1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.