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Appendix: Technical materials for the asymptotic results

In this appendix, we provide the detailed derivations for the asymptotic results under IRa.

The derivations for the asymptotic results under IRb and IRc as well as the theorems in Section 5 are delegated in the supplement. Throughout the appendix, we use ¯K to denote K+ 1 and ¯T to denote TK−1. To facilitate the analysis, we introduce the following the auxiliary identification condition (an intermediate step analysis)

AU1 The underlying parameter values θ = (Λ,Γ, F,Φ,Σee) satisfy: N1Λ∗′Σ−1eeΛ = Q,T1 PTt=1ftft∗′ =Ir1 and T1 PTt=1ftgt= 0, where Q is a diagonal matrix, whose diagonal elements are distinct and arranged in descending order.

Appendix A: The asymptotic results of the QMLE

In this appendix, we show that the QMLE ˜λi˜2i and ˜ftare respectively consistent estimator ofλi, σ2i and ft in AU1.

Proposition A.1 Under Assumptions A-D, together with AU1,

λ˜iλi =1 T

XT t=1

ftft∗′−11 T

XT t=1

fteit+Op(N1/2T1/2) +Op(T1), (A.1)

˜

γiγi =1 T

XT t=1

gtgt

−11 T

XT t=1

gteit

, (A.2)

f˜tft =1 N

XT i=1

1

σ2iλiλ∗′i 11 N

XN i=1

1

σ2iλieit+Op(N1/2T1/2) +Op(T1), (A.3)

˜

σ2iσ2i = 1 T

XT t=1

(e2itσ2i) +Op(N−1/2T−1/2) +Op(T−1), (A.4)

Proof of Proposition A.1Write zt= Λft+ Γgt+et into matrix form,

Z = ΛF∗′+ ΓG+e. (A.5)

Post-multiplyingMG=ITG(GG)1G on both sides, together withF∗′G= 0 by AU1, we have

ZMG= ΛF∗′+eMG.

LetY =ZMG and ytdenotes the t-th column ofY. The above equation is equivalent to yt= Λft+eteG(GG)−1gt (A.6) Bai and Li (2012) derive the asymptotic representations of ˜λi,f˜t,˜σi2 under the case that gt ≡ 1. However, when gt is a general random variable, as like in the present context, the derivation is the same since term eG(GG)1gt is essentially negligible. Using the arguments of Bai and Li (2012) under IC3, we obtain (A.1), (A.3) and (A.4). Consider (A.2). Substituting zit = λ∗′i ft+γi∗′gt+eit into ˜γi = (PTt=1gtgt)1 PTt=1gt(zit−˜λif˜t), we have

˜

γiγi = XT t=1

gtgt1 XT t=1

gteit XT t=1

gtgt1 XT t=1

gtft∗′λiλi)

XT t=1

gtgt−1 XT t=1

gt( ˜ftft)λ˜i

The second term of the right hand side is zero by PTt=1gtft∗′ =GF = 0. Consider the third term. Notice

f˜t= (˜ΛΣ˜−1eeΛ)˜ −1Λ˜Σ˜−1eeyt

= (˜ΛΣ˜ee1Λ)˜ 1Λ˜Σ˜ee1hzt XT s=1

zsgs XT s=1

gsgs1gti

= (˜ΛΣ˜−1eeΛ)˜ −1Λ˜Σ˜−1eehΛft+et XT s=1

esgs XT s=1

gsgs1gti Then it follows

f˜tft = −A∗′ft+ (˜ΛΣ˜−1eeΛ)˜ −1Λ˜Σ˜−1eeet

−(˜ΛΣ˜ee1Λ)˜ 1Λ˜Σ˜ee1 XT s=1

esgs XT s=1

gsgs−1gt

(A.7)

whereA = (˜Λ−Λ)Σ˜ee1Λ(˜˜ ΛΣ˜ee1Λ)˜ 1.

Given the above expression, together withPTt=1gtft∗′= 0, we have 1

T XT t=1

gt( ˜ftft) = 0 (A.8) Then (A.2) follows. This completes the proof.

Lemma A.1 Under Assumptions A-D, as Lemma C.1(e) of Bai and Li (2012). Given these results, the second term of J11 is Op(N1) +Op(T1). The last term can be proved to be the same magnitude by the similar arguments. Summarizing these results, we haveJ11=Op(N1) +Op(T1). TermsJ12and J21 can be proved to beOp(N−1/2T−1/2) +Op(T−1) similarly as J11. Then (a) follows.

