Startseite Bootstrap Prediction Intervals for Factor-Augmented Regressions with Cross-Section Averages
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Bootstrap Prediction Intervals for Factor-Augmented Regressions with Cross-Section Averages

  • Rui Chen , Yimeng Xie , Lizhi Tang EMAIL logo und Qing Tao
Veröffentlicht/Copyright: 10. Oktober 2025
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Studies in Nonlinear Dynamics & Econometrics
Aus der Zeitschrift Studies in Nonlinear Dynamics & Econometrics

Abstract

This paper revisits the problem of forecasting using factor augmented time series regressions. It is shown that the OLS estimator of the regression coefficients using cross-section average is asymptotically biased, when the time series dimension (T) and the cross-section dimension (N) satisfy T / N → c and 0 < c < ∞. We propose bootstrap algorithms using the OLS estimator to construct prediction intervals, and show they are asymptotically valid under Gaussianity of the innovations as well as under more general conditions. The effectiveness of the bootstrap prediction intervals is further justified by Monte Carlo simulations and an empirical application.

Keywords: bootstrap; cross-section average; factor-augmented regression models; forecast
JEL Classification: C10; C13; C15
Corresponding author: Lizhi Tang, Department of Statistics and Data Science at School of Economics, Xiamen University, 422 Siming South Road, Xiamen, 361005, Fujian, China, E-mail: 

Funding source: National Social Science Fund of China

Award Identifier / Grant number: 21BTJ037

Appendix A. Proofs of Main Results

Proof of Lemma 1:

Write F ̂ t = x ̄ t = λ ̄ ⊤ F t + e ̄ t in the matrix form,

(A.1) F ̂ = x ̄ = F λ ̄ + e ̄ ,

where x ̄ = x ̄ 1 , ⋯ , x ̄ T − h ⊤ is (T − h) × m, and F = F 1 , ⋯ , F T − h ⊤ is (T − h) × r. Let e ̄ = e ̄ r , e ̄ − r . Post-multiplying (A.1) by Λ ̄ , we have

F ̂ Λ ̄ = F λ ̄ Λ ̄ + e ̄ Λ ̄ = F , 0 T − h × m − r + e ̄ r λ ̄ r − 1 , e ̄ − r − e ̄ r λ ̄ r − 1 λ ̄ − r .

By (S.14) in Karabiyik and Westerlund (2021),

(A.2) Q N − 1 T δ ̂ − Q N δ 0 = T − 1 z ̂ 0 T z ̂ 0 + T − 1 / 2 z ̂ 0 ⊤ ε − e ̄ r 0 α ,

where ε = ε h + 1 , ⋯ , ε T ⊤ is (T − h) × 1, z ̂ 0 = z ̂ 1 0 , ⋯ , z ̂ T − h 0 ⊤ is (T − h) × (m + n), and e ̄ 0 = e ̄ r 0 , e ̄ − r 0 = e ̄ r λ ̄ r − 1 , N e ̄ − r − e ̄ r λ ̄ r − 1 λ ̄ − r is (T − h) × m. Define z t 0 = F t ⊤ , e ̄ − r , t 0 ⊤ , g t ⊤ ⊤ and z 0 = F , e ̄ − r 0 , g = z 1 0 , ⋯ , z T − h 0 ⊤ , which are (m + n) × 1 and (T − h) × (m + n), respectively. By (S.19) and (S.39) in Karabiyik and Westerlund (2021), T − 1 2 z ̂ 0 ⊤ ε = T − 1 2 z 0 ⊤ ε + O p N − 1 2 , and T − 1 z ̂ 0 ⊤ z ̂ 0 + → p Σ z 0 − 1 , where

Σ z 0 = Σ F 0 r × m − r Σ F g 0 m − r × r Λ − r ⊤ Σ e Λ − r 0 m − r × n Σ F g ⊤ 0 n × m − r Σ g ,

where Λ −r is from Λ = Λ r , Λ − r = lim N → ∞ Λ ̄ . Then we consider T − 1 / 2 z ̂ 0 ⊤ e ̄ r 0 ,

T − 1 2 z ̂ 0 ⊤ e ̄ r 0 = T − 1 2 z ̂ 0 − z 0 ⊤ e ̄ r 0 + T − 1 2 z 0 ⊤ e ̄ r 0 = T − 1 2 e ̄ r 0 ⊤ e ̄ r 0 0 n × m + T 0 r × 1 T − 1 e ̄ − r 0 ⊤ e ̄ r 0 0 n × 1 + O p N − 1 2 .

Also,

T − 1 2 e ̄ r 0 ⊤ e ̄ r 0 = T − 1 2 e ̄ Λ ̄ r ⊤ e ̄ Λ ̄ r = T − 1 2 Λ ̄ r ⊤ e ̄ ⊤ e ̄ Λ ̄ r = T N − 1 Λ ̄ r ⊤ N T − 1 e ̄ ⊤ e ̄ Λ ̄ r → p c Λ r ⊤ Σ e Λ r ,

Together with the above results, we have

T Q N − 1 ⁢ δ ̂ − Q N ⁢ δ 0 = T − 1 z ̂ 0 ⊤ z ̂ 0 + ⁢ T − 1 2 ⁢ z ̂ 0 ⊤ ⁢ ε − e ̄ r 0 ⁢ α = T − 1 z ̂ 0 ⊤ z ̂ 0 + ⁢ T − 1 2 ⁢ z ̂ 0 ⊤ ⁢ ε − T − 1 z ̂ 0 ⊤ z ̂ 0 + ⁢ T − 1 2 ⁢ z ̂ 0 ⊤ ⁢ e ̄ r 0 ⁢ α = Σ z 0 − 1 ⁢ T − 1 / 2 ⁢ z 0 ⊤ ⁢ ε − T 0 r × 1 T − 1 ⁢ e ̄ − r 0 ⊤ ⁢ e ̄ r 0 ⁢ α 0 n × 1 − T − 1 / 2 ⁢ e ̄ r 0 ⊤ ⁢ e ̄ r 0 ⁢ α 0 n × m + o p ⁢ 1 .

Finally, by Assumption 4(d), T Q N − 1 δ ̂ − Q N δ 0 → d N − c B 1 − B 2 , Σ z 0 − 1 Σ z 0 ε Σ z 0 − 1 , where

B 1 = Σ z 0 − 1 Λ r ⊤ Σ e Λ r α 0 n × 1 ,

B 2 = T N − 1 2 0 r × 1 Λ − r ⊤ Σ e Λ − r − 1 Λ − r ⊤ Σ e Λ r α 0 n × 1 ,

Σ z 0 ε = lim N , T → ∞ T − 1 ∑ t = 1 T − h E ε t + h 2 z t 0 z t 0 ⊤ .

