TECHNICAL APPENDIX FOR GENERALIZED LEAST SQUARES INFERENCE IN PANEL AND MULTILEVEL MODELS WITH SERIAL CORRELATION AND FIXED EFFECTS

Transcription

1 ECHNICAL APPENDIX FOR GENERALIZED LEA QUARE INFERENCE IN PANEL AND MULILEVEL MODEL WIH ERIAL CORRELAION AND FIXED EFFEC CHRIIAN B. HANEN Let C st = x st β + z st βs + v st, ) or, in vector notation, C s = X s β + Z s β s + V s, where C s = [C s,...,c s ] is, X s = [x s,...,x s ] is k, Z s = [z s,...,z s ] is k, and V s = [v s,...,v s ] is. Also, let x sth be the h th element of x st so that x st = [x st,...,x stk ], and define z sth similarly. Define v st = v st z st Z s Z s) Z s V s, ẍ st = x st z st Z s Z s) Z s X s, V s = [ v s,..., v s ], and Ẍ s = [ẍ s,...,ẍ s ]. Let v st be a vector with v st = [v st),...,v st) ], and define v st similarly. Assumtion, fixed). uose the data are generated by model ) and N. v st = vst α + η st, where η st is strictly stationary in t for each s, E[ηst ] = σ η, E[η st η sτ ] = 0 for t τ, and the roots of α z α z... α z = 0 have modulus greater than. N. {X s,v s,η s } are iid across s. {Z s } are nonstochastic and identical across s. N3. i) Rank t= E[ẍ stẍ st ]) = RankE[Ẍ sẍs]) = k. ii) RankZ s Z s) = k s. N4. E[V s X s ] = 0, E[V s V s X s ] = Γα). N5. E[ηst 4 ] = µ 4 < and E[x 4 sth ] < s, t, h. Proosition. If Assumtion is satisfied, α α α), where α α) = E[ t=+ v st v st ] E[ t=+ v st v st] = Γ α) + Γα)) Aα) + Aα)). Proosition. uose α α) is continuously differentiable in α and that H = Dα α) is invertible for all α such that N is satisfied, where Dα α) is the derivative matrix of α α) in α. Define α ) = α α). hen, if Assumtion is satisfied, α ) α 0 and α ) α) d H Γ α) + Γα)) N0,Ξ), where Ξ = E[ t =+ t =+ v st µ st µ st v st ]

2 CHRIIAN B. HANEN and µ st = v st v st α α). Assumtion, ). uose the data are generated by model ) and N. v st = v st α + η st, where η st is strictly stationary in t for each s, E[η st ] = σ η, E[η st η sτ ] = 0 for t τ, and the roots of α z α z... α z = 0 have modulus greater than. N. {X s,v s,η s } are iid across s. {Z s } are nonstochastic and identical across s. N3. i) [X,Z], where X = [X,...,X ] and Z = diagz,...,z ) has full rank. ii) Z s Z s is uniformly ositive definite with minimum eigenvalue λ s λ > 0 for all s. N4. E[V s X s ] = 0, E[V s V s X s] = Γα). N5. {X st,v st,η st } is α-mixing of size 3r r4, r > 4, and z ith <, E x ith r+δ <, and E η it r+δ < for some δ > 0 and all i,t,h. d Proosition 3. If Assumtion is satisfied, α α) NρBα),Γ ΞΓ ) if ρ 0, where Ξ = lim t =+ E[v st η st η st vst ]. In addition, if η st t =+ are indeendent for all s and t, Ξ = σ ηγ. Assumtion 3. For Bα,) defined in Lemma C.5, N6. Bα, ) is continuously differentiable in α with bounded derivative uniformly in for all α satisfying the stationarity condition given in N. Proosition 4. If Assumtion is satisfied, α ) α) d N0,Γ ΞΓ ) for Ξ defined in Proosition 3 if ρ 0. If Assumtion 3 is also satisfied, then α ) α) d N0,Γ ΞΓ ) if 0. 3 Proosition 5. If Assumtions and 3 are satisfied and H = Dα α) is invertible for all α satisfying the stationarity condition given in N uniformly in, where Dα α) is the derivative matrix of α α) in α, α ) α) d N0,Γ ΞΓ ) for Ξ defined in Proosition 3. Assumtion 4 Higher-Order Asymtotics). uose the data are generated by model ) and HO. v st = v st α + η st, where η st are iid N0,σ η ) random variables, and E[V sv s ] = Γα) is a ositive definite matrix with minimum eigenvalue bounded away from 0 and maximum eigenvalue bounded away from infinity uniformly in. HO. i) {X s,z s } are nonstochastic with {Z s } identical across s. ii) [X,Z], where X = [X,...,X ] and Z = diagz,...,z ) has full rank, and the minimum eigenvalue of X X X ZZ Z) Z X is bounded away from zero. iii) x sth and z sth.

