Testing Stationarity Through Finite-Dimensional Distributions
Strict stationarity is a statement about the joint distributions of a process, not about whether individual sample paths look regular or repeat. For a sequence, the practical test is to compare the distribution at any finite set of times with the distribution at those same times after a common shift. This makes iid sequences and randomly shifted periodic sequences useful examples: one is stationary because its joint laws are products of the same marginal law, and the other because shifting time simply changes a uniformly chosen phase.
Throughout, let \((\Omega,\mathcal{F},\mathbb{P})\) be a probability space and let \((X_n)_{n\ge0}\) be a process with state space \(S\), where \(S\) is equipped with a sigma-field. The time index is the nonnegative integers. For a random vector, equality in distribution means that the two vectors assign the same probability to every measurable set.
The condition applies to every finite collection of times, not just to one-dimensional marginal distributions or to a particular collection of sample paths. A sequence can have identical marginals at all times and still fail strict stationarity if its dependence structure changes with time. In the examples below, it is useful to compare the full finite-dimensional distributions directly.
Identically Distributed Independent Variables
Proof. Let \(F\) be the common distribution of each \(X_n\). Fix \(r\), times \(0\le n_1<\cdots<n_r\), and a shift \(h\ge0\). The variables \(X_{n_1},\ldots,X_{n_r}\) are independent and each has distribution \(F\), so the joint distribution of their vector is the product distribution \(F^{\otimes r}\). The shifted variables \(X_{n_1+h},\ldots,X_{n_r+h}\) are also distinct members of the iid sequence, so their joint distribution is the same product distribution \(F^{\otimes r}\). Thus the two vectors have the same distribution for every permitted choice of times and shift, which is strict stationarity. \(\square\)
Worked Example: An iid Bernoulli Sequence
Let \(S=\{0,1\}\), and let \((X_n)_{n\ge0}\) be iid on a probability space with \(\mathbb{P}(X_n=1)=\tfrac13\) and \(\mathbb{P}(X_n=0)=\tfrac23\) for every \(n\). The proposition shows that the sequence is strictly stationary. We can also check a finite-dimensional event explicitly.
For the event \(X_0=1,\ X_2=0,\ X_5=1\), independence gives
After shifting all three times by \(4\), the event is \(X_4=1,\ X_6=0,\ X_9=1\). These are again distinct iid variables, and substituting their probabilities gives
Check: The two event probabilities agree. The proposition supplies the full check for every finite set of times and every measurable event; this calculation illustrates one such comparison.
Randomizing the Phase of a Periodic Sequence
A deterministic periodic sequence does not automatically have stationary distributions: a shift can move its values to a different part of the cycle. A random phase can remove this dependence on the starting position. The construction uses a finite set of phases, each chosen with the same probability.
Proof. Fix \(r\), times \(0\le n_1<\cdots<n_r\), and \(h\ge0\). For each phase \(u\in\{0,\ldots,m-1\}\), the vector at the original times is \((a_{(n_1+u)\bmod m},\ldots,a_{(n_r+u)\bmod m})\). The vector at the shifted times, for that same phase, is \((a_{(n_1+h+u)\bmod m},\ldots,a_{(n_r+h+u)\bmod m})\). As \(u\) runs through the \(m\) phases, \((u+h)\bmod m\) also runs through every phase exactly once: addition by \(h\) modulo \(m\) is a bijection, with inverse addition by \(-h\) modulo \(m\). Since \(U\) assigns probability \(1/m\) to each phase, the shifted vector therefore takes the same list of values, with the same probabilities, as the original vector. Repeated values in the list do not affect this argument, because the phase map is still a bijection. The vectors have the same distribution for every choice of times and shift, proving strict stationarity. \(\square\)
Worked Example: A Uniformly Shifted Four-Phase Pattern
Take \(S=\{0,1,2\}\), \(m=4\), and the repeating list \(a_0=0,\ a_1=1,\ a_2=1,\ a_3=2\). On the four-point probability space, let \(\mathbb{P}(U=u)=\tfrac14\) for each \(u\in\{0,1,2,3\}\), and set \(X_n=a_{(n+U)\bmod4}\). The theorem guarantees strict stationarity. To see what a joint-distribution check looks like, list the possible vectors \((X_0,X_1,X_2)\):
Each displayed vector has probability \(\tfrac14\). At the times shifted by \(1\), the vectors \((X_1,X_2,X_3)\), in phase order \(U=0,1,2,3\), are respectively \((1,1,2)\), \((1,2,0)\), \((2,0,1)\), and \((0,1,1)\). They are the same four vectors, each with probability \(\tfrac14\), so these two three-dimensional distributions agree.
