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 ). 5

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