Semiparametric Estimation of Fixed
E¤ects Panel Data Models with Smooth
Coe¢ cients
Yiguo Sun
Department of Economics
University of Guelph
Guelph, ON, Canada N1G2W1
yisun@uoguelph.ca
Raymond J. Carroll
Department of Statistics, Texas A&M University
College Station, TX 77843-3134,
carroll@stat.tamu.edu
Qi Li
Department of Economics, Texas A&M University
College Station, TX 77843-4228,
qi@econmail.tamu.edu
Abstract
In this paper we consider the problem of estimating semiparametric …xede¤ects panel data models with smooth coe¢ cients by local linear regression
approach. We show that the proposed estimator has the usual nonparametric
convergence rate and is asymptotically normally distributed under regular conditions. A modi…ed least-squared cross-validatory method is used to …nd the
optimal bandwidth automatically. Moreover, we propose a test statistic for testing the null hypothesis of random e¤ects against …xed e¤ects for semiparametric
panel data regression models with smooth coe¢ cients. Monte Carlo simulations
are used to study …nite sample performance of the proposed estimator and test.
Some key words: Panel data; Local least squares; Smooth coe¢ cients; Bootstrap.
Short title: Fixed E¤ects Panel Data Models with Smooth Coe¢ cients
2
1
Introduction
Panel data traces information on each individual unit across time. Economists
often …nd that they could overcome econometrics di¢ culties and extract more
economic inferences using panel data, which would not be possible using pure
time-series data or cross-section data. With the increased availability of panel
data both theoretical and applied econometrics work in panel data analysis has
become ever popular in the recent years.
To avoid imposing inadequate parametric panel data model, many econometricians and statisticians have been working on theories of nonparametric and
semiparametric panel data regression models. Among many contributions, we
name a few here: Ke and Wang (2001), Li and Stengos (1996), and Ullah and
Roy (1998) for semiparametric panel data models with random e¤ects; Henderson and Ullah (2005), Lin and Carroll (2000, 2001, 2006), Lin, Wang, Welsh and
Carroll (2004), Lin and Ying (2001), Ruckstuhl, Welsh and Carroll (1999), Wu
and Zhang (2002) and Wang (2003) for estimation of nonparametric panel data
models with random e¤ects. Estimation of these types of models are appropriate
when the individual e¤ect is independent of the regressors.
However, random e¤ects estimators are inconsistent if the true model is one
with …xed e¤ects, i.e. individual e¤ects which are correlated with the regressors.
Indeed, economists often view the assumptions for the random e¤ects model as
being unsupported by the data. Su and Ullah (2006) consider a …xed e¤ects
partially linear panel data model. This paper considers …xed e¤ects panel data
models with smooth coe¢ cients. The model is assumed to apply to situations
1
where we see smooth transitive changes across sections and/or across time.
The remainder of the paper is organized as follows: Section 2 introduces
a semiparametric …xed-e¤ects panel data regression model with smooth coe¢ cients and estimation methodology. In Section 3, we propose a nonparametric
estimator for unknown smoothing coe¢ cients and derive its asymptotic results.
In Section 4 we propose a test statistic for testing for random e¤ects versus …xed
e¤ects in semiparametric panel data models with smooth coe¢ cients. Section 5
examines …nite sample properties with a small Monte Carlo study. Finally, Section 6 concludes the paper. We delay detailed mathematical proofs to Appendix
A for …xed-e¤ects estimation and Appendix B for random-e¤ects estimation.
2
Fixed E¤ects Semiparametric Panel Data Models with Smooth Coe¢ cients
We consider the following semiparametric …xed-e¤ects panel data regression
model with smooth coe¢ cients
T
Yi;t = Xi;t
(Zi;t ) +
i
+
i;t ; (i
= 1; :::; n; t = 1; :::; m)
(1)
where the covariate Zi;t = (Zi;t;1 ; :::; Zi;t;q )T is of dimension q, Xi;t = (Xi;t;1 ; :::; Xi;t;k )T
is of dimension k; ( ) = (
1
( );
;
T
k
( )) contains k unknown functions, and
all other variables are scalars. The random errors
with a zero mean, …nite variance
and t. Further,
i
2
v
i;t
are assumed to be i.i.d.
and independent of Zi;t and Xi;t for all i
has a …nite mean and variance. We allow
i
to be correlated
with Zi;t and/or Xi;t with an unknown correlation structure. Hence, model (1)
is a …xed-e¤ects model. Alternatively, when
2
i
is uncorrelated with Zi;t and
Xi;t model (1) is a random-e¤ects model.
For a given …xed e¤ects model, there are many ways of removing the unknown …xed e¤ects from the model.
Example 1 The usual …rst-di¤ erencing (FD) estimation method could deduct
one equation from another to remove the time-invariant …xed e¤ ects. For example, deducting equation for time t from that for time t
t = 2;
1; we have for
;m
y~i;t = yi;t
yi;t
1
T
= Xi;t
(Zi;t )
T
Xi;t
1
(Zi;t
1)
+ v~i;t ; v~i;t = vi;t
vi;t
or deducting equation for time t from that for time 1; we obtain for t = 2;
y~i;t = yi;t
T
yi1 = Xi;t
(Zi;t )
T
Xi;1
(Zi1 ) + v~i;t ; v~i;t = vi;t
vi;1 ;
1;
(2)
;m
(3)
and so on.
Example 2 The conventional …xed e¤ ects (FE) estimation method, on the
other hand, removes the …xed e¤ ects by deducting each equation from the crosstime average of the system, and it gives for t = 2;
m
T
y~i;t = Xi;t
(Zi;t )
where wi: = m
qt;s =
1
Pm
t=1
;m
m
X
1 X T
T
Xi;s (Zi;s ) + v~i;t =
qt;s Xi;s
(Zi;s ) + v~i;t
m s=1
s=1
wi;t and w
~i;t = wi;t wi: with w being y or v: In addition,
1=m if s 6= t and 1
1=m otherwise, and
Pm
s=1 qt;s
In general cases, we could introduce a constant (m
full rank m
(4)
1 such that P em = 0m
1
= 0 for all t.
1)
m matrix P with
where em is an m
1 vector of ones.
Premultiplying P to the m equations for each i will remove the unknown …xed
3
e¤ects
i.
For the FD estimator, each row of P only takes values of -1, 0,
and 1. Denote P = p01 ;
; p0m
vector. Model (2) assumes pt;t =
t = 1;
;m
m
1
0
m
; where pt = (pt;j )j=1 is an m
;m
is an (m
1; pt;t+1 = 1; and pt;j = 0 for
1: For the FE estimator, P = Sm
1)
1 row
1; pt;t+1 = 1; and pt;j = 0 for j 6= t; t + 1;
1: Model (3) assumes pt;1 =
j 6= 1; t + 1; t = 1;
where Sm
1
1
m matrix containing the last m
Im
1
0
m em em
;
1 rows of the
m identity matrix Im : When m = 2; FD estimator and FE estimator will
be exactly the same; this is not true for m > 2.
