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 T−K−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=1ft∗ft∗′ =Ir1 and T1 PTt=1ft∗g′t= 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,σ˜2i and ˜ftare respectively consistent estimator ofλ∗i, σ2i and ft∗ in AU1.
Proposition A.1 Under Assumptions A-D, together with AU1,
λ˜i−λ∗i =1 T
XT t=1
ft∗ft∗′−11 T
XT t=1
ft∗eit+Op(N−1/2T−1/2) +Op(T−1), (A.1)
˜
γi−γi∗ =1 T
XT t=1
gtgt
−11 T
XT t=1
gteit
, (A.2)
f˜t−ft∗ =1 N
XT i=1
1
σ2iλ∗iλ∗′i −11 N
XN i=1
1
σ2iλ∗ieit+Op(N−1/2T−1/2) +Op(T−1), (A.3)
˜
σ2i −σ2i = 1 T
XT t=1
(e2it−σ2i) +Op(N−1/2T−1/2) +Op(T−1), (A.4)
Proof of Proposition A.1Write zt= Λ∗ft∗+ Γ∗gt+et into matrix form,
Z = Λ∗F∗′+ Γ∗G′+e. (A.5)
Post-multiplyingMG=IT −G(G′G)−1G′ on both sides, together withF∗′G= 0 by AU1, we have
ZMG= Λ∗F∗′+eMG.
LetY =ZMG and ytdenotes the t-th column ofY. The above equation is equivalent to yt= Λ∗ft∗+et−eG(G′G)−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(G′G)−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=1gtg′t)−1 PTt=1gt(zit−˜λ′if˜t), we have
˜
γi−γi∗ = XT t=1
gtg′t−1 XT t=1
gteit− XT t=1
gtgt′−1 XT t=1
gtft∗′(˜λi−λ∗i)
− XT t=1
gtg′t−1 XT t=1
gt( ˜ft−ft∗)′λ˜i
The second term of the right hand side is zero by PTt=1gtft∗′ =G′F∗ = 0. Consider the third term. Notice
f˜t= (˜Λ′Σ˜−1eeΛ)˜ −1Λ˜′Σ˜−1eeyt
= (˜Λ′Σ˜−ee1Λ)˜ −1Λ˜′Σ˜−ee1hzt− XT s=1
zsg′s XT s=1
gsg′s−1gti
= (˜Λ′Σ˜−1eeΛ)˜ −1Λ˜′Σ˜−1eehΛ∗ft∗+et− XT s=1
esgs′ XT s=1
gsgs′−1gti Then it follows
f˜t−ft∗ = −A∗′ft∗+ (˜Λ′Σ˜−1eeΛ)˜ −1Λ˜′Σ˜−1eeet
−(˜Λ′Σ˜−ee1Λ)˜ −1Λ˜′Σ˜−ee1 XT s=1
esgs′ XT s=1
gsgs′−1gt
(A.7)
whereA∗ = (˜Λ−Λ∗)′Σ˜−ee1Λ(˜˜ Λ′Σ˜−ee1Λ)˜ −1.
Given the above expression, together withPTt=1gtft∗′= 0, we have 1
T XT t=1
gt( ˜ft−ft∗)′ = 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(N−1) +Op(T−1). The last term can be proved to be the same magnitude by the similar arguments. Summarizing these results, we haveJ11=Op(N−1) +Op(T−1). 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(N−1/2T−1/2) +Op(T−1) and (˜Λ′Σ˜−ee1Λ)˜ −1 =Op(N−1), 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˜ht−1+ Φ2˜ht−2+· · ·+ ΦK˜ht−K+error
By Lemma A.1(a) and (b), h1
T¯ XT t= ¯K
(˜ht−h∗t) ˜ψt′ih1 T¯
XT t= ¯K
ψ˜tψ˜t′i−1 =Op(N−1) +Op(T−1)
h1 T¯
XT t= ¯K
( ˜ψt−ψt∗) ˜ψ′tih1 T¯
XT t= ¯K
ψ˜tψ˜′ti−1 =Op(N−1) +Op(T−1) By Lemma A.1(a) and (c),
h1 T¯
XT t= ¯K
u∗tψ˜′tih1 T¯
XT t= ¯K
ψ˜tψ˜′ti−1 =h1 T¯
XT t= ¯K
u∗tψt∗′ih1 T¯
XT t= ¯K
ψ∗tψt∗′i−1 +Op(N−1) +Op(T−1)
Given this result, we have Φ˜ −Φ∗ =
XT t= ¯K
u∗tψt∗′1 T¯
XT t= ¯K
ψ∗tψt∗′−1+Op(N−1) +Op(T−1) Post-multiplyingik⊗Ir 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: N1Λ⋆′Σ−ee1Λ⋆ = Q⋆, E(ft⋆ft⋆′) =Ir1 and E(ft⋆g′t) = 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 Q′V 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˜i−λ⋆i = (˜λi−λ∗i) + (λ∗i −λ⋆i).
