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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.