 Poisson score linearization under proportional variance

With ${\mu }_i=e^{\beta x_i}$, define $U={\sum }_i{x}_i(Y_i-{\mu }_i)$ and $I={\sum }_i{x}_i^2{\mu }_i$. The estimating equation and a Taylor expansion give $\hat{\beta }-{\beta }_0\approx I({\beta }_0{)}^{-1}U({\beta }_0)$. If responses are independent with correct means and $\mathrm{Var}(Y_i)=\varphi {\mu }_i$, then $EU=0$ and $\mathrm{Var}U=\varphi I$. Consequently the first-order estimator variance is $\varphi /I$, even if the working Poisson distribution is wrong. Regular asymptotics require growing information and control of leverage and the Taylor remainder.