Appendices to the paper "Detecting Big Structural Breaks in Large Factor Models" (2013) by Chen, Dolado and Gonzalo.

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1 Appendices to the paper "Detecting Big Structural Breaks in Large Factor Models" 203 by Chen, Dolado and Gonzalo. A.: Proof of Propositions and 2 he proof proceeds by showing that the errors, factors and loadings in model 5 satisfy Assumptions A to D of Bai and g 2002 B 2002 hereafter. hen, once these results are proven, Propositions and 2 just follow immediately from application of heorems and 2 of B Define F t = [F t G t ], ɛ t = HG 2 t + e t, and Γ = [A Lemma. E Ft 4 < and Ft Ft Σ F. p Σ F Λ]. as for some positive matrix Proof. E F t 4 < follows from E F t 4 < by Assumption 2 and the definition of G t. o prove the second part, we partition the matrix Σ F = lim F t F t into: Σ Σ 2 Σ 2 Σ 22 where Σ = lim F t F t, Σ 22 = lim F 2 t F 2 t, Σ 2 = lim F t F 2 t, and F t is the k subvector of F t that has big breaks in their loadings, F 2 t is the k 2 subvector of F t that doesn t have big breaks in their loadings. By the definition of F t G t we have: and Ft Ft Ft Ft Ft 2 t=τ+ Ft Ft Ft = Ft 2 Ft Ft 2 Ft 2 t=τ+ Ft 2 Ft t=τ+ Ft Ft t=τ+ Ft Ft 2. t=τ+ Ft Ft By Assumption 2, the above matrix converges to Moreover, Σ Σ 2 π Σ Σ F = Σ 2 Σ 22 π Σ 2. π Σ π Σ 2 π Σ Σ Σ 2 0 detσ F = det Σ 2 Σ 22 π Σ 2 = detσ F detπ π Σ > π π Σ because Σ F is positive definite by assumption. his completes the proof.

2 Lemma 2. Γ i < for all i, and Γ Γ Σ Γ as for some positive definite matrix Σ Γ. Proof. his follows directly from Assumptions.a and 3. he following lemmae involve the new errors ɛ t. Let M and M denote some positive constants. Lemma 3. Eɛ it = 0, E ɛ it 8 M for all i and t. Proof. For t =,..., τ, E ɛ it 8 = E e it 8 < M by Assumption 4. For t = τ +,...,, E ɛ it 8 = E e it + η if t E e it 8 + E η if t 2 8 by Loève s inequality. ext, E η i F 2 t 8 η i 8 E F t 8 < by Assumptions.a and 2. hen the result follows. Lemma 4. Eɛ sɛ t / = E i= ɛ is ɛ it = γ s, t, γ s, s M for all s, and s= γ s, t2 M for all t and. Proof. γs, t = Eɛ is ɛ it i= = Ee is + η ig 2 see it + η ig 2 t i= = [ Eeis e it + Eη ig 2 sη ig 2 t ] i= Ee is e it + E η i 2E s G2 η i G 2 2. t i= Since i= Ee is e it = γ s, t by Assumption 4, and E η i 2 G2 t ηi 2 E F t 2 = O for all t by Assumptions.b and 2, we have γ s, t γ s, t + O. hen i= γ s, s γ s, s + O M 2

3 by Assumption 4. Moreover, γs, t 2 s= = s= s= γ s, t + O 2 γ s, t 2 + O M + O M by Assumption 4. hus, the proof is complete. Lemma 5. Eɛ it ɛ jt = τ ij,t with τ ij,t τ ij for some τ ij and for all t; and i= j= τ ij M. Proof. By Assumption 4, τ ij,t τ ij for some τ ij and all t, where τ ij,t = Ee it e jt. hen: for all t. herefore by Assumption 4. τ ij,t = Eɛ it ɛ jt = Ee it + η ig 2 t e jt + η jg 2 t Ee it e jt + E η i 2E s G2 η i G 2 t τ ij + O τij τ ij + O i= j= i= j= M + O Lemma 6. Eɛ it ɛ js = τ ij,ts and i= j= s= τ ij,ts M. M Proof. By Assumption 4, i= j= s= τ ij,ts M, where Ee it e js = τ ij,ts. hen: Eɛ it ɛ js = τ ij,ts = Ee it e js + Eη ig 2 t η jg 2 s = τ ij,ts + Eη ig 2 t η jg 2 s 2 3

