A TEST FOR INDIVIDUAL IN A ONE-WAY ERROR COMPONENTS MODEL

by Alberto Holly and Lucien Gardiol Cahier No. 9915 Version date : September 17, 1999

Preliminary version. Please do not quote without the authors’ explicit con- sent. Comments most welcome.

authors’ address

Département d’économétrie et d’économie politique Université de Lausanne Ecole des HEC 1015-Lausanne Switzerland Abstract

The purpose of this paper is to derive a Rao’s e¢cient score for testing for heteroscedasticity in an error components model with only indi- vidual e¤ects. We assume that the individual e¤ect exists and therefore do not test for it. In addition, we assume that the individual e¤ects, and not the white noise term may be heteroscedastic. Finally, we assume that the error components are normally distributed. We …rst establish, under a speci…c set of assumptions, the asymptotic distribution of the Score under contiguous alternatives. We then derive the expression for the Score test statistic for individual heteroscedasticity. Fi- nally, we discuss the asymptotic local power of this Score test statistic.

Key words :Panel , Error components model, Score test, Individual heteroscedasticity, Contiguous alternatives, Asymptotic local power. JEL classi…cation : C23, C12 A Score test for individual heteroscedasticity 1

1 Introduction

In the analysis of error-components models it is custumary to assume that the individual e¤ects are homoscedastic. In some situations, however, it may be appropriate to generalize the error components model context to the heteroscedastic case, as …rst suggested by Mazodier and Trognon (1978). Misspeci…cation errors in presence of heteroscedasticity can produce mis- leading results. However, if no heteroscedasticity exists, standard estima- tion and speci…cation test procedures can be applied straightforwardly. It would therefore simplify the analysis considerably if one were to test for het- eroscedasticity before implementing more elaborate inference procedures to deal with the possible heteroscedasticity situation. To this testing purpose, a natural procedure consists in using Rao’s e¢- cient score statistic [Rao (1948)], or it’s Lagrange Multipier (LM) interpreta- tion provided by Silvey (1959), as its computation is based on the usual error components model in the homoscedastic case. In another setting, Breusch and Pagan (1980) have considered the standard model with non–spherical disturbances and took the error-components model of Balestra and Nerlove (1966) as an example. They presented an LM test for the null hypothesis that the individual e¤ect is missing. Gourieroux, Holly and Mon- fort (1982) derived the asymptotic distribution of the LM test of Breusch and Pagan (1980) by taking into account the fact that the parameter de…ning the null hypothesis is on the boundary of the parameter set. They showed that the standard asymptotic distribution theory does not apply in this case and derived the appropiate nonstandard results.1 In a recent paper, Lejeune (1998) developed a pseudo-LM test procedure for jointly testing the null hypothesis of no individual e¤ects and homoscedas- ticity against the alternative of random individual e¤ects and heteroscedas-

1See also Baltagi, Chang and Li (1992) for an analysis of the behavior of one–sided LM tests. A Score test for individual heteroscedasticity 2 ticity in the white noise error term.2 The pseudo-LM test derived by Lejeune (1998) is distribution-free in the sense that it does not rely on any distribu- tion assumption such as normality. In this paper we consider a di¤erent setting than in Lejeune (1998). Firstly, we assume that the individual e¤ect exists and therefore do not test for it. Secondly, we assume that the individ- ual e¤ects, and not the white noise term may be heteroscedastic. Thirdly, we assume that the error components are normally distributed. In addition, not only the speci…cation considered in this paper di¤ers from that of Leje- une (1998) but also the method of derivation of the main results, which we believe to be useful in other contexts as well. The paper is organized as follows. The speci…cation of the model as well as some preliminary assumptions are presented in Section 2. The derivation of the asymptotic distribution of the Score under contiguous alternatives is contained in Section 3. The expression of the heteroscedaticity test statistic is derived in Section 4 and its asymptotic local power is discussed in Section 5. Throughout this paper, we tried to adhere to widely accepted set of no- tation in the context of Panel Data models. In particular, the unit vector = 1 T 1 ¶ (all elements ) of size £ is denoted by T and the unit matrix (all = 1 T T J = ¶ ¶ elements ) of size £ is denoted by T ( T T0 ). For a review of the main matrices used in this paper as well as their properties, see Crépon and Mairesse [(1996), Appendix].3 The notation D and AD are used throughout to the distribution and asymptotic distribution, respectively, of a or a random vector. The noncentral chi–square distribution with p degrees of freedom and 2 noncentrality parameter ± is de…ned as the distribution of the scalar product of a random p–variate normal vector with equal to the identity

