Re: [Gretl-users] arima + arch ?

Hi, Your script is confusing me a bit;) so if I had an arch process that was arch 24 and arima 1 1 would I change the script to this: open "C:\Documents and Settings\cquilley\Application Data\gretl\mydata.gdt" /* compute 1-step pred. errs + cond. variances */ function filter(series y, \ scalar arpar, scalar mapar, \ scalar om, scalar alpha1, scalar alpha2, \ scalar alpha3, scalar alpha4, scalar alpha5, \ scalar alpha6, scalar alpha7, scalar alpha8, \ scalar alpha9, scalar alpha10, scalar alpha11, \ scalar alpha12, scalar alpha13, scalar alpha14, \ scalar alpha15, scalar alpha16, scalar alpha17, \ scalar alpha18, scalar alpha19, scalar alpha20, \ scalar alpha21, scalar alpha22, scalar alpha23, \ scalar alpha24, \ series *e, series *h) scalar err = 0 scalar T = $nobs series e = 0 series h = om/(1-alpha1-alpha2-alpha3-alpha4-alpha5-alpha6-alpha7-alpha8-alpha9-alpha10-alpha11-alpha12-alpha13-alpha14-alpha15-alpha16-alpha17-alpha18-alpha19-alpha20-alpha21-alpha22-alpha23-alpha24) loop for i=2..T --quiet scalar e2lag = e[i-1]^2 scalar e3lag = e[i-2]^2 scalar e4lag = e[i-3]^2 scalar e5lag = e[i-4]^2 scalar e6lag = e[i-5]^2 scalar e7lag = e[i-6]^2 scalar e8lag = e[i-7]^2 scalar e9lag = e[i-8]^2 scalar e10lag = e[i-9]^2 scalar e11lag = e[i-10]^2 scalar e12lag = e[i-11]^2 scalar e13lag = e[i-12]^2 scalar e14lag = e[i-13]^2 scalar e15lag = e[i-14]^2 scalar e16lag = e[i-15]^2 scalar e17lag = e[i-16]^2 scalar e18lag = e[i-17]^2 scalar e19lag = e[i-18]^2 scalar e20lag = e[i-19]^2 scalar e21lag = e[i-20]^2 scalar e22lag = e[i-21]^2 scalar e23lag = e[i-22]^2 scalar e24lag = e[i-23]^2 scalar e25lag = e[i-24]^2 e[i] = y[i] - (arpar*y[i-1] + mapar*e[i-1]) h[i] = om + alpha1*e2lag + alpha2*e3lag + alpha3*e4lag \ + alpha4*e5lag + alpha5*e6lag + alpha6*e7lag + alpha7*e8lag \ + alpha8*e9lag + alpha9*e10lag + alpha10*e11lag + \ alpha11*e12lag + alpha12*e13lag + alpha13*e14lag + \ alpha14*e15lag + alpha15*e16lag + alpha16*e17lag + \ alpha17*e18lag +alpha18*e19lag + alpha19*e20lag + \ alpha20*e21lag + alpha21*e22lag +alpha22*e23lag + \ alpha23*e24lag + alpha24*e25lag end loop return scalar err end function /* now with a non-zero an MA part */ series y = TOTAL scalar phi = 0 scalar theta = 0 scalar a1 = 0.1 scalar a2 = 0.1 scalar a3 = 0.1 scalar a4 = 0.1 scalar a5 = 0.1 scalar a6 = 0.1 scalar a7 = 0.1 scalar a8 = 0.1 scalar a9 = 0.1 scalar a10 = 0.1 scalar a11 = 0.1 scalar a12 = 0.1 scalar a13 = 0.1 scalar a14 = 0.1 scalar a15 = 0.1 scalar a16 = 0.1 scalar a17 = 0.1 scalar a18 = 0.1 scalar a19 = 0.1 scalar a20 = 0.1 scalar a21 = 0.1 scalar a22 = 0.1 scalar a23 = 0.1 scalar a24 = 0.1 scalar omega = var(y)*(1-a1-a2-a3-a4-a5-a6-a7-a8-a9-a10-a11-a12-a13-a14-a15-a16-a17-a18-a19-a20-a21-a22-a23-a24) series ht = 1 series et = 0 mle loglik = -0.5*(log(ht) + ut) scalar err = filter(y, phi, theta, omega, a1, a2, a3, a4, a5, a6, a7, a8, a9, a10, a11, a12, a13, a15, a15, a16, a17, a18, a19, a20, a21, a22, a23, a24, &et, &ht) series ut = et^2/ht params phi, theta, omega, a1, a2, a3, a4, a5, a6, a7, a8, a9, a10, a11, a12, a13, a15, a15, a16, a17, a18, a19, a20, a21, a22, a23, a24 end mle --verbose ? When I run this I get the error "Bad character 'i' in date string" and the script halts. Also, the script only appears to output the estimated alphas and arima coefficients. How would I modify this to get a more garch like output with coefficients estimated for the function total = B1 + B2*X2 + B3*X3? i.e how do I get it to estimate the B's as well? Thanks, Chris On 6/5/07, Riccardo (Jack) Lucchetti <r.lucchetti(a)univpm.it&gt; wrote: