Department of Economics
University of Essex
Dr Gordon Kemp
Session 2012/2013
Spring Term
EC352 – Econometric Methods
Solutions for Exercises from Week 09
1
Question E07.2
Suppose that you are interested in the relationship between the number of letters of application
sent by a job seeker and the probability of being employed. You have data about a cohort of
Essex students in 2005 and know their job status (employed or unemployed) and the number of
letters of application they have sent in the last year of study. Let Ei denote a binary variable
that takes the value 1 if person i is employed and 0, otherwise; and let Li denote the number
of letters sent by person i.
1. Suppose that one student in the sample has sent 50 applications and is employed. Write
down the probability of that particular event as a function of (unknown) parameters.
Answer:
• The probability of this particular event would be:
exp(β0 + 50β1 )
1 + exp(β0 + 50β1 )
2. Derive formally the marginal effect on the probability of being employed of sending an
additional letter.
Answer:
• Treating number of letters as continuous:
d Pr {Ei = 1|Li }
dLi
dΛ (β0 + β1 Li )
= β1 λ (β0 + β1 Li )
dLi
)
(
exp (β0 + β1 Li )
= β1
[1 + exp (β0 + β1 Li )]2
= β1 Λ (β0 + β1 Li ) [1 − Λ (β0 + β1 Li )]
=
• Note that for a full answer you would need to derive this formally: it helps to use
property that d exp(z)/dz = exp(z). The formula is more complicated if you treat
Li as discrete.
3. Within the context of a logit model, is the marginal effect of an application letter on the
probability of being employed identical for all individuals? Explain.
Answer:
1
• In general no – the marginal effects would only be the same here when either β1 = 0
or when Li is the same for all individuals. This is because Λ(β0 + β1 Li ) varies with
Li provided that β1 6= 0.
• For a more complete answer, it is useful to graph λ(z) = dΛ(z)/dz = exp(z)/[1 +
exp(z)]2 = Λ(z)[1 − Λ(z)] against z.
2
Exercise C17.1
This question is about efficiency in spread betting markets. The data consists of 553 observations on games taken during the 1994-95 men’s college basketball season in the US and is
contained in the file pntsprd.dta (this data is also used in computing Exercise C8.5 crom
Wooldridge). The variable spread is the Las Vegas points spread for the day before the game
was played. In spread betting, the gambler usually wagers that the difference between the scores
of two teams will be less than or greater than the value specified by the bookmaker. The offered
bet consists of specifying one of the teams as the “favorite”, the other as the “underdog”, and
a points spread which is always positive. If the gambler bets $100 on the favorite and the favorite’s score minus the the underdog’s score is greater than the spread then the gambler wins
$100; if it less than the spread than the gambler loses $100; while if it is equal to spread no
money is won or lost. Often spreads are often offered in half-point fractions to aovid the last
outcome (in which case bookmaker would still have to cover the transactions costs of the bet).
2
1. The variable favwin is a binary variable if the team favored by the Las Vegas points spread
wins. A linear probability model (LPM) to estimate the probability that the favored team
wins is:
Pr(favwin = 1|spread) = β0 + β1 spread
Explain why, if the spread incorporates all relevant information, we expect β0 = 0.5.
Answer:
• Suppose that there are no overheads to betting, bookmakers are risk neutral, the
betting market is perfectly competitive, and pushes never occur. Then the bookmaker will offer a bet with an expected payout to the gambler of $0 which means
that the probability of the gambler winning should equal the probability of the
gambler losing and hence should equal 0.5. If the LPM is correct and the spread
was equal to zero that should mean that the probability of the favorite winning
should equal 0.5 which corresponds in the LPM to β0 = 0.5.
• In practice, there are overheads and pushes do occur. Furthermore, the LPM may
not be appropriate and in the data set bookmakers never offer zero spreads. All of
these factors put some doubt on the conclusion that β0 should equal 0.5.
