Why Account for Intermediate Observations?
In a time series, two observations can be associated simply because each is related to observations between them. For example, a monthly measure at time \(t\) may be correlated with the measure at \(t-2\) partly because both are related to the value at \(t-1\). To describe the association remaining after accounting for that intermediate value, use partial autocorrelation.
The word “accounting” has a specific meaning here: it refers to removing the linear association with the intermediate observations. It does not automatically mean that every possible nonlinear relationship, or every form of statistical dependence, has been removed.
Here, second-order stationarity means that the series has a finite, constant mean and finite variance, and its covariance between two observations depends only on the distance between their time indices. This assumption lets us describe the relevant correlations by lag. The definition concerns population quantities; a value calculated from observed data is an estimate and is subject to sampling variation.
Write \(\gamma_j=\operatorname{Cov}(X_t,X_{t-j})\) for the autocovariance at lag \(j\), and \(\rho_j=\gamma_j/\gamma_0\) for the autocorrelation, where \(\gamma_0=\operatorname{Var}(X_t)>0\). To define the partial autocorrelation at lag \(k\), the variance matrix of the intervening observations must be nonsingular, and both residual variances must be positive. These conditions ensure that the projections and their residual correlation are defined.
Residuals Give the Precise Meaning
A linear projection predicts a variable using a linear combination of specified observations. Its residual is the observed value minus that projection. Thus, to find the partial autocorrelation at lag \(k\), first remove from each endpoint the part linearly predictable from all \(k-1\) observations between the endpoints. Then calculate the correlation between what remains.
At lag \(2\), the only intermediate observation is \(X_{t-1}\). After removing the mean, project \(X_t\) and \(X_{t-2}\) separately on \(X_{t-1}\). Since the variance at each time is \(\gamma_0\) and the covariance of adjacent observations is \(\gamma_1\), each projection coefficient is \(\gamma_1/\gamma_0=\rho_1\). The two residuals are therefore \(X_t-\rho_1X_{t-1}\) and \(X_{t-2}-\rho_1X_{t-1}\), using centered observations.
The same residual variance applies to the other endpoint. Their residual covariance is \(\gamma_2-\gamma_1^2/\gamma_0=\gamma_0(\rho_2-\rho_1^2)\). Dividing this covariance by the common residual standard deviation gives the lag-two partial autocorrelation:
This expression requires \(1-\rho_1^2>0\), which is exactly the requirement that the residual variance after projecting on \(X_{t-1}\) be positive. The formula shows why \(\rho_2\) alone is not the lag-two partial autocorrelation: the term \(\rho_1^2\) accounts for the association explained through the intermediate observation.
Worked Example: Finding a Lag-Two Partial Autocorrelation
Suppose a hypothetical stationary monthly series describing a portfolio’s standardized claim activity has population autocorrelations \(\rho_1=0.6\) and \(\rho_2=0.5\). Find its lag-two partial autocorrelation. The task is to remove the linear contribution of the intervening month.
Method. Use the lag-two residual-correlation formula. First calculate the square of the lag-one autocorrelation: \(\rho_1^2=0.6^2=0.36\).
The numerator is \(\rho_2-\rho_1^2=0.5-0.36=0.14\). The denominator is \(1-\rho_1^2=1-0.36=0.64\). Therefore,
The lag-two partial autocorrelation is \(0.2188\), rounded to four decimal places. It is positive: after linearly accounting for the month between the endpoints, a positive association remains. The result is smaller than the ordinary lag-two autocorrelation of \(0.5\), because some of that ordinary association is accounted for by the shared relationship with the intermediate value.
Check. The denominator is positive, so the residual variance is positive. This example avoids the common mistake of reporting \(\rho_2=0.5\) as the partial autocorrelation.
Why the Definition Matches a Regression Coefficient
The residual-correlation definition is useful because it states directly what is being compared. There is also a closely related regression interpretation. Regress \(X_t\) on its previous \(k\) observations, \(X_{t-1},\ldots,X_{t-k}\). Under the stationary covariance assumptions and the nonsingularity conditions above, the coefficient of \(X_{t-k}\) in this regression equals the partial autocorrelation at lag \(k\).
The reason is that the coefficient on \(X_{t-k}\) measures the additional linear contribution of that observation after the other \(k-1\) predictors have been included. This can be seen by residualizing both the response and that predictor against the intermediate observations. The regression’s remaining coefficient is the slope relating those two residuals.
Let \(R_t\) be the residual from projecting \(X_t\) on \(X_{t-1},\ldots,X_{t-k+1}\).
Let \(R_{t-k}\) be the residual from projecting \(X_{t-k}\) on \(X_{t-1},\ldots,X_{t-k+1}\).
The least-squares coefficient of \(R_{t-k}\) in a regression of \(R_t\) is \(\operatorname{Cov}(R_t,R_{t-k})/\operatorname{Var}(R_{t-k})\). This follows from the least-squares condition that the fitted residual is uncorrelated with its predictor.
The endpoint residuals have equal variances: each uses the same covariance matrix for the intermediate observations, with the endpoint-to-intermediate covariances reversed. Reversing both the order of the intermediate observations and their covariance vector leaves the resulting residual variance unchanged. The residual slope is therefore their correlation, the partial autocorrelation.
