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Monotone risk calibration and finite-sample deployment guarantees — Examples

1 · Prerequisites

2 · Logical flowchart

3 · Definitions, theorems and proofs

None yet.

4 · Examples, counterexamples and false statements

ExampleConstruction: AI-generatedVerification: AI-generatedjudge pass (gpt-6-luna)Open item page →

The all-zero loss boundary

Example

Let $n, M \geq 1$ , $B > 0$ , $α \in [0, B)$ , and let every calibration and test loss be zero at every grid point. The rule passes at the first grid point exactly when $α \geq B / (n + 1)$ ; when $α < B / (n + 1)$ it returns the maximum-grid fallback. In either case the selected test loss is zero and its expected risk is zero.

Example references

Finite nested decision families and bounded risk: the bounded finite-grid risk setup.

Finite-grid conformal risk control: the B-corrected pass test and fallback.

Angelopoulos et al., Conformal Risk Control, §1.1

Facts & Assumptions

Given: The finite grid, the all-zero calibration and test losses, $B > 0$ , and $α \in [0, B)$ .

[F1]

The risk family uses a finite deterministic grid and bounded measurable loss rows; expected test risk is the expectation of the selected test loss. Finite nested decision families and bounded risk

[F2]

The algorithm selects the first grid point whose B-corrected calibration average is at most $α$ , and falls back to the maximum point if none passes. Finite-grid conformal risk control

Verification

1.1givenF2algebra

At every grid point, the calibration threshold is $(0 + B) / (n + 1) = B / (n + 1)$ . Hence, if $α \geq B / (n + 1)$ the first point passes, while if $α < B / (n + 1)$ every point fails and the algorithm returns the fallback.

2.1givenF1step 1.1algebra∎

The test loss is zero at every grid point, including the first point and the fallback. Its expectation is therefore $0$ , which is at most $α$ because $α \geq 0$ .

ExampleConstruction: AI-generatedVerification: AI-generatedjudge pass (gpt-6-luna)Open item page →

Fallback when the calibration correction exceeds the target

Example

Let $n, M \geq 1$ , $B > 0$ , and $0 \leq α < B / (n + 1)$ . Consider any loss-nested finite-grid family with calibration losses in $[0, B]$ and endpoint losses $L_{i} (λ_{M}) \leq α$ almost surely. The B-corrected calibration threshold is too large to pass at any grid point, so the rule falls back to $λ_{M}$ . The endpoint condition then bounds the realized test loss, and the expected test risk is at most $α$ .

Example references

Finite nested decision families and bounded risk: nonnegative bounded losses and the safe maximum-grid endpoint.

Finite-grid conformal risk control: the pass threshold includes the B correction and otherwise returns the maximum grid point.

Angelopoulos et al., Conformal Risk Control, §1.1

Facts & Assumptions

Given: A finite grid, $n \geq 1$ , $B > 0$ , $0 \leq α < B / (n + 1)$ , calibration losses in $[0, B]$ , and endpoint test loss at most $α$ almost surely.

[F1]

Losses are nonnegative and endpoint-safe, with $L_{n + 1} (λ_{M}) \leq α$ almost surely. Finite nested decision families and bounded risk

[F2]

At grid point $λ_{j}$ , the algorithm tests $(\sum_{i = 1}^{n} L_{i} (λ_{j}) + B) / (n + 1)$ and uses $λ_{M}$ when no point passes. Finite-grid conformal risk control

[F3]

Expectation preserves almost-sure inequalities for integrable real random variables. Linearity, monotonicity, and the modulus bound for expectation

Verification

1.1givenF2algebra

Since every calibration loss is nonnegative, for each $j$ the algorithm's threshold satisfies $(\sum_{i = 1}^{n} L_{i} (λ_{j}) + B) / (n + 1) \geq B / (n + 1) > α$ . No point passes, so the rule returns $λ_{M}$ .

2.1givenF1F3step 1.1∎

The endpoint condition gives $0 \leq L_{n + 1} (λ_{M}) \leq α$ almost surely. The loss is bounded and integrable, so [F3] implies $E [L_{n + 1} (λ_{M})] \leq α$ .

