\right)(-1)&=0 \\
\sum{\left[ Y_i- \hat{\beta}_1 - \hat{\beta}_2X_i -\hat{\beta}_{3} X_{3 i}\right]}  &=0 \\
\sum{( Y_i- \hat{Y}_i )} &=0 \\
\sum{e_i}  &=0 \\
\bar{e_i} &=0
\end{align}$$`

---

## OLS估计量的特征4

**特征4**：残差(
`$e_i$`)和
`$Y_i$`的拟合值(
`$\hat{Y_i}$`)不相关

`$$\begin{align}
Cov(e_i, \hat{Y_i}) &= E \left[ \left( e_i-E(e_i)\right )\cdot \left( \hat{Y_i}-E(\hat{Y_i})\right ) \right]\\
& = E(e_i \cdot \hat{y_i}) \\
& = \sum \mathrm{e}_{i}\left(\hat{y}_{i}+\bar{Y}\right)\\
& =\sum \mathrm{e}_{i} \hat{y}_{i}+\bar{Y} \sum e_{i}\\
& =0
\end{align}$$`

其中：

`$$\begin {align} 
\sum \hat{y}_{i} \mathrm{e}_{i} &=\hat{\beta}_{2} \sum x_{2 i} e_{i}+\hat{\beta}_{3} \sum x_{3 i} e_{i} && \leftarrow \left[ \hat{y}_{i}=\hat{\beta}_{2} x_{2 i}+\hat{\beta}_{3} x_{3 i} \right] \\ 
&=\hat{\beta}_{2} \sum\left(X_{2 i}-\overline{X}_{2}\right) e_{i}+\hat{\beta}_{3} \sum\left(X_{3 i}-\overline{X}_{3}\right) e_{i} \\ 
&=\hat{\beta}_{2} \sum X_{2 i} e_{i}-\hat{\beta}_{2} \overline{X}_{2} \sum e_{i}+\hat{\beta}_{3} \sum X_{3 i} e_{i}-\hat{\beta}_{3} \overline{X}_{3} \sum e_{i} \\ &=0 
 \end {align}$$`

---

## OLS估计量的特征5

**特征5**：残差(
`$e_i$`)和自变量(
`$X_{2i},X_{3i}$`)不相关

`$$\begin {align} 
\frac{\partial \sum e_{i}^{2}}{\partial \hat{\beta}_{2}} & =2 \sum\left(Y_{i}-\hat{\beta}_{1}-\hat{\beta}_{2} X_{2 i}-\hat{\beta}_{3} X_{3 i}\right)\left(-X_{2 i}\right)=0 \\ 
\frac{\partial \sum e_{i}^{2}}{\partial \hat{\beta}_{3}} &=2 \sum\left(Y_{i}-\hat{\beta}_{1}-\hat{\beta}_{2} X_{2 i}-\hat{\beta}_{3} X_{3 i}\right)\left(-X_{3 i}\right)=0\\
\sum{e_iX_{2i}} &=0 \\
\sum{e_iX_{3i}} &=0
\end {align}$$`

---

## OLS估计量的特征6

**特征6**：
`$var(\hat{\beta}_2)$`和
`$var(\hat{\beta}_3)$`的关系。

`$$\begin{align}
\operatorname{var}\left(\hat{\beta}_{2}\right)=\sigma_{\hat{\beta}_{2}}^{2} & =\frac{\sigma^{2}}{\sum x_{2 i}^{2}\left(1-r_{23}^{2}\right)} \\
\operatorname{var}\left(\hat{\beta}_{3}\right)=\sigma_{\hat{\beta}_{3}}^{2} &=\frac{\sigma^{2}}{\sum x_{3 i}^{2}\left(1-r_{23}^{2}\right)} \\
r_{23}^{2} &= \frac{\left(\sum x_{2 i} x_{3 i}\right)^{2}}{\sum x_{2 i}^{2} \sum x_{3 i}^{2}}
\end{align}$$`

`$$r_{23} \rightarrow 1, \operatorname{var}\left(\hat{\beta}_{2}\right) \rightarrow \infty ; \operatorname{var}\left(\hat{\beta}_{3}\right) \rightarrow \infty$$`

- 给定
`$\sum{x_{ki}^2},\sigma^2$`：真值
`$\beta_i$`的估计将变得很困难。

- 给定
`$\sum{x_{ki}^2},r_{23}^2$`：
`$var(\hat{\beta}_i)$`与总体方差呈正比。

- 给定
`$\sigma^2,r_{23}^2$`：
`$var(\hat{\beta}_i)$`与
`$\sum{x_{ki}^2}$`呈反比。表明
`$x_{ki}$`样本值变化越大，真值
`$\beta_i$`的估计精度越高！

---

## 多元判定系数

**多元判定系数**：在三变量（或者更多变量）的模型中，衡量Y的变异由变量
`$(X_{2i},X_{3i})$`等联合解释的比重，记作
`$R^2$`。

`$$\begin {align} 
Y_{i} &=\hat{\beta}_{1}+\hat{\beta}_{2} X_{2 i}+\hat{\beta}_{3} X_{3 i}+e_{i} \\ 
\overline{Y} &=\hat{\beta}_{1}+\hat{\beta}_{2} \overline{X}_{2}+\hat{\beta}_{3} \overline{X}_{3} \\ 
y_{i} &=\hat{\beta}_{2} x_{2 i}+\hat{\beta}_{3} x_{3 i}+e_{i} \\ &=\hat{y}_{i}+e_{i} 
\end {align}$$`

`$$\begin{alignedat}{2}
&&(Y_i - \bar{Y}) &&= (\hat{Y}_i - \bar{Y}) &&+ (Y_i - \hat{Y}_i ) \\
&&y_i &&= \hat{y}_i &&+ e_i \\
&&\sum{y_i^2} &&= \sum{\hat{y}_i^2} &&+ \sum{e_i^2} \\
&&TSS &&=ESS &&+RSS
\end{alignedat}$$`

---

## 多元判定系数

`$$\begin {align} 
RSS &= \sum e_{i}^{2}=\sum y_{i}^{2}-\hat{\beta}_{2} \sum y_{i} x_{2 i}-\hat{\beta}_{3} \sum y_{i} x_{3 i} \\
ESS &=\sum \hat{y}_{i}^{2}=\hat{\beta}_{2} \sum y_{i} x_{2 i}+\hat{\beta}_{3} \sum y_{i} x_{3 i} \\
TSS &= \sum y_{i}^{2}=\sum \hat{y}_{i}^{2}+\left(\sum y_{i}^{2}-\hat{\beta}_{2} \sum y_{i} x_{2 i}-\hat{\beta}_{3} \sum y_{i} x_{3 i}\right)
\end {align}$$`

`$$\begin {align} 
R^{2}=\frac{E S S}{T S S}=1-\frac{\mathrm{RSS}}{T S S}=1-\frac{\sum e_{i}^{2}}{\sum y_{i}^{2}}=\frac{\hat{\beta}_{2} \sum y_{i} x_{2 i}+\hat{\beta}_{3} \sum y_{i} x_{3 i}}{\sum y_{i}^{2}}
\end {align}$$`

比较一元回归下的判定系数：

`$$\begin {align} 
r^{2}=\frac{E S S}{T S S}=\frac{\hat{\beta}_{2}^{2} \sum x_{i}^{2}}{\sum y_{i}^{2}}=\hat{\beta}_{2}\left(\frac{\sum x_{i}^{2}}{\sum y_{i}^{2}}\right) 
\end {align}$$`

