线性回归误差项方差的估计

2023-11-19

线性回归误差项方差的估计

摘要

之前在文章线性回归系数的几个性质中，我们证明了线性回归系数项的几个性质。在这篇短文中，我们介绍线性回归模型中误差项方差的估计。

线性回归误差项概念的回顾

我们先来回忆一下什么是线性回归的误差项。在文章线性回归系数的几个性质中，我们指出，对于单变量的线性回归模型， y i = β 1 x i + β 0 + ϵ i , i = 1 , 2 , ⋯ , n y_i = \beta_1 x_i + \beta_0 + \epsilon_i, \, i = 1, \, 2, \, \cdots, n yi=β1xi+β0+ϵi,i=1,2,⋯,n，
其中 ϵ i \epsilon_i ϵi 为误差项 (error)，每个 ϵ i \epsilon_i ϵi 均是一个随机变量，独立且都服从一个均值为0, 方差为 σ 2 \sigma^2 σ2 的概率分布。这篇短文介绍的就是对这个 σ 2 \sigma^2 σ2 的估计。

残差平方和 (residual sum of squares)

我们定义 S S R e s = ∑ i = 1 n e i 2 = ∑ i = 1 n ( y i − y i ^ ) 2 \displaystyle SS_{\mathrm{Res}} = \sum_{i = 1}^n e_i^2 = \sum_{i = 1}^n (y_i - \hat{y_i})^2 SSRes=i=1∑nei2=i=1∑n(yi−yi^)2。

首先，我们证明，
S S R e s = ∑ i = 1 n y i 2 − n y ˉ 2 − β 1 ^ S x y SS_{\mathrm{Res}} = \sum_{i = 1}^n y_i^2 -n \bar{y}^2 - \hat{\beta_1} S_{xy} SSRes=i=1∑nyi2−nyˉ2−β1^Sxy。

我们把 y i ^ = β 0 ^ + β 1 ^ ⋅ x i \displaystyle \hat{y_i} = \hat{\beta_0} + \hat{\beta_1} \cdot x_i yi^=β0^+β1^⋅xi 代入 S S R e s \displaystyle SS_{\mathrm{Res}} SSRes 的表达式。

