Spatial autoregression model:strong consistency

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1 Statistics & Probability Letters 65 ( Spatial autoregression model:strong consistency B.B. Bhattacharyya a, J.-J. Ren b, G.D. Richardson b;, J. Zhang b a Department of Statistics, North Carolina State University, Raleigh, NC , USA b Department of Mathematics, University of Central Florida, Orlando, FL , USA Received February 2003; received in revised form March 2003 Abstract Let ( ˆ n ; ˆ n denote the Gauss Newton estimator of the parameter (; in the autoregression model Z ij = Z i 1;j + Z i;j 1 Z i 1;j 1 + ij. It is shown in an earlier paper that when = =1; {n 3=2 (ˆ n ; ˆ n } converges in distribution to a bivariate normal random vector. A two-parameter strong martingale convergence theorem is employed here to prove that n r (ˆ n ; ˆ n 0 almost surely when r 3 2. c 2003 Published by Elsevier B.V. MSC: 62F12; 60F05; 62M30 Keywords: Spatial autoregression; Unit roots; Two-parameter martingale 1. Introduction and preliminaries Martin (1979 introduced the spatial autoregression model Z ij = Z i 1;j + Z i;j 1 Z i 1;j 1 + ij (1.1 and indicated its applicability in a subsequent paper in The above model is used in the study of image processing by Jain (1981 and in the analysis of digital ltering and system theory by Tjostheim (1981. Basu and Reinsel (1994 illustrate the feasibility of (1.1 being nonstationary with a practical example. Moreover, Cullis and Gleeson (1991 include model (1.1 with = =1 in the class of models used to represent the error structure in linear regression to analyze eld data. When 1 and 1, various estimators of (; are known to converge in distribution to a bivariate normal, where the normalizing term is n (Basu, 1990; Khalil, A similar conclusion is proved by Bhattacharyya (1995 when = = 1 by using a Gauss Newton estimator ( ˆ n ; ˆ n and Corresponding author. Tel.: ; Fax: addresses: bhattach@stat.ncsu.edu (B.B. Bhattacharyya, garyr@pegasus.cc.ucf.edu (G.D. Richardson /$ - see front matter c 2003 Published by Elsevier B.V. doi: /j.spl

2 72 B.B. Bhattacharyya et al. / Statistics & Probability Letters 65 ( a normalizing term n 3=2. It is proved here that n r (ˆ n ; ˆ n 0 almost surely when = =1 and r 3 2. It is assumed without further mention that Z ij = 0 when i j 6 0. The following axioms are listed below for future reference. Assumptions. (A.1 = =1 (A.2 ij ; i;j 1, are i.i.d., each having mean zero, variance 2 and a nite fourth moment (A.3 E i k=1 k1 4+2 =O(i 2+ for some 0, where ij are i.i.d. and each has mean zero. (A.4 n and n are initial estimators obeying n r ( n 0 and n r ( n 0 almost surely when r 3 2. Remark 1.1. Assume model (1.1 obeys (A.1 (A.3. According to (1.4, X ij =X i 1;j + ij ; dene n to be the least squares estimator of. Since = =1;X ij = Z ij Z i;j 1 is observable and, moreover, when r 3 2 ; nr ( n =n r 3 n i;j=1 X i 1;j ij =n 3 n i;j=1 X i 1;j 2 0 a.s. by Theorem 2.2(i and Theorem 2.4(i. The estimator n is dened similarly and hence this establishes the existence of initial estimators in axiom (A.4 whenever (A.1 (A.3 are fullled. Let 0 =(; ; n =( n ; n ;f ij (a; b=az i 1;j + bz i;j 1 abz i 1;j 1 ;F ij(a; b=(@f ij (a; ij (a; b=@b =(Z i 1;j bz i 1;j 1 ;Z i;j 1 az i 1;j 1 and R ij (a; b = ( a( bz i 1;j 1. Denote ˆ n =[ i;j=1 F ij( n F ij( n ] 1 i;j=1 F ij( n (Z ij f ij ( n ; then ˆ n = ˆ n + n is said to be the Gauss Newton estimator of 0. It is shown by Bhattacharyya (1995 that ˆn obeys 1 n ˆ n 0 = F ij ( n F ij( n n F ij ( n (R ij ( n + ij : (1.2 i;j=1 i;j=1 Dene X ij = Z ij Z i;j 1 ;Y ij = Z ij Z i 1;j and observe that F ij( n =(X i 1;j +( n Z i 1;j 1 ;Y i;j 1 +( n Z i 1;j 1 : (1.3 Moreover, using model (1.1, i X ij = X i 1;j + ij and thus X ij = kj when = =1: (1.4 Likewise, Y ij = Y i;j 1 + ij and Y ij = k=1 j il when = =1: (1.5 l=1 The principal result of this work is stated below and proved in subsequent sections. Theorem 1.2. Assume that model (1.1 obeys axioms (A.1 (A.3; then n r ( ˆ n 0 0 almost surely when r 3 2.

