Explanation

For an ARMA(2,2) model to be covariance stationary, the roots of the characteristic equation must lie outside the unit circle. The AR part is:

$Y_t = 0.2 + 0.4Y_{t-1} - 0.3Y_{t-2} + \text{MA terms}$

The characteristic equation for the AR(2) part is:

$`$1` - 0.4z + 0.3z^2 = 0$$

Solving this quadratic equation:

$z = \frac{0.4 \pm \sqrt{(0.4)^2 - 4(1)(0.3)}}{2(0.3)} = \frac{0.4 \pm \sqrt{0.16 - 1.2}}{0.6} = \frac{0.4 \pm \sqrt{-1.04}}{0.6}$

The discriminant is negative (-1.04), which means the roots are complex. However, the magnitude of the roots can be calculated using:

$|z| = \sqrt{\frac{c}{a}} = \sqrt{\frac{0.3}{1}} = \sqrt{0.3} \approx 0.5477$

Since |z| < 1, the roots lie inside the unit circle, which violates the stationarity condition. For stationarity, we need |z| > 1.

Therefore, the model is not covariance stationary, making option D correct.

Analysis of other options:

• A: Incorrect - The AR coefficients sum to 0.4 + (-0.3) = 0.1, which is less than 1, but this is not sufficient for stationarity
• B: Incorrect - The MA coefficients sum to 0.6 + 0.1 = 0.7, which is less than 1, but this relates to invertibility, not stationarity
• C: Incorrect - The unconditional mean is not 0.2; for a stationary ARMA process, the mean is calculated differently
• E: Incorrect - The long-run variance would be finite if the process were stationary
• F: Incorrect - The model is not stationary as established above