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3 多元线性回归

  在2.3.1中,我们证得

\[\hat{\beta_1}=\beta_1+\frac{\sum_{i=1}^{n}(X_i-\bar{X})u_i}{\sum_{i=1}^{n}(X_i-\bar{X})^2}\]

因此有

\[\hat{\beta_1}\stackrel{d}{\rightarrow}\beta_1+\frac{\mathrm{cov}(X,u)}{\mathrm{var}(X)}=\beta_1+\rho_{Xu}\frac{\sigma_u}{\sigma_x}\]

如果模型具有内生性,即\(\mathrm{cov}(X,u)\neq0\)\(\hat{\beta_1}\)不是\(\beta_1\)的无偏和一致估计,此时简单线性回归出现了遗漏变量偏差. 为了解决这个问题,需要采用多元线性回归.

3.1 回归系数的估计

定义3.1.1 定义

\[Y_i=X_i^\prime\beta+u_i\]

多元线性回归模型.

  注意到样本多元回归中,

\[X_i=\left(\begin{array}{c} 1\\ X_{i1}\\ X_{i2}\\ \vdots\\ X_{ik} \end{array}\right)\]

而总体多元回归中,

\[X_{n\times(k+1)}=\left(\begin{array}{cccc} 1&X_{11}&\ldots&X_{1k}\\ 1&X_{21}&\ldots&X_{2k}\\ 1&X_{31}&\ldots&X_{3k}\\ \vdots&\vdots&\,&\vdots\\ 1&X_{n1}&\ldots&X_{nk} \end{array}\right)\]

其中,\(k\)表示解释变量的个数,\(n\)表示样本数.

  类似地,我们也可得到多元线性回归的正规方程组. 目标函数为

\[\min\sum_{i=1}^{n}(Y_i-X_i^\prime\hat{\beta})^2\]

\(\hat{\beta}\)进行求导,得

\[\nabla=-2\sum_{i=1}^{n}X_i^\prime\hat{u_i}=0\]

\[\sum_{i=1}^{n}X_i^\prime\hat{u_i}=0\]

将正规方程组写成标量形式,可得

\[\begin{cases} \sum_{i=1}^n\hat{u_i}=0\\ \sum_{i=1}^nX_{i1}\hat{u_i}=0\\ \sum_{i=1}^nX_{i2}\hat{u_i}=0\\ \ldots\\ \sum_{i=1}^nX_{ik}\hat{u_i}=0 \end{cases}\]

现对正规方程组进行求解:

\[X^\prime(Y-X\hat{\beta})=0\]

\[\hat{\beta}=(X^\prime X)^{-1}X^\prime Y\]

3.1.1 回归系数的期望和方差

  现验证\(\hat{\beta}\)的无偏性和一致性.

\[\mathbb{E}[\hat{\beta}|X]=\mathbb{E}[(X^\prime X)^{-1}X^\prime Y|X]=(X^\prime X)^{-1}X^\prime\mathbb{E}[Y|X]=(X^\prime X)^{-1}X^\prime X\beta=\beta\]

  由定理2.2.1可知,\(\beta=(\mathbb{E}[X^\prime X])^{-1}\mathbb{E}[X^\prime Y]\),根据连续映射定理,可得

\[\hat{\beta}=(X^\prime X)^{-1}X^\prime Y\stackrel{p}{\rightarrow}(\mathbb{E}[X_i X_i^\prime])^{-1}\mathbb{E}[X_i Y]=\beta\]

  接下来求\(\hat{\beta}\)的条件方差. 设误差方差矩阵为\(D(d_{ii}=\sigma_i^2,d_{ij}=0,i\neq j)\),则

\[\mathrm{var}(\hat{\beta}|X)=\mathrm{var}((X^\prime X)^{-1}X^\prime Y|X)=(X^\prime X)^{-1}X^\prime DX(X^\prime X)^{-1}\]

在同方差条件下,由于\(D=I_n\sigma^2\),则

\[\mathrm{var}(\hat{\beta})=\sigma^2(X^\prime X)^{-1}\]

3.1.2 回归系数的大样本抽样分布

  注意到

\[\hat{\beta}=(X^\prime X)^{-1}X^\prime Y=(X^\prime X)^{-1}X^\prime (X\beta+u)=\beta+(X^\prime X)^{-1}X^\prime u=\beta+\left(\sum_{i=1}^{n}X_iX_i^\prime\right)^{-1}\left(\sum_{i=1}^{n}X_iu_i\right)\]