我们有，

S S R e s = ∑ i = 1 n ( y i − y i ^ ) 2 = ∑ i = 1 n ( y i − ( β 0 ^ + β 1 ^ ⋅ x i ) ) 2 = ∑ i = 1 n y i 2 − 2 ∑ i = 1 n y i ( β 0 ^ + β 1 ^ ⋅ x i ) + ∑ i = 1 n ( β 0 ^ + β 1 ^ ⋅ x i ) 2 = ∑ i = 1 n y i 2 − 2 β 0 ^ ⋅ n y ˉ − 2 β 1 ^ ∑ i = 1 n x i y i + n β 0 ^ 2 + 2 β 0 ^ β 1 ^ ∑ i = 1 n x i + β 1 ^ 2 ∑ i = 1 n x i 2 = ∑ i = 1 n y i 2 − 2 ( y ˉ − β 1 ^ x ˉ ) n y ˉ − 2 β 1 ^ ∑ i = 1 n x i y i + n ( y ˉ − β 1 ^ x ˉ ) 2 + 2 ( y ˉ − β 1 ^ x ˉ ) β 1 ^ ∑ i = 1 n x i + β 1 ^ 2 ∑ i = 1 n x i 2 = ∑ i = 1 n y i 2 − 2 n y ˉ 2 + 2 n β 1 ^ x ˉ y ˉ − 2 β 1 ^ ∑ i = 1 n x i y i + n y ˉ 2 − 2 n β 1 ^ x ˉ y ˉ + n β 1 ^ 2 x ˉ 2 + 2 y ˉ β 1 ^ ∑ i = 1 n x i − 2 β 1 ^ 2 x ˉ ∑ i = 1 n x i + β 1 ^ 2 ∑ i = 1 n x i 2 = ∑ i = 1 n y i 2 − n y ˉ 2 − n β 1 ^ 2 x ˉ 2 + β 1 ^ 2 ∑ i = 1 n x i 2 − 2 β 1 ^ ∑ i = 1 n x i y i + 2 n β 1 ^ x ˉ y ˉ = ∑ i = 1 n y i 2 − n y ˉ 2 + β 1 ^ 2 ( ∑ i = 1 n x i 2 − n x ˉ 2 ) − 2 β 1 ^ ( ∑ i = 1 n x i y i − n x ˉ y ˉ ) = ∑ i = 1 n y i 2 − n y ˉ 2 + β 1 ^ 2 S x x − 2 β 1 ^ S x y = ∑ i = 1 n y i 2 − n y ˉ 2 + β 1 ^ ( S x y S x x ) S x x − 2 β 1 ^ S x y = ∑ i = 1 n y i 2 − n y ˉ 2 − β 1 ^ S x y \begin{aligned} \displaystyle SS_{\mathrm{Res}} &= \sum_{i = 1}^n (y_i - \hat{y_i})^2 = \sum_{i = 1}^n \big(y_i - (\hat{\beta_0} + \hat{\beta_1} \cdot x_i) \big)^2 \\ &= \sum_{i = 1}^n y_i^2 - 2 \sum_{i = 1}^n y_i (\hat{\beta_0} + \hat{\beta_1} \cdot x_i) + \sum_{i = 1}^n (\hat{\beta_0} + \hat{\beta_1} \cdot x_i)^2 \\ &= \sum_{i = 1}^n y_i^2 - 2 \hat{\beta_0} \cdot n \bar{y} - 2 \hat{\beta_1} \sum_{i = 1}^n x_i y_i + n \hat{\beta_0}^2 + 2 \hat{\beta_0} \hat{\beta_1} \sum_{i = 1}^n x_i + \hat{\beta_1}^2 \sum_{i = 1}^n x_i^2 \\ &= \sum_{i = 1}^n y_i^2 - 2 (\bar{y} - \hat{\beta_1} \bar{x}) n \bar{y} - 2 \hat{\beta_1} \sum_{i = 1}^n x_i y_i + n (\bar{y} - \hat{\beta_1} \bar{x})^2 + 2 (\bar{y} - \hat{\beta_1} \bar{x}) \hat{\beta_1} \sum_{i = 1}^n x_i + \hat{\beta_1}^2 \sum_{i = 1}^n x_i^2 \\ &= \sum_{i = 1}^n y_i^2 - 2 n \bar{y}^2 + 2n \hat{\beta_1} \bar{x} \bar{y} - 2 \hat{\beta_1} \sum_{i = 1}^n x_i y_i + n \bar{y}^2 - 2n \hat{\beta_1} \bar{x} \bar{y} + \\ & \hspace{5mm} n \hat{\beta_1}^2 \bar{x}^2 + 2 \bar{y} \hat{\beta_1} \sum_{i = 1}^n x_i - 2 \hat{\beta_1}^2 \bar{x} \sum_{i = 1}^n x_i + \hat{\beta_1}^2 \sum_{i = 1}^n x_i^2 \\ &= \sum_{i = 1}^n y_i^2 - n \bar{y}^2 - n \hat{\beta_1}^2 \bar{x}^2 + \hat{\beta_1}^2 \sum_{i = 1}^n x_i^2 - 2 \hat{\beta_1} \sum_{i = 1}^n x_i y_i + 2n \hat{\beta_1} \bar{x} \bar{y} \\ &= \sum_{i = 1}^n y_i^2 - n \bar{y}^2 + \hat{\beta_1}^2 \big( \sum_{i = 1}^n x_i^2 - n \bar{x}^2 \big) - 2 \hat{\beta_1} \big( \sum_{i = 1}^n x_i y_i - n \bar{x} \bar{y} \big) \\ &= \sum_{i = 1}^n y_i^2 - n \bar{y}^2 + \hat{\beta_1}^2 S_{xx} - 2 \hat{\beta_1} S_{xy} \\ &= \sum_{i = 1}^n y_i^2 - n \bar{y}^2 + \hat{\beta_1} \left( \frac{S_{xy} }{ S_{xx} } \right) S_{xx} - 2 \hat{\beta_1} S_{xy} \\ &= \sum_{i = 1}^n y_i^2 -n \bar{y}^2 - \hat{\beta_1} S_{xy} \end{aligned} SSRes=i=1∑n(yi−yi^)2=i=1∑n(yi−(β0^+β1^⋅xi))2=i=1∑nyi2−2i=1∑nyi(β0^+β1^⋅xi)+i=1∑n(β0^+β1^⋅xi)2=i=1∑nyi2−2β0^⋅nyˉ−2β1^i=1∑nxiyi+nβ0^2+2β0^β1^i=1∑nxi+β1^2i=1∑nxi2=i=1∑nyi2−2(yˉ−β1^xˉ)nyˉ−2β1^i=1∑nxiyi+n(yˉ−β1^xˉ)2+2(yˉ−β1^xˉ)β1^i=1∑nxi+β1^2i=1∑nxi2=i=1∑nyi2−2nyˉ2+2nβ1^xˉyˉ−2β1^i=1∑nxiyi+nyˉ2−2nβ1^xˉyˉ+nβ1^2xˉ2+2yˉβ1^i=1∑nxi−2β1^2xˉi=1∑nxi+β1^2i=1∑nxi2=i=1∑nyi2−nyˉ2−nβ1^2xˉ2+β1^2i=1∑nxi2−2β1^i=1∑nxiyi+2nβ1^xˉyˉ=i=1∑nyi2−nyˉ2+β1^2(i=1∑nxi2−nxˉ2)−2β1^(i=1∑nxiyi−nxˉyˉ)=i=1∑nyi2−nyˉ2+β1^2Sxx−2β1^Sxy=i=1∑nyi2−nyˉ2+β1^(SxxSxy)Sxx−2β1^Sxy=i=1∑nyi2−nyˉ2−β1^Sxy

