Test of hypotheses with panel data

Transcription

1 Stochastic modeling in economics and finance November 4, 2015

2 Contents 1 Test for poolability of the data 2 Test for individual and time effects 3 Hausman s specification test 4 Case study

3 Contents Test for poolability of the data 1 Test for poolability of the data 2 Test for individual and time effects 3 Hausman s specification test 4 Case study

4 Introduction I Test for poolability of the data Restricted model: Unrestricted model: represents a behavioral equation with the same parameters over time and across individuals. is the same behavioral equation but with different parameters across time or individuals. Note: we restrict to test poolability of the data for the case of pooling across individuals (pooling over time can be obtained in a similar fasion). The restricted panel data regression model: y it = α + Xit T it i = 1,..., N, t = 1,..., T, u it = µ i + ν it i = 1,..., N, t = 1,..., T. The unrestricted panel data regression model: y it = α i + Xit T i + u it i = 1,..., N, t = 1,..., T, u it = µ i + ν it i = 1,..., N, t = 1,..., T.

5 Introduction II Test for poolability of the data Why do we pool data? Pooling lead to widen database, and therefore we can obtain better and more reliable estimates of the parameters. Panel data allows to study individual and time effects. Panel data models are popular in applied economics (they allow to control for individual heterogeneity).

6 Test for poolability of the data Vector formulation of restricted and unrestricted model I The unrestricted panel data regression model: y i = α i + X i β i + u i = Z i δ i + u i, i = 1,..., N, where y T i = (y i1,..., y it ), Z i = (ι T, X i ) and δ i = (α i, β i ). y i : T 1, X i : T K, Z i : T (K + 1), δ i : (K + 1) 1, u i : T 1. The restricted panel data regression model: y = αι NT + Xβ + u = Z δ + u, where Z T = (Z T 1,..., Z T N ) and ut = (u T 1,..., ut N ). y : NT 1, X : NT K, Z : NT (K + 1), δ : (K + 1) 1, u : NT 1. We want to test the hypothesis H 0 : δ i = δ for all i = 1,..., N.

7 Test for poolability of the data Vector formulation of restricted and unrestricted model II The unrestricted model can be reformulated as: Z δ 1 0 Z 2 0 δ 2 y = u = Z δ + u. 0 0 Z N δ N In this case Z = Z I with I = (ι N I K ), K = K + 1. y : NT 1, Z : NT N(K + 1), δ : N(K + 1) 1, u : NT 1. We aim to compare restricted and unrestricted in the forms derived above: y = Z δ + u y = Z δ + u.

8 Test for poolability of the data Test for poolability under assumption u N(0, σ 2 I NT ) I For the restricted model under u N(0, σ 2 I NT ) the minimum variance unbiased (MVU) estimator for δ is: ˆδ ols = ˆδ mle = (Z T Z ) 1 Z T y and therefore y = Z ˆδ ols + e e = y Z ˆδ ols = (I NT Z (Z T Z ) 1 Z T )y = My = M(Z δ + u) = Mu. For the unrestricted model under u N(0, σ 2 I NT ) MVU estimator for δ i is: and therefore ˆδ i,ols = ˆδ i,mle = (Z T i Z i ) 1 Zi T y i y i = Z i ˆδ i,ols + e i e i = y i Z i ˆδi,ols = (I T Z i (Zi T Z i ) 1 Zi T )y i = M i y i = M i (Z i δ i + u i ) = M i u i.

9 Test for poolability of the data Test for poolability under assumption u N(0, σ 2 I NT ) II For the unrestricted model given as y = Z δ + u under u N(0, σ 2 I NT ) MVU estimator for δ i is: ˆδ ols = ˆδ mle = (Z T Z ) 1 Z T y and therefore y = Z ˆδ ols + e e = y Z ˆδ ols = (I NT Z (Z T Z ) 1 Z T )y = M y = M (Z δ + u) = M u. It can be shown that M M M 2 0 = , 0 0 M N M and M are idempotent and symmetric matrices with MM = M.

