This is, I guess, just a question of standard vocabulary.

All definitions I have seen for an ARIMA series just state that it is a series which becomes ARMA after differencing some number of times.

If you difference a time series model with linear trend

$$X_t = a + bt + w_t,$$

where $w_t$ is a white noise series, you get a stationary model

$$Y_t = b + (w_t - w_{t-1})$$

This would not be an ARMA model because of the nonzero mean, I guess, so $X_t$ doesn't fit the definition of ARIMA(0,1,1) I have seen. If you difference again, you get

$$Z_t = w_t -2 w_{t-1} + w_{t-2}$$

Thus, the series $X_t$ fits the definition I have seen of an ARIMA(0,2,2) model. Yet I see lots of websites and books informally suggest an ARIMA model should have only stochastic trends.

So my question is, do practitioners call a series like $X_t$ with a deterministic trend a type of ARIMA model, as would comport with the definitions I have seen? Or do the definitions I have seen need some revision to remove such series from consideration?

Also, would practitioners view the extra difference I applied to be a case of overdifferencing? I would think so, since a constant mean should be easy to deal with after that first difference.

  • $\begingroup$ Whenever a process does not have a unit root, differencing is overdifferencing. $\endgroup$ – Richard Hardy Feb 24 at 14:18
  • $\begingroup$ How are the ARMA polynomials modified in the presence of a constant term? $\endgroup$ – Barry Feb 24 at 14:26

No, $X_t$ is not considered an ARIMA model.

Walter Enders writes here in his textbook (3rd edition, p. 191)

We have shown that differencing can sometimes be used to transform a nonstationary model into a stationary model with an ARMA representation. This does not mean that all nonstationary models can be transformed into well-behaved ARMA models by appropriate differencing. Consider, for example, the a model that is the sum of of a deterministic trend and a pure noise component

$y_t = y_0 + a_1 t + \epsilon_t$

The first difference of $y_t$ is not well-behaved because

$\Delta y_t = a_1 + \epsilon_t - \epsilon_{t-1}$

Here $\Delta y_t$ is not invertible in the sense that $\Delta y_t$ cannot be expressed in the form of an autoregressive process. Recall that invertibility of a stationary process requires that the MA component does not have a unit root.

  • 3
    $\begingroup$ This quote does not actually say that $y_t$ is not an ARIMA process, but rather that $\Delta y_t$ isn't invertible (it is not a "well-behaved ARMA process"). Does Enders also require, in a different section, that every ARMA process be invertible? $\endgroup$ – Chris Haug Feb 24 at 21:54
  • $\begingroup$ Yes, p. 79, states "[...] the Box-Jenkins approach also neccessitates that the model be invertible. Formally, ${y_t}$ is invertible if it can be represented by a finite-order or convergent autoregressive process. Invertibility is important because the use of the ACF or PACF implicitly assumes that the ${y_t}$ seqeuence can be represented by an autoregressive model. $\endgroup$ – Arne Jonas Warnke Feb 25 at 7:16
  • 2
    $\begingroup$ ARMA processes exist independently of the Box-Jenkins approach for finding suitable ARMA models, and of ACF and PACF plots. It could be possible for a process to be technically ARMA but not one you would want to use in practice so, this quote still does not say that $y_t$ is not an ARIMA process. $\endgroup$ – Chris Haug Feb 25 at 17:21
  • 1
    $\begingroup$ Again I agree, and I do not find a more explicit statement, but the language from above's quote strongly suggests (in my view) that Walter Enders does not think of $y_t = y_o + a_1 t + \epsilon_t$ to be considered an ARIMA model. $\endgroup$ – Arne Jonas Warnke Feb 25 at 17:58

I personally consider Rob Hyndman's forecast package for R the gold standard in ARIMA modeling and forecasting. And this package will quite happily deal with time series of the form of your differenced series and call them "ARIMA with non-zero mean".

> set.seed(4); forecast::auto.arima(rnorm(100,5,1))
Series: rnorm(100, 5, 1) 
... snip ...

Similarly, a deterministic trend plus white noise is modeled as "ARIMA with drift":

> set.seed(4); forecast::auto.arima(1:100+rnorm(100,5,1))
Series: 1:100 + rnorm(100, 5, 1) 
ARIMA(5,1,0) with drift

So yes, I would consider deterministic trends ARIMA processes.

In addition, Brockwell & Davis' Introduction to Time Series and Forecasting (3rd ed., 2016) also consider "ARMA(p,q) processes with mean" on p. 74. I couldn't find an explicit discussion of trends that upon differencing turn into such ARMA(p,q) processes with (nonzero) mean, but I would say this extension is obvious enough to be accepted.

And I agree that this is a question of convention.

  • $\begingroup$ What about the title question Is a time series which is a deterministic linear trend + white noise considered an ARIMA model? $\endgroup$ – Richard Hardy Feb 24 at 17:29
  • $\begingroup$ @RichardHardy: per the question, such a series upon differencing becomes an ARMA process with a nonzero mean. forecast::auto.arima() considers this "ARIMA with drift". See my edit. $\endgroup$ – Stephan Kolassa Feb 24 at 17:38
  • $\begingroup$ Isn't rnorm(100,5,1) and i.i.d. Normal(5,1) sequence? How is it related to the processes described by the OP? Also, auto.arima fits only ARIMA models on whatever data you supply. This does not suggest the data follows an ARIMA process, hoewever. That I hit an item with a hammer does not prove the item is a nail. Similarly, a deterministic trend plus white noise is modeled as "ARIMA with drift": by the exact same argument, this is no proof the process is ARIMA. You could fit these processes with other models as long as this is technically possible, and your computer would not complain. $\endgroup$ – Richard Hardy Feb 24 at 20:09
  • $\begingroup$ I find the answer (in its current form) misleading, so -1. $\endgroup$ – Richard Hardy Feb 24 at 20:09
  • $\begingroup$ @RichardHardy: yes, rnorm(100,5,1) is 100 samples from a N(5,1) - white noise, just as specified by the question. I did not claim that this series is ARIMA(5,1,0) with drift - it isn't, by definition. What I am saying is that auto.arima() fits a deterministic trend plus ARMA, and then calls this fitted process an "ARIMA with drift". The question was about nomenclature as commonly used, and the probably most commonly used piece of ARIMA modeling software uses the terms as the OP asked. (con't) $\endgroup$ – Stephan Kolassa Feb 24 at 20:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.