笔记:OLS的线性性、无偏性和有效性的证明

  • 线性性
    定义 x i = X i X , y i = Y i Y x_i=X_i-\overline X,y_i=Y_i-\overline Y 。为了后续证明方便首先说明一些性质。
    x i = ( X i X ) = 0 \sum x_i=\sum (X_i -\overline X)=0
    x i y i = x i ( Y i Y ) = x i Y i Y x i = x i Y i \sum x_iy_i=\sum x_i(Y_i-\overline Y)\\=\sum x_iY_i-\overline Y\sum x_i =\sum x_i Y_i
    根据以上性质证明线性性:
    β ^ 1 = x i y i x i 2 = x i ( Y i Y ) x i 2 = x i x i 2 Y i = k i Y i \hat \beta_1=\frac{\sum x_i y_i}{\sum x_i^2}=\frac{\sum x_i(Y_i-\overline Y)}{\sum x_i^2}=\frac{\sum x_i}{\sum x_i^2}Y_i =\sum k_iY_i
    其中 k i = x i x i 2 k_i=\frac{x_i}{\sum x_i^2}
    β ^ 0 = Y β ^ 1 X = 1 n Y i k i Y i X = ( 1 n k i X ) Y i = w i Y i \hat \beta_0=\overline Y-\hat \beta_1\overline X=\frac{1}{n}\sum Y_i -\sum k_i Y_i \overline X\\=\sum(\frac{1}{n}-k_i\overline X)Y_i=\sum w_i Y_i
    其中 w i = 1 n X k i w_i=\frac{1}{n}-\overline X k_i

  • 无偏性
    无偏性的证明用到了 k i k_i 的一些性质:
    k i = x i x i 2 = 0 k i X i = x i X i x i 2 = x i X i x i X i X x i = 1 \sum k_i=\frac{\sum x_i}{\sum x_i^2}=0 \\\quad\\\sum k_iX_i=\frac{\sum x_i X_i}{\sum x_i^2} =\frac{\sum x_i X_i}{\sum x_iX_i-\overline X \sum x_i}=1
    利用上面的性质,
    β ^ 1 = k i Y i = k i ( β 0 + β 1 X i + μ i ) = β 0 k i + β 1 k i x i + k i μ i = β 1 + k i μ \hat \beta_1=\sum k_i Y_i=\sum k_i(\beta_0+\beta_1 X_i+\mu_i)\\=\beta_0\sum k_i+\beta_1\sum k_ix_i+\sum k_i \mu _i=\beta_1+\sum k_i\mu
    E ( β ^ 1 ) = E ( β 1 + k i μ ) = β 1 + k i E ( μ i ) = β 1 E(\hat \beta_1)=E(\beta_1+\sum k_i\mu)=\beta_1+\sum k_i E(\mu _i)=\beta_1
    同样地,
    E ( β ^ 0 ) = E ( β 0 + w i μ i ) = β 0 + w i E ( μ i ) = β 0 E(\hat\beta_0)=E(\beta_0+\sum w_i\mu_i)=\beta_0+\sum w_iE(\mu_i)=\beta_0

  • 有效性
    有效性是说在所有线性无偏估计中,OLS估计量具有最小方差。
    V a r ( β ^ 1 ) = V a r ( k i Y i ) = k i 2 ( β 0 + β 1 X i + μ i ) = k i 2 σ 2 = σ 2 x i 2 {\rm Var}(\hat\beta_1)={\rm Var}(\sum k_i Y_i)=\sum k_i^2(\beta_0+\beta_1X_i+\mu_i)\\=\sum k_i^2\sigma ^2=\frac{\sigma^2}{\sum x_i^2}
    β ^ 1 \hat\beta_1^* 是由其他方法得到的估计量:
    β ^ 1 = c i Y i \hat\beta_1^*=\sum c_iY_i
    其中 c i = k i + d i c_i=k_i+d_i
    由于 β ^ 1 \hat\beta_1^* 的无偏性, c i c_i k i k_i 一样具有性质:
    c i = 0 , c i X i = 1 \sum c_i=0,\sum c_iX_i=1
    d i d_i 有如下性质:
    k i d i = k i ( c i k i ) = x i c i x i 2 k i 2 = X i c i X c i x i 2 k i 2 = 1 x i 2 1 x i 2 = 0 \sum k_i d_i=\sum k_i(c_i-k_i)=\sum \frac{x_ic_i}{\sum x_i^2}-\sum k_i^2\\= \frac{\sum X_ic_i-\overline X\sum c_i}{\sum x_i^2}-\sum k_i^2 =\frac{1}{\sum x_i^2}-\frac{1}{\sum x_i^2}=0
    下面的证明利用了以上性质。
    V a r ( β ^ 1 ) = V a r ( c i Y i ) = σ 2 c i 2 = σ 2 ( k i + d i ) 2 = σ 2 k i 2 + σ 2 d i 2 + 2 σ 2 k i d i = σ 2 x i 2 + σ 2 d i 2 = V a r ( β ^ 1 ) + σ 2 d i 2 {\rm {Var}} (\hat \beta_1^*)={\rm{Var}}(\sum c_i Y_i)=\sigma^2\sum c_i^2 \\\quad\\=\sigma^2\sum(k_i+d_i)^2=\sigma^2\sum k_i^2+\sigma^2\sum d_i^2+2\sigma^2\sum k_id_i\\\quad\\=\frac{\sigma^2}{\sum x_i^2}+\sigma^2\sum d_i^2={\rm Var}(\hat \beta_1)+\sigma^2\sum d_i^2
    因为 d i 2 0 \sum d_i^2 \geq0 ,所以 V a r \rm Var ( β ^ 1 \hat \beta_1^* ) V a r ( β ^ 1 ) \geq{\rm Var}(\hat \beta_1)

Basic Econometrics,Damodar N.Gujarati
计量经济学,李子奈

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