 # Foreign Direct Investment and Growth Symbiosis: A Semiparametric System of Simultaneous Equations Analysis

• Nadine McCloud , and Subal C. Kumbhakar

## Abstract

We characterize the types of interactions between foreign direct investment (FDI) and economic growth, and analyze the effect of institutional quality on such interactions. To do this analysis, we develop a class of instrument-based semiparametric system of simultaneous equations estimators for panel data and prove that our estimators are consistent and asymptotically normal. Our new methodological tool suggests that across developed and developing economies, causal, heterogeneous symbiosis and commensalism are the most dominant types of interactions between FDI and economic growth. Higher institutional quality facilitates, impedes or has no effect on the interactions between FDI and economic growth.

## Acknowledgement

Elizabeth Byrd and Shamar Stewart provided excellent research assistance. We thank participants at the 2013 China Meeting of the Econometric Society, the 2013 Asian Meeting of the Econometric Society, and the Department of Economics Research Seminar at the University of the West Indies at Mona for helpful comments. McCloud thanks the support of the Mona Research Fellowship from the Office of the Principal at University of the West Indies at Mona. This research was supported in part by computational resources provided by Information Technology at Purdue – Rosen Center for Advanced Computing, Purdue University, West Lafayette, Indiana. This paper has been presented under its previous title “Are There Feedbacks Between Foreign Direct Investment And Economic Growth? A Semiparametric System Of Simultaneous Equations Analysis With Instrumental Variables”.

## Technical Appendix

In this appendix, we assume C ∈ (0, ∞) is an arbitrary bounded constant. Recall that nNT; we use these terms interchangeably. The integral symbol represents a multiple integral of varying dimensions depending on the context in which it is used. We provide the proofs for only Proposition 3.2, Proposition 3.3 and Proposition 3.9 and Theorem 3.4 because the proofs for Corollary 3.7 and Theorem 3.6 and Theorem 3.11 are less involved. Many of the ensuing proofs use convergence in mean square.

## Proof of Proposition Proposition 3.2:.

(i) Note that

(37)S~n=1n(Q1K11/2DθK11/2U1H11Q1K11/2DβK21/2U2H21Q2K21/2DβK11/2U1H11Q2K21/2DγK21/2U2H21):=(S~n,11S~n,12S~n,21S~n,22).

The proofs for S~n,11 and S~n,22 follow directly from Cai and Li (2008) [Proof of Proposition (i)], which yields S~n,11=f~1(z1)DθS1{1+oP(1)} and S~n,22=f~2(z2)DγS2{1+oP(1)}. To complete the proof of Proposition 3.2, it therefore remains to show that (ia) S~n,12=oP(1) and (ib) S~n,21=oP(1).

We now prove part (ib); the proof of part (ia) can be easily established using the approach below.

E[S~n,21]=E{1ni=1Nt=1TβQ2,itU~1,itKh11/2(Z1,itz1)Kh21/2(Z2,itz2)}=βE{Q2,itU~1,itKh11/2(Z1,itz1)Kh21/2(Z2,itz2)}=βE(W2X~1W2X~1(Z1z1)/h1X~1W2(Z1z1)/h1W2X~1(Z1z1)(Z1z1)/h12)×Kh11/2(Z1z1)Kh21/2(Z2z2)=βE(Ω21(Z1,Z2)Ω21(Z1,Z2)(Z1z1)/h1Ω21(Z1,Z2)(Z1z1)/h1Ω21(Z1,Z2)(Z1z1)(Z1z1)/h12)×Kh11/2(Z1z1)Kh21/2(Z2z2)=β(Ω21(u1,u2)Ω21(u1,u2)(u1z1)/h1Ω21(u1,u2)(u1z1)/h1Ω21(u1,u2)(u1z1)(u1z1)/h12)×Kh11/2(u1z1)Kh21/2(u2z2)f(u1,u2)du1du2=β(Ω21(z1+h1c1,z2+h2c2)Ω21(z1+h1c1,z2+h2c2)c1Ω21(z1+h1c1,z2+h2c2)c1Ω21(z1+h1c1,z2+h2c2)c1c1)×h1d1/2h2d2/2h1d1h2d2K11/2(c1)K21/2(c2)f(z1+h1c1,z2+h2c2)dc1dc2=O(h1d1/2h2d2/2)=o(1),

where the second equality is by virtue of Assumption A.1, the fourth equality follows from law of iterative expectations (LIE), the sixth equality uses a change of variable, and the remaining equalities are consequences of changes in the implied canonical differential form, Lebesgue Dominated Convergence Theorem, and Assumption A.2, Assumption A.3, and Assumption A.5.

