有关的证明过程
1． 线性特性
xiyi?xi(Yi?Y)???2??22xx?? ii
??Y?2i?xixiY?xi2x?i??KiYi
?1?Y???2X?Y?X?KiYi ?1?1???Yi??KiXYi????KiX?Yin?n?
2． 无偏性
?2??KiYi??Ki(?1??2Xi?ui) ???Ki?1??Ki?2Xi??Kiui ??1?Ki??2?KiXi??Kiui
xi?（Xi?X)???0?Ki??2?22?xi?xi?xi其中：
xi(Xi?X?X)??KiXi??2Xi?2xix?i
xixixi(Xi?X)??xiX??2x? i??xi2?X?xi
?xi2???11x?i
xi2?2??2??Kiui 故有：??2?E(?2??Kiui)??2??KiEui??2 E??1???1????KiX?Yi?n?
?1?????KiX???1??2Xi?ui??n?
???1n???2Xinui??n
???1KiX???2KiXXi??KiXui
??1??2X?u??1X?Ki??2X?KiXi?X?Kiui
1??1??(?XKi)ui n1??E?1??1??(?KiX)Eui??1 n3． 有效性
首先讨论参数估计量的方差。
?2)?E(??2?E(??2))2 Var(????)2?E((???E(?222???Kiui?2?(K1u1?K2u2???Knun)(K1u1?K2u2???Knun)?Kiui)??2)?E(2?Kiui)2??(Kiui)2???KiKjuiuji?j
?E(?Kiui)2?E?(K2iui)?E??KiKjuiuji?j
??K?2i2Eui2??2??x?i??2??x???i2??xi2 2Var(即：????
2)?xi2
同理有：
Var(??)??2Xi21?n?xi2
Var(??1)?E(??1?E(??1))2?E(???1?n?KiX???ui)2
?????22?1?n?KiX????ui??????1?n?KiX??2?ui
?1)??2Var(????2?1??1??KiX???KjX?uiujn??n?i?j?
??1??KX?i??n?
??2?KiX(2?2?Ki2X2)n n1
Ki22?2X??nn??22?Ki??2X2?
?n??2(?Xi)2n2?xi2
?2n??22?n(x)?(X)?2??i???i?xi2??
?n??(?2??Xi2?nX)?n221n(??Xi)??2?xi2
??2?n?xi2
?Xi2显然各自的标准误差为：
?)?se(?2?)??se(?1?xi2，
?n?xi2
Xi2标准差的作用：衡量估计值的精度。
由于σ为总体方差，也需要用样本进行估计。
?2???ei2n?2
证明过程如下：
回顾:Yi??1??2Xi?ui
因此有： Y??1??2X?u
那么：(Yi?Y)?yi?(?1??2Xi?ui)?(?1??2X?u) ??2xi?(ui?u)
?2xi， 根据定义：ei?yi??（实际观测值与样本回归线的差值） 则有：
?2xi?(ui?u)?(??2??2)xi ei?(?2xi?(ui?u))??两边平方，再求和：
?ei2??(ui?u)2???2??2)xi?2(ui?u)(???2??2)xi)2((?
?2??2)2?(??xi2???2??2)(ui?u)2?2(??(ui?u)xi
对上式两边取期望有：
E(?ei2)???2??)2xi2E(?2