---

## 多元判定系数

- 分母部分与
`$X_{ki}$`的变量数**无关**：

`$$\sum y_{i}^{2}=\sum\left(Y_{i}-\overline{Y}\right)^{2}$$`

- 分子部分与
`$X_{ki}$`的变量数**有关**：

`$$\sum e_{i}^{2}=\sum y_{i}^{2}-\hat{\beta}_{2} \sum y_{i} x_{2 i}-\hat{\beta}_{3} \sum y_{i} x_{3 i}$$`
- 如果
`$X_{ki}$`的变量数**增加**，RSS会**减小**，而TSS总是**不变**，因此判定系数
`$R^2$`自然会**变大**。一般而言，自变量数越多，判定系数约接近于1。

- 启示：模型选择时，较高的
`$R^2$`可能来自解释变量个数的增加，并不能说明模型就一定更好。

---

## 调整多元判定系数

**调整判定系数**（adjusted R square）： 利用相应的自由度对平方和进行校正，基于此计算得到的判定系数，记为
`$\bar{R}^2$`。对于如下的多元回归方程：

`$$\begin {align} 
Y_{i}=\beta_{1}+\beta_{2} X_{2 i}+\beta_{3} X_{3 i} + \cdots +\beta_{k} X_{k i}+u_{i} 
\end {align}$$`

则**调整判定系数**
`$\bar{R}^2$`可以计算为：

`$$\begin {align} 
\overline{R}^{2}= 1-\frac{\sum \mathrm{e}_{i}^{2} /(n-k)}{\sum y_{i}^{2} /(n-1)}
\end {align}$$`

---

## 调整多元判定系数

**调整判定系数**
`$\bar{R}^2$`还可以计算为：

`$$\begin {align} 
\bar{R}^{2}&=1-\frac{\hat{\sigma}^{2}}{S_{Y}^{2}}  
&& \leftarrow \left[\hat{\sigma}^{2} =\frac{\sum \mathrm{e}_{i}^{2}}{n-k}  \right] \\ 
\bar{R}^{2} &=1-\left(1-R^{2}\right) \frac{n-1}{n-k} 
&& \leftarrow \left[ S_{Y}^{2}=\frac{\sum(Y-\bar{Y})^{2}}{n-1} \right] 
\end {align}$$`

- 如果
`$k>1$`，则
`$R^2 > \bar{R}^2$`。表明随着变量数的增多，
`$\bar{R}^2$`相对要增大得慢一些。

- 
`$R^2 \geq 0$`，但
`$\bar{R}^2$`可以小于0

- 不能单凭最高的
`$\bar{R}^2$`之值来选择模型，可参考的标准还可以有AIC、APC等。

---

### 模型选择的标准：儿童死亡率案例

人均国民收入PGNP和妇女识字率FLR对儿童死亡率CM的二元回归**模型1**：可以在不同回归元之间进行分配吗？

.pull-left[
妇女识字率FLR对死亡率CM**回归2**：

]

.pull-right[
人均国民收入PGNP对死亡率CM**回归3**：

---

### 模型选择的标准：柯布道格拉斯生产曲线案例

以**柯布道格拉斯生产**模型为例：

`$$\begin {align} 
Y_{i}=\beta_{1} X_{2 i}^{\beta_{2}} X_{3 i}^{\beta_{3}} e^{u_{i}}
\end {align}$$`

`$$\begin {align} 
\ln Y_{i} &=\ln \beta_{1}+\beta_{2} \ln X_{2 i}+\beta_{3} \ln X_{3 i}+u_{i} \\ 
&=\beta_{0}+\beta_{2} \ln X_{2 i}+\beta_{3} \ln X_{3 i}+u_{i} && \leftarrow \left[ \beta_{0}=\ln \beta_{1} \right]
 \end {align}$$`

---

### 模型选择的标准：柯布道格拉斯生产曲线（数据）

<div class="figure" style="text-align: center">
<div id="htmlwidget-3767bf8eea42d81521f0" style="width:100%;height:auto;" class="datatables html-widget"></div>
<script type="application/json" data-for="htmlwidget-3767bf8eea42d81521f0">{"x":{"filter":"none","vertical":false,"data":[["1","2","3","4","5","6","7","8","9","10","11","12","13","14","15","16","17","18","19","20","21","22","23","24","25","26","27","28","29","30","31","32","33","34","35","36","37","38","39","40","41","42","43","44","45","46","47","48","49","50","51"],["Alabama","Alaska","Arizona","Arkansas","California","Colorado","Connecticut","Delaware","District of Columbia","Florida","Georgia","Hawaii","Idaho","Illinois","Indiana","Iowa","Kansas","Kentucky","Louisiana","Maine","Maryland","Massachusetts","Michigan","Minnesota","Mississippi","Missouri","Montana","Nebraska","Nevada","New Hampshire","New Jersey","New Mexico","New York","North Carolina","North Dakota","Ohio","Oklahoma","Oregon","Pennsylvania","Rhode Island","South Carolina","South Dakota","Tennessee","Texas","Utah","Vermont","Virginia","Washington","West Virginia","Wisconsin","Wyoming"],[38372840,1805427,23736129,26981983,217546032,19462751,28972772,14313157,159921,47289846,63015125,1809052,10511786,105324866,90120459,39079550,22826760,38686340,69910555,7856947,21352966,46044292,92335528,48304274,17207903,47340157,2644567,14650080,7290360,9188322,51298516,20401410,87756129,101268432,3556025,124986166,20451196,34808109,104858322,6541356,37668126,4988905,62828100,172960157,15702637,5418786,49166991,46164427,9185967,66964978,2979475],[424471,19895,206893,304055,1809756,180366,224267,54455,2029,471211,659379,17528,75414,963156,835083,336159,246144,384484,216149,82021,174855,355701,943298,456553,267806,439427,24167,163637,59737,96106,407076,43079,727177,820013,34723,1174540,201284,257820,944998,68987,400317,56524,582241,1120382,150030,48134,425346,313279,89639,694628,15221],[2689076,57997,2308272,1376235,13554116,1790751,1210229,421064,7188,2761281,3540475,146371,848220,5870409,5832503,1795976,1595118,2503693,4726625,415131,1729116,2706065,5294356,2833525,1212281,2404122,334008,627806,522335,507488,3295056,404749,4260353,4086558,184700,6301421,1327353,1456683,5896392,297618,2500071,311251,4126465,11588283,762671,276293,2731669,1945860,685587,3902823,361536]],"container":"<table class=\"display\">\n  <thead>\n    <tr>\n      <th> <\/th>\n      <th>state<\/th>\n      <th>Y<\/th>\n      <th>X2<\/th>\n      <th>X3<\/th>\n    <\/tr>\n  <\/thead>\n<\/table>","options":{"dom":"tip","columnDefs":[{"className":"dt-center","targets":"_all"},{"visible":false,"targets":0},{"orderable":false,"targets":0}],"pageLength":8,"order":[],"autoWidth":false,"orderClasses":false,"lengthMenu":[8,10,25,50,100]}},"evals":[],"jsHooks":[]}</script>
<p class="caption">美国51个地区的制造业投入产出数据</p>
</div>

---

### 模型选择的标准：柯布道格拉斯生产案例（回归）

双对数模型下：

`$$\begin{equation} \begin{alignedat}{999} &log(Y)=&& + \beta_{1} && + \beta_{2} log(X2)&& + \beta_{3} log(X3)&&+u_i\\ \end{alignedat} \end{equation}$$`