3 HO3. Let M z α) = I Γ α) )I Γ α) )ZZ I Γ α) )Z) Z I Γ α) ), M = IXX M z α)x) X M z α), A α) = X M z α)x, and Ψ α) = X M z α)mv. For i =,...,, j =,...,, and k =,...,, i) each matrix in Aα), A i α) = A α i α, and A ij = A α i α j α aroaches a limit as,, and limaα) is nonsingular, ii) the covariance matrices for the vectors in Ψ i α) = Ψ α i α, and Ψ ij α) = Ψ α i α j α aroach limits as,, and iii) all the matrices in A α), A i α), A ij α), A ijk α) = 3 A α i α j α k α, Ψ α), Ψ i α), Ψ ij α), and Ψ ijk α) = 3 Ψ α i α j α k α are bounded in robability uniformly for α A where A is a neighborhood of α. HO4. Bα, ) is three times continuously differentiable in α with the first three derivatives bounded uniformly in for all α satisfying the stationarity condition given in N. Proosition 6. Let α be the least squares estimator of α, α ) be the iteratively biascorrected estimator, and β α) and β α ) ) be the corresonding FGL estimators. If Assumtion 4 is satisfied and ρ as {, } jointly, the higher-order bias and variance of β α) and β α ) ) are Bias β α)) = Bias β α ) )) = 0 ) V ar β α) β)) = A + Υ/ 3) + ζ ij Bα,) ibα,) j + O/ ) i= j= V ar β α ) ) β)) = A + Υ/ + O/ ). 4) for ζ ij = E[Ψ i Ψ j ] and Υ = i= j= ζ ij ) Γ ) [i,j] where ) Γ ) [i,j] is the [i,j] element of matrix ) Γ. Also, V ar β α) β))v ar β α ) ) β)) = ζ ij Bα,) ibα,) j 0. i= j= Aendix A. Notation he following notation will be used throughout the aendix. uose the anel has a cross-sectional dimension s =,..., and a time-series dimension t =,..., with > where is the order of the autoregressive rocess. Let C st = x stβ + z stβ s + v st, 5) 3

4 or, in vector notation, C s = X s β +Z s β s +V s, where C s = [C s,..., C s ] is, X s = [x s,..., x s ] is k, Z s = [z s,..., z s ] is k, and V s = [v s,..., v s ] is. Also, let x sth be the h th element of x st so that x st = [x st,...,x stk ], and define z sth similarly. Let v st be a vector with v st = [v st),..., v st) ]. Define v st = v st z stz sz s ) Z sv s, ẍ st = x st z stz sz s ) Z sx s, Vs = [ v s,..., v s ], and Ẍ s = [ẍ s,..., ẍ s ]. hroughout, let A = [tracea A)] / be the Euclidean norm of a matrix A. Aendix B. Proof of Proosition and Proosition Proosition is verified by combining Lemmas B.3 and B.4 below, and Proosition follows from Proosition and Lemma B.6. All results resented below are for asymtotics where with fixed. Proof of Proosition. Immediate from Lemma B.3 and Lemma B.4. Proof of Proosition. hat α α) is continuously differentiable in α and that H = Dα α) is invertible for all α such that N is satisfied imly that α α) is invertible for all α such that N is satisfied by the Inverse Function heorem. ee, e.g. Fitzatrick 996) heorem 6.9.) α ) α 0 then follows immediately from the definition of α ) and Proosition. o verify the asymtotic normality, exand α ) about α = α α). his gives α ) = α α α)) + H α α) α α α)), where α α) is an intermediate value between α and α α). he conclusion then follows from continuity of H, Proosition, and Lemma B.6. B.. Lemmas. Lemma B.. Let β be the ordinary least squares estimate of β. hen if the conditions of Assumtion are satisfied, β β 0 and β β ) d N0, M ΩM ), where M = E[Ẍ sẍs] and Ω = E[Ẍ sγ α)ẍs]. Proof. Using AB A B and IZ s Z sz s ) Z s ositive semi-definite, E Ẍ sẍs E Ẍs = E[traceX s X s X s Z sz s Z s) Z s X s)] traceex s X s) = trace t= E[x stx st ]) < by N5. Also, Ẍ V s s E Ẍ s E V s ) / < from the Cauchy-chwarz inequality, N5, and the same argument as above. he Khinchin LLN then yields s= Ẍ sẍs M and s= Ẍ V s s 0, from which β β 0 follows. 4

5 o show asymtotic normality of β first note that Ẍ s V s is iid and has mean zero by N and N4. Also, E Ẍ s V s V s Ẍ s E X s 4 E V s 4 ) / < by N5, AB A B, the Cauchy- chwarz inequality, and E Ẍs 4 = E[traceX sx s )) tracex sx s )tracex sz s Z sz s ) Z sx s ) + tracex s Z sz s Z s) Z s X s)) ] E[traceX s X s)) ] = E X s 4, where the inequality follows from X sx s, X sz s Z sz s ) Z sx s, and I Z s Z sz s ) Z s ositive semi-definite. It then follows from the Lindeberg-Levy CL that s= Ẍ s V s d N0, Ω) since E[ Ẍ s V s V s Ẍ s ] = E[Ẍ sv s V sẍs] = E[Ẍ sγ α)ẍs], from which β β ) d N0, M ΩM ) is obtained. Lemma B.. Define ṽ st to be the residual from least squares regression of ); i.e. ṽ st = C st x β st z st β s = v stx st β β )z st β sβ ) = v st ẍ st β β ), where β and β s are least squares estimates of β and β. s Let ṽst be a vector with ṽst = [ṽ st),...,ṽ st) ], and let v st be a vector with v st = [ v st),..., v st) ]. Under the conditions of Assumtion, s= t=+ ṽ stṽ = s= s= t=+ t=+ Proof. ṽ stṽ st = v st v st +o / ), and s= t=+ ṽ stṽst = s= t=+ v st v st+o / ). s= t=+ v st v st ẍ st β β ) v st v stẍ st β β )+ẍ st β β ) β β ) ẍ st ), where ẍ st is a k matrix with ẍ st = [ẍ st),...,ẍ st) ]. Note vec s= t=+ ẍ st β β ) v st ) = { [ s= t=+ v st ẍ st)]} β β ) = o )O / ) by Lemma B., Assumtion, and the Khinchin LLN. imilarly, vec v stẍ st β β )) = o )O / ), and vec s= t=+ s= t=+ ẍ st β β ) β β ) ẍ st ) = O )O / )O / ). It then follows that s= t=+ ṽ stṽ st = s= t=+ v st v st + o / ) and, by a similar argument, s= t=+ ṽ stṽst = s= t=+ v st v st + o / ). Lemma B.3. Under the conditions of Assumtion, s= t=+ v st v st E[ t=+ v st v st ] = )Γ α) + Γα)), and s= t=+ v st v st E[ t=+ v st v st ] = )Aα) + Aα)), where Γ α) + Γα) is a matrix with i, j element E [ t=+ v st v st ] [i,j] = γ ij α) trace Z sγ i α)z s,j Z sz s ) ) + trace Z sγ j α)z s,i Z sz s ) ) 6) trace Z sγα)z s Z sz s ) Z s,iz s,j Z sz s ) ), 5

