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Alastair Hall
ECG 790: Topics in Advanced Econometrics
Fall 2006
Solutions to Practice Problem Set 2
1(a). Let xt = (n1,t , n2,t, pt) and α0 = (α0,1 , α0,2 , α0,3) . Suppose that the parameters of the
demand equation, α0 are to be estimated by GMM based on the following moment condition,
E[zt(qt − α0 xt )] = 0
(1)
where zt is a vector of instruments with zt ∈ Ωt. Recall that α0 is identified in this model if
rank{E[ztxt ]} = 3. Since,
(2)
E[ztxt ] = E[ztnt ], E[ztpt ]
we first derive E[ztpt ]. To do this, we need the reduced form associated with the system, that
is:
−1 −1 qt
1 −α0,3
α0,1 α0,2
1 −α0,3
uD
t
=
nt +
(3)
pt
1 −β0
0
0
1 −β0
uSt
This implies that
pt =
−1
1
(α0,1 n1,t + α0,2 n2,t) +
(uS − uD
t )
α0,3 − β0
α0,3 − β0 t
(4)
From (4) it follows that
E[ztpt ] =
−1
1
(α0,1 E[ztn1,t ] + α0,2 E[ztn2,t ]) +
E[zt(uSt − uD
t )]
α0,3 − β0
α0,3 − β0
(5)
Using E[uit|Ωt] = 0 for i = D, S and zt ∈ Ωt , it follows from (5) that
E[ztpt ] =
−1
(α0,1 E[ztn1,t ] + α0,2 E[ztn2,t ])
α0,3 − β0
(6)
Therefore, it can be seen from (2) that the third column of E[ztxt ] is a linear combination
of the first two columns, and so rank{E[ztxt ]} cannot exceed two. It follows that α0 is not
identified within this model.
1(b). Suppose that the parameters of the supply equation, β0 are to be estimated by GMM based
on the following moment condition,
E[zt(qt − β0 pt)] = 0
(7)
where zt is a vector of instruments. Recall that β0 is identified in this model if rank{E[ztxt ]} =
1. From (5), it follows that this condition is satisfied if zt contains variables that are correlated
with n1,t and/or n2,t . The obvious choices are n1,t , n2,t or nt . With any of these choices, β0
is identified.
1
1(c) Specializing the order condition to this model, we have: “if an equation is to have all its
parameters identified, at least one of its parameters must be believed to be zero” (Christ,
1994, p.39). Such a restriction is often refered to as an exclusion restriction because if a parameter is believed to be zero then the associated variable is excluded from the right hand side.
In this example there are two exogenous variables, n1,t and n2,t: neither are excluded from
the demand equation and so this equation fails the order condition; both are excluded from
the supply equation and so the order condition holds.
The order condition is only necessary; for further discussion see Sections 15.3.1-15.3.3 in
Greene, 2003, Econometric Analysis, fifth edition.
2(a) Recall that IV estimation in this model is GMM based on E[zt(yt − xt β0 )] = 0. Putting
zt = xt yields
(8)
E[xt(yt − xt β0 )] = 0
Notice that β0 is just identified by (8) and so the IV estimator is the Method of Moments
estimator based on (8), that is βˆT where
X (y − X βˆT ) = 0
(9)
Equation (9) is the first order condition associated OLS estimation and so βˆT is the OLS
estimator.
2(b) From Assumption 2.3, the condition for identification is rank{E[xtxt]} = p. In ECG 751,
you considered the linear regression model. In the analysis there, it was assumed that
limT →∞T −1 Tt=1 Et[xtxt ] = Q where Q is a nonsingular matrix. This assumption can be
viewed as a generalization of the condition in the classical regression model that the regressors
are not collinear and hence that rank(X X) = p. If xt is a stationary process then Q =
E[xtxt ]. Therefore, it can be seen that this rank condition in the linear regression analysis is
in fact an identification condition.
3(a). By definition, u
ˆt = yt − xt θˆT and so using yt = xt θ0 + ut , we have
u
ˆt = ut − xt(θˆT − θ0 )
(10)
= (Z X)−1 Z y
(11)
Since p = q, we have
θˆT
= θ0 + (Z X)
−1
Zu
(12)
where (12) is obtained by substituting y = Xθ0 + u into (11). Substituting (12) into (10) and
squaring, we obtain
u
ˆ2t = u2t − 2ut xt(Z X)−1 Z u + u Z(X Z)−1 xt xt (Z X)−1 Z u
2
(13)
3(b). By definition, we have
SˆT = T −1
T
u
ˆ2t zt zt
(14)
t=1
p
The proof strategy is as follows: we show that {SˆT }i,j → {S}i,j , where {.}i,j denotes the
i − j th element of the matrix in curly brackets, for all i, j.
By definition, we have:
{SˆT }i,j = T −1
T
u
ˆ2t zt,i zt,j
(15)
t=1
where zt,i is the ith element of zt . Substituting from (13) into (15), it follows that
{SˆT }i,j = T −1
T
u2t zt,i zt,j + u Z(X Z)−1 T −1
t=1
−2T −1
T
zt,i zt,j xt xt(Z X)−1 Z u
t=1
T
zt,i zt,j ut xt (Z X)−1 Z u
(16)
t=1
We now consider the terms on the right hand side. To this end recall from ECG751, the
following result:
Weak Law of Large Numbers: If {wt} is an i.i.d. scalar sequence with E[wt] = μ < ∞ then
p
T −1 Tt=1 wt → μ.