Proof of Theorem 1:

Similar as (A.2), it follows that

Q N * − 1 T δ ̂ * − Q N * δ 0 * = T − 1 z ̂ 0 * ⊤ z ̂ 0 * + T − 1 / 2 z ̂ 0 * ⊤ ε * − e ̄ r 0 * α ̂ .

We now evaluate T − 1 z ̂ 0 * ⊤ z ̂ 0 * , T − 1 / 2 z ̂ 0 * ⊤ ε * and T − 1 / 2 z ̂ 0 * ⊤ e ̄ r 0 * in turn.

Consider T − 1 z ̂ 0 * ⊤ z ̂ 0 * . First,

z ̂ 0 * = F ̂ 0 * , g = F ̂ * Λ ̄ * D N , g = F ̂ λ ̂ ̄ + e ̄ * Λ ̄ * D N , g = F ̂ + e ̄ 0 * , g ,

where e ̄ 0 * = e ̄ * Λ ̄ * D N . Then it follows that

1 T z ̂ 0 * ⊤ z ̂ 0 * = 1 T F ̂ + e ̄ 0 * ⊤ F ̂ + e ̄ 0 * F ̂ + e ̄ 0 * ⊤ g g ⊤ F ̂ + e ̄ 0 * g ⊤ g ,

where

1 T F ̂ + e ̄ 0 * ⊤ F ̂ + e ̄ 0 * = 1 T F ̂ ⊤ F ̂ + e ̄ 0 * ⊤ F ̂ + F ̂ ⊤ e ̄ 0 * + e ̄ 0 * ⊤ e ̄ 0 * .

Plug in F ̂ , we have

1 T F ̂ ⊤ F ̂ = 1 T F λ ̄ + e ̄ ⊤ F λ ̄ + e ̄ = 1 T λ ̄ ⊤ F ⊤ F λ ̄ + λ ̄ ⊤ F ⊤ e ̄ + e ̄ ⊤ F λ ̄ + e ̄ ⊤ e ̄ .

By (A.11) in Lemma 2 of Pesaran (2006), 1 T F ⊤ e ̄ = O p 1 N T . And by (S.27) in Karabiyik and Westerlund (2021), 1 T e ̄ ⊤ e ̄ = O p N − 1 . Thus, 1 T F ̂ ⊤ F ̂ = 1 T λ ̄ ⊤ F ⊤ F λ ̄ + O p 1 N T + O p N − 1 .

Also,

1 T F ̂ ⊤ e ̄ 0 * = 1 T F ̂ ⊤ e ̄ * Λ ̄ r * , N e ̄ * Λ ̄ − r * ,

1 T F ̂ ⊤ e ̄ * = 1 T ∑ t = 1 T − 1 F ̂ t e ̄ t * ⊤ , consider the l th row of 1 T F ̂ ⊤ e ̄ * ,

var * 1 T ∑ t = 1 T − 1 F ̂ l t e ̄ t * = var 1 T ∑ t = 1 T − 1 F ̂ l t 1 N ∑ i = 1 N e ̂ i t · η i t

= var 1 T ∑ t = 1 T − 1 F ̂ l t 1 N ∑ i = 1 N e ̂ i t .

Using (S.86) in Karabiyik and Westerlund (2021), E 1 N 2 ∑ i = 1 N e ̂ i t e ̂ i t ′ ⊤ = O p N − 1 . Therefore,

var * 1 T ∑ t = 1 T − 1 F ̂ l t e ̄ t * = O p * 1 N ∑ t = 1 T − 1 ∑ t ‘ = 1 T − 1 E F ̂ l t F ̂ l t ′ T 2 = O p * 1 N T .

Hence, 1 T F ̂ ⊤ e ̄ * = O p * 1 N T . 1 T F ̂ ⊤ e ̄ * Λ ̄ r * ≤ 1 T F ̂ ⊤ e ̄ * Λ ̄ r * = O p * 1 N T , 1 T F ̂ ⊤ N e ̄ * Λ ̄ − r * ≤ N 1 T F ̂ ⊤ e ̄ * Λ ̄ − r * = O p * 1 T .

In addition,

1 T e ̄ 0 * ⊤ e ̄ 0 * = 1 T e ̄ r 0 * ⊤ e ̄ r 0 * e ̄ r 0 * ⊤ e ̄ − r 0 * e ̄ − r 0 * ⊤ e ̄ r 0 * e ̄ − r 0 * ⊤ e ̄ − r 0 * ,

where 1 T e ̄ r 0 * ⊤ e ̄ r 0 * = 1 T Λ ̄ r * ⊤ e ̄ * ⊤ e ̄ * Λ ̄ r * ≤ 1 T e ̄ * ⊤ e ̄ * Λ ̄ r * 2 = O p * 1 N , 1 T e ̄ r 0 * ⊤ e ̄ − r 0 * = 1 T Λ ̄ r * ⊤ e ̄ * ⊤ N e ̄ * Λ ̄ − r * ≤ N 1 T e ̄ * ⊤ e ̄ * Λ ̄ r * Λ ̄ − r * = O p * 1 N . Consider 1 T e ̄ − r 0 * ⊤ e ̄ − r 0 * ,

1 T e ̄ − r 0 * ⊤ e ̄ − r 0 * = 1 T N Λ ̄ − r * ⊤ e ̄ * ⊤ N e ̄ * Λ ̄ − r * = Λ ̄ − r * ⊤ N T − 1 e ̄ * ⊤ e ̄ * Λ ̄ − r * .

According to Condition 2, 1 T e ̄ − r 0 * ⊤ e ̄ − r 0 * = Λ − r * ⊤ Σ e * Λ − r * + O p * 1 T , where Λ − r * is from Λ * = Λ r * , Λ − r * = lim N → ∞ Λ ̄ * .

It remains to consider

E * ⁢ 1 T ⁢ e ̄ * ⊤ ⁢ e ̄ * = E * ⁢ 1 N 2 T ∑ i = 1 N ∑ j = 1 N ∑ t = 1 T − 1 e i t * ⁢ e j t * ⊤ ⁢ E * ⁢ 1 N 2 T ∑ i = 1 N ∑ j = 1 N ∑ t = 1 T − 1 e ̂ i t · η i t ⁢ e ̂ j t · η j t ⊤ = 1 N 2 T ∑ i = 1 N ∑ j = 1 N ∑ t = 1 T − 1 e ̂ i t ⁢ e ̂ j t ⊤ ⁢ I ⁡ i = j .

