Identifying the Average Treatment E¤ect in a Two
Threshold Model - Supplemental Appendix
Arthur Lewbel and Thomas Tao Yang2
Boston College
This online appendix provides proofs for Theorem 2.4, 2.6, 2.8 and associated lemmas, which
are asymptotics for the following estimators in the paper,
2
b
n
1 X 4E
n
b
E
i=1
1
n
n
X
i=1
1
nT
1
nT
2
6
4
b
E
b
E
D i Yi
f (vi jxi )
xi
Di
f (vi jxi )
xi
D i Yi
fb(vi jxi )
xi
Di
fb(vi jxi )
T X
n
X
t=1 i=1
T X
n
X
t=1 i=1
b
E
(1 Di )Yi
fb(vi jxi )
b
E
1
nT
t
(1 Di )Yi
f (vi jxi )
b
E
xi
Dit Yit
fbv (vit )
b
E
1 Di
f (vi jxi )
1 Di
fb(vi jxi )
xi
xi
xi
T X
n
X
(1
t=1 i=1
T X
n
X
Dit
fbvt (vit )
xi
1
nT
t=1 i=1
Dit )Yit
fbvt (vit )
3
5;
(1)
3
(2)
:
(3)
7
5;
(1 Dit )
fbvt (vit )
Lemma 4.1 Suppose random variables si ; xi are i.i.d. across i = 1; 2; :::; n; and si is bounded:
1 1 k 1
r(xi ) is a bounded real function of xi : The following Lipschitz conditions hold:
jE(si jxi + ex )
E(si jxi )j
jr(xi + ex )
r(xi )j
jfx (xi + ex )
fx (xi )j
M1 kex k ;
M2 kex k ;
M3 kex k
for some positive number M1 ; M2 ; and M3 . ri (xi ), E(si jxi ); fx (xi ) are p-th order di¤ erentiable
and the p-th order derivatives are bounded.
2
Corresponding Author: Arthur Lewbel, Department of Economics, Boston College, 140 Commonwealth Avenue,
Chestnut Hill, MA, 02467, USA. (617)-552-3678, lewbel@bc.edu, https://www2.bc.edu/ lewbel/. Thomas Tao
Yang, Department of Economics, Boston College, 140 Commonwealth Avenue, Chestnut Hill, MA, 02467, USA.
yangta@bc.edu.
1
Let
n
Un =
=
1X
b (si jxi ) fb(xi ) E r(xi )sj 1 K xj xi
r(xi )E
n
h
hk
i=1
8
2
3
n
n
X
xj xi 5
1 X<
1
1
4
r(xi )
E r(xi )sj k K
sj K
k
:
n
h
(n 1)h
h
i=1
xj
h
j=1;j6=i
where K(:) is the kernel function de…ned in Assumption 4.3. n1
" h2k+2
xi
9
=
;
;
! 1; and nh2p ! 0; as
n ! 1; for a very small " > 0. De…ne term
n
X
en = 1
U
f[r(xi )si + r(xi )E(si jxi )] fx (xi )
n
i=1
p
2E [r(xi )E(si jxi )fx (xi )]g ;
en ) = op (1): Furthermore, since K xj xi is a p-th order kernel, and by assumption
n(Un U
h
h
i
xj xi
1
2p
that nh ! 0; E r(xi )sj hk K
in Un could be replaced by E[r(xi )E(si jxi )fx (xi )] and the
h
then
conclusion still holds.
The Lemma is just a modi…cation of Lemma 3.2 in Powell et al. (1988). It is pretty straightforward, given the property of U-statistics and uniformly convergence property of the nonparametric
regression. The Lipschitz condition is imposed to control the residual term; the existence and
boundedness conditions are imposed to control the biased term. Estimator (1) is consist of several
terms like Un and a residual term of the order op
p1
n
:
Proof of Lemma 4.1. Rewrite the …rst term in Un ; indexed by U1n as
U1n =
=
1
n(n
1)
2
n(n
1)
n
n
X
X
r(xi )sj
i=1 j=1;j6=i
n
X1
1
K
hk
xj
xi
n
X
1
1
(r(xi )sj + r(xj )si ) k K
2
h
i=1 j=i+1
Index zi = (si ; xi ); and de…ne p(zi ; zj ) =
1
2
(4)
h
[r(xi )sj + r(xj )si ] h1k K
xj
xi
h
xj xi
h
: Since the kernel
function we adopt is symmetric, p(zi ; zj ) = p(zj ; zi ): It is easy to see that E(kp(zi ; zj )k2 ) = o(n):
According to Lemma 3.2 in Powell et al. (1989),
U1n
e1n = op
U
2
1
p
n
;
(5)
where
n
X
e1n = E [p(zj ; zi )] + 2
[E (p(zj ; zi )jzi )
U
n
E (p(zj ; zi ))] :
(6)
i=1
E [p(zj ; zi )jzi ] is
E [p(zj ; zi )jzi ] =
=
Z
k
ZR
Rk
=
xj xi
1
1
fx (xj )dxj
[r(xi )E(sj jxj ) + r(xj )si ] k K
2
h
h
1
[r(xi )E(si jxi + hu) + r(xi + hu)si ] K (u) fx (xi + hu)du
2
1
[r(xi )E(si jxi ) + r(xi )si ] fx (xi ) + Ri ;
2
where second equality is obtained via variable transformation u =
xj xi
h ;
and Ri is the residual
term, which could be decomposed into two components R1i and R2i ,
Ri = R1i R2i
Z
1
[r(xi )E(si jxi + hu) + r(xi + hu)si r(xi )E(si jxi ) r(xi )si ] K (u) fx (xi )du
=
k 2
RZ
1
[r(xi )E(si jxi ) + r(xi )si ] K (u) [fx (xi + hu) fx (xi )] du:
Rk 2
E[p(zj ; zi )jzi ] E[p(zj ; zi )] in Equation (6) thus could be written as
E [p(zj ; zi )jzi ] E [p(zj ; zi )]
1
[r(xi )E(si jxi ) + r(xi )si ] fx (xi )
=
2
+R1i E(R1i ) [R2i E(R2i )]
1
=
[r(xi )E(si jxi ) + r(xi )si ] fx (xi )
2
(7)
E
1
[r(xi )E(si jxi ) + r(xi )si ] fx (xi )
2
E [r(xi )E(si jxi )fx (xi )] + R1i
E(R1i )
Then under Lipschitz conditions and bound conditions given in the Lemma,
2
E(R1i
) = O(h2 );
2
E R2i
= O(h2 ):
3
[R2i
E(R2i )] :
Thus by Lindberg Levy CLT
n
n
1 X
p
[R1i
n
1 X
p
[R2i
n
E(R1i )]
i=1
i=1
p
E(R2i )] ! 0:
(8)
Equation (5), (6), (7), and (8) imply that U1n E[p(zj ; zi )] is equal to
n
1X
f[r(xi )E(si jxi ) + r(xi )si ]fx (xi )
n
1
p
n
2E [r(xi )E(si jxi )fx (xi )]g + op
i=1
;
which is the …rst conclusion of the Lemma.
