Two canonical VARMA forms: SCM vis-à-vis Echelon form

Transcription

1 Two canonical VARMA forms: SCM vis-à-vis Echelon form George Athanasopoulos D.S. Poskitt Farshid Vahid

2 Introduction Outline of presentation Motivation for VARMA models SCM methodology Tiao and Tsay (1989) Athanasopoulos and Vahid (006a) Echelon form Hannan and Kavalieris (1984) Poskitt (199) Lütkepohl and Poskitt (1996) Comparison of the two forms: Theoretical - Theorem Experimental - Simulations Practical - Forecasting real macroeconomic variables Summary of findings and future research

3 Introduction Why VARMA? Closer approximations to Wold representation Parsimonious representations any invertible VARMA can be represented by an infinite order VAR Athanasopoulos and Vahid (006b) - VARMA outperformed VAR when forecasting macroeconomic data Difficult to Identify If univariate ARIMA modelling is difficult then VARMA modelling is even more difficult - some might say impossible! - Chatfield

4 Introduction Identification Problem Consider a bivariate VARMA(1,1)» y1,t y,t =»»»»» φ11 φ 1 y1,t 1 ε1,t θ11 θ + 1 ε1,t 1 φ 1 φ y,t 1 ε,t θ 1 θ ε,t 1 φ 1 = φ = θ 1 = θ = 0 nd row: y,t = ε,t y,t 1 = ε,t 1 φ 1 and θ 1 are not separately identifiable Set either coefficient to zero to achieve identification 1 Redundant parameters Exchangeable models

5 Canonical SCM Definition of a SCM: For a given K-dimensional VARMA(p, q) y t = Φ 1 y t Φ p y t p + η t Θ 1 η t 1... Θ q η t q (1) z r,t = α r y t SCM(p r, q r ) if α r satisfies α r Φ pr 0 T where 0 p r p α r Φ l = 0 T for l = p r + 1,..., p α r Θ qr 0 T where 0 q r q α r Θ l = 0 T for l = q r + 1,..., q SCM Methodology: Find K-linearly independent vectors A = (α 1,..., α K ) which transform (1) into Ay t = Φ 1y t Φ py t p + ε t Θ 1ε t 1... Θ qε t q () where Φ i = AΦ i, ε t = Aη t and Θ i = AΘ i A 1 Series of C/C tests: E [ y1,t 1 y,t 1 ] [ y1,t y,t ] [α1 ] = 0 α 1 y t SCM(0, 0)

6 Canonical SCM Example: K = α 1 y t SCM(1, 1) α y t SCM(1, 0) α y t SCM(0, 0) 4 α 11 α 1 α 1 α 1 α α α 1 α α 5 6 y t = 4 φ (1) 11 φ (1) 1 φ (1) 1 φ (1) 1 φ (1) φ (1) 7 5 y t 1 +ε t 4 θ(1) 11 θ (1) 1 0 Normalise diagonally (test for improper normalisations) Reduce parameters of A Canonical SCM α y t = 4 α 1 α 1 φ (1) 11 φ (1) 1 φ (1) 1 φ (1) 1 φ (1) φ (1) 5 ε t y t 1 + ε t 4 θ(1) 11 θ (1) ε t 1

7 Canonical reverse Echelon form Echelon Form Definition: A K dimensional VARMA representation (Φ 0 Φ 1 L... Φ p L p )y t = (Θ 0 Θ 1 L... Θ q L q )ε t is in Echelon form if [Φ(L) : Θ(L)] is left coprime and 1 Φ 0 = Θ 0 is lower triangular with unit diagonal elements, row r of the polynomial operators [Φ(L) : Θ(L)] is of maximum degree k r, the operators have the form: Xk r φ rr (L) = 1 φ (j) rr L j for r = 1,..., K, θ rc(l) = θ (0) rc j=1 Xk r φ rc(l) = Xk r j=1 j=k r k rc +1 φ (j) rc L j for r c, θ (j) rc L j with θ (0) rc = φ (0) rc for r, c = 1,..., K. where φ (j) rc specifies the element of Φ j in row r and column c.

8 Canonical reverse Echelon form Echelon Form Definition: A K dimensional VARMA representation (Φ 0 Φ 1 L... Φ p L p )y t = (Θ 0 Θ 1 L... Θ q L q )ε t is in Echelon form if [Φ(L) : Θ(L)] is left coprime and 1 Φ 0 = Θ 0 is lower triangular with unit diagonal elements, row r of the polynomial operators [Φ(L) : Θ(L)] is of maximum degree k r, the operators have the form: Xk r φ rr (L) = 1 φ (j) rr L j for r = 1,..., K, θ rc(l) = θ (0) rc j=1 Xk r φ rc(l) = Xk r j=1 j=k r k rc +1 φ (j) rc L j for r c, θ (j) rc L j with θ (0) rc = φ (0) rc for r, c = 1,..., K. where φ (j) rc specifies the element of Φ j in row r and column c.

9 Canonical reverse Echelon form Kronecker Indices (k r ): the maximum row degrees k = (k 1,..., k K ) which define the structure of the system, and k rc = { min(kr + 1, k c ) for r c min(k r, k c ) for r < c, () for r, c = 1,..., K, specifies the number of free parameters in the operator φ rc (L) for r c. Search through Kronecker indices for optimal model Poskitt(199) search procedure and BIC Canonical Reverse Echelon form representations Example: k = (k 1, k, k ) = (1, 1, 0) φ (0) 1 φ (0) y t = 4 φ (1) 11 φ (1) 1 φ (1) 1 φ (1) 1 φ (1) φ (1) y t 1 +Θ 0 ε t 4 θ (1) 11 θ (1) 1 0 θ (1) 1 θ (1) ε t 1.

