A supplement to Asymptotic Distributions of Quasi-Maximum Likelihood Estimators for Spatial Autoregressive. Appendix A: Some Useful Lemmas

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1 A supplemet to Asymptotic Distributios of Quasi-Maximum Likelihood Estimators for Spatial Autoregressive Models (for referece oly; ot for publicatio) Appedix A: Some Useful Lemmas A. Uiform Boudedess of Matrices i Row ad Colum Sums Lemma A. Suppose that the spatial weights matrix W is a o-egative matrix with its (i, j)th elemet beig w,ij = dij l= d il ad d ij 0 for all i, j. () If the row sums j= d ij are bouded away from zero at the rate h uiformly i i, ad the colum sums i= d ij are O(h ) uiformly i j, the {W } are uiformly bouded i colum sums. () If d ij = d ji for all i ad j ad the row sums j= d ij are O(h ) ad bouded away from zero at the rate h uiformly i i, the {W } are uiformly bouded i colum sums. Proof: () Let c ad c be positive costats such that c h j= d ij for all i ad i= d ij c h for all j, for large. It follows that i= w,ij = d ij i= d c il h i= d ij c c for all i. l= () This is a special case of () because l= d il = O(h ) ad i= d ij = i= d ji imply i= d ij = O(h ). Lemma A. Suppose that lim sup λ 0 W <, where is a matrix orm, the { S } is uiformly bouded i both row ad colum sums. Proof: For ay matrix orm, λ 0 W < implies that S 985, p.30). Let c = sup λ 0 W. The, S Lemma A.3 Suppose that { W } ad { S = k=0 (λ 0W ) k (Hor ad Johso k=0 λ 0W k = k=0 ck = c < for all. }, where is a matrix orm, are bouded. The { S (λ) }, where S (λ) =I λw, is uiformly bouded i a eighborhood of λ 0. Proof: Let c be a costat such that W c ad S c for all. We ote that S (λ) = (S (λ λ 0 )W ) = S (I (λ λ 0 )G ), where G = W S. By the submultiplicative property of a matrix orm, G W S c for all. Let B (λ 0 )={λ : λ λ 0 < /c }. It follows that, for ay λ B (λ 0 ), (λ λ 0 )G λ λ 0 G <. As (λ λ 0 )G <, I (λ λ 0 )G is ivertible ad (I (λ λ 0 )G ) = k=0 (λ λ 0) k G k. Therefore, (I (λ λ 0 )G ) k=0 λ λ 0 k G k k=0 λ λ 0 k c k = λ λ 0 c < for ay λ B (λ 0 ). The result follows by takig a close eighborhood B(λ 0 ) cotaied i B (λ 0 ). I B(λ 0 ),

2 sup λ B(λ0) λ λ 0 c <, ad, hece, sup λ B(λ 0) S (λ) S sup (I (λ λ 0 )G ) λ B(λ 0) sup λ B(λ 0) c λ λ 0 c <. Lemma A.4 Suppose that W for all, where is a matrix orm, the { S (λ) }, where S (λ) =I λw, are uiformly bouded i ay closed subset of (, ). Proof: For ay λ (, ), λw λ W < ad, hece, S (λ) = k=0 λk W k. It follows that, for ay λ <, S (λ) k=0 λ k W k k=0 λ k = λ. Hece, for ay closed subset B of (, ), sup λ B S (λ) sup λ B λ <. Lemma A.5 Suppose that elemets of the k matrices X are uiformly bouded; ad the limitig matrix of X X exists ad is osigular, the the projectors M ad (I M ), where M = I X (X X ) X, are uiformly bouded i both row ad colum sums. Proof: Let B =( X X ). From the assumptio of the lemma, B coverges to a fiite limit. Therefore, there exists a costat c b such that b,ij c b for all, where b,ij is the (i, j)th elemet of B. By the uiform boudedess of X, there exists a costat c x such that x,ij c x for all. Let A = X ( X X ) X, which ca be rewritte as A = k s= k r= b,rsx,r x,s, where x,r is the rth colum of X. It follows that j= a,ij j= k s= k r= b,rsx,ir x,js k c b c x, for all i =,,. Similarly, i= a,ij i= k s= k r= b,rsx,ir x,js k c b c x for all j =,,. That is, {X (X X ) X } are uiformly bouded i both row ad colum sums. Cosequetly, {M } are also uiformly bouded i both row ad colum sums. A. Orders of Some Relevat Quatities Lemma A.6 Suppose that the elemets of the sequeces of vectors P =(p,,p ) ad Q = (q,,q ) are uiformly bouded for all. ) If {A } are uiformly bouded i either row or colum sums, the Q A P = O(). ) If the row sums of {A } ad {Z } are uiformly bouded, z i, A P = O() uiformly i i, where z i, is the ith row of Z. Proof: Let costats c ad c such that p i c ad q i c. For ), there exists a costat such that i= j= a,ij c 3. Hece, Q A P = i= j= a,ijq i p j c c i= j= a,ij c c c 3. For ), let c 4 be a costat such that j= a,ij c 4 for all ad i. It follows that e i A P = j= a,ijp j c j= a,ij c c 4 where e i is the ith uit colum vector. Because {Z } is uiformly

