Economics 233
2 Semester 2009-2010
Problem Set 2
nd
7.2
a) The assumptions are:
 SOLS.1:
 SOLS.2:

is nonsingular (has rank k).
, where
The model is given by:
Under SOLS.1:
The estimator of
is thus defined as:
We write:
Since
from SOLS.1,
.
From the given assumption,
.
The Central Limit Theorem implies that:
*
Also,
. But
.
**
*, **, and the asymptotic equivalent lemma imply:
Therefore, the asymptotic variance of
is:
b) To estimate the asymptotic variance in (a), we find a consistent estimator for
. Let
and
.
and
A consistent estimator of A is:
In the case of B, we should estimate Ω first since the error is not observable. Hence, we
replace with the SOLS residuals.
Since
is a consistent estimator of
consistent estimator of . Then:
, we have
. As such,
is a
With the Law of Large Numbers,
. Hence,
is a consistent estimator of B.
Therefore, we estimate the asymptotic variance in (a) with:
c) Add the following assumptions:
 SGLS.1:
 SGLS.2: Ω is positive definite and
which is the unconditional variance matrix of
 SGLS.3:
is nonsingular (where
.
, where
.
We note that:
We will apply the following property: For any A and B that are invertible,
positive semi-definite if and only if
is positive semi-definite.
As such, to show that
that
is
is positive semi-definite, we show
is positive semi-definite.
We simplify the notations on expectations and N.
This expression is positive semi-definite since the projection matrix reduces it to a norm.
d) Under assumptions SGLS1- SGLS3 and given that
FGLS have the same asymptotic variance.
=
, show that SOLS and
e) As shown in (d), SOLS and FGLS have the same asymptotic variance if
.
However, even with this assumption FGLS is still more efficient than SOLS under
assumptions SGLS1, SOLS2, SGLS2 and (
)= (
. The advantage of FGLS
over SOLS is the homoskedasticity assumption because without this assumption SGLS1
will be even harder to meet.
7.8
Seemingly unrelated regression
---------------------------------------------------------------------Equation
Obs Parms
RMSE
"R-sq"
chi2
P
---------------------------------------------------------------------hrearn
616
12
4.3089
0.2051
158.93
0.0000
hrvacdays
616
12
.1389899
0.3550
339.01
0.0000
hrsicklve
616
12
.056924
0.2695
227.23
0.0000
hrinsur
616
12
.1573797
0.3891
392.27
0.0000
hrpension
616
12
.2500388
0.3413
319.16
0.0000
--------------------------------------------------------------------------------------------------------------------------------------------------|
Coef.
Std. Err.
z
P>|z|
[95% Conf. Interval]
-------------+---------------------------------------------------------------hrearn
|
educ |
.4588139
.068393
6.71
0.000
.3247662
.5928617
exper | -.0758428
.0567371
-1.34
0.181
-.1870455
.0353598
expersq |
.0039945
.0011655
3.43
0.001
.0017102
.0062787
tenure |
.1100846
.0829207
1.33
0.184
-.052437
.2726062
tenuresq | -.0050706
.0032422
-1.56
0.118
-.0114252
.0012839
union |
.8079933
.4034789
2.00
0.045
.0171892
1.598797
south | -.4566222
.5458508
-0.84
0.403
-1.52647
.6132258
nrtheast | -1.150759
.5993283
-1.92
0.055
-2.32542
.0239032
nrthcen | -.6362663
.5501462
-1.16
0.247
-1.714533
.4420005
married |
.6423882
.4133664
1.55
0.120
-.167795
1.452571
white |
1.140891
.6054474
1.88
0.060
-.0457639
2.327546
male |
1.784702
.3937853
4.53
0.000
1.012897
2.556507
_cons | -2.632127
1.215291
-2.17
0.030
-5.014054
-.2501997
-------------+---------------------------------------------------------------hrvacdays
|
educ |
.0201829
.0022061
9.15
0.000
.015859
.0245068
exper |
.0066493
.0018301
3.63
0.000
.0030623
.0102363
expersq | -.0001492
.0000376
-3.97
0.000
-.0002229
-.0000755
tenure |
.012386
.0026747
4.63
0.000
.0071436
.0176284
tenuresq | -.0002155
.0001046
-2.06
0.039
-.0004205
-.0000106
union |
.0637464
.0130148
4.90
0.000
.0382378
.0892549
south | -.0179005
.0176072
-1.02
0.309
-.05241
.016609
nrtheast | -.0169824
.0193322
-0.88
0.380
-.0548728
.0209081
nrthcen |
.0002511
.0177458
0.01
0.989
-.03453
.0350321
married |
.0227586
.0133337
1.71
0.088
-.0033751
.0488923
white |
.0084869
.0195296
0.43
0.664
-.0297905
.0467642
male |
.0569525
.0127021
4.48
0.000
.0320568
.0818482
_cons | -.1842348
.039201
-4.70
0.000
-.2610674
-.1074022
-------------+---------------------------------------------------------------hrsicklve
|
educ |
.0096054
.0009035
10.63
0.000
.0078346
.0113763
exper |
.002145
.0007495
2.86
0.004
.0006759
.0036141
expersq | -.0000383
.0000154
-2.48
0.013
-.0000684
-8.08e-06
tenure |
