Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 40 / 1 / b / Solution

For a discrete-time weakly stationary process, the spectral density of a stationary process, when it exists, is a nonnegative function satisfyingIf ${\sum }_h|\gamma (h)|<\infty$, thenA general second-order stationary process has a spectral measure of a stationary time series, which need not have a spectral density: a centered random constant has an atom at zero. Thus the spectral density calculation below assumes that $f_X$ exists; the corresponding measure identity holds without that extra assumption.

Absolute summability of the filter coefficients gives mean-square convergence becauseConsequently the bilateral filtered process is well defined, with constant mean $m{\sum }_s{a}_s$. Its covariance isThe sum is absolutely convergent, since $|{\gamma }_X(j)|\le {\gamma }_X(0)$ and $\sum |a_s|<\infty$. It depends only on lag, proving weak stationarity of $Y$.

Insert the spectral representation and interchange summation and integration using absolute summability and the finite total spectral mass. For real coefficients,Therefore the spectral density transformation under a linear filter isFor a bilateral sequence, $A$ is guaranteed to converge on the unit circle. Its negative powers need not converge inside the disk: for example, $a_{-j}=2^{-j}$ makes the series diverge at $z=1/4$. Only the unit-circle values are used here, so the unnecessarily broad domain printed for $A$ does not affect the proof.