3 bouded i row sums, j= z,ij c z for some costat c z. It follows that z i, A P j= z,ij e j A P ( j= z,ij )c c 4 c z c c 4. Lemma A.7 Suppose {A } are uiformly bouded either i row sums or i colum sums. The, ) elemets a,ij of A are uiformly bouded i i ad j, ) tr(a m )=O() for m, ad 3) tr(a A )=O(). Proof: If A is uiformly bouded i row sums, let c be the costat such that max i j= a,ij c for all. O the other had, if A is uiformly bouded i colum sums, let c be the costat such that max j i= a,ij c for all. Therefore, a,ij l= a,il c if A is uiformly bouded i row sums; otherwise, a,ij k= a,kj c if A is uiformly bouded i colum sums. The result ) implies immediately that tr(a )=O(). If A is uiformly bouded i row (colum) sums, the A m for m is uiformly bouded i row (colum) sums. Therefore, ) implies tr(a m )=O(). Fially, as tr(a A )= i= j= a,ij, tr(a A ) i= ( j= a,ij ) c if A is uiformly bouded i row sums; otherwise tr(a A ) j= ( i= a,ij ) c. Lemma A.8 Suppose that the elemets a,ij of the sequece of matrices {A }, where A =[a,ij ], are O( h ) uiformly i all i ad j; ad {B } is a sequece of coformable matrices. () If {B } are uiformly bouded i colum sums, the elemets of A B have the uiform order O( h ). () If {B } are uiformly bouded i row sums, the elemets of B A have the uiform order O( h ). For both cases () ad (), tr(a B )=tr(b A )=O( h ). Proof: Cosider (). Let a,ij = c,ij h. Because a,ij = O( h ) uiformly i i ad j, there exists a costat c so that c,ij c for all i, j ad. Because {B } is uiformly bouded i colum sums, there exists a costat c so that k= b,kj c for all ad j. Let a i, be the ith row of A ad b,l be the lth colum of B. It follows that a i, b,l h j= c,ijb,jl c h j= b,jl cc h, for all i ad l. Furthermore, tr(a B ) = i= a i,b,i i= a i,b,i c c h. These prove the results i (). The results i () follow from () because (B A ) = A B ad the uiform boudedess i row sums of {B } is equivalet to the uiform boudedess i colum sums of {B }. Lemma A.9 Suppose that A are uiformly bouded i both row ad colum sums. Elemets of the k matrices X are uiformly bouded; ad lim X X exists ad is osigular. Let M = I X (X X ) X. The 3

4 (i) tr(m A )=tr(a )+O(), (ii) tr(a M A )=tr(a A )+O(), (iii) tr[(m A ) ]=tr(a )+O(), ad (iv) tr[(a M A ) ]=tr[(m A A ) ]=tr[(a A ) ]+O(). Furthermore, if A,ij = O( h ) for all i ad j, the (a) tr (M A )=tr (A )+O( h ) ad (b) i= ((M A ) ii ) = i= (A,ii) + O( h ). Proof: The assumptios imply that elemets of the k k matrix ( X X ) are uiformly bouded for large eough. Lemma A.6 implies that elemets of the k k matrices X A X, X A A X ad X A X are also uiformly bouded. It follows that tr(m A )=tr(a ) tr[(x X ) X A X ]=tr(a )+O(), tr(a M A )=tr(a A ) tr[(x X ) X A A X ]=tr(a A )+O(), ad tr((m A ) )=tr(a ) tr[(x X ) X A X ]+tr([(x X ) X A X ] )=tr(a )+O(). By (iii), tr[(a M A ) ]=tr[(m A A ) ]=tr[(a A ) ]+O(). Whe A,ij = O( h ), from (i), tr (M A )=(tr(a )+O()) = tr (A )+tr(a ) O() + O() = tr (A )+O( h ). Because A is uiformly bouded i colum sums ad elemets of X are uiformly bouded, X A e i = O() for all i. Hece, i= (M A ) ii = i= (A,ii x i, (X X ) X A e i ) = i= (A,ii + O( )) = i= [(A,ii) +A,ii O( )+O( )] = i= (A,ii) + O( h ). Lemma A.0 Suppose that A is a matrix with its colum sums beig uiformly bouded ad elemets of the k matrix C are uiformly bouded. Elemets v i sofv =(v,,v ) are i.i.d.(0,σ ). The, C A V = O P (), Furthermore, if the limit of C A A C exists ad is positive defiite, the C A V D N(0,σ 0 lim C A A C ). Proof: This is Lemma A. i Lee (00). These results ca be established by Chebyshev s iequality ad Liapouov double array cetral limit theorem. A.3 First ad Secod Momets of Quadratic Forms ad Limitig Distributio For the lemmas i this subsectio, v s i V =(v,,v ) are i.i.d. with zero mea, variace σ ad fiite fourth momet µ 4. Lemma A. Let A =[a ij ] be a -dimesioal square matrix. The 4

5 ) E(V A V )=σ tr(a ), ) E(V A V ) =(µ 4 3σ 4 ) i= a ii + σ4 [tr (A )+tr(a A )+tr(a )], ad 3) var(v A V )=(µ 4 3σ 4 ) i= a ii + σ4 [tr(a A )+tr(a )]. I particular, if v s are ormally distributed, the E(V A V ) = σ 4 [tr (A )+tr(a A )+tr(a )] ad var(v A V )=σ 4 [tr(a A )+tr(a )]. Proof: The result i ) is trivial. For the secod momet, E(V A V ) = E( a ij v i v j ) = E( a ij a kl v i v j v k v l ). i= j= i= j= k= l= Because v s are i.i.d. with zero mea, E(v i v j v k v l ) will ot vaish oly whe i = j = k = l, (i = j) (k = l), (i = k) (j = l), ad (i = l) (j = k). Therefore, E(V A V ) = a ii E(v4 i )+ a ii a jj E(vi v j )+ a ij E(v i v j )+ a ij a ji E(vi v j ) i= i= i= j i =(µ 4 3σ 4 ) a ii + σ4 [ a ii a jj + =(µ 4 3σ 4 ) i= j= i= j i i= j= i= j= i= j i a ij + a ij a ji ] a ii + σ 4 [tr (A )+tr(a A )+tr(a )]. i= The result 3) follows from var(v A V )=E(V A V ) E (V A V ) ad those of ) ad ). Whe v s are ormally distributed, µ 4 =3σ. Lemma A. Suppose that {A } are uiformly bouded i either row ad colum sums, ad the elemets a,ij of A are O( h ) uiformly i all i ad j. The, E(V A V )=O( h ), var(v A V )=O( h ) ad V A V = O P ( h ). Furthermore, if lim h =0, h V A V h E(V A V )=o P (). Proof: E(V A V )=σ tr(a )=O( h ). From Lemma A., the variace of V A V is var(v A V )= (µ 4 3σ 4 ) i= a,ii + σ4 [tr(a A )+tr(a )]. Lemma A.8 implies that tr(a ) ad tr(a A ) are O( h ). As i= a,ii tr(a A ), it follows that i= a,ii = O( h ). Hece, var(v A V ) = O( h ). As E((V A V ) )=var(v A V )+E (V A V )=O(( h ) ), the geeralized Chebyshev iequality implies that P ( h V A V M) M ( h ) E((V A V ) )= M O() ad, hece, h V A V = O P (). Fially, because var( h V A V )=O( h )=o() whe lim h = 0, the Chebyshev iequality implies that h V A V h E(V A V )=o P (). Lemma A.3 Suppose that {A } is a sequece of symmetric matrices with row ad colum sums uiformly bouded ad {b } is a sequece of costat vectors with its elemets uiformly bouded. The momet E( v 4+δ ) for some δ>0 of v exists. Let σ Q be the variace of Q where Q = b V + V A V 5

