多元回归系数的标准误差？

我意识到这是一个非常基本的问题，但是我在任何地方都找不到答案。

我正在使用正态方程或QR分解计算回归系数。如何计算每个系数的标准误差？我通常认为标准错误的计算方式如下：

$SE_\bar{x}\ = \frac{\sigma_{\bar x}}{\sqrt{n}}$

什么是的每个系数？在OLS上下文中最有效的计算方法是什么？$\sigma_{\bar x}$

When doing least squares estimation (assuming a normal random component) the regression parameter estimates are normally distributed with mean equal to the true regression parameter and covariance matrix $\Sigma = s^2\cdot(X^TX)^{-1}$ where $s^2$ is the residual variance and $X^TX$ is the design matrix. $X^T$ is the transpose of $X$ and $X$ is defined by the model equation $Y=X\beta+\epsilon$ with $\beta$回归参数，是误差项。通过将（X T X ）− 1中的相应项乘以残余方差的样本估计值，然后取平方根，可以得出β参数的估计标准差。这不是一个非常简单的计算，但是任何软件包都会为您计算并在输出中提供。$\epsilon$$(X^TX)^{-1}$

例

上德雷珀和史密斯（在我的注释的参考）的第134页，它们提供以下数据通过最小二乘模型拟合，其中ε 〜Ñ （0 ，我σ 2）。$Y = \beta_0 + \beta_1 X + \varepsilon$$\varepsilon \sim N(0, \mathbb{I}\sigma^2)$

                      X                      Y                    XY
                      0                     -2                     0
                      2                      0                     0
                      2                      2                     4
                      5                      1                     5
                      5                      3                    15
                      9                      1                     9
                      9                      0                     0
                      9                      0                     0
                      9                      1                     9
                     10                     -1                   -10
                    ---                     --                   ---
Sum                  60                      5                    32
Sum of  Squares     482                     21                   528

看起来像一个斜率应接近0的示例。

$$X^t = \pmatrix{ 1 &1 &1 &1 &1 &1 &1 &1 &1 &1 \\ 0 &2 &2 &5 &5 &9 &9 &9 &9 &10 }.$$

所以

$$X^t X = \pmatrix{n &\sum X_i \\ \sum X_i &\sum X_i^2} = \pmatrix{10 &60 \\60 &482}$$

和

$$\eqalign{ (X^t X)^{-1} &= \pmatrix{ \frac{\sum X_i^2}{n \sum (X_i - \bar{X})^2} &\frac{-\bar{X}}{\sum (X_i-\bar{X})^2} \\ \frac{-\bar{X}}{\sum (X_i-\bar{X})^2} &\frac{1}{\sum (X_i-\bar{X})^2} } \\ &= \pmatrix{\frac{482}{10(122)} &-\frac{6}{122} \\ -\frac{6}{122} &\frac{1}{122}} \\ &= \pmatrix{0.395 &-0.049 \\ -0.049 &0.008} }$$

$\bar{X} = \sum X_i/n = 60/10 = 6$

Estimate for $β = (X^TX)^{-1} X^TY$ = ( b0 ) =(Yb-b1 Xb) b1 Sxy/Sxx

b1 = 1/61 = 0.0163 and b0 = 0.5- 0.0163(6) = 0.402

From $(X^TX)^{-1}$ above Sb1 =Se (0.008) and Sb0=Se(0.395) where Se is the estimated standard deviation for the error term. Se =√2.3085.

Sorry that the equations didn't carry subscripting and superscripting when I cut and pasted them. The table didn't reproduce well either because the spaces got ignored. The first string of 3 numbers correspond to the first values of X Y and XY and the same for the followinf strings of three. After Sum comes the sums for X Y and XY respectively and then the sum of squares for X Y and XY respectively. The 2x2 matrices got messed up too. The values after the brackets should be in brackets underneath the numbers to the left.