2We would like to thank B. Lejeune for making his unpublihed manuscripts available to us. 3Appendix based on an unpublished manuscript by Alain Trognon (1984). A Score test for individual heteroscedasticity 3

± Â2(±2) matrix and mean vector having a norm of , and is denoted by p .

2 Speci…cation of the model and preliminary assump- tions

We consider the one-way error components linear regression model

0 0 0 ynt = xnt¯ + unt n = 1; : : : ; N; t = 1; : : : ; T

y x K 1 where nt is the (scalar) observation of the dependent variable, nt a £ u0 vector of nonstochastic explanatory variables, and nt the unobservable error term which is decomposed as

0 0 0 unt = ¹n + vnt;

¹0 v0 where n is the unobservable random variable of individual e¤ects and nt the usual unobservable error term. We assume that

¹0 v0 n t v0 Assumption 1 n and nt are independent for all and ; the nt are inde- (0; ¾02) ¹0 pendent identically distributed asN v and the n are independent and ¡0; ¾02h ¡z µ0¢¢ z p 1 distributed as N ¹ n0 where n is a £ vector of explanatory z µ0 h : variables such that n0 does not contain a constant term; R ! R is a (s) strictly positive twice di¤erentiable function satisfying h(0) = 1 , h (0) 6= 0 (s) for s = 1 , 2 where h denotes the derivative of order s of h.

Let

0 0 0 02 02 0 ± = (¯ ; ¾v ; ¾¹ )

±0 ¢ ¢ K Assumption 2 2 where is a compact subset of R £ R+£R+ and 0 p µ 2 £ where £ is a compact subset of R . A Score test for individual heteroscedasticity 4

0 0 Assumption 3 (± ; µ ) is an interior point of ¢ £ £.

¾02 It is important to observe that we assume that ¹ is strictly positive ¾02 £ - in other words, that ¹ is not on the boundary of the parameter set . Therefore, we shall not question the existence of individual e¤ects. Instead, 0 we shall test for heteroscedasticity of the individual e¤ects by testing H : 0 a 0 µ = 0 against H : µ 6= 0: The T observations for individual n can be expressed in the following matrix form:

0 0 yn = Xn¯ + un;

y T 1 y X T K n where n is the £ vector of the nt, n is the £ matrix whose -th x u0 T 1 u0 row is n0 t and n is the £ vector of the nt. In this paper we deal with the so-called semi-asymptotic case where T is …xed and N goes to in…nity.

(X ; z ) F Assumption 4 The empiric distribution of n n , denoted by n, con- verges completely to a nondegenerate distribution function F (X; z). The z F marginal (limiting) distribution of will be denoted by z.

More assumptions will be introduced in the following section. Stacking the individuals one after the other, we have:

0 1 0 1 0 0 1 y1 X1 u1 B C B C B C B . C B . C B . C B . C B . C B . C B y C = B X C ¯0 + B u0 C B n C B n C B n C B C B C B C B . C B . C B . C @ . A @ . A @ . A 0 yN XN uN or more compactly: A Score test for individual heteroscedasticity 5

y = X¯0 + u0 where

¡ 0¢ D(u) = N 0; • and

¡ 0¢ •0 = ¾02I + ¾02 diag(h z0 µ ) J v NT ¹ n • T which may be conveniently written as

¡ 0¢ JT •0 = ¾02W + diag[¾02 + T ¾02h z0 µ ] v n v ¹ n • T where

µ J ¶ W = I I T n N • T ¡ T

3 Asymptotic distribution of the Score under contigu- ous alternatives

The log– is:

1 1 = constant ln det(•) u0•-1u L ¡2 ¡ 2 where

µ ¶ 1 1 JT •¡1 = W + diag 2 n 2 2 • (1) ¾v ¾v + T ¾¹h (zn0 µ) T A Score test for individual heteroscedasticity 6

