I am currently dealing with some monthly data on interest rates which goes back over thirty years.

The total 30 years worth of data is I(1). However when I shorten this data to go back only 10 years its I(2).

When running this data in a VECM for the 10 year interval, do I have to difference or log the variable it for it to be I(1) or is it more appropriate just to treat it like an I(1) variable (since the true nature of the process is I(1)) and put it in the VECM even though the 10 years worth of data is I(2)?

  • $\begingroup$ What about the Macroeconomic environment? Was there a crisis in the first 20 years? If so you should maybe check for structural changes and take the short time series. $\endgroup$ – Ferdi Nov 16 '17 at 13:28
  • $\begingroup$ How do you know that the data "is" integrated of a certain order? Did you simulate it? Typically, you would decide on an order of integration based on a statistical test (e.g., KPSS in auto.arima()), and the results can of course change if you shorten or lengthen your data. $\endgroup$ – S. Kolassa - Reinstate Monica Nov 16 '17 at 13:34
  • $\begingroup$ (Incidentally, I have not seen your notation I(d) elsewhere - "integrated of order d" is more common, if admittedly more cumbersome.) $\endgroup$ – S. Kolassa - Reinstate Monica Nov 16 '17 at 13:35
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    $\begingroup$ To say that data "is I(1)" is completely nonsensical. A stochastic process can be I(1), but what you actually have is some measure of evidence that your data could or could not come from a process of a specific order. Obviously, as you change the data, the balance of evidence will vary, especially if some of the assumptions for the test are wrong. For interest rates specifically, @Ferdi is correct, you have to deal with the structural changes. $\endgroup$ – Chris Haug Nov 16 '17 at 13:37
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    $\begingroup$ @StephanKolassa The I(d) notation is fairly widespread in econometrics (although its generality and terseness is indeed useless: in econometrics almost everything is either I(0) or I(1), we could just say "integrated" or "not integrated"). $\endgroup$ – Chris Haug Nov 16 '17 at 13:40

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