Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 38 / 1

Use the one-sided convention: integrating the one-sided spectral density over $[0,\pi ]$ gives the variance. A standard sufficient hypothesis is absolute summability of the autocovariance sequence:For a real weakly stationary process, extend the sequence by ${\gamma }_{-k}={\gamma }_k$. The Fourier pair isAbsolute summability makes the series uniformly convergent and continuous; cosine orthogonality gives the inverse relation. Nonnegativity follows from the spectral measure of a stationary time series, or from the nonnegative Fejér approximations obtained by taking variances of finite Fourier sums.

For precision, absolute summability is sufficient, not necessary. The exact general condition is that the spectral measure be absolutely continuous with respect to Lebesgue measure. For example, a density constant on $[0,\pi /2]$ and zero elsewhere has ${\gamma }_k$ proportional to $\sin (k\pi /2)/k$, which is not absolutely summable. Thus the displayed absolutely convergent Fourier formula answers the usual covariance-summability version; an arbitrary density need only have the integral inverse relation and Fejér-mean Fourier recovery in $L^1$.

The series defining the linear filter of a stationary time series converges in $L^2$, because the norm of a tail is at most $\Vert X_0{\Vert }_2$ times the corresponding tail sum of $|a_s|$. Its mean is the original mean times ${\sum }_s{a}_s$. Its covariance is time independent and equalsUnder the summability assumption in (i),We may therefore interchange all the sums. Setting $h=k-s+u$ in the Fourier series givesThe second factor is the conjugate of the first because the coefficients are real. HenceThis spectral density transformation under a linear filter also holds whenever the original spectral measure has a density, without covariance summability: insert its integral representation into the covariance double sum, justified by absolute summability of the filter coefficients.

In the usual causal stationary convention, a centered autoregressive moving-average model satisfiesSet $\Phi (z)=1-{\sum }_{j=1}^p{\varphi }_j{z}^j$ and $\Theta (z)=1+{\sum }_{j=1}^q{\theta }_j{z}^j$. After removal of common factors, the usual causality condition is that all zeros of $\Phi$ lie outside the closed unit disk. Then $\Theta (z)/\Phi (z)$ has an absolutely summable causal coefficient sequence, giving a linear-filter representation of $X$ in terms of white noise. Invertibility of the moving-average part is a separate issue and is not needed for its covariance spectrum.

White noise has one-sided density ${\sigma }^2/\pi$. Applying the filtering result, or equating the spectra of the two filtered sides of the defining equation, givesThe moving-average sign agrees with the positive $\theta$ convention used later. More generally, the same quotient applies to a stationary two-sided solution when the reduced denominator has no unit-circle zeros; causality selects the past-dependent realization. This is the spectral density of an ARMA process.

For $|\varphi |<1$, the causal representation is $X_t={\sum }_{j\ge 0}{\varphi }^j{\epsilon }_{t-j}$. ThusAlternatively the rational spectrum from (iii) is ${\sigma }^2/\{\pi (1-\varphi z)(1-\varphi z^{-1})\}$ on $|z|=1$. The absolutely convergent geometric sums giveConsequentlyThis is the Poisson-kernel expansion of an AR(1) spectrum, with the same one-sided normalization as (i).

For the causal case $|\varphi |<1$, expandHence the white noise coefficients are ${\psi }_0=1$ and ${\psi }_j=(\varphi +\theta ){\varphi }^{j-1}$ for $j\ge 1$. Orthogonality of white noise gives ${\gamma }_k={\sigma }^2{\sum }_{j\ge 0}{\psi }_j{\psi }_{j+k}$ for $k\ge 0$. Summing the geometric series yields the autocovariance of a causal ARMA(1,1) process:For $\varphi =0$, this means ${\gamma }_1={\sigma }^2\theta$ and ${\gamma }_k=0$ for $k\ge 2$, together with ${\gamma }_0={\sigma }^2(1+{\theta }^2)$. For $\theta =-\varphi$, all positive-lag covariances vanish and ${\gamma }_0={\sigma }^2$, as expected from cancellation of the two filter polynomials.