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

Rare-event reweighting, overlap and certified partition ratios

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

TheoremStatement: AI-adaptedProof: AI-generatedprecheck passverified 2026-10-22 (gpt-6.1-sol)Open item page →

Finite partition-ratio estimator and overlap variance

Statement

Let X be a finite nonempty set equipped with counting measure. For a∈{0,1}, let wa:X→(0,∞) be a finite positive weight, define Za=∑x∈Xwa(x), and define the probability mass function πa(x)=wa(x)/Za. For an integer n≥1, let X1,…,Xn be independent random variables with common law π0. Define R(x)=w1(x)/w0(x), r^n=n−1∑j=1nR(Xj), and χ2(π1∥π0)=∑x∈Xπ0(x)(π1(x)/π0(x)−1)2. Then

Eπ0[r^n]=Z1Z0,Var⁡π0(r^n)(Z1/Z0)2=χ2(π1∥π0)n.

The estimator uses only the finite raw-weight ratios w1(Xj)/w0(Xj); the partition sums define the target ratio and probability laws but are not inputs to its computation.

Facts & Assumptions

Given: The finite nonempty state set, strictly positive finite weights w0,w1, their finite normalizing sums Z0,Z1, and independent samples Xj∼π0.

[F1]

Ordinary importance sampling: for an unnormalized target p~, proposal density q, integrand f, and target normalizer Zp, the weighted value is fp~/(Zpq) on q>0 and zero otherwise. Its support and integrability conditions are explicit, and its normalizing constant Zp is a known input to that algorithm.

[F2]

Ordinary importance sampling has finite variance and a CLT under a weighted second moment: under the ordinary importance-sampling hypotheses and finite weighted second moment, the finite-sample variance is the variance of one weighted value divided by n; the finite-sample conclusion is choice-free.

Proof

Given: [F1], [F2], and the finite positive weights and IID π0 sample in the Statement.

1.1F1givenalgebra

Put R=w1/w0. Finite summation gives Eπ0R=∑x(w0(x)/Z0)(w1(x)/w0(x))=Z1/Z0. Since w0(x),w1(x)>0 on every point of X, the formal ordinary-importance-sampling setup with counting measure, p~=w1, q=π0, f≡1, and Zp=Z1 has full support and finite weighted second moment. Its weighted value is Y=(Z0/Z1)R=π1/π0, and finite summation gives Eπ0Y=(Z0/Z1)Eπ0R=1. Here Z1 is used only to identify this mathematical random variable; it is not supplied to or computed by the raw-weight ratio estimator.

2.1F2givenstep 1.1algebra

Apply [F2] to this formal weighted variable with mean 1 from step 1.1. It gives Var⁡(Y‾n)=n−1(Eπ0[Y2]−1). Since Y‾n=(Z0/Z1)r^n, rescaling gives Var⁡(r^n)/(Z1/Z0)2=n−1(∑xπ0(x)(π1(x)/π0(x))2−1).

3.1givenstep 2.1algebra∎

The probability masses sum to one, so expanding the square gives χ2(π1∥π0)=∑xπ0(x)(π1(x)/π0(x))2−2∑xπ1(x)+∑xπ0(x)=∑xπ0(x)(π1(x)/π0(x))2−1. Substitution into step 2.1 proves the variance identity.

Remarks

The proof uses the ordinary importance-sampling variance identity only after symbolically normalizing the target. It does not treat the ordinary importance-sampling algorithm, whose contract requires a known target normalizer, as a procedure for computing an unknown partition ratio.

AlgorithmAlgorithm: AI-adaptedProof: Not applicableprecheck passverified 2026-10-22 (gpt-6.1-sol)Open item page →

Independent finite bridge-ratio estimator

Definition

Input a finite nonempty state space X, an integer K≥1, positive finite weights w0,…,wK:X→(0,∞), positive integers n0,…,nK−1, and, on a common probability space (Ω,F,P), supplied random variables Xk,j:Ω→(X,P(X)) for 0≤k<K and 1≤j≤nk. For each k, set Zk=∑x∈Xwk(x) and πk(x)=wk(x)/Zk, a probability mass function on X and thus a probability measure on P(X). Require P(Xk,j=x)=πk(x) for every x∈X and require the entire finite family (Xk,j)k,j to be mutually independent. These exact draws are inputs, not outputs of this procedure; approximate Gibbs draws, correlated trajectories, and adaptively chosen bridges are outside its contract.

For each k=0,…,K−1, compute R‾k=nk−1∑j=1nkwk+1(Xk,j)/wk(Xk,j). Return R^=∏k=0K−1R‾k. The procedure evaluates only raw-weight ratios; it neither computes nor requires any Zk as a numeric input.

