Two Different Questions About an FCS Chain
An FCS chain can settle into a stable pattern of draws even when its specified conditional models do not agree with any single joint probability distribution. These are distinct properties. Convergence concerns the behavior of repeated applications of the chain’s transition rule. Compatibility concerns whether the conditional distributions used in that rule can all be derived from one joint distribution. Evidence for the first does not establish the second.
This distinction matters because a stable chain can produce reproducible summaries of its own stationary behavior while failing to sample from a joint model having the intended conditionals. The chain may still have a well-defined limiting distribution, but that distribution need not have every specified conditional as its corresponding conditional distribution.
The distinction is especially clear in a finite-state example. We will use two binary variables, \(X\) and \(Y\), and a systematic update that first draws \(X\) given the current \(Y\), then draws \(Y\) given the newly drawn \(X\). Every number below is hypothetical and chosen to make the distinction transparent.
A Compatibility Check for Two Binary Conditionals
Write \(p_y=P(X=1\mid Y=y)\) and \(q_x=P(Y=1\mid X=x)\). If a strictly positive joint distribution exists, the conditional models must imply the same joint odds ratio. From the \(X\)-given-\(Y\) conditionals, that ratio is the odds of \(X=1\) at \(Y=1\), divided by the odds of \(X=1\) at \(Y=0\). From the \(Y\)-given-\(X\) conditionals, it is the odds of \(Y=1\) at \(X=1\), divided by the odds at \(X=0\). Thus compatibility requires
This condition follows directly by writing the joint cell probabilities as \(P_{xy}=P(X=x,Y=y)\). The left side simplifies to \(P_{11}P_{00}/(P_{10}P_{01})\); the right side simplifies to the same quantity. If the two sides differ, no positive joint distribution can have both sets of conditionals.
Worked Example: Detecting Incompatible Binary Conditionals
Suppose the hypothetical FCS specification is \(P(X=1\mid Y=0)=0.20\), \(P(X=1\mid Y=1)=0.80\), \(P(Y=1\mid X=0)=0.10\), and \(P(Y=1\mid X=1)=0.90\). Check whether these conditionals are compatible.
For the \(X\)-given-\(Y\) models, the implied odds ratio is
For the \(Y\)-given-\(X\) models, it is
Because \(16\ne81\), the required equality fails. Therefore, no strictly positive joint distribution has all four specified conditional probabilities. This conclusion concerns compatibility only: it says nothing yet about whether the update algorithm converges.
Why the Chain Can Still Converge
Define one full sweep as follows: from the current state \((X,Y)\), draw \(X'\) from the specified distribution of \(X\mid Y\), then draw \(Y'\) from the specified distribution of \(Y\mid X'\). The new state is \((X',Y')\). In the example, every conditional probability and its complement is positive. Consequently, from any starting state, every possible new pair \((x',y')\) has positive probability:
For a finite-state Markov chain, positivity of every one-sweep transition makes the chain irreducible (every state can be reached) and aperiodic (the chain is not forced into a fixed cycle). A finite irreducible, aperiodic Markov chain has a unique stationary distribution, and the distribution of its state converges to that stationary distribution from any starting state. This result requires no compatibility assumption. It is a result about the transition matrix, not about the existence of a joint distribution generating all the conditionals.
Worked Example: Finding the Limiting Behavior of the Binary Chain
Use the conditionals from the preceding example. Track \(Y\) from one sweep to the next. If the current \(Y=0\), then \(X'=1\) with probability \(0.20\); if \(X'=1\), the next \(Y'=1\) with probability \(0.90\). If \(X'=0\), that probability is \(0.10\). Therefore,
If the current \(Y=1\), the corresponding calculation is
So the transition matrix for \(Y\), with states ordered \(0,1\), is
Its stationary probability of \(Y=1\), denoted \(m\), satisfies \(m=0.26(1-m)+0.74m\), giving \(m=0.50\). The other eigenvalue of the matrix is \(0.74-0.26=0.48\). Hence, from an initial probability \(r_0=P(Y_0=1)\), the probability after \(k\) sweeps is
For example, if the chain starts with \(Y_0=0\) with certainty, then after five sweeps \(P(Y_5=1)=0.50-0.50(0.48)^5\approx0.4873\), which is close to its limit of \(0.50\). The chain has a clear, rapidly approached limiting distribution despite the incompatibility established above.
What the Stationary Distribution Does—and Does Not—Mean
The full stationary distribution can also be calculated for this example. At stationarity, \(P(Y=0)=P(Y=1)=0.50\). During the next \(X\) update, \(P(X'=1)=0.50(0.20)+0.50(0.80)=0.50\). The second update draws \(Y'\) from the specified \(Y\mid X'\) model. Thus the joint probabilities at the end of a sweep are
These probabilities sum to one and define the stationary distribution at sweep endpoints. Under it, \(P(Y'=1\mid X'=1)=0.45/(0.45+0.05)=0.90\), and \(P(Y'=1\mid X'=0)=0.05/(0.05+0.45)=0.10\): the \(Y\mid X\) conditionals are recovered. But \(P(X'=1\mid Y'=1)=0.45/(0.45+0.05)=0.90\), not the specified \(0.80\). Likewise, \(P(X'=1\mid Y'=0)=0.05/(0.05+0.45)=0.10\), not the specified \(0.20\). The limiting distribution is real and stable; it simply does not have both prescribed conditionals.
