Exercise 3:
Bivariate regression analysis and model evaluation
𝑌𝑡 = 𝛼 + 𝛽𝑋𝑡 + 𝑢𝑡
a) Reasonable assumptions:
 Zero conditional mean: 𝐸(𝑢𝑡 \𝑋𝑡 ) = 0
 Constant variance: 𝑣𝑎𝑟(𝑢𝑡 \𝑋𝑡 ) = 𝜎2
 Autocorrelation:𝑢𝑡 = 𝜌𝑢𝑡−1 + 𝜀𝑡

Normal distributed: 𝑢𝑡 ~ 𝑁(0, 𝜎 2 )
b)
EQ( 1) Modelling Y by OLS
The dataset is: M:\ECON 4160\data\Sp8101.xls
The estimation sample is: 1980(1) - 2000(1)
Constant
X
Coefficient
23.8431
0.574115
sigma
R^2
Adj.R^2
no. of observations
mean(Y)
AR 1-5 test:
ARCH 1-4 test:
Normality test:
Hetero test:
Hetero-X test:
RESET23 test:
26.6013
0.91284
0.911737
81
209.948
F(5,74)
F(4,73)
Chi^2(2)
F(2,78)
F(2,78)
F(2,77)
=
=
=
=
=
=
Std.Error
7.113
0.01996
t-value
3.35
28.8
t-prob Part.R^2
0.0012
0.1245
0.0000
0.9128
RSS
55902.4978
F(1,79) =
827.4 [0.000]**
log-likelihood
-379.679
no. of parameters
2
se(Y)
89.5389
1473.0
4905.3
6.8954
27.065
27.065
901.70
[0.0000]**
[0.0000]**
[0.0318]*
[0.0000]**
[0.0000]**
[0.0000]**
Interpreting the result:
 We have to perform a t-test to say something about the 𝛽.
t-test:
(𝑒𝑠𝑡𝑖𝑚𝑎𝑡𝑒 − ℎ𝑦𝑝𝑦𝑒𝑧𝑖𝑠𝑒𝑑 𝑣𝑎𝑙𝑢𝑒)
𝑡=
𝑠𝑡𝑎𝑛𝑑𝑒𝑟 𝑒𝑟𝑟𝑜𝑟
To conclude that 𝛽 is not 1, the t-value in absolute value has to be larger than a critical value.
(0,57 − 1)
= −21,54
0.019
From this result we can conclude that 𝛽 is significantly different from 1.
High t-value gives very low p-value, which means
𝑡=

The model’s standard error is very low, and also from this we can say that the model
have a big certain. This is thus also the reason for the big t-value in absolute terms.
The t-value can though indicate statistical significance either because the estimated
𝛽 is “large” or the standard error is “small”. Too much focus on statistical significance
can thus lead to the false conclusion that a variable is important for explaining Y even
though its estimated effect is modest.

𝑅 2 = 0.91, this means that the variation in X explains approximately 90 % of the
variation in Y. This is a very high number.

c) Is this model an adequate conditional model of 𝑌𝑡 given𝑋𝑡 ?
Even though, as we stated in b), the 𝑅 2is large and the 𝛽 are significantly different from 1,
the test fail. We have autocorrelation, heteroskedasticity and the model is not normally
distributed. The assumptions we made in a) is not valid. We can’t perform a the t-test
correctly. So we can’t say for sure that the 𝛽 is significant.
When we correct for autocorrelation and heteroskedasticity in the model, we get robust
standard errors.
Robust standard errors
Coefficients
JHCSE
Constant
23.843
6.4130
X
0.57411
0.023236
t-JHCSE
Constant
3.7180
X
24.708
SE
HACSE
HCSE
7.1132
11.795
6.2624
0.019959
0.042387
0.022755
Coefficients
t-SE
t-HACSE
t-HCSE
23.843
3.3520
2.0214
3.8074
0.57411
28.764
13.545
25.231
The result is quite similar as in b). We have same values on the 𝛽, which is still significantly
different from 1, the standard error is still very small and there is no heteroskedasticity and
autocorrelation. But: it is still not normally distributed
The importance of normal distribution when we have robust standard error is a matter of
preferences to the person who is making the model.