OLS估计结果为：

`$$\begin{equation} \begin{alignedat}{999} &\widehat{log(Y)}=&&+3.89&&+0.47log(X2)&&+0.52log(X3)\\ &\text{(t)}&&(9.8115)&&(4.7342)&&(5.3803)\\&\text{(se)}&&(0.3962)&&(0.0989)&&(0.0969)\\&\text{(fitness)}&& R^2=0.9642;&& \bar{R^2}=0.9627\\& && F^{\ast}=645.93;&& p=0.0000 \end{alignedat} \end{equation}$$`

---

### 模型选择的标准：多项式回归

**多项式回归模型**(polynomial regression models)：

`$$\begin {align} 
Y &=\beta_{1}+\beta_{2} X+\beta_{3} X^{2} \\
Y_{i}&=\beta_{1}+\beta_{2} X_{i}+\beta_{3} X_{i}^{2}+u_{i} \\
Y_{i}&=\beta_{1}+\beta_{2} X_{i}+\beta_{3} X_{i}^{2}+\cdots+\beta_{k} X_{i}^{k}+u_{i} \\
\hat{Y}_{i}&=\hat{\beta}_{1}+\hat{\beta}_{2} X_{2 i}+\hat{\beta}_{3} X_{3 i}
 \end {align}$$`

- 思考1：上述模型是线性回归模型吗？

- 思考2： X与X的诸多幂函数之间是高度相关的吗？有没有违背自变量无多重共线性的CLRM假设？

---

### 模型选择的标准：总生产成本案例（数据）

<div id="htmlwidget-7687323615283ee40590" style="width:100%;height:auto;" class="datatables html-widget"></div>
<script type="application/json" data-for="htmlwidget-7687323615283ee40590">{"x":{"filter":"none","vertical":false,"data":[["1","2","3","4","5","6","7","8","9","10"],[1,2,3,4,5,6,7,8,9,10],[193,226,240,244,257,260,274,297,350,420],[1,4,9,16,25,36,49,64,81,100],[1,8,27,64,125,216,343,512,729,1000]],"container":"<table class=\"display\">\n  <thead>\n    <tr>\n      <th> <\/th>\n      <th>X<\/th>\n      <th>Y<\/th>\n      <th>XX<\/th>\n      <th>XXX<\/th>\n    <\/tr>\n  <\/thead>\n<\/table>","options":{"dom":"tip","columnDefs":[{"className":"dt-center","targets":"_all"},{"visible":false,"targets":0},{"orderable":false,"targets":0}],"pageLength":8,"order":[],"autoWidth":false,"orderClasses":false,"lengthMenu":[8,10,25,50,100]}},"evals":[],"jsHooks":[]}</script>

---

### 模型选择的标准：总生产成本案例（绘图）

产出X与总成本Y的散点图关系如下：

---

### 模型选择的标准：总生产成本案例（回归）

多项式模型下：

`$$\begin{equation} \begin{alignedat}{999} &Y=&& + \beta_{1} && + \beta_{2} X&& + \beta_{3} XX&& + \beta_{4} XXX&&+u_i\\ \end{alignedat} \end{equation}$$`

OLS估计结果为：

`$$\begin{equation} \begin{alignedat}{999} &\widehat{Y}=&&+141.77&&+63.48X&&-12.96XX&&+0.94XXX\\ &\text{(t)}&&(22.2368)&&(13.2837)&&(-13.1501)&&(15.8968)\\&\text{(se)}&&(6.3753)&&(4.7786)&&(0.9857)&&(0.0591)\\&\text{(fitness)}&& R^2=0.9983;&& \bar{R^2}=0.9975\\& && F^{\ast}=1202.22;&& p=0.0000 \end{alignedat} \end{equation}$$`

---

## 偏相关系数

**偏相关系数**（partial correlation coefficient）：
一个不依赖于
`$X_{2i}$`的，对
`$X_{3i}$`和
`$Y_i$`的影响的一种相关系数。

.pull-left[
- 保持
`$X_{3i}$`不变，
`$Y_i$`和
`$X_{2i}$`之间的相关系数：

`$$\begin {align} 
r_{12 \cdot 3}=\frac{r_{12}-r_{13} r_{23}}{\sqrt{\left(1-r_{13}^{2}\right)\left(1-r_{23}^{2}\right)}}
\end {align}$$`

]

.pull-right[

- 保持
`$X_{2i}$`不变，
`$Y_i$`和
`$X_{3i}$`之间的相关系数：

`$$\begin {align} 
r_{13.2}=\frac{r_{13}-r_{12} r_{23}}{\sqrt{\left(1-r_{12}^{2}\right)\left(1-r_{23}^{2}\right)}}
\end {align}$$`
]

- 保持
`$Y_i$`不变，
`$X_{2i}$`和
`$X_{3i}$`之间的相关系数：

`$$\begin {align} 
r_{23.1}=\frac{r_{23}-r_{12} r_{13}}{\sqrt{\left(1-r_{12}^{2}\right)\left(1-r_{13}^{2}\right)}}
\end {align}$$`

---

## 简单相关系数

**简单相关系数**（simple correlation coefficient）：

.pull-left[

`$Y_i$`和
`$X_{2i}$`之间的相关系数：

`$$\begin {align} 
r_{12}=\frac{\sum y_{i} x_{2 i}}{\sqrt{\sum y_{i}^{2}} \sqrt{\sum x_{2 i}^{2}}}
\end {align}$$`

]

.pull-right[

`$Y_i$`和
`$X_{3i}$`之间的相关系数：

`$$\begin {align} 
r_{13}=\frac{\sum y_{i} x_{3 i}}{\sqrt{\sum y_{i}^{2}} \sqrt{\sum x_{3 i}^{2}}}
\end {align}$$`
]

`$X_{2i}$`和
`$X_{3i}$`之间的相关系数：

`$$\begin {align} 
r_{23}=\frac{\sum x_{2 i} x_{3 i}}{\sqrt{\sum x_{2 i}^{2}} \sqrt{\sum x_{3 i}^{2}}}
\end {align}$$`

---

layout: false
class: center, middle, duke-softblue,hide_logo
name: inference

# 6.2 多元回归分析：推断问题

---

layout: true
  
<div class="my-header-h2"></div>
<div class="watermark1"></div>
<div class="watermark2"></div>
<div class="watermark3"></div>

<div class="my-footer"><span>huhuaping@   <a href="#chapter06a">第06章 多元回归I：代数部分</a>
&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;
&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;
<a href="#inference">6.2 多元回归分析：推断问题</a> </span></div>

---

## N-CLRM假设

**经典正态线性回归模型**(classical normal linear regression model , N-CLRM)：在经典线性回归模型(CLRM)假设中再增加干扰项
`$u_i$`服从正态性的相关假设。

- 均值为0：
`$E(u|X_i)=0$`

- 同方差：
`$Var(u_i) \equiv \sigma^2$`

- 无自相关：
`$E(u_i,u_j)=0$`

- 正态性分布：
`$u_i \sim N(0, \sigma^2)$`

以上几条也可以统写为：
`$u_i \sim iid. \  N(0, \sigma^2)$`

其中，iid表示独立同分布(Independent Identical Distribution, iid)。

---

## 个别回归系数的显著性检验（t检验理论）

`$$\begin {align} 
Y_{i}=\beta_{1}+\beta_{2} X_{2 i}+\beta_{3} X_{3 i}+u_{i} 
\end {align}$$`

二元线性回归模型，在N-CLRM假设容易得到：

`$$\begin{align} 
T_1&=\frac{\hat{\beta}_{1}-\beta_{1}}{S_{\hat{\beta}_{1}}} \sim t(n-3)  \\
T_2&=\frac{\hat{\beta}_{2}-\beta_{2}}{S_{\hat{\beta}_{2}}}
 \sim t(n-3) \\
T_3&=\frac{\hat{\beta}_{3}-\beta_{3}}{S_{\hat{\beta}_{3}}}
 \sim t(n-3)
\end{align}$$`

