ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck pass

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.

A one-leaf fit can project a nonzero gradient to zero

Example

Take $x_{1} = 0$ , $x_{2} = 1$ , labels $y_{1} = + 1$ , $y_{2} = - 1$ , squared loss $ℓ (y, z) = \frac{1}{2} (z - y)^{2}$ , initial scores $f_{0} (x_{1}) = f_{0} (x_{2}) = 0$ , and maximum tree depth $0$ . The pseudo-residual vector is $(1, - 1)$ , but its one-leaf least-squares tree fit predicts $0$ on both observations. Let $Φ (z_{1}, z_{2}) = \frac{1}{2} (z_{1} - 1)^{2} + \frac{1}{2} (z_{2} + 1)^{2}$ . A vector $u$ is a strict descent direction at $0$ when the derivative of $t \mapsto Φ (t u)$ at $t = 0$ is negative. The fitted vector $u = (0, 0)$ has derivative $0$ , although the unrestricted score gradient is $(- 1, + 1) \neq 0$ . The exact leaf loss minimizer is also $0$ , so the empirical loss stays at $1$ .

Facts & Assumptions

Given: The two-observation sample, squared loss, zero initial scores, and depth cap $0$ stated in the Example.

[F1]

Negative-gradient pseudo-residuals for a finite empirical loss: The pseudo-residual at observation $i$ is the negative second-argument derivative of its loss at the current score.

[F2]

Pseudo-residuals are the negative gradient in sample-score space: The pseudo-residual vector is the negative Euclidean gradient of the empirical score loss.

[F3]

Empirical axis-aligned binary decision trees: The root sample set contains the training indices, and a regression tree's terminal leaf sets partition those indices.

[F4]

Greedy axis-aligned CART growth on a finite sample: At a node whose depth equals the maximum depth, the procedure leaves the node unsplit; regression leaf predictions are sample means.

[F5]

Exact-line-search gradient tree boosting on a finite sample: Each leaf update uses an attained exact minimizer of the empirical loss over its real offset.

[F6]

Directional derivatives and partial derivatives of a map $U \subseteq R^{m} \to R^{n}$ : The directional derivative in direction $u$ is the derivative of the line map $t \mapsto Φ (0 + t u)$ at $t = 0$ .

Verification

1.1givenF1F2algebra

For any constant $c$ , the initial empirical loss is $\frac{1}{2} (c - 1)^{2} + \frac{1}{2} (c + 1)^{2} = c^{2} + 1$ , so $c_{0} = 0$ is an attained minimizer. At these scores, [F1] gives residuals $r_{1} = 1$ and $r_{2} = - 1$ , and [F2] gives $\nabla Φ (0) = (- 1, + 1) \neq 0$ .

2.1F3F4step 1.1algebra

With depth cap $0$ , [F4] leaves the root as a single terminal leaf; by [F3] it contains both observations. Its regression prediction is their residual mean $(1 + (- 1)) / 2 = 0$ , so the fitted direction on the sample is $u = (0, 0)$ .

3.1F2F6step 2.1

Since $t u = (0, 0)$ for every real $t$ , the line map in [F6] is constant and its derivative is $0$ . Thus $u$ is not a strict descent direction, even though the unrestricted gradient in step 1.1 is nonzero.

4.1F5step 1.1step 2.1algebra∎

The sole leaf's loss at offset $γ$ is $\frac{1}{2} (γ - 1)^{2} + \frac{1}{2} (γ + 1)^{2} = γ^{2} + 1$ , whose attained minimum is $1$ at $γ = 0$ . By [F5], the algorithm updates neither training score, so the loss remains $1$ .

Depends on

Used by

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