# ACF and PACF of AR(p)

Why does the PACF of AR(p) model cut off past the order of the series? Why does the ACF tail off to zero? What is the intuitive reason behind this?

Suppose you have an AR(1) process. (Higher orders are similar.)

$$y_t = \phi y_{t-1}+\epsilon_t.$$

Thus, the correlation between $y_t$ and $y_{t-1}$ is $\phi$. This is the first entry in the ACF plot. And of course, the correlation between $y_t$ and $y_{t-2}$ is $\phi^2$. This is the second entry in the ACF plot. And so forth.

Since $-1<\phi<1$, we get an ACF $(\phi, \phi^2, \phi^3, \dots)$ that decays towards zero - monotonically if $\phi>0$, and alternating between negative and positive values if $\phi<0$.

For the PACF, recall that it gives the correlation between $y_t$ and $y_{t-\ell}$ after accounting for all lower-order autocorrelations. If you have an AR(p) process, then there are only nonzero "true" autocorrelations of order up to $p$. So if we account for all of these, all higher partial autocorrelations are calculated by correlating $y_t$ with white noise, which should yield zero.

All this up to sampling variance, of course.

• I understood 1st one. But Im still unclear about PACF – Dominic Joseph Dec 26 '17 at 8:49