Testing for Cross-Sectional Dependence in a Random
Effects Model
Afees Salisu1, Sam Olofin1, Eugene Kouassi2
1Centre for Econometrics and Allied Research (CEAR), Department of Economics, University of Ibadan, Ibadan, Nigeria 2Department of Economics, University of Cocody, Abidjan, Cote d’Ivoire
Email: [email protected]
Received December 4,2011; revised December 28, 2011; accepted January 10, 2012
ABSTRACT
This paper extends and generalizes the works of [1,2] to allow for cross-sectional dependence in the context of a two-way error components model and consequently develops LM test. The cross-sectional dependence follows the first order spatial autoregressive error (SAE) process and is imposed on the remainder disturbances. It is important to note that this paper does not consider alternative forms of spatial lag dependence other than SAE. It also does not allow for endogeneity of the regressors and requires the normality assumption to derive the LM test.
Keywords: Cross-Sectional Dependence; Error Components Model; Lagrangian Multiplier (LM) Tests
1. Introduction
The standard error components model assumes, among others, spatial independence across cross-sectional units. However, this restrictive assumption may not hold for a lot of panel data applications. When one begins to look at a cross section of regions, states, countries, etc., these agg- regate units may exhibit cross-sectional correlation that has to be dealt with (see [3]). Ignoring cross-sectional dependence when in fact it exists, results in biased, in- consistent and inefficient estimates of regression coeffi- cients (see [1,4,5]).
In the literature, several test statistics have been de- veloped for spatial econometrics however in the context of either cross sectional framework or one-way error components model.1 The specification of cross-sectional
dependence in linear regression models by most of these works follows either spatial autoregressive (SAR) pro- cess often defined as spatial lag dependence (see [6-10]); spatial moving average processes (SMA) often called spatial error dependence (see [11]); spatial autoregressive error process (SAE) (see [6,12,13]); SARMA (a combi- nation of SAR and SMA) (see [2,4,14]); a combination of SAR and SARE (see [15]); direct representation form of cross-sectional dependence (see [16,17]) or spatial error component process (SEC) suggested by [18]. Con- sequently, various tests as well as estimators were de- rived against these different specification forms using ei- ther the Maximum Likelihood (ML) approach (see [2,19];
for a survey of the literature) or Instrumental Variables (IV) and Generalized Method of Moments (GMM) (see [9,20,21]).
The present study develops LM test for cross-sectional dependence in the context of panel data framework. The latter is a two-way random effects model where the cross-sectional dependence follows the SAE and is im- posed on the remainder disturbances. Prominent papers that have adopted the SAE include [2,4] in the context of cross-sectional framework, and [1,22,23] in the context of one-way error components framework. Thus, the main objective of this work is to extend and generalize the works of [1,2] to allow for cross-sectional dependence in the context of a two-way error components model. The panel data model considered here is the restricted two- way random effects model assuming no cross-sectional dependence in the remainder disturbances. Thus, the LM test will be similar to the one developed by [1] if we fur-ther modify the hypothesis to test for cross-sectional de-pendence assuming the presence of random individual effect only (while ignoring the presence of time effects). In the same vein, the LM test will be similar to [2,4] if the hypothesis is reconstructed to test for cross-sectional dependence ignoring the presence of both the random country and time effects.
In Section 2, the structure of the two-way error ran-dom effects model is described in the context of cross- sectional dependence in the remainder disturbance term. Analyses of the LM test are provided in Section 3 and Section 4 concludes the paper.
1A review of score test statistics for alternative specifications in spatial
2. The Model
k 1 We consider the following panel data regression model:; 1, , ; 1, ,
ytixtiu iti N t T
y
th
i tth
(1) where the index i denotes N regional units and the index t refers to the T observations of each region i. The i sub-script, therefore, denotes the cross-sectional dimension whereas t denotes time-series dimension. The total num-ber of observations is NT. ti is the observation on the
region over the time period; xti is the
observation on k explanatory variables and ti is the
regression disturbance term. The error term ti follows
a two-way random effects with both regional specific and temporal effects; that is,
th ti
u u
ti i t
uti v (2)
where i denotes regional specific effects, t denotes
temporal effects and ti
v
represents the remainder distur- bance term. Stacking the N observations of each time- period t, Equation (2) may be written as:
t t N t
u i v
u
iN
1, ,1
1, t2, , tN
(3)
where 1 2 tN , is a vector
of ones of N dimension,
, , ,
t t t
u u u
t t and
1, 2, , N
.
