IB9W00
UNIVERSITY OF WARWICK
January Examination 2019/20
Quantitative Methods for Financial Management
Time Allowed: 2 hours
Silent pocket calculators that are not capable of text storage or retrieval are permitted
but their instruction booklets, PDAs, mobile phones or any other hand-held devices that
facilitate wireless communication are NOT PERMITTED.
Closed Book examination
You should answer ALL questions
Read carefully the instructions on the answer book and make sure that the particulars
required are entered on each answer book
PLEASE TURN OVER
1
Question 1 [20 marks]
Suppose we want to model whether an application for a loan is successful as a
function of certain characteristics:
approve = β0 + β1unem+ β2cons+ β3mortg + β4term+ β5atotinc+ u, (1)
where approve = 1 if the loan application is successful, unem is the unemployment
rate by industry, cons is a score (the lower the better) on the credit history on con-
sumer payments, mortg is a score (the lower the better) on the credit history on
mortgage payments, term is the term of the loan in months, and atotinc is the total
monthly income. You estimate this equation by Logit using maximum likelihood (ML)
and obtain the following results:
a) Suppose the log-likelihood for the model with only the intercept is equal to
l0 = −740.34. Compute the McFadden-R2 as well as the pseudo-R2. What
would be the McFadden-R2 of a model with the intercept only? Explain. [7
marks]
Answer: For the Logit and Probit models, two goodness-of-fit measures are
• McFadden - R2 = 1− l1
l0
,
• Pseudo - R2 = 1− 1
1+2(l1−l0)/N ,
where l1 is the log-likelihood function for the unrestricted model, and l0 is the
log-likelihood function for the model with only an intercept. The idea is to com-
pare the log-likelihood of the unrestricted model with the one obtained from an
auxiliary model with only a constant term. Since l1 = −658.67 and l0 = −740.34,
we have
• McFadden - R2 = 1− l1
l0
= 1− −658.67−740.34 = 1− 0.8897 = 0.1103,
• Pseudo - R2 = 1 − 1
1+2(l1−l0)/N = 1 − 11+2(−658.67+740.34)/1989 = 1 − 11.082 =
1− 0.924 = 0.076.
2
From the above formulas, it is easy to see that for a model with only the inter-
cept l1 = l0, so that both R2 measures are equal to zero.
b) Suppose the sample means of the explanatory variables are
Compute the partial effect of each regressor on the probability of having the
loan application approved. Just focus on those explanatory variables whose
parameter estimates are statistically significant at the 1% confidence level. [7
marks]
Answer: Consider cons and mortg (both have a p-value equal to 0.000). The
partial effect of each explanatory variable can be approximated as
∂P (y = 1|x)
∂xj
= βˆj · g
(
βˆ0 + xβˆ
)
,
where dG(z)/dz = g(z) = exp(z)/(1 + exp(z))2, and x is the sample average of
the independent variables. Given the information in the text we can compute
the scaling factor as
g (4.0414− 0.0698 · 3.8823− 0.4066 · 2.1096− 0.4478 · 1.7084−
2.73e− 06 · 2351.46 + 0.000013 · 5195.55) = 0.0892.
Therefore, the partial effect of cons is given by
∂P (approve = 1|x)
∂cons
= βˆcons · g
(
βˆ0 + xβˆ
)
= −0.4066 · 0.0892 = −0.0363,
and the partial effect of mortg is given by
∂P (approve = 1|x)
∂mortg
= βˆmortg · g
(
βˆ0 + xβˆ
)
= −0.4478 · 0.0892 = −0.0399,
meaning that a bad consumer and mortgage payment history can negatively
affect the probability of obtaining a loan by about 4%.
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c) Consider the ML estimates of the Logit model without cons and term:
What is the value of the likelihood ratio test for the restriction β2 = β4 = 0?
Can we reject the null hypothesis H0 : β2 = β4 = 0 (against the alternative
H1 : H0 not true)? At what significance level? (Hint: the critical values of
the χ22 are equal to 4.61, 5.99, and 9.21 at the 10%, 5%, and 1% significance
levels, respectively.) [6 marks]
Answer: For the Logit model, linear restrictions on the parameters can be
tested by using the Likelihood-Ratio (LR) test. The LR test is twice the differ-
ence in the log-likelihoods:
LR = 2 (l1 − l0) ,
where l1 is the log-likelihood value for the unrestricted model and l0 is the log-
likelihood value for the restricted model. From the complete model (first esti-
mation output) we can see that l1 = −658.67. The text of the exercise provides
also the output for the restricted model without cons and term. Therefore, the
LR statistic is
LR = 2 (−658.67 + 723.79) = 130.24.
It is easy to see that a LR statistic of 130.24 is greater than the critical values
of the χ22 at the 10%, 5%, and 1% significance levels. Therefore, we strongly
reject the null hypothesis that cons and term are irrelevant.
As
Question 2 [20 marks]
Consider the fitted values from a simple linear regression model with intercept:
yˆ = 5 + 4x.
sume that R2=0.2, the total number of observations is n = 202, and that ∑n
i=1(xi − x¯)2 = 4, where x¯ is the sample mean of x. Assume that the classical
Gauss-Markov assumptions hold.
