The Geometric ACF Of An AR(1) Process

Why the AR(1) ACF Has a Geometric Pattern

An autocorrelation function (ACF) describes how strongly a time-series value is linearly related to values at different time lags. For a stationary AR(1) process, each additional lag applies the same factor, the autoregressive coefficient, to the preceding lag's correlation. This repeated multiplication creates a geometric pattern. If the coefficient is negative, the correlations alternate in sign as well as declining in magnitude.

The pattern is useful for calculating dependence at a specified lag and for understanding how a one-period relationship carries through multiple periods. The result applies to a stationary AR(1) process; it should not be used without checking that the model satisfies the stationarity condition.

Definition: A stationary AR(1) process with mean \(\mu\) can be written as \(X_t-\mu=\phi(X_{t-1}-\mu)+\varepsilon_t\), where \(\phi\) is the autoregressive coefficient. The innovations \(\varepsilon_t\) have mean zero and constant variance \(\sigma_\varepsilon^2\), and are uncorrelated across time and with past values of the process. For the stationary AR(1) solution, \(|\phi|<1\). Define the lag-\(k\) autocovariance by \(\gamma_k=\operatorname{Cov}(X_t,X_{t-k})\), and the lag-\(k\) autocorrelation by \(\rho_k=\gamma_k/\gamma_0\), for \(k\geq 0\).

Here \(\gamma_0=\operatorname{Var}(X_t)\), which is positive for a non-degenerate process. The condition \(|\phi|<1\) ensures the stationary solution has finite variance. The ACF formula below is a result for that stationary solution, not a general formula for every process that can be written as a first-order autoregression.

Deriving the AR(1) Autocorrelation Function

The AR(1) equation says that the centered value at time \(t\) is \(\phi\) times the centered value one period earlier, plus the new innovation. To relate \(X_t\) to \(X_{t-k}\), take the covariance of both sides of the equation with \(X_{t-k}\), where \(k\geq 1\).

\begin{aligned} \gamma_k &=\operatorname{Cov}(X_t,X_{t-k})\\ &=\operatorname{Cov}\bigl(\phi(X_{t-1}-\mu)+\mu+\varepsilon_t,\ X_{t-k}\bigr)\\ &=\phi\operatorname{Cov}(X_{t-1},X_{t-k}) +\operatorname{Cov}(\varepsilon_t,X_{t-k})\\ &=\phi\gamma_{k-1}. \end{aligned}

The constant \(\mu\) contributes no covariance. The innovation \(\varepsilon_t\) is uncorrelated with past process values, and \(X_{t-k}\) is a past value when \(k\geq 1\), so its covariance with that innovation is zero. The covariance recursion therefore holds at every positive lag. Repeated application gives \(\gamma_k=\phi^k\gamma_0\), and dividing by \(\gamma_0\) gives the ACF.

Theorem: For a stationary AR(1) process with \(|\phi|<1\), the autocorrelation at lag \(k\) is

\rho_k=\phi^k,\qquad k\geq 0.

In particular, \(\rho_0=1\), \(\rho_1=\phi\), and \(\rho_k=\phi\rho_{k-1}\) for \(k\geq 1\). For negative lags, stationarity gives \(\rho_{-k}=\rho_k\).

The first lag's autocorrelation equals the coefficient, but the coefficient is not itself the autocorrelation at every lag. At lag 2 the coefficient is squared; at lag 3 it is cubed; and the pattern continues. The autocovariances follow the corresponding scale relationship \(\gamma_k=\gamma_0\phi^k\). Thus, knowing the variance is necessary to calculate autocovariances, but not to calculate autocorrelations.

Reading the Size and Sign of the ACF

For \(0<\phi<1\), all positive-lag autocorrelations are positive and decrease in magnitude as the lag increases. Their ratio at consecutive lags is \(\rho_k/\rho_{k-1}=\phi\). For \(-1<\phi<0\), the magnitudes still decrease geometrically, but the sign alternates: odd-lag correlations are negative and even-lag correlations are positive.