Using (S.94) in Karabiyik and Westerlund (2021), we have 1 N T ∑ i = 1 N ∑ j = 1 N ∑ t = 1 T − 1 e ̂ i t e ̂ j t ⊤ = O p 1 . Hence, 1 T e ̄ * ⊤ e ̄ * = O p * 1 N . By adding the results, 1 T F ̂ + e ̄ * Λ ̄ * D N ⊤ F ̂ + e ̄ * Λ ̄ * D N = 1 T λ ̄ ⊤ F ⊤ F λ ̄ + 0 r × r 0 r × m − r 0 m − r × r Λ − r * ⊤ Σ e * Λ − r * + O p * 1 N . In addition, 1 T F ̂ 0 * ⊤ g = 1 T F ̂ + e ̄ 0 * ⊤ g = 1 T F ̂ ⊤ g + 1 T e ̄ 0 * ⊤ g , where the first term is

1 T F ̂ ⊤ g = 1 T F λ ̄ + e ̄ ⊤ g = 1 T λ ̄ ⊤ F ⊤ g + 1 T e ̄ ⊤ g .

Assumptions 2 and 4 imply that 1 T e ̄ ⊤ g = 1 T ∑ t = 1 T − 1 e ̄ t g t ⊤ = 1 T 1 N ∑ i = 1 N ∑ t = 1 T − 1 e i t g t ⊤ = O p 1 N T . Also, 1 T e ̄ 0 * ⊤ g = 1 T e ̄ * Λ ̄ r * , N e ̄ * Λ ̄ − r * ⊤ g , 1 T e ̄ * T g = 1 T ∑ t = 1 T − 1 e ̄ t * g t ⊤ , consider the lth row of 1 T e ̄ * ⊤ g ,

var * 1 T ∑ t = 1 T − 1 g l t e ̄ t * ⊤ = var 1 T ∑ t = 1 T − 1 g l t 1 N ∑ i = 1 N e ̂ i t · η i t = var 1 T ∑ t = 1 T − 1 g l t 1 N ∑ i = 1 N e ̂ i t ,

by (S.86) in Karabiyik and Westerlund (2021), E 1 N 2 ∑ i = 1 N e ̂ i t e ̂ i t ′ ⊤ = O p N − 1 . Thus,

var 1 T ∑ t = 1 T − 1 g l t e ̄ t * ⊤ = O p * 1 N ∑ t = 1 T − 1 ∑ t ′ = 1 T − 1 E g l t g l t ′ T 2 = O p * 1 N T ,

Therefore, 1 T e ̄ * ⊤ g = O p * 1 N T . This implies that 1 T Λ ̄ r * ⊤ e ̄ * ⊤ g ≤ 1 T e ̄ * ⊤ g Λ ̄ r * = O p * 1 N T , and 1 T N Λ ̄ − r * ⊤ e ̄ * ⊤ g ≤ N 1 T e ̄ * ⊤ g Λ ̄ − r * = O p * 1 T .

Together with the above results, we have

1 T ⁢ z ̂ 0 * ⊤ ⁢ z ̂ 0 * = 1 T ⁢ λ ̄ ⊤ ⁢ F ⊤ ⁢ F λ ̄ λ ̄ ⊤ ⁢ F ⊤ ⁢ g g ⊤ ⁢ F λ ̄ g ⊤ ⁢ g + 1 T ⁢ 0 r × r 0 r × m − r 0 r × n 0 m − r × r e ̄ − r 0 * ⊤ ⁢ e ̄ − r 0 * 0 m − r × n 0 n × r 0 n × m − r 0 n × n + O p * ⁢ 1 N + O p * ⁢ 1 T = S z * + S e * + O p ⁢ 1 N + O p * ⁢ 1 T ,

where

S z * = 1 T λ ̄ ⊤ F ⊤ F λ ̄ λ ̄ ⊤ F ⊤ g g ⊤ F λ ̄ g ⊤ g ,

S e * = 1 T 0 r × r 0 r × m − r 0 r × n 0 m − r × r e ̄ − r 0 * ⊤ e ̄ − r 0 * 0 m − r × n 0 n × r 0 n × m − r 0 n × n .

Note that

Q N = Λ ̄ D N 0 m × n 0 n × m I n = λ ̄ r − 1 − λ ̄ r − 1 λ ̄ − r 0 r × n 0 m − r × r N I m − r 0 m − r × n 0 n × r 0 n × m − r I n ,

Let F 0 = ( F , 0 (T−h)×(m−r)), then

Q N ⊤ ⁢ S z * ⁢ Q N = Λ ̄ D N 0 m × n 0 n × m I n ⊤ ⁢ 1 T ⁢ λ ̄ ⊤ ⁢ F ⊤ ⁢ F λ ̄ λ ̄ ⊤ ⁢ F ⊤ ⁢ g g ⊤ ⁢ F λ ̄ g ⊤ ⁢ g ⁢ Λ ̄ ⁢ D N 0 m × n 0 n × m I n = 1 T ⁢ D N ⁢ Λ ̄ ⊤ ⁢ λ ̄ ⊤ ⁢ F ⊤ ⁢ F λ ̄ ⁢ Λ ̄ ⁢ D N D N ⁢ Λ ̄ ⊤ λ ̄ ⊤ ⁢ F ⊤ ⁢ g g ⊤ ⁢ F λ ̄ ⁢ Λ ̄ ⁢ D N g ⊤ ⁢ g = 1 T ⁢ F 0 ⊤ ⁢ F 0 F 0 ⊤ ⁢ g g ⊤ ⁢ F 0 g ⊤ ⁢ g = 1 T ⁢ F ⊤ ⁢ F 0 r × m − r F ⊤ ⁢ g 0 m − r × r 0 m − r × m − r 0 m − r × n g ⊤ ⁢ F 0 n × m − r g ⊤ ⁢ g .

Equation (S.81) in Karabiyik and Westerlund (2021) yields

Λ ̄ D N + λ ̂ i = Λ ̄ r λ ̄ r λ i + o p 1 ,

then we obtain

Λ ̄ D N + λ ̂ ̄ = Λ ̄ r λ ̄ r λ ̄ + o p 1 ,

which implies that Λ ̄ * Λ ̄ D N = Λ ̄ + o p * 1 . Hence, Λ ̄ * = diag I r , N − 1 / 2 I m − r + o p * 1 , then we can get that 1 T N e ̄ − r 0 * ⊤ e ̄ − r 0 * = 0 m − r × r , I m − r 1 T e ̄ * ⊤ e ̄ * 0 m − r × r , I m − r ⊤ + o p * 1 = Λ ̄ − r ⊤ 1 T e ̄ ⊤ e ̄ Λ ̄ − r + o p * 1 . It follows that

Q N ⊤ S e * Q N = 1 T 0 r × r 0 r × m − r 0 r × n 0 m − r × r N e ̄ − r 0 * ⊤ e ̄ − r 0 * 0 m − r × n 0 n × r 0 n × m − r 0 n × n = 1 T 0 r × r 0 r × m − r 0 r × n 0 m − r × r e ̄ − r 0 ⊤ e ̄ − r 0 0 m − r × n 0 n × r 0 n × m − r 0 n × n + o p * 1 .