• Assuming that E[u2t zt,i zt,j ] = {E[u2 zz ]}i,j < ∞, it follows from the Assumptions 2.1,
2.4 and the WLLN above that
{SˆT }i,j = {E[u2zz ]}i,j
(17)
• For the second term in (16), note that
u Z(X Z)−1
T
zt,i zt,j xt xt (Z X)−1 Z u = T −1 u Z(T −1 X Z)−1 T −1
t=1
× (T
−1
Z X)
−1
T
−1
T
zt,i zt,j xt xt
t=1
Zu
(18)
p
Under Assumptions 2.1, 2.2 and 2.3, we have from the WLLN that T −1 Z u → 0. Under
p
Assumptions 2.1, 2.3 and 2.4, we have from the WLLN that T −1 Z X → E[ztxt ], a
p
nonsingular matrix and so using Slutsky’s Theorem we have (T −1 Z X)−1 → (E[ztxt ])−1 .
Under Assumptions 2.1 and 2.4 and assuming that E[zt,izt,j xt,r xt,s ] = m(i, j, r, s) < ∞
for all r, s = 1, 2, . . . p (where xt,r is the r th element of xt ), it follows from the WLLN that
p
p
T −1 Tt=1 zt,izt,j xt,r xt,s → m(i, j, r, s) and hence that T −1 Tt=1 zt,i zt,j xt xt → M (i, j)
3
where M (i, j) is the p × p matrix with r − sth element m(i, j, r, s). It follows from these
limiting results, (18) and Slutsky’s Theorem that
u Z(X Z)−1
T
p
zt,i zt,j xt xt(Z X)−1 Z u → 0
(19)
t=1
• For the third term in (20), note that
T −1
T
zt,i zt,j ut xt (Z X)−1 Z u = T −1
t=1
T
zt,i zt,j ut xt (T −1 Z X)−1 T −1 Z u
(20)
t=1
It has already been shown that (T −1 Z X)−1 = Op(1) and T −1 Z u = op (1). Assuming
that E[zt,izt,j ut xt,r ] = n(i, j, r) < ∞ for r = 1, 2, . . .p, it follows from the WLLN that
T −1 Tt=1 zt,izt,j ut xt = Op (1). It therefore follows from (21) that
T −1
T
p
zt,izt,j ut xt (Z X)−1 Z u → 0
(21)
t=1
p
Finally it follows from (16), (17), (19) and (21) that {SˆT }i,j → {S}i,j . For this to hold
p
for i, j = 1, 2, . . .q and hence SˆT → S, we require E[zt,izt,j xt,r xt,s] = m(i, j, r, s) < ∞ for
all i, j = 1, 2, . . . q, r, s = 1, 2, . . .p and E[zt,izt,j ut xt,r ] = n(i, j, r) < ∞ for i, j = 1, 2, . . .q,
r = 1, 2, . . .p in addition to Assumptions 2.1-2.4.
3(c). If E[u2 zz ] = σ02 E[ztzt ] then SˆCIV is not consistent for S. This scenario would mean that
−1
that does not
the second step GMM estimator is calculated with a weighting matrix SˆCIV
−1
converge to S and hence this second step estimator does not achieve the efficiency bound
for estimation based on the population moment condition in Assumption 2.2. This estimator
−1
satisfies the conditions of Assumption 3.7. The
would still be consistent, however, as SˆCIV
confidence intervals are no longer approximate 95% intervals - in the sense that the coverage probability is not 95% even in the limit - because the standard errors are estimated
inconsistently.
4(a). We have V ar[λ ht ] = λ Ωλ. Since Ω is positive definite, it follows that λ Ωλ > 0 for all
λ = 0. We now turn to the finiteness of λ Ωλ. Let ch = maxj V ar[ht,j ] and cλ = maxj λj
where λj is the j th element of λ; since V ar[ht,i] < ∞ for alli, ch < ∞; since λ = 1,
|cλ | ≤ 1. By the Cauchy-Schwarz inequality, |Cov[ht,i , ht,j ]| < V ar[ht,i]V ar[ht,j ] ≤ ch , and
so λ Ωλ ≤ λ Ch λ ≤ c2λ ιk Ch ιk , where Ch is a k×k matrix every element of which is ch , and ιk =
is k × 1 vector of ones. Now, ιk Ch ιk = k2 ch and so V ar[λ ht ] = λ Ωλ ≤ k2 ch c2λ ≤ k2 ch < ∞.
hT − μ) = T −1/2 Tt=1 vt where vt = λ Ω−1/2 (ht − μ). Now vt satisfies:
4(b). We have λ Ω−1/2 T 1/2 (¯
(i) {vt} is iid because {ht} is iid;
(ii) E[vt] = 0 because E[ht − μ] = 0;
4
(iii) V ar[vt ] = λ Ω−1/2 V ar[ht − μ]Ω−1/2 λ = λ Ω−1/2 ΩΩ−1/2λ = 1 because Ω = Ω1/2Ω1/2 ⇒
{Ω1/2}−1 Ω{Ω1/2}−1 = Ik and Ω−1/2 = {Ω1/2}−1 (which in turn implies Ω−1/2 =
{Ω1/2}−1 ).
Notice that (i)-(iii) hold for any λ such that λ = 1, and so for any λ it follows from
d
the Lindberg-Levy CLT that T −1/2 Tt=1 vt = λ Ω−1/2 T 1/2 (¯hT − μ) → N (0, 1). Now define
z ∼ N (0k , Ik ), and note that λ z ∼ N (0, 1) for any λ such that λ λ = 1. Therefore, using the
d
Cramer-Wold device, it follows that Ω−1/2 T 1/2 (¯hT − μ) → N (0k , Ik ).
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