Throughout the appendix, we useCto denote a generic finite constant large enough, which need not to be the same at each appearance. Denote the projection matrixPA=A(A′A)−1A′and the residual makerMA=I−PAfor a matrixA.
LetΞberx×rxdiagonal matrix that consist of the firstrxlargest eigenvalues of the T×T matrix (N T)−1PN
i=1XiX′i. Then by the definition of eigenvalues andF,b FΞb = (N T)−1PN
i=1XiX′iF. It’s easy to show thatb Ξ is invertible following the proof of Lemma A.3 inBai(2003). Then
Fb−F0R= 1 N T
XN i=1
F0Γ0iV′iFΞb −1+ 1 N T
XN i=1
ViΓ0i′F0′FΞb −1+ 1 N T
XN i=1
ViV′iFΞb −1 (A.1) and
ˆft−R′ft0= 1 N T
XN i=1
Ξ−1Fb′ViΓ0i′ft0+ 1 N T
XN i=1
Ξ−1Fb′F0Γ0ivit+ 1 N T
XN i=1
Ξ−1Fb′Vivit (A.2) whereR= (N T)−1PN
i=1Γ0iΓ0i′F0′FΞb −1. Following the proof of Lemma A.3 inBai(2003) again, we can show thatRis invertible.
LetΞLbe arx×rxdiagonal matrix that consists of the firstrxlargest eigenvalues of theT×Tmatrix (N T)−1PN
i=1Xi,−1X′i,−1. Then by the definition of eigenvalues andFb−1,Fb−1ΞL= (N T)−1PN
i=1Xi,−1X′i,−1Fb−1. It’s easy to show thatΞL is in-vertible following the proof of Lemma A.3 inBai(2003). Then
Fb−1−F0−1R
= 1 N T
XN i=1
F0−1Γ0iV′i,−1Fb−1Ξ−L1+ 1 N T
XN i=1
Vi,−1Γ0i′F0−′1Fb−1Ξ−L1+ 1 N T
XN i=1
Vi,−1V′i,−1Fb−1Ξ−L1 (A.3)
where R = (N T)−1PN
i=1Γ0iΓ0i′F0−′1Fb−1Ξ−L1. Following the proof of Lemma A.3 in Bai (2003) again, we can show thatRis invertible.
1
Lemma A.1 Under AssumptionsBtoD, we have
(a) T−1kFb−F0Rk2=Op(δNT−2), T−1kFb−1−F0−1Rk2=Op(δNT−2), (b) T−1(bF−F0R)′F0=Op(δNT−2), T−1(bF−F0R)′F0−1=Op(δNT−2),
T−1(Fb−1−F0−1R)′F0=Op(δNT−2), T−1(bF−1−F0−1R)′F0−1=Op(δ−NT2), (c) T−1(bF−F0R)′Fb=Op(δ−NT2), T−1(bF−F0R)′Fb−1=Op δNT−2
, T−1(Fb−1−F0−1R)′Fb =Op(δNT−2), T−1(bF−1−F0−1R)′Fb−1=Op(δNT−2), (d) T−1(bF−F0R)′H0=Op(δNT−2), T−1(bF−1−F0−1R)′H0=Op(δNT−2), (e) Ξ=Op(1) ,R=Op(1),Ξ−1=Op(1),R−1=Op(1),
ΞL=Op(1),R=Op(1),Ξ−L1=Op(1),R−1=Op(1)
(f) RR′−(T−1F0′F0)−1=Op(δ−NT2),RR′−(T−1F0−′1F0−1)−1=Op δ−NT2 , (g) MbF−MF0=Op(δ−NT1),MbF−1−MF0
−1 =Op(δ−NT1), (h) N−1/2T−1/2
XN ℓ=1
Γ00ℓ V′ℓ(bF−F0R) =Op(N−1/2T1/2) +Op(δ−NT2T1/2),
N−1/2T−1/2 XN ℓ=1
Γ00ℓ V′ℓ,−1(bF−1−F0−1R) =Op(N−1/2T1/2) +Op(δ−NT2T1/2),
(i) kN−1/2T−1 XN h=1
(VhV′h−E(VhV′h))F0k=Op(1),
(j) kN−1/2T−1 XN h=1
F0′(VhV′h−E(VhV′h))F0k=Op(1),
(k) kN−1/2T−1 XN h=1
(VhV′h−E(VhV′h))H0k=Op(1),
(l) kN−1/2T−1 XN h=1
F0′(VhV′h−E(VhV′h))H0k=Op(1),
(m) kN−1T−1/2 XN ℓ=1
E(VℓV′ℓ)k=O(1),
(n) N−1/2T−1 XN ℓ=1
[VℓV′ℓ−E(VℓV′ℓ)] =Op(1).
