3
$\begingroup$

Under the GARCH($m$,$s$) model, it can be shown that $ E(\eta_t\eta_{t-j}) = E[(a_t^{2}-\sigma_t^{2})(a_{t-j}^{2}-\sigma_{t-j}^{2})] = 0 $.

In my proof attempt, I came across $ E(\epsilon_t^{2}\epsilon_{t-j}^{2}) = 1 $ where $ \epsilon_t \sim^{iid} (0,1) $.

Is it indeed true that if $ \epsilon_t \sim^{iid} (0,1) $, then $ E(\epsilon_t^{2}\epsilon_{t-j}^{2}) = 1 $?

$\endgroup$
3
  • 3
    $\begingroup$ This is trivial. Use independence of $\epsilon$s, then relationship between raw and central second moments. Done. $\endgroup$
    – Glen_b
    May 3, 2014 at 11:36
  • $\begingroup$ "This is trivial." --> Okay, ouch. But thanks! I'm going to assume that independence of $\epsilon_t$s implies independence of $\epsilon_t^{2}$s $\endgroup$
    – BCLC
    May 3, 2014 at 11:40
  • 1
    $\begingroup$ Yes, that's right. If the squares were dependent, there's a form of dependence among the unsquared values. If you aren't aware of that fact, then the result is not so obvious. $\endgroup$
    – Glen_b
    May 3, 2014 at 11:43

1 Answer 1

3
$\begingroup$

Since the OP seems to have a handle on it now, I want to make sure this question has an answer.

First:

$$E(\epsilon_t^{2}) = \text{Var}(\epsilon_t) + [E(\epsilon_t)]^2 = 1+ 0^2 = 1.$$

and

\begin{eqnarray} E(\epsilon_t^{2}\epsilon_{t-j}^{2}) &=& E(\epsilon_t^{2})\,E(\epsilon_{t-j}^{2})\quad\quad \text{(*)}\\ &=&1\times 1\\ &=&1 \end{eqnarray}


(*) Independence of $\epsilon_t$ and $\epsilon_{t-j}$ implies independence of $\epsilon_t^2$ and $\epsilon_{t-j}^2$

$\endgroup$
1
  • $\begingroup$ Note: "independence of $ϵ_t$s implies independence of $ϵ_t^{2}$s" because "If the squares were dependent, there's a form of dependence among the unsquared values." hehe $\endgroup$
    – BCLC
    May 3, 2014 at 11:56

Your Answer

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

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