10 Test for poolability of the data Test for poolability under assumption u N(0, σ 2 I NT ) III Chow test extended to the case of N linear regressions (Baltagi (2005)) Under H 0 : δ i = δ for i = 1,..., N and u N(0, σ 2 I NT ), the statistic F obs given as F obs = (et e e T e )/(tr(m) tr(m ) e T e /tr(m ) = (et e e T 1 e 1... e T N en)/(k (N 1)) (e T 1 e e t N en)/n(t K ) is distributed as an F ((N 1)K, N(T K ). Hence the critical region for this test is defined as {F obs > F ((N 1)K, N(T K ; α 0 )} where α 0 denotes the level if significance of the test. Proof. Using properties of matrices M and M one can easily derive: e T e e T e = (Mu) T (Mu) (M u) T (M u) = u T Mu u T M u = u T (M M )u, e T e = u T M u. Since M and (M M ) are idempotent and u N(0, σ 2 I NT ), u T (M M )u/σ 2 follows χ 2 distribution with tr(m M ) degrees of freedom, and similarly, u T M u/σ 2 follows χ 2 distribution with tr(m ) degrees of freedom.

11 Test for poolability of the data Test for poolability under assumption u N(0, σ 2 I NT ) IV Matrices M and M are idempotent and therefore: tr(m) = r(m) = NT K, tr(m ) = r(m ) = NT K N = N(T K ), tr(m M ) = tr(m) tr(m ) = K (N 1). To finish the proof it remains to note that u T (M M )u and u T M u are independent variable, (M M )M = 0. Statistic F obs, as the ration of two independent random variables with χ 2 distribution both divided by their degrees of freedom, has to follow F distribution.

12 Test for poolability of the data Test for poolability under assumption u N(0, Ω) I In the general case u N(0, Ω) one seeks for a suitable transformation of the model so as the Chow test can be applied. Namely, consider that Ω = σ 2 Σ and multiply restricted as well as unrestricted models by Σ 1/2, then we get: ȳ = Z δ + ū, ȳ = Z δ + ū, where ȳ = Σ 1/2 y, Z = Σ 1/2 Z, ū = Σ 1/2 u and Z = Σ 1/2 Z. In this case E(ūū T ) = E(Σ 1/2 uu T Σ 1/2T ) = Sigma 1/2 E(uu T )Σ 1/2T = σ 2 I NT. For the restricted and unrestricted reformulated models we gain have: ˆ δ ols = ( Z T Z ) 1 Z T ȳ, ē = ȳ Z ˆ δ ols, ē = Mȳ = Mū, ˆ δ ols = ( Z T Z ) 1 Z T ȳ, ē = ȳ Z ˆ δ ols, ē = M ȳ = M ū.

13 Test for poolability of the data Test for poolability under assumption u N(0, Ω) II In order to apply Chow test, we need to verify: Z = Z I, M, M are symmetric and idempotent and M M = M, where M = I NT Z ( Z T Z ) 1 Z T and M = I NT Z ( Z T Z ) 1 Z T. Roy-Zellner test for poolability (Baltagi (2005)) Under H 0 : δ i = δ for i = 1,..., N and u N(0, Ω) the statistic F obs given as F obs = (ēt ē ē T ē )/(tr( M) tr( M ) ē T ē /tr( M ) is distributed as an F ((N 1)K, N(T K ). Hence the critical region for this test is defined as {F obs > F ((N 1)K, N(T K ; α 0 )} where α 0 denotes the level if significance of the test.

14 Contents Test for individual and time effects 1 Test for poolability of the data 2 Test for individual and time effects 3 Hausman s specification test 4 Case study

15 Introduction I Test for individual and time effects All presented tests are dedicated for two-way error component model given as: y it = α + Xit T it i = 1,..., N, t = 1,..., T, u it = µ i + λ t + ν it i = 1,..., N, t = 1,..., T, for which µ i IID(0, σ 2 µ ), λ IID(0, σ2 λ ) and ν it IID(0, σ 2 ν ). We want to test the hypotheses: H a 0 : σ2 µ = 0, H b 0 : σ2 λ = 0, H c 0 : σ2 µ = σ2 λ = 0 H d 0 : σ2 µ = 0 σ2 λ > 0 H e 0 : σ2 λ = 0 σ2 µ > 0.