We now show that Var[s~n,21]0 as Nhjdjhkdk. We define

s~n,21rs:=1ni=1Nt=1TβW2,itrX~1,itsKh11/2(Z1,itz1)Kh21/2(Z2,itz2)

where W2, itr is the r-th element of W2, it and X~1,its is the s-th element of X~1,it. Then, by Assumption A.1, we obtain

Var[s~n,21rs]=β21NT2Var{t=1TW2,itrX~1,itsKh11/2(Z1,itz1)Kh21/2(Z2,itz2)}=V1+V2,whereV1=β2NTVar{W2,i1rX~1,i1sKh11/2(Z1,i1z1)Kh21/2(Z2,i1z2)},V2=β22NTt=1T1(Tt)Cov(V21,V2(t+1)),V21=W2,i1rX~1,i1sKh11/2(Z1,i1z1)Kh21/2(Z2,i1z2),V2(t+1)=W2,i(t+1)rX~1,i(t+1)sKh11/2(Z1,i(t+1)z1)Kh21/2(Z2,i(t+1)z2).

Now, by Assumption A.1 and Assumption A.2, and for a fixed T, it is straightforward to show that V1CNTh1d1h2d2. By similar arguments and using the Cauchy-Schwarz result that |Cov(X,Y)|Var(X)Var(Y) yield |V2|CTNh1d1h2d2. Hence, we obtain Var[s~n,21rs]=O((Nh1d1h2d2)1)=o(1) as required. Therefore, we have

1ni=1Nt=1TβW2,itX~1,itKh11/2(Z1,itz1)Kh21/2(Z2,itz2)=oP(1).

By invoking similar steps to those above, we deduce that

1ni=1Nt=1TβW2,itX~1,it(Z1,itz1)/h1Kh11/2(Z1,itz1)Kh21/2(Z2,itz2)=oP(1),1ni=1Nt=1TβW2,itX~1,it(Z1,itz1)(Z1,itz1)/h12Kh11/2(Z1,itz1)Kh21/2(Z2,itz2)=oP(1).

Thus, the proof of part (i) is complete.

(ii) Note that

(38)Bn=1n(Q1K11/2DθK11/2X~1R¯1+Q1K11/2DβK21/2X~2R¯2Q2K21/2DβK11/2X~1R¯1+Q2K21/2DγK21/2X~2R¯2):=(Bn,11+Bn,12Bn,21+Bn,22).

The proofs for Bn,11 and Bn,22 follow directly from Cai and Li (2008) [Proof of Proposition (ii)], which yields Bn,11=(h12/2)f1(z1)B1(z1)+oP(h12) and Bn,22=(h22/2)f2(z2)B2(z2)+oP(h22). To complete the proof, it remains to show that (iia) Bn,12=oP(1) and (iib) Bn,21=oP(1).

For (iia),

E[Bn,12]=E{1ni=1Nt=1TβQ1,itX~2,itR¯2(Z2,itz2)Kh11/2(Z1,itz1)Kh21/2(Z2,itz2)}
h22E[Bn,12]=Eβ(W1,itX~2,itA2((Z2,itz2)/h2)W1,itX~2,itA2((Z2,itz2)/h2)(Z1,itz1)/h1)×Kh11/2(Z1,itz1)Kh21/2(Z2,itz2)=β(Ω12(u1,u2)A2((u2z2)/h2)Ω12(u1,u2)A2((u2z2)/h2)(u1z1)/h1)×Kh11/2(u1z1)Kh21/2(u2z2)f(u1,u2)du1du2=h1d1/2h2d2/2β(Ω12(z1+c1h1,z2+c2h2)A2(c2)Ω12(z1+c1h1,z2+c2h2)A2(c2)c1)×K11/2(c1)K21/2(c2)f(z1+c1h1,z2+c2h2)dc1dc2=O(h1d1/2h2d2/2).

