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Midterm Exam 2: ECON 141
UC Berkeley
Fall 2012
Suggested Solutions
November 6th, 2012
Please do not turn this page over until instructed to do so.
Instructions (PLEASE READ CAREFULLY before starting):
• This test has a total of 3 questions and 100 points. You have 1h 20m to solve it, that is, 80
minutes.
• Show your work, unless you are explicitly told not to! No credit will be given for correct answers
if you do not justify your argument.
• Please be sure that your handwriting is legible!
• Be precise but brief. If a correct reply is hidden among wrong, or irrelevant, arguments, you will
not get full credit.
• We will grade only what is written on your exam sheet. There should be plenty of space for all your
calculations. You will be provided extra sheets for auxiliary calculations, which you do not want/need
to include in your answer. Content in these sheets will not be graded. If you need extra
sheets, we will provide them..
• If time is running short, you should try to set up the problem without doing the final calculations.
Name:
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Question 1 (40 points)
Give a brief answer, explanation, and/or mathematical derivation to the five questions below. All the
parts within this question have equal weight.
Question 1.a: In a fixed effect regression model, Yit = αi + uit , where i = 1, ..., n and t = 1, ..., T and
n → ∞ but T is fixed. Are the fixed effects, αi , consistently estimatated? True or False? Explain.
Question 1.b: Let T Fit be the rate of traffic fatalities in state i at time t and BTit be the tax on beer
in state i at time t. Suppose you were considering the fixed effect regression model
T Fit = β0 + β1 BTit + αi + uit .
However, a researcher told you that traffic fatalities increase when roads are more icy. So states with more
snow will have more fatalities. She/He proposes the following two methods to estimate the effect of snow
on fatalities:
a. Collect data on the average (across time) snowfall for each state (i.e., Snowi ) and add it as a regressor.
b. Collect data on snowfall for each year and state (i.e., Snowit ) and add it as a regressor.
Which one would you prefer? And why?
Question 1.c: What two properties define an intrumental variable, Z (for the endogenous variable X,
where Y = βX + u)?
Question 1.d: Consider the following two models:
(a)
Yit = β1 Xit + αi + λt + uit .
(b)
Yit = β0 + β1 Xit + γ2 B2t + ... + γT BTt + δ2 D2i + ... + δn Dni + uit .
Where Bst = 1 it t = s and 0 otherwise with s = 2, ..., T and Dji = 1 if i = j and 0 otherwise with
j = 2, ..., n.
How are the coefficients (β1 , α1 , ..., αn , λ1 , ..., λT ) and β0 , β1 , γ2 , ..., γT , δ2 , ..., δn related?
Question 1.e: Consider the following production function Y = λK β1 Lβ2 eu where Y is output, K is
capital, L is labor. Suppose you have data on these variables, how would you use LS regression analysis
to estimate the parameters? Hint: Writing down the LS regression will suffice.
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Answers for Question 1
(a) The fixed effects cannot be consistently estimated, this is due to the fact that each estimate only relies
on T observations. In a dummy variable regression setup we have that α̂i is estimated as an individual
average i.e.
α̂i =
T
T
1X
1X
Yit = αi +
uit
T
T
t=1
t=1
with
M SE(α̂i ) = V ar(α̂i ) =
1
V ar(uit )
T
which doesn’t go to zero as n → ∞ and T is fixed.
(b) The question ask for an effect of snow on fatalities so uou would prefer a time varying estimate of snow
since the coefficient on a time invariant covariate such as Snowi can’t be estimated in a FE model.
(c) The two properties that an instrumental variable must satisfy are relevance and exogeneity i.e.
Cov(X, Z) 6= 0
and
Cov(u, Z) = 0
(d) The relation between the coefficients in the two models is
β0 = α1 + λ1
β1 = β1
γt = λ t − λ 1
for t = 2, · · · , T
δn = αn − α1
for n = 2, · · · , T
You can derive this if you consider i = t = 1 to find β0 = α1 + λ1 , then consider i = 1, t > 1 and use
the previous finding to derive γt = λt − λ1 and finally look at i, t > 1 together with the two previous
results to find δn = αn − α1 .
(e) Take log’s of the model, then
ln Y = ln λ + β1 ln K + β2 ln L + u
which can be estimated using least squares methods.
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Question 2 (40 points)
Consider the following regression
ln(Wi ) =β̂0 + β̂1 Y Si + β̂2 Di + ûi .
Where Wi is the hourly wage of individual i (in U$S dollars), and Y Si is the years of schooling of individual
i, and Di takes value 1 if i is female, 0 otherwise.
Let β̂0 = 3, SE(β̂0 ) = 1; β̂1 = 0.05, SE(β̂1 ) = 0.01; β̂2 = −0.21, SE(β̂2 ) = 0.1.
• 2.a What is the economic interpretation of β̂2 ? Be precise about the units.
• 2.b What is the economic interpretation of β̂1 ? Be precise about the units.
• 2.c Do you expect β̂1 to be consistent? Why or Why not?
