Stationary laws and circulating flow

What you will learn

Solve finite-chain stationary laws by conserving statewise probability flow, verify invariance independently and distinguish stationarity from uniqueness and convergence.

Prerequisites: Transition kernels and two-step laws

Ask for a law that survives a transition

A fictional inspection token occupies one of three workstations, A, B or C, at each integer tick. Its only possible move is clockwise, A to B to C to A; otherwise it stays put. Four equally likely fresh tickets determine each transition. At A one ticket moves clockwise, at B two move, and at C all four move. Assume the current workstation contains all relevant memory, ticket draws are fresh, ticks have equal duration and these rules do not change over time. These are stipulated mechanisms for learning, not fitted market regimes or a model of actual poker behavior. We use the state order A, B, C throughout.

The conditional transition rows are (3,1,0)/4, (0,2,2)/4 and (4,0,0)/4. Earlier work computed a future law from a supplied initial law. Here the unknown is the initial law itself: which probability vector will be reproduced after one transition? Write that row vector as π. The requirement is πP=π, together with nonnegative coordinates summing to one. Calling π stationary describes an unchanged distribution, not an unmoving token. It also does not make successive states independent. A token known to be at C must next be at A, even when its unconditional distribution is stationary.

Balance total flow without demanding reverse edges

For a proposed allocation q, the probability flowing from A to B in one tick is q(A)/4. From B to C it is q(B)/2, and from C to A it is q(C). Staying mass contributes to the next occupancy but cancels when computing a state’s net change. Thus the change at A is q(C)−q(A)/4; at B it is q(A)/4−q(B)/2; at C it is q(B)/2−q(C). These changes always sum to zero. Conservation alone is insufficient: a loss at one state can be hidden by an equal gain elsewhere.

Stationarity requires each change separately to vanish, giving a common clockwise flow J=q(A)/4=q(B)/2=q(C). Notice the unequal state masses implied by equal flows. A slow exit can sustain the same throughput only if more probability is waiting upstream of it. This is global balance. Pairwise detailed balance would additionally require q(A)P(A,B)=q(B)P(B,A). The right side is zero here while the left side will be positive. The circulation therefore gives an explicit stationary system without detailed balance. Do not delete legitimate motion simply because the occupancy totals are stable.

Solve the flow equations and audit the answer twice

Substitute q(A)=4J, q(B)=2J and q(C)=J into normalization. Their sum is 7J=1, so J=1/7 and π=(4/7,2/7,1/7). Normalization is essential: the unnormalized vector (4,2,1) balances algebraically but is not a probability law. Independently multiply by the conditional rows. The next A probability is (4/7)(3/4)+(1/7)(1)=3/7+1/7=4/7. Next B is (4/7)(1/4)+(2/7)(1/2)=1/7+1/7=2/7. Next C is (2/7)(1/2)=1/7. All three entries are reproduced, and their sum is one.

A second check counts elementary source-ticket combinations. Represent the proposed law using seven equally likely source tickets: four place the token at A, two at B and one at C. Pair each source ticket with one of four equally likely transition tickets, making twenty-eight pairs. The four A source tickets yield twelve A endpoints and four B endpoints. The two B tickets yield four B and four C endpoints. The C ticket yields four A endpoints. Endpoint totals are therefore sixteen, eight and four out of twenty-eight, reducing to the original law. This enumeration checks the multiplication without solving the flow equations again.

A fixed law does not prove attraction or uniqueness

The unequal-holding cycle can reach every state from every other state along positive-probability clockwise paths. This is irreducibility. Its self-loop at A permits return in one tick; combined with communication, this removes a common return-time cycle and gives aperiodicity. The finite-chain convergence theorem then supplies convergence to the unique stationary law from every start. We use that theorem with its conditions stated; a handful of numerical preview ticks is not its proof. In particular, the first few distributions may be far from π even though π is already an exact fixed point.

Now replace all exit rates by one. The token rotates every tick. Solving the flow equations still gives a unique stationary law, now (1/3,1/3,1/3). Yet starting at A produces A, B, C, A forever, so the sequence of time-specific laws has no limit. Starting with the stationary mixture reproduces that mixture at every tick. Next stop exits from A while keeping B and C exits positive. Only mass concentrated at A is stationary, and every start eventually reaches A. Finally stop exits from both A and B. Every mixture (t,1−t,0), for 0≤t≤1, is stationary. A start at A and a start at B now select different limits. These boundary constructions separate existence, uniqueness and attraction.

Construct a certificate by moving mass, then challenge it

The workshop gives you twenty-one mass chips, each representing probability 1/21. Choose a donor and receiver and transfer one chip at a time. The live ledger uses denominator eighty-four because each chip has four transition-ticket branches. For each state it shows outgoing mass, incoming mass, signed change and the next occupancy. Begin with seven chips everywhere. Under unequal holding this looks symmetric, but its outgoing numerators are seven, fourteen and twenty-eight, so it drifts. Build a balanced allocation instead, enter the common edge-flow probability and choose a justified long-run verdict.

Changing the mechanism clears the certificate and restores the equal allocation. In the forced-cycle mechanism that allocation already balances, but the preview starting specifically at A exposes cycling. Advance the preview several times and distinguish its starting law from your candidate stationary mixture. Under one absorbing state, move every chip to A and check zero circulating flow. Under two absorbing states, construct two different balanced allocations on A and B; no mass may remain at C. The twenty-one-chip grid samples this continuous family without exhausting it. Hints and reset are available, and drafts last for this visit only. A successful certificate is a checked consequence of this finite mechanism, not an empirical forecast or account credit.

Hands-on: Balance a circulating law

Move conserved probability mass, audit statewise flow and challenge convergence with periodic and absorbing mechanisms.

Common mistakes

• Equating state occupancies instead of statewise incoming and outgoing flow.
• Accepting an unnormalized fixed vector such as (4,2,1).
• Demanding detailed balance on a directed cycle and incorrectly concluding no stationary law exists.
• Reading a stationary distribution as an absorbing physical state or independent draws.
• Inferring convergence from uniqueness, or uniqueness from a single balanced candidate.

Keep these ideas

• A stationary law is a normalized nonnegative fixed point πP=π.
• Cancel staying mass to balance total incoming and outgoing flow at each state.
• Stationarity can sustain directed circulation without detailed balance.
• Finite irreducible aperiodic chains converge; periodic and absorbing boundaries need separate reasoning.
• Verify a candidate through both transition multiplication and an independent outcome count.

Further reading

• Data 140: Long Run Behavior, sections 10.3.1–10.3.5 and 10.3.7 ↗

  Read finite irreducible aperiodic convergence theorem, normalized balance equations and invariance under a stationary start. Original unequal-rate flow derivation and 28-pair audit; theorem used with conditions, not proved here. Checked 2026-10-24.

• Data 140: Deconstructing Chains, sections 10.2.1–10.2.2 ↗

  Read communication, irreducibility, return-time period and self-loop criterion. Original three-state forced-cycle and absorbing boundary constructions. Checked 2026-10-24.

• Data 140: Balance and Detailed Balance, sections 11.1.1–11.1.3 ↗

  Read total flow balance, pairwise detailed balance and its sufficient but stronger character. Original one-way unequal-holding cycle and mass-transfer activity. No video review or copied exercises. Checked 2026-10-24.