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Econ203C Homework #6 Anton Cheremukhin
June 1, 2006
Exercise 1 GMM and the Three Tests
The hardest part is to get a stable estimate for both the unrestricted and the restricted version.
We do this in six steps.
- We find unrestricted β
(1) using and identity weight matrix and "fminsearch".
- Given that, we compute the optimal weight matrix V
(2) and find the corresponding β
(2) .
- We Repeat step 2 and obtain V
(3) and the corresponding β
(3) .
- We Compute A and m n
β
(3)
- We find restricted β
(5)
using V
(3)
as the weight and "fmincon".
- We compute m n
β
(5)
and V
(5) .
After that we calculate the three tests: W = nr
0
μ
R
A
V
(3)
− 1
A
0
− 1
R
0
r,
LM = nm n
β
(5)
0
V
(3)
− 1
A
0
A
V
(5)
− 1
A
0
− 1
A
V
(3)
− 1
m n
β
(5)
LR = n
m n
β
(5)
0
V
(3)
− 1
m n
β
(5)
− m n
β
(3)
0
V
(3)
− 1
m n
β
(3)
where r = β 2
β 3
, R = [0, β 3
, β 2
, 1].
The reason why we use V
(3)
is because it is the weight matrix, we used when finding the minimum
- the one we regard as ’optimal’. Besides, otherwise we often get negative values for the LR test.
The whole procedure is very sensitive to V
(3) , that’s why we repeated the second step, to be sure
we are getting a good estimate of the ’optimal’ weight matrix.
The histograms are almost identical: one of them is depicted on the left. The graph on the
right compares the three empirical c.d.f.’s of the tests. They are almost indistinguishable. The
probabilities of type I errors for the 95% level are all around 11-12%. The means differ a little,
which can be clearly seen in the graph.
Wald LM LR
mean 1.47 1.62 1.
median 0.67 0.67 0.
P(type I error) 0.109 0.118 0.
Exercise 2 NLLS as an Example of GMM
y i
= g (x i
, θ 0
) + u i
E (u i
|x i
1) E
(yi − g (xi, θ))
2
→ minθ use law of iterated expectations:
FOC: E
∂
∂θ
g (x i
, θ) (y i
− g (x i
, θ))
= E
∂
∂θ
g (x i
, θ) E [(y i
− g (x i
, θ)) |x i
]
since E (ui|xi) = 0 ⇔ E (yi|xi) = g (xi, θ 0 ) , E
∂
∂θ
g (xi, θ) (g (xi, θ 0 ) − g (xi, θ))
If g (.) is strictly monotone in θ, this expression is satisfied iff θ = θ 0
- The usual conditions for existence are continuity of g (.) and compactness of Θ. For uniqueness
we need to add: θ 6 = θ 0
⇒ g (x, θ) 6 = g (x, θ 0
). Strict monotonicity in θ is one possibility.
b θn = arg minθ mn (θ)
0
(Vn)
− 1
mn (θ)
m n
(θ) =
1
n
Σϕ (y i
, x i
, θ) ϕ (y i
, x i
, θ) = z (y i
− g (x i
, θ))
Some simple ideas for instruments zji include: 1 , x 1 i, ..., xKi.
- Some additional simple ideas for instruments z ji
include: x
2
1 i
, ..., x
2
Ki
We can use higher powers, if needed. The justification to that would be a Taylor expansion of
∂
∂θ
g (x i
, θ) around θ 0
. These are all instruments since E [f (x i
) (y i
− g (x i
, θ))] = 0 only in θ 0
- An optimal GMM estimator uses an optimal weight matrix:
b V n
1
n
Σϕ
y i
, x i
, z,
b θ
0
ϕ
y i
, x i
, z,
b θ
0
0
Can use any consistent
b θ
0 , e.g.
b θ
0 = arg min θ
m n
(θ)
0
m n
(θ).
- Asymptotic distribution:
n
b θn − θ 0
d
→ N (0, Λ (θ 0 )) , where
A (θ 0
) = E
0
h
∂ϕ(y,x,z,θ 0
)
∂θ
i
, V (θ 0
) = E
0
ϕ (y, x, z, θ 0
) ϕ (y, x, z, θ 0
0
Λ (β 0
A (θ 0
) V (θ 0
− 1
A (θ 0
0
− 1
A consistent estimator for A (θ 0 ) is
b A =
P (^) ∂ϕ (yi,xi,zi,^
b θ
0 )
n∂β
A consistent estimator for V (θ 0
) is
b V n
1
n
P
z i
z
0
i
y i
− x
0
i
b θ
0
2
Proof is exactly the same as always.
- H 0 : θ 1 θ 2 ...θK = 1 H 1 : θ 1 θ 2 ...θK 6 = 1
r (θ) = θ 1
θ 2
...θ K
− 1 R (θ) = [θ 2
...θ K
, θ 1
θ 3
...θ K
, ..., θ 1
θ 2
...θ K− 1
]
e θn = arg minθ mn (θ)
0
e Vn
− 1
mn (θ) s.t. r (θ) = 0
e A =
P (^) ∂ϕ (yi,xi,zi,^
e θ
0 )
n∂β
e V n
1
n
Σϕ
y i
, x i
, z j
e θ
0
ϕ
y i
, x i
, z i
e θ
0
0
e θ
0 = arg min θ
m n
(θ)
0
m n
(θ) s.t. r (θ) = 0
W = nr
b θ
0
Ã
R
b θ
μ
b A
b V
− 1
b A
0
− 1
R
b θ
0
r
b θ
LM = nm
e θ
0
e V
− 1
e A
0
μ
e A
e V
− 1
e A
0
− 1
e A
e V
− 1
m
e θ
LR = n
μ
m
e θ
0
e V
− 1
m
e θ
− m
b θ
0
b V
− 1
m
b θ
All distributed as χ
2 (1).
