It is my understanding of the AIC formula that: $$ AIC = -2\log(L) + 2m $$ where $m =$ number of parameters.

I am wondering how to determine the value of $m$ in AIC for the following models:

  • AR(1)
  • MA(1)
  • ARMA(1,1)
  • ARIMA(1,1,1)
  • Seasonal ARIMA(1,1,1)(1,1,1)12

I believe for an AR(p) model $m=p$,
MA(q) model $m=q$,
ARMA(p,q) model $m = p+q$,
ARIMA(p,d,q) model $m=p+q$
and for a seasonal ARIMA(p,d,q)(P,D,Q)s model $m = p+q+P+Q+1$.

Could you please let me know if this is correct?
And if not, how would I determine the correct number of parameters ($m$)?

  • 1
    $\begingroup$ I think it's mostly right, although perhaps +1 should be added everywhere (except for SARIMA) because of the fact that also $\sigma^2_{\varepsilon}$ is being estimated extra to the AR and MA coefficients. But I am not sure. Also, what is your logic behind the +1 for SARIMA but not in the other cases? $\endgroup$ Apr 21, 2017 at 6:54
  • $\begingroup$ @RIchard The +1 is correct but nothing is lost by ignoring it with AIC (since they all have it). For BIC and AICc though, you can't just leave it out. Also you usually have a mean but for differenced models you often don't $\endgroup$
    – Glen_b
    Apr 21, 2017 at 7:00
  • $\begingroup$ @Glen, thanks. I presume you mean that nothing is lost in comparisons where such constants cancel out. But if one takes AIC of a given (as opposed to chosen) model and interprets it as the expected likelihood, then it starts to matter, right? $\endgroup$ Apr 21, 2017 at 7:02
  • $\begingroup$ Yes, nothing is lost in comparisons when it's common to both. $\endgroup$
    – Glen_b
    Apr 21, 2017 at 7:27
  • $\begingroup$ I actually thought the +1 was for the seasonal component of the seasonal ARIMA model. My interest is finding the exact number of parameters, so I do not want to leave out any additional parameters just for the ease of comparison. $\endgroup$ Apr 21, 2017 at 11:51

1 Answer 1


I think you are almost right. For a general SARIMA model, it would be $$ m=p+q+P+Q+1 $$ where $+1$ comes from the fact that also $\sigma^2_{\varepsilon}$ is being estimated extra to the AR and MA coefficients.

For submodels such as AR, MA or ARIMA, just set the appropriate coefficients to zero. Thus $m=p+1$ for AR; $m=q+1$ for MA; and $m=p+q+1$ for ARIMA.

(As @Glen_b notes, the $+1$ is not so important when we compare AICs across models; when we take a difference between two AICs, the $+1$ in each cancels out.)

  • $\begingroup$ Thank you for your answer, I was testing your answer on R by running the following code: arima(dowjones,c(1,0,0)). This resulted in R outputting a log likelihood of -49.86 and an aic of 105.72. Doing some basic maths with the AIC formula and the loglikelihood, R output is suggesting that the number of parameters is 3 rather than 2 (your suggested answer). $\endgroup$ Apr 21, 2017 at 11:43
  • $\begingroup$ I don't know if the reported likelihood is correct. It will be correct up to a constant but probably not beyond that. I find it annoying, but that is how most of the functions do it... $\endgroup$ Apr 21, 2017 at 11:55
  • $\begingroup$ I believe it is correct since -49.86*-2 = 99.72. Indicating exactly 6 needs to be added on to achieve the correct AIC. This indicates that the correct number of parameters is 3. Should I perhaps post this as a separate question with the R code included? $\endgroup$ Apr 21, 2017 at 15:24
  • $\begingroup$ Perhaps the constant counts as a parameter as well? I don't know... $\endgroup$ Apr 21, 2017 at 15:45

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