From what I understand, OLS gives consistent estimates for stationary AR(1) time series but not for unit-root ones. I am trying to illustrate this phenomenon with a small simulation in R but the OLS estimates in the unit-root case seem alright:

res <- c()

for (i in 1:1000) {
  ar <- c(0)
  for (j in 2:1000) ar[j] <- 1 * ar[j - 1] + rnorm(1)
  res[i] <- lm(ar[2:1000] ~ -1 + ar[1:999])$coef[1]


The above code gives results like 0.998191, 0.9980904, 0.998139, which are very close to the true coefficient of 1. In fact, the same simulation with a stationary AR(1) process often gives estimates that are more far-off. The only difference seems to be that the estimates in the unit-root case have a skewed distribution while the stationary ones do not.

What am I doing wrong? How can I show the inconsistency of the OLS estimator for unit-root AR(1) processes by simulation?

Let me clarify why I did think a unit-root AR(1) process could not be estimated by OLS: In Wooldridge’s Introductory Econometrics (2013), Section 11.3, it says: “The previous section shows that, provided the time series we use are weakly dependent, usual OLS procedures are valid under assumptions weaker than the classical linear model assumptions. [...] Using time series with strong dependence in regression analysis poses no problem, if the CLM assumptions in Chapter 10 hold. But the usual inference procedures are very susceptible to violation of these assumptions when the data are not weakly dependent, because then we cannot appeal to the law of large numbers and the central limit theorem.” Wooldridge then discusses the unit-root AR(1) model as an example of a strongly dependent time series. Although he does not explicitly claim that the model cannot be estimated by OLS, he also does not state that the CLM assumptions hold for the unit-root AR(1) model. I think this is very misleading…

  • $\begingroup$ Where did you read that the OLS estimator is inconsistent for non-stationary AR(1) ? And please explain what kind of non-stationarity you examine: for the mean? for the variance? A unit root?? $\endgroup$ Commented Mar 28, 2017 at 0:22
  • $\begingroup$ @AlecosPapadopoulos It's not stated explicitly but the code fragment reveals that here "non-stationary AR(1)" means $y_t = y_{t-1} + \varepsilon_t$ (i.e. yes, a unit root). $\endgroup$
    – Chris Haug
    Commented Mar 28, 2017 at 0:29
  • $\begingroup$ Thank you for your comments. I edited the question adding an explanation as to why I did think a unit-root AR(1) process could not be estimated by OLS. $\endgroup$
    – bbrot
    Commented Apr 3, 2017 at 15:11

2 Answers 2


You will not be able to show this result (by simulation or otherwise) because it does not hold. When the true AR parameter is unity, the OLS estimator is superconsistent, not inconsistent. See for example the discussion in Hamilton's Time Series Analysis, section "Asymptotic Properties of a First-Order Autoregression when the True Coefficient is Unity" (17.4).

What you can illustrate with simulation is this superconsistency. Simply repeat several Monte Carlo simulations of the OLS estimate (I did just 1000), similar to what was done above, but for a range of sample sizes. See plots below:

enter image description here enter image description here

What you should observe is that the sample bias of the OLS estimate gets closer to 0 faster in the nonstationary $\phi=1$ case (although it started out larger) and the variance shrinks to zero faster as well. OLS works just fine in that case.

Edit: In the text mentioned above, the following expression is derived in the case where the true coefficient is unity:

$$\sqrt{T}\left(\hat{\rho}_T-1\right) \to^P 0$$

Essentially, $\hat{\rho}_T$ (the OLS estimator) converges to 1 much faster than $\sqrt{T}$ goes to infinity. It's typical for consistent estimators to converge at a slower rate, so this behavior is termed "superconsistency".

  • $\begingroup$ (+1) Rightfully beat me to it. I would suggest to add the convergence in probability expression of Hamilton between eq. 17.4.8 and 17.4.9 (p. 488) so that it is clear what "superconsistent" means. $\endgroup$ Commented Mar 28, 2017 at 0:32
  • $\begingroup$ Sorry for the late reply, I have been on vacation for a couple of days. And thank you very much for this very helpful answer, especially for pointing me to the right section in Hamilton’s book. $\endgroup$
    – bbrot
    Commented Apr 3, 2017 at 14:34

The unit root issues in regression are usually associated with the presence of it in the dependent or independent variables. For instance, if you regress one variable on another and both of them have unit roots, then you'll likely to end up with a spurious regression.

Otherwise, the way the answers went on so war was on estimation of the autoregression coefficient $\phi$ in $x_t=c+\phi x_{t-1}+\varepsilon_t$, where the unit root is not an issue at all. It becomes an issue when you try to regress $y\sim x$, where $y$ has a unit root too.

To demonstrate the spurious regression you simply generate the pairs of independent processes $y,x$, regress $y\sim x$ and show how often the regression comes up significant. It should not be significant because the processes are not related, but you'll detect the slope where it shouldn't be.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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