We showλ∗i and λ⋆i 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 (λ⋆i′, 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∗ =R⋆11′ft⋆+R⋆21′gt. (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 followsR⋆11−1−→p Ir1. Let where Ndg{·} denotes the non-diagonal elements of the argument. Neglecting the terms U⋆Q⋆U⋆′ 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 ˜λi−λ⋆i. Notice λ˜i−λ⋆i = ˜λi−R⋆11λ∗i = (˜λi−λ∗i)−(R⋆11−Ir1)λ∗i By (A.1), the above result is equivalent to
λ˜i−λ⋆i =h1
ByR⋆′−111 =Ir1 +U⋆′, the above equation is equal to
f˜t−ft⋆ = ( ˜ft−ft∗)−U⋆′ft∗+R⋆11′−1R⋆′21gt Substituting (A.3) into the above result, we have
f˜t−ft⋆ = −U⋆′ft∗+R⋆11′−1R⋆21′gt+
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) andR⋆21=−W⋆+Op(T−1), we have
By Proposition A.2, we can rewrite the above result as Φ˜k−Φ⋆k=
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 (λ′i, γ′i)′. Then we have The last equation in (B.2) can be written as
ft⋆ =R11′ ft+R′21gt. (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(ft⋆gt′) = 0, we have R21=−∆−gg1∆gfR11.
Defineφt=R′−111ft⋆. From the above results,φhas an alternative expression
φt=ft−∆f g∆−gg1gt. (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 estimates Aˆ andBˆ, we have
AˆBˆ−1Aˆ′− AB−1A′ = ( ˆA − A)B−1A′+AB−1( ˆA − A)′− AB−1( ˆB − B)B−1A′+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−1Aˆ′− 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∗′MΨ∗H∗ =Op(N−1) +Op(T−1) (b) 1
T¯H⋆′MΨ⋆H⋆− 1
T¯H∗′MΨ∗H∗ =B⋆′Ω⋆+ Ω⋆B⋆+Op(T−1) (c) 1
T¯H˜′MΨ˜H˜ − 1
T¯H⋆′MΨ⋆H⋆ =−B⋆′Ω⋆−Ω⋆B⋆+Op(N−1) +Op(T−1) where 1¯
TH˜′MΨ˜H˜ is defined as 1
T¯
H˜′MΨ˜H˜ = 1 T¯
XT t= ¯K
˜ht˜h′t−1 T¯
XT t= ¯K
˜htψ˜′t1 T¯
XT t= ¯K
ψ˜tψ˜t′−11 T¯
XT t= ¯K
ψ˜t˜h′t,
and T1¯H∗′MΨ∗H∗ and T1¯H⋆′MΨ⋆H⋆ are defined similarly.
Proof of Lemma B.2. Consider (a). By Lemma A.1(a), we have 1
T¯ XT t= ¯K
h˜t˜h′t− 1 T¯
XT t= ¯K
h∗th∗′t =Op(N−1) +Op(T−1)
1 Given the above results, together with Lemma B.1, we have (a).
Consider (b). By h⋆t =R⋆′−1h∗t, we have ψt⋆ = (IK⊗R⋆′−1)ψt∗. 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) ˆft−ft=1 N
XN i=1
1
σi2λiλ′i−11 N
XT i=1
1
σi2λieit−V′ft−W′gt+Op(N−1) +Op(T−1) wherevec(V) =B−1
Q P1D+r1T1¯PTt= ¯K[εt⊗εt−vec(Ir1)];φt=ft−∆f g∆−1gggt;∆φφ=E(φtφ′t);
W = Ω−υυ1 1 T
PT
t= ¯Kυtε′t; ηt=gt−∆gf∆−f f1ft; ∆ηη =E(ηtηt′).