6 σ tr(a ). Assume that the variace σq is O( h ) with { h σ Q } bouded away from zero, the elemets of A are of uiform order O( h ) ad the elemets of b of uiform order O( h h ).Iflim + δ =0, the Q σ Q D N(0, ). Proof: The asymptotic distributio of the quadratic radom form Q ca be established via the martigale cetral limit theorem. Our proof of this Lemma follows closely the origial argumets i Kelejia ad Prucha (00). I their paper, σ Q is assumed to be bouded away from zero with the -rate. Our subsequet argumets modify theirs to take ito accout the differet rate of σq. The Q ca be expaded ito Q = i= b iv i + i= a,iivi + i i= j= a,ijv i v j σ tr(a )= i= Z i, where Z i = b i v i +a,ii (vi i σ )+v i j= a,ijv j. Defie σ-fields J i =<v,...,v i > geerated by v,...,v i. Because vs are i.i.d. with zero mea ad fiite variace, E(Z i J i )=b i E(v i )+a,ii (E(v i ) σ )+E(v i ) i j= a,ijv j = 0. The {(Z i, J i ) i, } forms a martigale differece double array. We ote that σ Q = i= E(Z i )asz i are martigale differeces. Also h σ Q = O(). Defie the ormalized variables Z i = Z i/σ Q. The {(Z i, J i) i } is a martigale differece double array ad Q σ Q = i= Z i. I order for the martigale cetral limit theorem to be applicable, we would show that there exists a δ > such that i= E Z i teds to zero as goes to ifiity. Secod, it will be show +δ that i= E(Z i J i ) p. For ay positive costats p ad q such that p + q =, Z i a,ii v i σ + v i ( b i + i j= a,ij v j )= a,ii p a,ii q v i σ + v i ( b i p bi q + i j= a,ij p a,ij q vj ). The Holder iequality for ier products applied to the last term implies that i p q i Z i q ( b i p ) p + ( a,ij p ) p ( b i q vi ) q +( a,ii q v i σ ) q + ( a,ij q vi v j ) q = b i + j= q i p a,ij b i v i q + a,ii vi i σ q + a,ij q v i q v j q. j= As {A } are uiformly bouded i row sums ad elemets of b are uiformly bouded, there exists a costat c such that j= a,ij c / ad b i c / for all i ad. Hece Z i q q q p c ( b i v i q + a,ii v i σ q + v i q i j= a,ij v j q ). Take q =+δ. Let c q > be a fiite costat such that E( v ) c q, E( v q ) c q ad E( v σ q ) c q. Such a costat exists uder the momet coditios of v. It follows that i= E Z i q q q p c c q i= ( b i + i j= a,ij ) =O(). As i= E Z i +δ = σ +δ i= E Z i +δ Q ad σ +δ Q =( h σ Q ) + δ ( h ) + δ c ( h ) + δ for some costat c>0 whe is large, i= E Z i +δ = 6 j= j= q

7 δ O( h+ δ )=O( h+ δ ) δ, which goes to zero as teds to ifiity. It remais to show that i= E(Z i J i ) p 0. As i i E(Zi J i )=(µ 4 σ 4 )a,ii + σ (b i + a,ij v j ) +µ 3 a,ii (b i + a,ij v j ), j= ad E(Zi )=(µ 4 σ 4 )a,ii +4σ4 i j= a,ij + σ b i +µ 3a,ii b i, because E(b i + i j= a,ijv j ) = b i +4σ i j= a,ij. Hece, i i i i E(Zi J i ) E(Zi) =4σ ( a,ij a,ik v j v k + a,ij(vj σ )) + 4(σ b i + µ 3 a,ii )( a,ij v j ) j= k j ad i= E(Z i J i ) = σq i= [E(Z i J i ) E(Zi )] = ad H 3 = h H = h i i a,ij a,ik v j v k, i= j= k j j= H = h j= j= 4σ (H h σ + H )+ 4 H h Q σ 3, where Q i a,ij(vj σ ), i= j= i= (σ b i + µ 3 a,ii ) i j= a,ijv j. We would like to show that H j for j =,, 3, coverge i probability to zero. It is obvious that E(H 3 ) = 0. By exchagig summatios, i= (σ b i + µ 3 a,ii ) i j= a,ijv j = j= ( i=j+ (σ b i + µ 3 a,ii )a,ij )v j. Thus, E(H3 h )=σ4 ( (b i + µ 3 σ a,ii)a,ij ) σ4 h ( max b i + µ 3 i σ a,ii ) ( j= i=j+ j= i=j+ a,ij ) = O( h ) because max i,j a,ij = O( h ), max i b i = O( h ) ad j= ( i=j+ a,ij ) = O(). E(H ) = 0 ad H ca be rewritte ito H = h j= ( i=j+ a,ij )(v j σ ). Thus E(H )=(h ) (µ 4 σ 4 ) ( a,ij ) ( h ) (µ 4 σ 4 ) max a,ij ( a,ij ) = O( i,j ). j= i=j+ We coclude that H 3 = o P () ad H = o P (). E(H ) = 0 but its variace is relatively more complex tha that of H ad H 3. By rearragig terms, H = h i i i= j= k j a,ija,ik v j v k = h S j= k j,jk v j v k, where S,jk = i=max{j,k}+ a,ija,ik. The variace of H is E(H ) =( h ) j= S,jk S,rs E(v j v k v r v s ). j= k j r= s r As k j ad s r, E(v j v k v r v s ) 0 oly for the cases that (j = r) (k = s) ad (j = s) (k = r). The variace of H ca be simplified ad E(H )=σ4 ( h ) σ 4 ( h ) σ 4 ( h j= k j i = j= k j S,jk σ4 ( h ) ( ( a,ija,ik ) max max ) j i = a,ij max i,k a,i k 7 j= k j i = i = j i = ( i = j= i=j+ a,ija,ika,ija,ik ) a,ij max i,k a,i k a,ij k= a,ik ) =O( h ),