6 Aα) + Aα) is a vector with i th element [ ] E v st v st = γ i α) trace Z sγ i α)z s,0 Z sz s ) ) t=+ [i,] + trace Z sγ 0 α)z s,i Z sz s ) ) 7) trace Z sγα)z s Z sz s ) Z s,iz s,0 Z sz s ) ), Γα) = E[V s V s ], Γ kα) = E[V s V s,k ], γ iα) = E[v st v sti) ], V s,k = v s+k), v s+k),..., v s k) ) and Z s,k is defined similarly. Proof. he roof is given for s= t=+ v st v st Γα)); s= t=+ v st v st similar argument. s= [ s= E[ t=+ v st v st ] = )Γ α) + E[ t=+ v st v st] = )Aα) + Aα)) follows by a t=+ v st v is a matrix with [i, j] element ] t=+ v st v st [i,j] = + = + s= s= s= s= s= s= s= s= E[ t=+ t=+ t=+ t=+ t=+ v sti) v stj) v sti) z stj) Z sz s ) Z sv s v stj) z sti) Z s Z s) Z s V s z sti) Z sz s ) Z sv s V sz s Z sz s ) z stj) v sti) v stj) V s,iz s,j Z sz s ) Z sv s V s,jz s,i Z sz s ) Z sv s 8) Z s,iz s Z s) Z s V sv s Z s Z s Z s) Z s,j t=+ v st v st ] [i,j] by the Khinchin LLN and reeated alication of the triangle and Cauchy-chwarz inequalities since 6

7 i) [ t=+ v st v ] [i,j] is iid by N, ii) E v sti) v stj) < by N, ) /, iii) E v sti) z stj) Z s Z s) Z s V s E z stj) Z s Z s) Z s V s E v sti) iv) E z sti) Z s Z s) Z s V sv s Z sz s Z s) z stj) ) /, E z sti) Z s Z s) Z s V s E z stj) Z s Z s) Z s V s and v) by N and N5, E z stj) Z s Z s) Z s V s = E z stj) Z s Z s) Z s Z s)z s Z s) Z s V s [ from which it follows that E z stj) Z sz s ) Z s E Z s Z sz s ) Z sv s = tracez stj) Z sz s ) z stj) ) E [ tracev s Z sz s Z s) Z s V s) ] k E[traceV s V s)] = k E[ ] t=+ v st v st <. [i,j] t=+ v st ] <, Lemma B.4. Let α = s= t=+ ṽ stṽst ) s= t=+ ṽ stṽ st ) be the least squares estimate of α using the least squares residuals, ṽ st from estimating β. If Assumtion is satisfied, α = Proof. By Lemma B., s= t=+ v st v st ) s= t=+ v st v st) + o / ). α = s= t=+ v st v st + M ) s= t=+ v st v st + M ), where M = o / ) and M = o / ). After some algebra, it then follows that α = + = s= t=+ s= t=+ s= t=+ s= t=+ v st v st ) s= t=+ v st v st + M ) M v st v st + M ) M v st v st ) s= t=+ v st v st ) s= t=+ v st v st ) s= t=+ v st v st ) v st v st) + O )o / ) + O )o / )O )O ) where the last equality is by Lemma B.3, which yields the conclusion. 7

8 Lemma B.5. Define µ st = v st v st α α). If Assumtion is satisfied, s= t=+ v st µ st) d N0, Ξ), where Ξ = E[ t =+ t =+ v st µ st µ st v st ]. Proof. t=+ v st µ st are iid by N. Also, E[ t=+ v st µ st] = E[ = E[ = 0, t=+ t=+ v st v st v st α α))] v st v st ] E{ t=+ v st v st E[ t=+ v st v st ]) E[ t=+ v st v st ]} where the second equality comes from α α) = E[ t=+ v st v ] E[ t=+ v st v st] from Proostion. Also E[ t =+ t =+ = E[ v sti) µ st µ st v stj)] t =+ t =+ E[ +E[ k= t =+ t =+ v sti) v st v st v stj)] E[ k = k = t =+ t =+ < k= t =+ t =+ v sti) v st v stk) v stj)α α) k ] v sti) v stk ) v stk ) v stj)α α) k α α) k ] v sti) v stk) v st v stj)α α) k ] since E v st v st v st3 v st4 E v st 4 E v st 4 E v st3 4 E v st4 4 ) /4 < by reeated alication of the Cauchy-chwarz inequality and E v st 4 = E e t V s Z s Z s Z s) Z s V s) 4 E e t 4 V s 4 I Z s Z s Z s) Z s 4 ) = k ) E V s 4 < by N5 and fixed, where e t is the t th unit vector. hen the conclusion follows by the Lindeberg-Levy CL since E[ t =+ t =+ v st µ st µ st v st ] [i,j] = E[ t =+ t =+ v sti) µ st µ st v stj)] < i, j. Lemma B.6. uose Assumtion holds, then αα α)) d Γ α)+ Γα)) N0, Ξ). Proof. For µ st defined in Lemma B.5, αα α) = s= t=+ v st v st ) s= t=+ v st µ st)+ o / ) by Lemma B.4. he conclusion is then immediate from Lemmas B.3 and B.5. 8

