Strict Stationarity

What Must Stay the Same Under a Time Shift?

A process can have the same distribution at every individual time and still change its dependence pattern as time passes. To describe time invariance fully, it is not enough to compare one-time distributions: we must compare the joint distributions at every finite collection of times. Strict stationarity is the requirement that those joint laws remain unchanged when all the times are shifted by the same amount.

Let \((\Omega,\mathcal{F},\mathbb{P})\) be a probability space, let \(S\) be a measurable state space, and let \((X_n)_{n\ge0}\) be an \(S\)-valued process in discrete time. For indices \(0\le n_1<\cdots<n_r\), the vector \((X_{n_1},\ldots,X_{n_r})\) has a finite-dimensional distribution: its joint probability law on \(S^r\). Equality in distribution of two such vectors means that their joint laws agree on every measurable subset of \(S^r\).

Definition: The process \((X_n)_{n\ge0}\) is strictly stationary if, for every integer \(r\ge1\), every choice \(0\le n_1<\cdots<n_r\), and every integer shift \(h\ge0\),

\((X_{n_1},\ldots,X_{n_r})\) and \((X_{n_1+h},\ldots,X_{n_r+h})\) have the same distribution.

The same shift \(h\) is applied to every index. The condition concerns the whole joint law, not merely the distributions of individual coordinates.

For \(r=1\), strict stationarity requires that \(X_n\) and \(X_{n+h}\) have the same distribution for all \(n,h\ge0\). Thus all one-time marginals must agree. For \(r=2\), however, the definition also compares the dependence between observations at different times. This extra information is essential: two processes may have identical marginals at every time while having different joint laws at different time locations.

Proposition: If a process is strictly stationary, then its one-time distributions are identical. More generally, the joint distribution at any fixed collection of relative time gaps is unchanged as the collection is shifted forward.

Proof. For the one-time statement, take \(r=1\) and \(n_1=n\) in the definition. For any \(h\ge0\), it gives \(X_n\) equal in distribution to \(X_{n+h}\). For the second statement, fix \(r\), indices \(n_1<\cdots<n_r\), and a shift \(h\ge0\). The definition directly gives equality in distribution of the vectors at indices \((n_1,\ldots,n_r)\) and \((n_1+h,\ldots,n_r+h)\). Their relative gaps are the same, since \((n_j+h)-(n_i+h)=n_j-n_i\). \(\square\)

Three Processes to Separate the Ideas

Worked Example: Independent Bernoulli Observations

Let \((X_n)_{n\ge0}\) be independent random variables on a probability space, each taking values in \(S=\{0,1\}\), with \(\mathbb{P}(X_n=1)=\tfrac13\) and \(\mathbb{P}(X_n=0)=\tfrac23\) for every \(n\). This process is strictly stationary.

To check the full condition, choose any distinct indices \(n_1<\cdots<n_r\) and any pattern \((x_1,\ldots,x_r)\in\{0,1\}^r\). Independence and the common Bernoulli distribution give

\mathbb{P}(X_{n_1}=x_1,\ldots,X_{n_r}=x_r) =\prod_{\ell=1}^r \left(\frac13\right)^{x_\ell} \left(\frac23\right)^{1-x_\ell}.

The right side depends on the pattern but not on the indices. Replacing every \(n_\ell\) by \(n_\ell+h\) therefore leaves the probability unchanged. Since every event in the finite state space \(\{0,1\}^r\) is a union of patterns, the entire joint distributions agree.

Check: For the pattern \((1,0,1)\), the probability at indices \((0,2,5)\) is \(\tfrac13\cdot\tfrac23\cdot\tfrac13=\tfrac2{27}\). At indices \((4,6,9)\), it is also \(\tfrac13\cdot\tfrac23\cdot\tfrac13=\tfrac2{27}\), as the definition requires.

Worked Example: Strict Stationarity Does Not Require Independence

Let \(Y\) be a fair \(\{0,1\}\)-valued random variable on a probability space, and set \(X_n=Y\) for every \(n\ge0\). The process is strictly stationary even though its observations are perfectly dependent.

For any \(r\) and any indices \(n_1<\cdots<n_r\), the vector is \((Y,\ldots,Y)\). It equals \((1,\ldots,1)\) with probability \(\tfrac12\), and \((0,\ldots,0)\) with probability \(\tfrac12\); every other vector in \(\{0,1\}^r\) has probability \(0\). These probabilities do not depend on the chosen indices, so shifting them by any \(h\ge0\) leaves the joint law unchanged.

Check: For two distinct times, \(\mathbb{P}(X_{n_1}=1,X_{n_2}=1)=\tfrac12\), while \(\mathbb{P}(X_{n_1}=1)\mathbb{P}(X_{n_2}=1)=\tfrac12\cdot\tfrac12=\tfrac14\). The unequal values confirm dependence without affecting strict stationarity.

Worked Example: Equal Marginals but Not Strictly Stationary

Let \(\Omega=\{(0,0),(0,1),(1,0),(1,1)\}\), with each outcome having probability \(\tfrac14\), and let \(U\) and \(V\) be the first and second coordinates. Define the process by \(X_0=X_1=X_2=U\), and \(X_n=V\) for \(n\ge3\).