Check the marginals. Across the four phases, \(X_0\) equals \(0\) once, \(1\) twice, and \(2\) once. Thus \(\mathbb{P}(X_0=0)=\tfrac14\), \(\mathbb{P}(X_0=1)=\tfrac24=\tfrac12\), and \(\mathbb{P}(X_0=2)=\tfrac14\). The same phase count applies at every time because adding a time index permutes the four phases. These probabilities sum to \(\tfrac14+\tfrac12+\tfrac14=1\). The theorem, rather than the marginal check alone, guarantees agreement of all joint distributions.
Why a Deterministic Alternating Sequence Fails
Worked Example: Alternation Without a Random Phase
Use the one-point probability space \(\Omega=\{\omega\}\), with \(\mathbb{P}(\{\omega\})=1\), and state space \(S=\{-1,1\}\). Define \(X_n(\omega)=(-1)^n\). The sequence is periodic with period \(2\), but it is not strictly stationary. Indeed, \(X_0=1\) with probability \(1\), whereas \(X_1=-1\) with probability \(1\). In terms of the event \(\{1\}\),
Strict stationarity requires equality of the one-dimensional distributions when the single time \(0\) is shifted by \(1\), so this difference is enough to disprove it. The same conclusion follows from the joint distributions: the vector \((X_0,X_1)\) equals \((1,-1)\) with probability \(1\), while \((X_1,X_2)\) equals \((-1,1)\) with probability \(1\).
For contrast, if a phase \(U\) is uniform on \(\{0,1\}\) and \(X_n=(-1)^{n+U}\), the random-phase theorem applies. The random phase makes the starting value \(1\) or \(-1\) equally likely, and shifting time swaps the two phases rather than changing their probabilities.
What Randomizing the Phase Does—and Does Not—Guarantee
Uniform phase is a reliable sufficient condition for stationarity in the periodic construction. Choosing a phase at random without choosing it uniformly is not enough in general. This distinction is visible already in a cycle with three distinct values.
Worked Example: A Nonuniform Phase Changes the Marginal Law
Let \(S=\{0,1,2\}\), use the repeating pattern \((0,1,2)\), and let \(U\) take values \(0,1,2\) with probabilities \(\tfrac12,\tfrac13,\tfrac16\), respectively. These probabilities sum to \(\tfrac12+\tfrac13+\tfrac16=1\). Define \(X_n=a_{(n+U)\bmod3}\) using \(a_0=0,\ a_1=1,\ a_2=2\). At time \(0\), the event \(X_0=0\) occurs exactly when \(U=0\), so \(\mathbb{P}(X_0=0)=\tfrac12\). At time \(1\), \(X_1=0\) occurs exactly when \(U=2\), so \(\mathbb{P}(X_1=0)=\tfrac16\).
Check: Since \(\tfrac12\ne\tfrac16\), even the one-dimensional distributions differ after a shift of \(1\). The sequence is not strictly stationary. This example shows why the uniformity in the random-phase theorem matters for a general repeating pattern.
Common pitfall. Periodicity is not the same as stationarity. A mistaken argument for the deterministic alternating sequence is: “It repeats every two steps, so shifting time does not change it.” The calculation \(X_{n+2}=X_n\) verifies invariance only for shifts by \(2\); it says nothing about the required shift by \(1\). In fact, \(X_0=1\) while \(X_1=-1\), so the distributions already differ at a single time. To test strict stationarity, compare distributions under every permitted common shift, not just shifts that preserve the pattern pointwise.
The examples also distinguish pathwise repetition from distributional invariance. Each random-phase sample path repeats the same list, but stationarity follows from the distribution of the phase: a time shift permutes phases without changing their probabilities. For iid sequences, the reason is different: every selected finite vector has the same product distribution. In either case, the defining check concerns finite-dimensional distributions; agreement of marginals alone is not sufficient.
Check Your Understanding
Use the definition to decide what must be compared, and support each conclusion with a distributional calculation or argument.
- Why does independence matter in the proof that an identically distributed sequence is strictly stationary?
- For the four-phase pattern \((0,1,1,2)\), what is the probability that \(X_0=1\), and why is it unchanged at time \(5\)?
- In the random-phase theorem, why does adding \(h\) modulo \(m\) preserve the phase distribution?
- Can the deterministic alternating sequence have the same one-dimensional distribution at times \(0\) and \(1\)? Explain using its specified probability space.
- For the nonuniform three-phase example, compare \(\mathbb{P}(X_0=2)\) and \(\mathbb{P}(X_1=2)\). Does this comparison also rule out strict stationarity?