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✓ 4 results · all verified · 0 also independently AI-judged

Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 4 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Verifiable generation, group-relative updates, and test-time sampling — Examples

1 · Prerequisites

2 · Logical flowchart

3 · Definitions, theorems and proofs

None yet.

4 · Examples, counterexamples and false statements

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passOpen item page →

Equal group rewards set every normalized advantage to zero

Example

For any prompt, suppose every response in the policy support receives reward $7$ and the group size is $G = 3$ . Then every sampled group has mean reward $7$ , population standard deviation $0$ , and normalized advantages $A_{i} = 0$ . The reward surrogate contribution to the finite group-relative objective is exactly zero for every candidate policy parameter; the separately specified KL penalty remains in the objective formula.

Example references

The population-standard-deviation zero branch and clipped group objective.

Facts & Assumptions

Given: A prompt with a finite supported response set, reward $r (y) = 7$ on every supported response, three independent group draws, and the finite group-relative update convention.

[F1]

Finite truncated response spaces, correctness predicates, and verifiers: a response policy has a finite fixed support and a finite real reward on each supported response.

[F2]

Finite-budget group-relative policy update: for $G \geq 2$ , the group uses the population standard deviation; when it is zero all advantages are set to zero, while the objective also includes its exact KL penalty.

Verification

Given: The objects and conventions in the Example.

1.1givenF1F2algebra

Since every reward is $7$ , the group mean is $\overline{R} = (7 + 7 + 7) / 3 = 7$ , and each difference $R_{i} - \overline{R}$ is zero. Thus the population standard deviation is $s = 0$ .

2.1F2step 1.1algebra∎

By the specified zero-standard-deviation branch in [F2], each $A_{i} = 0$ . For every sampled response and candidate parameter, both arguments of the clipped minimum are zero, so its average reward-surrogate contribution is zero. The objective's separately defined KL term remains.

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passOpen item page →

Perfect and false-positive verifiers have different rejection outcomes

Example

Let a prompt have one correct response $c$ with probability $3 / 4$ and one incorrect response $w$ with probability $1 / 4$ . With a perfect verifier $A = C$ , rejection search with budget $N = 2$ returns an accepted response with probability $15 / 16$ . With a false-positive verifier that accepts both $c$ and $w$ , rejection search always returns its first draw, so its false-acceptance probability is $1 / 4$ ; the per-draw union bound gives $2 (1 / 4) = 1 / 2$ .

Example references

Rejection search returns the first accepted response within its draw budget.

The independent success formula and the false-acceptance union bound.

Facts & Assumptions

Given: A finite two-response law with $\Pr (Y = c) = 3 / 4$ , $\Pr (Y = w) = 1 / 4$ , $C (c) = 1$ , $C (w) = 0$ , rejection budget $N = 2$ , and the two stated verifier choices.

[F1]

Finite truncated response spaces, correctness predicates, and verifiers: correctness and verifier outputs are binary functions on the finite response set.

[F2]

Budgeted verifier-guided generation and best-of-N sampling Budgeted verifier generation terminates with bounded response and selection calls: rejection search returns the first response accepted by $A$ , or failure after $N$ draws.

[F3]

Independent generation success and verifier false-acceptance bounds: with a perfect verifier, independent rejection succeeds with probability $1 - (1 - p)^{N}$ .

[F4]

Independent generation success and verifier false-acceptance bounds: if each draw has false-acceptance probability at most $ϵ$ , an accepted but incorrect selected output has probability at most $N ϵ$ .

Verification

Given: The objects and hypotheses in the Example.

1.1givenF1F2F3algebra

For the perfect verifier, $p = 3 / 4$ and rejection succeeds iff at least one of the two independent draws is $c$ . Its failure probability is $(1 - 3 / 4)^{2} = 1 / 16$ , so its success probability is $15 / 16$ .

2.1givenF1F2F4algebra∎

For the accept-all verifier, the first draw is always accepted. It is incorrect exactly when it is $w$ , an event of probability $1 / 4$ . Each draw has false-acceptance probability $ϵ = 1 / 4$ , so [F4] gives the valid bound $N ϵ = 1 / 2$ ; the exact first-draw probability $1 / 4$ is below it.

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passOpen item page →

Repeated dependent copies do not increase success probability

Example

Let a single response $Y$ be correct with probability $1 / 2$ , and define two candidates by $Y_{1} : = Y_{2} : = Y$ . With a perfect verifier, the probability that at least one candidate is correct is $1 / 2$ . The independent-draw formula at $N = 2$ would give $3 / 4$ , but these duplicated candidates are dependent and do not provide a second opportunity for success.

Example references

The success formula requires independent draws.

Facts & Assumptions

Given: A finite correct/incorrect response law with $\Pr (C (Y) = 1) = 1 / 2$ , and the repeated-copy construction $Y_{1} = Y_{2} = Y$ .

[F1]

Finite truncated response spaces, correctness predicates, and verifiers: correctness is a binary function on the response set.

[F2]

Independent generation success and verifier false-acceptance bounds: $N$ independent draws with one-draw success probability $p$ have success probability $1 - (1 - p)^{N}$ under a perfect verifier.

Verification

Given: The objects and construction in the Example.

