Answer Keys: Final Exam in Econometrics
ECON 837
April 15th, 2015
Exercise 1 []
see midterm
Exercise 2 []
(a) We want an estimator γ̂ such that E [γ̂] = µ2 . Consider X̄ 2 =
( 1 ∑n
n
i=1
Xi
)2
. We know
[ ]
[ ]
[ ]2 σ 2
E X̄ 2 = V ar X̄ + E X̄ =
+ µ2 .
n
)2
∑n (
1
We also know that an unbiased estimator of σ 2 is s2 = n−1
. So dene
i=1 Xi − X̄
γ̂ = X̄ 2 − n1 s2 , and we have
[ ] 1 [ ] σ2
σ2
E [γ̂] = E X̄ 2 − E s2 =
+ µ2 −
= µ2 .
n
n
n
p
(b) Since Xn → a, we know that
lim Pr [|Xn − a| < ε] = 1.
n→∞
Dene Yn =
√
Xn . Notice that we can write
lim Pr [|Xn − a| < ε]
[√
]
√ √
√ = lim Pr Xn − a Xn + a < ε
n→∞
[
]
√
√ √
√ Xn + a
ε
√
= lim Pr Xn − a
<√
n→∞
a
a
[√
]
√ ε
≤ lim Pr Xn − a < √
n→∞
a
1 =
and hence Yn =
n→∞
√
p √
Xn → a.
Exercise 3 []
1
We know that
∑ ∗
∑ ∗
∑ ∗
∑ ∗
(xi − x̄∗ ) yi∗
(xi − x̄∗ ) (yi + νi )
(xi − x̄∗ ) (α + βxi + εi )
(xi − x̄∗ ) νi
β̂ = ∑ ∗
=
=
+ ∑ ∗
∑ ∗
∑ ∗
(xi − x̄∗ )2
(xi − x̄∗ )2
(xi − x̄∗ )2
(xi − x̄∗ )2
∑
∑
∑
1
1
1
(x∗ − x̄∗ ) xi
(x∗i − x̄∗ ) εi
(x∗i − x̄∗ ) νi
n
n
= β n1 ∑ i ∗
+
+
.
∑ ∗
∑ ∗
1
1
∗ )2
∗ )2
∗ )2
(x
−
x̄
(x
−
x̄
(x
−
x̄
i
i
i
n
n
n
Applying a WLLN to each term, we have:
1∑ ∗
(xi − x̄∗ )2
n
1∑ ∗
plim
(xi − x̄∗ ) xi
n
1∑ ∗
plim
(xi − x̄∗ ) εi
n
1∑ ∗
plim
(xi − x̄∗ ) νi
n
plim
]
[
= E (x∗i − x̄∗ )2 = V ar [x∗i ] = σx2 + ση2
= E [(x∗i − x̄∗ ) xi ] = E [(xi + ηi − x̄) xi ] = V ar [xi ] + Cov [xi , ηi ] = σx2
= E [(x∗i − x̄∗ ) εi ] = E [(xi + ηi − x̄) εi ] = Cov [xi , εi ] + Cov [ηi , εi ] = 0
= E [(x∗i − x̄∗ ) νi ] = E [(xi + ηi − x̄) νi ] = Cov [xi , νi ] + Cov [ηi , νi ] = 0
(
)
so that plimβ̂ = βσx2 / σx2 + ση2
(a) If ση2 = 0, then plimβ̂ = βσx2 /σx2 = β (the value of σν2 is irrelevant here). Hence the least
squares estimator is consistent.
(
)
(b) If ση2 ̸= 0, then plimβ̂ = βσx2 / σx2 + ση2 ̸= β (again, the value of σν2 is irrelevant). Hence
the least squares estimator is inconsistent.
(c) Clearly, measurement error in xi is the problem. Measurement error in yi is irrelevant it just becomes part of the error term.
(d) Suppose we observe wi = xi + ξi . Notice that:
Cov [wi , x∗i ] = E [(xi + ξi − µ) (xi + ηi − µ)]
[
]
= E (xi − µ)2 + E [(xi − µ) ηi ] + E [(xi − µ) ξi ] + E [ηi ξi ]
= V ar [xi ] + Cov [xi , ηi ] + Cov [xi , ξi ] + Cov [ηi , ξi ]
= σx2
because the covariances are all zero by assumption. So we can easily estimate σx2 from this
covariance, and we can estimate σx2 + ση2 directly from the observed x∗i . Using these, we can
correct our estimate β̂ and obtain a consistent estimate of β (note: this is an IV estimator,
where wi is an instrument for xi ).
