Arima fitting

#1
I have obtained these two plots using R, I have to fit a model and the trouble is choosing between an ARMA(0,0), and an AR3. The main issue is the autocorrelation at lag 3, is it enough significant to be considered in the analysis? The BIC of the two models is very similar. Moreover Residuals are good for both models, altough AR3 removes the little correlation at lag 3 in the residuals ACF. Thank
s. 6fa57f99-d5a7-4fcb-9e83-9c78b0bd1a04.jpg
 

hlsmith

Not a robit
#2
The default given you have a representative and sufficiently powered sample is to select the simplest model, given they seem comparable in other regards (e.g., BIC, etc.).

Have you tried using the auto.arima (sp?) function in R?
 

hlsmith

Not a robit
#4
Well it all comes down to you, If you think there is autocorrelation there and it makes domain sense - then control for it. Would you have projected this autocorrelation in the model a priori before seeing it? All models are wrong, some are just informative - so it all comes down to what you think is correct and that you can defend if questioned about it.
 
#5
Thanks for the answer. How can I control if the autocorellation at lag 3 is significant enough? Moreover, looking at this graph do you think an ARCH/GARCH is necessary? Thanks!
Finale-squared residuals.png
 

hlsmith

Not a robit
#6
In all honesty, I am not sure I am savvy enough to provide any more information than I did. I am not really seeing that much lag 3 correlation.

Side question, how did you generate the graphs in post #5? Would you be willing to share that code? Thanks.
 
#7
After fitted the best model, so white noise, I search for volatility clustering (and so arch/garch)
The code is:
plot(fitWhiteNoise$residuals^2, ylab="Squared residuals")
pacf(fitWhiteNoise$residuals^2,main="Squared residuals")
acf(fitWhiteNoise$residuals^2,main="Squared residuals")
McLeod.Li.test(fitWhiteNoise)
 

noetsi

Fortran must die
#8
the rule of parsimony suggests using the simpler model. When you run the more complex model the parameters may not be statistically significant