Discussion
Let the time series \(\{Y_{t}\}\) be an ARIMA(2, 1, 2). Then \(\{Y_{t}\}\) is :
Answer: B
An ARIMA(2, 1, 2) process is having one unit root for the AR polynomial.
ARIMA(\(p, d, q\)) means that the \(d\)th difference is a stationary, invertible ARMA(\(p, q\)). So \((1 - B)Y_{t}\) is ARMA(2, 2), and the full autoregressive operator is \(\phi(B)(1 - B)\), of degree 3, with exactly one root at \(z = 1\) coming from the single difference.
Because of that unit root, \(Y_{t}\) itself is not stationary (A). The MA polynomial of an ARIMA model is taken to be invertible, with all roots outside the unit circle, so it has no unit root (C and D).
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