.0050021
.0010954
4.57
0.000
.002855
.0071491
tenuresq | -.0001391
.0000428
-3.25
0.001
-.0002231
-.0000552
union | -.0046655
.0053303
-0.88
0.381
-.0151127
.0057816
south |
-.011942
.0072111
-1.66
0.098
-.0260755
.0021916
nrtheast | -.0026651
.0079176
-0.34
0.736
-.0181833
.0128531
nrthcen | -.0222014
.0072679
-3.05
0.002
-.0364462
-.0079567
married |
.0038338
.0054609
0.70
0.483
-.0068694
.014537
white |
.0038635
.0079984
0.48
0.629
-.0118132
.0195401
male |
.0042538
.0052022
0.82
0.414
-.0059423
.01445
_cons | -.0937606
.016055
-5.84
0.000
-.1252278
-.0622935
-------------+---------------------------------------------------------------hrinsur
|
educ |
.0080042
.002498
3.20
0.001
.0031082
.0129002
exper |
.0054052
.0020723
2.61
0.009
.0013436
.0094668
expersq | -.0001266
.0000426
-2.97
0.003
-.00021
-.0000431
tenure |
.0116978
.0030286
3.86
0.000
.0057618
.0176338
tenuresq | -.0002466
.0001184
-2.08
0.037
-.0004787
-.0000146
union |
.1441536
.0147368
9.78
0.000
.11527
.1730372
south |
.0196786
.0199368
0.99
0.324
-.0193969
.0587541
nrtheast | -.0052563
.0218901
-0.24
0.810
-.0481601
.0376474
nrthcen |
.0242515
.0200937
1.21
0.227
-.0151315
.0636345
married |
.0365441
.0150979
2.42
0.016
.0069527
.0661355
white |
.0378883
.0221136
1.71
0.087
-.0054535
.0812301
male |
.1120058
.0143827
7.79
0.000
.0838161
.1401955
_cons | -.1180824
.0443877
-2.66
0.008
-.2050807
-.0310841
-------------+---------------------------------------------------------------hrpension
|
educ |
.0390226
.0039687
9.83
0.000
.031244
.0468012
exper |
.0083791
.0032924
2.55
0.011
.0019262
.0148321
expersq | -.0001595
.0000676
-2.36
0.018
-.0002921
-.000027
tenure |
.0243758
.0048118
5.07
0.000
.0149449
.0338067
tenuresq | -.0005597
.0001881
-2.97
0.003
-.0009284
-.0001909
union |
.1621404
.0234133
6.93
0.000
.1162513
.2080296
south | -.0130816
.0316749
-0.41
0.680
-.0751632
.049
nrtheast | -.0323117
.0347781
-0.93
0.353
-.1004755
.0358521
nrthcen | -.0408177
.0319241
-1.28
0.201
-.1033878
.0217525
married | -.0051755
.023987
-0.22
0.829
-.0521892
.0418381
white |
.0395839
.0351332
1.13
0.260
-.0292758
.1084437
male |
.0952459
.0228508
4.17
0.000
.0504592
.1400325
_cons | -.4928338
.0705215
-6.99
0.000
-.6310534
-.3546143
------------------------------------------------------------------------------
Marital status appears to have no effect on hourly earnings, hourly value of sick leave, and
hourly value of pension. On the other hand, marital status appears to affect hourly value of
vacation days and hourly value of employer-provided insurance.
To test whether another year of education increases expected pension value and expected
insurance by the same amount:
test [hrinsur]educ = [hrpension]educ
( 1)
[hrinsur]educ - [hrpension]educ = 0
chi2( 1) =
Prob > chi2 =
102.24
0.0000
Since the p-value is very small, we reject the null hypothesis that another year of education
increases expected pension value and expected insurance by the same amount.
7.12
If we add wealth at the beginning of year t to the saving equation, the strict exogeneity
assumption would not likely hold. Wealth at time t is likely to be affected by wealth in previous
periods, because we can think of present wealth as an accumulation of wealth acquired in
previous periods. Since wealth in previous periods is not included in the savings equation at time
t, the error term at time t is correlated with the independent variables (specifically with the
wealth at time s ≠ t) of the saving equation at time s ≠ t.
Likewise, it is also reasonable to assume that wealth at time t can be affected by savings in
previous periods, which would make the error term at time t to be correlated with wealth in other
time periods. Hence, the strict exogeneity assumption is violated.
8.4 Consider the system of equations (8.12), and let z be a row vector of variables exogenous in
every equation. Assume that the exogeneity assumption takes the stronger form E  u g | z   0 ,
g  0,1,..., G . This assumption means that z and nonlinear functions of z are valid instruments
in every equation.
a) Suppose that E  xg | z  is linear in z for all g. Show that adding nonlinear functions of z to
the instrument list cannot help in satisfying the rank condition. (Hint: Apply Problem 8.3)
Let Z   z z ... z  and X   x1 x2 ... xG  where Z is a G X M matrix, X is a G X K
matrix, z is a 1 X M vector, and xg is a 1 X Kg vector, g  1,2,..., G . Note that
K  K1  K 2  ...  K G .
z
z
 x

