程序代写代做代考 Alastair Hall ECON61001: Semester 1, 2020-21 Econometric Methods

Alastair Hall ECON61001: Semester 1, 2020-21 Econometric Methods
1 . ( a ) 1.(b)
1.(c)
Solutions to Problem Set for Tutorial 4
W e h a v e T 1 / 2 v ̄ T = T − 1 / 2 ι ′T v . U s i n g L e m m a 2 . 1 w i t h w = T 1 / 2 v ̄ T , c = 0 , C = T − 1 / 2 ι T , μ = 0 andΩ=IT,weobtainT1/2v ̄T ∼N(0,T−1ι′TιT). Sinceι′TιT =T,wehaveT1/2v ̄T ∼N(0,1).
Theevent|v ̄T|0,wehaveT1/2n→∞asT→∞;solimT→∞P(|v ̄T| 1/2 then the resulting statistic diverges.
ui and xi cannot be independent because of the discrete sample space for yi. Specifically, since the sample space of yi is {0, 1}, ui can only take one of two values given xi: 1 − x′iβ0 or −x′iβ0. Therefore, the sample space for ui depends on xi.
Since there only two possible outcomes for ui conditional on xi, the conditional distribution for ui given xi cannot be normal.
Recall that σˆN2 = e′e/(N − k) = u′(IN − P)u/(N − k) for P = X(X′X)−1X′. Multiplying out and for convenience multiplying and dividing by N, we obtain
σˆN2 = 􏰘 N 􏰙􏰋N−1u′u − N−1u′X(N−1X′X)−1N−1X′u􏰌 N−k
􏰘N􏰙􏰆NNNN􏰇
= N−k N−1􏰈u2i −N−1􏰈uix′i(N−1􏰈xix′i)−1N−1􏰈xiui (1)
i=1 i=1 i=1 i=1
We now analyze the large sample behaviour of the terms of the right hand side of the previous
equation. First note that as k is finite, we have limN →∞N/(N − k) = 1. For the remaining 1
2.(i)
2.(ii) 3.

terms, we note they all involve sums of i.i.d random variables. As in Lecture 4, we can apply the WLLN to deduce:
N
N−1􏰈xix′i →p Q,
i=1 N
N − 1 􏰈 x i u i →p 0 . t=1
Now consider, N−1􏰔Ni=1u2i. Since E[ui] = 0, we have E[u2i] = Var[ui] = σ02 under our assumptions. Therefore, it follows from the WLLN that
N
N−1􏰈u2i →p σ02.
i=1
From (1), it can be seen that σˆN2 is a continuous function of N−1 􏰔Ni=1 u2i , N−1 􏰔Ni=1 xix′i, N−1 􏰔Ni=1 xiui and N/(N − k). From Slutsky’s Theorem, it follows that σˆN2 converges in probability to the corresponding function of the limits of these terms, and so we have:
σˆN2 →p (1)􏰋σ02−0×Q−1×0􏰌=σ02.
4. From the solutions to Tutorial 2, Question 2, we have:
γˆN = β0,1 + (X1′ X1)−1X1′ X2β0,2 + (X1′ X1)−1X1′ u,
where the ith row of Xl is x′l,i for l = 1, 2, from which it follows that
γˆN = β0,1 + (N−1X1′X1)−1N−1X1′X2β0,2 + (N−1X1′X1)−1N−1X1′u. (2)
From (2), it can be seen that γˆN is a continuous function of N−1X1′X1, N−1X1′X2 and N−1X1′u. By similar arguments to Lecture 4, we have from the WLLN that N−1X1′u →p 0, and (decomposing X)
N−1X′X = 􏰄 N−1X1′X1 N−1X1′X2 􏰅 →p 􏰄 Q1,1 Q1,2 􏰅 = Q. N − 1 X 2′ X 1 N − 1 X 2′ X 2 Q 2 , 1 Q 2 , 2
Thus, using Slutsky’s Theorem, it follows from (2) and these applications of the WLLN that: γˆN →p β0,1 + Q−1Q1,2β0,2.
1,1
5. Recall from Lecture 4 that we can test H0 : g(β0) = 0 vs. H1 : g(β0) ̸= 0 using the statistic:
W(g) = Ng(βˆN)′ 􏰉G(βˆN)(N−1X′X)−1G(βˆN)′ 􏰊−1 g(βˆN)/σˆN2 N
where G(β ̄) = ∂g(β)/∂β′| ̄, and that under H0: W(g) →d χ2n where ng is the number of β=β Ng
restrictions. For this question, g(β0) = β0,2β0,3 −1. Therefore, we have G(β) = [0, β3, β2, 0, 0]. 2

Note that since we are given that β0,i ̸= 0 for i = 2, 3, it follows that rank{G(β0)} = 1 = ng. Using these results to specialize W(g) to this case, we obtain:
N
( g ) 􏰀 βˆ N , 2 βˆ N , 3 − 1 􏰁 2
W N = σˆ N2 􏰀 βˆ N2 , 3 m 2 , 2 + βˆ N2 , 2 m 3 , 3 + 2 βˆ N , 2 βˆ N , 3 m 2 , 3 􏰁
where mi,j is the (i,j)th element of (X′X)−1. The decision rule is to: reject H0 at the
approximate 100α% significance level if W (g) > c1(1 − α) where c1(1 − α) is the 100(1 − α)th
percentile of the χ21 distribution.
N
3