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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.

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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 $Z_t = Y_t - \mu$.

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  • $\begingroup$ How did you get $E[Z_t \mid \mathbb{H}_{t-1} ] = E[ \varepsilon_t \mid \mathbb{H}_{t-1}]$? $\endgroup$
    – ketchup
    May 15 '18 at 16:55
  • $\begingroup$ $E[Z_t|\mathbb{H}_{t-1}] = E[\varepsilon_t (\omega + \alpha Z_{t-1}^2)^{1/2}|\mathbb{H}_{t-1}] = (\omega + \alpha Z_{t-1}^2)^{1/2} E[\varepsilon_t]$. $\endgroup$
    – nestor556
    May 15 '18 at 19:17

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