Second conclusion follows by the assumption that nh2p ! 0 and those bound conditions and
by the fact that
E r(xi )sj
1
K
hk
xj
xi
h
E [r(xi )E(si jxi )fx (xi )] = op (hp );
the equality holds by the assumption that kernel function is of order p:
Proof of Theorem 2.4. The last conclusion of this theorem follows immediately after the …rst
conclusion via Lindeberg central limit theorem. For the …rst conclusion, it is enough to show that
n h
i
X
1
b (xi ) E(Y1 ) is equal to
1
n
i=1
n
1X
n
1 (xi )
i=1
+
h1i
E(g1i jxi )
E(h1i jxi )g1i
[E(g1i jxi )]2
E(Y1 ) + op
1
p
n
:
(9)
Note that
1 X hb
1 (xi )
n
n
i=1
1 (xi )
i
=
=
=
"
#
n
b 1i jxi ) E(h1i jxi )
1 X E(h
b
n
E(g1i jxi )
i=1 E(g1i jxi )
"
#
n
b e
1 X E(
h1i jxi ) E(e
h1i jxi ) E(e
h1i jxi ) E(e
h1i jxi )
+
b g1i jxi )
b g1i jxi )
b g1i jxi )
n
E(e
g1i jxi )
E(e
E(e
i=1 E(e
h
i
2
3
e
b
n
E(
h
jx
)
E(e
g
jx
)
E(e
g
jx
)
b e
1i i
1i i
1i i
1 X 4 E(
h1i jxi ) E(e
h1i jxi )
+ Rni 5 ;
n
E(e
g1i jxi )
[E(e
g1i jxi )]2
i=1
4
where the second equality follows by the fact that
Rni =
h
b e
E(
h1i jxi )
ih
E(e
h1i jxi ) E(e
g1i jxi )
b g1i jxi )
E(e
g1i jxi )E(e
Under Assumption 4.2, n1
" h2k+2
i=1
1
n
n
X
i=1
1
n
n h
X
i=1
2
E(e
h1i jxi )
E(e
g1i jxi ) ;
b g1i jxi )
E(e
b g1i jxi )
[E(e
g1i jxi )]2 E(e
h
E(e
g1i jxi ) = Op (n1
n
X
p1
n
for any arbitrary small " > 0: So
jRni j is op (1): Thus,
=
h
E(e
h1i jxi ) E(e
g1i jxi )
i
b g1i jxi )
E(e
h
E(e
h1i jxi ) = Op (n1
b g1i jxi )
sup E(e
n
X
b e
E(
h1i jxi ) E(h1i jxi )
b g1i jxi ) ; E(g1i jxi )
E(e
=
and
i2
:
! 1 and from Li and Racine (2007) Chapter 2.3
b e
sup E(
h1i jxi )
p1
n
b 1i jxi )
E(h
b 1i jxi )
E(g
i=1
b (xi )
1
e
b e
4 E(h1i jxi ) E(h1i jxi )
E(e
g1i jxi )
sup jRni j is Op (
"
h )
k
1
2
"
k
1
2
h )
1
1
n2
" k
h
i
i
;
;
): Since n1
" h2k+2
! 1 ,
i
(x
)
is equal to
1 i
h
b g1i jxi )
E(e
h1i jxi ) E(e
2
[E(e
g1i jxi )]
i3
E(e
g1i jxi )
5 + op
1
p
n
:
(10)
Note that
1
n
n
X
i=1
b e
E(
h1i jxi )
E(e
g1i jxi )
=
=
1
n
n
X
i=1
n
X
1
(n 1)hk
h1j K
j=1;j6=i
2
E(e
g1i jxi )
n
1X
1
r1 (xi ) 4
n
(n 1)hk
i=1
xj xi
h
n
X
s1j K
xj
j=1;j6=i
h
3
xi 5
:
Under Assumption 4.4, applying Lemma 4.1 by letting r(xi ) = r1 (xi ); sj = s1j ; one can get
n b e
n
h1i jxi ) E(e
h1i jxi )
1 X E(
1X
h1i
=
n
E(e
g1i jxi )
n
E(g1i jxi )
i=1
E(Y1 ) + op
i=1
1
p
n
:
(11)
Similarly, we can prove that
h
i
b g1i jxi ) E(e
n E(e
n
h
jx
)
E(e
g
jx
)
X
1i
i
1i
i
1
1 X E(h1i jxi )g1i
=
n
n
[E(e
g1i jxi )]2
[E(g1i jxi )]2
i=1
i=1
5
E(Y1 ) + op
1
p
n
:
(12)
Combining Equation (10) (11) (12) gives that
1 X hb
1 (xi )
n
n
i 1X
h1i
(x
)
=
i
1
n
E(g1i jxi )
n
i=1
i=1
Adding both sides of the above equation with
1
n
n
X
[
1
p
n
E(h1i jxi )g1i
+ op
[E(g1i jxi )]2
1 (xi )
E(Y1 )] gives that
1
n
i=1
= ( Xi Vi ); si ;
(k+1) 1
n h
X
i=1
is equal to Equation (9). The …rst conclusion is proved.
Lemma 4.2 Assume we observe Wi
:
Zi
1 1 (k+2) 1
k 1 1 1
= ( Wi
si );
(k+1) 1 1 1
b (xi )
1
E(Y1 )
which are i.i.d.
across i: r(xi ) is a real function of xi : si ; r(xi ) and density function fx ; fw are bounded: E(si jwi );
r(xi ); fx and fw are p-th order di¤ erentiable, and p-th order derivatives are bounded. E(si jwi ),
r(xi ); fx ; fw satisfy the Lipschitz condition
jE(si jwi + ew )
E(si jwi )j
jr(xi + ex )
r(xi )j
jfx (xi + ex )
fx (xi )j
jfw (wi + ew )
fw (wi )j
M1 kew k ;
M2 kex k ;
M3 kex k ;
M4 kew k
for some positive M1 , M2 ; M3 ; M4 . Under above assumptions, the following term
n
=
1X
b si fbw (wi ) xi fbx (xi )
r(xi )E
n
i=1
8
n
n
< 1
X
X
xj xi
1
1
r(xi )
sj K
k
:n 1
n
h
h
i=1
=
j=1;j6=i
1
n(n
1)2
n
n
X
X
n
X
r(xi )sj
i=1 j=1;j6=i l=1;l6=j
1
K
hk
xj
(13)
2
4
h
n
X
1
n
xi
1
l=1;l6=j
1
hk+1
1
hk+1
K
K
wl
wl
wj
h
wj
h
is equal to
n
1X
fr(xi )si fw (wi )fx (xi ) + r(xi )fx (xi )E [ si fw (wi )j xi ]
n
i=1
+r(xi )E (si jwi ) fw (wi )fx (xi )
2E [r(xi )si fw (wi )fx (xi )]g + op
6
1
p
n
:
39
=
5
;
i
After decomposing Estimator (2), the in‡uence function of our estimator is consist of several
terms like Equation (13). The proof of this lemma looks tedious but it only repeatedly uses the Ustatistics results from Powell et al. (1988) and uniformly convergence property of nonparametric
estimates.
Proof of Lemma 4.2. Rewrite Equation (13),
n
1X
b si fbw (wi ) xi fbx (xi )
r(xi )E
n
i=1
=
=
1
n
n
X
X
n(n
1)hk
n(n
1
1)2 h2k+1
i=1 j=1;j6=i
n
X
xj
r(xi )sj fbw (wj )K
n
X
r(xi )
i=1
n
X
xi
h
sj K
xj
xi
h
j=1;j6=i l=1;l6=j
K
wl
wj
h
:
Drop one term in the last summation to make it look more like a U-statistics,
n
1X
r(xi )
n
(n
i=1
n
X
1
1)(n 2)h2k+1
n
X
xj
sj K
xi
h
j=1;j6=i l=1;l6=i;j
wl
K
wj
h
:
(14)
Since all terms inside summation are bounded, one term dropped is of the order op ( n3 h12k+1 ); which
is of course op ( p1n ) by assumption nh2k+1 ! 1:
The structure of proof is as follows.