The regression interpretation does not change the definition into a causal effect. A nonzero coefficient describes a remaining linear association in the specified regression; by itself it does not show that one observation causes another.
Worked Example: A Lag-Three Association That Disappears
For a hypothetical stationary series of monthly exposure changes, suppose the autocorrelations are \(\rho_1=0.5\), \(\rho_2=0.25\), and \(\rho_3=0.125\). Find the lag-three partial autocorrelation by accounting for both intervening months, \(X_{t-1}\) and \(X_{t-2}\).
Method. Form each endpoint’s residual after projection on the two intermediate observations. For \(X_t\), its covariances with \((X_{t-1},X_{t-2})\), in units of \(\gamma_0\), are \((0.5,0.25)\). The projection coefficients are \((0.5,0)\): multiplying the intermediate covariance matrix by this vector gives \((0.5,0.25)\).
For \(X_{t-3}\), the covariances with those same intermediate observations are \((0.25,0.5)\). Its projection coefficients are \((0,0.5)\), which reproduce those covariances. The residuals are therefore \(R_t=X_t-0.5X_{t-1}\) and \(R_{t-3}=X_{t-3}-0.5X_{t-2}\), with centered observations.
Their covariance, divided by \(\gamma_0\), is
Each residual variance is \(\gamma_0(1-0.5^2)=0.75\gamma_0\). Thus both residual standard deviations are positive, but their covariance is zero. Their correlation, and hence the lag-three partial autocorrelation, is \(0\).
Check. The ordinary lag-three autocorrelation is \(0.125\), but the association vanishes after accounting for both intermediate observations. This example avoids interpreting a nonzero ordinary autocorrelation as proof of a nonzero partial autocorrelation.
Worked Example: The Partial Association Can Have the Opposite Sign
Suppose a hypothetical stationary monthly series of service utilization has \(\rho_1=0.5\) and \(\rho_2=0.1\). Find the lag-two partial autocorrelation and interpret its sign.
Method. Account for the intervening month using the lag-two formula. The squared lag-one autocorrelation is \(0.5^2=0.25\). The adjusted numerator is \(0.1-0.25=-0.15\), and the denominator is \(1-0.25=0.75\). Hence,
The partial autocorrelation is \(-0.2\). The ordinary lag-two autocorrelation is positive, but the residual association after accounting for the intermediate month is negative. This is possible because the two measures answer different questions. The result avoids the mistake of assuming the partial association must have the same sign as the ordinary autocorrelation.
Interpretation and Common Traps
A partial autocorrelation at lag \(k\) describes the linear association between the two endpoints separated by \(k\) time steps after removing their linear relationships with the \(k-1\) observations between them. A positive value indicates positive residual association; a negative value indicates negative residual association; a value near zero indicates little linear residual association. It does not imply that the endpoints are independent after conditioning on the intermediate observations.
The qualifier “linear” matters. Residualizing against intermediate observations removes the fitted linear component, not necessarily every possible dependence. Even a zero partial autocorrelation does not, without additional distributional assumptions, establish conditional independence. Also, a sample estimate near zero is not proof that the population partial autocorrelation is exactly zero.
- Using the ordinary autocorrelation instead. At lag two, \(\rho_2\) is not generally the partial autocorrelation; account for \(\rho_1\) using the derived formula.
- Accounting for too few observations. At lag \(k\), account for every observation strictly between the endpoints, not only the one immediately before \(X_t\).
- Calling the result a causal effect. The definition describes linear association after adjustment, not causation.
- Equating zero partial correlation with independence. Zero means no remaining linear correlation under the stated projection; it does not generally rule out nonlinear dependence.
Partial autocorrelation is useful because it gives a lag-specific description of what remains after intermediate observations are accounted for. Its interpretation depends on keeping the endpoints, the intervening observations, and the linear nature of the adjustment explicit. The definition and calculations here are aligned with the SOA Exam SRM (Statistics for Risk Modeling) syllabus.
Check Your Understanding
For each problem, identify what is being adjusted for before interpreting the requested partial autocorrelation.
- A stationary series has \(\rho_1=0.4\) and \(\rho_2=0.28\). Calculate its lag-two partial autocorrelation. (Answer: \(0.1429\), to four decimal places.)
- A stationary series has \(\rho_1=0.6\) and \(\rho_2=0.2\). Calculate its lag-two partial autocorrelation and state its sign. (Answer: \(-0.2500\); negative.)
- A stationary series has \(\rho_1=0.4\), \(\rho_2=0.16\), and \(\rho_3=0.064\). Its lag-three endpoint residuals, after projection on both intervening observations, have covariance \(0\) and positive variances. What is the lag-three partial autocorrelation? (Answer: \(0\).)
- For a stationary series, \(\gamma_0=25\), \(\gamma_1=10\), and \(\gamma_2=7\). First convert the autocovariances to autocorrelations, then find the lag-two partial autocorrelation. (Answer: \(\rho_1=0.4\), \(\rho_2=0.28\), and \(\operatorname{PACF}(2)=0.1429\), to four decimal places.)
- A lag-two partial autocorrelation is \(0.3\). State precisely what association this describes, and give one conclusion it does not establish. (Answer: It is the correlation between \(X_t\) and \(X_{t-2}\) residuals after each is linearly projected on \(X_{t-1}\); it does not establish causation or conditional independence.)