Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 47 / 2

For a weakly stationary process with mean $m$, the autocovariance function is ${\gamma }_k=\mathbb{E}[(X_{t+k}-m)(X_t-m)]$, independent of $t$. The positive definiteness of these covariances gives a unique finite positive spectral measure of a stationary time series on the frequency circle. Its cumulative function is the time-series spectral distribution $F$, with total mass ${\gamma }_0$, andUse one representative of the identified endpoints $-\pi$ and $\pi$ when specifying atoms. For a real-valued process the measure is symmetric, so the imaginary part integrates to zero. If it has a time-series spectral density, thenThe second identity is pointwise for an absolutely summable covariance sequence, as will apply to part (a). Existence of a density alone does not guarantee ordinary pointwise convergence of its Fourier series; in general the reconstruction can be interpreted by its Cesàro means in $L^1$. This fixes the frequency normalization for all three calculations below.

Only overlapping innovations contribute to the autocovariance function of the moving-average process. Set $c_0=1,c_1={\theta }_1,c_2={\theta }_2$. The overlap formula is ${\gamma }_k={\sigma }^2{\sum }_j{c}_j{c}_{j+|k|}$, with coefficients outside $0,1,2$ set to zero. ThusThe time-series spectral density is the Fourier sum of this finite covariance sequence:It also equals ${\sigma }^2|1+{\theta }_1{e}^{-i\lambda }+{\theta }_2{e}^{-2i\lambda }{|}^2/(2\pi )$, which directly verifies nonnegativity. Integrating from $-\pi$ gives the requested time-series spectral distribution:Extend it by zero below $-\pi$ and by ${\gamma }_0$ above $\pi$. In particular its total mass is the process variance, not necessarily one.

The even time-series spectral density makes the sine part vanish. At lag zero its integral is the area of the triangle, so ${\gamma }_0={\pi }^2$. For any nonzero integer $k$, integration by parts givesTherefore the triangular spectral density of a stationary time series has autocovariance functionNo additional factor $1/(2\pi )$ belongs in this inverse integral under the convention stated above.

The zero means of both amplitudes give $\mathbb{E}{X}_t=0$. Their unit variances and zero covariance give, for arbitrary integer $s,t$,Thus the variance is one and the covariance depends only on lag, proving weak stationarity. Independence or normality of the amplitudes is unnecessary. Sincethe spectral measure of a random harmonic oscillation assigns mass $1/2$ to each of $-{\omega }_0,{\omega }_0$. The corresponding right-continuous time-series spectral distribution isThese spectral atoms represent a persistent oscillation. There is no ordinary absolutely continuous spectral density for this process.