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✓ 3 results · all verified · 3 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; all 3 also cleared it.

CART trees and pruning

1 · Prerequisites

2 · Logical flowchart

3 · Definitions, theorems and proofs

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

Empirical axis-aligned binary decision trees

Definition

Fix a finite labeled sample (xi,yi)i=1n with n≥1, xi∈Rp and p≥1. In regression, yi∈R. In classification, the labels lie in a fixed ordered finite set C={c1<⋯<cK} with K≥1. A finite empirical axis-aligned binary decision tree has root sample-index set Iroot={1,…,n}. Each internal node v stores a coordinate j∈{1,…,p} and threshold t∈R; its left and right child sample sets are IvL={i∈Iv:xij≤t} and IvR={i∈Iv:xij>t}, and both are required to be nonempty. A leaf is a terminal node. The leaf sample sets partition {1,…,n}, and each node's feature region is the intersection of its path inequalities. Every input feature vector follows the unique path determined by the stored comparisons.

For a regression leaf with sample set I, put NI=∣I∣, yˉI=NI−1∑i∈Iyi, and QI=NI−1∑i∈I(yi−yˉI)2; its prediction is yˉI and its total squared-error loss is NIQI. For a classification leaf, put pIk=NI−1∣{i∈I:yi=ck}∣ and predict the class ck with largest pIk, choosing the least index in a tie. Its optional impurities are misclassification QImis=1−max⁡kpIk, Gini QIGini=1−∑k=1KpIk2, and entropy QIent=−∑k=1KpIklog⁡pIk, where log⁡ is the natural logarithm and 0log⁡0=0. A chosen classification criterion determines QI and the corresponding total leaf impurity is NIQI.

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

The sample mean uniquely minimizes squared error on a nonempty leaf

Statement

Let I be a nonempty finite set of sample indices and let yi∈R for every i∈I. Put N=∣I∣ and yˉI=N−1∑i∈Iyi. Among all constants c∈R, the unique minimizer of ∑i∈I(yi−c)2 is c=yˉI, and the minimum value is ∑i∈I(yi−yˉI)2.

Facts & Assumptions

Given: A nonempty finite sample-index set I and real responses (yi)i∈I.

[F1]

Empirical axis-aligned binary decision trees: A regression leaf with sample set I has NI=∣I∣≥1, mean yˉI=NI−1∑i∈Iyi, and squared-error impurity QI=NI−1∑i∈I(yi−yˉI)2.

[F2]

The standard formulas ⟨x,y⟩=∑k<nxkyk on Rn and ∑k<nxkyk‾ on Cn are inner products: For every n, the standard coordinate dot product on Rn is an inner product; in particular, identifying R with R1 supplies the standard inner products on both R and RN.

[F3]

The adjoint T∗:W→V is characterised by ⟨Tv,w⟩W=⟨v,T∗w⟩V: An adjoint satisfies ⟨Ta,v⟩=⟨a,T∗v⟩ for every a and v.

Proof

1.1givenconstructalgebra

Enumerate I in increasing sample-index order as i1,…,iN and define T(c)=(c,…,c) and b=(yi1,…,yiN). For all scalars a,c,d∈R, T(ac+d)=aT(c)+T(d) coordinatewise, so T is linear.

2.1F2step 1.1algebra

Identify R with R1. By [F2] for n=1 and n=N, the standard coordinate inner products make R and RN finite-dimensional inner-product spaces. Hence ∥Tc−b∥2=∑i∈I(yi−c)2.

3.1F2F3step 1.1step 2.1algebraconstruct

Define S:RN→R by S(v1,…,vN)=∑k=1Nvk. This map is linear, and for every c∈R and v∈RN, ⟨T(c),v⟩RN=c∑k=1Nvk=⟨c,S(v)⟩R. Thus S is an adjoint of T by [F3]. If another map also satisfied this identity, its value at each v would differ from S(v) by a scalar pairing to zero with every c∈R, hence would equal S(v) by positive definiteness. Therefore T∗=S.

4.1F4step 1.1step 2.1step 3.1algebraF1

Now [F4] applies to the linear map T between these finite-dimensional inner-product spaces, with its adjoint established in step 3.1. The normal equation T∗(Tc−b)=0 is Nc−∑i∈Iyi=0, whose unique solution is c=yˉI because N≥1. By [F4], this solution is the unique minimizer.

5.1givenstep 4.1algebra∎

Expanding around yˉI gives ∑i∈I(yi−c)2=∑i∈I(yi−yˉI)2+N(c−yˉI)2, since ∑i∈I(yi−yˉI)=0. Therefore the minimum value is ∑i∈I(yi−yˉI)2.

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

Cost-complexity criteria for subtrees

Definition

Let T0 be a fixed finite binary tree grown on the sample from an empirical decision tree. A subtree T⊆T0 is obtained by collapsing zero or more internal nodes and all their descendants to leaves. For each leaf m of T, let Nm be its sample count and let Qm(T) be its regression squared-error impurity or its chosen classification impurity, computed as in the empirical decision-tree definition. Define the total training impurity R(T)=∑m∈leaves(T)NmQm(T), the terminal-node count ∣T∣, and for α≥0 the cost-complexity criterion Cα(T)=R(T)+α∣T∣.

