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How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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

✓ 3 results · all verified · 2 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 1 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

AdaBoost and gradient boosting

1 · Prerequisites

2 · Logical flowchart

3 · Definitions, theorems and proofs

AlgorithmAlgorithm: Literature-sourcedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

Finite-sample AdaBoost.M1 with half-log coefficients

Definition

Fix a finite training sample (xi,yi)i=1N with N≥1, xi∈X, and yi∈{−1,+1}, a finite round limit M≥0, and a weak-learning routine which at each requested round returns a classifier hm:X→{−1,+1} for the current observation weights. Initialize wi(1)=1/N and F0(x)=0. At round m, compute the weighted error em=∑i=1Nwi(m)1{hm(xi)≠yi}. If em=0, halt and return hm itself as a perfect classifier with zero training error; do not form an infinite coefficient. If em≥1/2, halt and report that the weak-learning condition failed. For 0<em<1/2, set

am=12log⁡1−emem,Zm=2em(1−em),

update wi(m+1)=wi(m)exp⁡(−amyihm(xi))/Zm, and set Fm(x)=Fm−1(x)+amhm(x). After M completed weak rounds, return the classifier y^M(x)=sign⁡+(FM(x)), where sign⁡+(u)=+1 for u≥0 and −1 for u<0. Thus ties are assigned to +1. The coefficient convention is the half-log convention; multiplying every coefficient by the same positive factor does not change the sign classifier. The definition makes no claim about test error or generalization.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

Negative-gradient pseudo-residuals for a finite empirical loss

Definition

Let N≥1, let (xi,yi)i=1N be a finite sample with xi∈X and yi∈Y, let f:X→R be a score function, and let ℓ:Y×R→R be finite-valued. Assume that for each i, the one-variable function t↦ℓ(yi,t) is differentiable at zi=f(xi). For a score vector z=(z1,…,zN), define the empirical score loss Φ(z)=∑i=1Nℓ(yi,zi). The pseudo-residual at observation i for f is

ri(f)=−ddtℓ(yi,t)∣t=f(xi).

The sample-score gradient, when its coordinate partial derivatives exist, is the vector ∇Φ(z)=(∂z1Φ(z),…,∂zNΦ(z)). The sample pseudo-residual vector is (r1(f),…,rN(f)); the word “pseudo” distinguishes these score derivatives from derivatives with respect to parameters of a restricted model.

LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passOpen item page →

Pseudo-residuals are the negative gradient in sample-score space

Statement

Let N≥1, let (xi,yi)i=1N be a finite sample, let f:X→R, and let U⊆R be open with zi=f(xi)∈U for every i. Suppose each scalar loss t↦ℓ(yi,t) is defined on U and differentiable at zi. Write ∂2ℓ(yi,zi)=ddtℓ(yi,t)∣t=zi and define Φ:UN→R by Φ(w)=∑i=1Nℓ(yi,wi). Then Φ is totally differentiable at z, its derivative in a score increment v=(v1,…,vN) is

DΦ(z)v=∑i=1N∂2ℓ(yi,zi)vi,

and its Euclidean gradient is ∇Φ(z)=(∂2ℓ(yi,zi))i=1N. For these possibly local losses, define the pseudo-residual by ri(f)=−∂2ℓ(yi,zi); this agrees with the usual definition when the loss is finite-valued on all of Y×R. Then the pseudo-residual vector associated with f on this sample is exactly −∇Φ(z) in sample-score space.

Facts & Assumptions

Given: Sample scores zi=f(xi) in an open scalar domain U and scalar losses differentiable at those scores.

[F1]

Negative-gradient pseudo-residuals for a finite empirical loss: For a finite-valued loss on Y×R whose sampled sections are differentiable at the scores, the empirical score loss is Φ(z)=∑iℓ(yi,zi), and the pseudo-residual is ri=−∂2ℓ(yi,zi).

[F2]

The total (Fréchet) derivative Df(a) as the linear first-order approximation with o(∥h∥2) remainder: Total differentiability at a point means a linear first-order approximation whose remainder divided by the increment norm tends to zero.

[F3]

The chain rule for total derivatives: D(g∘f)(a)=Dg(f(a))∘Df(a): If f is totally differentiable at a and g at f(a), then g∘f is totally differentiable and D(g∘f)(a)=Dg(f(a))∘Df(a).

[F4]

Sums and scalar multiples of totally differentiable maps are totally differentiable with the expected derivatives: Finite sums of totally differentiable maps are totally differentiable and their derivatives add.