Proof of Proposition B.1. Consider the VAR expression under AU2:
h⋆t = Φ⋆1h⋆t−1+ Φ⋆2h⋆t−2+· · ·+ Φ⋆Kh⋆t−K+u⋆t. Pre-multiplyingR′−1 gives
ht= R′−1Φ⋆1R′ht−1+· · ·+ R′−1Φ⋆KR′ht−K+R′−1u⋆t. So we have Φi =R′−1Φ⋆iR′ fori= 1,2, . . . , K and ut=R′−1u⋆t. Then we have
εt=R′−111ε⋆t −R′−111R′21υ⋆
t, υt=υ⋆
t. (B.7)
Post-multiplyingυ′
t on both sides and taking the expectation, byE(εtυ′
t) = 0, we have R21= Ω⋆υυ−1Ω⋆υε, (B.8) Substituting the proceeding result into (B.7), byE(εtε′t) =Ir1, we have
Ω⋆εε·υ= Ω⋆εε−Ω⋆ευΩ⋆υυ−1Ω⋆υε=R′11R11. (B.9) where Ω⋆εε=E(ε⋆tε⋆t′),Ω⋆υυ =E(υ⋆tυ⋆t′) and Ω⋆ευ =E(ε⋆tυ⋆t′). In addition, the identification condition also requires that
Q= 1
NΛ′Σ−1eeΛ =R111
NΛ⋆′Σ−1eeΛ⋆R′11. This is equivalent to
Q⋆ = 1
NΛ⋆′Σ−1eeΛ⋆=R−111QR′−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) Rˆ′11Rˆ11= ˜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−Ω⋆εε·υ
−p
→0, where ˜Ωεε·υ and Ω⋆εε·υ
are defined in (B.9) and (B.12). Thus
Rˆ′11Rˆ11R−111R′−111 −→p Ir1, which, by the fact thatAB=I thenBA=I, leads to
( ˆR11R−111)′( ˆR11R−111)−→p Ir1 (B.19) Furthermore, by (B.13), we have
( ˆR11R11−1)hR111
NΛ˜′Σ˜−1eeΛ˜R′11i( ˆR11R−111)′ = diag
By N1Λ˜′Σ˜−ee1Λ˜ −N1Λ⋆′Σ−ee1Λ⋆ =op(1) andR11N1Λ⋆′Σ−ee1Λ⋆R′11= N1Λ′Σ−ee1Λ =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 ˆR11−R11 p
−
→ 0. Henceforth, we use ∆Rd11 to denote ˆR11−R11. Apparently
∆Rd11−→p 0. By (B.9) and (B.12), we have
Rˆ′11Rˆ11−R′11R11= ˜Ωεε−Ω⋆εε−( ˜ΩευΩ˜−1υυΩ˜υε−Ω⋆ευΩ⋆−1υυ Ω⋆υε)
Substituting (B.16)-(B.18) into the above equation, together with Lemma B.1, we have
∆Rd′11R11+R′11∆Rd11+∆Rd′11∆Rd11=−V⋆′Ω⋆εε·υ−Ω⋆εε·υV⋆ +1
T¯ XT t= ¯K
h(ε⋆t −Ω⋆ευΩ⋆υυ−1υ⋆
t)(ε⋆t −Ω⋆ευΩ⋆υυ−1υ⋆
t)′−Ω⋆εε·υ
i+Op(N−1) +Op(T−1).
