Why The PACF Cuts Off For AR(p)

Linear Prediction and the PACF

An autoregressive model predicts the current observation from a finite number of past observations. The partial autocorrelation function (PACF) describes the additional linear contribution of a particular lag after the nearer lags have already been included. This makes the PACF a natural way to express the finite prediction structure of an autoregressive process.

The key result is exact at the population level: for a stationary AR(p) process, the PACF at lag \(k\) is zero for \(k>p\). At lag \(p\), its value is the final AR coefficient, so it is nonzero precisely when the model truly needs that final lag.

Definition: Let \(X_t\) be a stationary process with mean \(\mu\), and write \(Z_t=X_t-\mu\). For a positive integer \(k\), linearly regress \(Z_t\) on \(Z_{t-1},\ldots,Z_{t-k}\). The population PACF at lag \(k\), denoted \(\alpha_{kk}\), is the coefficient of \(Z_{t-k}\) in this regression. Equivalently, it is the correlation between the residuals from linearly predicting \(Z_t\) and \(Z_{t-k}\) using \(Z_{t-1},\ldots,Z_{t-k+1}\). These definitions assume the relevant prediction coefficients are uniquely determined.

“Partial” is important: the PACF at lag \(k\) is not generally the ordinary correlation between \(X_t\) and \(X_{t-k}\). It measures the lag-\(k\) contribution after accounting linearly for the intervening lags. For example, an AR(1) process can have nonzero ordinary correlations at every positive lag, even though its PACF is zero at every lag greater than one.

Why the Cutoff Holds

Consider a zero-mean, stationary, causal AR(p) process

Z_t=\phi_1Z_{t-1}+\phi_2Z_{t-2}+\cdots+\phi_pZ_{t-p}+\varepsilon_t.

Assume the innovation sequence has mean zero and variance \(\sigma_\varepsilon^2>0\), and that each \(\varepsilon_t\) is uncorrelated with all observations from earlier times. A sufficient standard setup is independent, identically distributed innovations with finite variance and AR coefficients satisfying the stationarity condition. Causality and stationarity ensure the model describes a stable process rather than a nonstationary recursion. The coefficient vector in each regression must also be identifiable; in particular, the covariance matrix of the included lags must be nonsingular.

For any \(k\ge p\), the AR equation writes \(Z_t\) as a linear combination of the first \(p\) regressors, plus the innovation. The innovation is uncorrelated with each of \(Z_{t-1},\ldots,Z_{t-k}\), so the prediction error is orthogonal to every included regressor. Those are exactly the conditions for the population least-squares coefficients. Uniqueness then determines the coefficients.

Use the AR equation.
For a regression on \(k\ge p\) lags, propose coefficients \((\phi_1,\ldots,\phi_p,0,\ldots,0)\).

Identify the prediction error.
Subtracting this prediction from \(Z_t\) leaves \(\varepsilon_t\).

Check orthogonality.
By the innovation assumption, \(E[\varepsilon_t Z_{t-j}]=0\) for every \(j=1,\ldots,k\).

Read the PACF from the last coefficient.
The unique regression has zero coefficients after lag \(p\), so \(\alpha_{kk}=0\) for \(k>p\). At \(k=p\), \(\alpha_{pp}=\phi_p\).

This also explains a useful distinction. The final coefficient \(\phi_p\) cannot be made zero by cancellation with other coefficients in the regression: the regression coefficients are exactly \((\phi_1,\ldots,\phi_p)\). Therefore \(\alpha_{pp}=0\) if and only if \(\phi_p=0\). If \(\phi_p=0\), the same process can be represented using fewer AR lags.

Worked Examples

Worked Example: Checking the Cutoff for an AR(1)

A hypothetical monthly claim-count index, after centering, follows \(Z_t=0.6Z_{t-1}+\varepsilon_t\). Assume a stationary causal process with independent innovations \(\varepsilon_t\sim N(0,1)\). Find the PACF at lags 1 and 2, and verify the lag-2 result directly from its regression equations.

Method. At lag 1, the regression is the AR(1) equation itself, so \(\alpha_{11}=0.6\). To check lag 2, first obtain the needed autocorrelations. Write \(\rho_j=\operatorname{Corr}(Z_t,Z_{t-j})\). The AR(1) equation gives \(\rho_1=0.6\), and then \(\rho_2=0.6\rho_1=0.36\).

