Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 218 / 2 / d / ii

Under the intended assumption that the two white-noise sequences are mutually uncorrelated at every pair of times, $X$ and $W$ are uncorrelated. Their sum is therefore weakly stationary with

Strictly, the printed condition $\mathbb{E}[{\epsilon }_t{W}_t]=0$ only at equal times is insufficient. For example, $W_t=(-1{)}^t{\epsilon }_{t-1}$ is itself white noise and is contemporaneously uncorrelated with ${\epsilon }_t$, but the cross-covariance contribution can depend on $t$. The displayed answer therefore uses the standard intended cross-series white-noise assumption $\mathbb{E}[{\epsilon }_t{W}_s]=0$ for all $s,t$.