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ACF Versus PACF

Learn what the ACF and PACF measure, how to calculate a lag-2 PACF, and why the two values can tell different stories.

Beginner 10 min read

What You'll Learn

  • Distinguish the total linear association measured by the ACF from the adjusted association measured by the PACF
  • Calculate sample autocorrelations for a short series using a stated convention
  • Derive and apply the lag-2 PACF formula
  • Interpret PACF values as associations after accounting for intermediate observations
  • Avoid confusing an ACF value with a PACF value at the same lag
  • Recognize symmetry of the sample ACF at positive and negative lags

Why Compare ACF and PACF?

At lag \(k\), the autocorrelation function (ACF) describes the linear association between observations \(k\) time steps apart. That association can include pathways through observations at the intervening lags. The partial autocorrelation function (PACF) asks a more specific question: how strongly are the two observations associated after the linear contributions of those intermediate observations have been accounted for?

The distinction matters when reading time-series summaries. A large ACF at lag 2 does not, by itself, show that lag 2 contributes an association beyond what is already explained by lag 1. Conversely, a small ACF at a lag does not guarantee that the PACF there is small. This comparison is aligned with the SOA Exam SRM (Statistics for Risk Modeling) syllabus.

Definition: For a process with constant mean and finite, positive variance, the ACF at lag \(k\) is the correlation between observations \(k\) time steps apart, denoted \(\rho_k\). The PACF at lag \(k\) is the correlation between the residuals from linearly projecting each endpoint onto the observations at the intervening lags. Equivalently, under the usual stationary covariance setup, it is the coefficient on the lag-\(k\) observation in the linear projection of the current observation on lags 1 through \(k\). At lag 1, there are no intervening observations, so the ACF and PACF agree.

Sample ACF and PACF

For an observed series \(x_1,\ldots,x_n\), let \(\bar{x}\) be its sample mean. We will use the following sample ACF convention for nonnegative lag \(k\):

$$ r_k = \frac{\displaystyle\sum_{t=k+1}^{n}(x_t-\bar{x})(x_{t-k}-\bar{x})} {\displaystyle\sum_{t=1}^{n}(x_t-\bar{x})^2}. $$

The denominator uses all \(n\) centered observations, while the numerator uses the pairs available at lag \(k\). In particular, \(r_0=1\). For a real-valued series, the sample ACF is symmetric in the lag: \(r_{-k}=r_k\). The products in the lag-\(-k\) numerator pair the same observations as those in the lag-\(k\) numerator, with the factors reversed. Estimates at different positive lags need not follow any particular pattern.

A common way to estimate the PACF from sample ACF values is the Yule–Walker recursion. At lag 1, \(\widehat{\phi}_{11}=r_1\). At lag 2, the estimate simplifies to:

$$ \widehat{\phi}_{22} = \frac{r_2-r_1^2}{1-r_1^2}, \qquad |r_1|<1. $$

Here \(\widehat{\phi}_{22}\) estimates the lag-2 PACF. This plug-in calculation uses the sample ACF values and is not the only possible finite-sample estimation convention. When reporting a calculated PACF, identify the convention if the question or analysis specifies one. If the denominator is zero, this formula is undefined; it cannot be used to report a finite lag-2 PACF.

Worked Example: Comparing Lag 1 and Lag 2 in a Short Series

A six-period index of hypothetical claim-processing workload, after centering, is \(1,0,-1,-1,0,1\). Using the stated sample ACF convention, calculate \(r_1\), \(r_2\), and the Yule–Walker estimate of the lag-2 PACF. Interpret the comparison.

Method. First find the mean and the common denominator for the sample ACF. Then calculate the lagged-product sums and substitute the resulting ACF estimates into the lag-2 formula.

