I have a reduced form VAR

$$\begin{equation} y_t = c_o + A y_{t-1} + \epsilon_t \end{equation}$$

Where, $y_t \in \mathbb{R}^2$, $A$ is a $2$X$2$ matrix and

$$\begin{equation} E(\epsilon_t \epsilon_\tau ')=\left\{ \begin{array}{@{}ll@{}} \Omega, & \text{if}\ t=\tau \\ 0, & \text{otherwise} \end{array}\right. \end{equation} $$

$\Omega$ is not necessarily digonal. I have to show that I can use Least squares method. But, for this, I have to show that the errors $\epsilon_t$ are not correlated with the regressor $y_{t-1}$, i.e.

$$E(y_{t-1} \epsilon_t' ) = 0$$

I tried to use the Wald decomposition:

$$y_{t-1} = \mu + \sum_{i=0}^{\infty} A^{i} \epsilon_{t-1 - i} $$ But

$$E[ y_{t-1} \epsilon_t'] = E[\mu \epsilon_t'] + \sum_{i=0}^{\infty} A^{i} E[\epsilon_{t-1 - i} \epsilon_t'] =^{i =1} A \Omega$$

With this, I can not reach my goal. Some ideias?

  • $\begingroup$ What does $=^{i=1}$ mean in your last equation? $\endgroup$ – Richard Hardy Dec 9 '18 at 14:41
  • $\begingroup$ It was a mistake. I thought when i = 1, I could equal the indexes so I have $t-1-i = t$. In fact, it is true if $t+1-i = t$. Sorry $\endgroup$ – orrillo Dec 10 '18 at 4:44
  • $\begingroup$ If that is a mistake, could you fix it? $\endgroup$ – Richard Hardy Dec 10 '18 at 6:13
  • $\begingroup$ I can do this? The problem is trivial without this error. What should I do? Delete the question or make an update? Adding a note pointing out the mistake and the response given by the user below may be more interesting. What do you think of that? $\endgroup$ – orrillo Dec 10 '18 at 6:40

HIL you didn't write it down explicitly, but I'll assume $E(\epsilon_{t}) = 0$.

So, you're basically there already.

In your last equation, the first term is zero because $\mu \times E(\epsilon_{t}) = \mu \times 0 = 0 $.

For the second term, consider the expectation inside the sum. It's always taking expectations of epsilon_t and epsilon's which have subcripts that are earlier than t. But, by definition, the covariance $\Omega$ is only non-zero when the time subscripts of the $\epsilon$ are the same. Therefore, the second term is a summation of zero's so it's zero also. So both terms in your last expression are zero. I hope this helped but you did it all.

  • $\begingroup$ Sorry, you're right. I forgot to put the hypotheses. This is true, I made a mistake thinking that when it was i = 1, I would have $E[\epsilon_t \epsilon_t']$ , but in fact, this never happens. $\endgroup$ – orrillo Dec 7 '18 at 7:24
  • 1
    $\begingroup$ glad it helped. $\endgroup$ – mlofton Dec 8 '18 at 8:14

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