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Shrinkage GMM estimation in conditional moment restriction

models,

Supplementary material: Technical appendix

Ryo Okui∗ Kyoto University November 16, 2009

Abstract

This paper is a supplementary Material for “Shrinkage GMM estimation in conditional mo-ment restriction” and collects the proofs of the theorems in the paper.

Insitute of Economic Research, Kyoto University, Yoshida-Hommachi, Sakyo, Kyoto, Kyoto, 606-8501, Japan,

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1

Proofs

This appendix contains the proofs of the theorems. λmax(A) denotes the maximum eigenvalue of a matrix A. “CS” and “ME” are the abbreviations of the Cauchy-Schwartz inequality and the maximum eigenvalue inequality, respectively.1

1.1 The proof of Theorem 1

Define R(β) E(E(ρi(β)|xi)2/σi2), ˆR(β) ≡ gˆ(β)′Υ( ˜ˆ β)− 1

ˆ

g(β), ˆRs(β) ≡ gˆ1(β)′Υ11( ˜ˆ β)−1gˆ1(β) +

s˜g2(β)′Υ22( ˜˜ β)−1˜g2(β), and ˆβ ≡argmaxβRˆ(β). Note that ˆβs = argminβRˆs(β). In the view of the proof of Theorem 5.3 in Donald, Imbens and Newey (2003), we have:

sup β |

ˆ

R(β)R(β)| →p 0, Rˆ( ˆβ)R(β0)→p0,

asR(β) is uniquely minimized atβ0 ∈Θ, Θ is compact by Assumption 1, andR(β) is continuous by Assumptions 1 and 2. Since ˆRs(β)−Rˆ(β) =Op(K2/N)→p 0 and ˆRs( ˆβ)−Rˆ( ˆβ) =Op(K2/N)→p 0 by 1s=Op(K2/N), we get

sup β |

ˆ

Rs(β)−R(β)| →p 0, Rˆs( ˆβs)−R(β0)→p 0.

Therefore, ˆβ→pβ0 by Lemma A.1 in Donald, Imbens and Newey (2003). Next, we show asymptotic normality. By the mean value theorem,

∂2ˆ Rs( ¯β)

∂β∂β′ ( ˆβ−β0) +

∂Rˆs(β0)

∂β = 0, where ¯β is a mean value. ∂2ˆ

Rs( ¯β)/∂β∂β′ →p E(σi−2did′i) by Lemmas 1 and 2. √

N ∂Rˆs(β0)/∂β→d N(0, E(σi−2did′i)). Therefore,

N( ˆβs−β0)→dN(0, E(σi−2did′i)− 1

).

1.2 The proof of Theorem 2

In various steps of our proofs, we use the approach developed by Donald, Imbens and Newey (2009). The objective function of the shrinkage GMM can be written as:

ˆ

g(β)′Ψˆ−1

( ˜β)ˆg(β),

where

ˆ

Ψ−1

((

I 0

0 s−1/2I

)

ˆ

F′Υ( ˜ˆ β) ˆF

(

I 0

0 s−1/2I

))−1

ˆ F′,

and

ˆ F =

(

I Υ11( ˜ˆ β)−1Υ12( ˜ˆ β)

0 I

)

.

Tedious but straightforward algebra shows that

ˆ Ψ =

Υ11 Υ12ˆ

ˆ

Υ21 s−1

( ˆΥ22Υ21ˆ Υˆ−111Υ12) + ˆˆ Υ21Υˆ− 1 11Υ12ˆ

)

.

1

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The shrinkage GMM estimator together with the auxiliary vector parameter λsolves 1 N ∑ i ∂ ∂βρi(β)q

iλ= 0, 1 N

i

qiρi(β) + ˆΨλ= 0.

For simplicity, assume the true value of parameter is 0 without loss of generality. Also assume the dimension ofβ is 1. The multidimensional case is analyzed in a very similar fashion.

Let θ= (β, λ′)and ˆθ is an estimator ofθ obtained by solving the following equation

ˆ

m(θ) = 1 N

i

mi(θ) = 0.

Then, ˆθ has an expansion:

ˆ

θ = M−1mM−1( ˆMM)ˆθ1 2

j ˆ

θjM−1Ajθˆ+R,

R = 1

2

j ˆ

θjM−1(Aj−Aˆj)ˆθ− 1 6 ∑ j ∑ k ˆ

θjθˆkM−1Bjk∗ θ,ˆ

where

mmˆ(0), Mˆ 1 N

i

∂ ∂θmi(0)

= 1

N

i

Mi(0), Aˆj ≡ 1 N ∑ i ∂ ∂θj

Mi(0),

M and Aj are population values (conditional on xis) of ˆM and ˆAj, respectively, and the rows of

B∗

jk are the second derivatives of the corresponding rows of Mi evaluated at θ∗, which is between ˆ

θ and 0. Note that each row ofB∗

jk has a different mean valueθ∗. Then, for the SGMM estimator,

m= ( 0 ¯ g )