由于

\[\mathbb{E}\left[\sum_{i=1}^{n}X_iu_i|X_i\right]=0\]
\[\mathrm{var}(X_i u_i)=\mathbb{E}[X_i^\prime X_iu_i^2]\]

\[\sum_{i=1}^{n}X_iu_i\stackrel{d}{\rightarrow}N(0,n\mathbb{E}[X_i^\prime X_iu_i^2])\]

因为

\[\left(\sum_{i=1}^{n}X_iX_i^\prime\right)^{-1}\stackrel{p}{\rightarrow}\mathbb{E}[n(X_iX_i^\prime)]^{-1}=\frac{1}{n}\mathbb{E}[(X_iX_i^\prime)]^{-1}\]

\[(X^\prime X)^{-1}X^\prime u\stackrel{d}{\rightarrow}N\left( 0,\frac{1}{n}\mathbb{E}[(X_iX_i^\prime)]^{-1}\mathbb{E}[X_i^\prime X_iu_i^2]\mathbb{E}[(X_iX_i^\prime)]^{-1}\right) \]

即可得到\(\hat{\beta}\)的大样本抽样分布为

\[\hat{\beta}\stackrel{d}{\rightarrow}N\left( \beta,\frac{1}{n}\mathbb{E}[(X_iX_i^\prime)]^{-1}\mathbb{E}[X_i^\prime X_iu_i^2]\mathbb{E}[(X_iX_i^\prime)]^{-1}\right)\]

3.2 分块回归和偏回归

定义3.2.1 定义矩阵

\[P=X(X^\prime X)^{-1}X^\prime\]

投影矩阵.

  注意到\(\hat{Y}=X\hat{\beta}=X(X^\prime X)^{-1}X^\prime Y=PY\),用\(P\)左乘任何向量可得该向量在超平面\(X\)上的投影.

定理3.2.1 (投影矩阵的性质)(1)\(P^\prime=P\)

(2)\(P^2=P\)

(3)\(PX=X\)

(4)\(\mathrm{r}(P)=\mathrm{tr}(P)=k+1\).

证明:

\[P^\prime=X((X^\prime X)^\prime)^{-1}X^\prime=X^\prime(X^\prime X)^{-1}X=P\]
\[P^2=X(X^\prime X)^{-1}X^\prime X(X^\prime X)^{-1}X^\prime=X(X^\prime X)^{-1}X^\prime=P\]
\[PX=X(X^\prime X)^{-1}X^\prime X=X\]

由于\(P\)是实对称阵,可进行相似对角化,即

\[P=Q^\prime I_{k+1}Q\]

则有\(\mathrm{r}(P)=k+1\). 由幂等矩阵的性质可知,\(\mathrm{tr}(P)=\mathrm{r}(P)=k+1\).

定义3.2.2 定义矩阵

\[M=I_n-P\]

消灭矩阵.

  注意到\(\hat{u}=Y-\hat{Y}=Y-PY=MY\),用\(M\)左乘任何向量可得该向量投影后的残差向量.

定理3.2.2 (消灭矩阵的性质)(1)\(M^\prime=M\)

(2)\(M^2=M\)

(3)\(MX=0\)

(4)\(PM=0\)

(5)\(\mathrm{r}(M)=\mathrm{tr}(M)=n-k-1\).

证明:

\[M^\prime=I_n^\prime-P^\prime=I_n-P\]
\[M^2=(I_n-P)(I_n-P)=I_n-2P+P^2=I_n-P=M\]
\[MX=X-PX=X-X=0\]
\[PM=P(I_n-P)=P-P^2=0\]

\(PM=0\),可得\(\mathrm{r}(P)+\mathrm{r}(M)\leq n\);又由\(P+M=I_n\),可得\(n=\mathrm{r}(I_n)\leq \mathrm{r}(P)+\mathrm{r}(M)\),所以

\[\mathrm{r}(P)+\mathrm{r}(M)=n\]

则有\(\mathrm{r}(M)=n-k-1\). 由幂等矩阵的性质可知,\(\mathrm{tr}(M)=\mathrm{r}(M)=n-k-1\).

