Consider the following nonlinear regression model:
𝑦
𝑡
=
𝛼
𝑥
𝑡
𝛽
+
𝜀
𝑡
Assume i.i.d. data and
𝔼
[
𝜀
𝑡
|
𝑥
𝑡
]
=
0
. To estimate
𝛼
and
𝛽
by GMM, we use the following moment conditions:
𝔼
[
𝑦
𝑡
−
𝛼
𝑥
𝑡
𝛽
]
=
0
𝔼
[
(
𝑦
𝑡
−
𝛼
𝑥
𝑡
𝛽
)
𝑥
𝑡
]
=
0
We have an i.i.d. sample with
𝑇
=
1000
observations, with
∑
𝑡
=
1
𝑇
𝑥
𝑡
=
100
,
∑
𝑡
=
1
𝑇
𝑥
𝑡
2
=
200
and
∑
𝑡
=
1
𝑇
𝑥
𝑡
3
=
800
. We obtain point estimates
𝛼
̂
=
−
1
and
𝛽
̂
=
3
. To compute the variance of the estimates, we need to estimate the matrix
𝛤
0
,
𝛤
̂
0
=
[
𝛤
̂
11

𝛤
̂
12

𝛤
̂
21

𝛤
̂
22
]
Then, the value
𝛤
̂
11
is:

Question

Consider the following nonlinear regression model:
𝑦
𝑡
=
𝛼
𝑥
𝑡
𝛽
+
𝜀
𝑡
Assume i.i.d. data and 
𝔼
[
𝜀
𝑡
|
𝑥
𝑡
]
=
0
. To estimate 
𝛼
 and 
𝛽
 by GMM, we use the following moment conditions:
𝔼
[
𝑦
𝑡
−
𝛼
𝑥
𝑡
𝛽
]
=
0
𝔼
[
(
𝑦
𝑡
−
𝛼
𝑥
𝑡
𝛽
)
𝑥
𝑡
]
=
0
We have an i.i.d. sample with 
𝑇
=
1000
 observations, with 
∑
𝑡
=
1
𝑇
𝑥
𝑡
=
100
, 
∑
𝑡
=
1
𝑇
𝑥
𝑡
2
=
200
 and 
∑
𝑡
=
1
𝑇
𝑥
𝑡
3
=
800
. We obtain point estimates 
𝛼
̂
=
−
1
 and 
𝛽
̂
=
3
. To compute the variance of the estimates, we need to estimate the matrix 
𝛤
0
,
𝛤
̂
0
=
[
𝛤
̂
11
	
𝛤
̂
12

𝛤
̂
21
	
𝛤
̂
22
]
Then, the value 
𝛤
̂
11
 is:

BlackTom AI · Accepted Answer

To tackle this GMM set-up, start by identifying the first moment condition: E[y_t − α x_t^β] = 0. The derivative of this moment with respect to α is ∂/∂α [y_t − α x_t^β] = − x_t^β. The matrix Γ0 (the expectation of the Jacobian of the moment conditions) has its (1,1) entry as Γ11 = E[− x_t^β]. Since the given β-hat is 3, we interpret x_t^β as x_t^3.

Now, we estimate E[x_t^3] from the sample. With T = 1000 observations, the provided sum ∑ x_t^3 = 800 gives the sample average as (1/T) ∑ x_t^3 = 800 / 1000 = 0.8. Therefore, the estimated Γ11 is − E[x_t^3] ≈ −0.8.

Evaluating the answer options:

- Option 1: Γ̂11 = −0.8
  This matches the calculation above: the derivative with respect to α is −x_t^β, and using β = 3 we use E[x^3] ≈ 0.8, yielding Γ̂11 ≈ −0.8. This is consistent with the supplied sample moments.

- Option 2: There is not enough information to compute Γ̂11.
  Not correct. We have the necessary sample moments up to x^3, and the derivative with respect to α depends only on x^β. With β = 3 and the sum ∑ x^3 = 800, we can form an estimate of E[x^3]. Therefore, there is enough information to compute Γ̂11.

- Option 3: Γ̂11 = −0.1
  This would require E[x^3] ≈ 0.1, but the given sum ∑ x^3 = 800 over 1000 observations yields E[x^3] ≈ 0.8, not 0.1. Hence this value is inconsistent with the data.

- Option 4: Γ̂11 = 200
  This is inconsistent with the units and the derived form Γ11 = − E[x^3]. An answer of 200 would imply a positive, large expectation of x^3, which contradicts the observed average E[x^3] ≈ 0.8. Therefore this is incorrect.

- Option 5: Γ̂11 = 0.2
  This would correspond to E[x^3] ≈ −0.2 if we kept the negative sign, or misidentifying the sign altogether. The correct sign is negative, and the magnitude based on the sample is 0.8, not 0.2, so this option misstates both sign and magnitude.

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Similar Questions

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