What's a stationary VAR?

by Jase   Last Updated January 14, 2018 12:19 PM

  • What is a stationary VAR (vector autoregression)?
  • Can a VAR with non-stationary variables be stationary?
  • How do you test whether a VAR is stationary or non-stationary? (Example in R language if possible/applicable).


Answers 3


  • What is a stationary VAR?

I don't think the question is correct. VAR (Vector Autoregression) is an econometric technique used to model the relationship between time series variables. We cannot say that VAR is "stationary". You can have "stationary" time series, but not "stationary" VAR models. This is not correct to say! Anyway, a stationary time series variable is a variable which fluctuate around its mean (or its trend) over time. The series may deviate for a little while but it will definitely revert back to the mean or the trend later.

  • Can a VAR with non-stationary variables be stationary?

Again the question is not formulated correctly. But this is what you want to know. Non-stationary variables can have a stationary relationship. It means that they "move together" over time. We say that they are "cointegrated".

  • How do you test whether a VAR is stationary or non-stationary? (Example in R language if possible/applicable).

In order to test whether a variable is stationary, you can use a unit root test such as the Dickey-Fuller (DF) test. In R, the tseries package contains the adf.test function.

You may want to read this. It may be a little hard to digest, but there a few easy-to-understand illustrations that will help you understand.

user43899
user43899
April 16, 2014 15:51 PM

  1. VAR is actualy an equation. We say that process $X_t$ is VAR when it satisfies the following equation:

$$X_t=\alpha+\Phi_1X_{t-1}+...+\Phi_pX_{t-p}+\varepsilon_t,$$

where $\Phi_i$ are matrices and $\varepsilon_t$ is white noise process. If the process satistfying this equation is stationary we say that the VAR is stationary. Given matrices $\Phi_i$ you can test whether the solution is stationary or not. If the roots of the following equation are in modulo greater than 1, then the solution is stationary:

$$|I-\lambda\Phi_1-...-\lambda^p\Phi_p|=0,$$

where $|A|$ is the determinant of matrix $A$.

  1. Stationarity (both in weak and strong sense) is a property of a process. Whether it is a vector valued or scalar valued. For example the process is called stationary in weak sense, if it satisfies two conditions: $EX_t=c$ and $Cov(X_t,X_s)=r(t-s)$, where $c$ is a constant and $r$ is apropriate function. The definition is the same for vector and scalar processes. For vector processes it immediately implies that each individual element of the vector is stationary. Now if one of the elements is not stationary, then it is also immediately clear, that whole vector cannot be stationary too. The same reasoning applies for stationarity in strong sense. So the answer to the question would be no. You cannot get a stationary solution for VAR equation if one of the elements is not stationary.

  2. It is usual to test non-stationarity, or to be more precise unit-root non-stationarity for individual variables and then estimate VAR. Estimation assumes that you have either stationarity or cointegration. If the process is non-stationary and not cointegrated the estimation is not possible (it is possible to argue differently, but it is safe to assume that this holds for all the usual cases).

mpiktas
mpiktas
January 07, 2015 07:38 AM

When we use stationary variables in estimating VAR, the subsequent forecasting using such VAR model does not provide clear picture. The graphs that plot actual and forecast values of stationary variable generally show a straight line for forecast value and actual values trending around mean. The idea of whether variable is increasing or decreasing is not clear at all. How do we solve this issue

Harip
Harip
January 14, 2018 12:18 PM

Related Questions



Finding stationary values for an AR(2) process

Updated March 04, 2017 14:19 PM



Solve for inequality of AR model

Updated October 17, 2018 18:19 PM