  利用消灭矩阵的性质,我们可以将残差写成总体误差项的函数

\[\hat{u}=MY=M(X\beta+u)=Mu.\]

进而将残差平方和也写成总体误差项的函数

\[SSR=\hat{u}^\prime\hat{u}=u^\prime MMu=u^\prime Mu.\]

  在多元线性回归中,只要增加一个变量就会对所有的回归系数产生影响,然而仅从\(\hat{\beta}=(X^\prime X)^{-1}X^\prime Y\)这一表达式中很难看出不同变量的影响.为此,我们将\(X\)进行分块,即\(X=(X_1\quad X_2)\),分别对应于两组解释变量,此时多元线性回归模型可以改写为

\[Y=X_1\beta_1+X_2\beta_2+u\]

\(\hat{\beta}=(X^\prime X)^{-1}X^\prime Y\)可改写为

\[\begin{align*} \left(\begin{array}{c} \hat{\beta_1}\\ \hat{\beta_2} \end{array}\right)&=\left(\left(\begin{array}{c} X_1^\prime\\ X_2^\prime \end{array}\right)(X_1\quad X_2)\right)^{-1}\left(\begin{array}{c} X_1^\prime\\ X_2^\prime \end{array}\right)Y\\ &=\left(\begin{array}{cc} X_1^\prime X_1& X_1^\prime X_2\\ X_2^\prime X_1& X_2^\prime X_2 \end{array}\right)^{-1}\left(\begin{array}{c} X_1\\ X_2 \end{array}\right)Y\\ &=\left(\begin{array}{c} (X_1^\prime M_2X_1)^{-1}X_1^\prime M_2\\ (X_2^\prime M_1X_2)^{-1}X_2^\prime M_1 \end{array}\right)Y \end{align*}\]

\(r_1=M_1X_2\)\(\tilde{u}_1=M_1Y\),则

\[\hat{\beta_2}=(\hat{r_1}^\prime\hat{r_1})^{-1}\hat{r_1}^\prime\tilde{u}_1\]

这里可以看出,\(\hat{r_1}\)\(X_2\)\(X_1\)回归得到的残差矩阵,\(\tilde{u}_1\)\(Y\)\(X_1\)回归得到的残差矩阵,而\(\hat{\beta_2}\)恰恰是\(\tilde{u}_1\)\(\hat{r_1}\)进行回归所得的回归系数.这就是Frisch-Waugh-Lovell定理的内容.

  根据这一定理,我们就可对偏回归系数\(\beta_j\)进行估计.设\(X_j\)是解释变量\(X_2\)的样本组成的一维行向量. 首先由\(X_j\)对其他所有解释变量进行回归,得到残差列向量\(\hat{r}_j\);再由\(Y\)对其他所有解释变量进行回归,得到残差列向量\(\tilde{u}_j\),则

\[\hat{\beta_j}=\frac{\sum_{i=1}^{n}\hat{r}_{ij}\tilde{u}_{ij}}{\sum_{i=1}^{n}\hat{r}_{ij}^2}\]

由正规方程组易知

\[\sum_{i=1}^{n}\hat{r}_{ij}\tilde{u}_{ij}=\sum_{i=1}^{n}\hat{r}_{ij}(Y_i-X^*\tilde{\beta})\sum_{i=1}^{n}\hat{r}_{ij}Y_i\]

则有

\[\hat{\beta_j}=\frac{\sum_{i=1}^{n}\hat{r}_{ij}Y_i}{\sum_{i=1}^{n}\hat{r}_{ij}^2}\]

\(\beta_j\)是剔除了其他所有解释变量后的残差\(\hat{r}_{j}\)\(Y\)的样本协方差和\(\hat{r}_{j}\)方差之比.

注3.2.1 \(X^*\)为除\(X_j\)以外的其他解释变量组成的行向量,\(\tilde{\beta}\)\(Y\)对除\(X_j\)以外的其他解释变量的回归系数列向量,显然\(\hat{r}_{ij}\)\(X^*\)无关.

3.2.1 解释变量个数不同时偏回归系数之间的关系

  现在,我们用上述结论探究解释变量个数不同时偏回归系数之间的关系.

  先考察简单线性回归\(Y=\gamma_0+\gamma_1X_1+e\)和多元线性回归\(Y=\beta_0+\beta_1X_1+\cdots+\beta_kX_k+u\).

\[\begin{align*} \hat{\gamma_1}&=\frac{\sum_{i=1}^n(X_{i1}-\bar{X}_1)Y_i}{\sum_{i=1}^n(X_{i1}-\bar{X}_1)^2}\\ &=\frac{\sum_{i=1}^n(X_{i1}-\bar{X}_1)(\hat{\beta_0}+\hat{\beta_1}X_{i1}+\cdots+\hat{\beta_k}X_{ik}+\hat{u_i})}{\sum_{i=1}^n(X_{i1}-\bar{X}_1)^2}\\ &=\hat{\beta_1}+\hat{\beta_2}\hat{\gamma_{2,1}}+\cdots+\hat{\beta_k}\hat{\gamma_{k,1}} \end{align*}\]

\(\hat{\gamma_{j,1}}\)表示\(X_j\)\(X_1\)的回归系数.

