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Which Dickey-Fuller test should I apply to this time series?

  1. Nov 29, 2012 #1
    I have a time series of climate data that I'm testing for stationarity. Based on previous research, I expect the model underlying the data to have an intercept term, a positive linear time trend, and some normally distributed error term. In other words, I expect the underlying model to look something like this:

    yt = a0 + a1t + β yt-1 + ut $

    where ut is normally distributed. Since I'm assuming the underlying model has both an intercept and a linear time trend, I tested for a unit root with equation #3 of the simple Dickey-Fuller test, as shown:

    ∇yt = α01t+δ yt-1+ut

    This test returns a critical value that would lead me to reject the null hypothesis and conclude that the underlying model is non-stationary. However, I question if I'm applying this correctly, since even though the underlying model is assumed to have an intercept and a time trend, this does not imply that the first difference ∇yt will as well. Quite the opposite, in fact, if my math is correct.

    Calculating the first difference based on the equation of the assumed underlying model gives:
    ∇yt = yt - yt-1 = [a0 + a1 + β yt-1 + ut] - [a0 + a1(t-1) + β yt-2 + ut-1]

    ∇yt = [a0 - a0] + [a1t - a1(t-1)] + β[yt-1 - yt-2] + [ut - ut-1]

    ∇yt = a1 + β * ∇yt-1 + ut - ut-1$

    Therefore, the first difference ∇yt appears to only have an intercept, not a time trend.

    Because the underlying model has an intercept and a time trend, should I use the Dickey-Fuller test that includes an intercept and time trend when it tests for a unit root, or should I use the Dickey-Fuller test that only includes an intercept because the first difference of the original time series only has an intercept?
     
  2. jcsd
  3. Nov 29, 2012 #2

    Stephen Tashi

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    Science Advisor

    You are correct that the difference operation a linear trend term (k)(t) produces a constant, not another linear trend term.

    It's interesting to try to read that Wikipedia article you linked. It's about some hypothesis tests, but it never manages to state the null hypotheses and the test statistics. Apparently you understand the test statistics - which is more than I know.

    However, I did find this page: http://stats.stackexchange.com/questions/18133/selecting-regression-type-for-dickey-fuller-test where the reply mentions that using the test with term of the form (k)(t) implies we are investigating a model with a quadratic trend. So my guess is that you don't use that form to test your null hypothesis.
     
  4. Nov 29, 2012 #3
    As far as I know, the null hypothesis is that a unit root exists and that the process is therefore stationary. I had found that thread you linked, and I'm the user who started the short but ongoing discussion in the comments. The linked post (listed in the right-hand column) is also mine, and is quite similar to my post here.

    http://stats.stackexchange.com/q/44647

    I don't understand why using a test with a trend term a1t would be used with a quadratic term, however. Can you shed any light on that?
     
  5. Nov 30, 2012 #4

    Stephen Tashi

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    The only light I can shed is that if [itex] f(t) = kt^2 [/itex] then
    [itex] \triangle f(t) = f(t+1) - f(t) = k(t+1)^2 - kt^2 = k( t^2 + 2t + 1) - kt^2 = 2kt + k [/itex]
    So the [itex] \triangle [/itex] of a model with a [itex] t^2 [/itex] term has a term linear in [itex] t [/itex].
     
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