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LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-luna)
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The weighted centroid minimizes finite-cluster squared error

Statement

Let x1,…,xN∈Rd, let wi>0 with ∑i=1Nwi=1, and let S⊆{1,…,N} be nonempty. Put WS:=∑i∈Swi and μS:=WS−1∑i∈Swixi. For every c∈Rd,

∑i∈Swi∥xi−c∥22=∑i∈Swi∥xi−μS∥22+WS∥c−μS∥22.

Consequently, μS is the unique minimizer of the cluster's weighted squared error. If S is empty, its cost is zero for every center and no centroid is required.

Facts & Assumptions

Given: A finite list of vectors xi∈Rd, positive weights summing to one, and a nonempty index set S.

[A1]

The inner product and squared norm on Rd are ⟨x,y⟩=∑ℓ=1dxℓyℓ and ∥x∥22=⟨x,x⟩.

Proof

1.1givenA1algebra

Since S is nonempty and each wi>0, WS>0, so μS is defined and ∑i∈Swi(xi−μS)=0 by its formula.

2.1givenA1step 1.1algebra

Write xi−c=(xi−μS)+(μS−c) and expand the squared norm using [A1]. Summing over S, the cross term is 2⟨∑i∈Swi(xi−μS),μS−c⟩=0 by step 1.1; the remaining terms give the displayed identity.

3.1givenstep 2.1algebra∎

The last term in step 2.1 is nonnegative and vanishes exactly when c=μS, because WS>0 and a Euclidean squared norm vanishes exactly at the zero vector. Thus μS uniquely minimizes the cost. For empty S, the sum has no terms and equals zero for every c.

Used by

Dependency tree · 0 levels

Nothing. This result depends on no other item in the library.

Sources