Why Lag Sign and Zero Lag Matter
Autocovariance describes how values of a time series vary together when they are separated by a specified number of time steps. The sign of the lag tells us which time point is earlier under a chosen convention. It does not, by itself, tell us whether the covariance is positive or negative. Two basic facts make the interpretation reliable: reversing the sign of a lag leaves autocovariance unchanged, and autocovariance at lag zero is the variance.
These facts follow from the definition of covariance, together with the condition that covariance depends on the time separation rather than on the particular time origin. The condition matters: for an arbitrary time series, covariances may depend on both time points, so a single function of lag may not describe them.
The mean \(\mu\) is subtracted from each observation because covariance concerns variation around the mean. The same definition can be written using time-specific means if they are not constant, but a covariance function of lag alone then requires the relevant covariance structure to be invariant to shifts in time.
With this convention, \(h=2\) compares an observation with the one two time steps later, while \(h=-2\) compares it with the one two time steps earlier. The definition treats both signs as legitimate lags. Neither sign is a direction of correlation strength.
Deriving Symmetry Across Positive and Negative Lags
Covariance is symmetric in its two arguments: \(\operatorname{Cov}(U,V)=\operatorname{Cov}(V,U)\). Autocovariance symmetry follows by combining this property with the assumption that covariance depends only on lag.
By definition, \(\gamma(-h)=\operatorname{Cov}(X_t,X_{t-h})\).
Covariance is symmetric, so \(\operatorname{Cov}(X_t,X_{t-h})=\operatorname{Cov}(X_{t-h},X_t)\).
The ordered pair \(X_{t-h},X_t\) is separated by \(h\) steps. Because the covariance depends only on that separation, \(\operatorname{Cov}(X_{t-h},X_t)=\gamma(h)\).
This is a symmetry in the lag argument: the graph of the autocovariance function is mirrored around lag zero. The result does not say that the covariance is positive. It says that the values at matching positive and negative lags are equal. An autocovariance can be negative at both \(h\) and \(-h\), or positive at both.
The assumption that covariance depends only on lag is essential to the final step. Covariance symmetry alone always gives \(\operatorname{Cov}(X_t,X_{t-h})=\operatorname{Cov}(X_{t-h},X_t)\), but it does not guarantee that all pairs separated by \(h\) have the same covariance. Without that time-shift property, a single \(\gamma(h)\) may not be defined for the whole series.
Worked Example: A Sinusoidal Process at Positive and Negative Lags
A hypothetical index is modeled by \(X_t=A\cos(\pi t/3)+B\sin(\pi t/3)\), where \(A\) and \(B\) are independent, each has mean zero, and each has variance \(2\). Find its autocovariance at lags \(0\), \(1\), \(-1\), and \(2\). State why it is a function of lag alone.
Method. Expand the covariance using independence and then apply a trigonometric identity. This method avoids treating the negative sign on a lag as a sign on the covariance.
The mean of \(X_t\) is zero because both \(A\) and \(B\) have mean zero. For any \(t\) and \(h\), independence gives \(\operatorname{Cov}(A,B)=0\), so
The last equality uses \(\cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b)\). The resulting expression depends on \(h\) alone, as required for an autocovariance function.
At lag zero, \(2\cos(0)=2\), so \(\gamma(0)=2\). At lag \(1\), \(\gamma(1)=2\cos(\pi/3)=2(1/2)=1\). At lag \(-1\), \(\gamma(-1)=2\cos(-\pi/3)=2(1/2)=1\). At lag \(2\), \(\gamma(2)=2\cos(2\pi/3)=2(-1/2)=-1\).
The results are \(\gamma(0)=2\), \(\gamma(1)=\gamma(-1)=1\), and \(\gamma(2)=-1\). As a check, \(X_0=A\), so its variance is \(2\), agreeing with the zero-lag value. The negative covariance at lag \(2\) is also mirrored at lag \(-2\), not changed into a positive value.
Why Zero-Lag Autocovariance Is the Variance
At lag zero, the two time points in the covariance definition are the same observation. The result is therefore the variance of that observation, not a separate measure of dependence.
This derivation requires a finite second moment so that the variance exists. For a covariance function of lag alone, the variance is constant across time, and we may write \(\gamma(0)=\sigma^2\), where \(\sigma^2\) is that common variance. Even without a time-invariant covariance structure, the fixed-time identity \(\operatorname{Cov}(X_t,X_t)=\operatorname{Var}(X_t)\) remains true; what may fail is the claim that this variance has the same value at every \(t\).
Worked Example: A Shared Random Level and Alternating Component
A hypothetical monthly measure is \(X_t=12+C+(-1)^tD\), where \(C\) and \(D\) are independent, have mean zero, and have variances \(4\) and \(1\), respectively. Find \(\gamma(0)\), \(\gamma(1)\), and \(\gamma(-1)\), and verify that the covariance depends only on lag.
Method. Expand the covariance into the contribution from \(C\) and the contribution from \(D\). This avoids incorrectly assuming that the mean level \(12\) contributes to covariance.
The mean of \(X_t\) is \(12\), so \(X_t-12=C+(-1)^tD\). For any integer lag \(h\),
The cross-covariances involving one \(C\) and one \(D\) are zero because \(C\) and \(D\) are independent. Also, \((-1)^t(-1)^{t+h}=(-1)^{2t+h}=(-1)^h\), so the expression does not depend on \(t\).