Thus, E[Bn,12]=O(h1d1/2h2(d2+4)/2). In addition, any (r, s)-entry of the Var(Bn,12) converges to zero.

Similarly, for (iib), we can show that E[Bn,21]=O(h1(d1+4)/2h2d2/2), and any (r, s)-entry of the Var(Bn,21) converges to zero. Therefore, Bn,12=oP(1) and Bn,21=oP(1), which respectively do not statistically dominate Bn, 11 and Bn,22. Hence, the proof of part (ii) is complete.

(iii) Note that

(39)Rn=1n(Q1K11/2DθK11/2X~1R1+Q1K11/2DβK21/2X~2R2Q2K21/2DβK11/2X~1R1+Q2K21/2DγK21/2X~2R2):=(Rn,11+Rn,12Rn,21+Rn,22).

The proofs for Rn, 11 and Rn,22 follow directly from Cai and Li (2008) [Proof of Proposition (iii)], which yield that Rn,11=oP(h12) and Rn,22=oP(h22). To complete the proof, we now show that (iiia) Rn,12=oP(h1d1/2h2(d2+4)/2) and (iiib) Rn,21=oP(h1(d1+4)/2h2d2/2).

We prove part (iiib); by symmetry, the proof of part (iiia) easily follows.

E[Rn,21]=E{1ni=1Nt=1TβQ2,itX~1,itR1(Z1,itz1)Kh11/2(Z1,itz1)Kh21/2(Z2,itz2)}h12E[Rn,21]=βE(W2,itX~1,ith12R1(Z1,it,z1)W2,itX~1,ith12R1(Z1,it,z1)(Z1,itz1)/h1))×Kh11/2(Z1,itz1)Kh21/2(Z2,itz2)=β(Ω21(u1,u2)h12R1(u1,z1)Ω21(u1,u2)h12R1(u1,z1)(u1z1)/h1))×Kh11/2(u1z1)Kh21/2(u2z2)f(u1,u2)du1du2,

where the last equality is a consequence of LIE. Applying a change of variables, the result that h12R1(z1+c1h1,z1)=o(1), Lebesgue Dominated Convergence Theorem, and Assumption A.1 to Assumption A.3, we obtain Rn,21=oP(h1(d1+4)/2h2d2/2) as required. By symmetry, it is straightforward to derive the result that Rn,12=oP(h1d1/2h2(d2+4)/2). Furthermore, any (r, s)-entry of both the Var(Rn,21) and Var(Rn,12) converges to zero. In essence, the terms Rn,11 and Rn,22 stochastically dominate their counterparts. Therefore, we obtain the desired result. □

## Proof of Proposition Proposition 3.3:.

Since E[Tn]=0, we write nD~Var(Tn)=nD~E[TnTn]. Now

(40)E[TnTn]=E(Tn1Tn1Tn1Tn2Tn2Tn1Tn2Tn2),
whereTn1=1nQ1K11/2DθK11/2ϵ1+1nQ1K11/2DβK21/2ϵ2=Tn,11+Tn,12,Tn2=1nQ2K21/2DβK11/2ϵ1+1nQ2K21/2DγK21/2ϵ2=Tn,21+Tn,22.

To prove that

(41)nD~Var(Tn)=f~l~(z)Dθγ2S,

we will show that the off-diagonal block terms for E[TnTn] in A.4 are of smaller order than its (1, 1) and (2, 2) main-diagonal block terms, which are of orders O{(nh1d1)1} and O{(nh2d2)1} respectively.

(i) To compute E[Tn1Tn1], note that

(42)Tn1Tn1=Tn,11Tn,11+Tn,11Tn,12+Tn,12Tn,11+Tn,12Tn,12.

For the first term in (42), we have,

(43)E[Tn,11Tn,11]=Var(Tn,11)=Var{1ni=1Nt=1Tθ[Q1,itϵ1,itKh1(Z1,itz1)]}=V11,1+V11,2,
whereV11,1=θ2nVar{Q1,i1ϵ1,i1Kh1(Z1,i1z1)},V11,2=2θ2nTt=1T1(Tt)Cov(Q1,i1ϵ1,i1Kh1(Z1,i1z1),Q1,i(t+1)ϵ1,i(t+1)Kh1(Z1,i(t+1)z1)).

