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Problem auto.arima() in R

On 06/16/2011 03:04 AM, siddharth arun wrote:
I don't know.  Based on a brief glance at the manual page, it does
seem that it handles seasonal models.  I would recommend that you look
at the Journal of Statistical Software article referenced in
?auto.arima.  It's probably also worth browsing through Chap. 14 of
Venables and Ripley "Modern Applied Statistics with S" to get background
on time series analysis in R/S.

For auto.arima(), you need to know the frequency at which the data are
sampled (and assume that you are trying to fit annual patterns).  If you
are looking for patterns of an unknown frequency, spectral analysis is
probably a more appropriate approach.
ts(x,freq=1) ?? If each data point corresponds to a year then you
can't estimate seasonality, can you?  See ?ts.

  If (as mentioned above) you are trying to detect _periodicity_ rather
than _seasonality_ (i.e. regular patterns with an _unknown_ period) then
seasonal ARIMA is probably not the right way to go.
If you know that these are quarterly data, then specify it.

z <- ts(rep(4:1,3),frequency=4)
ARIMA(2,0,2)(1,0,1)[4] with non-zero mean : 1e+20 *
 ARIMA(0,0,0) with non-zero mean : 40.73225
 ARIMA(1,0,0)(1,0,0)[4] with non-zero mean : 1e+20 *
 ARIMA(0,0,1)(0,0,1)[4] with non-zero mean : 1e+20
 ARIMA(0,0,0)(1,0,0)[4] with non-zero mean : 1e+20 *
 ARIMA(0,0,0)(0,0,1)[4] with non-zero mean : 1e+20
 ARIMA(0,0,0)(1,0,1)[4] with non-zero mean : 1e+20 *
 ARIMA(1,0,0) with non-zero mean : 42.69388
 ARIMA(0,0,1) with non-zero mean : 1e+20
 ARIMA(1,0,1) with non-zero mean : 1e+20
 ARIMA(0,0,0) with zero mean     : 60.23336

 Best model: ARIMA(0,0,0) with non-zero mean

Series: z
ARIMA(0,0,0) with non-zero mean

Coefficients:
      intercept
         2.5000
s.e.     0.3228

sigma^2 estimated as 1.25:  log likelihood = -18.37
AIC = 40.73   AICc = 42.07   BIC = 41.7


  auto.arima() tries to fit seasonal models, but discards them.  This
data set is probably too short and too artificial to draw conclusions
from ...