Proof of Lemma A.1. For the proofs of (a) to (f), and (h), see Proof of Lemma 4 in Supplemental Material,Norkute et al.(2020). For (g), we decompose the left hand side term as
MbF−MF0 =− Fb
Fb−F0R′
T −
Fb−F0R R′F0′
T − 1
TF0 RR′−
F0′F0 T
−1! F0′ then it will be bounded in norm by
Fb
√T
Fb−F0R
√T +kRk
F0
√T
Fb−F0R
√T +
F0
√T
2RR′−
F0′F0 T
−1=Op δ−NT1
with LemmasA.1(a), (e), (f) and the facts that T−1/2Fb2 =rx andET−1/2F02≤C by AssumptionC. For (i), we have
2
√1 N T
XN h=1
(VhV′h−E(VhV′h))F0 =
1 T
XT s=1
√1 N T
XN h=1
XT t=1
[v′hsvht−E(v′hsvht)]ft0
2
1/2
=Op(1) sinceE√N T1 PN
h=1
PT
t=1[v′hsvht−E(v′hsvht)]ft0
2
≤C by AssumptionB7. For (j), it holds because
E
√1 N T
XN h=1
F0′(VhV′h−E(VhV′h))F0
2
≤C
which can be proved following the proof of Lemma A.2(i) in Bai(2009) with Assumption B8. The proofs of (k) and (l) are similar to those of (i) and (j), respectively.
Consider (m). Since
1 N
XN i=1
|E(v′isvit)| ≤ 1 N
XN i=1
(E(v′isvis)E(v′itvit))1/2
≤
1 N
XN i=1
E(v′isvis) 1 N
XN j=1
E v′jtvjt
1/2
≤C
we have
1 N√
T XN ℓ=1
E(VℓV′ℓ)
2
= 1 T
XT s=1
XT t=1
1 N
XN i=1
E(v′isvit)
2
≤1 T
XT s=1
XT t=1
1 N
XN i=1
|E(v′isvit)| 1 N
XN j=1
E v′jsvjt≤ C T
XT s=1
XT t=1
1 N
XN i=1
|E(v′isvit)| ≤ C T
XT s=1
XT t=1
˜
σst≤C2,
by AssumptionB3.
Consider (n), we can derive that
E
√1 N T
XN ℓ=1
[VℓV′ℓ−E(VℓV′ℓ)]
2
=E
1 T2
XT s=1
XT t=1
"
√1 N
XN ℓ=1
(v′ℓsvℓt−E(v′ℓsvℓt))
#2
= 1 T2
XT s=1
XT t=1
E
"
√1 N
XN ℓ=1
(v′ℓsvℓt−E(v′ℓsvℓt))
#2
≤C ,
by AssumptionB4. Then we can prove part (n) by Markov inequality. This completes the proof.
3
Lemma A.2 Under AssumptionsAtoD, for alli, we have
Proof of Lemma A.2. (a) can be derived easily with AssumptionA.
Consider (b). with AssumptionsAandC, we have E
Consider (c). By Cauchy-Schwardz inequality, we have
E by AssumptionsA, B3, andC.
Consider (d). We have
E
by AssumptionsA, B8, andC.
Consider (e). We have
E
√1 N T
XN ℓ=1
ε′i[VℓV′ℓ−E(VℓV′ℓ)]
2
= 1 T
XT t=1
E
√1 N T
XN ℓ=1
XT s=1
εis[v′ℓsvℓt−E(v′ℓsvℓt)]
2
≤C
Consider (f). We have E
1
√TV′iF
2
= 1 T
XT s=1
XT t=1
tr [E(visv′it)]E fs0′ft0
≤ 1 T
XT s=1
XT t=1
kEvisv′itk
Efs02Eft021/2
≤C T
XT s=1
XT t=1
˜ σst≤C2
by AssumptionsB3andC. This completes the proof.