First we will show that
(n
n
X
1
1)(n 2)h2k+1
n
X
sj K
xj
xi
h
j=1;j6=i l=1;l6=i;j
K(
wl
wj
h
)
(15)
is equal to
E
+
sj
K
h2k+1
xj
xi
h
K(
wl
wj
h
) xi +
1
n
1
n
X
j=1;j6=i
1
sj K
hk
xj
xi
h
xj xi
xj xi
1
1
E(sj jwj )K
fw (wj ) E k sj K
fw (wj ) xi
k
h
h
h
h
xj xi
1
1
E k E(sj jwj )K
fw (wj ) xi + Rnij E(Rnij jxi ) + op p
h
h
n
fw (wj ) (16)
;
where Rnij is some residual term that will be de…ned later.
Then substitute Equation (16) into Equation (14), the new expression becomes a standard
7
U-statistics. We then could show the new expression is equal to
n
1X
[r(xi )si fw (wi )fx (xi ) + r(xi )E (si jwi ) fw (wi )fx (xi )
E [r(xi )si fw (wi )fx (xi )] +
n
i=1
1
p
n
2E [r(xi )si fw (wi )fx (xi )]] + op
;
by which the conclusion of the lemma follows immediately.
Step 1: As discussed, we consider …rst the following term,
(n
=
(n
n
X
1
1)(n 2)h2k+1
n
X
xj
sj K
2)
n
X1
n
X
j=1;j6=i l=j+1;l6=i
xj
1
sj K
2
xi
wl
K
h
j=1;j6=i l=1;l6=i;j
2
1)(n
xi
(17)
h
xl
+ sl K
h
wj
xi
1
K
h2k+1
h
wl
wj
h
:
Let
P1 (zj ; zl ; xi ) =
xj
1
sj K
2
xi
h
+ sl K
xl
1
xi
h2k+1
h
wl
K
wj
h
;
then Equation (17) becomes
(n
2
1)(n
2)
n
X1
n
X
P1 (zj ; zl ; xi );
(18)
j=1;j6=i l=j+1;l6=i
following Powell et al. (1989), we …rst verify E P1 (zj ; zl ; xi )2 jxi = op (n):
E P1 (zj ; zl ; xi )2 jxi
(
Z Z
1
sj K
=
E
2
wj ;w
l
xj
xi
h
+ sl K
xl
xi
h
1
h2k+1
fw (wj )fw (wl )dwj dwl :
(
Z Z
1
1
=
E
[sj K (ui ) + sl K (uj + hul )] K(ul )
2k+1
2
h
uj ;u
K
wl
wj
h
2
wj ; wl
)
2
(xi + huj ; vj ); (xi + huj + hul ; vj + hul )
l
fw (xi + huj ; vj )fw (xi + huj + hul ; vj + hul ) duj dvj dul
1
= Op
= op (n);
h2k+1
where the second equality holds by variable changes ul =
wl wj
h ;
uj =
xj xi
h ;
the third equality
holds because of those bound conditions, and the last equality holds by assumption that nh2k+1 !
8
)
1: According to Lemma 3.2 in Powell et al. (1989), Equation (18) is equal to3
n
E [P1 (zj ; zl ; xi )jxi ] +
2X
fE [P1 (zj ; zl ; xi )jzj ; xi ]
n
E [P1 (zj ; zl ; xi )jxi ]g + op
j=1
1
p
n
:
(19)
Term inside the summation in Equation (19) could be analyzed as follows,
E [P1 (zj ; zl ; xi )jzj ; xi ] E [P1 (zj ; zl ; xi )jxi ]
xj xi
wj wl
wj wl
sj
1
1
xl xi
sl
=
K
zj ; xi + E 2k+1 K
K
zj ; xi
E 2k+1 K
2
h
h
2
h
h
h
h
wj wl
wj wl
sj
xj xi
1
1
sl
xl xi
K
xi
K
xi :
E 2k+1 K
E 2k+1 K
2
h
h
2
h
h
h
h
Since
sj
wj wl
xj xi
E 2k+1 K
K
zj ; xi
h
h
h
Z
sj
xj xi
wj wl
=
K
K
fw (wl )dwl
2k+1
h
h
h
wl
Z
xj xi
xj xi
sj
sj
K (ul ) [fw (wj + hul )
=
K
fw (wj ) + k K
h
h
hk
h
u
fw (wj )] dul ;
l
and similarly
sl
K
2k+1
h
xl
E [sj jwj ]
K
hk
xj
E
=
E [sj jwj ] K
xi
h
xi
h
xj
xi
h
K
wj
wl
h
fw (wj ) +
1
hk
zj ; xi
Z
ul
E [sj jwj + hul ] K
xj + hul
h
xi
fw (wj + hul )
fw (wj ) K (ul ) dul ;
the following holds
E [P1 (zj ; zl ; xi )jzj ; xi ] E [P1 (zj ; zl ; xi )jxi ]
xj xi
xj xi
sj
1 sj
1 E [sj jwj ]
1
=
K
K
fw (wj ) +
fw (wj )
E kK
k
k
2h
h
2
h
2
h
h
E [sj jwj ]
xj xi
1
K
E
fw (wj ) xi + Rnij E (Rnij jxi ) ;
2
h
hk
3
xj
xi
h
fw (wj ) xi
The argument in Powell et al. (1989) is without conditional expectation. With conditional expectation,
everything follows very similarly to the case without conditional expectation.
9
(20)
where
Rnij
Z
xj xi
sj
K
K (ul ) [fw (wj + hul ) fw (wj )] dul
=
h
hk
Z
xj + hul xi
1
fw (wj + hul )
+ k
E [sj jwj + hul ] K
h
h
u
(21)
l
E [sj jwj ] K
xj
xi
fw (wj )K (ul ) dul :
h
(22)
Equation (20) is the same as the term inside the summation of Equation (16), together with the
fact that p(zj ; zl ; xi ) are symmetric for zj ; zl conditional on xi and thus
E [P1 (zj ; zl ; xi )jxi ] = E sj K
xj
xi
h
K(
wl
wj
h
) xi ;
Equation (19) is equal to (16). By the assumption that r(xi ) is bounded, the op ( p1n ) term is still
op ( p1n ) after summing over i:
To accomplish the second part of proof, we will …rst discuss the residual term of the order
op ( p1n ) and then the in‡uence term.
