# 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.
• 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. – Jarle Tufto Feb 28 '19 at 10:40