Gini Splits

Equal Gain Needs a Tie Policy

When two candidate splits have the same exact gain, a stated tie policy decides which one is displayed as chosen.

highlighted = computed this step

Two candidates can tie

Start with parent counts [2,2]. Feature L and Feature R both split into [1,1] and [1,1].

Gw(L)=Gw(R)=1/2G_w(L)=G_w(R)=1/2
Equal gain tieTwo candidate questions have the same exact gain.Equal gain tie policyparent counts=[2, 2]; parent G=1/2candidatechildrenweighted Ginigainpolicy resultFeature L[1, 1] | [1, 1]1/20chosen firstFeature R[1, 1] | [1, 1]1/20same scoretie policy: first listed candidate -> Feature Lequal exact scores need a stated tie policynot a claim of accuracy, optimality, or generalizationnot proof Feature L is better

The gains are equal too

The parent Gini is 1/2, and each candidate weighted Gini is also 1/2. So each gain is 0.

gain=Gparent−Gw=1/2  −  1/2=0\text{gain}=G_{\text{parent}}-G_w=1/2\;-\;1/2=0
Equal gain tieTwo candidate questions have the same exact gain.Equal gain tie policyparent counts=[2, 2]; parent G=1/2candidatechildrenweighted Ginigainpolicy resultFeature L[1, 1] | [1, 1]1/20chosen firstFeature R[1, 1] | [1, 1]1/20same scoretie policy: first listed candidate -> Feature Lequal exact scores need a stated tie policynot a claim of accuracy, optimality, or generalizationnot proof Feature L is better

State the tie policy

The displayed policy chooses the first listed candidate, so Feature L is selected. The policy chooses among equal exact scores; it does not make an accuracy, optimality, or generalization claim.

first listed candidate→Feature L\text{first listed candidate}\rightarrow\text{Feature L}
Equal gain tieTwo candidate questions have the same exact gain.Equal gain tie policyparent counts=[2, 2]; parent G=1/2candidatechildrenweighted Ginigainpolicy resultFeature L[1, 1] | [1, 1]1/20chosen firstFeature R[1, 1] | [1, 1]1/20same scoretie policy: first listed candidate -> Feature Lequal exact scores need a stated tie policynot a claim of accuracy, optimality, or generalizationnot proof Feature L is better