 Solution

Write $a$ for the scalar intercept denoted $\mu$ in the question, to distinguish it from the individual means ${\mu }_i$. The log-likelihood of the independent Poisson observations isDifferentiating gives the score equations for an interior maximum-likelihood estimator:In particular, when $m={\sum }_i{y}_i>0$,The left side is a weighted covariate mean whose derivative is its weighted variance, so it is strictly increasing when the covariates are not all equal. An interior solution is unique if the observed count-weighted mean is strictly between the smallest and largest covariates. At an extreme it may require a limiting slope, and if all counts are zero the maximum occurs as $a\to -\infty$, rather than at finite parameters. Constant covariates make the slope and intercept unidentifiable.

Put $S_j={\sum }_i{x}_i^j{\mu }_i$. The negative Hessian matrix is deterministic at given parameters and is also the Fisher information matrix:Under the usual interior, increasing-information regularity for asymptotic normality of a maximum likelihood estimator, replace the unknown means by their fitted values and read the lower-right entry:Equivalently, Poisson slope information after eliminating an intercept is ${\sum }_i{\mu }_i(x_i-{\bar{x}}_{\mu }{)}^2$, with ${\bar{x}}_{\mu }=S_1/S_0$. This explains both the positive denominator for varying covariates and the loss of slope information when their spread is small.