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Uniformized finite-generator stationary residual certificate

Statement

Let Q be a finite column generator, let ν>0 satisfy ν≥max⁡jrj, and put P=I+Q/ν. Let m≥1 and q=δ(Pm)<1. Then the finite master equation p′=Qp has a unique stationary probability π, characterized among column probability vectors by Qπ=0. For any raw vector y∈Rn and column probability vector x, set c=∥y−x∥1. Then

∥y−π∥1≤c+m∥Qx∥1ν(1−q).

The identity Qx/ν=Px−x makes the generator residual directly usable by the discrete certificate. In the exact algorithm branch, y+=0 returns candidate-unavailable, and q≥1 returns certificate-unavailable; neither status supplies an algorithmic accuracy certificate.

In particular, when P and y are rational and the exact branch of Correct and certify a finite-chain stationary candidate returns certified with candidate x and correction c, those outputs satisfy the conditions above.

Facts & Assumptions

Given: A finite column generator Q, an admissible ν>0, its uniformized matrix P, an m-step coefficient q<1, a raw vector y, and a column probability vector x.

[F1]

Finite column generator: Q has nonnegative off-diagonal entries, zero column sums, exit rates rj=−Qjj, and acts on column probability vectors.

[F2]

The finite uniformized matrix is column stochastic: if ν≥max⁡jrj and ν>0, then P=I+Q/ν is column-stochastic.

[F3]

Finite master equation and uniformization formula: the solution of p′=Qp, p(0)=p0, is the Poisson-weighted uniformization series with matrix P, and each Pkp0 is a probability vector.

[F4]

m-step Dobrushin certificate for a stationary probability: if q=δ(Pm)<1, then P has a unique stationary probability πP and ∥x−πP∥1≤m∥Px−x∥1/(1−q) for every column probability vector x.

[F5]

Correct and certify a finite-chain stationary candidate: in its exact rational domain, a certified output has a column probability candidate x and correction distance c=∥y−x∥1.

Proof

Given: Q, ν, P, m, q, y, x, and c=∥y−x∥1 as above.

Proof technique: identify the generator-stationary law with the stationary law of its uniformized matrix, then substitute the generator residual.

1.1F1F2F3F4given

By [F1] and [F2], Q=ν(P−I) and P is column-stochastic. Since q=δ(Pm)<1, [F4] gives a unique stationary probability πP for P. The identity QπP=ν(P−I)πP=0 makes the constant trajectory a solution of the master equation; [F3] identifies it with the unique solution for that initial law. Conversely, any time-independent probability solution π of p′=Qp satisfies Qπ=0, hence Pπ=π, so it equals πP.

2.1F1F2F4F5step 1.1algebra∎

For the probability x, Px−x=Qx/ν follows from P=I+Q/ν. By [F4] and the triangle inequality, ∥y−πP∥1≤∥y−x∥1+∥x−πP∥1≤c+m∥Qx∥1/(ν(1−q)). If P and y are rational and the exact algorithm returns certified, [F5] supplies precisely such x and c.

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Sources