Stationarity, Convergence, and Compatibility

Three Different Questions About an FCS Chain

An FCS chain can appear stable while still being influenced by its starting values. Even if its distribution no longer depends on those values, that fact alone does not establish that the conditional models used to update variables can all arise from one joint probability distribution. These are three separate questions: Is there a stationary distribution? Does the chain converge to it from its initial state? Are the specified conditional models compatible with a joint distribution?

The earlier material on stationary distributions of FCS chains introduced the idea of an invariant distribution for a full update sweep. Here the emphasis is on distinctions and implications. Let $X_t$ denote the state of an FCS chain after sweep $t$, including the current imputed values; observed values are held fixed. If the update protocol is unchanged across sweeps, the chain has a transition kernel $K$, where $K(x,A)$ is the probability that the next state is in a set $A$, given current state $x$. The probability law of the next state is obtained by applying $K$ to the current law.

Definition: A distribution $\pi$ is stationary for a transition kernel $K$ if starting the chain with $X_0\sim\pi$ gives $X_t\sim\pi$ for every $t$. Equivalently, $\pi K=\pi$. The chain converges to $\pi$ from an initial state $x$ if the distribution of $X_t$ approaches $\pi$ as $t$ increases. For a finite state space, a common measure of this approach is total variation distance: $$ \|\mathcal{L}(X_t\mid X_0=x)-\pi\|_{\mathrm{TV}} =\frac{1}{2}\sum_{y}|\Pr(X_t=y\mid X_0=x)-\pi(y)|. $$

Stationarity describes what happens if the chain is already distributed as $\pi$. Convergence describes whether the chain reaches that distribution, at least approximately, when it starts elsewhere. Neither definition says that an individual chain settles at one completed data set. A stochastic chain normally continues to move, even after its marginal distribution has become nearly stationary.

When Does a Chain Forget Its Starting State?

A useful standard result applies when the chain has a finite state space and a fixed transition matrix. If the chain is irreducible (every state can eventually be reached from every other state with positive probability) and aperiodic (returns to a state are not restricted to multiples of an integer greater than one), it has a unique stationary distribution and converges to it from every starting state. The two conditions address different obstacles: irreducibility rules out separate closed regions, while aperiodicity rules out persistent cycles.

This finite-state result is a guide to the concepts, not an automatic convergence guarantee for every FCS application. Imputed values may have a continuous state space, and the behavior of the transition kernel depends on the conditional models and update procedure. The relevant conditions for a general-state chain must be justified for that chain; observing a stable trace is not a substitute for such justification.

Worked Example: Stationarity and Convergence in a Two-State Chain

Consider a hypothetical chain whose state is A or B, with transition matrix $$ P= \begin{pmatrix} 0.8 & 0.2\\ 0.3 & 0.7 \end{pmatrix}. $$ Rows indicate the current state and columns the next state. Find its stationary distribution and calculate how the probability of state A changes when the chain starts in A.

Write the stationary probabilities as $\pi(A)=a$ and $\pi(B)=1-a$. Stationarity requires that the probability of A after one transition equal $a$: $$ a=0.8a+0.3(1-a)=0.5a+0.3. $$ Therefore $0.5a=0.3$, so $a=0.6$, and $\pi(B)=0.4$. Both transitions between states have positive probability, so the chain is irreducible. Each state also has a positive probability of remaining where it is, so it is aperiodic. The finite-state result therefore implies convergence to this unique stationary distribution from either starting state.

Let $p_t=\Pr(X_t=\mathrm{A}\mid X_0=\mathrm{A})$. At the next step, the chain is in A either by moving from A to A or from B to A, giving $$ p_{t+1}=0.8p_t+0.3(1-p_t)=0.5p_t+0.3. $$ The fixed point is $0.6$. Subtracting it gives $p_{t+1}-0.6=0.5(p_t-0.6)$. Since $p_0=1$, repeated substitution yields $p_t=0.6+0.4(0.5)^t$. Thus $p_1=0.8$, $p_2=0.7$, and $p_4=0.625$, approaching $0.6$. The chain does not become fixed at A or B; its distribution gradually loses its initial excess probability of A.

The calculation illustrates a general point: convergence is about the changing probability distribution, not a requirement that the realized sequence of states stop changing. In practice, a single observed path cannot display its entire distribution at each iteration. Diagnostics therefore use indirect evidence, such as comparing summaries across chains started from different initial states and checking whether those summaries continue to drift.

Stationarity Does Not Guarantee Convergence

A stationary distribution may exist even when the chain fails to converge from a particular starting state. Periodicity gives a simple example. The chain can preserve a distribution when initialized from it, while a chain started at one state keeps cycling and its distribution oscillates rather than approaching the stationary distribution.

Worked Example: A Stationary Distribution With No Pointwise Convergence

Consider a hypothetical chain with two states, A and B, that switches state with probability 1: $$ P= \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}. $$ The distribution $\pi(A)=\pi(B)=0.5$ is stationary: after one transition, the probability of A is the previous probability of B, also $0.5$, and likewise for B. But starting in A gives

\Pr(X_t=\mathrm{A}\mid X_0=\mathrm{A}) = \begin{cases} 1, & t \text{ even},\\ 0, & t \text{ odd}. \end{cases}

This probability does not approach $0.5$; it alternates between 1 and 0. The chain is irreducible, but it is periodic, so the convergence result above does not apply. This example separates existence of a stationary distribution from convergence to it. It also shows why checking only whether a candidate distribution is invariant cannot establish that initialization has been forgotten.