Lemma A.3 Under AssumptionsAtoD, we have (a) N−1T−1
XN i=1
X′iMbFXi−N−1T−1 XN i=1
X′iMF0Xi=Op(δ−NT1),
(b) N−1T−1 XN
i=1
XN j=1
XN ℓ=1
wijwiℓX′jMbFXℓ−N−1T−1 XN i=1
XN j=1
XN ℓ=1
wijwiℓX′jMF0Xℓ=Op(δ−NT1),
(c) N−1T−1 XN i=1
XN j=1
wijX′jMbFMbF−1Xi,−1−N−1T−1 XN i=1
XN j=1
wijX′jMF0MF0
−1Xi,−1=Op(δ−NT1) (d) N−1T−1
XN i=1
XN j=1
wijX′jMbFXi−N−1T−1 XN i=1
XN j=1
wijX′jMF0Xi=Op(δ−NT1),
(e) N−1T−1 XN i=1
X′i,−1MbF−1Xi,−1−N−1T−1 XN i=1
X′i,−1MF0
−1Xi,−1=Op(δNT−1), (f) N−1T−1
XN i=1
X′i,−1MbF−1MbFXi−N−1T−1 XN i=1
X′i,−1MF0
−1MF0Xi=Op(δNT−1), (g) N−1T−1
XN i=1
XN j=1
wijw′iY′MbFXj−N−1T−1 XN i=1
XN j=1
wijw′iY′MF0Xj =Op(δNT−1),
(h) N−1T−1 XN i=1
w′iY′MbF−1Xi,−1−N−1T−1 XN i=1
w′iY′MF0
−1Xi,−1=Op(δ−NT1), (i) N−1T−1
XN i=1
w′iY′MbFXi−N−1T−1 XN i=1
w′iY′MF0Xi=Op(δNT−1),
(j) N−1T−1 XN i=1
XN j=1
wijy′i,−1MbFXj−N−1T−1 XN i=1
XN j=1
wijy′i,−1MF0Xj=Op(δNT−1),
(k) N−1T−1 XN i=1
y′i,−1MbF−1Xi,−1−N−1T−1 XN i=1
y′i,−1MF0
−1Xi,−1=Op(δ−NT1), (l) N−1T−1
XN i=1
y′i,−1MbFXi−N−1T−1 XN
i=1
y′i,−1MF0Xi=Op(δNT−1),
(m) N−1T−1 XN i=1
X′i,−1MbF−1Xi−N−1T−1 XN i=1
X′i,−1MF0
−1Xi=Op(δ−NT1).
Proof of Lemma A.3. For (a), please see Lemma 6 in Norkute et al.(2020). For (b), the left hand side is bounded in
5
norm by
by LemmaA.1(g). With AssumptionsB, CandD, it’s easy to show that
Ekxitk4≤C (A.4)
for alli andt. In addition, we can derive that
E
by AssumptionE3and the equation (A.4). Then by Markov inequality, we derive that 1
Combining the above equations, we can derive that (b). Consider (c), we can bound the left hand side term as
Similar to the equation (A.5), we can show that 1 N
XN i=1
XN j=1
|wij| Xj
√T Xi,−1
√T
=Op(1)
with Markov inequality. Then we have (c). The cases (d) to (m) can be proved similarly. This completes the proof.
Lemma A.4 Under AssumptionsAtoE, we have (a) 1
N XN i=1
XN j=1
|wij|Γ0jϕ0i=Op(1)
(b) 1 N
XN i=1
XN j=1
|wij|Γ0jϕ0i=Op(1)
(c) 1 N
XN i=1
XN j=1
|wij|Γ0j εi
√T
=Op(1)
(d) 1 N
XN i=1
XN j=1
|wij|Γ0j ui
√T
=Op(1)
(e) 1 N
XN i=1
XN j=1
|wij|Γ0j 1
√TF0′εi
=Op(1), XN j=1
|wij|Γ0j 1
√TH0′εi
=Op(1)
(f) 1 N
XN i=1
XN j=1
|wij|Γ0j
1 N T
XN ℓ=1
ε′iE(VℓV′ℓ)F0
=Op(1) (g) 1
N XN i=1
XN j=1
|wij|Γ0j
√1 N T
XT ℓ=1
ε′i[VℓV′ℓ−E(VℓV′ℓ)]F0
=Op(1) (h) 1
N XN i=1
XN j=1
|wij|Γ0j
√1 N T
XT ℓ=1
ε′i[VℓV′ℓ−E(VℓV′ℓ)]
=Op(1) (i) 1
N XN i=1
XN j=1
|wij|Γ0j1 T
F0−FRb −1′ εi
=Op δ−NT2 Proof of Lemma A.4. Consider (a). By Cauchy-Schwardz inequality, we have
E 1 N
XN i=1
XN j=1
|wij|Γ0jϕ0i = 1
N XN i=1
XN j=1
|wij|E Γ0jϕ0i
≤1 N
XN i=1
XN j=1
|wij|
EΓ0j2Eϕ0i21/2
≤ C N
XN i=1
XN j=1
|wij| ≤C2
Then we derive that P
1 N
XN i=1
XN j=1
|wij|Γ0jϕ0i≥M
≤ EN1
PN i=1
PN
j=1|wij|Γ0jϕ0i
M ≤ C2
M
by the Markov inequality. Thus, for anyǫ >0, there existsM =C2/ǫ, such that
P
1 N
XN i=1
XN j=1
|wij|Γ0jϕ0i≥M
≤ǫ .
Then we complete the proof of (a). Following the proof of part (a), we can prove part (b) with Ekϕik2≤C, and prove part (c) with Lemma A.2(a).
7
Consider (d). Note that T−1/2F0 = Op(1) and T−1/2H0 =Op(1). With LemmasA.4 (a) to (c), we can prove (d) easily with the definition ofui.
Following the proof part of (a), we can prove part (e) with Lemma A.2(b), and we can prove part (f) with Lemma A.2(c).
Consider (g). We can show that
E
by AssumptionD and LemmaA.2(d).
Following the proof of (g), we can prove part (h) with AssumptionDand Lemma A.2(e).