Assumption 1: Both i and vt are assumed
inde-pendent and normally distributed according to,
2
; ~
0, 2
t N v
~ 0,
i N v
( )
(4)
The remainder disturbance term t is assumed to
follow the first order spatial error correlation (see [2,3]), that is:
t W t et
(5) where t t1, ,tN and et
et1, ,etN
. Theterm is the scalar spatial autoregressive coefficient with 1. The matrix W is an spatial weight matrix which represents the degree of potential interac-tion between neighboring locainterac-tions whose diagonal ele-ments are zero and off-diagonal eleele-ments are non-zero. Equation (5) can be further simplified as:
NN
1t IN W et
(6)
Given Equation (6), the weight matrix W also satisfies the condition that IN W is nonsingular for all
1
. t is also assumed to be independent and
nor-mally distributed as: e
2
~ 0,
t e
e IN (7) The eti process is also independent of the i and
terms.
t
v
The model (1) can be re-written in matrix notation as:
y Xu
1 NT
(8) where y is of dimension vector, X is an NTk
matrix, is vector and u is vector. The matrix X is assumed to be of full column rank and its elements are assumed to be asymptotically bounded in absolute value. Given Equation (6), Equation (3) can be re-written as:
1
NT
1t N t N t
u I W e i v (9) We can write Equation (8) in vector from as:
1
T T N T N
u I B e i I I i v
(Ω)
(10)
The variance-covariance (VCV) matrix of Equa-tion (10) (that is, the unrestricted model) can be ex- pressed as:
1
2 2
2
Ω T e T N
T v N N
I B B J I
I i i
(11)
2
where JT i iT T and it is a matrix of ones of dimension
T. To obtain the spectral decomposition of Equation (11), we use the [24] method. Essentially, we replace JT by
T
TJ and IT by ET JT where ET IT JT and
T T
J J T
and consequently, we obtain3:
1
2 2
1
2 2 2
Ω T e v N N
T e N v N N
E B B i i
J B B T I i i
(12)
Also, using the [25] method of inversion, Equation (12) can be expressed as:
1 1
1 2 2
1 1
2 2 2
1 2
Ω T e v N N
T e N v N N
T T
E B B i i
J B B T I i i
E A J A
(13)
1
12 2
1 e v N N
A B B i i
where
1
12 2 2
2 e N v N N
A B B TI i i
and
2
.
3. Derivation of the LM Test
In this section, we derive the LM test for testing for no cross-sectional dependence in a two-way random effects model. We employ the Maximum Likelihood (ML) ap-proach and consequently, the log-likelihood function. The LM test derived is based on the idea that the score of the likelihood function evaluated under the null is equal to zero when the null hypothesis is true, so that a test based on the square of the score divided by the ap-propriate element of the information matrix (since this is the variance of the score) can be constructed. The use of the normal likelihood function requires the assumption of
2See the appendix for the derivation. 3Note that
T
normality of the error term.
Essentially, the derivation of the LM test involves the following steps:
Step 1: Derive the VCV matrix for the unrestricted model;
Step 2: Derive the VCV matrix for the restricted model;
Step 3: Derive the spectral decomposition for the ma-trices obtained in steps 1 and 2;
Step 4: Derive the inverse of the matrices obtained in steps 1 and 2 using the results from step 3;
Step 5: Derive the general log-likelihood function; Step 6: Use the information in steps 1 - 5 to derive the score functions of the likelihood evaluated from the re-stricted ML under 0
a H ;
Step 7: Derive the information matrix and its inverse; Step 8: Use the results obtained in steps 6 and 7 to de-velop the LM test.
The log likelihood function, L under normality of dis-turbances is given as:
ogΩ 1 Ω 12 2u u
u y X
1
, l
L c (14)
where
2
, , ,
and the vector of parameters is de-
noted as
, 2, 2e v
2 2 2
, , ,
where
e v
.