Z
X

Then,
  1
 
z
x2
 zx1
 zx
... xG    1
 zx1

 zx1
zx2
zx2
zx2
zx2
... zxG 
... zxG 
.
... zxG 

... zxG 
All rows of the block matrix are similar. Thus, if we assume that M  K and
rank E  z xg   K g , and xg   x f for some   , f , g  1,2,..., G , the rank E  Z X  is equal
to K  K1  K 2  ...  KG .
Suppose that we add nonlinear functions of z. Let w   z h  z  and W be the matrix that
contains the vector, w in each row.
From previous problem, 8.3, we can have rank E  zxg   rank E  wxg  , g  1,2,..., G .
Then, rank E W X   rank E  Z X  .
Therefore, adding nonlinear functions of z to the instrument list does not necessarily affect
the rank of E  Z X  . This implies that it also does not necessarily help in satisfying the rank
condition.
b) What happens if E  xg | z  is a nonlinear function of z for some g?
Consider the following case where for only one x say, x1 , its conditional expectation
given z is a nonlinear function of z; i.e., E  xg | z   h  z   .
h  z 


z 
Then, Z X  
x

 1


 z 
x2
 h  z  x1

0
... xG   
 0

 0
0
zx2
zx2
zx2
... 0 

... zxG 
.
... zxG 

... zxG 
All rows of the block matrix are similar except for the first row.
Again, if we assume that M  K and rank E  zxg   K g , and xg   x f for some



, f , g  2,..., G , the rank E  Z X  is equal to K 2  ...  KG plus rank E h  z  x1 .
When two of the x ’s, say have E  x | z  that is nonlinear function of z, then we have




rank E  Z X   K3  ...  KG  rank E h  z  x1  rank E h  z  x2 .
Increasing the number of xg ’s up to G, in which E  xg | z  is a nonlinear function of z
G


with the same function h, we have rank E  Z X    rank E h  z  xg .
g 1
If the nonlinear function of z are distinct for every g, then as the number of xg ’s increases
G


to G, then the rank condition is satisfied if rank E  Z X    rank E hg  z  xg  K .
g 1
Here, nonlinearity of E  xg | z  with respect to z may have an effect in satisfying the rank
condition.
8.6 Consider the system (8.12) in the G = 2 case, with an i subscript added:
yi1  xi11  ui1
yi 2  xi1 2  ui 2
z
0
The instrument matrix is Zi   i1
.
 0 zi 2 
 11  12 
Let  be the 2x2 variance matrix of ui  ui1 ui 2  and write 1   12
.
 22 

a) Find E  Zi1ui  and show that it is not necessarily zero under orthogonality conditions
E  zi1ui1   0 and E  zi2ui 2   0 .
0   11  12   ui1 


zi 2   12  22  ui 2 
  11 zi1  12 zi1   ui1 
  12
 
22
 zi 2  zi 2  ui 2 
z
Zi1ui   i1
0
  11 zi1ui1   12 zi1ui 2 
  12

22
 zi 2ui1   zi 2ui 2 
Then taking the expectation:
  11 z u   12 zi1ui 2 
E  Zi1ui   E  12 i1 i1