Coefficient	First few positive-lag correlations	Pattern
\(\phi=0.6\)	\(\rho_1=0.6,\ \rho_2=0.36,\ \rho_3=0.216\)	Positive, with decreasing magnitude
\(\phi=-0.6\)	\(\rho_1=-0.6,\ \rho_2=0.36,\ \rho_3=-0.216\)	Alternating signs, with decreasing magnitude

For either sign, \(|\rho_k|=|\phi|^k\). This means that the sign and the magnitude should be considered separately: the sign comes from whether the power \(k\) is odd or even, while the magnitude comes from raising \(|\phi|\) to \(k\). A negative correlation does not mean the dependence is increasing in strength; it means the linear relationship is inverse at that lag.

Worked Example: Positive Coefficient and Lagged Covariances

A hypothetical stationary monthly index follows an AR(1) model with \(\phi=0.6\) and variance \(\gamma_0=25\) index-units squared. Calculate \(\rho_1,\rho_2,\rho_4\) and \(\gamma_2,\gamma_4\). State what happens to the correlation magnitude from lag 2 to lag 4.

Method and assumptions. Use \(\rho_k=\phi^k\) and \(\gamma_k=\gamma_0\rho_k\). The process is specified as stationary, so the AR(1) ACF result applies.

Autocorrelations. At lag 1, \(\rho_1=0.6^1=0.6\). At lag 2, \(\rho_2=0.6^2=0.36\). At lag 4, \(\rho_4=0.6^4=(0.6^2)^2=0.36^2=0.1296\).

Autocovariances. At lag 2, \(\gamma_2=25(0.36)=9\) index-units squared. At lag 4, \(\gamma_4=25(0.1296)=3.24\) index-units squared.

Interpretation and check. The lag-4 correlation is positive and smaller in magnitude than the lag-2 correlation. In two more periods, the correlation is multiplied by \(0.6^2=0.36\): \(0.36(0.36)=0.1296\). The covariance calculation is consistent with the ACF because \(3.24/25=0.1296\). The common mistake avoided here is multiplying the lag-2 correlation by \(\phi\) only once when moving to lag 4; there are two additional lags, so the multiplier is \(\phi^2\).

Worked Example: Negative Coefficient and Alternating Signs

A hypothetical stationary weekly measurement has an AR(1) coefficient \(\phi=-0.8\) and variance \(\gamma_0=16\). Calculate the autocorrelations at lags 1 through 5, and calculate the lag-4 autocovariance. Give the correlations to four decimal places.

Method and assumptions. Apply \(\rho_k=(-0.8)^k\) for each lag. The sign is determined by whether the lag is odd or even. Use \(\gamma_4=16\rho_4\) for the autocovariance.

Calculate each power before rounding. \(\rho_1=(-0.8)^1=-0.8\); \(\rho_2=(-0.8)^2=0.64\); \(\rho_3=(-0.8)^3=-0.512\); \(\rho_4=(-0.8)^4=0.4096\); and \(\rho_5=(-0.8)^5=-0.32768\).

To four decimal places, these are \(-0.8000,\ 0.6400,\ -0.5120,\ 0.4096,\ -0.3277\), respectively. The lag-4 autocovariance is \(\gamma_4=16(0.4096)=6.5536\) measurement-units squared.

Interpretation and check. The sign alternates, while the magnitudes are \(0.8,\ 0.64,\ 0.512,\ 0.4096,\ 0.32768\), a decreasing sequence. Also, \(\gamma_4/\gamma_0=6.5536/16=0.4096=\rho_4\). The common mistake avoided is treating a negative coefficient as making every positive-lag correlation negative. Even powers of \(-0.8\) are positive.

Using the ACF to Recover the Coefficient

The ACF formula can also be used in reverse. In a theoretical stationary AR(1) process, \(\rho_1=\phi\), so a specified lag-1 autocorrelation directly gives the coefficient. If a different lag is given, the power relationship may identify the coefficient, but sign information must be handled carefully. For example, a positive \(\rho_2\) alone cannot distinguish a positive coefficient from a negative one because \(\rho_2=\phi^2\).

