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I am currently working on moving average process, and not very clear about the invertibility concept. For me it seems like the concept arose from the need of one-to-one relationship between MA model coefficients and autocorrelation function. However, most definition focusing on whether MA can be expressed as AR($\infty$), others relate to the root location.

I can see the importance of invertibility. But how does the definition link to the one-to-one relationship? Yes, If MA can be considered as AR($\infty$), it should have the one-to-one relationship, since all AR are invertible. But if I check why AR is invertible, it goes back to the definition of AR. I think I missed a critical proof if there is one.

Richard Hardy
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yuhao
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  • @JarleTufto The 5th paragraph seems crucial but I still cannot infer how root outside unit circle is equivalent to one-to-one relationship. "This particular convention is preferred since the model then can be put directly in AR(∞)..." Yes, in AR(∞), but why is that one-to-one? – yuhao Oct 19 '21 at 10:22
  • I'm not sure if this is what you're asking but if you consider the equations relating the MA coefficients with the coefficient for the infinite AR representation (obtained by https://en.wikipedia.org/wiki/Rational_function#Taylor_series), you will see that changing any of the MA coefficients leads to a change in the AR coefficients. Also, for given MA coefficients there is a single solution for the AR coefficients. The relationship is therefore one-to-one. Constraining the parameter space to invertible models, also makes the relationship to the distribution of the data one-to-one. – Jarle Tufto Oct 19 '21 at 11:24

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