最小平方法~如何證明B0的OLS是B0線性不偏變異最小者

Question

想請教大大~
我只會B1的證明,
那B0如何證哩?
(即BLUS)
謝謝!

Answer 1

(i)∵Βo = Ybar - Β1* Xbar = ΣYi / n -ΣKi*Xbar*Yi = Σ(1/n - Ki*Xbar)*Yi
where Ki = (Xi - Xbar) / Σ(Xi - Xbar)^2
∴Βo is a linear combination of Yi，i = 1…n

(ii)∵E[Βo]=E[Ybar] - E[Β1*Xbar] = βo +β1*Xbar - β1*Xbar =βo
∴Βo is an unbiased estimator of βo

(iii) for any unbiased estimator of βo，
let Βo'=ΣCi*Yi
then E[Βo'] = E[ΣCi*Yi] = ΣCi*E[Yi] = ΣCi*(βo+β1*Xi)
= ΣCi*βo +ΣCi*Xi*β1= βo
so that，ΣCi= 1 and ΣCi*Xi= 0；
Var(Βo) = Σ(1/n - Ki*Xbar) ^2 *σ^2
Var(Βo' ) = ΣCi^2 * σ^2；
what we want to show is ΣCi^2 >= Σ(1/n - Ki*Xbar) ^2，
let Ci - (1/n - Ki*Xbar) = Di，
then ΣCi^2 = Σ [ (1/n - Ki*Xbar) + Di ] ^2
= Σ (1/n - Ki*Xbar) ^2 +2Σ(1/n - Ki*Xbar)*Di + ΣDi^2
since Σ(1/n - Ki*Xbar)*Di
= Σ(1/n - Ki*Xbar)*[ Ci - (1/n - Ki*Xbar) ]
= Σ(1/n - Ki*Xbar)*Ci - Σ(1/n - Ki*Xbar)^2
= (ΣCi) / n - Σ Ki*Ci*Xbar - Σ[1/n^2 - 2*Ki*Xbar / n + (Ki*Xbar)^2]
= 1 / n - Σ(Xi - Xbar)*Ci*Xbar / Σ(Xj - Xbar)^2 -
(1 / n - 0 + Xbar^2 / Σ(Xj - Xbar)^2)
(∵ΣCi = 1，we have proved；Ki = (Xi - Xbar) / Σ(Xi - Xbar)^2；
ΣKi = 0；ΣKi^2 = 1 / Σ(Xi - Xbar)^2 )
= 0
(∵ΣCi*Xi= 0 =>
Σ(Xi - Xbar)*Ci*Xbar / Σ(Xj - Xbar)^2 = Xbar^2 / Σ(Xj - Xbar)^2 )
we have ΣCi^2 >= Σ(1/n - Ki*Xbar) ^2
and the equation holds when Di = 0 for each i = 1…n

conclusion：Βo is the BLUE of βo

2005-11-24 09:04:51 補充：
不客氣^^

Answer 2

太強了，初學者怎麼打出Σ的？