22
 zi 2ui1   zi 2ui 2 
 E  11 zi1ui1   12 zi1ui 2  


 E  12 zi 2ui1   22 zi 2ui 2 


  11 E  zi1ui1    12 E  zi1ui 2  
  12
.
22
 E  zi 2ui1    E  zi 2ui 2  
 12 E  zi1ui 2  
By the orthogonality conditions, we have E  Z i 1ui    12
.
 E  zi 2ui1  
Since E  zi1ui 2  and E  zi2ui1  are not necessarily zero, then E  Zi1ui  is not necessarily
zero.
b) What happens if  is diagonal (so that 1 is diagonal)?
 * 11
0 
.
 22 
Let * 1  
*
 0
 zi1
0
Then, Zi1ui  
0   * 11

zi 2   0
0   ui1   * 11 zi1ui1 
.

* 22  
  ui 2   * 22 zi 2ui 2 
Taking the expectation:
 * 11 zi1ui1 
1

E  Zi  ui   E  * 22

  zi 2ui 2 
 * 11 E  zi1ui1  
  * 22

  E  zi 2ui 2  
0 
 .
0 
Thus, when  is a diagonal, E  Zi1ui  is zero.
c) What if zi1  zi 2 (without restrictions on  )?
 12 E  zi1ui 2  
From (a) we have: E  Z i 1ui    12
.
 E  zi 2ui1  
 12 E  z u  
Since zi1  zi 2 , then E  Z i1ui    12 i 2 i 2  .
  E  zi1ui1  
0 
But with the orthongonality conditions, we get E  Z i 1ui     .
0
Thus, if zi1  zi 2 , E  Zi ui  is zero.
1
9.2 Write a two-equation system in the form
(i) y1   1 y2  z1 1  u1
(ii) y2   2 y1  z 2  2  u2
a) Show that reduced forms exist if and only if  1 2  1 .
Substitute (ii) into (i) and derive the reduce form of y1 :
 


y1   1  2 y1  z 2 2  u2  z11  u1
 y1   1 2 y1  z 2 1  2   1u2  z1 1  u1
 1   1 2  y1  z11  z 2 1  2  u1   1u2
(i’)  y1  z1 1 1   1 2   z 2  1 2 1   1 2    u1   1u2  1   1 2  .
Substitute (i’) into (ii) to get the reduce form of y2 . Then, we get
(ii’) y2  z1  11 1   1 2   z 2  2 1   1 2    2u1  u2  1   1 2  .
Since the vectors of coefficients of the exogenous variables and the error terms are
multiplied by the inverse of 1   1 2 , then for the reduce forms of y1 and y2 to exist,  1 2
should be not equal to 1.
b) State in words the rank condition for identifying each equation.
For the rank condition to be satisfied, the order condition should also be satisfied.
Assuming that the linear two-equation system above has exclusion restrictions, under
Theorem 9.1 (Order Condition with Exclusion Restrictions), each equation will be
identified if the number of excluded exogenous variables from the equation must be as
large as the number of included right-hand-side endogenous variables in the equation.
Since each equation in the given system has one endogenous variable, then the number of
excluded exogenous variables should be at least one for the order condition to be
satisfied. Thus, for the rank condition should be satisfied, there should be exactly one
excluded exogenous variable in each equation.
9.4 Given:
(1) y1 = γ12y2 + δ11z1 + δ12z2 + δ13z3 + u1
(2) y1 = γ22y2 + γ23y3 +δ21z1 + u2
(3) y3 = δ31z1 + δ32z2 + δ33z3 + u3
a. Show that a well-defined reduced form exists as long as γ12 ≠ γ22.
Suppose that γ12 ≠ γ22.
Subtracting the first equation from the second and solving for y2,
0 = (γ12 − γ22)y2 − γ23 y3 + (δ11 − δ21)z1 + δ12z2 + δ13z3 + (u1 − u2)
(γ22 − γ12)y2 = − γ23y3 + (δ11 − δ21)z1 + δ12z2 + δ13z3 + (u1 − u2)
Substitute for y3 from the third equation
(γ22 − γ12)y2 = − γ23(δ31z1 + δ32z2 + δ33z3 + u3) + (δ11 − δ21)z1 + δ12z2 + δ13z3 + (u1 − u2)
(γ22 − γ12)y2 = (δ11 − δ21 − γ23δ31)z1 − γ23δ32z2 + δ12z2 + δ13z3 − γ23δ33z3 + (u1 − u2 − γ23u3)
Finally, divide by (γ22 − γ12) to get y2
y2 = π1z1 + π2z2 + π3z3 + v2,
11   21   23 31
  
