Why Weak Stationarity Matters
A time series can have observations that vary over time without its statistical behavior changing over time. Weak stationarity gives a precise way to express this stability using three features: the mean, the variance, and the covariance between observations. These conditions help distinguish a series whose fluctuations have a stable structure from one whose moments change with time.
The key is to check all three features together. In particular, the covariance at lag zero is the variance, so a proposed covariance function must agree with the stated variance at that lag. A specification that fails this check cannot describe any process.
- The mean is constant: \(E[X_t]=\mu\) for every \(t\).
- The variance is finite and constant: \(\operatorname{Var}(X_t)=\gamma(0)<\infty\) for every \(t\).
- The covariance between observations depends only on their separation in time. Equivalently, there is a function \(\gamma(h)\), defined for integer lags \(h\), such that \(\operatorname{Cov}(X_t,X_{t+h})=\gamma(h)\) for every \(t\) and \(h\).
Time and Lag Are Different
The time index \(t\) identifies an observation. The lag \(h\) identifies how far apart two observations are. For example, \(\operatorname{Cov}(X_7,X_9)\) is a covariance at lag \(2\), and \(\operatorname{Cov}(X_3,X_5)\) is also at lag \(2\). For weak stationarity, these covariances must be equal, regardless of which time points are used.
The function \(\gamma(h)\) is the autocovariance function. Weak stationarity requires that the covariance be determined by \(h\), not by the starting time \(t\). It also requires a common finite mean and variance; having a covariance that depends only on lag is not enough by itself.
There are useful consequences of this definition. Since the variance is \(\gamma(0)\), it is the same at every time. Also, covariance is symmetric: \(\operatorname{Cov}(X_t,X_{t+h})=\operatorname{Cov}(X_{t+h},X_t)\). The second covariance is the first one viewed at lag \(-h\), so \(\gamma(-h)=\gamma(h)\). These properties are checks on a proposed autocovariance function, not replacements for checking all three stationarity conditions.
Why the Covariance Must Be a Function of Lag
The covariance condition can be understood as invariance under a common shift of the time indices. Suppose \(s=t+h\). If both times are shifted forward by the same integer \(k\), their separation is unchanged: \((t+k)-(s+k)=t-s\). For a weakly stationary process, this shift must leave their covariance unchanged.
Conversely, if the covariance is unchanged by every common time shift, then any two pairs of observations with the same separation have the same covariance. Taking one observation at time \(t\) and the other at \(t+h\) defines that common covariance as \(\gamma(h)\). This is why a lag-based covariance function expresses stability of the dependence structure over time.
Checking the Three Conditions
A reliable test checks each condition explicitly, including whether the quantities are well-defined and finite. A constant mean alone does not guarantee weak stationarity, and neither does a covariance formula that uses only the lag.
Calculate or identify \(E[X_t]\) as a function of \(t\). It must be the same finite value at every time.
Verify that \(\operatorname{Var}(X_t)\) exists, is finite, and does not change with \(t\).
For arbitrary \(t\) and \(h\), show that \(\operatorname{Cov}(X_t,X_{t+h})\) can be written using \(h\) alone.
Set \(h=0\). The covariance formula must give the same value as the variance.
Worked Examples
Worked Example: Independent Observations With a Common Distribution
A hypothetical monitoring series is defined by \(X_t=6+\varepsilon_t\), where the \(\varepsilon_t\) are independent and identically distributed \(N(0,9)\) random variables for all integer \(t\). Determine whether \(\{X_t\}\) is weakly stationary and give its autocovariance function.
Method. Check the mean, variance, and covariance for arbitrary times. Here each innovation has mean \(0\), variance \(9\), and distinct innovations are independent.
Mean. For every \(t\), \[ E[X_t]=E[6+\varepsilon_t]=6+0=6. \] The mean is constant.
Variance. Adding a constant does not change variance, so \[ \operatorname{Var}(X_t)=\operatorname{Var}(\varepsilon_t)=9. \] This is finite and constant.
Covariance. At lag \(h=0\), the covariance is the variance, \(9\). If \(h\ne0\), then \(t\) and \(t+h\) are distinct indices. Independence gives \[ \operatorname{Cov}(X_t,X_{t+h}) =\operatorname{Cov}(\varepsilon_t,\varepsilon_{t+h})=0. \] Therefore \[ \gamma(h)= \begin{cases} 9,&h=0,\\ 0,&h\ne0. \end{cases} \] This depends only on \(h\), and \(\gamma(0)=9\) agrees with the variance. The process is weakly stationary.
Common mistake avoided: Independence at different times makes the covariance zero only when \(h\ne0\). It does not make the zero-lag covariance zero; at lag zero it equals the variance.
Worked Example: A Random-Amplitude Oscillation
Consider \(X_t=5+U\cos(\pi t/3)+V\sin(\pi t/3)\), where \(U\) and \(V\) are independent \(N(0,4)\) random variables. The same pair \(U,V\) is used at every integer time \(t\). Determine whether the process is weakly stationary and calculate its covariance at lags \(1\), \(2\), and \(3\).
Method. Calculate the first two moments and then the covariance between \(X_t\) and \(X_{t+h}\). The shared random coefficients mean observations at different times need not be independent.
Mean. Since \(E[U]=E[V]=0\), \[ E[X_t]=5+0\cos(\pi t/3)+0\sin(\pi t/3)=5. \]
Variance. Independence gives \(\operatorname{Cov}(U,V)=0\). Using \(\operatorname{Var}(U)=\operatorname{Var}(V)=4\), \[ \operatorname{Var}(X_t) =4\cos^2(\pi t/3)+4\sin^2(\pi t/3) =4. \] The variance is finite and constant.