, Mˆ =

(

0 Γ¯′

0 ¯

Γ0 Ψ( ˜ˆ β)

)

, M =

(

0 Γ′

0 Γ0 Ψ

)

,

ˆ Aj =

     (

0 Γ¯′

1 ¯ Γ1 0

)

ifj= 1,

0 otherwise,

Aj =

     (

0 Γ′

1 Γ1 0

)

ifj= 1,

0 otherwise,

B∗jk =

     (

0 Γ¯∗′

2 ¯ Γ∗

2 0

)

ifj= 1, k= 1

0 otherwise, where

¯ g 1

N

i

qiρi(0), Υ≡

1 N

i

σi2qiqi′, Γ0¯ ≡ 1 N ∑ i qi ∂ ∂βρi(0),

Γ1 1 N

i

qiE

(

∂2

∂β2ρi(0)|xi

)

, Γ1¯ 1 N

i

qi

∂2

∂β2ρi(0), Γ¯

∗ 2 ≡ 1 N ∑ i qi ∂3 ∂β3ρi(β

).

Noting that

M−1 =

(

−Ω−1

s Ω−s1Γ′0Ψ−1 Ψ−1

Γ0Ω−1

s Σ

)

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where Ωs≡Γ′0Ψ− 1

Γ0 and ΣΨ−1 −Ψ−1

Γ0Ω−1 s Γ′0Ψ−

1

, we have the following expansion. ˆ

β = Ω−s1Γ′0Ψ− 1

¯

g+Tβ +Rβ1,

Ω−s1(¯Γ0Γ0)′λˆΩ−s1Γ′0Ψ−1( ˆΨ( ˜β)Ψ)ˆλs−1Γ′0Ψ−1(¯Γ0Γ0) ˆβ +1

2Ω

−1

s Γ′1βˆλˆ− 1 2Ω

−1 s Γ′0Ψ−

1 Γ1βˆ2,

and

ˆ

λ = Σ¯g+Tλ+Rλ1,

≡ −Ψ−1Γ0Ω−s1(¯Γ0Γ0)′λˆΣ(¯Γ0Γ0) ˆβΣ( ˆΨ( ˜β)Ψ)ˆλ,

where Rβ1 = op(K 2

/N2/3

) by Lemma 4 and Rλ

1 = Op(1/N) by Lemma 6. Also, we define R β 2 ≡ Tβ+Rβ1 so that ˆβ=−Ω−

1

s Γ′0Ψ−1g¯+Rβ2 andR β

2 =Op(K/N) by Lemmas 3 and 4 andRλ2 ≡Tλ+Rλ1 so that ˆλ=Σ¯g+Rλ2 and Rλ2 =Op(ζ(K)

K/N) by Lemmas 5 and 6. Furthermore, Lemma 7 gives

ˆ

Ψ( ˜β)Ψ = ˆΥv1+ 2 ˜βΥρη+RΨ,

where ˆ Υv1=

1 N

i

(ρ2i σi)qiqi′, Υρη = 1 N

i

σρη,iqiqi′, and ||R Ψ

||=Op(ζ(K) √

K/N +K/N).

We rewrite ˆλby using ˆβ =Ω−1

s Γ′0Ψ−1¯g+Rβ2, ˆλ=−Σ¯g+Rλ2 and ˆΨ( ˜β)−Ψ = ˆΥv1+2 ˜βΥρη+RΨ so that

ˆ

λ = Σ¯g+ Ψ−1Γ0Ω−s1(¯Γ0Γ0)′Σ¯g+ Σ(¯Γ0Γ0)Ω−s1Γ′0Ψ−1¯g +Σ ˆΥv1Σ¯g+ 2 ˜βΣΥρηΣ¯g+Rλ3,

whereRλ

3 =op(K 2

/N3/2

) by Lemma 8.

Using these results, we can write each term ofTβ as

Ω−s1(¯Γ0Γ0)′λˆ = Ω−s1(¯Γ0Γ0)′Σ¯g+ Ω−s1(¯Γ0Γ0)′Σ ˆΥv1Σ¯g+Rβ3, Ωs−1Γ′0Ψ−1( ˆΨ( ˜β)Ψ)ˆλ = s−1Γ′0Ψ−1Υˆv1Σ¯g+ Ω−s1Γ′0Ψ−

1ˆ

Υv1Σ ˆΥv1Σ¯g −2 ˜βΩ−s1Γ′0Ψ−1ΥρηΣ¯g+Rβ4,

Ω−s1Γ′0Ψ−1(¯Γ0Γ0) ˆβ = Ω−s1Γ0Ψ′ −1(¯Γ0Γ0)Ω−s1Γ′0Ψ−1g¯+Rβ5, Ω−s1Γ′1βˆλˆ = Ω−

1

s Γ′1Σ¯gΩ− 1 s Γ′0Ψ−

1 ¯ g+Rβ6, Ωs−1Γ′0Ψ−1Γ1βˆ2 = Ω−s1Γ′0Ψ−1Γ1(Ω−s1Γ′0Ψ−1¯g)2+Rβ7,

where Rβ3 =op(K2/N3/2) by Lemma 9, R4β =op(K2/N3/2) by Lemma 10, Rβ5 =op(K2/N3/2) by Lemma 11, Rβ6 =op(K