8 because A is uiformly bouded i row ad colum sums ad a,ij = O( h ) uiformly i i ad j. Thus, h H = o P () as lim h = 0 implied by the coditio lim (+ ) δ =0. As H j, j =,, 3, are o P () ad lim h σ Q > 0, i= E(Z i J i ) coverges i probability to. The cetral limit theorem for the martigale differece double array is thus applicable (see, Hall ad Heyde, 980; Potscher ad Prucha, 997) to establish the result. Lemma A.4 Suppose that A is a costat matrix uiformly bouded i both row ad colum sums. Let c be a colum vector of costats. If h c h c = o(), the c A V = o P (). O the other had, if h c c h = O(), the c A V = O P (). h Proof: The first result follows from Chebyshev s iequality if var( c A V )=σ0 h c A A c goes to zero. Let Λ be the diagoal matrix of eigevalues of A A ad Γ be the orthoormal matrix of eigevectors. As eigevalues i absolute values are bouded by ay orm of the matrix, eigevalues i Λ i absolute value are uiformly bouded because A (or A ) are uiformly bouded. Hece, h c A A c h c Γ Γ c λ,max = h c c λ,max = o(), where λ,max is the eigevalue of A A with the largest absolute value. Whe h c h c = O(), c A A c h c c λ,max = O(). I this case, var( h c A V )= σ h c A A c = O(). Therefore, h c A V = O P (). Appedix B: Detailed Proofs: Proof of Cosistecy (Theorem 3. ad Theorem 4.) We shall prove that l L (λ) Q (λ) coverges i probability to zero uiformly o Λ, ad the idetificatio uiqueess coditio holds, i.e., for ay ɛ>0, lim sup [max λ Nɛ(λ 0) (Q (λ) Q (λ 0 )] < 0 where N ɛ (λ 0 ) is the complemet of a ope eighborhood of λ 0 i Λ with radius ɛ. For the proof of these properties, it is useful to establish some properties for l S (λ) ad σ (λ), where σ (λ) = σ o tr(s S (λ)s (λ)s )=σ 0 [ + (λ 0 λ) tr(g )+(λ 0 λ) tr(g G )]. There is also a auxiliary model which has useful implicatios. Deote Q p, (λ) = (l(π) +) l σ (λ) +l S (λ). The log likelihood fuctio of a SAR process Y = λw Y + V, where V N(0,σ 0 I ), is l L p, (λ, σ )= l(π) l σ +l S (λ) σ Y S (λ)s (λ)y. It is apparet that Q p, (λ) = max σ E p (l L p, (λ, σ )), where E p is the expectatio uder this SAR process. By the Jese iequality, Q p, (λ) E p (l L p, (λ 0,σ 0 )) = Q p,(λ 0 ) for all λ. This implies that (Q p,(λ) Q p, (λ 0 )) 0 for all λ. 8

9 Let λ ad λ be i Λ. By the mea value theorem, (l S (λ ) l S (λ ) ) = tr(w S ( λ ))(λ λ ) where λ lies betwee λ ad λ. By the uiform boudedess of Assumptio 7, Lemma A.8 implies that tr(w S ( λ )) = O( h ). Thus, l S (λ) is uiformly equicotiuous i λ i Λ. As Λ is a bouded set, (l S (λ ) l S (λ ) ) =O() uiformly i λ ad λ i Λ. The σ (λ) is uiformly bouded away from zero o Λ. This ca be established by a couter argumet. Suppose that σ (λ) were ot uiformly bouded away from zero o Λ. The, there would exist a sequece {λ } i Λ such that lim σ (λ ) = 0. We have show that (Q p,(λ) Q p, (λ 0 )) 0 for all λ, ad (l S (λ 0 ) l S (λ) ) =O() uiformly o Λ. This implies that l σ (λ) l σ 0 + (l S (λ 0 ) l S (λ) ) =O(). That is, l σ (λ ) is bouded from above, a cotradictio. Therefore, σ (λ) must be bouded always from zero uiformly o Λ. (uiform covergece) Show that sup λ Λ l L (λ) Q (λ) = o P (). Note that l L (λ) Q (λ) = (l ˆσ (λ) l σ (λ)). Because M S (λ)y =(λ 0 λ)m G X β 0 + M S (λ)s V, ˆσ (λ) = Y S (λ)m S (λ)y =(λ 0 λ) (G X β 0 ) M (G X β 0 )+(λ 0 λ)h (λ)+h (λ), (B.) where H (λ) = (G X β 0 ) M S (λ)s V ad H (λ) = V S S (λ)m S (λ)s V. As H (λ) = (G X β 0 ) M V +(λ 0 λ) (G X β 0 ) M G V, Lemma A.0 ad the liearity of H (λ) iλ imply H (λ) =o P () uiformly i λ Λ. Note that H (λ) σ (λ) = V S S (λ)s (λ)s V σ 0 where T (λ) = V S tr(s S (λ)x (X X ) X S (λ)s V. By Lemma A.0, S (λ)s (λ)s ) T (λ), X S (λ)s V = X S V λ X G V = O p (). Thus, T (λ) = ( X S (λ)s V ) ( X X ) ( X S (λ)s V ) = o P (). By Lemma A., [V S S (λ)s (λ)s V σ0 tr(s S (λ)s (λ)s )] = o P () uiformly i λ Λ. These covergeces are uiform o Λ because λ appears simply as liear or quadratic factors i those terms. That is, H (λ) σ (λ) =o P () uiformly o Λ. Therefore, ˆσ (λ) σ (λ) =o P () uiformly o Λ. By the Taylor expasio, l ˆσ (λ) l σ (λ) = ˆσ (λ) σ (λ) / σ (λ), where σ (λ) lies betwee ˆσ (λ) ad σ (λ). Note that σ (λ) σ (λ) because σ (λ) =(λ 0 λ) (G X β 0 ) M (G X β 0 )+ 9

