1
$\begingroup$

I noticed that sometimes the moving average process is written in two different ways. I would like to know which one is correct.

$y_t = \epsilon_t + \theta \epsilon_{t-1}$

$y_t = \epsilon_t - \theta \epsilon_{t-1}$

$\endgroup$
  • $\begingroup$ This is a matter of notational conventions. You can see immediately that the model is the same for $\tilde\theta=-\theta$. $\endgroup$ – Richard Hardy Jan 8 '16 at 16:35
1
$\begingroup$

Both are correct, where the first is, I would say, more common. They are equivalent because you can always transform $\theta$ to $-\theta$ and get the same process, with the same dynamics.

The variance is \begin{eqnarray} E(Y_t-\mu)^2&=&E(\epsilon_t+\theta\epsilon_{t-1})^2\notag\\ &=&E(\epsilon_t^2+2\epsilon_t\theta\epsilon_{t-1}+\theta^2\epsilon_{t-1}^2)\notag\\ &=&\sigma^2+0+\theta^2\sigma^2 \end{eqnarray} and the first autocovariance is \begin{eqnarray*} E(Y_t-\mu)(Y_{t-1}-\mu)&=&E(\epsilon_t+\theta\epsilon_{t-1})(\epsilon_{t-1}+\theta\epsilon_{t-2})\\ &=&E(\epsilon_t\epsilon_{t-1}+\theta\epsilon_{t-1}^2+\theta\epsilon_t\epsilon_{t-2}+\theta^2\epsilon_{t-1}\epsilon_{t-2})\\ &=&0+\theta\sigma^2+0+0, \end{eqnarray*} whether you define $\epsilon_t+\theta\epsilon_{t-1}$ or $\epsilon_t-(-\theta)\epsilon_{t-1}$.

$\endgroup$
  • $\begingroup$ You are of course correct, but spelling out the variance and autocovariance equations may be more of a distraction rather than a useful illustration, IMHO. $\endgroup$ – Richard Hardy Jan 8 '16 at 16:37
  • $\begingroup$ That is possible indeed. Maybe OP will weigh in. $\endgroup$ – Christoph Hanck Jan 8 '16 at 16:38

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.