给定一组样本(Xi,Yi)(X_i,Y_i)(Xi,Yi),模型Yi=β0Xi+μiY_i=\beta_0X_i+\mu_iYi=β0Xi+μi的离差形式为:
yi=Yi−Y‾=β1xi+(μi−μ‾)y_i=Y_i-\overline Y=\beta_1x_i+(\mu_i-\overline\mu)yi=Yi−Y=β1xi+(μi−μ)
样本回归函数的离差形式:
y^i=β^1xi\hat y_i=\hat \beta_1x_iy^i=β^1xi
所以
∑ei2=∑(yi−y^i)2=∑[(β1−β^1)xi+(μi−μ‾i)]2=∑(β1−β^1)2xi2+∑(μi−μ‾)2+2∑(β1−β^1)xi(μi−μ‾i)]=∑(β1−β^1)2xi2+∑(μi−μ‾)2−2∑(∑kiμi)xi(ui−μ‾)=∑(β1−β^1)2xi2+∑(μi−μ‾)2−∑xiμi∑kiμi+2μ‾∑xi∑kiμi=∑(β1−β^1)2xi2+∑(μi−μ‾)2−2∑xiμi∑xiμi∑xi2 \sum e_i^2=\sum(y_i-\hat y_i)^2=\sum[(\beta_1-\hat \beta_1)x_i+(\mu_i-\overline \mu_i)]^2\\\quad\\=\sum(\beta_1-\hat\beta_1)^2x_i^2+\sum(\mu_i-\overline \mu)^2+2\sum(\beta_1-\hat\beta_1)x_i(\mu_i-\overline \mu_i)]\\\quad\\=\sum(\beta_1-\hat\beta_1)^2x_i^2+\sum(\mu_i-\overline \mu)^2-2\sum (\sum k_i\mu_i)x_i(u_i-\overline \mu)\\\quad\\= \sum(\beta_1-\hat\beta_1)^2x_i^2+\sum(\mu_i-\overline \mu)^2-\sum x_i\mu_i\sum k_i\mu_i+2\overline\mu\sum x_i\sum k_i\mu_i\\\quad\\= \sum(\beta_1-\hat\beta_1)^2x_i^2+\sum(\mu_i-\overline \mu)^2-2\sum x_i\mu_i\frac{\sum x_i\mu_i}{\sum x_i^2}∑ei2=∑(yi−y^i)2=∑[(β1−β^1)xi+(μ
笔记:干扰项方差的无偏估计
最新推荐文章于 2024-05-26 22:35:57 发布