TheoremStatement: AI-adaptedProof: AI-generatedprecheck passverified 2026-10-22 (gpt-6.1-sol)Open item page →

Mean and relative second moment of independent bridge ratios

Statement

Let X be a finite nonempty set, let K≥1, and let w0,…,wK:X→(0,∞) be finite positive weights. For each k, set Zk=∑x∈Xwk(x) and πk(x)=wk(x)/Zk. For strictly positive probability mass functions ν,μ:T→(0,1] on any finite nonempty set T (so ∑x∈Tν(x)=∑x∈Tμ(x)=1), define χ2(ν∥μ)=∑x∈Tμ(x)(ν(x)/μ(x)−1)2. On a common probability space (Ω,F,P), let Xk,j:Ω→(X,P(X)) for 0≤k<K and 1≤j≤nk be mutually independent random variables with P(Xk,j=x)=πk(x) for each x∈X, where nk≥1; these are the supplied stage samples in alg-csm-frontier-independent-bridge-ratios. Then

E[R^]=ZKZ0,E[R^2](ZK/Z0)2=∏k=0K−1(1+χ2(πk+1∥πk)nk).

Consequently, Var⁡(R^)/(ZK/Z0)2 is the displayed product minus 1. This is an exact finite-sample result under independent exact draws; it gives no claim for reused trajectories, approximate Gibbs draws, or adaptively selected bridges.

Facts & Assumptions

Given: The finite positive weights, stage laws, sample counts, and mutually independent stage samples in the Statement.

[F1]

Finite partition-ratio estimator and overlap variance: for a finite positive pair of weights and nk independent draws with common law πk, the adjacent raw-ratio sample mean has mean Zk+1/Zk and relative variance χ2(πk+1∥πk)/nk.

[F2]

Independent finite bridge-ratio estimator: the procedure returns the product of those adjacent sample means and requires all draws across stages to be mutually independent.

Proof

Given: [F1], [F2], and the finite positive weights and mutually independent stage samples.

1.1F1givenalgebra

For each k, apply [F1] to (wk,wk+1) and the nk independent samples with common law πk. With rk=Zk+1/Zk, the stage mean R‾k satisfies E[R‾k]=rk and Var⁡(R‾k)/rk2=χ2(πk+1∥πk)/nk. Since E[R‾k2]=Var⁡(R‾k)+(E[R‾k])2, it follows that E[R‾k2]/rk2=1+χ2(πk+1∥πk)/nk. All moments are finite because X is finite and the weights are positive and finite.

2.1F1F2givenstep 1.1algebra∎

Write Bk=(Xk,1,…,Xk,nk)∈Xnk. The blocks are independent finite-valued random elements, and R‾k is a function of Bk. If μk(b)=P(Bk=b), then for any functions Gk of the respective blocks, independence and finite summation give E[∏k=0K−1Gk(Bk)]=∑b0∈Xn0,…,bK−1∈XnK−1∏k=0K−1μk(bk)Gk(bk)=∏k=0K−1E[Gk(Bk)]. Apply this first with Gk=R‾k and then with Gk=R‾k2. Thus E[R^]=∏kE[R‾k]=∏k(Zk+1/Zk)=ZK/Z0, while E[R^2]/(ZK/Z0)2=∏kE[R‾k2]/rk2. Substitution from step 1.1 gives the stated product, and subtracting one yields the relative variance.

AlgorithmAlgorithm: AI-adaptedProof: Not applicableprecheck passverified 2026-10-22 (gpt-6.1-sol)Open item page →

Certified finite partition and expectation enclosure

Definition

Input a finite nonempty set X with an explicit enumeration (x1,…,xm) listing each element exactly once, functions w:X→(0,∞) and f:X→R, and, for each x∈X, rational numbers ax≤bx and cx≤dx such that w(x)∈Wx:=[ax,bx] and f(x)∈Fx:=[cx,dx]. Here a closed rational input interval [a,b] means a,b∈Q and a≤b; its endpoints need not be dyadic, so before conversion it is distinct from the finite-dyadic real interval defined in def-intervals-rectangles-and-complex-balls. Choose a finite dyadic format F as in def-dyadic-floating-point-format-and-directed-rounding, with its directed roundings. The interval assertions are certified input premises; this procedure does not derive them from energies or evaluate an exponential or logarithm.

For i=1,…,m in increasing order, convert Wxi and then Fxi by rounding each lower rational endpoint down and each upper rational endpoint up into F; denote the converted intervals by W^i and F^i. If any conversion overflows, return unresolved-overflow and no finite enclosure claim. Initialize Z0=N0=[0,0]. For i=1,…,m in increasing order, compute the following three operations in the displayed order:

Pi=W^iF^i,Zi=Zi−1+W^i,Ni=Ni−1+Pi.

