2
$\begingroup$

In the Multiplicative Error Model (MEM) specification by Engle in "New Frontiers for ARCH models", he wrote that MEM specified an error that is multipled times the mean and the specification is:

$$ x_t = \mu_t \epsilon_t $$

$$ \epsilon_t | \Im_{t-1} \sim D(1,\phi_t^2) $$

I am not sure I understand what epsilon is supposed to be, is it a normal of mean 1, but then what is the variance supposed to be ?

Engle mentions that the MEM is good to model the process which have to be positive. From my understanding this is only true if $\epsilon_t$ is positive, so unless $D$ is a log-normal distribution or another distribution guaranteeing positivity, you cannot guarantee that $x_t$ itself will be positive, is my understanding correct ?

$\endgroup$
5
  • 1
    $\begingroup$ what is $\Im_{t-1}$ and what does $D(1,\phi^{2}_{t})$ denote? $\endgroup$
    – Macro
    Jun 25, 2012 at 23:19
  • $\begingroup$ He doesn't know. $\endgroup$ Jun 26, 2012 at 1:12
  • $\begingroup$ Engle wrote et distributed D(0,σt$^2$) where D is some general disturbance distribution for an additive error and for a multiplicative model D(1.φt$^2$). $\endgroup$ Jun 26, 2012 at 1:19
  • $\begingroup$ Engle states that the variance of the error term is proportional to the square of the mean. $\endgroup$ Jun 26, 2012 at 1:27
  • $\begingroup$ @Michael Chernick: thank you, the reason I am asking is that Engle says that MEM is a good model to represent entities which are always positive, I am not sure why since unless you use a lognormal $\epsilon_t$ and $x_t$ can be negative ? (by the way how can I give you points for your comment) $\endgroup$
    – BlueTrin
    Jun 26, 2012 at 8:03

1 Answer 1

0
$\begingroup$

For a multiplicative error Engle picks a disturbance distribution for a multiplicative error term that is nonnegative with mean $1$. There are many distributions that are nonnegative. The lognormal is just one of them. The gamma distribution including the $\chi^2$ is another.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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