Many nonparametric local smoothing methods can be used to estimate the
unknown function ( ). However, for each i; the right-hand of equations (2)-(4)
T
contain linear combination of Xi;t
(Zi;t )
m
t=1
over time. If X contains an in-
tercept term, back…tting algorithm has to be used to recover the unknown functions,which brings not only computational burden but also complicate mathematical proofs.
Therefore, this paper proposes a new way of removing the unknown …xed
e¤ects, motived by least square dummy variable (LSDV) model in parametric
panel data analysis. As the text goes along, we will describe how the proposed
method removes …xed e¤ects by deducting a smoothed version of average across
time from each observation.
Rewriting model (1) in matrix format yields
Y = B (X; (Z)) + D0
where Y = Y1T ;
; YnT
T
and V = v1T ;
T
B (X; (Z)) stacks all Xi;t
(Zi;t ) into an (nm)
4
0
+ V;
; vnT
T
(5)
are (nm)
1 vectors,
1 vector in ascending order of
i …rst, then of t;
is a (nm)
0
=[
1;
T
n]
;
is an n
1 vector, and D0 = In
n matrix with main diagonal blocks being em ; where
em
n
refers to
Kroneker product operation. However, we can not work on this model without
T
further restriction on the …xed e¤ects for cases like Xi;t
(Zi;t ) =
Xi;t;2
2
(Zi;t ) +
; since we can not identify
1
which has been used by Su and Ullah (2006). De…ne
Pn
i=1
where D =
i
=[
2;
Pn
i=1
;
T
n]
1
= 0,
: With
I(n
T
em is a (nm)
1) (n 1)
(6)
(n
1) matrix.
De…ne an m m diagonal matrix KH (Zi ; z) = diag (KH (Zi;1 ; z) ;
; KH (Zi;m ; z))
; n; and a (nm) (nm) diagonal matrix WH (z) = diag(KH (Z1 ; z) ;
where KH (Zi;t ; z) = K H
is a q
i
= 0; we can rewrite (5) as
Y = B (X; (Z)) + D + V;
en
for i = 1; 2;
(Zi;t ) +
( ) in the presents of unknown
…xed e¤ects. Therefore, we impose an identi…cation condition like
restriction
1
1
(Zi;t
z) for all i and t; and H = diag (h1 ;
; hq )
q diagonal smoothing matrix. We then solve the following optimization
problem
min (Y
(Z);
B (X; (Z))
T
D ) WH (z) (Y
B (X; (Z))
D ):
(7)
We use the local weight matrix WH (z) to take into account localness of our
nonparametric …tting, and place no weight matrix for data variation since fvit g
are i.i.d. across equations. Taking …rst-order condition with respect to
DT WH (z) (Y
B (X; (Z))
D^ (z)) = 0;
gives
(8)
which yields
^ (z) = DT WH (z) D
1
DT WH (z) (Y
5
B (X; (Z))) :
(9)
; KH (Zn ; z)),
Replacing
in (7) by ^ (z) ; we obtain the concentrated weighted least
squares
T
min (Y
B (X; (Z))) SH (z) (Y
(Z)
B (X; (Z))) ;
(10)
T
where we de…ne SH (z) = MH (z) WH (z) MH (z) and MH (z) = I(nm)
1
D DT WH (z) D
DT WH (z) : Apparently, MH (z) D
0(nm)
1
for all z ef-
fectively removes unknown …xed e¤ects from model (1).
How does MH (z) transform the observed data? Simple calculations give
MH (z) = I(nm)
D A
(nm)
1
A
1
en 1 eTn 1 A 1 =
n
X
!
cH (Zi ; z) DT WH (z) ;
i=1
where cH (Zi ; z)
1
=
Pm
t=1
KH (Zi;t ; z) for i = 1;
We use the formula (A + BCD)
1
1
=A
A
1
; n and A = diag cH (Z2 ; z)
B DA
1
B+C
1
1
DA
1
1
;
; cH (Zn ; z)
to
derive the inverse matrix, see Appendix B in Poirier (1995).
3
Nonparametric Estimator and Asymptotic Theory
Local linear regression approach is usually used to estimate non-/semi-parametric
models. The basic idea of this method is to apply Taylor expansion up to the
second-order derivative. That is, for each l = 1;
; k; we have the following
Taylor expansion around a point z :
l
(zi;t )
l
(z)+ H
0
l
(z)
T
H
1
1
z) + rH;l (zi;t ; z) ; H
2
(zi;t
1
(zi;t
z) = O(1) a.s.,
(11)
1
where rH;l (zi;t ; z) = H
proximates
l
(zi;t ) and
0
l
(zi;t
z)
T
h
2
l (z)
H @@z@z
T H
(z) approximates
6
0
l
i
H
1
(zi;t
z) :
l
(z) ap-
(zi;t ) when zi;t is close to z: De…ne
1
.
l (z) =
and
h
l (z) ; H
(z) = [
umn of
1
0
l
(z) ;
(z)
;
i
T T
k
; a (q + 1)
T
(z)] , a k
1 column vector for l = 1; 2;
(q + 1) parameter matrix. The …rst col-
(z) is (z) : Therefore, we will replace (Zi;t ) in (1) by
1
for each i and t; where Gi;t (z; H) = [1; fH
(z) Gi;t (z; H)
z)gT ]T is a (q +1) 1 vector.
(Zi;t
To make matrix operation simpler, we stack the parameter matrix
a [k (q + 1)]
CT
; k;
(z) into
1 column vector and denote it by vec( (z)). Since vec (ABC) =
T
B) = AT
A vec (B) and (A
B T ; where
refers to Kronecker prodT
T
uct, we have Xi;t
(z) Gi;t (z; H) = (Gi;t (z; H)
Xi;t ) vec ( (z)) for all i and
t. Thus, we consider the following minimization problem
T
min [Y
R (z; H) vec ( (z))] SH (z) [Y
(z)
where
Ri (z; H)
R (z; H)
2
(Gi;1 (z; H)
6
..
= 4
.
=
h
(Gi;m (z; H)
T
R1 (z; H) ;
T
Xi;1 )
T
Xi;m )
R (z; H) vec ( (z))]
3
7
5 is an m
T
; Rn (z; H)
iT
(12)
[k (q + 1)] matrix, (13)
is a (nm)
[k (q + 1)] matrix.
(14)
Simple calculations give
vec b (z)
=
h
i
T
R (z; H) SH (z) R (z; H)
T
R (z; H) SH (z) Y
i
T
= vec ( (z)) + R (z; H) SH (z) R (z; H)
T
where An = R (z; H) SH (z)
element of the column vector
[rH;1 ( ; ) ;
1
h
(15)
1
(An =2 + Bn )(16)
;
T
(z; H) and Bn = R (z; H) SH (z) V: The (t + (i
th
1)m)
T
(z; H) is Xi;t
rH Z~i;t ; z ; where rH ( ; ) =
T
; rH;k ( ; )] and rH;l Z~i;t ; z = H
1
with Z~i;t lying between Zi;t and z for each i and t.
