Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-generatedPipeline-generated
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.

Conditional moments of one Poisson tau-leap increment

Statement

Fix a state x∈E and step h≥0 in Explicit Poisson tau-leaping with an invalid-state status. Let ΔY=∑j=1RνjKj, where conditionally independent Kj∼Pois⁡(aj(x)h). Then E[ΔY∣x]=h∑j=1Raj(x)νj, Cov⁡(ΔY∣x)=h∑j=1Raj(x)νjνjT, and E[∥ΔY∥2∣x]=h∑j=1Raj(x)∥νj∥2+h2∥∑j=1Raj(x)νj∥2.

Facts & Assumptions

Given: A state x∈E, a step h≥0, the product count law in Explicit Poisson tau-leaping with an invalid-state status, and the Poisson count laws below.

[F1]

The scalar Poisson count law: for λ≥0, a Pois⁡(λ) count has masses e−λλk/k! (with a point mass at zero when λ=0).

[F2]

Normalization, moments, and generating function of the Poisson count law: if K∼Pois⁡(λ), then EK=λ and Var⁡(K)=λ.

[F3]

Explicit Poisson tau-leaping with an invalid-state status: the channel counts are conditionally independent and the state increment is ΔY=∑jνjKj.

Proof

Given: The inputs above.

1.1givenF1F2F3algebra

Each parameter aj(x)h is finite, so [F1] defines the corresponding Poisson law and [F2] gives E[Kj∣x]=aj(x)h. By [F3], conditional expectation is linear, hence E[ΔY∣x]=∑jνjaj(x)h.

1.2F2F3algebra

Conditional independence in [F3] makes cross-channel covariances zero. The covariance of νjKj is νjνjTVar⁡(Kj∣x)=haj(x)νjνjT by [F2]. Summing proves the covariance formula.

2.1step 1.1step 1.2algebra∎

For a square-integrable vector, E∥ΔY∥2=tr⁡Cov⁡(ΔY)+∥EΔY∥2. Substitution of steps 1.1 and 1.2 gives the asserted identity.

Depends on

Used by

Dependency tree · two levels

14 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