Since our test statistic requires information only on the vector of parameters , consequently, information due to is ignored. Following [26], the gradient of the log likelihood with respect to can be expressed as:
1 1
Ω Ω
Ω
u u
1
1 2 1
Ω Ω
2 L
tr
(15)
2
i j L I E
(16)
By further simplification, it is easy to show that:
1 Ω 1 Ω
Ω Ω
i j
1 2
I tr
, 1, 2,3, 4.
i j
I
For Equations (15) and (16) represent
the score function and the information matrix respec- tively. The information matrix- is block diagonal.
The LM statistic can, therefore, be written generally as:
1D I D
LM
D
(17) where and I are the score function and informa-
tion matrix respectively evaluated at the null hypothesis. The LM test statistic expressed in (17) is distributed as
2
k
(i.e. chi-square distributed) with k degrees of
freedom, k being the number of parameters in the vector . Based on Equation (17), therefore, the fol-lowing hypotheses can be tested in relation to cross-sec-tional dependence:
2 2 2
0 : 0 0; 0; 0
a
e v
H (18)
This is a test of no cross-sectional dependence assume- ing the presence of random individual and time effects. This is the null hypothesis this study sets out to test.
2 2 2
0: 0 0; 0; 0
b
e v
H (19)
This hypothesis tests for cross-sectional dependence assuming the presence of random individual effect only (while ignoring the presence of time effects). This test is similar to [1] LM test for spatial error correlation as well as random country effects.
2 2 2
0: 0 0; 0; 0
c
e v
H (20)
This hypothesis tests for cross-sectional dependence ignoring the presence of both the random country and time effects. This is similar to the LM test by [2,4].
We derive below the score function for the null hypo- thesis expressed in (18) above which is the focus of this paper; that is:
2 2 2
0 : 0 0; 0; 0
a
e v
H
Under the null hypothesis in (18), the VCV matrix re-duces to:4
2 2
2 2 2
Ω T e N v N N
T e N N v N N
E I i i
J I T I i i
(21)
0
Given that ; then ti eti and, therefore,
Var Var e
ti ti . The Equation (21) is the VCV
ma-trix for the restricted model. Using [25] Lemma 2.1, the inverse of Equation (21) can be expressed as:
1
1 2 2
1
2 2 2
1 2
Ω T e N v N N
T e N N v N N
a a
T T
E I i i
J I T I i i
E A J A
(22)
12 2
1 e N v N N A I i i
where
12 2 2
2 e N N v N N
A I TI i i
and
.
The Equation (22) is the reduced form of Equation (13) and is also the VCV matrix for the familiar two-way random effects error components model. In addition, it is a principal component required in the log-likelihood function to derive the LM test. In particular, both Equa-tions (21) and (22) are required to derive the partial de-rivatives and information matrix for the LM test.
Using the general formulas on log likelihood differen-tiation, we derive its gradients evaluated at the restricted
ML under 0
a
H as follows:
Recall Equation (15):
Ω
Ω
Ω 1u 1 1 1 1 Ω Ω 2 2 L tr u
Assumption 2: Let M
B B
1
, then
1
M B B W B B W . Recall, BIN W a
and since under H0, 0; then BIN and MWW.
Assumption 3: If T, Tand T are idempotent and
symmetric matrices, we can write that E I J IT ETJT where ET IT JT. Then, ET and JT are orthogonal
(see [3]).
