The below is an ARCH(1) model
$$
Y_t = \mu + \sigma_t \varepsilon_t
$$
where the $\varepsilon_t$ are iid with $E[\varepsilon_t] = 0$, $E[\varepsilon_t^2] = 1$, and finite fourth moments. Additionally,
$$
\sigma_t^2 = \omega + \alpha (Y_{t-1} – \mu)^2
$$
I'm not sure how to derive the autocorrelation function
$$
\rho_X(k) = \frac{Cov(X_t, X_{t+k})}{\sqrt{V(X_t)V(X_{t+k})}}
$$
for $Y_t$ and for $\sigma_t^2 \varepsilon_t^2$.
I know for the autocorrelation for $Y_t$, the covariance between $Y_t$ and $Y_{t+k}$ becomes
$$E[\sigma_t \sigma_{t+k} \varepsilon_t \varepsilon_{t+k}]$$
but I get stuck here.
Best Answer
For an ARCH(1) process given by $Z_t = \varepsilon_t (\omega + \alpha Z_{t-1}^2)^{1/2}$, let $\mathbb{H}_{t}$ be the history up to time $t$, then using the law of total expectation:
$\text{cov}(Z_t, Z_{t-h}) = E [Z_t Z_{t-h}] = E [E[Z_t Z_{t-h}|\mathbb{H}_{t-1}]] = E [Z_{t-h} E[Z_t|\mathbb{H}_{t-1}]] = \\ E[Z_{t-h} (\omega + \alpha Z_{t-1}^2)^{1/2} E[\varepsilon_t|\mathbb{H}_{t-1}]] = E[\varepsilon_t|\mathbb{H}_{t-1}] E[Z_{t-h} (\omega + \alpha Z_{t-1}^2)^{1/2}] = 0$
I've also used the fact that $E[\varepsilon_t|\mathbb{H}_{t-1}] = E[\varepsilon_t] = 0$.
In your problem, you would have $Z_t = Y_t - \mu$.