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!!


1 Answer 1


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.

  • $\begingroup$ given that ARMA process $\{y_t\}_{t=1}^n$ is stationary, why $z_t = 1.1 y_{t-1}$ will be explosive? $\endgroup$
    – Stephen Ge
    Jun 5, 2020 at 7:00
  • $\begingroup$ OK, I may not have understood your notation correctly. I was basically reading it as removing the first lag, to get $Y_t=1.1Y_{t-1}+\epsilon_t$ $\endgroup$ Jun 5, 2020 at 7:07
  • $\begingroup$ 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. $\endgroup$
    – Stephen Ge
    Jun 5, 2020 at 7:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.