2 /N2/3

) by Lemma 12, andRβ7 =op(K 2

/N2/3

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Summing up, we have the following expansion of β

Nβˆ = Ω−s1

h+

4

j=1

Tjh+Zh

,

h = √NΓ′0Ψ−1¯g, T1h = −

N(¯Γ0Γ0)′Σ¯g, T2h = −

NΓ′0Ψ−1Υˆv1Σ¯g,

T3h = 2 √

Nβ˜Γ′0Ψ−1ΥρηΣ¯g+ √

NΓ′0Ψ−1(¯Γ0Γ0)Ωs−1Γ′0Ψ−1¯g +1

2 √

NΓ′1Σ¯gΩ−s1Γ′0Ψ−1¯g1 2

NΓ′0Ψ−1Γ1(Ω−s1Γ′0Ψ−1g¯)2, T4h =

N(¯Γ0Γ0)′Σ ˆΥv1Σ¯g− √

NΓ′0Ψ−1Υˆv1Σ ˆΥv1Σ¯g,

Zh = √NΩs(Rβ1 +R β 3 −R

β 4 −R

β

5 + 1/2R β

6 −1/2R β 7), andh=Op(1),T1h=Op(K/

N) by Lemma 14,Th 2 =Op(

K/N) by Lemma 15,Th

3 =Op(1/ √

N) by Lemma 16, T4h =Op(ζ(K)K/

N) by Lemma 17 and Zh =op(K2/N). This means that

(√Nβˆ)2

= Ω−1 s hh′Ω−

1 s + Ω−

1 s

T1hT1h′+ 2 4

j=1 Tjhh′

Ω

−1

s +op(γK,N).

E(Tjhh′|X) =op(γK,N) for j= 1, . . . ,4 by Lemma 18, 19, 20, and 21.

Now Ωs−1E(hh′)Ωs−1 = Ω∗−1+ Ω∗−1(Ω∗Ω + (1s)2Ω2)Ω∗−1+op(γK,N). To show this, first it is easy to show that

E(hh′|X) = Ω1+s2Ω2= Ω∗2(Ω∗Ωs) + Ω∗−Ω + (1−s)2Ω2. Second, a simple matrix algebra shows that

Ω−s1= (Ω∗(Ω∗Ωs))−1 = Ω∗−1−Ω∗−1(Ω∗−Ωs)Ω∗−1+op(γK,N), since||(Ω∗

s)||2 =op(γK,N). Then, we have

Ω−s1E(hh′|X)Ω−s1 = (Ω∗−1Ω∗−1(Ω∗Ωs)Ω∗−1+op(γK,N)) ×(Ω∗2(Ω∗Ωs) + Ω∗−Ω + (1−s)2Ω2) ×(Ω∗−1Ω∗−1(Ω∗Ωs)Ω∗−1+op(γK,N))

= Ω∗−12Ω∗−1(Ω∗Ωs)Ω∗−1+ Ω∗−1(Ω∗−Ω + (1−s)2Ω2)Ω∗−1 +Ω∗−1(Ω∗Ωs)Ω∗−1+ Ω∗−1(Ω∗−Ωs)Ω∗−1+op(γK,N)

= Ω∗−1+ Ω∗−1(Ω∗Ω + (1s)2Ω2)Ω∗−1+op(γK,N). Then, by Lemma 22, the decomposition (3) in the main paper holds with

S(s) = s 2

Π22

N + Ω

∗−1

(Ω∗Ω + (1s)2Ω2)Ω∗−1.

2

Lemmas

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Lemma 1. ||ΓiΨ−1Γj||=Op(1) and ||ΓiΣΓj||=Op(1) for i, j= 0,1,2. Proof.

ΓiΨ−1Γj = ΓiΥ−1Γj+ (1−s)Γi,2Υ˜−221Γj,2.

||ΓiΥ−1Γj|| = Op(1) and ||Γi,2Υ−221Γj,2|| = Op(1) by Lemma A.3 in Donald, Imbens and Newey (2003). (1s) =Op(1) by the assumption. A similar argument shows that||ΓiΣΓj||=Op(1).

Lemma 2. ||ΓiΨ−1ΓjΓj)||=Op(1/N)and ||ΓiΣ(¯ΓjΓj)||=Op(1/N). Also, ||Γ0Ψ−1g¯||=

Op(1/√N) and ||Γ′0Σ¯g||=Op(1/ √

N) for i, j= 0,1,2. Proof.