The standardized population regression of \(Z_t\) on \(Z_{t-1}\) and \(Z_{t-2}\) has covariance matrix and right-hand side

\begin{pmatrix}1&0.6\\0.6&1\end{pmatrix} \begin{pmatrix}b_1\\b_2\end{pmatrix} = \begin{pmatrix}0.6\\0.36\end{pmatrix}.

The proposed coefficients are \((b_1,b_2)=(0.6,0)\). Multiplication checks both equations:

\begin{pmatrix}1&0.6\\0.6&1\end{pmatrix} \begin{pmatrix}0.6\\0\end{pmatrix} = \begin{pmatrix}0.6\\0.36\end{pmatrix}.

Thus \(\alpha_{22}=b_2=0\), while the ordinary lag-2 correlation is \(\rho_2=0.36\). The model’s lag-2 association is explained by lag 1; it adds no partial linear contribution. The covariance matrix is nonsingular since its determinant is \(1-0.6^2=0.64>0\), so the solution is unique. The calculation avoids the common mistake of confusing the ACF value \(0.36\) with the PACF at lag 2.

Worked Example: An AR(2) Has Its PACF at Lag 2

A hypothetical quarterly reserve-change series, centered about its mean, follows \(Z_t=0.5Z_{t-1}-0.2Z_{t-2}+\varepsilon_t\), with independent innovations \(\varepsilon_t\sim N(0,1)\). The coefficient magnitudes sum to \(0.7<1\), a sufficient condition for a causal stationary solution. Find the PACF at lag 2 and verify that the PACF at lag 3 is zero.

Method. At lag 2, the regression is the AR equation, so its coefficients are \(0.5\) and \(-0.2\). In particular, \(\alpha_{22}=-0.2\). To verify the next lag numerically, calculate the required autocorrelations from the model. Multiplying the AR equation by the appropriate past observations and taking expectations gives

\rho_1=0.5-0.2\rho_1,\qquad \rho_2=0.5\rho_1-0.2,\qquad \rho_3=0.5\rho_2-0.2\rho_1.

Solving in order,

\rho_1=\frac{0.5}{1.2}=\frac{5}{12},\qquad \rho_2=0.5\left(\frac{5}{12}\right)-0.2 =\frac{5}{24}-\frac{1}{5} =\frac{1}{120},

and

\rho_3=0.5\left(\frac{1}{120}\right)-0.2\left(\frac{5}{12}\right) =\frac{1}{240}-\frac{1}{12} =-\frac{19}{240}.

For a regression on three lags, test the coefficient vector \((0.5,-0.2,0)\). The three rows of its covariance equations give

\begin{aligned} 0.5-0.2\rho_1&=\frac{5}{12}=\rho_1,\\ 0.5\rho_1-0.2&=\frac{1}{120}=\rho_2,\\ 0.5\rho_2-0.2\rho_1&=-\frac{19}{240}=\rho_3. \end{aligned}

So the vector satisfies all three equations. The lag covariance matrix is positive definite for this nondegenerate stationary process, making the regression solution unique. Therefore the lag-3 coefficient, and hence \(\alpha_{33}\), is zero. The trap is to assume the lag-3 ACF must also be zero: it is \(-19/240\), not zero. The cutoff applies to the PACF, not the ACF.

Worked Example: Verifying an AR(3) Cutoff at Lag 4

A hypothetical monthly policy-retention index, centered at its mean, follows \(Z_t=0.4Z_{t-1}+0.2Z_{t-2}-0.1Z_{t-3}+\varepsilon_t\), with independent innovations \(\varepsilon_t\sim N(0,1)\). Since \(0.4+0.2+0.1=0.7<1\), the coefficient absolute-sum condition ensures a causal stationary model. Find \(\alpha_{33}\) and verify \(\alpha_{44}=0\).

Method. At lag 3 the regression coefficients are the AR coefficients, giving \(\alpha_{33}=-0.1\). For the lag-4 check, compute the correlations needed in the four regression equations. The covariance identities are

\begin{aligned} \rho_1&=0.4+0.2\rho_1-0.1\rho_2,\\ \rho_2&=0.4\rho_1+0.2-0.1\rho_1,\\ \rho_3&=0.4\rho_2+0.2\rho_1-0.1,\\ \rho_4&=0.4\rho_3+0.2\rho_2-0.1\rho_1. \end{aligned}

From the second equation, \(\rho_2=0.3\rho_1+0.2\). Substitution in the first gives \(0.83\rho_1=0.38\), so \(\rho_1=38/83\). Then

\rho_2=\frac{28}{83},\qquad \rho_3=\frac{21}{166},\qquad \rho_4=\frac{6}{83}.