CounterexampleConstruction: AI-generatedVerification: AI-generatedjudge pass (gpt-6-luna)Open item page →

A nonmonotone loss can defeat first-pass calibration

Statement refuted

For every deterministic grid $Λ = {λ_{1} < \dots < λ_{M}}$ with $M \geq 1$ , calibration size $n \geq 1$ , bound $B > 0$ , and target $α \in [0, B)$ , suppose a probability space carries $n + 1$ exchangeable random elements $W_{i} = (L_{i} (λ_{j}))_{j = 1}^{M}$ of the Borel space $[0, B]^{M}$ , with $L_{i} (λ_{M}) \leq α$ almost surely for every $i$ . The B-corrected first-passing rule then has expected test loss $E [L_{n + 1} (\hat{λ})] \leq α$ , even when the losses are not nonincreasing along the grid.

Statement references

Finite-grid conformal risk control: the first-passing rule and maximum-grid fallback.

Exchangeable random elements: joint-law invariance under all row permutations.

Angelopoulos et al., Conformal Risk Control, §2.4

Facts & Assumptions

Given: The refuted assertion and the following finite model: $n = 1$ , $M = 2$ , $B = 1$ , $α = 3 / 4$ , and $λ_{1} < λ_{2}$ . Let $Z_{1}, Z_{2}$ be independent with $P (Z_{i} = a) = 1 / 10$ and $P (Z_{i} = b) = 9 / 10$ . Define the losses by $L_{i} (λ_{1}) = 0$ and $L_{i} (λ_{2}) = 3 / 4$ when $Z_{i} = a$ , and by $L_{i} (λ_{1}) = 1$ and $L_{i} (λ_{2}) = 3 / 4$ when $Z_{i} = b$ .

[F1]

Exchangeability means invariance of the joint law under every coordinate permutation. Exchangeable random elements

[F2]

The algorithm tests $(L_{1} (λ_{j}) + B) / (n + 1)$ at each grid point in increasing order and returns the maximum point if no test passes. Finite-grid conformal risk control

[A1]

The two rows are independent draws from the same two-point law specified above, so their joint law is exchangeable.

[A2]

The loss table has values in $[0, 1]$ and endpoint losses $L_{i} (λ_{2}) = 3 / 4 = α$ ; it is nonmonotone on type $a$ because $0 < 3 / 4$ as the grid index increases.

Counterexample

1.1A1A2F1

By [A1] and [F1], the two loss rows are exchangeable. By [A2], their values are bounded by $B = 1$ , the endpoint is safe, and the family violates the required nonincreasing-loss condition on an event of probability $1 / 10$ .

1.2A2F2algebra

If the calibration row is type $a$ , the first threshold is $(0 + 1) / 2 = 1 / 2 \leq 3 / 4$ , so the algorithm selects $λ_{1}$ . If it is type $b$ , the first threshold is $(1 + 1) / 2 = 1 > 3 / 4$ and the second is $(3 / 4 + 1) / 2 = 7 / 8 > 3 / 4$ , so the algorithm falls back to $λ_{2}$ .

2.1A1A2step 1.2algebra

Conditional on a type- $a$ calibration row, independence makes the expected test loss at $λ_{1}$ equal to $(1 / 10) 0 + (9 / 10) 1 = 9 / 10$ . Conditional on a type- $b$ calibration row, the selected endpoint loss is $3 / 4$ for either test type.

3.1A1A2step 2.1algebra∎

Averaging over the calibration row gives $E [L_{2} (\hat{λ})] = (1 / 10) (9 / 10) + (9 / 10) (3 / 4) = 153 / 200 = 3 / 4 + 3 / 200 > 3 / 4 = α$ . Thus the asserted guarantee fails despite exchangeability, boundedness, and a safe endpoint.