Consider (i). With the equation (A.1), we have 1
Lemma A.5 Under AssumptionsAtoD, we have
√1 N T
XN i=1
XN j=1
wijΓ0j′F0′MbFui
=− 1
N3/2T1/2 XN i=1
XN ℓ=1
XN j=1
wijΓ0j′(Υ0)−1Γ0ℓV′ℓMbFui
− 1
N3/2T3/2 XN i=1
XN ℓ=1
XN j=1
wijΓ0j′(Υ0)−1 Fb′F0 T
!−1
Fb′VℓV′ℓMbFui+Op
1 δNT2
+Op
r T N3
!
Proof of Lemma A.5. First, we have
√1 N T
XN i=1
XN j=1
wijΓ0j′(F0−FRb −1)′MbFui
=− 1
√N T XN i=1
XN j=1
wijΓ0j′ 1 N T
XN ℓ=1
R−1′Ξ−1Fb′VℓΓ0ℓ′F0′MbFui
− 1
√N T XN i=1
XN j=1
wijΓ0j′ 1 N T
XN ℓ=1
R−1′Ξ−1Fb′F0Γ0ℓV′ℓMbFui
− 1
√N T XN i=1
XN j=1
wijΓ0j′ 1 N T
XN ℓ=1
R−1′Ξ−1Fb′VℓV′ℓMbFui
=B1+B2+B3
We consider the termB1. By LemmaA.1(h), we have 1
N T XN ℓ=1
Fb′VℓΓ0ℓ′ =R′ 1 N T
XN ℓ=1
F0′VℓΓ0ℓ′+ 1 N T
XN ℓ=1
Fb−F0R′
VℓΓ0ℓ′
=Op
1
√N T
+Op
1 N
+Op
√ 1 N δNT2
! .
(A.6)
Note that thatMbFF0=MbF
F0−FRb −1
andMbF=IT −T−1FbbF′. Given the equation (A.6), we can derive that
B1=− 1
√N T XN i=1
XN j=1
wijΓ0j′ 1 N T
XN ℓ=1
R−1′Ξ−1Fb′VℓΓ0ℓ′
F0−FRb −1′
ui (A.7)
− 1
√N T XN i=1
XN j=1
wijΓ0j′ 1 N T2
XN ℓ=1
R−1′Ξ−1Fb′VℓΓ0ℓ′
F0−FRb −1′
FRb ′F0′ui (A.8)
− 1
√N T XN i=1
XN j=1
wijΓ0j′ 1 N T2
XN ℓ=1
R−1′Ξ−1Fb′VℓΓ0ℓ′
F0−FRb −1′
Fb
Fb−F0R′
ui (A.9)
=B1.1+B1.2+B1.3.
kB1.1k ≤
√1 N T
XN i=1
XN j=1
wijΓ0j′ 1 N T
XN ℓ=1
R−1′Ξ−1Fb′VℓΓ0ℓ′
F0−FRb −1′
H0ϕ0i
(A.10)
+
√1 N T
XN i=1
XN j=1
wijΓ0j′ 1 N T
XN ℓ=1
R−1′Ξ−1Fb′VℓΓ0ℓ′
F0−FRb −1′
εi
(A.11)
9
≤√ Consider the termB2. With the definition ofR, B2can be reformulated as
B2=− 1
Consider the termB3. With the definition ofR, B3can be written as B3=− 1
N3/2T3/2 XN i=1
XN ℓ=1
XN j=1
wijΓ0j′(Υ0)−1 Fb′F0 T
!−1
Fb′VℓV′ℓMbFui
Combining the above three terms, we can complete the proof.
Lemma A.6 Under AssumptionsAtoD, we have
− 1
N3/2T1/2 XN i=1
XN ℓ=1
XN j=1
wijΓ0j′(Υ0)−1Γ0ℓV′ℓMbFui
=− 1
N3/2T1/2 XN i=1
XN ℓ=1
XN j=1
wijΓ0j′(Υ0)−1Γ0ℓV′ℓMF0ui
+ rT
N 1 N
XN i=1
XN j=1
wijΓ0j′(Υ0)−1 1 N
XN ℓ=1
XN h=1
Γ0ℓV′ℓVh
T Γ0h′
! (Υ0)−1
F0′F0 T
−1
F0′ui
T +Op
1 δNT
+Op
√T N
!
