# If I split a stationary ARMA process into two parts, are they also stationary?

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 $$1 and $$1, 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!!

Not an answer, but maybe related:

Brockwell and Davis (Introduction to Time Series and Forecasting, 2016), Proposition 2.2.1 says the following:

Let $$Z_t$$ be a stationary time series with expectation zero and acf $$\gamma_Z$$. If $$\sum_{j=-\infty}^{\infty}|\psi_j|<\infty$$, then the series $$Y_t=\sum_{j=-\infty}^{\infty}\psi_j Z_{t-j}$$ is stationary with expectation $$0$$ and acf $$\gamma_Y(h)=\sum_{j=-\infty}^{\infty}\sum_{k=-\infty}^{\infty}\psi_j\psi_k\gamma_Z(h+k-j).$$

If, however, one were to just delete some lags, the statement could not be proven.

E.g., having a look at the "stationarity triangle" stated here, reveal that $$Y_t=1.1Y_{t-1}-0.2Y_{t-2}+\epsilon_t$$ is stationary (in the sense of admitting a causal representation), while omitting the second lag and just keeping the explosive first lag $$1.1Y_{t-1}$$ clearly isn't.

• given that ARMA process $\{y_t\}_{t=1}^n$ is stationary, why $z_t = 1.1 y_{t-1}$ will be explosive? Jun 5, 2020 at 7:00
• OK, I may not have understood your notation correctly. I was basically reading it as removing the first lag, to get $Y_t=1.1Y_{t-1}+\epsilon_t$ Jun 5, 2020 at 7:07
• I probably think the same thing, if I were you :). I feel that both parts will be stationary, but I don’t know how to prove it. Jun 5, 2020 at 7:12