Online Appendix B: Proofs of the main theoretical results for the two-step IV estimator

LetΞbery×ry diagonal matrix that consist of the firstrylargest eigenvalues of theT×T matrix (N T)⁻¹PN i=1ubiub^′_i. Then by the definition of eigenvalues andH,b HΞb = (N T)⁻¹PN

i=1ubiub^′_iH. It’s easy to show thatb Ξis invertible following the proof of Proposition A.1 (i) inBai(2009) given the covergence rate of ˆθto θ is√

N T. Then Hb −H⁰R

= 1 N T

XN i=1

Ci

θ−θˆ θ−θˆ_′

C^′_iHΞb ⁻¹+ 1 N T

XN i=1

Ci

θ−θˆ

u^′_iHΞb ⁻¹+ 1 N T

XN i=1

ui

θ−θˆ_′

C^′_iHΞb ⁻¹ + 1

N T XN i=1

H⁰ϕ⁰_iε^′_iHb + 1 N T

XN i=1

εiϕ⁰_i^′H^0′HΞb ⁻¹+ 1 N T

XN i=1

εiε^′_iHΞb ⁻¹

(B.1)

whereR= _{N T}¹ PN

i=1ϕ⁰_iϕ⁰_i^′H^0′HΞb ⁻¹. Following the proof of Proposition A.1 (ii) in Bai(2009), we can show that Ris invertible.

Lemma B.1 Under AssumptionsBtoD, we have (a) T⁻¹ bH−H⁰R²=Op δ_NT⁻²

,

(b) T⁻¹

Hb −H⁰R′

H⁰=Op δ_NT⁻² , T⁻¹

Hb −H⁰R′

F⁰=Op δ⁻_NT² ,

(c) T⁻¹

Hb −H⁰R′

Hb =Op δ_NT⁻² ,

(e) Ξ=Op(1),R=Op(1),Ξ⁻¹=Op(1),R⁻¹=Op(1). (f) RR^′−

H^0′H⁰ T

−1

=Op δ_NT⁻² ,

(g) M^Hb−M^H0 =Op δ⁻_NT¹ ,

(h) 1

√N T XN ℓ=1

ϕ_ℓε^′_ℓ

Hb −H⁰R

=Op

rT N

! +Op

√T δ²_NT

! ,

(i)

√1 N T

XN h=1

(εhε^′_h−E(εhε^′_h))H⁰

=Op(1), (j)

√1 N T

XN h=1

H^0′(εhε^′_h−E(εhε^′_h))H⁰

=Op(1), (m)

1 N√

T XN ℓ=1

E(εℓε^′_ℓ)

=O(1),

(n) 1

√N T XN ℓ=1

[εℓε^′_ℓ−E(εℓε^′_ℓ)] =Op(1).

Proof of Lemma B.1. We can follow the way of the proof of Lemma A.1to prove this lemma. Thus, we omitted the details.

Lemma B.2 Under AssumptionsBtoD, we have 1

N T XN

i=1

Zb^′_iM^HbCi− 1 N T

XN i=1

Z^′_iM^H0Ci=Op δ_NT⁻¹

1 N T

XN i=1

Zb^′_iM^HbZbi− 1 N T

XN i=1

Z^′_iM^H0Zi=Op δ_NT⁻¹

34

Proof of Lemma B.2. With LemmaB.1 (g), we can follow the way of the proof of LemmaA.3 to prove this lemma.

Thus, we omitted the details.

Lemma B.3 Under AssumptionsA toD, we have (a) 1

Proof of LemmaB.3. Following the proof of LemmaA.9(a), (b) and (d), we can easily prove parts (a)-(c), respectively.

35

Consider (d). Since HbR⁻¹−H⁰

= 1 N T

XN i=1

Ci

θ−θˆ θ−ˆθ′

C^′_iHb H^0′Hˆ T

!−1

(Υ⁰_ϕ)⁻¹+ 1 N T

XN i=1

Ci

θ−ˆθ

u_i^′Hb H^0′Hˆ T

!−1

(Υ⁰_ϕ)⁻¹

+ 1 N T

XN i=1

ui

θ−ˆθ′

C^′_iHb H^0′Hˆ T

!−1

(Υ⁰_ϕ)⁻¹+ 1 N T

XN i=1

H⁰ϕ⁰_iε^′_iHb H^0′Hˆ T

!−1

(Υ⁰_ϕ)⁻¹

+ 1 N

XN i=1

εiϕ⁰_i^′(Υ⁰_ϕ)⁻¹+ 1 N T

XN i=1

εiε^′_iHb H^0′Hˆ T

!₋1

(Υ⁰_ϕ)⁻¹

36

We have

Consider the first result in (f). We have

by LemmasA.1(i), (m) and (n) and LemmaB.3(a). The other variant is derived in a similar manner.

Replacing ϕ⁰_i by _T¹

Hb −H⁰R′

ε_i, with Lemma B.3 (d) and (e), we can prove (g). Following the way of proof of

38

LemmaA.4(i), we can prove (h). Now consider (i).

1 N

XN ℓ=1

XN h=1

Γ⁰_ℓV^′_ℓVh

T Γ⁰_h^′

≤

1 N T

XN ℓ=1

XN h=1

XT t=1

Γ⁰_ℓE(VℓtV^′_ht)Γ⁰_h^′ +

1 N T

XN ℓ=1

XN h=1

XT t=1

Γ⁰_ℓ

VℓtV^′_ht−E(VℓtV^′_ht) Γ⁰_h^′

≤1 N

XN ℓ=1

XN h=1

kΓ⁰_ℓkkΓ⁰_hk¯σℓh+ 1

√T

1 N√

T XN ℓ=1

XN h=1

XT t=1

Γ⁰_ℓ

VℓtV^′_ht−E(VℓtV^′_ht) Γ⁰_h^′

=Op(1) since

E 1 N

XN ℓ=1

XN h=1

kΓ⁰_ℓkkΓ⁰_hkσ¯ℓh

!

≤ 1 N

XN ℓ=1

XN h=1

EkΓ⁰_ℓk²EkΓ⁰_hk^21/2

¯ σℓh≤ C

N XN ℓ=1

XN h=1

¯

σℓh≤C² by AssumptionB3.