10 σ (λ). As σ (λ) is uiformly bouded away from zero o Λ, σ (λ) will be so too. It follows that, because ˆσ (λ) σ (λ) =o P () uiformly o Λ, ˆσ (λ) will be bouded away from zero uiformly o Λ i probability. Hece, l ˆσ (λ) l σ (λ) = o P () uiformly o Λ. Cosequetly, sup λ Λ l L (λ) Q (λ) = o P (). (uiform equicotiuity) We will show that l Q (λ) = (l(π)+) l σ (λ)+ l S (λ) is uiformly equicotiuous o Λ. The σ (λ) is uiformly cotiuous o Λ. This is so, because σ (λ) isa quadratic form of λ ad its coefficiets, (G X β 0 ) M (G X β 0 ), tr(g ) ad tr(g G ) are bouded by Lemmas A.6 ad A.8. The uiform cotiuity of l σ (λ) o Λ follows because σ (λ) is uiformly bouded o Λ. Hece l Q (λ) is uiformly equicotiuous o Λ. (idetificatio uiqueess) At λ 0, σ (λ 0)=σ 0. Therefore, Q (λ) Q (λ 0 )= (l σ (λ) l σ 0)+ (l S (λ) l S (λ 0 ) ) [l σ (λ) l σ (λ)] = (Q p,(λ) Q p, (λ 0 )) [l σ (λ) l σ (λ)]. Suppose that the idetificatio uiqueess coditio would ot hold. The, there would exist a ɛ>0 ad a sequece λ i N ɛ (λ 0 ) such that lim [ Q (λ ) Q (λ 0 )] = 0. Because N ɛ (λ 0 ) is a compact set, there would exist a coverget subsequece {λ m } of {λ }. Let λ + be the limit poit of λ m i Λ. As Q (λ) is uiformly equicotiuous i λ, lim m [ m Q m (λ + ) m Q m (λ 0 )] = 0. Because (Q p, (λ) Q p, (λ 0 )) 0 ad [l σ (λ) l σ (λ)] 0, this is possible oly if lim m (σ m (λ + ) σ m (λ + )) = 0 ad lim m ( m Q p,m (λ + ) m Q p,m (λ 0 )) = 0. The lim m (σ m (λ + ) σ m (λ + )) = 0 is a cotradictio whe lim (G X β 0 ) M (G X β 0 ) 0. I the evet that lim (G X β 0 ) M (G X β 0 ) = 0, the cotradictio follows from the relatio lim ( Q p,(λ + ) Q p,(λ 0 )) = 0 uder Assumptio 9. This is so, because, i this evet, Assumptio 9 is equivalet to that lim [ (l S (λ) l S ) (l σ (λ) l σ 0)] = lim [Q p,(λ) Q p, (λ 0 )] 0 for λ λ 0. Therefore, the idetificatio uiqueess coditio must hold. The cosistecy of ˆλ ad, hece, ˆθ follow from this idetificatio uiqueess ad uiform covergece (White 994, Theorem 3.4). Proof of Theorem 3. (Show that Σ θ is osigular): Let α =(α,α,α 3 ) be a colum vector of costats such that Σ θ α =0. It is sufficiet to show that α = 0. From the first row block of the liear equatio system Σ θ α = 0, oe X has lim X X α + lim GXβ0 α = 0 ad, therefore, α = lim (X X ) X G X β 0 α. From the last equatio of the liear system, oe has α 3 = σ0 lim tr(g) α. By elimiatig α ad 0

11 α 3, the remaiig equatio becomes { lim σ 0 [ (G X β 0 ) M (G X β 0 ) + lim tr(g G )+tr(g ) ]} tr (G ) α =0. (B.) Because tr(g G )+tr(g ) tr (G )= tr[(c + C )(C + C ) ] 0 ad Assumptio 8 implies that lim (G X β 0 ) M (G X β 0 ) is positive defiite, it follows that α = 0 ad, so, α =0. (the limitig distributio of l L (θ 0) θ ): The matrix G is uiformly bouded i row sums. As the elemets of X are bouded, the elemets of G X β 0 for all are uiformly bouded by Lemma A.6. With the existece of high order momets of v i Assumptio, the cetral limit theorem for quadratic forms of double arrays of Kelejia ad Prucha (00) ca be applied ad the limitig distributio of the score vector follows. (Show that l L ( θ ) θ θ l L (θ 0) θ θ p 0): The secod-order derivatives are l L (θ) β β = σ X X, l L (θ) β = σ X W Y, l L (θ) β σ = σ 4 X V (δ), l L (θ) = tr([w S (λ)] ) σ Y W W Y, l L (θ) σ = σ 4 Y W V (δ), ad l L (θ) σ σ = σ 4 σ V (δ)v 6 (δ). As X X = O(), X WY = O P () ad σ p σ0, it follows that l L ( θ ) β β l L (θ 0 ) β β =( σ0 σ ) X X = o p (), ad l L ( θ ) l L (θ 0 ) =( β β σ0 σ ) X W Y = o p (). As V ( δ )=Y X β λ W Y = X (β 0 β )+(λ 0 λ )W Y + V, l L ( θ ) β σ l L (θ 0 ) β σ =( σ0 4 σ 4 ) X V + X X σ 4 ( β β 0 )+ X W Y ñσ 4 ( λ λ 0 )=o p (). Let G (λ) =W S (λ). By the mea value theorem, tr(g ( λ )) = tr(g )+tr(g 3 ( λ )) ( λ λ 0 ), therefore, l L ( θ ) l L (θ 0) = tr(g3 ( λ )) ( λ λ 0 )+( σ 0 σ ) Y W WY = o p (), because tr(g 3 ( λ )) = O( h ) ad Y W W Y = O P ( h ). Note that G ( λ ) is uiformly bouded i row ad colum sums uiformly i a eighborhood of λ 0 by Lemma A.3 uder Assumptio 5. Therefore, tr(g 3 ( λ )) = O( h ). O the other had, because Y W V ( δ )= Y W X (β 0 β )+(λ 0 λ ) Y W W Y + Y W V = Y W V + o P ()

