例题:异方差导致问题及其修正

对一元回归模型
Y i = β 0 + β 1 X i + μ i Y_i=\beta_0+\beta_1 X_i+\mu_i
若异方差情况为Var( μ i \mu_i )= σ i 2 σ 2 \sigma_i^2\not = \sigma^2

  • 估计量有偏吗?
    我们知道
    β ~ 1 = x i y i x i 2 = x i ( β 1 X i + μ i ) x i 2 = β 1 + x i μ i x i 2 \tilde \beta_1=\frac{\sum x_iy_i}{\sum x_i^2}=\frac{\sum x_i(\beta_1 X_i+\mu_i)}{\sum x_i^2}\\=\beta_1+\frac{\sum x_i\mu_i}{\sum x_i^2}
    那么
    E ( β ~ 1 ) = β 1 + x i x i 2 E ( μ i ) = β 1 E(\tilde \beta_1)=\beta_1+\sum\frac{x_i}{\sum x_i^2}E(\mu_i)=\beta_1

  • 方差怎样变化呢?
    V a r ( β ~ 1 ) = V a r ( x i μ i x i 2 ) = ( x i x i 2 ) 2 V a r ( μ i ) + i j x i x i 2 x j x j 2 C o v ( μ i , μ j ) = x i 2 σ 2 ( x i 2 ) 2 {\rm Var}(\tilde \beta_1)={\rm Var}(\frac{\sum x_i\mu_i}{\sum x_i^2})\\\quad\\=\sum (\frac{x_i}{\sum x_i^2})^2 {\rm Var}(\mu_i)+\sum _{i\not =j}\frac{x_i}{\sum x_i^2}\frac{x_j}{\sum x_j^2}{\rm Cov}(\mu_i,\mu_j)\\\quad\\=\frac{\sum x_i^2\sigma^2}{(\sum x_i^2)^2}

  • 加权最小二乘法的估计量
    对原模型除以权重 σ i \sigma_i :
    Y i σ i = β 0 1 σ i + β 1 X i σ i + μ i σ i \frac{Y_i}{\sigma_i}=\beta_0\frac{1}{\sigma_i}+\beta_1\frac{X_i}{\sigma_i}+\frac{\mu_i}{\sigma_i}
    对以上模型用OLS估计就是加权最小二乘。
    容易看出变换后的模型是同方差的。
    记变换后的样本函数为
    Y i σ i = β 0 1 σ i + β 1 X i σ i + e i \frac{Y_i}{\sigma_i}=\beta_0^*\frac{1}{\sigma_i}+\beta_1^*\frac{X_i}{\sigma_i}+e_i^*
    最小化: ( e i ) 2 = 1 σ 2 ( Y i β 0 β i X i ) 2 \sum (e_i^*)^2=\sum\frac{1}{\sigma^2}(Y_i-\beta_0^*-\beta_i^*X_i)^2
    F O C \rm FOC
    β 0 w i + β 1 w i X i = w i Y i β 0 w i X i + β 1 w i X i 2 = w i X i Y i \beta_0^*\sum w_i+\beta_1^*\sum w_i X_i=\sum w_iY_i\\ \beta_0^*\sum w_i X_i+\beta_1^*\sum w_iX_i^2=\sum w_iX_iY_i
    解得
    β 1 = w i w i X i Y i w i X i w i Y i w i w i X i 2 ( w i X i ) 2 \beta_1^*=\frac{\sum w_i\sum w_iX_iY_i-\sum w_iX_i \sum w_i Y_i}{\sum w_i\sum w_iX_i^2-(\sum w_iX_i)^2}

其中 w i = 1 / σ i 2 w_i=1 /\sigma_i^2
进一步,令
X = w i X i w i , Y = w i Y i w i y i = Y i Y , x i = X i X \overline X^*=\frac{\sum w_i X_i}{\sum w_i},\overline Y^*=\frac{\sum w_i Y_i}{\sum w_i}\\ y_i=Y_i-\overline Y^*,x_i=X_i-\overline X^*
则估计式可简化为
β 1 = w i x i y i w i ( x i ) 2 \beta_1^*=\frac{\sum w_ix_i^*y_i^*}{\sum w_i(x_i^*)^2}

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转载自blog.csdn.net/weixin_39174856/article/details/104049879