---

## 个别回归系数的显著性检验（t检验理论）

其中：

`$$\begin {align} 
S_{\hat{\beta}_{2}}^{2} 
&  =\frac{\hat{\sigma}^{2}}{\sum x_{2 i}^{2}\left(1-r_{23}^{2}\right)} && \leftarrow \left[ r_{23}^{2}=\frac{\left(\sum x_{2 i} x_{3 i}\right)^{2}}{\sum x_{2 i}^{2} \sum x_{3 i}^{2}} \right] 
\end {align}$$`

`$$\begin {align} 
S_{\hat{\beta}_{3}}^{2} 
& =\frac{\hat{\sigma}^{2}}{\sum x_{3 i}^{2}\left(1-r_{23}^{2}\right)} && \leftarrow \left[ r_{23}^{2}=\frac{\left(\sum x_{2 i} x_{3 i}\right)^{2}}{\sum x_{2 i}^{2} \sum x_{3 i}^{2}} \right] 
\end {align}$$`

`$$\begin {align} 
\hat{\sigma}^{2}=\frac{\sum \mathrm{e}_{i}^{2}}{n-3}
= \frac{1}{n-3}\left(\sum y_{i}^{2}-\hat{\beta}_{2} \sum y_{i} x_{2 i}-\hat{\beta}_{3} \sum y_{i} x_{3 i}\right)
\end {align}$$`

---

## 个别回归系数的显著性检验（t检验理论）

假设：

`$$H_0: \beta_i =0; \quad H_1: \beta_i \neq 0, \quad i  \in (1,2,3)$$`
基于
`$H_0$`可以得到：

`$$\begin{align} 
t^{\ast}_{\hat{\beta}_1}&=\frac{\hat{\beta}_{1}}{S_{\hat{\beta}_{1}}} \\
t^{\ast}_{\hat{\beta}_2}&=\frac{\hat{\beta}_{2}}{S_{\hat{\beta}_{2}}}
\\
t^{\ast}_{\hat{\beta}_3}&=\frac{\hat{\beta}_{3}}{S_{\hat{\beta}_{3}}}
\end{align}$$`

---

## 个别回归系数的显著性检验（t检验理论）

给定显著性水平
`$\alpha=0.05$`下，查出统计量的**理论分布值**。
`$t_{1-\alpha/2}(n-3)$`。

得到显著性检验的判断结论。

- 若
`$|t^{\ast}_{\hat{\beta}_2}| > t_{1-\alpha/2}(n-3)$`，则
`$\beta_i$`的t检验结果**显著**。换言之，在显著性水平
`$\alpha=0.05$`下，应**显著**地拒绝原假设
`$H_0$`，接受备择假设
`$H_1$`，认为回归参数
`$\beta_i \neq 0$`。
    
- 若
`$|t^{\ast}_{\hat{\beta}_i}| < t_{1-\alpha/2}(n-3)$`，则
`$\beta_i$`的t检验结果**不显著**。换言之，在显著性水平
`$\alpha=0.05$`下，不能**显著**地拒绝原假设
`$H_0$`，只能暂时接受原假设
`$H_0$`，认为回归参数
`$\beta_i = 0$`。

---

## 个别回归系数的显著性检验（t检验案例）

儿童死亡率的二元回归模型如下：

`$$\begin{equation} \begin{alignedat}{999} &CM=&& + \beta_{1} && + \beta_{2} PGNP&& + \beta_{3} FLR&&+u_i\\ \end{alignedat} \end{equation}$$`

以上二元回归模型的OLS估计结果如下：

---

## 样本回归模型的整体显著性检验（F检验）

`$$\begin {align} 
Y_{i}=\beta_{1}+\beta_{2} X_{2 i}+\beta_{3} X_{3 i}+u_{i} 
\end {align}$$`

<table class="table" style="font-size: 20px; margin-left: auto; margin-right: auto;">
 <thead>
  <tr>
   <th style="text-align:center;"> 变异来源 </th>
   <th style="text-align:center;"> 平方和符号SS </th>
   <th style="text-align:center;"> 平方和计算公式 </th>
   <th style="text-align:center;"> 自由度df </th>
   <th style="text-align:center;"> 均方和符号MSS </th>
   <th style="text-align:center;"> 均方和计算公式 </th>
  </tr>
 </thead>
<tbody>
  <tr>
   <td style="text-align:center;"> 回归平方和 </td>
   <td style="text-align:center;"> ESS </td>
   <td style="text-align:center;"> $\hat{\beta}_2\sum{y_ix_{2i}}+\hat{\beta}_3\sum{y_ix_{3i}}$ </td>
   <td style="text-align:center;"> 2 </td>
   <td style="text-align:center;"> $MSS_{ESS}$ </td>
   <td style="text-align:center;"> $ESS/df_{ESS}=\sum{\hat{y}_i^2}/2 $ </td>
  </tr>
  <tr>
   <td style="text-align:center;"> 残差平方和 </td>
   <td style="text-align:center;"> RSS </td>
   <td style="text-align:center;"> $\sum{(Y_i-\hat{Y_i})^2}=\sum{e_i^2}$ </td>
   <td style="text-align:center;"> n-3 </td>
   <td style="text-align:center;"> $MSS_{RSS}$ </td>
   <td style="text-align:center;"> $RSS/df_{RSS}=\frac{\sum{e_i^2}}{n-3}$ </td>
  </tr>
  <tr>
   <td style="text-align:center;"> 总平方和 </td>
   <td style="text-align:center;"> TSS </td>
   <td style="text-align:center;"> $\sum{(Y_i-\bar{Y_i})^2}=\sum{y_i^2}$ </td>
   <td style="text-align:center;"> n-1 </td>
   <td style="text-align:center;"> $MSS_{TSS}$ </td>
   <td style="text-align:center;"> $TSS/df_{TSS}=\frac{\sum{y_i^2}}{n-1}$ </td>
  </tr>
</tbody>
</table>

---

## 样本回归模型的整体显著性检验（F检验）

在原假设下：

`$$H_0: \beta_2 = \beta_3= 0; \quad H_1: \text{not all} \quad \beta_j = 0, \quad j \in 2, 3$$`
有：

`$$\begin {align} 
F^{*}=\frac{MSS_{ESS}}{MSS_{RSS}}=\frac{ESS / df_{\mathrm{Ess}}}{RSS /d f_{\mathrm{RSS}}}=\frac{\left(\hat{\beta}_{2} \sum y_{i} x_{2 i}+\hat{\beta}_{3} \sum \mathrm{y}_{i} x_{3 i}\right) / 2}{\sum \mathrm{e}_{i}^{2} /(n-3)} \sim \mathrm{F}(2, \mathrm{n}-3)
\end {align}$$`

---

## k变量回归模型的F检验

k变量回归模型下：

`$$\begin{align}
Y_i = \beta_1 + \beta_2X_{2i} + \beta_3X_{3i}+ \cdots + \beta_kX_{ki}+ u_i
\end{align}$$`

给出如下假设：
`$$H_0: \beta_2 = \beta_3 =\cdots= \beta_k= 0; \quad H_1: \text{not all} \quad \beta_j = 0, \quad j \in 2, 3, \cdots, k$$`

