Representation of ARMA processes

Question

Question

Consider the following process: $$2y_t-3y_{t-1}+y_{t-2}=\epsilon_t-\theta\epsilon_{t-1}$$ What is the model for the process $w_t=\Delta y_t = y_t-y_{t-1}$?

Attempt

I have solved the question in two ways, which seem to be giving conflicting answers.

1. Substitute the form of $y_t$ into the expression of $w_t$: $$w_t=\frac{3}{2}y_{t-1}-\frac{1}{2}y_{t-2}+\frac{1}{2}\epsilon_t-\frac{\theta}{2}\epsilon_{t-1}-y_{t-1} \\ \implies w_t=\frac{1}{2}y_{t-1}-\frac{1}{2}y_{t-2}+\frac{1}{2}\epsilon_t-\frac{\theta}{2}\epsilon_{t-1} \\ \implies w_t = \frac{1}{2}w_{t-1} +\frac{1}{2}\epsilon_t-\frac{\theta}{2}\epsilon_{t-1} $$

   This is an ARMA(1,1) model.

2. Use the expanded forms of $y_t$ and $y_{t-1}$, and then substitute into the expression for $w_t$: $$y_t=\frac{3}{2}y_{t-1}-\frac{1}{2}y_{t-2}+\frac{1}{2}\epsilon_t-\frac{\theta}{2}\epsilon_{t-1} \\ y_{t-1} =\frac{3}{2}y_{t-2}-\frac{1}{2}y_{t-3}+\frac{1}{2}\epsilon_{t-1}-\frac{\theta}{2}\epsilon_{t-2} \\ \implies w_t = \frac{3}{2}w_{t-1} - \frac{1}{2}w_{t-2}+\frac{1}{2}\epsilon_t-\frac{\theta+1}{2}\epsilon_{t-1}+\frac{\theta}{2}\epsilon_{t-2}$$ where the facts that $w_{t-1}=\Delta y_{t-1}= y_{t-1} - y_{t-2}$ and $w_{t-2}=\Delta y_{t-2}= y_{t-2} - y_{t-3}$ were used. This is an ARMA(2,2) model.

As can be seen, both these approaches yield different models. The first is stationary, while the second is not (by checking the roots of the AR characteristic polynomials). Is this possible, or is there a mistake in one approach?

Answer 1

The second representation $$w_t = \frac{3}{2}w_{t-1} - \frac{1}{2}w_{t-2}+\frac{1}{2}\epsilon_t-\frac{\theta+1}{2}\epsilon_{t-1}+\frac{\theta}{2}\epsilon_{t-2}$$

is an artifact. Write it like this

$$w_t = \Big(\frac{1}{2}w_{t-1} + \frac{1}{2}\epsilon_t- \frac{\theta}{2}\epsilon_{t-1}\Big) +\Big(w_{t-1}- \frac{1}{2}w_{t-2}-\frac{1}{2}\epsilon_{t-1}+\frac{\theta}{2}\epsilon_{t-2}\Big)$$

The first big parenthesis is the first representation. The second big parenthesis equals identically zero, since it is the first representation (lagged once), subtracted from itself.