2. Estimate the model from Part (1) by OLS. Test H0 : β0 = 0.5 against a two-sided
alternative. Use both the usual and heteroskedasticity-robust standard errors.
Answer:
• An standard OLS regression of f avwin on spread gives the results:
. regress favwin spread
Source |
SS
df
MS
-------------+-----------------------------Model | 11.0636261
1 11.0636261
Residual | 88.9038241
551 .161349953
-------------+-----------------------------Total | 99.9674503
552 .181100453
Number of obs
F( 1,
551)
Prob > F
R-squared
Adj R-squared
Root MSE
=
=
=
=
=
=
553
68.57
0.0000
0.1107
0.1091
.40168
-----------------------------------------------------------------------------favwin |
Coef.
Std. Err.
t
P>|t|
[95% Conf. Interval]
-------------+---------------------------------------------------------------spread |
.0193655
.0023386
8.28
0.000
.0147718
.0239593
_cons |
.5769492
.0282345
20.43
0.000
.5214888
.6324097
------------------------------------------------------------------------------
Then we can perform a test of H0 : β0 = 0.5 against a two-sided alternative based
on the usual standard errors generating the results:
. test _cons=0.5
( 1)
_cons = .5
F(
1,
551) =
Prob > F =
7.43
0.0066
so this would lead us to reject the null hypothesis at all conventional significance
levels.
3
• However, we know that binary response models (including the LPM) exhibit heteroskedasticity. Redoing the regression with robust standard errors generates the
output:
. regress favwin spread, vce(robust)
Linear regression
Number of obs
F( 1,
551)
Prob > F
R-squared
Root MSE
=
=
=
=
=
553
101.54
0.0000
0.1107
.40168
-----------------------------------------------------------------------------|
Robust
favwin |
Coef.
Std. Err.
t
P>|t|
[95% Conf. Interval]
-------------+---------------------------------------------------------------spread |
.0193655
.0019218
10.08
0.000
.0155905
.0231405
_cons |
.5769492
.0316568
18.23
0.000
.5147664
.6391321
------------------------------------------------------------------------------
and then the test of H0 : β0 = 0.5 against a two-sided alternative based on
heteroskedasticity-robust standard errors generates the results:
. test _cons=0.5
( 1)
_cons = .5
F(
1,
551) =
Prob > F =
5.91
0.0154
This still rejects the null hypothesis although.
3. Is spread statistically significant? What is the estimated probability that the favored team
wins when spread = 10?
Answer:
• With the usual standard errors the t-stat on the spread is 8.28 with a p-value of 0
to 3s.f. while with the heteroskedasticity-robust standard errors the t-stat is 10.08
with a p-value of 0 to 3s.f. Clearly the spread is highly significant for the LPM
regardless of which standard errors are used.
• The estimated probability that the favored team wins when spread = 10 is:sma
0.5769 + (0.0193 × 10) = 0.7699
i.e. about 77%.
4. Now estimate a probit model for Pr(favwin = 1|spread). Interpret and test the null
hypothesis that the intercept is 0.
[Hint: Remember that Φ(0) = 0.5.]
Answer:
• Fitting a probit model gives:
4
. probit favwin spread
Iteration
Iteration
Iteration
Iteration
Iteration
0:
1:
2:
3:
4:
log
log
log
log
log
likelihood
likelihood
likelihood
likelihood
likelihood
=
=
=
=
=
-302.74988
-266.49244
-263.62542
-263.56223
-263.56219
Probit regression
Number of obs
LR chi2(1)
Prob > chi2
Pseudo R2
Log likelihood = -263.56219
=
=
=
=
553
78.38
0.0000
0.1294
-----------------------------------------------------------------------------favwin |
Coef.