Proposition 1: Based on assumptions 2 and 3, we can
write the derivatives
0 Ω a H
for the parameters, ,
2
e
, 2
and 2
v
, respectively, as:
2 e T 0 Ω a H I M 0 2 Ω a T N e H I I 0 2 Ω a T N H TJ I 0 2 Ω aT N N
v H
I i i
Proof:
2 1 1 N NN v N N
i i
I i i
B B 0 1 2 2 1 2 2 1 2 1 2 2 Ω (A) a H
T e v
T e
T e
T e
e T T
E B B
J B B T
E B B
J B B
E B B J
(24)
Based on assumption 1, it is easy to establish from Equation (24) that:
E JT
M
0 2 2 Ω a e T H e IT M (25)5
0 2 22 2 2 2 2
2 2
2
2 2 2
2 Ω B a e H T e
e N v N N T e N N v N N
T e N v N N
e
T e N N v N N
e
T N T N T T N
E
I i i J I T I i i
E I i i
J I T I i i
E I J I E J I
0 2 Ω a T N e H I I
(26)
0 2 2 2 22 2 2
2 2
2
2 2 2
2
2 2 2
2 2 2 Ω C a H
T e N v N N
T e N N v N N
T e N v N N
T e N N v N N
T e N N v N N
T N
E I i i
J I T I i i
E I i i
J I T I i i
J I T I i i
J T I
0 2 Ω a T N H TJ I
(27)
0 2 2 2 22 2 2
2 2
2
2 2 2
2 2 2 2 2 Ω D a v H
T e N v N N
v
T e N N v N N
T e N v N N
v
T e N N v N N
v
T v N N T v N N
v v
T N N T N N T T N N
E I i i
J I T I i i
E I i i
J I T I i i
E i i J i i
E i i J i i E J i i
ETJT
5We also replace by
T
0a
T N N
2
Ω
v H
I i i
(28)
Proposition 2: Based on proposition 1 and assumptions
2 and 3, we can write the derivations of
0 1 Ω Ω
a
H
for
the parameters, , 2
e
, 2
and 2
v
, respectively, as:
1 JT A M2
0
1 Ω 2
Ω
a
e T
H
E A
T 1JT A2
0
1 2
Ω Ω
a
e H
E A
TJT A2
0 1
2 Ω Ω
a
H
1
1 2
Ω
N N
A i i
0
2
Ω
a
T T
v H
E A J
Proof:
These derivatives are quite straightforward to show particularly using the information in proposition 1.
Proposition 3: Based on propositions 1 and 2 and
as-sumptions 2 and 3, we can write the derivations of
0 1 Ω 1
Ω Ω
a
H
for the parameters, ,
2
e
, 2
and
2
v
, respectively, as:
1 1 2
2 2 Ω
Ω Ω E A MA JT A MA
0
1 1
a
e T H
2
21 1
1 2
Ω
Ω Ω ET A JT A
0
2
a
e H
21 1
1 Ω
Ω Ω TJT A
0 2
a
H
1 1
1 1 2 2
2
Ω
Ω Ω ET A i iN N A JT A i iN N A
0
a
v H
Proof:
These derivatives are straightforward to show using the information in proposition 2.
Proposition 4: Following propositions 1 - 3, we can
easily calculate the partial derivates L
for,
2
e
, 2
and 2
v
, respectively, evaluated at the restricted MLE:
0
2 2 2
1 Ξ Ξ
2 2
a
e v e
D H
L
D T
M M
where
2 2 2
N N
e e v
i Mi M
N
;
1 1 2 1
1
N N
D
v i i D MD i M
i
g
Ξu ET M u;Ξu
JTM u
1
1
Ω
T N T
u E I u E A u
1
2 Ω
T N T
u E I u J A u
0
2
** 1 2
2
1
tr 1
2 a
e e H
L
D T A A
u u
where ; ** u u and 12 e2Nv2.
0
2
** 2
2 2 tr
a
H
L T
D A
0
2 2
1
1 2
tr 1
2 a
v v H
L
D
N
N T A A
T N N
u E i i u
where and
JT N N
u i i u
2, 2, 2,
e v.
Proof:
See the appendix for further simplifications and proofs of the partial derivatives.