However, by (B.7) and (B.8), we haveR′11εt=ε⋆t−Ω⋆ευΩ⋆υυ−1υ⋆t. Given this result, together with (B.9), we have
∆Rd′11R11+R′11∆Rd11+∆Rd′11∆Rd11=−V⋆′R′11R11−R′11R11V⋆ +R′11h1
T¯ XT t= ¯K
εtε′t−Ir1
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∆Rd′11∆Rd11R−111, we have
∆Rd11R−111 +R11V⋆R11−1+∆Rd11R11−1+R11V⋆R−111′
= 1 T¯
XT t= ¯K
(εtε′t−Ir1) +Op(N−1) +Op(T−1).
(B.22)
Now consider 1
NΛ˜′Σ˜−ee1Λ˜− 1
NΛ⋆′Σ−ee1Λ⋆ = 1 N
XN i=1
1
˜
σ2i(˜λi−λ⋆i)˜λ′i+ 1 N
XN i=1
1
˜
σ2iλ˜i(˜λi−λ⋆i)′
− 1 N
XN i=1
1
˜
σi2(˜λi−λ⋆i)(˜λi−λ⋆i)′+ 1 N
XN i=1
λ⋆iλ⋆′i ( 1
˜ σ2i − 1
σi2).
The last term is Op(N−1/2T−1/2) +Op(T−1) which is shown in Bai and Li (2012). The third term isOp(T−1). The first two terms areV⋆Q⋆+Q⋆V⋆′+Op(N−1/2T−1/2)+Op(T−1) by Proposition A.3. Then it follows
1
NΛ˜′Σ˜−ee1Λ˜− 1
NΛ⋆′Σ−ee1Λ⋆ =V⋆Q⋆+Q⋆V⋆′+Op(N−1/2T−1/2) +Op(T−1). (B.23) Given the above results, (B.13) is equivalent to
Ndg
Rˆ11(Q⋆+V⋆Q⋆+Q⋆V⋆′) ˆR′11
=Op(N−1/2T−1/2) +Op(T−1).
Substituting (B.10) into the proceeding equation, we have Ndg
Rˆ11(R−111QR′−111 +V⋆R−111QR′−111 +R−111QR′−111 V⋆′) ˆR′11
=Op(N−1/2T−1/2) +Op(T−1).
Replace ˆR11=∆Rd11+R11, the left hand side is (neglecting Ndg)
Q+∆Rd11R11−1Q+Q(∆Rd11R−111)′+∆Rd11R−111Q(∆Rd11R−111)′Q +R11V⋆R−111Q+∆Rd11V⋆R11−1Q+ ˆR11V⋆R−111Q(∆Rd11R−111)′ +QR′−111 V⋆′R′11+QR′−111 V⋆′∆Rd′11+ (∆Rd11R11−1)QR′−111 V⋆′Rˆ′11
By neglecting the terms of smaller magnitude and noticing that Ndg(Q) = 0, we have Ndgn∆Rd11R−111+R11V⋆R−111Q (B.24) +Q∆Rd11R11−1+R11V⋆R−111′o=Op(N−1/2T−1/2) +Op(T−1).
Let V = ∆Rd11R−111 +R11V⋆R11−1. 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
(εtε′t−Ir1)i+Op(N−1) +Op(T−1).
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
(εtε′t−Ir1)i+Op(N−1) +Op(T−1). (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(N−1/2T−1/2) +Op(T−1).
implying
D1[Q⊗Ir1+ (Ir1⊗Q)Kr1]vec(V) =Op(N−1/2T−1/2) +Op(T−1). (B.26) The preceding two equations imply
"
2Dr+1
D1[Q⊗Ir1+ (Ir1 ⊗Q)Kr1]
#
vec(V) =
"
Dr+1vec1¯
T
PT
t= ¯K(εtε′t−Ir1) 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
(εtε′t−Ir1)i+Op(N−1) +Op(T−1).
DefineV by
vec(V) =B−1Q P1Dr+1vech1 T¯
XT t= ¯K
(εtε′t−Ir1)i. Then by the definition ofV,
∆Rd11R−111+R11V⋆R11−1 =V + +Op(N−1) +Op(T−1). (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(T−1/2) and V =Op(T−1/2).
