ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-6.1-sol)audited 2026-10-02

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Uniform law for a finite doubly stochastic matrix

Example

Let $E = {1, \dots, n}$ with $n \geq 1$ and let $p$ be a transition matrix on $E$ (Transition matrices and n-step probabilities) whose columns also sum to one: $\sum_{x \in E} p (x, y) = 1$ for every $y \in E$ . Then the uniform probability $π (x) = 1 / n$ is invariant for $p$ (Invariant and stationary distribution for a Markov kernel). No irreducibility hypothesis is needed, and no uniqueness is asserted: the identity matrix on $E$ is doubly stochastic with the same uniform invariant law.

Facts & Assumptions

Given: A nonempty finite set $E = {1, \dots, n}$ , a transition matrix $p$ on $E$ with $\sum_{y \in E} p (x, y) = 1$ for all $x$ , and with the extra hypothesis $\sum_{x \in E} p (x, y) = 1$ for all $y$ .

[F1]

The entries satisfy $p (x, y) \geq 0$ , rows sum to one, and the one-step matrix entries are the kernel masses $p (x, y) = K (x, {y})$ . (Transition matrices and n-step probabilities)

[F2]

On a countable state space with transition matrix $p$ , a probability vector $π$ is invariant exactly when $π (y) = \sum_{x \in E} π (x) p (x, y)$ for every $y \in E$ ; a finite set is countable. (Invariant and stationary distribution for a Markov kernel)

Verification

technique · direct computation of the measure-matrix product column by column

1.1givenalgebra

Define $π (x) : = 1 / n$ for $x \in E$ . Since $n \geq 1$ , each entry satisfies $π (x) \geq 0$ and $\sum_{x \in E} π (x) = n \cdot \frac{1}{n} = 1$ , so $π$ is a probability vector.

1.2givenalgebra

For every $y \in E$ , $(π p) (y) = \sum_{x \in E} π (x) p (x, y) = \frac{1}{n} \sum_{x \in E} p (x, y) = \frac{1}{n} \cdot 1 = \frac{1}{n} = π (y)$ , where the second equality factors the finite constant $\frac{1}{n}$ out of a finite sum and the third is the column-sum hypothesis.

2.1F1F2step 1.2given

By [F2] the identity of step 1.2 says exactly that $π$ is an invariant probability vector, i.e. a stationary distribution for $p$ .

3.1F1F2step 2.1given∎

Irreducibility is not used: the identity matrix on a finite $E$ with $n \geq 2$ is doubly stochastic, has $π (x) = 1 / n$ invariant by step 1.2, and is reducible, so the hypothesis cannot be weakened to a uniqueness statement; the uniform law is one invariant law among possibly several, and for $n = 1$ it is the only one since $p (1, 1) = 1$ .

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