3 B.B. Bhattacharyya et al. / Statistics & Probability Letters 65 ( Even though (A.3 implies the existence of fourth moments of ij, condition (A.2 is listed separately since some results proved here are valid under (A.1 (A.2. Note that (A.3 is satised when ij ; i;j 1, are i.i.d. and each is distributed as N (0; 2, which leads to the corollary below. More generally, axiom (A.3 is fullled with = 1 when (A.2 is valid and E Corollary 1.3. Suppose that ==1 in model (1.1 and ij ;i;j 1, are i.i.d. and each is distributed as N(0; 2. Then n r ( ˆ n 0 0 almost surely when r Strong martingale Almost sure convergence results are established in this section by appealing to a martingale convergence theorem for doubly indexed processes due to Walsh (1979, Let N 2 denote the set of all ordered pairs of positive integers and dene s =(s 1 ;s 2 6 (t 1 ;t 2 =t i s i 6 t i and s t provided s i t i ; i =1; 2. Suppose that (; F;P denotes the underlying probability space and F t ;t N 2, is an increasing sequence of sub--elds of F; that is, F s F t F when s 6 t. Moreover, assume that V t ; t N 2,isanF t -measurable random variable. Then {V t ; F t ; t N 2 } is called a strong martingale if it obeys the following conditions: (a E[V t F s ]=V s when s 6 t (martingale (b E[V (s;t] F s ] = 0 provided s t, where V (s;t]=v t V s1t 2 V t1s 2 + V s and F s denotes the smallest -eld containing each F ij with either i 6 s 1 or j 6 s 2. Walsh (1979, 1986 proved that an L 1 -bounded strong martingale converges almost surely. An extension of Kronecker s lemma to two indices is given below. Lemma 2.1. Assume that {x ij : i; j 1} and {c ij : i; j 0} are sequences of real numbers, where c ij =0 when i j =0, satisfying the following conditions: (a c ij = c ij c i 1;j c i;j 1 + c i 1;j 1 0 when i j 1 (b c Mn =c nn 0 and c nm =c nn 0 as n, for each xed M 1 (c y mn = m;n k;l=1 x kl=c kl y as m n and {y mn : m; n 1} is bounded. Then (1=c nn i;j=1 x ij 0 as n. Proof. Let y ij = i j k=1 l=1 x kl=c kl when i j 1 and y ij = 0 elsewhere. Denote y ij = y ij y i 1;j y i;j 1 + y i 1;j 1 and thus y ij = x ij =c ij when i j 1. Then i;j=1 x ij = i;j=0 c ij y ij i;j=0 c ijy i 1;j i;j=0 c ijy i;j 1 + i;j=0 c ijy i 1;j 1 = i;j=0 c ijy ij 1 i=0 j=0 c i+1;jy ij 1 i=0 j=0 c i;j+1y ij + 1 i;j=0 c i+1;j+1y ij = 1 i;j=0 y ij c i+1;j+1 1 i=0 (c i+1;n c in y in 1 j=0 (c n;j+1 c nj y nj + c nn y nn. Dene U n =(1=c nn 1 i;j=0 y ij c i+1;j+1 ;V n =(1=c nn 1 i=0 (c i+1;n c in y in and W n =(1=c nn 1 j=0 (c n;j+1 c nj y nj.

