A one-observation Bernoulli-logit Fisher-scoring step

Given: The scalar design $X=[1]$, response $y_1=1$, current $\beta =0$, and the Bernoulli-logit family and link in the Example.

Generalized linear model with explicit conditioning: the canonical link is $g=\tau$, equivalently ${\theta }_i={\eta }_i$.

Canonical-link score and Hessian: under a common scale the score is $X^{\mathsf{T}}(y-\mu )\mathrm{/}a(\varphi )$.

General-link score and Fisher information: the expected score outer-product information has weights $w_i={\dot{\mu }}_i^2\mathrm{/}(a(\varphi )V_i)$.

One minimum-norm IRLS step: the working response and weights are $z_i={\eta }_i+(y_i-{\mu }_i)\mathrm{/}{\dot{\mu }}_i$ and $w_i={\dot{\mu }}_i^2\mathrm{/}(a(\varphi )V_i)$.

IRLS equals the Fisher-scoring step when information is invertible: when information is invertible, ${\beta }_{\mathrm{n}\mathrm{e}\mathrm{w}}=\beta +I(\beta {)}^{-1}U(\beta )$.

The Bernoulli-logit component mass on $\{0,1\}$ relative to counting measure is $f(y;\eta )=\exp \{y\eta -\log (1+e^{\eta })\}$. It sums to one; its support is fixed, its second moment is finite, and its first two $\eta$ derivatives are bounded on the finite support. (Bernoulli random variables and binomial random variables as sums of independent Bernoulli trials, The exponential function is smooth and $(\exp {)}^{\mathrm{\prime }}=\exp$, The natural logarithm has derivative 1/x and equals the integral from 1 to x of 1/t)