By Assumption A.1 and Assumption A.2, and invoking similar steps to Cai and Li (2008) [Proof of Proposition 2], we obtain nh1d1V11,1θ2f~1(z1)S1 and

Cov(Q1,i1ϵ1,i1Kh1(Z1,i1z1),Q1,i(t+1)ϵ1,i(t+1)Kh1(Z1,i(t+1)z1))=E{Q1,i1Q1,i(t+1)ϵ1,i1ϵ1,i(t+1)Kh1(Z1,i1z1)Kh1(Z1,i(t+1)z1)}f1,1(t+1)(z1,z1)(G1,t+1(11,11)(z1,z1)00G1,t+1(11,11)(z1,z1)μ1,2(K12)).

Hence, V11,2 = O(n−1) and therefore by virtue of Assumption A.2 we obtain

nh1d1Var[Tn,11]θ2f~1(z1)S1.

For the second term in (42), and using Assumption A.1 we have,

E[Tn,11Tn,12]=1n2Ei=1Nt=1Tθβ[Q1,itQ1,itϵ1,itϵ2,itKh13/2(Z1,itz1)Kh21/2(Z2,itz2)]+1n2Ei=1Nts=1Tθβ[Q1,itQ1,isϵ1,itϵ2,isKh1(Z1,itz1)Kh11/2(Z1,isz1)Kh21/2(Z2,isz2)]+1n2Eil=1Nt=1Tθβ[Q1,itQ1,ltϵ1,itϵ2,ltKh1(Z1,itz1)Kh11/2(Z1,ltz1)Kh21/2(Z2,ltz2)]+1n2Eil=1Nts=1Tθβ[Q1,itQ1,lsϵ1,itϵ2,lsKh1(Z1,itz1)Kh11/2(Z1,lsz1)Kh21/2(Z2,lsz2)]=O(N1)=o(1).

To see this observe the following. The third and fourth summands in E[Tn,11Tn,12] are zero by Assumption A.1. For the first summand in E[Tn,11Tn,12], note that

1n2Ei=1Nt=1Tθβ[Q1,itQ1,itϵ1,itϵ2,itKh13/2(Z1,itz1)Kh21/2(Z2,itz2)]=θβnE(W1,itW1,itW1,itW1,it(Z1,itz1)/h1W1,itW1,it(Z1,itz1)/h1W1,itW1,it(Z1,itz1)(Z1,itz1)/h12)×ϵ1,itϵ2,itKh13/2(Z1,itz1)Kh21/2(Z2,itz2)=θβn(Ω1112(u1,u2)Ω1112(u1,u2)(u1z1)/h1Ω1112(u1,u2)(u1z1)/h1Ω1112(u1,u2)(u1z1)(u1z1)/h12)×Kh13/2(u1z1)Kh21/2(u2z2)f(u1,u2)du1du2=h1d1/2h2d2/2θβn(Ω1112(z1+h1c1,z2+h2c2)Ω1112(z1+h1c1,z2+h2c2)c1Ω1112(z1+h1c1,z2+h2c2)c1Ω1112(z1+h1c1,z2+h2c2)c1c1/h12)×K13/2(c1)K21/2(c2)f(z1+h1c1,z2+h2c2)dc1dc2.

For a fixed T and by invoking Assumption A.2, Assumption A.3 and Assumption A.5, this first summand is O(N−1). Similarly, the second summand in E[Tn,11Tn,12] is o(1).

For the fourth term in (42),

E[Tn,12Tn,12]=Var(Tn,12)=Var{1ni=1Nt=1Tβ[Q1,itϵ2,itKh11/2(Z1,itz1)Kh21/2(Z2,itz2)]}=β2nV12,1+V12,2,whereV12,1=Var(Q1,i1ϵ2,i1Kh11/2(Z1,i1z1)Kh21/2(Z2,i1z2)),V12,2=2β2nTt=1T1(Tt)Cov(Q1,i1ϵ2,i1Kh11/2(Z1,i1z1)Kh21/2(Z2,i1z2),Q1,i(t+1)ϵ2,i(t+1)Kh11/2(Z1,i(t+1)z1)Kh21/2(Z2,i(t+1)z2)).
(44)V12,1=E(Q1,i1Q1,i1ϵ2,i12Kh1(Z1,i1z1)Kh2(Z2,i1z2))=E(W1,i1W1,i1W1,i1W1,i1(Z1,i1z1)/h1W1,i1W1,i1(Z1,i1z1)/h1W1,i1W1,i1(Z1,i1z1)(Z1,i1z1)/h12)×ϵ2,it2Kh1(Z1,i1z1)Kh2(Z2,i1z2)=(Ω1122(u1,u2)Ω1122(u1,u2)(u1z1)/h1Ω1122(u1,u2)(u1z1)/h1Ω1122(u1,u2)(u1z1)(u1z1)/h12)×Kh1(u1z1)Kh2(u2z2)f(u1,u2)du1du2f(z1,z2)(Ω1122(z1,z2)00Ω1122(z1,z2)μ1,2(K1)).