9 Aendix C. Proof of Proositions 3, 4, and 5 he roofs of Proositions 3, 4, and 5 are collected below. All results resented below are for asymtotics where,. Proof of Proosition 3. α α) = α α α) + α α) α) = α α α)) + Bα, ) by Lemma C.5, from which the conclusion follows by Lemmas C.5 and C.8. Proof of Proosition 4. α ) α) = [ α α + α α)) α α) + α α) α] = α α α)) + Bα, ) B α, )). he first conclusion then follows from Lemmas C.5, C.8, C.9, and the Continuous Maing heorem, and the second conclusion follows from Lemmas C.5, C.8, C.9, and a aylor exansion of B α, ) about α = α. Proof of Proosition 5. Recall α ) = α α) = α α α)) + H α α α α) α)). From Lemma C.5, α α) = α + Bα, ) which imlies H I as by N6. he conclusion then follows from Lemma C.8. C.. Lemmas. Lemma C.. Let β be the ordinary least squares estimate of β. hen if the conditions of Assumtion are satisfied, β β 0 and β β ) d N0, M ΩM ), where M = M XX M XZ M ZZ M XZ, Ω = lim E[ X sγα)x s ] M XZ M ZZ lim E[ Z sγα)x s ]) lim E[ X s Γα)Z s])m ZZ M XZ + M XZM ZZ lim [ Z s Γα)Z s])m ZZ M XZ, with M XX = lim E[ X s X s], M XZ = lim E[ X s Z s], and M ZZ = lim [ Z s Z s]. Proof. Let Q s = I Z s Z sz s ) Z s. hen E X sq s V s +δ E Q s X s +δ Q s V s +δ ) E X s +δ E V s +δ ) / k +δ, 9

10 where the first inequality is from AB A B, the second from the Cauchy-chwarz inequality and Q s A A, and the third from E X s +δ = E[ tracex s X s)] +δ = +δ E[ k t= h= +δ [ k t= h= x ith ]+δ E x ith +δ ) +δ ] +δ +δ k +δ ) +δ = k +δ, where the first inequality follows from Minkowski s inequality and the second from N5, and E V s +δ by a similar argument. It also follows that E X sq s X s +δ k +δ by the same reasoning. o E X s Q sx s and E X s Q sv s are uniformly integrable in. ee, e.g. Billingsley 995).) Also, under N3-N5, x st x st t= t= t= z st z st lim E[x st x st ] 0, t= z st z st, t= x st z st t= E[x st z st ] 0 t= x st v st 0, and z st v st 0 by a LLN, e.g. White 00) Corollary It then follows from Phillis and Moon 999) Corollary that s= X s Q sx s M and s= X s Q sv s 0. Hence, β β 0. Let K denote a generic constant. o verify asymtotic normality, note that E X s V s + δ C + δ t= t= t= + max{ E x st v st + δ +ǫ δ + ) δ +ǫ, [ E x st v st +ǫ ) + δ +ǫ ] CK < 9) by N4-N5 and Doukhan 994) heorem. Also, E X sz s ) Z sz s ) Z sv s ) + δ k M) + δ E X sz s ) 4+δ E Z sv s ) 4+δ ) / } CK <, 0) where the second inequality follows from AB A B, the Cauchy-chwarz inequality, and N3, and the third inequality is from N4-N5 and Doukhan 994) heorem. 9) and 0) imly E X s Q sv s + δ < since E X sq s V s + δ [E X sv s + δ ) + δ + E X sz s Z sz s ) Z sv s + δ ) + δ ] + δ by Minkowski s inequality which yields E X sq s V s uniformly integrable in. ee, e.g. Billings- ley 995).) X s Q d sv s N0, Ω) from White 00) heorem 5.0, from which it follows that Ω = E[X sq s Γα)Q s X s ] Ω. E.g. Billingsley 995) Corollary 5..) hen Phillis and 0

11 Moon 999) heorem 3 gives s= t= X s Q sv s d N0, Ω) which imlies β β ) d N0, M ΩM ). Lemma C.. Define ṽ st to be the residual from least squares regression of ); i.e. ṽ st = C st x β st z st β s = v st x st β β ) z st β s β ) = v st ẍ st β β ), where β and β s are least squares estimates of β and β s. Let ṽ st be a vector with ṽst = [ṽ st),...,ṽ st) ], and let v st be a vector with v st = [ v st),..., v st) ]. Under the conditions of Assumtion, s= t=+ ṽ stṽ st = s= t=+ v st v st + o ) / ), and s= t=+ ṽ stṽ st = s= t=+ v st v st + o ) / ). Proof. Using calculations similar to those found in the roof of Lemma C., it can be shown that ) s= t=+ v stẍ stj)h = o ) and ) s= t=+ ẍstẍ stj)h = O ) for j =,..., and h =,...,k. he conclusion then follows by Lemma C. and the same arguments used in roving Lemma B.. Lemma C.3. Under the conditions of Assumtion, i) ii) ) s= ) s= ) s= t=+ v st v st Γ α), and ) s= t=+ v st v st Aα). t=+ v st v st = ) s= t=+ v stvst + O ), and t=+ v st v st = ) s= t=+ v stv st + O ). Proof. he roof is given for ) s= t=+ v st v Aα) follows by a similar argument. Γ α); ) s= t=+ v st v st

