程序代写代做代考 Alastair Hall ECON61001: Autumn 2020 Econometric Methods

Alastair Hall ECON61001: Autumn 2020 Econometric Methods
Solutions to Problem Set for Tutorial 3
1.(a) Recall from Lecture 2 that under Assumptions CA1-CA6, we have:
βˆT,i − β0,i ∼ Student’s t distribution with T-k df
σˆT √mi,i
where σˆT2 = e′e/(T − k) and e = y − XβˆT . As discussed in the lecture, we base our decision rule on this statistic with β0,i replaced by parameter value that satisfies H0 but is closest to H1 which in this case is β∗,i. Our decision rule is then as follows: reject H0 in favour of H1 at the 100α% significance level if
βˆ T , i − β ∗ , i > τ ( 1 − α )
T − k
σˆ √ m
T i,i
where τT −k (1 − α) is the 100(1 − α)th percentile of the Student’s t distribution with T − k degrees of freedom.
1.(b) The test statistic is
βˆT,i = 0.067 = 3.190 σˆT √mi,i 0.021
Since T − k = 91, the critical value is τ91(0.95) = 1.662 to 3dp, and so we reject H0 at the 5% significance level. This evidence supports the view that increasing the speed limit lead to a higher percentage of traffic accidents involving fatalities.
2.(a) Put R = [0,1,0,−1,0], r = 0.
2.(b) Put
R=
non–zero then R is full rank. For (a): det(RR′) = 2; for (b) det(RR′) = 10/3.
3.(a) The ith row of Ik is a 1×k vector whose ith element is 1 and whose remaining elements are all zero. So in this case, Rβ0 = β0,i. It then follows from the definition of r that for the stated choices of R, r, Rβ0 = r is equivalent to β0,i = β∗,i.
1
􏰄 1 1 −1 0 0 􏰅 00 0 1/3−1
􏰄0􏰅 −1
r=
The easiest way to check rank(R) = nr is to calculate det(RR′). If this determinant is

3.(b) Recall from Lecture 3 that the generic structure of the F-statistic for testing H0 : Rβ0 = r is F = (RβˆT − r)′[R(X′X)−1R′]−1(RβˆT − r).
n r σˆ T2 Frompart(a)itfollowsimmediatelythat: RβˆT −r=βˆT,i−β∗,i. NowconsiderR(X′X)−1R′.
Given the definition of R in part (a), it follows that
R(X′X)−1R′ = [0,0,…1,0,…0](X′X)−1[0,0,…1,0,…0]′
Now[0,0,…1,0,…0](X′X)−1 =m′i wherem′i istheith rowof(X′X)−1,andm′i[0,0,…1,0,…0]′ = mi,i. Therefore in this case, we have (using nr = 1):
(βˆT,i − β∗,i)′[mi,i]−1(βˆT,i − β∗,i) 􏰆 βˆT,i − β∗,i 􏰇2
m. F= σˆ2 =σˆ√
T T i,i
4.(a) First consider E[βˆR,T]. Using the formula for RLS estimator, we have: E[βˆR,T] = E􏰉βˆT − (X′X)−1R′{R(X′X)−1R′}−1(RβˆT − r)􏰊.
Since E[·] is a linear operator and X, R and r are constants (the former from Assumption CA2), it follows that
E[βˆR,T] = E[βˆT] − (X′X)−1R′{R(X′X)−1R′}−1(RE[βˆT] − r). (1) In lecture 2, it is shown that under Assumptions CA1 – CA4 we have E[βˆT] = β0 and
substituting this result into (1),
E[βˆR,T] = β0 − (X′X)−1R′{R(X′X)−1R′}−1(Rβ0 − r). (2)
Finally, using the fact that the restrictions hold, Rβ0 = r, it follows form (2) that E[βˆR,T] = β0. Now consider V ar[βˆR,T ]. Recall that since βˆR,T = β0, we have V ar[βˆR,T ] = E[(βˆR,T − β0)(βˆR,T −β0)′], and so we first obtain an expression for βˆR,T −β0. Using the formula for the RLS estimator, we have:
βˆR,T − β0 = βˆT − β0 − (X′X)−1R′{R(X′X)−1R′}−1(RβˆT − r). (3) Notice that if the restrictions hold, Rβ0 = r, then (3) can be re-written as:
βˆR,T − β0 = βˆT − β0 − (X′X)−1R′{R(X′X)−1R′}−1R(βˆT − β0), (4)
and so we have: where
βˆR,T − β0 = C(βˆT − β0), (5)
C = Ik − (X′X)−1R′{R(X′X)−1R′}−1R = Ik − C1, say. 2

Therefore, using (5), C is a constant (as both X and R are), and E[βˆT ] = β0, it follows that
V ar[βˆR,T ] = E[(βˆR,T − β0)(βˆR,T − β0)′] = E[C(βˆT − β0)(βˆT − β0)′C′],
= CE[(βˆT − β0)(βˆT − β0)′]C′ = CE[(βˆT − E[βˆT ])(βˆT − E[βˆT ])′]C′,
= CVar[βˆT]C′. (6)
In lectures, it is shown that if Assumptions CA1 – CA5 hold then Var[βˆT] = σ02(X′X)−1. Using this result in (6), we obtain:
Var[βˆR,T] = σ02C(X′X)−1C′. (7) Multiplying out and using C = Ik − C1 , we have:
C(X′X)−1C′ = (X′X)−1 − (X′X)−1C1′ − C1(X′X)−1 + C1(X′X)−1C1′ = (X′X)−1 −D1 −D1 +D1
where
= D, (8) D1 = (X′X)−1R′{R(X′X)−1R′}−1R(X′X)−1.
Combining (7) and (8), we obtain V ar[βˆR,T ] = σ02D. (b) Using (1) and E[βˆT ] = β0, we have:
E[βˆR,T] = β0 − (X′X)−1R′{R(X′X)−1R′}−1(Rβ0 − r) = β0 + (X′X)−1R′Aa, where A = {R(X′X)−1R′}−1 and a = Rβ0 − r. The bias of RLS is thus:
E[βˆR,T] − β0 = −(X′X)−1R′Aa = μ, say.
We now show that Rβ ̸= r implies μ ̸= 0. Clearly, Rβ0 ̸= r implies that a ̸= 0. Now consider b = Aa: since A is nonsingular, it follows that a ̸= 0 implies b ̸= 0. Under the assumptions hereR′ isak×nr matrixwithfullcolumnrankandsoifb̸=0thenwehavec=R′b̸=0. Finally, as (X′X)−1 is nonsingular, it follows that c ̸= 0 implies μ = −(X′X)−1c ̸= 0.
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