Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 36 / 5

For independent Poisson counts, the log-likelihood at mean vector $\mu$ isThe saturated model maximizes each cell separately at ${\hat{\mu }}_i=y_i$, including the boundary value zero. If $e_i$ is the fitted mean under the log-linear regression, twice the saturated-minus-fitted log-likelihood is the full Poisson deviancewith $0\log (0/e_i)=0$ understood by continuity. The fitted intercept has score function $\partial \ell /\partial \mu ={\sum }_i(y_i-e_i)$, since differentiating the logarithmic mean with respect to that intercept gives one. Its maximum-likelihood equation therefore gives ${\sum }_i{e}_i={\sum }_i{y}_i$. This is why Poisson deviance simplifies when an intercept is fitted:Without the fitted unpenalized intercept, one cannot generally discard the two linear terms. The individual logarithmic terms in the simplified sum need not each be positive; nonnegativity belongs to the full deviance, or to the full cell contributions before cancellation.

For the approximation, write $y_i=e_i+r_i$ and expand $\log (1+r_i/e_i)$. When relative residuals are small,Consequently the quadratic Pearson approximation to the Poisson deviance isthe Pearson chi-squared statistic for unit Poisson dispersion. Adequate expected counts and small relative residuals are what make the quadratic approximation useful; it is not an exact identity and can be poor for sparse cells, especially a zero count with positive fitted mean. Any further chi-squared goodness-of-fit reference needs its own large-count regularity assumptions.

With the negative-binomial size $\theta >0$ held fixed, collect all mean-independent terms into $c(y_i,\theta )$. The cell log-likelihood isIts derivative isFor $y_i>0$ it is positive below $y_i$ and negative above $y_i$, so the saturated maximum is at ${\mu }_i=y_i$. For $y_i=0$, the maximum is the boundary limit ${\mu }_i\downarrow 0$. Evaluating the saturated-minus-fitted log-likelihood gives the negative binomial devianceAll gamma-function and factorial terms cancel because the same known $\theta$ is used in both models. The zero-count logarithmic term is interpreted as zero; its remaining cell contribution is $2\theta \log (1+e_i/\theta )\ge 0$. No intercept score cancellation is required for this formula, so it applies to the specified design whether or not its covariates include an intercept.

For completeness, the fitted coefficients obey the score equations of the fixed-size negative binomial generalized linear model,The logarithmic link is not the canonical link for this family, but it produces the requested positive means. As a check, when $\theta \to \infty$,so the formula tends to the full Poisson deviance, including the linear terms when an intercept-balance equation is unavailable. The size $\theta$ used here is the negative-binomial size parameter; it is distinct from the random-multiplier variance denoted by the same letter in Question 4.