We define S z 0 * as S z * + S e * . Next, we consider T − 1 z ̂ 0 * ⊤ z ̂ 0 * + − S z 0 * − 1 . Using Assumptions 1 and 4, we yield rank λ ̄ = r , rank 1 T F ⊤ F = r and rank 1 T g ⊤ g = n . Since 1 T e ̄ − r 0 * ⊤ e ̄ − r 0 * is random, we have rank S z 0 * = m + n . We also have rank T − 1 z ̂ 0 * ⊤ z ̂ 0 * = m + n , so rank T − 1 z ̂ 0 * ⊤ z ̂ 0 * = rank S z 0 * . According to Andrews (1987),

T − 1 z ̂ 0 * ⊤ z ̂ 0 * + − S z 0 * − 1 = O p * N − 1 / 2 + O p * T − 1 / 2 .

Define

Σ z 0 * = 1 T λ ⊤ F ⊤ F λ λ ⊤ F ⊤ g g ⊤ F λ g ⊤ g + 0 r × r 0 r × m − r 0 r × n 0 m − r × r Λ − r * ⊤ Σ e * Λ − r * 0 m − r × n 0 n × r 0 n × m − r 0 n × n .

And we also have rank Σ z 0 * = m + n . Therefore, rank S z 0 * = rank Σ z 0 * . According to Andrews (1987),

S z 0 * − 1 − Σ z 0 * − 1 = o p * 1 .

Finally,

T − 1 z ̂ 0 * ⊤ z ̂ 0 * + − Σ z 0 * − 1 ≤ T − 1 z ̂ 0 * ⊤ z ̂ 0 * + − S z 0 * − 1 + S z 0 * − 1 − Σ z 0 * − 1 = o p * 1 .

Consider T − 1 2 z ̂ 0 * ⊤ ε * ,

T − 1 2 z ̂ 0 * ⊤ ε * = T − 1 2 F ̂ + e ̄ 0 * g ⊤ ε * = T − 1 2 z 0 * ⊤ ε * + T − 1 2 e ̄ r 0 * ⊤ ε * 0 m + n − r × 1 ,

where z 0 * = F ̂ + 0 T − h × m − r , e ̄ − r 0 * , g , Condition 1 implies that T − 1 2 z 0 * ⊤ ε * → d * N 0 , Σ z ε * . In addition, T − 1 2 e ̄ * ⊤ ε * = 1 T ∑ t = 1 T − 1 ε t + 1 * e ̄ t * = 1 N 1 T ∑ t = 1 T − 1 ε t + 1 * ∑ i = 1 N e i t * . Since

E * T − 1 2 e ̄ * ⊤ ε * 2 = 1 N 2 1 T E * ∑ t = 1 T − 1 ∑ s = 1 T − 1 ε t + 1 * ε s + 1 * ∑ i = 1 N ∑ j = 1 N e i t * ⊤ e j s * = 1 N 1 T ∑ t = 1 T − 1 ε ̂ t + 1 2 1 N ∑ i = 1 N e ̂ i t ⊤ e ̂ i t = O p * 1 N ,

T − 1 2 e ̄ * ⊤ ε * = O p * 1 N . Thus T − 1 2 e ̄ r 0 * ⊤ ε * = T − 1 2 Λ ̄ r * ⊤ e ̄ * ⊤ ε * ≤ Λ ̄ r * T − 1 2 e ̄ * ⊤ ε * = O p * 1 N . Then we have that T − 1 2 z ̂ 0 * ⊤ ε * = T − 1 2 z 0 * ⊤ ε * + O p * 1 N → d * N 0 , Σ z ε * . Moreover, T − 1 2 z ̂ 0 * ⊤ e ̄ r 0 * = T − 1 2 z ̂ 0 * − z 0 * ⊤ e ̄ r 0 * + T − 1 2 z 0 * ⊤ e ̄ r 0 * . We have shown that 1 T F ̂ ⊤ e ̄ r 0 * = O p * 1 N T and 1 T e ̄ r 0 * ⊤ g = O p * 1 N T , then T − 1 2 z ̂ ⊤ e ̄ r 0 * = O p * 1 N T , and

T − 1 2 z 0 * ⊤ e ̄ r 0 * = T − 1 2 z ̂ ⊤ e ̄ r 0 * + 0 r × r T − 1 2 e ̄ − r 0 * ⊤ e ̄ r 0 * 0 n × r = 0 r × r T − 1 2 e ̄ − r 0 * ⊤ e ̄ r 0 * 0 n × r + O p * 1 N T .

It remains to consider T − 1 / 2 z ̂ 0 * − z 0 * ⊤ e ̄ r 0 * , T − 1 / 2 z ̂ 0 * − z 0 * ⊤ e ̄ r 0 * = T − 1 / 2 e ̄ r 0 * ⊤ e ̄ r 0 * 0 m + n − r × r .

Note that

T − 1 / 2 ⁢ e ̄ r 0 * ⊤ e ̄ r 0 * = T − 1 2 ⁢ e ̄ * Λ ̄ r * ⊤ ⁢ e ̄ * ⁢ Λ ̄ r * = T − 1 2 ⁢ Λ ̄ r * ⊤ ⁢ e ̄ * ⊤ ⁢ e ̄ * ⁢ Λ ̄ r * = T N − 1 ⁢ Λ ̄ r * ⊤ ⁢ N T − 1 ⁢ e ̄ * ⊤ ⁢ e ̄ * ⁢ Λ ̄ r * = T N − 1 ⁢ Λ ̄ r * ⊤ ⁢ 1 N T ∑ i = 1 N ∑ j = 1 N ∑ t = 1 T − 1 e i t * ⁢ e j t * ⊤ ⁢ Λ ̄ r * .

Condition 2 yields

T − 1 / 2 e ̄ r 0 * ⊤ e ̄ r 0 * = T N − 1 Λ ̄ r * ⊤ 1 N T ∑ i = 1 N ∑ j = 1 N ∑ t = 1 T − 1 e i t * e j t * ⊤ Λ ̄ r * → p * c Λ r * ⊤ Σ e * Λ r * .