  再将条件放宽,考察短回归\(Y=\gamma_0+\gamma_1X_1+\cdots+\gamma_pX_p+e\)和长回归\(Y=\beta_0+\beta_1X_1+\cdots+\beta_pX_p+\beta_{p+1}X_{p+1}+\cdots+\beta_kX_k+u\).

\[\begin{align*} \hat{\gamma_1}&=\frac{\sum_{i=1}^n\hat{r_i}Y_i}{\sum_{i=1}^n\hat{r_i}^2}\\ &=\frac{\sum_{i=1}^n\hat{r_i}(\hat{\beta_0}+\hat{\beta_1}X_{i1}+\cdots+\hat{\beta_k}X_{ik}+\hat{u_i})}{\sum_{i=1}^n\hat{r_i}^2}\\ &=\hat{\beta_1}+\hat{\beta}_{p+1}\frac{\sum_{i=1}^n\hat{r_i}X_{ip+1}}{\sum_{i=1}^n\hat{r_i}^2}+\cdots+\hat{\beta_k}\frac{\sum_{i=1}^n\hat{r_i}X_{ik}}{\sum_{i=1}^n\hat{r_i}^2}\\ &=\hat{\beta_1}+\hat{\beta}_{p+1}\hat{\delta}_{p+1,1}+\cdots+\hat{\beta_k}\hat{\delta}_{k,1} \end{align*}\]

\(\hat{\delta}_{j,1}(p+1\leq j\leq k)\)表示\(X_j\)\(X_1\)的回归系数.

3.2.2 偏回归系数的期望和方差

  现验证\(\hat{\beta_j}\)的无偏性.

\[\hat{\beta_j}=\frac{\sum_{i=1}^{n}\hat{r}_{ij}Y_i}{\sum_{i=1}^{n}\hat{r}_{ij}^2}=\beta_j+\frac{\sum_{i=1}^{n}\hat{r}_{ij}u_i}{\sum_{i=1}^{n}\hat{r}_{ij}^2}\]

\(\hat{\beta_j}-\beta_j\)取条件期望,得

\[\mathbb{E}[\hat{\beta_j}-\beta_j|X_{1j},X_{2j},\cdots,X_{nj}]=\frac{\sum_{i=1}^{n}\hat{r}_{ij}\mathbb{E}[u_i|X_{1j},X_{2j},\cdots,X_{nj}]}{\sum_{i=1}^{n}\hat{r}_{ij}^2}=0\]

\[\mathbb{E}[\hat{\beta_j}]=\beta_j\]

\(\hat{\beta_j}\)\(\beta_j\)的无偏估计量.

  在异方差条件下,

\[\mathrm{var}(\hat{\beta_j}|X_{1j},X_{2j},\cdots,X_{nj})=\frac{\sum_{i=1}^{n}\hat{r}_{ij}^2\mathbb{E}[u_i^2|X_{1j},X_{2j},\cdots,X_{nj}]}{\sum_{i=1}^{n}\hat{r}_{ij}^2}=\frac{\sum_{i=1}^n\hat{r}_{ij}^2\sigma_i^2}{(\sum_{i=1}^n\hat{r}_{ij}^2)^2}\]

  在同方差条件下,

\[\mathrm{var}(\hat{\beta_j})=\frac{\sigma^2}{\sum_{i=1}^n\hat{r}_{ij}^2}=\frac{\sigma^2}{TSS_j(1-R_j^2)}\]

其中,\(TSS_j=\sum_{i=1}^{n}(X_{ij}-\bar{X_j})^2\)\(X_j\)的总样本变异,\(R_j^2\)则是将\(X_j\)对所有其他自变量(包括截距项)进行回归得到的\(R^2\).

