 Solution

The autocovariance vanishes beyond lag $q$. For an ordinary consecutive sum, variance of a sum givesThe last sum is finite. For $h\ge 0$, $\gamma (h)={\sigma }^2{\sum }_{j=0}^{q-h}{\theta }_j{\theta }_{j+h}$. Counting all coefficient pairs givesFor the odd-indexed sample the lag-$h$ autocovariance is $\gamma (2h)$, so the same finite-sum argument gives ${\sum }_h\gamma (2h)$. Only coefficient pairs of the same parity contribute. If $E={\sum }_{j\mathrm{even}}{\theta }_j$ and $O={\sum }_{j\mathrm{odd}}{\theta }_j$, the odd-subsample limit isThis is the odd-subsample long-run variance of a moving average. Equivalently it is $[f_X(0)+f_X(1/2)]/2$ in the cycles convention. Under the printed invertibility assumption and positive noise variance, the polynomial is nonzero at both $1$ and $-1$, so both limits are positive. The same formulas also hold without invertibility, when cancellations can make a limit zero.