Consider (b). The left hand side of (b) is equal to

"1

The two none-zero terms of the above are Op(N−1) +Op(T−1), which are shown in (a).

Then (b) follows.

Consider (c). The left hand side of (c) is equal to

"1 So it suffices to consider term 1¯

T +Op(T−1) similarly as Lemma C.1(e) of Bai and Li (2012). Given these results, together withA=Op(N1/2T1/2) +Op(T1) and (˜ΛΣ˜ee1Λ)˜ 1 =Op(N1), we have

Proposition A.2 Under Assumptions A-D, together with the identification condition AU1, for each k= 1,2, . . . , K, we have Φ is obtained by running the regression˜

˜ht= Φ1˜ht1+ Φ2˜ht2+· · ·+ ΦK˜htK+error

By Lemma A.1(a) and (b), h1

T¯ XT t= ¯K

htht) ˜ψtih1 T¯

XT t= ¯K

ψ˜tψ˜ti1 =Op(N1) +Op(T1)

h1 T¯

XT t= ¯K

( ˜ψtψt) ˜ψtih1 T¯

XT t= ¯K

ψ˜tψ˜ti−1 =Op(N1) +Op(T1) By Lemma A.1(a) and (c),

h1 T¯

XT t= ¯K

utψ˜tih1 T¯

XT t= ¯K

ψ˜tψ˜ti1 =h1 T¯

XT t= ¯K

utψt∗′ih1 T¯

XT t= ¯K

ψtψt∗′i1 +Op(N1) +Op(T1)

Given this result, we have Φ˜ −Φ =

XT t= ¯K

utψt∗′1 T¯

XT t= ¯K

ψtψt∗′1+Op(N1) +Op(T1) Post-multiplyingikIr on both sides gives Proposition A.2.

Now we consider the following condition (denoted by AU2), in which the loading re-strictions are the same as AU1 but factor rere-strictions are imposed on the population.

AU2 The underlying parameter values θ = (Λ,Γ, F,Φ,Σee) satisfy: N1ΛΣee1Λ = Q, E(ftft⋆′) =Ir1 and E(ftgt) = 0, whereQ is a diagonal matrix, whose diagonal elements are distinct and arranged in descending order.

Note that the superscript “stars” in θ and θ are different. Different identification restrictions imply different notations. Because AU1 and AU2 are asymptotically the same (the former with sample moment restriction T1 Ptftft =Ir1 and the latter with population moment restriction E(ftft) = Ir1), θ and θ are also asymptotically the same. That is why the MLE is also consistent forθ, which will be proved below.

The following lemma is useful to our analysis.

Lemma A.2 Let Q be anr×r matrix satisfying QQ=Ir QV Q=D

whereV is anr×r diagonal matrix with strictly positive and distinct elements, arranged in decreasing order, andDis also diagonal. Then Qmust be a diagonal matrix with elements either−1 or 1 and V =D.

Lemma A.2 is proved in Bai and Li (2012). The following proposition summarizes the asymptotic results under AU2. It shows that the limiting distributions under AU2 have been changed.

Proposition A.3 Under Assumptions A-D, together with the identification condition AU2, when N, T → ∞, we have

Proof of Proposition A.3. Notice that

λ˜iλi = (˜λiλi) + (λiλi).

We showλi and λi are close to each other because AU1 and AU2 are asymptotically the same. Different identification restrictions imply different rotations. LetR be the rotation matrix, which transform (λ∗′i , γi∗′) to (λi, γi). Then we have

As mentioned in the main text, due to the fact that the factors gt are observed, matrix R⋆′12 is fixed to 0 and matrix R⋆′22 is fixed toIr2. So equation (A.12) reduces to The last equation of (A.13) can also be written as

ft =R11ft+R21gt. (A.14)

Post-multiplyinggt on both sides and taking summation overt, byPTt=1gtft∗′= 0, we have elements either 1 or−1. Since we assume that the sign problem is precluded in our analysis, it followsR111−→p Ir1. Let where Ndg{·} denotes the non-diagonal elements of the argument. Neglecting the terms UQU⋆′ and U⋆′U since they are of smaller order than U, we can uniquely determine matrixU by solving the equation system (A.20) and (A.21). Let V be the leading term

of U. It is easy to see that U =Op(T−1/2), V =Op(T−1/2) and U = V+Op(T−1).

