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 EF_z. (∙) 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

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

as M →∞. Then, as (N, T →∞),

N ^X ^(QiT ^- E^QiT) →p 0.

i

Proof of Lemma 13

Let F_z = σ (F_z1 , ..., F_zN) and E_Fz denote the conditional expectation on

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

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

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

To show (38), define:

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

Then, almost surely,

<<   <   30   31   32   33   34   35   36   >   >>