For a current nonterminal node t, let It be its sample set and let Tt be its current descendant subtree. Let R(t)=NtQt be the training impurity if t is collapsed to a leaf, using the same response or class rule as for a leaf; let R(Tt) be the sum of the impurities of the current descendant leaves. Define the weakest-link quotient g(t)=[R(t)−R(Tt)]/(∣Tt∣−1). The denominator is the number of leaves removed by collapsing t. This definition makes no claim that a weakest-link sequence globally minimizes Cα for every α or that validation selection guarantees test-risk performance.

Remark

Choosing α by validation or cross-validation is a data-selection procedure; by itself, that choice does not guarantee test-risk performance.

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

Greedy axis-aligned CART growth on a finite sample

Definition

The inputs are observations (xi,yi)i=1n with n≥1 and xi∈Rp for an integer p≥1, a response mode (regression with yi∈R or classification with labels in a fixed finite ordered class set), a fixed classification impurity criterion when in classification mode, and a finite integer maximum depth dmax⁡≥0. The root has sample set {1,…,n} and depth 0. At a current leaf with nonempty sample set I and depth d<dmax⁡, perform the following procedure.

  1. For each coordinate j∈{1,…,p}, sort the distinct observed values in that coordinate as u1<⋯<ur. For each k=1,…,r−1, form the candidate threshold t=(uk+uk+1)/2 and its child sets IL={i∈I:xij≤t} and IR={i∈I:xij>t}. These are the finite candidates at the node; if no coordinate has at least two distinct values, the candidate set is empty.

  2. For each candidate, compute its child score. In regression the score is ∑i∈IL(yi−yˉIL)2+∑i∈IR(yi−yˉIR)2. In classification it is ∣IL∣QIL+∣IR∣QIR for the fixed chosen impurity criterion. Select the candidate with the smallest score; when scores tie, select the least pair (j,k) by comparing coordinate first and cut index second.

  3. Store the selected coordinate and threshold at the current node, create its left and right children, and apply this procedure to the left child and then the right child at depth d+1. If d=dmax⁡ or the candidate set is empty, leave the current node as a leaf.

The output is the finite tree whose leaves use the regression means or classification majority labels from the empirical decision-tree definition. No strictly positive loss reduction is required for a split below the depth cap, and the returned tree is not asserted to be globally optimal among all trees.

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

Greedy CART minimizes the listed split loss and terminates

Statement

Fix a finite labeled sample, response mode, classification impurity criterion when needed, and finite maximum depth as in greedy CART growth. At any node where the procedure splits, its selected coordinate and midpoint threshold minimize the declared child-weighted empirical loss among all nonempty coordinate-threshold partitions of that node's sample, with the specified lexicographic tie rule. The procedure produces a finite tree of depth at most the maximum depth, with nonempty children at every split. This is a nodewise candidate-optimality result only; it gives no global tree-optimality claim.

Facts & Assumptions

Given: A finite labeled sample with n≥1, a regression or classification mode, the fixed classification impurity criterion when applicable, and a finite integer maximum depth dmax⁡≥0.

[F1]

Empirical axis-aligned binary decision trees: Each split compares the nonempty child sample sets formed by one coordinate threshold; regression leaf loss is the sum of squared deviations from its sample mean, and classification leaf loss is sample count times the selected impurity.

[F2]

The sample mean uniquely minimizes squared error on a nonempty leaf: On a nonempty regression leaf, the unique constant minimizing the sum of squared errors is the sample mean.

[F3]

Greedy axis-aligned CART growth on a finite sample: At each eligible node, the algorithm forms midpoint thresholds between adjacent distinct observed coordinate values and selects the least child score, resolving score ties by coordinate and cut-index order.

Proof

1.1F1F3construct

At a node with finite nonempty sample set I, each coordinate has finitely many distinct observed values. Arrange them in increasing order u1<⋯<ur. For every k<r, the midpoint between uk and uk+1 gives two nonempty children. Conversely, any real threshold with two nonempty children has a unique gap uk≤s<uk+1 and induces the same partition as that midpoint. Thus the algorithm's candidate list is finite and contains every nonempty partition produced by one coordinate threshold.

2.1F1F2F3step 1.1algebrachoose

For each candidate the child score is a finite real sum. If the candidate list is nonempty, repeatedly comparing its finite list of scores produces a minimum, and the coordinate/cut-index order selects a unique first minimizer among ties. By [F2], the regression child means attain the least squared error on each child; by [F1], the classification score is the declared weighted impurity. The chosen candidate therefore minimizes the declared child loss over all such partitions.

3.1F1F3step 1.1algebra∎

Every recursive call increases depth by one, and calls at depth dmax⁡ stop. Each split creates two disjoint nonempty sample sets from its parent, so every terminal leaf contains at least one of the n sample indices. Each split increases the number of leaves by one; hence there are at most n−1 splits. The recursion terminates with depth at most dmax⁡ and all created children nonempty.

4 · Examples, counterexamples and false statements

None yet.

Sources