Assignment 1

Yang Liu
May 5, 2015

Assignment 1, Part 1
(1)

Table 0.1: Estimate a logit using solver
Product of probability

Log likelihood

Intercept

Eduation coefficient

Age coefficient

2.06641E-11

-24.60262143

-11.15550863

0.531907452

0.113507304

(2)
M EE d =
M E Ag e =

βE d e X β
(1 + e X β )2 β Ag e e X β
(1 + e X β )2

=

1 N βE d (i ) e X βi
Σ
N 1 (1 + e X βi )2

=

1 N β Ag e(i ) e X βi
Σ
N 1 (1 + e X βi )2

1

Result:

Table 0.2: Add caption
Marginal Edu effect

Marginal age effect

0.085732302

0.018294992

(3)

Table 0.3: linear probability model
Coefficients

t Stat

-1.541134276
0.097414096
0.020646694

Intercept
EDUC
AGE

Std. Err.
0.429125046
0.025225082
0.005153578

-3.591340779
3.861794961
4.00628341

(4)
H = −Σi p i (1 − p i )x i x i = X ΩX
The diagonal of matrix Ω equal to p i (1 − p i )
The Hessian matrix


8.058949138 105.838015 300.4357025
H =  105.838015 1434.435526 3840.728111
300.4357025 3840.728111 12190.99391

(0.1)

The variance-covariance matrix equals to the inverse of Hessian matrix
Table 0.4: variance-covariance matrix
Constant
Constant
Edu
Age

Edu

Age

10.2699517
-0.511915922
-0.091816163

-0.511915922
0.029972725
0.003172894

-0.09182
0.003173
0.001345

2

(5)
According to Student’s t distribution, under 49 freedom degree, the probabilities that these coefficients are 0s are less than 0.005. We can say these coefficients are significant under 99.5% confident level.
Table 0.5: Add caption
Coeff
Constant
EDUC
AGE

SE

T-ratios

-11.15550863
0.531907452
0.113507304

3.204676536
0.173126328
0.036676191

-3.481009239
3.072366051
3.094849818

(6)
Likelihood ratio LR = −2[l nL R − l nLU ] From the table above, we know that the probability
Table 0.6: Likelihood ratio test
Intercept
Alternative
Null
LR prob-value Eduation coefficient

Age coefficient

log likelihood

-11.15550863
-4.318996308
13.49019143
0.000239814

0.531907
0.329524

0.113507
0

-24.6026
-31.3477

that the coefficient of Age is zero is very small. The coefficient of AGE is different from zero under 99.9% confident level.

(7)
From the below figure about propensity score for the treated group and control group, we can see not all the controls are on the "support of X". On one hand, from the first two bins in this figure, we can see that a lot of units’ propensity scores in control group are smaller than the minimum propensity score of the treated group. On the other hand, from last two bins, there is nearly no comparison units. Both of these situations would lead to poor matches.

(8)
Follow the two formulas, I get AT E = 1.539489921,T OT = 1.775639145

Assignment 1, Part 2
Q UESTION 1
I put the detail procedure in another file called Assignment1.smcl. Here, I just show my overall results of replication.
Table 0.7: Replication Results
Unadjusted (1)
NSW
CPS-3

Adjusted(2)

886.304
(472.0863)
-1007.818
(539.346)

798.3512
(472.1283)
-1.446362
(681.2072)

4

Q UESTION 2: I NVERSE P ROPENSITY S CORE W EIGHTING

Table 0.8: Results
ATE
a b TOT

-846.1722
1480.339

1399.907
1476.243

From the table above, we can see there is obvious difference between before trimming and after trimming, especially for ATE. Before trimming, ATE is a negative number which is significantly different from TOT. After trimming, the ATE become a positive number and there is only a slight difference between ATE and TOT.
To figure out why it is so different before and after trimming, I did the following two histograms.
The first histogram is propensity scores for treated group and control group before trimming while the second histogram is the propensity scores after trimming.
From the first figure, it is obvious that quite a lot units’ propensity scores in the control group are smaller than the minimum value of the treated group. Actually more than 25% of the units’ propensity scores in the control group are smaller than the minimum value of the treated group. I think it is the main reason which lead to the results of a. After trimming, only the overlapping part of the two groups left. In addition, from the second figure, nearly no bin only contains one group units. We can say the controls are on the "support of X" after trimming
Before Trimming (a)

After Trimming (b)

C

Table 0.9: Add caption
IPW
Treat

800.647
(573.9948)

No weighting

DW 1999

1176.843
(809.3973)

662
(776)

From the table above, we can see the coefficient of treat by inverse propensity score weighting is smaller than the no weighting regression. Weighting by inverse of propensity score, the treated group with high propensity score get smaller weight while the control group with high propensity score get higher weight. Some of the treatment effect may relate to

6