Lemma B.4 Under AssumptionsA toD, we have

√1 N T

XN i=1

XN j=1

wijV^′_jMb^FM^Hbui= 1

√N T XN i=1

XN j=1

wijV^′_jM^F0M^H0εi+Op

1 δNT

+Op

√N T δ_NT³

!

Proof of Lemma B.4. SinceMH⁰ui=MH⁰εi and Mb^FM^Hb−MF⁰MH⁰=MF⁰

M^Hb −MH⁰

+

M^Fb−MF⁰

MH⁰+

Mb^F−MF⁰ M^Hb −MH⁰

we have

√1 N T

XN i=1

XN j=1

wijV^′_jMb^FM^Hbui− 1

√N T XN i=1

XN j=1

wijV^′_jM^F0M^H0εi

= 1

√N T XN i=1

XN j=1

wijV^′_jM^F0

M^Hb −M^H0

ui+ 1

√N T XN i=1

XN j=1

wijV^′_j

Mb^F−M^F0 M^H0ui

+ 1

√N T XN i=1

XN j=1

wijV^′_j

Mb^F−MF⁰ M^Hb−MH⁰

ui

=H₁+H₂+H₃

39

Now we consider the term H₁. Since M^Hb −MH⁰ = −T⁻¹(Hb −H⁰R)R^′H^0′−T⁻¹H⁰R(Hb −H⁰R)^′ −T⁻¹(Hb − H⁰R)(Hb −H⁰R)^′−T⁻¹H⁰

RR^′− T⁻¹H^0′H⁰−1

H^′0, we have

√1 N T

XN i=1

XN j=1

wijV^′_jM^F0

M^Hb −M^H0 ui

=− 1

√N T XN i=1

XN j=1

wijV^′_j1

T(HbR⁻¹−H⁰)

H^0′H⁰ T

−1

H^0′ui

− 1

√N T XN i=1

XN j=1

wijV^′_j1

T(HbR⁻¹−H⁰) RR^′−

H^0′H⁰ T

−1! H^0′ui

− 1

√N T XN i=1

XN j=1

wijV^′_j1

TH⁰R(Hb −H⁰R)^′ui− 1

√N T XN i=1

XN j=1

wijV^′_j1

T(Hb −H⁰R)(Hb −H⁰R)^′ui

− 1

√N T XN i=1

XN j=1

wijV^′_j1

TH⁰ RR^′−

H^0′H⁰ T

−1! H^0′ui

+ 1

√N T XN i=1

XN j=1

wijV^′_jF⁰ F^0′F⁰−1

F^0′1

T(HbR⁻¹−H⁰)

H^0′H⁰ T

−1

H^0′ui

+ 1

√N T XN i=1

XN j=1

wijV^′_jF⁰ F^0′F⁰−1

F^0′1

T(HbR⁻¹−H⁰) RR^′−

H^0′H⁰ T

−1! H^0′ui

+ 1

√N T XN i=1

XN j=1

wijV^′_jF⁰ F^0′F⁰₋1

F^0′1

TH⁰R(Hb −H⁰R)^′ui

+ 1

√N T XN i=1

XN j=1

wijV^′_jF⁰ F^0′F⁰−1

F^0′1

T(Hb −H⁰R)(Hb −H⁰R)^′ui

+ 1

√N T XN i=1

XN j=1

wijV^′_jF⁰ F^0′F⁰−1

F^0′1

TH RR^′−

H^0′H⁰ T

−1! H^0′ui

=H_1.1+H_1.2+H_1.3+H_1.4+H_1.5+H_1.6+H_1.7+H_1.8+H_1.9

We first consider the termsH_1.2 toH_1.9. Note thatui=H⁰ϕ⁰_i +εi. ForH_1.2, we have

kH_1.2k=

√1 N T

XN i=1

XN j=1

wijV^′_j1

T(Hb −H⁰R)R⁻¹ RR^′−

H^0′H⁰ T

−1! H^0′ui

≤√

N T× 1 N

XN i=1

XN j=1

|wij| 1

TV^′_j(Hb −H⁰R)

ϕ⁰_i1 TH^0′H⁰

RR^′−

H^0′H⁰ T

−1 R⁻¹

+√ N× 1

N XN

i=1

XN j=1

|wij| 1

TV^′_j(Hb −H⁰R) 1

√TH^0′εi

RR^′−

H^0′H⁰ T

−1 R⁻¹

=Op

√N T δ_NT³

!

by LemmaA.1(f) and LemmasA.9(g), (h).

40

ForH_1.3, we have

by Lemma B.1 (b), Lemmas A.9 (b)-(c) and Lemma B.3 (c). Similarly, we can show that H_1.7 = Op Now we consider the termH_1.1. Since

HbR⁻¹−H⁰ we can decompose the termH_1.1 as follows

− 1

We consider the last five termsH_1.1.2 toH_1.1.6. ForH_1.1.2, we can derive that

For the termH_1.1.5, we have

For the termH_1.1.6, we have

Combining the above terms, we can show that H₁=− 1

− 1

by Lemma B.3 (e). In similar manner,

kH_2.2k=

by Lemma B.1 (b) and Lemma B.3 (e) (g).

kH_2.4k=

√1 N T

XN i=1

XN j=1

wijV^′_jT⁻¹F⁰

RR^′− T⁻¹F^0′F⁰−1

F^0′M^H0ui

(B.18)

≤

√1 N T

XN i=1

XN j=1

wijV^′_jT⁻¹F⁰

RR^′− T⁻¹F^0′F⁰−1 F^0′εi

(B.19)

+

√1 N T

XN i=1

XN j=1

wijV^′_jT⁻¹F⁰

RR^′− T⁻¹F^0′F⁰−1

F^0′H⁰ H^0′H⁰−1

H^0′εi

(B.20)

≤ rN

T 1 N

XN i=1

XN j=1

|wij|

V^′_jF⁰

√T

RR^′− T⁻¹F^0′F⁰−1 F^0′εi

√T

(B.21)

+ rN

T 1 N

XN i=1

XN j=1

|wij|

V^′_jF⁰

√T

RR^′− T⁻¹F^0′F⁰−1 F^0′H⁰

T

H^0′H⁰ T

−1 H^0′εi

√T

(B.22)

=Op

rN T

1 δ²_{N T}

!

