I may be mixing up my time series and non time series concepts, but what is the difference between a regression model that exhibits serial correlation and a model that exhibits a unit root?

In addition, why is it that you can use a Durbin-Watson test to test for serial correlation, but must use a Dickey-Fuller test for unit roots? (My textbook says this is because the Durbun Watson Test cannot be used in models that include lags in the independent variables.)


A simpler explanation can be this: if you have an AR(1) process $$y_t = \rho y_{t-1} + \epsilon_t,$$ where $\epsilon_t$ is white noise, then testing for autocorrelation is $H_{0;\mbox{AC}}: \rho=0$ (and you can run OLS which behaves properly under the null), while testing for the unit root is $H_{0;\mbox{UR}}: \rho=1$. Now, with the unit root, the process is non-stationary under the null, and OLS utterly fails, so you have to go into the Dickey-Fuller trickery of taking the differences and such.


If you have, say, an autoregressive process, and you look at what is called the characteristic polynomial, that polynomial has complex roots (maybe some or all are real roots). If all the roots are inside the unit circle the process is stationary otherwise it is non-stationary. A test for unit roots is looking to see if the specific process is stationary based on the observed data (parameters unknown).

A test for serial correlation is entirely different. It looks at the autocorrelation function, testing to see whether or not all correlations are zero (sometimes referred to as a test for white noise).

The answer to the second question is that different problems require different tests. I don't understand what your book is describing. I see these tests as tests on individual time series. I don't see where independent and dependent variables enter into it.

  • $\begingroup$ I think this answer would be improved by (a) specifying which "characteristic polynomial" you are considering since there are at least two common forms with one of them broadly fitting your description and the other one not (b) clarifying that for your particular choice of characteristic polynomial you are looking for roots strictly inside the unit circle and (c) essentially what a unit-root test is doing is precisely what it states, i.e., testing for a root which lies exactly on the unit circle. That said, one needs a little more than stated to get a fully wide-sense stationary process. $\endgroup$ – cardinal May 6 '12 at 22:11
  • $\begingroup$ Thanks for clarifying the unit root test for the OP. As far as ambiguity about the characteristic polynomial, I was not aware of it. It shold be clear from the time series literature what polynomial i am referring to. Check the definition in Box and Jenkins book if you are unsure. Any AR process with at least one root of the characteristic polynomial on or outside the unit circle is nonstationary. Of course the unit root test is testing for roots on the unit circle. But keep in mind that the coefficients for the AR process is not known. $\endgroup$ – Michael R. Chernick May 15 '12 at 10:32
  • $\begingroup$ So the data only provides us with estimated coefficients and so we are looking for characteristic polynomials close to the one with the sample estimates of the coefficients. Testing the hypothesis that the mean of a distribution is 0 doesn't really test that the mean is exactly 0 but practically speaking that it is very close to 0. Similarly a unit root test really is testing whether the characteristic polynomial for the model has a root near the unit circle and hence the process is close to at or outside the boundary of stationarity. It is a statistical hypothesis testing problem. $\endgroup$ – Michael R. Chernick May 15 '12 at 10:51
  • 1
    $\begingroup$ Michael, I raised the points in my first comment precisely because what is stated in this answer is the opposite of the usual presentation in the majority of the time-series literature. If your characteristic equation is $1 - \phi_1 B - \phi_2 B^2 - \ldots - \phi_p B^p = 0$, then the roots must lie outside the unit circle to ensure stationarity. (cont.) $\endgroup$ – cardinal May 15 '12 at 13:01
  • 2
    $\begingroup$ I checked Box - Jenkins and Reinsel. We can close this here. On page 56 they define the characteristic equation (same characteristic polynomial that I intended)The complex factorization gives terms 1-Gi B. They say for stationarity that Gi must lie in the unit circle. But it is the inverse (in the sense of complex numbers) that is the root of the equation . So the roots all do lie outside the unit circle for stationarity. That was my confusion. $\endgroup$ – Michael R. Chernick May 15 '12 at 18:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.