I have the following time series and need to show that the ACF is zero except at lag one.

$$X_t=\frac{0.8\epsilon_{t-1}^2}{1+\epsilon_{t-1}^2}+ \epsilon_t, \text{ and that } \{ \epsilon_t\} \sim_{i.i.d} N(0,\sigma^2)$$ My guess is that I need to calculate the covariance for the lags and show they don't share the same components and such are not correlated since I have i.i.d. variables. However I'm not reaching the desired result. Any insights?


Yes. That is exactly the idea. Using the independence assumption it is easy to see that $Cov(X_t,X_{t-h})$ for $h>1$ is zero.

$$Cov(X_t,X_{t-1}) = Cov(0.8 \varepsilon_{t-1}^2/(1+\varepsilon_{t-1}^2),0.8 \varepsilon_{t-2}^2/(1+\varepsilon_{t-2}^2)) + Cov(0.8 \varepsilon_{t-1}^2/(1+\varepsilon_{t-1}^2),\varepsilon_{t-1}) + \\Cov(\varepsilon_t,0.8 \varepsilon_{t-2}^2/(1+\varepsilon_{t-2}^2)) + Cov(\varepsilon_t,\varepsilon_{t-1})$$

Then using that $\varepsilon_t$ is iid, we obtain

\begin{align} Cov(X_t,X_{t-1}) &= Cov(0.8 \varepsilon_{t-1}^2/(1+\varepsilon_{t-1} ^2),\varepsilon_{t-1}) \\ &= E \bigg[\frac{0.8 \varepsilon_{t-1}^2}{(1+\varepsilon_{t-1}^2)}\varepsilon_{t-1} \bigg] + E[0.8 \varepsilon_{t-1}^2/(1+\varepsilon_{t-1}^2)]E[\varepsilon_{t-1} ] \\ &=E[0.8 \varepsilon_{t-1}^3/(1+\varepsilon_{t-1}^2) ] \end{align} Then you should only evaluate the expectation.


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