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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.)

chl
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hgcrpd
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    Related: [Intuitive explanation of unit root](http://stats.stackexchange.com/questions/29121/intuitive-explanation-of-unit-root). – whuber Feb 09 '13 at 15:45

2 Answers2

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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.

StasK
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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.

naught101
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Michael R. Chernick
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    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. – cardinal May 06 '12 at 22:11
  • 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. – Michael R. Chernick May 15 '12 at 10:32
  • 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. – Michael R. Chernick May 15 '12 at 10:51
  • The sample estimate of the charactersitice polynomial is a fixed polynomial and it can be determined algebraically whether or not it has a root on the unit circle. So maybe although it is looking for unit roots it is really a test as to where or not an autoregressive process is stationary because all its roots are well within the unit circle versus the alternative that one is close to or outside the unit circle (a one-sided test). – Michael R. Chernick May 15 '12 at 10:55
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    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.) – cardinal May 15 '12 at 13:01
  • (cont.) In the signal-processing literature, a related polynomial in terms of $z = B^{-1}$ is considered, the roots of which must be *inside* the unit circle to get the same conclusion. This was also discussed in [the comments to this answer](http://stats.stackexchange.com/questions/19788#comment35796_19822). – cardinal May 15 '12 at 13:02
  • My memory is that Box and Jenkins define it as is given by Irishstat in the referenced asnwer. I would think that although Box and Jenkins address their book to signal processing and engineering disciplines you might call their definition the one used in the signal processing literature. Now George Box and Gwilym Jenkins are and were respectively famous statisticians. So I would think that this definition could be considered part of the statistical literature in time series. – Michael R. Chernick May 15 '12 at 14:45
  • When you say they do it oppositely in the time series literature are you referring to specific journals like the British journal Time Series Analysis that Priestley started? I just have not seen it and I used to do a lot of time series analysis and some publishing in the field. – Michael R. Chernick May 15 '12 at 14:46
  • Michael: Box, Jenkins and Reinsel, for example, primarily use the form I provided which is the more common one in the statistical literature, though they briefly discuss both forms. Note that IrishStat's answer is at odds with this answer and is in-line with BJ&R, so there appears to be some confusion here. – cardinal May 15 '12 at 16:22
  • Maybe I am confused. I looked at his answer and I thought I saw him write that the AR process is stationary if the roots are inside the unit circle. I work with the 1970 and 1976 editions of Box and Jenkins. I also have the more recent version with Reinsel. I will take a look and see if it goes along with what you are saying. I very well could be wrong as it has been many years since I have looked seriously through the book and I have not looked closely at the edition with Reinsel. If there is any mor eto say on this I thinkit would be more appropriate to take it to discussion. – Michael R. Chernick May 15 '12 at 17:22
  • Michael: Maybe where IrishStat says *outside*, you read *inside*? I know my brain regularly plays this trick on me, especially for the words that I am *expecting* to see. I've never been able to convince it that that's a very mean trick to play on me. – cardinal May 15 '12 at 17:43
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    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. – Michael R. Chernick May 15 '12 at 18:47