Solution (source code)

= Solution

The working <score function> and its derivative are
$$
U(\beta)=\sum_i x_i\{Y_i-e^{\beta x_i}\},\qquad
U'(\beta)=-\sum_i x_i^2e^{\beta x_i}=-I(\beta).
$$
At an interior working maximizer, $U(\widehat\beta)=0$. A <Taylor expansion> at $\beta_0$ yields
$$
0=U(\beta_0)-I(\beta_0)(\widehat\beta-\beta_0)
-\frac12\left(\sum_i x_i^3e^{\beta^*x_i}\right)(\widehat\beta-\beta_0)^2
$$
for an intermediate value $\beta^*$. When the last term is negligible relative to the linear one, this gives
$$
\boxed{\widehat\beta-\beta_0\approx I(\beta_0)^{-1}U(\beta_0).}
$$
This is the <Poisson score linearization under proportional variance>. Correct specification of the mean alone implies $EU(\beta_0)=0$. The stipulated independence and <variance> relation give
$$
\operatorname{Var}U(\beta_0)=\sum_i x_i^2\operatorname{Var}(Y_i)=\phi\sum_i x_i^2e^{\beta_0x_i}=\phi I(\beta_0).
$$
Taking the mean and <variance> of the linear approximation therefore proves
$$
\boxed{E\widehat\beta\approx\beta_0,\qquad
\operatorname{Var}(\widehat\beta)\approx\frac{\phi}{I(\beta_0)}.}
$$
These are first-order conclusions, not exact finite-sample identities. The mean and <variance> assumptions alone do not guarantee that the approximation is accurate: an interior consistent solution, increasing information, control of leverage, and a suitable central limit condition are needed for normal inference.

The <quasi-likelihood> score for <variance> $\phi\mu_i$ is $U/\phi$, so its zero is unchanged by the unknown common <dispersion parameter>. Thus the working Poisson estimating equation remains appropriate for the mean parameter, but its unadjusted Poisson <variance> estimate is wrong when $\phi\ne1$. A usual <Pearson dispersion estimator> for this one-parameter model is
$$
\widehat\phi=\frac1{n-1}\sum_i\frac{(Y_i-\widehat\mu_i)^2}{\widehat\mu_i},\qquad
\widehat{\operatorname{Var}}(\widehat\beta)=\frac{\widehat\phi}{I(\widehat\beta)}.
$$
In particular <overdispersion> multiplies coefficient <standard errors> by approximately $\sqrt\phi$, even though it leaves the coefficient estimating equation unchanged.