by Lemma B.1 (f) and B.3 (c). To sum up,kH₂k=Op

√ N δ_{N T}²

+Op

√ N T δ³_{N T}

. Now we consider the termH₃. Using

Mb^F−MF⁰ =−T⁻¹(bF−F⁰R)R^′F^0′−T⁻¹F⁰R(bF−FR)^0′−T⁻¹(bF−F⁰R)(bF−F⁰R)^′−T⁻¹F⁰

RR^′− T⁻¹F^0′F⁰−1 F^0′ andM^Hb−M^H0 =−T⁻¹(Hb−H⁰R)R^′H^0′−T⁻¹H⁰R(Hb−H⁰R)^′−T⁻¹(Hb−H⁰R)(Hb−H⁰R)^′−T⁻¹H⁰

RR^′− T⁻¹H^0′H⁰−1 H we have

H₃= 1

√N T XN i=1

XN j=1

wijV^′_j

Mb^F−M^F0 M^Hb−M^H0

ui (B.23)

= 1

√N T XN i=1

XN j=1

wijV^′_jT⁻¹(bF−F⁰R)R^′F^0′T⁻¹(Hb −H⁰R)R^′H^0′ui (B.24)

+ 1

√N T XN i=1

XN j=1

wijV^′_jT⁻¹(bF−F⁰R)R^′F^0′T⁻¹H⁰R(Hb −H⁰R)^′ui (B.25)

+ 1

√N T XN i=1

XN j=1

wijV^′_jT⁻¹(bF−F⁰R)R^′F^0′T⁻¹(Hb −H⁰R)(Hb −H⁰R)^′ui (B.26)

+ 1

√N T XN i=1

XN j=1

wijV^′_jT⁻¹(bF−F⁰R)R^′F^0′T⁻¹H⁰

RR^′− T⁻¹H^0′H−1 H^0′ui

+ 1

√N T XN i=1

XN j=1

wijV_j^′T⁻¹F⁰R(Fb−F⁰R)^′T⁻¹(Hb −H⁰R)R^′H^0′ui (B.27)

+ 1

√N T XN i=1

XN j=1

wijV_j^′T⁻¹F⁰R(bF−F⁰R)^′T⁻¹H⁰R(Hb −H⁰R)^′ui (B.28)

+ 1

√N T XN i=1

XN j=1

wijV_j^′T⁻¹F⁰R(Fb−F⁰R)^′T⁻¹(Hb −H⁰R)(Hb −H⁰R)^′ui (B.29)

+ 1

√N T XN i=1

XN j=1

wijV_j^′T⁻¹F⁰R(bF−F⁰R)^′T⁻¹H⁰

RR^′− T⁻¹H^0′H−1 H^0′ui

47

+ 1

√N T XN

i=1

XN j=1

wijV^′_jT⁻¹(Fb−F⁰R)(Fb−F⁰R)^′T⁻¹(Hb −H⁰R)R^′H^′ui (B.30)

+ 1

√N T XN

i=1

XN j=1

wijV^′_jT⁻¹(bF−F⁰R)(bF−F⁰R)^′T⁻¹H⁰R(Hb −H⁰R)^′ui (B.31)

+ 1

√N T XN

i=1

XN j=1

wijV^′_jT⁻¹(Fb−F⁰R)(Fb−F⁰R)^′T⁻¹(Hb −H⁰R)(Hb −H⁰R)^′ui (B.32)

+ 1

√N T XN

i=1

XN j=1

wijV^′_jT⁻¹(bF−F⁰R)(bF−F⁰R)^′T⁻¹H⁰

RR^′− T⁻¹H^0′H⁰−1 H^0′ui

+ 1

√N T XN

i=1

XN j=1

wijV^′_jT⁻¹F⁰

RR^′− T⁻¹F^0′F⁰₋1

F^0′T⁻¹(Hb −H⁰R)R^′H^0′ui (B.33)

+ 1

√N T XN

i=1

XN j=1

wijV^′_jT⁻¹F⁰

RR^′− T⁻¹F^0′F⁰−1

F^0′T⁻¹H⁰R(Hb −H⁰R)^′ui (B.34)

+ 1

√N T XN

i=1

XN j=1

wijV^′_jT⁻¹F⁰

RR^′− T⁻¹F^0′F⁰−1

F^0′T⁻¹(Hb −H⁰R)(Hb −H⁰R)^′ui (B.35)

+ 1

√N T XN

i=1

XN j=1

wijV^′_jT⁻¹F⁰

RR^′− T⁻¹F^0′F⁰−1

F^0′T⁻¹H⁰

RR^′− T⁻¹H^0′H⁰−1 H^0′ui

=H_3.1.1+H_3.1.2+H_3.1.3+H_3.1.4+ (B.36)

H_3.2.1+H_3.2.2+H_3.2.3+H_3.2.4+ (B.37)

H_3.3.1+H_3.3.2+H_3.3.3+H_3.3.4+ (B.38)

H_3.4.1+H_3.4.2+H_3.4.3+H_3.4.4.

kH_3.1.1k=

√1 N T

XN i=1

XN j=1

wijV^′_jT⁻¹(bF−F⁰R)R^′F^0′T⁻¹(Hb −H⁰R)R^′H^0′ui

(B.39)

≤

√1 N T

XN i=1

XN j=1

wijV^′_jT⁻¹(bF−F⁰R)R^′F^0′T⁻¹(Hb −H⁰R)R^′H^0′H⁰ϕ⁰_i

(B.40)

+

√1 N T

XN i=1

XN j=1

wijV^′_jT⁻¹(bF−F⁰R)R^′F^0′T⁻¹(Hb −H⁰R)R^′H^0′εi

≤√ N T 1

N XN i=1

XN j=1

|wij|T⁻¹V^′_j(Fb−F⁰R)kR^′kT⁻¹F^0′(Hb −H⁰R)R^′H^0′H⁰ T

ϕ⁰_i (B.41)

+√ N 1

N XN i=1

XN j=1

|wij|T⁻¹V^′_j(bF−F⁰R)kR^′kT⁻¹F^0′(Hb −H⁰R) H^0′εi

√T

(B.42)

=Op

√N T δ_{N T}³

!

by Lemma B.1 (b) and Lemma B.3 (e) (f).

kH_3.1.2k=

√1 N T

XN i=1

XN j=1

wijV^′_jT⁻¹(bF−F⁰R)R^′F^0′T⁻¹H⁰R(Hb −H⁰R)^′ui

(B.43)

48

≤

+√

+

≤√

+

=Op Then, we have

√1

Now we consider the termH_1.1.1, then we have H_1.1.1=− 1

The termH_1.1.1.2 is bounded in norm by

This completes the proof.