Proof of Lemma A.6. Note thatMF0−MbF =PbF−PF0 andPbF =T−1FbbF′. We can derive that
− 1
N3/2T1/2 XN i=1
XN ℓ=1
XN j=1
wijΓ0j′(Υ0)−1Γ0ℓV′ℓMbFui− − 1 N3/2T1/2
XN i=1
XN ℓ=1
XN j=1
wijΓ0j′(Υ0)−1Γ0ℓV′ℓMF0ui
= 1
N3/2T3/2 XN i=1
XN ℓ=1
XN j=1
wijΓ0j′(Υ0)−1Γ0ℓV′ℓ(bF−F0R)R′F0′ui
+ 1
N3/2T3/2 XN i=1
XN ℓ=1
XN j=1
wijΓ0j′(Υ0)−1Γ0ℓV′ℓF0R(bF−F0R)′ui
+ 1
N3/2T3/2 XN i=1
XN ℓ=1
XN j=1
wijΓ0j′(Υ0)−1Γ0ℓV′ℓ(bF−F0R)(bF−F0R)′ui
+ 1
N3/2T3/2 XN i=1
XN ℓ=1
XN j=1
wijΓ0j′(Υ0)−1Γ0ℓV′ℓF0 RR′−(T−1F0′F0)−1 F0′ui
=C1+C2+C3+C4
We first consider the last three terms. Consider the termC2. By Lemmas A.1(b) and (d), LemmasA.4(a), (b) and (i), we can derive that
kC2k ≤
1 N
XN i=1
XN j=1
|wij|Γ0jϕ0i
√1 N T
XN ℓ=1
Γ0ℓV′ℓF0 1
T
Fb−F0R′
H0
(Υ0)−1kRk
+
1 N
XN i=1
XN j=1
|wij|Γ0j1 T
Fb−F0R′
εi
√1 N T
XN ℓ=1
Γ0ℓV′ℓF0
(Υ0)−1kRk
=Op δNT−2 given the fact that (N T)−1/2PN
ℓ=1Γ0ℓV′ℓF0=Op(1) andui =H0ϕ0i +εi.
11
Consider the termC3. Sinceui=H0ϕ0i +εi, we can derive that
We first consider the termC1.2. We have
We consider the term C1.1. By the equation (A.1), we have The termC1.1.1 can be decomposed as
1
The termC1.1.1.1is bounded in norm by
√1
by LemmaA.1(e) and LemmaA.4(d). Similarly, we can show the termC1.1.1.2is bounded in norm by
√1
For the termC1.1.3, we can decompose it as 1
N3/2T1/2 XN i=1
XN ℓ=1
XN j=1
wijΓ0j′(Υ0)−1Γ0ℓV′ℓ 1 N T
XN h=1
E(VhV′h)F00R F0′Fˆ T
!−1
(Υ0)−1
F0′F0 T
−1
F0′ui
T
+ 1
N3/2T1/2 XN i=1
XN ℓ=1
XN j=1
wijΓ0j′(Υ0)−1Γ0ℓV′ℓ 1 N T
XN h=1
(VhV′h−E(VhV′h))F0R F0′Fˆ T
!−1
(Υ0)−1
F0′F0 T
−1
F0′ui
T
+ 1
N3/2T1/2 XN i=1
XN ℓ=1
XN j=1
wijΓ0j′(Υ0)−1Γ0ℓV′ℓ 1 N T
XN h=1
E(VhV′h)
Fb−F0R F0′Fˆ T
!−1
(Υ0)−1
F0′F0 T
−1
F0′ui
T
+ 1
N3/2T1/2 XN i=1
XN ℓ=1
XN j=1
wijΓ0j′(Υ0)−1Γ0ℓV′ℓ 1 N T
XN h=1
(VhV′h−E(VhV′h))
Fb−F0R F0′Fˆ T
!−1
(Υ0)−1
F0′F0 T
−1
F0′ui
T
which are bounded by 1 N
XN i=1
XN j=1
|wij|Γ0j 1
√Tui F0′Fb
T
!−1
F0′F0 T
−1 F0
√T
(Υ0)−12kRk
× 1
√T 1
T XT s=1
XT t=1
1
√N XN ℓ=1
Γ0ℓvℓs
ft0′E1 N
XN h=1
v′hsvht + 1
√N 1
√N T XN ℓ=1
Γ0ℓV′ℓ 1
√N T XN h=1
(VhV′h−E(VhV′h))F0
+ 1
√N T XN ℓ=1
Γ0ℓV′ℓ 1 N√
T XN h=1
E(VhV′h)Fb−F0R
√T +
rT N
1
√N T XN ℓ=1
Γ0ℓV′ℓ
√1 N T
XN
h=1
VhV′h−E(VhV′h)
Fb−F0R
√T
!
=Op
√T δ2NT
! +Op
1
√N
by LemmasA.1(a), (i), (m) and (n) and LemmaA.4(d), and the fact that
1 T
XT s=1
XT t=1
√1 N
XN ℓ=1
Γ0ℓvℓs
!
ft0′E 1 N
XN h=1
v′hsvht
!=Op(1)
14
because
by AssumptionsB3,CandD.
Combining the above terms, we can derive that
− 1
This completes the proof.
