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
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