Some would say that the modell is still not valid, and some would think that this is an ok
model.
d) The criteria we use to evaluate this model:
 R-squared


T-test/significance level
The test: AR, ARCH, normality test, hetero test,
e) The inverted model:
𝛼
𝑋𝑡 = − 𝛽 +
1
𝑌
𝛽 𝑡
+ 𝑢𝑡
EQ( 2) Modelling X by OLS
The dataset is: M:\ECON 4160\data\Sp8101.xls
The estimation sample is: 1980(1) - 2000(1)
Coefficient
-9.65664
1.59000
Constant
Y
sigma
R^2
Adj.R^2
no. of observations
mean(X)
AR 1-5 test:
ARCH 1-4 test:
Normality test:
Hetero test:
Hetero-X test:
RESET23 test:
44.2691
0.91284
0.911737
81
324.16
F(5,74)
F(4,73)
Chi^2(2)
F(2,78)
F(2,78)
F(2,77)
Std.Error
12.60
0.05528
t-prob Part.R^2
0.4459
0.0074
0.0000
0.9128
RSS
154820.562
F(1,79) =
827.4 [0.000]**
log-likelihood
-420.935
no. of parameters
2
se(X)
149.008
=
=
=
=
=
=
Robust standard errors
Coefficients
JHCSE
Constant
-9.6566
4.8894
Y
1.5900
0.040972
t-value
-0.766
28.8
7880.4
8133.5
29.254
6.2583
6.2583
2.9361
[0.0000]**
[0.0000]**
[0.0000]**
[0.0030]**
[0.0030]**
[0.0590]
SE
HACSE
HCSE
12.605
9.3902
4.8329
0.055277
0.078550
0.040839
Coefficients
t-SE
t-HACSE
t-HCSE
-9.6566
-0.76612
-1.0284
-1.9981
1.5900
28.764
20.242
38.934
t-JHCSE
Constant
-1.9750
Y
38.807
If 𝛽 = 1, we have causality, which means that we don’t know if X is explained by Y or if Y is explained
by X.
Have to run a t-test: 𝑡 =
(1,59−1)
0,05528
= 10,77
The coefficient is significant different from 1
The standard error is very small and the R-squared is very high, 0.91.
The tests still fail: the is autocorrelation, heteroskedasticity and not normally distributed.
f)
Evaluation criteria:
Reasonable: the test still fail, the t-test haven’t changes, except the t-value of the constant is
no longer significant.
g)
𝑌𝑡 = 𝛼 + 𝛽0 𝑋𝑡 + 𝛽1 𝑋𝑡−1 + 𝛾1 𝑌𝑡−1 + 𝑢𝑡
EQ( 3) Modelling Y by OLS
The dataset is: M:\ECON 4160\data\Sp8101.xls
The estimation sample is: 1980(2) - 2000(1)
Y_1
Constant
X
X_1
Coefficient
1.05790
-0.611499
0.886688
-0.930794
sigma
R^2
Adj.R^2
no. of observations
mean(Y)
AR 1-5 test:
ARCH 1-4 test:
Normality test:
Hetero test:
Hetero-X test:
RESET23 test:
1.37508
0.999764
0.999754
80
212.206
F(5,71)
F(4,72)
Chi^2(2)
F(6,73)
F(9,70)
F(2,74)
=
=
=
=
=
=
Std.Error
0.006588
0.8666
0.08276
0.08366
t-value
161.
-0.706
10.7
-11.1
t-prob Part.R^2
0.0000
0.9971
0.4826
0.0065
0.0000
0.6016
0.0000
0.6196
RSS
143.704417
F(3,76) =
1.072e+005 [0.000]**
log-likelihood
-136.944
no. of parameters
4
se(Y)
87.7537
6.0258
1.9184
1.1521
1.3187
1.5844
21.630
[0.0001]**
[0.1166]
[0.5621]
[0.2598]
[0.1369]
[0.0000]**
R-squared adjusted is very high
𝛽1 is not significantly different from 1. Can not reject that this value I 1.
We see from the test that there is still autocorrelation. The other tests are ok. There is no longer
heteroskedasticity, and the normal distribution is valid.
i)
Strategy for evaluating the tree models against each other: model 2 explains more than
model 1 because the R-squared is bigger.