于是，我们就证明了 S S R e s = ∑ i = 1 n y i 2 − n y ˉ 2 − β 1 ^ S x y SS_{\mathrm{Res}} = \sum_{i = 1}^n y_i^2 -n \bar{y}^2 - \hat{\beta_1} S_{xy} SSRes=i=1∑nyi2−nyˉ2−β1^Sxy。

残差平方和的期望

下面我们来看残差平方和的期望，即 E [ S S R e s ] \displaystyle \mathbb{E} [ SS_{\mathrm{Res}} ] E[SSRes] 。

E [ S S R e s ] = E [ ∑ i = 1 n y i 2 ] − n E [ y ˉ 2 ] − E [ β 1 ^ S x y ] \displaystyle \mathbb{E} [ SS_{\mathrm{Res}} ] = \mathbb{E} \big[ \sum_{i = 1}^n y_i^2 \big] - n \mathbb{E} \big[ \bar{y}^2 \big] - \mathbb{E} \big[ \hat{\beta_1} S_{xy} \big] E[SSRes]=E[i=1∑nyi2]−nE[yˉ2]−E[β1^Sxy]。

我们分项来求。

我们先来求 E [ ∑ i = 1 n y i 2 ] \displaystyle \mathbb{E} \big[ \sum_{i = 1}^n y_i^2 \big] E[i=1∑nyi2]。我们知道，
E [ y i ] = E [ β 1 x i + β 0 + ϵ i ] = β 1 x i + β 0 \displaystyle \mathbb{E} \big[ y_i \big] = \mathbb{E} \big[ \beta_1 x_i + \beta_0 + \epsilon_i \big] = \beta_1 x_i + \beta_0 E[yi]=E[β1xi+β0+ϵi]=β1xi+β0。注意这里我们用到了 E [ ϵ i ] = 0 \displaystyle \mathbb{E} \big[ \epsilon_i \big] = 0 E[ϵi]=0。
而 V a r [ ϵ i ] = σ 2 \displaystyle \mathrm{Var} \big[ \epsilon_i \big] = \sigma^2 Var[ϵi]=σ2。于是，我们知道 V a r [ y i ] = σ 2 \displaystyle \mathrm{Var} \big[ y_i \big] = \sigma^2 Var[yi]=σ2。从而， E [ y i 2 ] = V a r [ y i ] + ( E [ y i ] ) 2 = ( β 1 x i + β 0 ) 2 + σ 2 \displaystyle \mathbb{E} \big[ y_i^2 \big] = \mathrm{Var} \big[ y_i \big] + (\mathbb{E} \big[ y_i \big])^2 = (\beta_1 x_i + \beta_0)^2 + \sigma^2 E[yi2]=Var[yi]+(E[yi])2=(β1xi+β0)2+σ2。 E [ ∑ i = 1 n y i 2 ] = ∑ i = 1 n ( ( β 1 x i + β 0 ) 2 + σ 2 ) = ∑ i = 1 n ( β 1 x i + β 0 ) 2 + n σ 2 \displaystyle \mathbb{E} \big[ \sum_{i = 1}^n y_i^2 \big] = \sum_{i = 1}^n \big( (\beta_1 x_i + \beta_0)^2 + \sigma^2 \big) = \sum_{i = 1}^n (\beta_1 x_i + \beta_0)^2 + n \sigma^2 E[i=1∑nyi2]=i=1∑n((β1xi+β0)2+σ2)=i=1∑n(β1xi+β0)2+nσ2。