Let

° = ¡¯; ¾2; ¾2 ; µ¢0 v ¹ (2) and consider contiguous alternatives of the form :

°a = °0 + N ¡1=2°a N (3) where, since we consider contiguous alternatives only for the heteroscedas- ticity coe¢cients,

a ³ a ´0 ° = 0; 0; 0; µ0 (4)

The purpose of this section is to show that under speci…c regularity as- N 1=2@ (°a )=@° sumptions, ¡ L N is asymptotically normaly distributed. This is the key result for the asymptotic distribution of the heteroscedasticity test to be derived in the following section. The …rst di¤erential of L is:

1 ¡ ¢ 1 d = tr •-1 d • u0•-1 d u + u0•-1 d ••-1u L ¡2 ¡ 2 (5) where

d u = ¡X d ¯ µ ¶ JT d • = I d ¾2 + T diag (h (z0 µ)) d ¾2 NT v n • T ¹ µ ¶ JT + T ¾2 diag (h0 (z0 µ) z0 d µ) ¹ n n • T W J By using the properties of the matrices n and T , it is not di¢cult to show that d L may be written as A Score test for individual heteroscedasticity 7

@ @ @ @ d = L(°)d¯+ L (°)d¾2 + L (°)d¾2 + L(°)dµ L 2 v 2 ¹ @¯ @¾v @¾¹ @µ where

@ L(°) = X0•¡1u; @¯ (6)

( N Ã ! @ 1 N(T 1) X 1 1 L (°) = ¡ + u0W u 2 ¡ 2 ¡ £ ¤ 4 n (7) @¾ 2 ¾ ¾2 + T ¾2 h (z0 µ) ¾ v v n=1 v ¹ n v " Ã ! # ) 1 JT +u0 diag u ; £ 2 2 ¤2 • T ¾v + T ¾¹h (zn0 µ)

( @ T N µ h (z µ) ¶ L (°) = X n0 @¾2 2 ¡ ¾2 + T ¾2 h (z0 µ) (8) ¹ n=1 v ¹ n " Ã ! # ) h (zn0 µ) JT +u0 diag u £ 2 2 ¤2 • T ¾v + T ¾¹h (zn0 µ)

N ( @ 1 T ¾2 h(1) (z0 µ) z L X ¹ n n (°) = £ ¤+ (9) @µ 2 ¾2 + T ¾2 h (z0 µ) n=1 v ¹ n " Ã 2 (1) ! # ) 1 T ¾¹h (zn0 µ) zn JT + u0 diag u 2 £ 2 2 ¤2 • T ¾v + T ¾¹h (zn0 µ)

N 1=2@ (°a )=@° In order to derive the asymptotic distribution of ¡ L N , it is necessary to evaluate the second di¤erential of L. Using the fact that the second di¤erential of u is equal to zero, the second di¤erential of L is equal to: A Score test for individual heteroscedasticity 8

1 ¡ ¢ 1 ¡ ¢ d2 = tr •¡1 d ••¡1 d • tr •¡1 d2 • L 2 ¡ 2 1 u0•¡1 d ••¡1 d ••¡1u + u0•¡1 d2 ••¡1u ¡ 2 (10) 1 1 1 + 2u0•¡ d ••¡ d u ¡ d u0•¡ d u

2 By taking the expectation of d L, we obtain after obvious simpli…cation, that:

¡ ¢ 1 ¡ ¢ d2 = tr •¡1 d ••¡1 d • + d ¯0X0•¡1X d ¯ E ¡ L 2 (11)

It is not di¢cult to show that

à N ! N(T 1) X 1 ¡ ¡1 ¡1 ¢ 2 ¡ 2 tr • d •• d • = d ¾v + d ¾v ¾4 £ 2 2 ¤2 v n=1 ¾v + T ¾¹h (zn0 µ)

à ! N h (z µ) 2 X n0 2 + d ¾v 2T d ¾¹ £ 2 2 ¤2 n=1 ¾v + T ¾¹h (zn0 µ)

à N ! X h(1) (z0 µ) 2 2 n 0 + d ¾v 2T ¾¹ zn d µ £ 2 2 ¤2 n=1 ¾v + T ¾¹h (zn0 µ)

