Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 38 / 4

Use $\alpha$ for the intercept, to avoid confusing it with the observation means. Let $e_i=\exp (\hat{\alpha }+{\hat{\beta }}^T{x}_i)$ be the fitted Poisson regression mean. Apart from constants, its log-likelihood isAt a finite interior maximum-likelihood estimate, the intercept score equation gives the Poisson regression margin-matching score equations:The statistical saturated model fits each mean to $y_i$, taking the boundary mean zero if $y_i=0$. Subtracting fitted from saturated log-likelihood gives the Poisson devianceThe intercept balance cancels the sum of linear terms. ConsequentlyThe unsimplified formula makes nonnegativity transparent; individual terms in the simplified formula need not each be positive.

For the deviance goodness-of-fit test, if the fitted design has rank $r$, compare a suitably calibrated $D$ with ${\chi }_{n-r}^2$, rejecting for large values. This approximation applies in a regular large-count regime, such as a fixed set of cells with increasing exposure and positive fitted proportions. It is not automatically accurate for many sparse Poisson observations. Check residual patterns and overdispersion; where the approximation is doubtful, simulate independent Poisson random variables with fitted means, refit each sample, and use a parametric bootstrap distribution of the same statistic. A large p-value means no detected lack of fit, not proof of the model. If a finite interior estimate fails to exist, the score proof needs a boundary-limit interpretation; for example all-zero data give a supremum with every fitted mean tending to zero and limiting deviance zero.

Here the negative binomial distribution has mean parameter $\mu >0$ and size parameter $\theta >0$. Its support in the original PDF is all nonnegative integers; a finite endpoint appearing in the TeX aid would be incorrect. The probability generating function, obtained from the generalized binomial series, isDifferentiating at one gives $G^{\prime }(1)=\mu$ and $G^{\prime \prime }(1)={\mu }^2(1+1/\theta )$. Since the variance is $G^{\prime \prime }(1)+G^{\prime }(1)-G^{\prime }(1{)}^2$, this provesThe extra positive term explains its overdispersion relative to a Poisson distribution.

For one observation, write the log-likelihood asThe mean score function and its cross derivative areTaking expectations gives $I_{\mu \theta }=-\mathbb{E}({\partial }_{\theta }{u}_{\mu })=0$. In addition,For reference, with $\psi$ the digamma function, the size score function isIts variance $I_{\theta \theta }$ is finite and positive for finite positive $\mu ,\theta$. Thus negative binomial mean-size parameter orthogonality makes the expected Fisher information matrix diagonal. For an independent identically distributed sample and a regular identifiable interior true parameter, asymptotic normality of a maximum likelihood estimator givesHence the asymptotic correlation is zero. This is first-order asymptotic independence, not a claim of exact finite-sample independence. Indeed the mean score yields $\hat{\mu }=\bar{y}$ at an interior fit, while the size fit also depends on the observed dispersion. The argument excludes a true boundary such as the Poisson limit $\theta =\infty$.