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



(iii) Uniformly in i and r [0, 1],

D31T EFzi x31,it - Ex31,it 0k31×1 a.s.;

D32τ (EFzix32,it Ex32,it) → √1rIk32g32,i (zi) a.s.;

D33T EFzi x33,it - Ex33,it τ33 (r) g33,i (zi) a.s.,
where

g32,i = (g1,32,i, ..., gk32,32,i)0 ; g33,i = (g1,33,i, ..., gk33,33,i)0 ,
and
g32,i (zi) and g33,i (zi) are zero-mean functions of zi with

0 <Esup kg3k,i (zi)k4q ∞, for some q>1,

i

and g3k,i 6= g3k,j for i 6= j, and τ33 (r) = diag (r-m1,33, ...r-mk33,33).

(iv) There exist τ (r) and Gi (zi) such that

∖∖D(Er,.x3,it Ex3,it) k ≤ e(r)Gei(zi),

where R τ (r)4q dr < and E supi (Gi (zi)4q for some q > 1.

(v) Uniformly in (i, j) and r [0, 1];

D31T (Ex31,it Ex31,jt) 0k31 ×1,

D32T (Ex32,it Ex32,jt) rlk32 μgg32iμfl32j) ,

D33T (Ex33,it Ex33,jt) τ33 (r)g33i Mg33j) ,

with suPi kμg32ik, suPi ∖∖μg33ik .

Some remarks would be useful to understand Assumption 6. First, to have
an intuition about what the assumPtion imPlies, we consider, as an illustrative
example, the simple model in CASE 3 in Section 2.2, in which x
3,it = ∖∖zz∕lm +
e
it, where eit is indePendent of zi and i.i.d. across i. For this case,

D3τ (EFzi X3,it Exз,it¢ = D3τi (zi Ezi) /^tm;

D3T (Ex3,it Ex3,jt) = D3T iEzi ΠjEzj) /tm.

Thus,

g3k,i (zi)  = Πi (zi Ezi);

μg3k,i = niEzi.

Second, Assumption 6(iii) makes the restriction that E supi g3k,i (zi)4q is
strictly positive, for k =2, 3. This restriction is made to warrant that g
3k,i (zi) 6=
0 a.s. If g
3k,i (zi) = 0 a.s.,15then

D3kTEFzi (x3k,itEx3k,it) ~ τ3k (r) g32,i (zi) = 0 a.s.,

15An example is the case in which x3,it = eit Πizi /tm ,where eit is independent of zi with
mean zero.

22



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