F样本统计量可以表达为：

`$$\begin {align} 
F^{\ast}=\frac{ESS / df_{ESS}}{RSS / df_{RSS}}=\frac{ESS /(k-1)}{RSS /(n-k)}=\frac{MSS_{ESS}}{MSS_{\mathrm{RSS}}} \sim \mathrm{F}(k-1, n-k)
\end {align}$$`

---

## k变量回归模型的F检验

k变量回归模型下，F样本统计量也可以用判定系数
`$R^2$`表达为：

`$$\begin {align} 
F &=\frac{(n-k) ESS}{(k-1) RSS} \\ 
&=\frac{n-k}{k-1} \cdot \frac{ESS}{TSS-ESS} \\
&=\frac{n-k}{k-1} \cdot \frac{ESS / TSS}{1-(ESS / TSS)} \\ &=\frac{n-k}{k-1} \cdot \frac{R^{2}}{1-R^{2}} \\ 
&=\frac{R^{2} /(k-1)}{\left(1-R^{2}\right) /(n-k)} 
\end {align}$$`

---

## k变量回归模型的F检验

k变量回归模型下，方差分析表（ANOVA）理论上可以写成：

<table class="table" style="font-size: 20px; margin-left: auto; margin-right: auto;">
 <thead>
  <tr>
   <th style="text-align:center;"> 变异来源 </th>
   <th style="text-align:center;"> 平方和符号SS </th>
   <th style="text-align:center;"> 平方和计算公式 </th>
   <th style="text-align:center;"> 自由度df </th>
   <th style="text-align:center;"> 均方和符号MSS </th>
   <th style="text-align:center;"> 均方和计算公式 </th>
  </tr>
 </thead>
<tbody>
  <tr>
   <td style="text-align:center;"> 回归平方和 </td>
   <td style="text-align:center;"> ESS </td>
   <td style="text-align:center;"> $ R^2\sum{y_i^2} $ </td>
   <td style="text-align:center;"> k-1 </td>
   <td style="text-align:center;"> $MSS_{ESS}$ </td>
   <td style="text-align:center;"> $  R^2\sum{y_i^2}/(k-1) $ </td>
  </tr>
  <tr>
   <td style="text-align:center;"> 残差平方和 </td>
   <td style="text-align:center;"> RSS </td>
   <td style="text-align:center;"> $ (1-R^2)\sum{y_i^2} $ </td>
   <td style="text-align:center;"> n-3 </td>
   <td style="text-align:center;"> $MSS_{RSS}$ </td>
   <td style="text-align:center;"> $ \frac{(1-R^2)\sum{y_i^2}}{n-3} $ </td>
  </tr>
  <tr>
   <td style="text-align:center;"> 总平方和 </td>
   <td style="text-align:center;"> TSS </td>
   <td style="text-align:center;"> $ \sum{y_i^2}$ </td>
   <td style="text-align:center;"> n-1 </td>
   <td style="text-align:center;"> $MSS_{TSS}$ </td>
   <td style="text-align:center;"> $TSS/df_{TSS}=\frac{\sum{y_i^2}}{n-1}$ </td>
  </tr>
</tbody>
</table>

---

## k变量回归模型的F检验（案例）

研究关注儿童死亡率（`CM`，千分数）与人均GNP（`PGNP`，1980年的人均GNP）和妇女识字率（`FLR`，百分数）的关系，并构建如下PRM：

`$$\begin{equation} \begin{alignedat}{999} &CM=&& + \beta_{1} && + \beta_{2} PGNP&& + \beta_{3} FLR&&+u_i\\ \end{alignedat} \end{equation}$$`

可以计算得到如下方差分析表（AVOVA）：

<div id="htmlwidget-e36547cd57ab0ca63f48" style="width:100%;height:auto;" class="datatables html-widget"></div>
<script type="application/json" data-for="htmlwidget-e36547cd57ab0ca63f48">{"x":{"filter":"none","vertical":false,"caption":"<caption>儿童死亡率案例的ANOVA分析表<\/caption>","data":[["1","2","3"],["回归平方和ESS","残差平方和RSS","总平方和TSS"],[[257362.373050769],[106315.626949231],[363678]],[2,61,63],[[128681.186525384],[1742.87913031527],[5772.66666666667]]],"container":"<table class=\"display\">\n  <thead>\n    <tr>\n      <th> <\/th>\n      <th>source<\/th>\n      <th>SS<\/th>\n      <th>df<\/th>\n      <th>MSS<\/th>\n    <\/tr>\n  <\/thead>\n<\/table>","options":{"dom":"t","columnDefs":[{"targets":2,"render":"function(data, type, row, meta) {\n    return type !== 'display' ? data : DTWidget.formatRound(data, 2, 3, \",\", \".\", null);\n  }"},{"targets":4,"render":"function(data, type, row, meta) {\n    return type !== 'display' ? data : DTWidget.formatRound(data, 2, 3, \",\", \".\", null);\n  }"},{"className":"dt-center","targets":"_all"},{"visible":false,"targets":0},{"orderable":false,"targets":0}],"pageLength":3,"order":[],"autoWidth":false,"orderClasses":false,"lengthMenu":[3,10,25,50,100]}},"evals":["options.columnDefs.0.render","options.columnDefs.1.render"],"jsHooks":[]}</script>

---

## k变量回归模型的F检验（案例）

因此可以计算得到样本F统计量值（
`$F^{\ast}$`）为：

`$$\begin {align}
F^{\ast} = \frac{ESS / df_{ESS}}{RSS / df_{RSS}}
=\frac{MSS_{ESS}}{MSS_{RSS}}
=\frac{128681.1865}{1742.8791}=73.8325
\end {align}$$`

给定显著性水平
`$\alpha=0.05$`下，查出F分布的**理论值**
`$F_{1-\alpha}(1,n-2)=F_{0.95}$`(2,61)=3.1478

得到显著性检验的判断结论。因为
`$F^{\ast}=$` 73.8325 .red[**大于**] `$F_{1-\alpha}(1,n-2)=F_{0.95}$`(2,61)=3.1478，所以模型整体显著性的F检验结果**显著**。

换言之，在显著性水平
`$\alpha=0.05$`下，应**显著**地拒绝原假设
`$H_0$`，接受备择假设
`$H_1$`，认为斜率参数
`$\beta_2=\beta_3 \neq 0$`。

---
layout: false
class: center, middle, duke-softblue,hide_logo
name: restricted-reg

# 6.3 受约束的最小二乘法

---
layout: true
  
<div class="my-header-h2"></div>
<div class="watermark1"></div>
<div class="watermark2"></div>
<div class="watermark3"></div>

<div class="my-footer"><span>huhuaping@   <a href="#chapter06a">第06章 多元回归I：代数部分</a>
&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;
&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;
<a href="#restricted-reg">6.3 受约束的最小二乘法</a>  </span></div>

---

## 线性等式约束条件

**线性等式约束条件**：根据已有的经济理论，某一回归模型中的系数需满足一些线性等式约束条件。

柯布道格拉斯生产函数(the Cobb-Douglas production function):

`$$\begin {align} 
Y_{i}=\beta_{1} X_{2 i}^{\beta_{2}} X_{3 i}^{\beta_{3}} e^{u_{i}}
\end {align}$$`

其中，
`$Y_i$`表示产出；
`$X_{2i}$`表示劳动投入；
`$X_{3i}$`表示资本投入。

可以将指数模型转换成如下线性模型：

假设所描述的生产是规模报酬不变，由经济理论可得如下的**线性等式约束条件**：

`$$\begin {align}
\beta_{2}+\beta_{3}=1
\end {align}$$`

---

## 线性等式约束的t检验法

**步骤1**：先做无约束的或**无限制的回归**（unrestricted or unconstrained  regression）

`$$\begin{equation} \begin{alignedat}{999} &log(Y)=&& + \hat{\beta}_{1} && + \hat{\beta}_{2} log(X2)&& + \hat{\beta}_{3} log(X3)&&+e_i\\ \end{alignedat} \end{equation}$$`

