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Intercept Homogeneity Test for Fixed Effect Models under Cross-sectional Dependence: Some Insights

Gopal K. Basak and Samarjit Das


This paper develops a test for intercept homogeneity in fixed-effects one-way error component models assuming slope homogeneity. We show that the proposed test works equally well when intercepts are assumed to be either fixed (non-stochastic) or random. Moreover, this test can also be used to test for random effect vs. fixed effect although in the restrictive sense. The test is shown to be robust to cross-sectional dependence; for both weak and strong dependence. The proposed test is shown to have a standard χ2 limiting distribution and is free from nuisance parameters under the null hypothesis. Monte Carlo simulations also show that the proposed test delivers more accurate finite sample sizes than existing tests for various combinations of N and T. Simulation study shows that F-test is either over-sized or under-sized depending on the pattern of cross-sectional dependence. The performance of Hausman test (1978), on the other hand, is quite unstable across various DGPs; and empirical size varies from 0% to the nominal sizes depending on the structure of error variance-covariance matrix. The power of the proposed test outperforms the other two tests. It is worthwhile to mention that the power of our proposed test increases with T in contrast to that of Hausman test which is known to have no power as T→∞. An empirical illustration to examine the Kuznets’ U curve hypothesis with balanced panel data of Indian states is also provided. This empirical illustration points out the efficacy and the necessity of our robust test.

JEL Classification: C33; C12


The authors are grateful to Aman Ullah for his careful reading and constructive suggestions. The authors would also like to thank an anonymous referee for his/her insightful and constructive suggestions.


Proof of Theorem 1

Here we will use maximum eigenvalue norm.

Result 1:β^FE is consistent under both H0A and H0B.



from Assumption 1, we claim that plim(XMX)/NT=R, i.e we claim that the second order moment of MX exists. Now note that, ||XMX||=||XMMX||≤||M|| ||XX||=||XX||. Hence the matrix R is a finite matrix.

Note that, (XMX)−1X′MD̅=0, where D̅=[11111…1]′=1NT. This is because D̅ is a linear combination of the columns of D and hence belongs to the column space of D and we know that MD=0.



Now we take plim on both sides and we have


Now limE(XMϵ/NT)=0 by Assumption 2. Again, by Assumption 5,


Hence lim(V(XMϵ/NT))→0 as T→∞, both under weak and strong dependence. For weak dependence, lim(V(XMϵ/NT))→0 as either T and / or N→∞.

So we have plimβ^FE=β under both the Null.

Result 2:β^FE is consistent under both H1A and H1B.



Here (XMX)−1X′MD=0, since MD=0 and plim(XMX)−1XMϵ=0, by similar arguments as stated in the previous proof both under weak and strong dependence. For weak dependence, lim(V(XMϵ/NT))→0 as either T and/or N→∞.

Hence plimβ^FE=β under the alternative hypotheses.

Result 3:β^ols is consistent under both H0A and H0B.



Now it can be easily seen from the form of M̅=INTD̅(D̅′D̅)−1D̅′ that M̅D̅=0.

Now, β^ols=β+(XM¯X)1XM¯ϵ. It is also easy to see from Assumption 1 that W=plim[(XM̅X)/NT] is a finite matrix.


Now limE(XM̅ε/NT)=0, by Assumption 2. Again, by Assumption 5,


Hence lim(V(Xϵ/NT))→0 as T→∞, both under weak and strong dependence. For weak dependence, lim(V(XM̅ϵ/NT))→0 as either T and / or N→∞.

Hence plim(XM̅ϵ/NT)=0. So we have plimβ^ols=β under both the Null.

Result 4:β^ols is inconsistent under both H1A and H1B.




Now plim(XM̅ϵ/NT)=0 as we have already seen in Result 1. But plim(XM¯Dμ/NT)=plim1N(X¯iX¯¯)(μiμ¯)0, hence the proof.

Proof of Theorem 2

We will apply conditional Liapounov CLT. In both the cases, conditioning is on {Xt}. Define a matrix based on fourth moments and cross moments whose dimension is N2×N2. It is given by VF=E((εtεt)(εtεt)). Note, trace(VF)λmax(VF)i,jΩij2λmax(Ω2)=λmax2(Ω). Hence for strong dependence O(λmax(VF))=O(N2), whereas, for weak dependence, in general, O(λmax(VF))≥O(N).