ΓiΨ−1(¯Γj−Γj) = ΓiΥ−1(¯Γj−Γj) + (1−s)Γi,2Υ˜22−1(¯Γj,2−Γj,2). ||ΓiΥ−1(¯Γj−Γj)||=Op(1/

N) and||Γi,2Υ˜−221(¯Γj,2−Γj,2)||=Op(1/ √

N) by Lemma A.4 in Donald, Imbens and Newey (2003). 1s=Op(1) by the assumption. Others can be shown similarly.

Lemma 3. Tβ =Op(K/N).

Proof. Recall that Tβ has the form:

Tβ = Ω−s1(¯Γ0Γ0)′λˆΩ−s1Γ′0Ψ−1( ˆΨ( ˜β)Ψ)ˆλs−1Γ′0Ψ−1(¯Γ0Γ0) ˆβ +1

2Ω

−1

s Γ′1βˆλˆ− 1 2Ω

−1 s Γ′0Ψ−

1 Γ1βˆ2.

First, since ||(¯Γ0Γ0)′λˆ|| = O

p(K/N) by CS, ||Ω−s1(¯Γ0 −Γ0)′ˆλ|| = Op(K/N) = Op(K/N). For the second term, using ||Γ′−1

( ˆΨ( ˜β) Ψ)|| = Op(

K/N) by Lemma A.4 in Donald, Imbens and Newey (2003) and Lemma 7, we get ||Ω−1

s Γ′0Ψ− 1

( ˆΨ( ˜β)Ψ)ˆλ|| = Op(K/N) by CS. Next, observe that ||Γ′−1

(¯Γ0 Γ0)|| = Op(1/ √

N) by Lemma 2. Hence, ||Ω−1

s Γ′0Ψ−1(¯Γ0 −Γ0) ˆβ|| =

Op(1/ √

N)Op(1/ √

N) = Op(K/N). For the fourth term, ||Ω−s1Γ′1βˆλˆ|| = ||Ω− 1

s || · ||Γ′1λˆ|| · ||βˆ|| = Op(1/

N)Op(1/ √

N) =Op(K/N). For the last term,||Ω−s1Γ′0Ψ−1Γ1βˆ2|| =Op(1/N) =Op(K/N), as||Γ′

0Ψ− 1

Γ1||=Op(1) by Lemma 1.

Lemma 4. Rβ1 =op(K2/N3/2), where

Rβ1 = 1 2Ω

−1

s (¯Γ1−Γ1)′βˆλˆ− 1 2Ω

−1 s Γ′0Ψ−

1

(¯Γ1Γ1) ˆβ21 6Ω

−1 s Γ′0Ψ−

1 Γ∗2βˆ

3 +1

6Ω

−1 s Γ∗′2βˆ

2ˆ λ.

Proof. First, since||(¯Γ1Γ1)′ˆλ||=Op(K/N) by CS,||Ω−s1(¯Γ1−Γ1)′βˆλˆ||=Op(K/N)Op(1/√N) =

op(K2/N3/2). For the second term,||Ω−s1Γ′0Ψ− 1

(¯Γ1Γ1) ˆβ2

||=Op(1/√N)Op(1/N) =op(K2/N3/2), where||Γ′−1

(¯Γ1Γ1)||=Op(1/ √

N) by Lemma 2. Next,||Γ′−1 Γ∗

2||=Op(1) by Lemma 1, which implies that ||Ω−1

s Γ′0Ψ−1Γ∗2βˆ 3

|| =Op(1/N3/2) = op(K2/N3/2). For the last term,||Ω−s1Γ∗′2βˆ 2ˆ

λ||= ||Ω−1

s || · ||Γ∗′2ˆλ|| · ||βˆ 2

||=Op(1/√N)Op(1/N) =op(K2/N3/2).

Lemma 5. Tλ=Op(ζ(K)K/N).

Proof. Recall that Tλ has the form:

Tλ =Ψ−1Γ0Ω−s1(¯Γ0Γ0)′λˆΣ(¯Γ0Γ0) ˆβΣ( ˆΨ( ˜β)Ψ)ˆλ. First, note that||Ψ−1

Γ0||=Op(1) by ME and||(¯Γ0−Γ0)′λˆ||=Op(K/N) by CS. Then||Ψ−1Γ0Ω−s1(¯Γ0− Γ0)′λˆ|| = O

p(K/N). For the second term, ||Σ(¯Γ0 −Γ0) ˆβ|| =Op(

K/N)Op(1/ √

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where ||Σ(¯Γ0 Γ0)|| = Op(

K/N) by ME and ||Γ0¯ Γ0)|| = Op(

K/N). For the third term, first note that||Σ( ˆΨ( ˜β)Ψ)ˆλ|| ≤ ||Σ ˆΥv1λˆ||+||2 ˜βΣΥρηλˆ||+||ΣRΨλˆ||by Lemma 7. Now,||Σ ˆΥv1λˆ||=

