ECON-AD 101 Midterm: ecn-1-fall-05-midterm-answers
Econometrics I
Midterm Examination
Fall 2005
Answer Key
Please answer all of the questions and show all of your work. If you think that a question is
ambiguous, clearly state how you interpret it before providing an answer. All question parts have
equal weight.
1. Consider the linear regression specification
yi=Xiβ+εi,
where εihas the following distribution for individual i. With probability πithe disturbance
term is drawn from the uniform distribution defined on [−a, a],where a>0is an unknown
parameter, and with probability (1 −πi)the disturbance is drawn from a mean zero normal
distribution with unknown variance σ2
u. Thus the marginal density of εiis
fi(εi)=πi
1
2a+(1−πi)σ−1
uφ(εi
σu
),
where φdenotes the standard normal probability density function. The unknown parameter
vector βis of dimension K.You have access to a random sample consisting of Nobservations
with information on yand X. Assume that the rank of X0Xis K.
1. Define the ordinary least squares (OLS) estimator of β, ˆ
β.
Answer:Thisissimply ˆ
β=(X0X)−1X0y.
2. Is ˆ
βan unbiased estimator of β?Why?
Answer: Yes, this is unbiased since, conditional on πi,the disturbance term is i.i.d.with
E(εi|πi)=πi×0+(1−πi)×0=0
for all πi.
Thus E(εi|Xi)=0for any πi.
3. If πi=πfor all i, is ˆ
βthe best linear unbiased estimator (BLUE) of β?Why?
Answer: Yes, since we know that ˆ
βis unbiased for any pattern of πi,and now we know
also that the variance is constant, being given by
V(εi)=π(2a)2
12 +(1−π)σ2
u.(0.1)
Under mean independence and heteroskedasticity, the Gauss-Markov theorem applies and
ˆ
βis best linear unbiased.
4. If πi=πfor all i, provide a discussion of whether it would be possible to separately
estimate the parameters a, σu,and π.
Answer: The estimator of the variance of ε, given by s2,will be an unbiased estimator
of the quantity given in (0.1), which is a function of all three parameters. Thus from this
information alone these three parameters are not estimable. However, by deriving the
higher order moments of εidentification may be possible. A natural way to proceed would
be to work out the third and fourth population moments, and then determine whether
these three moments (second, third, and fourth) were linearly independent functions of
π, a, and σ2
u.If so, one could replace the population moments with the sample analogs
and solve for the values of these three parameters. These estimators, if they exist, would
be consistent.
5. If πivaries in the population, is the OLS estimator BLUE? If not, under what conditions
would it be possible to find a more efficient linear estimator?
Answer: No, because now we face a case of heteroskedastic disturbances. We could find
amoreefficent esitmator if πiwas a function of observable individual characteristics
and a finite dimensional parameter vector, such as
πi=exp(Ziδ)
1+exp(Ziδ),
where Ziis a vector of observable characteristics of i(could be equal to Xi,for example).
Could then consider performing OLS in the first stage, and recovering the residuals ri.
Then form the nonlinear regression function
r2
i=πi
(2a)2
12 +(1−πi)σ2
u+ξi
=σ2
u+·(2a)2
12 −σ2
u¸πi+ξi,
and estimate the parameters using nonlinear least squares. With consistent estimates of
δ(which determines πiin these parametric assumptions), a, and σ2
u,we can perform
Feasible GLS.
2. Once again, consider the linear regression model
yi=Xiβ+εi,
where
E(εi|Xi)=0,for all i.
In the population the variance of εiis given by
Var(εi|Xi)=Xiδ.
We assume that, at the true value of δ, Xiδ>0for all i.
1. If δis known, define the Generalized Least Squares (GLS) estimator of β, which we will
denote by ˜
β, as well as the covariance matrix of ˜
β.
Answer: ˜
β=(X0A−1X)−1X0A−1y,
where Ais a diagonal element with Xiδas element (i, i).The covariance matrix of the
estimator is
Cov(˜
β)=(X0A−1X)−1.
2
Document Summary
Please answer all of the questions and show all of your work. ambiguous, clearly state how you interpret it before providing an answer. All question parts have equal weight: consider the linear regression speci cation yi = xi + i, where i has the following distribution for individual i. U where denotes the standard normal probability density function. The unknown parameter vector is of dimension k. you have access to a random sample consisting of n observations with information on y and x. Assume that the rank of x 0x is k: de ne the ordinary least squares (ols) estimator of , . Answer: yes, this is unbiased since, conditional on i, the disturbance term is i. i. d. with. E( i| i) = i 0 + (1 i) 0 = 0 for all i. Answer: yes, since we know that is unbiased for any pattern of i, and now we know also that the variance is constant, being given by.