12 t=+ v st v is a matrix with i, j element [ ] t=+ v st v st [i,j] = + = + t=+ t=+ t=+ t=+ t=+ v sti) v stj) v sti) z stj) Z sz s ) Z sv s v stj) z sti) Z s Z s) Z s V s z sti) Z sz s ) Z sv s V sz s Z sz s ) z stj) v sti) v stj) V s,iz s,j Z sz s ) Z sv s V s,jz s,i Z sz s ) Z sv s ) Z s,iz sz s ) Z sv s V sz sz sz s ) Z s,j, where V s,k = v s+k), v s+k),..., v s k) ) and Z s,k is defined similarly. Consider E V s Z sz s Z s) Z s,i Z s,jz s Z s) Z s V s + δ = E V s Z s) Z s Z s) Z s,i Z s,j) Z s Z s) Z s V s) + δ. ) Note that by N5. Also, E Z sv s ) +δ = Z s,i Z s,j k < 3) +δ +δ E z st v st ) +δ t= C max{ E z st v st +δ+ǫ ) +δ +δ+ǫ, [ E z st v st +ǫ ) t= t= +δ +ǫ ] } K < 4) by Doukhan 994) heorem and N5. It then follows from the Holder s inequality, the Cauchy- chwarz inequality, AB A B, 3), 4), and N3-N5 that ) is bounded away from, which gives V sz s Z sz s ) Z s,i Z s,jz sz s ) Z sv s uniformly integrable in. imilarly, it can be

13 shown that V s,i Z s,jz sz s ) Z sv s and t=+ v sti)v stj) are uniformly integrable in. Also, as,. t=+ v sti)v stj) γ ij α),. V s,i Z s,jz sz s ) Z sv s d Ψ ij M ZZ Ψ, 3. V sz s Z sz s ) Z s,i Z s,jz sz s ) Z sv s d Ψ M ZZ M ijm ZZ Ψ, where M ZZ = lim [Z s Z s], M ij = lim [Z s,i Z s,j], Ψ N0, lim [Z s Γα)Z s], Ψ ij N0, lim [Z s,j Γα)Z s,j), Γ k = E[v s v s,k ], and E[ΨΨ ij ] = lim [Z sγ i α)z s,j ]. ee White 00) Corollary 3.48 and heorem 5.0.) Combining.,., and 3. and ) gives t=+ v sti) v stj) γ ij α). It then follows from Phillis and Moon 999) Corollary that s= t=+ v sti) v stj) γ ij α), which yields the first conclusion. Also, using Phillis and Moon 999) Corollary and considering only., 3. and ) yields the second conclusion. Lemma C.4. Let α = ) s= t=+ ṽ stṽst ) ) s= t=+ ṽ stṽ st ) be the least squares estimate of α using the least squares residuals, ṽ st from estimating β. If Assumtion is satisfied, α = = s= t=+ v st v st ) vst v st ) s= t=+ s= t=+ s= t=+ v st v st ) + o ) / ) v st v st) + O ) + o ) / ). Proof. Follows immediately from Lemmas C. and C.3 using the same argument as in Lemma B.4. Lemma C.5. For α α) defined in Lemma, α α) α = Bα, ), where Bα, ) Bα) as, and α α) α = B α, ) if the conditions of Assumtion are met. Proof. Recall α α) = Γ α)+ Γα)) Aα)+ α α) α = Γ α) + Γα)) A α) + Γ α)α) = as follows from the roof of Lemma C.3. α α) α = fashion. 3 Aα)) and that α = Γ α) Aα). hen Bα, ). hat Bα, ) Bα) B α, ) follows in the same

14 Lemma C.6. Define µ st = v st v st α α). If Assumtion is satisfied, s= t=+ v st µ st = s= t=+ v st η st + o ). Proof. ) s= t=+ v st µ st = ) s= t=+ v st µ st E[ v st µ st]) since E[ t=+ v st µ st ] = E[ = E[ = 0, t=+ t=+ v st v st v st α α))] v st v st ] E{ t=+ v st v st E[ t=+ v st v st ]) E[ t=+ v st v st ]} where the second equality comes from α α) = E[ t=+ v st v ] E[ t=+ v st v st] from Proostion. Also note that µ st = η st v st v st ) vst α α) α) v st v st ) α α) and vst = vst ZstZ s Z s) Z s V s where Zst = [z st),...,z st) ]. o, v st µ st E[ v st µ st ] = v st Z stz sz s ) Z sv s ) η st z stz sz s ) Z sv s v st α α) α) + V sz s Z sz s ) Z st α α)) E[v st Z st Z s Z s) Z s V s) η st z stz sz s ) Z sv s v st α α) α) + V sz s Z sz s ) Z st α α))] = v st η st v st z st Z s Z s) Z s V s E[v st z st Z s Z s) Z s V s]) v stv st E[v stv st ])α α) α) +v st V s Z sz s Z s) Z st E[v st V s Z sz s Z s) Z st ])α α) Z stz sz s ) Z sv s η st E[Z stz sz s ) Z sv s η st ]) +Z stz s Z s) Z s V sz st Z s Z s) Z s V s 5) E[Z st Z s Z s) Z s V sz st Z s Z s) Z s V s]) +Z stz sz s ) Z sv s v st E[Z stz sz s ) Z sv s v st ])α α) α) Zst Z s Z s) Z s V sv s Z sz s Z s) Zst E[Z stz sz s ) Z sv s V sz s Z sz s ) Z st ])α α)). Now consider Y, = s= t=+ Zst Z s Z s) Z s V sv s Z sz s Z s) Zst E[Z stz sz s ) Z sv s V sz s Z sz s ) Z st ]). 4