Both \(U\) and \(V\) are fair: exactly two of the four equally likely outcomes have first coordinate \(1\), and exactly two have second coordinate \(1\). Consequently, \(\mathbb{P}(X_n=1)=\tfrac12\) and \(\mathbb{P}(X_n=0)=\tfrac12\) at every time \(n\). Thus all one-time marginals agree.

Strict stationarity fails. At the indices \((0,2)\), the pair is \((U,U)\), so \(\mathbb{P}(X_0=1,X_2=1)=\mathbb{P}(U=1)=\tfrac12\). Shifting both indices by \(1\) gives \((1,3)\), where the pair is \((U,V)\). Only the outcome \((U,V)=(1,1)\) among the four equally likely outcomes has both coordinates \(1\), so \(\mathbb{P}(X_1=1,X_3=1)=\tfrac14\). Since \(\tfrac12\ne\tfrac14\), these two joint distributions differ, violating the definition.

Check: The failure is in the joint distribution, not in the marginals: the two compared pairs each have fair coordinates, but the first pair is identical in both coordinates and the second pair consists of independent coordinates.

Consequences and a Useful Closure Property

The definition also gives simple consequences for real-valued processes. Moment statements require moment assumptions: equality in distribution alone does not guarantee that expectations or covariances are finite. When those moments exist, strict stationarity makes them invariant under common shifts.

Proposition: Suppose \((X_n)_{n\ge0}\) is a strictly stationary real-valued process. If \(\mathbb{E}|X_0|<\infty\), then \(\mathbb{E}[X_n]=\mathbb{E}[X_0]\) for every \(n\ge0\). If \(\mathbb{E}[X_0^2]<\infty\), then \(\operatorname{Cov}(X_n,X_{n+r})=\operatorname{Cov}(X_0,X_r)\) for every \(n,r\ge0\).

Proof. Strict stationarity with \(r=1\) shows that \(X_n\) and \(X_0\) have the same distribution. Since \(\mathbb{E}|X_0|<\infty\), their expectations exist and agree. For the covariance statement, strict stationarity applied to the indices \((0,r)\) and shift \(n\) shows that \((X_0,X_r)\) and \((X_n,X_{n+r})\) have the same joint distribution. The second-moment assumption makes the products integrable: \(2|X_nX_{n+r}|\le X_n^2+X_{n+r}^2\), and each term on the right has finite expectation by equality of the one-time distributions. Thus the two pairs have equal product expectations and equal coordinate expectations. Substituting these equalities into the definition of covariance proves the result. \(\square\)

Strict stationarity is also preserved by applying the same measurable transformation at each time. This lets us change how process values are recorded without losing the time-shift invariance.

Proposition: Let \((X_n)_{n\ge0}\) be strictly stationary with values in a measurable space \(S\), and let \(g:S\to T\) be measurable, where \(T\) is a measurable space. Then the process \((g(X_n))_{n\ge0}\) is strictly stationary.

Proof. Fix \(r\), indices \(n_1<\cdots<n_r\), and \(h\ge0\). By strict stationarity, the \(S^r\)-valued vectors \((X_{n_1},\ldots,X_{n_r})\) and \((X_{n_1+h},\ldots,X_{n_r+h})\) have the same distribution. The coordinatewise map \(g_r:S^r\to T^r\), defined by \(g_r(x_1,\ldots,x_r)=(g(x_1),\ldots,g(x_r))\), is measurable. Applying the same measurable map to equal-in-distribution vectors gives equal-in-distribution image vectors: for any measurable \(C\subseteq T^r\), the probability that the image lies in \(C\) is the probability that the original vector lies in \(g_r^{-1}(C)\), which is the same for both vectors. These image vectors are precisely the two collections of transformed process values required by the definition. \(\square\)

A Common Mistake

A tempting but incorrect argument is: “Every \(X_n\) has the same distribution, so the process is strictly stationary.” In the third example, this argument would claim stationarity because every marginal is fair. But strict stationarity also requires the joint laws to agree. At the same threshold event \(\{(x_1,x_2)=(1,1)\}\), the pair at \((0,2)\) has probability \(\tfrac12\), whereas the pair at \((1,3)\) has probability \(\tfrac14\). The correction is to compare finite-dimensional distributions at arbitrary index sets after a common shift, rather than checking only one-time distributions.

Keep the two levels distinct. Equal one-time marginals are a necessary consequence of strict stationarity, obtained by considering one index at a time. They are not sufficient, because they say nothing about how values at different times are jointly distributed. Conversely, strict stationarity does not mean that observations are independent: the constant process has unchanged joint laws under shifts but is maximally dependent across time.

Check Your Understanding

Use the definition to answer each question. When comparing distributions, specify the finite collection of indices being used.

What does strict stationarity require for a collection of indices \((n_1,\ldots,n_r)\) and a shift \(h\)?
Why does strict stationarity imply that all one-time marginal distributions agree?
For the constant process \(X_n=Y\), what are the probabilities of the all-zero and all-one vectors at \(r\) distinct times when \(Y\) is fair?
In the third worked example, which two index sets demonstrate that equal marginals do not imply strict stationarity?
Why is an integrability assumption needed before asserting that a strictly stationary real-valued process has a finite, constant expectation?

What You'll Learn