A post here: Distribution of integral with respect to Brownian motion described a really weird stationary process.
Let us make it simpler, by only considering for $n\in\mathbb{Z}$, $$X_{n}:=\int_{0}^{2\pi}\cos(nx)Z(dx),$$ where the orthogonal stochastic measure $Z(dx)$ as described in the post has the property $$\mathbb{E}Z=0\ \text{and}\ Var(Z(dx))=\rho(dx)=dx,$$ where $\rho$ is the structure measure of $Z$, which coincides with the spectral measure of $X_{n}$.
Then, following the post, I tried to compute the variance function, covariance function, mean, etc.
For the mean, recall that $\mathbb{E}[Z(d\lambda)]=0$, so we have $$\mathbb{E}X_{n}=\mathbb{E}\Big(\int_{0}^{2\pi}\cos(n\lambda)Z(d\lambda)\Big)=\int_{0}^{2\pi}\cos(n\lambda)\mathbb{E}[Z(d\lambda)]=0.$$
However, the covariance function is really weird. Below is my computation:
\begin{align*} \mathbb{E}(X_{n}\overline{X_{n+k}})&=\int_{0}^{2\pi}\cos(n\lambda_{1})\cos((n+k)\lambda_{2})\mathbb{E}[Z(d\lambda_{1})\overline{Z(d\lambda_{2})}]\\ &=\int_{0}^{2\pi}\cos(n\lambda)\cos[(n+k)\lambda]d\lambda\\ &=\dfrac{1}{2}\dfrac{\sin[2\pi(k+2n)]}{k+2n}+\dfrac{1}{2}\dfrac{\sin(2\pi k)}{k}. \end{align*}
The final answer can be easily computed using WolframAlpha.
So, the covariance function depends not only on $k$ but also on $n$ ???? Then the process cannot be stationary.. right?
To verify my computation, I changed a way to compute it by using Herglotz Theorem: $$c(n)=\int_{0}^{2\pi}e^{in\lambda}\rho(d\lambda),$$ but $\rho(d\lambda)=d\lambda$, so that for $n>0$, $$c(n)=\int_{0}^{2\pi}e^{in\lambda}d\lambda=\int_{0}^{2\pi}\cos(n\lambda)d\lambda+i\int_{0}^{2\pi}\sin(n\lambda)d\lambda=\dfrac{\sin(2\pi n)}{n}+i\dfrac{2\sin^{2}(\pi n)}{n}.$$
This times, $c(n)$ depends on $n$ but we create some complex number...Also, you can see that the case has to be discussed differently if $n=0$, or $n<0$.
So the process is complex-valued?? What is the exactly this process? I am really confused....
Thank you!
Note that $\sin 2\pi m = 0$ for integer $m$, so $$ \mathbb{E}(X_{n}\overline{X_{n+k}}) = 0, k\neq 0, $$ and $$ \mathbb{E}(|X_{n}|^2) = \pi. $$ Therefore, the process is (wide-sense) stationary.
The answer depends on whether your $Z$ is complex-valued or not (which is not completely clear from your question).