**步骤2**：构建T统计量：

`$$\begin {align} 
T=\frac{\left(\hat{\beta}_{2}+\hat{\beta}_{3}\right)-\left(\beta_{2}+\beta_{3}\right)}{S_{\left(\hat{\beta}_{2}+\hat{\beta}_{3}\right)}} \sim \mathrm{t}(n-3)
\end {align}$$`

---

## 线性等式约束的t检验法

其中：

`$$\begin {align} 
S^2_{\left(\hat{\beta}_{2}+\hat{\beta}_{3}\right)} &= S^2_{\hat{\beta}_{2}}+S^2_{\hat{\beta}_{3}}+2 S^2_{\hat{\beta}_{2}\hat{\beta}_{3}}
\end {align}$$`

---

## 线性等式约束的t检验法

提出理论假设：

`$$H_0: \beta_{2}+\beta_{3}=1; H_1: \beta_{2}+\beta_{3} \neq 1$$`

在原假设
`$H_0$`下，可以计算得到如下样本t统计量：

`$$\begin {align} 
t^{\ast}=\frac{\left(\hat{\beta}_{2}+\hat{\beta}_{3}\right)-1}{S_{\left(\hat{\beta}_{2}+\hat{\beta}_{3}\right)}} 
\end {align}$$`

给定显著性水平
`$\alpha=0.05$`下，查出统计量的**理论分布值**。
`$t_{1-\alpha/2}(n-3)$`。

得到显著性检验的判断结论。

- 若
`$|t^{\ast}| > t_{1-\alpha/2}(n-2)$`，则
`$\beta_i$`的t检验结果**显著**。换言之，在显著性水平
`$\alpha=0.05$`下，应**显著**地拒绝原假设
`$H_0$`，接受备择假设
`$H_1$`，认为
`$\beta_{2}+\beta_{3} \neq 1$`，也即**规模报酬可变**。
    
- 若
`$|t^{\ast}_{\hat{\beta}_i}| < t_{1-\alpha/2}(n-2)$`，则
`$\beta_i$`的t检验结果**不显著**。换言之，在显著性水平
`$\alpha=0.05$`下，不能**显著**地拒绝原假设
`$H_0$`，只能暂时接受原假设
`$H_0$`，认为回归参数
`$\beta_{2}+\beta_{3} = 1$`，也即**规模报酬不变**。

---

## 线性等式约束的F检验

**无约束模型**（Unrestricted model）：
`$$\begin {align} 
\ln Y_{i}&=\beta_{0}+\beta_{2} \ln X_{2 i}+\beta_{3} \ln X_{3 i}+u_{i} && \leftarrow \left[ \beta_{0}=\ln \beta_{1} \right]
 \end {align}$$`

在线性等式约束条件下：
`$$\begin {align}
\beta_{2} =1- \beta_{3}
\end {align}$$`

可以将原模型变换为如下的**受约束模型**（Restricted model）：

`$$\begin {align}
\ln Y_{i} &=\beta_{0}+\left(1-\beta_{3}\right) \ln X_{2 i}+\beta_{3} \ln X_{3 i}+u_{i} \\ 
\ln Y_{i} &=\beta_{0}+\ln X_{2 i}+\beta_{3}\left(\ln X_{3 i}-\ln X_{2 i}\right)+u_{i} \\
\left(\ln Y_{i}-\ln X_{2 i}\right) & =\beta_{0}+\beta_{3}\left(\ln X_{3 i}-\ln X_{2 i}\right)+u_{i} \\
\ln \left(Y_{i} / X_{2 i}\right) &= \beta_{0}+\beta_{3} \ln \left(X_{3 i} / X_{2 i}\right)+u_{i}
\end {align}$$`

---

## 线性等式约束的F检验

给出原假设
`$H_0: \beta_2 + \beta_3 =1$`下，可以得到如下样本F统计量：

`$$\begin {align}
F^{*}&=\frac{\left(RSS_{R}-RSS_{UR}\right) / m}{RSS_{UR} /(n-k)}
=\frac{\left(\sum e_{R}^{2}-\sum e_{U \mathrm{R}}^{2}\right) / m}{\sum e_{U \mathrm{R}}^{2} /(n-k)} \sim \mathrm{F}(\mathrm{m}, \mathrm{n}-\mathrm{k})\\
F^{*}&=\frac{\left(R_{UR}^{2}-R_{R}^{2}\right) / m}{\left(1-R_{UR}^{2}\right) /(n-k)}
\end {align}$$`

其中：
- 
`$RSS_UR$`表示**无约束回归模型**的RSS；

- 
`$RSS_R$`表示**受约束回归模型**的RSS；

- 
`$R_{UR}^{2}$`表示**无约束回归模型**的判定系数；

- 
`$R_{R}^{2}$`表示**受约束回归模型**的判定系数；

- 
`$m$`表示线性约束条件的个数；
`$n$`表示样本数；
`$k$`表示无约束回归模型中回归系数个数（包括截距）；

---

## 线性等式约束的F检验

给定显著性水平
`$\alpha=0.05$`下，查出统计量的**理论分布值**。
`$F_{1-\alpha}(m,n-k)$`

得到显著性检验的判断结论。

- 若
`$F^{\ast} > F_{1-\alpha}(m,n-k)$`，则**显著**地拒绝原假设
`$H_0$`，接受备择假设
`$H_1$`，认为**规模报酬可变**。

- 若
`$F^{\ast} < F_{1-\alpha}(m,n-k)$`，则不能**显著**地拒绝原假设
`$H_0$`，暂时接受原假设
`$H_0$`，认为**规模报酬不变**。