1.1givenF1algebra

Since $Y_{1} = Y_{2} = Y$ , the event that at least one candidate is correct equals ${C (Y) = 1}$ , so its probability is $1 / 2$ .

2.1givenF1F2step 1.1algebra∎

The copies are not independent because $\Pr (Y_{1} correct, Y_{2} correct) = 1 / 2 \neq (1 / 2)^{2}$ . Applying [F2] to genuinely independent draws with the same marginal would give $1 - (1 - 1 / 2)^{2} = 3 / 4$ , which differs from the repeated-copy probability in step 1.1.

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passOpen item page →

Pass-at-one, oracle pass-at-N, and verifier-selected accuracy differ

Example

For one prompt let the finite response law assign mass $1 / 4$ to a correct response $c$ and mass $3 / 8$ to each of two incorrect responses $w_{1}, w_{2}$ . Let the verifier accept $c$ and $w_{1}$ but reject $w_{2}$ . The single-response pass@1 rate is $1 / 4$ ; for any integer $M \geq 1$ , the empirical pass@1 over $M$ independent replicas from this same law has expectation $1 / 4$ . Oracle pass@2, the event that at least one of two independent candidates is correct, has probability $7 / 16$ . For the same two candidates, use rejection search with budget $N = 2$ ; its verifier-selected accuracy, the probability its output is correct conditional on being accepted, is $2 / 5$ .

Example references

The finite law, correctness predicate, and verifier.

Rejection search returns the first accepted response or explicit failure.

The perfect-verifier rejection formula and general false-acceptance bound.

Facts & Assumptions

Given: The response probabilities, $C (c) = 1$ , $C (w_{1}) = C (w_{2}) = 0$ , $A (c) = A (w_{1}) = 1$ , $A (w_{2}) = 0$ , an integer $M \geq 1$ with $M$ independent replicas from this response law when computing empirical pass@1, two independent candidates, and rejection search with budget $N = 2$ on those candidates.

[F1]

Finite truncated response spaces, correctness predicates, and verifiers: correctness and acceptance are separate binary functions on the finite response set.

[F2]

Budgeted verifier-guided generation and best-of-N sampling Budgeted verifier generation terminates with bounded response and selection calls: rejection search returns the first verifier-accepted response, or failure if all $N$ draws are rejected.

[F4]

Finite sums and finite products, by recursion: the empirical pass@1 rate is the finite average of the single-sample correctness indicators, with sums defined recursively. Field supplies distributivity for finite expansions over the response list.

[F5]

Expectation of a real random variable on a finite probability space: expectation on the finite sample space of the $M$ replicas is the probability-weighted sum over its outcomes.

[F6]

Linearity, monotonicity, and the modulus bound for expectation: expectation is linear for integrable random variables on a common probability space.

[F7]

Finite truncated response spaces, correctness predicates, and verifiers: the response law is a normalized mass function on the finite response set.

[F8]

Finite truncated response spaces, correctness predicates, and verifiers: independent responses from that law have the product joint mass.

Verification

Given: The objects and sampling rule in the Example.

1.1givenF1F4F5F6F7F8algebrainduction

A single sample is correct exactly when it is $c$ , so its pass@1 probability is $1 / 4$ . The $M$ replicas have joint mass $\prod_{m = 1}^{M} π (y_{m})$ on the finite product $Y_{q}^{M}$ by [F8]. For one coordinate the total mass is $\sum_{y} π (y) = 1$ by [F7]. When one more coordinate is added, distributivity in [F4] factors its finite sum as another factor $\sum_{y} π (y) = 1$ ; induction therefore shows the total mass on $Y_{q}^{M}$ is $1$ . If one coordinate is fixed at $y$ , summing each of the other $M - 1$ coordinates contributes a factor $1$ , so that coordinate has marginal mass $π (y)$ . Thus this is a finite probability space and each correctness indicator has expectation $1 / 4$ . The indicators and their average are bounded and integrable; by [F5]–[F6] the average's expectation is the average of their expectations, also $1 / 4$ .

1.2givenF1F4F8algebra

Neither candidate is correct exactly when both lie in ${w_{1}, w_{2}}$ . Summing the four pair masses from [F8] and factoring by distributivity in [F4] gives probability $(3 / 8 + 3 / 8)^{2} = 9 / 16$ . The complementary oracle pass@2 event therefore has probability $1 - 9 / 16 = 7 / 16$ .

1.3givenF1F2F8algebra

A response is accepted with probability $1 / 4 + 3 / 8 = 5 / 8$ . By the product law for independent candidates [F8], rejection search fails only if both candidates are $w_{2}$ , which has probability $(3 / 8)^{2} = 9 / 64$ , so the probability it returns an accepted response is $55 / 64$ .

2.1givenF1F2F8step 1.3algebra∎

By [F8], it returns a correct response if the first draw is $c$ , or if the first is rejected $w_{2}$ and the second is $c$ , an event of probability $1 / 4 + (3 / 8) (1 / 4) = 11 / 32 = 22 / 64$ . Conditional on an accepted output, the selected accuracy is therefore $(22 / 64) / (55 / 64) = 2 / 5$ , distinct from the pass@1 and oracle pass@2 values above.