Superconsistency Of Cointegrating Regression

Why Cointegrating Regressions Behave Differently

In a regression with stationary variables, estimation typically learns about a slope by accumulating information at the square-root-of-sample-size rate. A cointegrating regression has a different structure: its regressor is integrated, so its level wanders over a range that grows with the sample. The resulting variation in the regressor can dominate a stationary regression error and make the slope estimate converge faster.

This result is called superconsistency. It is a statement about the rate at which the estimated long-run coefficient approaches its population value, not a claim that every finite-sample estimate is accurate. It also does not require the regressor and regression error to be independent. Endogeneity can affect the limiting distribution and finite-sample behavior, even though the coefficient remains consistent under the conditions below.

Model and Assumptions

Consider the cointegrating regression

Here \(y_t\) and \(x_t\) are integrated of order one, while \(u_t\) is stationary. Thus, the linear combination \(y_t-\alpha-\beta x_t=u_t\) is stationary even though the two series being related are not. This is the defining feature of a cointegrating relation. For the rate derivation, use the following regularity conditions:

• \(x_t\) has no deterministic drift and can be written as \(x_t=x_{t-1}+v_t\), where \(v_t\) is a stationary, short-memory process with positive long-run variance.
• \(u_t\) is stationary, has finite moments, and is sufficiently weakly dependent for its partial sums to grow at order \(\sqrt n\).
• The joint partial sums of \(v_t\) and \(u_t\) obey a functional central limit theorem. They may be correlated, including contemporaneously.
• The regression includes an intercept, and the integrated component of \(x_t\) is nondegenerate.

“Short-memory” here rules out additional unit-root behavior and ensures that partial sums have the usual square-root scale. The joint limit condition allows dependence between \(v_t\) and \(u_t\); independence is not required. The OLS slope with an intercept is

Why the Denominator Grows Like \(n^2\)

The increments of an integrated regressor have stationary-scale variation, but its levels accumulate those increments. For a zero-drift random walk, a typical \(x_t\) is of order \(\sqrt n\) when \(t\) is near \(n\). Consequently, the squared deviations in the OLS denominator are typically of order \(n\) each, and summing \(n\) such terms gives order \(n^2\).

The functional central limit theorem makes that argument precise. Let \(W(r)\) denote the Brownian-motion limit of the scaled partial sums of \(v_t\), and let \(\sigma_v^2>0\) be the long-run variance of \(v_t\). With an intercept, centering the regressor corresponds in the limit to centering the Brownian path by its average. Under the stated conditions,

The random integral is positive with probability one: a Brownian path is not constant over the interval. Thus the denominator, divided by \(n^2\), converges to a finite positive random quantity. In particular, the denominator is of order \(n^2\), not \(n\).

Why the Numerator Grows Like \(n\)

Substitute the regression model into the OLS formula. Since \(y_t-\bar{y}=\beta(x_t-\bar{x})+(u_t-\bar{u})\), the intercept disappears after centering, leaving the exact decomposition

The numerator is of order \(n\) under the joint weak-dependence conditions. One way to see the scale is to write \(U_t=\sum_{j=1}^t(u_j-E[u_j])\). Stationarity and the partial-sum limit imply that \(U_t\) is typically of order \(\sqrt n\) when \(t\) is of order \(n\); the integrated regressor is also of order \(\sqrt n\). Summing their increments against the regressor therefore produces terms on the \(n\) scale. A joint functional limit justifies this order even when increments and errors are correlated. Centering by sample means does not increase the order.

This argument is an order argument, not a claim that the numerator has a particular expectation or equals a fixed constant times \(n\). Its scaled limit can depend on the joint dependence between \(v_t\) and \(u_t\). Combining an \(O_p(n)\) numerator with an \(O_p(n^2)\) denominator yields the result.

Worked Examples

Why Superconsistency Matters—and What It Does Not Say

The contrast with a stationary regression is useful. With a stationary regressor, the centered sum of squares is ordinarily order \(n\). If the stationary error is weakly dependent, the cross-product with the regressor is typically order \(\sqrt n\), giving a slope error of order \(1/\sqrt n\). In the cointegrating regression, the integrated regressor instead gives a denominator of order \(n^2\) and a cross-product of order \(n\). This difference is the source of the faster rate.

Superconsistency helps explain why a long-run coefficient can be estimated consistently despite endogeneity between the regressor’s innovations and the stationary error. It does not mean that endogeneity is irrelevant. It can affect the limiting distribution, and it does not justify using ordinary stationary-regression standard errors or reference distributions without further analysis. Nor does consistency alone show that \(x_t\) and \(y_t\) are cointegrated: the stationary-error assumption is part of the model, not a consequence of obtaining a high \(R^2\) or a plausible-looking slope.

Common Pitfall

A common mistake is to reason that correlation between \(v_t\) and \(u_t\) must make OLS inconsistent, just as endogeneity can cause inconsistency in a stationary regression. That conclusion ignores the different orders in the cointegrating regression. Under the conditions given here, the dependence can change the order-\(n\) numerator’s limiting behavior, but the denominator remains order \(n^2\). Their ratio still converges to zero. This correction is conditional: if the regression error is not stationary, or if the integrated regressor is degenerate or the regularity conditions fail, this argument does not establish superconsistency.