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In case I dont want to "pre-process" the time series. I do a unit root test, and if it gives that is a non-stationary time series, then I will stay away from ARIMA models. Is this correct?

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You probably are talking about stationarity of the mean. In this case the statement is incorrect because ARIMA handles so called integrated models, such as those with unit roots. Integrated means that it's on differences like $\Delta y_t$. ARIMA(0,1,0) is your standard random walk model.

So, the statement is too generic to be true. However, in some special cases it holds. For instance, if your process is nonstationary in variance, i.e. the error variance is not conditionally stationary. An example is exponential random walk. In this case a simple log-transform would make it ARIMA(0,1,0) but you indicated that no pre-processing is allowed, so your statement would be correct.

Summarizing, depending on circumstances your statement could be true, but it is not true always.

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  • $\begingroup$ Thank you @Aksakal. What test can i do to know that a process is or is not conditionally stationary? Because as i can understand from your great answer. It would be the test that will make this statement correct. $\endgroup$ – donpresente Jun 2 '15 at 17:03
  • $\begingroup$ @donpresente To see whether the error variance is stationary, you can look at the series of squared differences. If they're growing, then the error must be increasing. I don't use any formal tests for this. $\endgroup$ – Aksakal Jun 2 '15 at 17:40
  • $\begingroup$ so you mean (R_t - R_{t-1})^2, where R is the value that takes the time series at time t. Or you mean the errors? Thanks! $\endgroup$ – donpresente Jun 3 '15 at 7:00
  • $\begingroup$ Yes, that's what I meant. You can regress this over time to see if there is significant trend $\endgroup$ – Aksakal Jun 3 '15 at 11:05
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No. For instance, a random-Walk has a unit root, and it's an ARIMA(0,1,0).

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    $\begingroup$ (+1) But you might expand on what the "I" in "ARIMA" means. $\endgroup$ – Scortchi - Reinstate Monica Jun 2 '15 at 15:44
  • $\begingroup$ The I means integrated, so in fact, you should typically expect ARIMA processes to have unit roots. $\endgroup$ – Arthur B. Jun 2 '15 at 18:44

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