Then, for a fixed T, V12,1 = O(N−1) = o(1). In a similar manner, we obtain V12,2 = o(1).

(ii) Note that by symmetry, E[Tn1Tn2]=E[Tn2Tn1]. To compute E[Tn1Tn2], we use the decomposition

(45)Tn1Tn2=Tn,11Tn,21+Tn,11Tn,22+Tn,12Tn,21+Tn,12Tn,22.

For the first term in (45), and by Assumption A.1 and Assumption A.2,

E[Tn,11Tn,21]=1n2Ei=1Nt=1Tθβ[Q1,itQ2,itϵ1,it2Kh13/2(Z1,itz1)Kh21/2(Z2,itz2)]+1n2Ei=1Nts=1Tθβ[Q1,itQ2,isϵ1,itϵ1,isKh1(Z1,itz1)Kh11/2(Z1,isz1)Kh21/2(Z2,isz2)]+1n2Eil=1Nt=1Tθβ[Q1,itQ2,ltϵ1,itϵ1,ltKh1(Z1,itz1)Kh11/2(Z1,ltz1)Kh21/2(Z2,ltz2)]+1n2Eil=1Nts=1Tθβ[Q1,itQ2,lsϵ1,itϵ1,lsKh1(Z1,itz1)Kh11/2(Z1,lsz1)Kh21/2(Z2,lsz2)]=O(N1)=o(1).

Similarly, for the second term in (45),

E[Tn,11Tn,22]=1n2Ei=1Nt=1Tθγ[Q1,itQ2,itϵ1,itϵ2,itKh1(Z1,itz1)Kh2(Z2,itz2)]+1n2Ei=1Nts=1Tθγ[Q1,itQ2,isϵ1,itϵ2,isKh1(Z1,itz1)Kh2(Z2,isz2)]+1n2Eil=1Nt=1Tθγ[Q1,itQ2,ltϵ1,itϵ2,ltKh1(Z1,itz1)Kh2(Z2,ltz2)]+1n2Eil=1Nts=1Tθγ[Q1,itQ2,lsϵ1,itϵ2,lsKh1(Z1,itz1)Kh2(Z2,lsz2)]=o(1).

For the third term in (45),

E[Tn,12Tn,21]=1n2Ei=1Nt=1Tβ2[Q1,itQ2,itϵ1,itϵ2,itKh1(Z1,itz1)Kh2(Z2,itz2)]+1n2Ei=1Nts=1Tβ2[Q1,itQ2,isϵ1,isϵ2,itKh11/2(Z1,itz1)Kh21/2(Z2,itz2)×Kh11/2(Z1,isz1)Kh21/2(Z2,isz2)]+1n2Eil=1Nt=1Tβ2[Q1,itQ2,ltϵ1,ltϵ2,itKh11/2(Z1,itz1)Kh21/2(Z2,itz2)×Kh11/2(Z1,ltz1)Kh21/2(Z2,ltz2)]+1n2Eil=1Nts=1Tβ2[Q1,itQ2,lsϵ1,lsϵ2,itKh11/2(Z1,itz1)Kh21/2(Z2,itz2)×Kh11/2(Z1,lsz1)Kh21/2(Z2,lsz2)]=o(1).

