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In order to show that not all auto covariance functions have a spectral density, I am considering the trigonometric stationary time serie \begin{align} X_t = A \cos(\omega t) + B \sin(\omega t), \end{align} where $A$ and $B$ are uncorrelated random variables with mean $0$ and variance $1$. It could be shown that the auto covariance function is given by \begin{align} \gamma(h) = \cos(\omega h), \end{align} which is not expressible as \begin{align} \int_{-\pi}^{\pi} e^{ih\lambda} f(\lambda)\ d\lambda, \end{align} with $f$ a function on $(-\pi, \pi]$. Therefore, we are interested in the Spectral Representation theorem. It says that $\gamma(\cdot)$ can be written as the Fourier transform of a discrete distribution function. (right-continuous, non-decreasing, bounded on $[-\pi,\pi]$) However, I do not get why \begin{align} F(\lambda) = \begin{cases} 0 \qquad \text{if $\lambda < -\omega$} \\ \frac{1}{2} \qquad \text{if $ -\omega \leq \lambda < \omega$} \\ 1 \qquad \text{if $\lambda \geq \omega$}, \end{cases} \end{align} i.e. \begin{align} \cos(\omega h) = \int_{(-\pi,\pi]} e^{ih\lambda}\ dF(\lambda). \end{align} Any help is appreciated!

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  • $\begingroup$ Your $F(\lambda)$ is not increasing, it should that $F(\lambda)=1$, if $\lambda\ge \omega$. $\endgroup$ – JGWang Apr 5 '17 at 2:24
  • $\begingroup$ Yes, you are right. Do you know how $F$ is found explicitly? $\endgroup$ – iJup Apr 7 '17 at 18:22
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For your case, the $X$ may be expressed as the sum of finite terms of $A_i\cos(\omega_it)$ and $B_i\sin(\omega_it)$, from the intuition, you could consider the spectral distribution with the form $F(\lambda)=\sum_{i}[c_i\eta_{\{\omega_i\}}(\lambda)+d_i\eta_{\{-\omega_i\}}(\lambda)]$, where $$\eta_{\{\omega\}}(\lambda)=\begin{cases}0,&\lambda< \omega,\\ 1,&\lambda\ge \omega.\end{cases} $$ And determine the $c_i,d_i$ by using $\gamma(h)=\int_{-\pi}^{\pi} e^{ih\lambda}\,dF(\lambda)$
For general case(c.f. Blackwell &Davis, Time Series: Theory and Methods, 2nd Ed. Springer(1991), p.151, Th4.9.1), there is following formula for $F(\lambda)$(but it is only useful for the uniqueness of $F$) $$\lim_{n\to\infty}\frac1{2\pi}\sum_{|j|\le n}\gamma(j)\frac{e^{-ij\lambda_2}-e^{-ij\lambda_1}}{-ij}=F(\lambda_2)-F(\lambda_1). $$

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