Hello, I have a question about function arima.sim I tried to somulate a AR(1) process, with no innovation, no error term. I used this code:
library(forecast) e=rnorm(100,mean=0,sd=0) series=arima.sim(model=list(ar=0.75),n=100,innov=e)+20 Then I tried to applicate ti this series auto.arima function: mod1<-auto.arima(series,stepwise=FALSE,trace=TRUE,ic='aicc') The best model returned is ARIMA (3,2,1), that is different from (1,0,0) that I used to generate the series... Besides, I tried to force a model (1,0,0), and then I compared the fitted. I attached my plot The strange thing is that the ARIMA(3,2,1) fitted my series better than ARIMA (1,0,0) that I used to generate this... The code is: mod2<-arima(series,c(1,0,0)) plot(series,main='black=series - red=ARIMA(3,2,1) - green=ARIMA(1,0,0)') lines(fitted(mod1),col='red') lines(fitted(mod2),col='green') Did I do somothing wrong? Anyone can give me explanation? Thank you very much http://www.nabble.com/file/p25663200/plot.PNG plot.PNG -- View this message in context: http://www.nabble.com/Incoherence-between-arima.sim-and-auto.arima-tp25663200p25663200.html Sent from the R help mailing list archive at Nabble.com. ______________________________________________ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