Recall that we define
0
a
, therefore,
2, 2, 2, 0
e v can be defined as the
H ,
under
solution obtained after maximization of the first order
condition and MLE is the corresponding
re-sidual under
u y X
0
a
H . Note that all the parameters
2 2 2
. . e, , v,0
i e
were evaluated when
0D at the restricted MLE except D
. This isbecause we are testing whether is statistically dif-ferent from zero. Thus, the partial derivatives under H0a
are rewritten in vector form as:
2 2 2
2 2 2
0 0 0
1 Ξ Ξ
2 2
e
v
e v e
D
D
D D
D
D
T M M
0
a
the information matrix under H . The information
ma-trix is given by:
2 1
2
i j L
I E tr Ω 1 ΩΩ1 Ω
i j
(29)
Proposition 5: Using the formular expressed in
Equa-tion (29) and informaEqua-tion in proposiEqua-tion 2, we can derive respective elements in I under a
0
H for the vector of
parameters
2, 2,e
2, v
as follows:
2
21 2
2
1 tr
2 e T T
L
E E A J A
2
21 2
T A JT A
2 2 2
1 tr 2
e L
E E
22
T
E TJ A
2 2 2
1 tr 2
L
2 2
1 1 tr t
2 N N
v L
E T A i i
2
2 2
1 r A i i1N N
2 2 1 tr
e e
E T A M
2 2
2 2
1 tr 2
L
A M
2 2
2 2
tr 2
e T L
E 2 A M
2
1 1
1 1 tr tr
2
v
N N T A i i A M
2
1N N 1
L E
A i i A M
2
2 2
1 2
2 2
1 1 tr tr
N N N N
L
E T A i i A i i
2
e v
2
2 2 2 2 2tr
e
L T
E A
2 2
1 tr
2 N N
E A i i A
0
a
2 2 2
v L
Given these information under H , the LM statistic is
given by,6
1
LM D I D
,
a
(30)
Under 0
Decision Criteria:
The LM statistic is a scalar and the value obtained when the test is performed on the two-way error compo-nents model is compared with the critical value for the chi-squared distribution— 1
2
. The intention is to ascer-tain whether to reject the null hypothesis, 0
a
H , that there
is no cross-sectional dependence problem in a two-way random effects model. Essentially, if LM is less than
the critical value for the chi-squared distribution, then, we do not reject the null hypothesis implying that there is no cross-sectional dependence; otherwise, we reject it.
4. Concluding Remarks
Thispaper provides a framework for testing for no cross- sectional dependence assuming the presence of random individual and time effects. Thus, several important is- sues have not been incorporated. These include testing other hypothesesearlier specified, that is;
2 2 2
: 0 0; 0; 0
b
H0 e v which tests for cross-
sectional dependence assuming the presence of random individual effect only (while ignoring the presence of time
effects; and c: 0 2 0; 2 0; 2 0
H0 e v which
tests for cross-sectional dependence ignoring the pres-ence of both the random country and time effects. Also, the empirical applications section involving Monte Carlo experiments is also not yet considered. These are some of the suggestions for future research.
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[25] J. R. Magnus, “Multivariate Error Components Analysis of Linear and Nonlinear Regression Models by Maximum Likelihood,” Journal of Econometrics, Vol.19, No. 2-3, 1982, pp. 239-285. doi:10.1016/0304-4076(82)90005-7 [26] J. R. Magnus and H. Neudecker, “Matrix Differential
Calculus with Applications in Statistics and Economet-rics,”Wiley Series in Probability and Statistics, Chiches-ter, 1988.
[27] D. A. Harville, “Maximum Likelihood Approaches to Variance Component Estimation and to Related Prob-lems,” Journal of the American Statistical Association, Vol.72, No. 358, 1977, pp. 320-338.
Appendix
(A) Derivation of the VCV Matrix for the Unrestricted Model
Here, 0
N ITiN v
1 1
T T
T N
T N
I B
i I
v I i
and the VCV matrix of u can be derived as follows.
Recall Equation (10),
