Two Different Ways to Compare a Process Across Time
A process can look the same after shifting time in more than one sense. Strict stationarity compares the joint distributions of the process values themselves. Stationary increments compare the distributions of changes in those values. The distinction matters whenever the level of a process has a meaningful starting point: a process may have time-homogeneous changes while its values continue to drift away from that starting point.
We use processes with real-valued states and either discrete time \(n\ge0\) or continuous time \(t\ge0\), defined on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). For a one-sided time index, a time shift is nonnegative and must keep every index in the process's domain.
This definition compares increments jointly, including when the intervals overlap. A common weaker usage asks only that the distribution of \(X_{t+h}-X_t\) depend on the interval length \(h\), not on its starting time \(t\). The joint definition is the useful one here; it also makes the implication from strict stationarity precise.
Strict Stationarity Implies Stationary Increments
Proof. Fix a finite collection of intervals and a valid shift \(u\). List all endpoints \(t_j\) and \(t_j+h_j\), including any repeated endpoints only once, in a vector of process times. Strict stationarity says that the vector of process values at these endpoints has the same distribution after every time is shifted by \(u\). Each requested increment vector is a fixed function of this endpoint-value vector: for each interval, subtract the value at its start from the value at its end. Applying the same function to random vectors with the same distribution gives random vectors with the same distribution. These two resulting vectors are precisely the original increments and the shifted increments. Thus the process has stationary increments. \(\square\)
The implication does not require independent increments. It follows just from equality of the finite-dimensional distributions and the fact that increments are differences of process values. To see why the converse fails, it is enough to find a process whose increments have time-invariant laws but whose levels do not.
Worked Example: A Random Walk Has Stationary Increments, Not Strict Stationarity
Let \((\Omega,\mathcal{F},\mathbb{P})\) carry independent random variables \(\xi_1,\xi_2,\ldots\) with \(\mathbb{P}(\xi_k=1)=\mathbb{P}(\xi_k=-1)=\tfrac12\) for every \(k\). Define the discrete-time process on the state space \(S=\mathbb{Z}\) by \(X_0=0\) and \(X_n=\sum_{k=1}^{n}\xi_k\) for \(n\ge1\).
For any \(n\ge0\) and integer length \(r\ge0\), its increment is \(X_{n+r}-X_n=\sum_{k=n+1}^{n+r}\xi_k\). When \(r=0\), this is the empty sum \(0\). For \(r\ge1\), the sum uses \(r\) independent variables, each with the same distribution as every other \(\xi_k\). Its distribution therefore does not depend on \(n\). More generally, for any finite set of interval starts and lengths, the increment vector is formed from a finite segment of the independent sequence by the same sums and differences after shifting all the indices. The shifted segment has the same joint distribution, since the \(\xi_k\) are independent and identically distributed. The random walk has stationary increments.
It is not strictly stationary. At time zero, \(\mathbb{P}(X_0=0)=1\). At time one, \(X_1=\xi_1\), so \(\mathbb{P}(X_1=0)=0\) and \(\mathbb{P}(X_1=1)=\mathbb{P}(X_1=-1)=\tfrac12\). The one-time distributions differ, which already violates strict stationarity. In particular, stationary increments do not imply stationarity of the levels.
Check: The one-step increment \(X_{n+1}-X_n=\xi_{n+1}\) has probability \(\tfrac12\) of being \(1\) and probability \(\tfrac12\) of being \(-1\), for every \(n\ge0\). But \(X_0\) equals \(0\) with probability \(1\), while \(X_1\) equals \(0\) with probability \(0\). The two tests agree: stationary increments hold and strict stationarity fails.
In this example the failure is especially direct because the walk starts at a fixed level. In fact, any strictly stationary process with deterministic \(X_0=0\) must satisfy \(X_n=0\) almost surely for every \(n\): strict stationarity makes \(X_n\) have the same distribution as \(X_0\). Thus a nonconstant random walk started at zero cannot be strictly stationary.
A Process That Is Both
Strict stationarity and stationary increments are compatible. A process that remains at a random level provides a simple example. It also emphasizes that a process can have randomness in its level while having no changes over time.
Worked Example: A Constant Random Value
Take \(\Omega=\{0,1\}\), \(\mathcal{F}\) its power set, and probabilities \(\mathbb{P}(\{1\})=\tfrac13\) and \(\mathbb{P}(\{0\})=\tfrac23\). Let \(Z(\omega)=\omega\), and define the continuous-time process on \(S=\{0,1\}\) by \(X_t=Z\) for every \(t\ge0\).
For any \(m\) and any times \(t_1,\ldots,t_m\), the vector of process values is \((Z,\ldots,Z)\). Shifting all the times leaves that vector unchanged, so its distribution is unchanged. This verifies strict stationarity for every finite collection of times.
For every \(t,h\ge0\), \(X_{t+h}-X_t=Z-Z=0\). Thus any finite vector of increments is the zero vector before and after a time shift, so the process has stationary increments as well.
