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  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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Conditional stopping bounds for prime-tested Rz candidates

Statement

For the candidate sequence and exponents of Exponent growth of the Rz scaled grid, let Ej be success events and J=min⁡{j≥1:Ej}. Suppose there are numbers pj∈[0,1] with pj≥c/(kj+1), for a constant 0<c≤1 independent of the input, such that Pr⁡(J>j)≤∏i=1j(1−pi)(j≥1). Independent success events of probabilities pj are a sufficient special case; independence of the events is not otherwise required. Then J<∞ almost surely and, for every fixed positive integer r, E[Jr]=O((k2+1)r)=O((1+log⁡(1/ε))r). In particular E[J]=O(log⁡(1/ε)) and E[log⁡2J]=O(1+log⁡log⁡(1/ε)). The implied constants can depend on c and r, but not on θ or ε. For every fixed nonnegative integer d, E[J(1+k2+log⁡J)d] is polynomially bounded in 1+k2.

This is the explicit quantitative probabilistic assumption needed when Ross–Selinger Hypothesis 8.3 is used for a complexity claim. It is a conditional model, not an assertion of independence for a deterministic input sequence.

Facts & Assumptions

Given: The success events, joint-failure product bound, constant c, and random index J in the statement.

[F1]

Exponent growth of the Rz scaled grid gives kj≤k2+2⌈log⁡2j⌉ and k2=O(1+log⁡(1/ε)).

Proof

1.1F1givenalgebra

By [F1], choose constants a≥1 and b>0, with a=O(k2+1), so that pj≥1/(a+bln⁡j) for all j≥1. The assumed joint-failure bound and 1−x≤e−x imply Pr⁡(J>j)≤∏h=1j(1−ph)≤exp⁡(−j/(a+bln⁡j)). This upper bound tends to zero, which proves almost sure finiteness.

2.1step 1.1algebra

For j≥a, put x=j/a≥1. Since ln⁡a≤a and ln⁡x≤2x, one has a+bln⁡j≤(1+3b)ax. Hence Pr⁡(J>j)≤exp⁡(−j/a/(1+3b)) for j≥a. For an integer-valued positive random variable, summing the identity Jr=∑j=0J−1((j+1)r−jr) and using nonnegative sums gives E[Jr]=∑j≥0((j+1)r−jr)Pr⁡(J>j). The part j<a is O(ar). The remaining part is bounded by a constant times ∑j≥a(j+1)r−1e−j/a/(1+3b)=O(ar): comparison with the integral and substitution y=t/a reduce it to ar∫1∞y2r−1e−y/(1+3b) dy, which is finite by repeated integration by parts. This proves all the moment bounds.

3.1F1step 2.1algebra∎

The elementary tangent bound ln⁡x≤ln⁡μ+(x−μ)/μ for x,μ>0, with μ=E[J], gives E[log⁡2J]≤log⁡2E[J]=O(1+log⁡(k2+1)). Finally log⁡J≤J, and expansion of J(1+k2+J)d reduces its expectation to the finitely many moments in step 2.1; these are polynomial in k2+1. The stated bounds in ε follow from [F1].

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