Lemma B.5 Under AssumptionsA toD, we have

√1 N T

XN i=1

XN j=1

wijΓ⁰_j^′F^0′Mb^FM^Hbui

=− 1

N^3/2T^1/2 XN i=1

XN ℓ=1

XN j=1

wijΓ⁰_j^′(Υ⁰)⁻¹Γ⁰_ℓV^′_ℓMb^FM^Hbui

− 1

N^3/2T^3/2 XN i=1

XN ℓ=1

XN j=1

wijΓ⁰_j^′(Υ⁰)⁻¹ Fb^′F⁰ T

!−1

Fb^′VℓV^′_ℓMb^FM^Hbui+Op

1 δ²_NT

+Op

r T N³

!

Proof of Lemma B.5. We can follow the way of the proof of Lemma A.5to prove this lemma. Thus, we omitted the details.

Lemma B.6 Under AssumptionsA toD, we have 1

N^3/2T^3/2 XN i=1

XN ℓ=1

XN j=1

wijΓ⁰_j^′(Υ⁰)⁻¹ Fb^′F⁰ T

!₋1

Fb^′VℓV^′_ℓMb^FM^Hbui=Op

1 δNT

+Op

√N T

! +Op

√T N

!

Proof of Lemma B.6. We can follow the way of the proof of Lemma A.7to show that 1

N^3/2T^3/2 XN i=1

XN ℓ=1

XN j=1

wijΓ⁰_j^′(Υ⁰)⁻¹ Fb^′F⁰ T

!−1

Fb^′VℓV^′_ℓMb^FM^Hbui

= 1

N^3/2T^3/2 XN i=1

XN ℓ=1

XN j=1

wijΓ⁰_j^′(Υ⁰)⁻¹ Fb^′F⁰ T

!−1

Fb^′E(VℓV^′_ℓ)Mb^FM^Hbui+Op

1 δNT

+Op

√T N

!

To proceed, we can follow the way of the proof of LemmaA.8to show that 1

N^3/2T^3/2 XN i=1

XN ℓ=1

XN j=1

wijΓ⁰_j^′(Υ⁰)⁻¹ Fb^′F⁰ T

!₋1

Fb^′E(VℓV^′_ℓ)Mb^FM^Hbui (B.103)

= 1

N^3/2T^3/2 XN i=1

XN ℓ=1

XN j=1

wijΓ⁰_j^′(Υ⁰)⁻¹

F^0′F⁰ T

−1

F^0′E(VℓV^′_ℓ)M^F0M^H0εi+Op

1

√T

+Op

√N T

!

(B.104)

We write

H₄= 1 N^3/2T^3/2

XN i=1

XN ℓ=1

XN j=1

wijΓ⁰_j^′(Υ⁰)⁻¹

F^0′F⁰ T

−1

F^0′E(VℓV^′_ℓ)MF⁰MH⁰εi (B.105)

= 1

N^3/2T^3/2 XN i=1

XN ℓ=1

XN j=1

wijΓ⁰_j^′(Υ⁰)⁻¹

F^0′F⁰ T

−1

F^0′E(VℓV^′_ℓ) (IT −P^H0−P^F0+P^H0P^F0)εi (B.106)

=H_4.1+H_4.2+H_4.3+H_4.4. First considerH_4.1.

vec (H_4.1) = vec



 1 N^3/2T^3/2

XN i=1

XN ℓ=1

XN j=1

wijΓ⁰_j^′Γ⁰_ℓ

F^0′F⁰ T

−1

F^0′E(VℓV^′_ℓ)εi



 (B.107)

= vec



 1 N^3/2T^3/2

XN i=1

XN ℓ=1

XN j=1

wijΓ⁰_j^′Υ⁰)⁻¹

F^0′F⁰ T

−1XT s=1

XT t=1

fsE(v^′_ℓsvℓt)εit



 (B.108)

= 1

N^3/2T^3/2 XN i=1

XN ℓ=1

XN j=1

XT s=1

XT t=1

wijE(v^′_ℓsvℓt)εitf_s^0′⊗Γ⁰_j^′vec (Υ⁰)⁻¹ F^0′F

T

−1!

(B.109)

55

= 1

by LemmaA.1(m) and AssumptionB3. Then we have 1

. Next, in a similar manner we obtain

vec (H_4.2) = vec

Similar toH_4.1we can show that

1

. Combining the above facts, we derive thatH₄=Op 1

T

,which completes the proof.

56

Lemma B.7 Under AssumptionsA toD, we have

− 1

N^3/2T^1/2 XN

i=1

XN ℓ=1

XN j=1

wijΓ⁰_j^′(Υ⁰)⁻¹Γ⁰_ℓV^′_ℓMb^FM^Hbui

=− 1

N^3/2T^1/2 XN

i=1

XN ℓ=1

XN j=1

wijΓ⁰_j^′(Υ⁰)⁻¹Γ⁰_ℓV^′_ℓMF⁰MH⁰εi+Op

1 δNT

+Op

√T N

!

Proof of Lemma B.7. We can follow the way of the proof of Lemma A.6to show that 1

N^3/2T^1/2 XN i=1

XN ℓ=1

XN j=1

wijΓ⁰_j^′(Υ⁰)⁻¹Γ⁰_ℓV^′_ℓM^F0M^Hbui− 1 N^3/2T^1/2

XN i=1

XN ℓ=1

XN j=1

wijΓ⁰_j^′(Υ⁰)⁻¹Γ⁰_ℓV^′_ℓMb^FM^Hbui

= rT

N 1 N

XN i=1

XN j=1

wijΓ⁰_j^′(Υ⁰)⁻¹ 1 N

XN ℓ=1

XN h=1

Γ⁰_ℓV^′_ℓVh

T Γ⁰_h^′

! (Υ⁰)⁻¹

F^0′F⁰ T

−1F^0′M^Hbui

T +Op

1 δNT

+Op

√T N

!

