2
$\begingroup$

I just started out with financial time series and I'm a bit stuck with ARMA models. I have the following ARMA process:

$-4X_t + X_{t-2} = Z_t + 0.2 Z_{t-1}$

Now I am being asked for the polynomials of $\Phi$ and $\Theta$ so we can write the model as: $\Phi (B) X_t = \Theta (B) Z_t$.

This is how I am deriving my solution:

$ \Phi(B) = 1-\phi_1 B - \phi_2 B^2 - ... - \phi_p B^p$

$=-4 +0B --1B^2 $

$= -4 +1B^2$

However, I'm not convinced that this answer is legit. Shouldn't this polynomial always start with 1?

$\endgroup$
4
  • $\begingroup$ To put it in the standard form just multiply the original equation by -1/4 $\endgroup$
    – nbbo2
    Oct 26, 2015 at 20:22
  • $\begingroup$ You could ask purely statistical questions here: stats.stackexchange.com $\endgroup$
    – Richi Wa
    Oct 27, 2015 at 7:11
  • $\begingroup$ Shouldn't this polynomial always start with 1? Why? What`s your reasoning for this assumption? $\endgroup$
    – Carol.Kar
    Oct 27, 2015 at 7:52
  • $\begingroup$ You can not have the coefficient $1$ in both polynomials (you need one on the lhs (AR) and one on the lhs (MA) ) $\endgroup$
    – Richi Wa
    Oct 27, 2015 at 10:19

1 Answer 1

2
$\begingroup$

There is no particular issue with your polynomials. However if you really want them to both start with a 1, you can apply a change of variable by defining : \begin{equation}Y_t = -\frac{1}{4}X_t\end{equation} Then your polynomials $\Phi_y(B)$ and $\Theta(B)$ such that : \begin{equation}\Phi_y(B)Y_t=\Theta(B)Z_t\end{equation} will both start with a $1$.

It is indeed often more convenient for the economic intuition to have both of them starting with $1$ with the idea that you want to explain the value of $X_t$ and not the value of $3X_t$ or $\lambda\cdot X_t$ with $\lambda\in\mathbf{R}$.

$\endgroup$
3
  • $\begingroup$ Are you sure that the rhs of the defining equation is not affected? Maybe you have to define another white noise process too. $\endgroup$
    – Richi Wa
    Oct 28, 2015 at 12:13
  • $\begingroup$ Yes I am sure, you actually get $X_t$ from the data, so you can always define another variable, say $Y_t=\alpha X_t$ and then multiply and divide the original equation by $\alpha$, or in other words replace $X_t$ by $Y_t / \alpha$, there is nothing wrong with that, it will always be true. It is a sort of normalization. $\endgroup$
    – Louis. B
    Oct 29, 2015 at 6:13
  • $\begingroup$ But what about the right hand side? at least the factor will influence the variance of the white noise process and you could call it another WN $\tilde{Z}$ with variance $1/16 \sigma^2$ where $\sigma^2$ is the original variance. $\endgroup$
    – Richi Wa
    Oct 29, 2015 at 7:36

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.