15 Y, is a matrix with i, j element [Y, ] [i,j] = = s= t=+ z sti) Z s Z s) Z s V sv s Z sz s Z s) z stj) E[z sti) Z sz s ) Z sv s V sz s Z sz s ) z stj) ]) V s Z sz s Z s) Z s,i Z s,jz s Z s) Z s V s s= E[V s Z sz s Z s) Z s,i Z s,jz s Z s) Z s V s]), and E [Y, ] [i,j] = E V sz s Z sz s ) Z s,iz s,j Z sz s ) Z sv s E[V sz s Z sz s ) Z s,iz s,j Z sz s ) Z sv s ] 0, where the equality is from indeendence assumtion N, since E A E[A] and. E V s Z sz s Z s) Z s,i Z s,jz s Z s) Z s V s E V s Z s 4 Z s Z s) Z s,i Z s,j) Z s Z s) 4 E Z s V s 4 ) / by the Cauchy-chwarz inequality.. Z s Z s) Z s,i Z s,j) Z s Z s) k M) Z s,i Z s,j ) k M) k by N3 and N5. 3. E Z s V s 4 C max{ t= E z stv st 4+ǫ ) 4 4+ǫ, [ t= E z stv st +ǫ ) +ǫ ] } K by Doukhan 994) heorem and N5. Hence, [Y, ] [i,j] 0 by Chebychev s inequality, and it follows that Y, = o ). hat all other terms, excet ) s= t=+ v st v st E[v st v st ])α α) α), are o ) follows similarly. o show ) s= t=+ v stvst E[vstv st ])α α) α) = o ), note that α α) α) = O ) from Lemma C.5. Also, vec{ t=+ v st v st E[v st v st ])} d Ψ N0, Ω) where Ω = lim [Evst vst vst vst ) Evst vst )Evst v t =+ t =+ st )] as follows from a CL e.g. White 00) heorem 5.0) and the Cramer-Wold device, and t=+ vecv stv st E[v stv st ]) d Ψ by the Continuous Maing heorem. In 5

16 addition, E = trace{e[ = s= t=+ t =+ t =+ v st v t =+ t =+ st E[v st v st ]) v st v st ) Ev st v st ))v st v st ) Ev st v st )) ]} [Evst vst vst vst ) Evst vst )Evst vst )] traceω) = E Ψ as. It follows that ) s= t=+ v st v st E[v st v st ]) is uniformly integrable in. For examle, Billingsley 995) 6.4.) hen, using Phillis and Moon 999) heorem 3, s= t=+ v st v and the conclusion follows immediately. st E[v st v st ]) d N0, Ω) s= t=+ v st v Lemma C.7. If Assumtion is satisfied, ) s= t=+ v st η d st N0, Ξ), where Ξ = lim t =+ t =+ In addition, if η st are indeendent for all s and t, Ξ = σ η Γ. E[v st η st η st v st ]. st E[v st v st ]) = O ), Proof. As, t=+ v stη st d Ψ N0, Ξ) by a CL e.g. White 00) heorem 5.0) and the Cramer-Wold device, and t=+ v st η st d Ψ by the Continuous Maing heorem. Also, E t=+ v st η st = trace t =+ t =+ Ev st η st η st vst ) traceξ) = E Ψ, which imlies that E t=+ v st η st is uniformly integrable in. For examle, Billingsley 995) 6.4.) hen, using Phillis and Moon 999) heorem 3, s= t=+ v stη st d N0, Ξ). Also, if the η st are indeendent for all s and t, Ev st η st η st v st = 0 t t, and Ev stη stv st = σ ηγ. Lemma C.8. α α α)) d N0, Γ ΞΓ ) for Ξ defined in Lemma C.7 if Assumtion is satisfied. Proof. From Lemma C.4, αα α) = ) s= t=+ v st v st ) ) s= t=+ v st v st ) α α)+o ) / ) = ) s= t=+ v st v st ) ) s= t=+ v st µ st )+o ) / ). he conclusion then follows immediately from Lemmas C.3, C.6, and C.7. Lemma C.9. Under Assumtion, α α = max{o ), O ) / )}. 6

17 Proof. Lemma C.4 gives α = = s= t=+ s= t=+ v stv v stv st ) st ) s= t=+ s= t=+ v stv st ) + O ) + o ) / ) v stη st ) + α + O ) + o ) / ). ) s= t=+ v stη st = O ) / ) from Lemma C.8, and ) s= t=+ v stvst = O ) follows immediately from a LLN, e.g. White 00) Corollary Aendix D. Proof of Proosition 6 he roof of Proosition 6 is quite similar to the roof rovided in Rothenberg 984) and is sketched below. hroughout, let M z α) = I Γ α) ) I Γ α) )ZZ I Γ α) )Z) Z I Γ α) ). Denote the GL estimator of β as β α) = X M z α)x) X M z α)c and the FGL estimator of β corresonding to α as β α) = X M z α)x) X M z α)c. Also, define and A α) = X M z α)x, Ψ α) = X M z α)mv for M = I XX M z α)x) X M z α). hen for i =,...,, j =,...,, and k =,...,, let A i = A α i α, A ij = A α i α j α, A ijk = 3 A α i α j α k α, Ψ i = Ψ α, Ψ ij = Ψ 3 Ψ α, Ψ ijk = α. α i α i α j α i α j α k te. As in Rothenberg 984), note that the residual vector Ṽ = C [X, Z][X, Z] [X, Z]) [X, Z] C = V [X, Z][X, Z] [X, Z]) [X, Z] V does not deend on β and that α and α ) are even functions of Ṽ and hence V. For a given α, β α) is a comlete sufficient statistic for β since V is normally distributed. Also, by Basu s heorem ee, for examle, Lehmann 983). 46), any statistic whose distribution does not deend on β must be distributed indeendently of β α), so β α) is indeendent of α, α ), β α) β α), and β α ) ) β α). It then follows, for α equal to either α or α ), that E[ β α) β ) β α) β ) ] = E[ β α) β ) β α) β ) ] + E[ β α) β α)) β α) β α)) ]. 7

