Exam #1 Answer Key
Economics 435: Quantitative Methods
Spring 2006
1
Fitted values and residuals (33 points)
a) No it isn’t.
plim ŷi
=
plim (β̂0 + β̂1 xi )
= plim β̂0 + plim β̂1 plim xi
= β0 + β1 xi
= yi − u i =
6 yi
b) Yes it is:
plim ûi
= plim (yi − ŷi )
= plim yi − plim ŷi
= yi − (yi − ui )
= ui
c) Yes. A proof was not required, just some sort of explanation, but here’s a proof. Let zi = u2i and wi = wi2 :
Pn
1
(xi − x̄)(wi − w̄)
Pn
α̂1 = n 1i=1
2
i=1 (xi − x̄)
n
P
Pn
n
1
1
i=1 (xi − x̄)wi − n
i=1 (xi − x̄)w̄
n
Pn
=
1
2
i=1 (xi − x̄)
n
P
n
1
(xi − x̄)wi
= n1 Pi=1
n
2
i=1 (xi − x̄)
n
Pn
Pn
1
1
i=1 (xi − x̄)zi + n
i=1 (xi − x̄)(wi − zi )
n
Pn
=
1
2
i=1 (xi − x̄)
n
P
P
n
n
1
(xi − x̄)(zi − z̄) + n1 i=1 (xi − x̄)(wi − zi )
P
= n i=1
n
1
2
i=1 (xi − x̄)
n
Taking probability limits
α̂1
Pn
− x̄)(zi − z̄) + n1 i=1 (xi − x̄)(wi − zi )
Pn
= plim
1
2
i=1 (xi − x̄)
n
Pn
Pn
1
1
plim n i=1 (xi − x̄)(zi − z̄) + plim n i=1 (xi − x̄)(wi − zi )
P
=
n
plim n1 i=1 (xi − x̄)2
1
n
Pn
i=1 (xi
1
ECON 435, Spring 2006
=
=
=
=
2
Pn
− x̄)(zi − z̄) + n1 i=1 plim (xi − x̄)plim (wi − zi )
Pn
plim n1 i=1 (xi − x̄)2
Pn
Pn
plim n1 i=1 (xi − x̄)(zi − z̄) + n1 i=1 (xi − x̄)0
P
n
plim n1 i=1 (xi − x̄)2
cov(x, z)
var(x)
α1
plim
1
n
Pn
i=1 (xi
Your answer does not need to be this elaborate!
2
Averages (33 points)
a) The condition is a1 + a2 + · · · + an = 1.s
b) The answer is:
var(Wn ) = (a21 + a22 + · · · + a2n )σ 2
c) If Wn is an unbiased estimator then
(a1 + a2 + · · · + an )2
12
1
=
= ≤ a21 + a22 + · · · + a2n
n
n
n
Multiply both sides of the above by σ 2 and we get:
var(Wn ) = (a21 + a22 + · · · + a2n )σ 2 ≥
3
σ2
= var(X̄n )
n
Best linear predictors (34 points)
a)
2
E (y − b0 − b1 x)
= E(y 2 − b0 y − b1 xy − b0 y + b20 + b0 b1 x − b1 xy + b0 b1 x + b21 x2 )
= E(y 2 ) − 2b0 E(y) − 2b1 E(xy) + b20 + 2b0 b1 E(x) + b21 E(x2 )
Taking derivatives:
∂E/∂b0
∂E/∂b1
= −2E(y) + 2b0 + 2b1 E(x) = 0
= −2E(xy) + 2b0 E(x) + 2b1 E(x2 )
Solving, we get:
b0
b1
= E(y) − b1 E(x)
E(xy) − b0 E(x)
E(xy) − E(x)E(y)
=
=
= cov(x, y)/var(x)
2
E(x )
E(x2 ) − E(x)2
b) They are consistent estimators.
plim β̂1
=
plim
cov(x,
ˆ
y)
var(x)
ˆ
ECON 435, Spring 2006
3
plim cov(x,
ˆ
y)
plim var(x)
ˆ
cov(x, y)
=
var(x)
= b1
= plim (ȳ − β̂1 x̄)
=
plim β̂0
= plim ȳ − plim β̂1 plim x̄
= E(y) − b1 E(x)
= b0
c) No, because we never used either in proving consistency.