,  2  12 23 32 , and
 22   12
 22   12
  
u u  u
 3  13 23 33 , while v2  1 2 23 3 .
 22   12
 22   12
the reduced form of y2, where  1 
b. Identify which of the three equations is identified.
Note that in this system, G = 3, M = 3. We then have
1   1  12  13 11 12 13 
 2   1  22  23  21  22  23 
3   1  32  33 31 32 33  and
 1

  12

B   13
 11
 12

 13
1  31 
 22  32 
 23 1 
.
 21  31 
 22  32 

 23  33 
For the first equation, γ13 = 0, and therefore J1 = 1. Since G – 1 = 2, J1 < G – 1. Equation
(1) does not satisfy the Order condition which is necessary for identification; therefore,
(1) is not identified.
For the second equation, δ22 = 0 and δ23 = 0. Thus, J2 = 2 = G – 1. We must then check
the rank condition sufficient to determine if (2) is identified. Using the exclusion
restrictions for (2), we have
0 0 0 0 1 0
R2  
.
0 0 0 0 0 1
The rank condition requires that
rankR2B = G – 1.
 1

  12
 0 0 0 0 1 0    13
R 2B  

 0 0 0 0 0 1   11
 12

 13
1  31 
 22  32 
 23 1   12  22  32 
.

 21  31   13  23  33 
 22  32 

 23  33 
Imposing all the system restrictions to R2B, namely δ22 = 0, δ23 = 0, and γ13 = 0, we get
0  32 

rankR 2B  rank  12
  2 . Equation (2) is therefore identified because rankR2B

0

13
33


= G – 1.
There is no need to determine if (3) is identified because it is already in reduced form.
9.6. Given:
(1) y1 = δ10 + γ12y2 + γ13y2z1 + δ11z1 + δ12z2 + u1
(2) y2 = δ20 + γ21y1 + δ21z1 + δ23z3 + u2
a) Initially assume γ13 = 0. Discuss identification of each equation in this case.
When γ13 = 0,
y1 = δ10 + γ12y2 + γ13y2z1 + δ11z1 + δ12z2 + u1
(1′)
Hence G = 2 and M = 3. We also have
1   1  12 10 11 12 13 
 2    21 1  20  21  22  23 
 1