Now consider ˆλi−λi. By ˆλi = ˆR11λ˜i and λi=R11λ⋆i, we have
λˆi−λi = ˆR11λ˜i−R11λ⋆i =∆Rd11λ⋆i +R11(˜λi−λ⋆i) +∆Rd11(˜λi−λ⋆i).
The last term of right hand side isOp(T−1) +Op(N−2) by ˜λi−λ⋆i =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 =R11λ⋆i, we have
λˆi−λi =V λi+R111 T
XT t=1
ft⋆ft⋆′−11 T
XT t=1
ft⋆eit+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= ˆR21−R21. Notice Rˆ21−R21= ˜Ω−υυ1Ω˜υε−Ω⋆υυ−1Ω⋆υε=−Ω⋆υυ−1( ˜Ωυυ−Ω⋆υυ)Ω⋆υυ−1Ω⋆υε+ Ω⋆υυ−1( ˜Ωυε−Ω⋆υε)
−( ˜Ω−1υυ−Ω⋆−1υυ )( ˜Ωυυ−Ω⋆υυ)Ω⋆−1υυ Ω⋆υε+ ( ˜Ω−1υυ−Ω⋆−1υυ )( ˜Ωυε−Ω⋆υε).
The last two terms of the right hand side are Op(N−2) +Op(T−1). Substituting (B.17) and (B.18) into the above result, we have
∆Rd21= Ω⋆υυ−1 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(ft⋆g′t) = 0, we have
R21R−111=−∆−gg1∆gf.
The preceding two results imply that the third expression of (B.30) is equal to
−∆−1gg∆gf∆−1φφh1 T
XT t=1
φteiti+Op(T−1).
Let Ξt=∆gf∆−φφ1φt. Given the above result, the asymptotic representation of ˆγi−γi can be rewritten as
ˆ
γi−γi =∆−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=gt−E(gtft′)[E(ftft′)]−1ft=gt−∆gf∆−f f1ft. (B.32) which implies that
∆ηη =∆gg−∆gf∆−1f f∆f g By the Woodbury formula, we have
∆−gg1 =∆−ηη1−∆−ηη1∆gf(∆f f +∆f g∆−ηη1∆gf)−1∆f g∆−ηη1 (B.33) With (B.33) and the relation thatgt−Ξt=ηt+∆gf∆−f f1ft−∆gf∆−φφ1φt, 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+∆−1ηη∆gf1 T
XT t=1
(∆−1f fft−∆−1φφφt)eit
−∆−1ηη∆gf(∆f f+∆f g∆−1ηη∆gf)−1∆f g∆−1ηη 1 T
XT t=1
ηteit
−∆−1ηη∆gf(∆f f +∆f g∆−1ηη∆gf)−1∆f g∆−1ηη∆gf 1 T
XT t=1
(∆−1f fft−∆−1φφφt)eit
Consider the term (∆−1f fft−∆−1φφφt). From the definition ofφt=ft−∆f g∆−gg1gt, we have
∆φφ=∆f f−∆f g∆−gg1∆gf (B.34) which can be used to derive
φt=ft−∆f g∆−1gg(ηt+∆gf∆−1f fft) =∆φφ∆−1f fft−∆f g∆−1ggηt Then
(∆−f f1ft−∆−φφ1φt) =∆−φφ1∆f g∆−gg1ηt
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+∆−ηη1∆gf∆−φφ1∆f g∆−gg11 T
XT t=1
ηteit (B.35)
−∆−ηη1∆gf(∆f f+∆f g∆−ηη1∆gf)−1∆f g∆−ηη11 T
XT t=1
ηteit
−∆−ηη1∆gf(∆f f+∆f g∆−ηη1∆gf)−1∆f g∆−ηη1∆gf∆−φφ1∆f g∆−gg11 T
XT t=1
ηteit
From the two basic facts that
∆−1φφ =∆−1f f +∆−1f f∆f g∆−ηη1∆gf∆−1f f, and
∆−1f f∆f g∆−1ηη =∆−1φφ∆f g∆−1gg.
we can rewrite the 2nd, 3rd and 4th terms on the right hand side of (B.35) as
∆−1ηη∆gf∆−φφ1−∆−f f1−∆−f f1∆f g∆−1gg∆−gf1∆−φφ1∆f g∆−1gg 1 T
XT t=1
ηteit
which equals zero by (B.34). So we can alternatively write the asymptotic representation of ˆγi−γi as
ˆ
γi−γi =∆−1ηηh1 T
XT t=1
ηteiti+W λi+Op(N−1) +Op(T−1).
