Introduction to Econometrics
Chapter 6
Ezequiel Uriel Jiménez
University of Valencia
Valencia, September 2013
6 Relaxing the assumptions in the linear
classical model
6.1 Relaxing the assumptions in the linear classical model: an
overview
6.2 Misspecification
6.3 Multicollinearity
6.4 Normality test
6.5 Heteroskedasticity
6.6 Autocorrelation
Exercises
Appendix
6 Relaxing the assumptions in the linear classical
model
6.2 Misspecification
[3]
TABLE 6.1. Summary of bias in  when x2 is omitted in estimating equation.
2
Corr(x 2 ,x 3 )>0
Corr(x 2 ,x 3 )<0
 3 >0
Positive bias
Negative bias
 3 <0
Negative bias
Positive bias
6 Relaxing the assumptions in the linear classical
model
6.2 Misspecification
[4]
EXAMPLE 6.1 Misspecification in a model for determination of wages
(file wage06sp)
Initial model
wage = b1 + b2 educ + b3tenure + u
 = 4.679 + 0.681 educ + 0.293 tenure
wage
i
i
i
(1.55)
(0.146)
(0.071)
2
Rinit
= 0.249 n = 150
Augmented model
 + a wage
 +u
wage = b1 + b2 educ + b3tenure + a1 wage
1
2
2
Raugm
= 0.289
F
2
2
( Raugm
 Rinit
)/r
2
(1  Raugm
) / ( n  h)
 4.18
3
6 Relaxing the assumptions in the linear classical
model
6.3 Multicollinearity
[5]
EXAMPLE 6.2 Analyzing multicollinearity in the case of labor absenteeism
(file absent)
TABLE 6.2. Tolerance and VIF.
age
tenure
wage
Collinearity statistics
Tolerance
VIF
0.2346
42.634
0.2104
47.532
0.7891
12.673
6 Relaxing the assumptions in the linear classical
model
6.3 Multicollinearity
[6]
EXAMPLE 6.3 Analyzing the multicollinearity of factors determining time
devoted to housework (file timuse03)
houswork  1   2 educ   3 hhinc   4 age   5 paidwork  u

max
542.14

 8782
min
7.06 E  06
TABLE 6.3. Eigenvalues and variance decomposition proportions.
Eigenvalues
7.03E-06
0.000498
0.025701
1.861396
542.1400
Variance decomposition proportions
Variable
1
C
EDUC
HHINC
AGE
PAIDWORK
0.999995
0.295742
0.064857
0.651909
0.015405
Associated Eigenvalue
2
3
4
4.72E-06
0.704216
0.385022
0.084285
0.031823
8.36E-09
4.22E-05
0.209016
0.263805
0.007178
1.23E-13
2.32E-09
0.100193
5.85E-07
0.945516
5
1.90E-15
3.72E-11
0.240913
1.86E-08
7.80E-05
6 Relaxing the assumptions in the linear classical
model
6.4 Normality test
[7]
EXAMPLE 6.4 Is the hypothesis of normality acceptable in the model to
analyze the efficiency of the Madrid Stock Exchange?
(file bolmadef)
n=247
TABLE 6.4. Normality test in the model on the Madrid Stock Exchange.
skewness
coefficient
kurtosis coefficient
Bera and Jarque
statistic
-0.0421
4.4268
21.0232
y
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x
FIGURE 6.1. Scatter diagram
corresponding to a model with
homoskedastic disturbances.
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[8]
y

