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

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 $E(\epsilon_t^{2}\epsilon_{t-j}^{2}) = 1$?

• This is trivial. Use independence of $\epsilon$s, then relationship between raw and central second moments. Done. May 3, 2014 at 11:36
• "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
– BCLC
May 3, 2014 at 11:40
• 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. May 3, 2014 at 11:43

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 $$\epsilon_{t-j}^2$$

• 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
– BCLC
May 3, 2014 at 11:56