Now test coefficients \((0.4,0.2,-0.1,0)\) in the four-lag regression. The covariance matrix equations reduce row by row to

\begin{aligned} 0.4+0.2\rho_1-0.1\rho_2&=\frac{38}{83}=\rho_1,\\ 0.3\rho_1+0.2&=\frac{28}{83}=\rho_2,\\ 0.4\rho_2+0.2\rho_1-0.1&=\frac{21}{166}=\rho_3,\\ 0.4\rho_3+0.2\rho_2-0.1\rho_1&=\frac{6}{83}=\rho_4. \end{aligned}

Thus the proposed vector satisfies all four regression equations. Uniqueness gives \(\alpha_{44}=0\). This check also shows why the fourth autocorrelation need not vanish: \(\rho_4=6/83\), even though the fourth partial autocorrelation is zero. The example avoids treating “cutoff” as a claim that all dependence ends after the AR order.

What the Cutoff Does—and Does Not—Say

The result is about population linear projections under the stated stationary AR assumptions. It does not say that all observations more than \(p\) periods apart are independent, or that their ordinary correlations vanish. An AR process generally transmits dependence through intermediate observations, so its ACF may persist beyond lag \(p\). The PACF is zero at those higher lags because the first \(p\) observations already provide the complete linear prediction from the past.

The lag-\(p\) value is also informative about the order. For the specified AR(p) representation, \(\alpha_{pp}=\phi_p\). If \(\phi_p\ne0\), the lag-\(p\) PACF cannot be zero through cancellation. If \(\phi_p=0\), the model has a shorter effective AR order, and the PACF is zero at lag \(p\) as well as beyond it.

In an observed finite sample, estimated PACF values are random and will not usually be exactly zero beyond the true order. Sampling variation, model misspecification, and departures from stationarity can all affect the estimated pattern. The theoretical cutoff is a property of the population model, not a promise that every sample plot will display exact zeros.

Key takeaway: For a stationary causal AR(p) process with uncorrelated innovations and uniquely determined linear projections, the regression of \(Z_t\) on its first \(k\) lags has coefficients \((\phi_1,\ldots,\phi_p,0,\ldots,0)\) whenever \(k\ge p\). Consequently, \(\alpha_{pp}=\phi_p\), and \(\alpha_{kk}=0\) for \(k>p\).

Check Your Understanding

For each problem, identify the population PACF values requested and keep the distinction between partial and ordinary autocorrelation clear.

A centered stationary AR(2) process has coefficients \(\phi_1=0.7\) and \(\phi_2=-0.25\), with independent, finite-variance innovations. State its PACF at lags 2 and 5. (Answer: \(\alpha_{22}=-0.25\), \(\alpha_{55}=0\).)
A causal AR(3) process has coefficients \((0.2,0,0.15)\). What are its PACF values at lags 3 and 4, and what does the lag-3 value imply about the final coefficient? (Answer: \(\alpha_{33}=0.15\), \(\alpha_{44}=0\); the final coefficient is nonzero.)
A stationary AR(1) process has coefficient \(0.45\). Its lag-3 ACF is \(\rho_3=0.45^3\). Find its lag-3 PACF and distinguish it from that ACF. (Answer: \(\rho_3=0.091125\), \(\alpha_{33}=0\).)
A stationary AR(2) model has coefficients \((0.3,0.2)\). Give the population regression coefficients when predicting \(Z_t\) from its first four lags, and identify \(\alpha_{44}\) and \(\alpha_{22}\). (Answer: \((0.3,0.2,0,0)\); \(\alpha_{44}=0\), \(\alpha_{22}=0.2\).)
A sample PACF plot from a series fitted by a stationary AR(1) has a small estimated value at lag 4 rather than exactly zero. Does that contradict the population cutoff, and what is the correct interpretation? (Answer: No. The population PACF is zero at lag 4; a nonzero sample estimate can result from sampling variation.)

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