Lemma A.7 Under AssumptionsAtoD, we have
1
Proof of Lemma A.7. Denote the left-hand-side of the above equation
C5= 1
− 1
kC5.3.1k ≤
+
+ 1
×
Fb−F0R
√T
2
kRk
Fb−F0R′
H0 T
ϕ0i (A.81)
+√ T 1
N XN i=1
XN j=1
|wij|Γ0j′(Υ0)−1
Fb′F0 T
!−1
PN
ℓ=1(VℓV′ℓ−E(VℓV′ℓ))
√N T
(A.82)
×
Fb−F0R
√T
2
kRk
Fb−F0R′ εi
T
(A.83)
=Op
√
T /δ−N T2
by Lemma A.1 (a), Lemma A.2 (n) and Lemma A.4 (a) (e) (i). Thus we concludeC5 =Op
√
T /δN T−2
+Op T−1/2 as required.
Lemma A.8 Under AssumptionsAtoD, we have 1
N3/2T3/2 XN i=1
XN ℓ=1
XN j=1
wijΓ0j′(Υ0)−1 Fb′F0 T
!−1
Fb′E(VℓV′ℓ)MbFui
= 1
N3/2T3/2 XN i=1
XN ℓ=1
XN j=1
wijΓ0j′(Υ0)−1
F0′F0 T
−1
F0′E(VℓV′ℓ)MF0ui+Op
1
√T
+Op
√N T
!
ifT /N2→0 asN, T → ∞.
Proof of Lemma A.8. Note that by AssumptionsB5, we have
E(VℓV′ℓ)MbFui
≤µmax(E(VℓV′ℓ))MbFui
≤Ckuik and
E(VℓV′ℓ)
MbF−MF0 ui
≤µmax(E(VℓV′ℓ))
MbF−MF0 ui
≤CMbF−MF0kuik=Op(δNT−1)kuik In addition,
Fb′F0 T
!−1
Fb
√T
′
−
F0′F0 T
−1
F0
√T
′=
Fb′F0 T
!−1
Fb
√T
′
− Fb′F0 T
!−1
Fb
√T
′
PF0
=
Fb′F0 T
!−1
Fb
√T
′
MF0
=
Fb′F0 T
!−1
Fb−F0R′
√T MF0
≤
Fb′F0 T
!−1
Fb−F0R
√T
=Op δNT−1
20
By substracting and adding terms, we have 1
N3/2T3/2 XN
i=1
XN ℓ=1
XN j=1
wijΓ0j′(Υ0)−1 Fb′F0 T
!−1
Fb′E(VℓV′ℓ)MbFui
− 1
N3/2T3/2 XN i=1
XN ℓ=1
XN j=1
wijΓ0j′(Υ0)−1
F0′F0 T
−1
F0′E(VℓV′ℓ)MF0ui
= 1
N3/2T3/2 XN
i=1
XN ℓ=1
XN j=1
wijΓ0j′(Υ0)−1
Fb′F0 T
!−1
Fb′−
F0′F0 T
−1
F0′
E(VℓV′ℓ)MFbui
+ 1
N3/2T3/2 XN i=1
XN ℓ=1
XN j=1
wijΓ0j′(Υ0)−1
F0′F0 T
−1
F0′E(VℓV′ℓ)
MbF−MF0
ui
which are bounded in norm by rN
T 1 N2
XN i=1
XN j=1
XN ℓ=1
|wij|Γ0j
E(VℓV′ℓ)MbFui
√T
Fb′F0 T
!−1
Fb′
√T −
F0′F0 T
−1
F0′
√T
(Υ0)−1
+ rN
T 1 N2
XN i=1
XN j=1
XN ℓ=1
|wij|Γ0j
E(VℓV′ℓ)
MbF−MF0 ui
√T
F0′F0 T
−1 F0
√T
(Υ0)−1
≤ rN
T 1 N
XN i=1
XN j=1
|wij|Γ0j ui
√T
Fb′F0 T
!−1
Fb′
√T −
F0′F0 T
−1
F0′
√T
(Υ0)−1
+ 1 δNT
rN T
1 N
XN i=1
XN j=1
|wij|Γ0j ui
√T
F0′F0 T
−1 F0
√T
(Υ0)−1
=Op
√N δNT
√T
!
by the above three facts and LemmaA.4(d). This completes the proof.
21
Lemma A.9 Under AssumptionsAtoD, we have (a) 1
N XN
i=1
XN j=1
|wij| 1
√TVj
ϕ0i=Op(1)
(b) 1 N
XN i=1
XN j=1
|wij| 1
√TV′jF0
ϕ0i=Op(1)
(c) 1 N
XN i=1
XN j=1
|wij| 1
√TV′jF0
ϕ0i=Op(1)
(d) 1 N
XN i=1
XN j=1
|wij| 1
√TV′jF0 1
√Tε′iF0
=Op(1)
(e) 1 N
XN i=1
XN j=1
|wij| 1
√TVj
1
T
Fb−F0R′
εi
=Op
1 δ2NT
(f) 1 N
XN i=1
XN j=1
|wij| 1
√TV′jF0 1
T
Fb−F0R′ εi
=Op
1 δNT2
(g) 1 N
XN i=1
XN j=1
|wij| 1
TV′j
Fb−F0R
ϕ0i=Op
1 δNT
(h) 1 N
XN i=1
XN j=1
|wij| 1
TV′j
Fb−F0R
ϕ0i=Op
1 δNT
(i) 1 N
XN i=1
XN j=1
|wij| 1
TV′j
Fb−F0R 1
T
Fb−F0R′
εi
=Op
1 δ3NT
Proof of LemmaA.9. With AssumptionB1and LemmaA.2(f), we can easily prove parts (a)-(d) by following the proof of LemmaA.4(a).