à ! N [h (z µ)]2 2 2 X n0 2 + d ¾¹ T d ¾¹ £ 2 2 ¤2 n=1 ¾v + T ¾¹h (zn0 µ)

à N ! X h (z0 µ) h(1) (z0 µ) 2 2 2 n n 0 + d ¾¹ 2T ¾¹ zn d µ £ 2 2 ¤2 n=1 ¾v + T ¾¹h (zn0 µ) A Score test for individual heteroscedasticity 9

à N £ (1) ¤2 ! X h (zn0 µ) 0 2 4 + d µ T ¾¹ znzn0 d µ £ 2 2 ¤2 n=1 ¾v + T ¾¹h (zn0 µ)

We shall …rst prove the following result:

Lemma 1 The loglikelihood L is regular with respect to its …rst and second derivatives, i.e.

¡ 2 ¢ 2 E ¡ d L = E (d L)

Proof. 4 From (??) we have

2 1 £ ¤2 (d ) = u0•-1 d ••-1u (u0•-1 d ••-1u) + d u0•¡1uu0•¡1 d u L 4 ¡ E £ ¤ 1 ¡ u0•-1 d ••-1u ¡ E(u0•-1 d ••-1u) u0•¡ d u

The expectation of the third term of the right-hand side is equal to zero. Thus,

2 1 (d ) = V (u0•-1 d ••-1u) + d u0•¡1 d u E L 4 1 ¡ ¢ = tr •¡1 d ••¡1 d • + d u0•¡1 d u 2 ¡ 2 ¢ = E ¡ d L as stated. z j z j = 1; : : : ; p Let j be the ¡th component of , . We introduce the following additional assumptions.

R h(1)(z0µ) zj d Fz(z) < j = 1; : : : ; p Assumption 5 j h(z0µ) j 1 for every .

4This result holds in a more general situation than the speci…c one considered in this paper. The proof we provide is simpler than the corresponding proof in Magnus (1978). A Score test for individual heteroscedasticity 10

R h(2)(z0µ) zjzk d Fz(z) < j; k = 1; : : : ; p Assumption 6 j h(z0µ) j 1 for every .

R h(1)2(z0µ) zjzk d Fz(z) < j; k = 1; : : : ; p Assumption 7 h2(z0µ) j j 1 for every .

Proposition 1 Let Assumptions 1 through ?? hold. Then 2 a) E [¡(1=N)@ L(°)=@°@°0] converges uniformly on ¡ to the asymptotic in- formation matrix I(°); 2 b) ¡(1=N)@ L(°)=@°@°0converges almost surely and uniformly on ¡ to I(°).

0 0 1=2 0 Proof. Let e = • ¡ u and suppose that Assumptions 1 through ?? hold. Then, by inspection, one may easily verify that all the elements 2 of ¡(1=N)@ L(°)=@°@°0 which, to save space, are not reproduced here, are (1=N) PN f(X ; z ; e0 ; °) f(X; z; e0; °) of the form n=1 n n n where the functions are either uniformly bounded or dominated by a function independent of ° which is integrable with respect to the product measure Z Z 0 0 º(A) = 1IA(X; z; e ) d F (X; z) d ©(e )

©(e0) (0; 1) 1I (X; z; e0) = 1 (X; z; e0) A where is the N distribution, and A if 2 , 0 otherwise. The assertion of the proposition follows from the version of the Uniform strong law of large numbers proved in Gallant [(1987), Theorem 1, p. 159– 162]. Note that the asymptotic information matrix I(°) is of the form 0 1 (°) 0 0 0 I¯¯ B C 0 2 2 (°) 2 2 (°) 2 (°) B I¾v¾v I¾v¾¹ I¾v µ C I(°) = B C B 0 ¾2 ¾2 (°) ¾2 ¾2 (°) ¾2 µ(°) C @ I ¹ v I ¹ ¹ I ¹ A 0 2 (°) 2 (°) (°) Iµ¾v Iµ¾¹ Iµµ 0 where, when evaluated at ° , 1 (°0) = lim X0•0¡1X; I¯¯ N!1 N A Score test for individual heteroscedasticity 11