“simple-to-general”: you go from a model with few variables to many variables
“general-to-simple”: the opposite.
From (6) to (8), simple to general.
From (8) to (6) general to simple.
Exercise 4: Bivariate regressions with autocorrelated errors
Model 1:
𝑌𝑡 = 𝛼 + 𝛽𝑋𝑡 + 𝜀𝑡
EQ( 4) Modelling Y by OLS
The dataset is: M:\ECON 4160\data\Sp8101.xls
The estimation sample is: 1980(2) - 2000(1)
Constant
X
Coefficient
25.1914
0.570766
sigma
R^2
Adj.R^2
no. of observations
mean(Y)
AR 1-5 test:
ARCH 1-4 test:
Normality test:
Hetero test:
Hetero-X test:
RESET23 test:
26.6688
0.908811
0.907642
80
212.206
F(5,73)
F(4,72)
Chi^2(2)
F(2,77)
F(2,77)
F(2,76)
=
=
=
=
=
=
Std.Error
7.340
0.02047
t-value
3.43
27.9
t-prob Part.R^2
0.0010
0.1312
0.0000
0.9088
RSS
55475.3342
F(1,78) =
777.4 [0.000]**
log-likelihood
-375.182
no. of parameters
2
se(Y)
87.7537
1361.1
4733.9
6.9911
26.463
26.463
882.91
[0.0000]**
[0.0000]**
[0.0303]*
[0.0000]**
[0.0000]**
[0.0000]**
EQ( 5) Modelling Y by OLS
The dataset is: M:\ECON 4160\data\Sp8102.xls
The estimation sample is: 1980(1) - 2000(1)
Constant
X
Coefficient
-3.29461
0.987515
sigma
R^2
Adj.R^2
no. of observations
mean(Y)
AR 1-5 test:
ARCH 1-4 test:
Normality test:
Hetero test:
Hetero-X test:
RESET23 test:
2.49631
0.446567
0.439562
81
3.02321
F(5,74)
F(4,73)
Chi^2(2)
F(2,78)
F(2,78)
F(2,77)
=
=
=
=
=
=
Std.Error
0.8385
0.1237
t-value
-3.93
7.98
t-prob Part.R^2
0.0002
0.1635
0.0000
0.4466
RSS
492.294488
F(1,79) =
63.75 [0.000]**
log-likelihood
-188.021
no. of parameters
2
se(Y)
3.33453
41.391
24.357
16.133
1.6170
1.6170
1.3818
[0.0000]**
[0.0000]**
[0.0003]**
[0.2051]
[0.2051]
[0.2573]
EQ( 6) Modelling Y by OLS
The dataset is: M:\ECON 4160\data\Sp8103.xls
The estimation sample is: 1980(1) - 2000(1)
Constant
X
Coefficient
-0.0514192
0.00987416
Std.Error
0.09848
0.09443
t-value
-0.522
0.105
t-prob Part.R^2
0.6030
0.0034
0.9170
0.0001
sigma
R^2
Adj.R^2
0.885816
0.000138391
-0.0125181
RSS
61.9889045
F(1,79) =
0.01093 [0.917]
log-likelihood
-104.101
no. of observations
81
mean(Y)
-0.0510664
AR 1-5 test:
ARCH 1-4 test:
Normality test:
Hetero test:
Hetero-X test:
RESET23 test:


F(5,74)
F(4,73)
Chi^2(2)
F(2,78)
F(2,78)
F(2,77)
no. of parameters
se(Y)
2
0.880323
= 0.68673 [0.6350]
= 0.69229 [0.5997]
=
1.9979 [0.3683]
= 0.068104 [0.9342]
= 0.068104 [0.9342]
= 0.30157 [0.7405]
The first data-set is the same as in 3a)
Sp8102:
o

R-squared are about 0,44, so the variables explains less here than in the previous
data set.
o The standard error is a bit bigger, more variation in the observations
o the coefficient on X is not significantly different from 1.
o The test show that there is less probability for heteroskedasticity, but there is still
autocorrelation.