The values sum to \(1+0-1-1+0+1=0\), so \(\bar{x}=0\). The denominator is:

$$ \sum_{t=1}^{6}(x_t-\bar{x})^2 =1^2+0^2+(-1)^2+(-1)^2+0^2+1^2 =4. $$

At lag 1, the numerator is:

$$ (0)(1)+(-1)(0)+(-1)(-1)+(0)(-1)+(1)(0)=1, \qquad r_1=\frac{1}{4}=0.25. $$

At lag 2, the numerator is:

$$ (-1)(1)+(-1)(0)+(0)(-1)+(1)(-1)=-2, \qquad r_2=\frac{-2}{4}=-0.5. $$

Now apply the lag-2 PACF formula:

$$ \widehat{\phi}_{22} =\frac{-0.5-(0.25)^2}{1-(0.25)^2} =\frac{-0.5-0.0625}{1-0.0625} =\frac{-0.5625}{0.9375} =-0.6. $$

Interpretation and check. The estimated lag-2 ACF is \(-0.5\), while the estimated lag-2 PACF is \(-0.6\). Both indicate a negative association in this calculation; the PACF describes the lag-2 association after accounting linearly for lag 1. The denominator is nonzero, and substituting \(-0.6\) into the formula gives \(( -0.5-0.0625)/0.9375=-0.6\), as required. A common mistake is to report \(-0.5\), the ACF, as the lag-2 PACF.

Why the Lag-2 Formula Holds

The formula follows by removing the linear contribution of the middle observation. The derivation below uses a process with constant mean, finite positive variance, and autocorrelation depending only on lag, so the lag-1 correlation is the same for adjacent pairs. Center the process by subtracting its mean and write the centered observations as \(Z_t\). Let \(\rho_1=a\), \(\rho_2=b\), and let the variance be \(\gamma_0>0\).

1
Project each endpoint on the middle observation.
The coefficient in each projection is \(a\), because the covariance of an endpoint with the middle observation is \(a\gamma_0\), and the middle observation has variance \(\gamma_0\). The residuals are \(U=Z_t-aZ_{t-1}\) and \(V=Z_{t-2}-aZ_{t-1}\).
2
Calculate the residual covariance.
Expanding the covariance gives \(\operatorname{Cov}(U,V)=\gamma_0 b-a\gamma_0 a-a\gamma_0 a+a^2\gamma_0=\gamma_0(b-a^2)\).
3
Calculate the residual variances and correlation.
Each residual has variance \(\gamma_0(1-a^2)\). Dividing the residual covariance by the product of the residual standard deviations gives \(\{ \gamma_0(b-a^2)\}/\{\gamma_0(1-a^2)\}=(b-a^2)/(1-a^2)\).

Replacing the population correlations \(a,b\) by the sample ACF values \(r_1,r_2\) gives the stated Yule–Walker estimate. The derivation also shows why the adjustment is not simply “subtract \(r_1\) from \(r_2\)”: both the association removed and the residual scale matter.

Using the PACF Beyond Lag 2

At a higher lag, the PACF accounts for all intervening lags, not just the immediately preceding observation. Under the Yule–Walker convention, a useful recursion expresses the order-\(k\) projection in terms of the order-\((k-1)\) projection and the ACF values. For illustration, write \(\phi_{k,j}\) for the coefficient on lag \(j\) in the order-\(k\) projection. The PACF at lag \(k\) is \(\phi_{k,k}\), calculated by:

$$ \phi_{k,k} = \frac{\rho_k-\displaystyle\sum_{j=1}^{k-1}\phi_{k-1,j}\rho_{k-j}} {1-\displaystyle\sum_{j=1}^{k-1}\phi_{k-1,j}\rho_j}. $$

This recursion comes from solving the linear projection equations: each coefficient is chosen so the projection residual is uncorrelated with each included lag. Extending the projection by one lag and solving those equations gives the displayed final coefficient. For a sample estimate using this convention, replace the \(\rho\) values with the corresponding \(r\) values.

Worked Example: A Lag-3 PACF After Two Intermediate Lags

Suppose a hypothetical stationary series has autocorrelations \(\rho_1=0.6\), \(\rho_2=0.5\), and \(\rho_3=0.2\). Find the lag-2 and lag-3 PACFs using the recursion, then compare the lag-3 ACF and PACF.