In real FCS work, a chain that has not moved much is not necessarily close to stationarity, and a chain that moves is not necessarily far from it. Poor mixing, long transients, and multiple regions that are difficult to move between can make apparent stability misleading. Conversely, visible random fluctuation is expected after convergence and does not by itself indicate failure.

Convergence Is Not Compatibility

Compatibility concerns the conditional models, not the chain’s dependence on its initial state. A collection of conditional models is compatible if there exists a joint probability distribution whose corresponding full conditional distributions are exactly the specified models. Compatibility asks whether one joint distribution can underlie all of the conditionals. Convergence asks whether repeated application of a transition kernel approaches an invariant distribution.

Definition: A set of conditional models is compatible if there is a joint distribution for the variables such that conditioning that joint distribution on the other variables gives each specified conditional model. A stationary distribution of the full FCS transition kernel is an invariant law for that chain; its existence alone does not show that the update conditionals are the full conditionals of that law.

This difference matters because a systematic FCS sweep applies several conditional update steps in sequence. Invariance under the full sweep does not, by definition, mean that a single joint distribution generates every conditional model used in those steps. Thus, an FCS chain can have a stable limiting distribution without the specified conditionals being compatible. Conversely, compatibility does not imply that a particular implementation will reach the corresponding distribution quickly: convergence speed also depends on the transition behavior.

The compatibility question requires examining whether the specified conditionals can come from a common joint distribution. It is not answered by plotting traces, running more iterations, or finding that several chains have similar summaries. The next material on compatibility develops how that question is assessed; here the essential point is not to treat stability as evidence that compatibility has been established.

Worked Diagnostic Example: What Multiple Chains Can Show

Worked Example: Comparing Chains Started From Different Imputations

Suppose an analyst runs four hypothetical FCS chains and records the mean of an imputed continuous variable, in arbitrary units, at selected sweeps. The chains start from deliberately different completed data sets. Their recorded summaries are:

Sweep	Chain 1	Chain 2	Chain 3	Chain 4
10	12.1	15.0	9.8	14.2
50	13.4	13.8	13.1	13.6
100	13.5	13.6	13.4	13.7

At sweep 10, the across-chain range is $15.0-9.8=5.2$ units; at sweep 50 it is $13.8-13.1=0.7$ units; and at sweep 100 it is $13.7-13.4=0.3$ units. The summaries have moved closer together, and their central values are near $13.5$ units. This is evidence that the initial differences in this particular summary have diminished.

It is not proof that the full chain has converged. Chains could agree on this mean while differing in other features of their imputed-data distributions, or all could be trapped in the same region. A useful assessment checks several relevant summaries and patterns, and considers whether chains continue to show systematic movement. Even convincing evidence of convergence would establish neither compatibility of the conditionals nor that the limiting distribution is appropriate for the scientific analysis.

Practical Interpretation and Common Pitfalls

For an FCS analysis, the chain state includes all currently imputed missing values, so the full state can be much higher-dimensional than any one diagnostic summary. A trace plot or summary comparison is a practical diagnostic, not a direct observation of total variation distance between the chain’s distribution and an unknown stationary distribution. Assessment is necessarily limited, and apparent agreement among a finite number of chains cannot rule out every failure mode.

A common mistake is to reason: “The chains’ summaries look stable, therefore the conditionals are compatible and the imputations are valid.” The conclusion does not follow. Stability is evidence relevant to convergence for the summaries examined. Compatibility is a separate property of the conditional models, and validity also depends on whether the models and assumptions are appropriate for the data and analysis. Another mistake is to discard a fixed number of early sweeps and assume that this guarantees the starting state has been forgotten. Discarding early draws can help after adequate assessment, but no universal number of sweeps ensures convergence for every kernel.

Key takeaway: Stationarity is invariance, convergence is approach to an invariant distribution from an initial state, and compatibility is existence of a joint distribution that generates the specified conditionals. Evidence for one does not establish the others.

Check Your Understanding

Answer each question, then compare with the solutions below.

In one sentence each, distinguish stationarity from convergence.
A finite-state chain is irreducible but periodic. Does the finite-state convergence result stated here guarantee convergence to its stationary distribution? Explain.
For the first two-state chain, what is the probability of A after three transitions when starting in A? Show the calculation.
Four FCS chains have similar means for one imputed variable. What does that suggest, and what does it not establish?
Explain why convergence of an FCS chain does not prove compatibility of its conditional models.

Solutions. (1) Stationarity means that a chain initialized from a distribution retains that distribution after every transition; convergence means that the distribution from a specified initial state approaches a stationary distribution over time. (2) No. The result requires both irreducibility and aperiodicity; periodicity can cause oscillation despite stationarity. (3) Using $p_t=0.6+0.4(0.5)^t$, $p_3=0.6+0.4(0.125)=0.65$, so the probability is $0.650$. (4) Similar means are evidence that the initial differences in that particular summary may have diminished. They do not establish convergence of the full state distribution, compatibility, or model validity. (5) Convergence concerns the behavior of the full transition kernel; compatibility asks whether a joint distribution has the specified conditionals. An invariant distribution of the full sweep need not, merely by being invariant, generate each update conditional.