4 74 B.B. Bhattacharyya et al. / Statistics & Probability Letters 65 ( Observe that since c ij 0 and c ij =0 when i j=0, it follows that c ij 0 provided i j 0 and c ij c st if i 6 s; j 6 t and (i; j is distinct from (s; t. First, it is shown that U n y as n. Since 1 i;j=0 c i+1;j+1=c nn ; U n y =(1=c nn 1 i;j=0 (y ij y c i+1;j+1 6 (1=c nn N 1 i;j=0 y ij y c i+1;j+1 + (1=c nn N 1 i=0 1 j=n y ij y c i+1;j+1 +(1=c nn 1 i=n N 1 j=0 y ij y c i+1;j+1 +(1=c nn 1 i;j=n y ij y c i+1;j+1. Assumption (c implies that for 0; y ij y when i j N and also since {y ij } is bounded, y ij y 6 M for all i j 1. Then U n y 6 (M=c nn c NN +(M=c nn c Nn + (M=c nn c nn +(=c nn (c nn c Nn c nn + c NN and thus by assumption (b, U n y as n. Likewise, V n y =(1=c nn 1 i=0 (c i+1;n c in (y in y 6 (1=c nn N 1 i=0 (c i+1;n c in (y in y + (1=c nn 1 i=n (c i+1;n c in (y in y 6 (M=c nn c Nn +(=c nn (c nn c Nn and thus V n y as n. Similarly, W n y as n and thus by assumption (c, (1=c nn i;j=1 x ij 0asn. Recall from (1.4 and (1.5, when = = 1 in model (1.1, X ij = i k=1 kj;y ij = j l=1 il and, moreover, Z ij = i;j k;l=1 kl. Theorem 2.2. Suppose that model (1.1 and axioms (A.1 (A.2 are fullled. Then (i 1 n r i;j=1 X i 1;j ij 0 a.s. when r 3 2 (ii 1 n r i;j=1 Y i;j 1 ij 0 a.s. when r 3 2 (iii 1 n r i;j=1 Z i 1;j 1 ij 0 a.s. when r 2. Proof. (i Dene, for each t =(t 1 ;t 2 N 2 ;V t = t 1;t 2 i;j=1 X i 1;j ij and let F t be the smallest -eld for which each ij is measurable, i 6 t 1 ;j6 t 2. First, it is shown that {V t ; F t ; t N 2 } is a martingale. Fix s =(s 1 ;s 2 6 (t 1 ;t 2 =t; employing standard properties of conditional expectation, E[V t F s ]=V s + = V s + t 1;t 2 i=1;j=s 2+1 t 1;s 2 i=s 1+1;j=1 E(X i 1;j E( ij + [ s1 kj E( ij F s + k=1 t 1;s 2 i=s 1+1;j=1 i 1 k=s 1+1 E[X i 1;j ij F s ] ] E( kj ij = V s by (A.2. Hence {V t ; F t ; t N 2 } is a martingale. It remains to show that the above is a strong martingale. Assume that s=(s 1 ;s 2 (t 1 ;t 2 =t; then V (s;t]= t 1;t 2 i=s X 1+1;j=s 2+1 i 1;j ij. Hence E[V (s;t] F s ]= t 1;t 2 i=s E[X 1+1;j=s 2+1 i 1;j ij F s ]= t 1;t 2 [ i=s 1+1;j=s 2+1 s1 k=1 kje( ij + ] i 1 k=s E( 1+1 kje( ij =0, and thus {V t ; F t ; t N 2 } is a strong martingale. Dene W t = t 1;t 2 i;j=1 X i 1;j ij =i p j q, where p 1 and q 1 2. Then {W t; F t ; t N 2 } is also a strong martingale and, moreover, E(Wt 2 = t 1;t 2 i;j=1 E(X i 1;j 2 2 =i 2p j 2q = O(1. According to Walsh (1986, Corollary 2.8, W t W almost surely as t 1 t 2, for some W. Employing Lemma 2.1 with c ij = i p j q,it follows that (1=n r i;j=1 X i 1;j ij 0 almost surely when r 3. Parts (ii (iii are veried using 2 a similar argument. The next result is due to Hu et al. (1989.

