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Is the time series $\{Yt\}$ given by $Y_{t} = Z_{t} - \frac{1}{2}Z_{t-1}Z_{t-2}$

With $Z_{t} \sim{N}(0\,1)$, weakly stationary?

I do not know how to check if the above stated formula is stationary? I am struggling how to interpret the product of the same white noise time series with different lag?

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Assume $k> 0$ without loss of generality, and assuming white $Z_t$: $$\begin{align}\operatorname{cov}(Y_t,Y_{t-k})&=\operatorname{cov}(Z_t-\frac{1}{2}Z_{t-1}Z_{t-2},Z_{t-k}-\frac{1}{2}Z_{t-k-1}Z_{t-k-2})\\&=-\frac{1}{2}\operatorname{cov}(Z_{t-1}Z_{t-2},Z_{t-k})+\frac{1}{4}\operatorname{cov}(Z_{t-1}Z_{t-2},Z_{t-1-k}Z_{t-2-k})\end{align}$$

Check this expression for $k=1,2$ and greater values. You'll see that the covariance is always $0$. Also, the mean process will be constant.

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  • $\begingroup$ should second line be: cov(Zt,,Zt−k)−1/2cov(Zt,Zt−1Zt−2)−1/2cov(Zt−1Zt−2,Zt−k)+1/4cov(Zt−1Zt−2,Zt−1−kZt−2−k) $\endgroup$
    – Berecht
    Feb 15, 2021 at 11:24
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    $\begingroup$ Yes, I shortened that a bit because covariance of $Z_t$ with any of the past values will be $0$ $\endgroup$
    – gunes
    Feb 15, 2021 at 11:26

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