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?

  • $\begingroup$ To put it in the standard form just multiply the original equation by -1/4 $\endgroup$
    – noob2
    Oct 26 '15 at 20:22
  • $\begingroup$ You could ask purely statistical questions here: stats.stackexchange.com $\endgroup$
    – Ric
    Oct 27 '15 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 '15 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) ) $\endgroup$
    – Ric
    Oct 27 '15 at 10:19

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 $\lambda\in\mathbf{R}$.

  • $\begingroup$ Are you sure that the rhs of the defining equation is not affected? Maybe you have to define another white noise process too. $\endgroup$
    – Ric
    Oct 28 '15 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 '15 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. $\endgroup$
    – Ric
    Oct 29 '15 at 7:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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