4 and we have τij,ts τ ij,ts + Eη ig 2 t η jg 2 s i= j= i= j= i= j= M + O M following the same arguments as above. Lemma 7. For every t, s, E /2 i= [ɛ is ɛ it Eɛ is ɛ it ] 4 M. Proof. Since ɛ it = e it + η i G2 t, we have: ɛ it ɛ is Eɛ it ɛ is = e it e is Ee it e is + e it η ig 2 s + e is η ig 2 t + η ig 2 t η ig 2 s Eη ig 2 t η ig 2 s. Since E e it η i G 2 s 4 η i 4 E e it 4 E F t 4 = O p 2 2, and E η i G2 t η i G2 s 4 η i 8 E F t 4 = O p 4 4, the result follows from Loève s inequality and that E /2 [e is e it Ee is e it ] 4 M. i= Lemma 8. E i= F t ɛ it 2 M. Proof. By the definition of ɛ it we have: E i= Ft 2 ɛ it E i= Ft 2 e it + E i= Ft η ig 2 t 2 then by the definition of F t and G 2 t, F t e it 2 = F t e it 2 + Ft 2 e it, t=τ+ Ft η ig 2 t 2 = First, by Assumption 4 we have t=τ+ F t η if t t=τ+ F t η if 2 t 2. E i= F t e it 2 M. 4

5 Second, and = = E r k= r t=τ+ F t η if 2 t 2 E t=τ+ 2 F kt η if t 2 p= t=τ+ s=τ+ E F kt F ks η if t η if s E F kt F ks η if t 2 η if s 2 η i 2 E F t 4 = O, so we have E t=τ+ F t η i F t 2 2 = O/. he result then follows by noting that t=τ+ Ft 2 e it t=τ+ 2 F t e it and t=τ+ Ft η i F t 2 2 t=τ+ F t η i F t 2 2. As mentioned before, once it has been shown that the new factors: Ft, the new loadings: Γ and the new errors: ɛ t all satisfy the necessary conditions of B 2002, Propositions and 2 just follow directly from their heorems and 2, with r replaced by r + k and F t replaced by Ft. A.2: Proof of heorem We only derive the limiting distributions for the two versions of the LM test, since the proof for the Wald tests is very similar. Let ˆF t define the r vector of estimated factors. Under the null: k = 0, when r = r we have ˆF t = DF t + o p. Let D i denote the ith row of D, and D j denote the jth column of D. Define ˆF t = DF t, and ˆF kt = D k F t as the kth element of ˆFt. Let ˆF t be the first element of ˆFt, and ˆF t = [ ˆF 2t,..., ˆF rt ], while ˆF t and ˆF t can be defined in the same way. ote that ˆF t depends on and. For simplicity, let π denote [ π]. ote that under H 0, we allow for the existence of small breaks, so that the model can be written as X it = α i F t + e it + η i G 2 t. However, since η i G 2 t is O p / by Assumption, we can use similar methods as in Appendix A. to show that an error term of this order 5

6 can be ignored and that the asymptotic properties of ˆF t will not be affected See Remark 5 of Bai, herefore, for simplicity in the presentation below, we eliminate the last term and consider instead the model X it = α i F t + e it in the following lemmae 9 to 3 required to prove Lemma 4 which is the key one in the proof of heorem. Lemma 9. sup π [0,] π Proof. Following Bai 2003 we have: ˆF t ˆF t F t = O p δ 2,. π ˆF t ˆF t F t = 2 π s= π + 2 ˆF s F tξ st s= = I + II + III + IV π ˆF s F tγ s, t + 2 s= π ˆF s F tζ st + 2 s= ˆF s F tκ st where ζ st = e se t γ s, t. κ st = F sa e t /. ξ st = F ta e s /. First, note that: π I = 2 ˆF s DF s F tγ s, t + 2 π D F s F tγ s, t. s= s= Consider the first part of the right hand side, we have π 2 ˆF s DF s F tγ s, t s= = 2 ˆF π s DF s F tγ s, t s= /2 ˆFs DF s 2 π F t 2 s= π γ s, t 2. s= s= ˆFs DF s 2 is Op δ, 2 by heorem of B 2002, sup π [0,] F t 2 = Op by Assumption 2, and sup π [0,] π F t 2 s= π γ s, t 2 s= γ s, t 2 = 6

