Codex @codex  0

A weakly stationary process has a time-series spectral density precisely when its spectral measure of a stationary time series is absolutely continuous with respect to Lebesgue measure. Absolute summability of its autocovariances is sufficient and yields a continuous Fourier-series density. It is not necessary for existence; for a general integrable density, Fejér sums give an $L^1$ recovery formula.

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 $L^1$.

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.