16 Introduction II Test for individual and time effects All presented test statistics stem from the likelihood function, which under assumption of normality has the following form: L(δ, θ) = constant 1 2 log Ω 1 2 ut Ω 1 u, where θ T = (σ 2 µ, σ 2 λ, σ2 ν) and Ω is given as Ω = σ 2 µ(i N J T ) + σ 2 λ (J N I T ) + σ 2 νi NT. Breusch and Pagan (1980) derived a Lagrange multiplier (LM) statistic to test H0 c : σµ = σλ 2 = 0 based on the Fisher score and the Fisher information matrix. In the following text we denote as θ mle MLE of θ under H c 0, similarly Ω MLE of Ω under H c 0. Note that Ω = σ 2 ν I NT where σ 2 ν = ũt ũ/nt and ũ are the OLS residuals.

17 Introduction III Test for individual and time effects Derivation of the Fisher score and the Fisher information matrix. It has been shown that In our specific case: L θ r = 1 2 tr(ω 1 ( Ω/ θ r )) (ut Ω 1 ( Ω/ θ r )Ω 1 u). Ω/ θ 1 = (I N J T ), Ω/ θ 2 = (J N I T ), Ω/ θ 3 = I NT. Using tr(i N J T ) = tr(j N I T ) = tr(i NT ) = NT, one gets: [ ] L D( θ) = = NT θ θ 2 σ 2 mle ν 1 ũt (I N J T )ũ ũ T ũ 1 ũt (J N I T )ũ ũ T ũ 0.

18 Introduction IV Test for individual and time effects In order to calculate the information matrix for this model we need ( ) 2 L E = 1 ) (Ω θ s θ r 2 tr 1 ( Ω/ θ r )Ω 1 ( Ω/ θ s) Using tr((i N J T )(J N I T )) = tr(j NT ) = NT, tr(i N J T ) 2 = NT 2 and tr(j N I T ) 2 = N 2 T, the information matrix for this model is: [ ] J( θ) = E 2 L = NT T N 1 θ r θ s 2 σ θ ν 4 mle with J 1 ( θ) = σ 4 ν NT (N 1)(T 1) (N 1) 0 (1 N) 0 (T 1) (1 T ) (1 N) (1 T ) (NT 1).

19 Test for individual and time effects The Breusch-Pagan test I 1. The Breusch-Pagan test (Breusch and Pagan (1980)) Under H0 c : σ2 µ = σ2 λ = 0, the Breusch-Pagan test statistic LM given as: LM = ( ) 2 ( ) 2 NT 1 ũt (I N J T )ũ NT 2(T 1) ũ T + 1 ũt (J N I T )ũ ũ 2(N 1) ũ T ũ is asymptotically distributed as χ 2 2. Proof. The statistic can be obtain as: LM = D T J 1 D.

20 Test for individual and time effects The Breusch-Pagan test II Notes on the Breusch-Pagan test: The test is very popular (it requires only calculation of OLS residuals ũ. The test statistic LM 1, LM 1 = ( ) 2 NT 1 ũt (I N J T )ũ 2(T 1) ũ T ũ is asymptotically distributed as χ 2 1 and can be used to test Ha 0 : σ2 µ = 0. The test statistic LM 2, LM 2 = ( ) 2 NT 1 ũt (J N I T )ũ 2(N 1) ũ T ũ is asymptotically distributed as χ 2 1 and can be used to test Hb 0 : σ2 λ = 0. Both test statistic LM 1 and LM 2 can be applied on condition σ 2 λ = 0 and σ2 µ = 0.

21 Test for individual and time effects Other test for individual and time effects I The problem with the Breusch-Pagan test is that it assumes that the alternative hypothesis is two-sided, but variance are nonnegative, thus the alternative hypothesis should be one-sided. The Honda tests (Honda (1985)) Under hypothesis H a 0 : σ2 µ = 0, the Honda test statistic HO given as: ( ) NT HO A = 1 ũt (I N J T )ũ 2(T 1) ũ T ũ has the asymptotic normal distribution N(0, 1).