By adding the results, we obtain

Q N * − 1 ⁢ T δ ̂ * − Q N * ⁢ δ 0 * = T − 1 z ̂ 0 * ⊤ z ̂ 0 * + ⁢ T − 1 / 2 ⁢ z ̂ 0 * ⊤ ⁢ ε * − e ̄ r 0 * ⁢ α ̂ = T − 1 z ̂ 0 * ⊤ z ̂ 0 * + ⁢ T − 1 / 2 ⁢ z ̂ 0 * ⊤ ⁢ ε * − T − 1 z ̂ 0 * ⊤ z ̂ 0 * + ⁢ T − 1 / 2 ⁢ z ̂ 0 * ⊤ ⁢ e ̄ r 0 * ⁢ α ̂ = Σ z 0 * − 1 ⁢ T − 1 / 2 ⁢ z 0 * ⊤ ⁢ ε * − T 0 r × 1 T − 1 ⁢ e ̄ − r 0 * ⊤ ⁢ e ̄ r 0 * ⁢ α ̂ 0 n × 1 − T − 1 2 ⁢ e ̄ r 0 * ⊤ ⁢ e ̄ r 0 * ⁢ α ̂ 0 n × 1 + o p * ⁢ 1 .

Finally, using Condition 1, Q N * − 1 T δ ̂ * − Q N * δ 0 * → d * N − c B 1 * − B 2 * , Σ z 0 * − 1 Σ z ε * Σ z 0 * − 1 , where

B 1 * = Σ z 0 * − 1 Λ r * ⊤ Σ e * Λ r * α ̂ 0 n × 1 , B 2 * = T N − 1 2 0 r × 1 Λ − r * ⊤ Σ e * Λ − r * − 1 Λ − r * ⊤ Σ e * Λ r * α ̂ 0 n × 1 , Σ z ε * = lim N , T → ∞ T − 1 ∑ t = 1 T − 1 E * ε t + 1 2 z t 0 * z t 0 * ⊤ .

Proof of Corollary 1.

We have shown Σ z 0 * = Q 0 − 1 ⊤ Σ z 0 Q 0 − 1 in Lemma 1, where Q 0 = lim N → ∞ Q N . Also,

Σ z ε * = lim N , T → ∞ ⁢ var * ⁢ 1 T ∑ t = 1 T − 1 z t 0 * ⁢ ε t + 1 * = lim N , T → ∞ ⁢ var * ⁢ 1 T ∑ t = 1 T − 1 Q N − 1 ⁢ z ̂ t 0 ⁢ ε t + 1 * = Q 0 − 1 ⊤ ⁢ lim N , T → ∞ var 1 T ∑ t = 1 T − 1 z t 0 ⁢ ε t + 1 ⁢ Q 0 − 1 = Q 0 − 1 ⊤ ⁢ Σ z 0 ε ⁢ Q 0 − 1 .

Thus,

Σ z 0 * − 1 Σ z ε * Σ z 0 * − 1 = Q 0 − 1 ⊤ Σ z 0 Q 0 − 1 − 1 Q 0 − 1 ⊤ Σ z 0 ε Q 0 − 1 Q 0 − 1 ⊤ Σ z 0 Q 0 − 1 − 1 = Q 0 Σ z 0 − 1 Q 0 ⊤ Q 0 − 1 ⊤ Σ z 0 ε Q 0 − 1 Q 0 Σ z 0 − 1 Q 0 ⊤ = Q 0 Σ z 0 − 1 Σ z 0 ε Σ z 0 − 1 Q 0 ⊤ .

Proof of Theorem 2.

Using the fact that

z ̂ T 0 * = F ̂ T 0 * g T = D N Λ ̄ * ⊤ F ̂ T * g T = D N Λ ̄ * ⊤ λ ̂ ̄ ⊤ F ̂ T + e ̄ T * g T = F ̂ T g T + D N Λ ̄ * ⊤ e ̄ T * 0 n × 1 = z T 0 * + e ̄ r , T 0 * 0 m + n − r × 1 .

It follows that

y ̂ T + 1 T * − y T + 1 T * = T − 1 / 2 z ̂ T 0 * ⊤ Q N * − 1 T δ ̂ * − Q N * δ 0 * + N − 1 / 2 α ̂ ⊤ N e ̄ r , T 0 * = T − 1 / 2 z ̂ T 0 * ⊤ T Q N * − 1 δ ̂ * − δ 0 * + N − 1 / 2 α ̂ ⊤ N e ̄ r , T 0 * = T − 1 / 2 ⁢ z T 0 * + e ̄ r , T 0 * 0 m + n − r × 1 ⊤ ⁢ T Q N * − 1 δ ̂ * − δ 0 * + N − 1 / 2 ⁢ α ̂ ⊤ ⁢ N e ̄ r , T 0 * = T − 1 / 2 ⁢ z T 0 * ⊤ ⁢ T Q N * − 1 δ ̂ * − δ 0 * + N − 1 / 2 ⁢ α ̂ ⊤ ⁢ N e ̄ r , T 0 * + r T * ,

where

r T * = T − 1 2 e ̄ r , T 0 * 0 m + n − r × 1 ⊤ T Q N * − 1 δ ̂ * − δ 0 * = T − 1 2 Λ ̄ r * ⊤ e ̄ t * 0 m + n − r × 1 ⊤ T Q N * − 1 δ ̂ * − δ 0 * = O P * 1 N T .

The last equality holds for that T Q N * − 1 δ ̂ * − δ 0 * = O P * 1 , and T − 1 2 e ̄ t * = O P * 1 N T , which have been verified in the proof of Theorem 1.

Finally, we obtain

y ̂ T + 1 T * − y T + 1 T * T − 1 z ̂ T ⊤ Σ z 0 − 1 Σ z 0 ε Σ z 0 − 1 z ̂ T + N − 1 α ̂ ⊤ Σ e α ̂ → d * N 0 , 1 .

Proof of Corollary 2.

