TheoremStatement: Literature-sourcedProof: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Finite master equation and uniformization formula

Statement

Let $Q$ be an $n \times n$ finite column generator, let $p_{0}$ be a column probability vector, and let $t \geq 0$ . The unique differentiable solution of

$p^{'} (t) = Q p (t), p (0) = p_{0},$

is $p (t) = e^{t Q} p_{0}$ . If $ν > 0$ and $ν \geq \max_{j} r_{j}$ , put $P = I + Q / ν$ ; then

$p (t) = \sum_{k = 0}^{\infty} e^{- ν t} \frac{(ν t)^{k}}{k!} P^{k} p_{0} .$

Each term $P^{k} p_{0}$ is a probability vector, and the displayed Poisson mixture is a probability vector. If $ν = 0$ , then $Q = 0$ and $p (t) = p_{0}$ .

Facts & Assumptions

Given: A finite column generator $Q$ , a column probability vector $p_{0}$ , time $t \geq 0$ , and an admissible uniformization rate $ν \geq 0$ .

[F1]

Finite column generator: $Q$ has nonnegative off-diagonal entries, zero column sums, and exit rates $r_{j} = - q_{j j}$ .

[F2]

The finite uniformized matrix is column stochastic: if $ν > 0$ dominates every exit rate, $P = I + Q / ν$ is column stochastic; if $ν = 0$ , $Q = 0$ .

[F3]

The scalar Poisson count law: for $λ \geq 0$ , the Poisson mass at $k$ is $e^{- λ} λ^{k} / k!$ when $λ > 0$ and the law is concentrated at zero when $λ = 0$ .

[F4]

Normalization, moments, and generating function of the Poisson count law: the finite-parameter Poisson masses sum to one.

[F5]

Finite matrix exponential series and its identities: $e^{t Q}$ has its locally convergent factorial series, derivative $Q e^{t Q}$ , and the commuting-matrix product identity.

[F6]

The mean value theorem, as the case $g (x) = x$ of Cauchy's: for $f$ continuous on $[a, b]$ with $a < b$ and differentiable on $(a, b)$ there is $c \in (a, b)$ with $f (b) - f (a) = f^{'} (c) (b - a)$ : a real function continuous on a closed interval and differentiable on its interior has endpoint difference equal to its derivative at some interior point times the interval length.

[F7]

Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ : finite sums and products of differentiable real-valued functions obey the sum and product rules, applied entrywise to finite matrix-vector products.

Proof

technique · direct

1.1givenF1F2algebrainduction

Suppose $ν > 0$ . By [F2], $P$ is nonnegative and column stochastic. If $x$ is a column probability vector, then $P x$ is nonnegative and $\sum_{i} (P x)_{i} = \sum_{j} (\sum_{i} P_{i j}) x_{j} = \sum_{j} x_{j} = 1$ . Induction gives that $P^{k} p_{0}$ is a probability vector for every $k \geq 0$ .

1.2F2F3F5algebra

By [F3] and the matrix series of [F5], the uniformization sum equals $e^{- ν t} e^{ν t P} p_{0}$ . The power series gives $e^{- ν t I} = e^{- ν t} I$ ; since $- ν t I$ commutes with $ν t P$ , [F5] gives $e^{- ν t I} e^{ν t P} = e^{ν t (P - I)} = e^{t Q}$ using $Q = ν (P - I)$ . This proves the displayed identity.

1.3givenF5algebra

By [F5], $p (t) = e^{t Q} p_{0}$ has derivative $Q e^{t Q} p_{0} = Q p (t)$ and initial value $p_{0}$ .

1.4givenF5F6F7algebra

Let $u$ be any other differentiable solution with the same initial vector and put $z (t) = e^{- t Q} u (t)$ . The product rule [F7], the equation $u^{'} = Q u$ , and commutation of $Q$ with $e^{- t Q}$ from [F5] give $z^{'} (t) = 0$ . If $t > 0$ , [F6] applied to each coordinate of $z$ on $[0, t]$ gives $z (t) = z (0) = p_{0}$ ; if $t = 0$ this follows from the initial value directly. Multiplying by $e^{t Q}$ and using [F5] gives $u (t) = e^{t Q} p_{0}$ . Thus the solution is unique.

2.1givenF3F4step 1.1algebra

The weights in [F3] at parameter $ν t$ sum to one by [F4]. Hence the nonnegative weighted sum of the probability vectors from step 1.1 is itself a probability vector.

3.1givenF2step 1.4algebra∎

If $ν = 0$ , [F2] gives $Q = 0$ . The constant function $p (t) = p_{0}$ solves the equation. The uniqueness argument in step 1.4 applied with $Q = 0$ shows that every solution with this initial value equals it. This proves the zero-rate case.

Depends on

Used by

Dependency tree · two levels

35 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Ward Whitt, Continuous-Time Markov Chains, full lecture notes (2006) (standard reference, not scraped)
Bennett L. Fox and Peter W. Glynn, Computing Poisson Probabilities (standard reference, not scraped)
Brian Munsky and Mustafa Khammash, The Finite State Projection Algorithm for the Solution of the Chemical Master Equation (standard reference, not scraped)