Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 29 / 2 / Solution

For a real weakly stationary process, extend its autocovariance by ${\gamma }_{-k}={\gamma }_k$. The exact existence of a time-series spectral density condition is that its spectral measure of a stationary time series be absolutely continuous with respect to Lebesgue measure. Absolute summability ${\gamma }_0+2{\sum }_{k\ge 1}|{\gamma }_k|<\infty$ is a useful sufficient condition, not a necessary one.

We use the conventional angular-frequency density on $[-\pi ,\pi ]$, restricted to $[0,\pi ]$ by symmetry. Under the absolute-summability condition, the Fourier series and its inverse relation areThus $2{\int }_0^{\pi }f={\gamma }_0$. If a one-sided density is instead normalized to integrate to the full variance, use $g=2f$ and omit the factor two in the inverse formula. This is the positive-frequency spectral normalization convention difference. With merely an integrable spectral density, the inverse relation remains valid; one must not assume pointwise convergence of the unweighted Fourier series. Its Fejér sums recover the density in $L^1$:

If $X$ and $Y$ are independent stationary processes, the cross covariances vanish, soLinearity of the inverse Fourier series relation, or addition of the spectral measures of a stationary time series, gives $f_Z=f_X+f_Y$. Independence can in fact be weakened to zero cross-covariances at every lag.

For the ARMA$(1,1)$ representation, away from an uncancelled unit root the transfer function givesThe usual causal case has $|\varphi |<1$; the same expression holds for the two-sided stationary solution when $|\varphi |>1$. We work in the nondegenerate case $v,w>0$ and $\varphi \ne \pm 1$; cancellations and zero-noise cases are obtained by the appropriate reduced representation or limits. Since independent white noise of variance $w$ contributes $w/(2\pi )$, putThenThe white-noise addition to an ARMA(1,1) process problem is therefore the factorizationDefineBoth are positive under the stated nondegeneracy conditions. An invertible choice isThese satisfy $|\alpha |<1$, $\lambda (1+\alpha {)}^2=P$, and $\lambda (1-\alpha {)}^2=Q$. HenceIn particular, it is the autoregressive coefficient $\varphi$ that enters the terms from the added observation noise.

To obtain a representation of the actual $Z$, defineThe stable inverse exists because $|\alpha |<1$, and spectral filtering gives $f_{\xi }=\lambda /(2\pi )$. Thus $\xi$ is weak white noise, andIts variance can equivalently be writtenWhen $\alpha \ne 0$, also $\lambda =(v\theta -w\varphi )/\alpha$; when $\alpha =0$, necessarily $C=0$ and $\lambda =A$, so the latter quotient should not be used. The model may reduce in order through cancellation. Boundary limits with $P=0$ or $Q=0$ give $|\alpha |=1$ and need not be stably invertible; the displayed stable reconstruction applies to the nondegenerate case above.