Covariance. Write \(\theta=\pi t/3\) and \(\phi=\pi(t+h)/3\). The constant 5 does not affect covariance. The cross-covariances involving \(U\) and \(V\) are zero, so \[ \begin{aligned} \operatorname{Cov}(X_t,X_{t+h}) &=4\cos(\theta)\cos(\phi)+4\sin(\theta)\sin(\phi)\\ &=4\cos(\theta-\phi)\\ &=4\cos(\pi h/3). \end{aligned} \] The result depends only on \(h\). At lag zero, it is \(4\cos(0)=4\), matching the variance.
At lag \(1\), \(\gamma(1)=4\cos(\pi/3)=4(1/2)=2\). At lag \(2\), \(\gamma(2)=4\cos(2\pi/3)=4(-1/2)=-2\). At lag \(3\), \(\gamma(3)=4\cos(\pi)=-4\). The process is weakly stationary, even though its covariance can be negative at some lags.
Common mistake avoided: A visibly oscillating formula does not by itself establish that the process is nonstationary. Here the random coefficients make the mean and variance constant, and the covariance calculation reduces to a function of the lag.
Worked Example: A Changing Mean Despite Lag-Based Covariance
Let \(X_t=2+0.5t+\varepsilon_t\), where the \(\varepsilon_t\) are independent and identically distributed \(N(0,1)\) random variables for integer \(t\). Check the three weak-stationarity conditions.
Mean. Since \(E[\varepsilon_t]=0\), \[ E[X_t]=2+0.5t. \] At \(t=0\), the mean is \(2\); at \(t=4\), it is \(2+0.5(4)=4\). The mean changes, so the process is not weakly stationary.
Variance and covariance. The deterministic term does not affect variance or covariance. Thus \(\operatorname{Var}(X_t)=1\) for every \(t\), and \[ \operatorname{Cov}(X_t,X_{t+h})= \begin{cases} 1,&h=0,\\ 0,&h\ne0. \end{cases} \] The variance is constant and the covariance depends only on lag, with the correct zero-lag value. Nevertheless, failure of the constant-mean condition is enough to rule out weak stationarity.
Common mistake avoided: Do not conclude that a process is weakly stationary just because its variance and covariance have the required forms. All three conditions must hold.
Worked Example: Detecting an Impossible Covariance Specification
A proposed process is said to satisfy \(E[X_t]=4\), \(\operatorname{Var}(X_t)=9\), and \(\operatorname{Cov}(X_t,X_{t+h})=2/(1+|h|)\) for every integer \(t\) and \(h\). Determine whether these statements describe a weakly stationary process.
Method. First check the covariance at lag zero against the stated variance. Substituting \(h=0\) into the proposed formula gives \[ \operatorname{Cov}(X_t,X_t)=\frac{2}{1+|0|}=2. \] But by the definition of variance, \[ \operatorname{Cov}(X_t,X_t)=\operatorname{Var}(X_t)=9. \] The same quantity is required to equal both \(2\) and \(9\), which is impossible. Therefore no process can satisfy all the stated specifications; this is not merely a process that fails weak stationarity.
Common mistake avoided: It is incorrect to accept the covariance formula as a lag-only function and declare the process weakly stationary. A valid covariance formula must also agree with the stated variance at lag zero.
What Weak Stationarity Does—and Does Not—Say
Weak stationarity concerns the first two moments and the covariance structure. It does not, by definition, require the full probability distribution of the series to remain unchanged after shifting time. A series may satisfy the weak-stationarity conditions without its entire collection of distributions being invariant under shifts. Conversely, a proposed process with no finite variance cannot meet the definition, even if its mean is constant.
For a practical check, write the three requirements separately rather than relying on an informal impression that a series “looks stable.” In particular:
- A changing mean rules out weak stationarity, even when the variance is constant.
- A changing variance rules it out, even when the mean is constant.
- A covariance that depends on the calendar time \(t\), rather than only on the lag, rules it out.
- A covariance formula with \(\gamma(0)\ne\operatorname{Var}(X_t)\) is internally inconsistent.
Check Your Understanding
For each problem, check the stated moment conditions together rather than relying on just one of them.
- A process has \(E[X_t]=3\), \(\operatorname{Var}(X_t)=5\), and covariance \(5\) at lag zero and \(0\) at every nonzero lag. Is it weakly stationary? (Answer: Yes; the mean and variance are constant, and the covariance depends only on lag with \(\gamma(0)=5\).)
- A process has \(E[X_t]=t\), variance \(2\), and covariance \(2\) at lag zero and \(0\) at every nonzero lag. Which condition fails? (Answer: The mean is not constant; the process is not weakly stationary.)
- A process has constant mean \(1\), variance \(6\), and \(\operatorname{Cov}(X_t,X_{t+h})=6/(1+|h|)\). Is the specification consistent? (Answer: Yes; its zero-lag covariance is \(6\), matching the variance, and the stated moments depend on lag as required.)
- A process has constant mean \(0\), variance \(4\), and covariance \(3\) at lag zero, with covariance \(1\) at every nonzero lag. What conclusion follows? (Answer: No such process can satisfy all the specifications, because its zero-lag covariance must equal its variance, but \(3\ne4\).)
- Suppose \(\operatorname{Cov}(X_t,X_{t+h})=2\cos(\pi h/2)\) for every integer \(t,h\), and the mean is constant. What variance is required by this covariance formula, and what additional variance condition must be checked? (Answer: The required variance is \(\gamma(0)=2\cos(0)=2\); it must be finite and equal to \(2\) at every time.)