7
(Zi;t
z)
T
H
~i;t )
@ 2 l (Z
H
@z@z T
H
1
(Zi;t
z)
To derive the asymptotic distribution of vec b (z) , we …rst give some reg-
ularity conditions. We use M > 0 to stand for a …nite constant.
Assumption 1: The random variables (Yi;t ; Xi;t ; Zi;t ) are independent and
identically distributed (iid) across the i index, and E
T
T
X1;t X1;s
X1;t X1;s
M < 1 for all t and s. Zi;t are continuos random variables with a pdf ft ( );
conditional pdfs ft ( jxi;t ) and ft (zjxi;t ; xi;s ) : Denote ft;s (z; zjxi;t ; xi;s ) to be the
joint pdf of (Zi;t ; Zi;s ) conditional on (Xi;t ; Xi;s ) = (xi;t ; xi;s ). Also, all these
(marginal or conditional) pdfs and ( ) are twice continuously di¤erentiable
functions and their second-order derivatives satis…es Lipschitz conditions. Let
St denote the support of Zi;t ; then ft (z) is bounded away from zero in its
domain S t .
T
0
0
Assumption 2: X has full rank k, and Xi;t;l 6= Xi;t;l
0 Zi;t;j for any l 6= l (l; l =
1;
; k), and j = 1;
; q: In addition, n
1 out of n variables { i }ni=1 are
independently distributed with zero mean and variance
determined by
Pn
i=1
i
= 0. If Xi;t;l
2
and the other one is
Xi;l for some l, we assume
This is one identi…cation condition. For cases that X1;t
is estimable if MH (z) (en
MH (z) (en
em ) = n
1
Pn
i=1
Xi;l 6= 0:
1; then
1
(z)
em ) 6= 0 for a given z. Simple calculations give
Pn
i=1 cH
(Zi ; z)
1
(cH (Z1 ; z) ;
T
; cH (Zn ; z))
the proof of Lemma 6 in Appendix A can be used to show that MH (z) (en
em ;
em ) 6=
0 for any given z with probability one.
In addition, for cases that Xi;t;l
Xi;l and Xi;l is not constant for some l, i.e.
the lth variable in X is time invariant,
l
8
(z) is estimable if MH (z) (b
em ) 6= 0
for a given z, where b is a n
MH (z) (b
Pn
i=1
em ) =
b1 +
+bn
n
1 vector with bi = Xi;l : Simple calculations give
MH (z) (en
em ). Therefore,
l
(z) is identi…able if
Xi;l 6= 0:
There could be other ways of introducing identi…cation conditions. We use
Su and Ullah’s (2006) assumption,
Pn
i=1
i
= 0, this restriction simpli…es esti-
mation and proofs and there is no need to do back…tting iteration.
Assumption 3: K(v) =
Qq
s=1
k(vs ) is a product kernel, the univariate
kernel function k( ) is a uniformly bounded, symmetric (around zero) probability
density function and its …rst four moments are …nite. In addition, de…ne jHj =
qP
q
2
h1
hq and kHk =
j=1 hj : As n ! 1, kHk ! 0 and n jHj ! 1.
(z) is to be estimated by b(z) which is the …rst column of b (z).
Theorem 3 Under Assumptions 1-3, we have the following bias and variance
for b(z):
Bais b(z)
=
V ar b(z)
where
and
=
Pm
1
2
Z
2
K (u) uuT du
;
H (z) + o kHk
Z
1
1
2 2v n 1 jHj
K 2 (u) du
+o n
q
1
Pm
ft (z)
f (z)
T
E ft (zjX1;t ) X1;t X1;t
; f (z) =
iT
h
2
@ 2 1 (z)
k (z)
; tr H @@z@z
:
H (z) = tr H @z@z T H ;
T H
=
t=1
1
1
jHj
t=1
;
ft (z) ;
Moreover, we have derived the following asymptotic normality results for
b(z):
2+
Theorem 4 Under Assumptions 1-3, E kXi;t k
some
> 0; and
2+
< 1; E jvi;t j
p
p
2
n jHj kHk = o (1) as n ! 1, we have n jHj b(z)
9
< 1 for
d
(z) !
N 0;
(z)
; where
timator for
(z)
^
(z)
(z)
=2
R
2
v
K 2 (u) du
1
. Moreover, a consistent es-
is given as follows:
= Sk ^ (z; H)
1
J^ (z; H) ^ (z; H)
1
p
SkT !
(z) ;
^ (z; H)
= n
1
jHj
1
R (z; H) SH (z) R (z; H)
T
J^ (z; H)
= n
1
jHj
1
T
R (z; H) SH (z) V^ V^ T SH (z) R (z; H)
where V^ is the vector of estimated residuals and Sk includes the …rst k rows of
the [k (q + 1)]
[k (q + 1)] identify matrix.
When calculating J^ (z; H) ; we do not need to estimate , since SH (z) V^ V^ T SH (z)
= WH (z) MH (z) V^
MH (z) V^
T
WH (z) and MH (z) D = 0; MH (z) V^ are
the estimated residuals from the local least squares estimation (12).
4
A Nonparametric Test: Random E¤ects vs
Fixed E¤ects
In this section we discuss how to test for the presence of random e¤ects versus
…xed e¤ects in the semiparametric panel data model with smoothing coe¢ cients.
The model remains as (1). De…ne ui;t =
tion assumes that
…xed e¤ects case,
i
i
i+
i;t .
The random e¤ects speci…ca-
is uncorrelated with the regressors X and Z, while for the
is allowed to be correlated with X and/or Z in an unknown
way.
We are interested in testing the null hypothesis (H0 ) that
e¤ect versus the alternative hypothesis (H1 ) that
10
i
i
is a random
is a …xed e¤ect. The null
and alternative hypotheses can be written as
H0
:
Pr [E( i jZi;1 ; :::; Zi;m ; Xi;1 ;
; Xi;m )
0] = 1 for all i,
(17)
H1
:
Pr [E( i jZi;1 ; :::; Zi;m ; Xi;1 ;
; Xi;m ) 6= 0] > 0 for some i , (18)
while we keep the same setup given in model (1) under both H0 and H1 .
