# Interpret $R^2$ for a long-run equilibrium model (2 stage OLS)

I've built an error correction model using two stage OLS - first an OLS on the cointegrated I(1) variables in levels to get the cointegration coefficients, and then an ARDL in differences with the lagged ECT using the residuals of the cointegration model and lagged differences of the predictor variables.

The long-run OLS model has an $$R^2$$ of .97 which seems pretty optimistically high but actually makes sense theoretically. However, the short run model in differences with the lagged ECT and lagged differences of the target variable have an $$R^2$$ of about .33.

Is it typical to interpret the fit metrics for both long and short run models in this kind of process and if so, does the $$R^2$$ have the typical interpretation? Do I need to apply any sort of adjustment to the $$R^2$$ in either model for it to be unbiased?

• Regarding the close vote, what about this post is off topic? I find it pretty much on topic. Mar 14 at 17:38

## 1 Answer

I do not know how typical it is to focus on $$R^2$$ values in cointegrating regressions and error correction model in general; this must depend on the application.
I doubt the first-stage $$R^2$$ would be of much interest in most applications. For cointegrated series, it will usually be close to 1, so there is not much nuance to look for there.
The second-stage $$R^2$$ is more interesting. If you care about the model's goodness of fit for first-differenced data, it is a sensible (even if biased) metric.

Regarding unbiasedness, $$R^2$$ is upward biased. This is true not only for time series (cointegrated or otherwise) but also for a simple cross section. $$R^2_{adjusted}$$ is sometimes used and is unbiased under $$H_0$$ that the slope coefficients are jointly zero, so the true model only contains a constant. However, this is not a very interesting hypothesis, so I am not sure if you should prefer $$R^2_{adjusted}$$ to the vanilla $$R^2$$. Also, the bias disappears asymptotically, so unless your sample is fairly short, your $$R^2$$ should be fine.

• Thank you for the insight! Mar 15 at 14:13
• @Jared, you are welcome! Mar 15 at 14:50