Suppose I have a linear model with strongly correlated residuals. Suppose further that after adding one or more lags of the dependent variable, the residuals no longer appear to be autocorrelated under the usual tests (DW, etc.). I know that if the residuals in the autoregressive model remained correlated, OLS yields biased estimates. My question is, if the addition of an autoregressive term or terms cures the autocorrelation of the residuals, will OLS now yield unbiased results?

Also, I assume that if the estimates are unbiased, confidence intervals on coefficients and predictive intervals on forecasts can be constructed in the usual way. If this is not the case even though the estimator is unbiased, I would like a description of or a pointer to the literature on how these should be adjusted.

  • $\begingroup$ I know that if the residuals in the autoregressive model remained correlated, OLS yields biased estimates: more precisely, unbiased point estimates and biased standard errors. $\endgroup$ Nov 18, 2020 at 18:19
  • $\begingroup$ This should answer (at least part of) your question. $\endgroup$
    – Dayne
    Nov 19, 2020 at 4:48
  • $\begingroup$ @RichardHardy Thanks Richard! That will make my life considerably easier. I have to say though that this result runs counter to my intuitions as applied to the lagged dependent variable. The fact that my data fails a Durban-Watson when run without a lagged dependent variable but appears to have no autocorrelation in the residuals at all when that term is included made me think the coefficient on the lagged term and the correlation of the residuals were tangled together. $\endgroup$
    – andrewH
    Nov 20, 2020 at 6:44
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    $\begingroup$ @Dayne Thanks Dayne! There is a lot of meat in that posting (including, I notice, another post by Richard). I've printed the posts and downloaded the references and am reading them now. $\endgroup$
    – andrewH
    Nov 20, 2020 at 6:49
  • $\begingroup$ @andrewH, I think I made a mistake, very sorry about that! While you start by saying you have a linear model, you later say it is an AR model, and I missed that point. An AR model has biased coefficient estimates regardless of whether its errors are autocorrelated. However, I would not worry much about that as the bias vanishes asymptotically. Under uncorrelated residuals, confidence intervals in an AR model can be constructed in the usual way. Under correlated residuals, use autocorrelation-robust standard errors. $\endgroup$ Nov 20, 2020 at 6:49

1 Answer 1


I am going to take a shot at answering this, even though I don't really deserve the credit for it, because I found the answer, or at least, an answer under particular circumstances, so interesting. Also because I think it applies to the case I am looking at.

One of the more common causes of autocorrelated disturbances is model misspecification, where one has omitted some explanatory variable from the true model, or incorrectly specified the functional form, in such a way that the inclusion of the variable or the correction of the functional form corrects the misspecification and eliminates the autocorrelation. In the presence of autocorrelated residuals, OLS estimation of the coefficient on autoregressive term or terms is biased (but, as @RichardHardy observes above, still consistent). But absent autocorrelated disturbances, the AR portion of the model can be estimated by OLS without any special adjustment required.

It is not quite enough that a specification change makes the autocorrelation disappear. You still have to satisfy yourself that the addition of, e.g., an autoregressive term is not masking autocorrelation in the error process of the true model. Not sure how best to test this -- perhaps by nesting your model in a model with both autoregression and autocorrelation and seeing if an F test lets you drop one but not the other. That may not be the best possible test, but should be valid provided the models are properly nested. Otherwise, you might use a Cox test via AIC, which has weaker requirements for valid model comparison.


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