Our statistic is based on the squared di¤erence between the FE and RE
estimators, which is asymptotically zero under H0 and positive under H1 : To
simplify the proofs and save computing time, we use local constant estimator
instead of local linear estimator for the test. Then following the argument in
Section 2 and Appendix B, we have
^F E (z)
=
X T SH (z) X
^RE (z)
=
X T WH (z) X
where X is a (nm)
(Xi;1 ;
1
X T SH (z) Y
1
k matrix with X = X1T ;
X T WH (z) Y
; XnT ; and for each i, Xi =
T
; Xi;m ) is an m k matrix with Xi;t = [Xi;t;1 ;
T
; Xi;t;k ] : Motivated
by Li, Huang, Li, and Fu (2002), we remove the random denominator of ^F E (z)
by multiplying X T SH (z) X and the test statistic is de…ned as
Tn
Z h
iT
^F E (z) ^RE (z)
X T SH (z) X
=
Z
~ (z)T SH (z) XX T SH (z) U
~ (z) dz
=
U
since X T SH (z) X
h
^F E (z)
T
X T SH (z) X
i
^RE (z) = X T SH (z) Y
h
^F E (z)
i
^RE (z) dz
4
~ (z) :
X ^RE (z) = X T SH (z) U
To simplify the statistic, we make several changes in Tn . Firstly, we simplify the
~ (z) by U
^ , where U
^ =Y
integration calculation by replacing U
and B X; ^RE (Z)
B X; ^RE (Z)
T ^
stacks up Xi;t
RE (Zi;t ) in the increasing order of i …rst
11
then of t: Secondly, to overcome the complexity caused by the random denominator in MH (z), we replace MH (z) by MD = Inm
m
1
em eTm , and
In
apparently, MD D = 0. Now removing the term of j = i, we obtain
T~n =
n X
X
^iT Qm
U
i=1 j6=i
where Qm = Im
m
1
Z
^j ;
KH (Zi ; z) XiT Xj KH (Zj ; z) dzQm U
4
em eTm : If jHj ! 0 as n ! 1 and E kXi;t k < M < 1,
we obtain
jHj
1
Z
KH (Zi ; z) XiT Xj KH (Zj ; z) dz
where KH (Zi;t ; Zj;s ) =
R
K H
1
(Zi;t
2
T
KH (Zi;1 ; Zj;1 ) Xi;1
Xj;1
6
..
4
.
T
KH (Zi;m ; Zj;1 ) Xi;m
Xj;1
3
T
KH (Zi;1 ; Zj;m ) Xi;1
Xj;m
7
..
..
5,
.
.
T
KH (Zi;m ; Zj;m ) Xi;m Xj;m
(19)
Zj;s ) + ! K (!) d!. We then replace
KH (Zi;t ; Zj;s ) by KH (Zi;t ; Zj;s ); this replacement will not a¤ect the essence of
the test statistic since the local weight is untouched. Now, our proposed test
statistic is given as
n
T^n =
1 X X ^T
^j
Ui Qm Ai;j Qm U
2
n jHj i=1
(20)
j6=i
where Ai;j equals to the right-hand side of equation (19) after replacing KH (Zi;t ; Zj;s )
by KH (Zi;t ; Zj;s ). Finally, to remove the asymptotic bias term of the proposed
test statistic, the random-e¤ects estimator of (Zi;t ) are leave-one-unit-out random e¤ect estimators; for a given pair of (i; j), i 6= j, ^RE (Zi;t ) is calculated
m
without using observations on f(Xj;t ; Zj;t ; Yj;t )gt=1 :
We present the asymptotic properties of this test below and delay the proofs
to Appendix C.
12
4
Theorem 5 Under Assumptions 1-3, and n jHj kHk ! 0 as n ! 1; we have
under H0 as n ! 1
Jn = n
p
d
jHjT^n =^ 0 ! N (0; 1)
(21)
where
n
^ 20 =
1 XX ^T
Vi Qm Ai;j Qm V^j
2
n jHj i=1
2
(22)
j6=i
is a consistent estimator of
2
0
1
m
=2 1
2
4
v
Z
K 2 (u) du
t
m X
X
t=1 s=1
T ^
Xi;t
F E (Zi;t )
where V^i;t = Yi;t
h
i
2
T
E f2;tj1;s (Z1;s jZ1;s ; X1;s ; X2;t ) X1;s
X2;t
;
^ i and ^F E (Zi;t ) is a leave-two-unit-out
FE estimator without using the observations from the ith and jth units and
^ i = Yi
m
1
Pm
t=1
T ^
Xi;t
F E (Zi;t ) : Under H1 ; Pr [Jn > Bn ] ! 1 as n ! 1;
where Bn is any nonstochastic sequence with Bn = o n
Theorem 5 states that the test statistic Jn = n
p
p
jHj :
jHjT^n =^ 0 is a consistent
test for testing H0 against H1 . It is a one-sided test. If Jn is greater than
the critical values from the standard normal distribution, we will reject the null
hypothesis at the corresponding signi…cance levels.
5
Monte Carlo Simulations
In this section we report some Monte Carlo simulation results to examine the
…nite sample performance of the proposed estimator. The following data generating process is used:
Yi;t =
1 (Zi;t )
+
2 (Zi;t )Xi;t
13
+
i
+ vi;t ;
(23)
where
= 1 + z + z2,
1 (z)
2 (z)
= sin(z ), Zi;t = wi;t + wi;t
uniformly distributed in [0; =2], Xi;t = 0:5Xi;t
addition,
i
= c0 Zi +ui for i = 1; 2;
; n 1 and
0:5;and 1.0, ui is i.i.d. N (0; 1): When c0 6= 0,
c0 to control the correlation between
vi;t is i.i.d. N (0; 1), wi;t ,
i;t ,
1
i
i
+
n
i;t ,
wi;t is i.i.d.
is i.i.d. N (0; 1). In
i;t
Pn
1
i=1
=
1,
i
where c0 = 0;
and Zi;t are correlated; we use
and Zi = m
1
Pm
t=1
Zi;t . Moreover,
ui and vi;t are independent of each other.
We report estimation results for both the proposed …xed-e¤ects (FE) estimator and the random-e¤ects (RE) estimator (see Appendix B for the asymptotic
results of the RE estimator). To learn how the two estimators perform when
we have …xed-e¤ects model and when we have random-e¤ects model, we use the
integrated squared error as a standard measure of estimation accuracy:
ISE ^l =
Z h
^l (z)
l
i2
(z) f (z) dz;
(24)
which can be approximated by the average mean squared error AM SE(^l ) =
nm)
1
Pn
i=1
Pm ^
t=1 [ j;l (Zi;t )
2
j;l (Zi;t )]
for l = 1; 2: Finally, in Table 1 we
present the average value of AM SE(^l ) from 1000 Monte Carlo experiments.
We take m = 3 and the sample size n=50, 100, and 200.
Since the bias and variance of the proposed FE estimator do not depend on
the values of the …xed e¤ects, our estimates are the same for di¤erent values of
c0 ; however, it is not true if random-e¤ects model is true. Therefore, the results
derived from the FE estimator are only reported once for each c0 in Table 1.