Check: The one-time distribution at every \(t\) assigns probability \(\tfrac13\) to \(1\) and \(\tfrac23\) to \(0\). For any pair of times, the joint vector is \((1,1)\) with probability \(\tfrac13\) and \((0,0)\) with probability \(\tfrac23\), whether or not the times are shifted. Every increment is exactly \(0\). These direct calculations confirm both properties.
Here the random level \(Z\) is shared across every time. Consequently, the values at different times are generally dependent. Strict stationarity does not mean that values at different times are independent, and stationary increments do not mean that increments are nonzero.
The Poisson Counting Process: Stationary Increments Without Stationary Levels
For a continuous-time example, let \((N_t)_{t\ge0}\) be a Poisson counting process of rate \(\lambda>0\), with state space \(S=\{0,1,2,\ldots\}\). The properties needed here are \(N_0=0\), independent increments, and \(N_{t+h}-N_t\) having a Poisson distribution with parameter \(\lambda h\). In particular, for \(k=0,1,2,\ldots\),
The law depends on the interval length \(h\), not on its starting time \(t\), so each increment has a time-invariant distribution. For a collection of disjoint intervals, the increments are independent and each has a distribution determined by its interval length. Shifting all those intervals preserves their lengths and disjointness, so it preserves the joint distribution of their increments. This gives stationary increments.
Worked Example: A Poisson Process Is Not Strictly Stationary
Consider a Poisson counting process with rate \(\lambda=2\), on continuous time \(t\ge0\) and state space \(S=\{0,1,2,\ldots\}\). Its value at time zero is \(N_0=0\) with probability \(1\). At time \(t=1\), the count has a Poisson distribution with parameter \(2\), so the probability of zero events is \(e^{-2}\), which is strictly less than \(1\). Thus \(N_0\) and \(N_1\) do not have the same distribution, and strict stationarity fails.
For an interval of length \(h\), the increment distribution is obtained by substituting \(\lambda=2\) into the Poisson increment formula: \(\mathbb{P}(N_{t+h}-N_t=k)=e^{-2h}(2h)^k/k!\). The starting time \(t\) does not appear. For instance, the probability of no events on that interval is \(e^{-2h}\), the same for an interval starting at \(t=0\) or at any other nonnegative time. The independent-increment property gives the corresponding joint statement for disjoint intervals.
Check: At \(t=0\), \(\mathbb{P}(N_0=0)=1\). At \(t=1\), \(\mathbb{P}(N_1=0)=e^{-2}\). Since \(2>0\), \(0<e^{-2}<1\), so the marginal distributions differ. Meanwhile the increment law on every interval of length \(h\) has parameter \(2h\), regardless of its start. The process has stationary increments but is not strictly stationary.
The source of the contrast is the initial level. The Poisson process starts at zero and accumulates events, so the distribution of its level changes over time. Its increments, by contrast, count events in intervals of a given length and have a distribution that does not depend on where the interval begins.
How to Diagnose Which Property Holds
When checking a process, keep levels and changes separate. Strict stationarity is a statement about every finite vector of values. A difference between even two one-time marginal distributions is enough to disprove it, as in both the random walk and the Poisson process. To establish strict stationarity, however, matching marginal distributions alone is not enough: the joint laws at every finite set of times must also be invariant under shifts.
Stationary increments instead asks whether the joint law of changes across intervals is unchanged by a common shift. Checking only one increment at a time establishes only the marginal version of that property. When increments over several intervals are involved, use their joint distribution; independence, when available, can help establish it, as it does for the random walk and Poisson process.
Common mistake. A tempting but incorrect argument is, “The distribution of \(X_{t+h}-X_t\) does not depend on \(t\), so \(X_t\) is strictly stationary.” This checks only increments, not the level distributions or the joint laws of process values. For the rate-\(2\) Poisson process, the probability of zero events over an interval of length \(h\) is \(e^{-2h}\) for every starting time, but \(\mathbb{P}(N_0=0)=1\) and \(\mathbb{P}(N_1=0)=e^{-2}<1\). The correction is to test the two definitions separately: use finite-dimensional distributions of levels for strict stationarity and joint distributions of increments for stationary increments.
| Property | What is compared after a shift? | Example from this tutorial |
|---|---|---|
| Strict stationarity | All finite-dimensional distributions of process values | The constant random-value process has this property |
| Stationary increments | Joint distributions of increments across intervals | The random walk and Poisson counting process have this property |
| Implication | Strict stationarity implies stationary increments | The converse fails for the random walk and Poisson process |
Check Your Understanding
For each question, distinguish statements about the process values from statements about its increments.
- What must be compared to disprove strict stationarity? Why can a difference between two one-time marginal distributions suffice?
- For a random walk with independent, identically distributed increments, explain why the distribution of \(X_{n+r}-X_n\) does not depend on \(n\).
- For the constant random-value process, what is the joint vector of values at any finite collection of times? What is every increment?
- For a Poisson counting process of rate \(\lambda>0\), write the probability that an interval of length \(h\) contains zero events. Does this probability depend on the interval's starting time?
- State the implication between strict stationarity and stationary increments, and give one example showing why the converse fails.