These recurrences fix the finite expression DAG: both sums are accumulated from left to right in the supplied enumeration, with rounding after each displayed operation and no reassociation. On completing the loop, set Z=Zm and N=Nm.

Every addition and multiplication uses the outward-rounded interval operations; division is used only for a denominator interval excluding zero. If any operation returns a non-finite status, return unresolved-operation and no finite enclosure claim. Otherwise return the finite intervals Z and N. If the lower endpoint of Z is strictly positive, also return the outward interval quotient E=N/Z; if that endpoint is nonpositive, return unresolved-denominator and no normalized-expectation enclosure.

This is finite enumeration and can require exponentially many operations in the model size. It certifies arithmetic from supplied weight intervals; it does not certify how those intervals were obtained.

TheoremStatement: AI-adaptedProof: AI-generatedprecheck passverified 2026-10-22 (gpt-6.1-sol)Open item page →

Soundness and termination of finite partition enclosures

Statement

Let X be finite and nonempty, let w:X→(0,∞) be an exact finite positive weight, and let f:X→R be a finite observable. Suppose the finite rational input intervals in the definition of alg-csm-frontier-interval-finite-partition contain every exact w(x) and f(x). The algorithm terminates after finitely many operations. If it returns finite intervals Z and N, then they contain Z=∑x∈Xw(x) and N=∑x∈Xw(x)f(x), respectively. If it also returns an expectation interval E, then its denominator interval has positive lower endpoint and E contains the exact normalized expectation ∑xπ(x)f(x)=N/Z, where π(x)=w(x)/Z. An overflow, non-finite operation, or nonpositive denominator lower endpoint produces an unresolved status and carries no normalized-expectation enclosure claim.

Facts & Assumptions

Given: The exact finite positive weights and finite observable on the enumerated nonempty state space, the certified rational input intervals, and the algorithm's finite or unresolved return branch.

[F1]

Certified finite partition and expectation enclosure: the procedure converts rational input endpoints outward, constructs finite interval sums for weights and weighted observables, and divides only when the partition interval's lower endpoint is positive.

[F2]

Dyadic floating-point format and directed rounding: for an exact rational intermediate within the finite range, directed rounding returns a greatest representable lower endpoint and a least representable upper endpoint; overflow is reported and has no finite-error claim.

[F3]

Real intervals, complex rectangles and ball enclosures: a finite expression DAG is a finite directed acyclic graph with input and constant nodes; sharing a node reuses the same exact intermediate value.

[F4]

Outward-rounded interval, rectangle and ball operations: addition, subtraction, multiplication, division away from zero, and squaring use the stated exact endpoint hulls followed by downward and upward endpoint rounding; division by an interval containing zero returns a split or unresolved status rather than a finite quotient.

[F5]

Directed dyadic rounding encloses with explicit underflow and overflow bounds: each finite directed endpoint rounding brackets the exact real endpoint; overflow is explicitly not a finite enclosure result.

[F6]

Finite real interval operations enclose their exact values: for a finite acyclic expression DAG whose input and constant intervals enclose their values, each finite computed node interval encloses its exact value, including reused shared nodes.

Proof

Given: [F1]–[F6] and the certified finite rational input intervals in the Statement.

1.1F1F2F3F5givenconstruct

The finite enumeration supplies one input interval for each exact weight and observable. By [F2] and [F5], when endpoint conversion is finite, rounding the lower endpoint down and upper endpoint up preserves containment of each input interval; if conversion overflows, the algorithm returns unresolved and the theorem makes no finite-enclosure claim. The partition sum and weighted sum are finite acyclic expression DAGs by [F1] and [F3].

2.1F3F4F6step 1.1given

Order the nodes of each finite DAG topologically. Each input node encloses its exact value by step 1.1. At every operation node with a finite result, [F4] gives the interval extension prescribed for that operation and [F6] proves that its value encloses the exact operation on the predecessor values. Induction over the finite node order therefore places Z in Z and N in N whenever those intervals are returned.

3.1F1F3F4F6step 2.1givenalgebraconstruct∎

The exact partition sum satisfies Z>0 because X is nonempty and every w(x)>0. If the returned lower endpoint of Z is positive, then 0∉Z and [F4] defines outward interval division; applying [F6] to this final operation gives N/Z∈N/Z=E. If the lower endpoint is nonpositive, or any operation is non-finite, [F1] returns an unresolved status and no quotient enclosure is claimed. Finally, the procedure performs finitely many conversions and finitely many operations because X is finite, and every operation returns a finite interval or a status; hence it terminates.

5 · Examples, experiments and counterexamples

None yet.

Sources