LemmaStatement: AI-adaptedProof: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)

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Jensen's inequality for pairs of simplex-valued forecasts

Statement

For an integer $K \geq 2$ , let $Δ_{K} = {p \in [0, 1]^{K} : \sum_{j = 1}^{K} p_{j} = 1} \subset R^{K}$ with its subspace Borel sigma-algebra, and let $C = Δ_{K} \times Δ_{K} \subset R^{2 K}$ with its subspace Borel sigma-algebra. Let $(Ω, F, P)$ be a probability space, let $Z = (U, V) : Ω \to C$ be measurable, and let $ψ : C \to R$ be continuous and jointly convex, meaning $ψ (λ z + (1 - λ) w) \leq λ ψ (z) + (1 - λ) ψ (w)$ for all $z, w \in C$ and $λ \in [0, 1]$ . Define $E Z$ coordinatewise. Then $E Z = (E U, E V) \in C$ and $ψ (E U, E V) \leq E [ψ (U, V)] .$

Statement references

Durrett, Theorem 1.6.2: scalar Jensen inequality for an integrable random variable.

Vaicenavicius et al., supplementary Theorem 2: applies convexity to the pair of conditional mean forecasts in each fixed bin.

Facts & Assumptions

Given: A probability space, a measurable $C$ -valued random vector $Z = (U, V)$ , and a continuous jointly convex function $ψ : C \to R$ .

[F1]

$C$ is convex, closed, and bounded in finite-dimensional Euclidean space; it is therefore compact by Heine–Borel. Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line

[F2]

A continuous map from a compact metric space to a metric space is uniformly continuous. Heine-Cantor: a continuous map from a compact metric space to any metric space is uniformly continuous

[F3]

Probability is total mass one, expectation is integration, and the integral of a nonnegative simple function is the finite weighted sum of its values. Probability measures and probability spaces Expectation of a nonnegative or integrable random variable The integral of a nonnegative simple function

[F4]

A measurable sequence converging almost everywhere under one integrable majorant has convergent integrals. Dominated convergence

[F5]

A measurable map is one whose preimages of measurable sets are measurable. A measurable function between measurable spaces

Proof

1.1F1F3givenalgebra

If $Z$ has finite range $z_{1}, \dots, z_{m} \in C$ with probabilities $p_{1}, \dots, p_{m}$ , then each coordinate of $Z$ is a nonnegative simple function, so [F3] gives $E Z = \sum_{i} p_{i} z_{i}$ , a convex combination in $C$ . The real finite-range random variable $ψ (Z)$ is integrable. Its positive and negative parts are nonnegative simple functions, so applying [F3] to each and subtracting their finite expectations gives $E [ψ (Z)] = \sum_{i} p_{i} ψ (z_{i})$ . Joint convexity now proves $ψ (E Z) \leq E [ψ (Z)]$ by induction on the number of positive weights: combine the first $m - 1$ points into their normalized convex average, apply the induction hypothesis, then apply two-point convexity with weights $1 - p_{m}$ and $p_{m}$ .

1.2F1F5construct

For $n \geq 1$ set $D_{n} = {(m_{1} / n, \dots, m_{K} / n) : m_{i} \in N_{0}, \sum_{i} m_{i} = n} \subseteq Δ_{K}$ . This is a finite set with a canonical lexicographic order. For $p \in Δ_{K}$ , put $m_{i} = ⌊ n p_{i} ⌋$ for $i < K$ and $m_{K} = n - \sum_{i < K} m_{i}$ . Then $m_{K} \geq 0$ , the coordinates sum to $n$ , and the first $K - 1$ coordinate errors are below $1 / n$ while the last is below $(K - 1) / n$ . Thus some $q \in D_{n}$ has $∥ p - q ∥_{2} < \sqrt{K (K - 1)} / n$ . The product grid $D_{n} \times D_{n}$ therefore approximates every point of $C$ within $2 K / n$ . List its finitely many points canonically as $c_{n, 1}, \dots, c_{n, m_{n}}$ . Their relative open balls of radius $2 K / n$ cover $C$ , and the cells obtained by removing all earlier balls from each ball are Borel in the stated subspace sigma-algebra. Assign each point to its first covering ball; replacing $Z$ on each cell by its center defines a measurable finite-range $Z_{n} \in C$ with $∥ Z_{n} - Z ∥_{2} < 2 K / n$ everywhere. The grid, order and assignment are explicit, so no choice axiom is used.

2.1F1F3F4step 1.1

Step 1.1 gives $E Z_{n} \in C$ for every $n$ . Each coordinate of $Z_{n}$ and $Z$ is bounded by $1$ , so dominated convergence [F4] gives $E Z_{n} \to E Z$ coordinatewise; because the coordinate constraints defining $C$ are closed, $E Z \in C$ .

2.2F1F2F4step 1.2

By [F2], $ψ$ is uniformly continuous on $C$ . Choose $δ > 0$ so that points within distance $δ$ have $ψ$ -values within $1$ , and then choose $n$ with $2 K / n < δ$ . The finite grid in step 1.2 is a $δ$ -net, so $∣ ψ (z) ∣ \leq 1 + \max_{c \in D_{n} \times D_{n}} ∣ ψ (c) ∣$ for every $z \in C$ . This is a common finite bound. Since $∥ Z_{n} - Z ∥_{2} \to 0$ , continuity gives $ψ (Z_{n}) \to ψ (Z)$ pointwise; [F4] then gives $E ψ (Z_{n}) \to E ψ (Z)$ .

3.1F1step 1.1step 2.1step 2.2∎

Applying step 1.1 to each $Z_{n}$ gives $ψ (E Z_{n}) \leq E ψ (Z_{n})$ . By continuity of $ψ$ and steps 2.1 and 2.2, taking limits proves $ψ (E Z) \leq E ψ (Z)$ .

Depends on

Used by

Fixed-bin calibration limits underestimate for jointly convex discrepancies; refining bins can be consistent Theorem

Dependency tree · two levels

53 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

Rick Durrett, Probability: Theory and Examples, 5th ed., Theorem 1.6.2 (standard reference, not scraped)
Vaicenavicius et al., Evaluating Model Calibration in Classification, supplementary material, Theorem 2 and proof (standard reference, not scraped)