Alphabeta Math
LemmaStatement: 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.

A geometric upper bound for a Poisson right tail

Statement

For N∼Pois⁡(μ) with μ≥0 and integer K≥0, write τK(μ):=P(N>K). If μ=0, then τK(μ)=0. If μ>0 and μ<K+2, then

τK(μ)≤e−μμK+1(K+1)!(1−μ/(K+2)).

The right side is a geometric majorant computed from a Poisson mass, not a floating-point subtraction of a rounded partial sum from one.

Facts & Assumptions

Given: A finite parameter μ≥0 and integer K≥0.

[F1]

For μ>0, P(N=k)=e−μμk/k!; at μ=0 the law is concentrated at zero. (The scalar Poisson count law)

[F2]

The masses in The scalar Poisson count law normalize to one, so P(N>K)=∑k>KP(N=k). (Normalization, moments, and generating function of the Poisson count law)

[F3]

Factorials satisfy (k+1)!=(k+1)k! for k≥0. (The factorial n! and the falling factorial nk‾, defined by recursion in N)

[F4]

For 0≤ρ<1, the geometric series satisfies ∑j≥0ρj=1/(1−ρ). (For ∣r∣<1, ∑k≥0rk=1/(1−r), and for ∣r∣≥1 the series diverges)

Proof

technique · direct
1.1givenF1F3algebra

If μ=0, [F1] gives N=0 almost surely and K≥0, so the tail is zero. Now suppose μ>0 and put wk=e−μμk/k! and ρ=μ/(K+2). The hypothesis gives 0<ρ<1, and [F1]–[F3] yield wk+1/wk=μ/(k+1)≤ρ for every k≥K+1.

2.1step 1.1induction

Iterating the ratio bound gives wK+1+j≤wK+1ρj for each integer j≥0.

3.1F2F4step 2.1algebra∎

By [F2] and step 2.1, τK(μ)=∑j≥0wK+1+j≤wK+1∑j≥0ρj=wK+1/(1−ρ) by [F4]; substituting the values of wK+1 and ρ proves the stated bound.

Depends on

Used by

Dependency tree · two levels

37 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