Parts (e) and (f) can be proved similar to the proof ofA.4(i).
22
Consider (g). With the equation (A.1), we have LemmaA.9(e) and LemmaA.9(f). This completes the proof.
Lemma A.10 Under Assumptions AtoD, we have
√1
T−1F0
RR′− T−1F0′F0−1
F0′, we have
√1 N T
XN i=1
XN j=1
wijV′j
MbF−MF0 ui
=− 1
√N T XN i=1
XN j=1
wijV′j1
T(bFR−1−F0)
F0′F0 T
−1
F0′ui
− 1
√N T XN i=1
XN j=1
wijV′j1
T(FRb −1−F0) RR′−
F0′F0 T
−1! F0′ui
− 1
√N T XN i=1
XN j=1
wijV′j1
TF0R(bF−F0R)′ui− 1
√N T XN i=1
XN j=1
wijV′j1
T(bF−F0R)(bF−F0R)′ui
− 1
√N T XN i=1
XN j=1
wijV′j1
TF0 RR′−
F0′F0 T
−1! F0′ui
=D1+D2+D3+D4+D5
We first consider the last four terms. By LemmaA.1(f) and LemmasA.9(g), (h), we have
kD2k=
√1 N T
XN i=1
XN j=1
wijV′j1
T(bF−F0R)R−1 RR′−
F0′F0 T
−1! F0′ui
≤√
N T × 1 N
XN i=1
XN j=1
|wij| 1
TV′j(Fb−F0R)
ϕ0i1 TF0′H0
RR′−
F0′F0 T
−1 R−1
+√ N× 1
N XN i=1
XN j=1
|wij| 1
TV′j(Fb−F0R) 1
√TF0′εi RR′−
F0′F0 T
−1 R−1
=Op
√N T δNT3
!
ForD3, we have
kD3k=
√1 N T
XN i=1
XN j=1
wijV′j1
TF0R(bF−F0R)′ H0ϕ0i +εi
≤√ N× 1
N XN i=1
XN j=1
|wij| 1
√TV′jF0
ϕ0i1
T(bF−F0R)′H0 kRk
+√ N× 1
N XN i=1
XN j=1
|wij| 1
√TV′jF0 1
T(Fb−F0R)′εi
kRk
=Op
√N δNT2
!
by LemmasA.1(b), (d) and (e) and LemmasA.9(b), (c) and (f).
24
ForD4, we have
kD4k ≤√
N T × 1 N
XN i=1
XN j=1
|wij| 1
TV′j(bF−F0R)
ϕ0i1
T(bF−F0R)′H0 +√
N T × 1 N
XN i=1
XN j=1
|wij| 1
TV′j(bF−F0R) 1
T(bF−F0R)′εi
=Op
√N T δNT3
!
by LemmasA.1(b), (d) and LemmasA.9(g), (h) and (i).
ForD5, we have
kD5k=
√1 N T
XN i=1
XN j=1
wijV′j1
TF0 RR′−
F0′F0 T
−1!
F0′ H0ϕ0i +εi
≤√ N× 1
N XN i=1
XN j=1
|wij| 1
√TV′jF0
ϕ0i1 TF0′H0
RR′−
F0′F0 T
−1
+ rN
T × 1 N
XN i=1
XN j=1
|wij| 1
√TV′jF0 1
√TF0′εi
RR′−
F0′F0 T
−1
=Op
√N δNT2
!
by LemmaA.1(f) and LemmasA.9(b), (c) and (d).