3.2.3 偏回归系数的大样本抽样分布

  易知

\[\sum_{i=1}^{n}\hat{r}_{ij}u_i\stackrel{d}\rightarrow N\left(0,n\mathrm{var}(\hat{r}_{ij}u_i)\right)\]

\(\frac{1}{n}\sum_{i=1}^n\hat{r}_{ij}^2\stackrel{p}\rightarrow a_j^2\)

\[\hat{\beta_j}\stackrel{d}\rightarrow N\left(\beta_j,\frac{\mathrm{var}(\hat{r}_{ij}u_i)}{n(a_j^2)^2}\right)\]

在同方差假设下,

\[\hat{\beta_j}\stackrel{d}\rightarrow N\left(\beta_j,\frac{\sigma^2}{na_j^2}\right)\]

3.3 拟合优度和误差方差的估计

3.3.1 拟合优度

  在简单线性回归中,我们在一元条件下证明了定理2.4.1,现在我们将该定理推广到多元情况.

\[\begin{align*} TSS&=Y^\prime Y\\ &=(\hat{Y}+\hat{u})^\prime(\hat{Y}+\hat{u})\\ &=\hat{Y}^\prime\hat{Y}+2\hat{Y}^\prime\hat{u}+\hat{u}^\prime\hat{u} \end{align*}\]

由于

\[\hat{Y}^\prime\hat{u}=Y^\prime PMY=0\]

则有

\[TSS=\hat{Y}^\prime\hat{Y}+\hat{u}^\prime\hat{u}=SSR+ESS.\]

3.3.2 误差方差的估计

  考察残差平方和

\[SSR=u^\prime Mu=\mathrm{tr}(u^\prime Mu)=\mathrm{tr}(Muu^\prime)\]

则有

\[\mathbb{E}[SSR|X]=\mathrm{tr}(\mathbb{E}[Muu^\prime|X])=\mathrm{tr}(M\mathbb{E}[uu^\prime|X])=\mathrm{tr}(MD)\]

在同方差假设下,\(D=I_n\sigma^2\),则

\[\mathbb{E}[SSR]=\sigma^2(n-k-1)\]

整理可得

\[\mathbb{E}\left[\frac{\hat{u}^\prime\hat{u}}{n-k-1}\right]=\sigma^2\]

\[\hat{\sigma}^2=\frac{1}{n-k-1}\sum_{i=1}^n\hat{u_i}^2\]
\[SER=\sqrt{\frac{1}{n-k-1}\sum_{i=1}^n\hat{u_i}^2}=\sqrt{\frac{SSR}{n-k-1}}\]

3.4 正态回归

定义3.4.1\(u_i|X_i\sim N(0,\sigma^2)\)时,定义

\[Y_i=X_i^\prime\beta+\mu_i\]

正态回归模型.

  该模型在零条件均值、独立同分布、有限峰度、无完全共线假设的基础上,加入了正态假设,即\(\mathbb{E}[u|X]=0\)的假设加强为\(u|X\sim N(0,\sigma^2I_n)\). 由此,我们可以得到\(\hat{\beta}\)\(\hat{u}\)的条件分布:

  因为\(\hat{\beta}=(X^\prime X)^{-1}X^\prime Y=\beta+(X^\prime X)^{-1}X^\prime u\),则有\(\mathrm{var}(\hat{\beta}|X)=\sigma^2(X^\prime X)^{-1}X^\prime((X^\prime X)^{-1}X^\prime)^\prime=\sigma^2(X^\prime X)^{-1}\),那么

\[\hat{\beta}|X\sim N(\beta,\sigma^2(X^\prime X)^{-1})\]

  类似地,由于\(\hat{u}=Mu\),则

\[\hat{u}|X\sim N(0,\sigma^2M)\]

定理3.4.1 在正态回归模型中,\(\hat{\beta}\)\(\hat{u}\)相互独立.

证明:考察\(\hat{\beta}\)\(\hat{u}\)的联合正态分布.

\[\left(\begin{array}{c}\hat{\beta}-\beta\\\hat{u}\end{array}\right)=\left(\begin{array}{c}(X^\prime X)^{-1}X^\prime\\M\end{array}\right)u\]
\[\mathrm{var}\left(\left.\left(\begin{array}{c} \hat{\beta}-\beta\\ \hat{u} \end{array}\right)\right|X\right)=\sigma^2\left(\begin{array}{c} (X^\prime X)^{-1}X^\prime\\ M \end{array}\right)(X(X^\prime X)^{-1}\quad M)=\left(\begin{array}{cc} \sigma^2(X^\prime X)^{-1}&0\\ 0&\sigma^2M \end{array}\right) \]

由于\(\mathrm{cov}(\hat{\beta}-\beta,\hat{u})=0\),则\(\mathrm{cov}(\hat{\beta},\hat{u})=0\),在正态条件下\(\hat{\beta}\)\(\hat{u}\)相互独立.

