Why the Core Properties Matter
An autocorrelation summarizes the linear association between values of a time series separated by a specified lag. Because it is expressed without measurement units, it is easier to compare across lags than an autocovariance. Two basic checks apply to every theoretical autocorrelation function: its value at lag zero is one, and no value can be less than minus one or greater than one. These properties are useful both for interpreting dependence and for checking whether proposed calculations are possible.
The results below concern a process whose mean and variance are constant over time and whose covariance between two observations depends only on their separation. These are the moment conditions needed for an autocovariance function indexed by lag. A later discussion develops stationarity in more detail. We also require finite, positive variance; if the variance is zero, the autocorrelation is not defined.
The denominator is the variance at one time point. Consequently, the ACF has no units: the units of covariance and variance cancel. This makes \(\rho(k)\) a standardized measure of linear association, not a measure of every possible type of dependence. In particular, an autocorrelation of zero does not, by itself, imply independence.
Deriving the Value at Lag Zero and the Bounds
At lag zero, both entries in the covariance are the same random variable. Thus \(\gamma(0)=\operatorname{Cov}(X_t,X_t)=\operatorname{Var}(X_t)\). Provided this variance is positive, dividing it by itself gives the ACF at zero:
To establish the bounds, write \(X_t-\mu\) and \(X_{t-k}-\mu\) for the centered observations, where \(\mu\) is the common mean. Each has variance \(\gamma(0)\), and their covariance is \(\gamma(k)\). A variance cannot be negative. Apply that fact to both the difference and the sum of the centered observations:
The first inequality implies \(\gamma(k)\leq\gamma(0)\); the second implies \(\gamma(k)\geq-\gamma(0)\). Since \(\gamma(0)>0\), dividing all parts by \(\gamma(0)\) gives the required result.
This is a consequence of the nonnegativity of variance, rather than an extra modeling assumption. It applies at every lag for which the ACF is defined.
For the endpoint interpretation, the variance of the difference is zero when \(\gamma(k)=\gamma(0)\). A random variable with zero variance equals its mean with probability one, so the centered observations must be equal. At the other endpoint, the variance of their sum is zero, so the centered observations must be negatives of each other. Values strictly between the endpoints indicate that neither exact relationship holds.
Worked Examples
Worked Example: An Alternating Process Reaches Both Bounds
Suppose a hypothetical monthly index is \(X_t=40+(-1)^t Z\), where \(t\) is an integer month, \(E[Z]=0\), and \(\operatorname{Var}(Z)=4\). Find its mean, variance, autocovariance at lag \(k\), and ACF. Check the value at lag zero and explain the lag-one result.
Method. Center the process by subtracting its mean, then compute the covariance between months \(t\) and \(t-k\). This example also shows why the bound can be attained exactly.
Mean and variance. Since \(E[Z]=0\), \(E[X_t]=40+(-1)^tE[Z]=40\). The centered value is \(X_t-40=(-1)^tZ\), so \(\operatorname{Var}(X_t)=((-1)^t)^2\operatorname{Var}(Z)=4\). Therefore \(\gamma(0)=4\).
Autocovariance. The centered value \(k\) months earlier is \(X_{t-k}-40=(-1)^{t-k}Z\). Hence \[ \begin{aligned} \gamma(k) &=E\big[((-1)^t Z)((-1)^{t-k}Z)\big]\\ &=(-1)^{2t-k}E[Z^2] =(-1)^k(4). \end{aligned} \] Here \(E[Z^2]=\operatorname{Var}(Z)=4\), since \(E[Z]=0\). Thus \(\gamma(k)=4(-1)^k\).
ACF and check. Dividing by \(\gamma(0)=4\) gives \(\rho(k)=4(-1)^k/4=(-1)^k\). At \(k=0\), \(\rho(0)=1\). At \(k=1\), \(\rho(1)=-1\), which is within the bounds and reflects exact sign reversal around 40: \(X_t-40=-(X_{t-1}-40)\). The result avoids the mistake of treating a negative autocorrelation as an impossible value.
Worked Example: Interpreting a Negative Autocorrelation
A hypothetical series of monthly benefit amounts, measured in dollars, has variance \(\gamma(0)=36\) dollars squared. Its covariance at lag 3 is \(\gamma(3)=-12\) dollars squared. Calculate \(\rho(3)\), check the bound, and interpret the sign and magnitude.
Method. Divide the lag-3 covariance by the variance at a single time point. The denominator is the variance, not its square root.
Calculation. \[ \rho(3)=\frac{\gamma(3)}{\gamma(0)} =\frac{-12}{36}=-\frac{1}{3}\approx-0.3333. \] The calculation uses no rounding until the final step; the decimal is rounded to four places.
Check and interpretation. Since \(-1\leq-1/3\leq1\), the value satisfies the required bound. The negative sign indicates an inverse linear association between amounts three months apart: observations above their mean tend to be paired with observations below their mean. A value of \(-1/3\) is not perfect inverse association, so it does not imply that every pair follows an exact sign reversal. It also does not establish a causal relationship.
The units cancel: both \(-12\) and \(36\) are measured in dollars squared. Reporting an autocorrelation in dollars or dollars squared would therefore be incorrect.
Worked Example: Detecting an Impossible Covariance
A proposed model for a hypothetical quarterly reserve index states that \(\gamma(0)=9\) and \(\gamma(2)=12\), in squared index units. Determine whether these values can describe a process with a valid ACF. Show the tempting calculation that leads to the problem and correct it.