Step 2 (residual term): Here we claim that
1
n(n
1)
n
n
X
X
[r (xi ) Rnij
i=1 j=1;j6=i
1
r (xi ) E (Rnij jxi )] = op ( p ):
n
Let Rnij = R1nij + R2nij , where
R1nij =
R2nij
=
xj xi
sj
K(
)
h
hk
1
hk
Z
ul
Z
K (ul ) [fw (wj + hul )
E [sj jwj + hul ] K(
E [sj jwj ] K(
fw (wj )] dul ;
ul
xj
xi
h
xj + hul
h
xi
)fw (wj + hul )
)fw (wj ) K (ul ) dul :
So the object of this small session is
1
n(n
1)
n
n
X
X
i=1 j=1;j6=i
[r (xi ) R1nij
r (xi ) E (R1nij jxi ) + r (xi ) R2nij
10
r (xi ) E (R2nij jxi )] :
(23)
For
n
n
X
X
r (xi ) R1nij ; it is a standard U-statistics after being written as
i=1 j=1;j6=i
Z
n
n
xj xi
1 X X r (xi ) sj
K(
)
n
h
hk
=
1
n
i=1 j=1;j6=i
n X
n
X
i=1 j=i+1
K (ul ) [fw (wj + hul )
fw (wj )] dul
(24)
ul
n
n
2X X
[P21 (zi ; zj ) + P22 (zi ; zj )] =
P2 (zi ; zj );
n
i=1 j=i+1
where
P21 (zi ; zj ) = r (xj ) si
P22 (zi ; zj ) = r (xi ) sj
Z
Z
K (ul ) [fw (wi + hul )
fw (wi )] dul
1
K
hk
xj
K (ul ) [fw (wj + hul )
fw (wj )] dul
1
K
hk
xj
ul
ul
P2 (zi ; zj ) =
xi
h
xi
h
;
;
1
[P21 (zi ; zj ) + P22 (zi ; zj )] :
2
For the similar reason as in Step 1, easy to show that E P2 (zi ; zj )2 = op (n): Then we could apply
Lemma 3.2 in Powell et al. (1989) again to get that
1
n(n
1)
n
n
X
X
n
r (xi ) R1nij = E [P2 (zi ; zj )]+
i=1 j=1;j6=i
2X
[E [P2 (zi ; zj )jzi ]
n
E [P2 (zi ; zj )]]+op
i=1
1
p
n
:
(25)
So
1
n(n
n
=
[r (xi ) R1nij
i=1 j=1;j6=i
1X
[2E [P2 (zi ; zj )jzi ]
n
i=1
=
1)
n
n
X
X
r (xi ) E (R1nij jxi )]
r (xi ) E (R1nij jxi )
n
1X
[E [P21 (zi ; zj )jzi ] + E [P22 (zi ; zj )jzi ]
n
1
p
n
E [P2 (zi ; zj )]] + op
r (xi ) E (R1nij jxi )
i=1
(26)
;
E [P2 (zi ; zj )]] + op
1
p
n
:
Since zi and zj are identically distributed;
E [P21 (zi ; zj )] = E [P22 (zi ; zj )] = E [P2 (zi ; zj )] :
11
(27)
By changing variables uj =
E [P21 (zi ; zj )jzi ] = si
= si
1
hk
Z
Z
xj xi
h
in the integral,
K (ul ) [fw (wi + hul )
fw (wi )] dul
ul
K (ul ) [fw (wi + hul )
fw (wi )] dul
ul
Z
Z
xj
r(xj )
K
hk
xj
xi
h
fx (xj )dxj
r (xi + huj ) K (uj ) fx (xi + huj )duj
uj
has been cancelled out above, then by the assumptions that the …rst derivative of fw and
each term above are bounded, we know that
h
i
E E [P21 (zi ; zj )jzi ]2 = O(h2 ):
(28)
Note that
"
xj xi
sj
r (xi ) E (R1nij jxi ) = r (xi ) E
K(
)
k
h
h
Z
K (ul ) [fw (wj + hul )
fw (wj )] dul xi
ul
#
which is exactly E[P22 (zi ; zj )jzi ] : Therefore, Equation (26) is equal to the following
n
1X
[E [P21 (zi ; zj )jzi ]
n
E [P2 (zi ; zj )]] + op
i=1
n
1X
[E [P21 (zi ; zj )jzi ]
n
=
E [P21 (zi ; zj )]] + op
i=1
1
p
n
1
p
n
;
;
(29)
where the last equality follows by Equation (27). By Equation (28) and Lindeberg–Levy central
limit theorem,
n
1 X
p
[E [P21 (zi ; zj )jzi ]
n
i=1
p
E [P21 (zi ; zj )]] ! 0;
which implies that Equation (26) is op ( p1n ):
Similarly
1
n(n
1)
n
n
X
X
[r (xi ) R2nij
i=1 j=1;j6=i
1
r (xi ) E (R2nij jxi )] = op ( p ):
n
Then Equation (23) is proved.
Step 3 (in‡uence term): Taking results from Step 1 and 2, we know that the in‡uence
12
term for Equation (13) is
n
sj
1X
r(xi )E 2k+1 K
n
h
xj
h
i=1
+
1
n(n
1)
n
n
X
X
xj
xi
wl
wj
h
xj
xi
E
n
1
fw (wj ) +
xj
i=1
2
=
1
n(n
1)
n
n
X
X
i=1 j=1;j6=i
3
1
=
n(n
E
xj
1
sj K
hk
r(xi )
1)
xi
i=1 j=1;j6=i
xj
Easy the see that the in‡uence term is
xi
1
+
wl
E
fw (wj ) xi
h
2
+
3;
xi
wj
h
xi
h
fw (wj )
h
:
) xi ;
1
sj K
hk
xj
xi
fw (wj ) xi
h
1
E(sj jwj )K
hk
r(xi )
1
E(sj jwj )K
hk
xj
K(
fw (wj )
h
n
n
X
X
xi
h
xj
1
E(sj jwj )K
hk
1
E(sj jwj )K
hk
sj
1X
r(xi )E 2k+1 K
n
h
=
) xi
h
fw (wj ) xi
h
Let
K(
1
sj K
hk
r(xi )
i=1 j=1;j6=i
1
sj K
hk
E
xi
xj
xi
h
fw (wj ) xi
fw (wj )
:
we will discuss the property of them one
by one.
For
1;
the proof is very similar to the proof of the second conclusion in Lemma 4.1. Note
that
1
E(
1)
=
E
1
r(xi )sj
K
h2k+1
xj
xi
K(
h
wl
wj
h
)
(30)
n
=
1X
fr(xi )fx (xi )E [ si fw (wi )j xi ]
n
E [r(xi )si fx (xi )fw (wi )] + Rni
E(Rni )g ;
i=1
where
Rni = E
r(xi )sj
K
h2k+1
xj
xi
h
K(
wl
wj
h
13
) xi
E [ r(xi )si fx (xi )fw (wi )j xi ] :
;
Easy to see that E jRni j2 = O(h2 ); since xi is i.i.d. across i;
n
1 X
p
[Rni
n
p
E(Rni )] ! 0:
i=1
So the in‡uence term for
xj
r(xi )sj
K
h2k+1
E
xi
h
is
1
K(
wl
wj
h
n
) +
1X
fr(xi )fx (xi )E [ si fw (wi )j xi ]
n
E [r(xi )si fx (xi )fw (wi )]g :
i=1
(31)
Since the kernel function used here is of order p; and by assumption that fw ;E(si jxi ) are p-th
order di¤erentiable and p-th derivatives are bounded, we know that
E
r(xi )sj
K
h2k+1
xj
xi
K(
h
wl
wj
h
) = E [r(xi )si fx (xi )fw (wi )] + O(hp ):
1
Therefore, by assumption n 2 hp ! 0; Equation (31) is equal to