(B.114) Now we consider the first term on the right hand side in (B.114). Note thatMH⁰ui=MH⁰εiandMH⁰−M^Hb =P^Hb−PH⁰. We can derive that

− 1

N^3/2T^1/2 XN

i=1

XN ℓ=1

XN j=1

wijΓ⁰_j^′(Υ⁰)⁻¹Γ⁰_ℓV^′_ℓMF⁰M^Hbui−



− 1 N^3/2T^1/2

XN i=1

XN ℓ=1

XN j=1

wijΓ⁰_j^′(Υ⁰)⁻¹Γ⁰_ℓV^′_ℓMF⁰MH⁰εi





= 1

N^3/2T^1/2 XN i=1

XN ℓ=1

XN j=1

wijΓ⁰_j^′(Υ⁰)⁻¹Γ⁰_ℓV^′_ℓM^F0P^Hbui− 1 N^3/2T^1/2

XN i=1

XN ℓ=1

XN j=1

wijΓ⁰_j^′(Υ⁰)⁻¹Γ⁰_ℓV^′_ℓM^F0P^H0ui

= 1

N^3/2T^1/2 XN i=1

XN ℓ=1

XN j=1

wijΓ⁰_j^′(Υ⁰)⁻¹Γ⁰_ℓV^′_ℓ1

TMF⁰HbHb^′ui− 1 N^3/2T^1/2

XN i=1

XN ℓ=1

XN j=1

wijΓ⁰_j^′(Υ⁰)⁻¹Γ⁰_ℓV^′_ℓMF⁰PH⁰ui

= 1

N^3/2T^1/2 XN i=1

XN ℓ=1

XN j=1

wijΓ⁰_j^′(Υ⁰)⁻¹Γ⁰_ℓV^′_ℓ1 TM^F0

Hb −H⁰R

R^′H^0′ui

+ 1

N^3/2T^1/2 XN

i=1

XN ℓ=1

XN j=1

wijΓ⁰_j^′(Υ⁰)⁻¹Γ⁰_ℓV^′_ℓ1

TM^F0H⁰R

Hb −H⁰R′

ui

+ 1

N^3/2T^1/2 XN

i=1

XN ℓ=1

XN j=1

wijΓ⁰_j^′(Υ⁰)⁻¹Γ⁰_ℓV^′_ℓ1 TM^F0

Hb −H⁰R Hb −H⁰R′

ui

+ 1

N^3/2T^1/2 XN

i=1

XN ℓ=1

XN j=1

wijΓ⁰_j^′(Υ⁰)⁻¹Γ⁰_ℓV^′_ℓ1

TM^F0H⁰ RR^′−

H^0′H⁰ T

−1! H^0′ui

=I₁+I₂+I₃+I₄

We first consider the last three terms. Consider the termI₂. By LemmasA.1(b) and (d), Lemmas A.4(a), (b) and (i),

57

we can derive that

Consider the termI₃. We can derive that

kI₃k ≤

by LemmaA.1(f) and LemmaA.4(d).

We first consider the termI_1.2. We have

We consider the term I_1.1. By the equation (B.1), we have 1

For the termI_1.1.1, we have

The term I_1.1.1.1 is bounded in norm by

√1

by LemmaA.1(e) and LemmaA.4(d). Similarly, we can show the termI_1.1.1.2is bounded in norm by

√1

For the termI_1.1.3, with the fact thatE(εhε^′_h) =σ_ε²IT, we can decompose it as which are bounded in norm by

1

For the termI_1.1.5, with the fact thatkMF⁰Chk ≤ kChk, we have

Similarly, we can derive thatI_1.1.6=Op

√

which is bounded in norm by 1

given the facts that

by AssumptionB3.

Combining the above terms, for the first term on the right hand side in (B.114), we can derive that

− 1

Next, we consider the second term on the right hand side in (B.114). It can be written as rT

+

by Lemma A.4 (a) (b), Lemma B.1 (b) and Lemma B.3 (h) (i).

kI_6.3k ≤

rT N

1 N

XN i=1

XN j=1

wijΓ⁰_j^′(Υ⁰)⁻¹ 1 N

XN ℓ=1

XN h=1

Γ⁰_ℓV^′_ℓVh

T Γ⁰_h^′

! (Υ⁰)⁻¹

F^0′F⁰ T

−1

T⁻²F^0′

Hb −H⁰R Hb −H⁰R′

H⁰ϕ⁰_i +

rT N

1 N

XN i=1

XN j=1

wijΓ⁰_j^′Γ⁰_ℓ 1 N

XN ℓ=1

XN h=1

Γ⁰_ℓV^′_ℓVh

T Γ⁰_h^′

! (Υ⁰)⁻¹

F^0′F⁰ T

−1

T⁻²F^0′

Hb −H⁰R Hb −H⁰R′

εi

≤ rT

N 1 N

XN i=1

XN j=1

|wij|Γ⁰_j^′(Υ⁰)⁻¹ 1 N

XN ℓ=1

XN h=1

Γ⁰_ℓV^′_ℓVh

T Γ⁰_h^′

(Υ⁰)⁻¹

F^0′F⁰ T

−1

F^0′

Hb −H⁰R T

×

Hb −H⁰R_′ H⁰ T

ϕ⁰_i

+ rT

N 1 N

XN i=1

XN j=1

|wij|Γ⁰_j^′(Υ⁰)⁻¹

1 N

XN ℓ=1

XN h=1

Γ⁰_ℓV^′_ℓVh

T Γ⁰_h^′

(Υ⁰)⁻¹

F^0′F⁰ T

−1

F^0′

Hb −H⁰R T

×

Hb −H⁰R′

εi

T

=Op

rT N

1 δ⁴_{N T}

!

by Lemma A.4 (a) (b), Lemma B.1 (b) and Lemma B.3 (h) (i).

65

kI_6.4k

by Lemma A.4 (a) (b) (e), Lemma B.1 (f) and Lemma B.3 (i). Therefore, noting that T /N tends to a finite positive constant we concludeI₆=Op δ_{N T}⁻² with (B.115), we complete the proof.

