 Solution

The null model has only an intercept:For $\bar{Y}>0$, maximizing its log-likelihood gives $\hat{\mu }=\bar{Y}$ and ${\hat{\beta }}_0=\log \bar{Y}$. The saturated Poisson regression gives each observation its own fitted mean ${\hat{\mu }}_i=Y_i$, allowing zero as a limiting fitted mean. By the definition of Poisson deviance,Since ${\sum }_i(Y_i-\bar{Y})=0$,Use the continuous convention $0\log (0/\bar{Y})=0$. If all counts are zero, both the saturated and null likelihood maxima are attained as limits at zero mean and the deviance is zero.