 Negative binomial deviance

With known size $\theta >0$, the mean-dependent log mass of a negative binomial distribution is $y\log \mu -(y+\theta )\log (\mu +\theta )$. Its derivative is $\theta (y-\mu )/[\mu (\mu +\theta )]$, so the saturated mean is $y$, including its boundary value zero. Subtracting the fitted log mass from that maximum gives the displayed deviance. This formula needs no fitted-intercept cancellation. As $\theta \to \infty$, its second term tends to $-2(y_i-e_i)$ and it becomes the full Poisson deviance.