22 Test for individual and time effects Other test for individual and time effects II Under hypothesis H0 b : σ2 λ, the Honda test statistic HO given as: ( HO B = NT 2(N 1) has the asymptotic normal distribution N(0, 1). 1 ũt (J N I T )ũ ũ T ũ Under hypothesis H0 c : σ2 µ = σ2 λ = 0, the Honda test statistic HO given as: HO = (A + B)/ 2 has the asymptotic normal distribution N(0, 1). )

23 Test for individual and time effects Other test for individual and time effects III King and Wu (1997) suggested the alternative test statistic to testing H c 0 : σ2 µ = σ2 λ = 0. The King and Wu test (King and Wu (1997)) Under hypothesis H0 c : σ2 µ = σ2 λ = 0, the King and Wu test statistic KW given as: KW = has the asymptotic normal distribution N(0, 1). T 1 N 1 A + B N + T 2 N + T 2

24 Test for individual and time effects Other test for individual and time effects IV Moulton and Randolph (1989) suggested an alternative standardized Lagrange multiplier test, because they revealed poor performance of Honda tests (especially if the number of regressors is high). The standardized Lagrange multiplier test (Moulton and Randolph (1989)) Under hypothesis H0 a : σ2 µ = 0 or H0 b : σ2 λ = 0, the standardized Lagrange multiplier test statistic SLM given as: HO E(HO) SLM = var(ho) has the asymptotic normal distribution N(0, 1).

25 Test for individual and time effects Other test for individual and time effects V Gourieroux, Holly, and Monfort (1982) note that A and B can be negative for a specific application and suggest corrected test for testing H c 0 : σ2 µ = σ 2 λ = 0. The Gourieroux, Holly and Monfort test (Gourieroux, Holly, and Monfort (1982)) Under hypothesis H0 c : σ2 µ = σλ 2 = 0, the Gourieroux, Holly and Monfort test statistic GLM is given as: A 2 + B 2 if A > 0, B > 0 χ 2 A 2 if A > 0, B 0 m = B 2 if A 0, B > 0 0 if A 0, B 0 χ 2 m denotes the mixed χ2 distribution. Under the null hypothesis, χ 2 m 1 4 χ2 (0) χ2 (1) χ2 (2), where χ 2 (0) equals 0 with probability one.

26 Test for individual and time effects Other test for individual and time effects VI When using above tests for H0 a : σ2 µ = 0, one implicitly assumes that σ2 λ = 0. This may lead to incorrect decisions especially when the variance σλ 2 is large. the conditional LM tests (Baltagi and Li (1992)) Under hypothesis H0 d : σ2 µ = 0 (allowing σλ 2 > 0), the conditional LM test given as: LM µ = 2 σ 2 2 σ 2 ν T (T 1)( σ 4 ν + (N 1) σ42 ) Dµ, where ( D µ = T ũt ( J N J ) T )ũ 2 σ 2 2 σ T (N 1) 2 σ 2 ν ( ũt (E N J ) T )ũ (N 1) σ ν 2 1 with σ 2 2 = ũt ( J N I T )ũ/t and σ 2 ν = ũt (E N I T )ũ/t (N 1), is asymptotically distributed as N(0, 1).

27 Test for individual and time effects Other test for individual and time effects VII Under hypothesis H e 0 : σ2 λ = 0 (allowing σ2 µ > 0), the conditional LM test given as: LM λ = 2 σ 2 1 σ 2 ν N(N 1)( σ 4 ν + (T 1) σ41 ) Dλ, where ( D λ = N ũt ( J N J ) T )ũ 2 σ 1 2 σ N(T 1) 2 σ 2 ν ( ) ũt ( J N E T )ũ (T 1) σ ν 2 1 with σ 2 1 = ũt (I N J T )ũ/n and σ 2 ν = ũ T (I N E T )ũ/n(t 1), is asymptotically distributed as N(0, 1).

28 Test for individual and time effects Other test for individual and time effects VIII ANOVA F tests can be used as universal tests for testing each of the considered hypotheses. The ANOVA F tests The ANOVA F test statistics have the following form: F = y T MD(D T MD) 1 D T My/(p r), y T Gy/(NT ( k + p r)) where M = Z (Z T Z ) 1 Z T and G, D, k, p, r are chosen depending on hypothesis tested. Under the null hypothesis, this statistic has a F distribution with p r and NT ((k + p r) degrees of freedom.