For the fourth term in (45),

E[Tn,12Tn,22]=1n2Ei=1Nt=1Tβγ[Q1,itQ2,itϵ2,it2Kh11/2(Z1,itz1)Kh23/2(Z2,itz2)]+1n2Ei=1Nts=1Tβγ[Q1,itQ2,isϵ2,itϵ2,isKh11/2(Z1,itz1)Kh21/2(Z2,itz2)Kh21/2(Z2,isz2)]+1n2Eil=1Nt=1Tβγ[Q1,itQ2,ltϵ2,itϵ2,ltKh11/2(Z1,itz1)Kh21/2(Z2,itz2)Kh21/2(Z2,ltz2)]+1n2Eil=1Nts=1Tβγ[Q1,itQ2,lsϵ2,itϵ2,lsKh11/2(Z1,itz1)Kh21/2(Z2,itz2)Kh21/2(Z2,lsz2)]=o(1).

(iii) To compute E[Tn2Tn2], note that

(46)Tn2Tn2=Tn,21Tn,21+Tn,21Tn,22+Tn,22Tn,21+Tn,22Tn,22.

For the first term in (46), we proceed as follows,

E[Tn,21Tn,21]=Var(Tn,21)=1n2Var{i=1Nt=1TβQ2,itϵ1,itKh11/2(Z1,itz1)Kh21/2(Z2,itz2)}=β2NT2Var{t=1TQ2,itϵ1,itKh11/2(Z1,itz1)Kh21/2(Z2,itz2)}=β2nV21,1+V21,2,
whereV21,1=Var{Q2,itϵ1,itKh11/2(Z1,itz1)Kh21/2(Z2,itz2)},V21,2=β2nTt=1T1(Tt)Cov(Q2,i1ϵ1,i1Kh11/2(Z1,i1z1)Kh21/2(Z2,i1z2),Q2,i(t+1)ϵ1,i(t+1)Kh11/2(Z1,i(t+1)z1)Kh21/2(Z2,i(t+1)z2)).

Using the steps in (44), we can show that

V21,1f(z1,z2)(Ω2211(z1,z2)00Ω2211(z1,z2)μ2,2(K2)).

Hence, for a fixed T, V21,1 = O(N−1). Similarly, we obtain V21,2 = o(1).

For the second term in (46),

E[Tn,21Tn,22]=1n2Ei=1Nt=1Tβγ[Q2,itQ2,itϵ1,itϵ2,itKh11/2(Z1,itz1)Kh23/2(Z2,itz2)]+1n2Ei=1Nts=1Tβγ[Q2,itQ2,isϵ1,itϵ2,isKh11/2(Z1,itz1)Kh21/2(Z2,itz2)Kh2(Z2,isz2)+1n2Eil=1Nt=1Tβγ[Q2,itQ2,ltϵ1,itϵ2,ltKh11/2(Z1,itz1)Kh21/2(Z2,itz2)Kh2(Z2,ltz2)]+1n2Eil=1Nts=1Tβγ[Q2,itQ2,lsϵ1,itϵ2,lsKh11/2(Z1,itz1)Kh21/2(Z2,itz2)Kh2(Z2,lsz2)]=o(1).

For the fourth term in (46),

E[Tn,22Tn,22]=Var(Tn,22)=1n2Var{i=1Nt=1Tγ[Q2,itϵ2,itKh2(Z2,itz2)]}.

Similar to the above proof of E[Tn,11Tn,11], we can easily show that

nh2d2Var(Tn,22)γ2f~2(z2)S2.

In essence, the off-diagonal block terms for E[TnTn] in A.4 are of smaller order than its (1, 1) and (2, 2) main-diagonal block terms, which are of orders O{(nh1d1)1} and O{(nh2d2)1} respectively. Therefore, this completes the proof of Proposition 3.3. □

## Proof of Theorem Theorem 3.4:.

We apply the Cramér-Wold device to assist in establishing asymptotic normality, given the multivariate nature of our semiparametric system estimator. We introduce some additional notations for ease of exposition. We define

A~it1:=(Kh11/2(Z1,itz1)00Kh21/2(Z2,itz2))(θββγ)(Kh11/2(Z1,itz1)00Kh21/2(Z2,itz2))=(θKh1(Z1,itz1)βKh11/2(Z1,itz1)Kh21/2(Z2,itz2)βKh11/2(Z1,itz1)Kh21/2(Z2,itz2)γKh2(Z2,itz2))

For any λRl~ such that λ=1, we set ηit=λD~1/2QitA~it1ϵit, where Qit = block diag (Q1,it, Q2,it) and ϵit=(ϵ1,it,ϵ2,it) for i = 1, …, N and t = 1, …, T. Thus, we have

n1/2λD~1/2Tn=1ni=1Nt=1Tηit.

