Fourth moment of arch(1) process I have an ARCH(1) process
\begin{align*}
Y_t &= \sigma_t \epsilon_t, \\
\sigma_t^2 &= \omega + \alpha Y_{t-1}^2,
\end{align*}
and I am trying to express the fourth moment $\mathbb{E}[Y_t^4]$ in terms of $\omega$, $\alpha$ and $\mathbb{E}[\epsilon_t^4]$.
 A: For \begin{align*}
Y_t = \sigma_t \epsilon_t, \qquad \sigma^2_t = \omega + \alpha Y^2_{t-1}, \qquad \omega>0, \alpha \geq 0,
\end{align*}
we assume $\sigma_t$ and $\epsilon_t$ to be independent. I also assume standard normality for $\epsilon_t$, so that $E(\epsilon_t^4)=3$. (You will see from the proof what needs to happen for convergence when the fourth moment is different.)
Consider a recursion for the 4th moment.
\begin{align*}
E[Y^4_t] &= E[\sigma^4_t \epsilon^4_t] = E[\sigma^4_t] E[\epsilon^4_t] \\
&= 3 E[\sigma^4_t] = 3 E[(w + \alpha Y^2_{t-1})^2] \\
&= 3 E[\omega^2 + 2\omega \alpha Y^2_{t-1} + \alpha^2 Y^4_{t-1}] \\
&= 3 \omega^2 + 6 \omega \alpha E[Y^2_{t-1}] + 3 \alpha^2 E[Y^4_{t-1}] \\
&= \underbrace{ 3 \omega^2 + \frac{6 \omega^2 \alpha}{1 - \alpha}}_{=:c} + 3 \alpha^2 E[Y^4_{t-1}], \\
\end{align*}
where the last line uses results for the variance of an ARCH(1)-process.
Repeated substitution yields
\begin{align*}
E[Y^4_t] &= c + 3 \alpha^2 E[Y^4_{t-1}] \\
&= c + 3 \alpha^2 (c + 3 \alpha^2 E[Y^4_{t-2}]) \\
&= c + 3 \alpha^2 c + (3 \alpha^2)^2 E[Y^4_{t-2}] \\
&= c + 3 \alpha^2c + (3 \alpha^2)^2 (c + 3 \alpha^2 E[Y^4_{t-3}]) \\
&= c + 3 \alpha^2 c + (3 \alpha^2)^2 c + (3 \alpha^2)^3 E[Y^4_{t-3}]\\
& \qquad \qquad \qquad \qquad \vdots \\
&= c \sum^n_{i=0} (3 \alpha^2)^i + (3 \alpha^2)^{n+1} E[Y^4_{t-(n+1)}] \\
\end{align*}
For $E[Y^4_t]$ to be finite we hence need $3 \alpha^2 < 1$. In this case, we obtain
\begin{align*}
E[Y^4_t] &= c \sum^\infty_{i=0} (3 \alpha^2)^i \quad\overset{x:=3 \alpha^2}{=} c \sum^\infty_{i=0} x^i= \frac{c}{1 - x} \\
&= \frac{c}{1 - 3 \alpha^2} \\
& = \frac{3 w^2 (1 + \alpha)}{(1 - \alpha) (1 - 3 \alpha^2)}. \\
\end{align*}
A: I am not entirely comfortable with having the fourth moments of the process on both sides of the equation if we are not sure that they are finite. I propose an alternative derivation where we do not have the fourth moments on both sides of the equation.
Assume that $0<\alpha<1$ and that $\epsilon_t$'s are iid $N(0,1)$ random variables so that $\operatorname E\epsilon_0^4=3$. Then the square of the unique stationary causal solution of the ARCH equations is given by
$$
Y_t^2=\omega\sum_{j=0}^\infty\alpha^j\epsilon_t^2\epsilon_{t-1}^2\cdots\epsilon_{t-j}^2
$$
for $t\in\mathbb Z$ (see, for example, Equation (7.2.3) of Brockwell and Davis (2016)). We have that
\begin{align}
 &\operatorname EY_t^4=\\
 &=\omega^2\operatorname E\Bigl|\,\sum_{j=0}^\infty\alpha^j\epsilon_t^2\epsilon_{t-1}^2\cdots\epsilon_{t-j}^2\Bigr|^2\\
 &=\omega^2\lim_{n\to\infty}\operatorname E\Bigl|\,\sum_{j=0}^n\alpha^j\epsilon_t^2\epsilon_{t-1}^2\cdots\epsilon_{t-j}^2\Bigr|^2\\
 &=\omega^2\lim_{n\to\infty}\operatorname E\Bigl[\,\sum_{j=0}^n\alpha^{2j}\epsilon_t^4\epsilon_{t-1}^4\cdots\epsilon_{t-j}^4+2\sum_{j=0}^{n-1}\sum_{k=j+1}^n\alpha^j\epsilon_t^2\epsilon_{t-1}^2\cdots\epsilon_{t-j}^2\alpha^k\epsilon_t^2\epsilon_{t-1}^2\cdots\epsilon_{t-k}^2\Bigr]\\
 &=\omega^2\lim_{n\to\infty}\Bigl[\,3\sum_{j=0}^n(3\alpha^2)^j
 +6\sum_{j=0}^{n-1}(3\alpha)^j\sum_{k=j+1}^n\alpha^k\Bigr]\\
 &=3\omega^2\lim_{n\to\infty}\Bigl[\,\sum_{j=0}^n(3\alpha^2)^j
 +2\sum_{j=0}^{n-1}(3\alpha)^j\Bigl[\frac{\alpha^{j+1}-\alpha^{n+1}}{1-\alpha}\Bigr]\Bigr]\\
 &=\label{limit}\tag{#}3\omega^2\lim_{n\to\infty}\Bigl[\,\sum_{j=0}^n(3\alpha^2)^j
 +\frac{2\alpha}{1-\alpha}\Bigl\{\,\sum_{j=0}^{n-1}(3\alpha^2)^j-\alpha^n\sum_{j=0}^{n-1}(3\alpha)^j\Bigr\}\Bigr].
\end{align}
We can use the monotone convergence theorem to justify the interchange of the limit and the expected value. Limit \eqref{limit} is finite if and only if $0<\alpha<3^{-1/2}$ and if it is finite, it is equal to
$$
\operatorname EY_t^4
=\frac{3\omega^2(1+\alpha)}{(1-\alpha)(1-3\alpha^2)}.
$$
I hope this is useful.