  12
 10
B
 11
 12

 13
 21 

1 
 20 

 21 
 22 

 23 
Since G – 1 = 1, the order condition for (1′) is satisfied because there is only one
exclusion restriction, δ13 = 0. Thus, we must check the rank condition to determine
identification of (1′).
We have R1   0 0 0 0 0 1 and hence, R1B  13  23  .
Imposing all exclusion restrictions, we get rankR1B  rank  0  23   1  G 1 . The first
equation is identified.
The exclusion restriction for the second equation is δ22 = 0, i.e., J2 = 1. Equation (2) also
satisfies the order condition. To test for the rank condition, we have
R2   0 0 0 0 1 0  and R2B  12  22  .
Imposing all system restrictions results in rankR2B  rank 12
(1) is also identified.
b) For any value of γ13, find the reduced form for y1.
To get the reduced form for y1, plug (2) into (1):
Where
c) Assuming
we get,
0  1  G 1 . Equation
d) If  13  0 , then the first equation is identified.
If  13  0 , then there are effectively three endogenous variables, y1 , y2 , y2 z1 , hence G = 3.
But since only z3 is excluded from (1), we have J = 1 < 2 = G – 1 and (1) will not be
identified. However, if we had more information on the other parameters, such as crossequation restrictions, we could use such information to identify the equation.
e) Suggest a 2SLS procedure for estimating the first equation.
Stage 1: Regress
to get
Stage 2: Regress
to get
f) Define a matrix of instruments suitable for 3SLS estimation
Let
using 2SLS (see item e)
For 1st equation,
For 2nd equation,
Where
is from 2SLS for 2nd equation
Stage 1: Regress
to get
Stage 2: Regress
to get
Where
and
The matrix of instruments is
g) The parameters in the 1st equation will be unidentifiable if
using the order
st
condition, which means the 1 equation cannot be consistently estimated. BUT if we
have additional information on the parameters (e.g. using cross equation restrictions or
covariance restrictions), then we can achieve identification of 1st equation and can now
use estimation methods to consistently estimate the 1st equation.
Can
be tested?
Once we have estimated the parameters of the 1st equation, we can test the significance of
using standard or robust t-tests.
9.8
a)
Instrumental variables (2SLS) regression
Number of obs
Wald chi2(16)
Prob > chi2
R-squared
Root MSE
=
=
=
=
=
3010
810.17
0.0000
0.2716
.37869
-----------------------------------------------------------------------------lwage |
Coef.
Std. Err.
z
P>|z|
[95% Conf. Interval]
-------------+---------------------------------------------------------------educ |
.2620171
.1663848
1.57
0.115
-.0640911
.5881253
educsq | -.0062667
.0065139
-0.96
0.336
-.0190336
.0065003
exper |
.0676065
.0342299
1.98
0.048
.0005172
.1346959
expersq | -.0008376
.0015668
-0.53
0.593
-.0039085
.0022333
south | -.1441741
.0265757
-5.43
0.000
-.1962615
-.0920868
black | -.1764539
.0351076
-5.03
0.000
-.2452635
-.1076444
smsa |
.1261733
.0248384
5.08
0.000
.0774909
.1748558
reg668 | -.1811498
.0480325
-3.77
0.000
-.2752918
-.0870078
reg667 |
.0134968
.0428959
0.31
0.753
-.0705775
.0975711
reg666 |
.0413164
.0448594
0.92
0.357
-.0466063
.1292391
reg665 |
.0254031
.0401821
0.63
0.527
-.0533524
.1041585
reg664 | -.0678702
.0372212
-1.82
0.068
-.1408224
.005082
reg663 |
.0316705
.0290702
1.09
0.276
-.025306
.0886469
reg662 | -.0085347
.030534
-0.28
0.780
-.0683804
.0513109
reg661 | -.1179468
.0403077
-2.93
0.003
-.1969485
-.0389451
smsa66 |
.0153976
.0214269
0.72
0.472
-.0265982
.0573935
_cons |
3.414059
1.004698
3.40
0.001
1.444886
5.383232
-----------------------------------------------------------------------------Instrumented: educ educsq
Instruments:
exper expersq south black smsa reg668 reg667 reg666 reg665
reg664 reg663 reg662 reg661 smsa66 nearc4reg668 nearc4reg667
nearc4reg666 nearc4reg665 nearc4reg664 nearc4reg663
nearc4reg662 nearc4reg661 nearc4smsa66 nearc4smsa nearc4south
nearc4exp nearc4black
The 2SLS estimation, using as instruments interactions of nearac4 with all exogenous
variables, results in a negative coefficient for educsq which is not statistically significant
at any level. Therefore there is no need to include educsq into the log(wage) equation.
b) The two conditions for a good instrumental variable are 1) that it is uncorrelated with the
error term and 2) that it is partially correlated with the endogenous variable in question.
One reason for black∙zj to be a potential IV for black∙educ given any exogenous variable
zj is that black and zj are both exogenous and therefore are both uncorrelated with the
error. As a result, black∙zj must also be uncorrelated with the error term. Moreover, the
fact that black∙zj and black∙educ are two variables derived from the variable black means
that they must be partially correlated to each other. Conditions 1) and 2) are thus satisfied
by black∙zj, for any zj exogenous to the system.
c)
Instrumental variables (2SLS) regression
Number of obs
Wald chi2(16)
Prob > chi2
R-squared
Root MSE
=
3010
= 1306.81
= 0.0000
= 0.2986
= .37161
-----------------------------------------------------------------------------lwage |
Coef.
Std. Err.
z
P>|z|
[95% Conf. Interval]
-------------+---------------------------------------------------------------blackeduc |
.0407135
.0090967
4.48
0.000
.0228843
.0585427
exper |
.0787306
.0067514
11.66
0.000
.0654982
.091963
expersq | -.0019402
.0003254
-5.96
0.000
-.002578
-.0013024
south | -.1393783
.0260037
-5.36
0.000
-.1903447
-.0884119
black | -.7007531
.1135754
-6.17
0.000
-.9233567
-.4781495
smsa |
.1311069
.0200989
6.52
0.000
.0917139
.1704999
educ |
.0657696
.0040212
16.36
0.000
.0578883
.0736509
reg668 | -.1766328
.0462573
-3.82
0.000
-.2672953
-.0859702
reg667 | -.0163362
.0394655
-0.41
0.679
-.0936871
.0610147
reg666 |
.006394
.0401767
0.16
0.874
-.0723508
.0851388
reg665 | -.0046576
.0361899
-0.13
0.898
-.0755885
.0662733
reg664 | -.0707684
.035653
-1.98
0.047
-.1406471
-.0008897
reg663 |
.0194164
.0273543
0.71
0.478
-.034197
.0730297
reg662 | -.0246772
.028212
-0.87
0.382
-.0799717
.0306174
reg661 | -.1267875
.0388035
-3.27
0.001
-.202841
-.050734
smsa66 |
.023371
.0194232
1.20
0.229
-.0146978
.0614398
_cons |
4.893009
.0792223
61.76
0.000
4.737736
5.048282
-----------------------------------------------------------------------------Instrumented: blackeduc
Instruments:
exper expersq south black smsa educ reg668 reg667 reg666
reg665 reg664 reg663 reg662 reg661 smsa66 blackeduchat
The variable black∙educ is highly significant (even at 1%) when black∙educhat is used as
an instrument.
d) Let Var(u1|z) = σ2. The structural equation from Example 6.2 is
log(wage)  1educ   2black  educ  z  u
where z is the vector of exogenous variables including nearc4.
ˆ :
For the instrument black  educ
ˆ  zˆ  ˆ nearc4  ˆ exper  ˆ exper 2  ˆ south  ...
(1) educ
1
2
3
4
ˆ as an instrument is
The first stage regression using black  educ
ˆ  zˆ  ˆ educ
(2) black  educ  ˆblack  educ
1
Note that the asymptotic variance of ̂ 2 is
A var ˆ  