对于 E [ y ˉ 2 ] \displaystyle \mathbb{E} \big[ \bar{y}^2 \big] E[yˉ2]，我们采用一样的方法。因为 E [ y ˉ ] = 1 n ∑ i = 1 n ( β 1 x i + β 0 ) = β 0 + β 1 x ˉ \displaystyle \mathbb{E} \big[ \bar{y} \big] = \frac{1}{n} \sum_{i = 1}^n (\beta_1 x_i + \beta_0) = \beta_0 + \beta_1 \bar{x} E[yˉ]=n1i=1∑n(β1xi+β0)=β0+β1xˉ。 V a r [ y ˉ ] = 1 n 2 ∑ i = 1 n V a r [ y i ] = 1 n 2 ⋅ n σ 2 = σ 2 n \displaystyle \mathrm{Var} \big[ \bar{y} \big] = \frac{1}{n^2} \sum_{i = 1}^n \mathrm{Var} \big[ y_i \big] = \frac{1}{n^2} \cdot n \sigma^2 = \frac{\sigma^2}{n} Var[yˉ]=n21i=1∑nVar[yi]=n21⋅nσ2=nσ2。
于是， E [ y ˉ 2 ] = V a r [ y ˉ ] + ( E [ y ˉ ] ) 2 = σ 2 n + ( β 0 + β 1 x ˉ ) 2 \displaystyle \mathbb{E} \big[ \bar{y}^2 \big] = \mathrm{Var} \big[ \bar{y} \big] + \big( \mathbb{E} \big[ \bar{y} \big] \big)^2 =\frac{\sigma^2}{n} + ( \beta_0 + \beta_1 \bar{x} )^2 E[yˉ2]=Var[yˉ]+(E[yˉ])2=nσ2+(β0+β1xˉ)2。

对于最后一项， E [ β 1 ^ S x y ] \mathbb{E} \big[ \hat{\beta_1} S_{xy} \big] E[β1^Sxy]，因为 β 1 ^ = S x y S x x \displaystyle \hat{\beta_1} = \frac{S_{xy} }{ S_{xx} } β1^=SxxSxy，而 S x x S_{xx} Sxx 是常数，所以我们须要求 E [ S x y 2 ] \displaystyle \mathbb{E} \big[ S_{xy}^2 \big] E[Sxy2]。
注意到， S x y = ∑ i = 1 n ( x i − x ˉ ) ( y i − y ˉ ) = ∑ i = 1 n ( x i − x ˉ ) y i \displaystyle S_{xy} = \sum_{i = 1}^n (x_i - \bar{x}) (y_i - \bar{y}) =\sum_{i = 1}^n (x_i - \bar{x}) y_i Sxy=i=1∑n(xi−xˉ)(yi−yˉ)=i=1∑n(xi−xˉ)yi。