We proceed to consider ˆft−ft. Notice ˆft = ˆR11′−1f˜t−Rˆ′−111Rˆ′21gt and ft = R′−111ft⋆ − R′−111R′21gt. Then
fˆt−ft= ˆR′−111 f˜t−Rˆ′−111 Rˆ′21gt−R′−111 ft⋆−R′−111 R′21gt
=−R′−111( ˆR11′ −R′11)R′−111ft⋆+R′−111( ˜ft−ft⋆)−R′−111( ˆR′21−R′21)gt+R′−111( ˆR′11−R11′ )R′−111R′21gt
−( ˆR′−111 −R′−111 )( ˆR′11−R′11)R′−111 ft⋆+ ( ˆR′−111 −R′−111 )( ˜ft−ft⋆)−( ˆR′−111 −R′−111 )( ˆR21−R21)gt +( ˆR′−111 −R′−111 )( ˆR′11−R′11)R′−111 R′21gt
The last four terms of the above expression areOp(N−2) +Op(T−1). Given this result, by (B.28), (B.29) and Proposition A.3, we have
fˆt−ft=−V′(R′−111 ft⋆−R′−111 R′21gt)−W′gt +h 1
N XN i=1
1
σi2λiλ′ii−11 N
XT i=1
1
σ2iλieit+Op(N−1) +Op(T−1)
Byft=R′−111 ft⋆−R′−111 R′21gt, we have fˆt−ft=1
N XN i=1
1
σi2λiλ′i−11 N
XT i=1
1
σi2λieit−V′ft−W′gt+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
ψtψ′t−1(ik⊗Ir)−B′Φk+ ΦkB′+Op(N−1) +Op(T−1)
Proof of Proposition B.2. Consider ˆΦk −Φk. Notice ˆΦk = R′−1ΦˆkRˆ′ and Φk = R′−1Φ⋆kR′. Thus
Φˆk−Φk=R′−1ΦˆkRˆ′−R′−1Φ⋆kR′ =R′−1Φ⋆k∆Rd′−R′−1∆Rd′R′−1Φ⋆kR′+R′−1( ˆΦk−Φ⋆k)R′+V where
V = (R′−1−R′−1) ˆΦk∆Rd′+( ˆR′−1−R′−1)( ˆΦk−Φ⋆k)R′−( ˆR′−1−R′−1)∆Rd′R′−1+R′−1( ˆΦk−Φ⋆k)∆Rd′ However, notice
∆Rd = ˆR−R=
"
∆Rd11 0
∆Rd21 0
#
=
"
−R11V⋆+V R11 0 W R11−W⋆−R21V⋆ 0
#
+Op(N−1) +Op(T−1)
=BR−RB⋆+Op(N−1) +Op(T−1) (B.36)
where
B =
"
V 0 W 0
#
, B⋆ =
"
V⋆ 0 W⋆ 0
#
and W = (PTt=1υtυ′t)−1(PTt=1υtε′t). Then ∆Rd is Op(T−1/2) since both B and B⋆ are Op(T−1/2). This result together with ˆΦk −Φ⋆k = Op(T−1/2) +Op(N−1) implies V = Op(N−2) +Op(T−1). Given this result, together with Φk=R′−1Φ⋆kR′, we have
Φˆk−Φk= ΦkR′−1∆Rd′−R′−1∆Rd′Φk+R′−1( ˆΦk−Φ⋆k)R′+Op(N−2) +Op(T−1) (B.37) Substituting (B.36) into the above equation, together withut=R′−1u⋆t, ht =R′−1h⋆t and Proposition A.4, we have
Φˆk−Φk= XT t= ¯K
utψt′ XT t= ¯K
ψtψ′t−1(ik⊗Ir)−B′Φk+ ΦkB′+Op(N−1) +Op(T−1) This completes the proof.
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