Midterm Answer Key
Economics 435: Quantitative Methods
Fall 2010
1
Cluster samples
a) Yes.
E(x̄)
= E
n
S
1 XX
xis
nS s=1 i=1
!
=
n
S
1 XX
E(xis )
nS s=1 i=1
=
n
S
1 XX
E(µ + vs + uis )
nS s=1 i=1
=
n
S
1 XX
µ
nS s=1 i=1
1
(Snµ)
nS
= µ
=
b)
var(xis )
=
var(µ + vs + uis )
=
var(vs ) + var(uis ) + 2cov(vs , uis )
=
σv2 + σu2 + 0
=
σv2 + σu2
c) First we need to find the covariance:
cov(xis , xjs )
= cov(µ + vs + uis , µ + vs + ujs )
= cov(vs , vs ) + cov(vs , ujs ) + cov(uis , vs ) + cov(uis , ujs )
= σv2 + 0 + 0 + 0
= σv2
1
ECON 435, Fall 2010
2
and then we plug into the usual formula for the correlation:
corr(xis , xjs )
cov(xis , xjs )
var(xis )var(xjs )
=
p
=
p
=
σv2
σv2 + σu2
σv2
(σv2 + σu2 )(σv2 + σu2 )
d)
var(x̄)
!
= var
S
n
1 XX
xis
nS s=1 i=1
= var
n
S
1 XX
µ + vs + uis
nS s=1 i=1
=
=
=
=
=
!
n
S X
n
S X
n
S X
X
X
X
1
uis
v
+
µ
+
var
s
(nS)2
s=1 i=1
s=1 i=1
s=1 i=1
!
n
S X
S
X
X
1
uis
nvs +
var nSµ +
(nS)2
s=1 i=1
s=1
1
(nS)2
var(nSµ) +
S
X
var(nvs ) +
n
S X
X
!
var(uis )
s=1 i=1
s=1
1
0 + n2 Sσv2 + nSσu2
2
(nS)
σv2
σ2
+ u
S
nS
e) Yes. I didn’t ask for proof, but here it is. In a random sample of size nS:
var(x̄)
var(x)
nS
σv2 + σu2
nS
σv2
σ2
+ u
nS
nS
σv2
σu2
+
S
nS
=
=
=
<
f) Yes. To see this note that:
lim var(x̄)
S→∞
=
=
This implies that x̄ →m.s. µ, so x̄ →p µ as well.
lim
S→∞
0
σv2
σ2
+ u
S
nS
!
ECON 435, Fall 2010
3
g) No. To see this note that:
lim var(x̄)
n→∞
=
lim
n→∞
σv2
6= 0
S
=
2
σv2
σ2
+ u
S
nS
Permanent income and the black-white test score gap
a)
GAP 1(C)
= E(s|b = 1, c = C) − E(s|b = 0, c = C)
=
(γ0 + γ1 + γ2 C) − (γ0 + γ2 C)
= γ1
b)
GAP 2(P )
= E(s|b = 1, p = P ) − E(s|b = 0, p = P )
=
(β0 + β1 + β2 P ) − (β0 + β2 P )
= β1
c) Yes.
d)
\1(C)
GAP
=
γ̂1
e) Let’s try the hint out:
s = β0 + β1 b + β2 p + u
= β0 + β1 b + β2 (δ0 + δ1 b + δ2 c + w) + u
=
(β0 + β2 δ0 ) + (β1 + β2 δ1 )b + β2 δ2 c + (β2 w + u)
Now E(β2 w + u|b, c) = β2 E(w|b, c) + E(u|b, c) = β2 ∗ 0 + 0 = 0. This means that the model above fits
equation (2) where γ0 = (β0 + β2 δ0 ), γ1 = (β1 + β2 δ1 ), γ2 = β2 δ2 , and v = (β2 w + u). Solving for β1 we get:
β1 = γ1 − δ1 γ2 /δ2
This yields the estimator:
\2(P )
GAP
= β̂1
= γ̂1 − δ̂1 γ̂2 /δ̂2
f)
\1(C)
GAP
= γ̂1
= −0.426
\2(P )
GAP
= γ̂1 − δ̂1 γ̂2 /δ̂2
= −0.426 − (−0.306 ∗ 0.205/0.524)
= −0.306
\2(P ) and δ̂2 are both -0.306, but that’s just a coincidence.
By the way, you might have noticed that GAP
g) Yes.