Alastair Hall ECON61001: Semester 1 2020-21 Econometric Methods
Solutions to Problem Set for Tutorial 8
1.(a) We have:
ui(β) = yi −x′iβ = x′iβ0 +ui −x′iβ = ui +x′i(β0 −β). Therefore, it follows that
E[ziui(β)] = E[ziui(β0)] + E[zix′i](β0 − β).
Since the population moment condition is valid and so E[ziui(β0)] = 0 it follows that
E[ziui(β)] = E[zix′i](β0 − β). (1)
We need conditions under which E[ziui(β)] ̸= 0 for all β ̸= β0. Notice that (1) represents a set of linear equations in E[ziui(β)] that is, they are of the form “Av = b” where A = E[zix′i], v = (β0 − β) and b = E[ziui(β)]. From linear algebra theory, the condition for b ̸= 0 for v ̸= 0 iff A is full column rank. Since E[zix′i] is 2 × 2 here, the condition is that rank{E[zix′i]} = 2.
1.(b) Using the hint, E[zix′i] is full rank if it is nonsingular or equivalently if det{E[zix′i]} ≠ 0.
Since
it follows that
E[zix′i]) = E 1 x2,i , z2,i z2,ix2,i
det{E[ztx′t]}=E[z2,ix2,i]−E[x2,i]E[z2,i] = Cov(z2,i,x2,i). Therefore, the relevance condition is that Corr(z2,i, x2,i) ̸= 0.
1.(c) The relevance condition is satisfied if the instrument z2,i is linearly related to the variable for which it is used as an instrument x2,i.
2.(a) Multiplying out and using , we have:
QIV (β) = y′Z(Z′Z)−1Z′y − y′Z(Z′Z)−1Z′Xβ − β′X′Z(Z′Z)−1Z′y + β′X′Z(Z′Z)−1Z′Xβ. Since β′X′Z(Z′Z)−1Z′y is a scalar and so equal to its transpose, it follows that:
β′X′Z(Z′Z)−1Z′y = (β′X′Z(Z′Z)−1Z′y)′ = y′Z(Z′Z)−1Z′Xβ, (using (AB)′ = B′A′). Therefore, we can write:
QIV (β) = y′Z(Z′Z)−1Z′y − 2y′Z(Z′Z)−1Z′Xβ + β′X′Z(Z′Z)−1Z′Xβ. 1
Using the the matrix differentiation results in Tutorial 1 Question 3, we have:
The FOC are and so imply:
∂(y′Z(Z′Z)−1Z′Xβ) ∂β
∂QIV (β) = −2X′Z(Z′Z)−1Z′y + 2X′Z(Z′Z)−1Z′Xβ. ∂β
∂QIV(β) = 0, ∂ β β = βˆ I V
= X′Z(Z′Z)−1Z′y,
= 2X′Z(Z′Z)−1Z′Xβ, where the last result uses the symmetry of X′(Z′Z)−1X. Therefore, we have:
∂(β′X′Z(Z′Z)−1Z′Xβ) ∂β
−X′Z(Z′Z)−1Z′y + X′Z(Z′Z)−1Z′XβˆIV = 0
Given that X′Z and Z′Z are full rank, X′Z(Z′Z)−1Z′X is nonsingular, it follows that:
βˆIV = X′Z(Z′Z)−1Z′X −1 X′Z(Z′Z)−1Z′y.
(b) If q = k then X′Z is square matrix of full rank and so nonsingular. Using the result that
(ABC)−1 = C−1B−1A−1, we have:
X′Z(Z′Z)−1Z′X −1 = (Z′X)−1{(Z′Z)−1}−1(X′Z)−1 = (Z′X)−1Z′Z(X′Z)−1,
and substituting this result into the formula for βˆIV , we obtain:
βˆIV = (Z′X)−1Z′Z(X′Z)−1X′Z(Z′Z)−1Z′y.
Using in turn that (X′Z)−1X′Z = Ik and Z′Z(Z′Z)−1 = Ik, the formula simplifies to: βˆIV = (Z′X)−1Z′y.