It is well-known that the performance of non-/semi-parametric models depends on the choice of bandwidth. Therefore, we propose a leave-one-unit-out
cross validation method to automatically …nd the optimal bandwidth for es14
timating both the FE and RE models. Speci…cally, when estimating
( ) at a
m
point Zi;t ; we remove f(Xi;t ; Yi;t ; Zi;t )gt=1 from the data and only use the rest of
1) m observations to calculate ^(
(n
(Zi;t ). In computing the RE estimate,
i)
the leave-one-unit-out cross validation method is just a trivial extension of the
conventional leave-one-out cross validation method. However, such simple extension fails to provide satisfying result when we calculate a FE estimator due
to the existence of unknown …xed e¤ects. Therefore, based on the arguments
made in Section 2, when calculating the FE estimator, we use the following
modi…ed leave-one-unit-out cross validation method:
B X; ^(
^ opt = arg min Y
H
H
T
1)
T
MD
MD Y
(Z)
B X; ^(
1)
(Z)
;
(25)
where MD = I(nm)
(nm)
1
m
In
em eTm satis…es MD D = 0 (this is
n
used to remove the unknown …xed e¤ects) and B X; ^(
T ^
Xi;t
(
i)
1)
(Z)
stacks up
(Zi;t ) in the increasing order of i …rst then of t. Simple calculations
give
Y
B X; ^(
T
1)
(Z)
T
=
B (X; (Z))
B X; ^(
1)
(Z) + V
=
B (X; (Z))
B X; ^(
1)
(Z)
+2 B (X; (Z))
B X; ^(
T
MD
MD Y
T
B X; ^(
(Z)
(Z)
T
MD
MD B (X; (Z))
T
MD
MD B (X; (Z))
T
1)
1)
B X; ^(
B X; ^(
1)
; n; s = 1;
T
MD
MD V + V T MD MD V;
mg for all i and t, or (Xi;t ; Zi;t ) is
strictly exogenous variable, then the second term has zero expectation because
15
(Z) + V
(Z)
where the last term does not depend on the bandwidth at all. If vi;t is independent of f(Xj;s ; Zj;s ) : j = 1;
1)
(26)
the linear transformation matrix MD removes a cross-time not cross-sectional
average from each variable, e.g. Y~i;t = Yi;t
m
1
Pm
t=1
Yi;t for all i and t.
Therefore, the …rst term is the dominate term in large sample and (25) is used
to …nd an optimal smoothing matrix minimizing a weighted mean squared error
n
ot=1;m
of ^ (Zi;t )
: Of course, we could use other weight matrix in (25) instead
i=1;n
of MD as long as the weight matrix can remove the …xed e¤ects and does not
trigger non-zero expectation of the second term in (26).
Table 1 shows that the RE estimator performs much better than the FE
estimator when the true model is a random e¤ects model, and the opposite is
true when the true model is a …xed-e¤ects model. This observation is consistent
with what we have derived in a parametric panel data regression model analysis.
Therefore, our simulation results indicate that a test for random e¤ects against
…xed e¤ects will be always in demand when we analyze panel data models. In
Table 2 we give Monte Carlo simulations of the proposed nonparametric test of
random e¤ects against …xed e¤ects.
How to choose the bandwidth h for the test? For univariate case, Theorem
5 indicates that nh ! 1 and nh9=2 ! 0 as n ! 1; if we take h
Theorem 5 requires
2
2
9; 1
2=7
;
: If we balance the two conditions nh ! 1
and nh9=2 ! 0 as n ! 1, we have
2, we use h = c (nm)
n
= 2=7: Therefore, in producing Table
^ z to calculate the RE estimator with c being value
of :8, 1:0, and 1:2. The results in Table 2 are consistent with the …ndings in
the nonparametric tests literature in that a smaller bandwidth is good for size
approximation and a larger bandwidth is good for power approximation.
16
6
Conclusion
In this paper we proposed using a local least squares method to estimate a
semiparametric panel data model with …xed e¤ects and smooth coe¢ cients. In
addition, we suggested using a bootstrap procedure to test for random e¤ects
against …xed e¤ects. A data-driven method has been introduced to automatically …nd the optimal bandwidth for the proposed FE estimator. Monte Carlo
simulations indicate that the proposed estimator and test statistic have good
…nite sample performance. Finally, the choice of bandwidths for the test is
important when investigating the size and power of the test.
Acknowledgments
Sun’s research was supported from the Social Sciences and Humanities Research
Council of Canada (SSHRC). Carroll’s research was supported by a grant from
the National Cancer Institute (CA-57030), and by the Texas A&M Center for
Environmental and Rural Health via a grant from the National Institute of
Environmental Health Sciences (P30-ES09106). Li’s research was partially supported by the Private Enterprise Research Center, Texas A&M University.
Appendix A: Technical Sketch–Proof of Theorem
3
To make our mathematical formula short, we introduce some simpli…ed notaPm
1
tions …rst: for each i, t, and s; it = KH (Zit ; z) ; cH (Zi ; z) = t=1 it ; and
17
for any positive integers i; j; t; s
2
6
6
= Git (z; H) GTjs (z; H) = 6
4
[ ]it;js
"
=
1
H
th
where the (l + 1)
1
(Zit
1
Git1
..
.
Gjs1
Git1 Gjs1
..
.
Gitq
Gitq Gjs1
.
T
H 1 (Zjs z)
H 1 (Zit z) H
z)
..
1
7
7
7
5
Gitq Gjsq
#
T
(1)
zl ) =hl ; l = 1;
; q.
(Zjs
element of Gjs (z; H) is Gjsl = (Zjsl
3
Gjsq
Git1 Gjsq
..
.
z)
Simple calculations show that
0
= @1 +
[ ]i1 t1 ;i2 t2 [ ]j1 s1 ;j2 s2
T
Ri (z; H) KH (Zi ; z) em eTm KH (Zj ; z) Rj (z; H)
=
q
X
j=1
m X
m
X
1
;
Gj1 s1 j Gi2 t2 j A [ ]i1 t1 ;j2 s(2)
2
it js
T
Xit Xjs
(3)
[ ]it;js
s=1 t=1
In addition, we have for a …nite positive integer j
jHj
jHj
where
m
X
1
t=1
2
E4
St;j;1
St;j;2
where RK;j =
m
X
1
E
t=1
2j
it
q
X
j 0 =1
h
j
it
[ ]it;it jXit
i
m
X
=
t=1
3
m
X
G2itj 0 [ ]it;it jXit 5 =
t=1
#
R
(zjXit )
ft (zjXit ) K j (u) du @ft@z
HRK;j
T
=
it )
RK;j H @ft (zjX
ft (zjXit ) RK;j
@z
"
#
R 2j
(zjX1t )
H K;2j
ft (zjXit ) K (u) uT udu @ft @z
T
=
@ft (zjXit )
ft (zjXit ) K;2j
K;2j H
@z
K j (u) uuT du and
K;2j
=
R
(4)
;
2
(5)
;
E (St;j;2 jXit ) + Op kHk
"
R
2
E (St;j;1 jXit ) + Op kHk
K 2j (u) uT u
(6)
(7)
uuT du.
