Time series: $Y_t = X_t - \mu$ is $AR(2)$

Question

A problem says: "there is a time series $Y_t = X_t - \mu$ $AR(2)$ ..."

I'm a little bit lost here. If it's $AR(2)$, where are is the $x_{t-1}$ and $x_{t-2}$ terms?

Answer 1

It's simply a demeaned process. You have: $$ X_t=c+\phi_1 X_{t-1}+\phi_2 X_{t-2}+\epsilon_t $$ where $E(X_t)=\mu\neq0$. You can transform this to a zero-mean process by subtracting $\mu$: $$ X_t-\mu=c+\phi_1 X_{t-1}+\phi_2 X_{t-2}+\epsilon_t-\mu $$ which means that $E(X_t-\mu)=\mu-\mu=0$. Now we simply call the demeaned process $Y_t$: $$ Y_t\equiv X_t-\mu $$ and $Y_t$ is an AR(2) process as well.