A one-observation Poisson-log Fisher-scoring step

Given: The scalar design $X=[1]$, response $y_1=2$, current $\beta =0$, and the Poisson-log family and link in the Example.

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

Mean and variance from the cumulant function: the response mean is $b^{\mathrm{\prime }}({\theta }_i)$ and its variance is $a(\varphi )b^{\mathrm{\prime }\mathrm{\prime }}({\theta }_i)$.

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 Poisson response has support ${\mathbb{N}}_0$, counting base measure, and $f(k;\theta )=e^{-e^{\theta }}{e}^{k\theta }\mathrm{/}k!$. The exponential series and addition formula normalize the masses. Its first two $\theta$ derivatives are $f(k-e^{\theta })$ and $f\{(k-e^{\theta }{)}^2-e^{\theta }\}$. On any compact interval $[u,v]$ these are bounded by the summable sequence $e^{vk}\{1+(k+e^v{)}^2+e^v\}\mathrm{/}k!$; the ratio test gives summability. Reindexing gives $\mathbb{E}[Y^2]={\lambda }^2+\lambda <\mathrm{\infty }$ for $\lambda =e^{\theta }$. Thus the fixed-support, finite-moment and derivative-envelope hypotheses hold, and the component score has finite second moment. (The real exponential function and the number $e$ by a power series, The exponential addition formula $\exp (x+y)=\exp (x)\exp (y)$, The factorial $n!$ and the falling factorial $n^{\underset{‾}{k}}$, defined by recursion in $\mathbb{N}$, The exponential function is smooth and $(\exp {)}^{\mathrm{\prime }}=\exp$)