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Apr 23, 2021 at 13:27 comment added mark leeds Parseval: Even with the independence assumption, you still need the $h > q$ assumption, Otherwise, if $h$ was not necessarily greater than $q$ as steveoamerica pointed, you would have some terms that were the same in the two $Y_t$ and $Y_{t+h}$ which makes $Y_t$ and $Y_{t+h}$ "dependent" because they contain some of the same terms. But, "dependent" is maybe a misleading term in that case. It's more like "common term" dependent.
Apr 23, 2021 at 13:20 comment added mark leeds He did add later than it was an MA(q) so all the $X_{t}$ are independent. I like your proof with the block diagonal argument. It's more "visual" than ugly algebra.
Apr 22, 2021 at 20:26 comment added steveo'america We need not require all $X_t$ be independent, of course, only that the first $q+1$ elements of $\vec{X}$ be independent of the second $q+1$ elements, which would make $\Sigma$ block diagonal (2x2).
Apr 22, 2021 at 19:42 history answered steveo'america CC BY-SA 4.0