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I'm teaching an applied time series course and have come up with a question I'm not sure how to answer. Suppose we have a non-stationary time series and we try models using regular differencing, seasonal differencing, or both. All three transformations lead to data that appear to be stationary. An example like this appears in Chapter 12 of the book Stat 2, and involves vehicular traffic over the peace bridge. You can get the data using the following R block.

require(Stat2Data)
data(PeaceBridge2003)
plot(PeaceBridge2003$Traffic)

I considered several models and found three that leave residuals with no autocorrelation, namely

  • Model 1: SARIMA(1,1,0)x(1,0,0)[12], the model I got working "by hand" from ACFs and PACFs
  • Model 2: SARIMA(0,0,0)x(0,1,1)[12], the model chosen by auto.arima()
  • Model 3: SARIMA(0,1,1)x(1,1,0)[12], the model from the Stat 2 book

I hope this notation is ok. The first says to do first-order regular differencing then use an AR(1) term and a seasonal AR(1) term $x_{t-12}$. The second says to do first-order seasonal differencing, then include a seasonal MA(1) term $w_{t-12}$.

To choose between these models, my gut instinct would be to pick the simplest one, (which, to me, is model 1) since differencing is such a destructive transformation.

However, I am certain my students will suggest using AIC to compare the three models, and when you do that it picks model 3. My concern is that the three models represent three different transformations. The response variable can be either $\Delta x_t$ or $\Delta_{12} x_t = x_t - x_{t-12}$ or $\Delta_{12} (\Delta x_t)$.

It is well-known that you should not use AIC to compare models with different transformations on the response variable. For example, taking a log or square root transformation on the response variable will give wildly different AICs, as has been covered here before.

What is surprising to me is that, when I simulate data and use the function AIC() in the stats package, I don't get wildly different values for different kinds of differencing as in the example above. And AIC usually does select the correct model in simulations. I'm starting to suspect that the way it's implemented in R, it's ok to use it to compare the three models above, perhaps because, at the end of the day, the response variable in all three is still $x_t$ (e.g., R has built in the reverse transformation). Is that correct? Or, should we not use the AIC() function in the stats package to compare models like the above with different types of differencing?

Lastly, I am well-aware that the real decision-maker about what order of differencing to use is the question "is the data stationary after this kind of differencing?" And, thinking about the generating mechanism, only one of the three types of differencing should, in theory, lead to stationarity. And, often, seasonal differencing is not warranted. But the problem here is that we don't have enough data to be able to say for sure that two of the choices leave non-stationary data and the third succeeds (here $n = 156$; which is 13 years of monthly data). As far as we can tell, all three choices are valid. So, it would be helpful to know if AIC can be used to help decide in situations like this.

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I did some simulations and I am convinced that when stats::arima(), or forecast::auto.arima() or astsa::sarima() are used to fit models, R considers the response variable to be $x_t$ even if the models you fit involve regular or seasonal differencing. Essentially, R is doing the extra work to undo the differencing transformation. That's why, if you ask one of these models for the fitted_values, you will get values of $x_t$ not $\Delta x_t$ or $\Delta_{12} x_t$.

For that reason, AIC() can be used to compare all such models, even if they involve different types of differencing.

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  • $\begingroup$ This is an interesting question, but I think the answer is not as simple as that. It seems that depending on the model, the likelihood is calculated for a different set of observations with some initial ones dropped. See this recent thread. If that is the case, the likelihoods are not comparable, so information criteria are not comparable either. Having fitted a model, check the $nobs part of the resulting object. For seasonally integrated models, you will have lower values than for simply integrated ones. $\endgroup$ Commented Sep 30 at 5:44
  • $\begingroup$ I would also be curious to read your take on the linked question. $\endgroup$ Commented Sep 30 at 5:49

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