29 Test for individual and time effects Other test for individual and time effects IX The likelihood ratio tests can be used as universal tests for testing each of the considered hypotheses. The likelihood ratio tests The one-sided likelihood ratio LR tests have the following form: ( ) l(res) LR = 2 log l(unres) where l(res) denotes the restricted maximum likelihood value (under the null hypothesis), while l(unres) denotes the unrestricted maximum likelihood value. For H a 0, Hb 0, Hd 0 and He 0, LR is distributed as 1 2 χ2 (0) χ2 (1) and for H c 0 as 1 4 χ2 (0) χ2 (1) χ2 (2).

30 Test for individual and time effects Comparison of the tests I Baltagi and Li (1992) carried out a Monte Carlo simulation in order to compare performance of presented tests on two-way error component model. They obtained the following results: When H0 a : σ2 µ = 0 is true and σλ 2 is large, all usual tests (BP, HO, KW, SLM, GMH) preformed badly (they ignore the fact that σλ 2 > 0). When σµ 2 >> 0 all tests performed well in rejecting Ha 0, but for small σ2 µ the power of the tests decreases as σλ 2 increases. when testing H0 d : σ2 µ = 0 σ2 λ > 0, LMµ, LR and F performed well. Moreover, the power of the tests increases as σλ 2 increases. Overspecifying the model, i. e. assuming the model to be two-way error component when it is one-way, does not hurt the power of tests LM µ, LR or F. Therefore, one should not ignore σ 2 λ > 0 when testing σ2 µ = 0. When testing H0 c : σ2 µ = σλ 2 = 0 all tests are possible, but GHM and F are recommended.

31 Test for individual and time effects Comparison of the tests II H0 a H0 b H0 c H0 d H0 e σµ 2 = 0 σ2 λ = 0 σ2 µ = σ2 λ = 0 σ2 µ = 0 σ2 λ > 0 σ2 λ = 0 σ2 µ > 0 BP - - HO - - KW - - SLM GHM F LR LM µ LM λ Table: Suitability of the tests How to proceed when seeking for the most proper model: STEP 1 Testing H c 0 : σ2 µ = σ2 λ = 0 by GHM. If Hc 0 is not rejected, then use OLS. If Hc 0 is rejected, continue to STEP 2. STEP 2 Calculate LM µ and LM λ to test H0 d and He 0. Depending on the results apply one-way or two-way error component model to data.

32 Contents Hausman s specification test 1 Test for poolability of the data 2 Test for individual and time effects 3 Hausman s specification test 4 Case study

33 Introduction I Hausman s specification test Firstly, we will consider random one-way error component model whose critical assumption is that E(u it X it ) = 0. We intend to test the hypothesis H 0 : E(u it X it ) = 0. All test presented in this section are based on comparing estimates ˆβ GLS, ˆβ within and ˆβ between. ˆβ GLS is derived from the following model: y it = α + X T it β + u it i = 1,..., N, t = 1,..., T, Ω = E(u T u), Ω 1/2 y = Ω 1/2 αι NT + Ω 1/2 Xβ + Ω 1/2 Z µµ + Ω 1/2 ν ˆβ GLS = (X T Ω 1 X) 1 X T Ω 1 y. ˆβ within is derived from the following model: y it ȳ i = (X it X i ) T β + (ν it ν i ) i = 1,..., N, t = 1,..., T, Q = (I N E T ), Qy = QXβ + Qν ˆβ within = (X T QX) 1 X T Qy.

34 Introduction II Hausman s specification test ˆβ between is derived from the following model: ȳ i = α + X T i β + ū i i = 1,..., N, P = (I N J T ), Py = Pαι NT + PXβ + PZ µµ + Pν, ˆβ between = (X T PX) 1 X T Py.

35 Hausman s specification test Hausman s specification test I In the case E(u it X it ) 0, the GLM estimator ˆβ GLM becomes biased and inconsistent for β, whereas the Within transformation leaves the Within estimator ˆβ within unbiased and consistent for β. The test is based on the difference between ˆβ GLS and ˆβ within : ˆq 1 = ˆβ GLS ˆβ within = ( ˆβ GLS β) ( ˆβ within β) = (X T Ω 1 X) 1 X T Ω 1 u (X T QX) 1 X T Qu. To derive the test statistic we need to calculated the mean and variance of ˆq 1. Obviously, E(ˆq 1 ) = 0. In order to calculate the variance, we proceed as follows: cov( ˆβ GLS, ˆq 1 ) = cov( ˆβ GLS, ˆβ GLS ˆβ within ) = var( ˆβ GLS ) + cov( ˆβ GLS, ˆβ within ) = (X T Ω 1 X) 1 (X T Ω 1 X) 1 X T Ω 1 E(uu T )QX(X T QX) 1 = (X T Ω 1 X) 1 (X T Ω 1 X) 1 = 0