Now we consider the termD1. With (A.1), we can decompose the termD1 as follows
− 1
√N T XN i=1
XN j=1
wijV′j1
T(bFR−1−F0)
F0′F0 T
−1
F0′ui
=− 1
√N T XN i=1
XN j=1
wijV′j 1 N T
XN h=1
VhΓ0h′(Υ0)−1
F0′F0 T
−1
F0′ui
− 1
√N T XN i=1
XN j=1
wijV′j 1 N T2
XN h=1
F0Γ0hV′hFb F0′Fˆ T
!−1
(Υ0)−1
F0′F0 T
−1
F0′ui
− 1
√N T XN i=1
XN j=1
wij
1 N T2
XN h=1
E V′jVh
V′hFb F0′Fˆ T
!−1
(Υ0)−1
F0′F0 T
−1
F0′ui
− 1
√N T XN i=1
XN j=1
wij
1 N T2
XN h=1
V′jVh−E V′jVh
V′hFb F0′Fˆ T
!−1
(Υ0)−1
F0′F0 T
−1
F0′ui
=D1.1+D1.2+D1.3+D1.4
25
We consider the last three termsD1.2,D1.3, andD1.4. ForD1.2, we can derive that
ForD1.3, we can decompose it as follows
− 1
which are bounded in norm by
Easily, we can show that
1
Similarly, we can show that
Combining the above equations and the equation (A.84), we can derive that D1.3=Op
Consider the termD1.4.1. With LemmaA.2(f), we haveN−1PN Then we can derive that
E
Then from the equation (A.85), we have 1
by Markov inequality. Thus, for the termD1.4.1, we can derive that kD1.4.1k
= vec
1
√T 1 N
XN h=1
√1 N T
XN i=1
XN j=1
wij V′jVh−E V′jVh
× 1
√TV′hF0
×R F0′Fˆ T
!−1
(Υ0)−1
F0′F0 T
−1
F0′H0 T ×ϕ0i
=
√1 T
1 N
XN h=1
1
√N T XN i=1
XN j=1
XT t=1
ϕ0i′⊗wij(vjtv′ht−E(vjtv′ht))
×
R F0′Fˆ T
!−1
(Υ0)−1
F0′F0 T
−1
F0′H0 T
′
⊗Ik
×vec 1
√T XT s=1
vhsfs0′
!
≤ 1
√T 1 N
XN h=1
√1 N T
XN i=1
XN j=1
XT t=1
ϕ0i′⊗wij(vjtv′ht−E(vjtv′ht))
√1 T
XT s=1
vhsfs0′
× kRk
F0′Fˆ T
!−1
(Υ0)−1
F0′F0 T
−1 F0′H0
T kIkk
=Op
1
√T 1
N XN h=1
√1 N T
XN i=1
XN j=1
XT t=1
ϕ0i′⊗wij(vjtv′ht−E(vjtv′ht))
√1 T
XT s=1
vhsfs0′
=Op
1
√T
with the above equation, where the third and the fourth equations use the fact that vec (ABC) = (C′⊗A)vec(B) for any comfortable matricesA, BandC. Similarly, we can show that the termD1.4.2isOp T−1/2
.
30
For the termD1.4.3, we have The equality can be proved similar to the equation (A.85).
Similarly, we can derive that the termD1.4.2 andD1.4.4 areOp δNT−1
. Combining the above termsD1.4.1to D1.4.4, we derive thatD1.4=Op δNT−1
. This completes the proof.
Proof of Proposition 3.1. With the definition ofZbi, we have
√1
ByNorkute et al. (2020), we have
√1
where
Combining LemmasA.5,A.6,A.7,A.8andA.10, we can derive that
√1
whereZi=PN
j=1wijMF0Xj,MF0
−1
Xi,−1,MF0Xi
,b1= (b′11,b′12,b′13)′ andb2= (b′21,b′22,b′23)′. This completes the proof.
Proof of Theorem3.1 We have
√N T bθ−θ
=
Ab′Bb−1Ab−1
Ab′Bb−1 1
√N T XN
i=1
Zb′iui.
By LemmaA.3, Ab −A→p 0and Bb −B →p 0, andBb−1−B−1 →p 0by continuous mapping theorem, thus, Ab′Bb−1Ab − A′B−1A →p 0. Under Assumption F A′B−1A is positive definite which implies
Ab′Bb−1Ab−1
Ab′Bb−1 = Op(1). By Proposition3.1 √1
N T
PN
i=1Zb′iui=√1 N T
PN
i=1Z′i H0ϕ0i +εi +q
T
Nb1+q
N
Tb2+op(1). First, due to the independence betweenεiandZi, a suitable central limit theorem ensures that √1
N T
PN
i=1Z′iεi=Op(1). Consider the first component of
√1 N T
PN
i=1Z′iH0ϕ0i. RecallingZi=PN
j=1wijMF0Vj,MF0MF0
−1Vi,−1,MF0Vi
whereVi=Vi−N1
PN
ℓ=1VℓΓ0ℓ′(Υ0)−1Γ0i, Vi,−1=Vi,−1−N1
PN
ℓ=1Vℓ,−1Γ0ℓ′(Υ0)−1Γ0i, and notingMF0 =MF= FFT′, √1 N T
PN i=1
PN
j=1wijVj′F
√T
F′H0
T ϕ0i =Op(1) due to the independence betweenVjandFand cross-sectional independence betweenPN
j=1wijVjandϕ0i. In a similar manner, we can show that other components in √1
N T
PN
i=1Z′iH0ϕ0i and the bias termsb1 and b2are Op(1). Together with the conditionN/T →c where 0< c≤ ∞asN, T → ∞, we have√
N T θb−θ
=Op(1) as required.
33