定理3.4.2

\[\frac{(n-k-1)\hat{\sigma}^2}{\sigma^2}\sim\chi^2(n-k-1)\]

证明:

\[\begin{align*} \frac{(n-k-1)\hat{\sigma}^2}{\sigma^2}&=\frac{\hat{u}^\prime\hat{u}}{\sigma^2}\\ &=\left(\frac{u}{\sigma}\right)^\prime M\left(\frac{u}{\sigma}\right) \end{align*}\]

由定理3.2.2(5)可知\(\mathrm{r}(M)=n-k-1\),则

\[\frac{(n-k-1)\hat{\sigma}^2}{\sigma^2}\sim\chi^2(n-k-1)\]

3.4.1 置信区间

\(\beta_j\)的置信区间

构造\(t\)统计量

\[t=\frac{\hat{\beta}_j-\beta_j}{\hat{\sigma}\sqrt{[(X^\prime X)^{-1}]_{jj}}}\sim t(n-k-1)\]

\(\beta_j\)的置信水平为\(\alpha\)的置信区间为

\[\left(\hat{\beta}_j\pm t_{\frac{\alpha}{2}}(n-k-1)\hat{\sigma}\sqrt{[(X^\prime X)^{-1}]_{jj}}\right)\]

\(\sigma^2\)的置信区间

构造\(\chi^2\)统计量

\[\chi^2=\frac{\hat{u}^\prime\hat{u}}{\sigma^2}\sim\chi^2(n-k-1)\]

\(\sigma^2\)的置信水平为\(\alpha\)的置信区间为

\[\left(\frac{\hat{u}^\prime\hat{u}}{\chi_{\frac{\alpha}{2}}^2(n-k-1)},\frac{\hat{u}^\prime\hat{u}}{\chi_{1-\frac{\alpha}{2}}^2(n-k-1)}\right) \]

3.4.2 假设检验

偏回归系数\(\beta_j\)\(t\)检验

\[H_0: \beta_j=b,\,H_1: \beta_j\neq b\]

构造\(t\)统计量

\[t=\frac{\hat{\beta_j}-b}{SE(\hat{\beta_j})}\]

拒绝域为

\[|t|\geq t_{\frac{\alpha}{2}}(n-k-1)\]

对参数线性组合的\(t\)检验

\[H_0: \beta_1=\beta_2,\,H_1: \beta_1\neq\beta_2\]

构造\(t\)统计量

\[t=\frac{\hat{\beta_1}-\hat{\beta_2}}{SE(\hat{\beta_1}-\hat{\beta_2})}\]

其中

\[\hat{\beta_1}-\hat{\beta_2}=a^\prime\hat{\beta},\,a=\left( \begin{array}{c} 0\\ 1\\ -1\\ O \end{array}\right),\,SE(\hat{\beta_1}-\hat{\beta_2})=a^\prime\hat{\sigma}^2(X^\prime X)^{-1}a\]

拒绝域为

\[|t|\geq t_{\frac{\alpha}{2}}(n-k-1)\]

回归系数\(\beta\)\(F\)检验

\[H_0: A\beta=b,\,H_1: A\beta\neq b\]

\(\mathrm{r}(A)=m\),构造\(F\)统计量

\[F=\frac{(A\hat{\beta}-b)^\prime(\hat{\sigma}^2A(X^\prime X)^{-1}A^\prime)^{-1}(A\hat{\beta}-b)}{m}\sim F(m,n-k-1)\]

拒绝域为

\[F\geq F_{\alpha}(m,n-k-1)\]

对冗余变量的\(F\)检验

\[H_0: \beta_{k-q+1}=\cdots=\beta_k=0,\,H_1: H_0\,\mbox{为假}\]

无约束回归\(u\)\(Y=\beta_0+\beta_1X_1+\cdots+\beta_{k-q}X_{k-q}+\cdots+\beta_kX_k+u\)

有约束回归\(r\)\(Y=\beta_0+\beta_1X_1+\cdots+\beta_{k-q}X_{k-q}+u\)

基于残差平方和\(SSR\)\(F\)检验:

\[F=\frac{(SSR_r-SSR_u)/q}{SSR_u/(n-k-1)}\sim F(q,n-k-1)\]

基于\(R^2\)\(F\)检验:

\[F=\frac{(R_u^2-R_r^2)/q}{(1-R_u^2)/(n-k-1)}\sim F(q,n-k-1)\]

拒绝域为

\[F\geq F_{\alpha}(q,n-k-1)\]