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a) What is the value of the F -statistic? [7 marks]
Answer: F − stat = 50
b) What is the standard error of the regression? [6 marks]
Answer: s.e. regression =
√
2
5
c) What is the standard error of the estimated slope coefficient? [7 marks]
Answer: s.e. beta estimate = 2
√
2
5
Question 3 [20 marks]
Suppose that a time-series process {xt : t = 1, 2, . . . , T} is given by xt = z + et, for
all t = 1, 2, . . . , T, where et is an i.i.d. sequence with mean zero and variance σ2e .
The random variable z is constant over time, and it has mean zero and variance σ2z .
Furthermore, assume that et is uncorrelated with z.
a) Find the expected value and variance of xt. Do your answers depend on t? [5
marks]
Answer: The expected value can be computed as
E [xt] = E [z] + E [et] = 0,
while the unconditional variance can be computed as
V ar [xt] = V ar [z + et] = V ar [z] + V ar [et] + 2Cov [z, et] = σ
2
z + σ
2
e .
Neither of these depend on t.
b) Find Cov (xt, xt+h) for any t and h. Is xt weakly stationary? Explain. [5 marks]
Answer: When h = 0, we obtain V ar [xt]. When h > 0, in general, Cov (xt, xt+h)
can be written as
Cov (xt, xt+h) = E (xt, xt+h)− E (xt)E (xt+h) .
We saw in part a) that E (xt) = E (xt+h) = 0. Therefore,
Cov (xt, xt+h) = E (xt, xt+h) = E ((z + et) (z + et+h)) =
= E
(
z2
)
+ E (etz) + E (zet+h) + E (etet+h) = E
(
z2
)
= σ2z
because et is an i.i.d. sequence. From part a) we know that E(xt) and V ar(xt)
do not depend on t and we have shown that Cov(xt, xt+h) does not depend on
t. Therefore, xt is weakly (covariance) stationary.
c) Show that Corr (xt, xt+h) = σ2z/ (σ2z + σ2e) for all t and h. [5 marks]
Answer: We saw in parts a) and b) how to compute V ar (xt) and Cov (xt, xt+h).
From the definition of correlation it is easy to see that
Corr (xt, xt+h) =
Cov (xt, xt+h)
V ar (xt)
=
σ2z
σ2z + σ
2
e
> 0.
5
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d) Is xt weakly dependent? Explain. [5 marks]
Answer: No. The correlation between xt and xt+h is the same positive value
obtained in part c) no matter how large h is. In other words, no matter how
far apart xt and xt+h are, their correlation is always the same. Of course, the
persistent correlation over time is due to the presence of the time-constant
variable z.
Question 4 [20 marks]
a) Briefly describe the concepts of strict and weak exogeneity in time series anal-
ysis. Also discuss the concept of unobserved heterogeneity in the context of
panel data methods. [5 marks]
b) When do the Fixed Effects (FE) and the First-Differenced (FD) estimators pro-
vide the same answer? In general, which one is preferable and why? [5 marks]
c) Does the Random Effects (RE) estimator provide any advantage over the pooled
OLS estimator? [5 marks]
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d) Briefly describe the Hausman Test. [5 marks]
Question 5 [20 marks]
We would like to investigate the potential determinants of the total amount of time
spent sleeping at night. This is done by estimating the following model:
sleep = β0 + β1totwrk + β2educ+ β3age+ β4male+ u.
The variable sleep is total minutes per week spent sleeping at night, totwrk is the
total weekly minutes spent working, educ and age are measured in years, and male
is a gender dummy (equal to 1 if the individual is a male and 0 otherwise). You
estimate the model by OLS and obtain the following results:
a) All other factors being equal, is there any evidence that men sleep more than
women? How strong is the evidence? [6 marks]
Answer: The estimate of the regression coefficient, β4, is positive: βˆ4 = 87.99
and statistically significant (tβˆ4 = 2.56). Therefore, according to the estimation
output, we can safely say that male tend to sleep more than females in our
sample. This evidence is relatively strong since the corresponding p-value is
equal to 0.011, meaning we can reject the null hypothesis at the 5% signifi-
cance level, although not at the 1% significance level.
b) Is there a statistically significant tradeoff between working and sleeping? What
is the estimated tradeoff? [7 marks]
Answer: Yes, there is a tradeoff. Indeed, the OLS estimate of β1 is nega-
tive: βˆ1 = −0.166 and statistically significant (tβˆ1 = −9.23, p-value of 0.000).
Therefore, intuitively, the more you work, the less you sleep. It would be nice
to understand if this is correlated with the type of job, going beyond the simple
amount of work, but this is not part of the exercise.
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c) Suppose you now want to investigate the compounded effect of being male and
working hard on the amount of time spent to sleep. Then, you estimate a model
with an interaction effect:
sleep = β0 + β1totwrk + β2educ+ β3age+ β4male+ β5male× totwrk + u,
and obtain the following results:
How would you interpret the estimation results? [7 marks]
Answer: According to the model estimates, there is no statistically significant
effect of the amount of work within the group males. As a result, hard working
men still tend to sleep more than females. The OLS estimate of β5 is slightly
negative (βˆ5 = −0.042) and not statistically significant (tβˆ5 = −1.16, p-value of
0.248). More precisely, the interaction between being a man and working hard
does not significantly deflate the impact of being a man on the amount of sleep
time.