1
T T
u I B e i I
Using assumption (1), the VCV matrix can be ex-pressed as:
T N
T N
E uu I B E ee
i I E
I i E v
E uu Ω
2
T T N
(A.1)
Let in this case be represented by , and by further simplification, (A.1) becomes:
1 2
2
Ω T e
T v N N
I B B
I i i
i i I
2 2
T N
(A.2)
1 2 Ω T e
T v N N
I B B
I i i
J I
(A.3)
where JT i iT T
T
and it is a matrix of ones of dimension T. To obtain the spectral decomposition of (A.3), we use the [24] method which involves replacing J by TJT
and IT by ET JT where ET IT JT and
T T
J J T in (A.3). This is done as follows:
2
N N
T N N
T v N N
i
J i i
1 2
2 2
1 1
2
2 2
1
2 2
1
2 2
Ω e T T
T N v T T
e T T
T N v T N N
T e T v N N
T e T N
E J B B
TJ I E J i
E B B J B B
T J I E i i
E B B E i i
J B B J T I J
i i
2
T v N N 1
2 2
1
2 2
Ω T e v N N
T e N
E B B i i
J B B T I J
i i
(A.4)
Using the [25] method of inversion, therefore, the in-verse of Equation (A.5) can be expressed as:
1
1 2
N
N v N N
i i
I i i
1
1 2 2
1
2 2
1 2
Ω T e v N
T e
T T
E B B
J B B T
E A J A
(A.5)
where 2
1 2
11 e v N N
A B B i i
and
1
12 2 2
2 e N v N N
A B B TI i i
(B) Derivation of the VCV Matrix for the cted Model
Here, 0
Restri
and as a consequence,
2ti ti e
Var Var e . Given this assumption,
Equa-on (10) reduces to: ti
T N
T N
T N
u I I e i I I i v
n, using
(B.1)
V matrix can be ed as:
The assumption (1), the VC
express
T N T N
T N T N
T N T N
I E ee I I
i I E i I
E uu I
I i E vv I i
Thus, (A.4) under the unrestricted model reduces to: (B.2)
2 2
2
Ω T e N T T N T v N N
I I i i I
I i i
(B.3)
Ω T e T N
1
2 2
2
T v N N
I B B J I
(B.4)
I i i
Just as before, we use the [24] method to
spectral decomposition of (B.4) and following the same procedure as Appendix A, we have:
obtain the
T e N N T v N N
2 2
2 2 2
Ω ET eIN v i iN N
J I TI J i i
Similarly, using the [25] method of inversion, Ω 1
(B.5)
can be expressed as:
1
1 2 2
1
2 2 2
Ω T e N v N N
T e N N T v N N
E I i i
J I T I J i i
1
Ω ET A1JTA2 (B.6)
where 2 2
11 e N v N N
A I i i and
12 2 2
2 e N N T v N N
A I TI J i i
.
(C) Derivation of the Partial Derivatives
L
1 1 1
1 Ω 1
Ω Ω Ω
2 2
L
tr u u
Ω
0
1 1 1
1 Ω Ω 1 Ω ΩΩ
2 2
a
L
tr u u
2
1 2
2
1 1 2 2 2
1 2
2
1 1 2 2 2
1 2 2
1 1
1 2 1 2 1 2 1 2
1 2
2
H
e T T
e T T
e T T
e T T
e
e T
tr E A M J A M
u E A MA J A MA u
tr E A M J A M
u E A MA J A MA u
tr T A A M
u E A MA u
u J
T A MA u2 2
(C.1) Note that:
2 2 2 2v
e e N v
1 1
N tr A
2 2 2
1 g
; 1 1
g
v i i v i
i
i e
2
1 g
i i tr A
where g 2 T 2
.
Therefore,
M tr A M
2
By some algebraic simplifications, we can write that:
1 2 1
1 1
tr T A A M T tr A
1 2
2 2 2 2
2 1 1
1
1
1
v N N N N
v
T A A M
i Mi i D MD i
T
N
Note further that:
1 ET A1 u
in which case,
T 1
u E A u,
therefore;
2
1
g
e e v
v i i
tr
1 1
T
T T
u E A MA u
u E A E M
Tu E IN
Ω1
T 1 1
T M u
Ξu E A MA uu E
Similarly,
J A2
J M
JT A2
u
in which case,
2 2 T
T T
u J A MA u
u
1
1 Ω
T N T
J I u J A u
,
therefore; u
T 2 2
T
Ξu J A MA uu J M u .
ence, we can write (C.1) as: As a consequ
0aH
2 2 2
1 Ξ Ξ
2 2
e v e
D L
D
T M M
0
C.2)
where
(
2 2 2
N N
e e v
i Mi M
N
;
1 1 2 1
1
N N
D
v i i i D MD i M
g
0
1 1 1
2 2 2
1 2
2 2
1 2
1 2
2 2
1 2
1 Ω Ω 1 Ω Ω Ω
2 2
1 2 1 2 1 2
1 (C.3)
2 a
e H e e
T T
T T
T T
T T
L
tr u u
tr E A A
u E A J A u
tr E A J A
u E A J A u
Using the information leading to (C.2), we can prove that:
J
1 2
2
2 2
1
1 g 1
1 1
v i
v i
i
tr T A A
N T
2 2 2
2 1
g 1
g
i i
e e v
v i N
And also with the representations that:
2
1 1 1
T T T
u E A uu E A E A u,
in which case,
1
1
Ω
T N T
u E I u E A u,
therefore;
2
1 T
u E A uu u ;
ein, and in the same v
2
2 T
u J A uu u .