Moreover, for any …nite positive integer j1 and j2 ; we have
jHj
=
2
m X
X
t=1 s6=t
m X
X
t=1 s6=t
E
h
j1 j2
it is
(t;s)
[ ]it;is jXit ; Xis
i
E Tj1 ;j2 ;1 jXit ; Xis + Op kHk
18
(8)
2
jHj
=
m X
X
2
t=1 s6=t
m X
X
t=1 s6=t
E4
(t;s)
j1 j2
it is
0
@
q
X
j 0 =1
(t;s)
1
R
3
Gitj 0 Gisj 0 A [ ]it;is jXit ; Xis 5
E Tj1 ;j2 ;2 jXit ; Xis + Op kHk
where we de…ne bj1;j2 ;i1 ;i2 =
Tj1 ;j2 ;1 =
2
1
K j1 (u) u2i
1 du
R
(9)
2
K j2 (u) u12i2 du
ft;s (z; zjXit ; Xis ) bj1 ;j2 ;0;0
H 5t ft;s (z; zjXit ; Xis ) bj1 ;j2 ;1;0
5Ts ft;s (z; zjXit ; Xis ) Hbj1 ;j2 ;0;1
H 52t;s ft;s (z; zjXit ; Xis ) Hbj1 ;j2 ;1;1
tr H 52t;s ft;s (z; zjXit ; Xis ) H
H 5s ft;s (z; zjXit ; Xis )
5Tt ft;s (z; zjXit ; Xis ) H
ft;s (z; zjXit ; Xis ) Iq q
and
(t;s)
Tj1 ;j2 ;2 =
bj1 ;j2 ;1;1 ;
with 5s ft;s (z; zjXit ; Xis ) = @ft;s (z; zjXit ; Xis ) =@zs and 52t;s ft;s (z; zjXit ; Xis ) =
@ 2 ft;s (z; zjXit ; Xis ) = @zt @zsT .
The conditional bias and variance of vec b (z) are given as follows:
h
i
h
i 1
T
T
Bias vec b (z) j fXit ; Zit g = 21 R (z; H) SH (z) R (z; H)
R (z; H) SH (z) (z; H)
h
i
h
i 1 h
i
T
T 2
and V ar vec b (z) j fXit ; Zit g = 2v R (z; H) SH (z) R (z; H)
R (z; H) SH
(z) R (z; H)
h
i 1
T
R (z; H) SH (z) R (z; H)
:
Lemma 6 Under Assumptions 1-3, we have
h
n
1
jHj
where
1
=
i
T
R (z; H) SH (z) R (z; H)
Pm
t=1
1
q
ft (z)
f (z)
1
1
Op (kHk)
T
E ft (zjX1t ) X1t X1t
.
19
Op (kHk)
R
K (u) uuT du
1
1
;
Proof: First, simple calculation gives
An
T
T
= R (z; H) SH (z) R (z; H) = R (z; H) WH (z) MH (z) R (z; H)
n
X
T
=
Ri (z; H) KH (Zi ; z) Ri (z; H)
i=1
n X
n
X
T
qij Ri (z; H) KH (Zi ; z) em eTm KH (Zj ; z) Rj (z; H)
j=1 i=1
=
n X
m
X
i=1 t=1
n
X
i=1
qii
m X
X
it is
T
Xit Xit
[ ]it;it
n X
X
T
Xit Xis
[ ]it;is
qij
An4 ;
2
it js
T
Xit Xjs
[ ]it;js
em eTm WH (z), and the typical elements of Q
Q
(nm)
m X
m
X
s=1 t=1
j=1 i6=j
An3
are qii = cH (Zi ; z) cH (Zi ; z) =
Pn
i=1 cH
(Zi ; z) and qij =
for i 6= j: Simple calculations give
1=2
KH (Zit ; z)
= jHj f (z) + jHj
KH (Zit ; z)
= jHj ft (z) + jHj
t=1
2
it
s=1 t6=s
An2
where MH (z) = I(nm)
m
X
m
X
t=1
i=1
qii
= An1
n
X
T
Xit Xit
it [ ]it;it
f (z)
1=2
and it follows that cH (Zi ; z) = jHj
1=2
Z
ft (z)
cH (Zi ; z) cH (Zj ; z) =
1=2
K 2 (u) du
Z
the leading term of qii ; and that qij = Op n
1
i=1 cH
(Zi ; z)
1
+ Op jHj 2 kHk ;
1=2
K 2 (u) du
R
f (z) K 2 (u) du
Pn
1
+ Op jHj 2 kHk ;
1=2
+ op jHj
1
2
is
. As a result, it is easy to show
that An4 is dominated by the other three terms.
Applying (4), (5), (8), and (9) to An1 ; we have n
Op kHk
2
+ Op n
1
2
jHj
1
2
1
jHj
1
Pm
An1
t=1
T
Xit Xit
+
E St;1;1
if kHk ! 0 and n jHj ! 1 as n ! 1:
To cope with the di¢ culty caused by the random denominator cH (Zi ; z)
when taking expectations and variations of An2 and An3 ; we apply the technique
used for kernel curve estimation. Speci…cally, de…ne !
^ it (z) =
1=2
It then follows that !
^ it (z) = (f (z) =ft (z))
20
1
it
cH (Zi ; z)
1
:
1=2
+op (1) : De…ne ! t (z) = (f (z) =ft (z))
and g^it (z) =
it
T
Xit Xit
: Then
[ ]it;it
n X
n X
m
m
X
X
^ it (z)
g^it (z)
g^it (z) !
=
+ 1
!
^ (z)
!
^ (z) ! t (z)
i=1 t=1 it
i=1 t=1 it
An2
Pn
Since !
^ it (z) = ! t (z) + op (1) ; it can be shown that
!
^ it (z)
! t (z)
i=1
Pm
:
^it
t=1 g
(z) =! t (z)
is the leading term of An2 : Again applying (4), (5), (8), and (9), we have
Pm
2
1 Pn Pm
1
T
+Op kHk +
n 1 jHj
^it (z) =! t (z)
E St;1;1
Xit Xit
i=1
t=1 g
t=1 ! t (z)
Op n
1
2
jHj
1
2
:
Similarly, for An3 , we de…ne !
^ i (z) = jHj
R 2
1=2
f (z) K (u) du
, and we can show that n
Op n
1=2
jHj
1=2
1=2
1
cH (Zi ; z)
jHj
1
An3
1
and ! i (z) =
1=2
Op jHj
+
if kHk ! 0 and n jHj ! 1 as n ! 1: Taking all the
results together and will complete the proof of this lemma, where the inverse
matrix is derived by using Theorem A.4.4. in Poirier (1995, p.627) .