Sp8103:
o R-squared is zero and the coefficient is almost zero.
o The test are here positive though: no autocorrelation, no heteroskedastisity and
there is normality.
Model 2: 𝑌𝑡 = 𝛼′ + 𝛽 ′ 𝑋𝑡 + 𝛾′𝑋𝑡−1 + 𝛿′𝑌𝑡−1 + 𝜀𝑡
EQ( 7) Modelling Y by OLS
The dataset is: M:\ECON 4160\data\Sp8101.xls
The estimation sample is: 1980(1) - 2000(1)
Constant
X
Coefficient
23.8431
0.574115
sigma
R^2
Adj.R^2
no. of observations
mean(Y)
AR 1-5 test:
ARCH 1-4 test:
Normality test:
Hetero test:
Hetero-X test:
RESET23 test:
26.6013
0.91284
0.911737
81
209.948
F(5,74)
F(4,73)
Chi^2(2)
F(2,78)
F(2,78)
F(2,77)
EQ( 8) Modelling Y by OLS
=
=
=
=
=
=
Std.Error
7.113
0.01996
t-value
3.35
28.8
t-prob Part.R^2
0.0012
0.1245
0.0000
0.9128
RSS
55902.4978
F(1,79) =
827.4 [0.000]**
log-likelihood
-379.679
no. of parameters
2
se(Y)
89.5389
1473.0
4905.3
6.8954
27.065
27.065
901.70
[0.0000]**
[0.0000]**
[0.0318]*
[0.0000]**
[0.0000]**
[0.0000]**
The dataset is: M:\ECON 4160\data\Sp8102.xls
The estimation sample is: 1980(2) - 2000(1)
Y_1
Constant
X
X_1
Coefficient
0.910625
-0.698079
0.0476839
0.0943174
sigma
R^2
Adj.R^2
no. of observations
mean(Y)
AR 1-5 test:
ARCH 1-4 test:
Normality test:
Hetero test:
Hetero-X test:
RESET23 test:
0.85749
0.93712
0.934638
80
3.01193
F(5,71)
F(4,72)
Chi^2(2)
F(6,73)
F(9,70)
F(2,74)
=
=
=
=
=
=
Std.Error
0.04064
0.3294
0.09402
0.1007
t-value
22.4
-2.12
0.507
0.937
t-prob Part.R^2
0.0000
0.8685
0.0374
0.0558
0.6135
0.0034
0.3517
0.0114
RSS
55.8820059
F(3,76) =
377.5 [0.000]**
log-likelihood
-99.1637
no. of parameters
4
se(Y)
3.35402
0.33846
0.58851
1.3638
0.73998
1.0028
2.2272
[0.8880]
[0.6720]
[0.5057]
[0.6192]
[0.4462]
[0.1150]
EQ( 9) Modelling Y by OLS
The dataset is: M:\ECON 4160\data\Sp8103.xls
The estimation sample is: 1980(2) - 2000(1)
Y_1
Constant
X
X_1
Coefficient
0.0395480
-0.0732441
0.0208488
0.227191
Std.Error
0.1093
0.09597
0.09170
0.09151
t-value
0.362
-0.763
0.227
2.48
t-prob Part.R^2
0.7185
0.0017
0.4477
0.0076
0.8208
0.0007
0.0152
0.0750
sigma
0.856199
R^2
0.0769963
Adj.R^2
0.040562
no. of observations
80
mean(Y)
-0.0669558
RSS
55.7138403
F(3,76) =
2.113 [0.106]
log-likelihood
-99.0432
no. of parameters
4
se(Y)
0.87411
AR 1-5 test:
ARCH 1-4 test:
Normality test:
Hetero test:
Hetero-X test:
RESET23 test:
0.62517
0.70692
0.35271
0.68527
0.77278
0.99447
F(5,71)
F(4,72)
Chi^2(2)
F(6,73)
F(9,70)
F(2,74)
=
=
=
=
=
=
[0.6811]
[0.5898]
[0.8383]
[0.6620]
[0.6417]
[0.3748]
Sp8102:
 R-squared high
 The coefficients to the X’s is not significant
 Coefficient to the lagged Y is significant
 The tests are ok
Sp8103:
 R-squared low
 The coefficients to the X and lagged Y are not significant
 The coefficient to lagged X is significant
 The tests are ok
Model 3
RALS
EQ(10) Modelling Y by RALS
The dataset is: M:\ECON 4160\data\Sp8101.xls
The estimation sample is: 1980(2) - 2000(1)
Constant
X
Uhat_1
Coefficient
-6.92801
0.792874
1.05356
sigma
no. of observations
mean(Y)
1.37865
80
212.206
Std.Error
4.495
0.02379
0.005487
t-value
-1.54
33.3
192.
