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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?

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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.

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

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