" # 1 (T 1) 1 0 ¡ ¾2¾2 (° ) = + ; I v v 2 ¾04 ¡ 02 02¢2 v ¾v + T ¾¹

T 0 0 ¾2¾2 (° ) = ¾2 ¾2 (° ) = I v ¹ I ¹ v ¡ 02 02¢2 2 ¾v + T ¾¹

02 (1) T ¾¹ h (0) 1 (°0) = 0 (°0) = lim ¶0 Z ¾2µ µ¾2 N I v I v ¡ 02 02¢2 N N 2 ¾v + T ¾¹ !1

T 2 0 ¾2 ¾2 (° ) = I ¹ ¹ ¡ 02 02¢2 2 ¾v + T ¾¹

2 02 (1) T ¾¹ h (0) 1 (°0) = 0 (°0) = lim ¶0 Z ¾2 µ µ¾2 N I ¹ I ¹ ¡ 02 02¢2 N N 2 ¾v + T ¾¹ !1

T 2¾04 £h(1)(0)¤2 1 0 ¹ µµ(° ) = lim Z0Z I ¡ 02 02¢2 N N 2 ¾v + T ¾¹ !1

Let µ J ¶ Z = I N Z N ¡ N (12) be the matrix obtained by centering each column of Z. We introduce the following additional assumptions:

1 lim X0•0¡1X Assumption 8 N!1 N is nonsingular

lim 1 Z0Z Assumption 9 N!1 N is nonsingular

0 Lemma 2 Under Assumptions ?? and ??, I(° ) is nonsingular A Score test for individual heteroscedasticity 12

0 Proof. We may write I(° ) as à ! (°0) (°0) (°0) = I±± I±µ I (°0) (°0) Iµ± Iµµ

(°0) It is easy to verify that if Assumption ?? is satis…ed, then I±± is nonsingu- (°0) (°0) (°0) (°0) 1 lar. Therefore, I is nonsingular if and only if I±± ¡I±µ Iµµ ¡ £ (°0) Iµ± is nonsingular. In turn, this property is implied by Assumption ??

Proposition 2 Under Assumptions 1 through ??

[N ¡1=2@ (°a )=@°] = (0; (°0)) AD L N N I

Proof. If Assumptions 1 through ?? hold, then one can verify that, with- out additional assumptions, the for contiguous alter- N 1=2@ (°a )=@° natives proved in Gallant and Holly (1980) applies. Hence, ¡ L N converges in distribution to the stated normal distribution.

4 The heteroscedasticity Score test statistic

The necessary …rst–order conditions system for the maximization of the log– likelihood function subject to the constraint µ = 0 boils down to the familiar estimating equation for the homoscedastic one–way error components model; that is:

(c) ³ (c) ´-1 ³ (c)-1 ´ ¯e = X0•e -1X X0e• y u(c) W u(c) ¾2(c) = e 0 ne ev N (T ¡ 1) u(c) B u(c) u(c) u(c) ¾2(c) = e 0 ne e 0e e¹ ¡ N (T ¡ 1) NT (T ¡ 1) A Score test for individual heteroscedasticity 13 where

(c) (c) ue = y ¡ X¯e J B = I T n N • T

It is useful to note that

u(c) B u(c) ¾2(c) + T ¾2(c) = e 0 ne ev e¹ N (13)

Let

³ ´ °(c) = ¯(c); ¾2(c); ¾2(c); 0 0 e e ev e¹

All the components of the score vector @L(°)=@° evaluated at the constrained (c) (c) estimator e° are equal to zero, except @L(e° )=@µ which is equal to: @ T ¾2(c)h(1) (0) L(°(c)) = e¹ @µ e 2 ³ ´2 £ ¾2(c) + T ¾2(c) ev e¹ " N µ ¶ N # X JT ³ 2(c) 2(c)´ X u(c)0 u(c) z ¾ + T ¾ z £ n T n n ¡ ev e¹ n (14) n=1 n=1

(c) It is convenient to write @L(°e )=@° more compactly in matrix nota- (c) (c) (c) tion. To this purpose, let se be the N £ 1 vector of the sen where sen = u(c)0(J =T )u(c): en T en ¾2(c) + T ¾2(c) s(c) Using (??), it is easy to verify that ev e¹ is the mean of the en . We may thus write,