n
1X
E [r(xi )si fx (xi )fw (wi )] +
fr(xi )E [ fx (xi )si fw (wi )j xi ]
n
1
p
n
E [r(xi )si fx (xi )fw (wi )]g + op
i=1
n
1X
r(xi )E [ si fx (xi )fw (wi )j xi ] + op
n
=
1
p
n
i=1
Sum up the results for
1
so far,
1
:
is equal to
n
1X
r(xi )E [ si fx (xi )fw (wi )j xi ] + op
n
i=1
For
2
=
=
1
p
n
:
(32)
2;
1
n(n
1)
2
n(n
1)
n
n
X
X
r(xi )
i=1 j=1;j6=i
n
X1
n
X
xj
1
sj K
hk
xi
h
n
P3 (zi ; zj )
i=1 j=i+1
fw (wj )
1X
1
r(xi )E k sj K
n
h
i=1
E
xj
xj
1
sj K
hk
xi
h
h
fw (wj ) xi ;
where
P3 (zi ; zj ) =
1
[r(xi )sj fw (wj ) + r(xj )si fw (wi )] K
2hk
14
xi
xj
xi
h
:
fw (wj ) xi
(33)
Let
P31 (zi ; zj ) =
1
r(xj )si fw (wi )K
hk
xj
P32 (zi ; zj ) =
1
r(xi )sj fw (wj )K
hk
xi
xi
h
xj
h
;
;
then
P3 (zi ; zj ) =
Notice that the structure of
2
1
[P31 (zi ; zj ) + P32 (zi ; zj )] :
2
is the same as Equation (26) and P3 ; P31 ; P32 here are correspond-
ing to P2 ; P21 ; P22 respectively: By the results in Step 2 that Equation (26) is equal to Equation
(29), similarly,
2
is equal to
n
1X
fE [ P31 (zi ; zj )j zi ]
n
E [P31 (zi ; zj )]g + op
i=1
1
p
n
:
The above expression could be rewritten as
n
r(xj )
1X
K
si fw (wi )E
n
hk
=
1
n
i=1
n
X
i=1
fr(xi )si fw (wi )fx (xi )
xj
xi
h
zi
E [P31 (zi ; zj )] ;
E [r(xi )si fw (wi )fx (xi )] + Rni
E (Rni )g ;
where
Rni = si fw (wi )E
r(xj )
K
hk
xj
xi
h
zi
r(xi )si fw (wi )fx (xi ):
2 ) = O(h2 ); by i.i.d. assumption on z and Lindeberg Levy CLT;
Easy to see that E(Rni
i
n
1 X
p
[Rni
n
i=1
So
2
p
E (Rni )] ! 0:
is equal to
n
1X
fr(xi )si fw (wi )fx (xi )
n
E [r(xi )si fw (wi )fx (xi )]g + op
i=1
15
1
p
n
(34)
For
3;
it has the same structure as
2:
Similarly, one could get that
3
n
1X
fr(xi )E (si jwi ) fw (wi )fx (xi )
n
E [r(xi )si fw (wi )fx (xi )]g + op
i=1
At this stage, from Equation (32), (34), and (35), we could get that
1
+
is equal to
1
p
n
:
(35)
+
3
is equal to
2
n
1X
fr(xi )si fw (wi )fx (xi ) + r(xi )fx (xi )E [ si fw (wi )j xi ]
n
(36)
i=1
+r(xi )E (si jwi ) fw (wi )fx (xi )
1
p
n
2E [r(xi )si fw (wi )fx (xi )]g + op
:
Sum up the results in Step 1, 2, and 3, we know that the Equation (13) is equal to Equation (36),
which is the conclusion of this lemma.
Lemma 4.3 Adopt the same notation and assumptions as in Lemma 4.2: Then
n
X
1
(n
1)
j=1;j6=i
1
sj fbw (wj ) k K
h
xj
xi
E [si fw (wi )jxi ] fx (xi ) = Op (hp ) + Op
h
"
1
n1
" hk
1
2
#
:
(37)
This Lemma is a complement to Lemma 4.2. The result will be used to determine the rate of
convergence of the residual terms after decomposing our estimator.
Proof of Lemma 4.3. The proof of this lemma use some results in lemma 4.2.
First note that
1
(n
=
1)
1
(n
1)
n
X
j=1;j6=i
n
X
j=1;j6=i
1
sj fbw (wj ) k K
h
sj fbw (wj )
1
K
hk
b [si fw (wi )jxi ] fbx (xi )
+E
Under Assumption n1
" h2k+2
xj
xi
h
xj
xi
h
E [si fw (wi )jxi ] fx (xi )
b [si fw (wi )jxi ] fbx (xi )
E
b [si fw (wi )jxi ]
fx (xi ) + E
E [si fw (wi )jxi ] fx (xi ):
! 1; from Li and Racine (2007) Chapter 2.3,
b [si fw (wi )jxi ]
sup E
E [si fw (wi )jxi ] = Op
16
"
1
n1
" hk
1
2
#
:
From Silverman (1978), sup fbx (xi )
n
X
1
(n
1)
j=1;j6=i
fx (xi ) is also Op
xj
1
sj fbw (wj ) k K
h
h
1
n1
xi
" hk
1
2
i
. So we only need to focus on
b [si fw (wi )jxi ] fbx (xi )
E
h
(38)
which is
n
X
1
1)2 h2k+1
(n
n
X
sj K
xj
xi
h
j=1;j6=i l=1;l6=j
K(
wl
wj
h
1
(n 1)hk
)
n
X
xj
sj fw (wj )K
xi
h
j=1;j6=i
Drop l = i in the …rst part of the above expression to make it look more like a U-statistics. This
h
i
1
term is of the order Op n2 h12k+1 which is op ( n1 1" hk ) 2 by some simple calculation, so this term
could be ignored. After dropping one term, the above term becomes
(n
n
X
1
1)(n 2)h2k+1
n
X
xj
sj K
h
j=1;j6=i l=1;l6=i;j
1
(n 1)hk
n
X
sj fw (wj )K
xj
xi
K(
wl
wj
h
)
(39)
:
h
j=1;j6=i
xi
As already discussed in the Step 1 of Lemma 4.2,
(n
is
p
n
X
1
1)(n 2)h2k+1
n
X
sj K
j=1;j6=i l=1;l6=i;j
xj
xi
h
K(
wl
wj
h
)
n-equivalent to Equation (16). For the convenience of reading, Equation (16) is rewritten as
follows:
E
+
sj
h2k+1
K
xj
xi
h
K(
wl
wj
h
) xi +
1
n
1
n
X
j=1;j6=i
1
sj K
hk
xj
xi
h
xj xi
xj xi
1
1
E(sj jwj )K
fw (wj ) E k sj K
fw (wj ) xi
h
h
hk
h
xj xi
1
1
E k E(sj jwj )K
fw (wj ) xi + Rnij E(Rnij jxi ) + op p
h
h
n
17
fw (wj ) (40)
;
:
Substitute Equation (40) into Equation (39), then Equation (39) becomes
+
xj
sj
K
2k+1
h
E
n
1
wl
wj
xj
xi
xj
1
sj K
hk
fw (wj )
h
1
p
n
E(Rnij jxi ) + op
E
) xi
h
1
E(sj jwj )K
hk
j=1;j6=i
+Rnij
K(
h
n
X
1
xi
xi
h
1
E(sj jwj )K
hk
E
(41)
fw (wj ) xi
xj
xi
h
fw (wj ) xi
:
We will discuss each term in Equation (41).