Op(ζ(K)√K/N) by CS and Lemma 7,||2 ˜βΣΥρηλˆ||=Op(√K/N) by M, and||ΣRΨλˆ||=Op(ζ(K)K/N3/2+

K3/2/N3/2) by CS and Lemma 7. Hence,||Σ( ˆΨ( ˜β)Ψ)ˆλ||=Op(ζ(K)√K/N)

Lemma 6. Rλ1 =Op(1/N), where

Rλ1 = − 1 2Ψ

−1

Γ0Ω−s1Γ′1βˆλˆ− 1 2ΣΓ1βˆ

2

−12Ψ−1Γ0Ω−s1(¯Γ1Γ1)′βˆλˆ1

2Σ(¯Γ1−Γ1) ˆβ 2

−16ΣΓ∗2βˆ 3

−16Ψ−1Γ0Ω−s1Γ∗′2βˆ 2ˆ

λ.

Proof. First, note that||Ψ−1

Γ0||=Op(1) by ME. Hence,||Ψ−1Γ0Ωs−1Γ′1βˆλˆ||=Op(1/ √

N)Op(1/ √

N) = Op(1/N). For the second term, ||ΣΓ1βˆ2||=Op(1/N), where ||ΣΓ1||=Op(1) by ME. Third, since ||(¯Γ1Γ1)′λˆ||=O

p(K/N) by CS, it follows that||Ψ−1Γ0Ω−s1(¯Γ1−Γ1)′βˆλˆ||=Op(K/N)Op(1/ √

N) = Op(1/N). For the fourth term, noting that||Σ(¯Γ1−Γ1)||=Op(

K/N) by ME, we have||Σ(¯Γ1 Γ1) ˆβ2|| = Op(

K/N)Op(1/N) = Op(1/N). Fifth, applying ME gives ||ΣΓ∗2|| = Op(1). Hence, ||ΣΓ∗

2βˆ 3

||=Op(1/N3/2) =Op(1/N). Lastly,||Ψ−1Γ0Ω−s1Γ∗′2βˆ 2ˆ

λ||=Op(1/ √

N)Op(1/N) =Op(1/N).

Lemma 7. Ψ( ˜ˆ β)Ψ = ˆΥv1+ 2 ˜βΥρη+RΨ, where

ˆ Υv1 =

1 N

i

(ρ2i σ2i)qiqi′, Υρη= 1 N

i

σρη,iqiqi′, and ||R Ψ

||=Op(ζ(K) √

K/N +K/N)

Proof. Ψ( ˜ˆ β)Ψ can be written as: ˆ

Ψ( ˜β)Ψ

=

(ˆ

Υ11 Υ12ˆ

ˆ

Υ21 s−1

( ˆΥ22Υ21ˆ Υˆ−111Υ12) + ˆˆ Υ21Υˆ

−1 11Υ12ˆ

)

(

Υ11 Υ12

Υ21 s−1

(Υ22Υ21Υ−111Υ12) + Υ21Υ

−1 11Υ12

)

= ˆΥ( ˜β)Υ +

(

0 0

0 (s−1

−1)[( ˆΥ22Υ21ˆ Υˆ11−1Υ12)ˆ −(Υ22−Υ21Υ

−1 11Υ12)]

)

.

As in the proof of Proposition 1 in Donald, Imbens, and Newey (2009), the matrix ˆΥ( ˜β)Υ has a decomposition: ˆΥ( ˜β)Υ = ˆΥv1+ 2 ˜βΥρη +RΥ, where RΨ = Op(ζ(K)√K/N +K/N). Since 1s=Op(K2/N) and ( ˆΥ22−Υ21ˆ Υˆ−111Υ12)ˆ −(Υ22−Υ21Υ−

1

11Υ12) =op(1), ˆ

Ψ( ˜β)Ψ = ˆΥ( ˜β)Υ +op( (sK)2

N ) = ˆΥv1+ 2 ˜βΥρη+R Ψ

,

whereRΨ

=Op(ζ(K) √

K/N +K/N).

Lemma 8. Rλ3 =op(K2/N3/2), where

Rλ3 =−Ψ− 1

Γ0Ω−1

s (¯Γ0−Γ0)′Rλ2 −Σ(¯Γ0−Γ0)Rβ2 + ΣR Ψ

Σ¯gΣ( ˆΨ( ˜β)Ψ)Rλ2 +Rλ1.

