I am looking at an I(1) process for an AR(3) model: $$x_t=\theta_1x_{t-1}+\theta_2x_{t-2}+\theta_3x_{t-3}+\varepsilon_t$$ I rewrite it using the lag-operator: $$(1-\theta_1L-\theta_2L^2-\theta_3L^3)x_t=\varepsilon_t$$ And the I factorise the characteristic polynomial: $$(1-\theta_1z-\theta_2z^2-\theta_3z^3)=(1-\phi_1z)(1-\phi_2z)(1-\phi_3z)$$ Where $\phi_1,\phi_2$ and $\phi_3$ are the inverse roots.

Because it is an I(1) process there is obviously only one unit root such that $\phi_1=1$ and it must hold that $|\phi_2|<1$ and $|\phi_3|<1$.

Where I am stuck is that many textbooks then say that because $|\phi_2|<1$ is invertible is can then rewrite it as: $$\frac{1}{1-\phi_2z}=1+\phi_2z+\phi_2^2z^2+\phi_2^3z^3+...$$

I hope someone can help me out with this :)


  • $\begingroup$ This is the sum of a geometric series. See any beginning algebra textbook for a treatment with numbers: it works for polynomials (and more complicated objects) too. $\endgroup$
    – whuber
    Oct 11, 2023 at 16:32
  • $\begingroup$ I think you should ask @whuber first whether they want to turn their comment as an answer (as it seems it caters to your query) out of basic etiquette. After that, you can self-answer based on their comment. $\endgroup$ Oct 11, 2023 at 16:53
  • $\begingroup$ @User1865345 Sorry, I am new on StackExchange, so I don't really know the website all that well. I tried marking whuber 's comment as an answer, but was unable to $\endgroup$
    – Rstrobaek
    Oct 11, 2023 at 16:57
  • $\begingroup$ No problem. In any case you can't mark a comment as an answer. You could ask them to turn that into an answer. If they think that would add anything substantial, they would do. $\endgroup$ Oct 11, 2023 at 17:00

1 Answer 1


For any complex number $z$ for which $|\phi z|\lt 1,$ the right hand side is an absolutely convergent power series. Multiplying it by $1-\phi z$ gives

$$(\color{red}{1}-\color{blue}{\phi z})(1 + \phi z + (\phi z)^2 + \cdots) = \color{red}{ (1 + \phi z + (\phi z)^2 + \cdots) }- \color{blue}{(\phi z)(1 + \phi z + (\phi z)^2 + \cdots)}.$$

Because both terms (on either side of the "$-$") on the right are absolutely convergent series, we may subtract them term by term and we can do it in any order we please. So, compute the right hand side as

$$\color{red}{1}\ +\ \left(\color{red}{\phi z} - \color{blue}{(\phi z)(1)}\right)\ +\ \left(\color{red}{(\phi z)^2} - \color{blue}{(\phi z)(\phi z)}\right)\ +\ \cdots.$$

All the terms in parentheses after the initial $\color{red}{1}$ are of the form $\color{red}{(\phi z)^n} - \color{blue}{(\phi z)(\phi z)^{n-1}},$ which cancel, leaving $\color{red}{1}+0+0+\cdots = 1.$ Consequently, the right hand side is the multiplicative inverse of the left hand side in your equation: that's what is meant by the notation "$1/(1 - \phi z).$"

BTW, lest these appeals to properties of power series seem overly fussy, notice that if we didn't pay attention to them we would wind up with some truly strange identities such as Euler's infamous

$$-1 = \frac{1}{1-2} = 1 + 2 + 4 + 8 + \cdots$$

when $\phi z = 2.$ (For that to be correct, you would need to be working in the ring of 2-adic integers ;-).)


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