Why Use Both the ACF and PACF?
An autocorrelation pattern can suggest a time-series model, but it rarely identifies one by itself. A pure autoregressive model and a mixed autoregressive moving-average model can both have an ACF that decreases gradually. The partial autocorrelation function provides a second view: it measures the association at a lag after accounting for the intervening lags. Comparing the two functions is therefore useful for distinguishing model patterns.
The key pattern for a mixed ARMA model is that both the ACF and PACF generally tail off. Here, “tail off” means that correlations diminish with increasing lag rather than becoming exactly zero after a finite lag. The word “generally” matters: special parameter values can produce exceptions, and a sample correlogram will not display exact theoretical values.
The ACF records the correlation between observations \(k\) periods apart, including any association transmitted through the intervening observations. The PACF isolates the additional linear association at lag \(k\) after those intervening lags have been included. A PACF is not generally the same as a conditional correlation in an arbitrary non-Gaussian distribution; its definition here is in terms of linear projection.
Reference Patterns: Pure AR, Pure MA, and Mixed ARMA
For a stationary pure AR(\(p\)) model, the theoretical PACF is zero at lags greater than \(p\), while the ACF generally tails off. For a pure MA(\(q\)) model, the ACF is zero at lags greater than \(q\), while the PACF generally tails off. For a mixed ARMA model, neither function generally has a finite cutoff.
| Model type | Typical theoretical ACF | Typical theoretical PACF |
|---|---|---|
| Pure AR(\(p\)) | Tails off | Cuts off after lag \(p\) |
| Pure MA(\(q\)) | Cuts off after lag \(q\) | Tails off |
| Mixed ARMA | Generally tails off | Generally tails off |
These are theoretical patterns, not rules that every plotted sample must visibly follow. A finite data set introduces sampling variation: a true zero correlation may appear nonzero, and a true small correlation may be hard to distinguish from zero. The patterns are clues for model identification, not proof that a particular model generated the data.
Why the ACF of an ARMA(1,1) Tails Off
Consider a centered, stationary ARMA(1,1) model in the convention \(X_t-\mu=\phi(X_{t-1}-\mu)+\varepsilon_t+\theta\varepsilon_{t-1}\). Assume \(|\phi|<1\), and that the innovations \(\varepsilon_t\) have mean zero, variance \(\sigma^2\), and are uncorrelated across time. These conditions give the stationary causal ARMA(1,1) setting used in the calculations below. They are not a general formula for nonstationary models.
Taking the covariance with \(X_{t-k}-\mu\), for \(k\geq 2\), makes the innovation terms uncorrelated with that past observation. The resulting covariance recursion is \(\gamma_k=\phi\gamma_{k-1}\). Repeated application gives a geometric sequence of autocovariances and, after division by \(\gamma_0\), autocorrelations.
The first-lag correlation depends on both model parameters. For this parameterization, the stationary ARMA(1,1) moments are \(\gamma_0=\sigma^2(1+\theta^2+2\phi\theta)/(1-\phi^2)\) and \(\gamma_1=\sigma^2(\phi+\theta)(1+\phi\theta)/(1-\phi^2)\). Thus \(\rho_1=(\phi+\theta)(1+\phi\theta)/(1+\theta^2+2\phi\theta)\), when the denominator is positive. The formula shows why an MA contribution affects the initial correlation, while the later ACF values follow the AR coefficient’s geometric pattern.
Worked Example: Finding the ACF of a Mixed Model
A hypothetical quarterly claim-count index follows a centered ARMA(1,1) model with \(\phi=0.5\), \(\theta=0.5\), innovation variance \(\sigma^2=4\), and mean \(\mu=20\). Find \(\gamma_0\), \(\rho_1\), and \(\rho_2,\rho_3\). Use the positive-\(\theta\) convention in the model equation above.
Method. Apply the stationary ARMA(1,1) moment formulas, then use the lag recursion. The mean sets the level but does not enter the autocovariance or autocorrelation calculations.
First calculate the denominator shared by the moment formulas: \(1-\phi^2=1-(0.5)^2=0.75\). For the variance numerator, \(1+\theta^2+2\phi\theta=1+0.25+2(0.5)(0.5)=1.75\). Therefore \(\gamma_0=4(1.75)/0.75=7/0.75=28/3\).
Next calculate the first-lag numerator: \((\phi+\theta)(1+\phi\theta)=(0.5+0.5)(1+0.25)=1(1.25)=1.25\). So \(\gamma_1=4(1.25)/0.75=5/0.75=20/3\), and \(\rho_1=\gamma_1/\gamma_0=(20/3)/(28/3)=5/7\).
For later lags, \(\rho_k=\phi^{k-1}\rho_1\). Hence \(\rho_2=(0.5)(5/7)=5/14\), and \(\rho_3=(0.5)^2(5/7)=5/28\).
Check. The correlations decrease in magnitude by a factor of \(0.5\) at each subsequent lag, as the recursion requires. The autocovariance is positive, as expected from the positive values of both parameters. A common mistake is to use \(\rho_2=\phi^2\rho_1\); the recursion starts at lag 2 with one factor of \(\phi\), so that expression would incorrectly shift the lag exponent.