Proof. First, since ||(¯Γ0Γ0)′Rλ

2|| = Op(

K/N)Op(ζ(K)√K/N3/2) by CS and Lemmas 5 and 6, ||Ψ−1

Γ0Ω−1

s (¯Γ0−Γ0)′R2λ|| = op(K2/N3/2). For the second term, noting that ||Σ(¯Γ0−Γ0)|| =

Op(

K/N) and ||Rβ2|| = Op(K/N) by Lemmas 3 and 4, applying CS gives ||Σ(¯Γ0−Γ0)R β 2|| = Op(

K/N)Op(K/N) = op(K2/N3/2). For the third term, ||ΣRΨΣ¯g|| = op(K2/N3/2) by M and Lemma 7. For the fourth term, ||Σ( ˆΨ( ˜β)Ψ)R2λ|| = Op(ζ(K)

K/N)Op(ζ(K)√K/N) =

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Lemma 9. Rβ3 =op(K2/N3/2), where

Rβ3 = Ω

−1

s (¯Γ0−Γ0)′Ψ− 1

Γ0Ω−s1(¯Γ0Γ0)′Σ¯g+ Ω−s1(¯Γ0Γ0)′Σ(¯Γ0Γ0)Γ′0Ψ−1¯g +2 ˜βΩ−s1(¯Γ0Γ0)′ΣΥρηΣ¯g+ Ω−s1(¯Γ0−Γ0)′Rλ3.

Proof. Since||(¯Γ0Γ0)′Ψ−1

Γ0||=Op(1/√N) by Lemma 2 and||(¯Γ0−Γ0)′Σ¯g|| ≤λmax(Σ)||Γ0¯ −Γ0||· ||g¯||=Op(K/N), ||Ω−s1(¯Γ0−Γ0)′Ψ−1Γ0Ω−s1(¯Γ0−Γ0)′Σ¯g||=Op(1/

N)Op(K/N) =op(K2/N3/2). For the second term, note that ||(¯Γ0 Γ0)′Σ(¯Γ0 Γ0)|| ≤ λmax(Σ)||Γ0¯ Γ0||2

= Op(K/N) and that ||Γ′0Ψ−1

¯

g|| = Op(1/√N) by Lemma 2. Hence, ||Ω−s1(¯Γ0 − Γ0)′Σ(¯Γ0 − Γ0)||Γ′0Ψ−1¯g|| =

Op(K/N)Op(1/ √

N) =op(K2/N3/2). For the third term,||β˜Ωs−1(¯Γ0−Γ0)′ΣΥρηΣ¯g||=Op(1/ √

N)× Op(K/N3/2) =op(K2/N3/2), because||(¯Γ0−Γ0)′ΣΥρηΣ¯g|| ≤λmax(ΣΥρηΣ)||(¯Γ0−Γ0)||·||¯g||. Lastly, ||(¯Γ0Γ0)′Rλ3|| =Op((K/N)

3/2

) by CS and Lemma 8, which implies that ||Ω−1

s (¯Γ0−Γ0)′Rλ3||= op(K2/N3/2)

Lemma 10. Rβ4 =op(K2/N3/2), where

Rβ4 = Ω− 1 s Γ′0Ψ−

1ˆ

Υv1Ψ−1Γ0Ω−s1(¯Γ0−Γ0)′Σ¯g+ Ω− 1 s Γ′0Ψ−

1ˆ

Υv1Σ(¯Γ0−Γ0)Ω−s1Γ′0Ψ− 1

¯ g +2 ˜βΩ−s1Γ′0Ψ−1Υˆv1ΣΥρηΣ¯g+ Ω−s1Γ′0Ψ−

1ˆ Υv1Rλ3 +2 ˜βΩ−s1Γ′0Ψ−1ΥρηRλ2 + Ω−

1 s Γ′0Ψ−

1 RΨλ.ˆ

Proof. First, noting that||Γ′

0Ψ− 1ˆ

Υv1Ψ−1Γ0||=Op(1/√N) and||(¯Γ0−Γ0)′Σ¯g||=Op(K/N) by M, we get||Ω−1

s Γ′0Ψ−1Υˆv1Ψ−1Γ0Ω−s1(¯Γ0−Γ0)′Σ¯g||=Op(1/√N)Op(K/N) =op((sK)2/N3/2). Second, since||Γ′−1ˆ

Υv1Σ(¯Γ0−Γ0)||=Op(K/N) by M and||Γ′0Ψ−1¯g||=Op(1/ √

N) by Lemma 2, we get ||Ω−1

s Γ′0Ψ− 1ˆ

Υv1Σ(¯Γ0−Γ0)Ω−s1Γ′0Ψ− 1

¯

g||=Op(K/N)Op(1/√N) =op(K2/N3/2). The third term is:

||β˜Ω−s1Γ′0Ψ−1Υˆv1ΣΥρηΣ¯g||=||β˜|| · ||Ωs−1|| · ||Γ′0Ψ− 1ˆ

Υv1ΣΥρηΣ¯g||=Op(1/ √

N)Op(1)Op(K/N), where||Γ′

0Ψ− 1ˆ

Υv1ΣΥρηΣ¯g||=Op(K/N) by CS. Fourth, we have||Ω−s1Γ′0Ψ− 1ˆ

Υv1Rλ3||=op(K 2

/N3/2 ) by Lemma 8. Fifth,||β˜Ω−1

s Γ′0Ψ−1ΥρηRλ2||=Op(1/ √

N)Op(ζ(K)K/N) =op(K2/N3/2) by Lemmas 5 and 6. Lastly, applying CS gives||Γ′−1

RΨˆ

λ||=Op(ζ(K)K/N3/2+ (K/N)3/2). Hence, we have ||Ω−1

s Γ′0Ψ−1RΨλˆ||=op(K2/N3/2).

