You are given the time plots of the variables con, inv and inc.
cont
invt
inct
3
Graph 1: Impulse Responses of VAR(2) system for con, inv and inc
cont
cont
invt
cont
invt
invt
cont
inct
invt
cont
inct
cont
invt
inct
invt
inct
inct
inct
Table 1: Impulse Responses of VAR(2) system for con, inv and inc
Responses to a one-standard error shock in con
period
con
inv
inc
1
0.00954
0.01359
0.00617
2
-0.00089
0.00460
0.00274
3
0.00289
0.00410
0.00102
4
0.00082
0.00046
0.00123
5
0.00037
0.00219
0.00053
6
0.00043
0.00048
0.00024
7
0.00021
0.00026
0.00029
8
0.00008
0.00040
0.00009
9
0.00010
0.00012
0.00006
10
0.00004
0.00007
0.00006
11
0.00002
0.00008
0.00002
12
0.00002
0.00002
0.00002
13
0.00001
0.00002
0.00001
14
0.00001
0.00001
0.00000
15
0.00000
0.00001
0.00000
16
0.00000
0.00000
0.00000
17
0.00000
0.00000
0.00000
18
0.00000
0.00000
0.00000
19
0.00000
0.00000
0.00000
20
0.00000
0.00000
0.00000
4
Responses to a one-standard error shock in inv
period
con
inv
inc
1
0.00000
0.04029
-0.00067
2
-0.00008
-0.01121
0.00183
3
0.00228
-0.00191
0.00173
4
-0.00003
0.00438
-0.00026
5
0.00022
0.00064
0.00024
6
0.00014
-0.00060
0.00033
7
0.00015
0.00045
0.00003
8
0.00004
0.00021
0.00004
9
0.00002
0.00002
0.00005
10
0.00003
0.00003
0.00001
11
0.00001
0.00003
0.00001
12
0.00000
0.00002
0.00001
13
0.00001
0.00000
0.00000
14
0.00000
0.00000
0.00000
15
0.00000
0.00001
0.00000
16
0.00000
0.00000
0.00000
17
0.00000
0.00000
0.00000
18
0.00000
0.00000
0.00000
19
0.00000
0.00000
0.00000
20
0.00000
0.00000
0.00000
Responses to a one-standard error shock in inc
period
con
inv
inc
1
0.00000
0.00000
0.00881
2
0.00255
0.00297
-0.00109
3
0.00220
0.00209
0.00122
4
-0.00081
0.00220
0.00090
5
0.00079
0.00043
0.00000
6
0.00031
-0.00022
0.00038
7
-0.00005
0.00080
0.00010
8
0.00014
0.00006
0.00002
9
0.00005
-0.00004
0.00009
10
0.00001
0.00015
0.00001
11
0.00003
0.00002
0.00001
12
0.00001
0.00000
0.00002
13
0.00000
0.00003
0.00000
14
0.00001
0.00000
0.00000
15
0.00000
0.00000
0.00000
16
0.00000
0.00000
0.00000
17
0.00000
0.00000
0.00000
18
0.00000
0.00000
0.00000
19
0.00000
0.00000
0.00000
20
0.00000
0.00000
0.00000
5
Graph 2: Impulse Responses
6
Graph 1
Graph 2
10
Graph 1
Graph 2
13
Question 4
Suppose that we examine the effect of a teaching method known as PSI on the
performance of students in a course. The question is whether students exposed to the
method scored higher on exams in the class. Data from students in two classes, one in
which PSI was used and another in which a traditional teaching method was employed.
You are given the following variables:
Consider the following model:
GRADEi = 0 + 1GPAi + 2PSIi + 3TUCEi + ui
(a) You are given the following LPM estimation results:
Pˆ(GRADE i = 1) = −1.58 + 0.52* GPAi + 0.30 * PSIi + 0.006 * TUCEi
According to LPM, what is the chance of a student with a grade point average of 3,
taught by traditional methods, and scoring 20 on the TUCE exam to get an A?
Answer:
Pˆ(GRADEi = 1) = −1.58 + 0.52*3 + 0.30 * 0 + 0.006 *20 = 0.10
This student would have about a 10% chance of getting an A.
(b) According to LPM, what is the chance of a student with a grade point average of 3,
taught by PSI method, and scoring 20 on the TUCE exam to get an A?
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Pˆ(GRADEi = 1) = −1.58 + 0.52*3 + 0.30 *1 + 0.006 *20 = 0.40
Answer:
This student would have about a 40% chance of getting an A.
(c) What is the main disadvantage of LPM estimates?
Answer:
Negative probabilities can be predicted!
(d) Below, you are some logit model estimation results.
a. What is the chance of a student with a grade point average of 3, taught by
traditional methods, and scoring 20 on the TUCE exam to get an A?
b. What is the chance of a student with a grade point average of 3, taught by
PSI method, and scoring 20 on the TUCE exam to get an A?
c. Would getting into the PSI class increase the chances of getting an A or not?
How much the difference?
15
Answers:
a. The student who has a GPA of 3.0, is taught by traditional methods (PSI=0), and has
a score of TUCE has approximately 7.27% chance of getting an A
b. The student who has a GPA of 3.0, is taught by PSI method (PSI=1), and has a score
of TUCE has approximately 35.3% chance of getting an A.
c. The student would have about a 28.03% (=35.3-7.27) better chance.
16
**** Stata Commands *****
use logist.dta
*LPM
regress grade gpa psi tuce
* What is the chance of a student with a grade point average of 3, taught by traditional
methods, and scoring 20 on the TUCE exam to get an A?
* According to these results, a student with a grade point of 3.0, taught by
traditional methods, and scoring 20 on the TUCE exam would earn an A with probability
of 0.1. So, this person would have about a 10% chance of getting an A.
* logit estimation
logit grade gpa tuce psi
* margins
margins, at(gpa = 3 psi = 0 tuce = 20)
* How can you interpret the reported value for margin .0726861
* The student who has a GPA of 3.0, is taught by traditional methods (PSI=0), and has
a score of TUCE has approximately 7.27% chance of getting an A
margins, at(gpa = 3 psi = 1 tuce = 20)
* How can you interpret the reported value for margin .0726861
* The student who has a GPA of 3.0, is taught by PSI method (PSI=1), and has a score
of TUCE has approximately 35.3% chance of getting an A.
* Would getting into the PSI class increase the chances of getting an A or not? How
much the difference?
* Yes, the student would have about a 28.03% (=35.3-7.27) better chance.
17
Question 5
Consider the following simple regression model:
Yt = 0 + 1 X t + ut
a) Write the likelihood function to derive the Maximum Likelihood (ML) estimators for 0 ,
1 and  2 .
Answer:
b) Write the log likelihood function to derive the Maximum Likelihood (ML) estimators for
0 , 1 and  2 .
Answer:
c) Derive the Maximum Likelihood (ML) estimators for 0 , 1 and  2 . Are they unbiased
estimators?
Answer:
18
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