关于协整序列残差平稳性的技术问询：线性组合是否平稳？

Let's break down your two questions about cointegration with straightforward, practical explanations:

Question 1: Do two cointegrated sequences imply their residual is stationary? #

Short answer: Yes, but only when referring to the residual from the true cointegrating relationship.

Cointegration’s core definition hinges on this: If two time series (X_t) and (Y_t) are both integrated of order 1 (I(1)—non-stationary on their own, but stationary after first differencing), they’re cointegrated if there exists a non-trivial linear combination of them that is integrated of order 0 (I(0), i.e., stationary).

When we estimate a valid cointegrating regression like (Y_t = \alpha + \beta X_t + \epsilon_t), the residual (\epsilon_t = Y_t - \alpha - \beta X_t) is exactly that stationary linear combination—assuming we’ve identified the correct cointegrating coefficient (\beta).

A quick caveat: If you run a regression with arbitrary (incorrect) coefficients, the resulting "residual" won’t be stationary. But in the context of cointegration, we’re always talking about the residual tied to the true underlying stationary combination.

Question 2: If X and Y are cointegrated, is (Y - \beta X) always stationary? #

Let’s tie this directly to your regression setup: (Y = \alpha + \beta X + \epsilon).

From the definition of cointegration, we know the residual (\epsilon = Y - \alpha - \beta X) is I(0). Rearranging gives (Y - \beta X = \alpha + \epsilon).

Adding a constant ((\alpha)) to a stationary sequence doesn’t break its stationarity—stationarity only requires mean, variance, and autocovariances to be constant over time, and a fixed constant merely shifts the mean without violating those properties.

So yes, (Y - \beta X) will always be stationary here. The constant (\alpha) just adds a fixed offset to the stationary residual (\epsilon), leaving the entire combination as I(0).

内容的提问来源于stack exchange，提问作者Federico Caccia