# Cointegration where first differences are not jointly stationary

Note: This is a crosspost from this post on cross-validated, where it did not receive an answer. I thought I might have better luck here.

I am looking for a rigorous and general treatment of cointegration. Unfortunately, many of the econometrics textbooks and papers I have found in this area either place a lot of restrictions on the timeseries involved or tend to be sloppy (assuming the sum of stationary timeseries is stationary, for example).

Here is one question I am interested in, for example. Consider a set of timeseries $$x^i_t$$ such that, for each $$i$$, $$\Delta x^i_t$$ is strictly stationary. I do not assume, however, that the differenced timeseries are jointly stationary. Assemble these timeseries into a vector $$\mathbf{x}_t$$ and define a cointegration vector to be any vector $$\mathbf{v}$$ such that $$\mathbf{v}^{\top} \mathbf{x}_t$$ is strictly stationary.

The main question is, is the set of cointegration vectors closed under addition? If not, what is the minimal set of conditions one needs to ensure that this is true? Also, what is the situation when one considers weakly stationary timeseries rather than strictly stationary ones?

N.B. This question is based on the following question at mathoverflow, which has, as yet, only received my very non-expert interest.

• It seems $v_1^Tx_t$ and $v_2^Tx_t$ are jointly stationary, so their sum is also stationary. Hence it is closed under addition. – fesman Dec 1 '20 at 17:49
• Why are those quantities jointly stationary? – Aaron Bergman Dec 1 '20 at 17:56
• Hi Aaron: strict ( same as strong ) stationarity implies that the joint distribution of the series is time invariant. This is stronger than the notion of weak stationarity. As far as I know, cointegration assumes weak stationarity ( constant variance and constant mean ) because, in the gaussian case ( which is usually assumed ), weak implies strong. So, what is assumed is mean and variance constant over time. Then, this + gaussian assumption on noise implies strong. So,it's important to first define the type of stationarity one is referrring to and then more can be investigated after that. – mark leeds Dec 2 '20 at 2:36
• I am interested in the answer for either definition of stationarity. I believe you can have strongly stationary timeseries that are not weakly stationary, fwiw. Regardless, I think it is reasonable to consider the first differenced time series being separately stationary without being jointly stationary, and that is the basis of the question. – Aaron Bergman Dec 2 '20 at 2:40
• Hi Aaron: The term "joint stationarity" refers to one series, not multiple. Also, AFAIK, strong implies weak. Assuming, you are referring to weak stationarity in your question, I'll go back to your original question ( when I have more time ) and see if I can say something useful. – mark leeds Dec 2 '20 at 21:49