# Why do we care if an MA process is invertible?

I am having trouble understanding why we care if an MA process is invertible or not.

Please correct me if I'm wrong, but I can understand why we care whether or not an AR process is causal, ie if we can "re-write it," so to speak, as the sum of some parameter and white noise - ie a moving average process. If so, we can easily see that the AR process is causal.

However, I'm having trouble understanding why we care whether or not we can represent an MA process as an AR process by showing that it is invertible. I don't really understand why we care.

Any insight would be great.

Invertibility is not really a big deal because almost any Gaussian, non-invertible MA$$(q)$$ model can be changed to an invertible MA$$(q)$$ model representing the same process by changing the parameter values. This is mentioned in most textbooks for the MA(1) model but it is true more generally.

As an example, consider the MA(2) model $$z_t = (1-0.2B)(1-2B)w_t, \tag{1}$$ where $$w_t$$ is white noise with variance $$\sigma_w^2$$. This is not an invertible model because $$\theta(B)$$ has one root equal to 0.5 inside the unit circle. However, consider the alternative MA(2) model obtained by changing this root to its reciprocal value of 2 such that model takes the form $$z_t = (1-0.2B)(1-0.5 B)w_t' \tag{2}$$ where $$w_t'$$ has variance $$\sigma_w'^2=4\sigma_w^2$$. You can easily verify that models (1) and (2) both have the same autocovariance functions and hence specify the same distribution for the data if the process is Gaussian.

To make the model identifiable such that there is a one-to-one mapping from $$\theta_1,\theta_2,\dots,\theta_q,\sigma_w^2$$ to the distribution of the data, the parameter space is therefore by convention restricted to that of invertible models. This particular convention is preferred since the model then can be put directly in AR$$(\infty)$$ form with coefficients $$\pi_1,\pi_2,\dots$$ satisfying the simple difference equation $$\theta(B)\pi_i=0$$.

If we didn't impose this restriction on the parameter space, the likelihood function of an MA$$(q)$$ would in general have up to $$2^q$$ local optima (if the MA polynomial has $$q$$ distinct real roots) which is something we want to avoid.

You can always move roots from inside to outside the unit circle with a corresponding change in the white noise variance using the above technique, except in cases where the MA-polynomial has one or more roots exactly on the unit circle.

• Very interesting! Commented Feb 28, 2019 at 10:36
• Yes, I don't know why this isn't stated more clearly in textbooks. You can see this "trick" being used by function maInvert inside R's arima function to ensure that the parameter estimates correspond to a invertible model. Commented Feb 28, 2019 at 10:40
• @Cm7F7Bb If the white noise is for example iid unif$(-\sigma,\sigma)$, and $z_t$ is MA(1), then you can easily see (at least using simulations) that the joint density of e.g. $z_t,z_{t-1}$ will be different depending on whether $|\theta_1|<1$ or not, even if the autocovariance functions are identical. So in that case, you should in principle be able to infer from data if you have an "invertible" or "non-invertible" model. Commented Sep 27, 2023 at 13:51
• @Cm7F7Bb There seems to be some work on this sort of thing, see jstor.org/stable/4616635. Commented Sep 27, 2023 at 13:54
• @Cm7F7Bb Oh, what I mean by "almost" is that if the MA-polynomial has one or more unit roots, then you can't invert the model to pure AR$(\infty)$ form. Commented Sep 27, 2023 at 15:35