36 Hausman s specification test Hausman s specification test II ˆβ within = ˆβ GLS ˆq 1 var( ˆβ within ) = var( ˆβ GLS ) + var(ˆq 1 ) var(ˆq 1 ) = var( ˆβ within ) var( ˆβ GLS ) = σν 2 (X T QX) 1 (X T Ω 1 X) 1. Hausman s specification test (Hausman (1978)) Under H 0 : E(u it X it ) = 0, the Hausman s specification test statistic given as: m 1 = ˆq T 1 (var(ˆq 1)) 1ˆq 1 is asymptotically distributed as χ 2 K, where K denoted the dimension of slope vector β.

37 Hausman s specification test An alternative asymptotically equivalent test I Consider the following regression: σ νω 1/2 y = σ νω 1/2 Xβ + QXγ + ω y = X β + Xγ + ω. Then the Hausman s test ( ˆβ within = ˆβ GLS ) is equivalent to test whether γ = 0. To the later one can apply standard Wald test for omission of variables X. Performing OLS on the above stated mode, one gets the estimates: ( ˆβ ˆγ ) = ( X T (Q + φ 2 P)X X T QX X T QX X T QX ˆβ = ˆβ between = (X T PX) 1 X T Py ˆγ = (X T QX) 1 X T Qν (X T PX) 1 XPu = ˆβ within ˆβ between. ) 1 ( X T (Q + φ 2 P)y X T Qy ), where σ νω 1/2 = Q + φp and φ = σ ν/σ 1.

38 Hausman s specification test An alternative asymptotically equivalent test II The alternative statistic is based on ˆq 3 = ˆγ = ˆβ within ˆβ between for which E(ˆq 3 ) = 0 var(ˆq 3 ) = E(ˆq 3ˆq 3 T ) = σ2 ν(x T QX) 1 + σ1 2 (X T PX) 1 = var( ˆβ within ) + var( ˆβ between ) The alternative specification test Under H 0 : E(u it X it ) = 0, the alternative specification test statistic given as: m 3 = ˆq T 3 (var(ˆq 3)) 1ˆq 3 is asymptotically distributed as χ 2 K, where K denoted the dimension of slope vector β. The test m 1 and m 3 are numerically exactly identical and are also identical with the statistic m 2 = ˆq T 2 (var(ˆq 2)) 1ˆq 2, where ˆq 2 = ˆβ GLS ˆβ between. This follows from the relationship between the estimators: ˆβ GLS = W 1 ˆβwithin + W 2 ˆβbetween.

39 Hausman s specification test Hausman s test for the two-way error component model The Hausman s test for the two-way error component model is based on difference between the fixed effects estimator (with both time and individual dummies) and the two-way random effects GLS estimator, i. e. the Within and GLS estimators. The equivalent tests cannot be executed anymore, since there are two Between estimators. However there are other type of equivalences. Kang (1985) classifies five testable hypothesis, which consider between time periods estimator ˆβ T and between cross section estimator ˆβ C : Assume µ i fixed and test E(λ t X it ) = 0 based upon ˆβ within ˆβ T. Assume µ i random and test E(λ t X it ) = 0 based upon ˆβ T ˆβ GLS. Assume λ t fixed and test E(µ i X it ) = 0 based upon ˆβ within ˆβ C. Assume λ t random and test E(µ i X it ) = 0 based upon ˆβ C ˆβ GLS. Test E(µ i X it ) = E(λ t X it ) = 0 upon ˆβ GLS ˆβ within, where ˆβ GLS is the estimates assuming both µ i and λ t random and ˆβ within is the estimates assuming both µ i and λ t fixed.