2
black  educ

2
Substitute for black  educ from (2)
A var ˆ  
2
ˆ  zˆ  ˆ educ 
  ˆblack  educ
2
1
ˆ and rearranging, we get
Inserting (1) into educ
(3) A var ˆ  

2

2
ˆˆ
ˆ
ˆ
ˆ


ˆ ˆ
 1black  nearc 4  z   1educ    2exper   3exper   4 south  ...

2
.
For the instrument black  nearc4 , the first stage regression is
(4) black  educ  black  nearc4  z   1educ
Note that the asymptotic variance of ̂ 2 is
A var   
2
  black  educ 
2
Plugging in (4),
(5) A var   
2
  black  nearc4  z   1educ 
2
.
Comparing (4) and (5), we see that if   ˆ ,  1  ˆ1 , and ˆ1  0 , then denominator of
A var ˆ  is greater than that of A var   because of the presence of
(ˆ2exper  ˆ3exper 2  ˆ4 south  ...) , therefore A var ˆ  <A var   . This can be
explained by the fact that the variations in the other exogenous control variables that were
ˆ in (2) heightened the variation in the instrument black  educ
ˆ ,
used to explain educ
consequently making it an asymptotically more efficient IV than
black∙educ.
black  nearc4 for
9.10
a) When we estimate the first equation, y1, by 2SLS using all the exogenous variables as IVs
then it means that we are no longer imposing any exclusion restrictions on the reduced
form in the second equation, y2.
b) From (a) it follows that y2 is an unrestricted reduced form. And given that y1 is a
structural equation and y2 is the unrestricted reduced form, then the 3SLS estimate of y1
is the same as the 2SLS estimates.
c) In this case, even though 2SLS and 3SLS estimates are the same, 2SLS can be more
preferred because it is more robust than 3SLS. OLS is more robust than 2SLS which is in
turn more robust than 3SLS since single equation methods have lower standard errors and
are therefore more robust.