Weak and strict stationarity
Separate invariance of finite-dimensional laws from invariance of second moments, and identify the extra role of joint Gaussianity.
On this page
The question for today
What you’ll understand
- State strict stationarity and weak stationarity as different mathematical properties.
- Use joint Gaussianity and finite moments in the correct implication directions.
- Distinguish a homogeneous transition rule from a stationary initialized process.
A time series can preserve its mean and covariance while changing its distribution, and it can preserve its entire distribution without having a variance at all. Stationarity therefore needs a definition rather than a visual judgment about a plot. The weak definition supports linear prediction and covariance analysis; the strict definition concerns every finite collection of observations. An initialization, deterministic trend or changing measurement mechanism can break either property even when the same recursion is used at each date.
Distributional invariance involves every finite vector
A process indexed by integers is strictly stationary if shifting all times in any finite observation vector by the same integer leaves its joint distribution unchanged. Equal one-time marginals alone are insufficient, since the dependence between successive observations could change. An iid sequence is strictly stationary because its joint laws are products of one fixed marginal law. Strict stationarity does not itself impose integrability: an iid Cauchy sequence satisfies the definition despite lacking a finite mean and variance.
Weak stationarity instead requires a finite second moment at every time, one constant mean, and covariance depending only on time difference. We use gamma(h)=Cov(X at t+h, X at t). For a real scalar series, gamma is even. Weak stationarity permits covariance-based methods without assuming the whole process law is Gaussian. It is a statement about the population process, not a claim that sample means or sample covariances computed from two finite stretches will match exactly.
The implication needs the missing hypotheses
Strict stationarity with finite second moments implies weak stationarity by applying the distributional equality to one- and two-dimensional vectors. The reverse implication fails in general. However, if every finite vector is jointly Gaussian, its law is determined by its mean vector and covariance matrix. Weak stationarity then makes all those Gaussian laws shift invariant, proving strict stationarity. Saying only that each individual observation is marginally Gaussian does not supply that joint-Gaussian assumption.
White noise is another term requiring care. Weak white noise has mean zero, a constant finite variance and zero covariance at every nonzero lag. It need not have independent observations. Jointly Gaussian white noise is independent because its finite covariance matrices are diagonal, but a non-Gaussian process can conceal nonlinear dependence behind zero correlations. Later innovations will initially be defined by orthogonality, so independence must be justified separately when a likelihood calculation needs it.
Initialization and trend are part of the model
A time-homogeneous Markov transition law can generate a nonstationary process when its initial distribution differs from an invariant distribution. The transition probabilities stay fixed while marginal probabilities move toward equilibrium. Similarly, a stable autoregression started from a deterministic value has a transient variance before reaching its stationary law. A stationary representation extending into the infinite past and a finite recursion with a chosen initial value are related constructions, but they are not automatically the same process.
A deterministic trend added to a stationary error changes the mean with calendar time. Removing a known trend can reveal a stationary residual; estimating that trend introduces uncertainty and a data-availability question. A rolling plot that looks stable does not prove strict stationarity, ergodicity or independence. Those properties control different aspects of inference, and a structural change can make stationarity a useful local approximation rather than an exact global assumption.
Work it through
Second moments can hide a changing distribution
Let all observations be independent. At even integer times X_t is a fair sign taking ±1; at odd times it is standard normal. Classify weak and strict stationarity.
Both kinds of marginal have mean zero and variance one. Independence gives zero covariance between every pair of distinct times.
Thus gamma(0)=1 and gamma(h)=0 for nonzero h, independent of the date. The process is weakly stationary and is weak white noise.
Its one-time distributions change after a shift by one: at even times P(X_t=1)=1/2, while at odd times that probability is zero. Hence strict stationarity fails.
Constant means and covariance do not determine non-Gaussian laws. Independent observations need not be identically distributed.
The living notebook
Explore the model
Interactive lab · markov
Keep initialization in the stationarity question
Pause and reset, then set both controls to p = 0.2 and q = 0.3. Starting in state zero, predict probabilities of state one 0.2 after one Step and 0.3 after two, approaching invariant probability 0.4. These changing means show that this initialized process is not stationary. For a fresh comparison, pause and reset, then set p = 0 and q = 0.3; state zero is absorbing and the chosen initial law remains there. The instrument does not offer an arbitrary initial distribution or test stationarity from sampled data.
- P(state 1) at step 0
- 0%
- Stationary P(state 1)
- 60%
- Chain behavior
- Monotone convergence
Exact finite-step probabilities for a two-state chain starting in state 0, not sampled trajectories. Playback advances two transitions per second. When p = q = 0 every distribution is stationary; when p = q = 1 the chain alternates forever. A single zero transition creates an absorbing state. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forA fixed transition matrix does not guarantee that the chosen process has stationary marginals.
Pause & reason
Check your understanding
Your turn
Try it yourself
Classify an iid Cauchy sequence and the process Y_t=3t+epsilon_t with iid standard-normal epsilon. If the deterministic trend is known and removed from Y, what remains?
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Check moment existence before using the weak definition, and check the mean's time dependence before inspecting covariance.
Compare with the solution
The iid Cauchy sequence is strictly stationary but not weakly stationary because its second moment does not exist. The process Y has mean 3t, so it is neither weakly nor strictly stationary, even though its centered errors have constant covariance. Subtracting the known trend leaves iid standard-normal errors, which satisfy both notions. Estimating a trend from data is a different statistical operation and adds parameter uncertainty; a known subtraction should not be confused with proof that a fitted residual sequence has the assumed population law.
Keep the idea
Take it with you
- Strict stationarity concerns finite-dimensional laws, while weak stationarity requires finite second moments.
- Joint Gaussianity, rather than Gaussian marginals alone, supplies the reverse implication.
- Initialization and deterministic trends can break stationarity despite a fixed recursion.
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