# Can an ARMA(4,1) be the same as an ARMA(3,0)?

I would really appreciate it if someone could explain the reasoning behind whether this can be true or not.

• Check the definition of ARMA here and you'll have your answer: en.wikipedia.org/wiki/Autoregressive-moving-average_model – Moss Murderer Oct 30 '17 at 1:53
• I'm gonna say no because the 4 refers to the fourth order AR, and 1 refers to the 1st order of MA? – mistersunnyd Oct 30 '17 at 2:13
• Try writing out the mathematical expressions for each. Look at them. Do they look the same? – jbowman Oct 30 '17 at 3:07
• Why should they be the same? Do you have an argument for that? If they were, we would not use these indices. – Richard Hardy Oct 30 '17 at 6:25

No. And the reasoning why doesn't have anything to do with limits or approximations.

If you take an ARMA(4,1) model, the only way it could be an ARMA(3) model is if its AR polynomial had a root in common with its MA polynomial. If there was a common root, you could divide both sides of the model by it, and both $p$ and $q$ would reduce by $1$. Most of the definitions that I've seen restrict this behavior.

For example, in "Intro to Time Series and Forecasting," their definition is

$\{X_t\}$ is an ARMA(p,q) process if $\{X_t\}$ is stationary and if for every $t$, $$X_t - \phi_1 X_{t-1} - \cdots - \phi_p X_{t-p} = Z_t + \theta_1 Z_{t-1} + \cdots + \theta_q Z_{t-q},$$ where $\{Z_t\} \sim \text{WN}(0,\sigma^2)$ and the polynomials $(1 - \phi_1 z - \cdots - \phi_p z^p)$ and $(1 + \theta_1z + \cdots + \theta_q z^q)$ have no common factors.

Say, to the contrary, that your model had $z_1 = z_{1,\phi} = z_{1,\theta}$ as a common root. Then \begin{align*} \frac{\phi(z)}{\theta(z)}&= \frac{\phi_p(z-z_{1,\phi})(z-z_{2,\phi})\cdots(z-z_{p,\phi})}{\theta_q(z-z_{1,\theta})\cdots(z-z_{q,\theta})}\\ &= \frac{\phi_p(z-z_{2,\phi})\cdots(z-z_{p,\phi})}{\theta_q(z-z_{2,\theta})\cdots(z-z_{q,\theta})}. \end{align*} You could write down your larger model, divide both sides by the common factor, and the result would be a smaller model. And in your specific question, you are asking about the situation where $p=4$ and $q=1$.

Your question is the same as if we can convert an ARMA(4,1) process to an AR(3) process. Let's check it out!

A stochastic process $\{X_t\}_{t\in \mathbb{Z}}$ is an ARMA(p,q) if: $$X_t - \phi_1X_{t-1}-\cdots-\phi_pX_{t-p} = Z_t + \theta_1Z_{t-1}+\cdots+\theta_pZ_{t-q}$$ where $Z_{t}$ is white noise (WN) with mean 0 and variance $\sigma^2$. An ARMA(4,1) is therefore given by: $$X_t - \phi_1X_{t-1} - \phi_2X_{t-2}- \phi_3X_{t-3}- \phi_4X_{t-4}=Z_t+ \theta Z_{t-1}$$ Considering the lag operator $L^d(Z_t) = Z_{t-d}$, with the condition $|\theta|<1$, we have \begin{align} X_t - \phi_1X_{t-1} - \phi_2X_{t-2}- \phi_3X_{t-3}- \phi_4X_{t-4} & = Z_t(1+ \theta L)\\ (X_t - \phi_1X_{t-1} - \phi_2X_{t-2}- \phi_3X_{t-3}- \phi_4X_{t-4})(1+ \theta L)^{-1} & = Z_t\\ (X_t - \phi_1X_{t-1} - \phi_2X_{t-2}- \phi_3X_{t-3}- \phi_4X_{t-4})\left(1+ \sum_{k = 1}^\infty(-\theta L)^k\right)&=Z_t \end{align} with $\lim_{k\rightarrow\infty}\theta^k = 0$, because $|\theta|<1$, the process can be approximated to an AR(4). For $k=1$ we get an AR(4) - a process with $X_tZ_t$'s. If we don't approximate, then it can never be an AR(something). Note that a MA(1) can be converted to an infinite AR the same as an AR(1) is converted to an infinite MA.

• I modified your approach in my own answer. – Taylor Nov 6 '17 at 0:08