Worked Example: Recovering a Negative Coefficient and Checking the ACF

For a hypothetical stationary AR(1) process, the theoretical autocorrelations are \(\rho_2=0.25\) and \(\rho_3=-0.125\). The variance is \(\gamma_0=36\). Recover \(\phi\), calculate \(\rho_4\), and find \(\gamma_4\). Check that the given autocorrelations are consistent with the recovered coefficient.

Method and assumptions. The AR(1) relationships are \(\rho_2=\phi^2\) and \(\rho_3=\phi^3\). Divide the third-lag relationship by the second-lag relationship to obtain \(\phi\), since \(\rho_2\ne 0\).

Recover the coefficient. \(\phi=\rho_3/\rho_2=(-0.125)/0.25=-0.5\). This sign is also consistent with the negative value of \(\rho_3\).

Calculate lag 4. \(\rho_4=(-0.5)^4=0.0625\). Then \(\gamma_4=\gamma_0\rho_4=36(0.0625)=2.25\) squared units.

Check consistency. The recovered coefficient gives \((-0.5)^2=0.25=\rho_2\) and \((-0.5)^3=-0.125=\rho_3\), so both supplied values agree with the AR(1) pattern. A common mistake is to take only the positive square root of \(\rho_2=0.25\), obtaining \(0.5\). That choice would predict \(\rho_3=(0.5)^3=0.125\), contradicting the given negative \(\rho_3\). The odd-lag sign identifies the negative coefficient.

Common Exam Traps

The formula is compact, but small errors in the lag, sign, or quantity being requested can change the answer. Before calculating, identify whether the question asks for a correlation or covariance, and whether it specifies a positive lag or a negative lag.

Using \(\phi\) at every lag. The coefficient is \(\rho_1\), not the autocorrelation at every lag. At lag \(k\), use \(\phi^k\).
Ignoring the parity of a negative power. For negative \(\phi\), odd powers are negative and even powers are positive.
Confusing correlation with covariance. The ACF is dimensionless. Convert to autocovariance by multiplying by \(\gamma_0\): \(\gamma_k=\gamma_0\rho_k\).
Rounding powers too early. Keep the full value of \(\phi^k\) through subsequent calculations, then round the final result to the requested precision.
Applying the stationary formula without its condition. The geometric ACF result here assumes a stationary AR(1) solution with \(|\phi|<1\).

For a negative coefficient, do not take the absolute value before calculating the signed correlation. Absolute values are useful for describing the size of the correlation, but the signed ACF is \(\rho_k=\phi^k\). The sign can matter when describing whether observations at a given lag tend to move together or in opposite directions.

Key takeaway: For a stationary AR(1) process, \(\rho_k=\phi^k\). Its magnitude decreases geometrically because \(|\phi|<1\); if \(\phi<0\), the sign alternates between odd and even lags. Multiply by the process variance only when the requested quantity is an autocovariance.

Check Your Understanding

For each question, assume the stated process is a stationary AR(1) process. Keep signs when calculating correlations.

A process has \(\phi=0.4\) and \(\gamma_0=10\). Calculate \(\rho_4\) and \(\gamma_4\). (Answer: \(\rho_4=0.0256,\ \gamma_4=0.256\).)
A process has \(\phi=-0.6\). List \(\rho_1,\rho_2,\rho_3,\rho_4\), giving each to four decimal places. (Answer: \(-0.6000,\ 0.3600,\ -0.2160,\ 0.1296\).)
A theoretical AR(1) ACF has \(\rho_1=0.75\). Calculate \(\rho_5\) to four decimal places. (Answer: \(0.2373\).)
A theoretical AR(1) ACF has \(\rho_2=0.16\) and \(\rho_3=-0.064\). Recover \(\phi\) and calculate \(\rho_5\). (Answer: \(\phi=-0.4,\ \rho_5=-0.01024\).)
A process has \(\phi=-0.5\) and \(\gamma_0=20\). Find the smallest positive lag \(k\) for which \(|\rho_k|<0.02\), and calculate \(\rho_k\) and \(\gamma_k\) at that lag. (Answer: \(k=6,\ \rho_6=0.015625,\ \gamma_6=0.3125\).)

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