Now consider the asymptotic representation of ˜λiλi. Notice λ˜iλi = ˜λiR11λi = (˜λiλi)−(R11Ir1i By (A.1), the above result is equivalent to

λ˜iλi =h1

ByR⋆′−111 =Ir1 +U⋆′, the above equation is equal to

f˜tft = ( ˜ftft)−U⋆′ft+R11′−1R⋆′21gt Substituting (A.3) into the above result, we have

f˜tft = −Uft+R11′−1R21gt+

Given the above results, by (A.26), we have the last expression of Proposition A.3. This completes the proof of Proposition A.3.

Proposition A.4 Under Assumptions A-D, together with the identification condition AU2, we have

However, byR⋆−111 =Ir1+V+Op(T−1) andR21=−W+Op(T−1), we have

By Proposition A.2, we can rewrite the above result as Φ˜k−Φk=

Appendix B: The asymptotic results and their proofs under IRa

As in the main text, we use (Λ,Γ, F) to denote the underlying parameters satisfying IRa.

LetR be the rotation matrix which transforms (λ⋆′i , γi⋆′) into (λi, γi). Then we have The last equation in (B.2) can be written as

ft =R11 ft+R21gt. (B.3)

Note that the rotation matrix R is nonrandom. To see this, both AU2 and IRa impose restrictions on the loadings and the covariance of ht. So the rotation matrix R, which transform the underlying parameters from AU2 to IRa, only involves loadings and covari-ance ofht. Thus it is nonrandom. This is in contrast withR, which is random since AU1 involvesft.

Post-multiplyinggt on both sides and taking the expectation, byE(ftgt) = 0, we have R21=−gg1gfR11.

Defineφt=R′−111ft. From the above results,φhas an alternative expression

φt=ftf ggg1gt. (B.4) The following lemmas will be used in the subsequent proof.

Lemma B.1 For any compatible matrices A and B and their corresponding estimatesand, we have

AˆBˆ1− AB1A = ( ˆA − A)B1A+AB1( ˆA − A)− AB1( ˆB − B)B1A+R where

R=−( ˆA − A) ˆB−1( ˆB − B)B−1A+ ( ˆA − A) ˆB−1( ˆA − A)

+ABˆ−1( ˆB − B)B−1( ˆB − B)B−1− ABˆ−1( ˆB − B)B−1( ˆA − A). Lemma B.1 can be proved easily by matrix algebra.

Lemma B.2 Under Assumptions A-D, we have (a) 1

T¯

H˜MΨ˜H˜ − 1

T¯H∗′MΨH =Op(N−1) +Op(T−1) (b) 1

T¯H⋆′MΨH− 1

T¯H∗′MΨH =B⋆′+ ΩB+Op(T−1) (c) 1

T¯H˜MΨ˜H˜ − 1

T¯HMΨH =−B−ΩB+Op(N1) +Op(T1) where 1¯

TH˜MΨ˜H˜ is defined as 1

T¯

H˜MΨ˜H˜ = 1 T¯

XT t= ¯K

˜ht˜ht1 T¯

XT t= ¯K

˜htψ˜t1 T¯

XT t= ¯K

ψ˜tψ˜t−11 T¯

XT t= ¯K

ψ˜t˜ht,

and T1¯H∗′MΨH and T1¯HMΨH are defined similarly.

Proof of Lemma B.2. Consider (a). By Lemma A.1(a), we have 1

T¯ XT t= ¯K

h˜t˜ht− 1 T¯

XT t= ¯K

hth∗′t =Op(N1) +Op(T1)

1 Given the above results, together with Lemma B.1, we have (a).

Consider (b). By ht =R′−1ht, we have ψt = (IKR′−1t. This gives Substituting (B.6) into (B.5), we have (b).

Result (c) is a direct result of (a) and (b). This completes the proof.

