If $X_t = cos(\omega_0 t + \phi)$ where the constant $\omega_0 \in (0, \pi)$ and $\phi$ is a random variable with a uniform distribution on $(0, 2\pi)$ but is fixed for a single realisation of the process then how do I calculate the autocovariance function?
The book I am reading just states that $\gamma(k) = \frac{1}{2} cos(\omega_0 k)$ so it must be obvious but I can't see it. The book doesn't say but I also suspect in the definition of $X_t$, $t$ is defined only for integer values, i.e. $t = 0, 1, 2, 3, \dots$
$X_t = cos(\omega_0 t + \phi)$ and R.V. is $\phi$ with uniform distribution so its probability density function $p_\phi=\frac 1{2\pi}$
Now note that autocorrelation function is: $\begin{align}R_{X}(\tau)=&E[X_tX_{t+\tau}]\\=& \int_0^{2\pi}X_t X_{t+\tau}p_\phi d\phi\\=&\frac 1{2\pi}\int_0^{2\pi}\cos(\omega_0 t + \phi)\cos(\omega_0 (t+\tau) + \phi)d\phi\\=&\frac 1{4\pi}\int_0^{2\pi}\cos (\omega_0 t + \phi-\omega_0 (t+\tau) - \phi)-\cos (2\omega_0 t+\omega_0\tau+\phi)d\phi\\=&\frac 1{4\pi}\int_0^{2\pi}\cos(\omega_0\tau)d\phi=\frac 12\cos(\omega_0\tau)\end{align}$