[F5]

Directional derivatives and partial derivatives of a map U⊆Rm→Rn: A coordinate partial derivative is the directional derivative along the corresponding standard basis direction.

[F6]

A total derivative computes every directional derivative, and its matrix is the Jacobian: When a total derivative exists, its value on each coordinate direction is the corresponding partial derivative.

[F7]

The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case: For a scalar-valued function, the Euclidean gradient is the vector of coordinate partial derivatives in standard coordinate order.

Proof

1.1givenF2construct

For each i, let πi(z)=zi and gi(t)=ℓ(yi,t). The coordinate projection has derivative Dπi(z)v=vi because its linear approximation has zero remainder. Differentiability of gi at zi gives gi(zi+u)=gi(zi)+∂2ℓ(yi,zi)u+o(∣u∣), so Dgi(zi)u=∂2ℓ(yi,zi)u.

2.1F3step 1.1

By the chain rule, ψi=gi∘πi is totally differentiable and Dψi(z)v=∂2ℓ(yi,zi)vi.

3.1F4step 2.1induction

Since Φ=∑i=1Nψi, the finite-sum rule gives DΦ(z)v=∑i=1N∂2ℓ(yi,zi)vi.

4.1givenF1F5F6F7step 3.1construct∎

Evaluating step 3.1 on the coordinate direction that changes only zi gives ∂ziΦ(z)=∂2ℓ(yi,zi) by [F5] and [F6], so [F7] gives the asserted gradient. Extend the sampled loss sections by zero outside U to obtain a finite-valued loss on {y1,…,yN}×R. Since U is open, this extension agrees with each original section near zi and has the same derivative there. Thus [F1] applied to the extension yields exactly the locally defined ri(f), independently of the values outside U. Consequently (ri(f))i=1N=−∇Φ(z), also agreeing with [F1] whenever the original loss is globally defined.

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

AdaBoost's exponential training-loss and nonpositive-margin bound

Statement

Let M≥0 be a finite sequence of completed rounds of finite-sample AdaBoost.M1, with weighted errors 0<em<1/2, scores FM(x)=∑m=1Mamhm(x), and Zm=2em(1−em). Then

∑i=1Ne−yiFM(xi)=N∏m=1MZm

and

1N∑i=1N1{yiFM(xi)≤0}≤∏m=1MZm.

The empty product for M=0 is 1. Under the algorithm's +1 tie rule, the empirical classification error is also at most this product. This is a finite training-margin bound only and makes no test-generalization claim.

Facts & Assumptions

Given: A fixed finite binary-labeled sample and a finite sequence of completed rounds with 0<em<1/2.

[F1]

Finite-sample AdaBoost.M1 with half-log coefficients: The initialization is wi(1)=1/N and F0=0.

[F2]

Finite-sample AdaBoost.M1 with half-log coefficients: The weighted error is em=∑i=1Nwi(m)1{hm(xi)≠yi}.

[F3]

Finite-sample AdaBoost.M1 with half-log coefficients: For 0<em<1/2, am=12log⁡1−emem and Zm=2em(1−em).

[F4]

Finite-sample AdaBoost.M1 with half-log coefficients: The update is wi(m+1)=wi(m)exp⁡(−amyihm(xi))/Zm and Fm(x)=Fm−1(x)+amhm(x).

[F5]

Finite-sample AdaBoost.M1 with half-log coefficients: The classifier uses sign⁡+(u)=+1 for u≥0 and −1 for u<0, so ties are assigned to +1.

Proof

1.1givenF1F2F3F4algebrainduction

The initial weights sum to 1 by [F1]. Inductively assume the weights at round m sum to 1; by [F2], the total weights on correct and incorrect examples are 1−em and em. For a temporary coefficient a≥0, the exponential reweighting sum is qm(a)=(1−em)e−a+emea. Its second derivative is positive, and qm′(a)=emea−(1−em)e−a vanishes uniquely at a=12log⁡((1−em)/em)>0, which equals the algorithm's coefficient by [F3]. Substitution gives qm(am)=2em(1−em)=Zm. By [F4], the next weights divide the reweighted terms by this exact sum, so they also sum to 1. Induction proves normalization at every round and the stated minimization formula.

2.1F1F4step 1.1inductionalgebra

Put Sm=∑ie−yiFm(xi). By [F1], S0=N and wi(1)=e−yiF0(xi)/S0. If wi(m)=e−yiFm−1(xi)/Sm−1, then [F4] gives wi(m+1)=e−yiFm(xi)/(Sm−1Zm). Step 1.1 establishes that the weights normalize, so the update denominator is the sum of the reweighted terms and Sm=Sm−1Zm. Induction yields SM=N∏m=1MZm.