First, by Corollary 1 and z ̂ t * = Q N * ⊤ − 1 z ̂ t 0 * ,

φ ̂ * = z ̂ T * ⊤ Σ ̂ z * + Σ ̂ z ε * Σ ̂ z * + z ̂ T * = z ̂ T * ⊤ 1 T ∑ t = 1 T − 1 z ̂ t * z ̂ t * ⊤ + 1 T ∑ t = 1 T − 1 ε ̂ t + 1 * 2 z ̂ t * z ̂ t * ⊤ 1 T ∑ t = 1 T − 1 z ̂ t * z ̂ t * ⊤ + z ̂ T * = z ̂ T 0 * ⊤ 1 T ∑ t = 1 T − 1 z ̂ t 0 * z ̂ t 0 * ⊤ + 1 T ∑ t = 1 T − 1 ε ̂ t + 1 * 2 z ̂ t 0 * z ̂ t 0 * ⊤ 1 T ∑ t = 1 T − 1 z ̂ t 0 * z ̂ t 0 * ⊤ + z ̂ T 0 * = z T 0 * ⊤ 1 T ∑ t = 1 T − 1 z t 0 * z t 0 * ⊤ + 1 T ∑ t = 1 T − 1 ε ̂ t + 1 * 2 z t 0 * z t 0 * ⊤ 1 T ∑ t = 1 T − 1 z t 0 * z t 0 * ⊤ + z T 0 * = Q N − 1 ⊤ z ̂ T 0 ⊤ Σ z 0 * − 1 Σ z ε * Σ z 0 * − 1 Q N − 1 ⊤ z ̂ T 0 = z ̂ T 0 ⊤ Q N − 1 Σ z 0 * − 1 Σ z ε * Σ z 0 * − 1 Q N − 1 ⊤ z ̂ T 0 → p * z T 0 ⊤ Q 0 − 1 Q 0 Σ z 0 − 1 Σ z 0 ε Σ z 0 − 1 Q 0 − 1 ⊤ Q 0 ⊤ z T 0 = z T 0 ⊤ Σ z 0 − 1 Σ z 0 ε Σ z 0 − 1 z T 0 ⊤ = φ .

And Condition 2 implies N e ̄ r , T 0 * → d * N 0 m × 1 , Λ r * ⊤ Σ e * Λ r * . Then we have α ̂ * = Σ e * − 1 2 ⊤ M Σ e * 1 / 2 ⊤ Λ − r * Σ e * 1 / 2 ⊤ Λ r * α ̂ , Condition 2 and Λ r * = I r , 0 r × m − r ⊤ ensure that α ̂ * = α ̂ + o P * 1 . Hence, we have that α ̂ * ⊤ Σ ̂ e * α ̂ * → d * Φ + Δ ⊤ Σ e Φ + Δ .

Proof of Theorem 3.

Define the following empirical distribution function,

F T , ε ̂ − ε ̂ ̄ x = 1 T − h ∑ t = 1 T − h I ε ̂ t + h − ε ̂ ̄ ≤ x ,

F T , ε x = 1 T − h ∑ t = 1 T − h I ε t + h ≤ x ,

where ε ̂ ̄ = 1 T − h ∑ t = 1 T − h ε ̂ t + h . Note that F T , ε * x = F T , ε ̂ − ε ̂ ̄ x . It follows that

d 2 F T , ε ̂ − ε ̂ ̄ , F ε ≤ d 2 F T , ε ̂ − ε ̂ ̄ , F T , ε + d 2 F T , ε , F ε .

Using Lemma 8.4 in Bickel and Freedman (1981), d 2(F T,ε , F ε ) converges to 0 almost surely. It suffices to prove that d 2 F T , ε ̂ − ε ̂ ̄ , F T , ε = o p 1 . Let I represent the uniform distribution on {1, ⋯, T − h}, and define X I = ε ̂ I + h − ε ̂ ̄ and Y I  = ε I+h . We have that

d 2 F T , ε ̂ − ε ̂ ̄ , F T , ε 2 ≤ E X I − Y I 2 = E I ε ̂ I + h − ε ̂ ̄ − ε I + h 2 = 1 T − h ∑ t = 1 T − h ε ̂ t + h − ε ̂ ̄ − ε t + h 2 = 1 T − h ∑ t = 1 T − h ε ̂ t + h − ε t + h 2 − 2 T − h ∑ t = 1 T − h ε ̂ t + h − ε t + h ε ̂ ̄ + ε ̂ ̄ 2 .

We consider the first term of the above equality,

ε ̂ t + h − ε t + h = − T − 1 / 2 T δ ̂ − Q N δ 0 ⊤ Q N − 1 ⊤ z ̂ t 0 − N − 1 / 2 α ⊤ N e ̄ r , t 0 ,

This implies that

1 T − h ∑ t = 1 T − h ε ̂ t + h − ε ̂ ̄ − ε t + h 2 ≤ 2 T − h ∑ t = 1 T − h Q N − 1 δ ̂ − Q N δ 0 2 z ̂ t 0 2 + 2 T − h ∑ t = 1 T − h α 2 e ̄ r , t 0 2 = O p 1 T + O p 1 N = o p 1 .

Similarly,

ε ̂ ̄ = 1 T − h ∑ t = 1 T − h ε ̂ t + h = 1 T − h ∑ t = 1 T − h ε ̂ t + h − ε t + h + 1 T − h ∑ t = 1 T − h ε t + h = O p 1 T + O p 1 N + o p 1 = o p 1 .

This means that both 2 T − h ∑ t = 1 T − h ε ̂ t + h − ε t + h ε ̂ ̄ and ε ̂ ̄ 2 are o p (1). Then we obtain

d 2 F T , ε ̂ − ε ̂ ̄ , F T , ε = o p 1 .

Proof of Condition 1.

  1. Under the bootstrap measure by construction, we have E * ε t + 1 * = 0 ; 1 T ∑ t = 1 T − 1 E * ε t + 1 * 2

    = 1 T ∑ t = 1 T − 1 E * ε ̂ t + 1 2 v t + 1 2 = 1 T ∑ t = 1 T − 1 ε ̂ t + 1 2 ≤ 1 T ∑ t = 1 T − 1 ε ̂ t + 1 4 1 2 < ∞ .

  2. Let m t * = Σ z ε * − 1 / 2 z t 0 * ε t + 1 * , E * m t * = Σ z ε * − 1 / 2 z t 0 * E * ε t + 1 * = 0 ;

    var * 1 T ∑ t = 1 T − 1 m t * = var * 1 T ∑ t = 1 T − 1 Σ z ε * − 1 2 z t 0 * ε t + 1 * = Σ z ε * − 1 / 2 var * 1 T ∑ t = 1 T − 1 z t 0 * ε t + 1 * Σ z ε * − 1 / 2 = I m + n .

Proof of Condition 2.

E * 1 N T ∑ i = 1 N ∑ j = 1 N ∑ t = 1 T e i t * e j t * ⊤ − E * e i t * e j t * ⊤  

= 1 N T ∑ i = 1 N ∑ j = 1 N ∑ t = 1 T E * e i t * e j t * ⊤ − 1 N T ∑ i = 1 N ∑ j = 1 N ∑ t = 1 T E * e i t * e j t * ⊤ = 0 .

