# Does augmented Dicky-Fuller test test complex unit roots?

The augmented Dicky-Fuller test, as described in Tsay's Financial Time Series: The regression done in the test seems to only produce real value $\hat{\beta}$, so it seems to me the test only tests existence of root 1 or -1. (Am I missing something?)

A AR(p) model can be nonstationary because it has a complex unit root, so I wonder if the augmented Dicky-Fuller test tests existence of all unit roots (including complex ones, i.e. non-real unit roots)?

• I think you are mixing things up. An AR($p$) is stationary if the roots are inside the complex unit circle. It is non-stationary if there is at least one root on our outside of the complex unit circle, but it doesn't matter if the root has a complex part or if it's on the real line. In any case it will be a unit root, and that is what the test is testing. – hejseb May 7 '14 at 18:51
• you mean the test can test if the root is on the complex unit circle? not just if the root is 1 or -1? – Tim May 7 '14 at 18:53
• Are you especially interested in if you have a complex root? The test doesn't distinguish between complex and real roots, it simply tests if you're on the complex unit circle, not where on it you might be. – hejseb May 7 '14 at 18:58
• but $\beta$ is a root. in the test statistic, the estimate $\hat{\beta}$ is real, and the statistic takes the difference between $\hat{\beta}$ and $1$. It is small only when $\hat{\beta}$ is close to $1$, while a root on the complex unit circle can be far from $1$. That is why I think the test only tests if the root is 1. – Tim May 7 '14 at 19:04
• $\beta$ is not a root; if it were, $x_t$ could be complex which it is not. It is the coefficient that is used in the auxiliary model to test the presence of a unit root. But I think I see what you mean now, and yes the ADF test only tests for a traditional random walk, and not an oscillating one. – hejseb May 7 '14 at 19:10