RSS
no. of parameters
se(Y)
t-prob Part.R^2
0.1274
0.0299
0.0000
0.9352
0.0000
0.9979
146.35158
3
87.7537
NLS using analytical derivatives (eps1=0.0001; eps2=0.005):
Strong convergence
Roots of error polynomial:
real
imag
modulus
1.0536
0.00000
1.0536
ARCH 1-4 test:
Normality test:
Hetero test:
Hetero-X test:
F(4,72)
Chi^2(2)
F(2,77)
F(2,77)
=
=
=
=
2.5384
1.5593
0.56747
0.56747
[0.0472]*
[0.4586]
[0.5693]
[0.5693]
EQ(11) Modelling Y by RALS
The dataset is: M:\ECON 4160\data\Sp8102.xls
The estimation sample is: 1980(2) - 2000(1)
Constant
X
Uhat_1
Coefficient
0.186602
0.00911751
0.976263
sigma
no. of observations
mean(Y)
0.881768
80
3.01193
Std.Error
5.575
0.09517
0.03010
t-value
0.0335
0.0958
32.4
RSS
no. of parameters
se(Y)
t-prob Part.R^2
0.9734
0.0000
0.9239
0.0001
0.0000
0.9318
59.8686801
3
3.35402
NLS using analytical derivatives (eps1=0.0001; eps2=0.005):
Strong convergence
Roots of error polynomial:
real
imag
modulus
0.97626
0.00000
0.97626
ARCH 1-4 test:
Normality test:
Hetero test:
Hetero-X test:
F(4,72)
Chi^2(2)
F(2,77)
F(2,77)
=
=
=
=
0.99624
1.0621
0.23177
0.23177
[0.4153]
[0.5880]
[0.7937]
[0.7937]
EQ(12) Modelling Y by RALS
The dataset is: M:\ECON 4160\data\Sp8103.xls
The estimation sample is: 1980(2) - 2000(1)
Constant
X
Coefficient
-0.0679270
-0.00438685
Std.Error
0.1038
0.09414
t-value
-0.654
-0.0466
t-prob Part.R^2
0.5150
0.0055
0.9630
0.0000
Uhat_1
0.0469083
sigma
0.884456
no. of observations
80
mean(Y)
-0.0669558
0.1129
0.416
RSS
no. of parameters
se(Y)
0.6789
0.0022
60.2342082
3
0.87411
NLS using analytical derivatives (eps1=0.0001; eps2=0.005):
Strong convergence
Roots of error polynomial:
real
imag
modulus
0.046908
0.00000
0.046908
ARCH 1-4 test:
Normality test:
Hetero test:
Hetero-X test:
F(4,72)
Chi^2(2)
F(2,77)
F(2,77)
=
=
=
=
0.61528
1.7531
0.12131
0.12131
[0.6530]
[0.4162]
[0.8859]
[0.8859]
Sp8101


The autocorrelation-coefficient is bigger than 1. This means that the model is explosive and
therefore unstable.
The tests are ok. Which is logical because Rals-estimation take into account the
autocorrelation.
Sp8102


The autocorrelation-coefficient is less than 1, which means that the model is stable. But since
its value is 0,97, the speed of convergence is low.
The tests satisfy the classical assumptions
Sp8103:

The autocorrelation-coefficient is very low, 0,0469, which reflects a high speed of
convergence. This means that the model is very stable.
Exercise 10)
a) Using the equations given in this exercise;
pt = aqt + bst + ut1
qt = cpt + edt + ut2
X1 = qt = quantum
X2 = pt = pris
X3 = st = supply
X4 = dt = demand
Equation 1 consists of the variable, st, which is not represented in the demand equation. On
the other hand, we see that the demand function also consists of a variable, dt, that is not
represented in the supply equation. This is the requirement of exact identification of both
equations in this system.