By the assumption that fw ;E(si jxi ); fx are p-th order di¤erentiable and p-th derivatives are
bounded, and the kernel function we use is of p-th order, we know that
E
sj
K
2k+1
h
xj
h
sj
K
hk
E
xi
xj
K(
wl
wj
h
) xi = E [ si fx (xi )fw (wi )j xi ] + Op (hp );
xi
fw (wj ) xi = E [ si fx (xi )fw (wi )j xi ] + Op (hp );
wl
wj
h
so
E
sj
K
2k+1
h
xj
xi
h
K(
h
) xi
E
xj
sj
K
hk
xi
h
fw (wj ) xi = Op (hp ):
(42)
Let
=
1
n
1
n
X
j=1;j6=i
1
E(sj jwj )K
hk
xj
xi
h
fw (wj )
E
1
E(sj jwj )K
hk
xj
xi
h
fw (wj ) xi
By i.i.d. assumption, j-th observation is independent of l-th observation when j 6= l: So
E(
2
jxi ) =
1
n
1
var
"
1
E(sj jwj )K
hk
18
xj
xi
h
2
fw (wj )
#
xi :
:
By
E
=
=
Z
"
1
hk
1
E(sj jwj )K
hk
1
E(sj jwj )K
hk
wj
Z
uj ;vj
xj
2
xi
fw (wj )
h
xj
xi
h
xi
#
2
fw (wj )
fw (wj )dwj
[E(sj jxi + huj ; vj )K (uj ) fw (xi + huj ; vj )]2 fw (xi + huj ; vj )duj dvj = Op (
1
);
hk
we know
E(
2
According to Markov inequality,
= Op
Similarly, let
R=
"
1
1
(nhk ) 2
n
1 X
[Rnij
n
R = op
"
#
:
:
(43)
E(Rnij jxi )] ;
j=1;j6=i
and we can get that
1
nhk
jxi ) = Op
1
1
(nhk ) 2
#
:
(44)
Equation (38) is equal to Equation (41). The order
(41) could be obtained from
" of Equation
#
Equation (42), (43) and (44), which is Op (hp ) + Op
1
1
(nhk ) 2
:
Corollary 4.4 Adopt the same notation and assumptions as in Lemma 4.2: Then
b
E
"
si
fb(vi jxi )
jxi
#
"
#
s
1
i
p
fbx (xi ) E
jxi fx (xi ) = Op (h ) + Op
+ Op ( 1
1
f (vi jxi )
n
(nhk ) 2
1
" h2(k+1)
): (45)
This Corollary is a direct result of Lemma 4.3. It is useful, since our estimator includes a
nonparametric estimate of conditional density in the denominator.
19
Proof of Corollary 4.4.
b
E
b
= E
"
"
si
fb(vi jxi )
si
fb(vi jxi )
b
+ E
#
jxi fbx (xi )
E
#
jxi fbx (xi )
si
jxi
f (vi jxi )
E
b
E
si
jxi fx (xi )
f (vi jxi )
h
si
si
b
jxi fbx (xi ) + E
jxi fbx (xi )
f (vi jxi )
f (vi jxi )
si
jxi
f (vi jxi )
fx (xi ):
h
All terms except the …rst term are easily seen to be Op (n1
"
" hk )
1
2
#
i
: For the …rst term
si
b
jxi fbx (xi ) E
jxi fbx (xi )
b
f (vi jxi )
f (vi jxi )
n
n
X
X
sj fbx (xj ) 1
sj fx (xj ) 1
xj xi
xj xi
1
1
=
K
K
k
k
b
n 1
h
n 1
fw (wj ) h
h
h
j=1;j6=i fw (wj )
j=1;j6=i
h
i
n
sj fbx (xj ) fx (xj ) 1
X
xj xi
1
=
K
n 1
fw (wj )
h
hk
j=1;j6=i
h
i
bw (wj ) fw (wj )
n
s
f
(x
)
f
X
j
x
j
xj xi
1
1
+
K
n 1
fw2 (wj )
h
hk
j=1;j6=i
h
ih
i
bx (xj ) fx (xj ) fbw (wj ) fw (wj )
n
s
f
X
j
xj xi
1
1
+
K
n 1
fw2 (wj )
h
hk
j=1;j6=i
h
i2
b
b
n
s
f
(x
)
f
(w
)
f
(w
)
X
j x j
w
j
w
j
xj xi
1
1
+
K
:
k
n 1
h
h
fw2 (wj )fbw (wj )
b
E
si
i
fx (xi )
(46)
(47)
(48)
(49)
j=1;j6=i
According the results in Lemma 4.3, and by seeing that Equation (46) is exactly Equation (38),
h
i
1
Equation (46 is of order Op (hp ) + Op (nhk ) 2 : For the same reason, Equation (47) is also of
i
h
1
order Op (hp ) + Op (nhk ) 2 : From Silverman (1978),
sup fbx (xj )
sup fbw (wj )
h
fx (xj ) = O (n1
h
fx (wj ) = O (n1
1
2
" k
h )
" k+1
h
)
i
1
2
;
i
;
for any " > 0; so together with the bounds condition on each term in Equation (48) and (49), we
know that Equation (48) and (49) are of the order Op ( n1
20
1
" h2k
) and Op ( n1
1
" h2(k+1)
) respectively.
Combine results so far, the conclusion is proved.
We will …rst derive the property of
Proof of Theorem 2.6.
1
n
n
X
i=1
into several components as follows
b (xi ) =
1
=
b b
E
h1i xi
b ( gb1i j xi )
E
b
b e
E
h1i xi
E ( ge1i j xi )
where
Ri =
b
b e
E
h1i xi
=
E e
h1i xi
h
b
b e
E
h1i xi
b b
E
ge1i xi
E e
h1i xi
b b
E
ge1i xi
[E ( ge1i j xi )]2
So
1
nhk
n
X
1
n
i=1
+ Op ( n2
1
" h4(k+1)
h
b b
E
ge1i xi
[E ( ge1i j xi )]2
1
E( g
e1i jxi )
Notice that
+ Ri ;
8
b
>E
h1i xi
n >
<b e
X
> E ( ge1i j xi )
i=1 >
:
E e
h1i xi
(50)
is bounded, Ri is of order Op (h2p ) +
); which op ( p1n ); by Assumption that nh2p ! 0; and n1
b (xi ) is equal to
1
1
n
i
E ( ge1i j xi )
h
i
i2
b
b e
b b
h1i xi E
E ( ge1i j xi ) E
ge1i xi
E ( ge1i j xi )
+
:
[E ( ge1i j xi )]2
According to Corollary 4.4, and assumption that
Op
b (xi ): It could be divided
1
h
b b
E
ge1i xi
2
[E ( ge1i j xi )]
9
i>
=
E ( ge1i j xi ) >
>
>
;
+ op
1
p
n
" h4k+4
:
! 1.
(51)
b
b e
h1i xi
n E
1X
n
E ( ge1i j xi )
i=1
=
=
n
n
X
X
xj xi
Dj Yj 1
1
1
K
k
n (n 1)
E ( ge1i j xi ) fb(vj jxj ) h
h
i=1 j=1;j6=i
(
n
n
X
X
Dj Yj fbx (xj ) 1
xj xi
1
1
K
k
n (n 1)
E ( ge1i j xi ) fxv (xj ; vj ) h
h
i=1 j=1;j6=i
h
i
Dj Yj fx (xj ) fbxv (xj ; vj ) fxv (xj ; vj ) 1
xj xi
1
K
2 (x ; v )
k
E ( ge1i j xi )
fxv
h
h
j j
21
(52)
+ Rij
9
=
;
;
where
Rij
=
h
Dj Yj fbx (xj ) fbxv (xj ; vj )
fxv (xj ; vj )
i2
2 (x ; v )fb (x ; v )
E ( ge1i j xi ) fxv
j j xv j j
h
ih
Dj Yj fbx (xj ) fx (xj ) fbxv (xj ; vj )
2 (x ; v )
E ( ge1i j xi ) fxv
j j
1
K
hk
xj
xi
h
i
fxv (xj ; vj ) 1
K
hk
xj
xi
h
:
Consider the residual term …rst, again from Silverman (1978) and those bound conditions, we
know that
sup jRij j = Op
1
the second equality follows by Assumption that n1
1
p
n
= op
n1 " h2k
" h4k+4
;
(53)
! 1. Apply Lemma 4.2 (by letting
sj = s5j ; r(xi ) = r5 (xi )) on the …rst term in Equation (52)4 ,
n
n
X
X
Dj Yj fbx (xj ) 1
1
1
K
n (n 1)
E ( ge1i j xi ) fxv (xj ; vj ) hk
xj
xi
;
h
i=1 j=1;j6=i
we know that is equal to
n
1X
n
i=1
2E(h1i jxi )
h1i
+
E(g1i jxi )
E(g1i jxi )
2E
E(h1i jxi )
E(g1i jxi )
+ op
1
p
n
;
which by further simpli…cation is:
n
1X
h1i
2E (Y1 jxi ) +
n
E(g1i jxi )
2E (Y1 ) :
(54)
i=1
The second component in Equation (52) can be rewritten as
n
n
X
X
1
n (n 1)
i=1 j=1;j6=i
"
Dj Yj fx (xj )fbxv (xj ; vj ) 1
K
2 (x ; v ) hk
E ( ge1i j xi ) fxv
j j
Dj Yj fx (xj )
1
K
E ( ge1i j xi ) fxv (xj ; vj ) hk
xj
xi
h
xj
xi
h
(55)
:
Apply Lemma 4.2 (by letting sj = s6j ; r(xi ) = r6 (xi )) on the …rst term in Equation (55), we could
4
By assuming v in Lemma 4.2 is empty.
