Determine whether the following ARMA processes are stationary.zt = 25 + 0.4zt−1 − 0.3zt−2 + εt − 0.2εt−1 + 0.4εt−2

To determine if the ARMA process zt = 25 + 0.4zt−1 − 0.3zt−2 + εt − 0.2εt−1 + 0.4εt−2 is stationary, we need to check if it meets the following conditions:

1. The mean of the process is constant and finite.
2. The autocovariance function of the process depends only on the time lag and not on the specific time t.

Let's start by calculating the mean of the process:

E(zt) = E(25 + 0.4zt−1 − 0.3zt−2 + εt − 0.2εt−1 + 0.4εt−2)
= 25 + 0.4E(zt−1) − 0.3E(zt−2) + E(εt) − 0.2E(εt−1) + 0.4E(εt−2)

Since the process is assumed to be stationary, the mean should be constant over time. Therefore, we can set E(zt) = E(zt−1) = E(zt−2) = μ, where μ is a constant. This gives us:

μ = 25 + 0.4μ - 0.3μ + 0 - 0.2*0 + 0.4*0
μ = 25

So the mean of the process is constant and finite.

Next, let's calculate the autocovariance function of the process:

γ(h) = Cov(zt, zt−h)
= Cov(25 + 0.4zt−1 − 0.3zt−2 + εt − 0.2εt−1 + 0.4εt−2, 25 + 0.4zt−h−1 − 0.3zt−h−2 + εt−h − 0.2εt−h−1 + 0.4εt−h−2)
= Cov(0.4zt−1 − 0.3zt−2 + εt − 0.2εt−1 + 0.4εt−2, 0.4zt−h−1 − 0.3zt−h−2 + εt−h − 0.2εt−h−1 + 0.4εt−h−2)
= 0.4γ(h-1) - 0.3γ(h-2) + Cov(εt, εt−h) - 0.2Cov(εt−1, εt−h) + 0.4Cov(εt−2, εt−h)

Since the process is assumed to be stationary, the autocovariance function should depend only on the time lag h and not on the specific time t. Therefore, we can simplify the above expression as:

γ(h) = 0.4γ(h-1) - 0.3γ(h-2) + σ²(-1)^(h)

where σ² is the variance of the error term εt, and (-1)^(h) equals 1 if h is even and -1 if h is odd.

This expression shows that the autocovariance function depends on the time lag h, so the ARMA process is stationary.

In summary, the ARMA process zt = 25 + 0.4zt−1 − 0.3zt−2 + εt − 0.2εt−1 + 0.4εt−2 is stationary.