Proposition B.1 Under Assumption A-D, together with the identification condition IR1, we have

(c) ˆftft=1 N

XN i=1

1

σi2λiλi−11 N

XT i=1

1

σi2λieitVftWgt+Op(N1) +Op(T1) wherevec(V) =B1

Q P1D+r1T1¯PTt= ¯Ktεt−vec(Ir1)];φt=ftf g−1gggt;φφ=E(φtφt);

W = Ωυυ1 1 T

PT

t= ¯Kυtεt; ηt=gtgff f1ft; ηη =E(ηtηt).

Proof of Proposition B.1. Consider the VAR expression under AU2:

ht = Φ1ht1+ Φ2ht2+· · ·+ ΦKhtK+ut. Pre-multiplyingR′−1 gives

ht= R′−1Φ1Rht−1+· · ·+ R′−1ΦKRht−K+R′−1ut. So we have Φi =R′−1ΦiR fori= 1,2, . . . , K and ut=R′−1ut. Then we have

εt=R′−111εtR′−111R21υ

t, υt=υ

t. (B.7)

Post-multiplyingυ

t on both sides and taking the expectation, byE(εtυ

t) = 0, we have R21= Ωυυ1υε, (B.8) Substituting the proceeding result into (B.7), byE(εtεt) =Ir1, we have

εε·υ= Ωεε−Ωευυυ1υε=R11R11. (B.9) where Ωεε=E(εtεt),Ωυυ =E(υtυt) and Ωευ =E(εtυt). In addition, the identification condition also requires that

Q= 1

NΛΣ−1eeΛ =R111

NΛ⋆′Σ−1eeΛR11. This is equivalent to

Q = 1

NΛ⋆′Σ−1eeΛ=R111QR′−111. (B.10) However, our estimation procedure implies that the estimators of R11, R21, denoted by Rˆ11,Rˆ21, satisfy

Rˆ21= ˜Ω−1υυΩ˜υε (B.11) Rˆ11Rˆ11= ˜Ωεε·υ= ˜Ωεε−Ω˜ευΩ˜−1υυΩ˜υε (B.12)

Rˆ11

1

NΛ˜Σ˜ee1Λ˜Rˆ11 = diag (B.13) where ˜Ωεε,Ω˜υυ,Ω˜ευ are submatrices of ˜Ω, which is defined as

Ω =˜ 1 T¯

XT t= ¯K

˜ utu˜t

with ˜ut being the residuals of the regression

The above result can be rewritten as Ω˜ −Ω=

By (B.15), we have ˜Ω−Ω⋆ p−→0. Then it follows ˜Ωεε·υ−Ωεε·υ

p

→0, where ˜Ωεε·υ and Ωεε·υ

are defined in (B.9) and (B.12). Thus

Rˆ11Rˆ11R111R′−111 −→p Ir1, which, by the fact thatAB=I thenBA=I, leads to

( ˆR11R111)( ˆR11R111)−→p Ir1 (B.19) Furthermore, by (B.13), we have

( ˆR11R11−1)hR111

NΛ˜Σ˜−1eeΛ˜R11i( ˆR11R−111) = diag

By N1Λ˜Σ˜ee1Λ˜ −N1ΛΣee1Λ =op(1) andR11N1ΛΣee1ΛR11= N1ΛΣee1Λ =Q, we have ( ˆR11R−111)Q( ˆR11R−111) = diag (B.20) NoticeQis a diagonal matrix by identification. Applying Lemma A.2 to (B.19) and (B.20), we have ˆR11R11−1 converges to a diagonal matrix whose diagonal elements are either 1 or

−1. However, the possibility of−1 is precluded by our sign restrictions. Given this result, we have ˆR11R11 p

→ 0. Henceforth, we use ∆Rd11 to denote ˆR11R11. Apparently

∆Rd11−→p 0. By (B.9) and (B.12), we have

Rˆ11Rˆ11R11R11= ˜Ωεε−Ωεε−( ˜ΩευΩ˜−1υυΩ˜υε−Ωευ⋆−1υυυε)

Substituting (B.16)-(B.18) into the above equation, together with Lemma B.1, we have

∆Rd11R11+R11∆Rd11+∆Rd11∆Rd11=−V⋆′εε·υ−Ωεε·υV +1

T¯ XT t= ¯K

ht −Ωευυυ1υ

t)(εt −Ωευυυ1υ

t)−Ωεε·υ

i+Op(N1) +Op(T1).

