In the following vector autoregression model with lag polynomial representation:

$$\Phi (L) y_t= \epsilon_t$$

where $Y$ is the vector of endogenous variables, $\Phi$ is the parameters matrix, $\epsilon$ is the error term, and $L$ is the lag polynomial factor.

The basic assumption in the above model is that the residual follow a multivariate white noise, i.e.

$$E(\epsilon_t )=0$$

and $E(\epsilon_t \epsilon_s^{‚})$ equals either $0$ if $t \neq s$, or $\sum{\epsilon}$ if $t=s$.

My question is, what $t=s$ and $t \neq s$ really mean in the above condition and from where we get $s$?

  • $\begingroup$ Hi: $s$ is just a dummy variable used to denote a time instance other than $t$. So, in words, that condition is saying that, the noise terms of each equation are uncorrelated ( no covariance ) when the noise terms are at different time instances. When, the noise terms are at the same instant, then I think the sum should have $\epsilon^2$ rather than $\epsilon$ because the variance ( autocovariance at lag zero which is obviously non-zero since that is variance ) is the square of $\epsilon$. So, I'm pretty sure there's a typo in the summation. $\endgroup$
    – mark leeds
    Commented Sep 13, 2018 at 2:37
  • $\begingroup$ Also. note that the summation should also include the $t$ subscript on $\epsilon$ in order to keep things clear. ( even if $\epsilon$ is assumed to be independent of $t$ ). So, it should be written as $\sum \epsilon^2_{t}$. $\endgroup$
    – mark leeds
    Commented Sep 13, 2018 at 2:40
  • $\begingroup$ Hello, thank you for answering. What i meant by $\sum{\epsilon}$ is $\sum_{\epsilon}$ (it is a typo). So my guess it may indicate for existence of correlation in the residuals. $\endgroup$
    – Nord1
    Commented Sep 13, 2018 at 8:22
  • $\begingroup$ Hi: I don't think that's correct, unless I'm misunderstanding because, if you take the cross product of a vector $\epsilon$, your result is the sum of the squares of the elements in the vector. So, I think it should be the sum of the squared elements. $\endgroup$
    – mark leeds
    Commented Sep 14, 2018 at 9:21

1 Answer 1


I think the mistake is how to define $\ Y_t$. It is supposed to contain endogenous and exogenous variables. Hence, the multivariate white noise in the VAR analysis should full fill the following conditions:

$E(\epsilon_t )=0$ and $E(\epsilon_t \epsilon_s^{‚})$ equals either $0$ if $t \neq s$, or $\sum{\epsilon}$ if $t=s$.

In the case of $t=s$, this refer to multivariate covariance stationary condition for the endogenous variable. In the case of $t \neq s$, this refer to the exogenous condition (if the model have an exogenous variables). In this case, $E(\epsilon_t \epsilon_s^{‚})$ should equal zero (ie. $t \neq s$ refer to the existence of an exogenous variable).

reference: "Lecture note from Christopher F Baum"

  • $\begingroup$ I'm quite confident that you're correct but I'm clearly confused. Could you give a reference to a text or notes that explain it. Thanks. $\endgroup$
    – mark leeds
    Commented Sep 18, 2018 at 17:50
  • $\begingroup$ I added the reference above. $\endgroup$
    – Nord1
    Commented Sep 19, 2018 at 12:02

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