Computing Partial Autocorrelation In A Short Series

Why Compute a Partial Autocorrelation?

Two time points can be correlated partly because both are related to a time point between them. A partial autocorrelation measures the remaining linear association between the endpoints after accounting for the intervening observations. Computing it requires more than reading an ordinary autocorrelation at the same lag: the indirect association must be adjusted for.

For example, a value of \(0.6\) for the lag-two autocorrelation does not by itself tell us how strongly the present observation is associated with the observation two periods earlier after accounting for the observation one period earlier. The lag-two partial autocorrelation answers that adjusted question.

Definition: For a weakly stationary process with finite second moments, the lag-\(k\) partial autocorrelation is the correlation between the residuals from linearly predicting \(X_t\) and \(X_{t-k}\) using the intervening observations \(X_{t-1},\ldots,X_{t-k+1}\). Equivalently, it is the coefficient on \(X_{t-k}\) in the linear projection of \(X_t\) on \(X_{t-1},\ldots,X_{t-k}\). The lag-one partial autocorrelation equals the lag-one autocorrelation because there are no intervening observations to adjust for.

Here, weak stationarity means that the mean is constant over time and the covariance between two observations depends only on their separation in time. Write \(\rho_j\) for the population autocorrelation at lag \(j\), and \(r_j\) for its sample estimate. In practice, sample partial autocorrelations are often calculated by substituting sample autocorrelations into the same formulas. The correlations used must give nonzero residual variances; otherwise the relevant partial correlation is undefined.

The Lag-Two Calculation

At lag two, there is just one intervening observation, \(X_{t-1}\). The partial autocorrelation compares \(X_t\) with \(X_{t-2}\) after each has had its linear relationship with \(X_{t-1}\) removed. Under stationarity, both endpoints have the same correlation \(\rho_1\) with the middle observation.

\operatorname{PACF}(2) = \frac{\rho_2-\rho_1^2}{1-\rho_1^2}

The numerator is the covariance of the residualized endpoints when the original variables are standardized to variance one. The denominator is their common residual standard deviation squared. The following derivation shows why both parts appear.

Remove the middle observation's linear contribution.
For standardized observations, the linear projection of either endpoint on the middle observation has coefficient \(\rho_1\). Define residuals \(U=Z_t-\rho_1Z_{t-1}\) and \(V=Z_{t-2}-\rho_1Z_{t-1}\), where each \(Z\) has mean zero and variance one.

Find their covariance.
Expanding the covariance and using \(\operatorname{Corr}(Z_t,Z_{t-2})=\rho_2\) and both adjacent correlations equal to \(\rho_1\) gives \(\operatorname{Cov}(U,V)=\rho_2-\rho_1^2-\rho_1^2+\rho_1^2=\rho_2-\rho_1^2\).

Find each residual variance and standardize.
\(\operatorname{Var}(U)=1-2\rho_1^2+\rho_1^2=1-\rho_1^2\), and the same holds for \(V\). Dividing their covariance by the product of their residual standard deviations yields \((\rho_2-\rho_1^2)/(1-\rho_1^2)\).

This derivation uses linear projection and correlation; it does not require the process to be normally distributed. If \(|\rho_1|=1\), the residual variance is zero, so this formula does not define a partial correlation.

Worked Example: Adjusting the Lag-Two Correlation

A hypothetical monthly series has sample autocorrelations \(r_1=0.50\) and \(r_2=0.40\). Estimate its lag-two partial autocorrelation and explain what the result says about the adjustment.

Method. Substitute the sample autocorrelations into the lag-two formula:

\widehat{\operatorname{PACF}}(2) = \frac{r_2-r_1^2}{1-r_1^2}

First square the lag-one autocorrelation: \(r_1^2=(0.50)^2=0.25\). The adjusted covariance term is \(r_2-r_1^2=0.40-0.25=0.15\), and the residual variance is \(1-r_1^2=1-0.25=0.75\). Therefore,

\widehat{\operatorname{PACF}}(2) = \frac{0.15}{0.75} = 0.20.

The estimated partial autocorrelation is \(0.20\). The raw lag-two autocorrelation was \(0.40\); after adjusting for the intervening observation, the remaining linear association is smaller. Check: the denominator is positive, and the adjusted value lies between \(-1\) and \(1\). The common mistake to avoid is reporting \(r_2=0.40\) as the partial autocorrelation without making the adjustment.

Calculating a Lag-Three Partial Autocorrelation

At lag three, two intervening observations must be accounted for. The calculation can be organized with the Durbin–Levinson recursion. At order \(k\), let \(\phi_{k,j}\) be the coefficient on lag \(j\) in the linear projection using the first \(k\) lags. The new coefficient at order \(k\), \(\phi_{k,k}\), is the lag-\(k\) partial autocorrelation.

Recursion: Set \(\phi_{1,1}=\rho_1\). For \(k\geq 2\),

\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}, \qquad \phi_{k,j}=\phi_{k-1,j}-\phi_{k,k}\phi_{k-1,k-j} \quad (1\leq j<k).

Interpretation: The numerator adjusts the lag-\(k\) correlation for the projection using the shorter set of lags. The denominator is the corresponding residual variance factor. For sample calculations, replace each \(\rho_j\) with \(r_j\).

The recursion is a computational form of successive linear projection. At each order, it adds one more lag, adjusts the earlier projection coefficients, and calculates the new coefficient for the added lag. It avoids treating a lag-three correlation as if only one intermediate observation were present.

Worked Example: Recursing to Lag Three

A hypothetical quarterly series has sample autocorrelations \(r_1=0.60\), \(r_2=0.50\), and \(r_3=0.30\). Calculate its estimated partial autocorrelation at lag three. Keep enough digits in intermediate calculations to avoid premature rounding.

