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
Fzi
QiT) > ε → 0.
This follows from the dominated convergence theorem if we can show that
PF
z
I N X(QiT
EF
Fzi
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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