我们有 E [ ∑ i = 1 n ( x i − x ˉ ) y i ] = ∑ i = 1 n ( x i − x ˉ ) ⋅ E [ y i ] = ∑ i = 1 n ( x i − x ˉ ) ( β 0 + β 1 x i ) \displaystyle \mathbb{E} \big[ \sum_{i = 1}^n (x_i - \bar{x}) y_i \big] = \sum_{i = 1}^n (x_i - \bar{x}) \cdot \mathbb{E} \big[ y_i \big] = \sum_{i = 1}^n (x_i - \bar{x}) (\beta_0 + \beta_1 x_i) E[i=1∑n(xi−xˉ)yi]=i=1∑n(xi−xˉ)⋅E[yi]=i=1∑n(xi−xˉ)(β0+β1xi)。
另外， V a r [ ∑ i = 1 n ( x i − x ˉ ) y i ] = ∑ i = 1 n ( x i − x ˉ ) 2 ⋅ V a r [ y i ] = ∑ i = 1 n ( x i − x ˉ ) 2 σ 2 \displaystyle \mathrm{Var} \big[ \sum_{i = 1}^n (x_i - \bar{x}) y_i \big] = \sum_{i = 1}^n (x_i - \bar{x})^2 \cdot \mathrm{Var} \big[ y_i \big] = \sum_{i = 1}^n (x_i - \bar{x})^2 \sigma^2 Var[i=1∑n(xi−xˉ)yi]=i=1∑n(xi−xˉ)2⋅Var[yi]=i=1∑n(xi−xˉ)2σ2。这里我们用到了 y i , i = 1 , 2 , ⋯ n y_i, \, i = 1, \, 2, \, \cdots n yi,i=1,2,⋯n 是非相关的 (uncorrelated)。
从而，
E [ S x y 2 ] = E [ ( ∑ i = 1 n ( x i − x ˉ ) y i ) 2 ] = V a r [ ∑ i = 1 n ( x i − x ˉ ) y i ] + ( E [ ∑ i = 1 n ( x i − x ˉ ) y i ] ) 2 = ∑ i = 1 n ( x i − x ˉ ) 2 σ 2 + ( ∑ i = 1 n ( x i − x ˉ ) ( β 0 + β 1 x i ) ) 2 = S x x σ 2 + ( ∑ i = 1 n β 1 x i ( x i − x ˉ ) ) 2 = S x x σ 2 + β 1 2 S x x 2 \begin{aligned} \displaystyle \mathbb{E} \big[ S_{xy}^2 \big] &= \mathbb{E} \big[ \big( \sum_{i = 1}^n (x_i - \bar{x}) y_i \big)^2 \big] \\ &= \mathrm{Var} \big[ \sum_{i = 1}^n (x_i - \bar{x}) y_i \big] + \left( \mathbb{E} \big[ \sum_{i = 1}^n (x_i - \bar{x}) y_i \big] \right)^2 \\ &= \sum_{i = 1}^n (x_i - \bar{x})^2 \sigma^2 + \left( \sum_{i = 1}^n (x_i - \bar{x}) (\beta_0 + \beta_1 x_i) \right)^2 \\ &= S_{xx} \sigma^2 + \left( \sum_{i = 1}^n \beta_1 x_i (x_i - \bar{x}) \right)^2 \\ &= S_{xx} \sigma^2 + \beta_1^2 S_{xx}^2 \end{aligned} E[Sxy2]=E[(i=1∑n(xi−xˉ)yi)2]=Var[i=1∑n(xi−xˉ)yi]+(E[i=1∑n(xi−xˉ)yi])2=i=1∑n(xi−xˉ)2σ2+(i=1∑n(xi−xˉ)(β0+β1xi))2=Sxxσ2+(i=1∑nβ1xi(xi−xˉ))2=Sxxσ2+β12Sxx2

从而， E [ β 1 ^ S x y ] = S x x σ 2 + β 1 2 S x x 2 S x x = σ 2 + β 1 2 S x x \displaystyle \mathbb{E} \big[ \hat{\beta_1} S_{xy} \big] = \frac{ S_{xx} \sigma^2 + \beta_1^2 S_{xx}^2 }{ S_{xx}} = \sigma^2 + \beta_1^2 S_{xx} E[β1^Sxy]=SxxSxxσ2+β12Sxx2=σ2+β12Sxx。