12 ad V ( δ )V ( δ )=( β β 0 ) X X ( β β 0 )+( λ λ 0 ) Y W W Y it follows that +( λ λ 0 )( β β 0 ) X W Y = V V + o P (), +(β 0 β ) X V + V V +(λ 0 λ ) Y W V l L ( θ ) σ l L (θ 0 ) σ = Y W V ( δ ) σ 4 + Y W V σ0 4 = Y W X σ 4 ( β β 0 )+ Y W W Y σ 4 ( λ λ 0 )+( σ0 4 σ 4 ) Y W V = o p (), ad l L ( θ ) σ σ l L (θ 0 ) σ σ = σ 4 V ( δ )V ( δ ) σ 6 σ0 4 + V V σ0 6 = ( σ 4 σ0 4 )+( σ0 6 σ 6 ) V V + o P () = o p (). (Show l L (θ 0) θ θ E( l L (θ 0) θ θ ) p 0): By Lemma A.0, X G V = o P () ad X G G V = o P (). It follows that X W Y = X G X β 0 + o P (), Y W V = V G V + o P (), ad Y W W Y = (X β 0 ) G G X β 0 + V G G V + o P (). Lemmas A. ad A.8 imply E(V G V )=σ 0tr(G ) ad var( V G V )= (µ 4 3σ 4 0 ) Similarly, E(V G G V )=σ 0tr(G G ) ad i= var( V G G V )= (µ 4 3σ 4 0 ) G,ii + σ4 0 [tr(g G )+tr(g )] = O( ). h i= (G G ) ii +σ4 0 tr((g G ) )=O( h ). By the law of large umbers, V V p σ 0. With these properties, the covergece result follows. Fially, from the expasio ( ) (ˆθ θ 0 )= l L ( θ ) l L (θ 0) θ θ θ, the asymptotic distributio of ˆθ follows. Proof of Theorem 4. The osigularity of Σ θ will ow be guarateed by Assumptio 9 istead of Assumptio 8. With (B.) i the proof of Theorem 3., uder Assumptio 8, oe arrives at [ ] lim tr(g G )+tr(g ) (G ) tr α =0.

13 Because [tr(g G )+tr(g ) tr (G )] = tr[(c +C )(C +C ) ] > 0 for large implied by Assumptio 9, it follows that α = 0. Hece Σ θ is osigular. The remaiig argumets are similar as i the proof of Theorem 3.. Proof of Theorem 5. (Show that h [l L (λ) l L (λ 0 ) (Q (λ) Q (λ 0 )] the mea value theorem, p 0 uiformly o Λ): From (.7) ad (3.3), by h [l L (λ) l L (λ 0 ) (Q (λ) Q (λ 0 ))] = h [l ˆσ (λ) l ˆσ (λ 0 ) (l σ (λ) l σ (λ 0 ))] = h [(l ˆσ (λ) l σ (λ)) (l ˆσ (λ 0) l σ (λ 0))] = h [l ˆσ ( λ ) l σ ( λ )] (λ λ 0 ). With the expressios of ˆσ (λ) i (.6) ad σ (λ) i (3.), it follows that ˆσ (λ) = Y W M S (λ)y, ad σ (λ) = {(λ λ 0)(G X β 0 ) M (G X β 0 ) σ0 tr[g S (λ)s ]}. These imply that = = h [(l L (λ) l L (λ 0 )) (Q (λ) Q (λ 0 ))] h ˆσ ( λ ) {Y W M S ( λ )Y ˆσ ( λ ) σ ( λ ) [(λ 0 λ )(G X β 0 ) M (G X β 0 ) + σ0 tr(g S ( λ )S )]}(λ λ 0) h ˆσ ( λ ) {Y W M S ( λ )Y [(λ 0 λ )(G X β 0 ) M (G X β 0 )+σ0tr(g S ( λ )S )] By usig S (λ)s ˆσ ( λ ) σ ( λ ) [(λ σ 0 ( λ ) λ )(G X β 0 ) M (G X β 0 )+σ0tr(g S ( λ )S )]}(λ λ 0 ). = I +(λ 0 λ)g, the model implies that Y W M S (λ)y =(G X β 0 ) M S (λ)s + V G M S (λ)s V X β 0 +(G X β 0 ) M S (λ)s V + V G M S (λ)s X β 0 =(G X β 0 ) M (G X β 0 )(λ 0 λ)+(g X β 0 ) M V +(G X β 0 ) M G V (λ 0 λ) + V G M V + V G M G V (λ 0 λ). Lemma A.9 implies that tr(m G )=tr(g )+O() ad tr(g M G )=tr(g G )+O(). The law of large umbers i Lemma A. shows that h (V M G V σ 0 tr(g )) = o P () ad h (V G M G V 3

