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Complete separation gives an unattained logistic likelihood supremum

Statement

Let n,q≥1 be finite, let xi∈Rq be a fixed design row with an intercept coordinate when one is used, and let yi∈{0,1}. Suppose the unpenalized binary logistic log-likelihood is ℓ(β)=∑i[yixiTβ−log⁡(1+exiTβ)]. If there is a vector v∈Rq such that (2yi−1)xiTv>0 for every i, then sup⁡β∈Rqℓ(β)=0, and no finite coefficient vector attains this supremum.

Facts & Assumptions

Given: A finite fixed binary logistic data set with n≥1, yi∈{0,1}, the unpenalized log-likelihood, and a strict separating vector v satisfying the signed-margin inequalities in the Statement.

[F1]

The log-likelihood is the sum of the Bernoulli terms yiηi−log⁡(1+eηi) with ηi=xiTβ. Binary logistic log-likelihood and L1-penalized objective

[F2]

At every finite coefficient vector, the binary conditional probability is strictly between zero and one. Binary logistic and multiclass softmax conditional models

Proof

1.1F1givenconstructalgebra

Put β(t)=tv for t>0. If yi=1, the separation condition gives ai=xiTv>0 and the ith log-likelihood term is −log⁡(1+e−tai)→0. If yi=0, the condition gives ai=xiTv<0 and the term is −log⁡(1+etai)→0. Since the sample is finite, ℓ(β(t))→0, so the supremum is at least zero.

2.1F2givenstep 1.1algebra∎

For any finite β, each ηi is finite and [F2] gives 0<pi<1. The observed Bernoulli log-probability is log⁡pi<0 when yi=1 and log⁡(1−pi)<0 when yi=0. As n≥1, ℓ(β)<0 for every finite β. Thus the supremum is at most zero and is attained by no finite coefficient vector; step 1.1 shows that it equals zero.

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