ExampleConstruction: AI-generatedVerification: 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.

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Two-state gating through uniformization

Example

For an off/on gate, let the rate from off to on be $α \geq 0$ and the rate from on to off be $β \geq 0$ . With column probability vector $(p_{o f f}, p_{o n})^{T}$ and $s = α + β$ , the generator is

$Q = (\begin{array}{cc} - α & β \\ α & - β \end{array}) .$

Starting off, if $s > 0$ the exact on probability is $p_{o n} (t) = \frac{α}{s} (1 - e^{- s t})$ ; if $s = 0$ the gate remains off. The case $α = 0$ gives no opening, while $β = 0$ gives $p_{o n} (t) = 1 - e^{- α t}$ .

Facts & Assumptions

Given: The states are off and on; the transition rates are $α \geq 0$ from off to on and $β \geq 0$ from on to off; the initial vector is $e_{o f f} = (1, 0)^{T}$ .

[F1]

Finite column generator: the source state indexes the column, off-diagonal entries are nonnegative, and columns sum to zero.

[F2]

Finite master equation and uniformization formula: the unique master-equation solution equals the Poisson-weighted sum of $P^{k} e_{o f f}$ for any admissible $ν > 0$ .

[F3]

The scalar Poisson count law and Normalization, moments, and generating function of the Poisson count law: the Poisson weights sum to one and the mass at zero is $e^{- s t}$ .

Verification

technique · compute the two-state uniformization matrix and sum its nontrivial powers

1.1givenF1algebra

The matrix $Q$ in the Example has off-diagonal entries $α, β \geq 0$ and both columns sum to zero, so it satisfies [F1]; its exit rates are $α$ and $β$ .

1.2givenF1constructalgebrainduction

Suppose $s > 0$ and choose $ν = s$ . Then $P = I + Q / s = (\begin{array}{cc} β / s & β / s \\ α / s & α / s \end{array})$ . Both columns equal the probability vector $v = (β / s, α / s)^{T}$ . Therefore $P e_{o f f} = v$ and $P v = v$ , which by induction gives $P^{k} e_{o f f} = v$ for every $k \geq 1$ .

2.1givenF2F3step 1.2algebra

By [F2], the on-coordinate at time $t$ is $\sum_{k \geq 1} e^{- s t} (s t)^{k} (α / s) / k!$ . By [F3] the sum of the masses over $k \geq 1$ is $1 - e^{- s t}$ , so $p_{o n} (t) = \frac{α}{s} (1 - e^{- s t})$ . The off-coordinate is $1 - p_{o n} (t)$ because the solution is a probability vector.

3.1givenF1F2step 2.1algebra∎

If $s = 0$ , nonnegativity of $α, β$ implies $α = β = 0$ , so $Q = 0$ and the master equation keeps $e_{o f f}$ fixed. If $s > 0$ and $α = 0$ , the formula gives $p_{o n} = 0$ ; if $β = 0$ , it gives $p_{o n} = 1 - e^{- α t}$ . The endpoint cases agree with the stated rates.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

17 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)