Two Nonlinear Models for Time Series n David A. Dickey n North Carolina State University n (joint with S. Hwang – Bank of Korea)
Kinston Goldsboro
Model 1: Transfer Function Upstream: G t = log(Goldsboro flow) G t *: deviation from mean Downstream: K t = log(Kinston flow) Model: K t = 0 + 1 (G t-2 *) G t-1 *+ 2 (1 - (G t-2 *) )G t-2 *+ Z t (G t *) = exp( + G t *)/(1+exp( + G t *) ) Logistic 0 < (G t *) < 1 Straightforward nonlinear regression problem! Z t = 1 Z t-1 + 2 Z t-2 + 3 Z t-3 + 4 Z t-4 + e t
ANOVA Table SSE is 2.25 with 385 df, Corrected total SS is 366. Parameter Estimate Approx 95% CI beta beta beta delta gamma alpha alpha alpha alpha
Lag 1 Dynamic Constant Dynamic Lag 2
AR(1) Case 1: | |<1 “stationary” normal limits for estimator. Case 2: =1 “random walk,” “unit root process” limit distributions non-standard
Model 2: =f(Y t-1 ) Can we span -1< <1 or 0< <1 ?? Can we span -1< <1 or 0< <1 ?? n Logistic »exp( Y) / (exp( Y) +1) »exp( Y t-1 ) / (exp( Y t-1 ) +1) n Hyperbolic tangent »2(Logistic)-1 »(exp( Y) -1) / (exp( Y) +1) »(exp( Y t-1 ) -1) / (exp( Y t-1 ) +1) n Related to “smooth transition” models (Tong, 1990) n Harder problem than transfer function! - Why??
Can make progress under H 0 : =0 [ (Y)= 0 ] Can make progress under H 0 : =0 [ (Y t-1 )= 0 ] n Estimates: Use Taylor’s series F n = derivative matrix, hyperbolic tangent model. Estimates of N(0,G ) Estimates of N(0,G -1 2 )
Example 1: = 0, = 1
Example 2: =3, =3
600 obs. =0, =1 SAS, PROC NLIN 600 obs. =0, =1 PROC SAS is the registered trademark of SAS Institute, Cary, NC Approx 95% Parameter Estimate Confidence Limits A B MU (converged in 6 iterations) =3, =3 case did not even converge
Tong: Skeleton of Process * Recursion without the e’s * y t = (y t-1 ) y t-1 =0, =1 |y (y)|<|y| y (y) even function t
Usual path to normality: stationarity + ergodicity One of Tong’s conditions to show ergodicity: There exist K>0, 0< <1 such that from any y 0 the skeleton y t is bounded by |y 0 | K t Skeleton ratios y t /y 0 ( =0, =1) : Blue is y1 Green is y2 Etc.
Hyperbolic tangent case (increasing) Suppose (K, ) exist – eventually (T), must have K T <1. Let M=K T < 1 and B=M 1/T. Note M < B = M 1/T Pick y -1 with (y -1 )>B and y 0 = y -1 B -T > y -1 Note 1: y > y -1 (y)>B (monotonicity) so (y 0 ) > B Note 2: B t y 0 = y -1 B t-T > y -1 for t T (B t y 0 ) > B for t<=T Thus y 1 = y 0 (y 0 )>B y 0 > y -1, y 2 = y 1 (y 1 )>B 2 y 0 > y -1 etc. But y T > B T y 0 (=My 0 =K T y 0 ) is a contradiction!
Skeleton ratios y t /y 0 = 2.0, = 0.5 = -1.0, = -0.8
Where is “good” ( ) region? Symmetries: Generate hyperbolic tangent model with symmetric e t. Now use – and –e t s (Y t-1 ) = [exp( (-Y t-1 ))-1]/ [exp( (-Y t-1 ))+1] -Y t = (Y t-1 ) (-Y t-1 ) – e t Now –e t and e t have same distribution so ( ) in “good” region ( ) also in “good” region & distribution of - estimate is mirror image of estimate ( estimates are same). For (|Y t-1 |), symmetry is ( ) and (- ). (proper conf. int. coverage and good convergence rate)
Y t / = (Y t-1 ) Y t-1 / + e t / where (Y t-1 ) = [exp( (Y t-1 / ))-1]/ [exp( (Y t-1 / ))+1] can assume e~(0,1) and slope is Hwang’s simulations on (-4,4)x(-4,4) suggest (for hyperbolic tangent) approximately 0< <3 and -4< <3-7 /3 along with their symmetric counterparts | | -4
Example 1: N.C. Weekly Soybean Prices (Prof. Nick Piggott) AR(1) using hyperbolic tangent Sum of Mean Source DF Squares Square Model Error Uncor Tot Parm Estimate Approximate 95% CI a mu b
Example 2: Kinston log(flow) model 1:sinusoid & AR(2)
AR(2) lag 2 coefficient = product of roots Replace with -1< (Y t-2 )<1 Fitted Model Y t = log(flow) – S-.25C (sine-cos) Y t = 1.53Y t-1 + (Y t-2 ) Y t-2 + e t Parameter Estimate Approx 95% CI A B Mu S1b C1b D Plug-in forecasts
Roots of Forecasts & 95% intervals from m m- (Y t-2 ) 15,000 simulated futures --- plug in forecast --- simulation forecast
Thanks ! Questions ?