从“派生如下”开始:“第七个” =错误。
因为
∑i(xi−x¯)(ui−u¯)
=∑i(xi−x¯)ui−∑i(xi−x¯)u¯
=∑i(xi−x¯)ui−u¯∑i(xi−x¯)
=∑i(xi−x¯)ui−u¯(∑ixi−nx¯)
=∑i(xi−x¯)ui−u¯(∑ixi−∑ixi)
=∑i(xi−x¯)ui−u¯0
=∑i(xi−x¯)ui
因此,在第7个“ =”之后,应该是:
1(∑i(xi−x¯)2)2E[(∑i(xi−x¯)ui)2]
=1(∑i(xi−x¯)2)2E(∑i(xi−x¯)2u2i+2∑i≠j(xi−x¯)(xj−x¯)uiuj)
= 1(∑i(xi−x¯)2)2E(∑i(xi−x¯)2u2i)+2E(∑i≠j(xi−x¯)(xj−x¯)uiuj)
= 1(∑i(xi−x¯)2)2E(∑i(xi−x¯)2u2i)uiujE(uiuj)=0
= 1(∑i(xi−x¯)2)2(∑i(xi−x¯)2E(u2i))
σ2(∑i(xi−x¯)2)2