6 Relaxing the assumptions in the linear classical
model
6.5 Heteroskedasticity




x
FIGURE 6.2. Scatter diagram
corresponding to a model with
heteroskedastic disturbances.
6.5 Heteroskedasticity
6 Relaxing the assumptions in the linear classical
model
EXAMPLE 6.5 Application of the Breusch-Pagan-Godfrey test
[9]
TABLE 6.5. Hostel and inc data.
i
1
2
3
4
5
6
7
8
9
10
Step 1. Applying OLS to the model,
hostel
17
24
7
17
31
3
8
42
30
9
inc
500
700
250
430
810
200
300
760
650
320
hostel  b1 + b2 inc + u
using data from table 6.5, the following estimated model is obtained:
 = -7.427+ 0.0533 inc
hostel
i
i
(3.48)
(0.0065)
The residuals corresponding to this fitted model appear in table 6.6.
6.5 Heteroskedasticity
6 Relaxing the assumptions in the linear classical
model
EXAMPLE 6.5 Application of the Breusch-Pagan-Godfrey test. (Cont.)
[10]
TABLE 6.6. Residuals of the regression of hostel on inc.
i
1
2
3
4
5
6
7
8
9
10
uˆi
-2.226
-5.888
1.1
1.505
-4.751
-0.234
-0.565
8.913
2.777
-0.631
Step 2. The auxiliary regression
uˆi2  1   2 inci  i
uˆi2  23.93  0.0799inc
R 2  0.5045
Step 3. The BPG statistics is:
BPG = nRar2 = 10 (0.56) = 5.05
2(0.05)
Step 4. Given that 1
=3.84, the null hypothesis of homoskedasticity is
rejected for a significance level of 5%, but not for the significance level of 1%.
6.5 Heteroskedasticity
6 Relaxing the assumptions in the linear classical
model
EXAMPLE 6.6 Application of the White test
[11]
Step 1. This step is the same as in the Breusch-Pagan-Godfrey test.
Step 2. The regressors of the auxiliary regression will be
 1i  1 i
 2i  1 inci
 3i  inci2
uˆi2  1   2 inci   3inci2  i
uˆi2  14.29  0.10inci  0.00018inci2
R 2  0.56
Step 3. The W statistic:
W = nR 2 = 10 (0.56) = 5.60
Step 4. Given that  22(0.10) =4.61, the null hypothesis of homoskedasticity is
rejected for a 10% significance level because W=nR2>4.61, but not for
significance levels of 5% and 1%.
6.5 Heteroskedasticity
[12]
Heteroskedasticity in the linear model
  29.42+ 1.219 bookval
marktval
marktval = b1 + b2 bookval + u
(30.85)
(0.127)
n = 20
400
350
Residuals in absolute value
6 Relaxing the assumptions in the linear classical
model
EXAMPLE 6.7 Heteroskedasticity tests in models explaining the market value of the
Spanish banks (file bolmad95)
300
250
200
150
100
50
0
0
100
200
300
400
500
600
700
bookval
GRAPHIC 6.1. Scatter plot between the residuals in absolute value and the variable bookval in the linear model.
BPG  nRar2  20  0.5220  10.44
As
12(0.01)
=6.64<10.44, the null hypothesis of homoskedasticity is rejected
for a significance level of 1%, and therefore for =0.05 and for  =0.10.l
W  nRar2  20  0.6017  12.03
As
 22(0.01) =9.21<12.03, the null hypothesis of homoskedasticity is rejected for
a significance level of 1%.
[13]
EXAMPLE 6.7 Heteroskedasticity tests in models explaining the market value of the
Spanish banks (Cont.)
Heteroskedasticity in the log-log model

ln(
marktval )  0.676+ 0.9384 ln(bookval )
(0.265)
(0.062)
1.0
0.9
Residuals in absolute value
6 Relaxing the assumptions in the linear classical
model
6.5 Heteroskedasticity
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
6.0
6.5
7.0
ln(bookval)
GRAPHIC 6.2. Scatter plot between the residuals in absolute value and the variable bookval in the
log-log model.
TABLE 6.7. Tests of heteroskedasticity on the log-log model to explain the market value of Spanish
banks.
Test
Statistic
Table values
Breusch-Pagan
2
BP = nRra
=1.05
 22(0.10) =4.61
White
W=
nRra2
=2.64
 22(0.10) =4.61
6.5 Heteroskedasticity
[14]
ln(hostel)  b1 +b2ln(inc) +b3secstud +b4terstud +b5hhsize +u

ln(
hostel)i  -16.37+2.732ln(inc)i +1.398 secstudi +2.972terstudi -0.444 hhsizei
(2.26)
(0.324)
(0.258)
(0.088)
(0.333)
R2 = 0.921 n = 40
1.6
1.4
Residuals in absolute value
6 Relaxing the assumptions in the linear classical
model
EXAMPLE 6.8 Is there heteroskedasticity in demand of hostel services? (file hostel)
1.2
1
0.8
0.6
0.4
0.2
0
6.4
6.6
6.8
7
7.2
7.4
7.6
7.8
8
ln(inc)
GRAPHIC 6.3. Scatter plot between the residuals in absolute value and the variable ln(inc) in the hostel model.
TABLE 6.8. Tests of heteroskedasticity in the model of demand for hostel services.
Test
Breusch-PaganGodfrey
White
Statistic
2
BPG =nRra
=7.83
W=
nRra2
=12.24
Table values
 22(0.05)
=5.99
 22(0.01)
=9.21
6 Relaxing the assumptions in the linear classical
model
6.5 Heteroskedasticity
[15]
EXAMPLE 6.9 Heteroskedasticity consistent standard errors in the models
explaining the market value of Spanish banks (Continuation of example 6.7)
(file bolmad95)
Non consistent
  29.42+ 1.219 bookval
marktval
(30.85)
(0.127)

ln(
marktval )  0.676+ 0.9384 ln(bookval )
(0.265)
(0.062)
White procedure
  29.42+ 1.219 bookval
marktval
(18.67)
(0.249)

ln(
marktval )  0.676+ 0.9384 ln(bookval )
(0.3218)
(0.0698)
6 Relaxing the assumptions in the linear classical
model
6.5 Heteroskedasticity
[16]
EXAMPLE 6.10 Application of weighted least squares in the demand of hotel
services (Continuation of example 6.8) (file hostel)