Exercise 4 []
2
(a) By denition, β̃ = (X ′ Ω̂−1 X)−1 X ′ Ω̂−1 y . We want to test, H0 : β1 = 0. The asymptotic
distribution of the QGLS is:
 [
(
)]−1 
′ −1
√
X Ω̂ X
d

n(β̃ − β) → N 0, Plim
n
 [
(
)]−1 
′
−1
√
X Ω̂ X
d

n(β̃1 − β1 ) → N 0, Plim
⇒
n
11
[
( ′ −1 )]−1
To provide a (feasible) test statistic, we need a consistent estimator of Plim X Ω̂n X
.
11
Under the assumption that the 4th moment of X exists, a consistent estimator can be
dened as follows:
X ′ Ω̂X
n
As a result, I dene the following test statistic:
√
nβ̃1 d
t1 = √
→ N (0, 1) under H0
V̂11
Ŝ =
with
(
V̂11 = 
′
X Ω̂X
n
)−1 

11
Decision rule: for α the level of condence of the test and the associated quantile t∗α/2 dened
as:
P (|z| > t∗α/2 ) = α with z ∼ N (0, 1)
the decision rule is to "reject H0 " if |t1 | > t∗α/2 and to "not reject H0 " otherwise.
(b) Asymptotic power function of the test stated in (a) against H1 : β1 = c for some c ∈ R∗ .
Under H1 , we have:
√ β̃1 − c d
n√
→ N (0, 1)
V̂11
As a result, we have:
t1 =
√ β̃1 − c √
c
n√
+ n√
V̂11
V̂11
with |t1 | which grows towards innity as n → ∞ under H1 . Hence, the asymptotic power
function is 1 since:
P (|t1 | > t∗α/2 ) → 1
n
3
The above result is true for any nonzero real number c.
(c) Power function of the test stated in (a) against H2 : β1 =
√c
n
for some c ̸= 0. We have:
√
√ β̃1 − c/ n
c
t1 = n √
+√
V̂11
V̂11
√
which converges towards a normal distribution with mean µ̃ = c/Plim[ V̂11 ] under H2 . As
a result, the power function is:
K(c) = P (|z̃| > t∗α/2 ) with z̃ ∼ N (µ̃, 1)
and t∗α/2 as dened in (a).
(d) There is heteroskedasticity in the model, but the applied researcher uses the OLS estimator β̂ . As a result, the asymptotic distribution of β̂ is:
√
n(β̂ − β) → N (0, Q−1 SQ−1 )
d
with Q = Plim(X ′ X/n) and S = Plim(X ′ ΩX/n). As discussed in (a), a consistent estimator
of S is Ŝ = X ′ Ω̂X/n and of Q is Q̂ = X ′ X/n. The test statistic is then dened as:
√
nβ̂1
d
t̂1 = √[
] → N (0, 1)
Q̂−1 Ŝ Q̂−1
11
- Decision rule: for a level of condence α, the associated quantile t∗α/2 (see (a)), the decision
rule is to "reject H0 " is |t̂1 | > t∗α/2 and to "not reject H0 " otherwise.
- Asymptotic power function against H1 : β1 = c for some nonzero real number c: very
similar to (b). Under H1 , we have:
√
β̂1 − c
n √[
]
Q̂−1 Ŝ Q̂−1
d
→ N (0, 1)
11
As a result, the test statistic is such that:
√
√
β̂1 − c
t̂1 = n √[
]
−1
−1
Q̂ Ŝ Q̂
nc
+ √[
]
−1
−1
Q̂ Ŝ Q̂
11
11
and |t̂1 | increases towards innity as n → ∞ under H1 . We conclude that the asymptotic
power function is equal to 1 since:
P (|t̂1 | > t∗α/2 ) → 1
n
4
and this result holds for any nonzero real number c.
(e) Power function of the test stated in (d) against the local alternative H2 : very similar to
(c). The test statistic is such that:
√
√
β̂1 − c/ n
t̂1 = n √[
]
−1
−1
Q̂ Ŝ Q̂
c
+ √[
]
−1
−1
Q̂ Ŝ Q̂
11
11
√[
]
which converges towards a normal distribution with mean µ̂ = c/Plim[
Q̂−1 Ŝ Q̂−1 ]
11
under H2 . As a result, the power function is:
K̂(c) = P (|ẑ| > t∗α/2 ) with ẑ ∼ N (µ̂, 1)
and t∗α/2 as dened in (a).
