If $X_t$ is a AR(2)-process, what is $Y_t = X_t - X_{t-1}$?

Question

Q: If $X_t$ is a AR(2)-process, what is $Y_t = X_t - X_{t-1}$?

$X_t$ is a AR(2)-process if it's stationary and satisfies $X_t = \phi_1 X_{t-1} + \phi_2 X_{t-2} + \epsilon_t$, where $\epsilon_t$ is white noise.

One may elaborate \begin{equation}\begin{split} Y_t &= X_t - X_{t-1}\\ &= \phi_1 X_{t-1} + (\phi_2 - \phi_1) X_{t-2} - \phi_2 X_{t-3} + \epsilon_t - \epsilon_{t-1} \\ &=(\phi_1B +(\phi_2 - \phi_1)B^2 - \phi_2B^3)X_t + (1-B)\epsilon_t \end{split}\end{equation}

But I don't know the benefit.

I have the following options:

1. $Y_t$ is an ARMA(2,1)-process
2. $Y_t$ is an AR(2)-process
3. $Y_t$ is an MA(2)-process
4. $Y_t$ is an AR(3)-process

Answer 1

We have $$ \begin{align*} X_t &= \phi_1 X_{t - 1} + \phi_2 X_{t - 2} + \epsilon_t\\ X_{t - 1} &= \phi_1 X_{t - 2} + \phi_2 X_{t - 3} + \epsilon_{t - 1} \end{align*} $$

which implies $$ X_t - X_{t - 1} = \phi_1(X_{t - 1} - X_{t - 2}) + \phi_2(X_{t - 2} - X_{t - 3}) + \epsilon_t - \epsilon_{t - 1} $$ or $$ (1 - \phi_1 B - \phi_2 B^2)Y_t = (1 - B)\epsilon_t $$

Consider these cases:

• If $\phi_1 + \phi_2 = 1$,
  • If $\phi_1 = 1, \phi_2 = 0$, then $(Y_t)$ is the white noise
  • Otherwise, $(Y_t)$ is AR(1)
• If $\phi_1 + \phi_2 \neq 1$, then the model cannot be simplified and we have ARMA(2, 1)