Mean and Correlation of a First-Order ARCH(1) Process For a first-order ARCH(1) process 
$$
Y_t = \epsilon_t(\alpha_0 + \alpha_1Y_{t-1}^2)^{1/2}
$$
$$
t \in \mathbb{Z}
$$
$$
\alpha_0, \alpha_1 > 0
$$
$ \{\epsilon_t\}_{t \in \mathbb{Z}} $ and $Y_t$ is independent of $\epsilon_{t+i}$ for all $i \in \mathbb{Z}$. I need to show that the mean of this process is 0 and the correlation is that of white noise. I can already see that the mean of the process can be found as follows:
$$
E[Y_t]=E[\epsilon_t(\alpha_0 + \alpha_1Y_{t-1}^2)^{1/2}]=E[\epsilon_t]E[(\alpha_0 + \alpha_1Y_{t-1}^2)^{1/2}]
$$
$$
=0 \cdot E[(\alpha_0 + \alpha_1Y_{t-1}^2)^{1/2}]=0
$$
However, I am not sure how to go about determining the correlation. I know that for a process the correlation function is
$$
\rho(s, t) = \frac{\gamma(s,t)}{\{\gamma(s,s)\gamma(t,t)\}^{1/2}}
$$
I assume that there must be some what that only $\epsilon_t$ remains when calculating the correlation.

I also need to show that it is necessary for $ \alpha_1 <1 $ and that $var(Y_t)=\frac{\alpha_0}{1-\alpha_1}$ in order to make the process stationary, but I'm hoping that follows from the correlation above.
 A: The way to show that the correlation is zero: 
$$
C(Y_t,Y_{t-1}) = E[\varepsilon_t (\alpha_0 + \alpha_1 Y_{t-1}^2)^{1/2}\varepsilon_{t-1} (\alpha_0 + \alpha_1 Y_{t-2}^2)^{1/2}] - E[\varepsilon_t (\alpha_0 + \alpha_1 Y_{t-1}^2)^{1/2}]E[\varepsilon_{t-1} (\alpha_0 + \alpha_1 Y_{t-2}^2)^{1/2}]
$$
You have shown that the last product must be equal since the expected value of each component is equal to zero. Thus, 
$$
\begin{align*}
C(Y_t,Y_{t-1}) =& E[\varepsilon_t (\alpha_0 + \alpha_1 Y_{t-1}^2)^{1/2}\varepsilon_{t-1} (\alpha_0 + \alpha_1 Y_{t-2}^2)^{1/2}] \\
=& E[\varepsilon_t\varepsilon_{t-1} ] E[(\alpha_0 + \alpha_1 Y_{t-1}^2)^{1/2}(\alpha_0 + \alpha_1 Y_{t-2}^2)^{1/2}] \\
=& E[\varepsilon_t]E[\varepsilon_{t-1} ] E[(\alpha_0 + \alpha_1 Y_{t-1}^2)^{1/2}(\alpha_0 + \alpha_1 Y_{t-2}^2)^{1/2}] \\
=& 0 
\end{align*}
$$
where we have used the independence and zero mean assumption for the innovations. 
Assuming that the process is stationary, we will have 
$$
\begin{align*}
E[Y^2_t]=& \alpha_0 + \alpha_1 E[Y^2_{t-1}] \\
 =& \alpha_0 + \alpha_1 E[Y^2_t]
\end{align*}
$$
Thus, rewriting yields
$$
E[Y^2_t] = \frac{\alpha_0}{1-\alpha_1}
$$
The variance will be negative if $\alpha_1 > 1$, why it is not allowed. 