Ox Professional version 6.00 (Windows_64/U/MT) (C) J.A. Doornik, 19942009
---- PcGive 13.0 session started at 10:24:16 on
8-10-2010 ----
SYS( 1) Estimating the system by OLS (RF)
The dataset is: M:\ECON 4160\data\bfm101.xls
The estimation sample is: 1 - 400
URF equation for: X1
Coefficient
X3
0.376430
X4
0.146745
Constant
U
0.103629
sigma = 1.19601
t-value
9.11
3.44
1.72
t-prob
0.0000
0.0006
0.0856
t-value
-13.1
21.9
2.37
t-prob
0.0000
0.0000
0.0181
RSS = 567.8859679
URF equation for: X2
Coefficient
X3
-0.219346
X4
0.378187
Constant
U
0.0579151
sigma = 0.485192
Std.Error
0.04132
0.04260
0.06013
Std.Error
0.01676
0.01728
0.02440
RSS = 93.4582914
log-likelihood
-729.665772
|Omega|
0.13167411
R^2(LR)
0.862243
no. of observations
400
-T/2log|Omega|
405.485055
log|Y'Y/T|
-0.0451642009
R^2(LM)
0.49697
no. of parameters
6
F-test on regressors except unrestricted: F(4,792) = 335.467 [0.0000]
**
F-tests on retained regressors, F(2,396) =
X3
552.144 [0.000]**
X4
469.487
[0.000]**
Constant U
2.82915 [0.060]
correlation of URF residuals (standard deviations on diagonal)
X1
X2
X1
1.1960
0.77656
X2
0.77656
0.48519
correlation between actual and fitted
X1
X2
0.43589
0.79185
Single-equation diagnostics using reduced-form residuals:
X1
: Portmanteau(12): Chi^2(12) =
19.394 [0.0795]
X1
: AR 1-2 test:
F(2,395) =
3.0443 [0.0487]*
X1
: ARCH 1-1 test:
F(1,398) =
1.3745 [0.2417]
X1
: Normality test:
Chi^2(2) = 0.48019 [0.7866]
X1
X1
X2
X2
X2
X2
X2
X2
Vector
Vector
Vector
Vector
Vector
Vector
:
:
:
:
:
:
:
:
Hetero test:
Hetero-X test:
Portmanteau(12):
AR 1-2 test:
ARCH 1-1 test:
Normality test:
Hetero test:
Hetero-X test:
Portmanteau(12):
AR 1-2 test:
Normality test:
Hetero test:
Hetero-X test:
RESET23 test:
F(4,395)
F(5,394)
Chi^2(12)
F(2,395)
F(1,398)
Chi^2(2)
F(4,395)
F(5,394)
Chi^2(48) =
F(8,784) =
Chi^2(4) =
F(12,1040)=
F(15,1082)=
F(8,784) =
=
=
=
=
=
=
=
=
1.1420
1.1468
16.886
1.6114
0.11376
3.2973
1.1115
0.95466
47.614
1.0564
5.9089
0.80499
0.85315
0.82584
[0.3363]
[0.3350]
[0.1539]
[0.2009]
[0.7361]
[0.1923]
[0.3506]
[0.4455]
[0.4886]
[0.3919]
[0.2061]
[0.6456]
[0.6179]
[0.5799]
MOD( 2) Estimating the model by 1SLS (SF)
The dataset is: M:\ECON 4160\data\bfm101.xls
The estimation sample is: 1 - 400
Equation for: X1 (quantum)
Coefficient
X2
0.820108
X4
-0.180153
Constant
U
-0.00314324
Std.Error
0.1060
0.05974
0.06207
t-value
7.74
-3.02
-0.0506
t-prob
0.0000
0.0027
0.9596
Std.Error
0.02091
0.02287
0.02780
t-value
-17.8
16.6
-0.607
t-prob
0.0000
0.0000
0.5440
sigma = 1.22593
Equation for: X2 (price)
Coefficient
X3
-0.371772
X1
0.380681
Constant
U
-0.0168808
sigma = 0.553067
log-likelihood
-1126.41198
no. of observations
400
No restrictions imposed
-T/2log|Omega|
no. of parameters
8.7388452
6
correlation of structural residuals (standard deviations on diagonal)
