On the Identification of Misspecified Propensity Scores - School of ...

To check the second condition of the lemma, let δ > 0,
limn→∞ j
n 2 hn
a2 n (n−1)2
1
anh2 2
C+C1 E ε
n
2 j01 =limn→∞ hn
a2
1
n anh2 C+C1
n
≤limn→∞ hn
a 2 n
1
anh2 C+C1
n
2
j E
ε2 j01 2
j E
n √ hn|εj0 |
an(n−1)
n √ hn|εj0 |
an(n−1)
ε2 j01 √
n hn
an(n−1)
1
anh2
C+C1 >δ
n
1
anh2 C+C1
n
1
anh2 C+C1
n
>δ
>δ ,
where we use the fact that |εj0| ≤ 1 to write the last inequality. Note that limn→∞[1/(anh2 n )] <
∞ implies that 1/(a2 nh 3/2
n ) → 0. This in turn implies that for each δ > 0 only finitely many
terms in the last summation will be non-zero, and therefore the second condition of the lemma
is satisfied as well.
On the other hand, the same arguments show that the same conditions are satisfied for the
process
n[Q(Xj, θ) − Q(Xj, θ0)] Q(Xi, θ) − Q(Xi, θ0) Q(Xi, θ) − Q(Xj, θ)
E √ K
|Xj
n − 1
hn
hn
j
n[Q(Xj, θ) − Q(Xj, θ0)][Q(Xi, θ) − Q(Xi, θ0)]
− E
(n − 1) √
Q(Xi, θ) − Q(Xj, θ)
K
hn
hn
To verify the third condition note that since Θ is a compact subset of R l , its L 2 packing
number, D2(τ, Θ), is less than or equal to 2lτ −l . If K(·) is bounded and Lipschitz, Q(·, ·) is
Lipschitz, and n−1h −3/2
n = O(1), then D2(τ, Ψn) ≤ C · τ −l . Thus, the third condition of the
lemma is also satisfied.
We know that ψn0 ∈ Ψn. In addition, if
√ n( ˆ θ − θ0) d → N(0, Σθ),
then for any {an} sequence such that an/ √ n → 0, we will have an( ˆ θ − θ0) P → 0, and ˆ ψn will
belong to Ψn with probability approaching to 1.
j
The arguments given so far allow us to argue that
ˆψn(i, j) − E[
i,j
ˆ
ψn(i, j)|j] − ψn0(i, j) − E[ψn0 (i, j)|j]
=oP (1),
√
hnεj0
n − 1 E
(εi0 − ˆεi) 1
Q(Xi,
K
hn
ˆ θ) − Q(Xj, ˆ
θ)
|Xj
hn
=oP (1), and
n √ hn(εj0 −ˆε j )
n−1
E
(εi0 −ˆε i )
hn
i,j
K
Q(Xi , ˆ θ)−Q(X j , ˆ θ)
hn
|Xj
−E
17
n(εj0 −ε j )(ε i0 −ˆε i )
(n−1) √ hn
K
Q(Xi , ˆ θ)−Q(X j , ˆ θ)
hn
=oP (1).