Std. Err.
z
P>|z|
[95% Conf. Interval]
-------------+---------------------------------------------------------------spread |
.092463
.0121811
7.59
0.000
.0685885
.1163374
_cons | -.0105926
.1037469
-0.10
0.919
-.2139328
.1927476
------------------------------------------------------------------------------
When the spread is zero then the predicted probability that the favorite wins is
Φ(βb0 ). Since βb0 = −0.0105926 this gives:
. disp normal(-0.0105926)
.49577424
i.e. almost a 50% chance of winning (very slightly under). The z-stat on the
intercept is -0.10 with a p-value of 0.919 so we do not reject the null hypothesis
that the intercept is 0.
5. Use the probit model to estimate the probability that the favored team wins when spread =
10. Compare this with the LPM estimate from Part (3).
Answer:
• Plugging this in we get:
. disp normal(-.0105926 + (.092463 * 10))
.8196514
i.e. a predicted probability of winning of 82% (rather higher than with the LPM
estimates)
6. Add the variables f avhome, f av25 and und25 to the probit model and test the joint
significance of these variables using the Likelihood Ratio test. (How many df are in the
chi-square distribution?) Interpret this result focusing on the question of whether the
spread incorporates all observable information prior to a game.
Answer:
• Running this probit regression gives:
. probit favwin spread favhome fav25 und25
Iteration 0:
Iteration 1:
log likelihood = -302.74988
log likelihood = -265.47417
5
Iteration 2:
Iteration 3:
Iteration 4:
log likelihood = -262.70317
log likelihood = -262.64181
log likelihood = -262.64177
Probit regression
Number of obs
LR chi2(4)
Prob > chi2
Pseudo R2
Log likelihood = -262.64177
=
=
=
=
553
80.22
0.0000
0.1325
-----------------------------------------------------------------------------favwin |
Coef.
Std. Err.
z
P>|z|
[95% Conf. Interval]
-------------+---------------------------------------------------------------spread |
.0878845
.0129491
6.79
0.000
.0625047
.1132642
favhome |
.1485753
.1370571
1.08
0.278
-.1200517
.4172024
fav25 |
.003068
.15869
0.02
0.985
-.3079587
.3140946
und25 | -.2198082
.2505842
-0.88
0.380
-.7109443
.2713278
_cons | -.0551801
.128763
-0.43
0.668
-.3075509
.1971907
------------------------------------------------------------------------------
• To test the joint significance of these added variables using the Likelihood Ratio
test we compare:
h
i
bU R − L
bR
LR = 2 L
with a χ2 (q), i.e. a chi-square with q d.f. where q is the number of restrictions we
are testing.
• Here we want to test that three coefficients are equal to zero so q = 3. Then:
h
i
bU R − L
b R = 2 × [(−262.64177) − (−263.56219)]
LR = 2 L
= 1.84084
Now the 10% significance level critical value from a χ2 (3) test is 6.25139.4 so we
clearly do not reject the null. Thus adding these additional variables does not
seem to provide any significant additional information relevant to predicting the
probability of a favored team win that was not already accounted for by the spread.
3
Computing Exercise C17.2
This question is about discrimination in approval of loan applications. The data consists
of 1989 observations on loan applications in Boston in 1990 and are contained in the file
loanapp.dta (this data is also used in Computer Exercises C7.8, C8.7 and C9.7 from Wooldridge).
1. Estimate a probit model of approve on white. Find the estimated probability of loan
approval for both whites and non-whites. How do these compare with the linear probability
estimates?
Answer:
• Running the probit estimation gives:
. probit approve white
Iteration 0:
Iteration 1:
log likelihood = -740.34659
log likelihood = -701.33221
6
Iteration 2:
Iteration 3:
log likelihood = -700.87747
log likelihood = -700.87744
Probit regression
Number of obs
LR chi2(1)
Prob > chi2
Pseudo R2
Log likelihood = -700.87744
=
=
=
=
1989
78.94
0.0000
0.0533
-----------------------------------------------------------------------------approve |
Coef.
