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I am new to time series analysis, and I am self learner. I am using R language to learn how to do time series analysis. I started by studying the concepts and the theory behind such analysis, however I see a great concentration on the ARIMA method, whereas there is a very small attention for other methods.

Could somebody one tell me why ARIMA is preferred over the other methods.

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    $\begingroup$ Hi: It's not so much that it's preferred. It's more that ARIMA models are often a reduced form equivalent of many any other time series models. So, for example, the koyck distributed lag ( ADL in econometrics ) say can be written as an AR(1) model with exogenous regressor. It's often the case in time series that one model can be written in various ways. Another example is exponential smoothing and double exponential smoothing. These can be written as an ARIMA(01,1) and ARIMA(0,2,2) respectively. $\endgroup$
    – mlofton
    Feb 26, 2019 at 17:49
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    $\begingroup$ Some days, when I am particularly cynical, I believe that ARIMA models are so popular because you can actually prove mathematical theorems about them (e.g., involving characteristic polynomials, unit roots etc.), in contrast to some other methods like exponential smoothing that only recently got a firm mathematical foundation via state spaces. $\endgroup$ Feb 26, 2019 at 19:46
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    $\begingroup$ In actual practice, simpler models are more likely to be adopted, e.g. simple exponential smoothing of a deseasonalized series (although many exponential amoothing models can be expressed as ARIMA models). The concept that ARIMA models are somehow best may be an indirect measure of the huge influence of Box -- and his students. $\endgroup$
    – zbicyclist
    Feb 26, 2019 at 20:31
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    $\begingroup$ There is probably something related to ARIMA being a universal approximation that contributes to its popularity. $\endgroup$ Feb 27, 2019 at 9:48
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    $\begingroup$ Stephan Kolassa, exponential smoothing is equivalent to an ARIMA(0,1,1). Wold's decomposition theorem provides a solid theoretical reason for ARIMA modelling if you expect stationarity in some sense (e.g. after differencing the series) $\endgroup$
    – Stats
    Feb 4, 2023 at 14:48

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ARIMA models are not generally preferred over any other time series analysis method. There are certainly not preferred when the series demonstrate non-stationaries unable to be modelled using the ARIMA framework.

However, there is an important reason why the ARIMA might be preferred when the series are stationary (or gets so after differencing). And this reason is the Wold's decomposition theorem - any covariance stationary process has a linear representation: a linear deterministic component ($V_t$) and a linear indeterministic components ($\varepsilon_t$)

Suppose that ${X_t}$ is a covariance stationary process with $\mathbb{E}[X_t] = 0$ and covariance function, $\gamma(j) = \mathbb{E}[X_t X_{t−j}]$ , $ \forall j$. Then

$$X_t = \sum_{j=0}^{\infty} \psi_j \varepsilon_{t−j} + V_t$$

where

  • $\psi_0=1$, $\sum_{j=0}^{\infty} \psi_j^2<\infty$
  • $\varepsilon_{t−j} \sim WN(0, \sigma_{\varepsilon}^2)$
  • $\mathbb{E}[\varepsilon_t V_s] = 0, \forall s,t>0$
  • $\varepsilon_t = X_t - \mathbb{E}[X_t|X_{t-1},X_{t-2},...]$

As you may see, the first part of the representation looks like an $MA(\infty)$ process with square summable moving average terms. The second part is the deterministic part of $X_t$ because $V_t$ is perfectly predictable based on past observations on $X_t$. And we know that models of $MA(\infty)$ representations are in their most general form $ARMA(p,q)$ representations: as long as the roots of the autoregressive part of an ARMA process are less than unity in absolute value, the process has a $MA(\infty)$ representation.

However, note, while an ARMA process generates an $MA(\infty)$ with square summable weights, it is not the only form that does this. A process that is square summable is not necessarily absolutely summable. $ARMA(p,q)$ models have ‘short memory’ relative to the entire class representations envisioned by the Wold representation. But Wold representation - despite covering more general cases- provides us with a strong argument of why modelling with ARMA is justifiable on stationary, short memory series.

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    $\begingroup$ Comments are not for extended discussion. Instead, use comments constructive criticism and requests for clarification. $\endgroup$
    – Sycorax
    Mar 31, 2023 at 20:07
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ARIMA models account for a variety of possible problems which simpler models may have. For example, if you try an AR model, but there should be an MA component, this could make your estimates highly incorrect. Also the other way around where you model as MA but it should include AR.

But what if you don't need an ARIMA and an AR will suffice? In applied statistics, say economics, your plan to simply use an AR(p) model may be fantastic, and when you try an ARIMA you don't find any contribution of the -IMA - but there will always be people who say "Well I could come up with 5 scenarios where you should really have an ARIMA model instead of just AR," and so to appease the crowd you're forced to throw in everything they'll think of.

"Everything they'll think of" is another important point. In many disciplines, ARIMA is about as complex or outside-of-the-box as most people have even heard of. Though there are other models which may be outside the scope of an ARIMA model, ARIMA is the generic go-to, as well as the benchmark against which other time-series methods will be compared. And so, once again, you're forced to show an ARIMA version of the analysis.

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I am not sure they are preferred. They are popular because Box et el were well respected and they did a good job of presenting their material at a time there was not a lot of alternatives. To me they require judgement ('an art rather than a science') and knowledge of the data that may not exist in a workplace environment. They also are time consuming and (some argue) they are hard to identify correctly with real world data, particularly when mixed AR and MA exist.

Exponential smoothing models such as Holt or Winters have proven about equally accurate from what I have read -they are certainly a lot easier and faster to use.

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  • $\begingroup$ Thank you for your answer, could you please add a refernce for Holt or Winters models to prove that they are equaly accurate. Thank you in advance. $\endgroup$
    – Nizar
    Mar 11, 2019 at 4:45
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    $\begingroup$ I am not a statistician so you should take my comments with a grain of sand. You might look up the M competitions that found damp trend, a form of ESM, among the better methods. This is one example of where ESM was better (but no general rule is suggested). otexts.com/fpp2/arima-ets.html Here is another that found each better at different times aip.scitation.org/doi/10.1063/1.4801282 This found one form of ESM better in one case. matec-conferences.org/articles/matecconf/pdf/2016/44/… None suggest one is better in all cases. $\endgroup$
    – user54285
    Mar 12, 2019 at 21:45
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According to Box and Jenkins, ARIMA is a linear non-stationary model. It is an imporvement over ARMA.

In my experience, ARIMA might be favored over other methods because of its flexibility.

You can achieve far better results if you decompose your signal into simpler components and use simple linear models to forecast each time series and then combine them into one forecast.

So, it is initially easier to work with and it can handle non-stationarity if there is homogeneous nonstationary behavior in the time series.

There are many scenarios where ARIMA model fails. e.g. if there is heteroskedasticity in your data, it cannot be modeled with ARIMA.

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  • $\begingroup$ Perhaps a relevant quotation from the Box & Jenkins text would be helpful in providing context & justification for the claims made in the first sentence. $\endgroup$
    – Sycorax
    Mar 31, 2023 at 20:08

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