1
$\begingroup$

I once heard the following statement:

The PACF (partial autocorrelation function) for MA processes behaves much like the ACF for AR processes; the PACF for AR processes behaves much like the PACF for AR processes.

How to understand the logic underlying this statement?

In addition to the strict mathematical proof, are there any approaches to understand this statement, the inherent relationship of AR, MA along with their PACF/ACF, from time series properties, statistics, or any other high-level thoughts?

$\endgroup$
2
$\begingroup$

The statement is related to the fact that the ACF of a stationary AR process of order p goes to zero at an exponential rate, while the PACF becomes zero after lag p. For an MA process of order q the theoretical ACF and PACF show the reverse behaviour, the ACF truncates after lag q and the PACF goes to zero at an exponential rate.

These properties can be used as a guide to choose the orders of an ARMA model. See for instance, Chapter 3 in Time Series: Theory and Methods by Peter J. Brockwell and Richard A. Davis and this.

$\endgroup$
  • $\begingroup$ Wold you please explain why the ACF truncates after lag q and the PACF goes to zero at an exponential rate in case MA process is the true process. What is the concept or intuition behind using ACF for choosing MA order? I am asking for the mechanism, not observation. $\endgroup$ – Mumbo.Jumbo Jun 5 '17 at 6:27
  • $\begingroup$ @Mumbo.Jumbo by getting the expression of the autocovariances in an AR process, it can be checked the decaying ACF of an AR(1) model, see for example this. For the MA model, you can get its AR representation (assuming invertibility) and see that it has coefficients of the form $\theta^i$ ($i=1,2,\dots$ and $\theta$ is the MA coefficient), this suggests a decaying FACP of the MA model. $\endgroup$ – javlacalle Jun 6 '17 at 19:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.