Method. Calculate the order-2 coefficients first, since the order-3 calculation uses them. At order 1, \(\phi_{1,1}=0.6\). Thus:

$$ \phi_{2,2} =\frac{0.5-(0.6)^2}{1-(0.6)^2} =\frac{0.14}{0.64} =0.21875. $$

The other order-2 coefficient is \(\phi_{2,1}=\phi_{1,1}-\phi_{2,2}\phi_{1,1}\), so:

$$ \phi_{2,1}=0.6-(0.21875)(0.6)=0.46875. $$

Use these two coefficients in the order-3 recursion:

$$ \phi_{3,3} =\frac{0.2-(0.46875)(0.5)-(0.21875)(0.6)} {1-(0.46875)(0.6)-(0.21875)(0.5)} =\frac{-0.165625}{0.609375} =-\frac{53}{195} \approx -0.2718. $$

Interpretation and check. The lag-3 ACF is \(0.2\), a positive overall lag-3 correlation. The lag-3 PACF is approximately \(-0.2718\), a negative association after accounting linearly for lags 1 and 2. The difference is possible because the intervening observations contribute to the overall lagged association. The denominator is positive, and multiplying \(0.609375\) by \(-53/195\) gives \(-0.165625\), confirming the final division.

Common Mistakes and Interpretation

The ACF and PACF answer related but distinct questions. The ACF at lag \(k\) includes the association between observations \(k\) steps apart without removing the contributions of intermediate lags. The PACF at lag \(k\) measures their remaining linear association after those intermediate observations have been accounted for. Neither quantity, on its own, establishes a causal effect.

Common exam trap: If \(r_1=0.4\) and \(r_2=0.3\), it is wrong to call \(0.3\) the lag-2 PACF. The correct calculation is \((0.3-0.4^2)/(1-0.4^2)=0.14/0.84\approx0.1667\). The value \(0.3\) is the lag-2 ACF; approximately \(0.1667\) is the lag-2 PACF under the Yule–Walker convention.

A PACF estimate is not automatically a “more accurate” ACF. It is an adjusted measure answering a different question. In short series, estimates can also be sensitive to the sample and to the estimation convention. Interpret the requested lag and measure carefully, and do not infer a pattern merely from one value.

Key takeaway: The ACF describes overall linear association at a lag. The PACF describes association at that lag after linearly accounting for the intervening lags. At lag 1 they agree; at higher lags they need not.

Check Your Understanding

For numerical questions, use the lag-2 formula or Yule–Walker convention stated above. Give numerical answers to four decimal places where needed.

  1. A hypothetical series has \(r_1=0.4\) and \(r_2=0.12\). Calculate its estimated lag-2 PACF. (Answer: \(-0.0476\))
  2. A time-series summary gives \(\rho_1=-0.3\) and \(\rho_2=0.45\). Calculate the lag-2 PACF and state whether its sign agrees with the lag-2 ACF. (Answer: PACF \(=0.3956\); its sign is positive, agreeing with the positive lag-2 ACF.)
  3. For a hypothetical stationary series, \(\rho_1=0.4\), \(\rho_2=0.25\), and \(\rho_3=0.3\). Using the Yule–Walker recursion, calculate \(\phi_{2,2}\), \(\phi_{2,1}\), and \(\phi_{3,3}\). (Answer: \(\phi_{2,2}=0.1071\), \(\phi_{2,1}=0.3571\), and \(\phi_{3,3}\approx0.2022\).)
  4. A valid estimated summary has an ACF of \(0.02\) and a PACF of \(-0.45\) at lag 3. What does each value describe, and are the signs inconsistent? (Answer: The ACF describes the overall lag-3 association, while the PACF describes the lag-3 association after accounting for lags 1 and 2; the signs are not inconsistent.)
  5. Using the stated finite-sample ACF convention, a real-valued series has \(r_2=-0.31\). What is \(r_{-2}\), and does finite sample size remove this symmetry? (Answer: \(r_{-2}=-0.31\); no, the sample ACF remains symmetric in positive and negative lags.)

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