• Consider the following model
ln(Wit ) =β0 + β1 Y Sit + αi + uit .
Where Wit is the the hourly (average) wage of a state i (in U$S dollars) at time t, and Y Sit is the
(average) years of schooling in state i at time t.
• 2.d Suppose that an applied researcher suggests you to include another regressor, Zi that does not
change over time. Discuss the merits of his proposal, in the case you are running Fixed effect
estimators. Would your answer change if you were running Random Effect estimators?
• 2.e Suppose the Hausman test statistic equals 100. What does this tell you about the validity of
using random effects estimator vs. fixed effects estimator? Would the Random Effect be consistent?
What about the Fixed effects? Hint: The 99 percentile of a χ2q with q = 2 is (approx) 9.2.
Answers for Question 2
(a) The economic interpretation of β̂2 is the log-wage gap between men and women with the same amount of
education. Here the model is log-linear (and the size of β̂2 is small so we can rely on an approximation)
to conclude that women earn on average 21% less than men with the same amount of education.
(b) The causal interpretation of β̂1 is as the monetary returns to additional schooling. In this context a
person who obtains an extra year of schooling will increase hourly wages by 5%.
(c) You shouldn’t expect β̂1 to be consistent. An omitted variable such as cognitive ability which affects
both wages and schooling will bias the results.
(d) The inclusion of a time invariant covariate when using a FE estimator isn’t feasible since it is perfectly
collinear with the fixed effect and so will be differenced out. If you are using a RE estimator then this
will increase the reliability of your estimator since the assumptions of the RE model are more likely to
hold when controlling for multiple factors.
(e) A Hausman test statistic equal to 100 means that the RE model is rejected at any confidence level.
Hence you should use a FE estimator.
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Question 3 (20 points)
Consider the following panel data model
Yit = β0 + β1 Xit + αi + uit .
(1)
• 3.a (6 points) From equation (1) derive the following equation
Ȳi = β0 + β1 X̄i + αi + ūi .
(2)
• 3.b (8 points) Let β̃1 be the OLS estimate of β1 in equation (2). Suppose that X̄i have all the same
variance. Show that
σXα
plimn→∞ β̃1 = β1 +
V ar(X̄i )
with T fixed.
• 3.c (6 points) Suppose further that Xit are uncorrelated across time. What does the previous
σXα
) when T increases.
equation tell you about the behavior of the asymptotic bias (i.e., V ar(
X̄ )
i
Answers for Question 3
(a) Average of t for each individual to get the desired equation
Ȳi =
T
T
1X
1X
Yit =
(β0 + β1 Xit + αi + uit )
T
T
t=1
= β0 + β1
t=1
1
T
T
X
Xit + αi +
t=1
T
1X
uit
T
t=1
= β0 + β1 X̄i + αi + ūi .
(b) The OLS estimator in equation (2) is
1 Pn
(X̄i − X̄)Ȳi
n
= β1 +
β̃1 = 1 Pi=1
n
2
i=1 (X̄i − X̄)
n
1
n
Pn
i=1 (X̄i − X̄)(αi +
1 Pn
2
i=1 (X̄i − X̄)
n
ūi )
Analyzing the numerator and denominator separately and combining results is feasible due to Slutsky’s
theorem. For the denominator we have (using µX := E[Xit ])
n
n
1X
1X
(X̄i − X̄)2 =
(X̄i − µX + µX − X̄)2
n
n
i=1
i=1
n
n
n
1X
1X
2X
2
2
=
(X̄i − µX ) +
(µX − X̄) +
(µX − X̄)(X̄i − µX )
n
n
n
i=1
i=1
i=1
n
n
1X
2X
2
2
=
(X̄i − µX ) + (µX − X̄) + (µX − X̄)
(X̄i − µX )
n
n
i=1
i=1
n
1X
=
(X̄i − µX )2 − (µX − X̄)2
n
i=1
→p V ar(X̄i ) + 0
as n → ∞
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where convergence follows by the LLN and Slutsky’s theorem. Similarly for the numerator we have
n
n
n
i=1
i=1
n
X
1X
1X
1X
(X̄i − X̄)(αi + ūi ) =
(X̄i − µX )(αi + ūi ) +
(µX − X̄)(αi + ūi )
n
n
n
=
1
n
i=1
(X̄i − µX )αi +
i=1
1
n
n
X
n
(X̄i − µX )ūi + (µX − X̄)
i=1
1X
(αi + ūi )
n
i=1
→p Cov(X̄i , αi ) + Cov(X̄i , ūi ) + (µX − µX )E(αi + ūi )
=
T
1X
Cov(Xit , αi ) + 0 + 0 = σXα
T
t=1
Combining these two yields
plimn→∞ β̃1 = β1 +
σXα
V ar(X̄i )
(c) When Xit are uncorrelated across time we have V ar(X̄i ) =
T
1
TV
ar(Xit ) so the asymptotic bias is
σXα
V ar(Xit )
which grows larger in T when σXα 6= 0.
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