# Why if in OLS the autocorrelation between residuals is positive, it will lead to inflated t-stats?

Why if in OLS the autocorrelation between residuals is positive, it will lead to inflated t-stats for the coefficients? I've seen that statement as a given truth, but what is the intuition/proof between that result? Does it mean that if the autocorrelation between residuals is negative, it will lead to deflated t-stats in the regression?

## 1 Answer

Let me try to illustrate with the most simple regression example, that of a regression on a constant. The idea is similar in general regressions.

We then know that the OLS estimator $$\hat\beta=\bar{y}$$, the sample mean. The t-statistic is $$t=\frac{\sqrt{n}\bar{y}}{\sqrt{Var(\bar{y})}}$$ In the denominator of a t-statistic, we therefore require an estimator of $$Var(\bar{y})$$, related to the so-called long-run variance of $$Y_i$$.

As, for example, this answer demonstrates, this means the variance of a single $$Y_i$$, call it $$\sigma^2$$, plus two times autocovariances.

When we have positive serial correlation, these autocovariances will generally be positive, meaning that a good estimator (in the sense of yielding t-ratios that, asymptotically at least, behave like $$N(0,1)$$ random variables if the null is true) of $$Var(\bar{y})$$ will be larger than the default estimator that takes the average of the squared residuals and thus only estimates $$\sigma^2$$.

Put differently, if we nevertheless use the default estimator which is inappropriate under serial correlation, t-ratios will on average be too large ("inflated"), meaning tests reject too often.

For example, one can show that the long-run variance of an AR(1) process is $$\sigma^2/(1-\phi)^2$$ We see that this is larger than $$\sigma^2$$ when $$\phi>0$$, but smaller if $$\phi<0$$, so that "deflated" t-statistics are indeed also possible.