weighted and two stage least squares estimation of ... - Boston College

and applying (2.35) and (2.36) with h(y, x, z, e) = z(y − vα0)/f ∗ (v|z) gives
E[z(y − vα0)I(0 ≤ y ≤ k)/f ∗ (v|z)] = E∗ [z(y − vα0)I(0 ≤ y ≤ k)/f ∗ (v|z)]
prob(y ≥ 0)
= E∗ [z(x ′ β0 + e)I(0 ≤ y ≤ k)/f ∗ (v|z)]
prob(y ≥ 0)
= k[E∗ [zx ′ ]β + E ∗ [ze]]
|α0|prob(y ≥ 0)
where the last equality applies (2.35) with H(y, x, z, e, θ) = z(x ′ β0 + e)y. With E ∗ [ze] = 0,
it follows that E ∗ [zy ∗ ] = E ∗ [zx ′ ]β0.
We next provide an identification result for α0. Define η(k) by
η(k) =
∗ 2v I(0 ≤ y ≤ k)/f (v|z)
E [I(0 ≤ y ≤ k)/f ∗
(v|z)]
(2.38)
Corollary 2.3 Let Assumptions A.1’, A.2’, A.3’ A.4’ and A.5’ hold. Let k and k ∗ be any
constants that satisfy 0 < k ∗ < k ≤ k. Then
α0 =
k − k ∗
η(k) − η(k ∗ )
Proof of Corollary 2.3:
(2.39)
E[vI(0 ≤ y ≤ k)/f ∗ (v|z)] = E[α −1
0 (y − x ′ β − e)I(0 ≤ y ≤ k)/f ∗ (v|z)] (2.40)
2 k
=
2α0|α0| − kE∗ [x ′
β − e]
/prob(y ≥ 0) (2.41)
α0|α0|
where the second equality above applies (2.35) and (2.36) with h(y, x, v, z, e) = α −1
0 (y −
x ′ β − e)/f ∗ (v|z) which implies H(y, x, z, e) = α −1
0 [(y 2 /2) − y(x ′ β + e)]. Similarly, E[I(0 ≤
y ≤ k)/f ∗ (v|z)] is given by equation (2.37), so η(k) = (k/α0) − 2E ∗ [x ′ β − e], and equation
(2.39) follows immediately.
Equation (2.34) in Theorem 2.3 shows that β0 is identified, and can be estimated by
an ordinary linear two stage least squares regression of y ∗ on x, using instruments z. The
variable y ∗ depends on f ∗ (v|z), which we will estimate using a kernel density estimator, and
13