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 $\endgroup$
    – 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


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. $\endgroup$
    – 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

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