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 $\mathbb{E}[\epsilon_t^4]$.

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*}
