I would like to conduct an ARIMA forecast of an exchange rate with 3030 daily observations. I followed these steps;

  1. I looked for the stationary for the original series ($P_t$) and concluded that the series is not stationary,
  2. I took the natural log of the series and then differenced so that I came up with return series($r_t = \ln P_t-\ln P_{t-1})$ of the exchange rate.
  3. I used R functions auto.arima and arima on $r_t$ and got the same result as ARIMA(0,0,0).

Should I conclude that $r_t$ follows a random walk and the original series ($P_t$) fits ARIMA(0,1,0)?


1 Answer 1


auto.arima is not tool for testing whether the selected model is somehow "correct" or "true". What it does is tell you that among the candidate models it investigated, the one reported had the lowest AICc value. This suggests the model is the best among them when it comes to forecasting under square loss.

Thus you cannot make statements like $r_t$ follows ARIMA(0,0,0) ($p$-value<0.05); but you can take this as an indication that any ARIMA model tried by auto.arima is unlikely to provide a better forecast than ARIMA(0,0,0) for $r_t$ or ARIMA(0,1,0) for $\text{ln}(P_t)$.

And no, $r_t$ does not follow a random walk; $\text{ln}(P_t)$ does.

  • $\begingroup$ I see. Graph of rt is similar to this: link. So, do you think, it is possible to model this kind of series by ARIMA method? (While, because of the lnPt series is not stationary and it also have one root, it is not possible to model it by ARIMA without making lnPt stationary. ) $\endgroup$
    – Cenk
    Commented Mar 5, 2017 at 16:22
  • $\begingroup$ I don't see much in the graph, except for some volatility clustering. But daily FX rates are largely unpredictable, so the task of doing this with a relatively simple ARIMA model seems set for a failure. Regarding unit roots, ARIMA(p,d,q) deals with them just fine. The middle parameter d tells you how many unit roots there are. $\endgroup$ Commented Mar 5, 2017 at 16:35
  • $\begingroup$ ok! Does saying this " ARIMA model seems set for a failure." mean: in order to model lnPt more complicated methods are needed? $\endgroup$
    – Cenk
    Commented Mar 5, 2017 at 16:43
  • $\begingroup$ Sort of. It is not really about the level of complexity but more about the information you have. Just looking at the historical development is very unlikely to yield good forecasts, whatever model you take. Unless you have some relevant information that is not publicly available (e.g. you know what the FED chairman is going to say before she does that), you are doomed to failure, because there are so many smart people working night and day trying to beat the market that you need months of training and insider skills to get anywhere close to their abilities. $\endgroup$ Commented Mar 5, 2017 at 16:50
  • $\begingroup$ But as an academic exercise or a homework problem, you (or anyone) can always try to do that. Just do not think that ARIMA(p,1,q) will work better than ARIMA(0,1,0) in reality. $\endgroup$ Commented Mar 5, 2017 at 16:52

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