3.1F5step 2.1algebra∎

If yiFM(xi)≤0, then e−yiFM(xi)≥1; if the margin is positive, its indicator is zero. Hence each nonpositive-margin indicator is at most its exponential term, and summing and applying step 2.1 proves the stated bound. By [F5], a misclassified positive-label example has negative margin and a misclassified negative-label example has nonpositive margin, so training error is no larger than the bounded nonpositive-margin fraction.

AlgorithmAlgorithm: Literature-sourcedProof: Not applicableOpen item page →

Exact-line-search gradient tree boosting on a finite sample

Definition

Fix a finite regression sample (xi,yi)i=1N with N≥1, xi∈Rp for a finite p≥1, and yi∈Y. Let ℓ:Y×R→R be a finite-valued per-observation loss, differentiable in its second argument at every score used. Choose a finite round count M≥0 and finite integer tree-depth cap dmax⁡≥0. The algorithm is admissible only when the following required minima are attained: an initial constant minimizer and, at every round, a line minimizer in each terminal leaf. For any admitted run, choose any one such minimizer at each finite stage.

Initialize f0(x)=c0 with c0∈arg⁡min⁡c∈R∑i=1Nℓ(yi,c). For each round m=1,…,M, perform the following steps.

  1. For each observation compute the negative score derivative rim=−∂2ℓ(yi,fm−1(xi)).

  2. Run finite-depth CART regression-tree growth, using squared error and depth cap dmax⁡, on the responses rim at features xi. Let its nonempty terminal feature regions be R1m,…,RJmm; their training-index sets partition {1,…,N}. This tree fit approximates the unrestricted negative score gradient. Its fitted leaf means are used to determine the regression-tree fit and regions, but are not the boosting update offsets.

  3. For each leaf region choose an attained exact line minimizer γjm∈arg⁡min⁡γ∈R∑i:xi∈Rjmℓ(yi,fm−1(xi)+γ). In particular, zero is an admissible comparison offset.

  4. Update fm(x)=fm−1(x)+∑j=1Jmγjm1{x∈Rjm}.

The output is fM. Because the run has finitely many rounds and finitely many leaves, choosing one attained minimizer at each stage requires no axiom of choice. The defined variant uses the full training sample and exact leafwise line minimization; shrinkage and subsampling are separate variants. This definition asserts no convergence or generalization result.

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

Gradient tree boosting does not increase empirical loss

Statement

For any finite run of the defined gradient tree boosting algorithm for which the initial constant and every leaf line minimizer are attained, let Lm=∑i=1Nℓ(yi,fm(xi)). At every completed round m, Lm≤Lm−1, and hence LM≤L0. This is a finite-step empirical-loss statement only; it asserts no convergence or generalization result.

Facts & Assumptions

Given: One completed finite boosting round, the finite training sample, and the attained leafwise minimizers.

[F1]

Exact-line-search gradient tree boosting on a finite sample: The tree's terminal regions determine the update, and each γjm attains the minimum of the corresponding leaf loss over all real offsets.

[F2]

Empirical axis-aligned binary decision trees: A finite tree has nonempty terminal sample sets that partition the training indices, and each training observation follows its unique leaf path.

[F3]

Greedy CART minimizes the listed split loss and terminates: With finite depth, the CART procedure returns a finite tree with nonempty children at every split.

Proof

1.1givenF2F3

Fix a completed round m. By [F2] and [F3], its leaf index sets Ijm={i:xi∈Rjm} are finitely many nonempty sets whose disjoint union is {1,…,N}.

2.1F1step 1.1given

For every leaf j, its attained minimizer γjm has loss no greater than the admissible offset 0, so ∑i∈Ijmℓ(yi,fm−1(xi)+γjm)≤∑i∈Ijmℓ(yi,fm−1(xi)).

3.1F1F2step 1.1step 2.1algebra∎

By the update rule, each training score fm(xi) uses the offset for its unique leaf. Summing the inequalities in step 2.1 over the partition in step 1.1 gives Lm=∑j∑i∈Ijmℓ(yi,fm−1(xi)+γjm)≤∑j∑i∈Ijmℓ(yi,fm−1(xi))=Lm−1. Applying this to each of the finitely many rounds yields LM≤L0.

4 · Examples, counterexamples and false statements

None yet.

Sources