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.


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.

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