Under certain condition, an ARMA model is stationary.

  1. But I was wondering why an ARMA model can (always?) be used to model a stationary process in time series?

    • Is any stationary process an ARMA process? Or, for any stationary process, is there an ARMA process s.t. they are identical a.s., or have the same law, or they are the same in some other sense?
    • Or for any stationary process, is there a sequence of ARMA processes which converges to the stationary process in some sense?
  2. Or is there a stationary process which is not a ARMA process, or cannot be modelled as ARMA?

3 Answers 3


There is an important reason why the ARIMA might be preferred when the series are stationary. 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

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


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

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.

Note also, another example of a stationary process is the periodic processes.If $Z_1,Z_2$ independent $N(0,\sigma^2)$ and $\omega$ constant then the process

$$X_t = Z_1 \cos (t \omega) + Z_2 \sin (t \omega)$$

is second order stationary with mean zero and autocovariance $cov(X_t,X_{t+h}) = \sigma^2 cos (\omega h) $.


It's mainly by definition. You use ARMA if the series is stationary. If it is not stationary, you can convert the series into a stationary process by taking the nth difference, in this case the ARMA model becomes an ARIMA.

Hope this helps.


A series that contains a Level Shift can be made stationary by de-meaning. A series that has a level shift will appear to have significant acf structure. The remedy is NOT to build an ARIMA but to simply detect the point in time where the level shift occurs and the impact of the level shift. In practice there can be multiple level shifts and/or multiple time trends ... all possible obfuscated by Pulses and Seasonal Pulses.

  • $\begingroup$ Thanks! What does "level shift" mean? Is it same as jump? $\endgroup$
    – Tim
    Jul 23, 2013 at 13:14
  • 1
    $\begingroup$ A level shift is a jump . $\endgroup$
    – IrishStat
    Jul 23, 2013 at 15:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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