For strong dependence, if the eigenvector corresponding to the largest eigenvalue of Ω belongs to the row-space of (dtdt) then


and hence,


For weak dependence,


since λmin(Ω)=O(1).



as λmax(VF)=O(N).

Proof of Theorem 3

To prove this theorem, let us first concentrate on EX||dt(etϵt)||2.


Let us denote, for notational simplicity, X˜t as a N×k+1 matrix of all regressors including the intercept variable. Similarly, ψ is the k+1×1 vector of regression parameter including the intercept. First let us concentrate on the term ||et–εt||2.


where, ψ^ is the estimator of intercept and the β^FE. Note that ||dtdt||=Op(1NT2),||(X˜tX˜t)||e=Op(N). For strong dependence, ||(ψ^ψ)||2=Op(1T); and that for weak dependence is, Op(1NT), by Theorem 1.

Hence, for weak dependence,


and, for strong dependence,


Thus, if dtεtεtdt converges so is dtetetdt and to the same matrix.

Now let us concentrate on EX|l(dtϵtϵtdt)llVllVl|r for r≤2. Note that EX|l(dtϵtϵtdt)llVllVl|r=1(lVl)rEX|(ldt(ϵtϵtΩ)dtl)|r. By Burkholder inequality,

EX|l(dtϵtϵtdt)llVllVl|rCr(lVl)rEX({ldt(ϵtϵtΩ)dtl}2)r/2,Cr(lVl)r(EX{ldt(ϵtϵtΩ)dtl}2)r/2,   asr/21,=Cr(lVl)r(Var{ldtϵtϵtdtl})r/2,Cr(lVl)r(EX{ldtϵtϵtdtl}2)r/2,

where Cr is independent of T. For r=2, we have already shown in Theorem 2 that

Cr(lVl)r(EX{ldtϵtϵtdtl}2)r/2={O(1T),for strong dependenceO(NT),for weak dependence.

Hence dtϵtϵtdtV in probability, conditionally on {X}, in the sense that,

l(dtϵtϵtdt)llVllVl0   in probability.


l(dtetetdt)llVllVl=lV¯llVl10   in probability,

uniformly in l(ll=1)

which implies ||V−1/2V̅V−1/2I||e→0 in probability.


||V||=||dtΩdt||||dtdt||||Ω||{Op(1NT) for weak dependence,Op(1T) for strong dependence.


EX||V¯||=EX||dtetetdt||||dtdt||EX||etet||||dtdt||{EX{||(ψ^ψ)||2}||(X˜tX˜t)||e+||Ω||}{Op(1NT) for weak dependence,Op(1T) for strong dependence.

Hence by Theorem 2,

(β^FEβ^ols)(V1V¯1)(β^FEβ^ols)=(β^FEβ^ols)V1/2(IV1/2V¯1V1/2)V1/2(β^FEβ^ols)IV1/2V¯1V1/2e(β^FEβ^ols)V1(β^FEβ^ols)=V1/2V¯V1/2Ie×Op(1)=op(1)   for both weak and strong dependence.

Hence, MD and EMD have same limiting distribution.

Proof of Theorem 4

Let δμ=plim[(XM¯X/NT)1(XM¯Dμ/NT)]=W1plim1N(X¯iX¯¯)(μiμ¯) and δ^μ=(XM¯X)1XM¯Dμ. Under Alternative, by Result 3 of Theorem 1, β^FE=β+(XMX)1XMϵ, and by Result 4 of Theorem 1, β^ols=β+δ^μ+(XM¯X)1XM¯ϵ. Therefore, under Alternative,


as in the notation of equation (6).

Hence, (β^FEβ^ols+δ^μ)V1(β^FEβ^ols+δ^μ) is asymptotically χk2 as in Theorem 2.

Thus, as in Theorem 3, (β^FEβ^ols+δ^μ)V¯1(β^FEβ^ols+δ^μ) is asymptotically χk2. Therefore, (β^FEβ^ols)V¯1(β^FEβ^ols) is asymptotically non-central χk2 with the non-centrality parameter as δμV1δμ.


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Supplemental Material:

The online version of this article (DOI: 10.1515/jem-2015-0004) offers supplementary material, available to authorized users.

Published Online: 2016-4-19
Published in Print: 2017-1-1

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