I understand that an ARMA process is a linear combination of an AR and MA process, but why should the roots of the two characteristics polynomials be different? I'd like both an intuitive insight and maybe a reference to a formal proof... Thanks in advance!


1 Answer 1


If the roots are the same, the two cancel out. Consider an ARMA model of the form

$y_t = \frac{\Theta(B)}{\Phi(B)}\varepsilon_t$

that is both stationary and invertible.

Write the numerator and denominator characteristic polynomials as products of factors (i.e. factorize the polynomials). Divide through by a constant so the highest order coefficient in each is 1. Factorize the polynomials. If the numerator has a term like say $(B-a)$ and the denominator has the same term (which it will if they both share a root), then those two terms cancel out; this means that we could not distinguish a common root at $a$ from any other common root, at say $a'$. That is, the root at $a$ is not identifiable. You get the same fit as taking the common root to be $0$ - which is equivalent to simply omitting that term altogether.

Note that polynomials with real coefficients always have either linear factors or conjugate pairs of complex factors.

So imagine you have an ARMA(4,3) where the factored AR and MA polynomials are different, except that there's exactly one factor in each the same. Then the model is identical to an ARMA(3,2) model without that factor in either polynomial.

Similarly we can take any ARMA$(p,q)$ and construct an infinite sequence of models of order $(p+k,q+k),\,k=1,2,3,...$ that are identical to the ARMA$(p,q)$ in terms of the process (i.e. have $k$ factors that would cancel). Since the factors that cancel don't contribute anything (they yield the same infinite-MA), we take the model in its simplest terms, which means we must have no roots in common.

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    $\begingroup$ This was all about the representation of the process. But are there any issues with estimation? Can I get identical estimated roots, or would that be in conflict with principles of maximum likelihood estimation, optimization routines or something? What if I get very similar (although not identical) estimated roots? $\endgroup$ Commented Sep 29, 2015 at 8:36
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    $\begingroup$ @RichardHardy: the roots that cancel between the AR and the MA parts are not identifiable, so cannot be estimated. Finding if there are very similar roots in the AR and MA parts is equivalent to determining the order of the ARMA(p,q) process. $\endgroup$
    – Xi'an
    Commented Sep 29, 2015 at 9:24

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