(f) The power of a test is a "measure" of the capacity of the test to detect deviations of the
model with respect to the null hypothesis. For a given level of condence, the larger the
power is (when the null hypothesis does not hold), the better the test is. In questions (b)
and (d), we computed the asymptotic power function of the 2 tests under H1 and they were
both equal to 1. In questions (c) and (e), we computed the power function under sequence of
local alternatives to distinguish the 2 tests. The power function associated with the largest
mean (either µ̃ or µ̂) is preferred. Recall:
µ̃ = √
c
and
[Q̂−1 Ŝ Q̂−1 ]11
µ̂ = √
c
[Ŝ]−1
11
It is sucient to compare the denominators, or more specically their inverses, that is:
(
)
( ′ )−1 ( ′ −1 ) ( ′ )−1
′ −1
X
X
X
X
X
Ω̂
X
X
Ω̂
X
Q̂−1 Ŝ Q̂−1 =
and
Ŝ =
n
n
n
n
and we conclude that: µ̃ > µ̂ which means that the rst test should be preferred to the
second one.
Exercise 5:
(a) - The 2 moment conditions are:
E[yi ] = θ0
and
V ar[yi ] = θ0 ⇔ E[(yi − θ0 )2 ] = θ0
5
Dene the vector of moment conditions for the i-th observation:
(
)
yi − θ
g(yi , θ) =
(yi − θ)2 − θ
- By denition, the limiting expression for the variance matrix of these moment conditions
V is:
V
= E [g(yi , θ0 )g ′ (yi , θ0 )]
]
[(
)
)
(
yi − θ 0
= E
yi − θ0 (yi − θ0 )2 − θ0
(yi − θ0 )2 − θ0
[(
)]
(yi − θ0 )2
(yi − θ0 )3 − θ0 (yi − θ0 )
= E
(yi − θ0 )3 − θ0 (yi − θ0 ) [(yi − θ0 )2 − θ0 ]2
with
E[(yi − θ0 )2 ] = V ar[yi ] = θ0
E[(yi − θ0 )3 − θ0 (yi − θ0 )] = E[(yi − θ0 )3 ] − θ0 E[yi − θ0 ] = 0
E[(yi − θ0 )2 − θ0 ]2 = E[(yi θ0 )4 ] − 2θ0 E[(yi − θ0 )2 ] + θ02 = 2θ02
As a result:
(
V
)
θ0 0
=
0
2θ02
- The ecient (two-step) GMM estimator is dened as follows:
[
(
)
]
1/θ 0
′
θ̂ = arg min g n (θ)
g n (θ)
2
θ∈Θ
0
1/(2θ )
for some consistent estimator θ of θ0 with
( ∑
n
1
g n (θ) =
- The FOC are:
dg ′n (θ̂)
dθ
with
dg n (θ̂)
=
dθ
(
(
n
1
n
∑n
i=1 (yi
−1
− n2
− θ)2 − θ
)
1/θ 0
2
1/(2θ )
0
)
i=1 (yi − θ)
g n (θ̂) = 0
(
)
∑n
P
i=1 (yi
− θ̂) − 1
6
→−
1
1
)
where the last result follows after assuming that θ̂ is a consistent estimator of θ0 .
√
(b) To derive the asymptotic distribution of n(θ̂ − θ0 ), I perform a mean-value expansion
of the FOC written in (a):
dg ′n (θ̂)
dθ
(
)[
1/θ 0
0
2
1/(2θ )
]
dg n (θ̃)
g n (θ0 ) +
(θ̂ − θ0 ) = 0
dθ
with θ̃ between θ̂ and θ0 .
(
)
(
)
√
dg ′n (θ̂) 1/θ 0
dg n (θ̃) √
dg ′n (θ̂) 1/θ 0
⇔
n(θ̂ − θ0 ) = −
ng n (θ0 )
2
2
dθ
dθ
dθ
0
1/(2θ )
0
1/(2θ )
with
√
d
ng n (θ0 ) → N (0, V )
)
(
(
)
dg ′n (θ̂) 1/θ 0
P
1
1
→
2
2
θ0
2θ0
dθ
0
1/(2θ )
(
)
dg ′n (θ̂) 1/θ 0
dg n (θ̃) P 1
1
→
+ 2
2
dθ
dθ
θ0 2θ0
0
1/(2θ )
It follows that:
]−1 (
)
1
1
1
1
+
N 0, + 2
θ0 2θ02
θ0 2θ0
( [
]−1 )
√
1
1
d
⇔
n(θ̂ − θ0 ) → N 0,
+ 2
θ0 2θ0
(
)
√
2θ02
d
⇔
n(θ̂ − θ0 ) → N 0,
1 + 2θ0
√
d
n(θ̂ − θ0 ) →
[
7