However, by (B.7) and (B.8), we haveR11εt=εt−Ωευυυ1υt. Given this result, together with (B.9), we have

∆Rd11R11+R11∆Rd11+∆Rd11∆Rd11=−V⋆′R11R11R11R11V +R11h1

T¯ XT t= ¯K

εtεtIr1

iR11+Op(N−1) +Op(T−1). (B.21)

Pre-multiplying R′−111 and post-multiplying R−111 on both sides, and neglecting the smaller order termR′−111∆Rd11∆Rd11R111, we have

∆Rd11R−111 +R11VR11−1+∆Rd11R11−1+R11VR−111

= 1 T¯

XT t= ¯K

tεtIr1) +Op(N1) +Op(T1).

(B.22)

Now consider 1

NΛ˜Σ˜ee1Λ˜− 1

NΛΣee1Λ = 1 N

XN i=1

1

˜

σ2iλiλiλi+ 1 N

XN i=1

1

˜

σ2iλ˜iλiλi)

− 1 N

XN i=1

1

˜

σi2λiλi)(˜λiλi)+ 1 N

XN i=1

λiλ⋆′i ( 1

˜ σ2i − 1

σi2).

The last term is Op(N1/2T1/2) +Op(T1) which is shown in Bai and Li (2012). The third term isOp(T1). The first two terms areVQ+QV+Op(N1/2T1/2)+Op(T1) by Proposition A.3. Then it follows

1

NΛ˜Σ˜ee1Λ˜− 1

NΛΣee1Λ =VQ+QV+Op(N1/2T1/2) +Op(T1). (B.23) Given the above results, (B.13) is equivalent to

Ndg

Rˆ11(Q+VQ+QV⋆′) ˆR11

=Op(N−1/2T−1/2) +Op(T−1).

Substituting (B.10) into the proceeding equation, we have Ndg

Rˆ11(R−111QR′−111 +VR−111QR′−111 +R−111QR′−111 V⋆′) ˆR11

=Op(N−1/2T−1/2) +Op(T−1).

Replace ˆR11=∆Rd11+R11, the left hand side is (neglecting Ndg)

Q+∆Rd11R111Q+Q(∆Rd11R111)+∆Rd11R111Q(∆Rd11R111)Q +R11VR−111Q+∆Rd11VR11−1Q+ ˆR11VR−111Q(∆Rd11R−111) +QR′−111 VR11+QR′−111 V∆Rd11+ (∆Rd11R11−1)QR′−111 VRˆ11

By neglecting the terms of smaller magnitude and noticing that Ndg(Q) = 0, we have Ndgn∆Rd11R111+R11VR111Q (B.24) +Q∆Rd11R11−1+R11VR−111o=Op(N−1/2T−1/2) +Op(T−1).

Let V = ∆Rd11R−111 +R11VR11−1. Taking the half-vectorization operation vech(·) which stacks the elements on and below the diagonal of the argument into a vector on both sides of (B.22), we get

vech(V+V) = vechh1 T¯

XT t= ¯K

tεtIr1)i+Op(N1) +Op(T1).

By the definitions of duplication matrixDr1 and its Moore-Penrose inverseD+r1, and sym-metrizer matrixSr1 = (Ir2

1+Kr1)/2, the left hand side of the above equation can be written as

vech(V+V) =D+r1vec(V+V) = 2Dr+1Sr1vec(V) = 2Dr+1vec(V),

where the last equation is due toD+r1Sr1 =D+r1, we have 2Dr+1vec(V) =Dr+1vech1

T¯ XT t= ¯K

tεtIr1)i+Op(N1) +Op(T1). (B.25) Let veck(M) be the operation which stacks the elements below the diagonal into a vector.

Let D1 be the matrix such that veck(M) =D1vec(M) for any symmetric matrix M. By (B.24), we have

veck(VQ+QV) =Op(N1/2T1/2) +Op(T1).

implying

D1[Q⊗Ir1+ (Ir1Q)Kr1]vec(V) =Op(N1/2T1/2) +Op(T1). (B.26) The preceding two equations imply

"

2Dr+1

D1[Q⊗Ir1+ (Ir1Q)Kr1]

#

vec(V) =

"

Dr+1vec1¯

T

PT

t= ¯KtεtIr1) 0

#

+Op(N−1)+Op(T−1).

LetBQbe the matrix before vec(V) andP1 = [Ip,0p×q], then the above result is equivalent to

vec(V) =B−1Q P1D+r1vech1 T¯

XT t= ¯K

tεtIr1)i+Op(N−1) +Op(T−1).