7 O p by Lemma i of B herefore: sup 2 π π [0,] s= For the second part, note that: ˆF s DF s F tγ s, t = O p δ, /2. sup 2 π D F s F tγ s, t π [0,] s= 2 D F s F t γ s, t s= and E s= s= F s F t γ s, t E F t 2 γ s, t = E F t 2 γ s, t M s= by Assumptions 2 and 4, so the second part is O p given that D is O p. herefore, we have ext, II can be written as: Similarly, we have sup I = O p. A. π [0,] δ, π II = 2 ˆF s DF s F tζ st + 2 π D F s F tζ st. s= s= sup 2 π ˆF s DF s F tζ st π [0,] s= sup ˆFs DF s 2 π F t 2 π π [0,] s= 2 ζst 2 s= ˆFs DF s 2 F t 2 s= 2 ζst 2 s= = O p δ, 7

8 because E ζ st 2 = E /2 [e it e is Ee it e is ] 2 = O i= by Assumption 4. As for the second term of II, we have: where 2 π D F s F tζ st = π s= D q t F t q t = Since E q t 2 M by Assumption 4, we have = [e it e is Ee it e is ]F s. s= i= sup 2 π D F s F tζ st π [0,] s= sup π D q t F t π [0,] D sup π π [0,] O p = O p. q t 2 q t 2 π F t 2 F t 2 hen it follows that Regarding III, it can be written as: sup II = O p. A.2 π [0,] δ, π III = 2 ˆF s DF s F tκ st + 2 π D F s F tκ st s= s= and the second part on the right hand side can be written as D F s F s s= π α i F te it. i= 8

9 herefore: sup 2 π D F s F tκ st π [0,] s= D F s F s sup s= π [0,] = O p π α i F te it i= by Assumption 8. As for the first part on the right hand side of III, we have sup 2 π ˆF s DF s F tκ st π [0,] s= ˆFs DF s 2 sup π F t s= π [0,] κ st 2 s= sup F π [0,] π s s= i= O p δ, 2 F s sup π s= π [0,] i= = O p δ, = O p δ, by Assumption 8. hus, It can also be proved in the similar way that Finally we have: sup π [0,] π ˆF t ˆF t F t α i F t e it 2 α i F t e it sup III = O p. A.3 π [0,] sup IV = O p. A.4 π [0,] sup π [0,] I + sup II + sup π [0,] = O p δ, + O p δ, + O p = O p δ 2, π [0,]. 2 III + sup IV π [0,] 9

10 Lemma 0. Proof. ote that: sup π [0,] π ˆF t ˆF t π ˆF t ˆF t = Op δ 2,. = = = π π π π ˆF t ˆF t ˆF t ˆF t π π ˆF t ˆF t ˆF t ˆF t F td + DF t F td π ˆF t DF t F td ˆF t DF t ˆF t DF t + π D F t ˆF t DF t + π ˆF t DF t F td. hus, sup π [0,] sup π [0,] π π ˆF t ˆF t π ˆF t ˆF t π [0,] ˆF t DF t ˆF t DF t + 2 D sup π [0,] ˆFt DF t D sup π since ˆFt DF t 2 = Op δ, 2 and sup π [0,] Lemma 9, the proof is complete. ˆF t DF t F t π π ˆF t DF t F t ˆF t DF t F t is O p δ, 2 by he next two lemmae follow from Lemma 0 and Assumption 6: Lemma. sup π [0,] π Proof. See Lemma 0 and Assumption 6. ˆF t ˆFt π ˆF t ˆFt = op. Lemma 2. ˆF t ˆF t = op. Proof. By construction we have ˆF t ˆF t = 0, and then the result follows from Lemma. 0

