1. Two Nonlinear Models for Time Series n David A. Dickey n North Carolina State University n (joint with S. Hwang – Bank of Korea)

2. Kinston Goldsboro

3. 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

4. ANOVA Table SSE is 2.25 with 385 df, Corrected total SS is 366. Parameter Estimate Approx 95% CI beta0 7.5861 7.4143 7.7578 beta1 0.8933 0.6938 1.0929 beta2 0.5500 0.4211 0.6789 delta -0.1643 -0.5723 0.2437 gamma -0.4081 -0.6105 -0.2058 alpha1 1.3288 1.2291 1.4286 alpha2 -0.5278 -0.6945 -0.3610 alpha3 0.0320 -0.1366 0.2007 alpha4 0.1212 0.0182 0.2243

5. Lag 1 Dynamic Constant Dynamic Lag 2

6. AR(1) Case 1: |  |<1 “stationary” normal limits for  estimator. Case 2:  =1 “random walk,” “unit root process” limit distributions non-standard

7. 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??

8. 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 )

9. Example 1:  = 0,  = 1

10. Example 2:  =3,  =3

11. 600 obs.  =0,  =1 SAS, PROC NLIN 600 obs.  =0,  =1 SAS @, PROC NLIN @ SAS is the registered trademark of SAS Institute, Cary, NC Approx 95% Parameter Estimate Confidence Limits A -0.4188 -0.8727 0.0351 B 1.0695 0.8091 1.3299 MU 0.0437 -0.0555 0.1429 (converged in 6 iterations)  =3,  =3 case did not even converge

12. 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

13. 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.

14. 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!

15. Skeleton ratios y t /y 0  = 2.0,  = 0.5  = -1.0,  = -0.8

16. 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)

17. 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. -33 3 |  | -4 

18. Example 1: N.C. Weekly Soybean Prices (Prof. Nick Piggott) AR(1) using hyperbolic tangent Sum of Mean Source DF Squares Square Model 3 37652.0 12550.7 Error 895 37.2886 0.0417 Uncor Tot 898 37689.3 Parm Estimate Approximate 95% CI a 4.3006 2.1949 6.4063 mu 7.8189 6.8801 8.7578 b -1.2435 -1.9299 -0.5572

19. Example 2: Kinston log(flow) model 1:sinusoid & AR(2)

20. AR(2) lag 2 coefficient = product of roots Replace with -1<  (Y t-2 )<1 Fitted Model Y t = log(flow) – 7.4 -.95S-.25C (sine-cos) Y t = 1.53Y t-1 +  (Y t-2 ) Y t-2 + e t Parameter Estimate Approx 95% CI A -1.3267 -1.5837 -1.0698 B 0.0646 0.0120 0.1173 Mu 7.3838 7.1136 7.6540 S1b 0.9528 0.5856 1.3200 C1b 0.2498 -0.0606 0.5602 D 1.5278 1.4459 1.6097 Plug-in forecasts

21. Roots of Forecasts & 95% intervals from m 2 -1.53m-  (Y t-2 ) 15,000 simulated futures --- plug in forecast --- simulation forecast

22. Thanks ! Questions ?