Lemma 11. Rβ5 =op(K2/N3/2), where Rβ5 = Ω−1

s Γ′0Ψ− 1

(¯Γ0Γ0)Rβ2.

Proof. ||Γ′0Ψ−1(¯Γ0Γ0)||= Op(1/√N) by Lemma 2 and ||Rβ2|| =Op(K/N) by Lemma 3 and 4. ThereforeRβ5 =Op(1/

N)Op(K/N) =op(K2/N3/2).

Lemma 12. Rβ6 =op(K2/N3/2), where Rβ6 =−Ω− 1

s Γ′1Σ¯gR2β+ Ω− 1

s Γ′1Rλ2βˆ.

Proof. ||Γ′1Σ¯g|| = O

p(1/ √

N) and ||Rβ2|| = Op(K/N) by Lemma 2 and Lemma 6 respectively. Hence,||Ω−1

s Γ′1Σ¯gR β

2||=Op(1/ √

N)Op(K/N) =op(K2/N3/2). For the second term, since||Γ′1Rλ2||= Op(ζ(K)K/N) by CS and Lemma 5 and 6,||Ω−s1Γ′1Rλ2βˆ||=Op(ζ(K)K/N)Op(1/

N) =op(K2/N3/2).

Lemma 13. Rβ7 =op(K2/N3/2), whereRβ7 =2Ω−1

s Γ′0Ψ− 1

Γ1Ω−1 s Γ′0Ψ−

1 ¯

gRβ2+Ω− 1 s Γ′0Ψ−

1

Γ1(Rβ2) 2

.

Proof. Since||Γ′

0Ψ− 1

Γ1||=Op(1) by Lemma 1,||Γ′0Ψ− 1

¯

g||=Op(1/ √

N) by Lemma 2, and||Rβ2||= Op(K/N) by Lemmas 3 and 4, It follows that||Ω−s1Γ′0Ψ−

1 Γ1Ω−1

s Γ′0Ψ−1gR¯ β2||=Op(1/ √

N)Op(K/N) =

op(K2/N3/2) and||Ω−s1Γ′0Ψ− 1

Γ1(Rβ2) 2

||=Op(K2/N2) =op(K2/N3/2).

(9)

Proof. Applying M gives ||(¯Γ0Γ0)′Σ¯g||=O

p(K/N).

Lemma 15. T2h =Op(K/N). Proof. ||Γ′0Ψ−1ˆ

Υv1Σ¯g||=Op(1/√N)Op(

K/N) =Op(√K/N) by Lemma A.4 in Donald, Imbens and Newey (2003) and CS.

Lemma 16. T3h =Op(1/N).

Proof. For the first term, noting that ||Γ′0Ψ−1ΥρηΣ¯g|| = Op(1/√N) by by Lemma A.4 in Don-ald, Imbens and Newey (2003), we have ||β˜Γ′−1

ΥρηΣ¯g|| = Op(1/N). For the second term, as Lemma 2 implies that||Γ′−1

(¯Γ0Γ0)||=Op(1/ √

N) and that||Γ′−1 ¯

g||=Op(1/ √

N), we have ||Γ′

0Ψ− 1

(¯Γ0Γ0)Ω−1 s Γ′0Ψ−

1 ¯

g||=Op(1/N). Similarly, for the third term, as ||Γ′0Σ¯g||=Op(1/ √

N) and ||Γ′−1

¯

g|| =Op(1/ √

N) by Lemma 2, we have ||Γ′1Σ¯g−1

s Γ′0Ψ−1g¯|| =Op(1/N). For the last term, noting that ||Γ′−1

Γ1|| = Op(1) by Lemma 1 and ||Γ′0Ψ−1g¯|| = Op(1/ √

N) by Lemma 2, ||Γ′

0Ψ− 1

Γ1(Ω−1 s Γ′0Ψ−

1 ¯ g)2

||=Op(1/N).

Lemma 17. T4h =Op(ζ(K)K/N).

Proof. First, note that||Σ ˆΥv1Σ¯g||=Op(ζ(K) √

K/N). Then, CS gives that||(¯Γ0Γ0)′Σ ˆΥ

v1Σ¯g||=

Op(

K/N)Op(ζ(K) √

K/N) =Op(ζ(K)K/N3/2), and that||Γ′0Ψ− 1ˆ

Υv1Σ ˆΥv1Σ¯g||=Op(

K/N)× Op(ζ(K)√K/N) =Op(ζ(K)K/N3/2).

Lemma 18. E(T1hh|X) =op(γK,N).

