I use R to fit an ARIMA model to a time series (yearly granularity):


beer <- c(150,241,361,403,504,684,706,862,879,806,840,846,1024,1196,1239,1237,1281,1342)

ts_beer = ts(beer, start = c(1980), frequency = 1)

dif.ts_beer <- diff(ts_beer)


enter image description here

Based on the ACF and PACF, I fit an ARIMA(4,0,4) model.

dif.ts_beer.fit <- arima(dif.Gas, order = c(4,0,4))


It looks OK. But then I run auto.arima:


It gives:

Series: dif.ts_beer 
ARIMA(0,0,0) with non-zero mean 

s.e.  17.0359

sigma^2 estimated as 5242:  log likelihood=-96.4
AIC=196.81   AICc=197.67   BIC=198.48

So the manual ARIMA(4,0,4) is not a good choice for this case? If so, what ARIMA(p,d,q) model should I use?

  • 1
    $\begingroup$ While I agree fully with @Stephen_Kolassa, for academic interest you may want to explore EACF plots. EACF is extended ACF proposed by Tsay, R. and Tiao, G. (1984). It is available in TSA package in r. See here $\endgroup$ – Dayne Sep 19 '19 at 10:13

First, it is very hard to use (P)ACF plots to identify ARIMA(p,d,q) models if both p and q are nonzero. See Hyndman & Athanasopoulos:

If p and q are both positive, then the plots do not help in finding suitable values of p and q.

Second, your peaks at lag 4 only barely exceed the confidence bands.

I would always prefer auto.arima() over parsing (P)ACF plots myself, i.e., the Box-Jenkins approach. It is built by experts with a lot of experience, and it truly is a gold standard for ARIMA modeling, unless you are an expert yourself and you are working academically on the frontiers of knowledge.

In the present case, auto.arima() would prefer a simple mean model. I would recommend that you run with this.

  • 1
    $\begingroup$ @Stephan_Kolasa, thank you Mr. Klassa for the detail explanation and guidance on the matter. It benefits a lot! $\endgroup$ – Mark K Sep 19 '19 at 8:18
  • $\begingroup$ Great @StephanKolassa $\endgroup$ – Fr1 Sep 19 '19 at 9:42

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