AAE 637 Midterm
Answer Key
1. (a)
2
Q
   2  PEL   2571.74 
PEL
 EL 
Q PEL
 0.541 
 2571.74 
  1.0945
PEL Q
 1271.2 
Given that only β2 is only used in the calculation of the above elasticity, we only need
to use its variance in the calculation of the variance of this elasticity. As such, we
have: Std.Err.  ηEL  
1
PEL Q
Std.Err.  β 2  
1
164.87  0.2397
(0.541)(1271.2)
With the null hypothesis that the own price elasticity is -1.0, we have
t
1.0945  1
 0.3942
0.2397
(b) There is no question that there is heteroscedasticity associated with the use of
annual town electricity consumption as the dependent variable. The reason for this is
that by definition, for the ith town we have we have Qi =
1 Ni
 qij . Hence even if the
Ni j=1
electricity consumption of individual households in all towns has a common variance
V(qij)=σ2, the variance for Qi is nevertheless:
 1 Ni 
1 Ni
1
σ2
2
given that there is zero
V  Qi   V 
 qij   2  V qij  2 Nσ 
Ni
Ni
 Ni j=1  Ni j=1
 
covariance of the error terms by assumption. Note that the variance is different
across town given that the number of households per town is expected to be different.
This results implies that our CRM parameter estimates are unbiased but the standard
errors are incorrectly calculated. This also carries over to the estimates of the
variance of the standard error of the above elasticity estimate
(c) Given the above error structure we need to use a GLS estimator to obtain parameter
estimates. The weighted least squares regression would look like this:
-1
Qi Ni  0 Ni  1 Ni INCi   2 Ni PElec,i
 3 Ni PNG,i   4 Ni E_EQPi +
That is we would have
 N1
0

N2
 0
P

 0

 1
0
N
1

1

0

N2



 0


 i*  Ni εi ~ 0,σ2

 N1

0
0 
-1


where
P
P=






0
N 42 

 1
0 
0
N
1


1


0 
2 0
N2
 and   σ 




1 
 0


N 42 
0
0 
0 
or


N 42 

0 


0 



1 
N 42 

2. (a) We know that by definition that: R 2 =1 TSS 
0
N2
σ 2  T-1
σ 2u
 TSS= u 2
TSS  T-1
1-R
11.799(99)
 3583.132 .
0.326
Using the above definition of the TSS, we have for Model II
σ 2u 
0.202(3583.132)
 7.311 . For model III we have:
99
R2  1
6.043
 0.8330
3583.132 99
Note, for each model we have:
SSEI = 11.799*98 = 1156.302
SSEII = 7.311*97 = 709.167
SSEIII = 6.043*97 = 586.171
SSEIV = 6.095*96 = 585.12
(b) H0: Model I H1: Model IV
F
SSE I -SSE IV
σ2u,IV (2)
 46.857
Ni εi
(c) H0: Model I H1: Model II
F
SSE I -SSE II
F
SSE II -SSEIV
F
SSE I -SSE III
F
TSS-SSEIV
σ2u,II (1)
 61.59 (or use T-ratio in
Model II)
(d) H0: Model II H1: Model IV
σ2u,IV (1)
 20.352 (or use T-ratio
in Model IV)
(e) H0: Model I H1: Model III
σ 2u,III (1)
 94.35 (or use T-ratio in
Model III)
(f) H0: Model 0
H1: Model IV
σ2u,IV (3)
 163.960
3. (a) The sample likelihood function can be represented as:
T
1
l
t 1 

  
 1e  yt 
y
 t
1
    T 
T
 K 
T  
e
e
 Tt 1 yt   T y 1
t 1
t
 Tt 1 yt   T y 1
t 1
t
The total sample log-likelihood function can be represented as:
T
 yt
L  ln K  T  ln   t 1

T
    1 ln yt
t 1
The 1st and 2nd derivatives are:
T
 yt
L
T t 1

 2



T 
 y 2
 L T   t 1 t 


 2  2
4
2
Equating the FOC to 0 and solving yields the following solution:
T
 yt
y
 *  t 1 
T

Second-order conditions are satisfied at the above value for the unknown parameter.
(b) Taking expectations of the above optimal parameter value:
 
E *
T 
  yt  T 
 E  t 1  
   unbiased
 T  T


 
Var  *
(c)
T 
  yt 
1 T
 Var  t 1  
 Var  yt 
2
 T   T   t 1



1
T  
T  2 
2
2
T
The above implies the estimator is consistent since the estimator is unbiased and the
variances goes to zero as T →∞
4. (a)
The easiest way to proceed is to substitute the true, unknown relationship
for each observation: y1=β0+β1X1+e1, yn=β0+β1Xn+en
1 
0  1 X n  en  0  1 X1  e1
X n  X1
 1 

1  X n  X1   en  e1
X n  X1
en  e1
X n  X1
We then can take the expected value of both sides of the above:
 E  en   E  e1  

 e e 
e e 
E   E  1  n 1   1  E  n 1   1  
  1
X n  X1 

 X n  X1 
 X n  X1 
 
because E(ei)=0. Therefore this estimator is unbiased.

e  e  Var  en   Var  e1 
(b) Var 1  Var  1  n 1  
given that the errors are the
X n  X1 

 X n  X1  2
 
only random variables, the denominator is a constant, and we have homoescedastic
non-autocorrelated errors. This implies that
 
Var 1 
2 2
 X n  X1  2
(c) Given the above expression for the variance, the only way the above parameter
estimator generates a consistent parameter estimate is if the (Xn-X1)2 approaches
infinity as the sample size gets larger. In general, this estimator will not be consistent
since we are usually not able to find arbitrarily large or small values of X.
5. I was looking for the fact that the FOC for minimization of the SSE with respect to
the intercept terms generates the results in the non-stochastic component of the
equation shown in this question. Refer to equation (5.3.7) and (5.3.8) of JHGLL on
page 165. Sure, if you plug in the means you get an estimate of the intercept but the
question is whether this is the CRM estimate.