我们把， E [ ∑ i = 1 n y i 2 ] , E [ y ˉ 2 ] , [ S x y 2 ] \displaystyle \mathbb{E} \big[ \sum_{i = 1}^n y_i^2 \big], \, \mathbb{E} \big[ \bar{y}^2 \big], \, \big[ S_{xy}^2 \big] E[i=1∑nyi2],E[yˉ2],[Sxy2] 这三项代入 E [ S S R e s ] = E [ ∑ i = 1 n y i 2 ] − n E [ y ˉ 2 ] − E [ β 1 ^ S x y ] \displaystyle \mathbb{E} [ SS_{\mathrm{Res}} ] = \mathbb{E} \big[ \sum_{i = 1}^n y_i^2 \big] - n \mathbb{E} \big[ \bar{y}^2 \big] - \mathbb{E} \big[ \hat{\beta_1} S_{xy} \big] E[SSRes]=E[i=1∑nyi2]−nE[yˉ2]−E[β1^Sxy]。
我们有
E [ S S R e s ] = E [ ∑ i = 1 n y i 2 ] − n E [ y ˉ 2 ] − E [ β 1 ^ S x y ] = ∑ i = 1 n ( β 1 x i + β 0 ) 2 + n σ 2 − n ( σ 2 n + ( β 0 + β 1 x ˉ ) 2 ) − ( σ 2 + β 1 2 S x x ) = ( n − 2 ) σ 2 \begin{aligned} \displaystyle \mathbb{E} [ SS_{\mathrm{Res}} ] &= \mathbb{E} \big[ \sum_{i = 1}^n y_i^2 \big] - n \mathbb{E} \big[ \bar{y}^2 \big] - \mathbb{E} \big[ \hat{\beta_1} S_{xy} \big] \\ &= \sum_{i = 1}^n (\beta_1 x_i + \beta_0)^2 + n \sigma^2 - n \left( \frac{\sigma^2}{n} + ( \beta_0 + \beta_1 \bar{x} )^2 \right) - ( \sigma^2 + \beta_1^2 S_{xx} ) \\ &= (n - 2) \sigma^2 \end{aligned} E[SSRes]=E[i=1∑nyi2]−nE[yˉ2]−E[β1^Sxy]=i=1∑n(β1xi+β0)2+nσ2−n(nσ2+(β0+β1xˉ)2)−(σ2+β12Sxx)=(n−2)σ2

也就是说， S S R e s n − 2 \displaystyle \frac{ SS_{\mathrm{Res}} }{n - 2} n−2SSRes 是 σ 2 \sigma^2 σ2 的一个无偏估计。

实验验证

class sigmasqu_estimation:
    
    def __init__(self, arr_x: np.array, beta1: float, beta0: float, epsilon: float):
        #self.N = N
        self.X = arr_x
        self.beta1 = beta1
        self.beta0 = beta0
        self.epsilon = epsilon
        
        self.Sxx = ((self.X - self.X.mean()) ** 2).sum()
        self.X_bar = self.X.mean()
        
    def estimate_sigmasqu(self, N: int) -> tuple:
        res_sigmasqu_esti = []
        for i in range(N):
            #print(i)
            cur_error = np.random.normal(0, self.epsilon, arr_x.shape)
            cur_y = self.beta0 + self.beta1 * self.X + cur_error
            cur_y_bar = cur_y.mean()
            Sxy = ((self.X - self.X.mean()) * (cur_y - cur_y_bar)).sum()
            cur_beta1 = Sxy / self.Sxx
            cur_beta0 = cur_y_bar - cur_beta1 * self.X_bar
            cur_y_hat = cur_beta1 * self.X + cur_beta0 
            SS_res = ((cur_y - cur_y_hat) ** 2).sum()
            res_sigmasqu_esti.append(SS_res / (self.X.shape[0] - 2))
        return np.array(res_sigmasqu_esti)

arr_x = np.array(range(1, 11))
a = sigmasqu_estimation(arr_x, 2, 3, 1)
res = a.estimate_sigmasqu(10 ** 5)

res
np.mean(res)

1.0004110047596488

我们发现， S S R e s n − 2 \displaystyle \frac{ SS_{\mathrm{Res}} }{n - 2} n−2SSRes 的均值非常接近 σ 2 \sigma^2 σ2

事实上， ( n − 2 ) S S R e s / σ 2 \displaystyle (n - 2) SS_{\mathrm{Res}} / \sigma^2 (n−2)SSRes/σ2 服从的是 χ n − 2 2 \chi^2_{n - 2} χn−22 的分布 [1]。

plt.figure(figsize=(8, 6), dpi=100)
plt.hist(res, bins=50, density=True);
line_vert = [[1, c] for c in np.linspace(0, 1, 100)]
plt.plot([c[0] for c in line_vert], [c[1] for c in line_vert], '-', linewidth=4)
plt.xlabel("estimated $\sigma^2$ value", fontsize=20)
plt.ylabel("count", fontsize=20)
plt.xticks(fontsize=20)
plt.yticks(fontsize=20);

$对 \sigma^2 的估计值的统计$

参考文献

[1] Introduction to linear regression analysis, Douglas C. Montgomery, Elizabeth A. Peck, G. Geoffrey Vining, John Wiley & Sons (2021)

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