# Time series: ARMA characteristic polynomials have common roots

I have a question regarding the idea that if the roots of the characteristic polynomials of a time series (say some ARMA process) lie outside the unit circle, then the series will be invertible/causal (depending on which characteristic polynomial we're talking about). In addition to such roots lying outside the unit circle, the two characteristic polynomials must share no common roots. On the problems such a violation could cause, I found the following in Brockwell and Davis (pg 86 (1991). Time series: theory and methods. SpringerVerlag ) so with respect to point (a), say I had some ARMA(2,2) process which could be factorised in terms of the backshift operator as $$(1-\alpha_1B)(1-\beta_1B)X_t=(1-\alpha_1B)(1-\alpha_2B)Z_t$$ and all the roots of the characteristic polynomials lay outside the unit circle but we had the obvious common root $$(1-\alpha_1B)$$. What point (a) says, is that I can simply cancel such root to leave a causal and invertible ARMA(1,1) process? $$(1-\beta_1B)X_t=(1-\alpha_2B)Z_t$$

When writing down an ARMA process you should always eliminate common roots, as these roots are not identifiable (and thus not estimable). In your case you should indeed cancel out the common root $$(1-\alpha_1 B)$$ to be left with the $$\mathrm{ARMA}(1,1)$$ process $$(1-\beta_1 B)X_t = (1 - \alpha_2 B)Z_t$$