Questions
Single choice
Consider the following nonlinear regression model: 𝑦 𝑖 = 𝛼 + 𝑥 𝑖 𝛽 + 𝜀 𝑖 , Assume i.i.d. data and 𝔼 [ 𝜀 𝑖 | 𝑥 𝑖 ] = 0 . To estimate 𝛼 and 𝛽 by GMM, we use the two theoretical moment conditions 𝔼 [ 𝑦 𝑖 − 𝛼 − 𝑥 𝑖 𝛽 ] = 0 𝔼 [ ( 𝑦 𝑖 − 𝛼 − 𝑥 𝑖 𝛽 ) 𝑥 𝑖 ] = 0 To compute the variance of the GMM estimator we need the matrices 𝛤 0 and 𝛷 0 .
Options
A.The matrix
𝛷
0
is:
𝛷
0
=
𝔼
[
−
1
−
𝑥
𝑖
𝛽
log
(
𝑥
𝑖
)
−
𝑥
𝑖
−
𝑥
𝑖
𝛽
+
1
log
(
𝑥
𝑖
)
]
.
B.The matrix
𝛷
0
is:
𝛷
0
=
𝔼
[
1
𝑥
𝑖
𝛽
log
(
𝑥
𝑖
)
𝑥
𝑖
𝑥
𝑖
𝛽
+
1
log
(
𝑥
𝑖
)
]
.
C.There is not enough information to compute the matrix
𝛷
0
.
D.The matrix
𝛷
0
is:
𝛷
0
=
[
𝔼
[
(
𝑦
𝑖
−
𝛼
−
𝑥
𝑖
𝛽
)
2
]
𝔼
[
(
𝑦
𝑖
−
𝛼
−
𝑥
𝑖
𝛽
)
2
𝑥
𝑖
]
𝔼
[
(
𝑦
𝑖
−
𝛼
−
𝑥
𝑖
𝛽
)
2
𝑥
𝑖
]
𝔼
[
(
𝑦
𝑖
−
𝛼
−
𝑥
𝑖
𝛽
)
2
𝑥
𝑖
2
]
]
.
E.The matrix
𝛷
0
is:
𝛷
0
=
𝔼
[
𝛼
𝑥
𝑖
𝛽
𝑥
𝑖
𝛽
𝑥
𝑖
𝛽
log
(
𝑥
𝑖
)
]
.
View Explanation
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Step-by-Step Analysis
Let's break down what the question is asking and what each option is proposing about the matrix Φ0 in the context of GMM for a nonlinear regression with moment conditions. The setup uses two theoretical moment conditions: E[ y_i − α − x_i β ] = 0 and E[ (y_i − α − x_i β) x_i ] = 0, and we need the appropriate form of Φ0 to compute the variance of the GMM estimator.
Option 1: The matrix Φ0 is: Φ0 = E[ −1 − x_i β log(x_i) − x_i − x_i β + 1 log(x_i) ].
- This expression is trying to assemble Φ0 as a vector or matrix that contains expectations involving transformations like log(x_i) and products with x_i or β. However, Φ0 in the GMM context is typically the variance (or the expected outer product) of the moment conditions and their derivatives with respect to parameters, evaluated at the true parameter values. A single row vector with terms like −1, − x_i β log(x_i), etc., does not align with the standard construction of Φ0 for this model, which would involve E[ g_i g_i' ] or E[ ∂g_i/∂θ ... ] structures. This option appears to mix components in a way that does not correspond to the conventional Φ0 form, and the inclusion of log(x_i) in this mixed fashion is inappropriate unless motivated......Login to view full explanationLog in for full answers
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