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Linear classifiers, discriminants, and logistic models — Examples

1 · Prerequisites

2 · Logical flowchart

3 · Definitions, theorems and proofs

None yet.

4 · Examples, counterexamples and false statements

ExampleConstruction: AI-generatedVerification: AI-generatedjudge pass (gpt-6-luna)Open item page →

A Gaussian LDA model and a conditional logistic model can share a decision boundary

Example

Assume AC. In one dimension, take LDA classes $1$ and $2$ with equal priors and conditional laws $X ∣ Y = 1 \sim N_{1} (- 1, 1)$ and $X ∣ Y = 2 \sim N_{1} (1, 1)$ . Recode class $1$ as binary label $0$ and class $2$ as binary label $1$ in a conditional logistic model with $\Pr (Y = 1 ∣ x) = σ (2 x)$ . Both classifiers have tie boundary $x = 0$ and agree on both sides of that boundary; ties select class $1$ (binary label $0$ ). This equality of boundaries for these parameters does not identify the two estimation procedures or equate their coefficients in general.

The number line marks the two Gaussian means and their posterior tie boundary; the score calculation below establishes the boundary.

diagram failed to render:
\begin{tikzpicture}[x=1.25cm,y=1cm]
  \draw[->] (-2.1,0) -- (2.1,0) node[right] {$x$};
  \fill (-1,0) circle (2pt);
  \fill (1,0) circle (2pt);
  \draw[dashed] (0,-0.32) -- (0,0.42);
  \node[below] at (-1,0) {$\mu_1=-1$};
  \node[below] at (1,0) {$\mu_2=1$};
  \node[above] at (0,0.42) {$x=0$};
\end{tikzpicture}

Example references

Common-covariance LDA scores are affine and QDA scores are quadratic

Binary logistic and multiclass softmax conditional models

The Axiom of Choice

Hastie, Tibshirani, and Friedman, ESL, §4.4.5, equation (4.33)

Stanford CS229 Spring 2022 lecture notes, §4.1.3

Facts & Assumptions

Given: AC, equal LDA priors, $μ_{1} = - 1$ , $μ_{2} = 1$ , $Σ = 1$ , the stated class recoding, and the conditional logistic slope $2$ with zero intercept.

[F1]

Under common covariance, the LDA posterior log-score for class $k$ is $x^{T} Σ^{- 1} μ_{k} - \frac{1}{2} μ_{k}^{T} Σ^{- 1} μ_{k} + \log π_{k}$ , up to a class-independent term. Common-covariance LDA scores are affine and QDA scores are quadratic

[F2]

The binary conditional logistic probability is $σ (β_{0} + x β)$ , with $σ (t) = (1 + e^{- t})^{- 1}$ . Binary logistic and multiclass softmax conditional models

[A1]

AC is carried because the LDA Gaussian law and density theorem assume it. The Axiom of Choice

Verification

1.1givenF1A1algebra

For the LDA model, [F1] gives $δ_{2} (x) - δ_{1} (x) = x (1 - (- 1)) - \frac{1}{2} (1^{2} - (- 1)^{2}) + \log (1) = 2 x$ . The two posterior scores tie exactly at $x = 0$ , with class $1$ selected there by the smallest-label rule.

2.1givenF2step 1.1algebra∎

For the logistic model, [F2] gives $\Pr (Y = 1 ∣ x) = σ (2 x)$ . Since $σ^{'} (t) = σ (t) (1 - σ (t)) > 0$ and $σ (0) = 1 / 2$ , this probability is below, equal to, or above $1 / 2$ exactly when $x$ is below, equal to, or above zero. Under the stated recoding, the class decisions agree on each region and ties select class $1$ ; no coefficient-estimation equivalence is claimed.

ExampleConstruction: AI-generatedVerification: AI-generatedjudge pass (gpt-6-luna)Open item page →

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.

ExampleConstruction: AI-generatedVerification: AI-generatedOpen item page →

Two contradictory labels make the cyclic perceptron repeat a two-update cycle

Example

Use the ordered one-dimensional sample $(x_{1}, y_{1}) = (0, + 1)$ , $(x_{2}, y_{2}) = (0, - 1)$ . It is not separable by an affine score because both observations have the same augmented input $(0, 1)$ and require opposite strict signs. For the zero-initialized cyclic perceptron, every pass updates on both observations and returns the weight vector to zero, so no no-update pass occurs.

Example references

The cyclic perceptron update procedure

Hastie, Tibshirani, and Friedman, ESL, §4.5.1

Stanford CS229 Spring 2022 lecture notes, §2.2

Facts & Assumptions

Given: The ordered sample and the zero-initialized procedure with augmented input $(x, 1)$ .

[F1]

The procedure updates $w \leftarrow w + y_{i} (x_{i}, 1)$ whenever $y_{i} w^{T} (x_{i}, 1) \leq 0$ , and terminates only after a full pass with no updates. The cyclic perceptron update procedure

Verification

1.1givenF1algebra

Both augmented inputs equal $(0, 1)$ . Starting at $w_{0} = (0, 0)$ , the first observation has margin zero and triggers the update $w_{1} = (0, 1)$ .

2.1F1step 1.1algebra∎

The second observation then has signed margin $(- 1) (0, 1)^{T} (0, 1) = - 1$ , so [F1] updates to $w_{2} = (0, 1) - (0, 1) = (0, 0)$ . Thus each pass repeats the initial state and updates twice, so the stop condition never holds. The sample is nonseparable because a single score at the shared input cannot be strictly positive and strictly negative simultaneously.

Linear classifiers, discriminants, and logistic models — Examples

2 · Logical flowchart

3 · Definitions, theorems and proofs

4 · Examples, counterexamples and false statements

A Gaussian LDA model and a conditional logistic model can share a decision boundary

Example

Example references

Facts & Assumptions

Verification

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

Example

Example references

Facts & Assumptions

Verification

Two contradictory labels make the cyclic perceptron repeat a two-update cycle

Example

Example references

Facts & Assumptions

Verification

Sources