5 B.B. Bhattacharyya et al. / Statistics & Probability Letters 65 ( Theorem 2.3. Assume that {V nj :16 j 6 n; n 1} is an array of random variables satisfying: (i E(V nj =0 (ii V n1 ;V n2 ;:::;V nn are independent, for each xed n 1 (iii sup n;j E V nj 2p+, for some 1 6 p 2 and 0. Then (1=n 1=p j=1 V nj 0 almost surely. Theorem 2.4. Suppose that model (1.1 obeys axioms (A.1 (A.3. Then (i 1 n 3 i;j=1 X 2 i 1;j 2 2 a.s. (ii 1 n 3 i;j=1 Y 2 i;j a.s. Proof. (i Denote V nj =(1=n 2 i=1 [X i 1;j 2 E(X i 1;j 2 ]; Theorem 2.3 is used to show that (1=n j=1 V nj 0 a.s. Since X ij = i k=1 kj, axiom (A.2 implies that V n1 ;V n2 ;:::;V nn are i.i.d. random variables and thus Theorem 2.3(ii is satised. It follows from axiom (A.3 that E V nj 2+ 6 (C 1 =n 2(2+ n 1+ n i=1 E X i 1;j (C 2 =n 3+ i=1 i2+ =O(1. Employing Theorem 2.3 with p=1, (1=n 3 i;j=1 [Xi 1;j 2 E(X i 1;j 2 i 1;j 2 =2 and thus (1=n 3 i;j=1 X 2 ] 0 a.s. Since E(X 2 i 1;j =(i 12,(1=n 3 i;j=1 E(X 2 i j 2 =2 a.s. Part (ii is proved in a similar manner. 3. ProofofTheorem 1.2 Bhattacharyya (1995 proves that {n 3=2 ( ˆ n 0 } converges in distribution to a bivariate normal random vector. Under assumptions (A.1 (A.3, it is shown below that n r ( ˆ n 0 0 almost surely when r 3 2. Lemma 3.1. Suppose that model (1.1 and axioms (A.1 (A.2 are satised. Then (i var i;j=1 X i 1;j 2 =O(n 5, (ii var i;j=1 Y i;j 1 2 =O(n 5, (iii var =O(n 8, (iv var (v var (vi var i;j=1 Z i 1;j 1 2 i;j=1 X i 1;jY i;j 1 =O(n 4, i;j=1 X i 1;jZ i 1;j 1 =O(n 6, i;j=1 Y i;j 1Z i 1;j 1 =O(n 6. Proof. Only verication of (v is given here since the other arguments are similar. Since X i 1;j = i 1 k=1 kj and Z i 1;j 1 = i 1 j 1 k=1 l=1 kl are independent, mean zero random variables,

6 76 B.B. Bhattacharyya et al. / Statistics & Probability Letters 65 ( cov(x i 1;j Z i 1;j 1 ;X i 1;j Z i 1;j 1 = 0 unless j = j. Moreover, cov(x i 1;j ( Z i 1;j 1 ;X i 1;jZ i 1;j 1 = E(X i 1;j X i 1;jE(Z i 1;j 1 Z i 1;j 1 = O((i i 2 j and when i 6 i n, var i;j=1 X i 1;jZ i 1;j 1 = i ( n n i =1 i=1 j=1 O(i2 j=o(n 6. Therefore var i;j=1 X i 1;jZ i 1;j 1 =O(n 6. Let n =( n ; n denote the initial estimator dened in Remark 3.2, A n = diag(n 3=2 ;n 3=2, A = diag( 2 =2; 2 =2, G n = i;j=1 F ij( n F ij( n and R ij ( n = ( n ( n Z i 1;j 1, where F ij( n is given in (1.3. Lemma 3.3. Suppose that model (1.1 and axioms (A.1 (A.3 are fullled. Then (i A n G A n A a.s. (ii A n n i;j=1 F ij( n R ij ( n 0 a.s. (iii n A n n i;j=1 F ij( n ij 0 a.s. when 0. Proof. (i Denote [ ] bn c n A n G n A n = : c n d n Then b n =(1=n 3 i;j=1 [X i 1;j+( n Z i 1;j 1 ] 2 =(1=n 3 i;j=1 X i 1;j 2 +(2=n3 ( n i;j=1 X i 1;j Z i 1;j +(( n 2 =n 3 i;j=1 Z i 1;j 1 2 :=V n1+v n2 +V n3. According to Theorem 2.4(i, V n1 2 =2 a.s. It follows from Lemma 3.1(v and the Borel Cantelli Lemma that (1=n 7=2+ i;j=1 X i 1;jZ i 1;j 1 0 a.s. when 0, and hence by Remark 1.1, n 3=2 ( n (1=n 7=2+ i;j=1 X i 1;jZ i 1;j 1 0 a.s. In particular, V n2 0 a.s. A similar argument using Lemma 3.1(iii and the Borel Cantelli Lemma shows that V n3 0 a.s. Hence b n 2 =2 a.s.; likewise, c n 0 a.s. and d n 2 =2 a.s. (ii Verication is similar to part (i. (iii Appealing to (1.3, n A n n i;j=1 F ij( n ij =n 3=2 i;j=1 X i 1;j ij ; n +n i;j=1 Y 3=2 i;j 1 ij i;j=1 Z i 1;j 1 ij ( n ; n 0 a.s. by Theorem 2.2 and Remark 1.1. Verication of Theorem 1.2 now follows immediately from Lemma 3.3. Indeed, employing (1.2, n A 1 n ( ˆ n 0 =(A n G n A n 1 n A n n i;j=1 F ij( n (R ij ( n + ij A 1 0 =0 a.s. 4. Further reading Martin (1990 may also be of interest to the reader. Acknowledgements The authors are grateful to M. Amezziane for bringing Hu et al. s work to our attention.

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