Std. Err.
z
P>|z|
[95% Conf. Interval]
-------------+---------------------------------------------------------------white |
.7839465
.0867118
9.04
0.000
.6139946
.9538985
_cons |
.5469463
.075435
7.25
0.000
.3990964
.6947962
------------------------------------------------------------------------------
Running the LPM gives:
. regress approve white
Source |
SS
df
MS
-------------+-----------------------------Model | 10.4743407
1 10.4743407
Residual |
203.59303 1987 .102462521
-------------+-----------------------------Total | 214.067371 1988 .107679764
Number of obs
F( 1, 1987)
Prob > F
R-squared
Adj R-squared
Root MSE
=
=
=
=
=
=
1989
102.23
0.0000
0.0489
0.0485
.3201
-----------------------------------------------------------------------------approve |
Coef.
Std. Err.
t
P>|t|
[95% Conf. Interval]
-------------+---------------------------------------------------------------white |
.2005957
.01984
10.11
0.000
.1616864
.239505
_cons |
.7077922
.0182393
38.81
0.000
.6720221
.7435623
------------------------------------------------------------------------------
The coefficient estimates are quite different; however, they are not directly comparable. The estimate from the LPM suggests that the probability of approval in
a loan application for a white applciant is about 20 perecntage points higher than
for a non-white applicant.
• In both models the coefficient on white is positive and very strongly statistically
significant.
2. Now, add the variables hrat, obrat, loanprc, unem, male, married, dep, sch, cosign,
chist, pubrec, mortlat1, mortlat2, and vr to the probit model. Is there statistically
significant evidence of discrimination against non-whites?
Answer:
• Adding these variables to the probit generates the following results:
. probit approve white hrat obrat loanprc unem male married dep sch cosign
chist pubrec mortlat1 mortlat2 vr
Iteration
Iteration
Iteration
Iteration
Iteration
0:
1:
2:
3:
4:
log
log
log
log
log
likelihood
likelihood
likelihood
likelihood
likelihood
=
=
=
=
=
-737.97933
-603.5925
-600.27774
-600.27099
-600.27099
Probit regression
Number of obs
7
=
1971
LR chi2(15)
Prob > chi2
Pseudo R2
Log likelihood = -600.27099
=
=
=
275.42
0.0000
0.1866
-----------------------------------------------------------------------------approve |
Coef.
Std. Err.
z
P>|z|
[95% Conf. Interval]
-------------+---------------------------------------------------------------white |
.5202525
.0969588
5.37
0.000
.3302168
.7102883
hrat |
.0078763
.0069616
1.13
0.258
-.0057682
.0215209
(remaining output suppressed)
The estimated coefficient on white is smaller than in the earlier probit regression:
here the estimate is 0.520 while the earlier estimate was 0.784. However, again the
coefficients are not entirely comparable. We still seee that the coefficient is positive
and highly statistically significnat (though not as significant as previously).
• Note that the maximized log-likelihood has gone up by about 100 so that all this
additional stuff in the probit is jointly highly statistically significant.
3. Estimate the model from Part (2) by logit. Compare the coefficient on white to the probit
estimate.
Answer:
• Running the logit estimation of the model from Part (2) gives:
. logit approve white hrat obrat loanprc unem male married dep sch cosign chist
pubrec mortlat1 mortlat2 vr
Iteration
Iteration
Iteration
Iteration
Iteration
Iteration
0:
1:
2:
3:
4:
5:
log
log
log
log
log
log
likelihood
likelihood
likelihood
likelihood
likelihood
likelihood
=
=
=
=
=
=
-737.97933
-634.97536
-601.41194
-600.49724
-600.49616
-600.49616
Logistic regression
Number of obs
LR chi2(15)
Prob > chi2
Pseudo R2
Log likelihood = -600.49616
=
=
=
=
1971
274.97
0.0000
0.1863
-----------------------------------------------------------------------------approve |
Coef.
Std. Err.
z
P>|z|
[95% Conf. Interval]
-------------+---------------------------------------------------------------white |
.9377643
.1729043
5.42
0.000
.598878
1.27665
hrat |
.0132631
.0128802
1.03
0.303
-.0119816
.0385078
(remaining output suppressed)
The coefficient estimate on white is now 0.938 but again this is not directly comparable with the earlier estimates. It is both positive and highly statistically significant.