22
get that it is equal to
n
h1i
1X
E(h1i jxi ; vi )
E (Y1 jxi ) +
+
n
E(g1i jxi )
E(g1i jxi )
2E (Y1 ) + op
i=1
1
p
n
:
(56)
Apply Lemma 4.1 on the second term in Equation (55), we could get that it is equal to
n
E (Y1 ) +
h1i
1X
E (Y1 jxi ) +
n
E(g1i jxi )
2E (Y1 ) + op
i=1
1
p
n
:
(57)
From Equation (56), (57), we know that Equation (55) is equal to
n
1 X E(h1i jxi ; vi )
n
E(g1i jxi )
1
p
n
E (Y1 ) + op
i=1
Combine Equation (53), (54), (58), and (52), we can get that
1
n
n E
b
X
i=1
n
1X
E(Y1 ) +
n
i=1
h1i
E (g1i jxi )
E (h1i jxi ; vi )
+ 2 [E (Y1 jxi )
E (g1i jxi )
:
(58)
b
e
h1i xi
E( g
e1i jxi )
is equal to
E(Y1 )] + op
1
p
n
:
(59)
Use the same strategy on another term in Equation (51),
=
h
i
b b
n E e
h
x
E
g
e
x
E
(
g
e
j
x
)
X
1i
i
i
1i
i
1i
1
n
[E ( ge1i j xi )]2
i=1
8
9
e
b b
n
E e
h1i xi =
1 X < E h1i xi E ge1i xi
n
i=1
:
[E ( ge1i j xi )]2
from which we could get that it is equal to
n
1X
n
i=1
E (h1i jxi ) g1i
[E (g1i jxi )]2
Sum up the results for
E ( ge1i j xi ) ;
E (h1i jxi ) E (g1i jxi ; vi )
+ [E (Y1 jxi )
[E (g1i jxi )]2
1
n
n
X
i=1
E(Y1 )] + op
1
p
n
:
(60)
b (xi ): From Equation (50), (51), (59), (60), we know that
1
23
1
n
n
X
i=1
b (xi ) is equal to the di¤erence between Equation (59) and Equation(60),
1
n
1X
E(Y1 ) +
n
i=1
+
h1i
E (g1i jxi )
E (h1i jxi ; vi )
E (g1i jxi )
E (h1i jxi ) E (g1i jxi ; vi )
+ [E (Y1 jxi )
[E (g1i jxi )]2
E (h1i jxi ) g1i
[E (g1i jxi )]2
E(Y1 )] + op
1
p
n
(61)
:
Equation (61) is very similar to Equation (9). Compared to Equation (9), the additional term
E(h1i jxi ; vi ) exists is because we also estimate fxv nonparametrically.
n
X
b (xi ), it is equal to
Do the similar analysis for n1
2
i=1
n
E(Y0 ) +
1X
n
i=1
h2i
E (g2i jxi )
E (h2i jxi ; vi )
E (g2i jxi )
E (h2i jxi ) E (g2i jxi ; vi )
+
+ [E (Y0 jxi )
[E (g2i jxi )]2
E (h2i jxi ) g2i
[E (g2i jxi )]2
E(Y0 )] + op
1
p
n
(62)
:
From Equation (61) and (62), the conclusion is proved.
Lemma 4.5 ai ; bt are random vectors that satisfy Assumption 2.10. wit are random vectors and
wit ? wit0 jai ; for t 6= t0 ; wit ? wi0 t jbt for i 6= i0 ; wit ? wi0 t0 for i 6= i0 ; t 6= t0 : h(ai ; bt ; wit ) are a real
function that the …rst and second moment exist, and E h(ai ; bt ; wit )2 = o(n). E[h(ai ; bt ; wit )] =E[h(ai0 ; bt0 ; wi0 t0 )]
for any i; t; i0 ; t0 : T ! 1 as n ! 1: Then
n T
1 XX
h(ai ; bt ; wit )
nT
i=1 t=1
is equal to
E [h(ai ; bt ; wit )]+
n T
1 XX
[E [h(ai ; bt ; wit )jai ] + E [h(ai ; bt ; wit )jbt ]
nT
2E [h(ai ; bt ; wit )]]+op
i=1 t=1
1
p
T
wit are heterogeneous across t; but E(h) are assumed the same across t: It is not strange as it
looks; one typical case that satis…es those is that E(h) = 0 for any i; t:
24
:
Proof of Lemma 4.5. Let
n T
1 XX
[h(ai ; bt ; wit )
Q=
nT
E(h(ai ; bt ; wit )jai )
E(h(ai ; bt ; wit )jbt ) + E(h(ai ; bt ; wit ))] ; (63)
i=1 t=1
then it is equivalent to show that Q = op ( p1T ):
n T
n
T
1 XXX X
E[Q ] = 2 2
E [(h
n T
0
0
2
E(hjai )
E(hjbt ) + E(h)) (h
E(hjai0 )
E(hjbt0 ) + E(h))] :
i=1 t=1 i =1 t =1
For i 6= i0 ; t 6= t0 ; the term inside summation is zero. For the case when only one index is equal to
the other one, i.e., i = i0 , t 6= t0 ; since
E [h(ai ; bt ; wit )h(ai ; bt0 ; wit0 )] = E [E [h(ai ; bt ; wit )h(ai ; bt0 ; wit0 )jai ]]
= E [E [h(ai ; bt ; wit )jai ]E[h(ai ; bt0 ; wit0 )jai ]] ;
the term inside summation is zero again. So we can rewrite E[Q2 ] as
E[Q2 ] =
n T
1 XX h
E (h
n2 T 2
E(hjai )
i=1 t=1
By assumption E h2 = op (n); so E[Q2 ] = op
1
T
i
E(hjbt ) + E(h))2 :
; which implies Q = op
p1
T
:
Lemma 4.6 Under the same assumptions in Lemma 4.5 and Assumption 2.9. Further assume
var(E [h(ai ; bt ; wit )jai ])
M , for all i; where M is a …nite positive number. Then
n T
1 XX
[E [h(ai ; bt ; wit )jai ] + E [h(ai ; bt ; wit )jbt ]
nT
2E [h(ai ; bt ; wit )]]
i=1 t=1
is equal to
1
T
T
X
[E [h(ai ; bt ; wit )jbt ]
E [h(ai ; bt ; wit )]] + op
p1
T
:
t=1
Proof of Lemma 4.6.