ExampleConstruction: AI-generatedVerification: AI-generatedjudge pass (gpt-6-luna)Open item page →

Multilabel false-negative risk control

Example

Let $K, n \geq 1$ , let $(X, A)$ be a measurable space, give the finite set $2^{[K]}$ its full sigma-algebra, and let $0 \leq λ_{1} < \dots < λ_{M} = 1$ be a deterministic grid in $[0, 1]$ . Fix a measurable score map $f : X \to [0, 1]^{K}$ . Let $(X_{i}, Y_{i})$ , $1 \leq i \leq n + 1$ , be exchangeable random elements of the product measurable space $X \times 2^{[K]}$ , where $Y_{i} \subseteq [K]$ . Define $C_{λ} (x) = {k \in [K] : f_{k} (x) \geq 1 - λ}$ . For nonempty $Y_{i}$ , set $L_{i} (λ) = 1 - ∣ Y_{i} \cap C_{λ} (X_{i}) ∣ / ∣ Y_{i} ∣$ ; for $Y_{i} = \emptyset$ , define $L_{i} (λ) = 0$ . With $B = 1$ and any target $α \in [0, 1)$ , the finite-grid rule selects a threshold whose expected test false-negative loss is at most $α$ .

Example references

Finite nested decision families and bounded risk: exchangeable finite loss rows and endpoint-safe bounded risk.

Finite-grid conformal risk control: the finite-grid calibration rule.

Finite-sample expected-risk control on a finite grid: expected-risk control under loss nesting and endpoint safety.

Angelopoulos et al., Conformal Risk Control, §3.2

Facts & Assumptions

Given: The finite class set, fixed measurable score map, exchangeable example rows, nested threshold sets, explicit empty-label convention, and target $α \in [0, 1)$ .

[F1]

A finite family of bounded loss rows is loss-nested when each row is nonincreasing along the ordered grid, and endpoint-safe when every terminal loss is at most the target. Finite nested decision families and bounded risk

[F3]

For exchangeable, bounded, loss-nested rows with a safe endpoint, the finite-grid rule has expected test loss at most $α$ . Finite-sample expected-risk control on a finite grid

Verification

1.1givenF1construct

For each $k$ and grid point $λ_{j}$ , the event $f_{k} (X_{i}) \geq 1 - λ_{j}$ is measurable; finite intersections, unions, and sums in the displayed loss formula are therefore measurable. The same map is applied to every exchangeable example row, so the resulting loss vectors are exchangeable.

1.2givenalgebra

If $λ \leq λ^{'}$ , then $1 - λ^{'} \leq 1 - λ$ , so $C_{λ} (x) \subseteq C_{λ^{'}} (x)$ . For nonempty $Y_{i}$ , the number of included true labels can only increase, making $L_{i} (λ^{'}) \leq L_{i} (λ)$ ; for empty $Y_{i}$ the loss is constantly zero. The losses lie in $[0, 1]$ .

2.1givenstep 1.2

At $λ_{M} = 1$ , the threshold is zero and every score lies in $[0, 1]$ , so $C_{1} (x) = [K]$ . Hence every nonempty-label loss is zero, and the stated empty-label convention also gives zero. Thus $L_{i} (λ_{M}) = 0 \leq α$ .

3.1givenF3step 1.1step 1.2step 2.1∎

The finite loss rows are exchangeable, bounded by $B = 1$ , nonincreasing, and endpoint-safe by steps 1.1–2.1. Applying [F3] gives $E [L_{n + 1} (\hat{λ})] \leq α$ .

Monotone risk calibration and finite-sample deployment guarantees — Examples

2 · Logical flowchart

3 · Definitions, theorems and proofs

4 · Examples, counterexamples and false statements

The all-zero loss boundary

Example

Example references

Facts & Assumptions

Verification

Fallback when the calibration correction exceeds the target

Example

Example references

Facts & Assumptions

Verification

A nonmonotone loss can defeat first-pass calibration

Statement refuted

Statement references

Facts & Assumptions

Counterexample

Multilabel false-negative risk control

Example

Example references

Facts & Assumptions

Verification

Sources