18 β α) β is exactly normal with variance A, so the higher-order variance of β α) is A lus the higher-order variance of β α) β α). In addition, the bias of β α) is E[ β α) β ] = E[ β α) β ] + E[ β α) β α)] = 0 + E[ β α) β α)], so the higher-order bias is also determined by the higher-order bias of β α) β α). te. Exansion of β α) β α). For α equal to either α or α ), β α) β α) = A α) Ψ α). 6) Let d i = α α) i be the i th element of the vector α α). hen exanding 6) about α = α and noting Ψα) = 0 yields β α) β α)) = Aα) Ψ i α)d i / i= + i= j= Aα) Ψ ij α)d i d j Aα) A i α)aα) Ψ j α)d i d j 7) Aα) A j α)aα) Ψ i α)d i d j )/ + Rᾱ)/) 3/, where ᾱ is an intermediate value between α and α and Rᾱ) = [6Aᾱ) A i ᾱ)aᾱ) A j ᾱ)aᾱ) A k ᾱ)aᾱ) Ψᾱ) 6 i= j= k= + 3Aᾱ) A ij ᾱ)aᾱ) A k ᾱ)aᾱ) Ψᾱ) + 3Aᾱ) A i ᾱ)aᾱ) A jk ᾱ)aᾱ) Ψᾱ) + 6Aᾱ) A i ᾱ)aᾱ) A j ᾱ)aᾱ) Ψ k ᾱ) Aᾱ) A ijk ᾱ)aᾱ) Ψᾱ) 8) 3Aᾱ) A ij ᾱ)aᾱ) Ψ k ᾱ) 3Aᾱ) A i ᾱ)aᾱ) Ψ jk ᾱ) + Aᾱ) Ψ ijk ᾱ)]d i d j d k. te 3. Exansion of α α α). Recall that the least squares estimator of α using the residuals from the estimation of equation ), ṽ st, is ) ) α = ṽ stṽ ṽstṽst Ĝ Ĝ s= t=+ s= t=+ where ṽ st = ṽ st),..., ṽ st) ) and ṽ st = v st ẍ st β β ) for v st and ẍ st as defined in??) and??) in the text and β the least squares estimate of β. hen Ĝ = = D + M s= t=+ s= t=+ v st v st v st β β ) ẍ st ẍ st β β ) v st + ẍ st β β ) β β ) ẍ 8 st )

19 where D = ) s= t=+ v st v, and Ĝ = s= t=+ s= t=+ = D + D α α) + M v st v st v st β β ) ẍ st ẍ st β β ) v st + ẍ st β β ) β β ) ẍ st ) where D = ) s= t=+ v st µ st and µ st = v st v st α α). It then follows that α α α) = Ĝ Ĝ α α) = D + M ) D + M M α α)). It is shown below that M and M are O /), and it then follows from Lemmas C.3-C.9 that α and α ) are consistent and have the asymtotic distributions described in Proosition 3. A aylor series exansion then yields α α α)) = Γ D 9) + Γ M M α α)) Γ D Γ )Γ D 0) Γ M Γ D Γ D Γ )Γ M M α α)) D ) + Γ D Γ )Γ D Γ )Γ Γ M Γ M M α α)) D + Γ M Γ D Γ )Γ + Γ D Γ )Γ M Γ D + Γ M Γ M Γ D 6 Γ D Γ + M ) Γ D Γ + M ) Γ D Γ + M ) Γ D + M M α α)) ) ψ + ψ / + ψ 3 / + R/) 3/, 3) where Γ is a matrix of intermediate values that satisfy Γ Γ D Γ + M. It is shown in Lemmas C.3 and C.6 that D Γ + M 0 and that D Γ = O / ), and it follows from Lemmas C.6 and C.7 that D = O / ). 9

20 o show M = O /), let M Z = I ZZ Z) Z and define M Zs similarly. Recall that hen M = + s= t=+ s= t=+ v st V ẌẌ Ẍ) ẍ st s= t=+ ẍ st Ẍ Ẍ) Ẍ V vst ẍ st Ẍ Ẍ) Ẍ V V ẌẌ Ẍ) ẍ st. 4) E V ẌẌ Ẍ) E V M Z X ) X M Z X) ) k / λ min X M Z X) ) E V sm Zs X s s= ) k / = λ min X M Z X) ) tracex sm Zs ΩM Zs X s ) s= ) k / λmax Ω) λ min X M Z X) ) trace i k i i i k ) s= ) k / = λ min λ maxω)k /, X M Z X) where the first inequality follows from the definition of A and the second inequality follows from boundedness of x sth and c Ac λ max A)c c. Also, defining e t to be a unit vector with t th element equal to and all other elements equal to 0 and e t = [e t,..., e t ], E s= t=+ ẍ st v st = ) k M ) k M λ max Ω) ) s= t=+ τ=+ trace s= t=+ trace s= k M λ max Ω), traceeẍ stẍ sτ v st v sτ )) t=+ e t )M Z s ΩM Zs e t ) t=+ t=+ e t )) e t )) where the first inequality results from boundedness of x sth, the second from c Ac λ max A)c c, and the third from the definition of e t. It then follows that ) v st V ẌẌ Ẍ) ẍ st = ẍ st v st vecv ẌẌ Ẍ) ) s= t=+ s= t=+ = O / )O / ) = O /). Using similar arguments, it can be shown that the second and third terms of exression 4) are O /). It then follows that M = O /). hat M = O /) follows similarly. 0