• Note that the maximized log-likelihood from this logit estimation is −600.49616
while that from th earlier probit estimation was −600.27099 so these are not very
different.
4. Use Equation (17.17) from Wooldridge to estimate the sizes of the discrimination effects
for probit and logit.
8
Answer:
• Equation (17.17) from Wooldridge outline a way of calculating the average partial
effect (APE) of a discrete explanatory variable in a binary response model. This
can be done in Stata by using the margins command with the at option and
then combining the results. To do this for the probit model we re-run the probit
estimation and then we run the margins command at white=0 and at white=1 :
. probit approve white hrat obrat loanprc unem male married dep sch cosign chist
pubrec mortlat1 mortlat2 vr
(output suppressed)
. margins, at(white=0 white=1)
Predictive margins
Model VCE
: OIM
Number of obs
Expression
: Pr(approve), predict()
1._at
: white
=
0
2._at
: white
=
1
=
1971
-----------------------------------------------------------------------------|
Delta-method
|
Margin
Std. Err.
z
P>|z|
[95% Conf. Interval]
-------------+---------------------------------------------------------------_at |
1 |
.7923174
.0210981
37.55
0.000
.7509659
.8336689
2 |
.8965419
.0072597
123.49
0.000
.8823131
.9107707
-----------------------------------------------------------------------------. disp (.8965419 - .7923174)
.1042245
From this we see that for the probit model, the estimated APE on loan approval of
being white as compared to non-white is an increase of 0.104, i.e. 10.4 percentage
points. We can do the same for the logit model which gives:
. logit approve white hrat obrat loanprc unem male married dep sch cosign chist
pubrec mortlat1 mortlat2 vr
(output suppressed)
. margins, at(white=0 white=1)
Predictive margins
Model VCE
: OIM
Number of obs
Expression
: Pr(approve), predict()
1._at
: white
=
0
2._at
: white
=
1
=
1971
-----------------------------------------------------------------------------|
Delta-method
|
Margin
Std. Err.
z
P>|z|
[95% Conf. Interval]
-------------+----------------------------------------------------------------
9
_at |
1 |
.7955101
.0203441
39.10
0.000
.7556364
.8353838
2 |
.8963808
.0072951
122.87
0.000
.8820826
.9106789
-----------------------------------------------------------------------------. disp (.8963808-.7955101)
.1008707
From this we see that for the logit model, the estimated APE on loan approval of
being white as compared to non-white is an increase of 0.101, i.e. 10.1 percentage
points. We can see that these answers are fairly similar and are both noticeably
positive, suggesting tht there is discrimination in loan approval even after controlling for this additional collection of factors. Note that the corresponding estimated
APE from the LPM is simply the coefficient on white in the regression with the
other factors included:
. regress approve white hrat obrat loanprc unem male married dep sch cosign
chist pubrec mortlat1 mortlat2 vr
Source |
SS
df
MS
-------------+-----------------------------Model | 35.4004787
15 2.36003192
Residual | 178.393534 1955
.09124989
-------------+-----------------------------Total | 213.794013 1970
.10852488
Number of obs
F( 15, 1955)
Prob > F
R-squared
Adj R-squared
Root MSE
=
=
=
=
=
=
1971
25.86
0.0000
0.1656
0.1592
.30208
-----------------------------------------------------------------------------approve |
Coef.
Std. Err.
t
P>|t|
[95% Conf. Interval]
-------------+---------------------------------------------------------------white |
.1288196
.0197317
6.53
0.000
.0901223
.1675169
hrat |
.001833
.0012632
1.45
0.147
-.0006444
.0043104
(remaining output suppressed)
This gives an estimated APE of 0.129, i.e. about 12.9 percentage points which is
rather above the estimated APE’s from the probit and logit estimations.
10