3
$\begingroup$

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

Attempted solution:

$X_t = \phi_1 X_{t-1} + \phi_2 X_{t-2} + W_t$, where $W_t$ is white noise.

\begin{equation} \begin{split} Y_t &:= X_t - X_{t-1} = \phi_1 X_{t-1} + \phi_2 X_{t-2} + W_t - \phi_1 X_{t-2} - \phi_2 X_{t-3} - W_{t-1} \\ \end{split} \end{equation}

But what can we say about $Y_t$?

$\endgroup$
2
  • $\begingroup$ this sounds like a self study question. if you agree, consider adding self-study tag :-) $\endgroup$
    – Ute
    Commented May 29, 2023 at 13:23
  • $\begingroup$ Reorder your terms such that you have a linear combination of $X_{t-i}$ terms and a linear combination of $W_{t-j}$ terms. Then have a look at ARMA models $\endgroup$
    – Ute
    Commented May 29, 2023 at 13:28

1 Answer 1

3
$\begingroup$

If we write it with characteristic polynomials the relations will look like below $$Y_t=(1-B)X_t, \ \ \ \ \ \ \ X_t(1-\phi_1B-\phi_2B^2)=W_t$$

Then, the relation between $Y_t$ and $W_t$ can be written with $$Y_t=\frac{1-B}{1-\phi_1B-\phi_2B^2}W_t \rightarrow Y_t(1-\phi_1B-\phi_2B^2)=(1-B)W_t$$

Converting it back yields the following ARMA(2,1) model. $$Y_t-\phi_1 Y_{t-1}-\phi_2 Y_{t-2}=W_t-W_{t-1}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.