Large-N and Large-T Properties of Panel Data Estimators and the Hausman Test

Lemma 2 Under Assumptions 1-8, we obtain the following results as (N, T →
∞). For some positive semidefinite matrices Ψ_x and Ξ (defined in the Ap-
pendix),

(a) NN ΣiT ∑t Gχ,τXitXitGχ,τ →p Ψχ;

^(b) √1N Pi √T Pt Gx,T^xitvit ⇒ N (0, ^σv^ψx) ;

^(c) 11 Pi DTwiweDT →p ^ξ;

(d) √⅛∑i DTWivi →p 0(k+g)×ι.

Lemma 3 Under Assumptions 1-8 and Assumption 11 (local alternatives to
random effects), as (N, T →∞),

√N Pi Dt w^u ⇒ N (≡^{λ σ}U≡0.

Lemma 4 Under Assumptions 1-8 and Assumption 10 (fixed effects),

N Pi Dτ wiei →p^Ξλ,

as (N,T →∞) .

The following assumption is required for identification of the within and
between estimators of β and γ.

Assumption 12 The matrices Ψx and Ξ are positive definite.

Two remarks on this assumption follow. First, this assumption is also sufficient
for identification of the GLS estimation. Second, while the positive definiteness
of the matrix Ξ is required for identification of the between estimators, it is not
a necessary condition for the asymptotic distribution of the Hausman statistic
obtained below. We can obtain the same asymptotic results for the Hausman
test even if we alternatively assume that within estimation can identify β (pos-
itive definite Ψx) and between estimation can identify γ given β (the part of
Ξ corresponding to zei is positive definite).¹⁸ Nonetheless, we assume that Ξ is
invertible for convenience.

We now consider the asymptotic distributions of the within, between and
GLS estimators of β and γ:

we can show that

E(ui | Xi, Zi) = Xiλχ + Ziλz,
where d = (σf + σ2∕T)(σf + ση) — σf, λχ = — cσf ση/d, λz = c(σf + σ2∕T)ση/d. Observe
that λx 6=0, if λz 6=0(c 6= 0). Thus, λx is functionally related to λz . In addition, it is easy
to show that plim_N→∞β_b = β + λx 6= β, if λz 6=0(c 6= 0). Thus, nonzero λz biases β_b.

¹⁸ This claim can be checked with the following simple example. Consider a simple model
with one time-varying regressor xit and one time invariant regressor zi . Assume that xit =
azi + eit , where the eit are i.i.d. over different i and t. For this model, it is straightforward
to show that the matrix Ξ fails to be invertible. Nonetheless, under the random effects
assumption, the Hausman statistic can be shown to follow a χ² distribution with the degree
of freedom equal to one.

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