R
"
1
Lemma 7 Under Assumptions 1-3, we have n
where
H
h
2
1 (z)
;
(z) = tr H @@z@z
T H
2
jHj
1
k (z)
; tr H @@z@z
T H
Cn
iT
K (u) uuT du
Op kHk
H
(z)
3
#
,
:
Proof: Simple calculations give
Cn
T
= R (z; H) SH (z) (z; H)
n X
m
X
T
=
Xit ) Xit
rH Z~it ; z
it (Git
=
i=1 t=1
n X
m
X
i=1 t=1
n
X
qii
i=1
it
m X
X
is it
qij
j=1 i6=j
where
T
Xit ) Xit
rH Z~it ; z
s=1 t6=s
m X
m
X
n X
X
= Cn1
(Git
Cn2
(Git
js it
n X
n
X
qij
m X
m
X
s=1 t=1
j=1 i=1
n
m
X X
2
qii
it (Git
t=1
i=1
js it
(Git
T
Xit ) Xit
rH Z~it ; z
T
Xit ) Xis
rH Z~is ; z
(Git
T
Xit ) Xjs
rH Z~js ; z
s=1 t=1
Cn3
Cn4 ;
(z; H) is de…ned in Section 3. Similar to the proof of Lemma 6, it is
easy to show that Cn;3 and Cn;4 are dominated by the other two terms of Cn :
21
T
Xit ) Xjs
rH Z~js ; z
For l = 1;
; k we have
jHj
1
E[
it rH;l
(Zit ; z) jXit ] = ft (zjXit ) tr
jHj
1
E
it rH;l
(Zit ; z) H
1
1
and E n
jHj
1
n
Cn1
(Zit
R
1
ilarly we can show that V ar n
K (u) uuT du H
3
z) jXit = Op kHk
K (u) uuT du
jHj
Z
1
H
(z)
Cn1 = O n
1
T
jHj
4
+ Op kHk
;
3
; O kHk
1
@ 2 l (z)
H
@z@z T
4
kHk
oT
if E
. SimT
T
Xit Xis
Xit Xis
<
M < 1 for all t and s:
For Cn2 ; we use the same method to cope with the random denominator
problem as in the proof of Lemma 6. Then we can show that the dominant
Pn Pm q (z)
T
term of Cn2 is i=1 t=1 fft(z)
Xit ) Xit
rH (Zit ; z) :
it (Git
This will complete the proof of this lemma.
Lemma 8 Under Assumptions 1-3, we have
n
1
jHj
1
Bn
2
R
K 2 (u) du
Op (kHk)
ORp (kHk)
K 2 (u) uuT du
:
Proof: Simple calculations give
Bn
T
T
T
2
= R (z; H) SH
(z) R (z; H) = R (z; H) WH (z) MH (z) MH (z) WH (z) R (z; H)
n
n X
n
X
X
T
T
2
2
=
Ri (z; H) KH
(Zi ; z) Ri (z; H)
qji Rj (z; H) KH
(Zj ; z) em eTm KH (Zi ; z) Ri (z; H)
i=1
j=1 i=1
n X
n
X
T
2
qij Ri (z; H) KH (Zi ; z) em eTm KH
(Zj ; z) Rj (z; H)
j=1 i=1
+
n X
n X
n
X
T
2
qij qji0 Ri (z; H) KH (Zi ; z) em eTm KH
(Zj ; z) em eTm KH (Zi0 ; z) Ri0 (z; H)
j=1 i=1 i0 =1
= Bn1
Bn2
T
Bn2
+ Bn3 :
1
1
Similar to the proof of Lemma 6, we can show that n 1 jHj Bn 2n 1 jHj (Bn1
Pn Pm
T
and that the leading term of Bn1 is i=1 t=1 2it [ ]it;it
Xit Xit
and the
22
Bn2 )
leading term of Bn2 is
Pm
1
n 1 jHj Bn
t=1 1
Pn
Pm
i=1 qii
q
3
it
t=1
ft (z)
f (z)
T
Xit Xit
: Then we obtain
[ ]it;it
E St;2;1
2
T
Xit Xit
+Op kHk
+Op n
1=2
jHj
1=2
The three lemmas above are enough to give the result of Theorem 3. More-
over, applying Liaponuov’s CLT will give the result of Theorem 4. Since the
proof is a rather standard procedure, we drop the details for compactness of the
paper.
Appendix B: Technical Sketch–Random E¤ects
Estimator
The RE estimator, ^RE ( ), is the solution to the following optimization problem:
T
min [Y
R (z; H) vec ( (z))] WH (z) [Y
(z)
R (z; H) vec ( (z))] ;
that is, we have
i 1
T
T
R (z; H) WH (z) R (z; H)
R (z; H) WH (z) Y
h
i 1
T
T
R (z; H) WH (z) D
= vec ( (z)) + R (z; H) WH (z) R (z; H)
h
i 1
T
T
~n
R (z; H) WH (z) A~n =2 + B
+ R (z; H) WH (z) R (z; H)
vec ^ RE (z)
h
=
T
where A~n = R (z; H) WH (z)
~n = R (z; H)T WH (z) V .
(z; H) and B
Its
asymptotic results are as follows.
Lemma 9 Under Assumptions 1-3,
2+
E jvit j
2+
< 1; E j i j
all i and t and for some
where
(z);RE
=
< M < 1 and E kXit k
Pn
i=2
2+
< M < 1 for
> 0, we have under H0
p
n jHj bRE (z)
2
n
p
2
n jHj kHk = o (1) as n ! 1, and
2
i
+
2
v
d
(z) ! N 0;
1
R
23
(z);RE
K 2 (u) du and
=
;
(10)
Pm
t=1
T
E ft (zjX1;t ) X1;t X1;t
:
:
Under H1 , we have
Bias bRE (z)
Z
=
K (u) uuT du
n
H
(z)
m
1 XX
2
E [ft (zjXi;t ; i ) i Xi;t ] + o kHk
2n i=1 t=1
Z
1
2
1
1
jHj
K 2 (u) du:
vn
+
h
i
V ar ^RE (z) j fXit ; Zit g
=
Proof of Lemma 9: First, we have the following decomposition
p
h
n jHj ^RE (z)
i p
h
(z) = n jHj ^RE (z)
E ^RE (z)
i p
h
+ n jHj E ^RE (z)
i
(z) ;
where we are going to show that the …rst term converges to a normal distribution with mean zero and the second term contributes to the asymptotic bias.