At lag zero, \(\gamma(0)=4+1=5\). At lag \(1\), \(\gamma(1)=4-1=3\). At lag \(-1\), \(\gamma(-1)=4+(-1)^{-1}=4-1=3\), since \((-1)^{-1}=-1\). Thus \(\gamma(1)=\gamma(-1)=3\), and the zero-lag value is the variance of \(X_t\), namely \(4+1=5\).
Check. The constant 12 vanishes when centering, and the odd-lag values match for both signs. A common mistake is to read \((-1)^{-1}\) as a negative covariance; here it equals \(-1\) as a coefficient, producing the same covariance value at either odd lag.
Computing Several Lags from Shared Inputs
A useful way to calculate autocovariances is to identify which random inputs appear in both observations. If distinct inputs are independent and centered, their cross-covariances are zero. Only shared inputs contribute to the covariance. The following example applies that principle directly.
Worked Example: Finding the Zero, Unit, and Two-Step Covariances
Let \(Z_t\) be independent across time, with \(E[Z_t]=0\) and \(\operatorname{Var}(Z_t)=4\) for every \(t\). Define \(X_t=Z_t+0.5Z_{t-1}\). Find \(\gamma(0)\), \(\gamma(1)\), \(\gamma(-1)\), and \(\gamma(2)\).
Method. Expand each covariance and retain only terms involving the same \(Z\). This method avoids assuming that all nonzero lags must have positive covariance or that reversing a lag changes its sign.
The mean of \(X_t\) is zero. At lag zero,
The cross term is zero because \(Z_t\) and \(Z_{t-1}\) are independent. At lag \(1\), \(X_{t+1}=Z_{t+1}+0.5Z_t\). Expanding gives
The only shared input is \(Z_t\), with coefficients \(1\) and \(0.5\). By symmetry, \(\gamma(-1)=\gamma(1)=2\). For lag \(2\), the inputs in \(X_t\) are \(Z_t,Z_{t-1}\), while those in \(X_{t+2}\) are \(Z_{t+2},Z_{t+1}\). No input is shared, so every cross-covariance is zero and \(\gamma(2)=0\).
The answers are \(5,2,2,0\), respectively. The zero-lag result is also the direct variance of \(X_t\), and the unit-lag result agrees with the negative unit-lag result. Independence across time and equal input variances make the covariance depend only on the separation.
Common Mistakes and Interpretation
The most frequent sign error is to assert \(\gamma(-h)=-\gamma(h)\). That would describe an odd function, but autocovariance is even in the lag: \(\gamma(-h)=\gamma(h)\). For example, if \(\gamma(1)=2\), then \(\gamma(-1)\) is \(2\), not \(-2\). A negative value at one lag is mirrored as the same negative value at the opposite lag.
Another error is to assume that every covariance at a nonzero lag is positive, or that the zero-lag variance tells us the sign at other lags. Neither follows from the definition. The sinusoidal example had variance \(2\) but covariance \(-1\) at lag \(2\). The zero-lag value is nonnegative because it is a variance; nonzero-lag values can be positive, zero, or negative.
Finally, do not apply a lag-only formula without checking its condition. For a series whose covariance changes over calendar time, one might have \(\operatorname{Cov}(X_1,X_2)\ne\operatorname{Cov}(X_5,X_6)\), even though both pairs are separated by one step. Pairwise covariance is still symmetric, but there is no single value \(\gamma(1)\) for all such pairs unless covariance is invariant to the time origin.
Check Your Understanding
Work through each problem using the convention \(\gamma(h)=\operatorname{Cov}(X_t,X_{t+h})\).
- A process has a covariance function that depends only on lag, with \(\gamma(3)=-1.5\) and \(\operatorname{Var}(X_t)=6\). Find \(\gamma(-3)\) and \(\gamma(0)\). (Answer: \(\gamma(-3)=-1.5\), \(\gamma(0)=6\))
- Let \(X_t=C+(-1)^tD\), where \(C,D\) are independent, centered random variables with variances \(2\) and \(3\). Find \(\gamma(0)\), \(\gamma(1)\), and \(\gamma(-1)\). (Answer: \(5,-1,-1\))
- Let \(X_t=Z_t-2Z_{t-1}\), where the \(Z_t\) are independent, centered, and have variance \(3\). Find \(\gamma(0)\), \(\gamma(1)\), \(\gamma(-1)\), and \(\gamma(2)\). (Answer: \(15,-6,-6,0\))
- A process is defined by \(X_t=A\cos(\pi t/2)+B\sin(\pi t/2)\), where \(A,B\) are independent, centered, and each has variance \(4\). Calculate \(\operatorname{Cov}(X_0,X_1)\) and \(\operatorname{Cov}(X_1,X_0)\), then identify the corresponding positive- and negative-lag values. (Answer: both covariances are \(0\); \(\gamma(1)=\gamma(-1)=0\))
- A series has \(\operatorname{Cov}(X_2,X_3)=7\) and \(\operatorname{Cov}(X_8,X_9)=4\). Can one assign a single value to \(\gamma(1)\) for the entire series on the information given? What symmetry statement remains valid for each pair? (Answer: No, the one-step covariance depends on calendar time; for each pair, \(\operatorname{Cov}(X_s,X_{s+1})=\operatorname{Cov}(X_{s+1},X_s)\))