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A strictly separated two-point sample has an unattained logistic likelihood supremum

Example

Take two one-dimensional observations $(x_{1}, y_{1}) = (- 1, 0)$ and $(x_{2}, y_{2}) = (1, 1)$ with an intercept in the design. The augmented rows are $(1, - 1)$ and $(1, 1)$ , and the vector $(0, 1)$ strictly separates them. Along coefficients $β (t) = (0, t)$ with $t > 0$ , the log-likelihood is $- 2 \log (1 + e^{- t})$ , which tends to zero while remaining negative for every finite $t$ . The sample therefore has an unattained supremum at zero.

Example references

Complete separation gives an unattained logistic likelihood supremum

Binary logistic log-likelihood and L1-penalized objective

Hastie, Tibshirani, and Friedman, ESL, §4.4.5 and Exercise 4.5

Facts & Assumptions

Given: The two observations and the unpenalized binary logistic model with an intercept.

[F1]

The log-likelihood is $\sum_{i} [y_{i} η_{i} - \log (1 + e^{η_{i}})]$ , where $η_{i} = x_{i}^{T} β$ . Binary logistic log-likelihood and L1-penalized objective

[F2]

Strict separation of this finite design implies a zero supremum that no finite coefficient attains. Complete separation gives an unattained logistic likelihood supremum

Verification

1.1givenconstructalgebra

For $v = (0, 1)$ , the signed margins are $(2 y_{1} - 1) (1, - 1)^{T} v = (- 1) (- 1) = 1$ and $(2 y_{2} - 1) (1, 1)^{T} v = (1) (1) = 1$ . Hence the sample is strictly separated.

2.1F1F2step 1.1algebra∎

With $β (t) = (0, t)$ , the two predictors are $- t$ and $t$ . By [F1], the two log-likelihood terms are each $- \log (1 + e^{- t})$ , so $ℓ (β (t)) = - 2 \log (1 + e^{- t}) \to 0$ . Every finite coefficient has likelihood strictly below zero by [F2]; hence this is an unattained supremum.

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Used by

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Sources

Hastie, Tibshirani, and Friedman, The Elements of Statistical Learning, 2nd ed., corrected 12th printing (standard reference, not scraped)