40 Contents Case study 1 Test for poolability of the data 2 Test for individual and time effects 3 Hausman s specification test 4 Case study

41 Case study Grunfeld investment equation I Grunfeld (1958) considered the following investment equation for 10 large US manufacturing firms over 20 years, : I it = α + β 1 F it + β 2 C it + u it, i = 1,..., 10, t = 1,..., 20, I it : real gross investment for firm i in year t, F it : the real value of the firm (shares outstanding), C it : the real value of the capital stock In order to find a proper model we need to answer the following questions: Can we use the restricted model, can we pool the data across firms or/and time? If we choose the restricted model, are there any individual and time effects and should we use one-way or two-way error component model? If we choose the restricted model with random individual or/and time effects, do disturbances u it contain invariant effect which are unobservable and uncorrelated with explanatory variables?

42 Case study Tests for poolability of the data RRSS URSS F obs Distribution Quantile Across firms F(27, 170) 1.55 Across firms* F(18, 170) 1.62 Across time F(57, 140) 1.42 Across time* F(38, 140) 1.49 Table: Poolability of Grunfeld investment data across firms and time under assumption u N(0, σ 2 I NT ), (* denotes poolbility allowing varying intercept). F obs Distribution Quantile Across firms 4.35 F(27, 170) 1.55 Across time 2.72 F(57, 140) 1.42 Table: Poolability of Grunfeld investment data across firms and time under assumption u N(0, Ω).

43 Case study Tests for individual and time effects H0 a H0 b H0 c H0 d H0 e σµ 2 = 0 σ2 λ = 0 σ2 µ = σ2 λ = 0 σ2 µ = 0 σ2 λ > 0 σ2 λ = 0 σ2 µ > 0 BP (3.841) (3.841) (5.991) HO (1.645) (1.645) (1.645) KW (1.645) (1.645) (1.645) SLM (1.645) (1.645) GHM (4.231) F (1.930) (1.645) (1.543) (1.648) (1.935) LR (2.706) (2.706) (4.231) (2.706) (2.706) LM µ (2.706) LM λ (2.706) Table: Tests for individual and time effects for Grunfeld investment data

44 Case study Hausman s specification test m 1 m 2 m (5.9915) (5.9915) (5.9915) Table: Hausman s test Grunfeld investment data modelled as one-way error component model (with individual effects).

45 Conclusion Case study Based on the tests executed we arrive at the following conclusions: The tests reject poolability across firms as well as poolability across time. GHM test rejects hypothesis σ 2 µ = σ 2 λ = 0. LMµ and LM λ tests revealed that σ 2 λ = 0 and σ2 µ 0. Therefore if we decide to model Grunfeld data using panel data model, the model should be formulated as: I it = α + β 1 F it + β 2 C it + µ i + ν it, i = 1,..., 10, t = 1,..., 20, where µ i are random effects. Hausman s test and its alternative do not reject the hypothesis E(u it X it ) and for estimating the regression parameters we can use GLS estimator. Thus the coefficients of the model are: ˆα = ˆβ 1 = 0.11 ˆβ 2 = 0.31

46 Case study Thank you for your attention

47 References I Case study B. H. Baltagi. Econometric analysis of panel data. John Wiley & Sons, Ltd, Third edition. B. H. Baltagi and Q. Li. Prediction in the one-way error component model with serial correlation. Journal of Forecasting, 11: , T. S. Breusch and A.R. Pagan. The lagrange multiplier test and its applications to model specification in econometrics. Review of Economic Studies, 47: , C. Gourieroux, A. Holly, and A. Monfort. Likelihood ratio test, wald test, and kuhn-tucker test in linear models with inequality constraints on the regression parameters. Econometrica, 50:63 80, Y. Grunfeld. The determinants of corporate investment. PhD thesis, University of Chicago, Unpublished Ph.D. dissertation. J. A. Hausman. Specification tests in econometrics. Econometrica, 46: , Y. Honda. Testing the error components model with non-normal disturbances. Review of Economic Studies, 52: , S. Kang. A note on the equivalence of specification tests in the two-factor multivariate variance components model. Journal of Econometrics, 28: , 1985.

48 References II Case study M. L. King and p. x. Wu. Locally optimal one-sided tests for multiparameter hypotheses. Econometric Reviews, 16: , B. R. Moulton and W. C. Randolph. Alternative tests of the error components model. Econometrica, 57: , 1989.