---

## 线性等式约束的F检验（案例：数据）

<div id="htmlwidget-75b79541d036daa54a87" style="width:100%;height:auto;" class="datatables html-widget"></div>
<script type="application/json" data-for="htmlwidget-75b79541d036daa54a87">{"x":{"filter":"none","vertical":false,"data":[["1","2","3","4","5","6","7","8","9","10","11","12","13","14","15","16","17","18","19","20","21","22","23","24","25","26","27","28","29","30","31","32","33","34","35","36","37","38","39","40","41","42","43","44","45","46","47","48","49","50","51"],["Alabama","Alaska","Arizona","Arkansas","California","Colorado","Connecticut","Delaware","District of Columbia","Florida","Georgia","Hawaii","Idaho","Illinois","Indiana","Iowa","Kansas","Kentucky","Louisiana","Maine","Maryland","Massachusetts","Michigan","Minnesota","Mississippi","Missouri","Montana","Nebraska","Nevada","New Hampshire","New Jersey","New Mexico","New York","North Carolina","North Dakota","Ohio","Oklahoma","Oregon","Pennsylvania","Rhode Island","South Carolina","South Dakota","Tennessee","Texas","Utah","Vermont","Virginia","Washington","West Virginia","Wisconsin","Wyoming"],[38372840,1805427,23736129,26981983,217546032,19462751,28972772,14313157,159921,47289846,63015125,1809052,10511786,105324866,90120459,39079550,22826760,38686340,69910555,7856947,21352966,46044292,92335528,48304274,17207903,47340157,2644567,14650080,7290360,9188322,51298516,20401410,87756129,101268432,3556025,124986166,20451196,34808109,104858322,6541356,37668126,4988905,62828100,172960157,15702637,5418786,49166991,46164427,9185967,66964978,2979475],[424471,19895,206893,304055,1809756,180366,224267,54455,2029,471211,659379,17528,75414,963156,835083,336159,246144,384484,216149,82021,174855,355701,943298,456553,267806,439427,24167,163637,59737,96106,407076,43079,727177,820013,34723,1174540,201284,257820,944998,68987,400317,56524,582241,1120382,150030,48134,425346,313279,89639,694628,15221],[2689076,57997,2308272,1376235,13554116,1790751,1210229,421064,7188,2761281,3540475,146371,848220,5870409,5832503,1795976,1595118,2503693,4726625,415131,1729116,2706065,5294356,2833525,1212281,2404122,334008,627806,522335,507488,3295056,404749,4260353,4086558,184700,6301421,1327353,1456683,5896392,297618,2500071,311251,4126465,11588283,762671,276293,2731669,1945860,685587,3902823,361536],[90.4015586459381,90.7477758230711,114.726592973179,88.7404680074329,120.207382652689,107.906983577836,129.188743774162,262.843760903498,78.8176441596846,100.358111334413,95.5673823400503,103.209265175719,139.387726416846,109.353901133357,107.917966238087,116.253171862125,92.7374219968799,100.618855401005,323.436865310503,95.7918947586594,122.118132166652,129.446619492214,97.8858515548639,105.80211716931,64.2551063082978,107.731561783869,109.428849257252,89.5279185025392,122.040946147279,95.6061224065095,126.017048413564,473.581327328861,120.680561953967,123.496129939403,102.411225988538,106.412864610826,101.603684346495,135.009343728182,110.961422140576,94.8201255308971,94.0957441227827,88.2617118392187,107.907378559737,154.376058344386,104.663314003866,112.577097270121,115.592931401729,147.358830307809,102.477348029318,96.4040867917792,195.74765127127],[6.33512301193721,2.9151545614476,11.1568395257452,4.52626991827137,7.48947150886639,9.92842886131533,5.39637574854972,7.73232944633183,3.54263183834401,5.85996719091872,5.36940818558068,8.35069602921041,11.2475137242422,6.09497215404358,6.9843392812451,5.34263845382691,6.48042609204368,6.51182623984353,21.867438664995,5.06127699003914,9.88885648108433,7.60769578944113,5.6126017440936,6.20634406082098,4.5267133671389,5.4710384204885,13.8208300575164,3.83657730219938,8.74391080904632,5.28050277818242,8.09444919376234,9.39550593096404,5.85875653382876,4.98352830991704,5.31924084900498,5.36501183442028,6.59442876731385,5.65,6.23958145943166,4.31411715250699,6.24522815668583,5.50652820041044,7.08721130940624,10.3431534958612,5.08345664200493,5.7400797772884,6.42222802142256,6.21126854975916,7.64831156081616,5.61858001692992,23.7524472767886]],"container":"<table class=\"display\">\n  <thead>\n    <tr>\n      <th> <\/th>\n      <th>state<\/th>\n      <th>Y<\/th>\n      <th>X2<\/th>\n      <th>X3<\/th>\n      <th>Y.X2<\/th>\n      <th>X3.X2<\/th>\n    <\/tr>\n  <\/thead>\n<\/table>","options":{"dom":"tip","columnDefs":[{"targets":5,"render":"function(data, type, row, meta) {\n    return type !== 'display' ? data : DTWidget.formatRound(data, 4, 3, \",\", \".\", null);\n  }"},{"targets":6,"render":"function(data, type, row, meta) {\n    return type !== 'display' ? data : DTWidget.formatRound(data, 4, 3, \",\", \".\", null);\n  }"},{"className":"dt-center","targets":"_all"},{"visible":false,"targets":0},{"orderable":false,"targets":0}],"pageLength":7,"order":[],"autoWidth":false,"orderClasses":false,"lengthMenu":[7,10,25,50,100]}},"evals":["options.columnDefs.0.render","options.columnDefs.1.render"],"jsHooks":[]}</script>

- 
`$Y.X_2$`表示
`$Y/X_2$`；
`$X3.X_2$`表示
`$X_3/X_2$`

---

## 线性等式约束的F检验（案例：回归1）

**无约束模型**回归结果如下：

`$$\begin{equation} \begin{alignedat}{999} &log(Y)=&& + \beta_{1} && + \beta_{2} log(X2)&& + \beta_{3} log(X3)&&+u_i\\ \end{alignedat} \end{equation}$$`

其中：

- 
`$RSS_{UR}=$` 3.4155

- 
`$R^2_{UR}=$` 0.9642

---

## 线性等式约束的F检验（案例：回归2）

**受约束模型**回归结果如下：

`$$\begin{equation} \begin{alignedat}{999} &log(Y.X2)=&& + \beta_{1} && + \beta_{2} log(X3.X2)&&+u_i\\ \end{alignedat} \end{equation}$$`

`$$\begin{equation} \begin{alignedat}{999} &\widehat{log(Y.X2)}=&&+3.76&&+0.52log(X3.X2)\\ &\text{(t)}&&(20.2637)&&(5.4665)\\&\text{(se)}&&(0.1854)&&(0.0958)\\&\text{(fitness)}&& R^2=0.3788;&& \bar{R^2}=0.3661\\& && F^{\ast}=29.88;&& p=0.0000 \end{alignedat} \end{equation}$$`

其中：

- 
`$RSS_R=$` 3.4256

- 
`$R^2_R=$` 0.3788

---

## 线性等式约束的F检验（案例：F统计量）

.pull-left[
利用RSS计算样本F统计量：

`$$\begin{align}
F^{*}&=\frac{\left(RSS_{R}-RSS_{UR}\right) / m}{RSS_{UR} /(n-k)}\\
&=\frac{\left(3.4256-3.4155\right) /1}{3.4155/(51-3)} \\
&=0.1414
\end{align}$$`
]

.pull-right[
利用
`$R^2$`计算样本F统计量：

`$$\begin{align}
F^{*}&=\frac{\left(R_{UR}^{2}-R_{R}^{2}\right) / m}{\left(1-R_{UR}^{2}\right) /(n-k)}\\
&=\frac{\left(0.9642-0.3788\right) /1}{ (1-0.9642) /(51-3)} \\
&=784.2919
\end{align}$$`

]

给定显著性水平
`$\alpha=0.05$`下，查出F分布的**理论值**
`$F_{1-\alpha}(m,n-k)=F_{0.95}$`(1,48)=4.0427

因为
`$F^{\ast}=$` 0.1414 .red[**小于**] 
`$F_{1-\alpha}(m,n-k)=F_{0.95}$` (1,48)=4.0427，所以，认为
`$\beta_2+\beta_3 = 1$`，也即**规模报酬不变**。

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# 6.4 检验回归模型的结构或稳定性

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<div class="my-footer"><span>huhuaping@   <a href="#chapter06a">第06章 多元回归I：代数部分</a>
&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;
&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;
<a href="#structure-break">6.4 检验回归模型的结构或稳定性</a>  </span></div>