10 Theorem: Consider a VARMA process y t, represented in canonical reverse Echelon form with Kronecker indices k = (k 1,..., k K ). Now suppose that y t is also represented in a canonical SCM representation that consists of K SCMs of orders s r = (p r, q r ) for r = 1,..., K. k = (k 1,..., k K ) = s max = (s max 1,..., s max K ) where s max r = max (p r, q r ) for r = 1,..., K. Example 1: Consider y t SCM(1, 1), SCM(1, 1), SCM(0, 0) φ (1) φ (1) 1 φ (1) 1 θ (1) y t = 4 φ (1) 1 φ (1) φ (1) θ (1) y t 1 + ε t 4 θ (1) 1 θ (1) ε t 1. α 1 α 1 k = s max = (max (1, 1), max (1, 1), max (0, 0)) = (1, 1, 0) Example : Consider y t SCM(1, 1), SCM(1, 0), SCM(0, 0) φ (1) α φ (1) 1 φ (1) 1 y t = 4 φ (1) 1 φ (1) φ (1) 7 5 y t 1 + ε t 4 θ(1) 11 θ (1) ε t 1. α 1 α 1 k = s max = (max (1, 1), max (1, 0), max (0, 0)) = (1, 1, 0)

11 Theorem: Consider a VARMA process y t, represented in canonical reverse Echelon form with Kronecker indices k = (k 1,..., k K ). Now suppose that y t is also represented in a canonical SCM representation that consists of K SCMs of orders s r = (p r, q r ) for r = 1,..., K. k = (k 1,..., k K ) = s max = (s max 1,..., s max K ) where s max r = max (p r, q r ) for r = 1,..., K. Example 1: Consider y t SCM(1, 1), SCM(1,1), SCM(0, 0) φ (1) φ (1) 1 φ (1) 1 θ (1) y t = 4 φ (1) 1 φ (1) φ (1) θ (1) y t 1 + ε t 4 θ (1) 1 θ (1) ε t 1. α 1 α 1 k = s max = (max (1, 1), max (1, 1), max (0, 0)) = (1, 1, 0) Example : Consider y t SCM(1, 1), SCM(1,0), SCM(0, 0) φ (1) α φ (1) 1 φ (1) 1 y t = 4 φ (1) 1 φ (1) φ (1) 7 5 y t 1 + ε t 4 θ(1) 11 θ (1) ε t 1. α 1 α 1 k = s max = (max (1, 1), max (1, 0), max (0, 0)) = (1, 1, 0)

12 Identifying VARMA(p,q): SCM methodology q p 0 1 0? 1

13 Identifying VARMA(p,q): SCM methodology q p 0 1 0? 1?

14 Identifying VARMA(p,q): SCM methodology q p 0 1 0? 1??

15 Identifying VARMA(p,q): SCM methodology q p 0 1 0? 1???

16 Identifying VARMA(p,q): SCM methodology q p 0 1 0? 1??? Identifying VARMA(p,q): Echelon form q p 0 1 0? 1

17 Identifying VARMA(p,q): SCM methodology q p 0 1 0? 1??? Identifying VARMA(p,q): Echelon form q p ?

18 Identifying VARMA(p,q): SCM methodology q p 0 1 0? 1??? Identifying VARMA(p,q): Echelon form q p ?

19 Identifying VARMA(p,q): SCM methodology q p 0 1 0? 1??? Identifying VARMA(p,q): Echelon form q p ?

20 Monte Carlo Simulations: Identify some prespecified VARMA processes PANEL A: DGP of equation () PANEL D: DGP of equation (5) N SCM Echelon N SCM Echelon M.O. E.O. k SCM M.O. E.O. M.O. E.O. k SCM M.O. E.O SCMs - (1,0)(1,0)(1,0) SCMs - (1,1)(1,0)(0,0) PANEL G: DGP of equation (8) N SCM Echelon M.O. E.O. k SCM M.O. E.O SCMs - (1,1)(1,0)(1,0) PANEL H: DGP of equation (9) N SCM Echelon M.O. E.O. k SCM M.O. E.O SCMs - (1,1)(1,1)(1,1)

21 Forecasting: 40 monthly macroeconomic variables from 8 general categories of economic activity, 1959:1-1998:1 (N=480) 70 variable systems Test sample: N 1 = 00 Estimated canonical SCM VARMA Estimated canonical reverse Echelon form VARMA Hold-out sample: N = 180 Produced N h + 1 out-of-sample forecasts for each h=1 to 15 Forecast error measures: MSFE (tr(msfe)) Percentage Better: PB h Relative Ratios: RRdMSFE h = 1 70 MSFE ECH i 70 i=1 MSFEi SCM

22 Percentage Better: SCM Echelon form % Forecast horizon (h)

23 Relative Ratios: % Echelon form VARMA Forecast horizon (h)

24 Relative Ratios: % Echelon form VARMA VAR(AIC) VAR(BIC) Forecast horizon (h)

25 Summary of findings and future research Summary: 1 Theoretical: SCM more flexible than Echelon form Simulations: both perform well Forecasting: SCM outperformed Echelon form However both outperformed VARs Future Research: 1 Closer to automation via Echelon form than SCM Find automatic process to refine the model