With $X$ having rows $x_i^T=(1,g_i,m_i)$ and $\mu_i(\beta)=e^{x_i^T\beta}$, the <quasi-score equation> is
$$
X^T\{Y-\mu(\beta)\}=0.
$$
It is the same coefficient equation as for Poisson maximum likelihood, explaining why the two models have identical coefficient estimates.

Let $W=\operatorname{diag}(\widehat\mu_1,\ldots,\widehat\mu_n)$ and let coordinate $2$ denote gender. The model-based <standard error> is
$$
\operatorname{se}(\widehat\beta_1)
=\sqrt{\widehat\phi\,[(X^TWX)^{-1}]_{22}},
\qquad
\widehat\phi=\frac{X_P^2}{n-3}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 218 / 1 / c / Solution

For mean $\mu_i(\beta)$ and variance proportional to $V(\mu_i)$, the quasi-score is the sum of $(Y_i-\mu_i)V(\mu_i)^{-1}\partial\mu_i/\partial\beta$. Setting it to zero estimates $\beta$ without requiring a full response likelihood.


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