uˆi  0.0239+ 0.0003 inc
R 2  0.1638

uˆi  -0.4198+ 0.0235 inc
R 2  0.1733
1

uˆi  0.8857- 532.1
(5.39)
(-2.87) inc

uˆ  -2.7033+ 0.4389 ln(inc)
R 2  0.1780
(0.143)
(2.73)
(-1.34)
i
(-2.46)
(2.82)
(2.88)
R 2  0.1788
WLS estimation

ln(
hostel )i  -16.21+ 2.709 ln(inc)i + 1.401 secstudi +2.982 terstud i - 0.445 hhsizei
(2.15)
(0.309)
(0.247)
R 2 = 0.914
(0.326)
n = 40
(0.085)
6 Relaxing the assumptions in the linear classical
model
6.6 Autocorrelation
[17]
3
u
2
1
00
time
1
2
3
4
5
6
7
8
9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29
-1
-2
-3
FIGURE 6.3. Plot of non-autocorrelated disturbances.
6 Relaxing the assumptions in the linear classical
model
6.6 Autocorrelation
[18]
4
5
u
u
4
3
3
2
2
1
1
00
time
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29
00
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29
-1
-1
-2
-2
-3
-3
-4
-4
-5
FIGURE 6.4. Plot of positive
autocorrelated disturbances.
FIGURE 6.5. Plot of negative
autocorrelated disturbances.
time
6 Relaxing the assumptions in the linear classical
model
6.6 Autocorrelation
[19]
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y
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x
FIGURE 6.6. Autocorrelated disturbances due to a specification bias.
6.6 Autocorrelation
6 Relaxing the assumptions in the linear classical
model
EXAMPLE 6.11 Autocorrelation in the model to determine the efficiency of the
Madrid Stock Exchange (file bolmadef)
[20]
dL=1.664;
dU=1.684
Since DW=2.04>dU, we do not reject the null hypothesis that the
disturbances are not autocorrelated for a significance level of  =0.01, i.e.
of 1%.
4
3
2
1
0
-1
-2
-3
-4
GRAPHIC 6.4. Standardized residuals in the estimation of the model to determine the
efficiency of the Madrid Stock Exchange.
6 Relaxing the assumptions in the linear classical
model
6.6 Autocorrelation
[21]
EXAMPLE 6.12 Autocorrelation in the model for the demand for fish
(file fishdem)
For n=28 and k'=3, and for a significance level of 1%:
dL=0.969;
dU=1.415
Since dL<1.202<dU, there is not enough evidence to accept the null
hypothesis, or to reject it.
3
2
1
0
-1
-2
2
4
6
8
10
12
14
16
18
20
22
24
26
28
GRAPHIC 6.5. Standardized residuals in the model on the demand for fish.
6 Relaxing the assumptions in the linear classical
model
6.6 Autocorrelation
[22]
EXAMPLE 6.13 Autocorrelation in the case of Lydia E. Pinkham
(file pinkham)
h  rˆ
é dù
é 1.2012 ù
n
n
53
ê
ú
ú
 1 ê1 
2


ê
ú
ê
ú
ˆ
ˆ
2
2
1
53
0.0814
´
1- n var b j
ë
û 1- n var b j
ë
û
( )
( )
Given this value of h, the null hypothesis of no autocorrelation is
rejected for =0.01 or, even, for =0.001, according to the table of
the normal distribution.
5,0
4,0
3,0
2,0
1,0
0,0
-1,0
-2,0
-3,0
-4,0
-5,0
8
13
18
23
28
33
38
43
48
53
58
GRAPHIC 6.6. Standardized residuals in the estimation of the model of the Lydia E.
Pinkham case.
6 Relaxing the assumptions in the linear classical
model
6.6 Autocorrelation
[23]
EXAMPLE 6.14 Autocorrelation in a model to explain the expenditures of
residents abroad (file qnatacsp)

ln(
turimpt ) = -17.31+ 2.0155ln( gdpt )
(3.43)
R 2 = 0.531
(0.276)
DW = 2.055
n = 49
2.5
2.0
1.5
1.0
0.5
0.0
-0.5
-1.0
-1.5
-2.0
5
10
15
20
25
30
35
40
45
GRAPHIC 6.7. Standardized residuals in the estimation of the model explaining the expenditures of
residents abroad.
2
For a AR(4) scheme, is equal to BG =nRar =36.35. Given this value of BG, the
null hypothesis of no autocorrelation is rejected for =0.01, since  52( ) =15.09.
6 Relaxing the assumptions in the linear classical
model
6.6.4 HAC standard errors
[24]
EXAMPLE 6.15 HAC standard errors in the case of Lydia E. Pinkham
(Continuation of example 6.13) (file pinkham)
TABLE 6.9.The t statistics, conventional and HAC, in the case of Lydia E. Pinkham.
regressor
intercept
advexp
sales (-1)
d1
d2
d3
t conventional
2.644007
3.928965
7.45915
-1.499025
3.225871
-3.019932
t HAC
1.779151
5.723763
6.9457
-1.502571
2.274312
-2.658912
ratio
1.49
0.69
1.07
1
1.42
1.14