By Assumption A.2 and Proposition 3.3, and for any i = 1, …, N and t = 1, …, T, we obtain

Var(ηit)=η2(z)(1+o(1)),andt=2T|Cov(ηi1,ηit)|=o(1),

where η2(z):=λf~(z)Dθγ2Sλ. Thus, Var(n1/2λD~1/2Tn)=η2(z)(1+o(1)).

Continuing in this way, it remains to show that the Lyapounov condition holds. This is easily achieved by invoking the stipulated assumptions, Minkowski’s inequality and similar steps to Proof of Theorem 2 in Cai and Li (2008). □

## Proof of Proposition Proposition 3.9:.

Note that

(47)S~n=1n(Q1KU1~00Q2KU2~):=(S~n,1100S~n,22),
(48)Bn=1n(Q1KX~1R¯1Q2KX~2R¯2):=(Bn,11Bn,22),
(49)Rn=1n(Q1KX~1R1Q2KX~2R2):=(Rn,11Rn,22),
(50)Tn=1n(Q1Kϵ1Q2Kϵ2):=(Tn,1Tn,2).

Then E[S~n], E[Bn], and E[Rn] follow directly from the results in Cai and Li (2008), and we have the desired result for (i), (ii) and (iii) of Proposition 3.9. For Proposition 3.9 (iv), note that

(51)Var(Tn)=E(Tn,1Tn,1Tn,1Tn,2Tn,2Tn,1Tn,2Tn,2),

and Tn,1Tn,1=1n2Q1Kϵ1ϵ1KQ1, Tn,1Tn,2=1n2Q1Kϵ1ϵ2KQ2, and Tn,2Tn,2=1n2Q2Kϵ2ϵ2KQ2. Using the results in Cai and Li (2008), it is easy to show that

nhdE[Tn,1Tn,1]=f(z)S1andnhdE[Tn,2Tn,2]=f(z)S2.

Thus, it suffices to show that the off-diagonal block terms in (51) are also of the order of magnitude n−1hd. We only consider the (1,2) block-entry in (51), as the result for the (2,1) block-entry will follow by virtue of symmetry. To begin, we express E[Tn,1Tn,2] as

(52)E[Tn,1Tn,2]=1n2Ei=1Nt=1T[Q1,itQ2,itϵ1,itϵ2,itKh2(Zitz1)]+1n2Ei=1Nts=1T[Q1,itQ2,isϵ1,itϵ2,isKh(Zitz1)Kh(Zisz2)]+1n2Eil=1Nt=1T[Q1,itQ2,ltϵ1,itϵ2,ltKh(Zitz1)Kh(Zltz2)]+1n2Eil=1Nts=1T[Q1,itQ2,lsϵ1,itϵ2,lsKh(Zitz1)Kh(Zlsz2)].

The third and fourth terms in (52) are zero by Assumption A.1. For the first term in (52), we have

1n2Ei=1Nt=1T[Q1,itQ2,itϵ1,itϵ2,itKh2(Zitz1)]=1nE[Q1,itQ2,itϵ1,itϵ2,itKh2(Zitz1)]=1nE(W1W2W1W2(Zz)/hW2W1(Zz)/hW1W2(Zz)(Zz)/h2)ϵ1ϵ2Kh2(Zz)=1nE(Ω1212(Z)Ω1212(Z)(Zz)/hΩ1212(Z)(Zz)/hΩ1212(Z)(Zz)(Zz)/h2)Kh2(Zz)=1n(Ω1212(u)Ω1212(u)(uz)/hΩ1212(u)(uz)/hΩ1212(u)(uz)(uz)/h2)Kh2(uz)f(u)du=1nhdf(z)(Ω1212(z)ν000Ω1212(z)μ2(K2)):=1nhdS12

Hence, this completes the proof of Proposition 3.9. □

## Proof of Theorem Theorem 3.11:.

This is straightforward given the above results in the proofs of Theorem 3.4 and Proposition 3.9, and the results in Cai and Li (2008). □

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Published Online: 2017-8-19

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