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 ?