# Fourth moment of ARCH(2)

I am studying the ARCH(2) process given by

$$X_t = \sqrt{h_t} \varepsilon_t$$

where

$$h_t = \alpha_0 + \alpha_1 X_{t-1} ^2 + \alpha_2 X_{t-2} ^2$$

and $\varepsilon_t$ follows $N(0,1)$. Assuming the process is stationary both in variance and the fourth moment, such that

$$E(X_t ^2) = E(X_{t-1}^2) = E(X_{t-2} ^2),$$

how can I work on the term $E( X_{t-1} ^2 X_{t-2} ^2)$ in the $E(X_t ^4)$ expression?

What I have done so far is

\begin{align*} E(X_t ^4 ) &= E (E(X_t ^4 | \mathcal{F}_{t-1}))= E(\varepsilon_t ^4 | \mathcal{F}_{t-1}) E [(\alpha_0 + \alpha_1 X_{t-1} ^2 + \alpha_2 X_{t-2} ^2 ) ^2]\\ &= E(\varepsilon_t ^4 | \mathcal{F}_{t-1}) E[(\alpha_0 ^2 + \alpha_1 ^2 X_{t-1} ^4 + \alpha_2 ^2 X_{t-2} ^4 + 2 \alpha_0 \alpha_1 X_{t-1} ^2 + 2 \alpha_0 \alpha_2 X_{t-2} ^2 \\ &+ 2 \alpha_1 \alpha_2 X_{t-1} ^2 X_{t-2} ^2 ] \\ &= E(\varepsilon_t ^4 | \mathcal{F}_{t-1}) (\alpha_0 ^2 + \alpha_1 ^2 E(X_{t-1} ^4 )+ \alpha_2 ^2 E(X_{t-2} ^4) + 2 \alpha_0 \alpha_1 E(X_{t-1} ^2) + 2 \alpha_0 \alpha_2 E(X_{t-2} ^2) \\ & + 2 \alpha_1 \alpha_2E( X_{t-1} ^2 X_{t-2} ^2)) \\ &= E(\varepsilon_t ^4 | \mathcal{F}_{t-1}) (\alpha_0 ^2 + \alpha_1 ^2 E(X_{t-1} ^4 )+ \alpha_2 ^2 E(X_{t-2} ^4)) \\ & \left(+ 2 \alpha_0 \alpha_1 \frac{\alpha_0}{1- \alpha_1 - \alpha_2 } + 2 \alpha_0 \alpha_2 \frac{\alpha_0}{1- \alpha_1 - \alpha_2 } + 2 \alpha_1 \alpha_2E( X_{t-1} ^2 X_{t-2} ^2)\right) \end{align*}

since $E(X_t ^2 ) = \frac{\alpha_0}{1- \alpha_1 - \alpha_2}$.

How do I treat the treat $E(X_{t-1} ^2 X_{t-2} ^2)$?

• Is $\epsilon_t$ independent to $X_t$ and also independent for different $t$? – Gordon Oct 30 '15 at 14:46
• @Gordon, yes it is, if I got your question right! – KaRJ XEN Oct 31 '15 at 13:35