Property 1: For an AR(p) process yi = φ0 + φ1 yi-1 +…+ φp yi-p + εi, PACF(k) = φk
Thus, for k > p it follows that PACF(k) = 0
Example 1: Chart PACF for the data in Example 1 from Basic Concepts for Autoregressive Process
Using the PACF function and Property 1, we get the result shown in Figure 1.
Figure 1 – Graph of PACF for AR(1) process
Observation: We see from Figure 1 that the PACF values for lags > 1 are close to zero, as is expected, although there is some random fluctuation from zero.
Example 2: Repeat Example 1 for the AR(2) process
where εi ∼ N(0,1), and calculate ACF and PACF.
From Example 2 of Characteristic Equation of AR(p) Process, we know that this process is stationary.
Figure 2 – Simulated AR(2) process
This time we place the formula =5+0.4*0-0.1*0+B4 in cell C4, =5+0.4*C4-0.1*0+B5 in cell C5 and =5+0.4*C5-0.1*C4+B6 in cell C6, highlight the range C6:C103 and press Ctrl-D.
The ACF and PACF are shown in Figure 3.
Figure 3 – ACF and PACF for AR(2) process
As you can see, there isn’t a perfect fit between the theoretical and actual ACF and PACF values.
how to decide the upper limit and Lower Limit of ACF? is it always 0.2?
Hello Adam,
I don’t know what you are referring to. Where did you get the .2 value? Please explain in more detail.
Charles
+-2/sqrt(n)
Adam,
I don’t see lower and upper bounds on this webpage, nor do I see any reference to .2 or 2/sqrtn). Are you referencing a different webpage?
Charles