and in term of group means, =a+邓+E1+t1,i=1,2,…,N Therefore, taking deviation from the group means removes the heterogeneity =(xt-)+t-E2,i=1,2,…,N;t=1,2,…,T, cn-)2|=(r-1)2, B is observable(and therefore e is observed), then an unbiased estimator of o based on T observations in group i would be (Eit-Ei) Since B must be estimated, we may use the residual from the LSDV estimator (which is consistent and unbiased in general) and correct the degree of freedom to form the estimator ∑1(et-a)2 T-k-1 We have n such estimators so we average them to obtain 1(ea-)2 (et-E1)2 T一k-1 NT-NK-N The degree of freedom correction in $p is excessive because it assume that a and B are reestimated for each i. The unbiased estimator would be 02=2sr 1>1( where e, should be 0 It remains to estimate o Back to the original model vht=a+x1tB+t+t2,i=1,2,…,N;t=1,2, In spite of the correlation across observation, this is a classical regression model in which the OLS estimation slope and variance estimator are both consistentand in term of group means, y¯i = α + x¯ 0 iβ + ε¯i + ui , i = 1, 2, ..., N; (5) Therefore, taking deviation from the group means removes the heterogeneity: yit − y¯i = (xit − x¯i) 0β + εit − ε¯i , i = 1, 2, ..., N; t = 1, 2, ..., T, (6) Since E X T t=1 (εit − ε¯i) 2 # = (T − 1)σ 2 ε , if β is observable (and therefore ε is observed), then an unbiased estimator of σ 2 ε based on T observations in group i would be σˆ 2 ε (i) = PT t=1(εit − ε¯i) 2 T − 1 . Since β must be estimated, we may use the residual from the LSDV estimator (which is consistent and unbiased in general) and correct the degree of freedom to form the estimator: s 2 e (i) = PT t=1(eit − e¯i) 2 T − k − 1 . We have N such estimators, so we average them to obtain s¯ 2 e = 1 N X N i=1 s 2 e (i) = 1 N "PT t=1(eit − e¯i) 2 T − k − 1 # = PN i=1 PT t=1(eit − e¯i) 2 NT − Nk − N . The degree of freedom correction in s¯ 2 e is excessive because it assume that α and β are reestimated for each i. The unbiased estimator would be σˆ 2 ε = s 2 LSDV = PN i=1 PT t=1(eit − e¯i) 2 NT − N − K . where e¯i should be 0. It remains to estimate σ 2 u . Back to the original model: yit = α + x 0 itβ + εit + ui , i = 1, 2, ..., N; t = 1, 2, ..., T. (7) In spite of the correlation across observation, this is a classical regression model in which the OLS estimation slope and variance estimator are both consistent 13