n T
1 XX
[E [h(ai ; bt ; wit )jai ] + E [h(ai ; bt ; wit )jbt ]
nT
i=1 t=1
25
2E(h(ai ; bt ; wit ))]
First by assumption that ! it jbt is i.i.d across i; we know that
E [h(ai ; bt ; wit )jbt ] = E [h(ai0 ; bt ; wi0 t )jbt ] ;
which gives
n T
1 XX
[E [h(ai ; bt ; wit )jbt ]
nT
1
T
=
i=1 t=1
T
X
[E [h(ai ; bt ; wit )jbt ]
E(h(ai ; bt ; wit ))]
(64)
E(h(ai ; bt ; wit ))]
t=1
For the other part, note that
n T
1 XX
[E [h(ai ; bt ; wit )jai ] E(h(ai ; bt ; wit ))]
nT
i=1 t=1
" n
#
T
1X 1X
[E [h(ai ; bt ; wit )jai ] E(h(ai ; bt ; wit ))] ;
T
n
=
t=1
i=1
where E[h(ai ; bt ; wit )jai ] is independent across i:
2
=
!2 3
n
1X
E4
[E [h(ai ; bt ; wit )jai ]
n
E(h(ai ; bt ; wit ))]
i=1
n
1 X h
E ([E [h(ai ; bt ; wit )jai ]
n2
E(h(ai ; bt ; wit ))])2
i=1
i
5
M
;
n
by Markov’s inequality,
n
1X
[E [h(ai ; bt ; wit )jai ]
n
1
E(h(ai ; bt ; wit ))] = Op ( p );
n
i=1
which gives that
n T
1 XX
[E [h(ai ; bt ; wit )jai ]
nT
i=1 t=1
1
E(h(ai ; bt ; wit ))] = Op ( p ):
n
(65)
Combining Equation (64) and Equation (65), the lemma follows.
Lemma 4.7 Denote
n
= (A1n ; B1n ; A2n ; B2n )0 ; a 4-by-1 vector, where A1n ; B1n ; A2n ; B2n are
26
random variables that evolve as n goes to in…nity. Assume that
n
converge in probability to
= (0; B 1 ; 0; B 2 )0 ; where B 1 6= 0; B 2 6= 0; and
p
where
d
n[
] ! N (0; );
n
is a positive de…nite matrix
0
2
A1
B
B
B
=B
B
B
@
A1 B1
A1 A2
A1 B2
:
2
B1
B1 A2
B1 B2
:
:
2
A2
A2 B2
:
:
:
Then
p
n
A1n
B1n
A2n
B2n
d
0;
!N
2
B2
2
A1
2
B1
1
C
C
C
C:
C
C
A
2
2 A1 A2
+ A22
B1B2
B2
!
:
Proof. The Lemma follows immediately after delta method.
Proof of Theorem 2.8. First note that
T
n
1 XX
nT
1it
!
T
n
1 XX
nT
2it
!
p
t=1 i=1
p
t=1 i=1
1;
2:
we will show the Equation (3) is equal to
1
nT
1
nT
T X
n
X
t=1 i=1
T X
n
X
1it
1it
t=1 i=1
1
nT
1
nT
T X
n
X
t=1 i=1
T X
n
X
2it
1
+ op ( p ):
n
2it
t=1 i=1
It is enough to show that
T
n
1 X X Dit Yit
nT
t=1 i=1
E(e
ai + ebt + Y1 )
fbvt (vit )
27
T
n
1 XX
=
nT
t=1 i=1
1it
1
+ op ( p ):
n
To this end,
T
n
ai + ebt + Y1 )
1 X X Dit Yit E(e
nT
fbvt (vit )
t=1 i=1
(66)
T
n
ai + ebt + Y1 )
1 X X Dit Yit E(e
nT
fvt (vit )
=
t=1 i=1
E(e
ai + ebt + Y1 )
Dit Yit
fbvt (vit )
fv2t (vit )
where
Dit Yit
Rnit =
Again, by Silverman (1978),
1
nT
E(e
ai + ebt + Y1 )
fv2t (vit )fbvt (vit )
T X
n
X
t=1 i=1
fbvt (vit )
1
jRnit j = op
fvt (vit ) + Rnit ;
n1
" h2
fvt (vit )
; which is op
2
:
p1
n
; by n1
" h4
! 1:
Generalize Lemma 4.1 a little bit, by Assumption 4.6, not hard to see that,
n
ai + ebt + Y1 )
1 X Dit Yit E(e
n
fv2t (vit )
fbvt (vit )
i=1
is equal to
n
1XE
n
i=1
h
Yit
E(e
ai + ebt + Y1 ) Dit vit
fvt (vit )
Substitute this back to Equation (66), we could get
1
nT
i
fvt (vit )
+ op
T X
n
X
Dit (Yit
t=1 i=1
T
n
1 X X Yit
nT
t=1 i=1
which is
1
nT
T X
n
X
E(e
ai + ebt + Y1 ) Dit
1it
+ op
p1
n
E
h
Yit
fvt (vit )
1
p
n
:
E(e
ai +e
bt +Y1 ))
b
fvt (vit )
E(e
ai + ebt + Y1 ) Dit jvit
: For the same reason
t=1 i=1
T
n
T
n
1 X X (1 Dit )Yit
1 XX
=
nT
nT
fbvt (vit )
t=1 i=1
t=1 i=1
28
2it
+ op
1
p
n
:
i
is equal to
+ op
1
p
n
;
Therefore, we know the Equation (23) is equal to
1
nT
1
nT
T X
n
X
1it
t=1 i=1
T X
n
X
1it
T X
n
X
1
nT
2it
t=1 i=1
T X
n
X
1
nT
t=1 i=1
+ op
1
p
n
:
2it
t=1 i=1
Applying Lemma 4.6 on this expression, it is equivalent to
1
T
T
X
E
t=1
1
nT
h
e
1it j bt ; bt
T X
n
X
i
1
T
T
X
E
t=1
1
nT
1it
t=1 i=1
h
e
2it j bt ; bt
T X
n
X
i
1
p
T
+ op
;
2it
t=1 i=1
we have now shown that
1
nT
1
nT
1
nT
=
1
nT
1
T
=
T X
n
X
t=1 i=1
T X
n
X
t=1 i=1
T X
n
X
t=1 i=1
T X
n
X
t=1 i=1
T
h
X
E
t=1
1
nT
Dit Yit
fbvt (vit )
1
nT
t=1 i=1
T X
n
X
Dit
fbvt (vit )
1
nT
1
nT
1it
1
nT
1it
e
1it j bt ; bt
T X
n
X
t=1 i=1
1it
T X
n
X
(1
i
t=1 i=1
T
n
XX
Dit )Yit
fbvt (vit )
E(e
ai + ebt + Y1 ) + E(e
ai + ebt + Y0 )
(1 Dit )
fbvt (vit )
2it
t=1 i=1
T X
n
X
+ op
1
p
n
2it
t=1 i=1
T
h
i
X
1
ebt
E
j
b
;
2it
t
T
t=1
T X
n
X
1
2it
nT
t=1 i=1
+ op
1
p
T
;
which gives the conclusion by applying Lemma 4.7.
References
[1] Li, Q., and J. S. Racine (2007), "Nonparametric Econometrics: Theory and Practise," Princeton University Press.
29
[2] Powell, J. L., J. H. Stock, and T. M. Stoker (1989), "Semiparametric Estimation of Index
Coe¢ ents," Econometrica, 57, 1403-1430.
[3] Silverman, B. W. (1978), "Weak and Strong Uniform Consistency of the Kernel Estimate of
a Density Function and its Derivatives," Annals of Statistics, 6, 177-184.
30