Discussion
Consider the time series model :
\[ X_{t} = 0.2X_{t-2} - 0.6X_{t-1} + Z_{t} + 1.2Z_{t-1}, \]
where \(Z_{t} \sim\) iid Normal \((0, \sigma^{2})\). Which of the following statements is true ?
Answer: C
The process is causal but not invertible.
Write the model as \(\phi(B)X_{t} = \theta(B)Z_{t}\) with \(\phi(z) = 1 + 0.6z - 0.2z^{2}\) and \(\theta(z) = 1 + 1.2z\). Causality needs the roots of \(\phi\) outside the unit circle: \(0.2z^{2} - 0.6z - 1 = 0\) gives \(z^{2} - 3z - 5 = 0\), so
\[ z = \frac{3 \pm \sqrt{29}}{2} \approx 4.19,\ -1.19 \]
both of modulus greater than 1: causal (and stationary). Invertibility needs the root of \(\theta\) outside the unit circle, but \(1 + 1.2z = 0\) gives \(z = -0.83\), inside. So the process is not invertible, which rules out A, B and D. (The equivalent invertible representation would use the MA coefficient \(1/1.2\).)
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