DefineV by

vec(V) =B−1Q P1Dr+1vech1 T¯

XT t= ¯K

tεtIr1)i. Then by the definition ofV,

∆Rd11R111+R11VR111 =V + +Op(N1) +Op(T1). (B.27) Post-multiplyingR11on both sides of (B.27), we have

∆Rd11=−R11V+V R11+Op(N−1) +Op(T−1) =Op(T−1/2) +Op(N−1) (B.28) sinceV =Op(T1/2) and V =Op(T1/2).

Now consider ˆλiλi. By ˆλi = ˆR11λ˜i and λi=R11λi, we have

λˆiλi = ˆR11λ˜iR11λi =∆Rd11λi +R11λiλi) +∆Rd11λiλi).

The last term of right hand side isOp(T−1) +Op(N−2) by ˜λiλi =Op(T−1/2) +Op(N−1) and ∆Rd11=Op(T−1/2) +Op(N−1). By (B.28) and (A.24), together with λi =R11λi, we have

λˆiλi =V λi+R111 T

XT t=1

ftft⋆′−11 T

XT t=1

fteit+Op(N−1) +Op(T−1).

Using (B.4), the above expression can be rewritten as

To derive the remaining asymptotic results, we first consider∆Rd21= ˆR21R21. Notice Rˆ21R21= ˜Ωυυ1Ω˜υε−Ωυυ1υε=−Ωυυ1( ˜Ωυυ−Ωυυ)Ωυυ1υε+ Ωυυ1( ˜Ωυε−Ωυε)

−( ˜Ω−1υυ−Ω⋆−1υυ )( ˜Ωυυ−Ωυυ)Ω⋆−1υυυε+ ( ˜Ω−1υυ−Ω⋆−1υυ )( ˜Ωυε−Ωυε).

The last two terms of the right hand side are Op(N2) +Op(T1). Substituting (B.17) and (B.18) into the above result, we have

∆Rd21= Ωυυ1 Substi-tuting (A.25), (A.24) and (B.29) into the above result, we have

ˆ

Consider the last equation of (B.2). Post-multiplyinggt on both sides and taking expecta-tion, by E(ftgt) = 0, we have

R21R111=−gg1gf.

The preceding two results imply that the third expression of (B.30) is equal to

−1gggf−1φφh1 T

XT t=1

φteiti+Op(T−1).

Let Ξt=gfφφ1φt. Given the above result, the asymptotic representation of ˆγiγi can be rewritten as

ˆ

γiγi =−1ggh1 T

XT t=1

(gt−Ξt)eiti+W λi+Op(N−1) +Op(T−1). (B.31) The above asymptotic representation has an alternative expression. First, we define

ηt=gtE(gtft)[E(ftft)]1ft=gtgff f1ft. (B.32) which implies that

ηη =gggf−1f ff g By the Woodbury formula, we have

gg1 =ηη1ηη1gf(∆f f +f gηη1gf)1f gηη1 (B.33) With (B.33) and the relation thatgt−Ξt=ηt+∆gff f1ftgfφφ1φt, we can rewrite the first term of the right hand side of (B.31) as

−1ggh1 T

XT t=1

(gt−Ξt)eiti=−1ηη 1 T

XT t=1

ηteit+−1ηηgf1 T

XT t=1

(∆−1f fft−1φφφt)eit

−1ηηgf(∆f f+f g−1ηηgf)−1f g−1ηη 1 T

XT t=1

ηteit

−1ηηgf(∆f f +f g−1ηηgf)−1f g−1ηηgf 1 T

XT t=1

(∆−1f fft−1φφφt)eit

Consider the term (∆−1f fft−1φφφt). From the definition ofφt=ftf ggg1gt, we have

φφ=f ff ggg1gf (B.34) which can be used to derive

φt=ftf g−1ggt+gf−1f fft) =φφ−1f fftf g−1ggηt Then

(∆f f1ftφφ1φt) =φφ1f ggg1ηt

With the above equation, the first term of the right hand side of (B.31) can be further rewritten as

gg1h1 T

XT t=1

(gt−Ξt)eiti=ηη11 T

XT t=1

ηteit+ηη1gfφφ1f ggg11 T

XT t=1

ηteit (B.35)