3. A suitable test statistic is
W(IV) = N(RβˆIV −r)′[RVˆIVR′]−1(RβˆIV −r)
N
where Vˆ = Mˆ−1Qˆ Qˆ−1Ωˆ Qˆ−1Qˆ′ Mˆ−1, Mˆ = Qˆ Qˆ−1Qˆ′ , Qˆ = N−1 N z x′, Ωˆ =
IV xz zz h zz xz xz zz xz xz i=1 i i h N − 1 Ni = 1 u ˆ 2 i z i z i ′ a n d u ˆ i = y i − x ′ i β ˆ I V .
Aside: Under the null hypothesis, we have:
N1/2(RβˆIV − r) →d N(0,RVIVR′) 2
where V = M−1Q Q−1Ω Q−1Q′ M′, M = Q Q−1Q′ and Ω = E[h(z )z z′]. Using IV xz zzp h zz xz p xz zz xz h i i i
the WLLN, we have Qˆxz → Qxz and Qˆzz → Qzz. Using similar arguments to White’s het- eroscedasticity covariance matrix estimator, we have Ωˆh →p Ωh. Therefore, using Slutsky’s Theorem, we can deduce that RVˆIV R′ →p RVIV R′. So using Lemma 3.6 in the Lecture notes,
W I V →d χ 2n r .
A suitable decision rule is to reject H0 at the (approximate) 100α% significance level if
W (I V ) > cnr (α) where cnr (α) is the 100(1 − α)th of the χ2n distribution. Nr
4. First consider the orthogonality condition E[ztut] = 0. Here we have: E[ztut] = E[ut] .
and so we need to show that E[ut] = 0 and E[yt−2ut] = 0. From Tutorial 5 Question 3 part (a), E[ut] = 0. Now consider yt−2ut. Using the hint, it follows that E[yt−2ut] is a linear function E [ut] and E [utut−j ] for j = 2, 3, . . .. Using Tutorial 5 Question 3 parts (a) & (c), it follows that E[ut] = 0 and E[utut−j ] = 0 for j = 2, 3, . . .. Therefore, E[yt−2ut] = 0.
Now consider the relevance condition. Here the relevance condition is that rank{E[ztx′t]} = 2. Using Question 1 above, this condition is equivalent to Corr(yt−1, yt−2). Since we are given that the process is weakly stationary, we can use the hint to deduce that
Corr(yt−1,yt−2) = (φ+β0,2)(1+φβ0,2). 1 + 2φβ0,2 + φ2
It can be recognized that the relevance condition is satisfied because we are given that β0,2 and φ are both less than one in absolute value and φ ̸= −β0,2.
5.(a) The relevance condition is that rank{E[zixi]} = 1 and as E[zixi] is a scalar this is equivalent to E[zixi] ̸= 0. However, we are given that E[zixi] = 0, and so the relevance condition fails.
5.(b) Following the same steps as led to (1) above, we have E[ziui(β)] = E[zixi](β0 − β).
and so as E[zixi] = 0, it follows that E[ziui(β)] = 0 for all β.
5.(c) If E[ziui(β)] = 0 for β then being told that E[ziui(β0)] = tells us nothing unique about β0. In other words, the IV estimation is based on a moment condition that is uninformative about β0. Therefore intuition suggests that βˆIV is not consistent. (In fact, the IV estimator does not converge to a constant in this case but to a random variable. To see this, we write:
Ni=1 ziui
βIV =β0+N zx. (2)
i=1 i i
E[yt−2ut]
ˆ
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Recall that when the instruments satisfy the orthogonality and relevance conditions then we establish consistency by dividing the numerator and denominator of the ratio of the right- hand side of (2) by N and then applying the WLLN to N−1 Ni=1 ziui and N−1 Ni=1 zixi. However, this approach runs in to problems here because terms are converging to zero. Instead to develop an analysis of the large sample behaviour of the estimator, we divide the numerator and denominator of the fraction by N1/2. Then applying the CLT, we have:
N−1/2Ni=1ziui p ξ1
N−1/2 N z x → ξ ∼ N(0, Ψ),
i=1 i i 2
for some pd matrix Ψ. It then follows that βˆIV →p β0 + v where v = ξ1/ξ2.)
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