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 2002. 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 > 0 0 0 π π Σ because Σ F is positive definite by assumption. his completes the proof.
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 2 8 2 7 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
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
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 2 2 + t=τ+ F t η if 2 t 2. E i= F t e it 2 M. 4
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
can be ignored and that the asymptotic properties of ˆF t will not be affected See Remark 5 of Bai, 2009. 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
O p by Lemma i of B 2002. 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
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
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
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 2 + 2 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
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
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
= 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
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