ηη1gf(∆f f+f gηη1gf)1f gηη11 T

XT t=1

ηteit

ηη1gf(∆f f+f gηη1gf)1f gηη1gfφφ1f ggg11 T

XT t=1

ηteit

From the two basic facts that

−1φφ =−1f f +−1f ff gηη1gf−1f f, and

−1f ff g−1ηη =−1φφf g−1gg.

we can rewrite the 2nd, 3rd and 4th terms on the right hand side of (B.35) as

−1ηηgfφφ1f f1f f1f g−1gggf1φφ1f g−1gg 1 T

XT t=1

ηteit

which equals zero by (B.34). So we can alternatively write the asymptotic representation of ˆγiγi as

ˆ

γiγi =−1ηηh1 T

XT t=1

ηteiti+W λi+Op(N−1) +Op(T−1).

We proceed to consider ˆftft. Notice ˆft = ˆR11′−1f˜tRˆ′−111Rˆ21gt and ft = R′−111ftR′−111R21gt. Then

fˆtft= ˆR′−111 f˜tRˆ′−111 Rˆ21gtR′−111 ftR′−111 R21gt

=−R′−111( ˆR11R11)R′−111ft+R′−111( ˜ftft)−R′−111( ˆR21R21)gt+R′−111( ˆR11R11 )R′−111R21gt

−( ˆR′−111R′−111 )( ˆR11R11)R′−111 ft+ ( ˆR′−111R′−111 )( ˜ftft)−( ˆR′−111R′−111 )( ˆR21R21)gt +( ˆR′−111R′−111 )( ˆR11R11)R′−111 R21gt

The last four terms of the above expression areOp(N2) +Op(T1). Given this result, by (B.28), (B.29) and Proposition A.3, we have

fˆtft=−V(R′−111 ftR′−111 R21gt)−Wgt +h 1

N XN i=1

1

σi2λiλii11 N

XT i=1

1

σ2iλieit+Op(N−1) +Op(T−1)

Byft=R′−111 ftR′−111 R21gt, we have fˆtft=1

N XN i=1

1

σi2λiλi11 N

XT i=1

1

σi2λieitVftWgt+Op(N−1) +Op(T−1).

This completes the proof.

Proposition B.2 Under Assumptions A-D, together with the identification condition IR1, we have

Φˆk−Φk= XT t= ¯K

utψt XT t= ¯K

ψtψt−1(ikIr)−BΦk+ ΦkB+Op(N−1) +Op(T−1)

Proof of Proposition B.2. Consider ˆΦk −Φk. Notice ˆΦk = R′−1ΦˆkRˆ and Φk = R′−1ΦkR. Thus

Φˆk−Φk=R′−1ΦˆkRˆR′−1ΦkR =R′−1Φk∆RdR′−1∆RdR′−1ΦkR+R′−1( ˆΦk−Φk)R+V where

V = (R′−1R′−1) ˆΦk∆Rd+( ˆR′−1R′−1)( ˆΦk−Φk)R−( ˆR′−1R′−1)∆RdR′−1+R′−1( ˆΦk−Φk)∆Rd However, notice

∆Rd = ˆRR=

"

∆Rd11 0

∆Rd21 0

#

=

"

R11V+V R11 0 W R11WR21V 0

#

+Op(N−1) +Op(T−1)

=BRRB+Op(N−1) +Op(T−1) (B.36)

where

B =

"

V 0 W 0

#

, B =

"

V 0 W 0

#

and W = (PTt=1υtυt)1(PTt=1υtεt). Then ∆Rd is Op(T1/2) since both B and B are Op(T−1/2). This result together with ˆΦk −Φk = Op(T−1/2) +Op(N−1) implies V = Op(N−2) +Op(T−1). Given this result, together with Φk=R′−1ΦkR, we have

Φˆk−Φk= ΦkR′−1∆RdR′−1∆RdΦk+R′−1( ˆΦk−Φk)R+Op(N−2) +Op(T−1) (B.37) Substituting (B.36) into the above equation, together withut=R′−1ut, ht =R′−1ht and Proposition A.4, we have

Φˆk−Φk= XT t= ¯K

utψt XT t= ¯K

ψtψt1(ikIr)−BΦk+ ΦkB+Op(N1) +Op(T1) This completes the proof.

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