# Is an ARMA(2,2) process a weakly stationary process?

We have an ARMA process as below: $$X[n] = a_0 Z[n] + a_1 Z[n-1] + b_1 X[n-1] +b_2 X[n-2].$$ How can I find the mean and covariance of the process to show that it is/is not a weakly stationary process? I have found one procedure here:

Whether a AR(P) process is stationary or not?, but this does not answer the query. Any help is appreciated.

• What does WSS stand for? – Richard Hardy Nov 2 '16 at 10:24
• Weak Sense stationary process. – jhon_wick Nov 2 '16 at 14:26
• What about examining the roots of the AR and MA polynomials? E.g. for AR processes it is enough to show the roots of the characteristic polynomial lie outside the unit circle. – Richard Hardy Nov 2 '16 at 14:55

## 1 Answer

What you have called the "covariance of the process" is often called the autocovariance function $C_X(i,j) = \operatorname{cov}(X[i],X[j])$ (to distinguish it from the crosscovariance function $C_{X,Z}(i,j) = \operatorname{cov}(X[i],Z[j])$ which considers the covariance of variables drawn from two different processes, which you will also need in this problem). In this instance, use of the bilinearity property of the covariance operator allows us to expand out $C_X(n,n+m)$, which is just

$$\operatorname{cov}\left(\sum_{i=0}^1a_iZ[n-i]+ b_iX[n-i-1], \sum_{i=0}^1a_iZ[n+m-i]+ b_iX[n+m-i+1]\right),$$

into a sum of $16$ autocovariance and crosscovariance functions. Is the sum a function of just $m$, the difference between the arguments? You will see that it is not sufficient to assume that $X$ and $Z$ are WSS processes, but that it is also necessary to consider whether $X$ and $Z$ are jointly WSS processes.