Why do we distinguish between the terms "constant", "intercept" and "drift" in the ADF test? I understand them to be analogous yet the literature seems to define "intercept" separately? For example, here is a quote:

... if it looks like a random walk, include the constant term. If it looks like it might be a random walk with drift, include the constant term and the intercept."

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    $\begingroup$ Welcome Geoffrey. Can you provide some links/citations that include the interchangeable use you mention in your post? $\endgroup$ Oct 27, 2019 at 0:41
  • $\begingroup$ @Matt. Here is a section verbatim: "... if it looks like a random walk, include the constant term. If it looks like it might be a random walk with drift, include the constant term and the intercept." $\endgroup$ Oct 27, 2019 at 1:20

1 Answer 1


Hi: That's a very poor way that they describe the procedure. This topic can be extremely confusing if not presented clearly and in an organized way. If you haven't already looked at it, I highly suggest Hamilton's text for a really nice explanation.

What I think they are saying ( who wrote that ? ) based on your verbatim is the following:

If you think that the underlying process is

A) $y_t = y_{t-1} + \epsilon_t$, then do the DF test using

$y_t = \beta_0 + \beta_1 \times y_{t-1} + \epsilon_t$ as the model to be estimated.

On the other hand, you think that the underlying process is

B) $y_t = \mu + y_{t-1} + \epsilon_t$, then do the DF test using

$y_t = \beta_0 + \beta_1 \times t + \beta_2 \times y_{t-1} + \epsilon_t$ as the model to be estimated.

Essentially, the model to be estimated should always be one level more than what is assumed under the null hypothesis. They should have been more clear with their terminology. My experience is intercept and mean are $\beta_{0}$ and $\mu$. The drift term is $\beta_1 t$ and is often referred to as the "trend" term.

Note that I just realized that I wrote things with respect to the DF test. The concept is the same for the ADF test except that there are differenced lagged responses added on to the regression models. My mistake there.

ADDENDUM: Note that the way I wrote it is also not the way the actual DF test is implemented. The actual test subtracts $y_{t-1}$ from both sides so that the actual test ends up being a test that the $\beta$ coefficient being tested is zero. So, there are two issues going on in my answer. I wrote it for the DF and I didn't use the differencing trick. Neither of these really effect the answer but I strongly suggest that one look at Hamilton for the gory details.

SECOND ADDENDUM: For anyone reading this answer, the topic of ADF testing is quite complicated and the cases considered here are only two out of a possible four cases. I strongly recommend reading chapter 17 in Hamilton's Time Series Analysis text for a much more detailed and enlightening discussion. Page 502 summarizes the tests but I believe that one needs to go through the whole chapter for a clear understanding.

  • $\begingroup$ It all just clicked - many thanks! $\endgroup$ Oct 27, 2019 at 8:18
  • $\begingroup$ Are you sure this is correct? It has been a while since I used ADF test last, but your exposition looks counterintuitive to me. Why should the model specification in the test be different from the DGP? $\endgroup$ Oct 27, 2019 at 9:40
  • $\begingroup$ Hi Richard: I wrote it for the DF but it should be the same for the ADF. The way I've always thought about it intuitively speaking was that, if you don't give the actual regression the ability to estimate outside the null, then you're not being fair. By fair, I mean that you're sort of pre-supposing that the null is true. But it's from memory ( which could be getting worse with age ) so if you want to check Hamilton, feel free. I'm pretty sure it's correct. $\endgroup$
    – mlofton
    Oct 27, 2019 at 12:05
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    $\begingroup$ @Newwone, I do not have the time at the moment. If I were you, I would take a good textbook (mlofton recommends Hamilton) and read the relevant chapter/section carefully. $\endgroup$ Dec 9, 2020 at 13:36
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    $\begingroup$ @Newwone: Like Richard, I recommend going through Hamilton's chapter on this carefully because, as you can see from the discussion above between myself and Richard, the whole issue is quite confusing. There are 4 different cases where each of them corresponds to a different scenario. If I tried to explain it here, I would just end up making it more confusing. That being said, I still think Hamilton gives the best explanation on what is a confusing topic. I suggest reading his chapter carefully from start to finish. $\endgroup$
    – mlofton
    Dec 10, 2020 at 17:23

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