11 Let denote weak convergence. D, F t, F t, F t and S are defined as in the paper see Page 2. Similarly, let Di denote the ith row of D, and D j denote the jth column of D. hen: Lemma 3. π F t F t EF t F t S /2 W r π for π [0, ], where W r is a r vector of independent Brownian motions on [0, ]. Proof. F t F t is stationary and ergodic because F t is stationary and ergodic by Assumption 7. First, we show that {F kt F t EF kt F t, Ω t } is an adapted mixingale of size for k = 2,..., r. By definition, we have F kt F t = D k F td F t = r p= D kp F pt rp= D p F pt = rh= rp= D kp D h F ptf ht, and F kt F t EF kt F t = r h= rp= D kp D h F ptf ht EF pt F ht = rh= rp= D kp D h Y hp,t. hus: = E E 2 F kt F t EF kt F t Ω t m r r 2 E Dkp D h EY hp,t Ω t m h= p= r r DkpD h h= p= r r h= p= c hp t γm hp r 2 max c hp t maxγm hp E EY hp,t Ω t m 2 since maxγ hp m is Om δ for some δ > 0 by Assumption 7, we conclude that {F kt F t EF kt F t, Ω t } is an adapted mixingale of size for k = 2,..., r. ext, we prove the weak convergence using the Crame-Rao device. Define where a R r, and a a =. ote that z t = a S /2 F t F t EF t F t z t = r ã k [F kt F t EF kt F t ] k=2

12 where ã k is the k th element of a S /2. Ez 2 t r 2 2 E ã k [F kt F t EF kt F t ] k=2 r k=2 E F kt F t 2 EF kt F t 2 2 M because E F t 4 < and F kt = D k Ft. Moreover, z t is stationary and ergodic, and we can show {z t, Ω t } is an adapted mixingale sequence of size because: 2 E Ez t Ω t m = r E k=2 ã k E F 2 kt F t EF kt F t Ω t m k=2 r 2 ã k E E F kt F t EF kt F t Ω t m max ã k r c k t γ m. k By the results above we know that γ k m is Om δ for k = 2,..., r. Hence it follows that {z t, Ω t } is an adapted mixingale sequence of size. hen it follows from heorem 7.7 of White 200 that: k=2 π z t = a S /2 π F t F t EF t F t Wπ. Moreover, it can be shown that: a π 2 t= π F t F t EF t F t +a π 0 2 F t F t EF t F t d 0, π2 π a Sa +π 0 a 2Sa 2 by using Corollary 3. of Woodridge and White 988. he proof is completed by using Lemma A.4 of Andrews 993. A.3: More discussions on Remark 9 In Remark 9 of the paper we mention that, although our tests are designed for single break, they should also have power against multiple breaks. o see this, consider the simple example of a FM with one factor and two big breaks: X t = Af t t τ + Bf t τ < t < τ 2 + Df t t τ 2 + e t 2

13 = Ag t + Bh t + Ds t + e t where g t = f t t τ, h t = f t τ < t < τ 2, and s t = f t t τ 2. In view of Proposition 2, Bai and g s 2002 IC will lead to the choice of 3 factors which, when estimated by PCA, implies the following result: ˆf t d d 4 d 7 g t ˆf 2t = d 2 d 5 d 8 h t + o p. ˆf 3t d 3 d 6 d 9 s t hen, by the definition of g t, h t and s t we have: ˆf t d ˆf 2t = d 2 f t + o p for t =,..., τ, ˆf 3t d 3 ˆf t d 4 ˆf 2t = d 5 f t + o p for t = τ,..., τ 2, ˆf 3t d 6 ˆf t d 7 ˆf 2t = d 8 f t + o p for t = τ 2,...,. ˆf 3t d 9 Hence, we can find one vector [p, p 2, p 3 ] which is orthogonal to [d, d 2, d 3 ] and [d 4, d 5, d 6 ], plus another vector [p 4, p 5, p 6 ] which is orthogonal to [d 7, d 8, d 9 ]. It is easy to see that [p, p 2, p 3 ] a[p 4, p 5, p 6 ] for any a 0 otherwise the D matrix will be singular, and thus we can find a breaking relationship between the estimated factors and even use Bai and Perron s 998, 2003 to detect a second break. he simulation results about the power of our tests against multiple breaks are available upon request. 3

14 References Andrews, D ests for parameter instability and structural change with unknown change point. Econometrica 6 4, Bai, J Inferential theory for factor models of large dimensions. Econometrica 7, Bai, J Panel data models with interactive fixed effects. Econometrica 77 4, Bai, J. and S. g Determining the number of factors in approximate factor models. Econometrica 70, Bai, J. and P. Perron 998. Estimating and testing linear models with multiple structural changes. Econometrica 66, Bai, J. and P. Perron Computation and analysis of multiple structural change models. Journal of Applied Econometrics 8, 22. White, H Asymptotic theory for econometricians. Academic Press Orlando, Florida. Wooldridge, J. and H. White 988. Some invariance principles and central limit theorems for dependent heterogeneous processes. Econometric heory 4 2,