--- title: "Stationary Distributions and Long-Run Behaviour" description: "A stationary probability row vector satisfies pi P = pi and is a left eigenvector for eigenvalue 1. For a finite chain with fixed transitions, irreducibility and aperiodicity give convergence to a uni" canonical: https://lightmysky.com/learn/mathematics/stationary-distributions-and-long-run-behaviour-mt_MGfHvUGkeh source: https://lightmysky.com/learn/mathematics/stationary-distributions-and-long-run-behaviour-mt_MGfHvUGkeh.md retrieved: 2026-11-12 --- > **Agent view.** This is the Markdown twin of the page, for tools and assistants. > When to use this site, and the call that answers each job: https://lightmysky.com/agent-instructions.md > API description (OpenAPI 3.1): https://lightmysky.com/openapi.json · Authentication: https://lightmysky.com/auth.md > Pricing: https://lightmysky.com/pricing.md · Catalog: https://lightmysky.com/llms.txt · Full catalog: https://lightmysky.com/llms-full.txt > Every machine-readable file on this domain: https://lightmysky.com/.well-known/ai-catalog.json > Ask for Markdown with `Accept: text/markdown`, a `.md` address, or `?mode=agent`. # Stationary Distributions and Long-Run Behaviour A stationary probability row vector satisfies pi P = pi and is a left eigenvector for eigenvalue 1. For a finite chain with fixed transitions, irreducibility and aperiodicity give convergence to a unique stationary distribution from every starting distribution. Subject: Mathematics · Area: Probability · Ages 22 to 24 Page: https://lightmysky.com/learn/mathematics/stationary-distributions-and-long-run-behaviour-mt_MGfHvUGkeh ## Ready when they can - Solve pi P = pi and check nonnegative entries totaling one - Decide whether a chain is irreducible and aperiodic - Say what periodicity does to the long-run behaviour ## Lesson: Stationary weights and limiting distributions Markov updates weight and add, as expected value does. Multiply each transition to a target by its current-state probability and add: a row distribution u becomes uP. Use finite chains with fixed P and one order for rows (current) and columns (next). A stationary row vector pi has nonnegative entries totaling 1 and satisfies pi P = pi. It is a left eigenvector for eigenvalue 1. Transposing gives P^T pi^T = pi^T, where T means transpose. Check all three conditions. **Example.** Use state order (0, 1) and transition rows (0.9, 0.1), (0.4, 0.6). Write pi = (a, 1 - a). The state 0 equation is a = 0.9a + 0.4(1 - a), so 0.5a = 0.4 and a = 0.8. Thus pi = (0.8, 0.2). The state 1 update is 0.8 × 0.1 + 0.2 × 0.6 = 0.2. Both entries are nonnegative and total 1. *(drawing: For two states, the probability moving from 0 to 1 balances the probability moving back.)* A current row distribution u becomes uP after one step. For destination j, add u_i × P_ij over current states i. Repeat for further steps. With order (0, 1) and rows (0.6, 0.4), (0.1, 0.9), starting at 0 gives (0.6, 0.4) after one step. After two steps, the state 1 probability is 0.6 × 0.4 + 0.4 × 0.9 = 0.60. **Example.** For three states, write pi = (a, b, c). Form one equation per destination column and use a + b + c = 1. With order (0, 1, 2) and rows (0.6, 0.4, 0), (0.2, 0.4, 0.4), (0, 0.2, 0.8), the outer equations give 0.4a = 0.2b and 0.2c = 0.4b. Thus a = b/2 and c = 2b. Normalizing gives pi = (1/7, 2/7, 4/7). Substitute into every component equation to check. A chain is irreducible when every state can reach every other along a path of positive-probability transitions. The path may take several steps; a zero one-step entry does not prove reducibility. A state that stays at itself with probability 1 is absorbing. In a chain with other states, it cannot reach them, so the whole chain is reducible. A state's period is the greatest common divisor of its possible positive return times. Period 1 means aperiodic. A self-loop makes time 1 possible, so that state has period 1. All states of an irreducible chain share a period; one self-loop therefore makes that chain aperiodic. An absorbing state has period 1. Without self-loops, inspect longer return paths: possible times 2 and 3 already give greatest common divisor 1. The deterministic flip moves 0 to 1 and 1 to 0. It is irreducible, but returns occur at times 2, 4, 6 and so on, giving period 2. Its stationary distribution is (0.5, 0.5): exchanging equal weights leaves them unchanged. Starting at state 0 instead gives alternating distributions, so its matrix powers do not converge. Any stationary starting distribution stays fixed, even in a periodic chain. Every finite chain has a stationary distribution. Irreducibility makes it unique. If the chain is also aperiodic, uP^n approaches pi for every starting distribution u; every row of P^n approaches pi. Irreducibility and aperiodicity together are sufficient, not necessary, for a common limit. Rows (1, 0), (0.5, 0.5) give state 1 weight b × 0.5^n from starting weight b, so every start approaches (1, 0). Stationary distributions need not be unique. The identity transition matrix leaves every (a, 1 - a), with 0 <= a <= 1, unchanged. To turn a nonzero, nonnegative stationary row vector v into a probability distribution, divide every entry by the sum of its entries. If vP = v, this scaling preserves the equation and makes the total 1. For two states, let r = P_01 and s = P_10. Stationarity gives pi_0 × r = pi_1 × s. Derive this by cancelling the stay-in-state-0 term from pi_0 = pi_0(1 - r) + pi_1s. If pi_0 is positive, an unknown r equals pi_1 × s divided by pi_0. Check that the result lies between 0 and 1, then check the completed matrix and stationary update. In order (L, M, R), let L and R be absorbing and M's row be (r, t, s), with t < 1. From M, the limiting R probability h satisfies h = s + th. Solve: h = s/(1 - t). **Recap.** Check pi P = pi, nonnegative entries and total 1. Finite irreducible, aperiodic chains converge to one stationary distribution from every start. A stationary start stays fixed. *The lesson, practice questions and flashcards were written for this dataset.* ## Practice 18 questions on this page, each with its working shown. ## Needs first - [Markov Chains and the Memoryless Assumption](https://lightmysky.com/learn/mathematics/markov-chains-and-the-memoryless-assumption-mt_2wDa_JxJ_o) ## Optional connections - [Eigenvalues and Eigenvectors](https://lightmysky.com/learn/mathematics/eigenvalues-and-eigenvectors-mt_TVqqaw11qa) - [Diagonalisation and Powers of a Matrix](https://lightmysky.com/learn/mathematics/diagonalisation-and-powers-of-a-matrix-mt_w15uCVhdhX) ## Opens up - [Martingales and Optional Stopping](https://lightmysky.com/learn/mathematics/martingales-and-optional-stopping-mt_198uXX3jRJ)