66

Lemma B.8 Under AssumptionsAtoD, we have

√1 N T

XN i=1

Γ⁰_i^′F⁰₋^′₁Mb^F−1M^Hbui

=− 1

N^3/2T^1/2 XN i=1

XN ℓ=1

Γ⁰_i^′(Υ⁰)⁻¹Γ⁰_ℓV^′_ℓ,−1Mb^F−1M^Hbui

− 1

N^3/2T^3/2 XN i=1

XN ℓ=1

Γ⁰_i^′(Υ⁰)⁻¹ Fb^′₋₁F⁰₋₁ T

!−1

Fb^′₋₁Vℓ,−1V^′_ℓ,−1Mb^F−1M^Hbui+Op δ_NT⁻² +Op

r T N³

!

Proof of Lemma B.8. First, we have

√1 N T

XN i=1

Γ⁰_i^′(F⁰₋₁−Fb₋1R⁻¹)^′Mb^F−1M^Hbui

=− 1

√N T XN i=1

Γ⁰_i^′ 1 N T

XN ℓ=1

R^−1′Ξ⁻_L¹Fb^′₋₁Vℓ,−1Γ⁰_ℓ^′F⁰₋^′₁Mb^F−1M^Hbui

− 1

√N T XN i=1

Γ⁰_i^′ 1 N T

XN ℓ=1

R^−1′Ξ⁻_L¹Fb^′₋₁F⁰₋₁Γ⁰_ℓV^′_ℓ,−1Mb^F−1M^Hbui

− 1

√N T XN i=1

Γ⁰_i^′ 1 N T

XN ℓ=1

R^−1′Ξ⁻_L¹Fb^′₋₁Vℓ,−1V^′_ℓ,−1Mb^F−1M^Hbui

=J₁+J₂+J₃

We consider the term J₁. By LemmaA.1(h), we have 1

N T XN ℓ=1

Fb^′₋₁Vℓ,−1Γ⁰_ℓ^′=R^′ 1 N T

XN ℓ=1

F⁰₋^′₁Vℓ,−1Γ⁰_ℓ^′+ 1 N T

XN ℓ=1

Fb₋1−F⁰₋₁R′

Vℓ,−1Γ⁰_ℓ^′

=Op

1

√N T

+Op

1 N

+Op

√ 1 N δ²_NT

! .

(B.123)

Note that thatM^HbH⁰=M^Hb

H⁰−HbR⁻¹

. Given the equation (B.123), we can derive that

kJ₁k=

With the definition ofR,J₂can be reformulated as J₂=−N^−3/2T^−1/2

Combining the above three terms, we can complete the proof.

Lemma B.9 Under AssumptionsAtoD, we have

− 1

Proof of Lemma B.9. Note thatM^H0εi =M^H0ui, we can derive that

− 1

N^3/2T^1/2 XN i=1

XN ℓ=1

Γ⁰_i^′(Υ⁰)⁻¹Γ⁰_ℓV^′_ℓ,−1Mb^F−1M^Hbui− − 1 N^3/2T^1/2

XN i=1

XN ℓ=1

Γ⁰_i^′(Υ⁰)⁻¹Γ⁰_ℓV^′_ℓ,−1MF⁰

−1MH⁰εi

!

=− 1

N^3/2T^1/2 XN i=1

XN ℓ=1

Γ⁰_i^′(Υ⁰)⁻¹Γ⁰_ℓV^′_ℓ,−1

Mb^F−1−MF⁰

−1

MH⁰εi

− 1

N^3/2T^1/2 XN i=1

XN ℓ=1

Γ⁰_i^′(Υ⁰)⁻¹Γ⁰_ℓV^′_ℓ,−1MF⁰

−1

M^Hb−M^H0 ui

− 1

N^3/2T^1/2 XN i=1

XN ℓ=1

Γ⁰_i^′(Υ⁰)⁻¹Γ⁰_ℓV^′_ℓ,−1

Mb^F−1−MF⁰

−1 M^Hb−M^H0 ui

=K₁+K₂+K₃

Now we consider the termK₁. Then

− 1

N^3/2T^1/2 XN i=1

XN ℓ=1

Γ⁰_i^′(Υ⁰)⁻¹Γ⁰_ℓV^′_ℓ,−1(Mb^F−1−MF⁰

−1)M^H0εi

= 1

N^3/2T^3/2 XN i=1

XN ℓ=1

Γ⁰_i^′(Υ⁰)⁻¹Γ⁰_ℓV^′_ℓ,−1(Fb₋1−F⁰₋₁R)R^′F⁰₋^′₁M^H0εi

+ 1

N^3/2T^3/2 XN i=1

XN ℓ=1

Γ⁰_i^′(Υ⁰)⁻¹Γ⁰_ℓV^′_ℓ,−1F⁰₋₁R(bF₋1−F⁰₋₁R)^′MH⁰εi

+ 1

N^3/2T^3/2 XN i=1

XN ℓ=1

Γ⁰_i^′(Υ⁰)⁻¹Γ⁰_ℓV^′_ℓ,−1(bF₋1−F⁰₋₁R)(bF₋1−F⁰₋₁R)^′MH⁰εi

+ 1

N^3/2T^3/2 XN i=1

XN ℓ=1

Γ⁰_i^′(Υ⁰)⁻¹Γ⁰_ℓV^′_ℓ,−1F⁰₋₁ RR^′−(T⁻¹F⁰₋^′₁F⁰₋₁)⁻¹

F⁰₋^′₁M^H0εi

=K_1.1+K_1.2+K_1.3+K_1.4 The term K_1.1 is bounded in norm by

kK_1.1k ≤ 1 N

XN i=1

Γ⁰_i

√1 N T

XN ℓ=1

Γ⁰_ℓV^′_ℓ,−1

Fb₋1−F⁰₋₁R

√T

kRk(Υ⁰)⁻¹

F⁰₋^′₁M^H0εi

√T

≤1 N

XN i=1

Γ⁰_i

√1 N T

XN ℓ=1

Γ⁰_ℓV^′_ℓ,−1

Fb₋1−F⁰₋₁R

√T

kRk(Υ⁰)⁻¹

× F⁰₋^′₁εi

√T +

F⁰₋^′₁F⁰ T

F^0′F⁰ T

−1 F^0′ε_i

√T +

F⁰₋^′₁H⁰ T

H^0′H^0′ T

−1 H^0′ε_i

√T

!

