arima libretl/-x-12 contradicting results with no estimation
by oleg_komashko＠ukr.net

Dear all,
script below illustrates the problem
<hansl>
nulldata 100
setobs 4 1994:1 --time-series
set seed 13
include arima_sim.gfn
series y = seas_cum(normal())
# case 1
arima 0 0 0; 0 1 0; y --nc
eval $ess
eval $T
eval mean($uhat)
eval {$uhat}'{$uhat}
eval {$uhat}'{$uhat} - $ess
eval max(abs(sdiff(x) - $uhat))
eval max(abs(x - $yhat - $uhat))
yhat1 = $yhat
uhat1 = $uhat
gnuplot y yhat1 --time-series --with-lines --single-yaxis --output=display
## here evething looks as it should be
# case 2
arima 0 0 0; 0 1 0; y --nc --x-12-arima
eval $ess
eval $T
eval mean($uhat)
eval {$uhat}'{$uhat}
eval {$uhat}'{$uhat} - $ess
eval max(abs(sdiff(x) - $uhat))
eval max(abs(x - $yhat - $uhat))
yhat2 = $yhat
uhat2 = $uhat
gnuplot y yhat2 --time-series --with-lines --single-yaxis --output=display
## but not here
<hansl>
Oleh
5 years, 6 months

Re: [Gretl-devel] ADF p-value glitch?
by Ioannis A. Venetis

Eviews gives the same p-value.
I think there is no "problem" here as the asymptotic distribution is
highly asymmetric around zero including a band of positive ADF values
but with very small probability area (and it is an one-sided test).
Actually you need a test statistic of 2.8 ( urcpval(2.8, 0, 1, 3) ) to
reach an asymptotic p-value of exactly 1 according to the code (and the
response surface estimated by MacKinnon (1996)).
P-value here (in the ADF test) is interpreted as the integral of the
distribution to the left of the ADF statistic. If you get an estimate of
ADF=0 the probability to the left is 99.62%, that is under the null
hypothesis you will observe an ADF statistic of 0 or smaller (<= than 0)
with 99.62 percent probability. Why should this be 1 or why should be 1
if the ADF statistic is positive?
The estimated parameter could be higher than one - although "slightly"
higher in relevant economic applications. For example you estimated
AR=1.00390523 so the asymptotic p-value is 0.99749368 or 99.74% (is
higher than 99.62 etc). Under the null that AR=1, the probability that
you will estimate (with constant trend etc that is with all
specification "correct") AR<=1.00390523 is 99.74%.
Now in the left tail assume that you get ADF=-3.5 so that eval
urcpval(-3.5, 0, 1, 3) gives 0.039304454 or under the null the
probability to "draw" an ADF less than or equal to -3.5 is "only" 3.93%.
At 5% (singificance) you reject the null etc.
Yiannis
5 years, 6 months

a very strange case of arima (poor)convergence
by oleg_komashko＠ukr.net

Dear all,
The script below
illustrate the problem
Findings: extremely bad lnl
in comparison to --x-12-arima
zero values of the 2 last parameters
and gradients at all iterations
Strangely large scaling factor
Note that --x-12-arima gives
nice pol. roots and excellent Ljung-Box Q'
Also note that with the default $pd for
modtest --autocorr
obviously fails because zero df
open bad_data.gdt #attached
smpl 1 194
# note strange zeros for b[y_one] and b[y_two]
arima 3 0 0; 1 0 0; diff_series const y_one y_two
lnl1 = $lnl
modtest --autocorr 5
# compare
arima 3 0 0; 1 0 0; diff_series const y_one y_two --x-12-arima
lnl2 = $lnl
modtest --autocorr 5
eval lnl2 - lnl1
arima 3 0 0; 1 0 0; diff_series const y_one y_two --verbose
# Scaling y by 2.18989e+018 !!!
/*
Iteration 1: loglikelihood = -8292.74884180
Parameters: -2.8152e+016 0.65496 -0.090036 -0.16211 -0.44328 0.00000
0.00000
Gradients: 4.7608e-018 4.0881 1.2538 3.3817 -1.2644 0.00000
0.00000 (norm 7.59e-001)
Iteration 2: loglikelihood = -8292.71644305 (steplength = 0.0016)
Parameters: -2.8152e+016 0.66150 -0.088030 -0.15670 -0.44531 0.00000
0.00000
Gradients: 4.5624e-018 1.3507 -1.5003 1.0847 -1.4772 0.00000
0.00000 (norm 5.32e-001)
*/
#etc. etc
Oleh
5 years, 6 months