Suppose ${Y_{1t}}, {Y_{2t}},\cdots, {Y_{Nt}}$ are $N$ distinct strictly stationary time series processes, where $t\in Z$, and $Z$ is the set of all integers. We further assume that the $N$ time series processes ${Y_{1t}}, {Y_{2t}},\cdots, {Y_{Nt}}$ are independent. Define a new time series process ${Z_t}$ by putting $$Z_t =\prod_{k=1}^{N}{Y_{kt}},\ t\in Z.$$
Is the new time series process ${Z_t}$ weakly stationary? Explain.
From what I understand, we can show that this process is weakly stationary if we can show the mean and autocovariance do not vary with time.
Given the independence and stationarity of the distinct time series processes, it is straightforward to show that $E\left(Z_t\right)$ is time-independent, satisfying one part of the stationarity.
How can I show that the autocovariance is time-independent?
By independence, you have \begin{align} E(Z_t Z_s) &= E\left(\prod_{k=1}^N Y_{kt} \prod_{k=1}^N Y_{ks} \right)\\ &= E\left(Y_{1t}Y_{1s}Y_{2t}Y_{2s} \cdots Y_{Nt}Y_{Ns} \right)\\ &= E\left(Y_{1t}Y_{1s}\right)E\left(Y_{2t}Y_{2s}\right)\cdots E\left(Y_{Nt}Y_{Ns}\right)\\ &= \prod_{k=1}^N E(Y_{kt}Y_{ks}) \end{align} Now if each $Y_{kt}$ is stationary $E(Y_{kt}Y_{ks}) = E_k(t-s)$. Moreover, $\mu = E(Z_t)$ does not depend on time and therefore $$ \mathrm{Cov}(Z_t,Z_s )= \prod_{k=1}^N E_k(t-s) - \mu^2, $$ which shows that $E(Z_t Z_s)$ is a function of $t-s$ only and therefore that $Z_t$ is weakly stationary.