14 σ 0 tr(g G )) = o P (). Uder Assumptio 0, h (G X β 0 ) M V = o P () ad h (G X β 0 ) M G V = o P () by Lemma A.4. Therefore, h {Y W M S (λ)y (G X β 0 ) M (G X β 0 )(λ 0 λ) σ0tr(g S (λ)s )} = h {(G X β 0 ) M V +(λ 0 λ)(g X β 0 ) M G V + V G M V +(λ 0 λ)v G M G V σ 0tr(G ) σ 0(λ 0 λ)tr(g G )} = o P (). From (B.) ad (3.), ˆσ (λ) σ (λ) =(λ 0 λ) (G X β 0 ) M S (λ)s + {V S + σ 0 {tr[s = o P (), S (λ)m S (λ)s S (λ)m S (λ)s V V σ0tr[s S (λ)m S (λ)s ]} ] tr[s S (λ)s (λ)s ]} uiformly i λ by Chebyshev s LLN, Lemma A. ad Lemma A.9. Note that h (G X β 0 ) M (G X β 0 )= O() ad h tr(g S (λ)s )=O(). Whe h, σ(λ) =σ0[+(λ 0 λ) tr(g) +(λ λ 0 ) tr(g G ) ] σ 0 uiformly o Λ. As σ (λ) σ (λ) ad σ 0 > 0, σ ( λ is O() ad ) ˆσ ( λ is O ) P (). I coclusio, h [(l L (λ) l L (λ 0 )) (Q (λ) Q (λ 0 ))] = o P () uiformly i λ Λ. (Show the uiform equicotiuity of h (Q (λ) Q (λ 0 ))): Recall that h (Q (λ) Q (λ 0 )) = h (l σ (λ) l σ 0)+ h (l S (λ) l S (λ 0 ) ). As tr(s S (λ)s (λ)s ) =(λ 0 λ)tr(g + G )+(λ λ 0 ) tr(g G ), h (σ (λ) σ 0 ) =(λ λ 0 ) h (G X β 0 ) M (G X β 0 )+σ0 h {tr[s S (λ)s (λ)s ] } =(λ λ 0 ) h (G X β 0 ) M (G X β 0 )+σ0(λ 0 λ) h tr(g + G )+σ0(λ 0 λ) h tr(g G ) is uiformly equicotious i λ Λ. By the mea value theorem, h (l σ (λ) l σ 0 )= h σ (λ)(σ (λ) σ 0 ) where σ (λ) lies betwee σ (λ) ad σ 0.Asσ (λ) is uiformly bouded away from zero o Λ ad σ 0 > 0, σ (λ) is uiformly bouded from above. Hece, h (l σ (λ) l σ 0) is uiformly equicotiuous o Λ. The h (l S (λ) l S (λ 0 ) ) = h tr(w S ( λ ))(λ λ 0 ) is uiformly equicotiuous o Λ because h tr(w S ( λ )) = O P (). I coclusio, h (Q (λ) Q (λ 0 )) is uiformly equicotiuous o Λ. 4

15 (uiqueess idetificatio): For idetificatio, let D (λ) = h (l σ (λ) l σ 0 )+ h (l S (λ) l S (λ 0 ) ). The, h (Q (λ) Q (λ 0 )) = D (λ) h (l σ (λ) l σ (λ)). By the Taylor expasio, l σ (λ) l σ(λ) = (σ (λ) σ (λ)) σ (λ) = (λ λ 0) σ (λ) h (G X β 0 ) M (G X β 0 ), where σ lies betwee σ (λ) ad σ (λ). Because σ (λ) σ (λ) for all λ Λ, h (l σ (λ) l σ (λ)) h σ (λ) (λ λ 0) (G X β 0 ) M (G X β 0 ). For the situatio i Assumptio 8, σ (λ) σ (λ) =o P () uiformly o Λ. Thus, lim σ (λ) =σ 0. Therefore, uder Assumptio 0(a), lim h (l σ (λ) l σ (λ)) lim σ (λ) (λ λ 0) h (G X β 0 ) M (G X β 0 ) = (λ λ 0) σ 0 h lim (G X β 0 ) M (G X β 0 ) < 0, for ay λ λ 0. Furthermore, uder the situatio i Assumptio 0(b), D (λ) < 0 wheever λ λ 0. It follows that lim h (Q (λ) Q (λ 0 )) < 0 wheever λ λ 0. As h (Q (λ) Q (λ 0 )) is uiformly equicotiuous, the idetificatio uiqueess coditio holds ad θ 0 is idetifiably uique. The cosistecy of ˆλ follows from the uiform covergece ad the idetificatio uiqueess coditio. For the pure SAR process, β = 0 is imposed i the estimatio. The cosistecy of the QMLE of λ follows by similar argumets above. For the pure process, ˆσ (λ) = Y S (λ)s (λ)y ad the cocetrated log likelihood fuctio is l L (λ) = (l(π)+) l ˆσ (λ)+l S (λ). For the pure process, Q (λ) happes to have the same expressio as that of the case where G X β 0 is multicolliear with X. The simpler aalysis correspods to settig X = 0 ad M = I i the precedig argumets. Proof of Theorem 5. (Show h ( l L ( λ ) l L (λ 0) )=o P ()): The first ad secod order derivatives of the cocetrated log likelihood are l L (λ) = ˆσ (λ) Y W M S (λ)y tr(w S (λ)), ad l L (λ) = ˆσ (λ) 4 (Y W M S (λ)y ) ˆσ (λ) Y W M W Y tr([w S (λ)] ), where ˆσ (λ) = Y S (λ)m S (λ)y. For the pure SAR process, β 0 = 0 ad the correspodig derivatives are similar with M replaced by the idetity I. So it is sufficiet to cosider the regressive model. 5

16 Because M X = 0 ad S (λ) =S +(λ 0 λ)w, oe has Y W M W Y =(G X β 0 ) M (G X β 0 )+(G X β 0 ) M G V + V G M G V, ad Y W M S (λ)y = Y W M S Y +(λ 0 λ)y W M W Y = Y W M V +(λ 0 λ)y W M W Y =(G X β 0 ) M V + V G M V +(λ 0 λ)[(g X β 0 ) M (G X β 0 ) +(G X β 0 ) M G V + V G M G V ]. As show i the proof of Theorem 5., h (G X β 0 ) M G V = o P (). Hece, h Y W M W Y = h (G X β 0 ) M (G X β 0 )+ h V G M G V + o P () ad h Y W M S (λ)y = h V G M V +(λ 0 λ)[ h (G X β 0 ) M (G X β 0 )+ h V G M G V ]+o P (). Lemma A. implies that V G M V = O p ( h ) ad V G M G V = O p ( h ). Thus h Y W M S (λ)y = O P () uiformly o Λ. From the proof of Theorem 5., ˆσ (λ) =σ (λ) +o P () = σ 0 + o P () uiformly h o Λ, whe lim h =. Thus, ˆσ (λ) V G M G V = h V σ G M 0 G V + o P () uiformly o Λ. Therefore, oe has h l L (λ) = σ0 [ h (G X β 0 ) M (G X β 0 )+ h V G M G V ] h tr([w S (λ)] )+o P (), uiformly o Λ. By Lemma A.8, uder Assumptio 7, h tr(g 3 (λ)) = O() uiformly o Λ. Therefore, by the Taylor expasio, h ( l L ( λ ) l L (λ 0) )= h {tr([w S ( λ )] ) tr(g )} + o P () = h tr(g3 ( λ ))( λ λ 0 )+o P () = o P (), for ay λ which coverges i probability to λ 0. (Show h ( l L (λ 0) E(P (λ 0 ))) p 0): Defie P (λ 0 ) = [(G σ0 X β 0 ) M (G X β 0 )+V G M G V ] tr(g ). The h l L (λ 0) = h P (λ 0 )+o P (). From Lemma A.9, tr(g M G )=tr(g G )+O(), tr[(g M G ) ]=tr[(g G ) ]+ O() ad i= ((G M G ) ii ) = i= ((G G ) ii ) + O( h ). The tr(g M G ) ad tr((g M G ) ) are O( h ) ad i= ((G G ) ii ) = O( h ) from Lemma A.8. Therefore, E(P (λ 0 )) = σ0 (G X β 0 ) M (G X β 0 ) [tr(g G )+tr(g )] + O(). 6