=Op

1 δNT

Similarly, with the fact that √¹ N T

PN

ℓ=1Γ⁰_ℓV^′_ℓ,−1F⁰₋₁=Op(1), we can show thatK_1.4=Op T^−1/2δ_NT⁻² .

For the termK_1.2, we have

This completes the proof.

Lemma B.10 Under AssumptionsA toD, we have N^−3/2T^−3/2

XN i=1

XN ℓ=1

Γ⁰_i^′(Υ⁰)⁻¹(T⁻¹Fb^′₋₁F⁰₋₁)⁻¹Fb^′₋₁Vℓ,−1V^′_ℓ,−1Mb^F−1M^Hbui

=Op(δ_NT⁻¹) +Op(N^1/2δ⁻_NT²) +Op(T^1/2δ_NT⁻²)

Proof of Lemma B.10. Following the way of the proof of LemmaB.6, we can prove this lemma. Thus, we omit the details.

Lemma B.11 Under AssumptionsA toD, we have

√1 N T

XN i=1

V^′_iMb^F−1M^Hbui= 1

√N T XN i=1

V^′_iMF⁰

−1M^H0εi+Op

1 δNT

+Op

√N T δ³_NT

!

Proof of Lemma B.11. Following the way of the proof of LemmaB.4, we can show that

√1 N T

XN i=1

V^′_iMb^F−1M^Hbui= 1

√N T XN i=1

V^′_iM^Fb⁻¹MH⁰εi+Op

1 δNT

+Op

√N T δ³_NT

!

Then, similar to the proof of the termH₂, we can prove that

√1 N T

XN i=1

V^′_iM^Fb−1M^H0εi= 1

√N T XN i=1

V^′_iMF⁰

−1M^H0εi+Op

√N T δ_NT³

!

This completes the proof.

Proof of Proposition 3.2. The termN^−1/2T^−1/2PN

i=1Zb^′_iM^Hbui, is equal to





N^−1/2T^−1/2PN i=1

PN

j=1wijX^′_jMb^FM^Hbui

N^−1/2T^−1/2PN

i=1X^′_i,−1Mb^F−1M^Hbui

N^−1/2T^−1/2PN

i=1X^′_iMb^FM^Hbui



 (B.124)

Consider the first term in (B.124). By LemmasB.4,B.5,B.6 andB.7and the fact thatMH⁰Xj =MH⁰Vj, we can derive that

√1 N T

XN i=1

XN j=1

wijX^′_jMb^FM^Hbui

= 1

√N T XN i=1

XN j=1

wijV^′_jMH⁰εi− 1 N^3/2T^1/2

XN i=1

XN ℓ=1

XN j=1

wijΓ⁰_j^′(Υ⁰)⁻¹Γ⁰_ℓV^′_ℓMH⁰εi+Op

1 δNT

+Op

√N T δ³_NT

!

= 1

√N T XN i=1

XN j=1

wijX^′_jM^H0εi+Op

1 δNT

+Op

√N T δ³_NT

!

whereX_j=Xj−_N¹ PN

ℓ=1XℓΓ⁰_ℓ^′(Υ⁰)⁻¹Γ⁰_j.

Consider the second term in (B.124). By LemmasB.8,B.9,B.10andB.11, we have

√1 N T

XN i=1

X^′_i,−1Mb^F−1M^Hbui = 1

√N T XN i=1

X^′_i,

−1MF⁰

−1MH⁰εi+Op

1 δNT

+Op

√N T δ_NT³

!

Similarly, for the third term, we can show that

√1 N T

XN i=1

X^′_iMb^FM^Hbui= 1

√N T XN i=1

X^′_iM^H0εi+Op

1 δNT

+Op

√N T δ³_NT

!

then

To proceed, we consider another asymptotic representation. Since

E

by Assumptions B3andE3. Then, we can derive that 1

Similarly, we can show that √¹ N T

PN i=1

PN

j=1wijV^′_jP^H0εi=Op(T^−1/2). Then

√1 N T

XN i=1

XN j=1

wijX^′_jM^H0εi

= 1

√N T XN i=1

XN j=1

wijV^′_jεi− 1

√N T XN i=1

XN j=1

wijV^′_jP^H0εi− 1 N^3/2T^1/2

XN i=1

XN ℓ=1

XN j=1

wijΓ⁰_j^′(Υ⁰)⁻¹Γ⁰_ℓV^′_ℓM^H0εi

= 1

√N T XN i=1

XN j=1

wijV^′_jεi+Op

1 δNT

Following the way of the proof of the above term, we can prove that

√1 N T

XN i=1

X^′_i,

−1MF⁰

−1M^H0εi= 1

√N T XN i=1

V^′_i,−1εi+Op

1 δNT

√1 N T

XN i=1

X^′

iM^H0εi= 1

√N T XN i=1

V^′_iεi+Op

1 δNT

Thus, we have

√1 N T

XN i=1

Zb^′_iM^Hbui=





√1 N T

PN i=1

PN

j=1wijV^′_jεi

√1 N T

PN

i=1V^′_i,−1εi

√1 N T

PN i=1V^′_iε_i



+Op

1 δNT

+Op

√N T δ³_NT

!

With the above proof, we can also show that

√1 N T

XN i=1

Zb^′_iM^Hbui=





√1 N T

PN i=1

PN

j=1wijX^′_jM^F0εi

√1 N T

PN

i=1X^′_i,−1MF⁰

−1εi

√1 N T

PN

i=1X^′_iMF⁰εi



+Op

1 δNT

+Op

√N T δ_NT³

!

This completes the proof.

Proof of Theorem 3.2. Substituting yi=Ciθ+ui intoecy and multiplying by√

N T we have

√N T(eθ−θ) =(Ae^′Be⁻¹A)e ⁻¹Ae^′Be⁻¹·N^−1/2T^−1/2 XN i=1

Ze^′_iM^Hbui.

With LemmasA.3andB.2, we have Ae −A=N⁻¹T⁻¹

XN i=1

Zb^′_iM^HbCi−N⁻¹T⁻¹ XN i=1

Z^′_iM^H0Ci=Op(δ_NT⁻¹), Bb−B=N⁻¹T⁻¹

XN i=1

Zb^′_iM^HbZbi−N⁻¹T⁻¹ XN i=1

Z^′_iM^H0Zi=Op(δ_NT⁻¹).