Step 1: Calculate the lag-two coefficient. At order two,

\phi_{2,2} = \frac{0.50-(0.60)^2}{1-(0.60)^2} = \frac{0.14}{0.64} = 0.21875.

The adjusted coefficient on lag one at order two is \(\phi_{2,1}=r_1-\phi_{2,2}r_1=0.60-(0.21875)(0.60)=0.46875\).

Step 2: Calculate the lag-three numerator.

r_3-\left(\phi_{2,1}r_2+\phi_{2,2}r_1\right) = 0.30-\left((0.46875)(0.50)+(0.21875)(0.60)\right) = 0.30-(0.234375+0.13125) = -0.065625.

Step 3: Calculate the denominator.

1-\left(\phi_{2,1}r_1+\phi_{2,2}r_2\right) = 1-\left((0.46875)(0.60)+(0.21875)(0.50)\right) = 1-(0.28125+0.109375) = 0.609375.

Step 4: Divide.

\widehat{\operatorname{PACF}}(3) = \phi_{3,3} = \frac{-0.065625}{0.609375} = -\frac{7}{65} \approx -0.1077.

The estimated lag-three partial autocorrelation is approximately \(-0.1077\). Check: the denominator is positive and the result is a valid correlation. The negative sign indicates a small negative adjusted linear association. The common mistake to avoid is using \(r_3=0.30\) directly, which ignores the two shorter lags.

Computing From Three Time Points

A second way to understand the lag-two formula is to view it as a partial correlation among three standardized observations. If the correlation of each adjacent pair is \(\rho_1\) and the endpoint correlation is \(\rho_2\), then the partial correlation between the endpoints controlling for the middle point is

\frac{\rho_2-\rho_1\rho_1} {\sqrt{(1-\rho_1^2)(1-\rho_1^2)}}.

Because the two residual variances are equal under stationarity, the denominator simplifies to \(1-\rho_1^2\), giving the lag-two formula. This perspective emphasizes what is removed: the linear association with the middle observation, not every possible relationship involving that observation.

Worked Example: Interpreting Three Pairwise Correlations

For three consecutive standardized observations in a hypothetical series, suppose the correlation of each adjacent pair is \(0.70\), and the correlation of the two endpoints is \(0.55\). Find the endpoint partial correlation given the middle observation, then compare it with the raw endpoint correlation.

Let \(\rho_1=0.70\) and \(\rho_2=0.55\). The adjusted covariance term is \(0.55-(0.70)(0.70)=0.55-0.49=0.06\). Each endpoint residual variance is \(1-(0.70)^2=1-0.49=0.51\), so their residual standard deviation product is \(\sqrt{0.51(0.51)}=0.51\). Thus,

\operatorname{Corr}(Z_t,Z_{t-2}\mid Z_{t-1}\text{ linearly adjusted}) = \frac{0.06}{0.51} = 0.117647\ldots \approx 0.1176.

The raw endpoint correlation is \(0.55\), but the adjusted association is about \(0.1176\). Much of the raw association is accounted for by the strong adjacent correlations with the middle observation. Check: \(0.1176\) is less than \(0.55\) and has the expected positive sign because \(0.55\) exceeds \(0.49\). The common mistake is to subtract \(0.70\) once rather than the product \(0.70\times0.70\), which represents the linear association transmitted through the middle point.

Common Calculation Traps

Confusing autocorrelation with partial autocorrelation. The lag-\(k\) autocorrelation is an unadjusted correlation. The lag-\(k\) partial autocorrelation accounts for the intervening lags.
Forgetting the denominator. At lag two, subtracting \(r_1^2\) from \(r_2\) gives the adjusted covariance term, not the partial correlation. Divide by \(1-r_1^2\).
Using the lag-two formula at lag three. The lag-three calculation must adjust for both lag one and lag two; use the recursion or the equivalent multiple linear projection.
Rounding too early. Keep full precision in intermediate recursion steps, then round the final estimate. Small intermediate errors can be magnified if a denominator is small.
Overinterpreting the adjustment. Partial autocorrelation removes linear contributions of the specified intervening observations. It does not establish causation or imply conditional independence in a non-Gaussian process.

Correct computation matters because the adjusted association answers a different question from the raw ACF value. A candidate should identify the requested lag, include every intervening observation in the adjustment, check that residual variance is positive, and then interpret the sign and magnitude in context. These calculations are aligned with the SOA Exam SRM (Statistics for Risk Modeling) syllabus.

Check Your Understanding

For each question, calculate the requested value before checking the answer. Round final estimates to four decimal places where needed.

A hypothetical weekly series has \(r_1=0.30\) and \(r_2=0.21\). Calculate its lag-two partial autocorrelation. (Answer: \(0.1319\))
A hypothetical sensor series has \(r_1=0.50\) and \(r_2=0.10\). Calculate its lag-two partial autocorrelation and state whether the adjusted association is positive or negative. (Answer: \(-0.2000\), negative)
A hypothetical monthly series has \(r_1=0.40\) and \(r_2=0.25\). Calculate its lag-two partial autocorrelation and identify the residual variance factor used in the denominator. (Answer: \(0.1071\); denominator \(1-r_1^2=0.84\))
A hypothetical quarterly series has \(r_1=0.40\), \(r_2=0.30\), and \(r_3=0.20\). Use the recursion to calculate the lag-three partial autocorrelation. (Answer: \(2/49\approx0.0408\))
A hypothetical series has \(r_1=0.80\) and \(r_2=0.65\). Calculate the lag-two partial autocorrelation and explain why it is not \(0.65\). (Answer: \(0.0278\); the lag-one linear contribution is adjusted out.)

What You'll Learn