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



as M → ∞, where 1() is the indicator function that equals one if the argument
in the parenthesis is correct, and otherwise equals zero. Then, as
(N,T →∞),

N X (QiT - EFzi QiT) →p 0.

i

In fact, Lemma 13 still holds even if we replace the conditional mean oper-
ator
EFz. () by the unconditional operator E(). Thus, we have the following
corollary.

Corollary 14 Suppose that QiT (k × k) is a sequence of independent random
matrices across
i satisfying

(37)


supE kQiT k 1 {kQiT k >M} → 0 a.s.,
i,T

as M →∞. Then, as (N, T →∞),

N ^X (QiT - EQiT) p 0.

i

Proof of Lemma 13

Let Fz = σ (Fz1 , ..., FzN) and EFz denote the conditional expectation on

Fz . For any ε > 0, we need to show that

IN X(QiT


EF
F
zi


QiT) > ε → 0.


This follows from the dominated convergence theorem if we can show that

PF

z


I N X(QiT


EF
F
zi


QiT > ε → 0 a.s..


But, this in turn follows from the conditional Markov inequality if we can show
that

EF

z


N X(QiT
i


EFzi QiT


0 a.s..


(38)


To show (38), define:

PiT = QiT1 {kQiT k ≤ M} ; RiT = QiT1 {kQiT k >M} .

Then, almost surely,

EFz


EFz


EF

z


(Qi (QiT - EFzi QiT)
i


X (PiT - EFzi PiT) + EFz N X (RiT - EFzi RiT)


N X(PiT - EFzi PiT)
i


33


1

2 2

I  +2 N X EFzi kRiT k .

i




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