X1
X2
X1
1.2259
0.00000
X2
0.00000
0.55307
Single-equation diagnostics using reduced-form residuals:
X1
: AR 1-2 test:
F(2,395) =
238.62 [0.0000]**
X1
: ARCH 1-1 test:
F(1,398) =
2.0339 [0.1546]
X1
: Normality test:
Chi^2(2) =
4.3366 [0.1144]
X1
: Hetero test:
F(4,395) =
26.184 [0.0000]**
X1
: Hetero-X test:
F(5,394) =
34.393 [0.0000]**
X2
: AR 1-2 test:
F(2,395) =
559.43 [0.0000]**
X2
: ARCH 1-1 test:
F(1,398) = 0.075181 [0.7841]
X2
: Normality test:
Chi^2(2) = 0.36136 [0.8347]
X2
: Hetero test:
F(4,395) =
40.440 [0.0000]**
X2
: Hetero-X test:
F(5,394) =
97.155 [0.0000]**
Vector Normality test:
Chi^2(4)
=
2.2586 [0.6883]
Vector Hetero test:
Vector Hetero-X test:
F(12,1040)=
F(15,1082)=
41.372 [0.0000]**
56.822 [0.0000]**
MOD( 3) Estimating the model by 2SLS
The dataset is: M:\ECON 4160\data\bfm101.xls
The estimation sample is: 1 - 400
Equation for: X1
X2
X4
Constant
U
Coefficient
-1.71615
0.795769
0.203020
Std.Error
0.3017
0.1347
0.09915
t-value
-5.69
5.91
2.05
t-prob
0.0000
0.0000
0.0413
Coefficient
-1.18947
2.57716
-0.209154
Std.Error
0.2634
0.6609
0.1482
t-value
-4.52
3.90
-1.41
t-prob
0.0000
0.0001
0.1590
sigma = 1.91585
Equation for: X2
X3
X1
Constant
U
sigma = 2.72275
log-likelihood
-729.665772
no. of observations
400
No restrictions imposed
-T/2log|Omega|
no. of parameters
405.485055
6
correlation of structural residuals (standard deviations on diagonal)
X1
X2
X1
1.9158
-0.92495
X2
-0.92495
2.7228
Single-equation diagnostics using reduced-form residuals:
X1
: AR 1-2 test:
F(2,395) =
3.0443 [0.0487]*
X1
: ARCH 1-1 test:
F(1,398) =
1.3745 [0.2417]
X1
: Normality test:
Chi^2(2) = 0.48019 [0.7866]
X1
: Hetero test:
F(4,395) =
1.1420 [0.3363]
X1
: Hetero-X test:
F(5,394) =
1.1468 [0.3350]
X2
: AR 1-2 test:
F(2,395) =
1.6114 [0.2009]
X2
: ARCH 1-1 test:
F(1,398) = 0.11376 [0.7361]
X2
: Normality test:
Chi^2(2) =
3.2973 [0.1923]
X2
: Hetero test:
F(4,395) =
1.1115 [0.3506]
X2
: Hetero-X test:
F(5,394) = 0.95466 [0.4455]
Vector Normality test:
Vector Hetero test:
Vector Hetero-X test:


Chi^2(4) =
F(12,1040)=
F(15,1082)=
5.9089 [0.2061]
0.80499 [0.6456]
0.85315 [0.6179]
Each equation in the system has an endogenous variable along with the other exogenous
variables on the RHS. This violates the classical assumption about zero conditional mean in
the disturbance term, since we condition on all the RHS variables [𝐸(𝑢𝑡|𝑅𝐻𝑆)]. OLS on the
system gives biased and inconsistent estimators.
OLS regression result show that the estimate of X2 is positive on the demand function. This
contradict the economic theory about that an increase in price should dampen the demand.
The other estimates correspond to economic theory.

2SLS, ILS and IV gives the same result, when it is exact identification. This can be shown by
theory.