Proof. Define B =QΣQ′/N andA=QΨ−1

Γ0. Then,

T1hh′ =N(¯Γ0−Γ0)′Σ¯g¯g′Ψ− 1

Γ0 =N 1 Nη

QΣ1

NQ

ρ1

QΨ−1

Γ0 =η′Bρρ′A/N.

Therefore, N E(Th

1h′|X) is the sum of the terms: E(ηiBijρjρkAk|X) for i, j, k = 1, . . . , N. All of these terms are 0 by the independence and the third moment condition, which implies that E(T1hh′|X) = 0 =op(γK,N).

Lemma 19. E(T2hh|X) =op(γK,N).

Proof. Define B =QΣQ′/N andA=QΨ−1

Γ0. Then,

T2hh′ =NΓ′0Ψ− 1ˆ

Υv1Σ¯gg¯′Ψ−1Γ0 =NΓ′0Ψ−1 1 NQ

V QΣ1

NQ

ρ1

QΨ−1

Γ0 =A′V Bρρ′A/N.

Therefore,Th

2h′ is the sum of the terms: AiViiBijρjρkAkfori, j, k= 1, . . . N. By the independence and the third moment condition, we need to consider only the cases withi=j =l. Therefore,

E(T2hh′|X) =

i

E(Viiρ2i|xi)A2iBii=

i

(κi−1)A2iBii=Op(K/N) =op(γK,N).

(10)

Proof.

T3hh′ = −2Nβ˜Γ′0Ψ− 1

ΥρηΣ¯gg¯′Ψ−1Γ0−NΓ′0Ψ−1(¯Γ0−Γ0)Ω−s1Γ′0Ψ− 1

¯

gg¯′Ψ−1Γ0 −12NΓ′1Σ¯gΩ−s1Γ0Ψ′ −1¯gg¯′Ψ−1Γ0+ 1

2NΓ

0Ψ−1Γ1(Ω−s1Γ′0Ψ−1g¯)2g¯′Ψ−1Γ0.

Now, by the independence and the third moment condition, the expectations of the second, third and fourth terms on the right-hand side of the equation are 0. It is therefore enough to consider only the first term. Now, ˜β = Φ′ρ/N +o

p(N−1/2). Letting B = QΣQ′/N and A = QΨ−1Γ0, we can write

Nβ˜Γ′0Ψ−1ΥρηΣ¯gg¯′Ψ−1Γ0 = N 1 Nρ

ΦΓ−1 1 NQ

V

ρηQΣ 1 NQ

ρ1

QΨ−1

Γ0+op(γK,N) = ρ′ΦA′VρηBρρ′A/N2+op(γK,N).

Now, ρ′ΦAV

ρηBρρ′Aconsists of the terms ρiΦiAjVρη,jkBklρlρmAm fori, j, k, l, m= 1, . . . , N. The expectations of all of these are zeros by the independence and the third moment condition.

Lemma 21. E(T4hh|X) =op(γK,N).

Proof. Define B =QΣQ′/N andA=QΨ−1

Γ0. Then,

T4hh′ = −N(¯Γ0−Γ0)′Σ ˆΥv1Σ¯g¯g′Ψ−1Γ0+NΓ′0Ψ−1Υˆv1Σ ˆΥv1Σ¯gg¯′Ψ−1Γ0 = N 1

QΣ1

NQ

V QΣ1

NQρ 1 Nρ

QΨ−1

Γ0+NΓ′0Ψ−1 1 NQ

V QΣ1

NQ

V QΣ1

NQ

ρ 1

QΨ−1 Γ0 = η′BV Bρρ′A/N+A′V BV Bρρ′A/N.

E(η′BV BρρA) = 0 by the independence and the third moment condition. ByAV BV BρρA/N =

A′V BρρA/N+o

p(γK,N) and the proof of Lemma 19,E(A′V BV Bρρ′A/N|X) =op(γK,N).

Lemma 22. E(T1hT1h′|X) = (s

iE(ηiρi|xi)ξii,2)2+op(γK,N).

Proof.

T1hT1h′ =N(¯Γ0−Γ)′Σ¯gg¯′Σ(¯Γ0−Γ) =N 1 Nη

QΣ1

NQ

ρ 1

QΣ1

NQ

η =ηBρρBη/N,

whereB =QΣQ′/N. By the independence assumption,

E(T1hT1h′|X) =

i

E(η2iρ2i|xi)Bii2/N +

i6=j

E(ηi2|xi)E(ρ2j|xj)B2ij/N

+∑

i6=j

(E(ηiρj|xi))2BijBji/N +

i6=j

E(ηiρi|xi)E(ηjρj|xj)BiiBjj

=

( ∑

i

E(ηiρi|xi)Bii

)2

+op(γK,N).

Now,

( ∑

i

E(ηiρi|xi)Bii

)2

=s2

( ∑

i

E(ηiρi|xi)ξii,2

)2

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