Springer Texts in Business and Economics
Weighted Least Squares. This is based on Kmenta (1986)
a. From the first equation in (5.11), one could solve for a
n n n
5 ід2) = V°?) -" Xi/°2).
i=l i=1 i=1
i=1
n n n n
Yi/o? 1/* -" X,/ o2 1/o2)
Li=1 i=1 i=1 i=1
= Y* - "x*.
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Substituting a in the second equation of (5.11) one gets
2
Xi/o? 1/o?) C" X2/o?).
i= 1 i=1
n
Multiplying both sides by (1/o?) and solving for " one gets (5.12b)
i= 1
pOA? P(Y'X'/o? - Lj(4o? PAA)
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£ wi* (Xi — x*)2 i=1 |
Subtract this equation from the original regression equation to get Yi—Y* = "(Xi — X*) + (ui — it*). Substitute this in the expression for p in (5.12b), we get
n n
£ Wi* (Xi — X ) (ui — u*) £ Wi* (Xi — X )ui
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£ Wi* (Xi — X*)[1] i=1 |
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Ewi * (Xi — x*)2 i=1 |
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where the second equality uses the fact that n n /n/n Ew* (Xi—X*) = E w*Xi— (E w* E w*Xi /Ew? i=1 i=1 i=1 / i=1 Therefore, E(") = " as expected, and / n 2 I P wi* (Xi — X )u| |
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P wi* (Xi — X*)2 i=1 |
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P = P + ^------------------------------------- = P + ^-------------
b. From problem 5.3 part (b) we get
var(" blue) = —----------------------
P wi* (Xi - X*)2
i=1
where w* = (1/cr?) = (1/ct2X®) and X = P w’Xjpw* =
i=i! i=i
p (Xi/Xi8)
i=1 v______
n ’
1/Xi8
i=1
For Xi = 1,2,.., 10 and 8 = 0.5, 1, 1.5 and 2, this variance is tabulated below.
c. The relative efficiency of OLS is given by the ratio of the two variances computed in parts (a) and (b). This is tabulated below for various values
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of 8.
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As 8 increases from 0.5 to 2, the relative efficiency of OLS with respect to BLUE decreases from 0.9 for mild heteroskedasticity to 0.4 for more serious heteroskedasticity.
