Econometrics 532
Problem Set
NOTE: The solutions to problem 4 can be found with the midterm solutions.
Problem 1
A)
. reg y x2 x3 x4
Source |
SS
df
MS
-------------+-----------------------------Model | 1435.83073
3 478.610245
Residual | 3694.30641
11 335.846038
-------------+-----------------------------Total | 5130.13715
14 366.438368
Number of obs
F( 3,
11)
Prob > F
R-squared
Adj R-squared
Root MSE
=
=
=
=
=
=
15
1.43
0.2879
0.2799
0.0835
18.326
-----------------------------------------------------------------------------y |
Coef.
Std. Err.
t
P>|t|
[95% Conf. Interval]
-------------+---------------------------------------------------------------x2 |
.0209707
.0190552
1.10
0.295
-.0209695
.0629109
x3 |
15.43505
9.268231
1.67
0.124
-4.964192
35.83428
x4 |
.8213001
1.905179
0.43
0.675
-3.371971
5.014571
_cons | -35.57865
49.22479
-0.72
0.485
-143.9217
72.76438
------------------------------------------------------------------------------
B) We would expect that 2  0, 3  0, 4  0 . The last two of these require some thought.
Larger families may require the head of household to work more to support the family.
When the unemployment rate is higher, the head of household is more likely to have to
enter the labor force to attempt to support the family. This coefficient could go the other
way if the labor market is so poor that the head of household decides that it’s not worth
the effort to search for a job. Remember that an individual is counted in the labor force if
she is working or looking for work.
C) You use a t-statistic with n – k = 15 – 11 = 4 degrees of freedom. The t-statistic is 0.43
and we cannot reject the hypothesis the unemployment rate has no effect on labor force
participation.
D) The unemployment rate apparently could be dropped from the model. The t-statistic is
nearly zero for the unemployment rate.
E) One possibility is a variable that reflects the demographic composition of the household.
An example is the dependency ratio, the share of the household that is not of working
age.
Problem 2
 1 
 
 2
A)     3 
 
4 
 
 5
 1 
 
2
0
1
-1
0
0

     2  3  0 

B) R  
 3         1 
0
0
0
1
1

   4
 
5
4 
 
 5
0 1 0 0 0 
2


C) (i) R  0 0 1 0 0 ; r   2 


 
0 0 0 1 0 
 2 
0 1 -1 0 0 
0 
(ii) R  
;r   

0 0 0 1 -1
0 
(iii) R  0 1 -3 -5 0 ; r  0 
(iv) R  0 1 3 0 0; r  0
D) ˆ *  ˆ if r  R ˆ  0 , so that Rˆ  r
Problem 3
Consider our basic linear regression model, where we have n observations and k right-hand side
variables:
Y  X 
In class, we should have proven the following theorem by now:
( Rˆ  C ) '[ s 2 R( X ' X ) 1 R ']1 ( R ˆ  C )
~ F (J , n  k)
J
under H 0 : R  C and the standard assumptions on the error term, where J is the number of
restrictions in the matrix R.
Theorem:
We can use this fact to test H 0 .
Earlier in the class, we learned how to test restrictions by comparing residuals from the restricted
regression and residuals from the unrestricted regression.
Unrestricted regression: Y  X     ˆ  ( X ' X ) 1 X ' y
It can be shown that the coefficient vector estimated under the restrictions contained in R is:
Restricted regression:
Y  X  R   , s.t. R  C  ˆR  ˆ  ( X ' X )1 R '[ R( X ' X )1 R ']1 ( Rˆ  C)
A) Explain in three or four concise sentences why it makes sense to test restrictions by
comparing the restricted and unrestricted residuals.
If the restrictions are correct, then the residuals should be the same from the restricted and
unrestricted regressions. If the restrictions are wrong, though, the line that we fit under the
restrictions will be significantly worse than the unrestricted line. The residuals would then
be larger in the restricted regression.
B) Are the restricted residuals ever smaller than the unrestricted residuals? Explain.
No. The residuals will always be smallest when we allow regression to pick the best possible
values for the regression coefficients rather than restricting any of the coefficients.
C) It is possible to prove the following theorem:
F
(ˆR ' ˆR  ˆU ' ˆU ) / J
~ F (J , n  k)
ˆU ' ˆU /(n  k )
I am going to walk you through how to prove this theorem. Start by noticing that
ˆ ' ˆ
s 2  U U , so that we just need to show
nk
ˆR ' ˆR  ˆU ' ˆU  ( Rˆ  C)'[ R( X ' X )1 R ']1 ( Rˆ  C) to prove the theorem.
Step 1: Show that ˆR  ˆU  X (ˆ  ˆR )
ˆR  y  X ˆR and ˆU  y  X ˆ , so ˆR  ˆU  X (ˆ  ˆR )
Step 2: Use the definition of ˆ to show that ˆU ' X  0 . You need to use your matrix rules:
(A+B)’=A’+B’ and (AB)’=B’A’. (Hint: ˆ  ( X ' X ) 1 X ' y  ˆ '  y ' X ( X ' X ) 1 )

 

  y ' ˆ ' X '  X   y ' y ' X ( X ' X )
ˆU ' X  y  X ˆ ' X  y ' ( X ˆ ) ' X 
1
X ' X
 y ' X  y ' X ( X ' X ) 1 X ' X  y ' X  y ' X  0
Step 3: Use the previous two results to show that ˆR ' ˆR  ˆU ' ˆU  (ˆ  ˆR )'( X ' X )(ˆ  ˆR ) .
From Step 1, ˆR  ˆU  X (ˆ  ˆR ) .
Since Step 2 shows that ˆU ' X  0 , ˆR ' ˆR  ˆU ' ˆU  (ˆ  ˆR )'( X ' X )(ˆ  ˆR ) .
Step 4: Now use the definition of ˆR to substitute into the expression from step 3. After
simplifying, you should have ˆ ' ˆ  ˆ ' ˆ  ( Rˆ  C)'[ R( X ' X )1 R ']1 ( Rˆ  C) , which is
R
R
U
U
what you needed to show to prove the theorem.
You were given that ˆR  ˆ  ( X ' X )1 R '[ R( X ' X )1 R ']1 ( Rˆ  C) . Using rules for
transpose:
ˆR ' ˆR  ˆU ' ˆU  ( Rˆ  C)'[ R( X ' X )1 R ']1 R( X ' X )1 ( X ' X )( X ' X )1 R '[ R( X ' X )1 R ']1( Rˆ  C)
By cancelling matrices, we arrive at the desired result:
ˆR ' ˆR  ˆU ' ˆU  ( Rˆ  C)'[ R( X ' X )1 R ']1 ( Rˆ  C) //