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## 邹至庄检验

<div class="figure" style="text-align: center">
<div id="htmlwidget-bdcef755b0955e8de3d1" style="width:100%;height:auto;" class="datatables html-widget"></div>
<script type="application/json" data-for="htmlwidget-bdcef755b0955e8de3d1">{"x":{"filter":"none","vertical":false,"data":[["1","2","3","4","5","6","7","8","9","10","11","12","13","14","15","16","17","18","19","20","21","22","23","24","25","26"],[1970,1971,1972,1973,1974,1975,1976,1977,1978,1979,1980,1981,1982,1983,1984,1985,1986,1987,1988,1989,1990,1991,1992,1993,1994,1995],[61,68.6,63.6,89.6,97.6,104.4,96.4,92.5,112.6,130.1,161.8,199.1,205.5,167,235.7,206.2,196.5,168.4,189.1,187.8,208.7,246.4,272.6,214.4,189.4,249.3],[727.1,790.2,855.3,965,1054.2,1159.2,1273,1401.4,1580.1,1769.5,1973.3,2200.2,2347.3,2522.4,2810,3002,3187.6,3363.1,3640.8,3894.5,4166.8,4343.7,4613.7,4790.2,5021.7,5320.8],["spl1","spl1","spl1","spl1","spl1","spl1","spl1","spl1","spl1","spl1","spl1","spl1","spl2","spl2","spl2","spl2","spl2","spl2","spl2","spl2","spl2","spl2","spl2","spl2","spl2","spl2"]],"container":"<table class=\"display\">\n  <thead>\n    <tr>\n      <th> <\/th>\n      <th>Year<\/th>\n      <th>Saving<\/th>\n      <th>Income<\/th>\n      <th>sample<\/th>\n    <\/tr>\n  <\/thead>\n<\/table>","options":{"dom":"tip","columnDefs":[{"className":"dt-center","targets":"_all"},{"visible":false,"targets":0},{"orderable":false,"targets":0}],"pageLength":8,"order":[],"autoWidth":false,"orderClasses":false,"lengthMenu":[8,10,25,50,100]}},"evals":[],"jsHooks":[]}</script>
<p class="caption">1970-1995年间美国储蓄、可支配收入数据（n=26）</p>
</div>

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## 邹至庄检验

按时间分段绘制散点图如下：

其中：spl1表示1970-1981年；spl2表示1982-1995年。

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## 邹至庄检验

根据散点图的情况，我们可以构建如下三个模型：

`$$\begin {align} 
1970-1981 : & Y_{t}=\lambda_{1}+\lambda_{2} X_{t}+u_{1 t}&& {n_{1}=12} \\ 
1982-1995 : & Y_{t}=\gamma_{1}+\gamma_{2} X_{t}+u_{2 t} && {n_{2}=14} \\ 
1970-1995 : & Y_{t}=\alpha_{1}+\alpha_{2} X_{t}+u_{t} && n=\left(n_{1}+n_{2}\right)=26
\end {align}$$`

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## 邹至庄检验

.pull-left[
样本段spl1（1970-1981年）**回归1**：

`$$\begin{equation} \begin{alignedat}{999} &\widehat{Saving}=&&+1.02&&+0.08Income\\ &\text{(t)}&&(0.0873)&&(9.6016)\\&\text{(se)}&&(11.6377)&&(0.0084)\\&\text{(fitness)}&& R^2=0.9021;&& \bar{R^2}=0.8924\\& && F^{\ast}=92.19;&& p=0.0000 \end{alignedat} \end{equation}$$`
]

.pull-right[
样本段spl2（1982-1995年）**回归2**：

`$$\begin{equation} \begin{alignedat}{999} &\widehat{Saving}=&&+153.49&&+0.01Income\\ &\text{(t)}&&(4.6923)&&(1.7708)\\&\text{(se)}&&(32.7123)&&(0.0084)\\&\text{(fitness)}&& R^2=0.2072;&& \bar{R^2}=0.1411\\& && F^{\ast}=3.14;&& p=0.1020 \end{alignedat} \end{equation}$$`
]

全部样本（1970-1995年）**回归3**：

`$$\begin{equation} \begin{alignedat}{999} &\widehat{Saving}=&&+62.42&&+0.04Income\\ &\text{(t)}&&(4.8918)&&(8.8938)\\&\text{(se)}&&(12.7607)&&(0.0042)\\&\text{(fitness)}&& R^2=0.7672;&& \bar{R^2}=0.7575\\& && F^{\ast}=79.10;&& p=0.0000 \end{alignedat} \end{equation}$$`

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## 邹至庄检验

**邹至庄检验**的原理和过程如下：

**步骤1**：估计在全部样本下的**约束方程**（方程3），得到**约束残差平方和**（记为
`$RSS_R$`，此处也记为
`$RSS_3$`）。在全部样本下（1970-1995）的模型，如果参数是稳定的，也即可以认为约束了如下两个条件：

`$$\gamma_1=\lambda_1; \quad \gamma_2=\lambda_2$$`

**步骤2**：估计分段样本（spl1=1970-1981）下的子方程1，得到其残差平方和
`$RSS_1$`，其自由度为
`$df_{RSS_1}=n_1-k$`

**步骤3**：估计分段样本（spl2=1982-1995）下的子方程2，得到其残差平方和
`$RSS_2$`，其自由度为
`$df_{RSS_2}=n_2-k$`

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## 邹至庄检验

**步骤4**：计算得到**无约束残差平方和**（
`$RSS_{UR}=RSS_1 +RSS_2$`），其自由度为
`$df_{RSS_{UR}}=n_1 +n_2 -2k$`

**步骤5**：如果没有结构性变动（
`$H_0$`），则构造得到如下样本F统计量：

`$$\begin {align} 
F^{\ast}=\frac{\left(\mathrm{RSS}_{\mathrm{R}}-\mathrm{RSS}_{\mathrm{UR}}\right) / k}{\left(\mathrm{RSS}_{\mathrm{UR}}\right) /\left(n_{1}+n_{2}-2 k\right)} \sim F_{\left[k,\left(n_{1}+n_{2}-2 k\right)\right]}
\end {align}$$`

**步骤6**：得到显著性检验的判断结论。

- 若
`$F^{\ast} > F_{1-\alpha}(k,n_1+n_2-2k)$`，则**显著**地拒绝原假设
`$H_0$`，接受备择假设
`$H_1$`，认为**存在结构变动**（也即参数不稳定）。

- 若
`$F^{\ast} < F_{1-\alpha}(k,n_1+n_2-2k)$`，则不能**显著**地拒绝原假设
`$H_0$`，暂时接受原假设
`$H_0$`，认为**不存在结构变动**（也即参数稳定）。

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## 邹至庄检验

在本案例中：

- **约束残差平方和**（记为
`$RSS_R=$` 23248.2982）

- **无约束残差平方和**（
`$RSS_{UR}=RSS_1 +RSS_2=$` 1785.0321+10005=11790.2528），其自由度为
`$df_{RSS_{UR}}=n_1 +n_2 -2k=$` 12+14-2*2=22

`$$\begin{align}
F^{*}&=\frac{\left(RSS_{R}-RSS_{UR}\right) / k}{RSS_{UR} /(n_1+n_2-2k)}\\
&=\frac{\left(23248.2982-11790.2528\right) /2}{11790.2528/(12+14-2*2)} \\
&=10.6901
\end{align}$$`

- 因为
`$F^{\ast}=$` 10.6901 .red[**大于**]查表值
`$F_{1-\alpha}(k,n_1+n_2-2k)=F_{0.95}$`(2,12+14-2\*2)=3.4434，所以显著拒绝
`$H_0$`，接受备择假设
`$H_1$`，认为**存在结构性变动**，也即**参数不稳定**。

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## 邹至庄检验

牢记关于邹至庄检验的一些警告：

- 必须满足该检验背后的假定。

- 邹至庄检验只是告诉我们子样本模型之间是否有差别，并没有告诉我们差别是来自截距、斜率还是二者都有。

- 邹至庄检验假定我们知道结构转折点。

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