Strict Versus Weak Stationarity

Why There Are Two Notions of Stationarity

A time series can look stable in its average and covariance structure while changing in other ways. For example, its values might have the same mean and variance at every date, yet alternate between two different distribution shapes. Whether that process counts as stationary depends on what features of its distribution the definition requires to stay unchanged.

The distinction is between strict stationarity, which concerns entire finite-dimensional distributions, and weak stationarity, which concerns means and covariances. Weak stationarity is often convenient because it summarizes dependence with second moments. Strict stationarity imposes a stronger distributional invariance. Under finite second moments, strict stationarity implies weak stationarity, but the converse does not generally hold.

Definitions

Definition: A process $\{X_t:t\in\mathbb{Z}\}$ is strictly stationary if, for every positive integer $k$, every collection of time indices $t_1,\ldots,t_k$, and every integer shift $h$, $$ (X_{t_1},\ldots,X_{t_k}) \overset{d}{=} (X_{t_1+h},\ldots,X_{t_k+h}). $$ In words, shifting all observation dates by the same amount leaves every finite joint distribution unchanged.

The definition applies to one observation, any pair of observations, and every larger finite collection. It therefore preserves all features of those joint distributions, not just their means and covariances. The distribution at a given time does not change with the date, and the joint behavior of observations at specified relative lags does not change with the calendar position.

Definition: A real-valued process $\{X_t:t\in\mathbb{Z}\}$ is weakly stationary if its variables have finite second moments, its mean is constant, and its covariance depends only on the time difference. That is, for some constant $\mu$ and function $\gamma$, $$ E[X_t]=\mu \quad\text{and}\quad \operatorname{Cov}(X_t,X_{t+h})=\gamma(h) $$ for all $t$ and $h$.

Weak stationarity is also called covariance stationarity in many settings. Its requirements refer only to first and second moments. In particular, it does not directly require that the distribution of $X_t$ be unchanged over time, nor does it constrain all higher-order joint distributional features.

Worked Example: A Strictly and Weakly Stationary Process

Consider a hypothetical process in which the $X_t$ are independent and identically distributed, with $P(X_t=1)=P(X_t=-1)=0.5$ at every date. This is a non-Gaussian process with finite second moments.

For any finite set of dates, the variables have the same independent Rademacher distributions. Shifting the dates does not change their joint distribution, so the process is strictly stationary.

Its mean and variance are $$ E[X_t]=(1)(0.5)+(-1)(0.5)=0, \qquad \operatorname{Var}(X_t)=E[X_t^2]-E[X_t]^2=(1)(0.5)+(1)(0.5)-0^2=1. $$ At a nonzero lag $h$, $X_t$ and $X_{t+h}$ are independent, so their covariance is zero. Thus $\gamma(0)=1$ and $\gamma(h)=0$ for $h\ne 0$. The mean is constant and the covariance depends only on the lag; the process is weakly stationary as well.

Why Strict Stationarity Implies Weak Stationarity When Moments Exist

The implication follows by applying the definition of strict stationarity to one-variable and two-variable distributions. Suppose the process is strictly stationary and $E[X_0^2]$ is finite. Since each $X_t$ has the same distribution as $X_0$, every $X_t$ has finite second moments, and its mean and variance are the same as those of $X_0$.

For any fixed lag $h$, strict stationarity also says that the pair $(X_t,X_{t+h})$ has the same distribution as $(X_0,X_h)$. Therefore its covariance is the same for every starting time $t$:

\operatorname{Cov}(X_t,X_{t+h}) = E[(X_t-\mu)(X_{t+h}-\mu)] = E[(X_0-\mu)(X_h-\mu)] = \gamma(h).

This establishes a constant mean and a covariance that depends only on the lag, which are precisely the moment conditions for weak stationarity. The finite-second-moment condition matters: strict stationarity alone does not guarantee that the mean or covariance exists.

Worked Example: Strict Stationarity Without Weak Stationarity

Let $X_t$ be independent and identically distributed standard Cauchy variables. Because their joint distribution is unchanged by shifting the dates, the process is strictly stationary.

However, a Cauchy variable has no finite mean or variance. In particular, $E[X_t]$ is undefined, so it cannot meet the finite-second-moment and constant-mean requirements of weak stationarity. This example shows why the moment condition is needed when claiming that strict stationarity implies weak stationarity.