Also,

var * 1 N T ∑ i = 1 N ∑ j = 1 N ∑ t = 1 T e i t * e j t * ⊤ − E * e i t * e j t * ⊤ = ∑ t = 1 T ∑ s = 1 T 1 N 2 ∑ i , j , k , l Cov * e i t * e j t * ⊤ , e l s * e k s * ⊤ ≤ κ · 1 T 1 N 1 N T ∑ i = 1 N ∑ t = 1 T e ̂ i t 4 = O p 1 T N = o p 1 .

Hence, N T − 1 ∑ t = 1 T e ̄ t * e ̄ t * ⊤ → p * Σ e * , that is, N T − 1 ∑ t = 1 T e ̄ t * e ̄ t * ⊤ − Σ e * = o p * 1 .

Appendix B. Monte Carlo Simulations

See Tables B.1–B.4

Table B.1:

DGP5, m = 2 > r = 1, e it is homoskedastic and ε t+1 is normal.

N = 50 N = 100 N = 200
T = 50 T = 100 T = 200 T = 400 T = 50 T = 100 T = 200 T = 400 T = 50 T = 100 T = 200 T = 400
y T+1|T Coverage rate
F 0.93 0.95 0.94 0.95 0.95 0.95 0.92 0.96 0.93 0.95 0.95 0.94
PC 0.77 0.65 0.58 0.46 0.84 0.78 0.68 0.57 0.89 0.81 0.76 0.63
CA 0.96 0.94 0.94 0.95 0.98 0.96 0.97 0.95 0.97 0.97 0.97 0.96
PCBS 0.90 0.89 0.90 0.91 0.94 0.92 0.95 0.93 0.94 0.93 0.94 0.92
CABS 0.93 0.93 0.94 0.91 0.93 0.93 0.94 0.94 0.95 0.93 0.93 0.93
Length
F 0.75 0.53 0.38 0.26 0.77 0.53 0.37 0.27 0.76 0.52 0.39 0.26
PC 0.79 0.56 0.41 0.30 0.78 0.54 0.38 0.28 0.77 0.52 0.39 0.26
CA 1.44 1.21 1.05 0.99 1.28 1.03 0.87 0.78 1.19 0.92 0.74 0.62
PCBS 1.20 0.99 0.90 0.86 1.05 0.85 0.73 0.68 0.95 0.71 0.61 0.53
CABS 1.49 1.23 1.06 1.00 1.31 1.03 0.85 0.77 1.23 0.92 0.72 0.61
y T+1 Coverage rate
F 0.95 0.96 0.97 0.95 0.95 0.96 0.94 0.94 0.97 0.95 0.93 0.94
PC 0.96 0.96 0.97 0.94 0.95 0.95 0.94 0.96 0.97 0.96 0.93 0.93
CA 0.96 0.96 0.97 0.95 0.96 0.96 0.95 0.96 0.98 0.96 0.93 0.94
PCBS 0.95 0.95 0.96 0.95 0.96 0.93 0.94 0.95 0.95 0.95 0.93 0.94
CABS 0.95 0.95 0.96 0.95 0.96 0.95 0.94 0.95 0.96 0.95 0.92 0.93
Length
F 3.90 3.89 3.93 3.92 3.91 3.91 3.91 3.92 3.93 3.91 3.92 3.91
PC 4.02 4.01 4.04 4.03 3.97 3.97 3.97 3.98 3.97 3.94 3.95 3.94
CA 4.16 4.14 4.15 4.14 4.06 4.05 4.04 4.05 4.04 3.99 3.99 3.98
PCBS 4.14 4.03 4.03 3.98 4.13 4.02 3.95 3.95 4.08 3.98 3.93 3.90
CABS 4.16 4.09 4.05 4.02 4.14 4.05 3.98 3.94 4.14 4.01 3.95 3.94

  1. Notes : F, PC, CA, PCBS and CABS refer to the results based on the true factors, PC based on knowing the true number of factors, CA, PC + Bootstrap, and CA + Bootstrap, respectively.

Table B.2:

DGP6, m = 2 > r = 1, e it is homoskedastic and ε t+1 is mixture.

N = 50 N = 100 N = 200
T = 50 T = 100 T = 200 T = 400 T = 50 T = 100 T = 200 T = 400 T = 50 T = 100 T = 200 T = 400
y T+1|T Coverage rate
F 0.91 0.94 0.95 0.94 0.94 0.93 0.96 0.93 0.91 0.92 0.95 0.91
PC 0.73 0.68 0.55 0.43 0.85 0.77 0.70 0.53 0.87 0.80 0.77 0.62
CA 0.93 0.94 0.94 0.94 0.96 0.97 0.97 0.95 0.97 0.97 0.97 0.96
PCBS 0.87 0.90 0.88 0.91 0.93 0.93 0.93 0.93 0.91 0.92 0.95 0.93
CABS 0.90 0.91 0.93 0.92 0.95 0.93 0.95 0.93 0.95 0.92 0.95 0.95
Length
F 0.76 0.54 0.37 0.27 0.75 0.54 0.37 0.27 0.74 0.52 0.38 0.27
PC 0.80 0.57 0.40 0.30 0.77 0.55 0.39 0.28 0.75 0.52 0.38 0.27
CA 1.45 1.20 1.05 0.98 1.28 1.02 0.87 0.78 1.19 0.88 0.73 0.62
PCBS 1.25 1.03 0.91 0.86 1.11 0.86 0.76 0.67 0.99 0.73 0.62 0.54
CABS 1.51 1.22 1.06 1.00 1.36 1.03 0.86 0.77 1.28 0.89 0.72 0.61
y T+1 Coverage rate
F 0.90 0.92 0.92 0.87 0.90 0.93 0.91 0.88 0.93 0.92 0.89 0.88
PC 0.90 0.93 0.92 0.87 0.90 0.93 0.91 0.89 0.93 0.92 0.89 0.88
CA 0.91 0.93 0.93 0.87 0.91 0.93 0.91 0.89 0.93 0.92 0.90 0.89
PCBS 0.95 0.96 0.95 0.94 0.96 0.96 0.94 0.96 0.96 0.97 0.97 0.93
CABS 0.95 0.96 0.96 0.93 0.96 0.98 0.93 0.96 0.97 0.97 0.97 0.95
Length
F 3.86 3.88 3.90 3.93 3.88 3.91 3.92 3.90 3.83 3.86 3.91 3.92
PC 3.99 4.01 4.01 4.04 3.94 3.97 3.97 3.96 3.87 3.88 3.94 3.95
CA 4.13 4.13 4.12 4.15 4.03 4.04 4.04 4.02 3.94 3.93 3.98 3.99
PCBS 4.41 4.39 4.27 4.25 4.53 4.28 4.16 4.11 4.32 4.15 4.16 4.10
CABS 4.52 4.36 4.31 4.28 4.50 4.34 4.19 4.16 4.40 4.21 4.16 4.08
Table B.3:

DGP7, m = 2 > r = 1, e it is heteroskedastic and ε t+1 is normal.