(B.125)

SinceZi= (PN

j=1wijMF⁰Xj,MF⁰

−1Xi,−1,MF⁰Xi), we have N⁻¹T⁻¹

XN i=1

Z^′_iP^H0Ci=





N⁻¹T⁻¹PN i=1

PN

j=1wijV^′_j(PH⁰−PF⁰)Ci

N⁻¹T⁻¹PN

i=1V^′_i,−1

PH⁰−PF⁰

−1PH⁰

Ci

N⁻¹T⁻¹PN

i=1V^′_i(P^H0−P^F0)Ci





The first block in the above matrix isOp(T^−1/2), since it is bounded in norm by kN⁻¹T⁻¹

XN i=1

XN j=1

wijV^′_jP^H0Cik+kN⁻¹T⁻¹ XN i=1

XN j=1

wijV^′_jP^F0Cik

≤T^−1/2·N⁻¹ XN i=1

XN j=1

|wij|kT^−1/2V^′_jH⁰kkT^−1/2Cik · kT^−1/2H⁰kk(T⁻¹H^0′H⁰)⁻¹k

+T^−1/2·N⁻¹ XN

i=1

XN j=1

|wij|kT^−1/2V^′_jF⁰kkT^−1/2Cik · kT^−1/2F⁰kk(T⁻¹F^0′F⁰)⁻¹k

=Op(T^−1/2),

with the similar argument, the other blocks can be proved to beOp(T^−1/2). Then N⁻¹T⁻¹

XN i=1

Z^′_iP^H0Ci=Op(T^−1/2). (B.126) A similar derivation gives

N⁻¹T⁻¹ XN i=1

Z^′_iP^H0Zi=Op(T^−1/2). (B.127) By Proposition3.2, (B.125)(B.126), and (B.127), we obtain

√N T .(eθ−θ) = (A^′B⁻¹A)⁻¹A^′B⁻¹·N^−1/2T^−1/2 XN i=1

Z^′_iεi+Op(δ⁻_NT¹) +Op(N^1/2T^1/2δ_NT⁻³).

As N, T → ∞ with N/T² → 0 and T /N² → 0, the central limit theorem of the martingale difference in Kelejian and Prucha(2001) can be applicable. Then

√N T(eθ−θ)−→^d N(0,Ψ).

With the equation (B.125), it’s sufficient to prove ˜σ²_ε is the consistent estimator ofσ_ε². Note that M^Hb(yi−Ci˜θ) =Ci(θ−˜θ) +M^HbH⁰ϕ⁰_i +M^Hbεi,

then

˜ σ²_ε−σ²_ε

=N⁻¹T⁻¹ XN i=1

(θ−θ)˜ ^′C^′_iCi(θ−θ)˜

+ 2N⁻¹T⁻¹ XN i=1

(θ−˜θ)^′C^′_iM^HbH⁰ϕ⁰_i + 2N⁻¹T⁻¹ XN i=1

(θ−˜θ)^′C^′_iM^Hbεi

+N⁻¹T⁻¹ XN i=1

ϕ⁰_i^′H^0′M^HbH⁰ϕ⁰_i + 2N⁻¹T⁻¹

XN i=1

ϕ⁰_i^′H^0′M^Hbε_i−N⁻¹T⁻¹ XN i=1

ε^′_iP^Hbε_i+N⁻¹T⁻¹ XN i=1

XT t=1

(ε²_it−σ²_ε).

The first term can be easily shown both to beOp(δ_NT⁻⁴). The last term is Op(N^−1/2T^−1/2). SinceM^HbH⁰= M^Hb(H⁰−HbR⁻¹), the second term is bounded in norm by

N⁻¹ XN i=1

kT^−1/2Cikkϕ⁰_ik · kT^−1/2(H⁰−HbR⁻¹)kkθ−θ˜k=Op(N^−1/2T^−1/2δ⁻_NT¹) +Op(δ_NT⁻⁴),

the fourth term is bounded in norm by N⁻¹

XN i=1

kϕ⁰_ik²· kT^−1/2(H⁰−HbR⁻¹)k²=Op(δ_NT⁻²), the fifth term is bounded in norm by

2N⁻¹ XN

i=1

kT^−1/2ε_ikkϕ⁰_ik · kT^−1/2(H⁰−HbR⁻¹)k=Op(δ⁻_NT¹). The third term is bounded in norm by

2N⁻¹T⁻¹ XN i=1

kT^−1/2CikkT^−1/2εik · kθ−θ˜k=Op(N^−1/2T^−1/2) +Op(δ_NT⁻³), the sixth term is bounded in norm by

N⁻¹T⁻² XN i=1

kε^′_iHbk²≤N⁻¹T⁻¹ XN i=1

kT^−1/2ε^′_iH⁰k²· kRk²+N⁻¹T⁻¹ XN i=1

kε_ik²· kT^−1/2(Hb −H⁰R)k², which isOp δ⁻_NT²

. Collecting the above terms, we have ˜σ_ε²−σ_ε²=Op(δ_NT⁻¹). Thus, we complete the proof.

Proof of Theorem3.3. Notingeui=ui−Ci

θe−θ

we have √¹ N T

PN

i=1Zb^′_iM^Hbeui= √¹ N T

PN

i=1Zb^′_iM^Hbui− Ae√

N T e θ−θ

. Since √ N T

e θ−θ

= A^′B⁻¹A−1

A^′B⁻¹√¹ N T

PN

i=1Z^′_iM^H0ε_i+op(1) by Proposition 3.2 and definingL =Ω^−1/2Awe have Ωˆ^−1/2√¹

N T

PN

i=1Zb^′_iM^Hbeui =MLΩ^−1/2√¹ N T

PN

i=1Z^′_iM^H0εi+op(1) withML=I3k−L(L^′L)⁻¹L^′ whose rank isν= 3k−(k+ 2), which yields

1 N T

PN

i=1ue′M^HbZb

iΩˆ⁻_{N T}¹ PN

i=1Zb^′_iM^Hbue→^d χ²_ν as required. Thus, we complete the proof.