I am confused while trying to find a general expression for the mean and variance of a stationary VAR model. I am trying to do it for VAR(1). I also can't find it in the literature. Can anyone help me?

  • $\begingroup$ I was going through my old answers and noticed this one was not accepted. Do you perhaps need further clarification? $\endgroup$ Commented Feb 26, 2017 at 12:17

2 Answers 2


Taking the variance of both sides of the equation $$ y_t = \nu + A_1 y_{t-1} + u_t $$ leads to $$ \operatorname{Var}y_t = A_1\operatorname{Var}y_{t-1}A_1^T+\Sigma_u. $$ Stationary implies that $\operatorname{Var}y_t =\operatorname{Var}y_{t-1}=\Gamma_0$ so you need to solve the matrix equation $$ \Gamma_0 = A_1\Gamma_0 A_1^T+\Sigma_u. $$ Applying the vec-function, this can be rewritten (see wikipedia) as $$ \operatorname{vec}\Gamma_0 = (A_1\otimes A_1) \operatorname{vec}\Gamma_0 + \operatorname{vec}\Sigma_u $$ and solved using standard methods for the unknown covariances given by $$ \operatorname{vec}\Gamma_0 = (I-A_1\otimes A_1)^{-1} \operatorname{vec}\Sigma_u. $$ So you don't need to work out the infinite sum from the MA$(\infty)$-representation.

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    $\begingroup$ If I understand correctly, the result is the same but your representation is more convenient as it readily yields an empirically feasible solution (unlike the infinite sum). $\endgroup$ Commented Dec 15, 2016 at 10:19
  • $\begingroup$ Yes, it's just two different ways of expressing the same covariance matrix. $\endgroup$ Commented Dec 15, 2016 at 10:21
  • $\begingroup$ Thanks for the answer. Quite what I was looking for. Not sure why the answer is not accepted. It should be. $\endgroup$
    – Xbel
    Commented Jul 27, 2021 at 7:17

According to Lütkepohl (2005), p. 14-15, if we have a $K$-variate VAR(1) process of the form $$ y_t = \nu + A_1 y_{t-1} + u_t, $$ then the unconditional mean is $$ (I_K-A_1)^{-1}\nu $$ (where $I_K$ is an identity matrix of dimension $K\times K$) and the unconditional covariance for lag $h$ (i.e. $\text{Cov}(y_t,y_{t-h})$) is $$ \sum_{i=0}^\infty A_1^{h+i}\Sigma_u {A_1^i}' $$ where $\Sigma_u$ is the covariance matrix of the error term $u_t$. Then the unconditional variance can be obtained by taking $h=0$ in the above expression.

The same applies to VAR($p$) after having expressed the process in its alternative $Kp$-dimensional VAR(1) representation.

These results are obtained using the vector moving-average (VMA) representation of the VAR(1) process.


  • Lütkepohl, Helmut. New Introduction to Multiple Time Series Analysis. Springer Science & Business Media, 2005.
  • $\begingroup$ Is there a way to derive the unconditional variance? $\endgroup$ Commented Jan 28, 2017 at 9:22
  • $\begingroup$ @Bonsaibubble, the answer by Jarle Tufto does that. $\endgroup$ Commented Jan 28, 2017 at 9:51
  • $\begingroup$ But how does this correspond to what Lütkepohl defines? $\endgroup$ Commented Jan 28, 2017 at 10:07
  • $\begingroup$ @Bonsaibubble, both answers agree on the substance (they do not imply different things), they just use different approach and notation. $\endgroup$ Commented Jan 28, 2017 at 10:09
  • $\begingroup$ Ok, I dont understand Tufto's approach then $\endgroup$ Commented Jan 28, 2017 at 10:13

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