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TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passaudited 2026-10-02
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Every transition matrix on a nonempty finite state space has a stationary distribution

Statement

Let E be a nonempty finite set and let p be a transition matrix on E (Transition matrices and n-step probabilities). Then p has an invariant probability distribution (Invariant and stationary distribution for a Markov kernel), that is, some probability vector π on E satisfies πp=π. No irreducibility, aperiodicity or recurrence hypothesis is needed, and the argument uses no choice principle and no Markov-chain path law.

Facts & Assumptions

Given: A nonempty finite set E and a transition matrix p on E, with E={x1,…,xn} for some n≥1.

[F1]

The entries satisfy p(x,y)≥0 and ∑y∈Ep(x,y)=1 for every x∈E, and the matrix powers are the n-step probabilities p(k)(x,y) of the iterated kernel; the finite sums over E are ordinary finite sums. (Transition matrices and n-step probabilities)

[F2]

On a countable state space with transition matrix p, a probability vector π is invariant exactly when π(y)=∑x∈Eπ(x)p(x,y) for every y∈E; a finite set is countable. (Invariant and stationary distribution for a Markov kernel)

[F3]

Every bounded sequence of reals has a convergent subsequence: there is a strictly increasing nj and a real L with xnj→L. (Bolzano-Weierstrass: every bounded real sequence has a convergent subsequence)

Proof

Given: A nonempty finite set E={x1,…,xn} with n≥1 and a transition matrix p on E.

Proof technique: Cesàro-average a single point mass in the compact finite simplex and pass to a convergent subsequence, using the exact telescoping identity for the drift.

1.1F1given

Fix the state x1 and let μ be its point mass. For every N≥1 define the row vector νN:=1N∑k=0N−1μp(k), where p(k) is the k-th matrix power. Its entries are finite nonnegative sums of products of the entries of p, hence νN(y)∈[0,1] for every y∈E; and using [F1] twice, ∑y∈EνN(y)=1N∑k=0N−1∑y∈Eμp(k)(y)=1N∑k=0N−11=1, since a point mass is a probability vector and rows of every p(k) sum to one. So every νN lies in S:={ρ:E→[0,1]:∑yρ(y)=1}.

1.2F1step 1.1algebra

For every N≥1 the exact telescoping identity νNp−νN=1N∑k=0N−1(μp(k+1)−μp(k))=1N(μp(N)−μ) holds coordinatewise as an identity of finite real sums, its entries being differences of numbers in [0,1], so no infinite sum is rearranged.

2.1F3step 1.1given

There is a strictly increasing sequence Nj and a vector π:E→[0,1] such that νNj(y)→π(y) for every y∈E. Enumerate E={x1,…,xn} and argue by induction on the number i of coordinates already handled: the i-th coordinate sequence along the subsequence produced so far is bounded in [0,1], so [F3] supplies a further strictly increasing subsequence on which it converges; after finitely many successive subsequence choices, every coordinate converges. Finite induction on these existential choices requires no choice axiom.

2.2F1step 1.2given

Consequently νNp−νN→0 coordinatewise as N→∞: each entry of 1N(μp(N)−μ) has absolute value at most 2N, since every entry of μp(N) and of μ lies in [0,1].

3.1step 2.1step 1.1algebra

The limit π is a probability vector: π(y)≥0 for every y because a limit of nonnegative numbers is nonnegative; and ∑y∈Eπ(y)=1 because a finite sum of convergent sequences converges to the sum of the limits, applied to the constant sums 1 from step 2.1 and step 1.1.

3.2step 2.1step 2.2algebra

Passing to the subsequence of step 2.1, νNj→π coordinatewise and hence νNjp→πp coordinatewise, because each entry of the finite matrix product is a finite sum ∑y∈EνNj(y)p(y,z) of finitely many convergent sequences. By step 2.2 the left side of νNp−νN tends to 0 along Nj, so πp−π=0, that is, πp=π.

4.1F1F2F3step 3.1step 3.2given∎

By step 3.1, π is a probability vector and by step 3.2 it satisfies πp=π; [F2] then identifies it as an invariant probability distribution for p. The state space was required nonempty so that x1 exists; the empty matrix has no probability vector, so E=∅ is excluded by the hypothesis. If n=1, then p(x1,x1)=1 and π=δx1 is invariant, consistent with the construction. The argument uses only finite enumerations, finite sums and the subsequence theorem; no irreducibility, aperiodicity, recurrence, product-space path law or choice principle is used, and the conclusion is a one-way existence assertion rather than an equivalence.

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