A time series of yearly data, I want to compare the AIC and BIC values by auto.arima and manual ARIMA.


drink <- c(188,301,451,504,630,855,883,1078,1099,1008,1050,1058)

drink_ts <- ts(drink, frequency = 1, start=c(1950))


# Series: drink_ts
# ARIMA(0,1,0) with drift 
# Coefficients:
#         drift
#       79.0909
# s.e.  26.5245

# sigma^2 estimated as 8513:  log likelihood=-64.86
# AIC=133.71   AICc=135.21   BIC=134.51

I want to replicate it, so do a manual with same ARIMA(0,1,0):

drink.fit <- arima(drink_ts, order = c(0,1,0))

To get the AIC and BIC values:

BIC = AIC(drink.fit,k = log(length(drink_ts)))

# AIC: 138.7121
# BIC: 138.2272

The AIC and BIC values by auto.arima and manual ARIMA are slightly different.

Does it matter? Am I missing anything in the manual ARIMA?

(by the way, if there’s a direct way to get the AIC and BIC from the manual ARIMA)

Thank you.


1 Answer 1


You are missing one thing. Note that auto.arima() fits an ARIMA(0,1,0) model with drift. This is the following model:

$$ (1-B)(y_t-\mu t) = e_t, $$

or after rearranging,

$$ y_t = y_{t-1}+\mu+e_t. $$

The estimated value of $\mu$ is the 79.0909 you get.

So if you wanted to reconstruct your auto.arima() results, you would first have to "correct" your time series drink_ts by this drift term. Unfortunately, it's not clear how the t parameter runs - does it start from 1950 (being the first year in your series), or from 1 (which sounds more reasonable - but would yield a nonzero intercept).

I played around a bit, but couldn't fully reconstruct your AIC values. Part of the problem is of course accounting for the degree of freedom expended in estimating the drift parameter, similarly whether to allow a mean or constrain it, and finally likely slight differences in estimating methods between arima() and auto.arima().

In general, I would trust auto.arima() more than a hand-crafted call to arima. (And regarding your parenthetical question, you can extract the AIC by arima(drink_ts)$aic, but I don't know of a way to get the BIC, or the AICc.)

If your question is still open, maybe you could either ask a new question that avoids the drift complications (possibly linking the two questions), or edit to explain why you are trying to do what you are trying to do?

  • $\begingroup$ @Stephan_Kolassa, Thank you for the insights and detail explanation. Your expertise and professionalism are highly appreciated. Just a matter of interest I raised this question, no other specific goals from this analysis. $\endgroup$
    – Mark K
    Commented Sep 25, 2019 at 8:52

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