Considering an Auto-Regressive Moving Average (ARMA) model, \begin{equation*} y_k = \phi_0 + \sum_{j=1}^{p} \phi_j y_{k-j} + \sum_{l=1}^{q} \theta_l \varepsilon_{k-l}+ \varepsilon_k, \qquad \text{for}\quad k=1,\cdots,n \end{equation*} where the noise term $\varepsilon_k$ follows the Normal distribution, such that $\varepsilon_k\sim\mathcal{N}(0,\sigma^2_{\varepsilon})$.
If we split ARMA process $\{y_k\}_{k=1}^n$ into two parts: \begin{equation*} x_k = \phi_0 + \sum_{j=1}^{r} \phi_j y_{k-j} + \sum_{l=1}^{s} \theta_l \varepsilon_{k-l}, \qquad \text{for}\quad k=1,\cdots,n \end{equation*} and \begin{equation*} z_k = \sum_{j=r+1}^{p} \phi_j y_{k-j} + \sum_{l=s+1}^{q} \theta_l \varepsilon_{k-l} + \varepsilon_k, \qquad \text{for}\quad k=1,\cdots,n \end{equation*} where $1<r<p$ and $1<s<q$, so that $y_k=x_k+z_k$.
If ARMA process $\{y_k\}_{k=1}^n$ is wide-sense stationary, can I say that both sequences $\{x_k\}_{k=1}^n$ and $\{z_k\}_{k=1}^n$ are stationary? How to prove it? Many thanks!!
In short, yes, $\{_\}_{}^{=1}$ and$\{z_\}_{}^{=1}$ are both weakly stationary (ie stationary in the wide sense). I won't give you the direct proof, but the way this is done is to directly calculate the mean and covariance, which you can do because you have the ARMA process in hand.
What's interesting about this problem is what happens if you were to actually simulate this ARMA process, $y_k$, in practice. Let's say you want to simulate an ARMA process with a specified mean and covariance, for $N=500$ time steps. You write up a little demo in python, R, MATLAB or whatever, and you run your model. You plot the mean and variance of your ARMA model, but it doesn't look WSS. While the mean appears to be constant, the variance appears to be changing with time. Huh, I wonder why that is?
So, what's happening is that there are 'transients' at the start of a simulated ARMA model, because it is just that, simulated. So, it takes additional time steps for your ARMA model to approach the theoretical variance. This is a phenomena called 'burn in' (yes, the same as the one you've heard of from MCMC and the like). So, a common practice would be to take your $N=500$ samples, throw away the first $100$ or so, which don't have the correct statistics, and keep remaining $400$ samples. So, in your problem, you would basically throw away $\{_\}_{}^{=1}$ and keep $\{z_\}_{}^{=1}$, which you are allowed to do because both are WSS (furthermore, both are ARMA).
Another question that may get you thinking on the right track. Does the converse hold? If I have two WSS signals, is their sum also WSS?
For further reference, I suggest Time Series Analysis and Its Applications , I've found it helpful in the past. cheers!