21 Finally, making use of M = O /), M = O /), D Γ = O / ), and D = O / ) yields that ψ, ψ, ψ 3, and R are all O ). te 4. Exansion of α α and α ) α. i. α α) = α α α))+ α α) α) = α α α))+ C.5. It then follows that Bα, ) by Lemma α α) = ψ + Bα, ) + ψ / + ψ 3 / + R/) 3/ 5) for ψ, ψ, ψ 3, and R defined above. ii. Recall α ) = α α) = α α α)) + H α α) α α α)) H + α α) α α α)) α α α)) i 6) α i + i= i= j= H α i α j α α) α α α)) α α α)) i α α α)) j where α α)) α α) α α α). From Lemma C.5, α α) = α + I + Bα,) α, H α i = Bα,) α α i O/) by Assumtion HO5. hus, H α i H H H α i H α j H +H Also, we have = O/) by Assumtion HO5, and H α i α j = Bα, ), so H = 3 Bα,) α α i α j = H α α) = H α i H α α) = O/) and H α i α j = α α) H H α j H α i H H H α i α j H α = O/) by Assumtion HO5. α) H = I Bα, ) + Bα, ) Bα, ) ) H Bα, ) H Bα, ) H Bα, ) H I + H + H + ) R 3/ H where H, H, and H 3 are O) by Assumtion HO5 and ρ. Finally lugging the exansion for H, the exressions for the derivatives of H and H, and 3) into 6) and collecting terms of the same orders yields α ) α) = ψ + ψ / + H α α)ψ / + ψ 3 / + H α α)ψ / + H α α)ψ / H Bα, ) H α α)ψ ψ ) i / + R ) /) 3/ α α i i= = ψ + ψ ) / + ψ ) 3 / + R ) /) 3/. 7)

22 te 5. Higher Order Bias and Variance. Recall β α) β = β α) β α) + β α) β. 8) i) Plugging 7) in for β α) β α) and 5) in for the d i in 8) yields β α)β ) = β α) β ) + A Ψ i ψ + Bα, )) i/ + + A Ψ i ψ ) i / i= ψ + i= j= i= A Ψ ij A A j A Ψ i A A i A Ψ j ) Bα, )) iψ + Bα, )) j/ + R/) 3/ where R = O ). ince the Ψ i, Ψ ij,... are odd functions of V and α is an even function of V, it follows that Bias β α)) = 0. Using the indeendence obtained in te, the higher order variance is V ar β α) β )) = A + E[ A Ψ i ψ + i= Bα, )) i/ ) A Ψ i ψ + i= Bα, )) i/ ) ]. Following Rothenberg 984), note that since α is an even function of V and i is an odd function of V, the ψ s and Ψ i are uncorrelated. Hence, E[ A Ψ i ψ + Bα, )) i/ ) A Ψ i ψ + Bα, )) i/ ) ] i= = i= j= i= E[Ψ i Ψ j ]E[ψ ) i ψ ) j ] + Bα, ) ibα, ) j ). Lemmas C.6 and C.7 and Assumtion 4 imly that E[ψ ψ ] = ) Γ + O/). hen defining ζ ij = E[Ψ i Ψ j ] and ) Γ + Γ ΘΓ ) = Υ = i= j= ζ ij Γ ) ij 9) yields V ar β α)β )) = A + Υ/ + i= j= ζ ij Bα, ) ibα, ) j + O/ ). 30)

23 ii) he derivation with α relaced with α ) is similar. In articular, the exansion for β α ) β ) is β α ) )β ) = β α) β ) + A Ψ i ψ ) i / + A Ψ i ψ ) ) i / + i= j= + R/) 3/. i= A Ψ ij A A j A Ψ i A A i A Ψ j )ψ ) i ψ ) j / i= It then follows as in i) that and that Bias β α ) ) = 0 V ar β α ) )β )) = A + Υ/ + O/ ). 3) Finally, the O/ ) term in 30) and 3) is the same, so V ar β α) β )) V ar β α ) ) β )) = is ositive semidefinite since ζ ij Bα, ) ibα, ) j = E[ i= j= References i= i= j= Bα, ) i Ψ i ) ζ ij Bα, ) ibα, ) j Bα, ) j Ψ j )]. Billingsley, P. 995): Probability and Measure. New York: Wiley, third edn. Doukhan, P. 994): Mixing: Proerties and Examles, vol. 85 of Lecture Notes in tatistics ringer- Verlag). New York: ringer-verlag. Fitzatrick, P. M. 996): Advanced Calculus: A Course in Mathematical Analysis. PW Publishing Comany: Boston, Massachussetts. Lehmann, E. L. 983): heory of Point Estimation. New York: Wiley. Phillis, P. C. B., and H. R. Moon 999): Linear Regression Limit heory for Nonstationary Panel Data, Econometrica, 675), 057. Rothenberg,. J. 984): Aroximate Normality of Generalized Least quares Estimates, Econometrica, 54), White, H. 00): Asymtotic heory for Econometricians. an Diego: Academic Press, revised edn. j= 3