Method. Use the variance at lag zero as the ACF denominator, then apply the bound. A covariance cannot have a larger absolute value than the variance when both time points have the same variance.
Calculation and check. \[ \rho(2)=\frac{\gamma(2)}{\gamma(0)} =\frac{12}{9}=\frac{4}{3}\approx1.3333. \] But \(4/3>1\), contradicting \(-1\leq\rho(2)\leq1\). Equivalently, the proposed values would give \(\operatorname{Var}(X_t-X_{t-2})=2(9)-2(12)=-6\), which cannot be a variance.
Common mistake and correction. Dividing \(12\) by the standard deviation \(3\) gives \(4\), but this is not an autocorrelation: the defining denominator is \(\gamma(0)=9\), not \(\sqrt{\gamma(0)}\). Using the correct denominator gives \(4/3\), which reveals that the supplied covariance values are impossible. This is not a case where the autocorrelation should be clipped to 1; at least one proposed moment value must be wrong.
Worked Example: Changing the Scale Does Not Change the ACF
A hypothetical series \(X_t\) has variance \(\gamma_X(0)=9\) and lag-2 covariance \(\gamma_X(2)=3\). A second analyst records \(Y_t=5-2X_t\). Find the lag-2 autocorrelation of \(Y_t\), and verify that it agrees with the ACF of \(X_t\).
Method. Calculate how adding a constant and multiplying by a fixed number change the variance and covariance. Then form the ratio defining the ACF.
Variance. Adding 5 does not change variance, while multiplying by \(-2\) multiplies variance by \((-2)^2=4\): \[ \gamma_Y(0)=\operatorname{Var}(5-2X_t)=4(9)=36. \]
Covariance. The constant disappears from covariance, and each of the two centered terms is multiplied by \(-2\): \[ \gamma_Y(2)=\operatorname{Cov}(5-2X_t,5-2X_{t-2}) =(-2)(-2)\gamma_X(2)=4(3)=12. \]
ACF and check. \[ \rho_Y(2)=\frac{12}{36}=\frac{1}{3}, \qquad \rho_X(2)=\frac{3}{9}=\frac{1}{3}. \] The results agree. The negative scale factor reverses the direction of each centered observation, but both entries in the covariance reverse together, so the covariance and variance are multiplied by the same positive factor. Their ratio is unchanged.
Practical Checks and Common Pitfalls
The ACF’s defining properties are useful safeguards when reading or calculating theoretical values. At lag zero, the answer must be one, not zero: an observation is perfectly correlated with itself when its variance is positive. At a nonzero lag, a proposed value outside \([-1,1]\) signals an error or incompatible moment assumptions. The appropriate response is to revisit the covariance, variance, or model assumptions—not to force the result back into range.
Several distinctions help avoid common errors:
- Do not confuse covariance and correlation. Covariance retains squared measurement units and can have a magnitude larger than one. The bounds apply to autocorrelation, after division by the variance.
- Use the variance in the denominator. For this lag-based ACF, \(\rho(k)=\gamma(k)/\gamma(0)\). Dividing by the standard deviation does not give the defined quantity.
- Do not interpret the sign as a guarantee about every pair. A negative value describes inverse linear association, not necessarily exact alternation. Only \(-1\) gives exact sign reversal of centered observations.
- Check that the denominator is positive. If \(\gamma(0)=0\), the ratio is undefined; it is not legitimate to assign the process an ACF of one at lag zero.
- Keep theoretical and observed quantities distinct. These results concern the process’s theoretical moments. A later topic considers estimates calculated from a finite observed series.
These properties matter because they make the ACF interpretable as a standardized measure and provide immediate checks on any proposed value. They do not identify a particular time-series model or establish independence; those questions require additional evidence and assumptions.
Check Your Understanding
For each problem, use the defining ratio and check whether the result is consistent with the ACF bounds.
- A hypothetical monthly index has \(\gamma(0)=49\) and \(\gamma(3)=14\). Find \(\rho(3)\) and state whether it is within the bounds. (Answer: \(\rho(3)=14/49=0.2857\), within the bounds.)
- A proposed quarterly series has \(\gamma(0)=16\) and \(\gamma(4)=-20\). Calculate the implied autocorrelation and decide whether the stated moments are possible. (Answer: \(\rho(4)=-20/16=-1.2500\); impossible because it is below \(-1\).)
- Let \(X_t=12+(-1)^tZ\), where \(E[Z]=0\) and \(\operatorname{Var}(Z)=2.25\). Find \(\gamma(0)\), \(\rho(0)\), and \(\rho(3)\). (Answer: \(\gamma(0)=2.25\), \(\rho(0)=1\), and \(\rho(3)=-1\).)
- A series has \(\gamma(0)=9\) and \(\gamma(2)=3\). For \(Y_t=5-2X_t\), calculate \(\gamma_Y(0)\), \(\gamma_Y(2)\), and \(\rho_Y(2)\). (Answer: \(\gamma_Y(0)=36\), \(\gamma_Y(2)=12\), and \(\rho_Y(2)=1/3\).)
- A hypothetical process has variance \(25\) and lag-5 autocorrelation \(-0.6\). Find its lag-5 autocovariance and check the bound on that covariance. (Answer: \(\gamma(5)=-0.6(25)=-15\); \(|-15|\leq25\), so it satisfies the covariance bound.)