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The Bernoulli identity-link boundary leaves the regular family

Example

For one Bernoulli response take x1=1, identity link g(μ)=μ, and a(ϕ)=1. The admissible interior parameter is β=η=μ∈(0,1). The boundary value μ=0 is a valid degenerate Bernoulli law in the closure, but it is not represented by a finite natural parameter.

Facts & Assumptions

Given: The Bernoulli cumulant b(θ)=log⁡(1+eθ), x1=1, identity link η=μ=β, and the closure boundary μ↓0.

[F1]

Generalized linear model with explicit conditioning: μi=g−1(ηi) and θi=τ(μi), where τ satisfies b′(τ(μ))=μ.

[F2]

General-link score and Fisher information: w=μ˙2/(a(ϕ)V(μ)) where V=b′′(θ).

[F3]

One minimum-norm IRLS step: the scoring step requires finite positive weights and a finite nonzero link derivative at the current iterate.

[F4]

In this Bernoulli GLM, Yi∈{0,1} and p(θ)=eθ/(1+eθ); relative to counting measure the mass is f(y;θ)=exp⁡{yθ−log⁡(1+eθ)}. It is normalized by summing over y=0,1. Writing the natural-parameter derivatives as f′=f(y−p) and f′′=f{(y−p)2−p(1−p)}, the first two derivatives have constant integrable bounds on the two-point support, and E[Y2]≤1. Thus the fixed-support, finite-moment, derivative-envelope, and score-moment hypotheses used above hold. (Bernoulli random variables and binomial random variables as sums of independent Bernoulli trials, The exponential function is smooth and (exp⁡)′=exp⁡, The natural logarithm has derivative 1/x and equals the integral from 1 to x of 1/t)

Verification

1.1F1F4givenalgebra

For Bernoulli, b′(θ)=eθ/(1+eθ), so the finite natural parameter corresponding to μ∈(0,1) is θ=log⁡(μ/(1−μ)). Under the identity link, η=μ=β; as μ↓0, θ→−∞, so μ=0 is outside the finite-natural-parameter interior.

2.1F2F3F4step 1.1algebra∎

On the interior, μ˙=dμ/dη=1 and V(μ)=b′′(θ)=μ(1−μ). Hence [F2] gives w=1/{μ(1−μ)}→+∞ as μ↓0. The boundary therefore violates the finite-weight admissibility in [F3], even though the limiting Bernoulli law itself is a probability measure.

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