Without confusing readers, we drop the subscription ‘RE’.
a. Under H0 , the conditional bias and variance of ^ (z) are as follows:
h
i
h
i 1
T
T
Bias0 ^ (z) j fXit ; Zit g = Sk R (z; H) WH (z) R (z; H)
R (z; H) WH (z) (z; H) =2
h
i
h
i 1 h
T
T
and V ar ^ (z) j fXit ; Zit g = Sk R (z; H) WH (z) R (z; H)
R (z; H) WH (z) V ar(U U T )WH (z) R (z; H)
h
i 1
T
R (z; H) WH (z) R (z; H)
SkT : It is simple to show that V ar(U U T ) = D DT +
; 2n is the covariance matrix of ( 2 ;
; n ).
h
i
h
i
b. Under H1 , we notice that Bias1 ^ (z) j fXit ; Zit g is the sum of Bias0 ^ (z) j fXit ; Zit g
h
i 1
T
T
plus an additional term Sk R (z; H) WH (z) R (z; H)
R (z; H) WH (z) D .
2
v I(nm) (nm) ,
where
= diag
2
2
;
2+
It is easy to show that under Assumptions 1-3, and E j i j
E kXit k
2+
n
< M < 1 for all i and t and for some
1
1
< M < 1 and
>0
T
jHj Sk R (z; H) WH (z) D
n X
m
X
2
= n 1
E [ft (zjXit ; i ) i Xit ] + Op kHk + Op
i=1 t=1
p
1
n jHj
!
(; 11)
which is op (1) under H0 and is a non-zero constant plus a term of op (1) under
T
H1 : Based on the facts that R (z; H) WH (z) R (z; H) is An1 in Lemma 6 and
24
T
R (z; H) WH (z)
(z; H) is Cn1 in Lemma 7 we have
Z
h
i
2
Bias0 ^ (z) j fXit ; Zit g =
K (u) uuT du
;
H (z) + o kHk
which is the same as the bias term of the FE estimator. Our results indicate
that under H1 the bias of the RE estimator will not vanish as n ! 1 and this
leads to the inconsistency of the RE estimator under H1 :
Pm
T
;
As for the conditional variance, if we denote = t=1 E ft (zjX1t ) X1t X1t
we can easily show that under H0
h
i
V ar ^ (z) j fXit ; Zit g = n
1
jHj
1
2n
1
n
X
2
ui
2
v
+
i=2
!
1
Z
K 2 (u) du;
(12)
and under H1
h
i
V ar ^ (z) j fXit ; Zit g =
2
1
vn
jHj
T
1
2
1
Z
K 2 (u) du;
(13)
where we have recognized that R (z; H) WH (z) R (z; H) is Bn1 in Lemma 8.
A
Appendix C: Technical Sketch–Proof of Theorem 5
T
De…ne 4i = (4i;1 ;
; 4i;m )
T
with 4i;t = Xi;t
(Zi;t )
^RE (Zi;t ) . Since
MD D = 0; we can decompose the proposed statistic into four terms
n
T^n
=
1 X X ^T
^j
Ui Qm Ai;j Qm U
2
n jHj i=1
j6=i
=
1
2
n jHj
+
n X
X
4Ti Qm Ai;j Qm 4j +
i=1 j6=i
n X
X
1
n2 jHj
n
2 XX T
4i Qm Ai;j Qm Vj
n2 jHj i=1
j6=i
ViT Qm Ai;j Qm Vj
i=1 j6=i
= Tn1 + 2Tn2 + Tn3
where Vi = (vi;1 ;
T
; vi;m )
1 error vector. Since ^RE (Zi;t ) is the
is the m
leave-one-unit-out estimator for a pair of (i; j); it is easy to see that E (Tn2 ) = 0:
25
The proofs fall into the standard procedures seen in the literature of nonparametric tests. We therefore give a very brief proof below.
Firstly, applying Hall’s (1984) CLT, we can show that under both H0 and
H1
n
by de…ning Hn
i;
j
p
d
jHjTn3 ! N 0;
= ViT Qm Ai;j Qm Vj with
2
0
i
(14)
= (Xi ; Zi ; Vi ), which a sym-
metric, centred and degenerate variable. We are able to show that
E G2n (
1;
2)
+n
fE [Hn2 ((
1
E Hn4 ((
1;
2 ))]g
1;
2 ))
2
3
=
O jHj
+O n
1
jHj
2
O jHj
if jHj ! 0 and n jHj ! 1 as n ! 1; where Gn (
2
0
1;
2)
=E
1 2
!0
[Hn (( 1 ; i )) Hn (( 2 ; i ))] :
h
i
Pm Pm
2
2
2
T
jHj
E
K
(Z
;
Z
)
X
X
:
1s
2t
2t
v
1s
H
t=1
s=1
i
= 2 jHj E Hn2 ( 1 ; 2 )
2 1 m
p
2
1=2
Secondly, we can show that n jHjTn2 = Op kHk + Op n 1=2 jHj
p
under H0 and n jHjTnj = Op (1) under H1 : Moreover, we have, under H0 ,
p
p
p
p
4
n jHjTn1 = Op n jHj kHk ; under H1 , n jHjTn1 = Op n jHj :
In addition, jHj
Finally, to estimate
2
0
consistently under both H0 and H1 , we replace the
unknown Vi and Vj in Tn3 by the estimated residual vectors from FE estimator. Simple calculations show that the typical element of V^i Qm is e
v~it =
P
m
T^
T^
yit Xit
yi m 1 t=1 Xit
vi = 4it (vit vi ),
F E (Zit ) vit
F E (Zit )
P
m
T
T
where 4it = Xit
(Zit ) ^F E (Zit )
m 1 t=1 Xit
(Zit ) ^F E (Zit ) =
Pm
T
(Zil ) ^F E (Zil ) with qtt = 1 1=m and qlt = 1=m for l 6= m:
l=1 qlt Xil
The leave-two-unit-out FE estimator does not use the observations from the ith
h
2
Pm Pm
2
T
and jth units, and this leads to E V^iT Qm Ai;j Qm V^j
t=1
s=1 E KH (Zis ; Zjt ) Xis Xjt
Pm
where v~it = vit vi and vi = m 1 t=1 vit :
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28
Table 1: Average mean squared errors (AMSE) of the …xed and random e¤ects
estimators when the data generation process is a random e¤ects model and when
it is a …xed e¤ects model.
Random E¤ects Estimator
Fixed E¤ects Estimator
Data Process
n = 50 n = 100 n = 200 n = 50 n = 100 n = 200
Estimating 1 ( ):
c0 = 0
.0951
.0533
.0277
c0 = 0:5
.6552
.5830
.5544
.1381
.1163
.1021
c0 = 1:0
2.2010
2.1239
2.2310
Estimating 2 ( ):
c0 = 0
.1562
.0753
.0409
c0 = 0:5
.8629
.7511
.7200
.1984
.1379
.0967
c0 = 1:0
2.8707
2.4302
2.5538
Table 2: Percentage Rejection Rate from
c=1.0
c0
n = 50
1%
5% 10%
0
.011 .023 .041
0.5 .682 .780 .819
1.0 .908 .913 .921
30
1000 Monte Carlo Simulations with
1%
.025
.935
.962
n = 100
5%
.04
.943
.966
10%
.062
.951
.967