N " Ã N !# X JT 1 X u(c)0 u(c) z z = Z0s(c) en T en n ¡ N n e n=1 n=1

(c) We may thus write @L(e° )=@° more compactly as: A Score test for individual heteroscedasticity 14

@ µ @ ¶0 L(°(c)) = 0; 0; 0; L(°(c)) @° e @µ e (15)

where

T ¾2(c)h(1) (0) @ (c) 1 ¹ L(° ) = e Z0s(c) @µ e 2 ³ ´2 e (16) ¾2(c) + T ¾2(c) ev e¹

As usual, the information matrix evaluated at ° is de…ned as ½ 2 ¾ @ L(°) N (°) = I ¡E @°@°0

¡1(°) We may write IN as

à ±± ±µ ! N (°) N (°) ¡1(°) = I I IN µ±(°) µµ(°) (17) IN IN

By using (??) and (??), we easily verify that

³ ´2 2 ¾2(c) + T ¾2(c) (c) ev e¹ 1 µµ(° ) = (Z0Z)¡ IN e T 2¾4(c) [h(1)(0)]2 (18) e¹

We are now in position to derive the expression for the Score test statistic S » given by:

S @ ³ (c)´0 (c) @ ³ (c)´ » = L ° ¡1(° ) L ° @° e IN e @° e (19)

Straightforward computation shows, by using (??), (??) and (??), that

S 1 1 » = s(c)0Z (Z0Z)¡ Z0s(c) 2(¾2(c) + T ¾2(c))2 e e (20) ev e¹ A Score test for individual heteroscedasticity 15

¾2(c) + T ¾2(c) s(c) Alternatively, by using the fact that ev e¹ is the mean of the en , S we may write the expression for the Score test statistic » as follows,

µ (c) ¶0 µ (c) ¶ 1 s 1 s »S = e ¶ Z (Z0Z)¡ Z0 e ¶ 2 s ¡ N s ¡ N (21)

(c) where s is the mean of se . S The Score test statistic » is thus one half of the explained sum of squares (c) of the OLS regression of se =s¡1 against Z as in Breusch and Pagan (1979).5

5 Asymptotic local power

[N 1=2@ (°a )=@°] = (0; (°0)) Since, according to Proposition ??, AD ¡ L N N I , one can show that, under contiguous alternatives, the distribution of the S Score test statistic » converges to the noncentral chi–square distribution a a with p degrees of freedom and noncentrality parameter ° 0A° , that is,

S ¡» ¢ = Â2 (°a0A°a) AD p where

0 1 0 0 0 0 B C B 0 0 0 0 C A = B C B 0 0 0 0 C @ A µµ 0 0 0 0 I (° )

For a proof see, for example, Holly (1987). The asymptotic power of the test is given by the noncentrality parameter a a ° 0A° . Its expression is given by:

5Notice also the di¤erence and similarity with the particular expression of the Pseudo- LM test in Lejeune (1998) when the normality assumption is assumed to hold. A Score test for individual heteroscedasticity 16

T 2¾4 ¡h(1)(0)¢2 1 ¹ a a °a0A°a = lim µ 0 (Z0Z=N)µ N 2 ¡ 2 2 ¢2 !1 ¾v + T ¾¹

The asymptotic power is in‡uenced by three factors. Firstly, not surpris- a ingly, the power is in‡uenced by µ , the test is more powerful to detect alter- natives which are away from the null hypothesis: Secondly, although the Score (1) test statistic itself does not depend on h(0) or h (0), the asymptotic power ¡ (1) ¢2 is an increasing function of h (0) for any given alternative. Thirdly, the T T ¾2 = ¡¾2 + T ¾2 ¢ power increases with , as the multiplicative constant ¹ v ¹ con- verges to 1 when T goes to in…nity. This last e¤ect shows that the test is improved when the number of observations for each individual sample in- creases. One could also note that the local power of the test tends to zero ¾2 T ¾2 when ¹ tends to zero and will tend to be small if ¹ is small compared to ¾2 v. Thus, as one should expect, the test will be powerful in situations where the individual heteroscedasticity is high. A Score test for individual heteroscedasticity 17

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