17 As h [P (λ 0 ) E(P (λ 0 ))] = σ0 +o(), where = h [V G M G V σ0 tr(g M G )], h [P (λ 0 ) E(P (λ 0 ))] = o P () if = o P (). By Lemma A. ad the orders of relevat terms, E( )=( h ) var(v G M G V )=( h ) [(µ 4 3σ 4 0) which goes to zero. Therefore, h (Show h l L (λ 0) Let q = V C M V. Thus, i= ( ) l L (λ 0) E(P (λ 0 )) p N(0,σ λ ) whe lim h = ): (G,iM G,i ) +σ 4 0tr((G M G ) )] = O( h ), = o P (). h l L (λ 0 ) = h ˆσ (λ 0 ) [(G X β 0 ) M V + q ]. The mea of q is E(q )=σ 0 tr(m C )= σ 0 tr[(x X ) X C X ]=O() because X X = O() ad X CX = O() from Lemma A.6. The variace of q from Lemma A. is σ q =(µ 4 3σ0 4 ) ((C M ) ii ) + σ0 4 [tr(c M C )+tr((c M ) )] =(µ 4 3σ 4 0) i= (C,ii ) + σ0[tr(c 4 C )+tr(c)] + O(), i= where the last expressio follows because ((C M ) ii ) = i= i= (C,ii ) +O( h ), tr(c M C )=tr(c C )+O() ad tr[(c M ) ]=tr(c )+O() from Lemma A.9. The covariace of a liear term Q V ad a quadratic form V P V is E(Q V V P V )= Q i= j= p,ije(v v i v j )=Q vec D (P )µ 3. Deote σ lq = var((g X β 0 ) M V + q ). Thus, As σ lq = σ 0(G X β 0 ) M (G X β 0 )+σ q +(G X β 0 ) M vec D (C M )µ 3. h h E(q )=O( ), which goes to zero, h l L (λ 0 ) h σ lq = ˆσ (λ 0) h σ lq = ˆσ (λ 0) [(G X β 0 ) M V + q E(q )] h + σ lq [(G X β 0 ) M V + q E(q )] σ lq E(q ) ˆσ (λ 0) + o P () p h N(0, lim σlq σ0 4 ). As (C,ii ) = O( h ), h tr (G )=O( h )=o(), h i= (C,ii) = O( h ) which goes to zero whe lim h =. Fially, as h lim [tr(c C h )+tr(c )] = lim [tr(g G )+tr(g ) h tr (G )] = lim [tr(g G )+tr(g )]. 7

18 The limitig distributio of (ˆλ λ 0 ) follows from the Taylor expasio ad the covergece results above. Proof of Theorem 5.3 As S (λ) =S +(λ 0 ˆλ )W, ˆβ β 0 =(X X ) X V (ˆλ λ 0 )(X X ) X W Y =(X X ) X V (ˆλ λ 0 )(X X ) X G X β 0 (ˆλ λ 0 )(X X ) X G V. As (ˆλ λ 0 )(X X ) X G V = O P ( h ) because X X = O() ad X GV = O P () ad ˆλ λ 0 = h O P ( ) by Theorem 5., ˆβ β 0 =(X X ) X V (ˆλ λ 0 )(X X ) X G X β 0 + O p ( h ). I geeral, ( h ˆβ β 0 )= ( X X h ) X V = (ˆλ λ 0 ) (X h X ) X G X β 0 + O P ( ) (ˆλ λ 0 ) (X h X ) X G X β 0 + O P ( ). h If β 0 is zero, ( ˆβ β 0 )=( X X ) X V h + O P ( ) D N(0,σ0 lim ( X X ) ). As ˆσ = Y S (ˆλ )M S (ˆλ )Y = Y S M S Y +(λ 0 ˆλ ) Y W M S Y +(λ 0 ˆλ ) Y W M W Y ad Y S M S Y = V M V, it follows that (ˆσ σ0 )= (V V σ0 ) V X (X X ) X V h (ˆλ λ 0 ) h Y W M S Y + (ˆλ λ 0 ) h h Y W M W Y. Because M X = 0, ad h (G X β 0 ) M V = O P () ad h (G X β 0 ) M G V = O P () uder Assumptio 0, h Y W M h S Y = (G X β 0 + G V ) h M V = V G M V + O P ( )=O P ( ) h ad h Y W M h W Y = (G X β 0 + G V ) M (G X β 0 + G V ) = h V G M G V + O P ( )=O P ( ) h h by Lemmas A. ad A.4. As E( V X (X X ) X V ) = σ 0 tr(x (X X ) X ) = σ 0 k goes to zero, the Markov iequality implies that V X (X X ) X V = o P (). Hece, as lim h =, (ˆσ σ 0 )= (V V σ 0 )+o P () D N(0,µ 4 σ 4 ). 8

19 Proof of Theorem 5.4 Let X =(X,X ), M = I X (X X ) X ad M = I X (X X ) X. Usig a matrix partitio for (X X ), ˆβ β 0 =(X M X ) X M V c (ˆλ λ 0 )+O P ( h ), ad ˆβ β 0 =(X M X ) X M V + O P ( h ). Therefore, ( h ˆβ β 0 )= ( h X M X ) X M V c (ˆλ λ 0 )+O P ( ) h = c (ˆλ λ 0 )+O P ( ), h h ad ( ˆβ β 0 )=( X M X ) X M h V + O P ( ). The asymptotic distributios of ˆβ ad ˆβ follow. 9