Why the Converse Can Fail

Mean and covariance restrictions do not generally determine a process's full distribution. Different distributions can share the same mean and variance, and a process can change its distribution over time while keeping those moments fixed. The following construction makes the distinction explicit.

Worked Example: Weakly but Not Strictly Stationary

Construct a hypothetical process with independent observations at all dates. At even dates, let $X_t$ equal $1$ or $-1$, each with probability $0.5$. At odd dates, let $X_t\sim N(0,1)$. Both types of observations have mean $0$ and variance $1$.

The mean is therefore $E[X_t]=0$ for every $t$. At lag zero, the covariance is the variance, so $\operatorname{Cov}(X_t,X_t)=1$. At every nonzero lag, the observations are independent and have mean zero, giving $$ \operatorname{Cov}(X_t,X_{t+h}) = E[X_tX_{t+h}]-E[X_t]E[X_{t+h}] = 0-(0)(0) = 0 \qquad (h\ne 0). $$ Thus $\gamma(0)=1$ and $\gamma(h)=0$ for $h\ne 0$, regardless of the starting date $t$. The process is weakly stationary.

It is not strictly stationary: the marginal distribution changes when the time index shifts by one. For instance, at an even date, $P(X_t=1)=0.5$, whereas at an odd date with a standard normal distribution, $P(X_t=1)=0$. Since even the one-dimensional distribution is not invariant under the shift, the strict stationarity condition fails.

When the Definitions Agree for Gaussian Processes

A Gaussian process is one for which every finite vector of observations has a multivariate normal distribution. A useful special feature of Gaussian distributions is that each such distribution is determined by its mean vector and covariance matrix. Consequently, for a Gaussian process, weak stationarity implies strict stationarity: weak stationarity keeps the mean vector and the covariance matrix of every finite collection unchanged when all dates are shifted, and those quantities determine that collection's joint distribution. The converse follows from the finite-moment implication established above. Thus, for Gaussian processes with finite second moments, the two notions agree.

Worked Example: A Gaussian Process Satisfying Both Definitions

Let $Z$ and $\{\varepsilon_t\}$ be mutually independent standard normal variables, and define the hypothetical process $X_t=Z+\varepsilon_t$. Each $X_t$ is normal with mean $0$ and variance $2$, since it is the sum of two independent $N(0,1)$ variables.

For two different dates $t$ and $s$, the shared $Z$ contributes covariance $1$, while the independent terms $\varepsilon_t$ and $\varepsilon_s$ contribute none: $$ \operatorname{Cov}(X_t,X_s) = \operatorname{Cov}(Z+\varepsilon_t,Z+\varepsilon_s) = \operatorname{Var}(Z) = 1 \qquad (t\ne s). $$ At the same date, the variance is $2$. The mean is constant, and the covariance is $2$ at lag zero and $1$ at every nonzero lag, so the process is weakly stationary.

Every finite vector from this process is jointly Gaussian. After any shift, its mean vector is still a vector of zeros; its covariance matrix still has $2$ on the diagonal and $1$ off the diagonal. Those quantities determine its multivariate normal distribution, so the shifted vector has the same distribution. The process is strictly stationary too.

What the Distinction Means in Practice

Weak stationarity is often the relevant starting point when an analysis uses means, variances, and covariances to describe a time series. But a conclusion based only on those quantities should not be interpreted as proof that every aspect of the process's distribution is stable. The alternating-distribution example has an unchanged covariance structure even though its marginal distribution changes over time.

Gaussianity changes what can be inferred from the moment conditions. For Gaussian processes, the finite-dimensional distributions are determined by their means and covariance matrices, so weak stationarity is enough to establish strict stationarity. Without Gaussianity, the same inference is not available in general. Strict stationarity also has a separate limitation: if moments do not exist, it need not imply weak stationarity.

Common mistake: “The mean and covariance are constant over time, so the process is strictly stationary.” This reasoning confuses invariance of second moments with invariance of every finite-dimensional distribution. The correction is to conclude only weak stationarity unless additional conditions, such as Gaussianity, justify the stronger conclusion.

Check Your Understanding

Use the definitions to decide what each condition does and does not establish.

State the condition that must hold for every finite vector under strict stationarity.
Why does strict stationarity imply weak stationarity when the process has finite second moments?
In the alternating-distribution example, which moment conditions hold, and what distributional feature shows that strict stationarity fails?
Why can weak stationarity imply strict stationarity for a Gaussian process?
Can a strictly stationary process fail to be weakly stationary? Give the condition that makes this possible.