Why the PACF of a Mixed Model Generally Tails Off
The PACF is not obtained by simply reading the ACF at the same lag. It requires accounting for the correlations among all the intermediate lags. For a nonsingular stationary series, the Durbin–Levinson recursion computes the PACF and the coefficients of the successive linear projections. Write \(a_{k,j}\) for the coefficient on lag \(j\) in the projection using lags \(1,\ldots,k\). Then
with coefficient updates
The numerator for \(\kappa_k\) is the part of the lag-\(k\) correlation not accounted for by the shorter-lag projection; the denominator adjusts for the residual variance from that projection. This explains why a nonzero ACF does not automatically mean a nonzero PACF at every lag. In a pure AR(\(p\)), the recursion yields zero PACF beyond \(p\). In a mixed ARMA model, the ACF typically keeps contributing at later lags, and the recursion generally leaves a nonzero partial correlation rather than a finite cutoff.
Worked Example: Calculating the First Three PACF Values
For the centered ARMA(1,1) process with \(\phi=0.5\) and \(\theta=0.5\) in the preceding example, calculate \(\kappa_1,\kappa_2,\kappa_3\). Use the ACF values \(\rho_1=5/7\), \(\rho_2=5/14\), and \(\rho_3=5/28\).
Method. Apply the Durbin–Levinson recursion successively. Keep fractions through the calculation so that the values can be checked exactly.
At lag 1, \(\kappa_1=\rho_1=5/7\). At lag 2, \[ \kappa_2=\frac{\rho_2-\rho_1^2}{1-\rho_1^2} =\frac{5/14-25/49}{1-25/49} =\frac{-15/98}{24/49} =-\frac{5}{16}. \] The lag-2 projection coefficient on lag 1 is \(a_{2,1}=a_{1,1}-\kappa_2a_{1,1} =(5/7)(1+5/16)=15/16\).
At lag 3, substitute \(a_{2,1}=15/16\), \(a_{2,2}=-5/16\): \[ \kappa_3= \frac{5/28-(15/16)(5/14)-(-5/16)(5/7)} {1-(15/16)(5/7)-(-5/16)(5/14)} =\frac{15/224}{99/224} =\frac{5}{33}. \] Thus, to four decimal places, \(\kappa_1=0.7143\), \(\kappa_2=-0.3125\), and \(\kappa_3=0.1515\).
Check. The PACF values do not become zero at lag 2 or 3, consistent with the general mixed-model pattern. A common error is to use \(\rho_3\) itself as \(\kappa_3\); the recursion subtracts the contribution already explained by the two shorter lags.
Reading the Patterns Without Overclaiming
For theoretical patterns, the combination of a tailing ACF and a tailing PACF is consistent with a mixed ARMA model. A tailing ACF and a PACF that cuts off after lag \(p\) is consistent with a pure AR(\(p\)); an ACF that cuts off after lag \(q\) and a tailing PACF is consistent with a pure MA(\(q\)). These clues are most useful when one function has a clear cutoff and the other has a clear tail.
Worked Example: Interpreting Three Theoretical Patterns
A model analyst is comparing three hypothetical stationary series. Series A has a nonzero ACF at lags 1, 2, and higher, with correlations decreasing steadily; its PACF is nonzero at lag 1 but exactly zero at every lag above 1. Series B has an ACF that is nonzero through lag 2 and exactly zero thereafter; its PACF remains nonzero at later lags. Series C has nonzero ACF and PACF values that both diminish without an exact cutoff. Identify the pattern each series supports.
Series A. The PACF cuts off after lag 1 while the ACF tails off. This is the theoretical signature of a pure AR(1) model.
Series B. The ACF cuts off after lag 2 while the PACF tails off. This supports a pure MA(2) model.
Series C. Neither function cuts off, and both diminish. This supports considering a mixed ARMA model rather than treating the pattern as a pure finite-order AR or MA signature.
Check. Each conclusion follows from which function cuts off, not from which function has the larger first-lag value. A common mistake is to call Series C a pure AR model because its ACF tails off; a tailing ACF alone does not distinguish pure AR from mixed ARMA.
In practice, correlograms come from finite samples, so exact zeros are rarely visible. The observed tail can also be obscured by sampling noise, especially at larger lags. Use the ACF and PACF together, consider whether the pattern is broadly compatible with the proposed model, and avoid treating one isolated nonzero spike as a model-order decision. This pattern-based interpretation is aligned with the SOA Exam SRM (Statistics for Risk Modeling) syllabus.
Check Your Understanding
For each problem, identify the relevant pattern or calculate the requested correlation.
- A stationary ARMA(1,1) model has \(\phi=0.4\) and \(\theta=0.2\), using the positive-\(\theta\) convention. Calculate \(\rho_1\) and \(\rho_2\). (Answer: \(\rho_1=0.5400\), \(\rho_2=0.2160\))
- A theoretical series has an ACF that is nonzero through lag 3 and zero at every higher lag, while its PACF tails off. Which model pattern does this support? (Answer: A pure MA(3) pattern)
- A theoretical series has a PACF that cuts off after lag 2 and an ACF that tails off. Which model pattern does this support? (Answer: A pure AR(2) pattern)
- A stationary ARMA(1,1) model has \(\phi=0.3\) and \(\rho_1=0.5\). Using the ARMA(1,1) ACF recursion, calculate \(\rho_4\). (Answer: \(\rho_4=0.0135\))
- A correlogram from a finite sample appears to have both an ACF and PACF that taper gradually. What model class is consistent with that pattern, and why is the conclusion not definitive? (Answer: A mixed ARMA model is consistent; sample variation means the pattern alone does not establish the model.)