I’m hoping for some help with my understanding of the ACF and the PACF as some aspects have got me rather confused: 1) I understand the PACF to be an indication of how much of the autocorrelation can be attributed to specific lags after we have stripped out the effects of where this autocorrelation is brought in via intervening lags. Is this interpretation correct? 2) Given that the current value of an MA(q) process will only be significantly autocorrelated with the most recent q values in the process, how do I interpret the meaning of the PACF being non-zero at lags greater than q? This seems counterintuitive as there is no autocorrelation to explain at these higher lags. 3) For an MA(1) process, Chapter 12 states that the graph of the ACF cuts off after 1 lag and the PACF declines approximately geometrically over many lags. However, it also states that an invertible MA(1) process can be expressed as an AR process of infinite order. An AR process of infinite order would not have an ACF that cut off after 1 lag. As this is still the same process, how can it be that the ACF is now completely different? Thanks in advance.
Sorry, I can't answer your query but I was amused to see your username. I well remember that mnemonic from geometry over 50 years ago but have never had occasion to use it since then!
Yes - it measures the direct correlation (and strips out the indirect correlation) between the \(x_t\)'s. The ACF measures the total connection (direct or indirect) between the \(X_t\)'s. But when you fix the intermediate values it does affect it. eg If \(X_t = e_t + e_{t-1}\) then when I fix \(X_{t-1} = e_{t-1} + e_{t-2}\) then changing \(X_{t-2} = e_{t-2} + e_{t-3}\) will affect \(X_t\). You can rewritten \(e_t\) as an infinite series of \(X_t\)'s but the ACF still calculates the correlation between the \(X_t\)'s not the \(e_t\)'s.
Thanks for your reply John My thinking still seems to be going wrong somewhere though! Re query 2, if we have a typical MA(q) process, the ACF for lag (q+1) will be around zero while the PACF will be non-zero. In your reply, you mentioned that the ACF measures the total connection (direct or indirect) between the \(X_t\)'s. If the PACF for lag (q+1) is non-zero, this suggests to me that there is a "connection" between the \(X_t\) value and the \(X_{t-(q+1)}\) value, so I don't understand how this can be consistent with the ACF for lag (q+1) being zero. Can you tell me what is incorrect in this line of thinking? Re query 3, after we have rewritten \(e_t\) as an infinite series of \(X_t\), I thought that we could rearrange the equation so that we have \(X_t\) on the left-hand side by itself and a series of historical \(X_t\)'s and \(e_t\) on the right-hand side i.e. it looks similar in style to the usual ARs that the course covers. Therefore I had been thinking we were considering the autocorrelation of this process i.e. of the \(X_t\)'s rather than the \(e_t\)'s. It was the ACF of this expression that I was thinking would not suddenly cut off, meaning that this process has a different ACF to that of the MA process that it originates from. Please can you tell me what is incorrect in this line of thinking? Thank you
Th It is a bit hard to "see". If we have \(X_t = e_t + 0.5e_{t-1}\) then we have \(\rho_1 = 0.4\) and \(\rho_2 = 0\) and \(\phi_1 = 0.4\) and \(\phi_2 = -4/21\). The ACF at lag 2 takes into account the negative direct connection between \(X_t\)'s 2 apart and the sum of the positive direct lag 1 connections (eg \(X_{t-2}\) with \(X_{t-1}\) and also \(X_{t-1}\) with \(X_t\). But in the grand scheme of things in the exam you'll simply be asked to calculate them. Again if we have \(X_t = e_t + 0.5e_{t-1}\) then the infinite series will be: \(e_t = (1 + 0.5B)^{-1}X_t = X_t - 0.5X_{t-1} + 0.5^2X_{t-2} - 0.5^3X_{t-3} + ...\) Rearranging this gives: \(X_t = 0.5X_{t-1} - 0.5^2X_{t-2} + 0.5^3X_{t-3} - ... + e_t\) If you are brave enough to obtain the Yule_walker equations, you'll discover when you solve for \(\gamma_2\) lots of things "cancel out" and you get zero. Please don't ask me to do this though
