 Solution

Let ${\ell }_{\mathrm{sat}}$ denote the maximized log-likelihood in the saturated model, with one fitted mean per observation. With known dispersion, define the deviance of a fitted model as $D=2({\ell }_{\mathrm{sat}}-{\ell }_{\mathrm{fitted}})$; for an exponential-dispersion convention that scales deviance by the dispersion $\varphi$, divide that scaled quantity by $\varphi$ in the following test.

Let $D_0,D_1$ be the fitted deviances under the null and full models. Their difference is the likelihood-ratio test statisticUnder the regular identifiable interior null hypothesis, Wilks theorem gives the limiting ${\chi }_{p-p_0}^2$ distribution. Reject for an observed difference above its chosen upper-tail critical value. This concerns the difference of fitted model deviances, not simply the full-model deviance compared with zero.

For a Poisson regression, the means are ${\mu }_i=e^{x_i^T\beta }$ and $\ell ={\sum }_i(y_i\log {\mu }_i-{\mu }_i-\log y_i!)$. Saturation sets ${\mu }_i=y_i$, taking the limiting value at $y_i=0$. Consequently the Poisson deviance iswith $0\log 0=0$. The intercept score implies ${\sum }_i(y_i-{\hat{\mu }}_i)=0$ at a finite fit, although retaining that term makes the individual deviance contributions clear.

For $y_i={\hat{\mu }}_i+{\delta }_i$ with $|{\delta }_i|/{\hat{\mu }}_i$ small, expand $({\hat{\mu }}_i+{\delta }_i)\log (1+{\delta }_i/{\hat{\mu }}_i)-{\delta }_i$. The linear term cancels and the result is ${\delta }_i^2/(2{\hat{\mu }}_i)+O(|{\delta }_i{|}^3/{\hat{\mu }}_i^2)$. Thereforethe Pearson chi-squared statistic. The approximation requires adequately large fitted counts and small relative residuals; it is not an identity for sparse observations.