N = 50 N = 100 N = 200
T = 50 T = 100 T = 200 T = 400 T = 50 T = 100 T = 200 T = 400 T = 50 T = 100 T = 200 T = 400
y T+1|T Coverage rate
F 0.92 0.97 0.95 0.95 0.95 0.94 0.95 0.94 0.96 0.94 0.93 0.96
PC 0.78 0.65 0.56 0.47 0.84 0.77 0.67 0.58 0.90 0.86 0.76 0.64
CA 0.96 0.92 0.95 0.95 0.97 0.95 0.97 0.95 0.98 0.97 0.96 0.97
PCBS 0.91 0.90 0.91 0.91 0.93 0.91 0.94 0.91 0.97 0.94 0.93 0.94
CABS 0.96 0.91 0.94 0.94 0.93 0.93 0.95 0.93 0.96 0.93 0.94 0.93
Length
F 0.75 0.53 0.38 0.26 0.77 0.54 0.37 0.27 0.78 0.52 0.39 0.27
PC 0.79 0.56 0.41 0.29 0.78 0.55 0.39 0.28 0.79 0.53 0.40 0.27
CA 1.39 1.14 1.02 0.94 1.27 1.00 0.84 0.75 1.23 0.89 0.72 0.60
PCBS 1.20 1.00 0.93 0.88 1.06 0.86 0.75 0.69 0.97 0.72 0.61 0.53
CABS 1.49 1.18 1.06 0.99 1.33 1.03 0.85 0.78 1.27 0.89 0.72 0.60
y T+1 Coverage rate
F 0.96 0.95 0.95 0.95 0.95 0.95 0.94 0.96 0.96 0.95 0.94 0.96
PC 0.96 0.95 0.95 0.95 0.94 0.95 0.94 0.95 0.96 0.95 0.95 0.96
CA 0.97 0.96 0.96 0.96 0.96 0.95 0.94 0.95 0.96 0.96 0.95 0.97
PCBS 0.95 0.93 0.96 0.95 0.94 0.93 0.92 0.95 0.94 0.95 0.94 0.96
CABS 0.95 0.93 0.95 0.95 0.94 0.94 0.94 0.95 0.95 0.94 0.94 0.96
Length
F 3.93 3.92 3.92 3.91 3.90 3.93 3.91 3.92 3.91 3.90 3.92 3.90
PC 4.05 4.04 4.02 4.02 3.96 4.00 3.97 3.98 3.94 3.93 3.95 3.93
CA 4.17 4.15 4.13 4.12 4.05 4.07 4.03 4.04 4.02 3.98 3.99 3.97
PCBS 4.17 4.08 4.01 3.98 4.09 4.04 3.92 3.96 4.05 3.94 3.93 3.89
CABS 4.23 4.09 4.03 4.00 4.17 4.03 3.97 3.97 4.17 3.97 3.93 3.92
Table B.4:

DGP8, m = 2 > r = 1, e it is heteroskedastic and ε t+1 is mixture.

N = 50 N = 100 N = 200
T = 50 T = 100 T = 200 T = 400 T = 50 T = 100 T = 200 T = 400 T = 50 T = 100 T = 200 T = 400
y T+1|T Coverage rate
F 0.91 0.95 0.93 0.95 0.91 0.94 0.95 0.95 0.92 0.94 0.95 0.96
PC 0.75 0.65 0.58 0.46 0.79 0.76 0.60 0.52 0.86 0.82 0.75 0.67
CA 0.94 0.93 0.92 0.95 0.97 0.97 0.95 0.93 0.97 0.97 0.96 0.97
PCBS 0.91 0.89 0.92 0.92 0.94 0.92 0.92 0.92 0.94 0.92 0.95 0.94
CABS 0.94 0.93 0.91 0.94 0.94 0.94 0.93 0.91 0.92 0.92 0.93 0.94
Length
F 0.74 0.52 0.37 0.27 0.77 0.52 0.37 0.26 0.75 0.53 0.38 0.27
PC 0.77 0.55 0.40 0.30 0.79 0.54 0.38 0.27 0.76 0.54 0.39 0.27
CA 1.36 1.14 1.01 0.94 1.31 1.00 0.84 0.75 1.19 0.90 0.72 0.60
PCBS 1.25 1.03 0.93 0.89 1.14 0.85 0.74 0.69 1.00 0.75 0.62 0.54
CABS 1.45 1.19 1.04 0.98 1.40 1.03 0.84 0.76 1.25 0.93 0.72 0.61
y T+1 Coverage rate
F 0.93 0.92 0.93 0.91 0.91 0.90 0.89 0.91 0.91 0.92 0.91 0.91
PC 0.93 0.93 0.93 0.91 0.91 0.90 0.89 0.91 0.91 0.92 0.91 0.91
CA 0.93 0.93 0.93 0.91 0.92 0.90 0.89 0.91 0.92 0.93 0.91 0.91
PCBS 0.97 0.97 0.95 0.96 0.98 0.96 0.94 0.95 0.97 0.97 0.97 0.96
CABS 0.97 0.96 0.97 0.95 0.96 0.95 0.94 0.95 0.97 0.98 0.96 0.96
Length
F 3.80 3.88 3.90 3.92 3.87 3.86 3.89 3.90 3.83 3.89 3.90 3.91
PC 3.92 4.00 4.02 4.03 3.94 3.91 3.95 3.96 3.87 3.92 3.93 3.94
CA 4.04 4.10 4.11 4.13 4.04 3.98 4.02 4.02 3.94 3.97 3.97 3.98
PCBS 4.50 4.34 4.24 4.26 4.44 4.19 4.16 4.14 4.41 4.20 4.14 4.09
CABS 4.51 4.41 4.27 4.24 4.53 4.34 4.22 4.14 4.47 4.26 4.16 4.09

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Supplementary Material

This article contains supplementary material (https://doi.org/10.1515/snde-2024-0102).

Received: 2024-09-18
Accepted: 2025-09-25
Published Online: 2025-10-10

© 2025 Walter de Gruyter GmbH, Berlin/Boston

https://doi.org/10.1515/snde-2024-0102
Schlagwörter für diesen Artikel
bootstrap; cross-section average; factor-augmented regression models; forecast
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