# Is non-invertibility a problem for (AR)MA processes?

I'm reading Time Series Analysis: Forecasting and Control (3rd ed.) by Box, Jenkins and Reinsel.

There are some arguments about invertibility that I can't wrap my head around.

Considering a MA(1) model:

Here it says: "the current deviation $$z_t$$ in (3.1.4) depends on $$z_{t-1}$$,$$z_{t-2}$$,...,$$z_{t-k}$$ with weights that increase as $$k$$ increases." This seems in contrast with this answer which I agree with, it doesn't feel correct to state that $$z_t$$ depends more and more on past values, since there is actually statistical independence for lags greater that 1. The problem is that in (3.1.4) there are a lot of terms that cancel out (actual math in the cited answer). (1) I am missing something, why would the above statement be true?

Moreover:

Assuming the process is Gaussian, I get that invertibility assures the model is identifiable, but I don't understand why would it make a difference in how we associate past happenings with present. For almost each invertible Gaussian process there is a non-invertible one that defines the same distribution, so I would argue that the two processes associate the past with the present in the same manner, (2) where am I wrong?

If we don't assume Gaussian distribution, (3) then why would we exclude the possibility of a process being non-invertible, effectively excluding some distributions that could be closer (and maybe predict better) the real process underlying the phenomenon being studied?

Again, in chapter 6, about model identification: And

Earlier it is stated: “Identification methods are rough procedures applied to a set of data to indicate the kind of representational model worthy of further investigation. The specific aim here is to obtain some idea of the values of d, p, q ... and to obtain initial estimates for the parameters”.

So, to assure the identification procedure can logically be used to uniquely identify a model (why couldn’t more than one model be “identified” for further investigation?), they exclude non invertibile models, apparently making them “not worthy of further investigation”. To me this makes sense only if we assume normality. They seem to point that “expressing the current value $$w_t$$ exclusively in terms of previous history” is required for a model, (4) why would it be?

• @RichardHardy I am not sure either, do you have a suggestion? Commented Dec 24, 2021 at 4:58
• No, I do not, as I indicated above :) Nevermind. Commented Dec 24, 2021 at 7:37