Worked Example: Interpreting Stable Empirical Summaries
Imagine a long hypothetical run of the chain above. Its sweep-endpoint frequencies approach \(0.45, 0.05, 0.05, 0.45\) for \((X,Y)=(1,1),(1,0),(0,1),(0,0)\), respectively. The resulting marginal frequency for \(X=1\) is \(0.45+0.05=0.50\), and for \(Y=1\) it is also \(0.50\). These frequencies can become very stable as the run grows.
Now estimate the conditional probability \(P(X=1\mid Y=1)\) from those endpoint frequencies. The estimate approaches \(0.45/(0.45+0.05)=0.90\), whereas the FCS update specified \(0.80\). Similarly, the empirical \(P(X=1\mid Y=0)\) approaches \(0.10\), whereas its update specified \(0.20\). Stable marginal frequencies and stable joint cell frequencies therefore describe the chain’s limiting behavior; they do not certify that the chain’s limiting distribution reproduces every update model.
The arithmetic assumes the run has reached the stationary regime and that the empirical frequencies adequately represent it. Even granting both assumptions, the result is a statement about this algorithm’s stationary distribution, not a compatibility proof.
What Stability Checks Can Establish
Stability checks address whether the simulated chain appears to have settled into a consistent operating regime. For example, summaries from later portions of a run may be similar, and runs initialized differently may show broadly similar behavior. Such evidence can help assess whether the algorithm has reached and explored its own limiting behavior. It cannot establish that the conditionals are compatible: a chain generated by incompatible conditionals can pass these checks, as the finite example demonstrates.
Compatibility requires a different kind of argument. One can check necessary compatibility constraints where they are available, derive conditionals from a proposed joint model and compare them with the specified models, or establish compatibility analytically for the model class. A simulation-based stability check is not a substitute for these structural checks. In the example, checking whether the two implied odds ratios agree answers the compatibility question directly; observing stable state frequencies does not.
The main alternatives therefore answer different questions:
| Approach | Question it addresses | What it cannot establish by itself |
|---|---|---|
| Check compatibility constraints or derive from a joint model | Can the specified conditionals come from one joint distribution? | That a particular simulation has converged in practice |
| Examine chain behavior across iterations or runs | Does the algorithm appear to have settled into stable behavior? | That the specified conditionals are compatible |
| Analyze the transition kernel mathematically | Does this transition process have a stationary distribution and converge under stated conditions? | That the stationary distribution has all the specified full conditionals |
The finite-state positivity argument also has limits. It applies because there are finitely many states and every state can be reached with positive probability in one sweep. In other settings, these conditions may fail, and convergence requires a suitable argument for that transition process. Even when convergence is established, compatibility remains a separate question.
Common Mistake
Wrong reasoning: “The chains from different starting values settle to similar distributions, so the conditional models must be compatible.” The conclusion does not follow. Similar stable behavior can arise because the same transition kernel has a unique stationary distribution, even if the conditional models violate compatibility constraints.
Correction: State precisely what the evidence supports. Chain stability may support the claim that the simulation is approaching the stationary behavior of its transition process. To claim compatibility, check whether a common joint distribution exists. Do not turn a numerical convergence assessment into a structural compatibility result.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions below.
- For the binary conditionals in the example, what odds ratio is implied by each set of conditionals, and what does their difference establish?
- Why does positivity of every one-sweep transition imply convergence for the finite binary chain?
- At stationarity, the endpoint frequency for \((X,Y)=(1,1)\) is \(0.45\), and for \((X,Y)=(0,1)\) it is \(0.05\). What conditional probability does this imply for \(P(X=1\mid Y=1)\), and how does it compare with the specified update?
- A chain’s marginal means appear stable over a long run. Does that prove either convergence or compatibility? State the more limited conclusion supported by the observation.
Solutions.
1. The \(X\mid Y\) conditionals imply \([0.80/0.20]/[0.20/0.80]=16\). The \(Y\mid X\) conditionals imply \([0.90/0.10]/[0.10/0.90]=81\). Since these differ, no strictly positive joint distribution has both sets of conditionals.
2. Positivity means every state can reach every state in one sweep. The finite chain is therefore irreducible; positive self-transition probabilities also make it aperiodic. A finite irreducible, aperiodic Markov chain has a unique stationary distribution and converges to it from any starting distribution. This argument does not require compatibility.
3. The implied conditional probability is \(0.45/(0.45+0.05)=0.90\). The specified update gives \(P(X=1\mid Y=1)=0.80\), so the stationary endpoint distribution does not reproduce that conditional.
4. Stable marginal means alone do not prove convergence or compatibility. They are limited evidence that those particular summaries have stopped changing appreciably in the observed run. They do not establish convergence of the full chain, and they provide no compatibility proof.